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id stringlengths 77 537	text stringlengths 101 1.47M	source stringclasses 1 value	added stringdate 2025-03-17 19:52:06 2025-03-17 22:27:57	created timestamp[s]date 2013-05-16 21:27:22 2025-03-14 14:40:33	metadata dict
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.04%3A_Conservative_Forces_and_Potential_Energy	7.4: Conservative Forces and Potential Energy Learning Objectives By the end of this section, you will be able to: - Define conservative force, potential energy, and mechanical energy. - Explain the potential energy of a spring in terms of its compression when Hooke’s law applies. - Use the work-energy theorem to show how having only conservative forces implies conservation of mechanical energy. Work is done by a force, and some forces, such as weight, have special characteristics. A conservative force is one, like the gravitational force, for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken. We can define a potential energy (PE) for any conservative force, just as we did for the gravitational force. For example, when you wind up a toy, an egg timer, or an old-fashioned watch, you do work against its spring and store energy in it. (We treat these springs as ideal, in that we assume there is no friction and no production of thermal energy.) This stored energy is recoverable as work, and it is useful to think of it as potential energy contained in the spring. Indeed, the reason that the spring has this characteristic is that its force is conservative . That is, a conservative force results in stored or potential energy. Gravitational potential energy is one example, as is the energy stored in a spring. We will also see how conservative forces are related to the conservation of energy. Potential Energy and Conservative Forces Potential energy is the energy a system has due to position, shape, or configuration. It is stored energy that is completely recoverable. A conservative force is one for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken. We can define a potential energy (PE) for any conservative force. The work done against a conservative force to reach a final configuration depends on the configuration, not the path followed, and is the potential energy added. Potential Energy of a Spring First, let us obtain an expression for the potential energy stored in a spring $(PE_s$. We calculate the work done to stretch or compress a spring that obeys Hooke’s law. (Hooke’s law was examined in Elasticity: Stress and Strain , and states that the magnitude of force $F$ on the spring and the resulting deformation $\Delta L$ are proportional, $F = k\Delta L$). (See Figure.) For our spring, we will replace $\Delta L$ (the amount of deformation produced by a force $F$) by the distance $x$ that the spring is stretched or compressed along its length. So the force needed to stretch the spring has magnitude $F = kx$, where $k$ is the spring’s force constant. The force increases linearly from 0 at the start to $kx$ in the fully stretched position. The average force is $kx/2$. Thus the work done in stretching or compressing the spring is \[W_s = Fd = (\frac{kx}{2})x = \frac{1}{2}kx^2. \] Alternatively, we noted in Kinetic Energy and the Work-Energy Theorem that the area under a graph of $F$ vs. $x$ is the work done by the force. In Figure (c) we see that this area is also $\frac{1}{2}kx^2$. We therefore define the potential energy of a spring , $PE_s\|) to be \[PE_s = \dfrac{1}{2}kx^2,\] where \(k$ is the spring’s force constant and $x$ is the displacement from its undeformed position. The potential energy represents the work done on the spring and the energy stored in it as a result of stretching or compressing it a distance $x$. The potential energy of the spring $PE_s$ does not depend on the path taken; it depends only on the stretch or squeeze $x$ in the final configuration. The equation $PE_s = \frac{1}{2}kx^2$ has general validity beyond the special case for which it was derived. Potential energy can be stored in any elastic medium by deforming it. Indeed, the general definition of potential energy is energy due to position, shape, or configuration. For shape or position deformations, stored energy is $PE_s = \frac{1}{2}kx^2$, where $k$ is the force constant of the particular system and $x$ is its deformation. Another example is seen in Figure for a guitar string. Conservation of Mechanical Energy Let us now consider what form the work-energy theorem takes when only conservative forces are involved. This will lead us to the conservation of energy principle. The work-energy theorem states that the net work done by all forces acting on a system equals its change in kinetic energy. In equation form, this is \[W_{net} = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mv_0^2 = \Delta KE.\] If only conservative forces act, then \[W_{net} = W_c\] where $W_c$ is the total work done by all conservative forces. Thus, \[W_c = \Delta KE.\] Now, if the conservative force, such as the gravitational force or a spring force, does work, the system loses potential energy. That is, \[W_c = -\Delta PE\] Therefore, \[-\Delta PE = KE\] or \[\Delta PE + \Delta KE = 0. \] This equation means that the total kinetic and potential energy is constant for any process involving only conservative forces. That is, \[KE + PE = {constant}\] or \[KE_i + PE_i = KE_f + PE_f \] \[(conservative \, forces \, only),\] where i and f denote initial and final values. This equation is a form of the work-energy theorem for conservative forces; it is known as the conservation of mechanical energy principle . Remember that this applies to the extent that all the forces are conservative, so that friction is negligible. The total kinetic plus potential energy of a system is defined to be its mechanical energy , $(KE + PE)$. In a system that experiences only conservative forces, there is a potential energy associated with each force, and the energy only changes form between $KE$ and the various types of $PE$, with the total energy remaining constant. Example $\PageIndex{1}$: Using Conservation of Mechanical Energy to Calculate the Speed of a Toy Car A 0.100-kg toy car is propelled by a compressed spring, as shown in Figure. The car follows a track that rises 0.180 m above the starting point. The spring is compressed 4.00 cm and has a force constant of 250.0 N/m. Assuming work done by friction to be negligible, find (a) how fast the car is going before it starts up the slope and (b) how fast it is going at the top of the slope. Strategy The spring force and the gravitational force are conservative forces, so conservation of mechanical energy can be used. Thus, \[KE_i + PE_i = KE_f + PE_f\] or \[\dfrac{1}{2} mv_i^2 + mgh_i + \dfrac{1}{2}kx_i^2 = \dfrac{1}{2}mv_f^2 + mgh_f + \dfrac{1}{2}kx_f^2 \nonumber\] where $h$ is the height (vertical position) and $x$ is the compression of the spring. This general statement looks complex but becomes much simpler when we start considering specific situations. First, we must identify the initial and final conditions in a problem; then, we enter them into the last equation to solve for an unknown. Solution for (a) This part of the problem is limited to conditions just before the car is released and just after it leaves the spring. Take the initial height to be zero, so that both $h_i$ and $h_f$ are zero. Furthermore, the initial speed $v_i$ is zero and the final compression of the spring $x_f$ is zero, and so several terms in the conservation of mechanical energy equation are zero and it simplifies to \[\dfrac{1}{2}kx_i^2 = \dfrac{1}{2}mv_f^2. \nonumber\] In other words, the initial potential energy in the spring is converted completely to kinetic energy in the absence of friction. Solving for the final speed and entering known values yields \[ \begin{align} v_f &= \sqrt{\dfrac{k}{m}}x_i \\[5pt] &= \sqrt{\dfrac{250N/m}{0.100kg}}(0.0400\space m) \\[5pt] &= 2.00 m/s \end{align}\] Solution for (b) One method of finding the speed at the top of the slope is to consider conditions just before the car is released and just after it reaches the top of the slope, completely ignoring everything in between. Doing the same type of analysis to find which terms are zero, the conservation of mechanical energy becomes \[\dfrac{1}{2}kx_i^2 = \dfrac{1}{2}mv_f^2 + mgh_f. \nonumber\] This form of the equation means that the spring’s initial potential energy is converted partly to gravitational potential energy and partly to kinetic energy. The final speed at the top of the slope will be less than at the bottom. Solving for $v_f$ and substituting known values gives \[ \begin{align} v_f = \sqrt{\dfrac{kx^2}{m} - 2gh_f} \\[5pt] &= \sqrt{\left(\dfrac{250N/m}{0.100kg}\right)(0.0400m)^2 - 2(9.80 m/s^2)(0.180m)} \\[5pt] &= 0.687 \, m/s \end{align}\] Discussion Another way to solve this problem is to realize that the car’s kinetic energy before it goes up the slope is converted partly to potential energy—that is, to take the final conditions in part (a) to be the initial conditions in part (b). Note that, for conservative forces, we do not directly calculate the work they do; rather, we consider their effects through their corresponding potential energies, just as we did in Example. Note also that we do not consider details of the path taken—only the starting and ending points are important (as long as the path is not impossible). This assumption is usually a tremendous simplification, because the path may be complicated and forces may vary along the way. PHET Explorations: Energy Skate Park Learn about conservation of energy with a skater dude! Build tracks, ramps and jumps for the skater and view the kinetic energy, potential energy and friction as he moves. You can also take the skater to different planets or even space! Summary - A conservative force is one for which work depends only on the starting and ending points of a motion, not on the path taken. - We can define potential energy $(PE$ for any conservative force, just as we defined $PE_g$ for the gravitational force. - The potential energy of a spring is $PE_s = \frac{1}{2}kx^2$, where $k$ is the spring’s force constant and \|(x\) is the displacement from its undeformed position. - Mechanical energy is defined to be $KE = PE$ for conservative force. - When only conservative forces act on and within a system, the total mechanical energy is constant. In equation form, \[ KE + PE = \, constant \] or \[KE_i +PE_i = KE_f + PE_f \] where i and f denote initial and final values. This is known as the conservation of mechanical energy. Glossary - conservative force - a force that does the same work for any given initial and final configuration, regardless of the path followed - potential energy - energy due to position, shape, or configuration - potential energy of a spring - the stored energy of a spring as a function of its displacement; when Hooke’s law applies, it is given by the expression $\frac{1}{2}kx^2$ where $x$ is the distance the spring is compressed or extended and $k$ is the spring constant - conservation of mechanical energy - the rule that the sum of the kinetic energies and potential energies remains constant if only conservative forces act on and within a system - mechanical energy - the sum of kinetic energy and potential energy	libretexts	2025-03-17T19:53:26.342529	2015-11-01T04:57:57	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.04%3A_Conservative_Forces_and_Potential_Energy", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "7.4: Conservative Forces and Potential Energy", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.05%3A_Nonconservative_Forces	7.5: Nonconservative Forces Learning Objectives By the end of this section, you will be able to: - Define nonconservative forces and explain how they affect mechanical energy. - Show how the principle of conservation of energy can be applied by treating the conservative forces in terms of their potential energies and any nonconservative forces in terms of the work they do. Nonconservative Forces and Friction Forces are either conservative or nonconservative. A nonconservative force is one for which work depends on the path taken. Friction is a good example of a nonconservative force. As illustrated in Figure $\PageIndex{1}$, work done against friction depends on the length of the path between the starting and ending points. Because of this dependence on path, there is no potential energy associated with nonconservative forces. An important characteristic is that the work done by a nonconservative force adds or removes mechanical energy from a system . Friction , for example, creates thermal energy that dissipates, removing energy from the system. Furthermore, even if the thermal energy is retained or captured, it cannot be fully converted back to work, so it is lost or not recoverable in that sense as well. How Nonconservative Forces Affect Mechanical Energy Mechanical energy may not be conserved when nonconservative forces act. For example, when a car is brought to a stop by friction on level ground, it loses kinetic energy, which is dissipated as thermal energy, reducing its mechanical energy. Figure compares the effects of conservative and nonconservative forces. We often choose to understand simpler systems such as that described in Figure $\PageIndex{2a}$ first before studying more complicated systems as in Figure $\PageIndex{2b}$. How the Work-Energy Theorem Applies Now let us consider what form the work-energy theorem takes when both conservative and nonconservative forces act. We will see that the work done by nonconservative forces equals the change in the mechanical energy of a system. As noted in Kinetic Energy and the Work-Energy Theorem , the work-energy theorem states that the net work on a system equals the change in its kinetic energy, or $ W_{net} = \Delta KE$. The net work is the sum of the work by nonconservative forces plus the work by conservative forces. That is, \[W_{net} = W_{nc} + W_c,\] so that \[W_{nc} + W_c = \Delta KE,\] where $W_{nc}$ is the total work done by all nonconservative forces and $W_c$ is the total work done by all conservative forces. Consider Figure $\PageIndex{3}$, in which a person pushes a crate up a ramp and is opposed by friction. As in the previous section, we note that work done by a conservative force comes from a loss of gravitational potential energy, so that $W_c = -\Delta PE$. Substituting this equation into the previous one and solving for $W_{nc}$ gives \[W_{nc} = \Delta KE + \Delta PE.\label{eq4}\] This equation means that the total mechanical energy $(KE + PE)$ changes by exactly the amount of work done by nonconservative forces. In Figure $\PageIndex{3}$, this is the work done by the person minus the work done by friction. So even if energy is not conserved for the system of interest (such as the crate), we know that an equal amount of work was done to cause the change in total mechanical energy. We rearrange Equation \ref{eq5} to obtain \[KE_i + PE_i + W_{nc} = KE_f + PE_f \label{eq5}\] This means that the amount of work done by nonconservative forces adds to the mechanical energy of a system. If $W_{nc}$ is positive, then mechanical energy is increased, such as when the person pushes the crate up the ramp in Figure. If $W_{nc} $ is negative, then mechanical energy is decreased, such as when the rock hits the ground in Figure $\PageIndex{2b}$. If $W_{nc}$ is negative, then mechanical energy is decreased, such as when the rock hits the ground in Figure $\PageIndex{2b}$. If $W_{nc}$ is zero, then mechanical energy is conserved, and nonconservative forces are balanced. For example, when you push a lawn mower at constant speed on level ground, your work done is removed by the work of friction, and the mower has a constant energy. Applying Energy Conservation with Nonconservative Forces When no change in potential energy occurs, applying Equation \ref{eq5} amounts to applying the work-energy theorem by setting the change in kinetic energy to be equal to the net work done on the system, which in the most general case includes both conservative and nonconservative forces. But when seeking instead to find a change in total mechanical energy in situations that involve changes in both potential and kinetic energy, the previous equation $KE_i + PE_i + W_{nc} = KE_f + PE_f $ says that you can start by finding the change in mechanical energy that would have resulted from just the conservative forces, including the potential energy changes, and add to it the work done, with the proper sign, by any nonconservative forces involved. Example $\PageIndex{1}$: Calculating Distance Traveled: How Far a Baseball Player Slides Consider the situation shown in Figure $\PageIndex{4}$, where a baseball player slides to a stop on level ground. Using energy considerations, calculate the distance the 65.0-kg baseball player slides, given that his initial speed is 6.00 m/s and the force of friction against him is a constant 450 N. Strategy Friction stops the player by converting his kinetic energy into other forms, including thermal energy. In terms of the work-energy theorem, the work done by friction, which is negative, is added to the initial kinetic energy to reduce it to zero. The work done by friction is negative, because f is in the opposite direction of the motion (that is, $\theta = 180^o $, and so $cos \, \theta = -1$). Thus, $W_{nc} = -fd $. The equation simplifies to \[\dfrac{1}{2} mv_i^2 - bfd = 0 \nonumber\] or \[fd = \dfrac{1}{2}mv_i^2. \nonumber\] This equation can now be solved for the distance $d$. Solution Solving the previous equation for $d$ and substituting known values yields \[\begin{align} d &= \dfrac{mv_i^2}{2f} \\[5pt] &= \dfrac{65.0 \, kg)(6.00 \, m/s)^2}{(2)(450 \, N)} \\[5pt] &= 2.60 \, m. \end{align}\] Discussion The most important point of this example is that the amount of nonconservative work equals the change in mechanical energy. For example, you must work harder to stop a truck, with its large mechanical energy, than to stop a mosquito. Example $\PageIndex{2}$: Calculating Distance Traveled: Sliding Up an Incline Suppose that the player from Example $\PageIndex{1}$ is running up a hill having a $5.00^o$ incline upward with a surface similar to that in the baseball stadium. The player slides with the same initial speed, and the frictional force is still 450 N. Determine how far he slides. Strategy In this case, the work done by the nonconservative friction force on the player reduces the mechanical energy he has from his kinetic energy at zero height, to the final mechanical energy he has by moving through distance $d$ to reach height $h$ along the hill, with $h = d \, \sin \, 5.00^o$. This is expressed by Equation \ref{eq5} \[KE_i + PE_i + W_{nc} = KE_f + PE_f . \nonumber\] Solution The work done by friction is again $W_{nc} = -fd$; initially the potential energy is $PE_i = mg \cdot 0 = 0$ and the kinetic energy is $KE_i = \frac{1}{2} \, mv^2$; the final energy contributions are $KE_f = 0$ for the kinetic energy and $PE_f = mgh = mgd \, sin \, \theta$ for the potential energy. Substituting these values gives \[ \nonumber \dfrac{1}{2} mv_i^2 + 0 + (-fd) = 0 + mgd \, \sin \, \theta.\] Solve this for $d$ to obtain \[\begin{align} d &= \dfrac{\frac{1}{2} mv_i^2}{f + mg \, \sin \, \theta} \\[5pt] &= \dfrac{(0.5)(65.0 \, kg)(6.00 \, m/s)^2}{450 \, N + (65.0 \, kg)(9.80 \, m/s^2)sin \, (5.00^o)} \\[5pt] &= 2.31 \, m. \end{align}\] Discussion As might have been expected, the player slides a shorter distance by sliding uphill. Note that the problem could also have been solved in terms of the forces directly and the work energy theorem, instead of using the potential energy. This method would have required combining the normal force and force of gravity vectors, which no longer cancel each other because they point in different directions, and friction, to find the net force. You could then use the net force and the net work to find the distance $d$ that reduces the kinetic energy to zero. By applying conservation of energy and using the potential energy instead, we need only consider the gravitational potential energy $mgh$, without combining and resolving force vectors. This simplifies the solution considerably. Making Connections: Take-Home Investigation - Determining Friction from the Stopping Distance This experiment involves the conversion of gravitational potential energy into thermal energy. Use the ruler, book, and marble from Take-Home Investigation—Converting Potential to Kinetic Energy . In addition, you will need a foam cup with a small hole in the side, as shown in Figure $\PageIndex{6}$. From the 10-cm position on the ruler, let the marble roll into the cup positioned at the bottom of the ruler. Measure the distance $d$ the cup moves before stopping. What forces caused it to stop? What happened to the kinetic energy of the marble at the bottom of the ruler? Next, place the marble at the 20-cm and the 30-cm positions and again measure the distance the cup moves after the marble enters it. Plot the distance the cup moves versus the initial marble position on the ruler. Is this relationship linear? With some simple assumptions, you can use these data to find the coefficient of kinetic friction $\mu_k$ of the cup on the table. The force of friction $f$ on the cup is $\mu_k N$, where the normal force $N$ is just the weight of the cup plus the marble. The normal force and force of gravity do no work because they are perpendicular to the displacement of the cup, which moves horizontally. The work done by friction is $fd$. You will need the mass of the marble as well to calculate its initial kinetic energy. It is interesting to do the above experiment also with a steel marble (or ball bearing). Releasing it from the same positions on the ruler as you did with the glass marble, is the velocity of this steel marble the same as the velocity of the marble at the bottom of the ruler? Is the distance the cup moves proportional to the mass of the steel and glass marbles? Phet Explorations: The Ramp Explore forces, energy and work as you push household objects up and down a ramp . Lower and raise the ramp to see how the angle of inclination affects the parallel forces acting on the file cabinet. Graphs show forces, energy and work. Summary - A nonconservative force is one for which work depends on the path. - Friction is an example of a nonconservative force that changes mechanical energy into thermal energy. - Work $W_{nc}$ done by a nonconservative force changes the mechanical energy of a system. In equation form, $W_{nc} = \Delta KE + \Delta PE $ or, equivalently, $KE_i + PE_i + W_{nc} = KE_f + PE_f .$ - When both conservative and nonconservative forces act, energy conservation can be applied and used to calculate motion in terms of the known potential energies of the conservative forces and the work done by nonconservative forces, instead of finding the net work from the net force, or having to directly apply Newton’s laws Glossary - nonconservative force - a force whose work depends on the path followed between the given initial and final configurations - friction - the force between surfaces that opposes one sliding on the other; friction changes mechanical energy into thermal energy	libretexts	2025-03-17T19:53:26.441572	2015-11-01T04:58:19	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.05%3A_Nonconservative_Forces", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "7.5: Nonconservative Forces", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.06%3A_Conservation_of_Energy	7.6: Conservation of Energy Learning Objectives By the end of this section, you will be able to: - Explain the law of the conservation of energy. - Describe some of the many forms of energy. - Define efficiency of an energy conversion process as the fraction left as useful energy or work, rather than being transformed, for example, into thermal energy. Energy, as we have noted, is conserved, making it one of the most important physical quantities in nature. The law of conservation of energy can be stated as follows: We have explored some forms of energy and some ways it can be transferred from one system to another. This exploration led to the definition of two major types of energy—mechanical energy $(KE + PE)$ and energy transferred via work done by nonconservative forces $(W_{nc})$ But energy takes many other forms, manifesting itself in many different ways, and we need to be able to deal with all of these before we can write an equation for the above general statement of the conservation of energy. Other Forms of Energy than Mechanical Energy At this point, we deal with all other forms of energy by lumping them into a single group called other energy $(OE)$. Then we can state the conservation of energy in equation form as \[KE_i + PE_i + W_{nc} + OE_i = KE_f + PE_f + OE_f .\] All types of energy and work can be included in this very general statement of conservation of energy. Kinetic energy is $KE$, work done by a conservative force is represented by $PE$, work done by nonconservative forces is $W_{nc}$ and all other energies are included as $OE$. This equation applies to all previous examples; in those situations $OE$ was constant, and so it subtracted out and was not directly considered. Usefulness of the Energy Conservation Principle The fact that energy is conserved and has many forms makes it very important. You will find that energy is discussed in many contexts, because it is involved in all processes. It will also become apparent that many situations are best understood in terms of energy and that problems are often most easily conceptualized and solved by considering energy. When does $OE$ play a role? One example occurs when a person eats. Food is oxidized with the release of carbon dioxide, water, and energy. Some of this chemical energy is converted to kinetic energy when the person moves, to potential energy when the person changes altitude, and to thermal energy (another form of $OE$). Some of the Many Forms of Energy What are some other forms of energy? You can probably name a number of forms of energy not yet discussed. Many of these will be covered in later chapters, but let us detail a few here. Electrical energy is a common form that is converted to many other forms and does work in a wide range of practical situations. Fuels, such as gasoline and food, carry chemical energy that can be transferred to a system through oxidation. Chemical fuel can also produce electrical energy, such as in batteries. Batteries can in turn produce light, which is a very pure form of energy. Most energy sources on Earth are in fact stored energy from the energy we receive from the Sun. We sometimes refer to this as radiant energy , or electromagnetic radiation, which includes visible light, infrared, and ultraviolet radiation. Nuclear energy comes from processes that convert measurable amounts of mass into energy. Nuclear energy is transformed into the energy of sunlight, into electrical energy in power plants, and into the energy of the heat transfer and blast in weapons. Atoms and molecules inside all objects are in random motion. This internal mechanical energy from the random motions is called thermal energy , because it is related to the temperature of the object. These and all other forms of energy can be converted into one another and can do work. Table gives the amount of energy stored, used, or released from various objects and in various phenomena. The range of energies and the variety of types and situations is impressive. Problem-Solving Strategies for Energy You will find the following problem-solving strategies useful whenever you deal with energy. The strategies help in organizing and reinforcing energy concepts. In fact, they are used in the examples presented in this chapter. The familiar general problem-solving strategies presented earlier—involving identifying physical principles, knowns, and unknowns, checking units, and so on —continue to be relevant here. Step 1. Determine the system of interest and identify what information is given and what quantity is to be calculated. A sketch will help. Step 2. Examine all the forces involved and determine whether you know or are given the potential energy from the work done by the forces. Then use step 3 or step 4. Step 3. If you know the potential energies for the forces that enter into the problem, then forces are all conservative, and you can apply conservation of mechanical energy simply in terms of potential and kinetic energy. The equation expressing conservation of energy is \[KE_i + PE_i = KE_f + PE_f.\] Step 4. If you know the potential energy for only some of the forces, possibly because some of them are nonconservative and do not have a potential energy, or if there are other energies that are not easily treated in terms of force and work, then the conservation of energy law in its most general form must be used. \[KE_i + PE_i + W_{nc} + OE_i = KE_f + PE_f + OE_f.\] In most problems, one or more of the terms is zero, simplifying its solution. Do not calculate $W_c$, the work done by conservative forces; it is already incorporated in the $PE$ terms. Step 5. You have already identified the types of work and energy involved (in step 2). Before solving for the unknown, eliminate terms wherever possible to simplify the algebra. For example, choose $h = 0$ at either the initial or final point, so that $PE_g$ is zero there. Then solve for the unknown in the customary manner. Step 6. Check the answer to see if it is reasonable . Once you have solved a problem, reexamine the forms of work and energy to see if you have set up the conservation of energy equation correctly. For example, work done against friction should be negative, potential energy at the bottom of a hill should be less than that at the top, and so on. Also check to see that the numerical value obtained is reasonable. For example, the final speed of a skateboarder who coasts down a 3-m-high ramp could reasonably be 20 km/h, but not 80 km/h. Transformation of Energy The transformation of energy from one form into others is happening all the time. The chemical energy in food is converted into thermal energy through metabolism; light energy is converted into chemical energy through photosynthesis. In a larger example, the chemical energy contained in coal is converted into thermal energy as it burns to turn water into steam in a boiler. This thermal energy in the steam in turn is converted to mechanical energy as it spins a turbine, which is connected to a generator to produce electrical energy. (In all of these examples, not all of the initial energy is converted into the forms mentioned. This important point is discussed later in this section.) Another example of energy conversion occurs in a solar cell. Sunlight impinging on a solar cell (Figure 7.7.1) produces electricity, which in turn can be used to run an electric motor. Energy is converted from the primary source of solar energy into electrical energy and then into mechanical energy. \| Object/phenomenon \| Energy in joules \| \|---\|---\| \| Big Bang \| $10^{68}$ \| \| Energy released in a supernova \| $10^{44}$ \| \| Fusion of all the hydrogen in Earth’s oceans \| $10^{34}$ \| \| Annual world energy use \| $4 \times 10^{20}$ \| \| Large fusion bomb (9 megaton) \| $3.8 \times 10^{16}$ \| \| 1 kg hydrogen (fusion to helium) \| $6.4 \times 10^{14}$ \| \| 1 kg uranium (nuclear fission) \| $8.0 \times 10^{13}$ \| \| Hiroshima-size fission bomb (10 kiloton) \| $4.2 \times 10^{13}$ \| \| 90,000-ton aircraft carrier at 30 knots \| $1.1 \times 10^{10}$ \| \| 1 barrel crude oil \| $5.9 \times 10^9$ \| \| 1 ton TNT \| $4.2 \times 10^9$ \| \| 1 gallon of gasoline \| $1.2 \times 10^8$ \| \| Daily home electricity use (developed countries) \| $7n\times 10^7$ \| \| Daily adult food intake (recommended) \| $1.2 \times 10^7$ \| \| 1000-kg car at 90 km/h \| $3.1 \times 10^5$ \| \| 1 g fat (9.3 kcal) \| $3.9 \times 10^4$ \| \| ATP hydrolysis reaction \| $3.2 \times 10^4$ \| \| 1 g carbohydrate (4.1 kcal) \| $1.7 \times 10^4$ \| \| 1 g protein (4.1 kcal) \| $1.7 \times 10^4$ \| \| Tennis ball at 100 km/h \| $22$ \| \| Mosquito 10 –2 g at 0.5 m/s ) \| $1.3 \times 10^{-6}$ \| \| Single electron in a TV tube beam \| $4.0 \times 10^{-15}$ \| \| Energy to break one DNA strand \| $4.0 \times 10^{-19}$ \| Efficiency Even though energy is conserved in an energy conversion process, the output of useful energy or work will be less than the energy input. The efficiency $E_{ff}$ of an energy conversion process is defined as \[Efficiency \, (E_{ff}) = \dfrac{useful \, energy \, or \, work \, output}{total \, energy \, input} = \dfrac{W_{out}}{E_{in}}.\] Table lists some efficiencies of mechanical devices and human activities. In a coal-fired power plant, for example, about 40% of the chemical energy in the coal becomes useful electrical energy. The other 60% transforms into other (perhaps less useful) energy forms, such as thermal energy, which is then released to the environment through combustion gases and cooling towers. \| Activity/device \| Efficiency (%) \| \|---\|---\| \| Cycling and climbing \| 20 \| \| Swimming, surface \| 2 \| \| Swimming, submerged \| 4 \| \| Shoveling \| 3 \| \| Weightlifting \| 9 \| \| Steam engine \| 17 \| \| Gasoline engine \| 30 \| \| Diesel engine \| 35 \| \| Nuclear power plant \| 35 \| \| Coal power plant \| 42 \| \| Electric motor \| 98 \| \| Compact fluorescent light \| 20 \| \| Gas heater (residential) \| 90 \| \| Solar cell \| 10 \| Efficiency of the Human Body and Mechanical Devices PhET Explorations: Masses and Springs A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can even slow time. Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energies for each spring. Summary - The law of conservation of energy states that the total energy is constant in any process. Energy may change in form or be transferred from one system to another, but the total remains the same. - When all forms of energy are considered, conservation of energy is written in equation form as \[KE_i + PE_i + W_{nc} + OE_i = KE_f + PE_f + OE_f ,\] where $OE$ is all other forms of energy besides mechanical energy. - Commonly encountered forms of energy include electric energy, chemical energy, radiant energy, nuclear energy, and thermal energy. - Energy is often utilized to do work, but it is not possible to convert all the energy of a system to work. The efficiency $E_{ff}$ of a machine or human is defined to be $E_{ff} = \frac{W_{out}}{E_{in}},$ where $W_{out}$ is useful work output and $E_{in}$ s the energy consumed. Glossary - law of conservation of energy - the general law that total energy is constant in any process; energy may change in form or be transferred from one system to another, but the total remains the same - electrical energy - the energy carried by a flow of charge - chemical energy - the energy in a substance stored in the bonds between atoms and molecules that can be released in a chemical reaction - radiant energy - the energy carried by electromagnetic waves - nuclear energy - energy released by changes within atomic nuclei, such as the fusion of two light nuclei or the fission of a heavy nucleus - thermal energy - the energy within an object due to the random motion of its atoms and molecules that accounts for the object's temperature - efficiency - a measure of the effectiveness of the input of energy to do work; useful energy or work divided by the total input of energy	libretexts	2025-03-17T19:53:26.538731	2015-11-01T04:58:46	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.06%3A_Conservation_of_Energy", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "7.6: Conservation of Energy", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.07%3A_Power	7.7: Power Learning Objectives By the end of this section, you will be able to: - Calculate power by calculating changes in energy over time. - Examine power consumption and calculations of the cost of energy consumed. What is Power? Power —the word conjures up many images: a professional football player muscling aside his opponent, a dragster roaring away from the starting line, a volcano blowing its lava into the atmosphere, or a rocket blasting off, as in Figure. These images of power have in common the rapid performance of work, consistent with the scientific definition of power $P$ as the rate at which work is done. Power Power is the rate at which work is done. \[P = \dfrac{W}{t}\] The SI unit for power is the watt $W$, where 1 watt equals 1 joule/second $(1 \, W = 1 \, J/s)$. Because work is energy transfer, power is also the rate at which energy is expended. A 60-W light bulb, for example, expends 60 J of energy per second. Great power means a large amount of work or energy developed in a short time. For example, when a powerful car accelerates rapidly, it does a large amount of work and consumes a large amount of fuel in a short time. Calculating Power from Energy Example $\PageIndex{1}$: Calculating the Power to Climb Stairs What is the power output for a 60.0-kg woman who runs up a 3.00 m high flight of stairs in 3.50 s, starting from rest but having a final speed of 2.00 m/s? (See Figure.) Strategy and Concept The work going into mechanical energy is $W = KE + PE$. At the bottom of the stairs, we take both $KE$ and $PE$ as initially zero; thus $W = KE_f + PE_g = \frac{1}{2}mv_f^2 + mgh,$ where $h$ is the vertical height of the stairs. Because all terms are given, we can calculate $W$ and then divide it by time to get power. Solution Substituting the expression for $W$ into the definition of power given in the previous equation, $P = W/t$ yields \[W = \dfrac{W}{t} = \dfrac{\frac{1}{2}mv_f^2 + mgh}{t}.\] Entering known values yields \[P = \dfrac{0.5(60 \, kg)(2.00 m/s^2) + (60.0 \, kg)(9.80 m/s^2)(3.00 \, m)}{3.50 \, s}\] \[= \dfrac{120 \, J + 1764 \, J}{3.50 \, s}\] \[= 538 \, W.\] Discussion The woman does 1764 J of work to move up the stairs compared with only 120 J to increase her kinetic energy; thus, most of her power output is required for climbing rather than accelerating. It is impressive that this woman’s useful power output is slightly less than 1 horsepower $(1 \, hp = 746 \, W)$. People can generate more than a horsepower with their leg muscles for short periods of time by rapidly converting available blood sugar and oxygen into work output. (A horse can put out 1 hp for hours on end.) Once oxygen is depleted, power output decreases and the person begins to breathe rapidly to obtain oxygen to metabolize more food—this is known as the aerobic stage of exercise. If the woman climbed the stairs slowly, then her power output would be much less, although the amount of work done would be the same. Making Connections: Take-Home Investigation —Measure Your Power Rating - Determine your own power rating by measuring the time it takes you to climb a flight of stairs. We will ignore the gain in kinetic energy, as the above example showed that it was a small portion of the energy gain. Don’t expect that your output will be more than about 0.5 hp. Examples of Power Examples of power are limited only by the imagination, because there are as many types as there are forms of work and energy. (See Table for some examples.) Sunlight reaching Earth’s surface carries a maximum power of about 1.3 kilowatts per square meter $kW/m^2$ A tiny fraction of this is retained by Earth over the long term. Our consumption rate of fossil fuels is far greater than the rate at which they are stored, so it is inevitable that they will be depleted. Power implies that energy is transferred, perhaps changing form. It is never possible to change one form completely into another without losing some of it as thermal energy. For example, a 60-W incandescent bulb converts only 5 W of electrical power to light, with 55 W dissipating into thermal energy. Furthermore, the typical electric power plant converts only 35 to 40% of its fuel into electricity. The remainder becomes a huge amount of thermal energy that must be dispersed as heat transfer, as rapidly as it is created. A coal-fired power plant may produce 1000 megawatts; 1 megawatt (MW) is $10^6$ of electric power. But the power plant consumes chemical energy at a rate of about 2500 MW, creating heat transfer to the surroundings at a rate of 1500 MW. (See Figure.) \| Object or Phenomenon \| Power in Watts \| \|---\|---\| \| Supernova (at peak) \| $5 \times 10^{37}$ \| \| Milky Way galaxy \| $10^{37}$ \| \| Crab Nebula pulsar \| $10^{28}$ \| \| The Sun \| $4 \times 10^{26}$ \| \| Volcanic eruption (maximum) \| $4 \times 10^{15}$ \| \| Lightning bolt \| $2 \times 10^{12}$ \| \| Nuclear power plant (total electric and heat transfer) \| $3 \times 10^9$ \| \| Aircraft carrier (total useful and heat transfer) \| $10^8$ \| \| Dragster (total useful and heat transfer) \| $2 \times 10^6$ \| \| Car (total useful and heat transfer) \| $8 \times 10^4$ \| \| Football player (total useful and heat transfer) \| $5 \times 10^3$ \| \| Clothes dryer \| $4 \times 10^3$ \| \| Person at rest (all heat transfer) \| $100$ \| \| Typical incandescent light bulb (total useful and heat transfer) \| $60$ \| \| Heart, person at rest (total useful and heat transfer) \| $8$ \| \| Electric clock \| $3$ \| \| Pocket calculator \| $10^{-3}$ \| Power and Energy Consumption We usually have to pay for the energy we use. It is interesting and easy to estimate the cost of energy for an electrical appliance if its power consumption rate and time used are known. The higher the power consumption rate and the longer the appliance is used, the greater the cost of that appliance. The power consumption rate is $P = \frac{W}{t} = \frac{E}{t}$, where $E$ is the energy supplied by the electricity company. So the energy consumed over a time $t$ is \[E = Pt.\] Electricity bills state the energy used in units of kilowatt-hours $(kW \cdot h)$, which is the product of power in kilowatts and time in hours. This unit is convenient because electrical power consumption at the kilowatt level for hours at a time is typical. Example $\PageIndex{2}$: Calculating Energy Costs What is the cost of running a 0.200-kW computer 6.00 h per day for 30.0 d if the cost of electricity is $0.120 per $kW \cdot h$? Strategy Cost is based on energy consumed; thus, we must find $E$ from $E = Pt$ and then calculate the cost. Because electrical energy is expressed in $kW \cdot h$ at the start of a problem such as this it is convenient to convert the units into $kW$ and hours. Solution The energy consumed in $kW \cdot h$ is \[E = Pt = (0.200 \, kW)(6.00 \, h/d)(30.0 \, d)\] \[= 36.0 \, kW \cdot h,\] and the cost is simply given by \[cost = (36.0 \, kW \cdot h)($0.120 \, per \, kW \cdot h) = $4.32 \, per \, month.\] Discussion The cost of using the computer in this example is neither exorbitant nor negligible. It is clear that the cost is a combination of power and time. When both are high, such as for an air conditioner in the summer, the cost is high. The motivation to save energy has become more compelling with its ever-increasing price. Armed with the knowledge that energy consumed is the product of power and time, you can estimate costs for yourself and make the necessary value judgments about where to save energy. Either power or time must be reduced. It is most cost-effective to limit the use of high-power devices that normally operate for long periods of time, such as water heaters and air conditioners. This would not include relatively high power devices like toasters, because they are on only a few minutes per day. It would also not include electric clocks, in spite of their 24-hour-per-day usage, because they are very low power devices. It is sometimes possible to use devices that have greater efficiencies—that is, devices that consume less power to accomplish the same task. One example is the compact fluorescent light bulb, which produces over four times more light per watt of power consumed than its incandescent cousin. Modern civilization depends on energy, but current levels of energy consumption and production are not sustainable. The likelihood of a link between global warming and fossil fuel use (with its concomitant production of carbon dioxide), has made reduction in energy use as well as a shift to non-fossil fuels of the utmost importance. Even though energy in an isolated system is a conserved quantity, the final result of most energy transformations is waste heat transfer to the environment, which is no longer useful for doing work. As we will discuss in more detail in Thermodynamics , the potential for energy to produce useful work has been “degraded” in the energy transformation. Summary - Power is the rate at which work is done, or in equation form, for the average power $P$ for work $W$ done over a time $t$, $P = W/t$. - The SI unit for power is the watt (W), where $1 \, W = 1 \, J/s$. - The power of many devices such as electric motors is also often expressed in horsepower (hp), where $1\space hp = 746 \, W.$ Glossary - power - the rate at which work is done - watt - (W) SI unit of power, with $1 W=1 J/s$ - horsepower - an older non-SI unit of power, with $1 hp=746 W$ - kilowatt-hour - ($kW⋅h$) unit used primarily for electrical energy provided by electric utility companies	libretexts	2025-03-17T19:53:26.679941	2015-11-01T04:59:04	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.07%3A_Power", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "7.7: Power", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.08%3A_Work_Energy_and_Power_in_Humans	7.8: Work, Energy, and Power in Humans Learning Objectives By the end of this section, you will be able to: - Explain the human body’s consumption of energy when at rest vs. when engaged in activities that do useful work. - Calculate the conversion of chemical energy in food into useful work. Energy Conversion in Humans Our own bodies, like all living organisms, are energy conversion machines. Conservation of energy implies that the chemical energy stored in food is converted into work, thermal energy, and/or stored as chemical energy in fatty tissue. (Figure 7.09.1.) The fraction going into each form depends both on how much we eat and on our level of physical activity. If we eat more than is needed to do work and stay warm, the remainder goes into body fat. Power Consumed at Rest The rate at which the body uses food energy to sustain life and to do different activities is called the metabolic rate . The total energy conversion rate of a person at rest is called the basal metabolic rate (BMR) and is divided among various systems in the body, as shown in Table. The largest fraction goes to the liver and spleen, with the brain coming next. Of course, during vigorous exercise, the energy consumption of the skeletal muscles and heart increase markedly. About 75% of the calories burned in a day go into these basic functions. The BMR is a function of age, gender, total body weight, and amount of muscle mass (which burns more calories than body fat). Athletes have a greater BMR due to this last factor. \| Organ \| Power consumed at rest (W) \| Oxygen consumption (mL/min) \| Percent of BMR \| \|---\|---\|---\|---\| \| Liver & spleen \| 23 \| 67 \| 27 \| \| Brain \| 16 \| 47 \| 19 \| \| Skeletal muscle \| 15 \| 45 \| 18 \| \| Kidney \| 9 \| 26 \| 10 \| \| Heart \| 6 \| 17 \| 7 \| \| Other \| 16 \| 48 \| 19 \| \| Totals \| 85 W \| 250 mL/min \| 100% \| Energy consumption is directly proportional to oxygen consumption because the digestive process is basically one of oxidizing food. We can measure the energy people use during various activities by measuring their oxygen use. (See Figure7.09.1.) Approximately 20 kJ of energy are produced for each liter of oxygen consumed, independent of the type of food. Table shows energy and oxygen consumption rates (power expended) for a variety of activities. Power of Doing Useful Work Work done by a person is sometimes called useful work , which is work done on the outside world , such as lifting weights. Useful work requires a force exerted through a distance on the outside world, and so it excludes internal work, such as that done by the heart when pumping blood. Useful work does include that done in climbing stairs or accelerating to a full run, because these are accomplished by exerting forces on the outside world. Forces exerted by the body are non-conservative, so that they can change the mechanical energy $(KE + PE)$ of the system worked upon, and this is often the goal. A baseball player throwing a ball, for example, increases both the ball’s kinetic and potential energy. If a person needs more energy than they consume, such as when doing vigorous work, the body must draw upon the chemical energy stored in fat. So exercise can be helpful in losing fat. However, the amount of exercise needed to produce a loss in fat, or to burn off extra calories consumed that day, can be large, as Example 7.09.1 illustrates. Example $\PageIndex{1}$: Calculating Weight Loss from Exercising If a person who normally requires an average of 12,000 kJ (3000 kcal) of food energy per day consumes 13,000 kJ per day, he will steadily gain weight. How much bicycling per day is required to work off this extra 1000 kJ? Solution Table states that 400 W are used when cycling at a moderate speed. The time required to work off 1000 kJ at this rate is then \[Time = \dfrac{energy}{\left(\frac{energy}{time} \right)} = \dfrac{1000 \, kJ}{400 \, W} = 2500 \, s = 42 \, min.\] Discussion If this person uses more energy than he or she consumes, the person’s body will obtain the needed energy by metabolizing body fat. If the person uses 13,000 kJ but consumes only 12,000 kJ, then the amount of fat loss will be \[ Fat \, loss = (1000 \, kJ) \left ( \frac{1 \, g \, fat}{30 \, kJ} \right) = 26 \, g,\] assuming the energy content of fat to be 39 kJ/g. Energy and Oxygen Consumption Rates: \| Activity \| Energy consumption in watts \| Oxygen consumption in liters O 2 /min \| \|---\|---\|---\| \| Sleeping \| 83 \| 0.24 \| \| Sitting at rest \| 120 \| 0.34 \| \| Standing relaxed \| 125 \| 0.36 \| \| Sitting in class \| 210 \| 0.60 \| \| Walking (5 km/h) \| 280 \| 0.80 \| \| Cycling (13–18 km/h) \| 400 \| 1.14 \| \| Shivering \| 425 \| 1.21 \| \| Playing tennis \| 440 \| 1.26 \| \| Swimming breaststroke \| 475 \| 1.36 \| \| Ice skating (14.5 km/h) \| 545 \| 1.56 \| \| Climbing stairs (116/min) \| 685 \| 1.96 \| \| Cycling (21 km/h) \| 700 \| 2.00 \| \| Running cross-country \| 740 \| 2.12 \| \| Playing basketball \| 800 \| 2.28 \| \| Cycling, professional racer \| 1855 \| 5.30 \| \| Sprinting \| 2415 \| 6.90 \| All bodily functions, from thinking to lifting weights, require energy. (See Figure 7.09.3.) The many small muscle actions accompanying all quiet activity, from sleeping to head scratching, ultimately become thermal energy, as do less visible muscle actions by the heart, lungs, and digestive tract. Shivering, in fact, is an involuntary response to low body temperature that pits muscles against one another to produce thermal energy in the body (and do no work). The kidneys and liver consume a surprising amount of energy, but the biggest surprise of all it that a full 25% of all energy consumed by the body is used to maintain electrical potentials in all living cells. (Nerve cells use this electrical potential in nerve impulses.) This bioelectrical energy ultimately becomes mostly thermal energy, but some is utilized to power chemical processes such as in the kidneys and liver, and in fat production. Summary - The human body converts energy stored in food into work, thermal energy, and/or chemical energy that is stored in fatty tissue. - The rate at which the body uses food energy to sustain life and to do different activities is called the metabolic rate, and the corresponding rate when at rest is called the basal metabolic rate (BMR) - The energy included in the basal metabolic rate is divided among various systems in the body, with the largest fraction going to the liver and spleen, and the brain coming next. - About 75% of food calories are used to sustain basic body functions included in the basal metabolic rate. - The energy consumption of people during various activities can be determined by measuring their oxygen use, because the digestive process is basically one of oxidizing food. Glossary - metabolic rate - the rate at which the body uses food energy to sustain life and to do different activities - basal metabolic rate - the total energy conversion rate of a person at rest - useful work - work done on an external system	libretexts	2025-03-17T19:53:26.778852	2015-11-01T04:59:29	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.08%3A_Work_Energy_and_Power_in_Humans", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "7.8: Work, Energy, and Power in Humans", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.09%3A_World_Energy_Use	7.9: World Energy Use Learning Objectives By the end of this section, you will be able to: - Describe the distinction between renewable and nonrenewable energy sources. - Explain why the inevitable conversion of energy to less useful forms makes it necessary to conserve energy resources. Energy is an important ingredient in all phases of society. We live in a very interdependent world, and access to adequate and reliable energy resources is crucial for economic growth and for maintaining the quality of our lives. But current levels of energy consumption and production are not sustainable. About 40% of the world’s energy comes from oil, and much of that goes to transportation uses. Oil prices are dependent as much upon new (or foreseen) discoveries as they are upon political events and situations around the world. The U.S., with 4.5% of the world’s population, consumes 24% of the world’s oil production per year; 66% of that oil is imported! Renewable and Nonrenewable Energy Sources The principal energy resources used in the world are shown in Figure 7.10.1. The fuel mix has changed over the years but now is dominated by oil, although natural gas and solar contributions are increasing. Renewable forms of energy are those sources that cannot be used up, such as water, wind, solar, and biomass. About 85% of our energy comes from nonrenewable fossil fuels —oil, natural gas, coal. The likelihood of a link between global warming and fossil fuel use, with its production of carbon dioxide through combustion, has made, in the eyes of many scientists, a shift to non-fossil fuels of utmost importance—but it will not be easy. The World’s Growing Energy Needs World energy consumption continues to rise, especially in the developing countries. (See Figure 7.10.1.) Global demand for energy has tripled in the past 50 years and might triple again in the next 30 years. While much of this growth will come from the rapidly booming economies of China and India, many of the developed countries, especially those in Europe, are hoping to meet their energy needs by expanding the use of renewable sources. Although presently only a small percentage, renewable energy is growing very fast, especially wind energy. For example, Germany plans to meet 20% of its electricity and 10% of its overall energy needs with renewable resources by the year 2020. (See 7.10.2.) Energy is a key constraint in the rapid economic growth of China and India. In 2003, China surpassed Japan as the world’s second largest consumer of oil. However, over 1/3 of this is imported. Unlike most Western countries, coal dominates the commercial energy resources of China, accounting for 2/3 of its energy consumption. In 2009 China surpassed the United States as the largest generator of $CO_2$ In India, the main energy resources are biomass (wood and dung) and coal. Half of India’s oil is imported. About 70% of India’s electricity is generated by highly polluting coal. Yet there are sizeable strides being made in renewable energy. India has a rapidly growing wind energy base, and it has the largest solar cooking program in the world. Table displays the 2006 commercial energy mix by country for some of the prime energy users in the world. While non-renewable sources dominate, some countries get a sizeable percentage of their electricity from renewable resources. For example, about 67% of New Zealand’s electricity demand is met by hydroelectric. Only 10% of the U.S. electricity is generated by renewable resources, primarily hydroelectric. It is difficult to determine total contributihttp://physwiki.ucdavis.ed..._Energy_Useons of renewable energy in some countries with a large rural population, so these percentages in this table are left blank. \| Country \| Consumption, in EJ (10 18 J) \| Oil \| Natural Gas \| Coal \| Nuclear \| Hydro \| Other Renewables \| Electricity Use per capita (kWh/yr) \| Energy Use per capita (GJ/yr) \| \|---\|---\|---\|---\|---\|---\|---\|---\|---\|---\| \| Australia \| 5.4 \| 34% \| 17% \| 44% \| 0% \| 3% \| 1% \| 10000 \| 260 \| \| Brazil \| 9.6 \| 48% \| 7% \| 5% \| 1% \| 35% \| 2% \| 2000 \| 50 \| \| China \| 63 \| 22% \| 3% \| 69% \| 1% \| 6% \| 1500 \| 35 \| \| \| Egypt \| 2.4 \| 50% \| 41% \| 1% \| 0% \| 6% \| 990 \| 32 \| \| \| Germany \| 16 \| 37% \| 24% \| 24% \| 11% \| 1% \| 3% \| 6400 \| 173 \| \| India \| 15 \| 34% \| 7% \| 52% \| 1% \| 5% \| 470 \| 13 \| \| \| Indonesia \| 4.9 \| 51% \| 26% \| 16% \| 0% \| 2% \| 3% \| 420 \| 22 \| \| Japan \| 24 \| 48% \| 14% \| 21% \| 12% \| 4% \| 1% \| 7100 \| 176 \| \| New Zealand \| 0.44 \| 32% \| 26% \| 6% \| 0% \| 11% \| 19% \| 8500 \| 102 \| \| Russia \| 31 \| 19% \| 53% \| 16% \| 5% \| 6% \| 5700 \| 202 \| \| \| U.S. \| 105 \| 40% \| 23% \| 22% \| 8% \| 3% \| 1% \| 12500 \| 340 \| \| World \| 432 \| 39% \| 23% \| 24% \| 6% \| 6% \| 2% \| 2600 \| 71 \| Energy and Economic Well-being The last two columns in this table examine the energy and electricity use per capita. Economic well-being is dependent upon energy use, and in most countries higher standards of living, as measured by GDP (gross domestic product) per capita, are matched by higher levels of energy consumption per capita. This is borne out in Figure 7.10.4. Increased efficiency of energy use will change this dependency. A global problem is balancing energy resource development against the harmful effects upon the environment in its extraction and use. Conserving Energy As we finish this chapter on energy and work, it is relevant to draw some distinctions between two sometimes misunderstood terms in the area of energy use. As has been mentioned elsewhere, the “law of the conservation of energy” is a very useful principle in analyzing physical processes. It is a statement that cannot be proven from basic principles, but is a very good bookkeeping device, and no exceptions have ever been found. It states that the total amount of energy in an isolated system will always remain constant. Related to this principle, but remarkably different from it, is the important philosophy of energy conservation. This concept has to do with seeking to decrease the amount of energy used by an individual or group through (1) reduced activities (e.g., turning down thermostats, driving fewer kilometers) and/or (2) increasing conversion efficiencies in the performance of a particular task—such as developing and using more efficient room heaters, cars that have greater miles-per-gallon ratings, energy-efficient compact fluorescent lights, etc. Since energy in an isolated system is not destroyed or created or generated, one might wonder why we need to be concerned about our energy resources, since energy is a conserved quantity. The problem is that the final result of most energy transformations is waste heat transfer to the environment and conversion to energy forms no longer useful for doing work. To state it in another way, the potential for energy to produce useful work has been “degraded” in the energy transformation. (This will be discussed in more detail in Thermodynamics .) Summary - The relative use of different fuels to provide energy has changed over the years, but fuel use is currently dominated by oil, although natural gas and solar contributions are increasing. - Although non-renewable sources dominate, some countries meet a sizeable percentage of their electricity needs from renewable resources. - The United States obtains only about 10% of its energy from renewable sources, mostly hydroelectric power. - Economic well-being is dependent upon energy use, and in most countries higher standards of living, as measured by GDP (Gross Domestic Product) per capita, are matched by higher levels of energy consumption per capita. - Even though, in accordance with the law of conservation of energy, energy can never be created or destroyed, energy that can be used to do work is always partly converted to less useful forms, such as waste heat to the environment, in all of our uses of energy for practical purposes. Glossary - renewable forms of energy - those sources that cannot be used up, such as water, wind, solar, and biomass - fossil fuels - oil, natural gas, and coal	libretexts	2025-03-17T19:53:26.894113	2015-11-01T04:59:53	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.09%3A_World_Energy_Use", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "7.9: World Energy Use", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.E%3A__Work_Energy_and_Energy_Resources_(Exercise)	7.E: Work, Energy, and Energy Resources (Exercise) Conceptual Questions 7.1: Work: The Scientific Definition 1. Give an example of something we think of as work in everyday circumstances that is not work in the scientific sense. Is energy transferred or changed in form in your example? If so, explain how this is accomplished without doing work. 2. Give an example of a situation in which there is a force and a displacement, but the force does no work. Explain why it does no work. 3. Describe a situation in which a force is exerted for a long time but does no work. Explain. 7.2: Kinetic Energy and the Work-Energy Theorem 4. The person in Figure does work on the lawn mower. Under what conditions would the mower gain energy? Under what conditions would it lose energy? 5. Work done on a system puts energy into it. Work done by a system removes energy from it. Give an example for each statement. 6. When solving for speed in Example, we kept only the positive root. Why? 7.3: Gravitational Potential Energy 7. In Example, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m below the start. We would find in that case that its final speed is the same as its initial speed. Explain in terms of conservation of energy. 8. Does the work you do on a book when you lift it onto a shelf depend on the path taken? On the time taken? On the height of the shelf? On the mass of the book? 7.4: Conservative Forces and Potential Energy 9. What is a conservative force? 10. The force exerted by a diving board is conservative, provided the internal friction is negligible. Assuming friction is negligible, describe changes in the potential energy of a diving board as a swimmer dives from it, starting just before the swimmer steps on the board until just after his feet leave it. 11. Define mechanical energy. What is the relationship of mechanical energy to nonconservative forces? What happens to mechanical energy if only conservative forces act? 12. What is the relationship of potential energy to conservative force? 7.6: Conservation of Energy 13. Consider the following scenario. A car for which friction is not negligible accelerates from rest down a hill, running out of gasoline after a short distance. The driver lets the car coast farther down the hill, then up and over a small crest. He then coasts down that hill into a gas station, where he brakes to a stop and fills the tank with gasoline. Identify the forms of energy the car has, and how they are changed and transferred in this series of events. (See Figure.) A car experiencing non-negligible friction coasts down a hill, over a small crest, then downhill again, and comes to a stop at a gas station. 14. Describe the energy transfers and transformations for a javelin, starting from the point at which an athlete picks up the javelin and ending when the javelin is stuck into the ground after being thrown. 15. Do devices with efficiencies of less than one violate the law of conservation of energy? Explain. 16. List four different forms or types of energy. Give one example of a conversion from each of these forms to another form. 17. List the energy conversions that occur when riding a bicycle. 7.7: Power 18. Most electrical appliances are rated in watts. Does this rating depend on how long the appliance is on? (When off, it is a zero-watt device.) Explain in terms of the definition of power. 19. Explain, in terms of the definition of power, why energy consumption is sometimes listed in kilowatt-hours rather than joules. What is the relationship between these two energy units? 20. A spark of static electricity, such as that you might receive from a doorknob on a cold dry day, may carry a few hundred watts of power. Explain why you are not injured by such a spark. 7.8: Work, Energy, and Power in Humans 21. Explain why it is easier to climb a mountain on a zigzag path rather than one straight up the side. Is your increase in gravitational potential energy the same in both cases? Is your energy consumption the same in both? 22. Do you do work on the outside world when you rub your hands together to warm them? What is the efficiency of this activity? 23. Shivering is an involuntary response to lowered body temperature. What is the efficiency of the body when shivering, and is this a desirable value? 24. Discuss the relative effectiveness of dieting and exercise in losing weight, noting that most athletic activities consume food energy at a rate of 400 to 500 W, while a single cup of yogurt can contain 1360 kJ (325 kcal). Specifically, is it likely that exercise alone will be sufficient to lose weight? You may wish to consider that regular exercise may increase the metabolic rate, whereas protracted dieting may reduce it. 7.9: World Energy Use 25. What is the difference between energy conservation and the law of conservation of energy? Give some examples of each. 26. If the efficiency of a coal-fired electrical generating plant is 35%, then what do we mean when we say that energy is a conserved quantity? Problems & Exercises 7.1: Work: The Scientific Definition 27. How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories. Solution $\displaystyle 3.00 J=7.17×10^{−4} kcal$ 28. A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. 29. (a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. (b) What is the work done on the lift by the gravitational force in this process? (c) What is the total work done on the lift? Solution (a) $\displaystyle 5.92×10^5J$ (b) $\displaystyle −5.88×10^5J$ (c) The net force is zero. 30. Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See [link] for the energy content of gasoline.) (a) What is the magnitude of the force exerted to keep the car moving at constant speed? (b) If the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s? 31. Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of $\displaystyle 20.0º$ with the horizontal. (See Figure.) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp. A man pushes a crate up a ramp. Solution $\displaystyle 3.14×10^3J$ 32. How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in Figure? Assume no friction acts on the wagon. The boy does work on the system of the wagon and the child when he pulls them as shown. 33. A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction $\displaystyle 25.0º$ below the horizontal. (a) What is the work done on the cart by friction? (b) What is the work done on the cart by the gravitational force? (c) What is the work done on the cart by the shopper? (d) Find the force the shopper exerts, using energy considerations. (e) What is the total work done on the cart? Solution (a) $\displaystyle −700J$ (b) 0 (c) 700 J (d) 38.6 N (e) 0 34. Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a $\displaystyle 60.0º$ slope at constant speed, as shown in Figure. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done? A rescue sled and victim are lowered down a steep slope. 7.2: Kinetic Energy and the Work-Energy Theorem 35. Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h. Solution $\displaystyle 1/250$ 36. (a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates. 37. Confirm the value given for the kinetic energy of an aircraft carrier in [link]. You will need to look up the definition of a nautical mile (1 knot = 1 nautical mile/h). Solution $\displaystyle 1.1×10^{10}J$ 38. (a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a). 39. A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.200 m while bringing a 900-kg car to rest from an initial speed of 1.1 m/s. Solution $\displaystyle 2.8×10^3N$ 40. Boxing gloves are padded to lessen the force of a blow. (a) Calculate the force exerted by a boxing glove on an opponent’s face, if the glove and face compress 7.50 cm during a blow in which the 7.00-kg arm and glove are brought to rest from an initial speed of 10.0 m/s. (b) Calculate the force exerted by an identical blow in the gory old days when no gloves were used and the knuckles and face would compress only 2.00 cm. (c) Discuss the magnitude of the force with glove on. Does it seem high enough to cause damage even though it is lower than the force with no glove? 41. Using energy considerations, calculate the average force a 60.0-kg sprinter exerts backward on the track to accelerate from 2.00 to 8.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him. Solution 102 N 7.3: Gravitational Potential Energy 42. A hydroelectric power facility (see Figure) converts the gravitational potential energy of water behind a dam to electric energy. (a) What is the gravitational potential energy relative to the generators of a lake of volume $\displaystyle 50.0 km^3 (mass=5.00×10^{13}kg)$, given that the lake has an average height of 40.0 m above the generators? (b) Compare this with the energy stored in a 9-megaton fusion bomb. Hydroelectric facility (credit: Denis Belevich, Wikimedia Commons) Solution (a) $\displaystyle 1.96×10^{16}J$ (b) The ratio of gravitational potential energy in the lake to the energy stored in the bomb is 0.52. That is, the energy stored in the lake is approximately half that in a 9-megaton fusion bomb. 43. (a) How much gravitational potential energy (relative to the ground on which it is built) is stored in the Great Pyramid of Cheops, given that its mass is about $\displaystyle 7 × 10^9 kg$ and its center of mass is 36.5 m above the surrounding ground? (b) How does this energy compare with the daily food intake of a person? 44. Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a) How much work did the bird do on the snake? (b) How much work did it do to raise its own center of mass to the branch? Solution (a) 1.8 J (b) 8.6 J 45. In Example, we found that the speed of a roller coaster that had descended 20.0 m was only slightly greater when it had an initial speed of 5.00 m/s than when it started from rest. This implies that $\displaystyle ΔPE >> KE_i$. Confirm this statement by taking the ratio of $\displaystyle ΔPE$ to $\displaystyle KE_i$. (Note that mass cancels.) 46. A 100-g toy car is propelled by a compressed spring that starts it moving. The car follows the curved track in Figure. Show that the final speed of the toy car is 0.687 m/s if its initial speed is 2.00 m/s and it coasts up the frictionless slope, gaining 0.180 m in altitude. A toy car moves up a sloped track. (credit: Leszek Leszczynski, Flickr) Solution $\displaystyle v_f=\sqrt{2gh+v_0^2}=\sqrt{2(9.80 m/s^2)(−0.180 m)+(2.00 m/s)^2}=0.687 m/s$ 47. In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy on even small hills.) To demonstrate this, find the final speed and the time taken for a skier who skies 70.0 m along a $\displaystyle 30º$ slope neglecting friction: (a) Starting from rest. (b) Starting with an initial speed of 2.50 m/s. (c) Does the answer surprise you? Discuss why it is still advantageous to get a running start in very competitive events. 7.4: Conservative Forces and Potential Energy 48. A $\displaystyle 5.00×10^5-kg$ subway train is brought to a stop from a speed of 0.500 m/s in 0.400 m by a large spring bumper at the end of its track. What is the force constant $\displaystyle k$ of the spring? Solution $\displaystyle 7.81×10^5N/m$ 49. A pogo stick has a spring with a force constant of $\displaystyle 2.50×10^4N/m$, which can be compressed 12.0 cm. To what maximum height can a child jump on the stick using only the energy in the spring, if the child and stick have a total mass of 40.0 kg? Explicitly show how you follow the steps in the Problem-Solving Strategies for Energy. 7.5: Nonconservative Forces 50. A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-m-high rise as shown in Figure. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.) The skier’s initial kinetic energy is partially used in coasting to the top of a rise. Solution 9.46 m/s 51. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h? (b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction? (c) What is the average force of friction if the hill has a slope $\displaystyle 2.5º$ above the horizontal? 7.6: Conservation of Energy 52. Using values from Table, how many DNA molecules could be broken by the energy carried by a single electron in the beam of an old-fashioned TV tube? (These electrons were not dangerous in themselves, but they did create dangerous x rays. Later model tube TVs had shielding that absorbed x rays before they escaped and exposed viewers.) Solution $\displaystyle 4×10^4$ molecules 53. Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water with an initial speed of 15.0 m/s strikes the water with a speed of 24.8 m/s independent of the direction thrown. Solution Equating $\displaystyle ΔPE_g$ and $\displaystyle ΔKE$, we obtain $\displaystyle v=\sqrt{2gh+v_0^2}=\sqrt{2(9.80 m/s^2)(20.0 m)+(15.0 m/s)^2}=24.8 m/s$ 54. If the energy in fusion bombs were used to supply the energy needs of the world, how many of the 9-megaton variety would be needed for a year’s supply of energy (using data from Table)? This is not as far-fetched as it may sound—there are thousands of nuclear bombs, and their energy can be trapped in underground explosions and converted to electricity, as natural geothermal energy is. 55. (a) Use of hydrogen fusion to supply energy is a dream that may be realized in the next century. Fusion would be a relatively clean and almost limitless supply of energy, as can be seen from Table. To illustrate this, calculate how many years the present energy needs of the world could be supplied by one millionth of the oceans’ hydrogen fusion energy. (b) How does this time compare with historically significant events, such as the duration of stable economic systems? Solution (a) $\displaystyle 25×10^6years$ (b) This is much, much longer than human time scales. 7.7: Power 56. The Crab Nebula (see Figure) pulsar is the remnant of a supernova that occurred in A.D. 1054. Using data from Table, calculate the approximate factor by which the power output of this astronomical object has declined since its explosion. Crab Nebula (credit: ESO, via Wikimedia Commons) Solution $\displaystyle 2×10^{−10}$ 57. Suppose a star 1000 times brighter than our Sun (that is, emitting 1000 times the power) suddenly goes supernova. Using data from Table: (a) By what factor does its power output increase? (b) How many times brighter than our entire Milky Way galaxy is the supernova? (c) Based on your answers, discuss whether it should be possible to observe supernovas in distant galaxies. Note that there are on the order of $\displaystyle 10^{11}$ observable galaxies, the average brightness of which is somewhat less than our own galaxy. 58. A person in good physical condition can put out 100 W of useful power for several hours at a stretch, perhaps by pedaling a mechanism that drives an electric generator. Neglecting any problems of generator efficiency and practical considerations such as resting time: (a) How many people would it take to run a 4.00-kW electric clothes dryer? (b) How many people would it take to replace a large electric power plant that generates 800 MW? Solution (a) 40 (b) 8 million 59. What is the cost of operating a 3.00-W electric clock for a year if the cost of electricity is $0.0900 per $\displaystyle kW⋅h$? 60. A large household air conditioner may consume 15.0 kW of power. What is the cost of operating this air conditioner 3.00 h per day for 30.0 d if the cost of electricity is $0.110 per $\displaystyle kW⋅h$? Solution $149 61. (a) What is the average power consumption in watts of an appliance that uses $\displaystyle 5.00 kW⋅h$ of energy per day? (b) How many joules of energy does this appliance consume in a year? 62. (a) What is the average useful power output of a person who does $\displaystyle 6.00×10^6J$ of useful work in 8.00 h? (b) Working at this rate, how long will it take this person to lift 2000 kg of bricks 1.50 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output here.) Solution (a) 208 W (b) 141 s 63. A 500-kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is its average power output in watts and horsepower if this takes 7.30 s? 64. (a) How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp = 746 W) to reach a speed of 15.0 m/s, neglecting friction? (b) How long will this acceleration take if the car also climbs a 3.00-m-high hill in the process? Solution (a) 3.20 s (b) 4.04 s 65. (a) Find the useful power output of an elevator motor that lifts a 2500-kg load a height of 35.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg—so that only 2500 kg is raised in height, but the full 10,000 kg is accelerated. (b) What does it cost, if electricity is $0.0900 per $\displaystyle kW⋅h$? 66. (a) What is the available energy content, in joules, of a battery that operates a 2.00-W electric clock for 18 months? (b) How long can a battery that can supply $\displaystyle 8.00×10^4J$ run a pocket calculator that consumes energy at the rate of $\displaystyle 1.00×10^{−3}W$? Solution (a) $\displaystyle 9.46×10^7J$ (b) $\displaystyle 2.54 y$ 67. (a) How long would it take a $\displaystyle 1.50×10^5$-kg airplane with engines that produce 100 MW of power to reach a speed of 250 m/s and an altitude of 12.0 km if air resistance were negligible? (b) If it actually takes 900 s, what is the power? (c) Given this power, what is the average force of air resistance if the airplane takes 1200 s? (Hint: You must find the distance the plane travels in 1200 s assuming constant acceleration.) 68. Calculate the power output needed for a 950-kg car to climb a $\displaystyle 2.00º$ slope at a constant 30.0 m/s while encountering wind resistance and friction totaling 600 N. Explicitly show how you follow the steps in the Problem-Solving Strategies for Energy. Solution Identify knowns: $\displaystyle m=950 kg$, slope angle $\displaystyle θ=2.00º, v=3.00 m/s, f=600 N$ Identify unknowns: power $\displaystyle P$ of the car, force $\displaystyle F$ that car applies to road Solve for unknown: $\displaystyle P=\frac{W}{t}=\frac{Fd}{t}=F(\frac{d}{t})=Fv,$ where $\displaystyle F$ is parallel to the incline and must oppose the resistive forces and the force of gravity: $\displaystyle F=f+w=600 N+mgsinθ$ Insert this into the expression for power and solve: $\displaystyle P=(f+mgsinθ)v$ $\displaystyle =[600 N+(950 kg)(9.80 m/s^2)sin 2º](30.0 m/s)$ $\displaystyle =2.77×10^4W$ About 28 kW (or about 37 hp) is reasonable for a car to climb a gentle incline. 69. (a) Calculate the power per square meter reaching Earth’s upper atmosphere from the Sun. (Take the power output of the Sun to be $\displaystyle 4.00×10^{26}W$.) (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of $\displaystyle 1.30 kW/m^2$ reaches Earth’s surface. Calculate the area in $\displaystyle km^2$ of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States’ energy needs $\displaystyle (1.05×10^{20}J)$? Australia’s energy needs $\displaystyle (5.4×10^{18}J)$? China’s energy needs $\displaystyle (6.3×10^{19}J)$? (These energy consumption values are from 2006.) 7.8: Work, Energy, and Power in Humans 70. (a) How long can you rapidly climb stairs (116/min) on the 93.0 kcal of energy in a 10.0-g pat of butter? (b) How many flights is this if each flight has 16 stairs? Solution (a) 9.5 min (b) 69 flights of stairs 71. (a) What is the power output in watts and horsepower of a 70.0-kg sprinter who accelerates from rest to 10.0 m/s in 3.00 s? (b) Considering the amount of power generated, do you think a well-trained athlete could do this repetitively for long periods of time? 72. Calculate the power output in watts and horsepower of a shot-putter who takes 1.20 s to accelerate the 7.27-kg shot from rest to 14.0 m/s, while raising it 0.800 m. (Do not include the power produced to accelerate his body.) Shot putter at the Dornoch Highland Gathering in 2007. (credit: John Haslam, Flickr) Solution 641 W, 0.860 hp 73. (a) What is the efficiency of an out-of-condition professor who does $\displaystyle 2.10×10^5 J$ of useful work while metabolizing 500 kcal of food energy? (b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of 20%? 74. Energy that is not utilized for work or heat transfer is converted to the chemical energy of body fat containing about 39 kJ/g. How many grams of fat will you gain if you eat 10,000 kJ (about 2500 kcal) one day and do nothing but sit relaxed for 16.0 h and sleep for the other 8.00 h? Use data from Table for the energy consumption rates of these activities. Solution 31 g 75. Using data from Table, calculate the daily energy needs of a person who sleeps for 7.00 h, walks for 2.00 h, attends classes for 4.00 h, cycles for 2.00 h, sits relaxed for 3.00 h, and studies for 6.00 h. (Studying consumes energy at the same rate as sitting in class.) 76. What is the efficiency of a subject on a treadmill who puts out work at the rate of 100 W while consuming oxygen at the rate of 2.00 L/min? (Hint: See Table.) Solution 14.3% 77. Shoveling snow can be extremely taxing because the arms have such a low efficiency in this activity. Suppose a person shoveling a footpath metabolizes food at the rate of 800 W. (a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process? 78. Very large forces are produced in joints when a person jumps from some height to the ground. (a) Calculate the magnitude of the force produced if an 80.0-kg person jumps from a 0.600–m-high ledge and lands stiffly, compressing joint material 1.50 cm as a result. (Be certain to include the weight of the person.) (b) In practice the knees bend almost involuntarily to help extend the distance over which you stop. Calculate the magnitude of the force produced if the stopping distance is 0.300 m. (c) Compare both forces with the weight of the person. Solution (a) $\displaystyle 3.21×10^4 N$ (b) $\displaystyle 2.35×10^3 N$ (c) Ratio of net force to weight of person is 41.0 in part (a); 3.00 in part (b) 79. Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet and legs. (a) Calculate the magnitude of the force needed to stop the downward motion of a jogger’s leg, if his leg has a mass of 13.0 kg, a speed of 6.00 m/s, and stops in a distance of 1.50 cm. (Be certain to include the weight of the 75.0-kg jogger’s body.) (b) Compare this force with the weight of the jogger. 80. (a) Calculate the energy in kJ used by a 55.0-kg woman who does 50 deep knee bends in which her center of mass is lowered and raised 0.400 m. (She does work in both directions.) You may assume her efficiency is 20%. (b) What is the average power consumption rate in watts if she does this in 3.00 min? Solution (a) 108 kJ (b) 599 W 81. Kanellos Kanellopoulos flew 119 km from Crete to Santorini, Greece, on April 23, 1988, in the Daedalus 88, an aircraft powered by a bicycle-type drive mechanism (see Figure). His useful power output for the 234-min trip was about 350 W. Using the efficiency for cycling from [link], calculate the food energy in kilojoules he metabolized during the flight. The Daedalus 88 in flight. (credit: NASA photo by Beasley) 82. The swimmer shown in Figure exerts an average horizontal backward force of 80.0 N with his arm during each 1.80 m long stroke. (a) What is his work output in each stroke? (b) Calculate the power output of his arms if he does 120 strokes per minute. Solution (a) 144 J (b) 288 W 83. Mountain climbers carry bottled oxygen when at very high altitudes. (a) Assuming that a mountain climber uses oxygen at twice the rate for climbing 116 stairs per minute (because of low air temperature and winds), calculate how many liters of oxygen a climber would need for 10.0 h of climbing. (These are liters at sea level.) Note that only 40% of the inhaled oxygen is utilized; the rest is exhaled. (b) How much useful work does the climber do if he and his equipment have a mass of 90.0 kg and he gains 1000 m of altitude? (c) What is his efficiency for the 10.0-h climb? 84. The awe-inspiring Great Pyramid of Cheops was built more than 4500 years ago. Its square base, originally 230 m on a side, covered 13.1 acres, and it was 146 m high, with a mass of about $\displaystyle 7×10^9kg$. (The pyramid’s dimensions are slightly different today due to quarrying and some sagging.) Historians estimate that 20,000 workers spent 20 years to construct it, working 12-hour days, 330 days per year. (a) Calculate the gravitational potential energy stored in the pyramid, given its center of mass is at one-fourth its height. (b) Only a fraction of the workers lifted blocks; most were involved in support services such as building ramps (see Figure), bringing food and water, and hauling blocks to the site. Calculate the efficiency of the workers who did the lifting, assuming there were 1000 of them and they consumed food energy at the rate of 300 kcal/h. What does your answer imply about how much of their work went into block-lifting, versus how much work went into friction and lifting and lowering their own bodies? (c) Calculate the mass of food that had to be supplied each day, assuming that the average worker required 3600 kcal per day and that their diet was 5% protein, 60% carbohydrate, and 35% fat. (These proportions neglect the mass of bulk and nondigestible materials consumed.) Ancient pyramids were probably constructed using ramps as simple machines. (credit: Franck Monnier, Wikimedia Commons) Solution (a) $\displaystyle 2.50×10^{12} J$ (b) 2.52% (c) $\displaystyle 1.4×10^4 kg$ (14 metric tons) 85. (a) How long can you play tennis on the 800 kJ (about 200 kcal) of energy in a candy bar? (b) Does this seem like a long time? Discuss why exercise is necessary but may not be sufficient to cause a person to lose weight. 7.9: World Energy Use 86. Integrated Concepts (a) Calculate the force the woman in Figure exerts to do a push-up at constant speed, taking all data to be known to three digits. (b) How much work does she do if her center of mass rises 0.240 m? (c) What is her useful power output if she does 25 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans. Forces involved in doing push-ups. The woman’s weight acts as a force exerted downward on her center of gravity (CG). Solution (a) 294 N (b) 118 J (c) 49.0 W 87. Integrated Concepts A 75.0-kg cross-country skier is climbing a $\displaystyle 3.0º$ slope at a constant speed of 2.00 m/s and encounters air resistance of 25.0 N. Find his power output for work done against the gravitational force and air resistance. (b) What average force does he exert backward on the snow to accomplish this? (c) If he continues to exert this force and to experience the same air resistance when he reaches a level area, how long will it take him to reach a velocity of 10.0 m/s? 88. Integrated Concepts The 70.0-kg swimmer in [link] starts a race with an initial velocity of 1.25 m/s and exerts an average force of 80.0 N backward with his arms during each 1.80 m long stroke. (a) What is his initial acceleration if water resistance is 45.0 N? (b) What is the subsequent average resistance force from the water during the 5.00 s it takes him to reach his top velocity of 2.50 m/s? (c) Discuss whether water resistance seems to increase linearly with velocity. Solution (a) $\displaystyle 0.500 m/s^2$ (b) $\displaystyle 62.5 N$ (c) Assuming the acceleration of the swimmer decreases linearly with time over the 5.00 s interval, the frictional force must therefore be increasing linearly with time, since $\displaystyle f=F−ma$. If the acceleration decreases linearly with time, the velocity will contain a term dependent on time squared ($\displaystyle t^2$). Therefore, the water resistance will not depend linearly on the velocity. 89. Integrated Concepts A toy gun uses a spring with a force constant of 300 N/m to propel a 10.0-g steel ball. If the spring is compressed 7.00 cm and friction is negligible: (a) How much force is needed to compress the spring? (b) To what maximum height can the ball be shot? (c) At what angles above the horizontal may a child aim to hit a target 3.00 m away at the same height as the gun? (d) What is the gun’s maximum range on level ground? 90. Integrated Concepts (a) What force must be supplied by an elevator cable to produce an acceleration of $\displaystyle 0.800 m/s^2$ against a 200-N frictional force, if the mass of the loaded elevator is 1500 kg? (b) How much work is done by the cable in lifting the elevator 20.0 m? (c) What is the final speed of the elevator if it starts from rest? (d) How much work went into thermal energy? Solution (a) $\displaystyle 16.1×10^3N$ (b) $\displaystyle 3.22×10^5J$ (c) $\displaystyle 5.66 m/s$ (d) 4.00 kJ 91. Unreasonable Results A car advertisement claims that its 900-kg car accelerated from rest to 30.0 m/s and drove 100 km, gaining 3.00 km in altitude, on 1.0 gal of gasoline. The average force of friction including air resistance was 700 N. Assume all values are known to three significant figures. (a) Calculate the car’s efficiency. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent? 92. Unreasonable Results Body fat is metabolized, supplying 9.30 kcal/g, when dietary intake is less than needed to fuel metabolism. The manufacturers of an exercise bicycle claim that you can lose 0.500 kg of fat per day by vigorously exercising for 2.00 h per day on their machine. (a) How many kcal are supplied by the metabolization of 0.500 kg of fat? (b) Calculate the kcal/min that you would have to utilize to metabolize fat at the rate of 0.500 kg in 2.00 h. (c) What is unreasonable about the results? (d) Which premise is unreasonable, or which premises are inconsistent? Solution (a) $\displaystyle 4.65×10^3kcal$ (b) 38.8 kcal/min (c) This power output is higher than the highest value on [link], which is about 35 kcal/min (corresponding to 2415 watts) for sprinting. (d) It would be impossible to maintain this power output for 2 hours (imagine sprinting for 2 hours!). 93. Construct Your Own Problem Consider a person climbing and descending stairs. Construct a problem in which you calculate the long-term rate at which stairs can be climbed considering the mass of the person, his ability to generate power with his legs, and the height of a single stair step. Also consider why the same person can descend stairs at a faster rate for a nearly unlimited time in spite of the fact that very similar forces are exerted going down as going up. (This points to a fundamentally different process for descending versus climbing stairs.) 94. Construct Your Own Problem Consider humans generating electricity by pedaling a device similar to a stationary bicycle. Construct a problem in which you determine the number of people it would take to replace a large electrical generation facility. Among the things to consider are the power output that is reasonable using the legs, rest time, and the need for electricity 24 hours per day. Discuss the practical implications of your results. 95. Integrated Concepts A 105-kg basketball player crouches down 0.400 m while waiting to jump. After exerting a force on the floor through this 0.400 m, his feet leave the floor and his center of gravity rises 0.950 m above its normal standing erect position. (a) Using energy considerations, calculate his velocity when he leaves the floor. (b) What average force did he exert on the floor? (Do not neglect the force to support his weight as well as that to accelerate him.) (c) What was his power output during the acceleration phase? Solution (a) 4.32 m/s (b) $\displaystyle 3.47×10^3N$ (c) 8.93 kW Contributors and Attributions - Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .	libretexts	2025-03-17T19:53:27.039702	2018-05-04T03:00:13	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/07%3A_Work_Energy_and_Energy_Resources/7.E%3A__Work_Energy_and_Energy_Resources_(Exercise)", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "7.E: Work, Energy, and Energy Resources (Exercise)", "author": null }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions	8: Linear Momentum and Collisions We use the term momentum in various ways in everyday language, and most of these ways are consistent with its precise scientific definition. We speak of sports teams or politicians gaining and maintaining the momentum to win. We also recognize that momentum has something to do with collisions. For example, looking at the rugby players in the photograph colliding and falling to the ground, we expect their momenta to have great effects in the resulting collisions. Generally, momentum implies a tendency to continue on course—to move in the same direction—and is associated with great mass and speed. - - 8.1: Linear Momentum and Force - The scientific definition of linear momentum is consistent with most people’s intuitive understanding of momentum: a large, fast-moving object has greater momentum than a smaller, slower object. Linear momentum is defined as the product of a system’s mass multiplied by its velocity. Momentum is directly proportional to the object’s mass and also its velocity. Thus the greater an object’s mass or the greater its velocity, the greater its momentum. - - 8.2: Impulse - The effect of a force on an object depends on how long it acts, as well as how great the force is. A very large force acting for a short time had a great effect on the momentum of the tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time. - - 8.3: Conservation of Momentum - Momentum is an important quantity because it is conserved. Yet it appears to not be conserved in the previous exampless, where large changes in momentum were produced by forces acting on the system of interest. Under what circumstances is momentum conserved? The answer to this question entails considering a sufficiently large system. It is always possible to find a larger system in which total momentum is constant, even if momentum changes for components of the system. - - 8.4: Elastic Collisions in One Dimension - An elastic collision is one that also conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the objects in the system. Truly elastic collisions can only be achieved with subatomic particles, such as electrons striking nuclei. Macroscopic collisions can be very nearly, but not quite, elastic—some kinetic energy is always converted into other forms of energy such as heat transfer due to friction and sound. - - 8.5: Inelastic Collisions in One Dimension - An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). This lack of conservation means that the forces between colliding objects may remove or add internal kinetic energy. Work done by internal forces may change the forms of energy within a system. For inelastic collisions, such as when colliding objects stick together, this internal work may transform some internal kinetic energy into heat transfer. - - 8.6: Collisions of Point Masses in Two Dimensions - One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin. - - 8.7: Introduction to Rocket Propulsion - Rockets range in size from fireworks so small that ordinary people use them to immense Saturn Vs that once propelled massive payloads toward the Moon. The propulsion of all rockets, jet engines, deflating balloons, and even squids and octopuses is explained by the same physical principle—Newton’s third law of motion. Matter is forcefully ejected from a system, producing an equal and opposite reaction on what remains. Thumbnail: A pool break-off shot. (CC-SA-BY; No-w-ay ).	libretexts	2025-03-17T19:53:27.107998	2015-10-27T19:22:41	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "8: Linear Momentum and Collisions", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.00%3A_Prelude_to_Linear_Momentum_and_Collisions	8.0: Prelude to Linear Momentum and Collisions We use the term momentum in various ways in everyday language, and most of these ways are consistent with its precise scientific definition. We speak of sports teams or politicians gaining and maintaining the momentum to win. We also recognize that momentum has something to do with collisions. For example, looking at the rugby players in the photograph colliding and falling to the ground, we expect their momenta to have great effects in the resulting collisions. Generally, momentum implies a tendency to continue on course—to move in the same direction—and is associated with great mass and speed. Momentum, like energy, is important because it is conserved. Only a few physical quantities are conserved in nature, and studying them yields fundamental insight into how nature works, as we shall see in our study of momentum.	libretexts	2025-03-17T19:53:27.186859	2015-11-01T05:02:58	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.00%3A_Prelude_to_Linear_Momentum_and_Collisions", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "8.0: Prelude to Linear Momentum and Collisions", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.01%3A_Linear_Momentum_and_Force	8.1: Linear Momentum and Force Learning Objectives By the end of this section, you will be able to: - Define linear momentum. - Explain the relationship between momentum and force. - State Newton’s second law of motion in terms of momentum. - Calculate momentum given mass and velocity. The scientific definition of linear momentum is consistent with most people’s intuitive understanding of momentum: a large, fast-moving object has greater momentum than a smaller, slower object. Linear momentum is defined as the product of a system’s mass multiplied by its velocity. Linear Momentum Linear momentum is defined as the product of a system’s mass multiplied by its velocity: \[p = mv \label{linearmomentum}\] Momentum is directly proportional to the object’s mass and also its velocity. Thus the greater an object’s mass or the greater its velocity, the greater its momentum. Momentum $p$ is a vector having the same direction as the velocity $v$. The SI unit for momentum is $kg \cdot m/s.$ Example $\PageIndex{1}$: Calculating Momentum: A Football Player and a Football - Calculate the momentum of a 110-kg football player running at 8.00 m/s. - Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s. Strategy No information is given regarding direction, and so we can calculate only the magnitude of the momentum, $p$ (As usual, a symbol that is in italics is a magnitude, whereas one that is italicized, boldfaced, and has an arrow is a vector.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum given in Equation \ref{linearmomentum}, which becomes \[p = mv \nonumber\] when only magnitudes are considered. Solution for (a) To determine the momentum of the player, substitute the known values for the player’s mass and speed into the equation. \[\begin{align} p_{player} &= (110 \, kg)(8.00 \, m/s) \\[5pt] &= 880 \, kg \cdot m/s \end{align}\] Solution for (b) To determine the momentum of the ball, substitute the known values for the ball’s mass and speed into the equation. \[\begin{align} p_{ball} &= (0.410 \, kg)(25.0 \, m/s) \\[5pt] &= 10.3 \, kg \cdot m/s \end{align}\] The ratio of the player’s momentum to that of the ball is \[\dfrac{p_{player}}{p_{ball}} = \dfrac{880}{10.3} = 85.0 \nonumber\] Discussion Although the ball has greater velocity, the player has a much greater mass. Thus the momentum of the player is much greater than the momentum of the football, as you might guess. As a result, the player’s motion is only slightly affected if he catches the ball. We shall quantify what happens in such collisions in terms of momentum in later sections. Momentum and Newton’s Second Law The importance of momentum, unlike the importance of energy, was recognized early in the development of classical physics. Momentum was deemed so important that it was called the “quantity of motion.” Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. Newton’s Second Law of Motion in Terms of Momentum The net external force equals the change in momentum of a system divided by the time over which it changes. \[F_{net} = \dfrac{\Delta p}{\Delta t}\] where $F_{net} $ is the net external force, $\Delta p$ is the change in momentum, and $\Delta t$ is the change in time. Making Connections: Force and Momentum Force and momentum are intimately related. Force acting over time can change momentum, and Newton’s second law of motion, can be stated in its most broadly applicable form in terms of momentum. Momentum continues to be a key concept in the study of atomic and subatomic particles in quantum mechanics. This statement of Newton’s second law of motion includes the more familiar $F_{net} = ma$ as a special case. We can derive this form as follows. First, note that the change in momentum $\Delta p$ is given by \[\Delta p = \Delta (mv)\] If the mass of the system is constant, then \[\Delta (mv) = m\Delta v.\] So that for constant mass, Newton’s second law of motion becomes \[F_{net} = \dfrac{\Delta p}{\Delta t} = \dfrac{m \Delta v}{\Delta t}.\] Because $\frac{\Delta v}{\Delta t} = a, $ we get the familiar equation \[F_{net} = ma\] when the mass of the system is constant . Newton’s second law of motion stated in terms of momentum is more generally applicable because it can be applied to systems where the mass is changing, such as rockets, as well as to systems of constant mass. We will consider systems with varying mass in some detail ; however, the relationship between momentum and force remains useful when mass is constant, such as in the following example. Example $\PageIndex{2}$: Calculating Force: Venus Williams’ Racquet During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What is the average force exerted on the 0.057-kg tennis ball by Venus Williams’ racquet, assuming that the ball’s speed just after impact is 58 m/s, that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)? Strategy This problem involves only one dimension because the ball starts from having no horizontal velocity component before impact. Newton’s second law stated in terms of momentum is then written as \[F_{net} = \dfrac{\Delta p}{\Delta t} \nonumber\] As noted above, when mass is constant, the change in momentum is given by \[\Delta p = m\Delta v = m(v_f - v_i). \nonumber\] In this example, the velocity just after impact and the change in time are given; thus, once $\Delta p$ s calculated, $F_{net} = \frac{\Delta p}{\Delta t}$ can be used to find the force. Solution To determine the change in momentum, substitute the values for the initial and final velocities into the equation above. \[\begin{align} \Delta p &= m(v_f - v_i) \\[5pt] &= (0.057 \, kg)(58 \, m/s - 0 \, m/s)\\[5pt] &= 3.306 \, kg \cdot m/s = 3.3 \, kg \cdot m/s \end{align} \] Now the magnitude of the net external force can determined by using $F_{net} = \frac{\Delta p}{\Delta t}$ \[\begin{align} F_{net} &= \dfrac{\Delta p}{\Delta t} = \dfrac{3.306 \, kg}{5.0 \times 10^{-3}}\\[5pt] &= 661 \, N,\end{align} \] where we have retained only two significant figures in the final step. Discussion This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact (note that the ball also experienced the 0.56-N force of gravity, but that force was not due to the racquet). This problem could also be solved by first finding the acceleration and then using $F = ma$ but one additional step would be required compared with the strategy used in this example. Summary - Linear momentum ( momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. - In symbols, linear momentum $p$ is defined to be \[p = mv \nonumber\] where $m$ is the mass of the system and $v$ is its velocity. - The SI unit for momentum is $kg \cdot m/s.$ - Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes. - In symbols, Newton’s second law of motion is defined to be \[F_{net} = \frac{\Delta p}{\Delta t} \nonumber \]where $F_{net}\nonumber $ is the net external force, $\Delta p$ s the change in momentum, and $\Delta t$ is change in time. Glossary - linear momentum - the product of mass and velocity - second law of motion - physical law that states that the net external force equals the change in momentum of a system divided by the time over which it changes	libretexts	2025-03-17T19:53:27.262448	2015-11-01T05:03:19	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.01%3A_Linear_Momentum_and_Force", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "8.1: Linear Momentum and Force", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.02%3A_Impulse	8.2: Impulse Learning Objectives By the end of this section, you will be able to: - Define impulse. - Describe effects of impulses in everyday life. - Determine the average effective force using graphical representation. - Calculate average force and impulse given mass, velocity, and time. The effect of a force on an object depends on how long it acts, as well as how great the force is. In [link] , a very large force acting for a short time had a great effect on the momentum of the tennis ball. A small force could cause the same change in momentum , but it would have to act for a much longer time. For example, if the ball were thrown upward, the gravitational force (which is much smaller than the tennis racquet’s force) would eventually reverse the momentum of the ball. Quantitatively, the effect we are talking about is the change in momentum $\Delta p$. By rearranging the equation $\Delta F_{net} = \frac{\Delta p}{\Delta t} $ to be \[\Delta p = F_{net} \Delta t,\] we can see how the change in momentum equals the average net external force multiplied by the time this force acts. The quantity $F_{net} \Delta t$ is given the name impulse . Impulse is the same as the change in momentum. Impulse: Change in Momentum Change in momentum equals the average net external force multiplied by the time this force acts. \[\Delta p = F_{net} \Delta t\] The quantity $ F_{net}\Delta t$ is given the name impulse. There are many ways in which an understanding of impulse can save lives, or at least limbs. The dashboard padding in a car, and certainly the airbags, allow the net force on the occupants in the car to act over a much longer time when there is a sudden stop. The momentum change is the same for an occupant, whether an air bag is deployed or not, but the force (to bring the occupant to a stop) will be much less if it acts over a larger time. Cars today have many plastic components. One advantage of plastics is their lighter weight, which results in better gas mileage. Another advantage is that a car will crumple in a collision, especially in the event of a head-on collision. A longer collision time means the force on the car will be less. Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could crumple or collapse in the event of an accident. Bones in a body will fracture if the force on them is too large. If you jump onto the floor from a table, the force on your legs can be immense if you land stiff-legged on a hard surface. Rolling on the ground after jumping from the table, or landing with a parachute, extends the time over which the force (on you from the ground) acts. Example $\PageIndex{1}$: Calculating Magnitudes of Impulses: Two Billiard Balls Striking a Rigid Wall Two identical billiard balls strike a rigid wall with the same speed, and are reflected without any change of speed. The first ball strikes perpendicular to the wall. The second ball strikes the wall at an angle of $30^o$ from the perpendicular, and bounces off at an angle of $30^o$ from perpendicular to the wall. - Determine the direction of the force on the wall due to each ball. - Calculate the ratio of the magnitudes of impulses on the two balls by the wall. Strategy for (a) In order to determine the force on the wall, consider the force on the ball due to the wall using Newton’s second law and then apply Newton’s third law to determine the direction. Assume the $x-$axis to be normal to the wall and to be positive in the initial direction of motion. Choose the $y-$axis to be along the wall in the plane of the second ball’s motion. The momentum direction and the velocity direction are the same. Strategy for (a) In order to determine the force on the wall, consider the force on the ball due to the wall using Newton’s second law and then apply Newton’s third law to determine the direction. Assume the $x-$axis to be normal to the wall and to be positive in the initial direction of motion. Choose the $y-$axis to be along the wall in the plane of the second ball’s motion. The momentum direction and the velocity direction are the same. Solution for (a) The first ball bounces directly into the wall and exerts a force on it in the +$x$ direction. Therefore the wall exerts a force on the ball in the -$y$ direction. The second ball continues with the same momentum component in the $y-$ direction, but reverses its $x$-component of momentum, as seen by sketching a diagram of the angles involved and keeping in mind the proportionality between velocity and momentum. These changes mean the change in momentum for both balls is in the $-x$ direction, so the force of the wall on each ball is along the $-x$ direction. Strategy for (b) Calculate the change in momentum for each ball, which is equal to the impulse imparted to the ball. Solution for (b) Let $\mu$ be the speed of each ball before and after collision with the wall, and $m$ the mass of each ball. Choose the $x-$axis and $y-$axis as previously described, and consider the change in momentum of the first ball which strikes perpendicular to the wall. \[p_{xi} = m\mu; \, p_{yi} = 0\] \[p_{xf} = -m\mu; \, p_{yf} = 0\] Impulse is the change in momentum vector. Therefore the $x-$component of impulse is equal to -$2m\mu$ and the $y-$component of impulse is equal to zero. Now consider the change in momentum of the second ball. \[p_{xi} = m/mu \, cos 30^o; \, p_{yi} = -m\mu \, 30^o\] \[p_{xf} = -m/mu \, cos 30^o; \, p_{yf} = -m\mu \, 30^o\] It should be noted here that while $p_x$ changes sign after the collision, $p_y$ does not. Therefore the -component of impulse is equal to $-2m\mu \, cos \, 30^o$ and the $y-$component of impulse is equal to zero. The ratio of the magnitudes of the impulse imparted to the balls is \[ \dfrac{2m\mu}{2m\mu \, 30^o} = \dfrac{2}{sqrt{3}} = 1.155.\] Discussion The direction of impulse and force is the same as in the case of (a); it is normal to the wall and along the negative $x-$direction. Making use of Newton’s third law, the force on the wall due to each ball is normal to the wall along the positive $x-$direction. Our definition of impulse includes an assumption that the force is constant over the time interval $\Delta t$. Forces are usually not constant . Forces vary considerably even during the brief time intervals considered. It is, however, possible to find an average effective force $F_{eff}$ that produces the same result as the corresponding time-varying force. Figure shows a graph of what an actual force looks like as a function of time for a ball bouncing off the floor. The area under the curve has units of momentum and is equal to the impulse or change in momentum between times $t_1$ and $t_2$. That area is equal to the area inside the rectangle bounded by $F_{eff}, \, t_1$, and $t_2$ Thus the impulses and their effects are the same for both the actual and effective forces. MAKING CONNECTIONS: Take-Home Investigation—Hand Movement and Impulse Try catching a ball while “giving” with the ball, pulling your hands toward your body. Then, try catching a ball while keeping your hands still. Hit water in a tub with your full palm. After the water has settled, hit the water again by diving your hand with your fingers first into the water. (Your full palm represents a swimmer doing a belly flop and your diving hand represents a swimmer doing a dive.) Explain what happens in each case and why. Which orientations would you advise people to avoid and why? MAKING CONNECTIONS: Constant Force anD Constant Acceleration The assumption of a constant force in the definition of impulse is analogous to the assumption of a constant acceleration in kinematics. In both cases, nature is adequately described without the use of calculus. Summary - Impulse, or change in momentum, equals the average net external force multiplied by the time this force acts: \[ \Delta p = F_{net}\Delta t.\] - Forces are usually not constant over a period of time. Glossary - change in momentum - the difference between the final and initial momentum; the mass times the change in velocity - impulse - the average net external force times the time it acts; equal to the change in momentum	libretexts	2025-03-17T19:53:27.334553	2015-11-01T05:03:38	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.02%3A_Impulse", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "8.2: Impulse", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.03%3A_Conservation_of_Momentum	8.3: Conservation of Momentum Learning Objectives By the end of this section, you will be able to: - Describe the principle of conservation of momentum. - Derive an expression for the conservation of momentum. - Explain conservation of momentum with examples. - Explain the principle of conservation of momentum as it relates to atomic and subatomic particles. Momentum is an important quantity because it is conserved. Yet it was not conserved in the examples in Impulse and Linear Momentum and Force , where large changes in momentum were produced by forces acting on the system of interest. Under what circumstances is momentum conserved? The answer to this question entails considering a sufficiently large system. It is always possible to find a larger system in which total momentum is constant, even if momentum changes for components of the system. If a football player runs into the goalpost in the end zone, there will be a force on him that causes him to bounce backward. However, the Earth also recoils —conserving momentum—because of the force applied to it through the goalpost. Because Earth is many orders of magnitude more massive than the player, its recoil is immeasurably small and can be neglected in any practical sense, but it is real nevertheless. Consider what happens if the masses of two colliding objects are more similar than the masses of a football player and Earth—for example, one car bumping into another, as shown in Figure $\PageIndex{1}$. Both cars are coasting in the same direction when the lead car (labeled $m_2$ is bumped by the trailing car (labeled $m_1$. The only unbalanced force on each car is the force of the collision. (Assume that the effects due to friction are negligible.) Car 1 slows down as a result of the collision, losing some momentum, while car 2 speeds up and gains some momentum. We shall now show that the total momentum of the two-car system remains constant. Using the definition of impulse, the change in momentum of car 1 is given by \[\Delta p_1 = F_1 \Delta t,\] is the force on car 1 due to car 2, and $\Delta t$ where $F_1$ is the time the force acts (the duration of the collision). Intuitively, it seems obvious that the collision time is the same for both cars, but it is only true for objects traveling at ordinary speeds. This assumption must be modified for objects travelling near the speed of light, without affecting the result that momentum is conserved. Similarly, the change in momentum of car 2 is \[ \Delta p_2 = F_2 \Delta t,\] where $F_2$ is the force on car 2 due to car 1, and we assume the duration of the collision $\Delta t$ is the same for both cars. We know from Newton’s third law that $F_2 = - F_1$, and so \[ \Delta p_2 = -F_1\Delta t = - \Delta p_1.\] Thus, the changes in momentum are equal and opposite, and \[\Delta p_1 + \Delta p_2 = 0.\] Because the changes in momentum add to zero, the total momentum of the two-car system is constant. That is, \[p_1 + p_2 = constant\] \[p_1 + p_2 = p_1' + p_2', \] where $p_1'$ and $p_2'$ are the momenta of cars 1 and 2 after the collision. (We often use primes to denote the final state.) This result—that momentum is conserved—has validity far beyond the preceding one-dimensional case. It can be similarly shown that total momentum is conserved for any isolated system, with any number of objects in it. In equation form, the conservation of momentum principle for an isolated system is written \[p_{tot} = constant,\] or \[p_{tot} = p_{tot},\] where $ p_{tot}$ is the total momentum (the sum of the momenta of the individual objects in the system) and $ p_{tot},$ is the total momentum some time later. (The total momentum can be shown to be the momentum of the center of mass of the system.) An isolated system is defined to be one for which the net external force is zero $(F_{net} = 0)$. Conservation of Momentum Principle \[ p_{tot} = constant\] \[p_{tot} = p_{tot}' \, (isolated \, system)\] Isolated System An isolated system is defined to be one for which the net external force is zero $(F_{net} = 0)$. Perhaps an easier way to see that momentum is conserved for an isolated system is to consider Newton’s second law in terms of momentum, $F_{net} = \frac{\Delta p_{tot}}{\Delta t}$. For an isolated system, $(F_{net} = 0)$; thus $\Delta p_{tot} = 0$ and $\Delta p$ is constant. We have noted that the three length dimensions in nature x, y and z are independent, and it is interesting to note that momentum can be conserved in different ways along each dimension. For example, during projectile motion and where air resistance is negligible, momentum is conserved in the horizontal direction because horizontal forces are zero and momentum is unchanged. But along the vertical direction, the net vertical force is not zero and the momentum of the projectile is not conserved (Figure $\PageIndex{2}$). However, if the momentum of the projectile-Earth system is considered in the vertical direction, we find that the total momentum is conserved. The conservation of momentum principle can be applied to systems as different as a comet striking Earth and a gas containing huge numbers of atoms and molecules. Conservation of momentum is violated only when the net external force is not zero. But another larger system can always be considered in which momentum is conserved by simply including the source of the external force. For example, in the collision of two cars considered above, the two-car system conserves momentum while each one-car system does not. MAKING CONNECTIONS: TAKE-HOME Investigation—Drop of Tennis Ball and a Basketball Hold a tennis ball side by side and in contact with a basketball. Drop the balls together. (Be careful!) What happens? Explain your observations. Now hold the tennis ball above and in contact with the basketball. What happened? Explain your observations. What do you think will happen if the basketball ball is held above and in contact with the tennis ball? MAKING CONNECTIONS: TAKE-HOME Investigation—Two Tennis Balls in a Ballistic Trajectory Tie two tennis balls together with a string about a foot long. Hold one ball and let the other hang down and throw it in a ballistic trajectory. Explain your observations. Now mark the center of the string with bright ink or attach a brightly colored sticker to it and throw again. What happened? Explain your observations. Some aquatic animals such as jellyfish move around based on the principles of conservation of momentum. A jellyfish fills its umbrella section with water and then pushes the water out resulting in motion in the opposite direction to that of the jet of water. Squids propel themselves in a similar manner but, in contrast with jellyfish, are able to control the direction in which they move by aiming their nozzle forward or backward. Typical squids can move at speeds of 8 to 12 km/h. The ballistocardiograph (BCG) was a diagnostic tool used in the second half of the 20th century to study the strength of the heart. About once a second, your heart beats, forcing blood into the aorta. A force in the opposite direction is exerted on the rest of your body (recall Newton’s third law). A ballistocardiograph is a device that can measure this reaction force. This measurement is done by using a sensor (resting on the person) or by using a moving table suspended from the ceiling. This technique can gather information on the strength of the heart beat and the volume of blood passing from the heart. However, the electrocardiogram (ECG or EKG) and the echocardiogram (cardiac ECHO or ECHO; a technique that uses ultrasound to see an image of the heart) are more widely used in the practice of cardiology. Making Connections: Conservation of Momentum and Collision Conservation of momentum is quite useful in describing collisions. Momentum is crucial to our understanding of atomic and subatomic particles because much of what we know about these particles comes from collision experiments. Subatomic Collisions and Momentum The conservation of momentum principle not only applies to the macroscopic objects, it is also essential to our explorations of atomic and subatomic particles. Giant machines hurl subatomic particles at one another, and researchers evaluate the results by assuming conservation of momentum (among other things). On the small scale, we find that particles and their properties are invisible to the naked eye but can be measured with our instruments, and models of these subatomic particles can be constructed to describe the results. Momentum is found to be a property of all subatomic particles including massless particles such as photons that compose light. Momentum being a property of particles hints that momentum may have an identity beyond the description of an object’s mass multiplied by the object’s velocity. Indeed, momentum relates to wave properties and plays a fundamental role in what measurements are taken and how we take these measurements. Furthermore, we find that the conservation of momentum principle is valid when considering systems of particles. We use this principle to analyze the masses and other properties of previously undetected particles, such as the nucleus of an atom and the existence of quarks that make up particles of nuclei. Figure $\PageIndex{3}$ below illustrates how a particle scattering backward from another implies that its target is massive and dense. Experiments seeking evidence that quarks make up protons (one type of particle that makes up nuclei) scattered high-energy electrons off of protons (nuclei of hydrogen atoms). Electrons occasionally scattered straight backward in a manner that implied a very small and very dense particle makes up the proton—this observation is considered nearly direct evidence of quarks. The analysis was based partly on the same conservation of momentum principle that works so well on the large scale. Summary - The conservation of momentum principle is written \[p_{tot} = constant\] or \[p_{tot} = p'_{tot} \, (isolated \, system),\] - $p_{tot}$ is the initial total momentum and $p'_{tot}$ is the total momentum some time later. An isolated system is defined to be one for which the net external force is zero $(F_{net} = 0$ - During projectile motion and where air resistance is negligible, momentum is conserved in the horizontal direction because horizontal forces are zero. - Conservation of momentum applies only when the net external force is zero. - The conservation of momentum principle is valid when considering systems of particles. Glossary - conservation of momentum principle - when the net external force is zero, the total momentum of the system is conserved or constant - isolated system - a system in which the net external force is zero - quark - fundamental constituent of matter and an elementary particle	libretexts	2025-03-17T19:53:27.483856	2015-11-01T05:03:59	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.03%3A_Conservation_of_Momentum", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "8.3: Conservation of Momentum", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.04%3A_Elastic_Collisions_in_One_Dimension	8.4: Elastic Collisions in One Dimension Learning Objectives By the end of this section, you will be able to: - Describe an elastic collision of two objects in one dimension. - Define internal kinetic energy. - Derive an expression for conservation of internal kinetic energy in a one dimensional collision. - Determine the final velocities in an elastic collision given masses and initial velocities. Let us consider various types of two-object collisions. These collisions are the easiest to analyze, and they illustrate many of the physical principles involved in collisions. The conservation of momentum principle is very useful here, and it can be used whenever the net external force on a system is zero. We start with the elastic collision of two objects moving along the same line—a one-dimensional problem. An elastic collision is one that also conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the objects in the system. Figure illustrates an elastic collision in which internal kinetic energy and momentum are conserved. Truly elastic collisions can only be achieved with subatomic particles, such as electrons striking nuclei. Macroscopic collisions can be very nearly, but not quite, elastic—some kinetic energy is always converted into other forms of energy such as heat transfer due to friction and sound. One macroscopic collision that is nearly elastic is that of two steel blocks on ice. Another nearly elastic collision is that between two carts with spring bumpers on an air track. Icy surfaces and air tracks are nearly frictionless, more readily allowing nearly elastic collisions on them. Elastic Collision An elastic collision is one that conserves internal kinetic energy. Internal Kinetic Energy Internal kinetic energy is the sum of the kinetic energies of the objects in the system. Now, to solve problems involving one-dimensional elastic collisions between two objects we can use the equations for conservation of momentum and conservation of internal kinetic energy. First, the equation for conservation of momentum for two objects in a one-dimensional collision is \[p_1 +p_2 = p'_1 + p'_2 \, (F_{net} = 0)\] or \[m_1v_1 + m_2 v_2 = m_1 v'_1 + m'_2v'_2 \, (F_{net} = 0),\] where the primes (') indicate values after the collision. By definition, an elastic collision conserves internal kinetic energy, and so the sum of kinetic energies before the collision equals the sum after the collision. Thus, \[ \dfrac{1}{2} m_1 v_1^2 + \dfrac{1}{2} m_2 v_2^2 = \dfrac{1}{2} m_1 v_1^{'2} + \dfrac{1}{2} m_2 v_2^{'2} \] expresses the equation for conservation of internal kinetic energy in a one-dimensional collision. Example $\PageIndex{1}$: Calculating Velocities Following an Elastic Collision Calculate the velocities of two objects following an elastic collision, given that \[m_1 = 0.500 \, kg, \, m_2 = 3.50 \, kg, \, v_1 = 4.00 \, m/s, \, and \, v_2 = 0,\] Strategy and Concept First, visualize what the initial conditions mean—a small object strikes a larger object that is initially at rest. This situation is slightly simpler than the situation shown in Figure where both objects are initially moving. We are asked to find two unknowns (the final velocities $v'_1$ and $ v'_2$). To find two unknowns, we must use two independent equations. Because this collision is elastic, we can use the above two equations. Both can be simplified by the fact that object 2 is initially at rest, and thus $v_2 = 0.$ Once we simplify these equations, we combine them algebraically to solve for the unknowns. Solution For this problem, note that $v_2 = 0$ and use conservation of momentum. Thus, \[p_1 = p'_1 + p'_2\] or \[m_1v_1 = m_1v'_1 + m_2v'_2.\] Using conservation of internal kinetic energy and that $v_2 = 0$, \[\dfrac{1}{2}m_1v_1^2 = \dfrac{1}{2}mv_1v_1^{'2} + \dfrac{1}{2}mv_2v_2^{'2}.\] Solving the first equation (momentum equation) for $v'_2$, we obtain \[v'_2 = \dfrac{m_1}{m_2}(v_1 - v'_1).\] Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable $v'_2$, leaving only $v'_1$ as an unknown (the algebra is left as an exercise for the reader). There are two solutions to any quadratic equation; in this example, they are as an unknown (the algebra is left as an exercise for the reader). There are two solutions to any quadratic equation; in this example, they are \[v'_1 = 4.00 \, m/s\] and \[v'_1 = -3.00 \, m/s.\] As noted when quadratic equations were encountered in earlier chapters, both solutions may or may not be meaningful. In this case, the first solution is the same as the initial condition. The first solution thus represents the situation before the collision and is discarded. The second solution $(v'_1 = -3.00 \, m/s$) is negative, meaning that the first object bounces backward. When this negative value of $v'_1$ is used to find the velocity of the second object after the collision, we get \[v'_2 = \dfrac{m_1}{m_2}(v_1 - v'_1) = \dfrac{0.500 \, kg}{3.50 \, kg}[4.00 - (-3.00)] \, m/s\] or \[v'_2 = 1.00 \, m/s.\] Discussion The result of this example is intuitively reasonable. A small object strikes a larger one at rest and bounces backward. The larger one is knocked forward, but with a low speed. (This is like a compact car bouncing backward off a full-size SUV that is initially at rest.) As a check, try calculating the internal kinetic energy before and after the collision. You will see that the internal kinetic energy is unchanged at 4.00 J. Also check the total momentum before and after the collision; you will find it, too, is unchanged. The equations for conservation of momentum and internal kinetic energy as written above can be used to describe any one-dimensional elastic collision of two objects. These equations can be extended to more objects if needed. ' 4 . 00 m/s Making Connections: Take-Home Investigation—Ice Cubes and Elastic Collision Find a few ice cubes which are about the same size and a smooth kitchen tabletop or a table with a glass top. Place the ice cubes on the surface several centimeters away from each other. Flick one ice cube toward a stationary ice cube and observe the path and velocities of the ice cubes after the collision. Try to avoid edge-on collisions and collisions with rotating ice cubes. Have you created approximately elastic collisions? Explain the speeds and directions of the ice cubes using momentum. PHET EXPLORATIONS: COLLISIONS LAB Investigate collisions on an air hockey table. Set up your own experiments: vary the number of discs, masses and initial conditions. Is momentum conserved? Is kinetic energy conserved? Vary the elasticity and see what happens. Summary - An elastic collision is one that conserves internal kinetic energy. - Conservation of kinetic energy and momentum together allow the final velocities to be calculated in terms of initial velocities and masses in one dimensional two-body collisions. Glossary - elastic collision - a collision that also conserves internal kinetic energy - internal kinetic energy - the sum of the kinetic energies of the objects in a system	libretexts	2025-03-17T19:53:27.557750	2015-11-01T05:04:18	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.04%3A_Elastic_Collisions_in_One_Dimension", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "8.4: Elastic Collisions in One Dimension", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.05%3A_Inelastic_Collisions_in_One_Dimension	8.5: Inelastic Collisions in One Dimension Learning Objectives By the end of this section, you will be able to: - Define inelastic collision. - Explain perfectly inelastic collision. - Apply an understanding of collisions to sports. - Determine recoil velocity and loss in kinetic energy given mass and initial velocity. We have seen that in an elastic collision, internal kinetic energy is conserved. An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). This lack of conservation means that the forces between colliding objects may remove or add internal kinetic energy. Work done by internal forces may change the forms of energy within a system. For inelastic collisions, such as when colliding objects stick together, this internal work may transform some internal kinetic energy into heat transfer. Or it may convert stored energy into internal kinetic energy, such as when exploding bolts separate a satellite from its launch vehicle. Definition: Inelastic Collisions An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). Figure $\PageIndex{1}$ shows an example of an inelastic collision. Two objects that have equal masses head toward one another at equal speeds and then stick together. Their total internal kinetic energy is initially \[\dfrac{1}{2}mv^2 + \dfrac{1}{2}mv^2 = mv^2.\] The two objects come to rest after sticking together, conserving momentum. But the internal kinetic energy is zero after the collision. A collision in which the objects stick together is sometimes called a perfectly inelastic collision because it reduces internal kinetic energy more than does any other type of inelastic collision. In fact, such a collision reduces internal kinetic energy to the minimum it can have while still conserving momentum. Definition: Perfectly Inelastic Collisions A collision in which the objects stick together is sometimes called “perfectly inelastic.” Example $\PageIndex{1}$: Calculating Velocity and Change in Kinetic Energy - Inelastic Collision of a Puck and a Goalie - Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. - How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible (Figure $\PageIndex{2}$) Strategy Momentum is conserved because the net external force on the puck-goalie system is zero. We can thus use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same. Once the final velocity is found, the kinetic energies can be calculated before and after the collision and compared as requested. Solution for (a) Momentum is conserved because the net external force on the puck-goalie system is zero. Conservation of momentum is \[p_1 + p_2 = p'_1 + p'_2 \nonumber\] or \[m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2. \nonumber\] Because the goalie is initially at rest, we know $v_2 = 0.$ Because the goalie catches the puck, the final velocities are equal, or $v'_1 = v'_2 = v'.$ Thus, the conservation of momentum equation simplifies to \[m_1v_1 = (m_1 + m_2)v'. \nonumber\] Solving for $v'$ yields \[v' = \dfrac{m_1}{m_1 + m_2}v_1. \nonumber\] Entering known values in this equation, we get \[\begin{align} v' &= \left( \dfrac{0.150 \, kg}{70.0 \, kg + 0.150 \, kg} \right)(35.0 \, m/s) \\[5pt] &= 7.48 \times 10^{-2} m/s)^2 \\[5pt] &= 0.196 \, J. \end{align}\] The change in internal kinetic energy is thus \[\begin{align} KE'_{int} - KE_{int} &= 0.196 \, J - 91.9 \, J \\[5pt] &= -91.7 \, J \end{align}\] where the minus sign indicates that the energy was lost. Discussion for (b) Nearly all of the initial internal kinetic energy is lost in this perfectly inelastic collision. $KE_{int} $ is mostly converted to thermal energy and sound. During some collisions, the objects do not stick together and less of the internal kinetic energy is removed—such as happens in most automobile accidents. Alternatively, stored energy may be converted into internal kinetic energy during a collision. Figure $\PageIndex{3}$ shows a one-dimensional example in which two carts on an air track collide, releasing potential energy from a compressed spring. Example $\PageIndex{2}$ deals with data from such a collision. Collisions are particularly important in sports and the sporting and leisure industry utilizes elastic and inelastic collisions. Let us look briefly at tennis. Recall that in a collision, it is momentum and not force that is important. So, a heavier tennis racquet will have the advantage over a lighter one. This conclusion also holds true for other sports—a lightweight bat (such as a softball bat) cannot hit a hardball very far. The location of the impact of the tennis ball on the racquet is also important, as is the part of the stroke during which the impact occurs. A smooth motion results in the maximizing of the velocity of the ball after impact and reduces sports injuries such as tennis elbow. A tennis player tries to hit the ball on the “sweet spot” on the racquet, where the vibration and impact are minimized and the ball is able to be given more velocity. Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations. Take-Home Experiment—Bouncing of Tennis Ball - Find a racquet (a tennis, badminton, or other racquet will do). Place the racquet on the floor and stand on the handle. Drop a tennis ball on the strings from a measured height. Measure how high the ball bounces. Now ask a friend to hold the racquet firmly by the handle and drop a tennis ball from the same measured height above the racquet. Measure how high the ball bounces and observe what happens to your friend’s hand during the collision. Explain your observations and measurements. - The coefficient of restitution $(c)$ is a measure of the elasticity of a collision between a ball and an object, and is defined as the ratio of the speeds after and before the collision. A perfectly elastic collision has a $c$ of 1. For a ball bouncing off the floor (or a racquet on the floor), $c$ can be shown to be $c = (h/H)^{1/2} $ where $h$ is the height to which the ball bounces and $H$ is the height from which the ball is dropped. Determine $c$ for the cases in Part 1 and for the case of a tennis ball bouncing off a concrete or wooden floor ($c = 0.85 $ for new tennis balls used on a tennis court). Example $\PageIndex{2}$: Calculating Final Velocity and Energy Release - Two Carts Collide In the collision pictured in Figure $\PageIndex{3}$, two carts collide inelastically. Cart 1 (denoted $m_1$ carries a spring which is initially compressed. During the collision, the spring releases its potential energy and converts it to internal kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and the cart and the spring together have an initial velocity of $-0.500 \, m/s$. After the collision, cart 1 is observed to recoil with a velocity of $-4.00 \, m/s$. - What is the final velocity of cart 2? - How much energy was released by the spring (assuming all of it was converted into internal kinetic energy)? Strategy We can use conservation of momentum to find the final velocity of cart 2, because $F_{net} = 0$ (the track is frictionless and the force of the spring is internal). Once this velocity is determined, we can compare the internal kinetic energy before and after the collision to see how much energy was released by the spring. Solution for (a) As before, the equation for conservation of momentum in a two-object system is \[m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2. \nonumber\] The only unknown in this equation is $v'_2.$ Solving for $v'_2$ and substituting known values into the previous equation fields \[\begin{align} v'_2 &= \dfrac{m_1v_1 + m_2v_2 - m_1v'_1}{m_2} \\[5pt] &= \dfrac{0.350 \, kg)(2.00 \, m/s) + (0.500 \, kg)(-0.500 \, m/s)}{0.500 \, kg} - \dfrac{(0.350 \, kg)(-4.00 \, m/s)}{0.500 \, kg} \\[5pt] &= 3.70 \, m/s.\end{align}\] Solution for (b) The internal kinetic energy before the collision is \[\begin{align} KE_{int} &= \dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2} m_2v_2^2 \\[5pt] &= \dfrac{1}{2}(0.350 \, kg)(2.00 \, m/s)^2 + \dfrac{1}{2}(0.500 \, kg)(-0.500 \, m/s)^2 \\[5pt] &= 0.763 \, J. \end{align}\] After the collision, the internal kinetic energy is \[\begin{align} KE'_{int} &= \dfrac{1}{2}m_1v_1^{'2} + \dfrac{1}{2} m_2v_2^{'2} \\[5pt] &= \dfrac{1}{2}(0.350 \, kg)(-4.00 \, m/s)^2 + \dfrac{1}{2}(0.500 \, kg)(0.370 \, m/s)^2 \\[5pt] &= 6.22 \, J. \end{align}\] The change in internal kinetic energy is thus \[\begin{align} KE' - KE &= 6.22 \, J - 0.763 \, J \\[5pt] &= 5.46 \, J. \end{align}\] Discussion The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. The internal kinetic energy in this collision increases by 5.46 J. That energy was released by the spring. Summary - An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). - A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision. - Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations. Glossary - inelastic collision - a collision in which internal kinetic energy is not conserved - perfectly inelastic collision - a collision in which the colliding objects stick together	libretexts	2025-03-17T19:53:27.634159	2015-11-01T05:04:36	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.05%3A_Inelastic_Collisions_in_One_Dimension", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "8.5: Inelastic Collisions in One Dimension", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.06%3A_Collisions_of_Point_Masses_in_Two_Dimensions	8.6: Collisions of Point Masses in Two Dimensions Learning Objectives By the end of this section, you will be able to: - Discuss two dimensional collisions as an extension of one dimensional analysis. - Define point masses. - Derive an expression for conservation of momentum along x -axis and y -axis. - Describe elastic collisions of two objects with equal mass. - Determine the magnitude and direction of the final velocity given initial velocity, and scattering angle. In the previous two sections, we considered only one-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and we shall see that their study is an extension of the one-dimensional analysis already presented. The approach taken (similar to the approach in discussing two-dimensional kinematics and dynamics) is to choose a convenient coordinate system and resolve the motion into components along perpendicular axes. Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously. One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses —that is, structureless particles that cannot rotate or spin. We start by assuming that $F_{net} = 0,$ so that momentum $p$ is conserved. The simplest collision is one in which one of the particles is initially at rest. (See Figure .) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure . Because momentum is conserved, the components of momentum along the $x-$ and $y-$ axes ($p_x $ and $p_y$) will also be conserved, but with the chosen coordinate system, $p_y$ is initially zero and $p_x$ is the momentum of the incoming particle. Both facts simplify the analysis. (Even with the simplifying assumptions of point masses, one particle initially at rest, and a convenient coordinate system, we still gain new insights into nature from the analysis of two-dimensional collisions.) Along the $x$-axis, the equation for conservation of momentum is \[p_{1x} + p_{2x} = p'_{1x} + p'_{2x} .\] Where the subscripts denote the particles and axes and the primes denote the situation after the collision. In terms of masses and velocities, this equation is \[m_1v_1 + m_2v_{2x} = m_1v'_{1x} + m_2v'_{2x}.\] But because particle 2 is initially at rest, this equation becomes \[m_1v_1 = m_1v'_{1x} + m_2v'_{2x}.\] The components of the velocities along the $x$-axis have the form $v \, cos \, \theta$. Because particle 1 initially moves along the $x$-axis, we find $v_{1x} = v_1$. Conservation of momentum along the $x$-axis gives the following equation \[ m_1v_1 = m_1v'_1 \, cos \, \theta_1 + m_2v'_2 \, cos \, \theta_2,\] where $\theta_1$ and $\theta_2$ are as shown in Figure . Conservation of Momentum Along the x-axis \[ m_1v_1 = m_1v'_1 \, cos \, \theta_1 + m_2v'_2 \, cos \, \theta_2\] Along the $y$-axis, the equation for conservation of momentum is \[p_{1y} + p_{2y} = p'_{1y} + p'_{2y},\] or \[m_1v_1 + m_2v_{2y} = m_1v'_{1y} + m_2v'_{2y}.\] But $v_{1y} $ is zero, because particle 1 initially moves along the $x$-axis. Because particle 2 is initially at rest, $v_{2y}$ is also zero. The equation for conservation of momentum along the $y$-axis becomes \[0 = m_1v'_{1y} + m_2v'_{2y}.\] The components of the velocities along the $y$-axis have the form $v \, sin \, \theta$. Thus, conservation of momentum along the $y$-axis gives the following equation: \[0 = m_1v'_{1y} \, sin \, \theta_1 + m_2v'_{2y} \, sin \, \theta_2.\] Conservation of Momentum Along y-axis \[0 = m_1v'_{1y} \, sin \, \theta_1 + m_2v'_{2y} \, sin \, \theta_2.\] The equations of conservation of momentum along the $x$-axis and $y$-axis are very useful in analyzing two-dimensional collisions of particles, where one is originally stationary (a common laboratory situation). But two equations can only be used to find two unknowns, and so other data may be necessary when collision experiments are used to explore nature at the subatomic level. Example $\PageIndex{1}$: Determining the Final Velocity of an Unseen Object from the Scattering of Another Object Suppose the following experiment is performed. A 0.250-kg object $(m_1)$ is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg $(m_2)$. The 0.250-kg object emerges from the room at an angle of $45^o$ with its incoming direction.The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity $(v'_2$ and $\theta_2)$ of the 0.400-kg object after the collision. Strategy Momentum is conserved because the surface is frictionless. The coordinate system shown in Figure is one in which $m_2$ is originally at rest and the initial velocity is parallel to the $x$-axis, so that conservation of momentum along the $x$- and $y$-axes is applicable. Everything is known in these equations except $v'_2$ and $\theta_2$, which are precisely the quantities we wish to find. We can find two unknowns because we have two independent equations: the equations describing the conservation of momentum in the $x$- and $y$-directions. Solution Solving $m_1v_1 = m_1v'_1 \, cos \, \theta_1 + m_2v'_2 \, cos \, \theta_2$ for $v'_2 \, cos \, \theta_2$ and $0 = m_1v'_{1y} \, sin \, \theta_1 + m_2v'_{2y} \, sin \, \theta_2$ for $v'_2 \, sin \, \theta_2$ and taking the ratio yields an equation (in which $\theta_2$ is the only unknown quantity. Applying the identity $\left(tan \, \theta = \frac{sin \, \theta}{cos \, \theta} \right) $, we obtain \[tan \, \theta_2 = \dfrac{v'_1 \, sin \, \theta_1}{v'_1 \, cos \, \theta_1 - v_1}.\] Entering known values into the previous equation gives \[tan \, \theta_2 = \dfrac{(1.50 \, m/s)(0.7071)}{(1.50 \, m/s)(0.7071) - 2.00 \, m/s} = -1.129.\] Thus, \[\theta_2 = tan^{-1}9-1.129) = 311.5^o \approx 312^o.\] Angles are defined as positive in the counter clockwise direction, so this angle indicates that $m_2$ is scattered to the right in Figure , as expected (this angle is in the fourth quadrant). Either equation for the $x$- or $y$-axis can now be used to solve for $v_2$, but the latter equation is easiest because it has fewer terms. \[v'_2 = - \left( \dfrac{0.250 \, kg}{0.400 \, kg} \right) (1.50 \, m/s) \left(\dfrac{0.7071}{-0.7485} \right).\] Thus, \[v'_2 = 0.886 \, m/s.\] Discussion It is instructive to calculate the internal kinetic energy of this two-object system before and after the collision. (This calculation is left as an end-of-chapter problem.) If you do this calculation, you will find that the internal kinetic energy is less after the collision, and so the collision is inelastic. This type of result makes a physicist want to explore the system further. Elastic Collisions of Two Objects with Equal Mass Some interesting situations arise when the two colliding objects have equal mass and the collision is elastic. This situation is nearly the case with colliding billiard balls, and precisely the case with some subatomic particle collisions. We can thus get a mental image of a collision of subatomic particles by thinking about billiards (or pool). (Refer to Figure for masses and angles.) First, an elastic collision conserves internal kinetic energy. Again, let us assume object 2 $m_2$ is initially at rest. Then, the internal kinetic energy before and after the collision of two objects that have equal masses is \[\dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_1^{'2} + \dfrac{1}{2}mv_2^{'2}.\] Because the masses are equal, $m_1 = m_2 = m$. Algebraic manipulation (left to the reader) of conservation of momentum in the $x$- and $y$-directions can show that \[\dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_1^{'2} + \dfrac{1}{2}mv_2^{'2} + mv'_1 v'_2 \, cos (\theta_1 - \theta_2).\] (Remember that $\theta_2$ is negative here.) The two preceding equations can both be true only if \[mv'_1 v'_2 \, cos (\theta_1 - \theta_2) = 0.\] There are three ways that this term can be zero. They are $v'_1 = 0$: head-on collision; incoming ball stops; $v'_2 = 0$: no collision; incoming ball continues unaffected All three of these ways are familiar occurrences in billiards and pool, although most of us try to avoid the second. If you play enough pool, you will notice that the angle between the balls is very close to 90º after the collision, although it will vary from this value if a great deal of spin is placed on the ball. (Large spin carries in extra energy and a quantity called angular momentum , which must also be conserved.) The assumption that the scattering of billiard balls is elastic is reasonable based on the correctness of the three results it produces. This assumption also implies that, to a good approximation, momentum is conserved for the two-ball system in billiards and pool. The problems below explore these and other characteristics of two-dimensional collisions. CONNECTIONS TO NUCLEAR AND PARTICLE PHYSICS Two-dimensional collision experiments have revealed much of what we know about subatomic particles, as we shall see in Medical Applications of Nuclear Physics and Particle Physics . Ernest Rutherford, for example, discovered the nature of the atomic nucleus from such experiments. Summary - The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the x-axis parallel to the velocity of the incoming particle. - Two-dimensional collisions of point masses where mass 2 is initially at rest conserve momentum along the initial direction of mass 1 (the x-axis), stated by $m_1v_1=m_1v′_1cosθ_1+m_2v′_2cosθ_2$ and along the direction perpendicular to the initial direction (the y-axis) stated by $0=m_1v′_1y+m_2v′2_y$. - The internal kinetic before and after the collision of two objects that have equal masses is $\frac{1}{2}mv_1^2=\frac{1}{2}mv′_1^2+\frac{1}{2}mv′_2^2+mv′_1v′_2cos(θ_1−θ_2)$. - Point masses are structureless particles that cannot spin. Glossary - point masses - structureless particles with no rotation or spin	libretexts	2025-03-17T19:53:27.710483	2015-11-01T05:04:57	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.06%3A_Collisions_of_Point_Masses_in_Two_Dimensions", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "8.6: Collisions of Point Masses in Two Dimensions", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.07%3A_Introduction_to_Rocket_Propulsion	8.7: Introduction to Rocket Propulsion Learning Objectives By the end of this section, you will be able to: - State Newton’s third law of motion. - Explain the principle involved in propulsion of rockets and jet engines. - Derive an expression for the acceleration of the rocket. - Discuss the factors that affect the rocket’s acceleration. - Describe the function of a space shuttle. Rockets range in size from fireworks so small that ordinary people use them to immense Saturn Vs that once propelled massive payloads toward the Moon. The propulsion of all rockets, jet engines, deflating balloons, and even squids and octopuses is explained by the same physical principle—Newton’s third law of motion. Matter is forcefully ejected from a system, producing an equal and opposite reaction on what remains. Another common example is the recoil of a gun. The gun exerts a force on a bullet to accelerate it and consequently experiences an equal and opposite force, causing the gun’s recoil or kick. Making Connections: Take-Home Experiment —Propulsion of a Balloon - Hold a balloon and fill it with air. Then, let the balloon go. In which direction does the air come out of the balloon and in which direction does the balloon get propelled? If you fill the balloon with water and then let the balloon go, does the balloon’s direction change? Explain your answer. Figure $\PageIndex{1}$ shows a rocket accelerating straight up. In part (a), the rocket has a mass $m$ and a velocity $v$ relative to Earth, and hence a momentum $mv$ In part (b), a time $\Delta t$ has elapsed in which the rocket has ejected a mass $\Delta m$ of hot gas at a velocity $v_e$ relative to the rocket. The remainder of the mass $(m - \Delta m)$ now has a greater velocity $(v + \Delta v)$. The momentum of the entire system (rocket plus expelled gas) has actually decreased because the force of gravity has acted for a time $\Delta t$, producing a negative impulse $\Delta p = -mg\Delta t$. (Remember that impulse is the net external force on a system multiplied by the time it acts, and it equals the change in momentum of the system.) So, the center of mass of the system is in free fall but, by rapidly expelling mass, part of the system can accelerate upward. It is a commonly held misconception that the rocket exhaust pushes on the ground. If we consider thrust; that is, the force exerted on the rocket by the exhaust gases, then a rocket’s thrust is greater in outer space than in the atmosphere or on the launch pad. In fact, gases are easier to expel into a vacuum.By calculating the change in momentum for the entire system over $\Delta t$, and equating this change to the impulse, the following expression can be shown to be a good approximation for the acceleration of the rocket. \[a = \dfrac{v_e}{m} \dfrac{\Delta m}{\Delta t} - g,\] where $a$ is the acceleration of the rocket, $v_e$ is the escape velocity, $m$ is the mass of the rocket, $\ Delta m$ is the mass of the ejected gas, and $\Delta t$ is the time in which the gas is ejected. A rocket’s acceleration depends on three major factors, consistent with the equation for acceleration of a rocket . First, the greater the exhaust velocity of the gases relative to the rocket, $v_e$, the greater the acceleration is. The practical limit for $v_e$ is about $2.5 \times 10^3 \, m/s$ for conventional (non-nuclear) hot-gas propulsion systems. The second factor is the rate at which mass is ejected from the rocket. This is the factor $(\Delta m/\Delta t)v_e$, with units of newtons, is called "thrust.” The faster the rocket burns its fuel, the greater its thrust, and the greater its acceleration. The third factor is the mass $m$ of the rocket. The smaller the mass is (all other factors being the same), the greater the acceleration. The rocket mass $m$ decreases dramatically during flight because most of the rocket is fuel to begin with, so that acceleration increases continuously, reaching a maximum just before the fuel is exhausted. Factors Affecting a Rocket’s Acceleration - The greater the exhaust velocity $v_e$ of the gases relative to the rocket, the greater the acceleration. - The faster the rocket burns its fuel, the greater its acceleration. - The smaller the rocket’s mass (all other factors being the same), the greater the acceleration. Example $\PageIndex{1}$ : Calculating Acceleration: Initial Acceleration of a Moon Launch A Saturn V’s mass at liftoff was $2.80 \times 10^6 \, kg$, its fuel-burn rate was $1.40 \times 10^4 \times kg/s$, and the exhaust velocity was $2.40 \times 10^3 m/s$. Calculate its initial acceleration. Strategy This problem is a straightforward application of the expression for acceleration because is the unknown and all of the terms on the right side of the equation are given. Solution Substituting the given values into the equation for acceleration yields \[\begin{align} a &= \dfrac{v_e}{m} \dfrac{\Delta m}{\delta t} - g \\[4pt] &= \dfrac{2.40 \times 10^3 \, m/s}{2.80 \times 10^6 \, kg}(1.40 \times 10^4 \, kg/s) - 9.8 \, m/s^2 \\[4pt] &= 2.20 \, m/s^2. \end{align}\] Discussion This value is fairly small, even for an initial acceleration. The acceleration does increase steadily as the rocket burns fuel, because $m$ decreases while $v_e$ and $\frac{\Delta m}{\Delta t} $ remain constant. Knowing this acceleration and the mass of the rocket, you can show that the thrust of the engines was $3.36 \times 10^7 \, N.$ To achieve the high speeds needed to hop continents, obtain orbit, or escape Earth’s gravity altogether, the mass of the rocket other than fuel must be as small as possible. It can be shown that, in the absence of air resistance and neglecting gravity, the final velocity of a one-stage rocket initially at rest is \[v = v_e \, ln \dfrac{m_0}{m_r},\] where $ln (m_0/m_r) $ is the natural logarithm of the ratio of the initial mass of the rocket $(m_0)$ to what is left $(m_r)$ after all of the fuel is exhausted. (Note that $v$ is actually the change in velocity, so the equation can be used for any segment of the flight. If we start from rest, the change in velocity equals the final velocity.) For example, let us calculate the mass ratio needed to escape Earth’s gravity starting from rest, given that the escape velocity from Earth is about $11.2 \times 10^3 \, m/s$, and assuming an exhaust velocity $v_e = 2.5 \times 10^3 \, m/s.$ \[ln \dfrac{m_0}{m_r} = \dfrac{v}{v_e} = \dfrac{11.2 \times 10^3 \, m/s}{2.5 \times 10^3 \, m/s} = 4.48 \] Solving for $m_0/m_r$ gives \[\dfrac{m_0}{m_r} = e^{4.48} = 88.\] Thus, the mass of the rocket is \[ m_r = \dfrac{m_0}{88}.\] This result means that only $1/88$ of the mass is left when the fuel is burnt, and $87/88$ of the initial mass was fuel. Expressed as percentages, 98.9% of the rocket is fuel, while payload, engines, fuel tanks, and other components make up only 1.10%. Taking air resistance and gravitational force into account, the mass $m_r$ remaining can only be about $m_0/180$. It is difficult to build a rocket in which the fuel has a mass 180 times everything else. The solution is multistage rockets. Each stage only needs to achieve part of the final velocity and is discarded after it burns its fuel. The result is that each successive stage can have smaller engines and more payload relative to its fuel. Once out of the atmosphere, the ratio of payload to fuel becomes more favorable, too.The space shuttle was an attempt at an economical vehicle with some reusable parts, such as the solid fuel boosters and the craft itself. (See Figure ) The shuttle’s need to be operated by humans, however, made it at least as costly for launching satellites as expendable, unmanned rockets. Ideally, the shuttle would only have been used when human activities were required for the success of a mission, such as the repair of the Hubble space telescope. Rockets with satellites can also be launched from airplanes. Using airplanes has the double advantage that the initial velocity is significantly above zero and a rocket can avoid most of the atmosphere’s resistance. Phet Explorations: Lunar Lander Can you avoid the boulder field and land safely, just before your fuel runs out, as Neil Armstrong did in 1969? Our version of this classic video game accurately simulates the real motion of the lunar lander with the correct mass, thrust, fuel consumption rate, and lunar gravity. The real lunar lander is very hard to control. Phet Explorations: Lunar Lander Can you avoid the boulder field and land safely, just before your fuel runs out, as Neil Armstrong did in 1969? Our version of this classic video game accurately simulates the real motion of the lunar lander with the correct mass, thrust, fuel consumption rate, and lunar gravity. The real lunar lander is very hard to control. Summary This page connects force to thrust and weight of a rocket and foxes on Newton's second law: $a=\frac{F}{mass}.$ - Newton’s third law of motion states that to every action, there is an equal and opposite reaction. - Acceleration of a rocket is $a = \frac{v_e}{m} \frac{\Delta m}{\Delta t} - g.$ - A rocket’s acceleration depends on three main factors. They are - The greater the exhaust velocity of the gases, the greater the acceleration. - The faster the rocket burns its fuel, the greater its acceleration. - The smaller the rocket's mass, the greater the acceleration.	libretexts	2025-03-17T19:53:27.781824	2015-11-01T05:05:20	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.07%3A_Introduction_to_Rocket_Propulsion", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "8.7: Introduction to Rocket Propulsion", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.E%3A_Linear_Momentum_and_Collisions_(Exercises)	8.E: Linear Momentum and Collisions (Exercises) - - Last updated - Save as PDF Conceptual Questions 8.1: Linear Momentum and Force 1. An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy? 2. An object that has a small mass and an object that has a large mass have the same kinetic energy. Which mass has the largest momentum? 3. Professional Application 4. Football coaches advise players to block, hit, and tackle with their feet on the ground rather than by leaping through the air. Using the concepts of momentum, work, and energy, explain how a football player can be more effective with his feet on the ground. 5. How can a small force impart the same momentum to an object as a large force? 8.2: Impulse 6. Professional Application Explain in terms of impulse how padding reduces forces in a collision. State this in terms of a real example, such as the advantages of a carpeted vs. tile floor for a day care center. 7. While jumping on a trampoline, sometimes you land on your back and other times on your feet. In which case can you reach a greater height and why? 8. Professional Application Tennis racquets have “sweet spots.” If the ball hits a sweet spot then the player's arm is not jarred as much as it would be otherwise. Explain why this is the case. 8.3: Conservation of Momentum 9. Professional Application If you dive into water, you reach greater depths than if you do a belly flop. Explain this difference in depth using the concept of conservation of energy. Explain this difference in depth using what you have learned in this chapter. 10. Under what circumstances is momentum conserved? 11. Can momentum be conserved for a system if there are external forces acting on the system? If so, under what conditions? If not, why not? 12. Momentum for a system can be conserved in one direction while not being conserved in another. What is the angle between the directions? Give an example. 13. Professional Application Explain in terms of momentum and Newton’s laws how a car’s air resistance is due in part to the fact that it pushes air in its direction of motion. 14. Can objects in a system have momentum while the momentum of the system is zero? Explain your answer. 15. Must the total energy of a system be conserved whenever its momentum is conserved? Explain why or why not. 8.5: Inelastic Collisions in One Dimension 17. What is an inelastic collision? What is a perfectly inelastic collision? 18. Mixed-pair ice skaters performing in a show are standing motionless at arms length just before starting a routine. They reach out, clasp hands, and pull themselves together by only using their arms. Assuming there is no friction between the blades of their skates and the ice, what is their velocity after their bodies meet? 19. A small pickup truck that has a camper shell slowly coasts toward a red light with negligible friction. Two dogs in the back of the truck are moving and making various inelastic collisions with each other and the walls. What is the effect of the dogs on the motion of the center of mass of the system (truck plus entire load)? What is their effect on the motion of the truck? 8.6: Collisions of Point Masses in Two Dimensions 19. Figure shows a cube at rest and a small object heading toward it. (a) Describe the directions (angle $\displaystyle θ_1$) at which the small object can emerge after colliding elastically with the cube. How does $\displaystyle θ_1$ depend on $\displaystyle b$, the so-called impact parameter? Ignore any effects that might be due to rotation after the collision, and assume that the cube is much more massive than the small object. (b) Answer the same questions if the small object instead collides with a massive sphere. A small object approaches a collision with a much more massive cube, after which its velocity has the direction $\displaystyle θ_1$. The angles at which the small object can be scattered are determined by the shape of the object it strikes and the impact parameter $\displaystyle b$. 8.7: Introduction to Rocket Propulsion 20. Professional Application Suppose a fireworks shell explodes, breaking into three large pieces for which air resistance is negligible. How is the motion of the center of mass affected by the explosion? How would it be affected if the pieces experienced significantly more air resistance than the intact shell? 21. Professional Application During a visit to the International Space Station, an astronaut was positioned motionless in the center of the station, out of reach of any solid object on which he could exert a force. Suggest a method by which he could move himself away from this position, and explain the physics involved. 22. Professional Application It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. When that is the case, the gas velocity and gas momentum are in the same direction as that of the rocket. How is the rocket still able to obtain thrust by ejecting the gases? Problems & Exercises 8.1: Linear Momentum and Force 23. (a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of 7.50 m/s size 12{7 "." "50"``"m/s"} {}. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of 600 m/s size 12{"600"``"m/s"} {}. (c) What is the momentum of the 90.0-kg hunter running at 7.40 m/s size 12{7 "." "40"``"m/s"} {} after missing the elephant? Solution (a) $\displaystyle 1.50×10^4kg⋅m/s$ (b) 625 to 1 (c) $\displaystyle 6.66×10^2kg⋅m/s$ 24. (a) What is the mass of a large ship that has a momentum of $\displaystyle 1.60×10^9kg⋅m/s$, when the ship is moving at a speed of $\displaystyle 48.0 km/h$? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of $\displaystyle 1200 m/s$. 25. (a) At what speed would a $\displaystyle 2.00×10^4-kg$ airplane have to fly to have a momentum of $\displaystyle 1.60×10^9kg⋅m/s$ (the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of $\displaystyle 60.0 m/s$? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship. Solution (a) $\displaystyle 8.00×10^4m/s$ (b) $\displaystyle 1.20×10^6kg⋅m/s$ (c) Because the momentum of the airplane is 3 orders of magnitude smaller than of the ship, the ship will not recoil very much. The recoil would be $\displaystyle −0.0100 m/s$, which is probably not noticeable. 26. (a) What is the momentum of a garbage truck that is $\displaystyle 1.20×10^4kg$ and is moving at $\displaystyle 10.0 m/s$ (b) At what speed would an 8.00-kg trash can have the same momentum as the truck? 27. A runaway train car that has a mass of 15,000 kg travels at a speed of $\displaystyle 5.4 m/s$ down a track. Compute the time required for a force of 1500 N to bring the car to rest. Solution 54 s 28. The mass of Earth is $\displaystyle 5.972×10^{24}kg$ and its orbital radius is an average of $\displaystyle 1.496×10^{11}m$. Calculate its linear momentum. 8.2: Impulse 29. A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms (milliseconds)? Solution $\displaystyle 9.00×10^3N$ 30. Professional Application A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg. 31. A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds from an initial speed of 4.00 m/s. (a) What is the average force exerted on the leg, taking the effective mass of the hand and forearm to be 1.50 kg? (b) Would the force be any different if the woman clapped her hands together at the same speed and brought them to rest in the same time? Explain why or why not. Solution a) $\displaystyle 2.40×10^3N$ toward the leg b) The force on each hand would have the same magnitude as that found in part (a) (but in opposite directions by Newton’s third law) because the change in momentum and the time interval are the same. 32. Professional Application A professional boxer hits his opponent with a 1000-N horizontal blow that lasts for 0.150 s. (a) Calculate the impulse imparted by this blow. (b) What is the opponent’s final velocity, if his mass is 105 kg and he is motionless in midair when struck near his center of mass? (c) Calculate the recoil velocity of the opponent’s 10.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer’s body. (d) Discuss the implications of your answers for parts (b) and (c). 33. Professional Application Suppose a child drives a bumper car head on into the side rail, which exerts a force of 4000 N on the car for 0.200 s. (a) What impulse is imparted by this force? (b) Find the final velocity of the bumper car if its initial velocity was 2.80 m/s and the car plus driver have a mass of 200 kg. You may neglect friction between the car and floor. Solution a) $\displaystyle 800 kg⋅m/s$ away from the wall b) $\displaystyle 1.20 m/s$ away from the wall 34. Professional Application One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of $\displaystyle 4.00×10^3m/s$ , given the collision lasts $\displaystyle 6.00×10^{–8}s$. 35. Professional Application A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) Calculate the average force on the person if he is stopped by a padded dashboard that compresses an average of 1.00 cm. (b) Calculate the average force on the person if he is stopped by an air bag that compresses an average of 15.0 cm. Solution (a) $\displaystyle 1.50×10^6N$ away from the dashboard (b) $\displaystyle 1.00×10^5N$ away from the dashboard 36. Professional Application Military rifles have a mechanism for reducing the recoil forces of the gun on the person firing it. An internal part recoils over a relatively large distance and is stopped by damping mechanisms in the gun. The larger distance reduces the average force needed to stop the internal part. (a) Calculate the recoil velocity of a 1.00-kg plunger that directly interacts with a 0.0200-kg bullet fired at 600 m/s from the gun. (b) If this part is stopped over a distance of 20.0 cm, what average force is exerted upon it by the gun? (c) Compare this to the force exerted on the gun if the bullet is accelerated to its velocity in 10.0 ms (milliseconds). 37. A cruise ship with a mass of $\displaystyle 1.00×10^7kg$ strikes a pier at a speed of 0.750 m/s. It comes to rest 6.00 m later, damaging the ship, the pier, and the tugboat captain’s finances. Calculate the average force exerted on the pier using the concept of impulse. (Hint: First calculate the time it took to bring the ship to rest.) Solution $\displaystyle 4.69×10^5N$ in the boat’s original direction of motion 38. Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalpost and experiences a backward force of $\displaystyle 1.76×10^4N$ for $\displaystyle 5.50×10^{–2}s$. 39. Water from a fire hose is directed horizontally against a wall at a rate of 50.0 kg/s and a speed of 42.0 m/s. Calculate the magnitude of the force exerted on the wall, assuming the water’s horizontal momentum is reduced to zero. Solution $\displaystyle 2.10×10^3N$ away from the wall 40. A 0.450-kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving the nail 1.00 cm into a board. (a) Calculate the duration of the impact. (b) What was the average force exerted on the nail? 41. Starting with the definitions of momentum and kinetic energy, derive an equation for the kinetic energy of a particle expressed as a function of its momentum. Solution $\displaystyle p=mv⇒p^2=m^2v^2⇒\frac{p^2}{m}=mv^2$⇒$\displaystyle \frac{p^2}{2m}=\frac{1}{2}mv^2=KE$ $\displaystyle KE=\frac{p^2}{2m}$ 42. A ball with an initial velocity of 10 m/s moves at an angle $\displaystyle 60º$ above the $\displaystyle +x$-direction. The ball hits a vertical wall and bounces off so that it is moving $\displaystyle 60º$ above the $\displaystyle −x$-direction with the same speed. What is the impulse delivered by the wall? 43. When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racquet exerts a force of 540 N on the ball for 5.00 ms, giving it a final velocity of 45.0 m/s. Using these data, find the mass of the ball. Solution 60.0 g 44. A punter drops a ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at an angle 55º size 12{"55"°} {} above the horizontal. What is the impulse delivered by the foot (magnitude and direction)? 8.3: Conservation of Momentum 45. Professional Application Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 150,000 kg and a velocity of 0.300 m/s, and the second having a mass of 110,000 kg and a velocity of $\displaystyle −0.120 m/s$. (The minus indicates direction of motion.) What is their final velocity? Solution 0.122 m/s 46. Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay model, which is initially motionless and has a mass of 0.350 kg. Both being soft clay, they naturally stick together. What is their final velocity? 47. Professional Application Consider the following question: A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seatbelt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg. Would the answer to this question be different if the car with the 70-kg passenger had collided with a car that has a mass equal to and is traveling in the opposite direction and at the same speed? Explain your answer. Solution In a collision with an identical car, momentum is conserved. Afterwards $\displaystyle v_f=0$ for both cars. The change in momentum will be the same as in the crash with the tree. However, the force on the body is not determined since the time is not known. A padded stop will reduce injurious force on body. 48. What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 12.0 m/s in the same direction? Assume the deer remains on the car. 49. A 1.80-kg falcon catches a 0.650-kg dove from behind in midair. What is their velocity after impact if the falcon’s velocity is initially 28.0 m/s and the dove’s velocity is 7.00 m/s in the same direction? Solution 22.4 m/s in the same direction as the original motion 8.4: Elastic Collisions in One Dimension 50. Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved. 51. Professional Application Two manned satellites approach one another at a relative speed of 0.250 m/s, intending to dock. The first has a mass of $\displaystyle 4.00×10^3kg$, and the second a mass of $\displaystyle 7.50×10^3kg$. If the two satellites collide elastically rather than dock, what is their final relative velocity? Solution 0.250 m/s 52. A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities be in this case? 8.5: Inelastic Collisions in One Dimension 53. A 0.240-kg billiard ball that is moving at 3.00 m/s strikes the bumper of a pool table and bounces straight back at 2.40 m/s (80% of its original speed). The collision lasts 0.0150 s. (a) Calculate the average force exerted on the ball by the bumper. (b) How much kinetic energy in joules is lost during the collision? (c) What percent of the original energy is left? Solution (a) 86.4 N perpendicularly away from the bumper (b) 0.389 J (c) 64.0% 54. During an ice show, a 60.0-kg skater leaps into the air and is caught by an initially stationary 75.0-kg skater. (a) What is their final velocity assuming negligible friction and that the 60.0-kg skater’s original horizontal velocity is 4.00 m/s? (b) How much kinetic energy is lost? 55. Professional Application Using mass and speed data from [link] and assuming that the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: (a) the final velocity if the ball and player are going in the same direction and (b) the loss of kinetic energy in this case. (c) Repeat parts (a) and (b) for the situation in which the ball and the player are going in opposite directions. Might the loss of kinetic energy be related to how much it hurts to catch the pass? Solution (a) 8.06 m/s (b) -56.0 J (c)(i) 7.88 m/s; (ii) -223 J 56. A battleship that is $\displaystyle 6.00×10^7kg$ and is originally at rest fires a 1100-kg artillery shell horizontally with a velocity of 575 m/s. (a) If the shell is fired straight aft (toward the rear of the ship), there will be negligible friction opposing the ship’s recoil. Calculate its recoil velocity. (b) Calculate the increase in internal kinetic energy (that is, for the ship and the shell). This energy is less than the energy released by the gun powder—significant heat transfer occurs. 57. Professional Application Two manned satellites approaching one another, at a relative speed of 0.250 m/s, intending to dock. The first has a mass of $\displaystyle 4.00×10^3kg$, and the second a mass of $\displaystyle 7.50×10^3kg$. (a) Calculate the final velocity (after docking) by using the frame of reference in which the first satellite was originally at rest. (b) What is the loss of kinetic energy in this inelastic collision? (c) Repeat both parts by using the frame of reference in which the second satellite was originally at rest. Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both. Solution (a) 0.163 m/s in the direction of motion of the more massive satellite (b) 81.6 J (c) $\displaystyle 8.70×10^{−2}m/s$ in the direction of motion of the less massive satellite, 81.5 J. Because there are no external forces, the velocity of the center of mass of the two-satellite system is unchanged by the collision. The two velocities calculated above are the velocity of the center of mass in each of the two different individual reference frames. The loss in KE is the same in both reference frames because the KE lost to internal forces (heat, friction, etc.) is the same regardless of the coordinate system chosen. 58. Professional Application A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal into it. (a) What is the final velocity of the loaded freight car? (b) How much kinetic energy is lost? 59. Professional Application Space probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a 4800-kg satellite uses this method to separate from the 1500-kg remains of its launcher, and that 5000 J of kinetic energy is supplied to the two parts. What are their subsequent velocities using the frame of reference in which they were at rest before separation? Solution 0.704 m/s –2.25 m/s 60. A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse if you hold the gun loosely a few centimeters from your shoulder rather than holding it tightly against your shoulder. (a) Calculate the recoil velocity of the rifle if it is held loosely away from the shoulder. (b) How much kinetic energy does the rifle gain? (c) What is the recoil velocity if the rifle is held tightly against the shoulder, making the effective mass 28.0 kg? (d) How much kinetic energy is transferred to the rifle-shoulder combination? The pain is related to the amount of kinetic energy, which is significantly less in this latter situation. (e) Calculate the momentum of a 110-kg football player running at 8.00 m/s. Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s. Discuss its relationship to this problem. Solution (a) 4.58 m/s away from the bullet (b) 31.5 J (c) –0.491 m/s (d) 3.38 J 61. Professional Application One of the waste products of a nuclear reactor is plutonium-239 ($\displaystyle ^{239}Pu$). This nucleus is radioactive and decays by splitting into a helium-4 nucleus and a uranium-235 nucleus ($\displaystyle ^4He+^{235}U$), the latter of which is also radioactive and will itself decay some time later. The energy emitted in the plutonium decay is (8.40×10^{–13}J\) and is entirely converted to kinetic energy of the helium and uranium nuclei. The mass of the helium nucleus is $\displaystyle 6.68×10^{–27}kg$, while that of the uranium is $\displaystyle 3.92×10^{–25}kg$ (note that the ratio of the masses is 4 to 235). (a) Calculate the velocities of the two nuclei, assuming the plutonium nucleus is originally at rest. (b) How much kinetic energy does each nucleus carry away? Note that the data given here are accurate to three digits only. 62. Professional Application The Moon’s craters are remnants of meteorite collisions. Suppose a fairly large asteroid that has a mass of $\displaystyle 5.00×10^{12}kg\(\displaystyle about a kilometer across) strikes the Moon at a speed of 15.0 km/s. (a) At what speed does the Moon recoil after the perfectly inelastic collision (the mass of the Moon is \(\displaystyle 7.36×10^{22}kg$) ? (b) How much kinetic energy is lost in the collision? Such an event may have been observed by medieval English monks who reported observing a red glow and subsequent haze about the Moon. (c) In October 2009, NASA crashed a rocket into the Moon, and analyzed the plume produced by the impact. (Significant amounts of water were detected.) Answer part (a) and (b) for this real-life experiment. The mass of the rocket was 2000 kg and its speed upon impact was 9000 km/h. How does the plume produced alter these results? Solution (a) $\displaystyle 1.02×10^{−6}m/s$ (b) $\displaystyle 5.63×10^{20}J$ (almost all KE lost) (c) Recoil speed is $\displaystyle 6.79×10^{−17}m/s$, energy lost is $\displaystyle 6.25×10^9J$. The plume will not affect the momentum result because the plume is still part of the Moon system. The plume may affect the kinetic energy result because a significant part of the initial kinetic energy may be transferred to the kinetic energy of the plume particles. 63. Professional Application Two football players collide head-on in midair while trying to catch a thrown football. The first player is 95.0 kg and has an initial velocity of 6.00 m/s, while the second player is 115 kg and has an initial velocity of –3.50 m/s. What is their velocity just after impact if they cling together? 64. What is the speed of a garbage truck that is $\displaystyle 1.20×10^4kg$ and is initially moving at 25.0 m/s just after it hits and adheres to a trash can that is 80.0 kg and is initially at rest? Solution 24.8 m/s 65. During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of 10.0 kg and the horizontal component of its velocity is 8.00 m/s when the 65.0-kg performer catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity? 66. (a) During an ice skating performance, an initially motionless 80.0-kg clown throws a fake barbell away. The clown’s ice skates allow her to recoil frictionlessly. If the clown recoils with a velocity of 0.500 m/s and the barbell is thrown with a velocity of 10.0 m/s, what is the mass of the barbell? (b) How much kinetic energy is gained by this maneuver? (c) Where does the kinetic energy come from? Solution (a) 4.00 kg (b) 210 J (c) The clown does work to throw the barbell, so the kinetic energy comes from the muscles of the clown. The muscles convert the chemical potential energy of ATP into kinetic energy. 8.6: Collisions of Point Masses in Two Dimensions 67. Two identical pucks collide on an air hockey table. One puck was originally at rest. (a) If the incoming puck has a speed of 6.00 m/s and scatters to an angle of $\displaystyle 30.0º$,what is the velocity (magnitude and direction) of the second puck? (You may use the result that $\displaystyle θ_1−θ_2=90º$ for elastic collisions of objects that have identical masses.) (b) Confirm that the collision is elastic. Solution (a) 3.00 m/s, $\displaystyle 60º$ below $\displaystyle x$-axis (b) Find speed of first puck after collision: $\displaystyle 0=mv'_1sin30º−mv'_2sin60º⇒v'_1=v′_2\frac{sin60º}{sin30º=}5.196 m/s$ Verify that ratio of initial to final KE equals one: $\displaystyle KE=\frac{1}{2}mv_1^2=18mJ$ $\displaystyle KE=\frac{1}{2}mv_{1}^{'2}+\frac{1}{2}mv_2^{'2}=18mJ$ $\displaystyle \frac{KE}{KE'}=1.00$ 68. Confirm that the results of the example Example do conserve momentum in both the $\displaystyle x$- and $\displaystyle y$-directions. 69. A 3000-kg cannon is mounted so that it can recoil only in the horizontal direction. (a) Calculate its recoil velocity when it fires a 15.0-kg shell at 480 m/s at an angle of $\displaystyle 20.0º$ above the horizontal. (b) What is the kinetic energy of the cannon? This energy is dissipated as heat transfer in shock absorbers that stop its recoil. (c) What happens to the vertical component of momentum that is imparted to the cannon when it is fired? Solution (a) $\displaystyle −2.26m/s$ (b) $\displaystyle 7.63×10^3J$ (c) The ground will exert a normal force to oppose recoil of the cannon in the vertical direction. The momentum in the vertical direction is transferred to the earth. The energy is transferred into the ground, making a dent where the cannon is. After long barrages, cannon have erratic aim because the ground is full of divots. 70. Professional Application A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered at an angle of $\displaystyle 85.0º$ to the initial direction of the bowling ball and with a speed of 15.0 m/s. (a) Calculate the final velocity (magnitude and direction) of the bowling ball. (b) Is the collision elastic? (c) Linear kinetic energy is greater after the collision. Discuss how spin on the ball might be converted to linear kinetic energy in the collision. 71. Professional Application Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei ($\displaystyle ^4He$) from gold-197 nuclei ($\displaystyle ^{197}Au$). The energy of the incoming helium nucleus was $\displaystyle 8.00×10^{−13}J$, and the masses of the helium and gold nuclei were $\displaystyle 6.68×10^{−27}kg$ and $\displaystyle 3.29×10^{−25}kg$, respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of $\displaystyle 120º$ during an elastic collision with a gold nucleus, calculate the helium nucleus’s final speed and the final velocity (magnitude and direction) of the gold nucleus. (b) What is the final kinetic energy of the helium nucleus? Solution (a) $\displaystyle 5.36×10^5m/s$ at $\displaystyle −29.5º$ (b) $\displaystyle 7.52×10^{−13}J$ 72. Professional Application Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and is approaching at $\displaystyle 8.00m/s$ due south. The second car has a mass of 850 kg and is approaching at $\displaystyle 17.0m/s$ due west. (a) Calculate the final velocity (magnitude and direction) of the cars. (b) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.) Note that because both cars have an initial velocity, you cannot use the equations for conservation of momentum along the $\displaystyle x$-axis and $\displaystyle y$-axis; instead, you must look for other simplifying aspects. 73. Starting with equations $\displaystyle m_1v_1=m_1v'_1cosθ_1+m_2v'_2cosθ_2$ and $\displaystyle 0=m_1v'_1sinθ_1+m_2v'_2sinθ_2$ for conservation of momentum in the $\displaystyle x-$ and $\displaystyle y-$directions and assuming that one object is originally stationary, prove that for an elastic collision of two objects of equal masses, $\displaystyle \frac{1}{2}mv_1^2=\frac{1}{2}mv_1^{'2}+\frac{1}{2}mv_2^{'2}+mv'_1v'_2cos(θ_1−θ_2)$ as discussed in the text. Solution We are given that $\displaystyle m_1=m_2≡m$. The given equations then become: $\displaystyle v_1=v_1cosθ_1+v_2cosθ_2$ and $\displaystyle 0=v'_1sinθ_1+v'_2sinθ_2.$ Square each equation to get $\displaystyle v_1^2=v_1^{'2}cos^2θ_1+v_2^{'2}cos^2θ_2+2v'_1v'_2cosθ_1cosθ_2$ $\displaystyle 0=v_1^{'2}sin^2θ_1+v_2^{'2}sin^2θ_2+2v'_1v'_2sinθ_1sinθ_2.$ Add these two equations and simplify: $\displaystyle v_1^2=v_1^{'2}+v_2^{'2}+2v'_1v'_2(cosθ_1cosθ_2+sinθ_1sinθ_2)$ $\displaystyle =v_1^{'2}+v_2^{'2}+2v'_1v'_2[\frac{1}{2}cos(θ_1−θ_2)+\frac{1}{2}cos(θ_1+θ_2)+\frac{1}{2}cos(θ_1−θ_2)−\frac{1}{2}cos(θ_1+θ_2)]$ $\displaystyle =v_1^{'2}+v_2^{'2}+2v'_1v'_2cos(θ_1−θ_2).$ Multiply the entire equation by $\displaystyle \frac{1}{2}m$ to recover the kinetic energy: $\displaystyle \frac{1}{2}mv_1^2=\frac{1}{2}mv_1^{'2}+\frac{1}{2}mv_2^{'2}+mv'_1v'_2cos(θ_1−θ_2)$ 74. Integrated Concepts A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 15.0 m away? 8.7: Introduction to Rocket Propulsion 75. Professional Application Antiballistic missiles (ABMs) are designed to have very large accelerations so that they may intercept fast-moving incoming missiles in the short time available. What is the takeoff acceleration of a 10,000-kg ABM that expels 196 kg of gas per second at an exhaust velocity of $\displaystyle 2.50×10^3m/s$? Solution $\displaystyle 39.2m/s^2$ 76. Professional Application What is the acceleration of a 5000-kg rocket taking off from the Moon, where the acceleration due to gravity is only $\displaystyle 1.6m/s^2$, if the rocket expels 8.00 kg of gas per second at an exhaust velocity of $\displaystyle 2.20×10^3m/s$? 77. Professional Application Calculate the increase in velocity of a 4000-kg space probe that expels 3500 kg of its mass at an exhaust velocity of $\displaystyle 2.00×10^3m/s$. You may assume the gravitational force is negligible at the probe’s location. Solution $\displaystyle 4.16×10^3m/s$ 78. Professional Application Ion-propulsion rockets have been proposed for use in space. They employ atomic ionization techniques and nuclear energy sources to produce extremely high exhaust velocities, perhaps as great as $\displaystyle 8.00×10^6m/s$. These techniques allow a much more favorable payload-to-fuel ratio. To illustrate this fact: (a) Calculate the increase in velocity of a 20,000-kg space probe that expels only 40.0-kg of its mass at the given exhaust velocity. (b) These engines are usually designed to produce a very small thrust for a very long time—the type of engine that might be useful on a trip to the outer planets, for example. Calculate the acceleration of such an engine if it expels $\displaystyle 4.50×10^{−6}kg/s$ at the given velocity, assuming the acceleration due to gravity is negligible. 79. Derive the equation for the vertical acceleration of a rocket. Solution The force needed to give a small mass $\displaystyle Δm$an acceleration $\displaystyle a_{Δm}$ is $\displaystyle F=Δma_{Δm}$. To accelerate this mass in the small time interval $\displaystyle Δt$ at a speed $\displaystyle v_e=a_{Δm}Δt$, so $\displaystyle F=v_e\frac{Δm}{Δt}$. By Newton’s third law, this force is equal in magnitude to the thrust force acting on the rocket, so $\displaystyle F_{thrust}=v_e\frac{Δm}{Δt}$, where all quantities are positive. Applying Newton’s second law to the rocket gives $\displaystyle F_{thrust}−mg=ma⇒a=\frac{v_e}{m}\frac{Δm}{Δt}−g$, where $\displaystyle m$ is the mass of the rocket and unburnt fuel. 80. Professional Application (a) Calculate the maximum rate at which a rocket can expel gases if its acceleration cannot exceed seven times that of gravity. The mass of the rocket just as it runs out of fuel is 75,000-kg, and its exhaust velocity is $\displaystyle 2.40×10^3m/s$. Assume that the acceleration of gravity is the same as on Earth’s surface $\displaystyle (9.80m/s^2)$. (b) Why might it be necessary to limit the acceleration of a rocket? Solution Given the following data for a fire extinguisher-toy wagon rocket experiment, calculate the average exhaust velocity of the gases expelled from the extinguisher. Starting from rest, the final velocity is 10.0 m/s. The total mass is initially 75.0 kg and is 70.0 kg after the extinguisher is fired. 81. How much of a single-stage rocket that is 100,000 kg can be anything but fuel if the rocket is to have a final speed of $\displaystyle 8.00km/s$, given that it expels gases at an exhaust velocity of $\displaystyle 2.20×10^3m/s$? Solution $\displaystyle 2.63×10^3kg$ 82. Professional Application (a) A 5.00-kg squid initially at rest ejects 0.250-kg of fluid with a velocity of 10.0 m/s. What is the recoil velocity of the squid if the ejection is done in 0.100 s and there is a 5.00-N frictional force opposing the squid’s movement. (b) How much energy is lost to work done against friction? Solution (a) 0.421 m/s away from the ejected fluid. (b) $\displaystyle 0.237J$ 83. Unreasonable Results Squids have been reported to jump from the ocean and travel $\displaystyle 30.0m$ (measured horizontally) before re-entering the water. (a) Calculate the initial speed of the squid if it leaves the water at an angle of $\displaystyle 20.0º$, assuming negligible lift from the air and negligible air resistance. (b) The squid propels itself by squirting water. What fraction of its mass would it have to eject in order to achieve the speed found in the previous part? The water is ejected at $\displaystyle 12.0m/s$; gravitational force and friction are neglected. (c) What is unreasonable about the results? (d) Which premise is unreasonable, or which premises are inconsistent? 84. Construct Your Own Problem Consider an astronaut in deep space cut free from her space ship and needing to get back to it. The astronaut has a few packages that she can throw away to move herself toward the ship. Construct a problem in which you calculate the time it takes her to get back by throwing all the packages at one time compared to throwing them one at a time. Among the things to be considered are the masses involved, the force she can exert on the packages through some distance, and the distance to the ship. 85. Construct Your Own Problem Consider an artillery projectile striking armor plating. Construct a problem in which you find the force exerted by the projectile on the plate. Among the things to be considered are the mass and speed of the projectile and the distance over which its speed is reduced. Your instructor may also wish for you to consider the relative merits of depleted uranium versus lead projectiles based on the greater density of uranium. Contributors and Attributions - Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .	libretexts	2025-03-17T19:53:27.919070	2017-04-20T14:36:29	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/08%3A_Linear_Momentum_and_Collisions/8.E%3A_Linear_Momentum_and_Collisions_(Exercises)", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "8.E: Linear Momentum and Collisions (Exercises)", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque	9: Statics and Torque How can we guarantee that a body is in equilibrium and what can we learn from systems that are in equilibrium? There are actually two conditions that must be satisfied to achieve equilibrium. These conditions are the topics of the first two sections of this chapter. - - 9.0: Prelude to Statics and Torque - Statics is the study of forces in equilibrium, a large group of situations that makes up a special case of Newton’s second law. We have already considered a few such situations; in this chapter, we cover the topic more thoroughly, including consideration of such possible effects as the rotation and deformation of an object by the forces acting on it. - - 9.2: The Second Condition for Equilibrium - The second condition necessary to achieve equilibrium involves avoiding accelerated rotation (maintaining a constant angular velocity. A rotating body or system can be in equilibrium if its rate of rotation is constant and remains unchanged by the forces acting on it. To understand what factors affect rotation, let us think about what happens when you open an ordinary door by rotating it on its hinges. - - 9.3: Stability - There are three types of equilibrium: stable, unstable, and neutral. A system is in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement. A system is in unstable equilibrium if, when displaced, it experiences a net force or torque in the same direction as the displacement from equilibrium. A neutral equilibrium is if its equilibrium is independent of displacements from its original position. - - 9.4: Applications of Statics, Including Problem-Solving Strategies - Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We begin with a discussion of problem-solving strategies specifically used for statics. Since statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Problem-Solving Strategies, still apply. - - 9.5: Simple Machines - Simple machines are devices that can be used to multiply or augment a force that we apply – often at the expense of a distance through which we apply the force. Levers, gears, pulleys, wedges, and screws are some examples of machines. Energy is still conserved for these devices because a machine cannot do more work than the energy put into it. Machines can reduce the input force that is needed to perform the job. The ratio of output to input force magnitudes is called its mechanical advantage. - - 9.6: Forces and Torques in Muscles and Joints - Muscles, bones, and joints are some of the most interesting applications of statics. There are some surprises. Muscles, for example, exert far greater forces than we might think. Figure shows a forearm holding a book and a schematic diagram of an analogous lever system. The schematic is a good approximation for the forearm, which looks more complicated than it is, and we can get some insight into the way typical muscle systems function by analyzing it. Thumbnails: Relationship between force (F), torque(τ), momentum (p), and angular momentum (L) vectors in a rotating system. (r) is the radius. (Public domain; Yawe).	libretexts	2025-03-17T19:53:27.988174	2015-10-27T19:23:49	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "9: Statics and Torque", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.00%3A_Prelude_to_Statics_and_Torque	9.0: Prelude to Statics and Torque What might desks, bridges, buildings, trees, and mountains have in common—at least in the eyes of a physicist? The answer is that they are ordinarily motionless relative to the Earth. Furthermore, their acceleration is zero because they remain motionless. That means they also have something in common with a car moving at a constant velocity, because anything with a constant velocity also has an acceleration of zero. Now, the important part—Newton’s second law states that net $F = ma$, and so the net external force is zero for all stationary objects and for all objects moving at constant velocity. There are forces acting, but they are balanced. That is, they are in equilibrium . Statics Statics is the study of forces in equilibrium, a large group of situations that makes up a special case of Newton’s second law. We have already considered a few such situations; in this chapter, we cover the topic more thoroughly, including consideration of such possible effects as the rotation and deformation of an object by the forces acting on it. How can we guarantee that a body is in equilibrium and what can we learn from systems that are in equilibrium? There are actually two conditions that must be satisfied to achieve equilibrium. These conditions are the topics of the first two sections of this chapter.	libretexts	2025-03-17T19:53:28.044987	2015-11-01T05:06:21	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.00%3A_Prelude_to_Statics_and_Torque", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "9.0: Prelude to Statics and Torque", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.01%3A_The_First_Condition_for_Equilibrium	9.1: The First Condition for Equilibrium Learning Objectives By the end of this section, you will be able to: - State the first condition of equilibrium. - Explain static equilibrium. - Explain dynamic equilibrium. The first condition necessary to achieve equilibrium is the one already mentioned: the net external force on the system must be zero. Expressed as an equation, this is simply \[net \, F = 0\] Note that if net $F$ is zero, then the net external force in any direction is zero. For example, the net external forces along the typical x - and y -axes are zero. This is written as \[net \, F_x \, and \, F_y = 0\] Figuress $\PageIndex{1}$ and $\PageIndex{2}$ illustrate situations where $net \, F = 0$ for both static equilibrium (motionless), and dynamic equilibrium (constant velocity). However, it is not sufficient for the net external force of a system to be zero for a system to be in equilibrium. Consider the two situations illustrated in Figures $\PageIndex{3}$ and $\PageIndex{4}$ where forces are applied to an ice hockey stick lying flat on ice. The net external force is zero in both situations shown in the figure; but in one case, equilibrium is achieved, whereas in the other, it is not. In Figure $\PageIndex{3}$, the ice hockey stick remains motionless. But in Figure $\PageIndex{4}$, with the same forces applied in different places, the stick experiences accelerated rotation. Therefore, we know that the point at which a force is applied is another factor in determining whether or not equilibrium is achieved. This will be explored further in the next section. PhET Explorations: Torque Investigate how torque causes an object to rotate . Discover the relationships between angular acceleration, moment of inertia, angular momentum and torque. Summary - Statics is the study of forces in equilibrium. - Two conditions must be met to achieve equilibrium, which is defined to be motion without linear or rotational acceleration. - The first condition necessary to achieve equilibrium is that the net external force on the system must be zero, so that $F = 0$. Glossary - static equilibrium - a state of equilibrium in which the net external force and torque acting on a system is zero - dynamic equilibrium - a state of equilibrium in which the net external force and torque on a system moving with constant velocity are zero	libretexts	2025-03-17T19:53:28.109574	2015-11-01T05:06:42	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.01%3A_The_First_Condition_for_Equilibrium", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "9.1: The First Condition for Equilibrium", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.02%3A_The_Second_Condition_for_Equilibrium	9.2: The Second Condition for Equilibrium Learning Objectives By the end of this section, you will be able to: - State the second condition that is necessary to achieve equilibrium. - Explain torque and the factors on which it depends. - Describe the role of torque in rotational mechanics. Definition: Torque The second condition necessary to achieve equilibrium involves avoiding accelerated rotation (maintaining a constant angular velocity. A rotating body or system can be in equilibrium if its rate of rotation is constant and remains unchanged by the forces acting on it. To understand what factors affect rotation, let us think about what happens when you open an ordinary door by rotating it on its hinges. Several familiar factors determine how effective you are in opening the door (Figure $\PageIndex{1}$). First of all, the larger the force, the more effective it is in opening the door—obviously, the harder you push, the more rapidly the door opens. Also, the point at which you push is crucial. If you apply your force too close to the hinges, the door will open slowly, if at all. Most people have been embarrassed by making this mistake and bumping up against a door when it did not open as quickly as expected. Finally, the direction in which you push is also important. The most effective direction is perpendicular to the door—we push in this direction almost instinctively. The magnitude, direction, and point of application of the force are incorporated into the definition of the physical quantity called torque. Torque is the rotational equivalent of a force. It is a measure of the effectiveness of a force in changing or accelerating a rotation (changing the angular velocity over a period of time). In equation form, the magnitude of torque is defined to be \[\tau = rF\, \sin \, \theta\] where $\tau$ (the Greek letter tau) is the symbol for torque, $r$ is the distance from the pivot point to the point where the force is applied, $F$ is the magnitude of the force, and $\theta$ is the angle between the force and the vector directed from the point of application to the pivot point, as seen in Figures $\PageIndex{1}$ and $\PageIndex{2}$. An alternative expression for torque is given in terms of the perpendicular lever arm $r_{\perp}$ as shown in Figures $\PageIndex{1}$ and $\PageIndex{2}$, which is defined as \[r_{\perp} = r \, \sin \, \theta\] so that \[\tau = r_{\perp}F.\] The perpendicular lever arm $r_{\perp}$ is the shortest distance from the pivot point to the line along which $F$ acts; it is shown as a dashed line in Figures $\PageIndex{1}$ and $\PageIndex{2}$. Note that the line segment that defines the distance $r_{\perp}$ is perpendicular to $F$, as its name implies. It is sometimes easier to find or visualize $r_{\perp}$ than to find both $r$ and $\theta$. In such cases, it may be more convenient to use $\tau = r_{perp}F$ rather than $\tau = rF \, \sin \, \theta$ for torque, but both are equally valid. The SI unit of torque is newtons times meters, usually written as $N \cdot m$. For example, if you push perpendicular to the door with a force of 40 N at a distance of 0.800 m from the hinges, you exert a torque of $32 \, N \cdot m(0.800 \, m \times 40 \, N \times sin \, 90^o)$ relative to the hinges. If you reduce the force to 20 N, the torque is reduced to $16 \, N \cdot m$, and so on. The torque is always calculated with reference to some chosen pivot point. For the same applied force, a different choice for the location of the pivot will give you a different value for the torque, since both $r$ and $\theta$ depend on the location of the pivot. Any point in any object can be chosen to calculate the torque about that point. The object may not actually pivot about the chosen “pivot point.” Note that for rotation in a plane, torque has two possible directions. Torque is either clockwise or counterclockwise relative to the chosen pivot point, as illustrated for points B and A, respectively, in Figure $\PageIndex{2}$. If the object can rotate about point A, it will rotate counterclockwise, which means that the torque for the force is shown as counterclockwise relative to A. But if the object can rotate about point B, it will rotate clockwise, which means the torque for the force shown is clockwise relative to B. Also, the magnitude of the torque is greater when the lever arm is longer. Now, the second condition necessary to achieve equilibrium is that the net external torque on a system must be zero . An external torque is one that is created by an external force. You can choose the point around which the torque is calculated. The point can be the physical pivot point of a system or any other point in space—but it must be the same point for all torques. If the second condition (net external torque on a system is zero) is satisfied for one choice of pivot point, it will also hold true for any other choice of pivot point in or out of the system of interest. (This is true only in an inertial frame of reference.) The second condition necessary to achieve equilibrium is stated in equation form as \[net \, \tau = 0\] where net means total. Torques, which are in opposite directions are assigned opposite signs. A common convention is to call counterclockwise (ccw) torques positive and clockwise (cw) torques negative. When two children balance a seesaw as shown in Figure $\PageIndex{3}$, they satisfy the two conditions for equilibrium. Most people have perfect intuition about seesaws, knowing that the lighter child must sit farther from the pivot and that a heavier child can keep a lighter one off the ground indefinitely. Example $\PageIndex{1}$: She Saw Torques On A Seesaw The two children shown in Figure $\PageIndex{3}$ are balanced on a seesaw of negligible mass. (This assumption is made to keep the example simple—more involved examples will follow.) The first child has a mass of 26.0 kg and sits 1.60 m from the pivot. - If the second child has a mass of 32.0 kg, how far is she from the pivot? - What is $F_p$, the supporting force exerted by the pivot? Strategy Both conditions for equilibrium must be satisfied. In part (a), we are asked for a distance; thus, the second condition (regarding torques) must be used, since the first (regarding only forces) has no distances in it. To apply the second condition for equilibrium, we first identify the system of interest to be the seesaw plus the two children. We take the supporting pivot to be the point about which the torques are calculated. We then identify all external forces acting on the system. Solution (a) The three external forces acting on the system are the weights of the two children and the supporting force of the pivot. Let us examine the torque produced by each. Torque is defined to be \[\tau = rF \, \sin \, \theta. \nonumber\] Here $\theta = 90^o,$ so that $sin \, \theta = 1$ for all three forces. That means $r_{\perp} = r$ for all three. The torques exerted by the three forces are first, \[\tau_1 = r_1w_1\nonumber\] second, \[\tau_2 = -r_2w_2\nonumber\] and third, \[ \begin{align} \tau_p &= r_pF_p \\[5pt] &= 0 \cdot F_p \\[5pt] &= 0. \end{align}\] Note that a minus sign has been inserted into the second equation because this torque is clockwise and is therefore negative by convention. Since $F_p$ acts directly on the pivot point, the distance $r_p$ is zero. A force acting on the pivot cannot cause a rotation, just as pushing directly on the hinges of a door will not cause it to rotate. Now, the second condition for equilibrium is that the sum of the torques on both children is zero. Therefore \[\tau_2 = -\tau_1,\nonumber\] or \[r_2w_2 = r_1w_1.\nonumber\] Weight is mass times the acceleration due to gravity. Entering $mg$ for $w$, we get \[r_2m_2g = r_1w_1g.\nonumber\] Solve this for the unknown $r_2$: \[r_2 = r_1\dfrac{m_1}{m_2}.\nonumber\] The quantities on the right side of the equation are known; thus, $r_2$ is \[ \begin{align} r_2 &= (1.60 \, m)\dfrac{26.0 \, kg}{32.0 \, kg} \\[5pt] &= 1.30 \, m \end{align}\] As expected, the heavier child must sit closer to the pivot (1.30 m versus 1.60 m) to balance the seesaw. Solution (b) This part asks for a force $F_p$. The easiest way to find it is to use the first condition for equilibrium, which is \[net \, F = 0.\nonumber\] The forces are all vertical, so that we are dealing with a one-dimensional problem along the vertical axis; hence, the condition can be written as \[net \, F_y = 0 \nonumber\] where we again call the vertical axis the y -axis. Choosing upward to be the positive direction, and using plus and minus signs to indicate the directions of the forces, we see that \[F_p - w_1 - w_2 = 0.\nonumber\] This equation yields what might have been guessed at the beginning: \[F_p = w_1 + w_2. \nonumber\] So, the pivot supplies a supporting force equal to the total weight of the system: \[F_p = m_1g + m_2g. \nonumber\] Entering known values gives \[ \begin{align} F_p &= (26.0 \, kg)(9.80 \, m/s^2) + (32.0 \, kg)(9.80 \, m/s^2) \\[5pt] &= 568 \, N. \end{align}\] Discussion The two results make intuitive sense. The heavier child sits closer to the pivot. The pivot supports the weight of the two children. Part (b) can also be solved using the second condition for equilibrium, since both distances are known, but only if the pivot point is chosen to be somewhere other than the location of the seesaw’s actual pivot! Several aspects of the preceding example have broad implications. First, the choice of the pivot as the point around which torques are calculated simplified the problem. Since $F_p$ is exerted on the pivot point, its lever arm is zero. Hence, the torque exerted by the supporting force $F_p$ is zero relative to that pivot point. The second condition for equilibrium holds for any choice of pivot point, and so we choose the pivot point to simplify the solution of the problem. Second, the acceleration due to gravity canceled in this problem, and we were left with a ratio of masses. This will not always be the case . Always enter the correct forces—do not jump ahead to enter some ratio of masses. Third, the weight of each child is distributed over an area of the seesaw, yet we treated the weights as if each force were exerted at a single point. This is not an approximation—the distances $r_{\perp}$ and $r_2$ are the distances to points directly below the center of gravity of each child. As we shall see in the next section, the mass and weight of a system can act as if they are located at a single point. Finally, note that the concept of torque has an importance beyond static equilibrium. Torque plays the same role in rotational motion that force plays in linear motion. We will examine this in the next chapter. Take-Home Experiment - Take a piece of modeling clay and put it on a table, then mash a cylinder down into it so that a ruler can balance on the round side of the cylinder while everything remains still. Put a penny 8 cm away from the pivot. Where would you need to put two pennies to balance? Three pennies? Summary - The second condition assures those torques are also balanced. Torque is the rotational equivalent of a force in producing a rotation and is defined to be \[\tau = rF \, \sin \, \theta \nonumber\] where $\tau$ is torque, $r$ is the distance from the pivot point to the point where the force is applied, $F$ is magnitude of the force, and $\theta$ is the angle between $F$ and the vector directed from the point where the force acts to the pivot point. The perpendicular lever arm $r_{\perp}$ is defined to be \[r_{\perp} = r \, \sin \, \theta \nonumber\] so that \[\tau = r_{\perp}F. \nonumber\] - The perpendicular lever arm $r_{\perp}$ is the shortest distance from the pivot point to the line along which $F$ acts. The SI unit for torque is newton-meter (N \cdot m). The second condition necessary to achieve equilibrium is that the net external torque on a system must be zero: \[ net \, \tau = 0 \nonumber\] By convention, counterclockwise torques are positive, and clockwise torques are negative. Glossary - torque - turning or twisting effectiveness of a force - perpendicular lever arm - the shortest distance from the pivot point to the line along which $F$ lies - SI units of torque - newton times meters, usually written as N·m - center of gravity - the point where the total weight of the body is assumed to be concentrated	libretexts	2025-03-17T19:53:28.188944	2015-11-01T05:07:01	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.02%3A_The_Second_Condition_for_Equilibrium", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "9.2: The Second Condition for Equilibrium", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.03%3A_Stability	9.3: Stability Learning Objectives By the end of this section, you will be able to: - State the types of equilibrium. - Describe stable and unstable equilibriums. - Describe neutral equilibrium. It is one thing to have a system in equilibrium; it is quite another for it to be stable. The toy doll perched on the man’s hand in Figure $\PageIndex{1}$, for example, is not in stable equilibrium. There are three types of equilibrium : stable , unstable , and neutral . Figures throughout this module illustrate various examples. Figure $\PageIndex{1}$ presents a balanced system, such as the toy doll on the man’s hand, which has its center of gravity (cg) directly over the pivot, so that the torque of the total weight is zero. This is equivalent to having the torques of the individual parts balanced about the pivot point, in this case the hand. The cgs of the arms, legs, head, and torso are labeled with smaller type. A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement. For example, a marble at the bottom of a bowl will experience a restoring force when displaced from its equilibrium position. This force moves it back toward the equilibrium position. Most systems are in stable equilibrium, especially for small displacements. For another example of stable equilibrium, see the pencil in Figure $\PageIndex{2}$. A system is in unstable equilibrium if, when displaced, it experiences a net force or torque in the same direction as the displacement from equilibrium. A system in unstable equilibrium accelerates away from its equilibrium position if displaced even slightly. An obvious example is a ball resting on top of a hill. Once displaced, it accelerates away from the crest. See the next several figures for examples of unstable equilibrium. A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. A marble on a flat horizontal surface is an example. Combinations of these situations are possible. For example, a marble on a saddle is stable for displacements toward the front or back of the saddle and unstable for displacements to the side. Figure $\PageIndex{7}$ shows another example of neutral equilibrium. When we consider how far a system in stable equilibrium can be displaced before it becomes unstable, we find that some systems in stable equilibrium are more stable than others. The pencil in Figure $\PageIndex{7}$ and the person in Figure $\PageIndex{8a}$ are in stable equilibrium, but become unstable for relatively small displacements to the side. The critical point is reached when the cg is no longer above the base of support. Additionally, since the cg of a person’s body is above the pivots in the hips, displacements must be quickly controlled. This control is a central nervous system function that is developed when we learn to hold our bodies erect as infants. For increased stability while standing, the feet should be spread apart, giving a larger base of support. Stability is also increased by lowering one’s center of gravity by bending the knees, as when a football player prepares to receive a ball or braces themselves for a tackle. A cane, a crutch, or a walker increases the stability of the user, even more as the base of support widens. Usually, the cg of a female is lower (closer to the ground) than a male. Young children have their center of gravity between their shoulders, which increases the challenge of learning to walk. Animals such as chickens have easier systems to control. Figure $\PageIndex{9}$ shows that the cg of a chicken lies below its hip joints and between its widely separated and broad feet. Even relatively large displacements of the chicken’s cg are stable and result in restoring forces and torques that return the cg to its equilibrium position with little effort on the chicken’s part. Not all birds are like chickens, of course. Some birds, such as the flamingo, have balance systems that are almost as sophisticated as that of humans. Figure $\PageIndex{9}$ shows that the cg of a chicken is below the hip joints and lies above a broad base of support formed by widely-separated and large feet. Hence, the chicken is in very stable equilibrium, since a relatively large displacement is needed to render it unstable. The body of the chicken is supported from above by the hips and acts as a pendulum between the hips. Therefore, the chicken is stable for front-to-back displacements as well as for side-to-side displacements. Engineers and architects strive to achieve extremely stable equilibriums for buildings and other systems that must withstand wind, earthquakes, and other forces that displace them from equilibrium. Although the examples in this section emphasize gravitational forces, the basic conditions for equilibrium are the same for all types of forces. The net external force must be zero, and the net torque must also be zero. Take-Home Experiment - Stand straight with your heels, back, and head against a wall. Bend forward from your waist, keeping your heels and bottom against the wall, to touch your toes. Can you do this without toppling over? Explain why and what you need to do to be able to touch your toes without losing your balance. Is it easier for a woman to do this? Summary - A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite the direction of the displacement. - A system is in unstable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in the same direction as the displacement from equilibrium. - A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. Glossary - neutral equilibrium - a state of equilibrium that is independent of a system’s displacements from its original position - stable equilibrium - a system, when displaced, experiences a net force or torque in a direction opposite to the direction of the displacement - unstable equilibrium - a system, when displaced, experiences a net force or torque in the same direction as the displacement from equilibrium	libretexts	2025-03-17T19:53:28.258111	2015-11-01T05:07:20	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.03%3A_Stability", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "9.3: Stability", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.04%3A_Applications_of_Statics_Including_Problem-Solving_Strategies	9.4: Applications of Statics, Including Problem-Solving Strategies Learning Objectives By the end of this section, you will be able to: - Discuss the applications of Statics in real life. - State and discuss various problem-solving strategies in Statics. Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We begin with a discussion of problem-solving strategies specifically used for statics. Since statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Problem-Solving Strategies , still apply. Problem-Solving Strategy: Static Equilibrium Situations - The first step is to determine whether or not the system is in static equilibrium . This condition is always the case when the acceleration of the system is zero and accelerated rotation does not occur . - It is particularly important to draw a free body diagram for the system of interest . Carefully label all forces, and note their relative magnitudes, directions, and points of application whenever these are known. - Solve the problem by applying either or both of the conditions for equilibrium (represented by the equations $net \, F = 0$ and $net \, \tau = 0 $, depending on the list of known and unknown factors. If the second condition is involved, choose the pivot point to simplify the solution . Any pivot point can be chosen, but the most useful ones cause torques by unknown forces to be zero. (Torque is zero if the force is applied at the pivot (then $r = 0$), or along a line through the pivot point (then \\theta = 0\)). Always choose a convenient coordinate system for projecting forces. - Check the solution to see if it is reasonable by examining the magnitude, direction, and units of the answer. The importance of this last step never diminishes, although in unfamiliar applications, it is usually more difficult to judge reasonableness. These judgments become progressively easier with experience. Now let us apply this problem-solving strategy for the pole vaulter shown in the three figures below. The pole is uniform and has a mass of 5.00 kg. In Figure $\PageIndex{1}$, the pole’s cg lies halfway between the vaulter’s hands. It seems reasonable that the force exerted by each hand is equal to half the weight of the pole, or 24.5 N. This obviously satisfies the first condition for equilibrium $(net \, F = 0)$. The second condition $(net \, \tau = 0)$ is also satisfied, as we can see by choosing the cg to be the pivot point. The weight exerts no torque about a pivot point located at the cg, since it is applied at that point and its lever arm is zero. The equal forces exerted by the hands are equidistant from the chosen pivot, and so they exert equal and opposite torques. Similar arguments hold for other systems where supporting forces are exerted symmetrically about the cg. For example, the four legs of a uniform table each support one-fourth of its weight. In Figure $\PageIndex{2}$, a pole vaulter holding a pole with its cg halfway between his hands is shown. Each hand exerts a force equal to half the weight of the pole, $F_r = F_L = w/2$. (b) The pole vaulter moves the pole to his left, and the forces that the hands exert are no longer equal. See Figure . If the pole is held with its cg to the left of the person, then he must push down with his right hand and up with his left. The forces he exerts are larger here because they are in opposite directions and the cg is at a long distance from either hand. Similar observations can be made using a meter stick held at different locations along its length. If the pole vaulter holds the pole as shown in Figure , the situation is not as simple. The total force he exerts is still equal to the weight of the pole, but it is not evenly divided between his hands. (If $F_L = F_R $, then the torques about the cg would not be equal since the lever arms are different.) Logically, the right hand should support more weight, since it is closer to the cg. In fact, if the right hand is moved directly under the cg, it will support all the weight. This situation is exactly analogous to two people carrying a load; the one closer to the cg carries more of its weight. Finding the forces $F_L$ and $F_R$ is straightforward, as the next example shows. If the pole vaulter holds the pole from near the end of the pole ( Figure ), the direction of the force applied by the right hand of the vaulter reverses its direction. Example $\PageIndex{1}$: What Force Is Needed to Support a Weight Held Near Its CG? For the situation shown in Figure , calculate: (a) $F_R$, the force exerted by the right hand, and (b) $F_L$, the force exerted by the left hand. The hands are 0.900 m apart, and the cg of the pole is 0.600 m from the left hand. Strategy Figure includes a free body diagram for the pole, the system of interest. There is not enough information to use the first condition for equilibrium $(net \, F = 0)$, since two of the three forces are unknown and the hand forces cannot be assumed to be equal in this case. There is enough information to use the second condition for equilibrium $(net \, \tau = 0)$ if the pivot point is chosen to be at either hand, thereby making the torque from that hand zero. We choose to locate the pivot at the left hand in this part of the problem, to eliminate the torque from the left hand. Solution for (a) There are now only two nonzero torques, those from the gravitational force $(\tau_W)$ and from the push or pull of the right hand $(\tau_R)$. Stating the second condition in terms of clockwise and counterclockwise torques, \[net \, \tau_{CW} = -net \, \tau_{CCW}.\] or the algebraic sum of the torques is zero. Here this is \[\tau_R = -\tau_W\] since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise toque. Using the definition of torque, $\tau = rF \, sin \, \theta$, noting that $\theta = 90^o$, and substituting known values, we obtain \[(0.900 \, m)(F_R) = (0.600 \, m)(mg).\] Thus, \[F_R = (0.667)(5.00 \, kg)(9.80 \, m/s^2)\]\[=32.7 \, N.\] Solution for (b) The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton’s second law: \[F_L + F_R - mg = 0\] From this we can conclude \[F_L + F_R = w = mg\] Solving for $F_L$, we obtain \[F_L = mg - F_R\] \]= mg - 43.7 \, N\] \[= (5.00 \, kg)(9.80 \, m/s^2) - 32.7 \, N\] \[= 16.3 \, N\] Discussion $F_L$ is seen to be exactly half of $F_R$, as we might have guessed, since $F_L$ is applied twice as far from for cg as $F_R$. If the pole vaulter holds the pole as he might at the start of a run, shown in Figure , the forces change again. Both are considerably greater, and one force reverses direction. Take-Home Experiment - This is an experiment to perform while standing in a bus or a train. Stand facing sideways. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Now stand facing forward. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Why is it easier and safer to stand facing sideways rather than forward? Note: For your safety (and those around you), make sure you are holding onto something while you carry out this activity! PhET Explorations: Balancing Act Play with objects on a teeter totter to learn about balance. Test what you've learned by trying the Balance Challenge game. Figure $\PageIndex{4}$: Balancing Act Summary - Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We have discussed the problem-solving strategies specifically useful for statics. Statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Problem-Solving Strategies , still apply. Glossary - static equilibrium - equilibrium in which the acceleration of the system is zero and accelerated rotation does not occur	libretexts	2025-03-17T19:53:28.330747	2015-11-01T05:07:40	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.04%3A_Applications_of_Statics_Including_Problem-Solving_Strategies", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "9.4: Applications of Statics, Including Problem-Solving Strategies", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.05%3A_Simple_Machines	9.5: Simple Machines Learning Objectives By the end of this section, you will be able to: - Describe different simple machines. - Calculate the mechanical advantage. Simple machines are devices that can be used to multiply or augment a force that we apply – often at the expense of a distance through which we apply the force. The word for “machine” comes from the Greek word meaning “to help make things easier.” Levers, gears, pulleys, wedges, and screws are some examples of machines. Energy is still conserved for these devices because a machine cannot do more work than the energy put into it. However, machines can reduce the input force that is needed to perform the job. The ratio of output to input force magnitudes for any simple machine is called its mechanical advantage (MA). \[MA = \dfrac{F_o}{F_i}\] One of the simplest machines is the lever, which is a rigid bar pivoted at a fixed place called the fulcrum. Torques are involved in levers, since there is rotation about a pivot point. Distances from the physical pivot of the lever are crucial, and we can obtain a useful expression for the MA in terms of these distances. Figure shows a lever type that is used as a nail puller. Crowbars, seesaws, and other such levers are all analogous to this one $F_i$ is the input force and $F_o$ is the output force. There are three vertical forces acting on the nail puller (the system of interest) – these are $F_i$, $F_o$, and $N.$ $F_n$ is the reaction force back on the system, equal and opposite to $F_o$. (note that $F_o$ is not a force on the system.) $N$ is the normal force upon the lever, and its torque is zero since it is exerted at the pivot. The torques due to $F_i$ and $F_n$ must be equal to each other if the nail is not moving, to satisfy the second condition for equilibrium $(net \, \tau = 0)$. (In order for the nail to actually move, the torque due to $F_i$ must be ever-so-slightly greater than torque due to $F_n$.) Hence, \[l_iF_i = l_oF_o\] where $l_i$ and $l_o$ are the distances from where the input and output forces are applied to the pivot, as shown in the figure. Rearranging the last equation gives \[ \dfrac{F_o}{F_i} = \dfrac{l_i}{l_o}.\] What interests us most here is that the magnitude of the force exerted by the nail puller,$F_o$, is much greater than the magnitude of the input force applied to the puller at the other end, $F_i$. For the nail puller, \[MA = \dfrac{F_o}{F_i} = \dfrac{l_i}{l_o}.\] This equation is true for levers in general. For the nail puller, the MA is certainly greater than one. The longer the handle on the nail puller, the greater the force you can exert with it. Two other types of levers that differ slightly from the nail puller are a wheelbarrow and a shovel, shown in Figure . All these lever types are similar in that only three forces are involved – the input force, the output force, and the force on the pivot – and thus their MAs are given by $MA = \frac{F_o}{F_i} $ and $MA = \frac{d_1}{d_2}$, with distances being measured relative to the physical pivot. The wheelbarrow and shovel differ from the nail puller because both the input and output forces are on the same side of the pivot. In the case of the wheelbarrow, the output force or load is between the pivot (the wheel’s axle) and the input or applied force. In the case of the shovel, the input force is between the pivot (at the end of the handle) and the load, but the input lever arm is shorter than the output lever arm. In this case, the MA is less than one. Example $\PageIndex{1}$: What is the Advantage for the Wheelbarrow? In the wheelbarrow of Figure , the load has a perpendicular lever arm of 7.50 cm, while the hands have a perpendicular lever arm of 1.02 m. (a) What upward force must you exert to support the wheelbarrow and its load if their combined mass is 45.0 kg? (b) What force does the wheelbarrow exert on the ground? Strategy Here, we use the concept of mechanical advantage. Solution (a) In this case, $\frac{F_o}{F_i} = \frac{l_i}{l_o}$ becomes \[F_i = F_o\dfrac{l_o}{l_i},\] Adding values into this equation yields \[F_i = (45.0 \, kg)(9.80 \, m/s^2)\dfrac{0.075\space m}{1.02 \, m} = 32.4 \, N.\] The free-body diagram (see Figure ) gives the following normal force: \[F_i = + N = W.\] Therefore, \[N = (45.0 \, kg)(9.80 \, m/s^2) - 32.4 \, N = 409 \, N.\] $N$ is the normal force acting on the wheel; by Newton’s third law, the force the wheel exerts on the ground is $409 \, N$. Discussion An even longer handle would reduce the force needed to lift the load. The MA here is $MA = 1.01/0.0750 = 13.6$ Another very simple machine is the inclined plane. Pushing a cart up a plane is easier than lifting the same cart straight up to the top using a ladder, because the applied force is less. However, the work done in both cases (assuming the work done by friction is negligible) is the same. Inclined lanes or ramps were probably used during the construction of the Egyptian pyramids to move large blocks of stone to the top. A crank is a lever that can be rotated $360^o$ about its pivot, as shown in Figure . Such a machine may not look like a lever, but the physics of its actions remain the same. The MA for a crank is simply the ratio of the radii $r_i/r_o$. Wheels and gears have this simple expression for their MAs too. The MA can be greater than 1, as it is for the crank, or less than 1, as it is for the simplified car axle driving the wheels, as shown. If the axle’s radius is $2.0 \, cm$ and the wheel's radius is $24.0 \, cm$, then $MA = 2.0/24.0 = 0.083$ and the axle would have to exert a force of $12,000 \, N$ on the wheel to enable it to exert a force of $1000 \, N$ on the ground. An ordinary pulley has an MA of 1; it only changes the direction of the force and not its magnitude. Combinations of pulleys, such as those illustrated in Figure , are used to multiply force. If the pulleys are friction-free, then the force output is approximately an integral multiple of the tension in the cable. The number of cables pulling directly upward on the system of interest, as illustrated in the figures given below, is approximately the MA of the pulley system. Since each attachment applies an external force in approximately the same direction as the others, they add, producing a total force that is nearly an integral multiple of the input force $T$. Summary - Simple machines are devices that can be used to multiply or augment a force that we apply – often at the expense of a distance through which we have to apply the force. - The ratio of output to input forces for any simple machine is called its mechanical advantage - A few simple machines are the lever, nail puller, wheelbarrow, crank, etc. Glossary - mechanical advantage - the ratio of output to input forces for any simple machine	libretexts	2025-03-17T19:53:28.399709	2015-11-01T05:08:04	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.05%3A_Simple_Machines", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "9.5: Simple Machines", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.06%3A_Forces_and_Torques_in_Muscles_and_Joints	9.6: Forces and Torques in Muscles and Joints Learning Objectives By the end of this section, you will be able to: - Explain the forces exerted by muscles. - State how a bad posture causes back strain. - Discuss the benefits of skeletal muscles attached close to joints. - Discuss various complexities in the real system of muscles, bones, and joints. Muscles, bones, and joints are some of the most interesting applications of statics. There are some surprises. Muscles, for example, exert far greater forces than we might think. Figure shows a forearm holding a book and a schematic diagram of an analogous lever system. The schematic is a good approximation for the forearm, which looks more complicated than it is, and we can get some insight into the way typical muscle systems function by analyzing it. Muscles can only contract, so they occur in pairs. In the arm, the biceps muscle is a flexor—that is, it closes the limb. The triceps muscle is an extensor that opens the limb. This configuration is typical of skeletal muscles, bones, and joints in humans and other vertebrates. Most skeletal muscles exert much larger forces within the body than the limbs apply to the outside world. The reason is clear once we realize that most muscles are attached to bones via tendons close to joints, causing these systems to have mechanical advantages much less than one. Viewing them as simple machines, the input force is much greater than the output force, as seen in Figure . Example $\PageIndex{1}$: Muscles Exert Bigger Forces Than You Might Think Calculate the force the biceps muscle must exert to hold the forearm and its load as shown in Figure , and compare this force with the weight of the forearm plus its load. You may take the data in the figure to be accurate to three significant figures. Strategy There are four forces acting on the forearm and its load (the system of interest). The magnitude of the force of the biceps is $F_B$; that of the elbow joint is $F_E$; that of the weights of the forearm is $w_a$, and its loadis $w_b$. Two of these are unknown $(F_B$ and $F_E)$ so that the first condition for equilibrium cannot by itself yield $F_B$. But if we use the second condition and choose the pivot to be at the elbow, then the torque due to $F_B$ is zero, and the only unknown becomes $F_B$. Solution The torques created by the weights are clockwise relative to the pivot, while the torque created by the biceps is counterclockwise; thus, the second condition for equilibrium $(net \, \tau = 0)$ becomes \[r_2w_a + r_3w_b = r_1F_B.\] Note that $sin \, \theta = 1$ for all forces, since $\theta = 90^o$ for all forces. This equation can easily be solved for $F_B$ in terms of known quantities, yielding \[F_B = \dfrac{r_2w_a + r_3w_b}{r_1}.\] Entering the known values gives \[F_B = \dfrac{(0.160 \, m)(2.50 \, kg)(9.80 \, m/s^2) + (0.380 \, m)(4.00 \, kg)(9.80 \, m/s^2)}{0.0400 \, m}\] which yields \[F_B = 470 \, N.\] Now, the combined weight of the arm and its load is $(6.50 \, kg)(9.80 \, m/s^2) = 63.7 \, N$, so that the ratio of the force exerted by the biceps to the total weight is \[\dfrac{F_B}{w_a + w_b} = \dfrac{470}{63.7} = 7.38.\] Discussion This means that the biceps muscle is exerting a force 7.38 times the weight supported. In the above example of the biceps muscle, the angle between the forearm and upper arm is 90°. If this angle changes, the force exerted by the biceps muscle also changes. In addition, the length of the biceps muscle changes. The force the biceps muscle can exert depends upon its length; it is smaller when it is shorter than when it is stretched. Very large forces are also created in the joints. In the previous example, the downward force $F_E$ exerted by the humerus at the elbow joint equals 407 N, or 6.38 times the total weight supported. (The calculation of $F_E$ is straightforward and is left as an end-of-chapter problem.) Because muscles can contract, but not expand beyond their resting length, joints and muscles often exert forces that act in opposite directions and thus subtract. (In the above example, the upward force of the muscle minus the downward force of the joint equals the weight supported—that is, $470 \, N - 470 \, N = 63 \, N$, approximately equal to the weight supported.) Forces in muscles and joints are largest when their load is a long distance from the joint, as the book is in the previous example. In racquet sports such as tennis the constant extension of the arm during game play creates large forces in this way. The mass times the lever arm of a tennis racquet is an important factor, and many players use the heaviest racquet they can handle. It is no wonder that joint deterioration and damage to the tendons in the elbow, such as “tennis elbow,” can result from repetitive motion, undue torques, and possibly poor racquet selection in such sports. Various tried techniques for holding and using a racquet or bat or stick not only increases sporting prowess but can minimize fatigue and long-term damage to the body. For example, tennis balls correctly hit at the “sweet spot” on the racquet will result in little vibration or impact force being felt in the racquet and the body—less torque as explained in Collisions of Extended Bodies in Two Dimensions . Twisting the hand to provide top spin on the ball or using an extended rigid elbow in a backhand stroke can also aggravate the tendons in the elbow. Training coaches and physical therapists use the knowledge of relationships between forces and torques in the treatment of muscles and joints. In physical therapy, an exercise routine can apply a particular force and torque which can, over a period of time, revive muscles and joints. Some exercises are designed to be carried out under water, because this requires greater forces to be exerted, further strengthening muscles. However, connecting tissues in the limbs, such as tendons and cartilage as well as joints are sometimes damaged by the large forces they carry. Often, this is due to accidents, but heavily muscled athletes, such as weightlifters, can tear muscles and connecting tissue through effort alone. The back is considerably more complicated than the arm or leg, with various muscles and joints between vertebrae, all having mechanical advantages less than 1. Back muscles must, therefore, exert very large forces, which are borne by the spinal column. Discs crushed by mere exertion are very common. The jaw is somewhat exceptional—the masseter muscles that close the jaw have a mechanical advantage greater than 1 for the back teeth, allowing us to exert very large forces with them. A cause of stress headaches is persistent clenching of teeth where the sustained large force translates into fatigue in muscles around the skull. Figure shows how bad posture causes back strain. In part (a), we see a person with good posture. Note that her upper body’s cg is directly above the pivot point in the hips, which in turn is directly above the base of support at her feet. Because of this, her upper body’s weight exerts no torque about the hips. The only force needed is a vertical force at the hips equal to the weight supported. No muscle action is required, since the bones are rigid and transmit this force from the floor. This is a position of unstable equilibrium, but only small forces are needed to bring the upper body back to vertical if it is slightly displaced. Bad posture is shown in part (b); we see that the upper body’s cg is in front of the pivot in the hips. This creates a clockwise torque around the hips that is counteracted by muscles in the lower back. These muscles must exert large forces, since they have typically small mechanical advantages. (In other words, the perpendicular lever arm for the muscles is much smaller than for the cg.) Poor posture can also cause muscle strain for people sitting at their desks using computers. Special chairs are available that allow the body’s CG to be more easily situated above the seat, to reduce back pain. Prolonged muscle action produces muscle strain. Note that the cg of the entire body is still directly above the base of support in part (b) of Figure . This is compulsory; otherwise the person would not be in equilibrium. We lean forward for the same reason when carrying a load on our backs, to the side when carrying a load in one arm, and backward when carrying a load in front of us, as seen in Figure . You have probably been warned against lifting objects with your back. This action, even more than bad posture, can cause muscle strain and damage discs and vertebrae, since abnormally large forces are created in the back muscles and spine. Example: $\PageIndex{1}$: Do Not Lift with Your Back Consider the person lifting a heavy box with his back, shown in Figure . (a) Calculate the magnitude of the force $F_B$ - in the back muscles that is needed to support the upper body plus the box and compare this with his weight. The mass of the upper body is 55.0 kg and the mass of the box is 30.0 kg. (b) Calculate the magnitude and direction of the force $F_V$ - exerted by the vertebrae on the spine at the indicated pivot point. Again, data in the figure may be taken to be accurate to three significant figures. Strategy By now, we sense that the second condition for equilibrium is a good place to start, and inspection of the known values confirms that it can be used to solve for $F_B$ - if the pivot is chosen to be at the hips. The torques created by $w_{ab}$ and $w_{box}$ - are clockwise, while that created by $F_B $ - is counterclockwise. Solution for (a) Using the perpendicular lever arms given in the figure, the second condition for equilibrium $(net \, \tau = 0)$ becomes \[(0.350 \, m)(55.0 \, kg)(9.80 \, m/s^2) + (0.500 \, m)(30.0 \, kg)(9.80 \, m/s^2) = (0.0800 \, m)F_B \] Solving for $F_B$ yields \[F_B = 4.20 \times 10^3 \, N.\] The ratio of the force the back muscles exert to the weight of the upper body plus its load is \[\dfrac{F_B}{w_{ub} + w_{box}} = \dfrac{4200 \, N}{833 \, N} = 5.04.\] This force is considerably larger than it would be if the load were not present. Solution for (b) More important in terms of its damage potential is the force on the vertebrae $F_V$ The first condition for equilibrium $(net \, F= 0)$ can be used to find its magnitude and direction. Using $y$ fo4r vertical and $x$ for horizontal, the condition for the net external forces along those axes to be zero \[net \, F_y = 0 \, and \, net \, F_x = 0\] Starting with the vertical $(y)$ components, this yields \[F_{V_y} - w_{ub} - w_{box} - F_B \, sin \, 29^o = 0.\] Thus, \[F_{V_y} = w_{ub} + w_{box} + F_B \, sin \, 29.0^o\] \[= 833 \, N + (4200 \, N) \, sin \, 29.0^o\] yielding \[F_{V_y} = 2.87 \times 10^3 \, N.\] Similarly, for the horizontal $(x)$ components, \[F_{V_x} - F_B \, cos \, 29.0^o = 0\] yielding \[F_{V_x} = 3.67 \times 10^3 \, N.\] The magnitude of $F_V $ is given by the Pythagorean theorem: \[F_V = \sqrt{F_{V_x}^2 + F_{V_Y}^2} = 4.66 \times 10^3 \, N.\] The direction of $F_V$ is \[\theta = tan^{-1} \left(\dfrac{F_{V_y}}{F_{V_x}} \right) = 38.0^o.\] Note that the ratio of $F_V$ to the weight supported is \]\dfrac{F_V}{w_{ub} + w_{box}} = \dfrac{4660 \, N}{833 \, N} = 5.59.\] Discussion This force is about 5.6 times greater than it would be if the person were standing erect. The trouble with the back is not so much that the forces are large—because similar forces are created in our hips, knees, and ankles—but that our spines are relatively weak. Proper lifting, performed with the back erect and using the legs to raise the body and load, creates much smaller forces in the back—in this case, about 5.6 times smaller. What are the benefits of having most skeletal muscles attached so close to joints? One advantage is speed because small muscle contractions can produce large movements of limbs in a short period of time. Other advantages are flexibility and agility, made possible by the large numbers of joints and the ranges over which they function. For example, it is difficult to imagine a system with biceps muscles attached at the wrist that would be capable of the broad range of movement we vertebrates possess. There are some interesting complexities in real systems of muscles, bones, and joints. For instance, the pivot point in many joints changes location as the joint is flexed, so that the perpendicular lever arms and the mechanical advantage of the system change, too. Thus the force the biceps muscle must exert to hold up a book varies as the forearm is flexed. Similar mechanisms operate in the legs, which explain, for example, why there is less leg strain when a bicycle seat is set at the proper height. The methods employed in this section give a reasonable description of real systems provided enough is known about the dimensions of the system. There are many other interesting examples of force and torque in the body—a few of these are the subject of end-of-chapter problems. Summary - Statics plays an important part in understanding everyday strains in our muscles and bones. - Many lever systems in the body have a mechanical advantage of significantly less than one, as many of our muscles are attached close to joints. - Someone with good posture stands or sits in such a way that their center of gravity lies directly above the pivot point in their hips, thereby avoiding back strain and damage to disks.	libretexts	2025-03-17T19:53:28.474804	2015-11-01T05:08:26	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.06%3A_Forces_and_Torques_in_Muscles_and_Joints", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "9.6: Forces and Torques in Muscles and Joints", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.E%3A_Statics_and_Torque_(Exercises)	9.E: Statics and Torque (Exercises) - - Last updated - Save as PDF Conceptual Questions 9.1: The First Condition for Equilibrium 1. What can you say about the velocity of a moving body that is in dynamic equilibrium? Draw a sketch of such a body using clearly labeled arrows to represent all external forces on the body. 2. Under what conditions can a rotating body be in equilibrium? Give an example. 9.2: The Second Condition for Equilibrium 3. What three factors affect the torque created by a force relative to a specific pivot point? 4. A wrecking ball is being used to knock down a building. One tall unsupported concrete wall remains standing. If the wrecking ball hits the wall near the top, is the wall more likely to fall over by rotating at its base or by falling straight down? Explain your answer. How is it most likely to fall if it is struck with the same force at its base? Note that this depends on how firmly the wall is attached at its base. 5. Mechanics sometimes put a length of pipe over the handle of a wrench when trying to remove a very tight bolt. How does this help? (It is also hazardous since it can break the bolt.) 9.3: Stability 6. A round pencil lying on its side as in Figure is in neutral equilibrium relative to displacements perpendicular to its length. What is its stability relative to displacements parallel to its length? 7. Explain the need for tall towers on a suspension bridge to ensure stable equilibrium. 9.4: Applications of Statics, Including Problem-Solving Strategies 8. When visiting some countries, you may see a person balancing a load on the head. Explain why the center of mass of the load needs to be directly above the person’s neck vertebrae. 9.5: Simple Machines 9. Scissors are like a double-lever system. Which of the simple machines in Figure and Figure is most analogous to scissors? 10. Suppose you pull a nail at a constant rate using a nail puller as shown in Figure. Is the nail puller in equilibrium? What if you pull the nail with some acceleration – is the nail puller in equilibrium then? In which case is the force applied to the nail puller larger and why? 11. Why are the forces exerted on the outside world by the limbs of our bodies usually much smaller than the forces exerted by muscles inside the body? 12. Explain why the forces in our joints are several times larger than the forces we exert on the outside world with our limbs. Can these forces be even greater than muscle forces (see previous Question)? 9.6: Forces and Torques in Muscles and Joints 13. Why are the forces exerted on the outside world by the limbs of our bodies usually much smaller than the forces exerted by muscles inside the body? 14. Explain why the forces in our joints are several times larger than the forces we exert on the outside world with our limbs. Can these forces be even greater than muscle forces? 15. Certain types of dinosaurs were bipedal (walked on two legs). What is a good reason that these creatures invariably had long tails if they had long necks? 16. Swimmers and athletes during competition need to go through certain postures at the beginning of the race. Consider the balance of the person and why start-offs are so important for races. 17. If the maximum force the biceps muscle can exert is 1000 N, can we pick up an object that weighs 1000 N? Explain your answer. 18. Suppose the biceps muscle was attached through tendons to the upper arm close to the elbow and the forearm near the wrist. What would be the advantages and disadvantages of this type of construction for the motion of the arm? 19. Explain one of the reasons why pregnant women often suffer from back strain late in their pregnancy. Problems & Exercises 9.2: The Second Condition for Equilibrium 20. (a) When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850m from the hinges. What torque are you exerting relative to the hinges? (b) Does it matter if you push at the same height as the hinges? Solution a) 46.8 N⋅m b) It does not matter at what height you push. The torque depends on only the magnitude of the force applied and the perpendicular distance of the force’s application from the hinges. (Children don’t have a tougher time opening a door because they push lower than adults, they have a tougher time because they don’t push far enough from the hinges.) 21. When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt. (a) How much torque are you exerting in newton × meters (relative to the center of the bolt)? (b) Convert this torque to footpounds. 22. Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible. Solution 23.3 N 23. Use the second condition for equilibrium ($\displaystyle net τ = 0$) to calculate $\displaystyle F_p$ in Example, employing any data given or solved for in part (a) of the example. 24. Repeat the seesaw problem in Example with the center of mass of the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0 kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium. Solution Given: $\displaystyle m_1=26.0 kg,m_2=32.0 kg,m_s=12.0 kg,$ $\displaystyle r_1=1.60 m,r_s=0.160 m,$ find $\displaystyle (a)r_2,(b)Fp$ a) Since children are balancing: net $\displaystyle τ_{cw}=–$net$\displaystyle τ_{ccw}⇒w_1r_1+m_sgr_s=w_2r_2$ So, solving for $r^2$gives: $\displaystyle r_2=\frac{w_1r_1+m_sgr_s}{w_2}=\frac{m_1gr_1+m_sgr_s}{m_2g}=\frac{m_1r_1+m_sr_s}{m_2}$ $\displaystyle =\frac{(26.0 kg)(1.60 m)+(12.0 kg)(0.160 m)}{32.0 kg}$ $\displaystyle =1.36 m$ b) Since the children are not moving: net $\displaystyle F=0=F_p–w_1–w_2–w_s⇒F_p=w_1+w_2+w_s$ So that $\displaystyle F_p=(26.0 kg+32.0 kg+12.0 kg)(9.80m/s^2)=686 N$ 9.3: Stability 25. Suppose a horse leans against a wall as in Figure. Calculate the force exerted on the wall assuming that force is horizontal while using the data in the schematic representation of the situation. Note that the force exerted on the wall is equal in magnitude and opposite in direction to the force exerted on the horse, keeping it in equilibrium. The total mass of the horse and rider is 500 kg. Take the data to be accurate to three digits. Solution $\displaystyle F_{wall}=1.43×10^3N$ 26. Two children of mass 20.0 kg and 30.0 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3.00 m, at what distance from the pivot point is the small child sitting in order to maintain the balance? 27. (a) Calculate the magnitude and direction of the force on each foot of the horse in Figure (two are on the ground), assuming the center of mass of the horse is midway between the feet. The total mass of the horse and rider is 500kg. (b) What is the minimum coefficient of friction between the hooves and ground? Note that the force exerted by the wall is horizontal. Solution a) $\displaystyle 2.55×10^3N, 16.3º$ to the left of vertical (i.e., toward the wall) b) 0.292 28. A person carries a plank of wood 2.00 m long with one hand pushing down on it at one end with a force $\displaystyle F_1$ and the other hand holding it up at .500 m from the end of the plank with force $\displaystyle F_2$. If the plank has a mass of 20.0 kg and its center of gravity is at the middle of the plank, what are the magnitudes of the forces $\displaystyle F_1$ and $\displaystyle F_2$? 29. A 17.0-m-high and 11.0-m-long wall under construction and its bracing are shown in Figure. The wall is in stable equilibrium without the bracing but can pivot at its base. Calculate the force exerted by each of the 10 braces if a strong wind exerts a horizontal force of 650 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths. Neglect the thickness of the wall. Solution $\displaystyle F_B=2.12×10^4N$ 30. (a) What force must be exerted by the wind to support a 2.50-kg chicken in the position shown in Figure? (b) What is the ratio of this force to the chicken’s weight? (c) Does this support the contention that the chicken has a relatively stable construction? 31. Suppose the weight of the drawbridge in Figure is supported entirely by its hinges and the opposite shore, so that its cables are slack. (a) What fraction of the weight is supported by the opposite shore if the point of support is directly beneath the cable attachments? (b) What is the direction and magnitude of the force the hinges exert on the bridge under these circumstances? The mass of the bridge is 2500 kg. A small drawbridge, showing the forces on the hinges ($\displaystyle F$), its weight ($\displaystyle w$), and the tension in its wires ($\displaystyle T$). Solution a) 0.167, or about one-sixth of the weight is supported by the opposite shore. b) $\displaystyle F=2.0×10^4N$, straight up. 32. Suppose a 900-kg car is on the bridge in Figure with its center of mass halfway between the hinges and the cable attachments. (The bridge is supported by the cables and hinges only.) (a) Find the force in the cables. (b) Find the direction and magnitude of the force exerted by the hinges on the bridge. 33. A sandwich board advertising sign is constructed as shown in Figure. The sign’s mass is 8.00 kg. (a) Calculate the tension in the chain assuming no friction between the legs and the sidewalk. (b) What force is exerted by each side on the hinge? A sandwich board advertising sign demonstrates tension. Solution a) 21.6 N b) 21.6 N 34. (a) What minimum coefficient of friction is needed between the legs and the ground to keep the sign in Figure in the position shown if the chain breaks? (b) What force is exerted by each side on the hinge? 35. A gymnast is attempting to perform splits. From the information given in Figure, calculate the magnitude and direction of the force exerted on each foot by the floor. A gymnast performs full split. The center of gravity and the various distances from it are shown. Solution 350 N directly upwards 9.4: Applications of Statics, Including Problem-Solving Strategies 36. To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom? 37. In Figure, the cg of the pole held by the pole vaulter is 2.00 m from the left hand, and the hands are 0.700 m apart. Calculate the force exerted by (a) his right hand and (b) his left hand. (c) If each hand supports half the weight of the pole in Figure, show that the second condition for equilibrium (net $\displaystyle τ=0$) is satisfied for a pivot other than the one located at the center of gravity of the pole. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium described above. 9.5: Simple Machines 38. What is the mechanical advantage of a nail puller—similar to the one shown in Figure —where you exert a force 45 cm from the pivot and the nail is 1.8 cm on the other side? What minimum force must you exert to apply a force of 1250 N to the nail? Solution 25 50 N 39. Suppose you needed to raise a 250-kg mower a distance of 6.0 cm above the ground to change a tire. If you had a 2.0-m long lever, where would you place the fulcrum if your force was limited to 300 N? 40. a) What is the mechanical advantage of a wheelbarrow, such as the one in Figure, if the center of gravity of the wheelbarrow and its load has a perpendicular lever arm of 5.50 cm, while the hands have a perpendicular lever arm of 1.02 m? (b) What upward force should you exert to support the wheelbarrow and its load if their combined mass is 55.0 kg? (c) What force does the wheel exert on the ground? Solution a) $\displaystyle MA=18.5$ b) $\displaystyle F_i=29.1 N$ c) 510 N downward 41. A typical car has an axle with $\displaystyle 1.10 cm$ radius driving a tire with a radius of $\displaystyle 27.5 cm$. What is its mechanical advantage assuming the very simplified model in Figure(b)? 42. What force does the nail puller in Exercise exert on the supporting surface? The nail puller has a mass of 2.10 kg. Solution $\displaystyle 1.3×10^3N$ 43. If you used an ideal pulley of the type shown in Figure(a) to support a car engine of mass 115 kg , (a) What would be the tension in the rope? (b) What force must the ceiling supply, assuming you pull straight down on the rope? Neglect the pulley system’s mass. 44. Repeat Exercise for the pulley shown in Figure(c), assuming you pull straight up on the rope. The pulley system’s mass is 7.00 kg . Solution a) $\displaystyle T=299 N$ b) 897 N upward 9.6: Forces and Torques in Muscles and Joints 45. Verify that the force in the elbow joint in Example is 407 N, as stated in the text. Solution $\displaystyle F_B=470 N;r_1=4.00 cm;w_a=2.50 kg;r_2=16.0 cm;w_b=4.00 kg;r_3=38.0 cm$ $\displaystyle F_E=w_a(\frac{r_2}{r_1}−1)+w_b(\frac{r_3}{r_1}−1)$ $\displaystyle =(2.50 kg)(9.80m/s^2)(\frac{16.0 cm}{4.0 cm}–1)+(4.00 kg)(9.80m/s^2)(\frac{38.0 cm}{4.00 cm}–1)$ $\displaystyle =407 N$ 46. Two muscles in the back of the leg pull on the Achilles tendon as shown in Figure. What total force do they exert? The Achilles tendon of the posterior leg serves to attach plantaris, gastrocnemius, and soleus muscles to calcaneus bone. 47. The upper leg muscle (quadriceps) exerts a force of 1250 N, which is carried by a tendon over the kneecap (the patella) at the angles shown in Figure. Find the direction and magnitude of the force exerted by the kneecap on the upper leg bone (the femur). The knee joint works like a hinge to bend and straighten the lower leg. It permits a person to sit, stand, and pivot. Solution $\displaystyle 1.1×10^3N$ $\displaystyle θ=190º$ ccw from positive x axis 48. A device for exercising the upper leg muscle is shown in Figure, together with a schematic representation of an equivalent lever system. Calculate the force exerted by the upper leg muscle to lift the mass at a constant speed. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium in Applications of Statistics, Including Problem-Solving Strategies. A mass is connected by pulleys and wires to the ankle in this exercise device. 49. A person working at a drafting board may hold her head as shown in Figure, requiring muscle action to support the head. The three major acting forces are shown. Calculate the direction and magnitude of the force supplied by the upper vertebrae $\displaystyle F_V$ to hold the head stationary, assuming that this force acts along a line through the center of mass as do the weight and muscle force. Solution $\displaystyle F_V=97N,θ=59º$ 50. We analyzed the biceps muscle example with the angle between forearm and upper arm set at 90º . Using the same numbers as in Example, find the force exerted by the biceps muscle when the angle is 120º and the forearm is in a downward position. 51. Even when the head is held erect, as in Figure, its center of mass is not directly over the principal point of support (the atlanto-occipital joint). The muscles at the back of the neck should therefore exert a force to keep the head erect. That is why your head falls forward when you fall asleep in the class. (a) Calculate the force exerted by these muscles using the information in the figure. (b) What is the force exerted by the pivot on the head? The center of mass of the head lies in front of its major point of support, requiring muscle action to hold the head erect. A simplified lever system is shown. Solution (a) 25 N downward (b) 75 N upward 52. A 75-kg man stands on his toes by exerting an upward force through the Achilles tendon, as in Figure. (a) What is the force in the Achilles tendon if he stands on one foot? (b) Calculate the force at the pivot of the simplified lever system shown—that force is representative of forces in the ankle joint. The muscles in the back of the leg pull the Achilles tendon when one stands on one’s toes. A simplified lever system is shown. Solution (a) $\displaystyle F_A=2.21×10^3N$ upward (b) $\displaystyle F_B=2.94×10^3N$ downward 53. A father lifts his child as shown in Figure. What force should the upper leg muscle exert to lift the child at a constant speed? A child being lifted by a father’s lower leg. 54. Unlike most of the other muscles in our bodies, the masseter muscle in the jaw, as illustrated in Figure, is attached relatively far from the joint, enabling large forces to be exerted by the back teeth. (a) Using the information in the figure, calculate the force exerted by the lower teeth on the bullet. (b) Calculate the force on the joint. A person clenching a bullet between his teeth. Solution (a) $\displaystyle F_{\text{teeth on bullet}}=1.2×10^2N$ upward (b) $\displaystyle F_J=84 N$ downward 55. Integrated Concepts Suppose we replace the 4.0-kg book in Exercise of the biceps muscle with an elastic exercise rope that obeys Hooke’s Law. Assume its force constant $\displaystyle k=600N/m$. (a) How much is the rope stretched (past equilibrium) to provide the same force $\displaystyle F_B$ as in this example? Assume the rope is held in the hand at the same location as the book. (b) What force is on the biceps muscle if the exercise rope is pulled straight up so that the forearm makes an angle of $\displaystyle 25º$with the horizontal? Assume the biceps muscle is still perpendicular to the forearm. 56. (a) What force should the woman in Figure exert on the floor with each hand to do a push-up? Assume that she moves up at a constant speed. (b) The triceps muscle at the back of her upper arm has an effective lever arm of 1.75 cm, and she exerts force on the floor at a horizontal distance of 20.0 cm from the elbow joint. Calculate the magnitude of the force in each triceps muscle, and compare it to her weight. (c) How much work does she do if her center of mass rises 0.240 m? (d) What is her useful power output if she does 25 pushups in one minute? A woman doing pushups. Solution (a) 147 N downward (b) 1680 N, 3.4 times her weight (c) 118 J (d) 49.0 W 57. You have just planted a sturdy 2-m-tall palm tree in your front lawn for your mother’s birthday. Your brother kicks a 500 g ball, which hits the top of the tree at a speed of 5 m/s and stays in contact with it for 10 ms. The ball falls to the ground near the base of the tree and the recoil of the tree is minimal. (a) What is the force on the tree? (b) The length of the sturdy section of the root is only 20 cm. Furthermore, the soil around the roots is loose and we can assume that an effective force is applied at the tip of the 20 cm length. What is the effective force exerted by the end of the tip of the root to keep the tree from toppling? Assume the tree will be uprooted rather than bend. (c) What could you have done to ensure that the tree does not uproot easily? 58. Unreasonable Results Suppose two children are using a uniform seesaw that is 3.00 m long and has its center of mass over the pivot. The first child has a mass of 30.0 kg and sits 1.40 m from the pivot. (a) Calculate where the second 18.0 kg child must sit to balance the seesaw. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent? Solution a) $\displaystyle \bar{x}_2=2.33 m$ b) The seesaw is 3.0 m long, and hence, there is only 1.50 m of board on the other side of the pivot. The second child is off the board. c) The position of the first child must be shortened, i.e. brought closer to the pivot. 59. Construct Your Own Problem Consider a method for measuring the mass of a person’s arm in anatomical studies. The subject lies on her back, extends her relaxed arm to the side and two scales are placed below the arm. One is placed under the elbow and the other under the back of her hand. Construct a problem in which you calculate the mass of the arm and find its center of mass based on the scale readings and the distances of the scales from the shoulder joint. You must include a free body diagram of the arm to direct the analysis. Consider changing the position of the scale under the hand to provide more information, if needed. You may wish to consult references to obtain reasonable mass values. Contributors and Attributions - Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .	libretexts	2025-03-17T19:53:28.587265	2017-11-27T02:03:13	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/09%3A_Statics_and_Torque/9.E%3A_Statics_and_Torque_(Exercises)", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "9.E: Statics and Torque (Exercises)", "author": null }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum	10: Rotational Motion and Angular Momentum In physics, angular momentum (rarely, moment of momentum or rotational momentum) is the rotational analog of linear momentum. It is an important quantity in physics because it is a conserved quantity – the angular momentum of a system remains constant unless acted on by an external torque. - - 10.0: Prelude to Rotational Motion and Angular Momentum - Why do tornadoes spin at all? And why do tornados spin so rapidly? The answer is that air masses that produce tornadoes are themselves rotating, and when the radii of the air masses decrease, their rate of rotation increases. An ice skater increases her spin in an exactly analogous manner. The skater starts her rotation with outstretched limbs and increases her spin by pulling them in toward her body. The same physics describes the exhilarating spin of a skater and the wrenching force of a torna - - 10.1: Angular Acceleration - Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer’s hard disk slows to a halt when switched off. In all these cases, there is an angular acceleration, in which ω changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration α is defined as the rate of change of angular velocity. - - 10.2: Kinematics of Rotational Motion - Just by using our intuition, we can begin to see how rotational quantities like θ,ω and α are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel’s angular acceleration α is large for a long period of time t then the final angular velocity ω and angle of rotation θ are large. - - 10.5: Angular Momentum and Its Conservation - Angular momentum is completely analogous to linear momentum. It has the same implications in terms of carrying rotation forward, and it is conserved when the net external torque is zero. Angular momentum, like linear momentum, is also a property of the atoms and subatomic particles. Thumbnail: The torque caused by the normal force – Fg and the weight of the top causes a change in the angular momentum L in the direction of that torque. This causes the top to precess. (CC-BY-SA-2.5; Xavier Snelgrove ) .	libretexts	2025-03-17T19:53:28.728779	2015-10-27T19:25:46	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "10: Rotational Motion and Angular Momentum", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.00%3A_Prelude_to_Rotational_Motion_and_Angular_Momentum	10.0: Prelude to Rotational Motion and Angular Momentum Why do tornadoes spin at all (Figure $\PageIndex{1}$)? And why do tornados spin so rapidly? The answer is that air masses that produce tornadoes are themselves rotating, and when the radii of the air masses decrease, their rate of rotation increases. An ice skater increases her spin in an exactly analogous manner as seen in Figure $\PageIndex{2}$. The skater starts her rotation with outstretched limbs and increases her spin by pulling them in toward her body. The same physics describes the exhilarating spin of a skater and the wrenching force of a tornado. Clearly, force, energy, and power are associated with rotational motion. These and other aspects of rotational motion are covered in this chapter. We shall see that all important aspects of rotational motion either have already been defined for linear motion or have exact analogs in linear motion. First, we look at angular acceleration—the rotational analog of linear acceleration.	libretexts	2025-03-17T19:53:28.786031	2015-11-01T05:25:04	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.00%3A_Prelude_to_Rotational_Motion_and_Angular_Momentum", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "10.0: Prelude to Rotational Motion and Angular Momentum", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.01%3A_Angular_Acceleration	10.1: Angular Acceleration Learning Objectives By the end of this section, you will be able to: - Describe uniform circular motion. - Explain non-uniform circular motion. - Calculate angular acceleration of an object. - Observe the link between linear and angular acceleration. Uniform Circular Motion and Gravitation discussed only uniform circular motion, which is motion in a circle at constant speed and, hence, constant angular velocity. Recall that angular velocity $\omega$ was defined as the time rate of change of angle $\theta$. \[ \omega = \dfrac{\Delta \theta}{\Delta t},\] where $\theta$ is the angle of rotation as seen in Figure $\PageIndex{1}$. The relationship between angular velocity $\omega$ and linear velocity $v$ was also defined in Rotation Angle and Angular Velocity as \[v = r \omega\] or \[\omega = \dfrac{v}{r}\] where $r$ is the radius of curvature, also seen in Figure $\PageIndex{1}$. According to the sign convention, the counter clockwise direction is considered as positive direction and clockwise direction as negative Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer’s hard disk slows to a halt when switched off. In all these cases, there is an angular acceleration , in which $\omega$ changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration $\alpha$ is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows: \[\alpha = \dfrac{\Delta \omega}{\Delta t},\] where $Δω$ is the change in angular velocity and $Δt$ is the change in time. The units of angular acceleration are (rad/s)/s, or rad/$s^2$. If $ω$ increases, then $α$ is positive. If $ω$ decreases, then $α$ is negative. Example $\PageIndex{1}$: Calculating the Angular Acceleration and Deceleration of a Bike Wheel Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the angular acceleration in $rad/s^2$. (b) If she now slams on the brakes, causing an angular acceleration of –87.3 $rad/s^2$, how long does it take the wheel to stop? Strategy for (a) The angular acceleration can be found directly from its definition in $α=\frac{Δω}{Δt}$ because the final angular velocity and time are given. We see that $Δω$ is 250 rpm and $Δt$ is 5.00 s. Solution for (a) Entering known information into the definition of angular acceleration, we get $α=\frac{Δω}{Δt}=\frac{250 rpm}{5.00 s}$. Because $Δω$ is in revolutions per minute (rpm) and we want the standard units of $rad/s^2$ for angular acceleration, we need to convert $Δω$ from rpm to rad/s: $Δω=250\frac{rev}{min}⋅\frac{2π rad}{rev}⋅\frac{1 min}{60 sec}=26.2rads$. Entering this quantity into the expression for $α$, we get $α=\frac{Δω}{Δt}=\frac{26.2 rad/s}{5.00 s}=5.24 rad/s^2$. Strategy for (b) In this part, we know the angular acceleration and the initial angular velocity. We can find the stoppage time by using the definition of angular acceleration and solving for Δt, yielding $Δt=\frac{Δω}{α}$. Solution for (b) Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that $Δω$ is –26.2 rad/s, and α is given to be –87.3 $rad/s^2$. Thus, $Δt=\frac{–26.2 rad/s}{–87.3rad/s^2}=0.300 s.$ Discussion Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are analogous to what happens with linear motion. For example, there is a large deceleration when you crash into a brick wall—the velocity change is large in a short time interval. If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then come to a stop. This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear and angular acceleration are related. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in Figure $\PageIndex{2}$. Thus, linear acceleration is called tangential acceleration $a_t$. Figure $\PageIndex{2}$: In circular motion, linear acceleration $a$, occurs as the magnitude of the velocity changes: $a$ is tangent to the motion. In the context of circular motion, linear acceleration is also called tangential acceleration $a_t$. Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. We know from Uniform Circular Motion and Gravitation that in circular motion centripetal acceleration, ac, refers to changes in the direction of the velocity but not its magnitude. An object undergoing circular motion experiences centripetal acceleration, as seen in Figure $\PageIndex{3}$. Thus, $a_t$ and $a_c$ are perpendicular and independent of one another. Tangential acceleration $a_t$ is directly related to the angular acceleration $α$ and is linked to an increase or decrease in the velocity, but not its direction. Now we can find the exact relationship between linear acceleration $a_t$ and angular acceleration $α$. Because linear acceleration is proportional to a change in the magnitude of the velocity, it is defined (as it was in One-Dimensional Kinematics) to be $a_t=\frac{Δv}{Δt}$. For circular motion, note that $v=rω$, so that $a_t=\frac{Δ(rω)}{Δt}$. The radius r is constant for circular motion, and so $Δ(rω)=r(Δω)$. Thus, $a_t=r\frac{Δω}{Δt}$. By definition, $α=\frac{Δω}{Δt}$. Thus, $a_t=rα$, or $α=\frac{a_t}{r}$. These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger the linear (tangential) acceleration is, and vice versa. For example, the greater the angular acceleration of a car’s drive wheels, the greater the acceleration of the car. The radius also matters. For example, the smaller a wheel, the smaller its linear acceleration for a given angular acceleration $α$. Exercise $\PageIndex{1}$: Calculating the Angular Acceleration of a Motorcycle Wheel A powerful motorcycle can accelerate from 0 to 30.0 m/s (about 108 km/h) in 4.20 s. What is the angular acceleration of its 0.320-m-radius wheels? (See Figure .) Strategy We are given information about the linear velocities of the motorcycle. Thus, we can find its linear acceleration $a_t$. Then, the expression $α=\frac{a_t}{r}$ can be used to find the angular acceleration. Solution The linear acceleration is $a_t=\frac{Δv}{Δt}=\frac{30.0 m/s}{4.20 s}=7.14m/s^2$. We also know the radius of the wheels. Entering the values for $a_t$ and $r$ into $α=\frac{a_t}{r}$, we get $α=\frac{a_t}{r}=\frac{7.14m/s^2}{0.320 m}=22.3rad/s^2$. Discussion Units of radians are dimensionless and appear in any relationship between angular and linear quantities. So far, we have defined three rotational quantities— $θ, ω,$ and $α$. These quantities are analogous to the translational quantities $x, v,$ and $a$. Table $\PageIndex{2}$ displays rotational quantities, the analogous translational quantities, and the relationships between them. \| Rotational \| Translational \| Relationship \| \|---\|---\|---\| \| $θ$ \| $x$ \| $θ=\frac{x}{r}$ \| \| $ω$ \| $v$ \| $ω=\frac{v}{r}$ \| \| $α$ \| $a$ \| $α=\frac{a_t}{r}$ \| MAKING CONNECTIONS: TAKE-HOME EXPERIMENT Sit down with your feet on the ground on a chair that rotates. Lift one of your legs such that it is unbent (straightened out). Using the other leg, begin to rotate yourself by pushing on the ground. Stop using your leg to push the ground but allow the chair to rotate. From the origin where you began, sketch the angle, angular velocity, and angular acceleration of your leg as a function of time in the form of three separate graphs. Estimate the magnitudes of these quantities. Exercise $\PageIndex{2}$ Angular acceleration is a vector, having both magnitude and direction. How do we denote its magnitude and direction? Illustrate with an example. - Answer - The magnitude of angular acceleration is $α$ and its most common units are $rad/s^2$. The direction of angular acceleration along a fixed axis is denoted by a + or a – sign, just as the direction of linear acceleration in one dimension is denoted by a + or a – sign. For example, consider a gymnast doing a forward flip. Her angular momentum would be parallel to the mat and to her left. The magnitude of her angular acceleration would be proportional to her angular velocity (spin rate) and her moment of inertia about her spin axis. PHET EXPLORATIONS: LADYBUG REVOLUTION Join the Ladybug Revolution in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs. Summary - Uniform circular motion is the motion with a constant angular velocity $ω=\frac{Δθ}{Δt}$. - In non-uniform circular motion, the velocity changes with time and the rate of change of angular velocity (i.e. angular acceleration) is $α=\frac{Δω}{Δt}$. - Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction, given as $a_t=\frac{Δv}{Δt}$. - For circular motion, note that $v=rω$, so that $a_t=\frac{Δ(rω)}{Δt}$. - The radius r is constant for circular motion, and so $Δ(rω)=rΔω$. Thus, $a_t=r\frac{Δω}{Δt}$. - By definition, $Δω/Δt=α$. Thus, $a_t=rα$ or $α=\frac{a_t}{r}$. Glossary - angular acceleration - the rate of change of angular velocity with time - change in angular velocity - the difference between final and initial values of angular velocity - tangential acceleration - the acceleration in a direction tangent to the circle at the point of interest in circular motion	libretexts	2025-03-17T19:53:28.870276	2015-11-01T05:25:23	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.01%3A_Angular_Acceleration", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "10.1: Angular Acceleration", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.02%3A_Kinematics_of_Rotational_Motion	10.2: Kinematics of Rotational Motion Learning Objectives By the end of this section, you will be able to: - Observe the kinematics of rotational motion. - Derive rotational kinematic equations. - Evaluate problem solving strategies for rotational kinematics. Just by using our intuition, we can begin to see how rotational quantities like $\theta, \omega$ and $\alpha$ are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a fairly long time, it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel’s angular acceleration $\alpha$ is large for a long period of time $t$ then the final angular velocity $\omega$ and angle of rotation $\theta$ are large. The wheel’s rotational motion is exactly analogous to the fact that the motorcycle’s large translational acceleration produces a large final velocity, and the distance traveled will also be large.Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating $\omega, \alpha$, and $t$. To determine this equation, we recall a familiar kinematic equation for translational, or straight-line, motion: \[v = v_0 + at \, (constant \, a)\] Note that in rotational motion $a = a_t$, and we shall use the symbol $a$ for tangential or linear acceleration from now on. As in linear kinematics, we assume $a$ is constant, which means that angular acceleration $\alpha$ is also a constant, because $a = r\alpha$. Now, let us substitute $v = r\omega$ and $a = r\alpha$ into the linear equation above: \[rw = r\omega_o + r\alpha t,\] The radius $r$ cancels in the equation, yielding \[\omega = \omega_o + at \, (constant \, a),\] where $\omega_o$ is the initial angular velocity. This last equation is a kinematic relationship among $\omega, \alpha$, and $t$ - that is, it describes their relationship without reference to forces or masses that may affect rotation. It is also precisely analogous in form to its translational counterpart. Making Connections. Kinematics for rotational motion is completely analogous to translational kinematics, first presented in One-Dimensional Kinematics . Kinematics is concerned with the description of motion without regard to force or mass. We will find that translational kinematic quantities, such as displacement, velocity, and acceleration have direct analogs in rotational motion. Starting with the four kinematic equations we developed in One-Dimensional Kinematics , we can derive the following four rotational kinematic equations (presented together with their translational counterparts): \| Rotational \| Translational \| \| \|---\|---\|---\| \| $\theta = \overline{\omega} t$ \| $x = \overline{v}t$ \| \| \| $\omega = \omega_o + \alpha t$ \| $v = v_o +at$ \| $(constant \, \alpha, a)$ \| \| $\Theta = \omega_ot + \frac{1}{2}\alpha t^2$ \| $x = v_0t + \frac{1}{2}at^2$ \| $(constant \, \alpha, a)$ \| \| $\omega^2 = \omega_o^2 + 2\alpha \theta$ \| $v^2 = v_0^2 + 2 ax$ \| $(constant \, \alpha, a)$ \| In these equations, the subscript 0 denotes initial values ($\theta_0, x_0$ and $t_0$ are initial values), and the average angular velocity $overline{\omega}$ and average velocity $\overline{v}$ are defined as follows: \[\overline{\omega} = \dfrac{\omega_0 + \omega}{2} \, and \, \overline{v} = \dfrac{v_0 + v}{2}.\] The equations given above in Table $\PageIndex{1}$ can be used to solve any rotational or translational kinematics problem in which $a$ and $\alpha$ are constant. Problem-Solving Strategy for Rotational Kinematics - Examine the situation to determine that rotational kinematics (rotational motion) is involved. Rotation must be involved, but without the need to consider forces or masses that affect the motion. - Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful. - Make a list of what is given or can be inferred from the problem as stated (identify the knowns). - Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog because by now you are familiar with such motion. - Substitute the known values along with their units into the appropriate equation, and obtain numerical solutions complete with units. Be sure to use units of radians for angles. - Check your answer to see if it is reasonable: Does your answer make sense? Example $\PageIndex{1}$: Calculating the Acceleration of a Fishing Reel A deep-sea fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of $110 \, rad/s^2$ for 2.00 s as seen in Figure 10.3.1. - What is the final angular velocity of the reel? - At what speed is fishing line leaving the reel after 2.00 s elapses? - How many revolutions does the reel make? - How many meters of fishing line come off the reel in this time? Strategy In each part of this example, the strategy is the same as it was for solving problems in linear kinematics. In particular, known values are identified and a relationship is then sought that can be used to solve for the unknown. Solution for (a) Here $\alpha$ and $t$ are given and $\omega$ needs to be determined. The most straightforward equation to use is $\omega = \omega_0 + \alpha t$ because the unknown is already on one side and all other terms are known. That equation states that \[\omega = \omega_0 + \alpha t.\] We are also given that $\omega_0 = 0$ (it starts from rest), so that \[\omega = 0 + (110 \, rad/s^2)(2.00s) = 220 \, rad/s.\] Solution for (b) Now that $\omega$ is known, the speed $v$ can most easily be found using the relationship \[v = r\omega,\] where the radius $r$ ofthe reel is given to be 4.50 cm; thus, \[ v = (0.0450 \, m)(220 \, rad/s) = 9.90 \, m/s.\] Note again that radians must always be used in any calculation relating linear and angular quantities. Also, because radians are dimensionless, we have $m \times rad = m$. Solution for (c) Here, we are asked to find the number of revolutions. Because $1\space rev = 2\pi \, rad$, we can find the number of revolutions by finding $\theta$ in radians. We are given $\alpha$ and $t$, and we know $\omega_o$ is zero, so that $\theta$ can be obtained using $\theta = \omega_0t + \frac{1}{2}\alpha t^2$. \[\theta = \omega_0t + \dfrac{1}{2} \alpha t^2\] \[= 0 + (0.500)(110 \, rad/s^2)(2.00s)^2 = 220 rad.\] Converting radians to revolutions gives \[\theta = (220 \, rad)\dfrac{1 \, rev}{2\pi \, rad} = 35.0 \, rev.\] Solution for (d) The number of meters of fishing line is $x$ which can be obtained through its relationship with $\theta$. \[x = r\theta = (0.0450 \, m)(220 \, rad) = 9.90 \, m.\] Discussion This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. We also see in this example how linear and rotational quantities are connected. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds.) The amount of fishing line played out is 9.90 m, about right for when the big fish bites. Example $\PageIndex{2}$: Calculating the Duration When the Fishing Reel Slows Down and Stops Now let us consider what happens if the fisherman applies a brake to the spinning reel, achieving an angular acceleration of - $300 \, rad/s^2$. How long does it take the reel to come to a stop? Strategy We are asked to find the time for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is $\omega_0 = 220 \, rad/s$ and the final angular velocity $\omega$ is zero. The angular acceleration is given to be $\alpha = - 300 \, rad/s^2.$ Examining the available equations, we see all quantities but t are known in $\omega = \omega_0 + \alpha t$, making it easiest to use this equation. Solution The equation states \[\omega = \omega_0 + \alpha t.\] We solve the equation algebraically for t , and then substitute the known values as usual, yielding \[t = \dfrac{\omega - \omega_0}{\alpha} = \dfrac{0 - 220 \, rad/s}{-300 \, rad/s^2} = 0.733 \, s.\] Discussion Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration. Example $\PageIndex{3}$: Calculating the Slow Acceleration of Trains and Their Wheels Large freight trains accelerate very slowly. Suppose one such train accelerates from rest, giving its 0.350-m-radius wheels an angular acceleration of $0.250 \, rad/s^2$. After the wheels have made 200 revolutions (assume no slippage): (a) How far has the train moved down the track? (b) What are the final angular velocity of the wheels and the linear velocity of the train? Strategy In part (a), we are asked to find $x$, and in (b) we are asked to find $\omega$ and $v$. We are given the number of revolutions $\theta$, the radius of the wheels $r$, and the angular accelerationn$\alpha$. Solution for (a) The distance $x$ is very easily found from the relationship between distance and rotation angle: \[\theta = \dfrac{x}{r}.\] Solving this equation for $x$ yields \[x = r\theta.\] Before using this equation, we must convert the number of revolutions into radians, because we are dealing with a relationship between linear and rotational quantities: \[\theta = (200 \, rev)\dfrac{2\pi \, rad}{1 \, rev} = 1257 \, rad.\] Now we can substitute the known values into $x = r\theta$ to find the distance the train moved down the track: \[x = r\theta = (0.350 \, m)(1257 \, rad) = 440 \, m.\] Solution for (b) We cannot use any equation that incorporates $t$ to find $\omega$, because the equation would have at least two unknown values. The equation $\omega^2 = \omega_0^2 + 2\alpha \theta$ will work, because we know the values for all variables except $\omega$. \[\omega^2 = \omega_0^2 + 2 \alpha \theta\] Taking the square root of this equation and entering the known values gives \[\omega = [0 + 2(0.250 \, rad/s^2)(1257 \, rad)]^{1/2}\] \[= 25.1 \, rad/s.\] We can find the linear velocity of the train, $v$, through its relationship to $\omega$: \[v = r\omega = (0.350 \, m)(25.1 \, rad/s) = 8.77 \, m/s.\] Discussion The distance traveled is fairly large and the final velocity is fairly slow (just under 32 km/h). There is translational motion even for something spinning in place, as the following example illustrates. Figure10.3.2 shows a fly on the edge of a rotating microwave oven plate. The example below calculates the total distance it travels. Example $\PageIndex{4}$: Calculating the Distance Traveled by a Fly on the Edge of a Microwave Oven Plate A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate and remains there. If the plate has a radius of 0.15 m and rotates at 6.0 rpm, calculate the total distance traveled by the fly during a 2.0-min cooking period. (Ignore the start-up and slow-down times.) Strategy First, find the total number of revolutions $\theta$, and then the linear distance $x$ traveled. $\theta = \overline{\omega}$ can be used to find $\theta$ because $\overline{\omega}$ is given to be 6.0 rpm. Solution Entering known values into $\theta = \overline{\omega}$ gives \[\theta = \overline{\omega} = (6.0 \, rpm)(2.0 \, min) = 12 \, rev.\] As always, it is necessary to convert revolutions to radians before calculating a linear quantity like $x$ from an angular quantity like $\theta$: \[\theta = (12 \, rev)\left(\dfrac{2\pi \, rad}{1 \, rev}\right) = 75.4 \, rad.\] Now, using the relationship between $x$ and $\theta$, we can determine the distance traveled: \[x = r\theta = (0.15 \, m)(75.4 \, rad) = 11 \, m.\] Discussion Quite a trip (if it survives)! Note that this distance is the total distance traveled by the fly. Displacement is actually zero for complete revolutions because they bring the fly back to its original position. The distinction between total distance traveled and displacement was first noted in One-Dimensional Kinematics . Exercise $\PageIndex{1}$ Rotational kinematics has many useful relationships, often expressed in equation form. Are these relationships laws of physics or are they simply descriptive? (Hint: the same question applies to linear kinematics.) - Answer - Rotational kinematics (just like linear kinematics) is descriptive and does not represent laws of nature. With kinematics, we can describe many things to great precision but kinematics does not consider causes. For example, a large angular acceleration describes a very rapid change in angular velocity without any consideration of its cause. Summary - Kinematics is the description of motion. - The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. - Starting with the four kinematic equations we developed in the One-Dimensional Kinematics , we can derive the four rotational kinematic equations (presented together with their translational counterparts) seen in Table . - In these equations, the subscript 0 denotes initial values $({x_0}$ and $t_o$ are initial values), and the average angular velocity $\overline{\omega}$ and average velocity $\overline{v}$ are defined as follows: \[ \overline{\omega} = \dfrac{\omega_0 + \omega}{2} \, and \, \dfrac{v_0 + v}{2}.\] Glossary - kinematics of rotational motion - describes the relationships among rotation angle, angular velocity, angular acceleration, and time	libretexts	2025-03-17T19:53:28.958166	2015-11-01T05:25:40	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.02%3A_Kinematics_of_Rotational_Motion", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "10.2: Kinematics of Rotational Motion", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.03%3A_Dynamics_of_Rotational_Motion_-_Rotational_Inertia	10.3: Dynamics of Rotational Motion - Rotational Inertia Learning Objectives By the end of this section, you will be able to: - Understand the relationship between force, mass and acceleration. - Study the turning effect of force. - Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angular acceleration. If you have ever spun a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity as seen in Figure 10.4.1. In fact, your intuition is reliable in predicting many of the factors that are involved. For example, we know that a door opens slowly if we push too close to its hinges. Furthermore, we know that the more massive the door, the more slowly it opens. The first example implies that the farther the force is applied from the pivot, the greater the angular acceleration; another implication is that angular acceleration is inversely proportional to mass. These relationships should seem very similar to the familiar relationships among force, mass, and acceleration embodied in Newton’s second law of motion. There are, in fact, precise rotational analogs to both force and mass. To develop the precise relationship among force, mass, radius, and angular acceleration, consider what happens if we exert a force $F$ on a point mass $m$ that is at a distance $r$ from a pivot point, as shown in Figure 10.4.2. Because the force is perpendicular to $r$, an accelerationn$a = frac{F}{m}$ is obtained in the direction of $F$. We can rearrange this equation such that $F = ma$ and then look for ways to relate this expression to expressions for rotational quantities. We note that $a = r\alpha$, and we substitute this expression into $F = ma$, yielding \[F = mr\alpha.\] Recall that torque is the turning effectiveness of a force. In this case, because $F$ is perpendicular to $r$, torque is simply $\tau = Fr$. So, if we multiply both sides of the equation above by $r$, we get torque on the left-hand side. That is, \[rF = mr^2\alpha\] or \[\tau = mr^2\alpha.\] This last equation is the rotational analog of Newton’s second law $F = ma$, where torque is analogous to force, angular acceleration is analogous to translational acceleration, and $mr^2$ is analogous to mass (or inertia). The quantity $mr^2$ is called the rotational inertia or moment of inertia of a point mass $m$ a distance $r$ from the center of rotation. Making Connections: Rotational Motion Dynamics Dynamics for rotational motion is completely analogous to linear or translational dynamics. Dynamics is concerned with force and mass and their effects on motion. For rotational motion, we will find direct analogs to force and mass that behave just as we would expect from our earlier experiences. Rotational Inertia and Moment of Inertia Before we can consider the rotation of anything other than a point mass like the one in Figure , we must extend the idea of rotational inertia to all types of objects. To expand our concept of rotational inertia, we define the moment of inertia $I$ of an object to be the sum of $mr^2$ for all the point masses of which it is composed. That is, $I = \sum mr^2$. Here $I$ is analogous to $m$ in translational motion. Because of the distance $r$, the moment of inertia for any object depends on the chosen axis. Actually, calculating $I$ is beyond the scope of this text except for one simple case—that of a hoop, which has all its mass at the same distance from its axis. A hoop’s moment of inertia around its axis is therefore $MR^2$, where $M$ is its total mass and $R$ its radius. (We use $M$ and $R$ for an entire object to distinguish them from $m$ and $r$ for point masses.) In all other cases, we must consult Figure 10.4.3 (note that the table is piece of artwork that has shapes as well as formulae) for formulas for $I$ that have been derived from integration over the continuous body. Note that $I$ has units of mass multiplied by distance squared $(kg \cdot m^2)$ as we might expect from its definition. The general relationship among torque, moment of inertia, and angular acceleration is \[net \, \tau = I \alpha\] or \[\alpha = \dfrac{net \, \tau}{I},\] where net $\tau$ is the total torque from all forces relative to a chosen axis. For simplicity, we will only consider torques exerted by forces in the plane of the rotation. Such torques are either positive or negative and add like ordinary numbers. The relationship in $\tau = I\alpha$, $\alpha = \frac{net \, \tau}{I}$ is the rotational analog to Newton’s second law and is very generally applicable. This equation is actually valid for any torque, applied to any object, relative to any axis. As we might expect, the larger the torque is, the larger the angular acceleration is. For example, the harder a child pushes on a merry-go-round, the faster it accelerates. Furthermore, the more massive a merry-go-round, the slower it accelerates for the same torque. The basic relationship between moment of inertia and angular acceleration is that the larger the moment of inertia, the smaller is the angular acceleration. But there is an additional twist. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates. For example, it will be much easier to accelerate a merry-go-round full of children if they stand close to its axis than if they all stand at the outer edge. The mass is the same in both cases; but the moment of inertia is much larger when the children are at the edge. Take-Home Experiment - Cut out a circle that has about a 10 cm radius from stiff cardboard. Near the edge of the circle, write numbers 1 to 12 like hours on a clock face. Position the circle so that it can rotate freely about a horizontal axis through its center, like a wheel. (You could loosely nail the circle to a wall.) Hold the circle stationary and with the number 12 positioned at the top, attach a lump of blue putty (sticky material used for fixing posters to walls) at the number 3. How large does the lump need to be to just rotate the circle? Describe how you can change the moment of inertia of the circle. How does this change affect the amount of blue putty needed at the number 3 to just rotate the circle? Change the circle’s moment of inertia and then try rotating the circle by using different amounts of blue putty. Repeat this process several times. Problem-Solving Strategy for Rotational Dynamics - Examine the situation to determine that torque and mass are involved in the rotation . Draw a careful sketch of the situation. - Determine the system of interest . - Draw a free body diagram . That is, draw and label all external forces acting on the system of interest. - Apply $net \, \tau = \alpha$, $\alpha = \frac{net \, \tau}{I},$ - the rotational equivalent of Newton’s second law, to solve the problem . Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation. - As always, check the solution to see if it is reasonable . Making Connections: Statics vs. Kinetics In statics, the net torque is zero, and there is no angular acceleration. In rotational motion, net torque is the cause of angular acceleration, exactly as in Newton’s second law of motion for rotation. Example $\PageIndex{1}$: Calculating the Effect of Mass Distribution Consider the father pushing a playground merry-go-round in Figure . He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction. Strategy Angular acceleration is given directly by the expression $\alpha = \frac{net \, \tau}{I}$ \[\alpha = \dfrac{\tau}{I}.\] To solve for $\alpha$, we must first calculate the torque (which is the same in both cases) and moment of inertia $I$ (which is greater in the second case). To find the torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that\[\tau = rF \, sin \, \theta = (1.50 \, m)(250 \, N) = 375 \, N \cdot m.\] Solution for (a) The moment of inertia of a solid disk about this axis is given in Figure to be \[\dfrac{1}{2}MR^2,\] where $M = 50.0 \, kg$ and $R = 1.50 \, m$, so that \[I = (0.500)(50.0 \, kg)(1.50 \, m)^2 = 56.25 \, kg \cdot m^2\] Now, after we substitute the known values, we find the angular acceleration to be \[\alpha = \dfrac{\tau}{I} = \dfrac{375 \, N \cdot m}{56.25 \, kg \cdot m^2} = 6.67 \dfrac{rad}{s^2}.\] Solution for (b) We expect the angular acceleration for the system to be less in this part, because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia $I$, we first find the child’s moment of inertia $I_c$ by considering the child to be equivalent to a point mass at a distance of 1.25 m from the axis. Then, \[I_c = MR^2 = (18.0 \, kg)(1.25 \, m)^2 = 28.13 \, kg \cdot m^2.\] The total moment of inertia is the sum of moments of inertia of the merry-go-round and the child (about the same axis). To justify this sum to yourself, examine the definition of $I$: \[I = 28.13 \, kg \cdot m^2 + 56.25 \, kg \cdot m^2 = 84.38 \, kg \cdot m^2.\] Substituting known values into the equation for $\alpha $ gives \[\alpha = \dfrac{\tau}{I} = \dfrac{375 \, N \cdot m}{84.38 \, kg \cdot m^2} = 4.44 \dfrac{rad}{s^2}.\] Discussion The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is on it. In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev/s, respectively. The father would end up running at about 50 km/h in the first case. Summer Olympics, here he comes! Confirmation of these numbers is left as an exercise for the reader. Exercise $\PageIndex{1}$:Check Your Understanding Torque is the analog of force and moment of inertia is the analog of mass. Force and mass are physical quantities that depend on only one factor. For example, mass is related solely to the numbers of atoms of various types in an object. Are torque and moment of inertia similarly simple? No. Torque depends on three factors: force magnitude, force direction, and point of application. Moment of inertia depends on both mass and its distribution relative to the axis of rotation. So, while the analogies are precise, these rotational quantities depend on more factors Summary - The farther the force is applied from the pivot, the greater is the angular acceleration; angular acceleration is inversely proportional to mass. - If we exert a force $F$ on a point mass $m$ that is a distance r from a pivot point and because the force is perpendicular to r and acceleration $a = F/m$ is obtained in the direction of $F$. We can rearrange this equation such that \[F = ma,\] and then look for ways to relate this expression to expressions for rotational quantities. We note that $a = r \alpha$, and we substitute this expression into $F = ma$, yielding \[F = mr\alpha\] - Torque is the turning effectiveness of a force. In this case, because $F$ is perpendicular to $r$, torque is simply $\tau = rF$. If we multiply both sides of the equation above by $r$, we get torque on the left-hand side. That is, \[rF = mr^2\alpha\] or \[\tau = mr^2\alpha.\] - The moment of inertia $I$ of an object is the sum of $MR^2$ for all the point masses of which it is composed. That is, \[I = \sum mr^2.\] - The general relationship among torque, moment of inertia, and angular acceleration is \[\tau = I \alpha\] or \[\alpha = \dfrac{net \, \tau}{I}.\] Glossary - torque - the turning effectiveness of a force - rotational inertia - resistance to change of rotation. The more rotational inertia an object has, the harder it is to rotate - moment of inertia - mass times the square of perpendicular distance from the rotation axis; for a point mass, it is I=mr2 and, because any object can be built up from a collection of point masses, this relationship is the basis for all other moments of inertia	libretexts	2025-03-17T19:53:29.036554	2015-11-01T05:25:58	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.03%3A_Dynamics_of_Rotational_Motion_-_Rotational_Inertia", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "10.3: Dynamics of Rotational Motion - Rotational Inertia", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.04%3A_Rotational_Kinetic_Energy_-_Work_and_Energy_Revisited	10.4: Rotational Kinetic Energy - Work and Energy Revisited Learning Objectives By the end of this section, you will be able to: - Derive the equation for rotational work. - Calculate rotational kinetic energy. - Demonstrate the Law of Conservation of Energy. In this module, we will learn about work and energy associated with rotational motion. Figure $\PageIndex{1}$ shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable rotational kinetic energy . Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in Uniform Circular Motion and Gravitation for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in Figure $\PageIndex{2}$) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled: \[net \, W = (net \, F)\Delta s.\] To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by $r$, and gather terms: \[net \, W = (r \, net \, F)\dfrac{\Delta s}{r}.\] We recognize that $r$ $net \, F = net \, \tau$ and $\Delta s/r = \theta$, so that \[net \, W = (net \, \tau)\theta. \label{netw}\] This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. Equation \ref{netw} is valid in general, even though it was derived for a special case. To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that \[net \, W = I \alpha \theta\] Work and energy in rotational motion are completely analogous to work and energy in translational motion, first presented in Uniform Circular Motion and Gravitation . Now, we solve one of the rotational kinematics equations for $\alpha \theta$. We start with the equation \[\omega^2 = \omega_0^2 + 2\alpha \theta.\] Next, we solve for $\alpha \theta$: \[\alpha \theta = \dfrac{\omega^2 - \omega_0^2}{2}.\] Substituting this into the equation for net $W$ and gathering terms yields \[net \, W = \dfrac{1}{2}I\omega^2 - \dfrac{1}{2}I\omega_0^2.\] This equation is the work-energy theorem for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term $\left(\frac{1}{2}\right)\omega^2$ to be rotational kinetic energy $KE_{rot}$ for an object with a moment of inertia $I$ and an angular velocity $\omega$: \[KE_{rot} = \dfrac{1}{2}I \omega^2.\] The expression for rotational kinetic energy is exactly analogous to translational kinetic energy, with $I$ being analogous to $m$ and $\omega$ to $v$. Rotational kinetic energy has important effects. Flywheels, for example, can be used to store large amounts of rotational kinetic energy in a vehicle, as seen in Figure $\PageIndex{3}$. Example $\PageIndex{1}$: Calculating the Work and Energy for Spinning a Grindstone Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in Figure . In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. - How much work is done if she exerts a force of 200 N through a rotation of $1.00 \, rad(57.3^o)$? The force is kept perpendicular to the grindstone’s 0.320-m radius at the point of application, and the effects of friction are negligible. - What is the final angular velocity if the grindstone has a mass of 85.0 kg? - What is the final rotational kinetic energy? (It should equal the work.) Strategy To find the work, we can use Equation \ref{netw}. We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships. In the last part, we can calculate the rotational kinetic energy from its expression in $KE_{rot} = \frac{1}{2}I\omega^2$. Solution for (a) The net work is expressed in the equation \[net \, W = (net \, \tau)\theta,\] where net $\tau$ is the applied force multiplied by the radius $(rF)$ because there is no retarding friction, and the force is perpendicular to $r$. The angle $\theta$ is given. Substituting the given values in the equation above yields \[\begin{align} net \, W &= rF\theta \\[5pt] &= (0.320 \, m)(200 \, N)(1.00 \, rad) \\[5pt] &= 64.0 \, N \cdot m. \end{align}\] Noting that $1 \, N \cdot m = 1 \, J.$, \[net \, W = 64.0 \, J.\] Figure $\PageIndex{4}$. A large grindstone is given a spin by a person grasping its outer edge. Solution for (b) To find $\omega$ from the given information requires more than one step. We start with the kinematic relationship in the equation \[\omega^2 = \omega_0^2 + 2 \alpha \theta.\] Note that $\omega_0 = 0$ because we start from rest. Taking the square root of the resulting equation gives \[\omega = (2\alpha \theta)^{1/2}.\] Now we need to find $\alpha.$ One possibility is \[\alpha = \dfrac{net \, \tau}{I},\] where the torque is \[net \, \tau = rF = (0.320 \, m)(200 \, N) = 64.0 \, N \cdot m.\] The formula for the moment of inertia for a disk is found in [link] : \[I = \dfrac{1}{2}MR^2 = 0.5(85.0 \, kg)(0.320 \, m)^2 = 4.352 \, kg \cdot m^2.\] Substituting the values of torque and moment of inertia into the expression for $\alpha$, we obtain \[\alpha = \dfrac{64.0 \, N\cdot m}{4.352 \, kg \cdot m^2} = 14.7 \dfrac{rad}{s^2}.\] Now, substitute this value and the given value for $\theta $ into the above expression for $\omega$: \[\omega = (2\alpha \theta)^{1/2} = [2(14.7 \dfrac{rad}{s^2})(1.00 \, rad)]^{1/2} = 5.42 \dfrac{rad}{s}.\] Solution for (c) The final rotational kinetic energy is \[KE_{rot} = \dfrac{1}{2}I\omega^2.\] Both $I$ and $\omega$ were found above. Thus, \[KE_{rot} = (0.5)(4.352 \, kg \cdot m^2)(5.42 \, rad/s)^2 = 64.0 \, J. Discussion The final rotational kinetic energy equals the work done by the torque, which confirms that the work done went into rotational kinetic energy. We could, in fact, have used an expression for energy instead of a kinematic relation to solve part (b). We will do this in later examples. Helicopter pilots are quite familiar with rotational kinetic energy. They know, for example, that a point of no return will be reached if they allow their blades to slow below a critical angular velocity during flight. The blades lose lift, and it is impossible to immediately get the blades spinning fast enough to regain it. Rotational kinetic energy must be supplied to the blades to get them to rotate faster, and enough energy cannot be supplied in time to avoid a crash. Because of weight limitations, helicopter engines are too small to supply both the energy needed for lift and to replenish the rotational kinetic energy of the blades once they have slowed down. The rotational kinetic energy is put into them before takeoff and must not be allowed to drop below this crucial level. One possible way to avoid a crash is to use the gravitational potential energy of the helicopter to replenish the rotational kinetic energy of the blades by losing altitude and aligning the blades so that the helicopter is spun up in the descent. Of course, if the helicopter’s altitude is too low, then there is insufficient time for the blade to regain lift before reaching the ground. Problem-Solving Strategy for Rotational Energy - Determine that energy or work is involved in the rotation. - Determine the system of interest. A sketch usually helps. - Analyze the situation to determine the types of work and energy involved. - For closed systems, mechanical energy is conserved. That is, $KE_i + PE_f = KE_f + PE_f$. Note that $KE_i $ and $KE_f$ may each include translational and rotational contributions.For open systems, mechanical energy may not be conserved, and other forms of energy (referred to previously as $OE$), - such as heat transfer, may enter or leave the system. Determine what they are, and calculate them as necessary. - Eliminate terms wherever possible to simplify the algebra. - Check the answer to see if it is reasonable. Example $\PageIndex{2}$: Calculating Helicopter Energies A typical small rescue helicopter, similar to the one in Figure $\PageIndex{5}$, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. - Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. - Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades - To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it? Strategy Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy. Solution for (a) The rotational kinetic energy is \[KE_{rot} = \dfrac{1}{2}I \omega^2 \nonumber\] We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find $KE_{rot}$, the angular velocity $\omega$ is \[\omega = \dfrac{300 \, rev}{1.00 \, min} \cdot \dfrac{2 \pi \, rad}{1 \, rev} \cdot \dfrac{1.00 \, min}{60.0 \, s} = 31.4 \dfrac{rad}{s}.\] The moment of inertia of one blade will be that of a thin rod rotated about its end, found in [link] . The total $I$ is four times this moment of inertia, because there are four blades. Thus, \[I = 4 \dfrac{Ml^2}{3} = 4 \times \dfrac{(50.0 \, kg)(4.00 \, kg)^2}{3} = 1067 \, kg \cdot m^2.\] Entering $\omega$ and $I$ into the expression for rotational kinetic energy gives \[KE_{rot} = 0.5(1067 \, kg \cdot m^2)(31.4 \, rad/s)^2\] \[= 5.26 \times 10^5 \, J\] Solution for (b) Translational kinetic energy was defined in U niform Circular Motion and Gravitation . Entering the given values of mass and velocity, we obtain \[KE_{trans} = \dfrac{1}{2} mv^2 = (0.5)(1000 \, kg)(20.0 \, m/s)^2 = 2.00 \times 10^5 \, J.\] To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is \[ \dfrac{2.00 \times 10^5 \, J}{5.26 \times 10^5 \, J} = 0.380.\] Solution for (c) At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies: \[KE_{rot} = PE_{grav}\] or \[\dfrac{1}{2} I\omega^2 = mgh.\] We now solve for $h$ and substitute known values into the resulting equation \[h = \dfrac{\frac{1}{2}I \omega^2}{mg} = \dfrac{5.26 \times10^5 \, J}{(1000 \, kg)(9.80 \, m/s^2)} = 53.7 \, m\] Discussion The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades—something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades. Conservation of energy includes rotational motion, because rotational kinetic energy is another form of $KE$. Uniform Circular Motion and Gravitation has a detailed treatment of conservation of energy. How Thick Is the Soup? Or Why Don’t All Objects Roll Downhill at the Same Rate? One of the quality controls in a tomato soup factory consists of rolling filled cans down a ramp. If they roll too fast, the soup is too thin. Why should cans of identical size and mass roll down an incline at different rates? And why should the thickest soup roll the slowest? The easiest way to answer these questions is to consider energy. Suppose each can starts down the ramp from rest. Each can starting from rest means each starts with the same gravitational potential energy $PE_{grav}$, which is converted entirely to $KE$, provided each rolls without slipping. $KE$ however, can take the form of $KE_{trans}$ or $KE_{rot}$, and total $KE$ is the sum of the two. If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. Thus, the can goes slower than it would if it slid down. Furthermore, the thin soup does not rotate, whereas the thick soup does, because it sticks to the can. The thick soup thus puts more of the can’s original gravitational potential energy into rotation than the thin soup, and the can rolls more slowly, as seen in Figure $\PageIndex{6}$. Assuming no losses due to friction, there is only one force doing work—gravity. Therefore the total work done is the change in kinetic energy. As the cans start moving, the potential energy is changing into kinetic energy. Conservation of energy gives \[PE_i = KE_f.\] More specifically, \[PE_{grav} = KE_{trans} + KE_{rot}\] or \[mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2.\] So, the initial $mgh$ is divided between translational kinetic energy and rotational kinetic energy; and the greater $I$ is, the less energy goes into translation. If the can slides down without friction, then $\omega = 0$ and all the energy goes into translation; thus, the can goes faster. TAKE-HOME EXPERIMENT Locate several cans each containing different types of food. First, predict which can will win the race down an inclined plane and explain why. See if your prediction is correct. You could also do this experiment by collecting several empty cylindrical containers of the same size and filling them with different materials such as wet or dry sand. Example $\PageIndex{3}$: Calculating the Speed of a Cylinder Rolling Down an Incline Calculate the final speed of a solid cylinder that rolls down a 2.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm. Strategy We can solve for the final velocity using conservation of energy, but we must first express rotational quantities in terms of translational quantities to end up with as the only unknown. Solution Conservation of energy for this situation is written as described above: \[mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2.\] Before we can solve for $v$, we must get an expression for $I$ from [link] . Because $v$ and $\omega$are related (note here that the cylinder is rolling without slipping), we must also substitute the relationship $\omega = v/R$ into the expression. These substitutions yield \[mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{1}{2}mR^2\right)\left(\dfrac{v^2}{R^2}\right).\] Interestingly, the cylinder’s radius $R$ and mass $m$ cancel, yielding \[gh = \dfrac{1}{2}v^2 + \dfrac{1}{4}v^2 = \dfrac{3}{4}v^2.\] Solving algebraically, the equation for the final velocity $v$ gives \[v = \left(\dfrac{4gh}{3}\right)^{1/2}.\] Substituting known values into the resulting expression yields \[v = \left[\dfrac{4(9.80 \, m/s^2)(2.00 \, m)}{3} \right]^{1/2} = 5.11 \, m/s.\] Discussion Because $m$ and $R$ cancel, the result $v = (\frac{4}{3}gh)^{1/2}$ is valid for any solid cylinder, implying that all solid cylinders will roll down an incline at the same rate independent of their masses and sizes. (Rolling cylinders down inclines is what Galileo actually did to show that objects fall at the same rate independent of mass.) Note that if the cylinder slid without friction down the incline without rolling, then the entire gravitational potential energy would go into translational kinetic energy. Thus, $\frac{1}{2}mv^2 = mgh$ and $v = (2gh)^{1/2}, which is 22% greater than \((4gh/3)^{1/2}$. That is, the cylinder would go faster at the bottom. Exercise $\PageIndex{1}$: Analogy of Rotational and Translational Kinetic Energy Is rotational kinetic energy completely analogous to translational kinetic energy? What, if any, are their differences? Give an example of each type of kinetic energy. Yes, rotational and translational kinetic energy are exact analogs. They both are the energy of motion involved with the coordinated (non-random) movement of mass relative to some reference frame. The only difference between rotational and translational kinetic energy is that translational is straight line motion while rotational is not. An example of both kinetic and translational kinetic energy is found in a bike tire while being ridden down a bike path. The rotational motion of the tire means it has rotational kinetic energy while the movement of the bike along the path means the tire also has translational kinetic energy. If you were to lift the front wheel of the bike and spin it while the bike is stationary, then the wheel would have only rotational kinetic energy relative to the Earth. Summary - The rotational kinetic energy $KE_{rot} $ for an object with a moment of inertia $I$ and an angular velocity $\omega$ is given by \[KE_{rot} = \dfrac{1}{2}I\omega^2.\] - Helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades. - Work and energy in rotational motion are completely analogous to work and energy in translational motion. - The equation for the work-energy theorem for rotational motion is, \[net \, W = \dfrac{1}{2}I \omega^2 - \dfrac{1}{2}I \omega_0^2.\] Glossary - work-energy theorem - if one or more external forces act upon a rigid object, causing its kinetic energy to change from $KE_1$ to $KE_2$, then the work $W$ done by the net force is equal to the change in kinetic energy - rotational kinetic energy - the kinetic energy due to the rotation of an object. This is part of its total kinetic energy	libretexts	2025-03-17T19:53:29.132271	2015-11-01T05:26:19	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.04%3A_Rotational_Kinetic_Energy_-_Work_and_Energy_Revisited", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "10.4: Rotational Kinetic Energy - Work and Energy Revisited", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.05%3A_Angular_Momentum_and_Its_Conservation	10.5: Angular Momentum and Its Conservation Learning Objectives By the end of this section, you will be able to: - Understand the analogy between angular momentum and linear momentum. - Observe the relationship between torque and angular momentum. - Apply the law of conservation of angular momentum. Why does Earth keep on spinning? What started it spinning to begin with? And how does an ice skater manage to spin faster and faster simply by pulling her arms in? Why does she not have to exert a torque to spin faster? Questions like these have answers based in angular momentum, the rotational analog to linear momentum. By now the pattern is clear—every rotational phenomenon has a direct translational analog. It seems quite reasonable, then, to define angular momentum $L$ as \[L = I\omega\] This equation is an analog to the definition of linear momentum as \[p = mv\] Units for linear momentum are $kg \cdot m/s$, while units for angular momentum are $kg \cdot m/s^2$. As we would expect, an object that has a large moment of inertia $I$, such as Earth, has a very large angular momentum. An object that has a large angular velocity $\omega$, such as a centrifuge, also has a rather large angular momentum. Angular momentum is completely analogous to linear momentum, first presented in Uniform Circular Motion and Gravitation . It has the same implications in terms of carrying rotation forward, and it is conserved when the net external torque is zero. Angular momentum, like linear momentum, is also a property of the atoms and subatomic particles. Example $\PageIndex{1}$: Calculating Angular Momentum of the Earth What is the angular momentum of the earth? Strategy No information is given in the statement of the problem; so we must look up pertinent data before we can calculate $L = I\omega$ First, the formula for the moment of inertia of a sphere is \[I = \dfrac{2MR^2}{5}\] so that \[L = I\omega = \dfrac{2MR^2\omega}{5}.\] Earth's mass $M$ is $5.979 \times 10^{24} \, kg$ and its radius $R$ is $6.376 \times 10^6 \, m$. The Earth’s angular velocity $\omega$ is, of course, exactly one revolution per day, but we must covert $\omega$ to radians per second to do the calculation in SI units. Solution Substituting known information into the expression for $L$ and converting $\omega$ to radians per second gives \[L = 0.4(5.979 \times 10^{24} \, kg)(6.376 \times 10^6 \, m)\left(\dfrac{1 \, rev}{d}\right)\] \[= 9.72 \times 10^{37} \, kg \cdot m^2 \cdot rev/d\] Substituting $2\pi$ for 1 rev and $8.64 \times 10^4 \, s$ for 1 day gives \[L = (9.72 \times 10^{37} \, kg \cdot m^2)\left(\dfrac{2\pi \, rad/rev}{8.64 \times 10^4 \, s/d}\right) (1 \, rev/d)\] \[= 7.07 \times 10^{33} \, kg \cdot m^2/s.\] Discussion This number is large, demonstrating that Earth, as expected, has a tremendous angular momentum. The answer is approximate, because we have assumed a constant density for Earth in order to estimate its moment of inertia. When you push a merry-go-round, spin a bike wheel, or open a door, you exert a torque. If the torque you exert is greater than opposing torques, then the rotation accelerates, and angular momentum increases. The greater the net torque, the more rapid the increase in $L$. The relationship between torque and angular momentum is \[net \, \tau = \dfrac{\Delta L}{\Delta t}.\] This expression is exactly analogous to the relationship between force and linear momentum, $F = \Delta p/\Delta t$. The equation $net \, \tau = \Delta L/ \Delta t$ is very fundamental and broadly applicable. It is, in fact, the rotational form of Newton’s second law. Example $\PageIndex{1}$: Calculating the Torque Putting Angular Momentum Into a Lazy Susan Figure $\PageIndex{1}$ shows a Lazy Susan food tray being rotated by a person in quest of sustenance. Suppose the person exerts a 2.50 N force perpendicular to the lazy Susan’s 0.260-m radius for 0.150 s. - What is the final angular momentum of the lazy Susan if it starts from rest, assuming friction is negligible? - What is the final angular velocity of the lazy Susan, given that its mass is 4.00 kg and assuming its moment of inertia is that of a disk? Figure $\PageIndex{1}$: A partygoer exerts a torque on a lazy Susan to make it rotate. The equation $net \, \tau = \dfrac{\Delta L}{\Delta t}$ gives the relationship between torque and the angular momentum produced. Strategy We can find the angular momentum by solving $net \, \tau = \dfrac{\Delta L}{\Delta t}$ for $\Delta L$, and using the given information to calculate the torque. The final angular momentum equals the change in angular momentum, because the lazy Susan starts from rest. That is, $\Delta L = L$ To find the final velocity, we must calculate $\omega$ from the definition of $L$ in $L = I\omega$. Solution for (a) Solving $net \, \tau = \dfrac{\Delta L}{\Delta t}$ for $\Delta L$ gives \[\Delta L = (net \, \tau)\Delta t.\] Because the force is perpendicular to $r$, we see that $net \, \tau = rF$ so that \[L = rF\Delta t = (0.260 \, m)(2.50 \, N)(0.150 \, s)\] \[= 9.75 \times 10^{-2} \, kg \cdot m^2 /s.\] Solution for (b) The final angular velocity can be calculated from the definition of angular momentum, \[L = I\omega.\] Solving for $\omega$ and substituting the formula for the moment of inertia of a disk into the resulting equation gives \[\omega = \dfrac{L}{I} = \dfrac{L}{\dfrac{1}{2}MR^2}.\] And substituting known values into the preceding equation yields \[\omega = \dfrac{9.75 \times 10^{-2} \, kg \cdot m^2 /s}{(0.500)(4.00 \, kg)(0.260 \, m)} = 0.721 \, rad/s.\] Discussion Note that the imparted angular momentum does not depend on any property of the object but only on torque and time. The final angular velocity is equivalent to one revolution in 8.71 s (determination of the time period is left as an exercise for the reader), which is about right for a lazy Susan. Example $\PageIndex{3}$: Calculating the Torque in a Kick The person whose leg is shown in $\PageIndex{1}$ kicks his leg by exerting a 2000-N force with his upper leg muscle. The effective perpendicular lever arm is 2.20 cm. Given the moment of inertia of the lower leg is $1.25 \, kg \cdot m^2$. - find the angular acceleration of the leg. - Neglecting the gravitational force, what is the rotational kinetic energy of the leg after it has rotated through $57.3^0$ (1.00 rad)? Strategy The angular acceleration can be found using the rotational analog to Newton’s second law, or $\alpha = net \, \tau/I$. The moment of inertia $I$ is given and the torque can be found easily from the given force and perpendicular lever arm. Once the angular acceleration $\alpha$ is known, the final angular velocity and rotational kinetic energy can be calculated. Solution to (a) From the rotational analog to Newton’s second law, the angular acceleration $\alpha$ is \[\alpha = \dfrac{net \, \tau}{I}.\] Because the force and the perpendicular lever arm are given and the leg is vertical so that its weight does not create a torque, the net torque is thus \[net \, \tau = r_{\perp}F\] \[= (0.0220 \, m)(2000 \, N)\] \[= 44.0 \, N \cdot m\] Substituting this value for the torque and the given value for the moment of inertia into the expression for $\alpha$ gives \[\alpha = \dfrac{44.0 \, N\cdot m}{1.25 \, kg \cdot m^2} = 35.2 \, rad/s^2\] Solution to (b) The final angular velocity can be calculated from the kinematic expression \[\omega^2 = \omega_0^2 + 2\alpha \theta\] or \[\omega^2 = 2\alpha \theta\] because the initial angular velocity is zero. The kinetic energy of rotation is \[KE_{rot} = \dfrac{1}{2}I \omega^2\] so it is most convenient to use the value of $\omega^2$ just found and the given value for the moment of inertia. The kinetic energy is then \[KE_{rot}= 0.5(1.25 \, kg \cdot m^2)(70.4 \, rad^2/s^2)\] \[= 44 \, J\] Discussion These values are reasonable for a person kicking his leg starting from the position shown. The weight of the leg can be neglected in part (a) because it exerts no torque when the center of gravity of the lower leg is directly beneath the pivot in the knee. In part (b), the force exerted by the upper leg is so large that its torque is much greater than that created by the weight of the lower leg as it rotates. The rotational kinetic energy given to the lower leg is enough that it could give a ball a significant velocity by transferring some of this energy in a kick. Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero. Conservation of Angular Momentum We can now understand why Earth keeps on spinning. As we saw in the previous example, $\Delta L = (net \, \tau)\Delta t$. This equation means that, to change angular momentum, a torque must act over some period of time. Because Earth has a large angular momentum, a large torque acting over a long time is needed to change its rate of spin. So what external torques are there? Tidal friction exerts torque that is slowing Earth’s rotation, but tens of millions of years must pass before the change is very significant. Recent research indicates the length of the day was 18 h some 900 million years ago. Only the tides exert significant retarding torques on Earth, and so it will continue to spin, although ever more slowly, for many billions of years. What we have here is, in fact, another conservation law. If the net torque is zero , then angular momentum is constant or conserved . We can see this rigorously by considering $net \, \tau = \dfrac{\Delta L}{\Delta t}$ for the situation in which the net torque is zero. In that case, \[net \, \tau = 0\] implying that \[\dfrac{\Delta L}{\Delta t} = 0.\] If the change in angular momentum $\Delta L$ is zero, then the angular momentum is constant; thus, \[L = constant \, (net \, \tau = 0)\] or \[L = L'(net \, \tau = 0).\] These expressions are the law of conservation of angular momentum . Conservation laws are as scarce as they are important. An example of conservation of angular momentum is seen in Figure $\PageIndex{3}$, in which an ice skater is executing a spin. The net torque on her is very close to zero, because there is relatively little friction between her skates and the ice and because the friction is exerted very close to the pivot point. (Both $F$ and $\tau$ are small, and so $\tau$ is negligibly small.) Consequently, she can spin for quite some time. She can do something else, too. She can increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that \[L = L'.\] Expressing this equation in terms of the moment of inertia, \[I\omega = I'\omega'\] where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because $I'$ is smaller, the angular velocity $\omega'$ must increase to keep the angular momentum constant. The change can be dramatic, as the following example shows. Example $\PageIndex{4}$: Calculating the Angular Momentum of a Spinning Skater Suppose an ice skater, such as the one in Figure $\PageIndex{3}$, is spinning at 0.800 rev/ s with her arms extended. She has a moment of inertia of $2.34 \, kg \cdot m^2$ with her arms extended and of $0.363 \, kg \cdot m^2$ with her arms close to her body. (These moments of inertia are based on reasonable assumptions about a 60.0-kg skater.) (a) What is her angular velocity in revolutions per second after she pulls in her arms? (b) What is her rotational kinetic energy before and after she does this? Strategy In the first part of the problem, we are looking for the skater’s angular velocity $\omega'$ after she has pulled in her arms. To find this quantity, we use the conservation of angular momentum and note that the moments of inertia and initial angular velocity are given. To find the initial and final kinetic energies, we use the definition of rotational kinetic energy given by \[KE_{rot} = \dfrac{1}{2} I\omega^2\] Solution for (a) Because torque is negligible (as discussed above), the conservation of angular momentum given in $I\omega = I'\omega'$ is applicable. Thus, \[L = L'\] or \[I\omega^2 = I'\omega'\] Solving for $\omega'$ and substituting known values into the resulting equation gives \[\omega' = \dfrac{I}{I'}\omega = \left(\dfrac{2.34 \, kg \cdot m^2}{0.363 \, kg \cdot m^2} \right)(0.800 \, rev/s)\] \[= 5.16 rev/s.\] Solution for (b) Rotational kinetic energy is given by \[KE_{rot} = \dfrac{1}{2}I\omega^2\] The initial value is found by substituting known values into the equation and converting the angular velocity to rad/s: \[KE_{rot} = (0.5)(2.34 \, kg \cdot m^2)((0.800 \, rev/s)(2\pi \, rad/rev))^2\] \[= 29.6 \, J.\] The final rotational kinetic energy is \[KE_{rot}' = \dfrac{1}{2}I'\omega^{'2}\] Substituting known values into this equation gives \[KE_{rot}' = (0.5)(0.363 \, kg \cdot m^2)[(5.16 \, rev/s)((2\pi \, rad/rev)]^2\] \[= 191 \, J.\] Discussion In both parts, there is an impressive increase. First, the final angular velocity is large, although most world-class skaters can achieve spin rates about this great. Second, the final kinetic energy is much greater than the initial kinetic energy. The increase in rotational kinetic energy comes from work done by the skater in pulling in her arms. This work is internal work that depletes some of the skater’s food energy. There are several other examples of objects that increase their rate of spin because something reduced their moment of inertia. Tornadoes are one example. Storm systems that create tornadoes are slowly rotating. When the radius of rotation narrows, even in a local region, angular velocity increases, sometimes to the furious level of a tornado. Earth is another example. Our planet was born from a huge cloud of gas and dust, the rotation of which came from turbulence in an even larger cloud. Gravitational forces caused the cloud to contract, and the rotation rate increased as a result (Figure $\PageIndex{3}$). Exercise $\PageIndex{1}$: Check Your Undestanding Is angular momentum completely analogous to linear momentum? What, if any, are their differences? Solution Yes, angular and linear momenta are completely analogous. While they are exact analogs they have different units and are not directly inter-convertible like forms of energy are. Summary - Every rotational phenomenon has a direct translational analog , likewise angular momentum $L$ can be defined as $L = I\omega$. - This equation is an analog to the definition of linear momentum as $p = mv$. The relationship between torque and angular momentum is $net \, \tau = \Delta L/\Delta t$. - Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero. Glossary - angular momentum - the product of moment of inertia and angular velocity - law of conservation of angular momentum - angular momentum is conserved, i.e., the initial angular momentum is equal to the final angular momentum when no external torque is applied to the system	libretexts	2025-03-17T19:53:29.225669	2015-11-01T05:26:39	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.05%3A_Angular_Momentum_and_Its_Conservation", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "10.5: Angular Momentum and Its Conservation", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.06%3A_Collisions_of_Extended_Bodies_in_Two_Dimensions	10.6: Collisions of Extended Bodies in Two Dimensions Learning Objectives By the end of this section, you will be able to: - Observe collisions of extended bodies in two dimensions. - Examine collision at the point of percussion. Bowling pins are sent flying and spinning when hit by a bowling ball—angular momentum as well as linear momentum and energy have been imparted to the pins. (See Figure $\PageIndex{1}$). Many collisions involve angular momentum. Cars, for example, may spin and collide on ice or a wet surface. Baseball pitchers throw curves by putting spin on the baseball. A tennis player can put a lot of top spin on the tennis ball which causes it to dive down onto the court once it crosses the net. We now take a brief look at what happens when objects that can rotate collide. Consider the relatively simple collision shown in Figure $\PageIndex{2}$, in which a disk strikes and adheres to an initially motionless stick nailed at one end to a frictionless surface. After the collision, the two rotate about the nail. There is an unbalanced external force on the system at the nail. This force exerts no torque because its lever arm is zero. Angular momentum is therefore conserved in the collision. Kinetic energy is not conserved, because the collision is inelastic. It is possible that momentum is not conserved either because the force at the nail may have a component in the direction of the disk’s initial velocity. Let us examine a case of rotation in a collision in Example $\PageIndex{1}$ . Example $\PageIndex{1}$: Rotation in a Collision Suppose the disk in Figure $\PageIndex{2}$ has a mass of 50.0 g and an initial velocity of 30.0 m/s when it strikes the stick that is 1.20 m long and 2.00 kg. - What is the angular velocity of the two after the collision? - What is the kinetic energy before and after the collision? - What is the total linear momentum before and after the collision? Strategy for (a) We can answer the first question using conservation of angular momentum as noted. Because angular momentum is $I\omega$, we can solve for angular velocity. Solution for (a) Conservation of angular momentum states \[L = L',\] where primed quantities stand for conditions after the collision and both momenta are calculated relative to the pivot point. The initial angular momentum of the system of stick-disk is that of the disk just before it strikes the stick. That is, \[L = I\omega,\] where $I$ is the moment of inertia of the disk and $\omega$ is its angular velocity around the pivot point. Now, $I = mr^2$ (taking the disk to be approximately a point mass) and $\omega = v/r$, so that \[L = mr^2\dfrac{v}{r} = mvr.\] After collision, \[L' = I'\omega'.\] It is $\omega'$ that we wish to find. Conservation of angular momentum gives \[I'\omega' = mvr.\] Rearranging the equation yields \[\omega' = \dfrac{mvr}{I'},\] where $I'$ is the moment of inertia of the stick and disk stuck together, which is the sum of their individual moments of inertia about the nail. [link] gives the formula for a rod rotating around one end to be $I = Mr^2/3$. Thus, \[I' = mr^2 + \dfrac{Mr^2}{3} = \left(m + \dfrac{M}{3}\right)r^2\] Entering known values in this equation yields, \[I' = (0.0500 \, kg + 0.667 \, kg)(1.20 \, m)^2 = 1.032 \, kg \cdot m^2\] The value of $I'$ is now entered into the expression for $\omega'$, which yields \[\omega' = \dfrac{mvr}{I'} = \dfrac{(0.0500 \, kg)(30.0 \, m/s)(1.20 \, m)}{1.032 \, kg \cdot m^2}\] \[= 1.744 \, rad/s \approx 1.74 \, rad/s.\] Strategy for (b) The kinetic energy before the collision is the incoming disk’s translational kinetic energy, and after the collision, it is the rotational kinetic energy of the two stuck together. Solution for (b) First, we calculate the translational kinetic energy by entering given values for the mass and speed of the incoming disk. \[KE = \dfrac{1}{2}mv^2 = (0.500)(0.0500 \, kg)(30.0 \, m/s)^2 = 22.5 \, J.\] After the collision, the rotational kinetic energy can be found because we now know the final angular velocity and the final moment of inertia. Thus, entering the values into the rotational kinetic energy equation gives \[KE' = \dfrac{1}{2} I'\omega^{'2} = (0.5)(1.032 \, kg \cdot m^2)(1.744 \, rad/s)^2\] \[= 1.57 \, J.\] Strategy for (c) The linear momentum before the collision is that of the disk. After the collision, it is the sum of the disk’s momentum and that of the center of mass of the stick. Solution of (c) Before the collision, then, linear momentum is \[p = mv = (0.0500 \, kg)(30.0 \, m/s) = 1.50 \, kg \cdot m/s.\] After the collision, the disk and the stick’s center of mass move in the same direction. The total linear momentum is that of the disk moving at a new velocity $v' = r\omega'$ plus that of the stick’s center of mass, which moves at half this speed because $v_{CM} = \left(\frac{r}{2}\right)\omega' = \frac{v'}{2}$. Thus, \[p' = mv' + Mv_{CM} = mv' + \dfrac{Mv'}{2}.\] Gathering similar terms in the equation yields, \[p' = \left(m + \dfrac{M}{2}\right)v'\] so that \[p' = \left(m + \dfrac{M}{2}\right)r\omega'\] Substituting known values into the equation, \[p' = (1.050 \, kg)(1.20 \, m)(1.744 \, rad/s) = 2.20 \, kg \cdot m/s.\] Discussion First note that the kinetic energy is less after the collision, as predicted, because the collision is inelastic. More surprising is that the momentum after the collision is actually greater than before the collision. This result can be understood if you consider how the nail affects the stick and vice versa. Apparently, the stick pushes backward on the nail when first struck by the disk. The nail’s reaction (consistent with Newton’s third law) is to push forward on the stick, imparting momentum to it in the same direction in which the disk was initially moving, thereby increasing the momentum of the system. The above example has other implications. For example, what would happen if the disk hit very close to the nail? Obviously, a force would be exerted on the nail in the forward direction. So, when the stick is struck at the end farthest from the nail, a backward force is exerted on the nail, and when it is hit at the end nearest the nail, a forward force is exerted on the nail. Thus, striking it at a certain point in between produces no force on the nail. This intermediate point is known as the percussion point . An analogous situation occurs in tennis as seen in Figure $\PageIndex{3}$. If you hit a ball with the end of your racquet, the handle is pulled away from your hand. If you hit a ball much farther down, for example, on the shaft of the racquet, the handle is pushed into your palm. And if you hit the ball at the racquet’s percussion point (what some people call the “sweet spot”), then little or no force is exerted on your hand, and there is less vibration, reducing chances of a tennis elbow. The same effect occurs for a baseball bat. Exercise $\PageIndex{1}$ Is rotational kinetic energy a vector? Justify your answer. Solution No, energy is always scalar whether motion is involved or not. No form of energy has a direction in space and you can see that rotational kinetic energy does not depend on the direction of motion just as linear kinetic energy is independent of the direction of motion. Summary - Angular momentum $L$ is analogous to linear momentum and is given by $L = I\omega$. - Angular momentum is changed by torque, following the relationship $net \, \tau = \frac{\Delta L}{\Delta t}$. - Angular momentum is conserved if the net torque is zero $L = constant \, (net \, \tau = 0)$ or $L = L' \, (net \, \tau = 0)$. This equation is known as the law of conservation of angular momentum, which may be conserved in collisions.	libretexts	2025-03-17T19:53:29.298256	2015-11-01T05:26:56	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.06%3A_Collisions_of_Extended_Bodies_in_Two_Dimensions", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "10.6: Collisions of Extended Bodies in Two Dimensions", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.07%3A_Gyroscopic_Effects-_Vector_Aspects_of_Angular_Momentum	10.7: Gyroscopic Effects- Vector Aspects of Angular Momentum Learning Objectives By the end of this section, you will be able to: - Describe the right-hand rule to find the direction of angular velocity, momentum, and torque. - Explain the gyroscopic effect. - Study how Earth acts like a gigantic gyroscope. Angular momentum is a vector and, therefore, has direction as well as magnitude . Torque affects both the direction and the magnitude of angular momentum. What is the direction of the angular momentum of a rotating object like the disk in Figure ? The figure shows the right-hand rule used to find the direction of both angular momentum and angular velocity. Both $L$ and $\omega$ are vectors—each has direction and magnitude. Both can be represented by arrows. The right-hand rule defines both to be perpendicular to the plane of rotation in the direction shown. Because angular momentum is related to angular velocity by $L = I\omega$, the direction of $L$ is the same as the direction of $\omega$. Notice in the figure that both point along the axis of rotation. Now, recall that torque changes angular momentum as expressed by \[net \, \tau = \dfrac{\Delta L}{\Delta t}.\] This equation means that the direction of $\Delta L$ is the same as the direction of the torque $\tau$ that creates it. This result is illustrated in Figure , which shows the direction of torque and the angular momentum it creates. Let us now consider a bicycle wheel with a couple of handles attached to it, as shown in Figure . (This device is popular in demonstrations among physicists, because it does unexpected things.) With the wheel rotating as shown, its angular momentum is to the woman's left. Suppose the person holding the wheel tries to rotate it as in the figure. Her natural expectation is that the wheel will rotate in the direction she pushes it—but what happens is quite different. The forces exerted create a torque that is horizontal toward the person, as shown in Figure (a). This torque creates a change in angular momentum $L$ in the same direction, perpendicular to the original angular momentum $L$, thus changing the direction of $L$ but not the magnitude of $L$. Figure shows how $\Delta L$ and $L$ add, giving a new angular momentum with direction that is inclined more toward the person than before. The axis of the wheel has thus moved perpendicular to the forces exerted on it , instead of in the expected direction. This same logic explains the behavior of gyroscopes. Figure shows the two forces acting on a spinning gyroscope. The torque produced is perpendicular to the angular momentum, thus the direction of the torque is changed, but not its magnitude. The gyroscope precesses around a vertical axis, since the torque is always horizontal and perpendicular to $L$. If the gyroscope is not spinning, it acquires angular momentum in the direction of the torque ($ L = \Delta L$), and it rotates around a horizontal axis, falling over just as we would expect. Earth itself acts like a gigantic gyroscope. Its angular momentum is along its axis and points at Polaris, the North Star. But Earth is slowly precessing (once in about 26,000 years) due to the torque of the Sun and the Moon on its nonspherical shape. Exercise $\PageIndex{1}$ Rotational kinetic energy is associated with angular momentum? Does that mean that rotational kinetic energy is a vector? - Answer - No, energy is always a scalar whether motion is involved or not. No form of energy has a direction in space and you can see that rotational kinetic energy does not depend on the direction of motion just as linear kinetic energy is independent of the direction of motion. Section Summary - Torque is perpendicular to the plane formed by $\tau$ and $F$ and is the direction your right thumb would point if you curled the fingers of your right hand in the direction of $F$. The direction of the torque is thus the same as that of the angular momentum it produces. - The gyroscope precesses around a vertical axis, since the torque is always horizontal and perpendicular to $L$. If the gyroscope is not spinning, it acquires angular momentum in the direction of the torque ($L = \Delta L$), and it rotates about a horizontal axis, falling over just as we would expect. - Earth itself acts like a gigantic gyroscope. Its angular momentum is along its axis and points at Polaris, the North Star. Glossary - right-hand rule - direction of angular velocity ω and angular momentum L in which the thumb of your right hand points when you curl your fingers in the direction of the disk’s rotation	libretexts	2025-03-17T19:53:29.365012	2015-11-01T05:27:13	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.07%3A_Gyroscopic_Effects-_Vector_Aspects_of_Angular_Momentum", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "10.7: Gyroscopic Effects- Vector Aspects of Angular Momentum", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.E%3A_Rotational_Motion_and_Angular_Momentum_(Exercises)	10.E: Rotational Motion and Angular Momentum (Exercises) - - Last updated - Save as PDF Conceptual Questions 10.1: Angular Acceleration 1. Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse. 2. Explain why centripetal acceleration changes the direction of velocity in circular motion but not its magnitude. 3. In circular motion, a tangential acceleration can change the magnitude of the velocity but not its direction. Explain your answer. 4. Suppose a piece of food is on the edge of a rotating microwave oven plate. Does it experience nonzero tangential acceleration, centripetal acceleration, or both when: (a) The plate starts to spin? (b) The plate rotates at constant angular velocity? (c) The plate slows to a halt? 10.3: Dynamics of Rotational Motion: Rotational Inertia 5. The moment of inertia of a long rod spun around an axis through one end perpendicular to its length is $ML^2/3$. Why is this moment of inertia greater than it would be if you spun a point mass $M$ at the location of the center of mass of the rod (at $L/2$)? (That would be $ML^2/4$.) 6. Why is the moment of inertia of a hoop that has a mass $M$ and a radius $R$ greater than the moment of inertia of a disk that has the same mass and radius? Why is the moment of inertia of a spherical shell that has a mass $M$ and a radius $R$ greater than that of a solid sphere that has the same mass and radius? 7. Give an example in which a small force exerts a large torque. Give another example in which a large force exerts a small torque. 8. While reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame? The image shows a side view of a racing bicycle. Can you see evidence in the design of the wheels on this racing bicycle that their moment of inertia has been purposely reduced? (credit: Jesús Rodriguez) 9. A ball slides up a frictionless ramp. It is then rolled without slipping and with the same initial velocity up another frictionless ramp (with the same slope angle). In which case does it reach a greater height, and why? 10.4: Rotational Kinetic Energy: Work and Energy Revisited 10. Describe the energy transformations involved when a yo-yo is thrown downward and then climbs back up its string to be caught in the user’s hand. 11. What energy transformations are involved when a dragster engine is revved, its clutch let out rapidly, its tires spun, and it starts to accelerate forward? Describe the source and transformation of energy at each step. 12. The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. Where did this energy come from? An immense cloud of rotating gas and dust contracted under the influence of gravity to form the Earth and in the process rotational kinetic energy increased. (credit: NASA) 10.5: Angular Momentum and Its Conservation 13. When you start the engine of your car with the transmission in neutral, you notice that the car rocks in the opposite sense of the engine’s rotation. Explain in terms of conservation of angular momentum. Is the angular momentum of the car conserved for long (for more than a few seconds)? 14. Suppose a child walks from the outer edge of a rotating merry-go round to the inside. Does the angular velocity of the merry-go-round increase, decrease, or remain the same? Explain your answer. A child may jump off a merry-go-round in a variety of directions. 15. Suppose a child gets off a rotating merry-go-round. Does the angular velocity of the merry-go-round increase, decrease, or remain the same if: (a) He jumps off radially? (b) He jumps backward to land motionless? (c) He jumps straight up and hangs onto an overhead tree branch? (d) He jumps off forward, tangential to the edge? Explain your answers. (Refer to Figure). 16. Helicopters have a small propeller on their tail to keep them from rotating in the opposite direction of their main lifting blades. Explain in terms of Newton’s third law why the helicopter body rotates in the opposite direction to the blades. 17. Whenever a helicopter has two sets of lifting blades, they rotate in opposite directions (and there will be no tail propeller). Explain why it is best to have the blades rotate in opposite directions. 18. Describe how work is done by a skater pulling in her arms during a spin. In particular, identify the force she exerts on each arm to pull it in and the distance each moves, noting that a component of the force is in the direction moved. Why is angular momentum not increased by this action? 19. When there is a global heating trend on Earth, the atmosphere expands and the length of the day increases very slightly. Explain why the length of a day increases. 20. Nearly all conventional piston engines have flywheels on them to smooth out engine vibrations caused by the thrust of individual piston firings. Why does the flywheel have this effect? 21. Jet turbines spin rapidly. They are designed to fly apart if something makes them seize suddenly, rather than transfer angular momentum to the plane’s wing, possibly tearing it off. Explain how flying apart conserves angular momentum without transferring it to the wing. 22. An astronaut tightens a bolt on a satellite in orbit. He rotates in a direction opposite to that of the bolt, and the satellite rotates in the same direction as the bolt. Explain why. If a handhold is available on the satellite, can this counter-rotation be prevented? Explain your answer. 23. Competitive divers pull their limbs in and curl up their bodies when they do flips. Just before entering the water, they fully extend their limbs to enter straight down. Explain the effect of both actions on their angular velocities. Also explain the effect on their angular momenta. The diver spins rapidly when curled up and slows when she extends her limbs before entering the water. 24. Draw a free body diagram to show how a diver gains angular momentum when leaving the diving board. 25. In terms of angular momentum, what is the advantage of giving a football or a rifle bullet a spin when throwing or releasing it? The image shows a view down the barrel of a cannon, emphasizing its rifling. Rifling in the barrel of a canon causes the projectile to spin just as is the case for rifles (hence the name for the grooves in the barrel). (credit: Elsie esq., Flickr) 10.6: Collisions of Extended Bodies in Two Dimensions 26. Describe two different collisions—one in which angular momentum is conserved, and the other in which it is not. Which condition determines whether or not angular momentum is conserved in a collision? 27. Suppose an ice hockey puck strikes a hockey stick that lies flat on the ice and is free to move in any direction. Which quantities are likely to be conserved: angular momentum, linear momentum, or kinetic energy (assuming the puck and stick are very resilient)? 28. While driving his motorcycle at highway speed, a physics student notices that pulling back lightly on the right handlebar tips the cycle to the left and produces a left turn. Explain why this happens. 10.7: Gyroscopic Effects: Vector Aspects of Angular Momentum 29. While driving his motorcycle at highway speed, a physics student notices that pulling back lightly on the right handlebar tips the cycle to the left and produces a left turn. Explain why this happens. 30. Gyroscopes used in guidance systems to indicate directions in space must have an angular momentum that does not change in direction. Yet they are often subjected to large forces and accelerations. How can the direction of their angular momentum be constant when they are accelerated? Problem & Exercises 10.1: Angular Acceleration 31. At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second? Solution $ω=0.737 rev/s$ 32. Integrated Concepts An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) What is its angular acceleration in $rad/s^2$? (b) What is the tangential acceleration of a point 9.50 cm from the axis of rotation? (c) What is the radial acceleration in $m/s^2$ and multiples of $g$ of this point at full rpm? 33. Integrated Concepts You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (b) How many turns will the stone make before coming to rest? Solution (a) $−0.26 rad/s^2$ (b) $27rev$ 34. Unreasonable Results You are told that a basketball player spins the ball with an angular acceleration of $100 rad/s^2$. (a) What is the ball’s final angular velocity if the ball starts from rest and the acceleration lasts 2.00 s? (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent? 10.2: Kinematics of Rotational Motion 35. With the aid of a string, a gyroscope is accelerated from rest to 32 rad/s in 0.40 s. (a) What is its angular acceleration in rad/s2? (b) How many revolutions does it go through in the process? Solution (a) $80rad/s^2$ (b) 1.0 rev 36. Suppose a piece of dust finds itself on a CD. If the spin rate of the CD is 500 rpm, and the piece of dust is 4.3 cm from the center, what is the total distance traveled by the dust in 3 minutes? (Ignore accelerations due to getting the CD rotating.) 37. A gyroscope slows from an initial rate of 32.0 rad/s at a rate of $0.700 rad/s^2$. (a) How long does it take to come to rest? (b) How many revolutions does it make before stopping? Solution (a) 45.7 s (b) 116 rev 38. During a very quick stop, a car decelerates at $7.00 m/s^2$. (a) What is the angular acceleration of its 0.280-m-radius tires, assuming they do not slip on the pavement? (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is $95.0 rad/s$? (c) How long does the car take to stop completely? (d) What distance does the car travel in this time? (e) What was the car’s initial velocity? (f) Do the values obtained seem reasonable, considering that this stop happens very quickly? Yo-yos are amusing toys that display significant physics and are engineered to enhance performance based on physical laws. (credit: Beyond Neon, Flickr) 39. Everyday application: Suppose a yo-yo has a center shaft that has a 0.250 cm radius and that its string is being pulled. (a) If the string is stationary and the yo-yo accelerates away from it at a rate of $1.50 m/s^2$, what is the angular acceleration of the yo-yo? (b) What is the angular velocity after 0.750 s if it starts from rest? (c) The outside radius of the yo-yo is 3.50 cm. What is the tangential acceleration of a point on its edge? Solution a) $600 rad/s^2$ b) 450 rad/s c) 21.0 m/s 10.3: Dynamics of Rotational Motion: Rotational Inertia 40. This problem considers additional aspects of example Calculating the Effect of Mass Distribution on a Merry-Go-Round. (a) How long does it take the father to give the merry-go-round an angular velocity of 1.50 rad/s? (b) How many revolutions must he go through to generate this velocity? (c) If he exerts a slowing force of 300 N at a radius of 1.35 m, how long would it take him to stop them? Solution (a) 0.338 s (b) 0.0403 rev (c) 0.313 s 41. Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater with arms extended is approximately a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends. 42. The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of $2.00×10^3N$ with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of $120rad/s^2$. What is the moment of inertia of the boxer’s forearm? Solution $0.50 kg⋅m^2$ 43. A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her leg. She produces an angular acceleration of $30.00 rad/s^2$ and her lower leg has a moment of inertia of $0.750 kg⋅m^2$. What is the force exerted by the muscle if its effective perpendicular lever arm is 1.90 cm? 44. Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk). (a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis? Solution (a) $50.4 N⋅m$ (b) $17.1rad/s^2$ (c) $17.0rad/s^2$ 45. Consider the 12.0 kg motorcycle wheel shown in Figure. Assume it to be approximately an annular ring with an inner radius of 0.280 m and an outer radius of 0.330 m. The motorcycle is on its center stand, so that the wheel can spin freely. (a) If the drive chain exerts a force of 2200 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? (b) What is the tangential acceleration of a point on the outer edge of the tire? (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s? A motorcycle wheel has a moment of inertia approximately that of an annular ring. 46. Zorch, an archenemy of Superman, decides to slow Earth’s rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Superman is not immediately concerned, because he knows Zorch can only exert a force of $4.00×10^7N$ (a little greater than a Saturn V rocket’s thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Superman time to devote to other villains.) Explicitly show how you follow the steps found in Problem-Solving Strategy for Rotational Dynamics. Solution $3.96×10^{18}s$ or $1.26×10^{11}y$ 47. An automobile engine can produce 200 N ∙ m of torque. Calculate the angular acceleration produced if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car, given the following information. The car is suspended so that the wheels can turn freely. Each wheel acts like a 15.0 kg disk that has a 0.180 m radius. The walls of each tire act like a 2.00-kg annular ring that has inside radius of 0.180 m and outside radius of 0.320 m. The tread of each tire acts like a 10.0-kg hoop of radius 0.330 m. The 14.0-kg axle acts like a rod that has a 2.00-cm radius. The 30.0-kg drive shaft acts like a rod that has a 3.20-cm radius. 48. Starting with the formula for the moment of inertia of a rod rotated around an axis through one end perpendicular to its length ($I=Mℓ^2/3$) , prove that the moment of inertia of a rod rotated about an axis through its center perpendicular to its length is $I=Mℓ^2/12$. You will find the graphics in Figure useful in visualizing these rotations. Solution $I_{end}=I_{center}+m(\frac{l}{2})^2$ Thus,$I_{center}=I_{end}−\frac{1}{4}ml^2=\frac{1}{3}ml^2−\frac{1}{4}ml^2=\frac{1}{12}ml^2$ 49. Unreasonable Results A gymnast doing a forward flip lands on the mat and exerts a 500-N ∙ m torque to slow and then reverse her angular velocity. Her initial angular velocity is 10.0 rad/s, and her moment of inertia is $0.050kg⋅m^2$. (a) What time is required for her to exactly reverse her spin? (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent? Solution (a) 2.0 ms (b) The time interval is too short. (c) The moment of inertia is much too small, by one to two orders of magnitude. A torque of $500 N⋅m$ is reasonable. 50. Unreasonable Results An advertisement claims that an 800-kg car is aided by its 20.0-kg flywheel, which can accelerate the car from rest to a speed of 30.0 m/s. The flywheel is a disk with a 0.150-m radius. (a) Calculate the angular velocity the flywheel must have if 95.0% of its rotational energy is used to get the car up to speed. (b) What is unreasonable about the result? (c) Which premise is unreasonable or which premises are inconsistent? Solution (a) 17,500 rpm (b) This angular velocity is very high for a disk of this size and mass. The radial acceleration at the edge of the disk is > 50,000 gs. (c) Flywheel mass and radius should both be much greater, allowing for a lower spin rate (angular velocity). 10.4: Rotational Kinetic Energy: Work and Energy Revisited 51. This problem considers energy and work aspects of [link]—use data from that example as needed. (a) Calculate the rotational kinetic energy in the merry-go-round plus child when they have an angular velocity of 20.0 rpm. (b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. (c) Again, using energy considerations, calculate the force the father must exert to stop the merry-go-round in two revolutions Solution (a) 185 J (b) 0.0785 rev (c) $W=9.81 N$ 52. What is the final velocity of a hoop that rolls without slipping down a 5.00-m-high hill, starting from rest? 53. (a) Calculate the rotational kinetic energy of Earth on its axis. (b) What is the rotational kinetic energy of Earth in its orbit around the Sun? Solution (a) $2.57×10^{29}J$ (b) $KE_{rot}=2.65×10^{33}J$ 54. Calculate the rotational kinetic energy in the motorcycle wheel ([link]) if its angular velocity is 120 rad/s. Assume M = 12.0 kg, $R_1$ = 0.280 m, and $R_2$ = 0.330 m. 55. A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 20.0 m/s at a distance of 0.480 m from the joint and the moment of inertia of the forearm is $0.500 kg⋅m^2$, what is the rotational kinetic energy of the forearm? Solution $KE_{rot}=434 J$ 56. While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is $3.75 kg⋅m^2$ and its rotational kinetic energy is 175 J. (a) What is the angular velocity of the leg? (b) What is the velocity of tip of the punter’s shoe if it is 1.05 m from the hip joint? (c) Explain how the football can be given a velocity greater than the tip of the shoe (necessary for a decent kick distance). 57. A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total mass of 10,000 kg. (a) Calculate the angular velocity the flywheel must have to contain enough energy to take the bus from rest to a speed of 20.0 m/s, assuming 90.0% of the rotational kinetic energy can be transformed into translational energy. (b) How high a hill can the bus climb with this stored energy and still have a speed of 3.00 m/s at the top of the hill? Explicitly show how you follow the steps in the Problem-Solving Strategy for Rotational Energy. Solution (a) $128 rad/s$ (b) $19.9 m$ 58. A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. Treating the ball as a spherical shell, calculate the vertical height it reaches. (b) Repeat the calculation for the same ball if it slides up the hill without rolling. 59. While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contacting the muscles in the back of the upper leg. (a) Find the angular acceleration produced given the mass lifted is 10.0 kg at a distance of 28.0 cm from the knee joint, the moment of inertia of the lower leg is $0.900 kg⋅m^2$, the muscle force is 1500 N, and its effective perpendicular lever arm is 3.00 cm. (b) How much work is done if the leg rotates through an angle of $20.0º$ with a constant force exerted by the muscle? Solution (a) $10.4 rad/s^2$ (b) net $W=6.11 J$ 60. To develop muscle tone, a woman lifts a 2.00-kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of $60.0º$. (a) What is the angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of $0.250 kg⋅m^2$, and the net force she exerts is 750 N at an effective perpendicular lever arm of 2.00 cm? (b) How much work does she do? 61. Consider two cylinders that start down identical inclines from rest except that one is frictionless. Thus one cylinder rolls without slipping, while the other slides frictionlessly without rolling. They both travel a short distance at the bottom and then start up another incline. (a) Show that they both reach the same height on the other incline, and that this height is equal to their original height. (b) Find the ratio of the time the rolling cylinder takes to reach the height on the second incline to the time the sliding cylinder takes to reach the height on the second incline. (c) Explain why the time for the rolling motion is greater than that for the sliding motion. 62. What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a final velocity of 6.00 m/s? Express the moment of inertia as a multiple of $MR^2$, where $M$ is the mass of the object and $R$ is its radius. 63. Suppose a 200-kg motorcycle has two wheels like, the one described in Problem 10.15 and is heading toward a hill at a speed of 30.0 m/s. (a) How high can it coast up the hill, if you neglect friction? (b) How much energy is lost to friction if the motorcycle only gains an altitude of 35.0 m before coming to rest? 64. In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km/h. (a) Find the rotational kinetic energy of the pitcher’s arm given its moment of inertia is $0.720 kg⋅m^2$ and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder. (b) What force did the muscles exert to cause the arm to rotate if their effective perpendicular lever arm is 4.00 cm and the ball is 0.156 kg? Solution (a) $1.49 kJ$ (b) $2.52×10^4N$ 65. Construct Your Own Problem Consider the work done by a spinning skater pulling her arms in to increase her rate of spin. Construct a problem in which you calculate the work done with a “force multiplied by distance” calculation and compare it to the skater’s increase in kinetic energy. 10.5: Angular Momentum and Its Conservation 66. (a) Calculate the angular momentum of the Earth in its orbit around the Sun. (b) Compare this angular momentum with the angular momentum of Earth on its axis. Solution (a) $2.66×10^{40}kg⋅m^2/s$ (b) $7.07×10^{33}kg⋅m^2/s$ The angular momentum of the Earth in its orbit around the Sun is $3.77×10^6$ times larger than the angular momentum of the Earth around its axis. 67. (a) What is the angular momentum of the Moon in its orbit around Earth? (b) How does this angular momentum compare with the angular momentum of the Moon on its axis? Remember that the Moon keeps one side toward Earth at all times. (c) Discuss whether the values found in parts (a) and (b) seem consistent with the fact that tidal effects with Earth have caused the Moon to rotate with one side always facing Earth. 68. Suppose you start an antique car by exerting a force of 300 N on its crank for 0.250 s. What angular momentum is given to the engine if the handle of the crank is 0.300 m from the pivot and the force is exerted to create maximum torque the entire time? Solution $22.5 kg⋅m^2/s$ 69. A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge? The child is initially at rest. 70. Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm? Solution 25.3 rpm 71. (a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is $0.400kg⋅m^2$. (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity decreases to 1.25 rev/s. (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 15.0 s? 72. Construct Your Own Problem Consider the Earth-Moon system. Construct a problem in which you calculate the total angular momentum of the system including the spins of the Earth and the Moon on their axes and the orbital angular momentum of the Earth-Moon system in its nearly monthly rotation. Calculate what happens to the Moon’s orbital radius if the Earth’s rotation decreases due to tidal drag. Among the things to be considered are the amount by which the Earth’s rotation slows and the fact that the Moon will continue to have one side always facing the Earth. 10.6: Collisions of Extended Bodies in Two Dimensions 73. Repeat Example in which the disk strikes and adheres to the stick 0.100 m from the nail. Solution (a) $0.156 rad/s$ (b) $1.17×10^{−2}J$ (c) $0.188 kg⋅m/s$ 74. Repeat Example in which the disk originally spins clockwise at 1000 rpm and has a radius of 1.50 cm. 75. Twin skaters approach one another as shown in Figure and lock hands. (a) Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate their moments of inertia to be that of point masses at this radius. (b) Compare the initial kinetic energy and final kinetic energy. Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin. Solution (a) 3.13 rad/s (b) Initial KE = 438 J, final KE = 438 J 76. Suppose a 0.250-kg ball is thrown at 15.0 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in Figure. (a) Calculate the final linear velocity of the person, given his mass is 70.0 kg. (b) What is his angular velocity if each arm is 5.00 kg? You may treat the ball as a point mass and treat the person's arms as uniform rods (each has a length of 0.900 m) and the rest of his body as a uniform cylinder of radius 0.180 m. Neglect the effect of the ball on his center of mass so that his center of mass remains in his geometrical center. (c) Compare the initial and final total kinetic energies. The figure shows the overhead view of a person standing motionless on ice about to catch a ball. Both arms are outstretched. After catching the ball, the skater recoils and rotates. 77. Repeat Example in which the stick is free to have translational motion as well as rotational motion. Solution (a) 1.70 rad/s (b) Initial KE = 22.5 J, final KE = 2.04 J (c) $1.50 kg⋅m/s$ 10.7: Gyroscopic Effects: Vector Aspects of Angular Momentum 78. Integrated Concepts The axis of Earth makes a 23.5° angle with a direction perpendicular to the plane of Earth’s orbit. As shown in Figure, this axis precesses, making one complete rotation in 25,780 y. (a) Calculate the change in angular momentum in half this time. (b) What is the average torque producing this change in angular momentum? (c) If this torque were created by a single force (it is not) acting at the most effective point on the equator, what would its magnitude be? The Earth’s axis slowly precesses, always making an angle of 23.5° with the direction perpendicular to the plane of Earth’s orbit. The change in angular momentum for the two shown positions is quite large, although the magnitude L size 12{L} {} is unchanged. Solution (a) $5.64×10^{33}kg⋅m^2/s$ (b) $1.39×10^{22}N⋅m$ (c) $2.17×10^{15}N$ Contributors and Attributions - Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .	libretexts	2025-03-17T19:53:29.492262	2017-10-05T14:58:46	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/10%3A_Rotational_Motion_and_Angular_Momentum/10.E%3A_Rotational_Motion_and_Angular_Momentum_(Exercises)", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "10.E: Rotational Motion and Angular Momentum (Exercises)", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics	11: Fluid Statics Fluid statics is the branch of fluid mechanics that studies incompressible fluids at rest. It encompasses the study of the conditions under which fluids are at rest in stable equilibrium as opposed to fluid dynamics, the study of fluids in motion. - - 11.4: Variation of Pressure with Depth in a Fluid - Pressure is the weight of the fluid $mg$ divided by the area $A$ supporting it (the area of the bottom of the container): \[P = \dfrac{mg}{A}. \nonumber\] Pressure due to the weight of a liquid is given by \[P = h\rho g, \nonumber\] where $P$ is the pressure, $h$ is the height of the liquid, $\rho$ is the density of the liquid, and $g$ is the acceleration due to the gravity. - - 11.5: Pascal’s Principle - Pressure is force per unit area. A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. A hydraulic system is an enclosed fluid system used to exert forces. Since atoms in a fluid are free to move about in an enclosed fluid, they transmit the pressure to all parts of the fluid and to the walls of the container. Remarkably, the pressure is transmitted undiminished. This phenomenon is called Pascal’s principle. - - 11.6: Gauge Pressure, Absolute Pressure, and Pressure Measurement - Gauge pressure is the pressure relative to atmospheric pressure. Absolute pressure is the sum of gauge pressure and atmospheric pressure. Aneroid gauge measures pressure using a bellows-and-spring arrangement connected to the pointer of a calibrated scale. Open-tube manometers have U-shaped tubes and one end is always open. It is used to measure pressure. A mercury barometer is a device that measures atmospheric pressure. - - 11.7: Archimedes’ Principle - Buoyant force is the net upward force on any object in any fluid. If the buoyant force is greater than the object’s weight, the object will rise to the surface and float. If the buoyant force is less than the object’s weight, the object will sink. If the buoyant force equals the object’s weight, the object will remain suspended at that depth. The buoyant force is always present whether the object floats, sinks, or is suspended in a fluid. Archimedes’ principle states that the buoyant force on an - - 11.8: Cohesion and Adhesion in Liquids - Surface Tension and Capillary Action - Attractive forces between molecules of the same type are called cohesive forces. Attractive forces between molecules of different types are called adhesive forces. Cohesive forces between molecules cause the surface of a liquid to contract to the smallest possible surface area. This general effect is called surface tension. Capillary action is the tendency of a fluid to be raised or suppressed in a narrow tube, or capillary tube which is due to relative strengths of cohesive and adhesive forces. - - 11.9: Pressures in the Body - Measuring blood pressure is among the most common of all medical examinations. The pressures in various parts of the body can be measured and often provide valuable medical indicators. The shape of the eye is maintained by fluid pressure, called intraocular pressure. When the circulation of fluid in the eye is blocked, it can lead to a buildup in pressure, a condition called glaucoma. Some of the other pressures in the body are spinal and skull pressures, bladder pressure, pressures in the skele Thumbnail: Surface tension preventing a paper clip from submerging. (CC-SA-BY 3.0; Alvesgaspar via Wikipedia )	libretexts	2025-03-17T19:53:29.561824	2015-10-27T19:34:27	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11: Fluid Statics", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.00%3A_Prelude_to_Fluid_Statics	11.0: Prelude to Fluid Statics Much of what we value in life is fluid: a breath of fresh winter air; the hot blue flame in our gas cooker; the water we drink, swim in, and bathe in; the blood in our veins. What exactly is a fluid? Can we understand fluids with the laws already presented, or will new laws emerge from their study? The physical characteristics of static or stationary fluids and some of the laws that govern their behavior are the topics of this chapter. Fluid Dynamics and Its Biological and Medical Applications explores aspects of fluid flow. Contributors and Attributions Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .	libretexts	2025-03-17T19:53:29.620480	2015-11-01T05:30:10	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.00%3A_Prelude_to_Fluid_Statics", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11.0: Prelude to Fluid Statics", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.01%3A_What_Is_a_Fluid	11.1: What Is a Fluid? Learning Objectives By the end of this section, you will be able to: - State the common phases of matter. - Explain the physical characteristics of solids, liquids, and gases. - Describe the arrangement of atoms in solids, liquids, and gases. Matter most commonly exists as a solid, liquid, or gas; these states are known as the three common phases of matter . Solids have a definite shape and a specific volume, liquids have a definite volume but their shape changes depending on the container in which they are held, and gases have neither a definite shape nor a specific volume as their molecules move to fill the container in which they are held (Figure $\PageIndex{1}$). Liquids and gases are considered to be fluids because they yield to shearing forces, whereas solids resist them. Note that the extent to which fluids yield to shearing forces (and hence flow easily and quickly) depends on a quantity called the viscosity which is discussed in detail in Viscosity and Laminar Flow; Poiseuille’s Law . We can understand the phases of matter and what constitutes a fluid by considering the forces between atoms that make up matter in the three phases. Atoms in solids are in close contact, with forces between them that allow the atoms to vibrate but not to change positions with neighboring atoms. (These forces can be thought of as springs that can be stretched or compressed, but not easily broken.) Thus a solid resists all types of stress. A solid cannot be easily deformed because the atoms that make up the solid are not able to move about freely. Solids also resist compression, because their atoms form part of a lattice structure in which the atoms are a relatively fixed distance apart. Under compression, the atoms would be forced into one another. Most of the examples we have studied so far have involved solid objects which deform very little when stressed. Submicroscopic Explanation of Solids and Liquids Atomic and molecular characteristics explain and underlie the macroscopic characteristics of solids and fluids. This submicroscopic explanation is one theme of this text and is highlighted in Conservation of Momentum . See, for example, microscopic description of collisions and momentum or microscopic description of pressure in a gas. This present section is devoted entirely to the submicroscopic explanation of solids and liquids. - In contrast, liquids deform easily when stressed and do not spring back to their original shape once the force is removed because the atoms are free to slide about and change neighbors—that is, they flow (so they are a type of fluid), with the molecules held together by their mutual attraction. When a liquid is placed in a container with no lid on, it remains in the container (providing the container has no holes below the surface of the liquid!). Because the atoms are closely packed, liquids, like solids, resist compression. Atoms in gases are separated by distances that are large compared with the size of the atoms. The forces between gas atoms are therefore very weak, except when the atoms collide with one another. Gases thus not only flow (and are therefore considered to be fluids) but they are relatively easy to compress because there is much space and little force between atoms. When placed in an open container gases, unlike liquids, will escape. The major distinction is that gases are easily compressed, whereas liquids are not. We shall generally refer to both gases and liquids simply as fluids , and make a distinction between them only when they behave differently. PhET Explorations: States of Matter - Basics Heat, cool, and compress atoms and molecules and watch as they change between solid, liquid, and gas phases. Summary - A fluid is a state of matter that yields to sideways or shearing forces. Liquids and gases are both fluids. Fluid statics is the physics of stationary fluids. Glossary - fluids - liquids and gases; a fluid is a state of matter that yields to shearing forces	libretexts	2025-03-17T19:53:29.684841	2015-11-01T05:30:45	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.01%3A_What_Is_a_Fluid", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11.1: What Is a Fluid?", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.02%3A_Density	11.2: Density Learning Objectives By the end of this section, you will be able to: - Define density. - Calculate the mass of a reservoir from its density. - Compare and contrast the densities of various substances. Which weighs more, a ton of feathers or a ton of bricks? This old riddle plays with the distinction between mass and density. A ton is a ton, of course; but bricks have much greater density than feathers, and so we are tempted to think of them as heavier (Figure $\PageIndex{1}$). Density , as you will see, is an important characteristic of substances. It is crucial, for example, in determining whether an object sinks or floats in a fluid. Definition: Density Density is mass per unit volume. \[\rho = \dfrac{m}{V},\label{density}\] where the Greek letter $\rho$ (rho) is the symbol for density, $m$ is the mass, and $V$ is the volume occupied by the substance. In the riddle regarding the feathers and bricks, the masses are the same, but the volume occupied by the feathers is much greater, since their density is much lower. The SI unit of density is $kg/m^3$, representative values are given in Table $\PageIndex{1}$. The metric system was originally devised so that water would have a density of $1 \, g/cm^3$, equivalent to $10^3 \, kg/m^3$. Thus the basic mass unit, the kilogram, was first devised to be the mass of 1000 mL of water, which has a volume of $1000 \, cm^3$. \| Substance \| $\rho(10^3\frac{kg}{m^3} \, or \, \frac{g}{mL})$ \| Substance \| $\rho(10^3\frac{kg}{m^3} \, or \, \frac{g}{mL})$ \| Substance \| $\rho(10^3\frac{kg}{m^3} \, or \, \frac{g}{mL})$ \| \|---\|---\|---\|---\|---\|---\| \| Solids \| Liquids \| Gases \| \|\|\| \| Aluminum \| 2.7 \| Water (4ºC) \| 1.000 \| Air \| $1.29 \times 10^{-3}$ \| \| Brass \| 8.44 \| Blood \| 1.05 \| Carbon dioxide \| $1.98 \times 10^{-3}$ \| \| Copper (average) \| 8.8 \| Sea water \| 1.025 \| Carbon monoxide \| $1.25 \times 10^{-3}$ \| \| Gold \| 19.32 \| Mercury \| 13.6 \| Hydrogen \| $0.090 \times 10^{-3}$ \| \| Iron or steel \| 7.8 \| Ethyl alcohol \| 0.79 \| Helium \| $0.18 \times 10^{-3}$ \| \| Lead \| 11.3 \| Petrol \| 0.68 \| Methane \| $0.72 \times 10^{-3}$ \| \| Polystyrene \| 0.10 \| Glycerin \| 1.26 \| Nitrogen \| $1.25 \times 10^{-3}$ \| \| Tungsten \| 19.30 \| Olive oil \| 0.92 \| Nitrous oxide \| $1.98 \times 10^{-3}$ \| \| Uranium \| 18.70 \| Oxygen \| $1.43 \times 10^{-3}$ \| \|\| \| Concrete \| 2.30–3.0 \| Steam $100^o$ \| $0.60 \times 10^{-3}$ \| \|\| \| Cork \| 0.24 \| \|\|\|\| \| Glass, common (average) \| 2.6 \| \|\|\|\| \| Granite \| 2.7 \| \|\|\|\| \| Earth’s crust \| 3.3 \| \|\|\|\| \| Wood \| 0.3–0.9 \| \|\|\|\| \| Ice (0°C) \| 0.917 \| \|\|\|\| \| Bone \| 1.7–2.0 \| As you can see by examining Table $\PageIndex{1}$, the density of an object may help identify its composition. The density of gold, for example, is about 2.5 times the density of iron, which is about 2.5 times the density of aluminum. Density also reveals something about the phase of the matter and its substructure. Notice that the densities of liquids and solids are roughly comparable, consistent with the fact that their atoms are in close contact. The densities of gases are much less than those of liquids and solids, because the atoms in gases are separated by large amounts of empty space. TAKE-HOME EXPERIMENT SUGAR AND SALT A pile of sugar and a pile of salt look pretty similar, but which weighs more? If the volumes of both piles are the same, any difference in mass is due to their different densities (including the air space between crystals). Which do you think has the greater density? What values did you find? What method did you use to determine these values? Example $\PageIndex{1}$: Calculating the Mass of a Reservoir From Its Volume A reservoir has a surface area of $50 \, km^2$ and an average depth of 40.0 m. What mass of water is held behind the dam? (See Figure $\PageIndex{2}$ for a view of a large reservoir—the Three Gorges Dam site on the Yangtze River in central China.) Strategy We can calculate the volume $V$ of the reservoir from its dimensions, and find the density of water $\rho$ in Table $\PageIndex{1}$. Then the mass $m$ can be found from the definition of density (Equation \ref{density}). Solution Solving Equation \ref{density} for $m$ gives \[m = \rho V. \nonumber\] The volume $V$ of the reservoir is its surface area $A$ times its average depth $h$: \[\begin{align} V &= Ah \\[5pt] &= (50.0 \, km^2)(40.0 \, m) \\[5pt] &= \left [ (50.0 \, km^2)\left(\frac{10^3 \, m}{1 \, km}\right )\right ](40.0 \, m) \\[5pt] &= 2.00 \times 10^9 \, m^3 \end{align}\] The density of water $\rho$ from Table $\PageIndex{1}$ is $1.000 \times 10^3 \, kg/m^3 $. Substituting $V$ and $\rho$ into the expression for mass gives \[\begin{align} m &= (1.00 \times 10^3 \, kg/m^3)(2.00 \times 10^9 \, m^3) \\[5pt] &= 2.00 \times 10^{12} \, kg.\end{align}\] Discussion A large reservoir contains a very large mass of water. In this example, the weight of the water in the reservoir is $mg = 1.96 \times 10^{13} \, N$, where $g$ is the acceleration due to the Earth’s gravity (about $9.80 \, m/s^2$). It is reasonable to ask whether the dam must supply a force equal to this tremendous weight. The answer is no. As we shall see in the following sections, the force the dam must supply can be much smaller than the weight of the water it holds back. Summary - Density is the mass per unit volume of a substance or object. In equation form, density is defined as \[\rho = \dfrac{m}{V}. n\nonumber\] - The SI unit of density is $kg/m^3$. Glossary - density - the mass per unit volume of a substance or object	libretexts	2025-03-17T19:53:29.769215	2015-11-01T05:31:09	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.02%3A_Density", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11.2: Density", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.03%3A_Pressure	11.3: Pressure Learning Objectives By the end of this section, you will be able to: - Define pressure. - Explain the relationship between pressure and force. - Calculate force given pressure and area. You have no doubt heard the word pressure being used in relation to blood (high or low blood pressure) and in relation to the weather (high- and low-pressure weather systems). These are only two of many examples of pressures in fluids. Definition: Pressure Pressure is defined as the force divided by the area perpendicular to the force over which the force is applied, or \[P = \dfrac{F}{A}. \label{pressure}\] where $F$ is a force applied to an area $A$ that is perpendicular to the force. A given force can have a significantly different effect depending on the area over which the force is exerted, as shown in Figure Figure $\PageIndex{1}$. The SI unit for pressure is the pascal , where \[1 \, Pa = 1 \, N/m^2.\] In addition to the pascal, there are many other units for pressure that are in common use. In meteorology, atmospheric pressure is often described in units of millibar (mb), where \[100 \, mb = 1 \times 10^4 \, Pa.\] Pounds per square inch $(lb/in^2 \, or \, psi)$ is still sometimes used as a measure of tire pressure, and millimeters of mercury (mm Hg) is still often used in the measurement of blood pressure. Pressure is defined for all states of matter but is particularly important when discussing fluids. Example $\PageIndex{1}$: Calculating Force Exerted by the Air - What Force Does a Pressure Exert? An astronaut is working outside the International Space Station where the atmospheric pressure is essentially zero. The pressure gauge on her air tank reads $6.9 \times 10^6 \, Pa$. What force does the air inside the tank exert on the flat end of the cylindrical tank, a disk 0.150 m in diameter? Strategy We can find the force exerted from the definition of pressure (Equation \red{pressure}) provided we can find the area $A$ acted upon. Solution By rearranging the definition of pressure (Equation \red{pressure}) to solve for force, we see that \[F = PA.\nonumber\] Here, the pressure $P$ is given, as is the area of the end of the cylinder $A$, given by $A = \pi r^2$. Thus \[ \begin{align} F &= (6.90 \times 10^6 \, Pa)(3.14)(0.0750 \, m)^2 \\[5pt] &= 1.22 \times 10^5 \, N. \end{align}\] Discussion Wow! No wonder the tank must be strong. Since we found $F = PA$, we see that the force exerted by a pressure is directly proportional to the area acted upon as well as the pressure itself. The force exerted on the end of the tank is perpendicular to its inside surface. This direction is because the force is exerted by a static or stationary fluid. We have already seen that fluids cannot withstand shearing (sideways) forces; they cannot exert shearing forces, either. Fluid pressure has no direction, being a scalar quantity. The forces due to pressure have well-defined directions: they are always exerted perpendicular to any surface. (See the tire in Figure $\PageIndex{2}$, for example.) Finally, note that pressure is exerted on all surfaces. Swimmers, as well as the tire, feel pressure on all sides (Figure $\PageIndex{3}$). PHET EXPLORATIONS: GAS PROPERTIES Pump gas molecules in this simulation to a box and see what happens as you change the volume, add or remove heat, change gravity, and more. Measure the temperature and pressure, and discover how the properties of the gas vary in relation to each other. Summary - Pressure is the force per unit perpendicular area over which the force is applied. In equation form, pressure is defined as \[F = PA. \nonumber\] - The SI unit of pressure is pascal and $1 \, Pa = 1 \, N/m^2.$ Glossary - pressure - the force per unit area perpendicular to the force, over which the force acts	libretexts	2025-03-17T19:53:29.834955	2015-11-01T05:31:30	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.03%3A_Pressure", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11.3: Pressure", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.04%3A_Variation_of_Pressure_with_Depth_in_a_Fluid	11.4: Variation of Pressure with Depth in a Fluid Learning Objectives By the end of this section, you will be able to: - Define pressure in terms of weight. - Explain the variation of pressure with depth in a fluid. - Calculate density given pressure and altitude. If your ears have ever popped on a plane flight or ached during a deep dive in a swimming pool, you have experienced the effect of depth on pressure in a fluid. At the Earth’s surface, the air pressure exerted on you is a result of the weight of air above you. This pressure is reduced as you climb up in altitude and the weight of air above you decreases. Under water, the pressure exerted on you increases with increasing depth. In this case, the pressure being exerted upon you is a result of both the weight of water above you and that of the atmosphere above you. You may notice an air pressure change on an elevator ride that transports you many stories, but you need only dive a meter or so below the surface of a pool to feel a pressure increase. The difference is that water is much denser than air, about 775 times as dense. Consider the container in Figure $\PageIndex{1}$. Its bottom supports the weight of the fluid in it. Let us calculate the pressure exerted on the bottom by the weight of the fluid. That pressure is the weight of the fluid $mg$ divided by the area $A$ supporting it (the area of the bottom of the container): \[P = \dfrac{mg}{A}.\] We can find the mass of the fluid from its volume and density: \[m = \rho V.\] The volume of the fluid $V$ is related to the dimensions of the container. It is \[V = Ah,\] where $A$ is the cross-sectional area and $h$ is the depth. Combining the last two equations gives \[m = \rho Ah.\] If we enter this into the expression for pressure, we obtain \[P = \dfrac{(\rho Ah)g}{A}.\] The area cancels, and rearranging the variables yields \[P = h\rho g. \label{eq10}\] This value is the pressure due to the weight of a fluid . Equation \ref{eq10} has general validity beyond the special conditions under which it is derived here. Even if the container were not there, the surrounding fluid would still exert this pressure, keeping the fluid static. Thus Equation \ref{eq10} represents the pressure due to the weight of any fluid of average density $\rho$ at any depth $h$ below its surface. For liquids, which are nearly incompressible, this equation holds to great depths. For gases, which are quite compressible, one can apply this equation as long as the density changes are small over the depth considered. Example $\PageIndex{1}$ illustrates this situation. Example $\PageIndex{1}$: Calculating the Average Pressure and Force Exerted: What Force Must a Dam Withstand? In [link] , we calculated the mass of water in a large reservoir. We will now consider the pressure and force acting on the dam retaining water (Figure $\PageIndex{2}$). The dam is 500 m wide, and the water is 80.0 m deep at the dam. - What is the average pressure on the dam due to the water? - Calculate the force exerted against the dam and compare it with the weight of water in the dam (previously found to be $1.96 \times 10^{13} \, N.$ Strategy for (a) The average pressure $\overline{P}$ due to the weight of the water is the pressure at the average depth $\overline{h}$ of 40.0 m, since pressure increases linearly with depth. Solution for (a) The average pressure due to the weight of a fluid (Equation \ref{eq10}) is \[\overline{P} = \overline{h}\rho g. \nonumber \] Entering the density of water from [link] and taking $\overline{h}$ to be the average depth of 40.0 m, we obtain \[ \begin{align} \overline{P} &= (40.0 \, m)\left(10^3 \, \frac{kg}{m^3} \right) \left(9.80\, \frac{m}{s^2} \right) \\[5pt] &= 3.92 \times 10^5 \, \frac{N}{m^2} = 392 \, kPa. \end{align} \] Strategy for (b) The force exerted on the dam by the water is the average pressure times the area of contact: \[F = \overline{P} A. \nonumber \] Solution for (b) We have already found the value for $\overline{P}$. The area of the dam is \[A = 80.0 \, m \times 500 \, m = 4.00 \times 10^4 \, m^2,\nonumber\] so that \[ \begin{align} F &= (3.92 \times 10^5 \, N/m^2)(4.00 \times 10^4 \, m^2) \\[5pt] &= 1.57 \times 10^{10} \, N.\end{align} \] Discussion Although this force seems large, it is small compared with the $1.96 \times 10^{13} \, N$ weight of the water in the reservoir—in fact, it is only $0.0800 \, \% $ of the weight. Note that the pressure found in part (a) is completely independent of the width and length of the lake—it depends only on its average depth at the dam. Thus the force depends only on the water’s average depth and the dimensions of the dam, not on the horizontal extent of the reservoir. In the diagram, the thickness of the dam increases with depth to balance the increasing force due to the increasing pressure.epth to balance the increasing force due to the increasing pressure. Atmospheric pressure is another example of pressure due to the weight of a fluid, in this case due to the weight of air above a given height. The atmospheric pressure at the Earth’s surface varies a little due to the large-scale flow of the atmosphere induced by the Earth’s rotation (this creates weather “highs” and “lows”). However, the average pressure at sea level is given by the standard atmospheric pressure $P_{atm}$, measured to be \[1 \, atmosphere \, (atm) = P_{atm} = 1.01 \times 10^5 \, N/m^2 = 101 \, kPa.\] This relationship means that, on average, at sea level, a column of air above $1.00 \, m^2 $ of the Earth’s surface has a weight of $1.01 \times 10^5 \, N,$ equivalent to 1 atm (Figure $\PageIndex{3}$). Example $\PageIndex{2}$: Calculating Average Density: How Dense Is the Air? Calculate the average density of the atmosphere, given that it extends to an altitude of 120 km. Compare this density with that of air listed in [link] . Strategy If we solve $P = h\rho g $ for density, we see that \[\overline{\rho} = \dfrac{P}{hg}.\nonumber\] We then take $P$ to be atmospheric pressure, $h$ is given, and $g$ is known, and so we can use this to calculate $\overline{\rho}$. Solution Entering known values into the expression for $\overline{\rho}$ yields \[ \begin{align} \overline{\rho} &= \dfrac{1.01 \times 10^5 \, N/m^2}{(120 \times 10^3 \, m)(9.80 \, m/s^2)} \\[5pt] &= 8.59 \times 10^{-2} \, kg/m^3. \end{align} \] Discussion This result is the average density of air between the Earth’s surface and the top of the Earth’s atmosphere, which essentially ends at 120 km. The density of air at sea level is given in [link] as $1.29 \, kg/m^3 $ - about 15 times its average value. Because air is so compressible, its density has its highest value near the Earth’s surface and declines rapidly with altitude. Example $\PageIndex{3}$: Calculating Depth Below the Surface of Water: What Depth of Water Creates the Same Pressure as the Entire Atmosphere? Calculate the depth below the surface of water at which the pressure due to the weight of the water equals 1.00 atm. Strategy We begin by solving the equation $P = h\rho g$ for depth $h$: \[h = \dfrac{P}{\rho g}.\nonumber\] Then we take $P$ to be 1.00 atm and $\rho $ to be the density of the water that creates the pressure. Solution Entering the known values into the expression for $h$ gives \[ \begin{align} h &= \dfrac{1.01 \times 10^5 \, N/m^2}{(1.00 \times 10^3 \, m)(9.80 \, m/s^2)} \\[5pt] &= 10.3 \, m. \end{align} \] Discussion Just 10.3 m of water creates the same pressure as 120 km of air. Since water is nearly incompressible, we can neglect any change in its density over this depth. What do you suppose is the total pressure at a depth of 10.3 m in a swimming pool? Does the atmospheric pressure on the water’s surface affect the pressure below? The answer is yes. This seems only logical, since both the water’s weight and the atmosphere’s weight must be supported. So the total pressure at a depth of 10.3 m is 2 atm—half from the water above and half from the air above. We shall see in Pascal’s Principle that fluid pressures always add in this way. Summary - Pressure is the weight of the fluid $mg$ divided by the area $A$ supporting it (the area of the bottom of the container): \[P = \dfrac{mg}{A}. \nonumber\] - Pressure due to the weight of a liquid is given by \[P = h\rho g, \nonumber\] where $P$ is the pressure, $h$ is the height of the liquid, $\rho$ is the density of the liquid, and $g$ is the acceleration due to the gravity. Glossary - pressure - the weight of the fluid divided by the area supporting it	libretexts	2025-03-17T19:53:29.989958	2015-11-01T05:31:50	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.04%3A_Variation_of_Pressure_with_Depth_in_a_Fluid", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11.4: Variation of Pressure with Depth in a Fluid", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.05%3A_Pascals_Principle	11.5: Pascal’s Principle Learning Objectives By the end of this section, you will be able to: - Define pressure. - State Pascal’s principle. - Understand applications of Pascal’s principle. - Derive relationships between forces in a hydraulic system. Pressure is defined as force per unit area. Can pressure be increased in a fluid by pushing directly on the fluid? Yes, but it is much easier if the fluid is enclosed. The heart, for example, increases blood pressure by pushing directly on the blood in an enclosed system (valves closed in a chamber). If you try to push on a fluid in an open system, such as a river, the fluid flows away. An enclosed fluid cannot flow away, and so pressure is more easily increased by an applied force. What happens to a pressure in an enclosed fluid? Since atoms in a fluid are free to move about, they transmit the pressure to all parts of the fluid and to the walls of the container. Remarkably, the pressure is transmitted undiminished . This phenomenon is called Pascal’s principle , because it was first clearly stated by the French philosopher and scientist Blaise Pascal (1623–1662): A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. Definition: Pascal’s Principle A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. Pascal’s principle, an experimentally verified fact, is what makes pressure so important in fluids. Since a change in pressure is transmitted undiminished in an enclosed fluid, we often know more about pressure than other physical quantities in fluids. Moreover, Pascal’s principle implies that the total pressure in a fluid is the sum of the pressures from different sources . We shall find this fact—that pressures add—very useful. Blaise Pascal had an interesting life in that he was home-schooled by his father who removed all of the mathematics textbooks from his house and forbade him to study mathematics until the age of 15. This, of course, raised the boy’s curiosity, and by the age of 12, he started to teach himself geometry. Despite this early deprivation, Pascal went on to make major contributions in the mathematical fields of probability theory, number theory, and geometry. He is also well known for being the inventor of the first mechanical digital calculator, in addition to his contributions in the field of fluid statics. Application of Pascal’s Principle One of the most important technological applications of Pascal’s principle is found in a hydraulic system , which is an enclosed fluid system used to exert forces. The most common hydraulic systems are those that operate car brakes. Let us first consider the simple hydraulic system shown in Figure $\PageIndex{1}$. Relationship Between Forces in a Hydraulic System We can derive a relationship between the forces in the simple hydraulic system shown in Figure $\PageIndex{1}$ by applying Pascal’s principle. Note first that the two pistons in the system are at the same height, and so there will be no difference in pressure due to a difference in depth. Now the pressure due to $F_1$ acting on area $A_1$ is simply $P_1 = \frac{F_1}{A_1}$, as defined by $P = \frac{F}{A}$. According to Pascal’s principle, this pressure is transmitted undiminished throughout the fluid and to all walls of the container. Thus, a pressure $P_2$ is felt at the other piston that is equal to $P_1$. That is \[P_1 = P_2\] But since $P_2 - \frac{F_2}{A_2}$, we see that \[\dfrac{F_1}{A_1} = \frac{F_2}{A_2}. \label{eq20}\] This equation relates the ratios of force to area in any hydraulic system, providing the pistons are at the same vertical height and that friction in the system is negligible. Hydraulic systems can increase or decrease the force applied to them. To make the force larger, the pressure is applied to a larger area. For example, if a 100-N force is applied to the left cylinder in Figure $\PageIndex{1}$ and the right one has an area five times greater, then the force out is 500 N. Hydraulic systems are analogous to simple levers, but they have the advantage that pressure can be sent through tortuously curved lines to several places at once. Example $\PageIndex{1}$: Calculating Force of Slave Cylinders: Pascal Puts on the Brakes Consider the automobile hydraulic system shown in Figure $\PageIndex{2}$. A force of 100 N is applied to the brake pedal, which acts on the cylinder—called the master—through a lever. A force of 500 N is exerted on the master cylinder. (The reader can verify that the force is 500 N using techniques of statics from Applications of Statics, Including Problem-Solving Strategies .) Pressure created in the master cylinder is transmitted to four so-called slave cylinders. The master cylinder has a diameter of 0.500 cm, and each slave cylinder has a diameter of 2.50 cm. Calculate the force $F_2$ created at each of the slave cylinders. Strategy We are given the force $F_1$ that is applied to the master cylinder. The cross-sectional areas $A_1$ and $A_2$ can be calculated from their given diameters. Then Equation \ref{eq20} can be used to find the force $F_2$. Manipulate this algebraically to get $F_2$ on one side and substitute known values: Solution Pascal’s principle applied to hydraulic systems is given by Equation \ref{eq20}: \[\begin{align} F_2 &= \dfrac{A_2}{A_1}F_1 \\[5pt] &= \dfrac{\pi r_2^2}{\pi r_1^2}F_1 \\[5pt] &= \dfrac{(1.25 \, cm)^2}{(0.250 \, cm)^2} \times 500 \, N \\[5pt] &= 1.25 \times 10^4 \, N. \end{align} \] Discussion This value is the force exerted by each of the four slave cylinders. Note that we can add as many slave cylinders as we wish. If each has a 2.50-cm diameter, each will exert $1.25 \times 10^4 \, N.$ A simple hydraulic system, such as a simple machine, can increase force but cannot do more work than done on it. Work is force times distance moved, and the slave cylinder moves through a smaller distance than the master cylinder. Furthermore, the more slaves added, the smaller the distance each moves. Many hydraulic systems—such as power brakes and those in bulldozers—have a motorized pump that actually does most of the work in the system. The movement of the legs of a spider is achieved partly by hydraulics. Using hydraulics, a jumping spider can create a force that makes it capable of jumping 25 times its length! Definition: Conservation of Energy Conservation of energy applied to a hydraulic system tells us that the system cannot do more work than is done on it. Work transfers energy, and so the work output cannot exceed the work input. Power brakes and other similar hydraulic systems use pumps to supply extra energy when needed. Summary - Pressure is force per unit area. - A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. - A hydraulic system is an enclosed fluid system used to exert forces. Glossary - Pascal’s Principle - a change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container	libretexts	2025-03-17T19:53:30.061098	2015-11-01T05:32:08	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.05%3A_Pascals_Principle", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11.5: Pascal’s Principle", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.06%3A_Gauge_Pressure_Absolute_Pressure_and_Pressure_Measurement	11.6: Gauge Pressure, Absolute Pressure, and Pressure Measurement Learning Objectives By the end of this section, you will be able to: - Define gauge pressure and absolute pressure. - Understand the working of aneroid and open-tube barometers. If you limp into a gas station with a nearly flat tire, you will notice the tire gauge on the airline reads nearly zero when you begin to fill it. In fact, if there were a gaping hole in your tire, the gauge would read zero, even though atmospheric pressure exists in the tire. Why does the gauge read zero? There is no mystery here. Tire gauges are simply designed to read zero at atmospheric pressure and positive when pressure is greater than atmospheric. Similarly, atmospheric pressure adds to blood pressure in every part of the circulatory system. (As noted in Pascal’s Principle , the total pressure in a fluid is the sum of the pressures from different sources—here, the heart and the atmosphere.) But atmospheric pressure has no net effect on blood flow since it adds to the pressure coming out of the heart and going back into it, too. What is important is how much greater blood pressure is than atmospheric pressure. Blood pressure measurements, like tire pressures, are thus made relative to atmospheric pressure. In brief, it is very common for pressure gauges to ignore atmospheric pressure—that is, to read zero at atmospheric pressure. We therefore define gauge pressure to be the pressure relative to atmospheric pressure. Gauge pressure is positive for pressures above atmospheric pressure, and negative for pressures below it. Definition: Gauge Pressure Gauge pressure is the pressure relative to atmospheric pressure. Gauge pressure is positive for pressures above atmospheric pressure, and negative for pressures below it. In fact, atmospheric pressure does add to the pressure in any fluid not enclosed in a rigid container. This happens because of Pascal’s principle. The total pressure, or absolute pressure , is thus the sum of gauge pressure and atmospheric pressure: \[P_{abs} = P_g + P_{atm} \] where $P_{abs}$ is absolute pressure, $P_g$ is gauge pressure, and $P_{atm} $ is atmospheric pressure. For example, if your tire gauge reads 34 psi (pounds per square inch), then the absolute pressure is 34 psi plus 14.7 psi ($P_{atm}$ in psi), or 48.7 psi (equivalent to 336 kPa). Definition: Absolute Pressure Absolute pressure is the sum of gauge pressure and atmospheric pressure. For reasons we will explore later, in most cases the absolute pressure in fluids cannot be negative. Fluids push rather than pull, so the smallest absolute pressure is zero. (A negative absolute pressure is a pull.) Thus the smallest possible gauge pressure is $P_g = -P_{atm} $ (this makes $P_{abs}\|) zero). There is no theoretical limit to how large a gauge pressure can be. There are a host of devices for measuring pressure, ranging from tire gauges to blood pressure cuffs. Pascal’s principle is of major importance in these devices. The undiminished transmission of pressure through a fluid allows precise remote sensing of pressures. Remote sensing is often more convenient than putting a measuring device into a system, such as a person’s artery. Figure shows one of the many types of mechanical pressure gauges in use today. In all mechanical pressure gauges, pressure results in a force that is converted (or transduced) into some type of readout. An entire class of gauges uses the property that pressure due to the weight of a fluid is given by \(P = h\rho g$. Consider the U-shaped tube shown in Figure , for example. This simple tube is called a manometer . In Figure (a), both sides of the tube are open to the atmosphere. Atmospheric pressure therefore pushes down on each side equally so its effect cancels. If the fluid is deeper on one side, there is a greater pressure on the deeper side, and the fluid flows away from that side until the depths are equal. Let us examine how a manometer is used to measure pressure. Suppose one side of the U-tube is connected to some source of pressure $P_{abs}$ such as the toy balloon in Figure (b) or the vacuum-packed peanut jar shown in Figure (c). Pressure is transmitted undiminished to the manometer, and the fluid levels are no longer equal. In Figure (b), $P_{abs}$ is greater than atmospheric pressure, whereas in Figure (c), $P_{abs}$ is less than atmospheric pressure. In both cases, $P_{abs}$ differs from atmospheric pressure by an amount $h\rho g$, where $\rho$ is the density of the fluid in the manometer. In Figure (b), $P_{abs}$ can support a column of fluid of height $h$, and so it must exert a pressure $h\rho g$ greater than atmospheric pressure (the gauge pressure $P_g$ is positive). In Figure (c), atmospheric pressure can support a column of fluid of height $h$, and so $P_{abs}$ is less than atmospheric pressure by an amount $h\rho g$ (the gauge pressure $P_g$ is negative). A manometer with one side open to the atmosphere is an ideal device for measuring gauge pressures. The gauge pressure is $P_g = h\rho g$ and is found by measuring $h$. Mercury manometers are often used to measure arterial blood pressure. An inflatable cuff is placed on the upper arm as shown in Figure . By squeezing the bulb, the person making the measurement exerts pressure, which is transmitted undiminished to both the main artery in the arm and the manometer. When this applied pressure exceeds blood pressure, blood flow below the cuff is cut off. The person making the measurement then slowly lowers the applied pressure and listens for blood flow to resume. Blood pressure pulsates because of the pumping action of the heart, reaching a maximum, called systolic pressure , and a minimum, called diastolic pressure , with each heartbeat. Systolic pressure is measured by noting the value of $h$ when blood flow first begins as cuff pressure is lowered. Diastolic pressure is measured by noting when blood flows without interruption. The typical blood pressure of a young adult raises the mercury to a height of 120 mm at systolic and 80 mm at diastolic. This is commonly quoted as 120 over 80, or 120/80. The first pressure is representative of the maximum output of the heart; the second is due to the elasticity of the arteries in maintaining the pressure between beats. The density of the mercury fluid in the manometer is 13.6 times greater than water, so the height of the fluid will be 1/13.6 of that in a water manometer. This reduced height can make measurements difficult, so mercury manometers are used to measure larger pressures, such as blood pressure. The density of mercury is such that $1 \, mm \, Hg = 133 \, Pa$. Definition: Systolic Pressure Systolic Pressure is the maximum blood pressure. Definition: Diastolic Pressure Diastolic pressure is the minimum blood pressure. Example $\PageIndex{1}$: Calculating Height of IV Bag: Blood Pressure and Intravenous Infusions Intravenous infusions are usually made with the help of the gravitational force. Assuming that the density of the fluid being administered is 1.00 g/ml, at what height should the IV bag be placed above the entry point so that the fluid just enters the vein if the blood pressure in the vein is 18 mm Hg above atmospheric pressure? Assume that the IV bag is collapsible. Strategy for (a) For the fluid to just enter the vein, its pressure at entry must exceed the blood pressure in the vein (18 mm Hg above atmospheric pressure). We therefore need to find the height of fluid that corresponds to this gauge pressure. Solution We first need to convert the pressure into SI units. Since $1.0 \, mm \, Hg = 133 \, Pa$, \[\begin{align}P = 18 \, mm \, Hg \times \dfrac{133 \, Pa}{1.0 \, mm \, Hg} = 2400 \, Pa \\[5pt] &= 0.24 \, Pa \end{align}\] Discussion The IV bag must be placed at 0.24 m above the entry point into the arm for the fluid to just enter the arm. Generally, IV bags are placed higher than this. You may have noticed that the bags used for blood collection are placed below the donor to allow blood to flow easily from the arm to the bag, which is the opposite direction of flow than required in the example presented here. A barometer is a device that measures atmospheric pressure. A mercury barometer is shown in Figure . This device measures atmospheric pressure, rather than gauge pressure, because there is a nearly pure vacuum above the mercury in the tube. The height of the mercury is such that $h\rho g = P_{atm}$. When atmospheric pressure varies, the mercury rises or falls, giving important clues to weather forecasters. The barometer can also be used as an altimeter, since average atmospheric pressure varies with altitude. Mercury barometers and manometers are so common that units of mm Hg are often quoted for atmospheric pressure and blood pressures. Table gives conversion factors for some of the more commonly used units of pressure. \| Conversion to N/m 2 (Pa) \| Conversion from atm \| \| $1.0 atm=1.013×10^5N/m^2$ \| $1.0 atm=1.013×10^5N/m^2$ \| \| $1.0dyne/cm^2=0.10N/m^2$ \| $1.0atm=1.013×10^6dyne/cm^2$ \| \| $1.0kg/cm^2=9.8×10^4N/m^2$ \| $1.0atm=1.013kg/cm^2$ \| \| $1.0lb/in.^2=6.90×10^3N/m^2$ \| $1.0atm=14.7lb/in.^2$ \| \| $1.0 mm Hg=133N/m^2$ \| $1.0atm=760 mm Hg$ \| \| $1.0 cm Hg=1.33×10^3N/m^2$ \| $1.0atm=76.0 cm Hg$ \| \| $1.0 cm water=98.1N/m^2$ \| $1.0atm=1.03×10^3cm water$ \| \| $1.0 bar=1.000×10^5N/m^2$ \| $1.0atm=1.013 bar$ \| \| $1.0 millibar=1.000×10^2N/m^2$ \| $1.0 atm=1013 millibar$ \| Conversion Factors for Various Pressure Units Summary - Gauge pressure is the pressure relative to atmospheric pressure. - Absolute pressure is the sum of gauge pressure and atmospheric pressure. - Aneroid gauge measures pressure using a bellows-and-spring arrangement connected to the pointer of a calibrated scale. - Open-tube manometers have U-shaped tubes and one end is always open. It is used to measure pressure. - A mercury barometer is a device that measures atmospheric pressure. Glossary - absolute pressure - the sum of gauge pressure and atmospheric pressure - diastolic pressure - the minimum blood pressure in the artery - gauge pressure - the pressure relative to atmospheric pressure - systolic pressure - the maximum blood pressure in the artery	libretexts	2025-03-17T19:53:30.142281	2015-11-01T05:32:28	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.06%3A_Gauge_Pressure_Absolute_Pressure_and_Pressure_Measurement", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11.6: Gauge Pressure, Absolute Pressure, and Pressure Measurement", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.07%3A_Archimedes_Principle	11.7: Archimedes’ Principle Learning Objectives By the end of this section, you will be able to: - Define buoyant force. - State Archimedes’ principle. - Understand why objects float or sink. - Understand the relationship between density and Archimedes’ principle. When you rise from lounging in a warm bath, your arms feel strangely heavy. This is because you no longer have the buoyant support of the water. Where does this buoyant force come from? Why is it that some things float and others do not? Do objects that sink get any support at all from the fluid? Is your body buoyed by the atmosphere, or are only helium balloons affected (Figure $\PageIndex{1}$)? Answers to all these questions, and many others, are based on the fact that pressure increases with depth in a fluid. This means that the upward force on the bottom of an object in a fluid is greater than the downward force on the top of the object. There is a net upward, or buoyant force on any object in any fluid (Figure $\PageIndex{2}$). If the buoyant force is greater than the object’s weight, the object will rise to the surface and float. If the buoyant force is less than the object’s weight, the object will sink. If the buoyant force equals the object’s weight, the object will remain suspended at that depth. The buoyant force is always present whether the object floats, sinks, or is suspended in a fluid. Defintion: Buoyant Force The buoyant force is the net upward force on any object in any fluid. Just how great is this buoyant force? To answer this question, think about what happens when a submerged object is removed from a fluid, as in Figure $\PageIndex{3}$. The space it occupied is filled by fluid having a weight $w_{fl}$. This weight is supported by the surrounding fluid, and so the buoyant force must equal $w_{fl}$, the weight of the fluid displaced by the object. It is a tribute to the genius of the Greek mathematician and inventor Archimedes (ca. 287–212 B.C.) that he stated this principle long before concepts of force were well established. Stated in words, Archimedes’ principle is as follows: The buoyant force on an object equals the weight of the fluid it displaces. In equation form, Archimedes’ principle is \[F_B = w_{fl},\] where $F_B$ is the buoyant force and $w_{fl}$ is the weight of the fluid displaced by the object. Archimedes’ principle is valid in general, for any object in any fluid, whether partially or totally submerged. Archimedes’ Principle According to this principle the buoyant force on an object equals the weight of the fluid it displaces. In equation form, Archimedes’ principle is \[F_B = w_{fl},\] where $F_B$ is the buoyant force and $w_{fl}$ is the weight of the fluid displaced by the object. Humm … High-tech body swimsuits were introduced in 2008 in preparation for the Beijing Olympics. One concern (and international rule) was that these suits should not provide any buoyancy advantage. How do you think that this rule could be verified? Making Connections: Take-Home Investigation The density of aluminum foil is 2.7 times the density of water. Take a piece of foil, roll it up into a ball and drop it into water. Does it sink? Why or why not? Can you make it sink? Floating and Sinking Drop a lump of clay in water. It will sink. Then mold the lump of clay into the shape of a boat, and it will float. Because of its shape, the boat displaces more water than the lump and experiences a greater buoyant force. The same is true of steel ships. Example $\PageIndex{1}$: Calculating buoyant force: dependency on shape - Calculate the buoyant force on 10,000 metric tons $(1.00 \times 10^7 \, kg)$ of solid steel completely submerged in water, and compare this with the steel’s weight. - What is the maximum buoyant force that water could exert on this same steel if it were shaped into a boat that could displace $1.00 \times 10^5 \, m^3$ of water? Strategy for (a) To find the buoyant force, we must find the weight of water displaced. We can do this by using the densities of water and steel given in [link] . We note that, since the steel is completely submerged, its volume and the water’s volume are the same. Once we know the volume of water, we can find its mass and weight. Solution for (a) First, we use the definition of density $\rho = \frac{m}{V}$ to find the steel’s volume, and then we substitute values for mass and density. This gives \[V_{st} = \dfrac{m_{st}}{\rho_{st}} = \dfrac{1.00 \times 10^7 \, kg}{7.8 \times 10^3 \, kg/m^3} = 1.28 \times 10^3 \, m^3.\] Because the steel is completely submerged, this is also the volume of water displaced, $V_W$. We can now find the mass of water displaced from the relationship between its volume and density, both of which are known. This gives \[m_W = \rho_WV_W = (1.000 \times 10^3 \, kg/m^3)(1.28 \times 10^3 \, m^3)\] \[= 1.3 \times 10^6 \, kg.\] By Archimedes’ principle, the weight of water displaced is $m_Wg$, so the buoyance force is \[F_B = w_W = m_Wg = (1.28 \times 10^6\space kg)(9.80 \, m/s^2)\] \[= 1.3 \times 10^7 \, N.\] The steel’s weight is $m_W g = 9.80 \times 10^7 \, N$, which is much greater than the buoyant force, so the steel will remain submerged. Note that the buoyant force is rounded to two digits because the density of steel is given to only two digits. Strategy for (b) Here we are given the maximum volume of water the steel boat can displace. The buoyant force is the weight of this volume of water. Solution for (b) The mass of water displaced is found from its relationship to density and volume, both of which are known. That is, \[m_W = \rho_WV_W = (1.000 \times 10^3 \, kg/m^3)(1.00 \times 10^5 \, m^3)\] \[= 9.80 \times 10^8 \, kg.\] The maximum buoyant force is the weight of this much water, or \[F_B = w_W = m_W g = (1.00 \times 10^8 \, kg)(9.80 \, m/s^2)\] \[= \times 10^8 \, N.\] Discussion The maximum buoyant force is ten times the weight of the steel, meaning the ship can carry a load nine times its own weight without sinking. Making Connections: Take-Home Investigation - A piece of household aluminum foil is 0.016 mm thick. Use a piece of foil that measures 10 cm by 15 cm. (a) What is the mass of this amount of foil? (b) If the foil is folded to give it four sides, and paper clips or washers are added to this “boat,” what shape of the boat would allow it to hold the most “cargo” when placed in water? Test your prediction. Density and Archimedes’ Principle Density plays a crucial role in Archimedes’ principle. The average density of an object is what ultimately determines whether it floats. If its average density is less than that of the surrounding fluid, it will float. This is because the fluid, having a higher density, contains more mass and hence more weight in the same volume. The buoyant force, which equals the weight of the fluid displaced, is thus greater than the weight of the object. Likewise, an object denser than the fluid will sink. The extent to which a floating object is submerged depends on how the object’s density is related to that of the fluid. In Figure $\PageIndex{4}$, for example, the unloaded ship has a lower density and less of it is submerged compared with the same ship loaded. We can derive a quantitative expression for the fraction submerged by considering density. The fraction submerged is the ratio of the volume submerged to the volume of the object, or \[fraction \, submerged = \dfrac{V_{sub}}{V_{obj}} = \dfrac{V_{fl}}{V_{obj}}.\] The volume submerged equals the volume of fluid displaced, which we call $V_{fl}$. Now we can obtain the relationship between the densities by substituting $\rho = \frac{m}{V}$ into the expression. This gives \[\dfrac{V_{fl}}{V_{obj}} = \dfrac{m_{fl}/\rho_{fl}}{m_{obj}/\overline{\rho}_{obj}},\] where $\overline{\rho}_{obj}$ is the average density of the object and $\rho_{fl}$ is the density of the fluid. Since the object floats, its mass and that of the displaced fluid are equal, and so they cancel from the equation, leaving \[fraction \, submerged = \dfrac{\overline{\rho}_{obj}}{\rho_{fl}}.\] We use this last relationship to measure densities. This is done by measuring the fraction of a floating object that is submerged—for example, with a hydrometer. It is useful to define the ratio of the density of an object to a fluid (usually water) as specific gravity : \[specific \, gravity = \dfrac{\overline{\rho}}{\rho_W},\] where $\overline{\rho}$ is the average density of the object or substance and $\rho_W$ is the density of water at 4.00°C. Specific gravity is dimensionless, independent of whatever units are used for $\rho$. If an object floats, its specific gravity is less than one. If it sinks, its specific gravity is greater than one. Moreover, the fraction of a floating object that is submerged equals its specific gravity. If an object’s specific gravity is exactly 1, then it will remain suspended in the fluid, neither sinking nor floating. Scuba divers try to obtain this state so that they can hover in the water. We measure the specific gravity of fluids, such as battery acid, radiator fluid, and urine, as an indicator of their condition. One device for measuring specific gravity is shown in Figure $\PageIndex{5}$. Definition: Specific Gravity Specific gravity is the ratio of the density of an object to a fluid (usually water). Example $\PageIndex{2}$: Calculating Average Density: Floating Woman Suppose a 60.0-kg woman floats in freshwater with $97.0\%$ of her volume submerged when her lungs are full of air. What is her average density? Strategy We can find the woman’s density by solving the equation \[fraction \, submerged = \dfrac{\overline{\rho}_{obj}}{\rho_{fl}}\] for the density of the object. This yields \[\overline{\rho}_{obj} = \overline{\rho}_{person} = (fraction \, submerged) \cdot \rho_{fl}.\] We know both the fraction submerged and the density of water, and so we can calculate the woman’s density. Solution Entering the known values into the expression for her density, we obtain \[\overline{\rho}_{person} = 0.970 \cdot \left(10^3 \, \dfrac{kg}{m^3}\right) = 970 \, \dfrac{kg}{m^3}.\] Discussion Her density is less than the fluid density. We expect this because she floats. Body density is one indicator of a person’s percent body fat, of interest in medical diagnostics and athletic training. (See Figure $\PageIndex{7}$) There are many obvious examples of lower-density objects or substances floating in higher-density fluids—oil on water, a hot-air balloon, a bit of cork in wine, an iceberg, and hot wax in a “lava lamp,” to name a few. Less obvious examples include lava rising in a volcano and mountain ranges floating on the higher-density crust and mantle beneath them. Even seemingly solid Earth has fluid characteristics. More Density Measurements One of the most common techniques for determining density is shown in Figure $\PageIndex{7}$. An object, here a coin, is weighed in air and then weighed again while submerged in a liquid. The density of the coin, an indication of its authenticity, can be calculated if the fluid density is known. This same technique can also be used to determine the density of the fluid if the density of the coin is known. All of these calculations are based on Archimedes’ principle. Archimedes’ principle states that the buoyant force on the object equals the weight of the fluid displaced. This, in turn, means that the object appears to weigh less when submerged; we call this measurement the object’s apparent weight . The object suffers an apparent weight loss equal to the weight of the fluid displaced. Alternatively, on balances that measure mass, the object suffers an apparent mass loss equal to the mass of fluid displaced. That is \[apparent \, weight \, loss = weight \, of \, fluid \, displaced\] or \[apparent \, mass \, loss = mass \, of \, fluid \, displaced.\] The next example illustrates the use of this technique. Example $\PageIndex{3}$: Calculating Density: Is the Coin Authentic? The mass of an ancient Greek coin is determined in air to be 8.630 g. When the coin is submerged in water as shown in Figure $\PageIndex{7}$, its apparent mass is 7.800 g. Calculate its density, given that water has a density of $1.000 \, g/m^3$ and that effects caused by the wire suspending the coin are negligible. Strategy To calculate the coin’s density, we need its mass (which is given) and its volume. The volume of the coin equals the volume of water displaced. The volume of water displaced $\rho = \frac{m}{V}$ for $V$. Solution The volume of water is $V_W = \frac{m_W}{\rho_W}$ where $m_W$ is the mass of water displaced. As noted, the mass of the water displaced equals the apparent mass loss, which is $m_W = 8.630 \, g - 7.800 \, g = 0.830 \, g$. Thus the volume of water is $V_W = \frac{0.830 \, g}{1.000 \, g/cm^3} = 0.830 \, cm^3$. This is also the volume of the coin, since it is completely submerged. We can now find the density of the coin using the definition of density: \[\rho_c = \dfrac{m_c}{V_c} = \dfrac{8.630 \, g}{0.830 \, cm^3} = 10.4 \, g/cm^3.\] Discussion You can see from [link] that this density is very close to that of pure silver, appropriate for this type of ancient coin. Most modern counterfeits are not pure silver. This brings us back to Archimedes’ principle and how it came into being. As the story goes, the king of Syracuse gave Archimedes the task of determining whether the royal crown maker was supplying a crown of pure gold. The purity of gold is difficult to determine by color (it can be diluted with other metals and still look as yellow as pure gold), and other analytical techniques had not yet been conceived. Even ancient peoples, however, realized that the density of gold was greater than that of any other then-known substance. Archimedes purportedly agonized over his task and had his inspiration one day while at the public baths, pondering the support the water gave his body. He came up with his now-famous principle, saw how to apply it to determine density, and ran naked down the streets of Syracuse crying “Eureka!” (Greek for “I have found it”). Similar behavior can be observed in contemporary physicists from time to time! PhET Explorations: Buoyancy When will objects float and when will they sink? Learn how buoyancy works with blocks. Arrows show the applied forces, and you can modify the properties of the blocks and the fluid. Summary - Buoyant force is the net upward force on any object in any fluid. If the buoyant force is greater than the object’s weight, the object will rise to the surface and float. If the buoyant force is less than the object’s weight, the object will sink. If the buoyant force equals the object’s weight, the object will remain suspended at that depth. The buoyant force is always present whether the object floats, sinks, or is suspended in a fluid. - Archimedes’ principle states that the buoyant force on an object equals the weight of the fluid it displaces. - Specific gravity is the ratio of the density of an object to a fluid (usually water). Glossary - Archimedes’ principle - the buoyant force on an object equals the weight of the fluid it displaces - buoyant force - the net upward force on any object in any fluid - specific gravity - the ratio of the density of an object to a fluid (usually water)	libretexts	2025-03-17T19:53:30.233504	2015-11-01T05:32:49	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.07%3A_Archimedes_Principle", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11.7: Archimedes’ Principle", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.08%3A_Cohesion_and_Adhesion_in_Liquids_-_Surface_Tension_and_Capillary_Action	11.8: Cohesion and Adhesion in Liquids - Surface Tension and Capillary Action Learning Objectives By the end of this section, you will be able to: - Understand cohesive and adhesive forces. - Define surface tension. - Understand capillary action. Children blow soap bubbles and play in the spray of a sprinkler on a hot summer day (Figure $\PageIndex{1}$). An underwater spider keeps his air supply in a shiny bubble he carries wrapped around him. A technician draws blood into a small-diameter tube just by touching it to a drop on a pricked finger. A premature infant struggles to inflate her lungs. What is the common thread? All these activities are dominated by the attractive forces between atoms and molecules in liquids—both within a liquid and between the liquid and its surroundings. Attractive forces between molecules of the same type are called cohesive forces . Liquids can, for example, be held in open containers because cohesive forces hold the molecules together. Attractive forces between molecules of different types are called adhesive forces . Such forces cause liquid drops to cling to window panes, for example. In this section we examine effects directly attributable to cohesive and adhesive forces in liquids. Definition: Cohesive Forces Attractive forces between molecules of the same type are called cohesive forces. Definition: Adhesive Forces Attractive forces between molecules of different types are called adhesive forces. Surface Tension Cohesive forces between molecules cause the surface of a liquid to contract to the smallest possible surface area. This general effect is called surface tension . Molecules on the surface are pulled inward by cohesive forces, reducing the surface area. Molecules inside the liquid experience zero net force, since they have neighbors on all sides. Definition: Surface Tension Cohesive forces between molecules cause the surface of a liquid to contract to the smallest possible surface area. This general effect is called surface tension. Surface Tension Forces between atoms and molecules underlie the macroscopic effect called surface tension. These attractive forces pull the molecules closer together and tend to minimize the surface area. This is another example of a submicroscopic explanation for a macroscopic phenomenon. The model of a liquid surface acting like a stretched elastic sheet can effectively explain surface tension effects. For example, some insects can walk on water (as opposed to floating in it) as we would walk on a trampoline—they dent the surface as shown in Figure $\PageIndex{2a}$. Figure $\PageIndex{2b}$ shows another example, where a needle rests on a water surface. The iron needle cannot, and does not, float, because its density is greater than that of water. Rather, its weight is supported by forces in the stretched surface that try to make the surface smaller or flatter. If the needle were placed point down on the surface, its weight acting on a smaller area would break the surface, and it would sink. Surface tension is proportional to the strength of the cohesive force, which varies with the type of liquid. Surface tension $\overline{\gamma}$ is defined to be the force F per unit length $L$ exerted by a stretched liquid membrane: \[\gamma = \dfrac{F}{L}.\] Table $\PageIndex{1}$ lists values of $\overline{\gamma}$ for some liquids. \| Liquid \| Surface tension γ(N/m) \| \|---\|---\| \| Water at $0^oC$ \| 0.0756 \| \| Water at $20^oC$ \| 0.0728 \| \| Water at $100^OC$ \| 0.0589 \| \| Soapy water (typical) \| 0.0370 \| \| Ethyl alcohol \| 0.0223 \| \| Glycerin \| 0.0631 \| \| Mercury \| 0.465 \| \| Olive Oil \| 0.032 \| \| Tissue fluids (typical) \| 0.050 \| \| Blood, whole at $37^oC$ \| 0.058 \| \| Blood plasma at $37^oC$ \| 0.073 \| \| Gold at $1070^oC$ \| 1.000 \| \| Oxygen at $-193^oC$ \| 0.0157 \| \| Helium at $-269^oC$ \| 0.00012 \| For the insect of Figure Figure $\PageIndex{1a}$, its weight $w$ is supported by the upward components of the surface tension force: $w = \gamma L \sin \theta$, where $L$ is the circumference of the insect’s foot in contact with the water. Figure $\PageIndex{3}$ shows one way to measure surface tension. The liquid film exerts a force on the movable wire in an attempt to reduce its surface area. The magnitude of this force depends on the surface tension of the liquid and can be measured accurately. Surface tension is the reason why liquids form bubbles and droplets. The inward surface tension force causes bubbles to be approximately spherical and raises the pressure of the gas trapped inside relative to atmospheric pressure outside. It can be shown that the gauge pressure $P $ inside a spherical bubble is given by \[P = \dfrac{4\gamma}{r},\] where $r$ is the radius of the bubble. Thus the pressure inside a bubble is greatest when the bubble is the smallest. Another bit of evidence for this is illustrated in Figure $\PageIndex{4}$. When air is allowed to flow between two balloons of unequal size, the smaller balloon tends to collapse, filling the larger balloon. Example $\PageIndex{1}$: Surface Tension: Pressure Inside a Bubble Calculate the gauge pressure inside a soap bubble $2.00 \times 10^{-4} m$ in radius using the surface tension for soapy water in Table . Convert this pressure to mm Hg. Strategy The radius is given and the surface tension can be found in Table , and so $P$ can be found directly from the equation $P = \frac{4\gamma}{r}$. Solution Substituding $r$ and $\gamma$ into this equation $P = \frac{4\gamma}{r}$, we obtaiin \[P = \dfrac{4\gamma}{r} = \dfrac{4(0.037 \, N/m)}{2.00 \times 10^{-4}m} = 740 \, N/m^2 = 740 \, Pa.\] We use a conversion factor to get this into units of mm Hg: \[P = (740 \, N/m^2)\dfrac{1.00 \, mm \, Hg}{133 \, N/m^2} = 5.56 \, mm \, Hg.\] Discussion Note that if a hole were to be made in the bubble, the air would be forced out, the bubble would decrease in radius, and the gauge pressure would reduce to zero, and the absolute pressure inside would decrease to atmospheric pressure (760 mm Hg). Our lungs contain hundreds of millions of mucus-lined sacs called alveoli , which are very similar in size, and about 0.1 mm in diameter. (See Figure .) You can exhale without muscle action by allowing surface tension to contract these sacs. Medical patients whose breathing is aided by a positive pressure respirator have air blown into the lungs, but are generally allowed to exhale on their own. Even if there is paralysis, surface tension in the alveoli will expel air from the lungs. Since pressure increases as the radii of the alveoli decrease, an occasional deep cleansing breath is needed to fully reinflate the alveoli. Respirators are programmed to do this and we find it natural, as do our companion dogs and cats, to take a cleansing breath before settling into a nap. The tension in the walls of the alveoli results from the membrane tissue and a liquid on the walls of the alveoli containing a long lipoprotein that acts as a surfactant (a surface-tension reducing substance). The need for the surfactant results from the tendency of small alveoli to collapse and the air to fill into the larger alveoli making them even larger (as demonstrated in Figure ). During inhalation, the lipoprotein molecules are pulled apart and the wall tension increases as the radius increases (increased surface tension). During exhalation, the molecules slide back together and the surface tension decreases, helping to prevent a collapse of the alveoli. The surfactant therefore serves to change the wall tension so that small alveoli don’t collapse and large alveoli are prevented from expanding too much. This tension change is a unique property of these surfactants, and is not shared by detergents (which simply lower surface tension). (See Figure .) If water gets into the lungs, the surface tension is too great and you cannot inhale. This is a severe problem in resuscitating drowning victims. A similar problem occurs in newborn infants who are born without this surfactant—their lungs are very difficult to inflate. This condition is known as hyaline membrane disease and is a leading cause of death for infants, particularly in premature births. Some success has been achieved in treating hyaline membrane disease by spraying a surfactant into the infant’s breathing passages. Emphysema produces the opposite problem with alveoli. Alveolar walls of emphysema victims deteriorate, and the sacs combine to form larger sacs. Because pressure produced by surface tension decreases with increasing radius, these larger sacs produce smaller pressure, reducing the ability of emphysema victims to exhale. A common test for emphysema is to measure the pressure and volume of air that can be exhaled. Making Connections: Take-Home Investigation - Try floating a sewing needle on water. In order for this activity to work, the needle needs to be very clean as even the oil from your fingers can be sufficient to affect the surface properties of the needle. - Place the bristles of a paint brush into water. Pull the brush out and notice that for a short while, the bristles will stick together. The surface tension of the water surrounding the bristles is sufficient to hold the bristles together. As the bristles dry out, the surface tension effect dissipates. - Place a loop of thread on the surface of still water in such a way that all of the thread is in contact with the water. Note the shape of the loop. Now place a drop of detergent into the middle of the loop. What happens to the shape of the loop? Why? - Sprinkle pepper onto the surface of water. Add a drop of detergent. What happens? Why? - Float two matches parallel to each other and add a drop of detergent between them. What happens? Note: For each new experiment, the water needs to be replaced and the bowl washed to free it of any residual detergent. Adhesion and Capillary Action Why is it that water beads up on a waxed car but does not on bare paint? The answer is that the adhesive forces between water and wax are much smaller than those between water and paint. Competition between the forces of adhesion and cohesion are important in the macroscopic behavior of liquids. An important factor in studying the roles of these two forces is the angle $\theta$ between the tangent to the liquid surface and the surface (Figure $\PageIndex{7}$). The contact angle $\theta$ is directly related to the relative strength of the cohesive and adhesive forces. The larger the strength of the cohesive force relative to the adhesive force, the larger $\theta$ is, and the more the liquid tends to form a droplet. The smaller $\theta$ is, the smaller the relative strength, so that the adhesive force is able to flatten the drop. Table lists contact angles for several combinations of liquids and solids. Definition: Contact Angle The angle $\theta$ between the tangent to the liquid surface and the surface is called the contact angle . One important phenomenon related to the relative strength of cohesive and adhesive forces is capillary action —the tendency of a fluid to be raised or suppressed in a narrow tube, or capillary tube . This action causes blood to be drawn into a small-diameter tube when the tube touches a drop. Capillary Action The tendency of a fluid to be raised or suppressed in a narrow tube, or capillary tube, is called capillary action. If a capillary tube is placed vertically into a liquid, as shown in Figure , capillary action will raise or suppress the liquid inside the tube depending on the combination of substances. The actual effect depends on the relative strength of the cohesive and adhesive forces and, thus, the contact angle $\theta$ given in the table. If $\theta$ is less that $90^o$, then the fluid will be raised; if $\theta$ is greater than $90^o$, it will be suppressed. Mercury, for example, has a very large surface tension and a large contact angle with glass. When placed in a tube, the surface of a column of mercury curves downward, somewhat like a drop. The curved surface of a fluid in a tube is called a meniscus . The tendency of surface tension is always to reduce the surface area. Surface tension thus flattens the curved liquid surface in a capillary tube. This results in a downward force in mercury and an upward force in water, as seen in Figure . \| Interface \| Contact angle $\phi$ \| \|---\|---\| \| Mercury-glass \| $140^o$ \| \| Water-glass \| $0^o$ \| \| Water-paraffin \| $107^o$ \| \| Water-silver \| $90^o$ \| \| Organic liquids (most)-glass \| $0^o$ \| \| Ethyl alcohol-glass \| $0^o$ \| \| Kerosene-glass \| $26^o$ \| Capillary action can move liquids horizontally over very large distances, but the height to which it can raise or suppress a liquid in a tube is limited by its weight. It can be shown that this heigh $h$ is given by \[h = \dfrac{2\gamma \cos \theta}{\rho gr}.\] If we look at the different factors in this expression, we might see how it makes good sense. The height is directly proportional to the surface tension $\gamma$, which is its direct cause. Furthermore, the height is inversely proportional to tube radius—the smaller the radius , the higher the fluid can be raised, since a smaller tube holds less mass. The height is also inversely proportional to fluid density , since a larger density means a greater mass in the same volume. (See Figure .) Example $\PageIndex{2}$: Calculating Radius of a Capillary Tube: Capillary Action: Tree Sap Can capillary action be solely responsible for sap rising in trees? To answer this question, calculate the radius of a capillary tube that would raise sap 100 m to the top of a giant redwood, assuming that sap’s density is $1050 \, kg/m^3$, its contact angle is zero, and its surface tension is the same as that of water at $20.0^oC$. Strategy The height to which a liquid will rise as a result of capillary action is given by $h = \frac{2\gamma \cos \theta}{\rho gr}$, and every quantity is known exceptfor $r$. Solution Solving for $r$ and substituting known values produces \[r = \dfrac{2\gamma \cos \theta}{\rho gh} = \dfrac{2(0.0728 \, N/m)cos(0^o)}{(1050 \, kg/m^3)(9.80 \, m/s^2)(100 \, m)}\] \[= 1.41 \times 10^{-7} \, m.\] Discussion This result is unreasonable. Sap in trees moves through the xylem , which forms tubes with radii as small as $2.5 \times 10^{-7} \, m.$ This value is about 180 times as large as the radius found necessary here to raise sap $100 m$. This means that capillary action alone cannot be solely responsible for sap getting to the tops of trees. How does sap get to the tops of tall trees? (Recall that a column of water can only rise to a height of 10 m when there is a vacuum at the top—see [link] .) The question has not been completely resolved, but it appears that it is pulled up like a chain held together by cohesive forces. As each molecule of sap enters a leaf and evaporates (a process called transpiration), the entire chain is pulled up a notch. So a negative pressure created by water evaporation must be present to pull the sap up through the xylem vessels. In most situations, fluids can push but can exert only negligible pull , because the cohesive forces seem to be too small to hold the molecules tightly together. But in this case, the cohesive force of water molecules provides a very strong pull. Figure shows one device for studying negative pressure. Some experiments have demonstrated that negative pressures sufficient to pull sap to the tops of the tallest trees can be achieved. Summary - Attractive forces between molecules of the same type are called cohesive forces. - Attractive forces between molecules of different types are called adhesive forces. - Cohesive forces between molecules cause the surface of a liquid to contract to the smallest possible surface area. This general effect is called surface tension. - Capillary action is the tendency of a fluid to be raised or suppressed in a narrow tube, or capillary tube which is due to the relative strength of cohesive and adhesive forces. Glossary - adhesive forces - the attractive forces between molecules of different types - capillary action - the tendency of a fluid to be raised or lowered in a narrow tube - cohesive forces - the attractive forces between molecules of the same type - contact angle - the angle $θ$ between the tangent to the liquid surface and the surface - surface tension - the cohesive forces between molecules which cause the surface of a liquid to contract to the smallest possible surface area	libretexts	2025-03-17T19:53:30.334212	2015-11-01T05:33:06	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.08%3A_Cohesion_and_Adhesion_in_Liquids_-_Surface_Tension_and_Capillary_Action", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11.8: Cohesion and Adhesion in Liquids - Surface Tension and Capillary Action", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.09%3A_Pressures_in_the_Body	11.9: Pressures in the Body Learning Objectives By the end of this section, you will be able to: - Explain the concept of pressure the in human body. - Explain systolic and diastolic blood pressures. - Describe pressures in the eye, lungs, spinal column, bladder, and skeletal system.	libretexts	2025-03-17T19:53:30.393126	2015-11-01T05:33:21	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.09%3A_Pressures_in_the_Body", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11.9: Pressures in the Body", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.E%3A_Fluid_Statics_(Exercises)	11.E: Fluid Statics (Exercises) Conceptual Questions 11.1: What Is a Fluid? 1. What physical characteristic distinguishes a fluid from a solid? 2. Which of the following substances are fluids at room temperature: air, mercury, water, glass? 3. Why are gases easier to compress than liquids and solids? 4. How do gases differ from liquids? 11.2: Density 5. Approximately how does the density of air vary with altitude? 6. Give an example in which density is used to identify the substance composing an object. Would information in addition to average density be needed to identify the substances in an object composed of more than one material? 7. Figure shows a glass of ice water filled to the brim. Will the water overflow when the ice melts? Explain your answer. 11.3: Pressure 8. How is pressure related to the sharpness of a knife and its ability to cut? 9. Why does a dull hypodermic needle hurt more than a sharp one? 10. The outward force on one end of an air tank was calculated in Example. How is this force balanced? (The tank does not accelerate, so the force must be balanced.) 11. Why is force exerted by static fluids always perpendicular to a surface? 12. In a remote location near the North Pole, an iceberg floats in a lake. Next to the lake (assume it is not frozen) sits a comparably sized glacier sitting on land. If both chunks of ice should melt due to rising global temperatures (and the melted ice all goes into the lake), which ice chunk would give the greatest increase in the level of the lake water, if any? 13. How do jogging on soft ground and wearing padded shoes reduce the pressures to which the feet and legs are subjected? 14. Toe dancing (as in ballet) is much harder on toes than normal dancing or walking. Explain in terms of pressure. 15. How do you convert pressure units like millimeters of mercury, centimeters of water, and inches of mercury into units like newtons per meter squared without resorting to a table of pressure conversion factors? 11.4: Variation of Pressure with Depth in a Fluid 16. Atmospheric pressure exerts a large force (equal to the weight of the atmosphere above your body—about 10 tons) on the top of your body when you are lying on the beach sunbathing. Why are you able to get up? 17. Why does atmospheric pressure decrease more rapidly than linearly with altitude? 18. What are two reasons why mercury rather than water is used in barometers? 19. Figure shows how sandbags placed around a leak outside a river levee can effectively stop the flow of water under the levee. Explain how the small amount of water inside the column formed by the sandbags is able to balance the much larger body of water behind the levee. Because the river level is very high, it has started to leak under the levee. Sandbags are placed around the leak, and the water held by them rises until it is the same level as the river, at which point the water there stops rising. 20. Why is it difficult to swim under water in the Great Salt Lake? 21. Is there a net force on a dam due to atmospheric pressure? Explain your answer. 22. Does atmospheric pressure add to the gas pressure in a rigid tank? In a toy balloon? When, in general, does atmospheric pressure not affect the total pressure in a fluid? 23. You can break a strong wine bottle by pounding a cork into it with your fist, but the cork must press directly against the liquid filling the bottle—there can be no air between the cork and liquid. Explain why the bottle breaks, and why it will not if there is air between the cork and liquid. 11.5: Pascal’s Principle 24. Suppose the master cylinder in a hydraulic system is at a greater height than the slave cylinder. Explain how this will affect the force produced at the slave cylinder. 11.6: Gauge Pressure, Absolute Pressure, and Pressure Measurement 25. Explain why the fluid reaches equal levels on either side of a manometer if both sides are open to the atmosphere, even if the tubes are of different diameters. 26. Figure shows how a common measurement of arterial blood pressure is made. Is there any effect on the measured pressure if the manometer is lowered? What is the effect of raising the arm above the shoulder? What is the effect of placing the cuff on the upper leg with the person standing? Explain your answers in terms of pressure created by the weight of a fluid. 27. Considering the magnitude of typical arterial blood pressures, why are mercury rather than water manometers used for these measurements? 11.7: Archimedes’ Principle 28. More force is required to pull the plug in a full bathtub than when it is empty. Does this contradict Archimedes’ principle? Explain your answer. 29. Do fluids exert buoyant forces in a “weightless” environment, such as in the space shuttle? Explain your answer. 30. Will the same ship float higher in salt water than in freshwater? Explain your answer. 31. Marbles dropped into a partially filled bathtub sink to the bottom. Part of their weight is supported by buoyant force, yet the downward force on the bottom of the tub increases by exactly the weight of the marbles. Explain why. 11.8: Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action 32. The density of oil is less than that of water, yet a loaded oil tanker sits lower in the water than an empty one. Why? 33. Is surface tension due to cohesive or adhesive forces, or both? 34. Is capillary action due to cohesive or adhesive forces, or both? 35. Birds such as ducks, geese, and swans have greater densities than water, yet they are able to sit on its surface. Explain this ability, noting that water does not wet their feathers and that they cannot sit on soapy water. 36. Water beads up on an oily sunbather, but not on her neighbor, whose skin is not oiled. Explain in terms of cohesive and adhesive forces. 37. Could capillary action be used to move fluids in a “weightless” environment, such as in an orbiting space probe? 38. What effect does capillary action have on the reading of a manometer with uniform diameter? Explain your answer. 39. Pressure between the inside chest wall and the outside of the lungs normally remains negative. Explain how pressure inside the lungs can become positive (to cause exhalation) without muscle action. Problems & Exercises 11.2: Density 40. Gold is sold by the troy ounce (31.103 g). What is the volume of 1 troy ounce of pure gold? Solution $1.610cm^3$ 41. Mercury is commonly supplied in flasks containing 34.5 kg (about 76 lb). What is the volume in liters of this much mercury? 42. (a) What is the mass of a deep breath of air having a volume of 2.00 L? (b) Discuss the effect taking such a breath has on your body’s volume and density. Solution (a) 2.58 g (b) The volume of your body increases by the volume of air you inhale. The average density of your body decreases when you take a deep breath, because the density of air is substantially smaller than the average density of the body before you took the deep breath. 43. A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 240-g rock that displaces $89.0cm^3$ of water? (Note that the accuracy and practical applications of this technique are more limited than a variety of others that are based on Archimedes’ principle.) Solution $2.70g/cm^3$ 44. Suppose you have a coffee mug with a circular cross section and vertical sides (uniform radius). What is its inside radius if it holds 375 g of coffee when filled to a depth of 7.50 cm? Assume coffee has the same density as water. 45. (a) A rectangular gasoline tank can hold 50.0 kg of gasoline when full. What is the depth of the tank if it is 0.500-m wide by 0.900-m long? (b) Discuss whether this gas tank has a reasonable volume for a passenger car. Solution (a) 0.163 m (b) Equivalent to 19.4 gallons, which is reasonable 46. A trash compactor can reduce the volume of its contents to 0.350 their original value. Neglecting the mass of air expelled, by what factor is the density of the rubbish increased? 47. A 2.50-kg steel gasoline can holds 20.0 L of gasoline when full. What is the average density of the full gas can, taking into account the volume occupied by steel as well as by gasoline? Solution $7.9×10^2kg/m^3$ 48. What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are parts by mass, not volume.) Assume that this is a simple mixture having an average density equal to the weighted densities of its constituents. Solution $15.6g/cm^3$ 49. There is relatively little empty space between atoms in solids and liquids, so that the average density of an atom is about the same as matter on a macroscopic scale—approximately $10^3kg/m^3$. The nucleus of an atom has a radius about $10^{−5}$ that of the atom and contains nearly all the mass of the entire atom. (a) What is the approximate density of a nucleus? (b) One remnant of a supernova, called a neutron star, can have the density of a nucleus. What would be the radius of a neutron star with a mass 10 times that of our Sun (the radius of the Sun is $7×10^8m$)? Solution (a) $10^{18}kg/m^3$ (b) $2×10^4m$ 11.3: Pressure 50. As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted on the floor by the heel if it has an area of $1.50cm^2$ and the woman’s mass is 55.0 kg. Express the pressure in Pa. (In the early days of commercial flight, women were not allowed to wear high-heeled shoes because aircraft floors were too thin to withstand such large pressures.) Solution $3.59×10^6Pa$; or $521lb/in^2$ 51. The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.200 mm in radius, what pressure is exerted on the record in $N/m^2$? 52. Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What force must be exerted on a nail with a circular tip of 1.00 mm diameter to create a pressure of $3.00×10^9N/m^2$?(This high pressure is possible because the hammer striking the nail is brought to rest in such a short distance.) Solution $2.36×10^3N$ 11.4: Variation of Pressure with Depth in a Fluid 53. What depth of mercury creates a pressure of 1.00 atm? Solution 0.760 m 54. The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure due to the ocean at the bottom of this trench, given its depth is 11.0 km and assuming the density of seawater is constant all the way down. 55. Verify that the SI unit of $hρg$ is $N/m^2$. $(hρg)_{units}=(m)(kg/m^3)(m/s^2)=(kg⋅m^2)/(m^3⋅s^2)$ $=(kg⋅m/s^2)(1/m^2)$ $=N/m^2$ 56. Water towers store water above the level of consumers for times of heavy use, eliminating the need for high-speed pumps. How high above a user must the water level be to create a gauge pressure of $3.00×10^5N/m^2$? 57. The aqueous humor in a person’s eye is exerting a force of 0.300 N on the $1.10-cm^2$ area of the cornea. (a) What pressure is this in mm Hg? (b) Is this value within the normal range for pressures in the eye? Solution (a) 20.5 mm Hg (b) The range of pressures in the eye is 12–24 mm Hg, so the result in part (a) is within that range 58. How much force is exerted on one side of an 8.50 cm by 11.0 cm sheet of paper by the atmosphere? How can the paper withstand such a force? 59. What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weight of the gasoline in it when it is full? Solution $1.09×10^3N/m^2$ 60. Calculate the average pressure exerted on the palm of a shot-putter’s hand by the shot if the area of contact is $50.0cm^2$ and he exerts a force of 800 N on it. Express the pressure in $N/m^2$ and compare it with the $1.00×10^6Pa$ pressures sometimes encountered in the skeletal system. 61. The left side of the heart creates a pressure of 120 mm Hg by exerting a force directly on the blood over an effective area of $15.0cm^2$. What force does it exert to accomplish this? Solution 24.0 N 62. Show that the total force on a rectangular dam due to the water behind it increases with the square of the water depth. In particular, show that this force is given by $F=ρgh^2L/2$, where $ρ$ is the density of water, $h$ is its depth at the dam, and $L$ is the length of the dam. You may assume the face of the dam is vertical. (Hint: Calculate the average pressure exerted and multiply this by the area in contact with the water. (See Figure.) 11.5: Pascal’s Principle 63. How much pressure is transmitted in the hydraulic system considered in Example? Express your answer in pascals and in atmospheres. Solution $2.55×10^7Pa$; or 251 atm 64. What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resting on the slave cylinder? The master cylinder has a 2.00-cm diameter and the slave has a 24.0-cm diameter. 65. A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force. Solution $5.76×10^3N$ extra force 66. A certain hydraulic system is designed to exert a force 100 times as large as the one put into it. (a) What must be the ratio of the area of the slave cylinder to the area of the master cylinder? (b) What must be the ratio of their diameters? (c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses to friction. 67. (a) Verify that work input equals work output for a hydraulic system assuming no losses to friction. Do this by showing that the distance the output force moves is reduced by the same factor that the output force is increased. Assume the volume of the fluid is constant. (b) What effect would friction within the fluid and between components in the system have on the output force? How would this depend on whether or not the fluid is moving? Solution (a) $V=d_iA_i=d_oA_o⇒d_o=d_i(\frac{A_i}{A_o})$. Now, using equation: $\frac{F_1}{A_1}=\frac{F_2}{A_2}⇒F_o=F_i(\frac{A_o}{A_i})$. Finally, $W_o=F_od_o=(\frac{F_iA_o}{A_i})(\frac{d_iA_i}{A_o})=F_id_i=W_i$. In other words, the work output equals the work input. (b) If the system is not moving, friction would not play a role. With friction, we know there are losses, so that $W_{out}=W_{in}−W_f$; therefore, the work output is less than the work input. In other words, with friction, you need to push harder on the input piston than was calculated for the nonfriction case. 11.6: Gauge Pressure, Absolute Pressure, and Pressure Measurement 68. Find the gauge and absolute pressures in the balloon and peanut jar shown in Figure, assuming the manometer connected to the balloon uses water whereas the manometer connected to the jar contains mercury. Express in units of centimeters of water for the balloon and millimeters of mercury for the jar, taking $h=0.0500 m$ for each. Solution Balloon: Pg=5.00 cmH2O, Pabs=1.035×103cmH2O. Jar: Pg=−50.0 mm Hg, Pabs=710 mm Hg. 69. (a) Convert normal blood pressure readings of 120 over 80 mm Hg to newtons per meter squared using the relationship for pressure due to the weight of a fluid ($P=hρg$) rather than a conversion factor. (b) Discuss why blood pressures for an infant could be smaller than those for an adult. Specifically, consider the smaller height to which blood must be pumped. 70. How tall must a water-filled manometer be to measure blood pressures as high as 300 mm Hg? Solution 4.08 m 71. Pressure cookers have been around for more than 300 years, although their use has strongly declined in recent years (early models had a nasty habit of exploding). How much force must the latches holding the lid onto a pressure cooker be able to withstand if the circular lid is $25.0 cm$ in diameter and the gauge pressure inside is 300 atm? Neglect the weight of the lid. 72. Suppose you measure a standing person’s blood pressure by placing the cuff on his leg 0.500 m below the heart. Calculate the pressure you would observe (in units of mm Hg) if the pressure at the heart were 120 over 80 mm Hg. Assume that there is no loss of pressure due to resistance in the circulatory system (a reasonable assumption, since major arteries are large). Solution $ΔP=38.7 mm Hg$, Leg blood pressure$=\frac{159}{119}.$ 73. A submarine is stranded on the bottom of the ocean with its hatch 25.0 m below the surface. Calculate the force needed to open the hatch from the inside, given it is circular and 0.450 m in diameter. Air pressure inside the submarine is 1.00 atm. 74. Assuming bicycle tires are perfectly flexible and support the weight of bicycle and rider by pressure alone, calculate the total area of the tires in contact with the ground. The bicycle plus rider has a mass of 80.0 kg, and the gauge pressure in the tires is $3.50×10^5Pa$. Solution $22.4cm^2$ 11.7: Archimedes’ Principle 75. What fraction of ice is submerged when it floats in freshwater, given the density of water at 0°C is very close to $1000 kg/m^3$? Solution $91.7%$ 76. Logs sometimes float vertically in a lake because one end has become water-logged and denser than the other. What is the average density of a uniform-diameter log that floats with 20.0% of its length above water? 77. Find the density of a fluid in which a hydrometer having a density of $0.750 g/mL$ floats with $92.0%$ of its volume submerged. Solution $815 kg/m^3$ 78. If your body has a density of $995 kg/m^3$, what fraction of you will be submerged when floating gently in: (a) Freshwater? (b) Salt water, which has a density of $1027 kg/m^3$? 79. Bird bones have air pockets in them to reduce their weight—this also gives them an average density significantly less than that of the bones of other animals. Suppose an ornithologist weighs a bird bone in air and in water and finds its mass is $45.0 g$ and its apparent mass when submerged is $3.60 g$ (the bone is watertight). (a) What mass of water is displaced? (b) What is the volume of the bone? (c) What is its average density? Solution (a) $41.4 g$ (b) $41.4cm^3$ (c) $1.09 g/cm^3$ 80. A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is displaced? (b) What is the volume of the rock? (c) What is its average density? Is this consistent with the value for granite? 81. Archimedes’ principle can be used to calculate the density of a fluid as well as that of a solid. Suppose a chunk of iron with a mass of 390.0 g in air is found to have an apparent mass of 350.5 g when completely submerged in an unknown liquid. (a) What mass of fluid does the iron displace? (b) What is the volume of iron, using its density as given in [link] (c) Calculate the fluid’s density and identify it. Solution (a) 39.5 g (b) $50cm^3$ (c) $0.79g/cm^3$ It is ethyl alcohol. 82. In an immersion measurement of a woman’s density, she is found to have a mass of 62.0 kg in air and an apparent mass of 0.0850 kg when completely submerged with lungs empty. (a) What mass of water does she displace? (b) What is her volume? (c) Calculate her density. (d) If her lung capacity is 1.75 L, is she able to float without treading water with her lungs filled with air? 83. Some fish have a density slightly less than that of water and must exert a force (swim) to stay submerged. What force must an 85.0-kg grouper exert to stay submerged in salt water if its body density is $1015kg/m^3$? Solution 8.21 N 84. (a) Calculate the buoyant force on a 2.00-L helium balloon. (b) Given the mass of the rubber in the balloon is 1.50 g, what is the net vertical force on the balloon if it is let go? You can neglect the volume of the rubber. 85. (a) What is the density of a woman who floats in freshwater with $4.00%$ of her volume above the surface? This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water (briefly). (b) What percent of her volume is above the surface when she floats in seawater? Solution (a) $960kg/m^3$ (b) $6.34%$ She indeed floats more in seawater. 86. A certain man has a mass of 80 kg and a density of $955kg/m^3$ (excluding the air in his lungs). (a) Calculate his volume. (b) Find the buoyant force air exerts on him. (c) What is the ratio of the buoyant force to his weight? 87. A simple compass can be made by placing a small bar magnet on a cork floating in water. (a) What fraction of a plain cork will be submerged when floating in water? (b) If the cork has a mass of 10.0 g and a 20.0-g magnet is placed on it, what fraction of the cork will be submerged? (c) Will the bar magnet and cork float in ethyl alcohol? Solution (a) $0.24$ (b) $0.68$ (c) Yes, the cork will float because $ρ_{obj}<ρ_{\text{ethyl alcohol}}(0.678g/cm^3<0.79g/cm^3)$ 88. What fraction of an iron anchor’s weight will be supported by buoyant force when submerged in saltwater? 89. Scurrilous con artists have been known to represent gold-plated tungsten ingots as pure gold and sell them to the greedy at prices much below gold value but deservedly far above the cost of tungsten. With what accuracy must you be able to measure the mass of such an ingot in and out of water to tell that it is almost pure tungsten rather than pure gold? Solution The difference is 0.006%. 90 . A twin-sized air mattress used for camping has dimensions of 100 cm by 200 cm by 15 cm when blown up. The weight of the mattress is 2 kg. How heavy a person could the air mattress hold if it is placed in freshwater? 91. Referring to Figure, prove that the buoyant force on the cylinder is equal to the weight of the fluid displaced (Archimedes’ principle). You may assume that the buoyant force is $F_2−F_1$ and that the ends of the cylinder have equal areas $A$. Note that the volume of the cylinder (and that of the fluid it displaces) equals $(h_2−h_1)A$. Solution $F_{net}=F_2−F_1=P_2A−P_1A=(P_2−P_1)A$ $=(h_2ρ_{fl}g−h_1ρ_{fl}g)A$ $=(h_2−h_1)ρ_{fl}gA$ where $ρ_{fl}$ = density of fluid. Therefore, $F_{net}=(h_2−h_1)Aρ_{fl}g=V_{fl}ρ_{fl}g=m_{fl}g=w_{fl}$ where is $w_{fl}$ the weight of the fluid displaced. 92. (a) A 75.0-kg man floats in freshwater with $3.00%$ of his volume above water when his lungs are empty, and $5.00%$ of his volume above water when his lungs are full. Calculate the volume of air he inhales—called his lung capacity—in liters. (b) Does this lung volume seem reasonable? 11.8: Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action 93. What is the pressure inside an alveolus having a radius of $2.50×10^{−4}m$ if the surface tension of the fluid-lined wall is the same as for soapy water? You may assume the pressure is the same as that created by a spherical bubble. Solution $592N/m^2$ 94. (a) The pressure inside an alveolus with a $2.00×10^{−4}$-m radius is $1.40×10^3Pa$, due to its fluid-lined walls. Assuming the alveolus acts like a spherical bubble, what is the surface tension of the fluid? (b) Identify the likely fluid. (You may need to extrapolate between values in Table.) 95. What is the gauge pressure in millimeters of mercury inside a soap bubble 0.100 m in diameter? Solution $2.23×10^{−2}mm Hg$ 96. Calculate the force on the slide wire in Figure if it is 3.50 cm long and the fluid is ethyl alcohol. 97. Figure(a) shows the effect of tube radius on the height to which capillary action can raise a fluid. (a) Calculate the height $h$ for water in a glass tube with a radius of 0.900 cm—a rather large tube like the one on the left. (b) What is the radius of the glass tube on the right if it raises water to 4.00 cm? Solution (a) $1.65×10^{−3}m$ (b) $3.71×10^{–4}m$ 98. We stated in Example that a xylem tube is of radius $2.50×10^{−5}m$. Verify that such a tube raises sap less than a meter by finding h for it, making the same assumptions that sap’s density is $1050kg/m^3$, its contact angle is zero, and its surface tension is the same as that of water at $20.0º C$. 99. What fluid is in the device shown in Figure if the force is $3.16×10^{−3}N$ and the length of the wire is 2.50 cm? Calculate the surface tension $γ$ and find a likely match from Table. Solution $6.32×10^{−2}N/m$ Based on the values in table, the fluid is probably glycerin. 100. If the gauge pressure inside a rubber balloon with a 10.0-cm radius is 1.50 cm of water, what is the effective surface tension of the balloon? 101. Calculate the gauge pressures inside 2.00-cm-radius bubbles of water, alcohol, and soapy water. Which liquid forms the most stable bubbles, neglecting any effects of evaporation? Solution $P_w=14.6N/m^2,$ $P_a=4.46N/m^2,$ $P_{sw}=7.40N/m^2.$ Alcohol forms the most stable bubble, since the absolute pressure inside is closest to atmospheric pressure. 102. Suppose water is raised by capillary action to a height of 5.00 cm in a glass tube. (a) To what height will it be raised in a paraffin tube of the same radius? (b) In a silver tube of the same radius? 103. Calculate the contact angle $θ$ for olive oil if capillary action raises it to a height of 7.07 cm in a glass tube with a radius of 0.100 mm. Is this value consistent with that for most organic liquids? Solution $5.1º$ This is near the value of $θ=0º$ for most organic liquids. 104. When two soap bubbles touch, the larger is inflated by the smaller until they form a single bubble. (a) What is the gauge pressure inside a soap bubble with a 1.50-cm radius? (b) Inside a 4.00-cm-radius soap bubble? (c) Inside the single bubble they form if no air is lost when they touch? 105. Calculate the ratio of the heights to which water and mercury are raised by capillary action in the same glass tube. Solution $−2.78$ The ratio is negative because water is raised whereas mercury is lowered. 106. What is the ratio of heights to which ethyl alcohol and water are raised by capillary action in the same glass tube? 11.9: Pressures in the Body 107. During forced exhalation, such as when blowing up a balloon, the diaphragm and chest muscles create a pressure of 60.0 mm Hg between the lungs and chest wall. What force in newtons does this pressure create on the $600cm^2$ surface area of the diaphragm? Solurion 479 N 108. You can chew through very tough objects with your incisors because they exert a large force on the small area of a pointed tooth. What pressure in pascals can you create by exerting a force of $500 N$ with your tooth on an area of $1.00mm^2$? 109. One way to force air into an unconscious person’s lungs is to squeeze on a balloon appropriately connected to the subject. What force must you exert on the balloon with your hands to create a gauge pressure of 4.00 cm water, assuming you squeeze on an effective area of $50.0cm^2$? Solution 1.96 N 110. Heroes in movies hide beneath water and breathe through a hollow reed (villains never catch on to this trick). In practice, you cannot inhale in this manner if your lungs are more than 60.0 cm below the surface. What is the maximum negative gauge pressure you can create in your lungs on dry land, assuming you can achieve $−3.00 cm$ water pressure with your lungs 60.0 cm below the surface? Solution $−63.0 cm$ $H_2O$ 111. Gauge pressure in the fluid surrounding an infant’s brain may rise as high as 85.0 mm Hg (5 to 12 mm Hg is normal), creating an outward force large enough to make the skull grow abnormally large. (a) Calculate this outward force in newtons on each side of an infant’s skull if the effective area of each side is $70.0cm^2$. (b) What is the net force acting on the skull? 112. A full-term fetus typically has a mass of 3.50 kg. (a) What pressure does the weight of such a fetus create if it rests on the mother’s bladder, supported on an area of $90.0cm^2$? (b) Convert this pressure to millimeters of mercury and determine if it alone is great enough to trigger the micturition reflex (it will add to any pressure already existing in the bladder). Solution (a) $3.81×10^3N/m^2$ (b) $28.7 mm Hg$, which is sufficient to trigger micturition reflex 113. If the pressure in the esophagus is $−2.00 mm Hg$ while that in the stomach is $+20.0 mm Hg$, to what height could stomach fluid rise in the esophagus, assuming a density of 1.10 g/mL? (This movement will not occur if the muscle closing the lower end of the esophagus is working properly.) 114. Pressure in the spinal fluid is measured as shown in Figure. If the pressure in the spinal fluid is 10.0 mm Hg: (a) What is the reading of the water manometer in cm water? (b) What is the reading if the person sits up, placing the top of the fluid 60 cm above the tap? The fluid density is 1.05 g/mL. A water manometer used to measure pressure in the spinal fluid. The height of the fluid in the manometer is measured relative to the spinal column, and the manometer is open to the atmosphere. The measured pressure will be considerably greater if the person sits up. Solution (a) 13.6 m water (b) 76.5 cm water 115. Calculate the maximum force in newtons exerted by the blood on an aneurysm, or ballooning, in a major artery, given the maximum blood pressure for this person is 150 mm Hg and the effective area of the aneurysm is $20.0cm^2$. Note that this force is great enough to cause further enlargement and subsequently greater force on the ever-thinner vessel wall. 116. During heavy lifting, a disk between spinal vertebrae is subjected to a 5000-N compressional force. (a) What pressure is created, assuming that the disk has a uniform circular cross section 2.00 cm in radius? (b) What deformation is produced if the disk is 0.800 cm thick and has a Young’s modulus of $1.5×10^9N/m^2$ ? Solution (a) $3.98×10^6Pa$ (b) $2.1×10^{−3}cm$ 117. When a person sits erect, increasing the vertical position of their brain by 36.0 cm, the heart must continue to pump blood to the brain at the same rate. (a) What is the gain in gravitational potential energy for 100 mL of blood raised 36.0 cm? (b) What is the drop in pressure, neglecting any losses due to friction? (c) Discuss how the gain in gravitational potential energy and the decrease in pressure are related. 118. (a) How high will water rise in a glass capillary tube with a 0.500-mm radius? (b) How much gravitational potential energy does the water gain? (c) Discuss possible sources of this energy. Solution (a) 2.97 cm (b) $3.39×10^{−6}J$ (c) Work is done by the surface tension force through an effective distance $h/2$ to raise the column of water. 119. A negative pressure of 25.0 atm can sometimes be achieved with the device in Figure before the water separates. (a) To what height could such a negative gauge pressure raise water? (b) How much would a steel wire of the same diameter and length as this capillary stretch if suspended from above? (a) When the piston is raised, it stretches the liquid slightly, putting it under tension and creating a negative absolute pressure $P=−F/A$ (b) The liquid eventually separates, giving an experimental limit to negative pressure in this liquid. 120. Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at $15.0 m/s$ size 12{"15" "." 0`"m/s"} {} and brought to rest in 2.80 mm. (a) What average force is exerted on the nail? (b) How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long? (c) What pressure is created on the 1.00-mm-diameter tip of the nail? Solution (a) $2.01×10^4N$ (b) $1.17×10^{−3}m$ (c) $2.56×10^{10}N/m^2$ 121. Calculate the pressure due to the ocean at the bottom of the Marianas Trench near the Philippines, given its depth is $11.0 km$ and assuming the density of sea water is constant all the way down. (b) Calculate the percent decrease in volume of sea water due to such a pressure, assuming its bulk modulus is the same as water and is constant. (c) What would be the percent increase in its density? Is the assumption of constant density valid? Will the actual pressure be greater or smaller than that calculated under this assumption? 122. The hydraulic system of a backhoe is used to lift a load as shown in Figure. (a) Calculate the force $F$ the slave cylinder must exert to support the 400-kg load and the 150-kg brace and shovel. (b) What is the pressure in the hydraulic fluid if the slave cylinder is 2.50 cm in diameter? (c) What force would you have to exert on a lever with a mechanical advantage of 5.00 acting on a master cylinder 0.800 cm in diameter to create this pressure? Hydraulic and mechanical lever systems are used in heavy machinery such as this back hoe. Solution (a) $1.38×10^4N$ (b) $2.81×10^7N/m^2$ (c) 283 N 123. Some miners wish to remove water from a mine shaft. A pipe is lowered to the water 90 m below, and a negative pressure is applied to raise the water. (a) Calculate the pressure needed to raise the water. (b) What is unreasonable about this pressure? (c) What is unreasonable about the premise? 124. You are pumping up a bicycle tire with a hand pump, the piston of which has a 2.00-cm radius. (a) What force in newtons must you exert to create a pressure of $6.90×10^5Pa$ (b) What is unreasonable about this (a) result? (c) Which premises are unreasonable or inconsistent? Solution (a) 867 N (b) This is too much force to exert with a hand pump. (c) The assumed radius of the pump is too large; it would be nearly two inches in diameter—too large for a pump or even a master cylinder. The pressure is reasonable for bicycle tires. 125. Consider a group of people trying to stay afloat after their boat strikes a log in a lake. Construct a problem in which you calculate the number of people that can cling to the log and keep their heads out of the water. Among the variables to be considered are the size and density of the log, and what is needed to keep a person’s head and arms above water without swimming or treading water. 126. The alveoli in emphysema victims are damaged and effectively form larger sacs. Construct a problem in which you calculate the loss of pressure due to surface tension in the alveoli because of their larger average diameters. (Part of the lung’s ability to expel air results from pressure created by surface tension in the alveoli.) Among the things to consider are the normal surface tension of the fluid lining the alveoli, the average alveolar radius in normal individuals and its average in emphysema sufferers. Contributors and Attributions - Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .	libretexts	2025-03-17T19:53:30.540717	2017-04-20T14:40:23	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/11%3A_Fluid_Statics/11.E%3A_Fluid_Statics_(Exercises)", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "11.E: Fluid Statics (Exercises)", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications	12: Fluid Dynamics and Its Biological and Medical Applications We have dealt with many situations in which fluids are static, but by their very definition, fluids flow. Examples come easily—a column of smoke rises from a camp fire, water streams from a fire hose, blood courses through your veins. Why does rising smoke curl and twist? How does a nozzle increase the speed of water emerging from a hose? How does the body regulate blood flow? The physics of fluids in motion—fluid dynamics—allows us to answer these and many other questions.	libretexts	2025-03-17T19:53:30.599968	2015-10-27T19:34:55	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "12: Fluid Dynamics and Its Biological and Medical Applications", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.00%3A_Prelude_to_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications	12.0: Prelude to Fluid Dynamics and Its Biological and Medical Applications We have dealt with many situations in which fluids are static. But by their very definition, fluids flow. Examples come easily—a column of smoke rises from a camp fire, water streams from a fire hose, blood courses through your veins. Why does rising smoke curl and twist? How does a nozzle increase the speed of water emerging from a hose? How does the body regulate blood flow? The physics of fluids in motion— fluid dynamics —allows us to answer these and many other questions. Glossary - fluid dynamics - the physics of fluids in motion	libretexts	2025-03-17T19:53:30.656829	2015-11-01T05:34:25	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.00%3A_Prelude_to_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "12.0: Prelude to Fluid Dynamics and Its Biological and Medical Applications", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.01%3A_Flow_Rate_and_Its_Relation_to_Velocity	12.1: Flow Rate and Its Relation to Velocity Learning Objectives By the end of this section, you will be able to: - Calculate flow rate. - Define units of volume. - Describe incompressible fluids. - Explain the consequences of the equation of continuity. Flow rate $Q$ is defined to be the volume of fluid passing by some location through an area during a period of time, as seen in Figure $\PageIndex{1}$. In symbols, this can be written as \[Q = \dfrac{V}{t},\] where $V$ is the volume ant $t$ is the elapsed time. The SI unit for flow rate is $m^3/s$, but a number of other units for $Q$ are in common use. For example, the heart of a resting adult pumps blood at a rate of 5.00 liters per minute (L/min). Note that a liter (L) is 1/1000 of a cubic meter or 1000 cubic centimeters ($10^{-3} \, m^3$ or $10^3 \, cm$). In this text we shall use whatever metric units are most convenient for a given situation. Example $\PageIndex{1}$: Calculating Volume from Flow Rate: The Heart Pumps a Lot of Blood in a Lifetime How many cubic meters of blood does the heart pump in a 75-year lifetime, assuming the average flow rate is 5.00 L/min? Strategy Time and flow rate $Q$ are given, and so the volume $V$ can be calculated from the definition of flow rate. Solution Solving $Q = V/t$ for volume gives \[V = Qt. \nonumber \] Substituting known values yields \[\begin{align}V &= \left(\dfrac{5.00 \, L}{1 \, min}\right)(75 \, y)\left(\dfrac{1 \, m^3}{10^3 \, L}\right)(5.26 \times 10^5 \, \left(\dfrac{min}{y}\right) \\[5pt] &= 2.0 \times 10^5 \, m^3 \end{align}\] Discussion This amount is about 200,000 tons of blood. For comparison, this value is equivalent to about 200 times the volume of water contained in a 6-lane 50-m lap pool. Flow rate and velocity are related, but quite different, physical quantities. To make the distinction clear, think about the flow rate of a river. The greater the velocity of the water, the greater the flow rate of the river. But flow rate also depends on the size of the river. A rapid mountain stream carries far less water than the Amazon River in Brazil, for example. The precise relationship between flow rate $Q$ and velocity $\overline{v}$ is \[Q = A \overline{v},\] where $A$ is the cross-sectional area and $ \overline{v}$ is the average velocity. This equation seems logical enough. The relationship tells us that flow rate is directly proportional to both the magnitude of the average velocity (hereafter referred to as the speed) and the size of a river, pipe, or other conduit. The larger the conduit, the greater its cross-sectional area. Figure $\PageIndex{1}$ illustrates how this relationship is obtained. The shaded cylinder has a volume \[V = Ad,\] which flows past the point $P$ in a time $t$. Dividing both sides of this relationship by $t$ gives \[\dfrac{V}{t} = \dfrac{Ad}{t}.\] We note that $Q = V\t$ and the average speed is $\overline{v} = d/t$. Thus the equation becomes $Q = A\overline{v}$. Figure $\PageIndex{2}$ shows an incompressible fluid flowing along a pipe of decreasing radius. Because the fluid is incompressible, the same amount of fluid must flow past any point in the tube in a given time to ensure continuity of flow. In this case, because the cross-sectional area of the pipe decreases, the velocity must necessarily increase. This logic can be extended to say that the flow rate must be the same at all points along the pipe. In particular, for points 1 and 2, \[Q_1 = Q_2\] \[A_1\overline{v}_1 = A_2\overline{v}_2\] This is called the equation of continuity and is valid for any incompressible fluid. The consequences of the equation of continuity can be observed when water flows from a hose into a narrow spray nozzle: it emerges with a large speed—that is the purpose of the nozzle. Conversely, when a river empties into one end of a reservoir, the water slows considerably, perhaps picking up speed again when it leaves the other end of the reservoir. In other words, speed increases when cross-sectional area decreases, and speed decreases when cross-sectional area increases. Since liquids are essentially incompressible, the equation of continuity is valid for all liquids. However, gases are compressible, and so the equation must be applied with caution to gases if they are subjected to compression or expansion. Example $\PageIndex{2}$: Calculating Fluid Speed: Speed Increases When a Tube Narrows A nozzle with a radius of 0.250 cm is attached to a garden hose with a radius of 0.900 cm. The flow rate through hose and nozzle is 0.500 L/s. Calculate the speed of the water (a) in the hose and (b) in the nozzle. Strategy We can use the relationship between flow rate and speed to find both velocities. We will use the subscript 1 for the hose and 2 for the nozzle. Solution for (a) First, we solve $Q = A\overline{v}$ for $v_1$ and note that the cross-sectional area is $A = \pi r^2$, yielding \[\overline{v}_1 = \dfrac{Q}{A_1} = \dfrac{Q}{\pi r_1^2}, \nonumber \] Substituting known values and making appropriate unit conversions yields \[\overline{v}_1 = \dfrac {(0.500 \, L/s)(10^{-3} m^3/L)}{\pi (9.00 \times 10^{-3}m)^2} = 1.96 \, m/s. \nonumber \] Solution for (b) We could repeat this calculation to find the speed in the nozzle $\overline{v}_2$, but we will use the equation of continuity to give a somewhat different insight. Using the equation which states \[A_1\overline{v}_1 = A_2\overline{v}_2 \nonumber \] solving for $\overline{v}_2$ and substituting $\pi r^2$ for the cross-sectional area yields \[\overline{v}_2 = \dfrac{A_1}{A_2}\overline{v}_1 = \dfrac{\pi r_1^2}{\pi r_2^2}\overline{v}_1 = \dfrac{r_1^2}{r_2^2}\overline{v}_1. \nonumber \] Substituting known values, \[\overline{v}_2 = \dfrac{ (0.900 \, cm)^2}{(0.250 \, cm)^2} 1.96 \, m/s = 25.5 \, m/s. \nonumber \] Discussion A speed of 1.96 m/s is about right for water emerging from a nozzleless hose. The nozzle produces a considerably faster stream merely by constricting the flow to a narrower tube. The solution to the last part of the example shows that speed is inversely proportional to the square of the radius of the tube, making for large effects when radius varies. We can blow out a candle at quite a distance, for example, by pursing our lips, whereas blowing on a candle with our mouth wide open is quite ineffective. In many situations, including in the cardiovascular system, branching of the flow occurs. The blood is pumped from the heart into arteries that subdivide into smaller arteries (arterioles) which branch into very fine vessels called capillaries. In this situation, continuity of flow is maintained but it is the sum of the flow rates in each of the branches in any portion along the tube that is maintained. The equation of continuity in a more general form becomes \[n_1A_1\overline{v}_1 = n_2A_2\overline{v}_2,\] where $n_1$ and $n_2$ are the number of branches in each of the sections along the tube. Example $\PageIndex{3}$: Calculating Flow Speed and Vessel Diameter: Branching in the Cardiovascular System The aorta is the principal blood vessel through which blood leaves the heart in order to circulate around the body. (a) Calculate the average speed of the blood in the aorta if the flow rate is 5.0 L/min. The aorta has a radius of 10 mm. (b) Blood also flows through smaller blood vessels known as capillaries. When the rate of blood flow in the aorta is 5.0 L/min, the speed of blood in the capillaries is about 0.33 mm/s. Given that the average diameter of a capillary is $8.0 \, \mu m$, calculate the number of capillaries in the blood circulatory system. Strategy We can use $Q = A\overline{v}$ to calculate the speed of flow in the aorta and then use the general form of the equation of continuity to calculate the number of capillaries as all of the other variables are known. Solution for (a) The flow rate is given by $Q = A\overline{v}$ or $\overline{v} = \frac{Q}{\pi r^2} $ for a cylindrical vessel. Substituting the known values (converted to units of meters and seconds) gives \[\begin{align} \overline{v} &= \dfrac{(5.0 \, L/min)(10^{-3} \, m^3/L)(1 \, min/60 \, s)}{\pi(0.010 \, m)^2} \\[5pt] &= 0.27 \, m/s. \end{align}\] Solution for (b) Using $n_1A_1\overline{v}_1 = n_2A_2\overline{v}_1$, assigning the subscript 1 to the aorta and 2 to the capillaries, and solving for $n_2$ (the number of capillaries) gives $n_2 = \frac{n_1A_1\overline{v}_1}{A-2\overline{v}_2}.$ Converting all quantities to units of meters and seconds and substituting into the equation above gives \[\begin{align} n_2 &= \dfrac{(1)(\pi)(10 \times 10^{-3} m)^2(0.27 \, m/s)}{(\pi)(4.0 \times 10^{-6} m)(0.33 \times 10^{-3} m/s)} \\[5pt] &= 5.0 \times 10^9 \, capillaries.\end{align}\] Discussion Note that the speed of flow in the capillaries is considerably reduced relative to the speed in the aorta due to the significant increase in the total cross-sectional area at the capillaries. This low speed is to allow sufficient time for effective exchange to occur although it is equally important for the flow not to become stationary in order to avoid the possibility of clotting. Does this large number of capillaries in the body seem reasonable? In active muscle, one finds about 200 capillaries per $mm^3$, or about $200 \times 10^6$ per 1 kg of muscle. For 20 kg of muscle, this amounts to about $4 \times 10^9$ capillaries. × 10 9 Summary - Flow rate $Q$ is defined to be the volume $V$ flowing past a point in time $t$, or $Q = \frac{V}{t}$ where $V$ is volume and $t$ is time. - The SI unit of volume is $m^3$. - Another common unit is the liter (L), which is $10^{-3}m^3$ - Flow rate and velocity are related by $Q = A\overline{v}$ where $A$ is the cross-sectional area of the flow and $v$ is its average velocity. - For incompressible fluids, flow rate at various points is constant. That is, \[Q_1 = Q_2\] \[A_1\overline{v}_1 = A_2\overline{v}_2\] \[n_1A_1\overline{v}_1 = n_2A_2\overline{v}_2\] Glossary - flow rate - abbreviated Q , it is the volume V that flows past a particular point during a time t , or Q = V/t - liter - a unit of volume, equal to 10 −3 m 3	libretexts	2025-03-17T19:53:30.737233	2015-11-01T05:34:42	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.01%3A_Flow_Rate_and_Its_Relation_to_Velocity", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "12.1: Flow Rate and Its Relation to Velocity", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.02%3A_Bernoullis_Equation	12.2: Bernoulli’s Equation Learning Objectives By the end of this section, you will be able to: - Explain the terms in Bernoulli’s equation. - Explain how Bernoulli’s equation is related to conservation of energy. - Explain how to derive Bernoulli’s principle from Bernoulli’s equation. - Calculate with Bernoulli’s principle. - List some applications of Bernoulli’s principle. When a fluid flows into a narrower channel, its speed increases. That means its kinetic energy also increases. Where does that change in kinetic energy come from? The increased kinetic energy comes from the net work done on the fluid to push it into the channel and the work done on the fluid by the gravitational force, if the fluid changes vertical position. Recall the work-energy theorem, \[W_{net} = \dfrac{1}{2} mv^2 - \dfrac{1}{2}mv_0^2.\] There is a pressure difference when the channel narrows. This pressure difference results in a net force on the fluid: recall that pressure times area equals force. The net work done increases the fluid’s kinetic energy. As a result, the pressure will drop in a rapidly-moving fluid , whether or not the fluid is confined to a tube. There are a number of common examples of pressure dropping in rapidly-moving fluids. Shower curtains have a disagreeable habit of bulging into the shower stall when the shower is on. The high-velocity stream of water and air creates a region of lower pressure inside the shower, and standard atmospheric pressure on the other side. The pressure difference results in a net force inward pushing the curtain in. You may also have noticed that when passing a truck on the highway, your car tends to veer toward it. The reason is the same—the high velocity of the air between the car and the truck creates a region of lower pressure, and the vehicles are pushed together by greater pressure on the outside (Figure $\PageIndex{1}$) This effect was observed as far back as the mid-1800s, when it was found that trains passing in opposite directions tipped precariously toward one another. Making Connections: Take Home Investigation with a Sheet of Paper: Hold the short edge of a sheet of paper parallel to your mouth with one hand on each side of your mouth. The page should slant downward over your hands. Blow over the top of the page. Describe what happens and explain the reason for this behavior. Bernoulli’s Equation The relationship between pressure and velocity in fluids is described quantitatively by Bernoulli’s equation , named after its discoverer, the Swiss scientist Daniel Bernoulli (1700–1782). Bernoulli’s equation states that for an incompressible, frictionless fluid, the following sum is constant: \[P + \dfrac{1}{2}\rho v^2 + \rho gh = constant \label{eq1}\] where $P$ is the absolute pressure, $\rho$ is the fluid density, $v$ is the velocity of the fluid, $h$ is the height above some reference point, and $g$ is the acceleration due to gravity. If we follow a small volume of fluid along its path, various quantities in the sum may change, but the total remains constant. Let the subscripts 1 and 2 refer to any two points along the path that the bit of fluid follows; Bernoulli’s equation becomes: \[P_1 + \underbrace{\dfrac{1}{2}\rho v_1^2}_{\text{kinetic energy}} + \underbrace{\rho gh_1}_{\text{potential energy}} = P_2 + \underbrace{\dfrac{1}{2}\rho v_2^2}_{\text{kinetic energy}} + \underbrace{\rho gh_2. \label{eq2}}_{\text{potential energy}} \] Bernoulli’s equation is a form of the conservation of energy principle. Note that the second and third terms are the kinetic and potential energy with $m$ replaced by $\rho$. In fact, each term in the equation has units of energy per unit volume. We can prove this for the second term by substituting $\rho = m/V$ into it and gathering terms: \[\dfrac{1}{2}\rho v^2 = \dfrac{\frac{1}{2}mv^2}{V} = \dfrac{KE}{V}. \label{eq3}\] So $\frac{1}{2}\rho v^2$ is the kinetic energy per unit volume. Making the same substitution into the third term of Equation \ref{eq3}, we find \[\rho gh = \dfrac{mgh}{V} = \dfrac{PE_g}{V}, \label{eq4}\] so $\rho gh$ is the gravitational potential energy per unit volume. Note that pressure $P$ has units of energy per unit volume, too. Since $P = F/A$, its units are $N/m^2$. If we multiply these by m/m, we obtain $N \cdot m/m^3 = J/m^3$, or energy per unit volume. Bernoulli’s equation is, in fact, just a convenient statement of conservation of energy for an incompressible fluid in the absence of friction . Making Connections: Conservation of Energy Conservation of energy applied to fluid flow produces Bernoulli’s equation. The net work done by the fluid’s pressure results in changes in the fluid’s $KE$ and $PE_g$ per unit volume. If other forms of energy are involved in fluid flow, Bernoulli’s equation can be modified to take these forms into account. Such forms of energy include thermal energy dissipated because of fluid viscosity. The general form of Bernoulli’s equation has three terms in it (Equation \ref{eq1}), and it is broadly applicable. To understand it better, we will look at a number of specific situations that simplify and illustrate its use and meaning. Bernoulli’s Equation for Static Fluids Let us first consider the very simple situation where the fluid is static—that is, $v_1 = v_2 = 0$. Bernoulli’s equation in that case is \[P_1 + \rho gh_1 = P_2 + \rho gh_2.\] We can further simplify the equation by taking $h_2 = 0$ (we can always choose some height to be zero, just as we often have done for other situations involving the gravitational force, and take all other heights to be relative to this). In that case, we get \[P_2 = P_1 + \rho gh_1.\] This equation tells us that, in static fluids, pressure increases with depth. As we go from point 1 to point 2 in the fluid, the depth increases by $h_1$, and consequently, $P_2$ is greater than $P_1$ by an amount $\rho gh_1$. In the very simplest case, $P_1$ is zero at the top of the fluid, and we get the familiar relationship $P = \rho gh$. (Recall that $P = \rho gh$ and $\Delta PE_g = mgh$.) Bernoulli’s equation includes the fact that the pressure due to the weight of a fluid is $\rho gh$. Although we introduce Bernoulli’s equation for fluid flow, it includes much of what we studied for static fluids in the preceding chapter. Bernoulli’s Principle—Bernoulli’s Equation at Constant Depth Another important situation is one in which the fluid moves but its depth is constant—that is, $h_1 = h_2$. Under that condition, Bernoulli’s equation becomes \[P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2.\] Situations in which fluid flows at a constant depth are so important that this equation is often called Bernoulli’s principle . It is Bernoulli’s equation for fluids at constant depth. (Note again that this applies to a small volume of fluid as we follow it along its path.) As we have just discussed, pressure drops as speed increases in a moving fluid. We can see this from Bernoulli’s principle. For example, if $v_2$ is greater than $v_1$ in the equation, then $P_2$ must be less than $P_1$ for the equality to hold. Example $\PageIndex{1}$: Calculating Pressure: Pressure Drops as a Fluid Speeds Up Previously , we found that the speed of water in a hose increased from 1.96 m/s to 25.5 m/s going from the hose to the nozzle. Calculate the pressure in the hose, given that the absolute pressure in the nozzle is $1.01 \times 10^5 \, N/m^2$ (atmospheric, as it must be) and assuming level, frictionless flow. Strategy Level flow means constant depth, so Bernoulli’s principle applies. We use the subscript 1 for values in the hose and 2 for those in the nozzle. We are thus asked to find $P_1$. Solution Solving Bernoulli’s principle for $P_1$ yields \[\begin{align} P_1 &= P_2 + \dfrac{1}{2}\rho v_2^2 - \dfrac{1}{2}\rho v_1^2 \\[5pt] &= \dfrac{1}{2}\rho(v_2^2 - v_1^2).\end{align}\] Substituting known values, \[\begin{align} P_1 &= 1.01 \times 10^5 \, N/m^2 + \dfrac{1}{2}(10^3 \, kg/m^3)[(25.5 m/s)^2 - (1.96 m/s)^2] \\[5pt] &= 4.24 \times 10^5 \, N/m^2.\end{align}\] Discussion This absolute pressure in the hose is greater than in the nozzle, as expected since $v$ is greater in the nozzle. The pressure $P_2$ in the nozzle must be atmospheric since it emerges into the atmosphere without other changes in conditions. Applications of Bernoulli’s Principle There are a number of devices and situations in which fluid flows at a constant height and, thus, can be analyzed with Bernoulli’s principle. Application: Entrainment People have long put the Bernoulli principle to work by using reduced pressure in high-velocity fluids to move things about. With a higher pressure on the outside, the high-velocity fluid forces other fluids into the stream. This process is called entrainment . Entrainment devices have been in use since ancient times, particularly as pumps to raise water small heights, as in draining swamps, fields, or other low-lying areas. Some other devices that use the concept of entrainment are shown in Figure $\PageIndex{2}$. Application: Wings and Sails The airplane wing is a beautiful example of Bernoulli’s principle in action. Figure $\PageIndex{1a}$ shows the characteristic shape of a wing. The wing is tilted upward at a small angle and the upper surface is longer, causing air to flow faster over it. The pressure on top of the wing is therefore reduced, creating a net upward force or lift. (Wings can also gain lift by pushing air downward, utilizing the conservation of momentum principle. The deflected air molecules result in an upward force on the wing — Newton’s third law.) Sails also have the characteristic shape of a wing. (See Figure $\PageIndex{1b}$.) The pressure on the front side of the sail, $P_{front}$ is lower than the pressure on the back of the sail, $P_{back}$. This results in a forward force and even allows you to sail into the wind. Making Connections: Take Home Investigation with Two Strips of Paper For a good illustration of Bernoulli’s principle, make two strips of paper, each about 15 cm long and 4 cm wide. Hold the small end of one strip up to your lips and let it drape over your finger. Blow across the paper. What happens? Now hold two strips of paper up to your lips, separated by your fingers. Blow between the strips. What happens? Application: Velocity Measurement Figure $\PageIndex{4}$ shows two devices that measure fluid velocity based on Bernoulli’s principle. The manometer in Figure $\PageIndex{1a}$ is connected to two tubes that are small enough not to appreciably disturb the flow. The tube facing the oncoming fluid creates a dead spot having zero velocity ($v_1 = 0$) in front of it, while fluid passing the other tube has velocity $v_2$. This means that Bernoulli’s principle as stated in \[P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2\] becomes \[P_1 = P_2 + \dfrac{1}{2}\rho v_2^2.\] Thus pressure $P_2$ over the second opening is reduced by $\frac{1}{2}\rho v_2^2$, and so the fluid in the manometer rises by $h$ on the side connected to the second opening, where \[h \propto \dfrac{1}{2} \rho v_2^2.\] (Recall that the symbol $\propto$ means “proportional to.”) Solving for $v_2$, we see that \[v_2 \propto \sqrt{h}.\] Figure $\PageIndex{1b}$ shows a version of this device that is in common use for measuring various fluid velocities; such devices are frequently used as air speed indicators in aircraft. Summary - Bernoulli’s equation states that the sum on each side of the following equation is constant, or the same at any two points in an incompressible frictionless fluid: \[P_1 + \dfrac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \dfrac{1}{2}\rho v_2^2 + \rho gh_2.\] - Bernoulli’s principle is Bernoulli’s equation applied to situations in which depth is constant. The terms involving depth (or height h ) subtract out, yielding \[P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2.\] - Bernoulli’s principle has many applications, including entrainment, wings and sails, and velocity measurement. Glossary - Bernoulli’s equation - the equation resulting from applying conservation of energy to an incompressible frictionless fluid: P + 1/2 pv 2 + pgh = constant , through the fluid - Bernoulli’s principle - Bernoulli’s equation applied at constant depth: P 1 + 1/2 pv 1 2 = P 2 + 1/2 pv 2 2	libretexts	2025-03-17T19:53:30.821978	2015-11-01T05:35:02	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.02%3A_Bernoullis_Equation", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "12.2: Bernoulli’s Equation", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.03%3A_The_Most_General_Applications_of_Bernoullis_Equation	12.3: The Most General Applications of Bernoulli’s Equation Learning Objectives By the end of this section, you will be able to: - Calculate using Torricelli’s theorem. - Calculate power in fluid flow. Torricelli’s Theorem Figure $\PageIndex{1}$ shows water gushing from a large tube through a dam. What is its speed as it emerges? Interestingly, if resistance is negligible, the speed is just what it would be if the water fell a distance $h$ from the surface of the reservoir; the water’s speed is independent of the size of the opening. Let us check this out. Bernoulli’s equation must be used since the depth is not constant. We consider water flowing from the surface (point 1) to the tube’s outlet (point 2). Bernoulli’s equation as stated in previously is \[P_1 + \dfrac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \dfrac{1}{2}\rho v_2^2 + \rho gh_2.\] Both $P_1$ and $P_2$ equal atmospheric pressure ($P_1$ is atmospheric pressure because it is the pressure at the top of the reservoir. $P_2$ must be atmospheric pressure, since the emerging water is surrounded by the atmosphere and cannot have a pressure different from atmospheric pressure.) and subtract out of the equation, leaving \[\dfrac{1}{2}\rho v_1^2 + \rho gh_1 = \dfrac{1}{2}\rho v_2^2 + \rho gh_2.\] Solving this equation for $v_2^2$ noting that the density $\rho$ cancels (because the fluid is incompressible), yields \[v_2^2 = v_1^2 + 2g(h_1 - h_2).\] We let $h = h_1 - h_2$, the equation then becomes \[v_2^2 = v_1^2 + 2gh\] where $h$ is the height dropped by the water. This is simply a kinematic equation for any object falling a distance $h$ with negligible resistance. In fluids, this last equation is called Torricelli’s theorem . Note that the result is independent of the velocity’s direction, just as we found when applying conservation of energy to falling objects. All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change. (See Figure .) Example $\PageIndex{1}$: Calculating Pressure: A Fire Hose Nozzle Fire hoses used in major structure fires have inside diameters of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of $ 1.62 \times 10^6 \, N/m^2$. The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Assuming negligible resistance, what is the pressure in the nozzle? Strategy Here we must use Bernoulli’s equation to solve for the pressure, since depth is not constant. Solution Bernoulli’s equation states \[P_1 + \dfrac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \dfrac{1}{2}\rho v_2^2 + \rho gh_2,\nonumber\] where the subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds $v_1$ and $v_2$. Since $Q = A_1v_1$, we get \[\begin{align}v_1 &= \dfrac{Q}{A_1} \\[5pt] &= \dfrac{40.0 \times 10^{-3} m^3/s}{\pi(3.20 \times 10^{-2} m)^2} \\[5pt] &= 12.4 \, m/s. \end{align}\] Similarly, we find \[v_2 = 56.6 \, m/s.\nonumber\] (This rather large speed is helpful in reaching the fire.) Now, taking $h_1$ to be zero, we solve Bernoulli’s equation for $P_2$: \[P_2 = P_1 + \dfrac{1}{2}\rho(v_1^2 - v_2^2) - \rho gh_2. \nonumber\] Substituting known values yields \[\begin{align} P_2 &= 1.62 \times 10^6 N/m^2 + \dfrac{1}{2}(1000 \, kg/m^3)[(12.4 \, m/s)^2 - (56.6 \, m/s)^2] - (1000 \, kg/m^3)(9.80 m/s^2)(10.0 \, m) \\[5pt] &= 0 \end{align}\] Discussion This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus the nozzle pressure equals atmospheric pressure, as it must because the water exits into the atmosphere without changes in its conditions. Power in Fluid Flow Power is the rate at which work is done or energy in any form is used or supplied. To see the relationship of power to fluid flow, consider Bernoulli’s equation: \[P + \dfrac{1}{2}\rho v^2 + \rho gh = constant.\] All three terms have units of energy per unit volume, as discussed in the previous section. Now, considering units, if we multiply energy per unit volume by flow rate (volume per unit time), we get units of power. That is $(E/V)(V/t) = E/t$. This means that if we multiply Bernoulli’s equation by flow rate $Q$, we get power. In equation form, this is \[\left(P + \dfrac{1}{2}\rho v^2 + \rho gh \right)Q = power.\] Each term has a clear physical meaning. For example, $PQ$ is the power supplied to a fluid, perhaps by a pump, to give it its pressure $P$. Similarly, $\frac{1}{2}\rho v^2Q$ is the power supplied to a fluid to give it its kinetic energy. And $\rho ghQ$ is the power going to gravitational potential energy. Making Connections: Power Power is defined as the rate of energy transferred, or $E/t$. Fluid flow involves several types of power. Each type of power is identified with a specific type of energy being expended or changed in form. Example $\PageIndex{2}$: Calculating Power in a Moving Fluid Suppose the fire hose in the previous example is fed by a pump that receives water through a hose with a 6.40-cm diameter coming from a hydrant with a pressure of $0.700 \times 10^6 \, N/m^2$. What power does the pump supply to the water? Strategy Here we must consider energy forms as well as how they relate to fluid flow. Since the input and output hoses have the same diameters and are at the same height, the pump does not change the speed of the water nor its height, and so the water’s kinetic energy and gravitational potential energy are unchanged. That means the pump only supplies power to increase water pressure by $0.92 \times 10^6 \, N/m^2$ (from $0.700 \times 10^6 \, N/m^2 $ to $1.62 \times 10^6 \, N/m^2)$. Solution As discussed above, the power associated with pressure is \[\begin{align} power &= PQ \\[5pt] &= (0.920 \times 10^6 \, N/m^2)(40.0 \times 10^{-3} m^3/s).\\[5pt] &= 3.68 \times 10^4 \, W \\[5pt] &= 36.8 \, kW \end{align} \] Discussion Such a substantial amount of power requires a large pump, such as is found on some fire trucks. (This kilowatt value converts to about 50 hp.) The pump in this example increases only the water’s pressure. If a pump—such as the heart—directly increases velocity and height as well as pressure, we would have to calculate all three terms to find the power it supplies. Summary - Power in fluid flow is given by the equation $(P_1 + \frac{1}{2}\rho v^2 + \rho gh)Q = power$, where the first term is power associated with pressure, the second is power associated with velocity, and the third is power associated with height.	libretexts	2025-03-17T19:53:30.894017	2015-11-01T05:35:16	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.03%3A_The_Most_General_Applications_of_Bernoullis_Equation", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "12.3: The Most General Applications of Bernoulli’s Equation", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.04%3A_Viscosity_and_Laminar_Flow_Poiseuilles_Law	12.4: Viscosity and Laminar Flow; Poiseuille’s Law Learning Objectives By the end of this section, you will be able to: - Define laminar flow and turbulent flow. - Explain what viscosity is. - Calculate flow and resistance with Poiseuille’s law. - Explain how pressure drops due to resistance. Laminar Flow and Viscosity When you pour yourself a glass of juice, the liquid flows freely and quickly. But when you pour syrup on your pancakes, that liquid flows slowly and sticks to the pitcher. The difference is fluid friction, both within the fluid itself and between the fluid and its surroundings. We call this property of fluids viscosity . Juice has low viscosity, whereas syrup has high viscosity. In the previous sections we have considered ideal fluids with little or no viscosity. In this section, we will investigate what factors, including viscosity, affect the rate of fluid flow. The precise definition of viscosity is based on laminar , or nonturbulent, flow. Before we can define viscosity, then, we need to define laminar flow and turbulent flow. Figure shows both types of flow. Laminar flow is characterized by the smooth flow of the fluid in layers that do not mix. Turbulent flow, or turbulence , is characterized by eddies and swirls that mix layers of fluid together. Figure shows schematically how laminar and turbulent flow differ. Layers flow without mixing when flow is laminar. When there is turbulence, the layers mix, and there are significant velocities in directions other than the overall direction of flow. The lines that are shown in many illustrations are the paths followed by small volumes of fluids. These are called streamlines . Streamlines are smooth and continuous when flow is laminar, but break up and mix when flow is turbulent. Turbulence has two main causes. First, any obstruction or sharp corner, such as in a faucet, creates turbulence by imparting velocities perpendicular to the flow. Second, high speeds cause turbulence. The drag both between adjacent layers of fluid and between the fluid and its surroundings forms swirls and eddies, if the speed is great enough. We shall concentrate on laminar flow for the remainder of this section, leaving certain aspects of turbulence for later sections. Making Connections: Take-Home Experiment: Go Down to the River Try dropping simultaneously two sticks into a flowing river, one near the edge of the river and one near the middle. Which one travels faster? Why? Figure shows how viscosity is measured for a fluid. Two parallel plates have the specific fluid between them. The bottom plate is held fixed, while the top plate is moved to the right, dragging fluid with it. The layer (or lamina) of fluid in contact with either plate does not move relative to the plate, and so the top layer moves at while the bottom layer remains at rest. Each successive layer from the top down exerts a force on the one below it, trying to drag it along, producing a continuous variation in speed from to 0 as shown. Care is taken to insure that the flow is laminar; that is, the layers do not mix. The motion in Figure is like a continuous shearing motion. Fluids have zero shear strength, but the rate at which they are sheared is related to the same geometrical factors $A$ and $L$ as is shear deformation for solids. A force $F$ is required to keep the top plate in Figure moving at a constant velocity $v$, and experiments have shown that this force depends on four factors. First, $F$ is directly proportional to $v$ (until the speed is so high that turbulence occurs—then a much larger force is needed, and it has a more complicated dependence on $v$). Second, $F$ is proportional to the area $A$ of the plate. This relationship seems reasonable, since $A$ is directly proportional to the amount of fluid being moved. Third, $F$ is inversely proportional to the distance between the plates $L$. This relationship is also reasonable, $L$ is like a lever arm, and the greater the lever arm, the less force that is needed. Fourth, $F$ is directly proportional to the coefficient of viscosity , $\eta$. The greater the viscosity, the greater the force required. These dependencies are combined into the equation \[F = \eta \dfrac{\nu A}{L},\] which gives us a working definition of fluid viscosity $\eta$. Solving for $\eta$ gives \[\eta = \dfrac{FL}{\nu A},\] which defines viscosity in terms of how it is measured. The SI unit of viscosity is \[N \cdot m/[(m/s)m^2] = (N/m^2)s \, or \, Pa \cdot s\), Table lists the coefficients of viscosity for various fluids. Viscosity varies from one fluid to another by several orders of magnitude. As you might expect, the viscosities of gases are much less than those of liquids, and these viscosities are often temperature dependent. The viscosity of blood can be reduced by aspirin consumption, allowing it to flow more easily around the body. (When used over the long term in low doses, aspirin can help prevent heart attacks, and reduce the risk of blood clotting.) Laminar Flow Confined to Tubes—Poiseuille’s Law What causes flow? The answer, not surprisingly, is pressure difference. In fact, there is a very simple relationship between horizontal flow and pressure. Flow rate $Q$ is in the direction from high to low pressure. The greater the pressure differential between two points, the greater the flow rate. This relationship can be stated as \[Q = \dfrac{P_2 - P_1}{R},\] where $P_1$ and $P_2$ are the pressures at two points, such as at either end of a tube, and $R$ is the resistance to flow. The resistance $R$ includes everything, except pressure, that affects flow rate. For example, $R$ is greater for a long tube than for a short one. The greater the viscosity of a fluid, the greater the value of $R$. Turbulence greatly increases $R$, whereas increasing the diameter of a tube decreases $R$. If viscosity is zero, the fluid is frictionless and the resistance to flow is also zero. Comparing frictionless flow in a tube to viscous flow, as in Figure , we see that for a viscous fluid, speed is greatest at midstream because of drag at the boundaries. We can see the effect of viscosity in a Bunsen burner flame, even though the viscosity of natural gas is small. The resistance $R$ to laminar flow of an incompressible fluid having viscosity $\eta$ through a horizontal tube of uniform radius $r$ and length $l$ such as the one in Figure , is given by \[R = \dfrac{8\eta l}{\pi r^4}.\] This equation is called Poiseuille’s law for resistance after the French scientist J. L. Poiseuille (1799–1869), who derived it in an attempt to understand the flow of blood, an often turbulent fluid. Let us examine Poiseuille’s expression for $R$ to see if it makes good intuitive sense. We see that resistance is directly proportional to both fluid viscosity $\eta$ and the length $l$ of a tube. After all, both of these directly affect the amount of friction encountered—the greater either is, the greater the resistance and the smaller the flow. The radius $r$ of a tube affects the resistance, which again makes sense, because the greater the radius, the greater the flow (all other factors remaining the same). But it is surprising that $r$ is raised to the fourth power in Poiseuille’s law. This exponent means that any change in the radius of a tube has a very large effect on resistance. For example, doubling the radius of a tube decreases resistance by a factor of $2^4 = 16$. Taken together, $Q = \frac{P_2 - P_1}{R}$ and $R = \frac{8\eta l}{\pi r^4}$ give the following expression for flow rate: \[Q = \dfrac{(P_2 - P_1)\pi r^4}{8\eta l}.\] This equation describes laminar flow through a tube. It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law . Example $\PageIndex{1}$: Using Flow Rate: Plaque Deposits Reduce Blood Flow Suppose the flow rate of blood in a coronary artery has been reduced to half its normal value by plaque deposits. By what factor has the radius of the artery been reduced, assuming no turbulence occurs? Strategy Assuming laminar flow, Poiseuille’s law states that \[Q = \dfrac{(P_2 - P_1)\pi r^4}{8\eta l}.\] We need to compare the artery radius before and after the flow rate reduction. Solution With a constant pressure difference assumed and the same length and viscosity, along the artery we have \[\dfrac{Q_1}{r_1^4} = \dfrac{Q_2}{r_2^4}\] So, given that $Q_2 = 0.5Q_1$, we find that $r_2^4 = 0.5r_1^4$. Therefore, $r_2 = (0.5)^{0.25}r_1$ a decrease in the artery radius of 16%. Discussion This decrease in radius is surprisingly small for this situation. To restore the blood flow in spite of this buildup would require an increase in the pressure difference $(P_2 - P_1)$ of a factor of two, with subsequent strain on the heart. \| Fluid \| Temperature (ºC) \| Viscosity (mPa⋅s) \| \|---\|---\|---\| \| Gases \| \|\| \| Air \| 0 \| 0.0171 \| \| 20 \| 0.0181 \| \| \| 40 \| 0.0190 \| \| \| 100 \| 0.0218 \| \| \| Ammonia \| 20 \| 0.00974 \| \| Carbon dioxide \| 20 \| 0.0147 \| \| Helium \| 20 \| 0.0196 \| \| Hydrogen \| 0 \| 0.0090 \| \| Mercury \| 20 \| 0.0450 \| \| Oxygen \| 20 \| 0.0203 \| \| Steam \| 100 \| 0.0130 \| \| Liquids \| \|\| \| Water \| 0 \| 1.792 \| \| 20 \| 1.002 \| \| \| 37 \| 0.6947 \| \| \| 40 \| 0.653 \| \| \| 100 \| 0.282 \| \| \| Whole blood 1 \| 20 \| 3.015 \| \| 37 \| 2.084 \| \| \| Blood plasma 2 \| 20 \| 1.810 \| \| 37 \| 1.257 \| \| \| Ethyl alcohol \| 20 \| 1.20 \| \| Methanol \| 20 \| 0.584 \| \| Oil (heavy machine) \| 20 \| 660 \| \| Oil (motor, SAE 10) \| 30 \| 200 \| \| Oil (olive) \| 20 \| 138 \| \| Glycerin \| 20 \| 1500 \| \| Honey \| 20 \| 2000–10000 \| \| Maple Syrup \| 20 \| 2000–3000 \| \| Milk \| 20 \| 3.0 \| \| Oil (Corn) \| 20 \| 65 \| The circulatory system provides many examples of Poiseuille’s law in action—with blood flow regulated by changes in vessel size and blood pressure. Blood vessels are not rigid but elastic. Adjustments to blood flow are primarily made by varying the size of the vessels, since the resistance is so sensitive to the radius. During vigorous exercise, blood vessels are selectively dilated to important muscles and organs and blood pressure increases. This creates both greater overall blood flow and increased flow to specific areas. Conversely, decreases in vessel radii, perhaps from plaques in the arteries, can greatly reduce blood flow. If a vessel’s radius is reduced by only 5% (to 0.95 of its original value), the flow rate is reduced to about $(0.95)^4 = 0.81$ of its original value. A 19% decrease in flow is caused by a 5% decrease in radius. The body may compensate by increasing blood pressure by 19%, but this presents hazards to the heart and any vessel that has weakened walls. Another example comes from automobile engine oil. If you have a car with an oil pressure gauge, you may notice that oil pressure is high when the engine is cold. Motor oil has greater viscosity when cold than when warm, and so pressure must be greater to pump the same amount of cold oil. Example $\PageIndex{2}$: What Pressure Produces This Flow Rate? An intravenous (IV) system is supplying saline solution to a patient at the rate of $0.120 \, cm^3/s$ through a needle of radius 0.150 mm and length 2.50 cm. What pressure is needed at the entrance of the needle to cause this flow, assuming the viscosity of the saline solution to be the same as that of water? The gauge pressure of the blood in the patient’s vein is 8.00 mm Hg. (Assume that the temperature is $20^o \, C$. Strategy Assuming laminar flow, Poiseuille’s law applies. This is given by \[Q = \dfrac{(P_2 - P_1)\pi r^4}{8\eta l},\] where $P_2$ is the pressure at the entrance of the needle and $P_1$ is the pressure in the vein. The only unknown is $P_2$. Solution Solving for $P_2$ yields \[P_2 = \dfrac{8\eta l}{\pi r^4}Q + P_1\] $P_1$ is given as 8.00 mm Hg, which converts to $1.066 \times 10^3 \, N/m^2$. Substituting this and the other known values yields \[ P_2 = \left[ \dfrac{8(1.00 \times 10^{-3} N\cdot s/m^2)(2.50 \times 10^{-2}m)}{\pi (0.150 \times 10^{-3} \, m^4)} \right] (1.20 \times 10^{-7} \, m^3/s) + 1.066 \times 10^3 \, N/m^2 \] \[= 1.62 \times 10^4 \, N/m^2\] Discussion This pressure could be supplied by an IV bottle with the surface of the saline solution 1.61 m above the entrance to the needle (this is left for you to solve in this chapter’s Problems and Exercises), assuming that there is negligible pressure drop in the tubing leading to the needle. Flow and Resistance as Causes of Pressure Drops You may have noticed that water pressure in your home might be lower than normal on hot summer days when there is more use. This pressure drop occurs in the water main before it reaches your home. Let us consider flow through the water main as illustrated in Figure $\PageIndex{6}$. We can understand why the pressure $P_1$ to the home drops during times of heavy use by rearranging \[ Q = \dfrac{P_2 - P_1}{R}\] to \[P_2 - P_1 = RQ,\] where, in this case, $P_2$ is the pressure at the water works and $R$ is the resistance of the water main. During times of heavy use, the flow rate $Q$ is large. This means that $P_2 - P_1 = RQ$ is valid for both laminar and turbulent flows. We can use $P_2 - P_1 = RQ$ to analyze pressure drops occurring in more complex systems in which the tube radius is not the same everywhere. Resistance will be much greater in narrow places, such as an obstructed coronary artery. For a given flow rate $Q$, the pressure drop will be greatest where the tube is most narrow. This is how water faucets control flow. Additionally, $R$ is greatly increased by turbulence, and a constriction that creates turbulence greatly reduces the pressure downstream. Plaque in an artery reduces pressure and hence flow, both by its resistance and by the turbulence it creates. Figure $\PageIndex{7}$ is a schematic of the human circulatory system, showing average blood pressures in its major parts for an adult at rest. Pressure created by the heart’s two pumps, the right and left ventricles, is reduced by the resistance of the blood vessels as the blood flows through them. The left ventricle increases arterial blood pressure that drives the flow of blood through all parts of the body except the lungs. The right ventricle receives the lower pressure blood from two major veins and pumps it through the lungs for gas exchange with atmospheric gases – the disposal of carbon dioxide from the blood and the replenishment of oxygen. Only one major organ is shown schematically, with typical branching of arteries to ever smaller vessels, the smallest of which are the capillaries, and rejoining of small veins into larger ones. Similar branching takes place in a variety of organs in the body, and the circulatory system has considerable flexibility in flow regulation to these organs by the dilation and constriction of the arteries leading to them and the capillaries within them. The sensitivity of flow to tube radius makes this flexibility possible over a large range of flow rates. Each branching of larger vessels into smaller vessels increases the total cross-sectional area of the tubes through which the blood flows. For example, an artery with a cross section of $1 \, cm^2$ may branch into 20 smaller arteries, each with cross sections of $0.5 \, cm^2$, with a total of $10 \, cm^2$ In that manner, the resistance of the branchings is reduced so that pressure is not entirely lost. Moreover, because $Q = A \overline{v}$ and $A$ increases through branching, the average velocity of the blood in the smaller vessels is reduced. The blood velocity in the aorta $(diameter = 1 \, cm)$ is about 25 cm/s, while in the capillaries ($20 \, \mu m$ in diameter) the velocity is about 1 mm/s. This reduced velocity allows the blood to exchange substances with the cells in the capillaries and alveoli in particular. Section Summary - Laminar flow is characterized by smooth flow of the fluid in layers that do not mix. - Turbulence is characterized by eddies and swirls that mix layers of fluid together. - Fluid viscosity $\eta$ is due to friction within a fluid. Representative values are given in Table . Viscosity has units of $(N/m^2)$s or $Pa \cdot s$. - Flow is proportional to pressure difference and inversely proportional to resistance: \[Q = \dfrac{P_2 - P_1}{R}.\] - For laminar flow in a tube, Poiseuille’s law for resistance states that \[R = \dfrac{8\eta l}{\pi r^4}.\] - Poiseuille’s law for flow in a tube is \[Q = \dfrac{(P_2 - P_1) \pi r^4}{8 \eta l}.\] - The pressure drop caused by flow and resistance is given by \[P_2 - P_1 = RQ.\] Footnotes 1 The ratios of the viscosities of blood to water are nearly constant between 0°C and 37° 2. See note on Whole Blood. Glossary - laminar - a type of fluid flow in which layers do not mix - turbulence - fluid flow in which layers mix together via eddies and swirls - viscosity - the friction in a fluid, defined in terms of the friction between layers - Poiseuille’s law for resistance - the resistance to laminar flow of an incompressible fluid in a tube: R = 8 ηl / πr 4 - Poiseuille’s law - the rate of laminar flow of an incompressible fluid in a tube: Q = ( P 2 − P 1 ) πr 4 /8 ηl	libretexts	2025-03-17T19:53:31.000804	2015-11-01T05:35:32	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.04%3A_Viscosity_and_Laminar_Flow_Poiseuilles_Law", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "12.4: Viscosity and Laminar Flow; Poiseuille’s Law", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.05%3A_The_Onset_of_Turbulence	12.5: The Onset of Turbulence Learning Objectives By the end of this section, you will be able to: - Calculate Reynolds number. - Use the Reynolds number for a system to determine whether it is laminar or turbulent. Sometimes we can predict if flow will be laminar or turbulent. We know that flow in a very smooth tube or around a smooth, streamlined object will be laminar at low velocity. We also know that at high velocity, even flow in a smooth tube or around a smooth object will experience turbulence. In between, it is more difficult to predict. In fact, at intermediate velocities, flow may oscillate back and forth indefinitely between laminar and turbulent. An occlusion, or narrowing, of an artery, such as shown in Figure $\PageIndex{1}$, is likely to cause turbulence because of the irregularity of the blockage, as well as the complexity of blood as a fluid. Turbulence in the circulatory system is noisy and can sometimes be detected with a stethoscope, such as when measuring diastolic pressure in the upper arm’s partially collapsed brachial artery. These turbulent sounds, at the onset of blood flow when the cuff pressure becomes sufficiently small, are called Korotkoff sounds . Aneurysms, or ballooning of arteries, create significant turbulence and can sometimes be detected with a stethoscope. Heart murmurs, consistent with their name, are sounds produced by turbulent flow around damaged and insufficiently closed heart valves. Ultrasound can also be used to detect turbulence as a medical indicator in a process analogous to Doppler-shift radar used to detect storms. An indicator called the Reynolds number $N_R$ can reveal whether flow is laminar or turbulent. For flow in a tube of uniform diameter, the Reynolds number is defined as \[ \underbrace{N_R = \dfrac{2\rho vr}{\eta}}_{\text{flow in tube}}\] where $\rho$ is the fluid density, $v$ its speed, $\eta$ its viscosity, and $r$ the tube radius. The Reynolds number is a unitless quantity. Experiments have revealed that $N_R$ is related to the onset of turbulence. For $N_R$ below about 2000, flow is laminar. For $N_R$ above about 3000, flow is turbulent. For values of $N_R$ between about 2000 and 3000, flow is unstable—that is, it can be laminar, but small obstructions and surface roughness can make it turbulent, and it may oscillate randomly between being laminar and turbulent. The blood flow through most of the body is a quiet, laminar flow. The exception is in the aorta, where the speed of the blood flow rises above a critical value of 35 m/s and becomes turbulent. Example $\PageIndex{1}$: Is This Flow Laminar or Turbulent? Calculate the Reynolds number for flow in the needle considered in Example 12.8 to verify the assumption that the flow is laminar. Assume that the density of the saline solution is $1025 \, kg/m^3$. Strategy We have all of the information needed, except the fluid speed $v$, which can be calculated from $\overline{v} = Q/A = 1.70 \, m/s$ (verification of this is in this chapter’s Problems and Exercises). Solution Entering the known values into $N_R = \frac{2\rho vr}{\eta}$ gives \[N_R = \dfrac{2\rho vr}{\eta}\] \[= \dfrac{2(1025 \, kg/m^3)(1.70 \, m/s)(0.150 \times 10^{-3} \, m)}{1.00 \times 10^{-3} \, N\cdot s/m^2}\] \[= 523.\] Discussion Since $N_R$ is well below 2000, the flow should indeed be laminar. Take-Home Experiment: Inhalation Under the conditions of normal activity, an adult inhales about 1 L of air during each inhalation. With the aid of a watch, determine the time for one of your own inhalations by timing several breaths and dividing the total length by the number of breaths. Calculate the average flow rate $Q$ of air traveling through the trachea during each inhalation. The topic of chaos has become quite popular over the last few decades. A system is defined to be chaotic when its behavior is so sensitive to some factor that it is extremely difficult to predict. The field of chaos is the study of chaotic behavior. A good example of chaotic behavior is the flow of a fluid with a Reynolds number between 2000 and 3000. Whether or not the flow is turbulent is difficult, but not impossible, to predict—the difficulty lies in the extremely sensitive dependence on factors like roughness and obstructions on the nature of the flow. A tiny variation in one factor has an exaggerated (or nonlinear) effect on the flow. Phenomena as disparate as turbulence, the orbit of Pluto, and the onset of irregular heartbeats are chaotic and can be analyzed with similar techniques. Summary - The Reynolds number $N_R$ can reveal whether flow is laminar or turbulent. It is \[N_R = \dfrac{2\rho vr}{\eta}.\] - For $N_R$ below about 2000, flow is laminar. For $N_R$ above about 3000, flow is turbulent. For values of $N_R$ between 2000 and 3000, it may be either or both. Glossary - Reynolds number - a dimensionless parameter that can reveal whether a particular flow is laminar or turbulent	libretexts	2025-03-17T19:53:31.067287	2015-11-01T05:35:56	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.05%3A_The_Onset_of_Turbulence", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "12.5: The Onset of Turbulence", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.06%3A_Motion_of_an_Object_in_a_Viscous_Fluid	12.6: Motion of an Object in a Viscous Fluid Learning Objectives By the end of this section, you will be able to: - Calculate the Reynolds number for an object moving through a fluid. - Explain whether the Reynolds number indicates laminar or turbulent flow. - Describe the conditions under which an object has a terminal speed. A moving object in a viscous fluid is equivalent to a stationary object in a flowing fluid stream. (For example, when you ride a bicycle at 10 m/s in still air, you feel the air in your face exactly as if you were stationary in a 10-m/s wind.) Flow of the stationary fluid around a moving object may be laminar, turbulent, or a combination of the two. Just as with flow in tubes, it is possible to predict when a moving object creates turbulence. We use another form of the Reynolds number $N_R'$, defined for an object moving in a fluid to be \[N_R' = \dfrac{\rho vL}{\eta}(object \, in \, fluid),\] where $L$ is a characteristic length of the object (a sphere’s diameter, for example), $\rho$ the fluid density, $\eta$ its viscosity, and $v$ the object’s speed in the fluid. If $N_R'$ is less than about 1, flow around the object can be laminar, particularly if the object has a smooth shape. The transition to turbulent flow occurs for $N_R'$ between 1 and about 10, depending on surface roughness and so on. Depending on the surface, there can be a turbulent wake behind the object with some laminar flow over its surface. For an $N_R'$ between 10 and $10^6$, the flow may be either laminar or turbulent and may oscillate between the two. For $N_R'$ greater than about $10^6$, the flow is entirely turbulent, even at the surface of the object (Figure $\PageIndex{1}$). Laminar flow occurs mostly when the objects in the fluid are small, such as raindrops, pollen, and blood cells in plasma. Example $\PageIndex{1}$: Does a Ball Have a Turbulent Wake? Calculate the Reynolds number $N_R'$ for a ball with a 7.40-cm diameter thrown at 40.0 m/s. Strategy We can use $N_R' = \frac{\rho vL}{\eta}$ to calculate $N_R'$, since all values in it are either given or can be found in tables of density and viscosity. Solution Substituting values into the equation for $N_R'$ yields \[N_R' = \dfrac{(1.20 \, kg/m^3)(40.0 \, m/s)(0.0740 \, m)}{1.81 \times 10^5 \, 1.00 \, Pa-s}\] \[= 2.11 \times 10^5.\] Discussion This value is sufficiently high to imply a turbulent wake. Most large objects, such as airplanes and sailboats, create significant turbulence as they move. As noted before, the Bernoulli principle gives only qualitatively-correct results in such situations. One of the consequences of viscosity is a resistance force called viscous drag $F_V$ that is exerted on a moving object. This force typically depends on the object’s speed (in contrast with simple friction). Experiments have shown that for laminar flow ( $N_R'$ less than about one) viscous drag is proportional to speed, whereas for $N_R'$ between about 10 and $10^6$, viscous drag is proportional to speed squared. (This relationship is a strong dependence and is pertinent to bicycle racing, where even a small headwind causes significantly increased drag on the racer. Cyclists take turns being the leader in the pack for this reason.) For $N_R'$ greater than $10^6$, drag increases dramatically and behaves with greater complexity. For laminar flow around a sphere, $F_V$ is proportional to fluid viscosity $\eta$, the object’s characteristic size $L$, and its speed $v$. All of which makes sense—the more viscous the fluid and the larger the object, the more drag we expect. Recall Stoke’s law $F_S = 6\pi r\eta v$. For the special case of a small sphere of radius $R$ moving slowly in a fluid of viscosity $\eta$, the drag force $F_S$ is given by \[F_S = 6\pi R\eta v.\] An interesting consequence of the increase in $F_V$ with speed is that an object falling through a fluid will not continue to accelerate indefinitely (as it would if we neglect air resistance, for example). Instead, viscous drag increases, slowing acceleration, until a critical speed, called the terminal speed , is reached and the acceleration of the object becomes zero. Once this happens, the object continues to fall at constant speed (the terminal speed). This is the case for particles of sand falling in the ocean, cells falling in a centrifuge, and sky divers falling through the air. Figure $\PageIndex{2}$ shows some of the factors that affect terminal speed. There is a viscous drag on the object that depends on the viscosity of the fluid and the size of the object. But there is also a buoyant force that depends on the density of the object relative to the fluid. Terminal speed will be greatest for low-viscosity fluids and objects with high densities and small sizes. Thus a skydiver falls more slowly with outspread limbs than when they are in a pike position—head first with hands at their side and legs together. Take-Home Experiment: Don’t Lose Your Marbles By measuring the terminal speed of a slowly moving sphere in a viscous fluid, one can find the viscosity of that fluid (at that temperature). It can be difficult to find small ball bearings around the house, but a small marble will do. Gather two or three fluids (syrup, motor oil, honey, olive oil, etc.) and a thick, tall clear glass or vase. Drop the marble into the center of the fluid and time its fall (after letting it drop a little to reach its terminal speed). Compare your values for the terminal speed and see if they are inversely proportional to the viscosities as listed in [link] . Does it make a difference if the marble is dropped near the side of the glass? Knowledge of terminal speed is useful for estimating sedimentation rates of small particles. We know from watching mud settle out of dirty water that sedimentation is usually a slow process. Centrifuges are used to speed sedimentation by creating accelerated frames in which gravitational acceleration is replaced by centripetal acceleration, which can be much greater, increasing the terminal speed. Summary - When an object moves in a fluid, there is a different form of the Reynolds number $N_R' = \frac{\rho vL}{\eta}(object \, in \, fluid)$, which indicates whether flow is laminar or turbulent. - For $N_R'$ less than about one, flow is laminar. - For $N_R'$ greater than $10^6$, flow is entirely turbulent. Glossary - viscous drag - a resistance force exerted on a moving object, with a nontrivial dependence on velocity - terminal speed - the speed at which the viscous drag of an object falling in a viscous fluid is equal to the other forces acting on the object (such as gravity), so that the acceleration of the object is zero	libretexts	2025-03-17T19:53:31.209328	2015-11-01T05:36:11	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.06%3A_Motion_of_an_Object_in_a_Viscous_Fluid", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "12.6: Motion of an Object in a Viscous Fluid", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.07%3A_Molecular_Transport_Phenomena-_Diffusion_Osmosis_and_Related_Processes	12.7: Molecular Transport Phenomena- Diffusion, Osmosis, and Related Processes Learning Objectives Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes - Define diffusion, osmosis, dialysis, and active transport. - Calculate diffusion rates. Diffusion There is something fishy about the ice cube from your freezer—how did it pick up those food odors? How does soaking a sprained ankle in Epsom salt reduce swelling? The answer to these questions are related to atomic and molecular transport phenomena—another mode of fluid motion. Atoms and molecules are in constant motion at any temperature. In fluids they move about randomly even in the absence of macroscopic flow. This motion is called a random walk and is illustrated in Figure . Diffusion is the movement of substances due to random thermal molecular motion. Fluids, like fish fumes or odors entering ice cubes, can even diffuse through solids. Diffusion is a slow process over macroscopic distances. The densities of common materials are great enough that molecules cannot travel very far before having a collision that can scatter them in any direction, including straight backward. It can be shown that the average distance $x_{rms}$ that a molecule travels is proportional to the square root of time: \[x_{rms} = \sqrt{2Dt,}\] where $x_{rms}$ stands for the root-mean-square distance and is the statistical average for the process. The quantity $D$ is the diffusion constant for the particular molecule in a specific medium. Table $\PageIndex{1}$ lists representative values of $D$ for various substances, in units of $m^2/s$. \| Diffusing molecule \| Medium \| D ($m^2/s$) \| \|---\|---\|---\| \| Hydrogen $(H_2)$ \| Air \| $6.4 \times 10^{-5}$ \| \| Oxygen $(O_2)$ \| Air \| $1.8 \times 10^{-5}$ \| \| Oxygen $(O_2)$ \| Water \| $1.0 \times 10^{-9}$ \| \| Glucose $(C_6H_{12}O_6)$ \| Water \| $6.7 \times 10^{-10}$ \| \| Hemoglobin \| Water \| $6.9 \times 10^{-11}$ \| \| DNA \| Water \| $1.3 \times 10^{-12}$ \| Note that $D$ gets progressively smaller for more massive molecules. This decrease is because the average molecular speed at a given temperature is inversely proportional to molecular mass. Thus the more massive molecules diffuse more slowly. Another interesting point is that $D$ for oxygen in air is much greater than $D$ for oxygen in water. In water, an oxygen molecule makes many more collisions in its random walk and is slowed considerably. In water, an oxygen molecule moves only about $40 \, \mu m$ in 1 s. (Each molecule actually collides about $10^{10}$ times per second!). Finally, note that diffusion constants increase with temperature, because average molecular speed increases with temperature. This is because the average kinetic energy of molecules, $\frac{1}{2}mv^2$, is proportional to absolute temperature. Example $\PageIndex{1}$: Calculating Diffusion: How Long Does Glucose Diffusion Take? Calculate the average time it takes a glucose molecule to move 1.0 cm in water. Strategy We can use $x_{tms} = \sqrt{2Dt},$ the expression for the average distance moved in time $t$, and solve it for $t$. All other quantities are known. Solution Solving for $t$ and substituting known values yields \[t = \frac{x_{rms}^2}{2D} = \frac{(0.010 \, m)^2}{2(6.7 \times 10^{-10} m^2/s)}\] \[ = 7.5 \times 10^4 s = 21 \, h.\] Discussion This is a remarkably long time for glucose to move a mere centimeter! For this reason, we stir sugar into water rather than waiting for it to diffuse. Because diffusion is typically very slow, its most important effects occur over small distances. For example, the cornea of the eye gets most of its oxygen by diffusion through the thin tear layer covering it. The Rate and Direction of Diffusion If you very carefully place a drop of food coloring in a still glass of water, it will slowly diffuse into the colorless surroundings until its concentration is the same everywhere. This type of diffusion is called free diffusion, because there are no barriers inhibiting it. Let us examine its direction and rate. Molecular motion is random in direction, and so simple chance dictates that more molecules will move out of a region of high concentration than into it. The net rate of diffusion is higher initially than after the process is partially completed (Figure $\PageIndex{1}$). The net rate of diffusion is proportional to the concentration difference. Many more molecules will leave a region of high concentration than will enter it from a region of low concentration. In fact, if the concentrations were the same, there would be no net movement. The rate of diffusion is also proportional to the diffusion constant $D$, which is determined experimentally. The farther a molecule can diffuse in a given time, the more likely it is to leave the region of high concentration. Many of the factors that affect the rate are hidden in the diffusion constant $D$. For example, temperature and cohesive and adhesive forces all affect values of $D$. Diffusion is the dominant mechanism by which the exchange of nutrients and waste products occur between the blood and tissue, and between air and blood in the lungs. In the evolutionary process, as organisms became larger, they needed quicker methods of transportation than net diffusion, because of the larger distances involved in the transport, leading to the development of circulatory systems. Less sophisticated, single-celled organisms still rely totally on diffusion for the removal of waste products and the uptake of nutrients. Osmosis and Dialysis—Diffusion across Membranes Some of the most interesting examples of diffusion occur through barriers that affect the rates of diffusion. For example, when you soak a swollen ankle in Epsom salt, water diffuses through your skin. Many substances regularly move through cell membranes; oxygen moves in, carbon dioxide moves out, nutrients go in, and wastes go out, for example. Because membranes are thin structures (typically $6.5 \times 10^{-9}$ to $10 \times 10^{-9} $ m across) diffusion rates through them can be high. Diffusion through membranes is an important method of transport. Membranes are generally selectively permeable, or semipermeable (Figure $\PageIndex{3}$). One type of semipermeable membrane has small pores that allow only small molecules to pass through. In other types of membranes, the molecules may actually dissolve in the membrane or react with molecules in the membrane while moving across. Membrane function, in fact, is the subject of much current research, involving not only physiology but also chemistry and physics. Osmosis is the transport of water through a semipermeable membrane from a region of high concentration to a region of low concentration. Osmosis is driven by the imbalance in water concentration. For example, water is more concentrated in your body than in Epsom salt. When you soak a swollen ankle in Epsom salt, the water moves out of your body into the lower-concentration region in the salt. Similarly, dialysis is the transport of any other molecule through a semipermeable membrane due to its concentration difference. Both osmosis and dialysis are used by the kidneys to cleanse the blood. Osmosis can create a substantial pressure. Consider what happens if osmosis continues for some time, as illustrated in Figure $\PageIndex{4}$. Water moves by osmosis from the left into the region on the right, where it is less concentrated, causing the solution on the right to rise. This movement will continue until the pressure $\rho gh$ created by the extra height of fluid on the right is large enough to stop further osmosis. This pressure is called a back pressure . The back pressure $\rho gh$ that stops osmosis is also called the relative osmotic pressure if neither solution is pure water, and it is called the osmotic pressure if one solution is pure water. Osmotic pressure can be large, depending on the size of the concentration difference. For example, if pure water and sea water are separated by a semipermeable membrane that passes no salt, osmotic pressure will be 25.9 atm. This value means that water will diffuse through the membrane until the salt water surface rises 268 m above the pure-water surface! One example of pressure created by osmosis is turgor in plants (many wilt when too dry). Turgor describes the condition of a plant in which the fluid in a cell exerts a pressure against the cell wall. This pressure gives the plant support. Dialysis can similarly cause substantial pressures. Reverse osmosis and reverse dialysis (also called filtration) are processes that occur when back pressure is sufficient to reverse the normal direction of substances through membranes. Back pressure can be created naturally as on the right side of Figure $\PageIndex{4}$. (A piston can also create this pressure.) Reverse osmosis can be used to desalinate water by simply forcing it through a membrane that will not pass salt. Similarly, reverse dialysis can be used to filter out any substance that a given membrane will not pass. One further example of the movement of substances through membranes deserves mention. We sometimes find that substances pass in the direction opposite to what we expect. Cypress tree roots, for example, extract pure water from salt water, although osmosis would move it in the opposite direction. This is not reverse osmosis, because there is no back pressure to cause it. What is happening is called active transport , a process in which a living membrane expends energy to move substances across it. Many living membranes move water and other substances by active transport. The kidneys, for example, not only use osmosis and dialysis—they also employ significant active transport to move substances into and out of blood. In fact, it is estimated that at least 25% of the body’s energy is expended on active transport of substances at the cellular level. The study of active transport carries us into the realms of microbiology, biophysics, and biochemistry and it is a fascinating application of the laws of nature to living structures. Summary - Diffusion is the movement of substances due to random thermal molecular motion. - The average distance $x_{rsm}$ a molecule travels by diffusion in a given amount of time is given by \[ x_{rsm} = \sqrt{2Dt},\] where $D$ is the diffusion constant, representative values of which are found in Table . - Osmosis is the transport of water through a semipermeable membrane from a region of high concentration to a region of low concentration. - Dialysis is the transport of any other molecule through a semipermeable membrane due to its concentration difference. - Both processes can be reversed by back pressure. - Active transport is a process in which a living membrane expends energy to move substances across it. Footnotes 1. At 20°C and 1 atm Glossary - diffusion - the movement of substances due to random thermal molecular motion - semipermeable - a type of membrane that allows only certain small molecules to pass through - osmosis - the transport of water through a semipermeable membrane from a region of high concentration to one of low concentration - dialysis - the transport of any molecule other than water through a semipermeable membrane from a region of high concentration to one of low concentration - relative osmotic pressure - the back pressure which stops the osmotic process if neither solution is pure water - osmotic pressure - the back pressure which stops the osmotic process if one solution is pure water - reverse osmosis - the process that occurs when back pressure is sufficient to reverse the normal direction of osmosis through membranes - reverse dialysis - the process that occurs when back pressure is sufficient to reverse the normal direction of dialysis through membranes - active transport - the process in which a living membrane expends energy to move substances across	libretexts	2025-03-17T19:53:31.291873	2015-11-01T05:36:28	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.07%3A_Molecular_Transport_Phenomena-_Diffusion_Osmosis_and_Related_Processes", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "12.7: Molecular Transport Phenomena- Diffusion, Osmosis, and Related Processes", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.E%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications_(Exercises)	12.E: Fluid Dynamics and Its Biological and Medical Applications (Exercises) - - Last updated - Save as PDF Conceptual Questions 12.1: Flow Rate and Its Relation to Velocity 1. What is the difference between flow rate and fluid velocity? How are they related? 2. Many figures in the text show streamlines. Explain why fluid velocity is greatest where streamlines are closest together. (Hint: Consider the relationship between fluid velocity and the cross-sectional area through which it flows.) 3. Identify some substances that are incompressible and some that are not. 12.2: Bernoulli’s Equation 4. You can squirt water a considerably greater distance by placing your thumb over the end of a garden hose and then releasing, than by leaving it completely uncovered. Explain how this works. 5. Water is shot nearly vertically upward in a decorative fountain and the stream is observed to broaden as it rises. Conversely, a stream of water falling straight down from a faucet narrows. Explain why, and discuss whether surface tension enhances or reduces the effect in each case. 6. Look back to Figure. Answer the following two questions. Why is $\displaystyle P_o$ less than atmospheric? Why is $\displaystyle P_o$ greater than $\displaystyle P_i$? 7. Give an example of entrainment not mentioned in the text. 8. Many entrainment devices have a constriction, called a Venturi, such as shown in Figure. How does this bolster entrainment? A tube with a narrow segment designed to enhance entrainment is called a Venturi. These are very commonly used in carburetors and aspirators. 9. Some chimney pipes have a T-shape, with a crosspiece on top that helps draw up gases whenever there is even a slight breeze. Explain how this works in terms of Bernoulli’s principle. 10. Is there a limit to the height to which an entrainment device can raise a fluid? Explain your answer. 11. Why is it preferable for airplanes to take off into the wind rather than with the wind? 12. Roofs are sometimes pushed off vertically during a tropical cyclone, and buildings sometimes explode outward when hit by a tornado. Use Bernoulli’s principle to explain these phenomena. 13. Why does a sailboat need a keel? 14. It is dangerous to stand close to railroad tracks when a rapidly moving commuter train passes. Explain why atmospheric pressure would push you toward the moving train. 15. Water pressure inside a hose nozzle can be less than atmospheric pressure due to the Bernoulli effect. Explain in terms of energy how the water can emerge from the nozzle against the opposing atmospheric pressure. 16. A perfume bottle or atomizer sprays a fluid that is in the bottle. (Figure.) How does the fluid rise up in the vertical tube in the bottle? Atomizer: perfume bottle with tube to carry perfume up through the bottle. (credit: Antonia Foy, Flickr) 17. If you lower the window on a car while moving, an empty plastic bag can sometimes fly out the window. Why does this happen? 12.3: The Most General Applications of Bernoulli’s Equation 18. Based on Bernoulli’s equation, what are three forms of energy in a fluid? (Note that these forms are conservative, unlike heat transfer and other dissipative forms not included in Bernoulli’s equation.) 19. Water that has emerged from a hose into the atmosphere has a gauge pressure of zero. Why? When you put your hand in front of the emerging stream you feel a force, yet the water’s gauge pressure is zero. Explain where the force comes from in terms of energy. 20. The old rubber boot shown in Figure has two leaks. To what maximum height can the water squirt from Leak 1? How does the velocity of water emerging from Leak 2 differ from that of leak 1? Explain your responses in terms of energy. Water emerges from two leaks in an old boot. 21. Water pressure inside a hose nozzle can be less than atmospheric pressure due to the Bernoulli effect. Explain in terms of energy how the water can emerge from the nozzle against the opposing atmospheric pressure. 12.4: Viscosity and Laminar Flow; Poiseuille’s Law 22. Explain why the viscosity of a liquid decreases with temperature—that is, how might increased temperature reduce the effects of cohesive forces in a liquid? Also explain why the viscosity of a gas increases with temperature—that is, how does increased gas temperature create more collisions between atoms and molecules? 23. When paddling a canoe upstream, it is wisest to travel as near to the shore as possible. When canoeing downstream, it may be best to stay near the middle. Explain why. 24. Why does flow decrease in your shower when someone flushes the toilet? 25. Plumbing usually includes air-filled tubes near water faucets, as shown in Figure. Explain why they are needed and how they work. The vertical tube near the water tap remains full of air and serves a useful purpose. 12.5: The Onset of Turbulence 26. Doppler ultrasound can be used to measure the speed of blood in the body. If there is a partial constriction of an artery, where would you expect blood speed to be greatest, at or nearby the constriction? What are the two distinct causes of higher resistance in the constriction? 27. Sink drains often have a device such as that shown in Figure to help speed the flow of water. How does this work? You will find devices such as this in many drains. They significantly increase flow rate. 28. Some ceiling fans have decorative wicker reeds on their blades. Discuss whether these fans are as quiet and efficient as those with smooth blades. 12.6: Motion of an Object in a Viscous Fluid 29. What direction will a helium balloon move inside a car that is slowing down—toward the front or back? Explain your answer. 30. Will identical raindrops fall more rapidly in 5º C air or 25º C air, neglecting any differences in air density? Explain your answer. 31. If you took two marbles of different sizes, what would you expect to observe about the relative magnitudes of their terminal velocities? 12.7: Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes 32. Why would you expect the rate of diffusion to increase with temperature? Can you give an example, such as the fact that you can dissolve sugar more rapidly in hot water? 33. How are osmosis and dialysis similar? How do they differ? Problem & Exercises 12.1: Flow Rate and Its Relation to Velocity 34. What is the average flow rate in $\displaystyle cm^3/s$ of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L? Solution $\displaystyle 2.78cm^3/s$ 35. The heart of a resting adult pumps blood at a rate of 5.00 L/min. (a) Convert this to $\displaystyle cm^3/s$. (b) What is this rate in $\displaystyle m^3/s$? 36. Blood is pumped from the heart at a rate of 5.0 L/min into the aorta (of radius 1.0 cm). Determine the speed of blood through the aorta. Solution 27 cm/s 37. Blood is flowing through an artery of radius 2 mm at a rate of 40 cm/s. Determine the flow rate and the volume that passes through the artery in a period of 30 s. 38. The Huka Falls on the Waikato River is one of New Zealand’s most visited natural tourist attractions (see Figure). On average the river has a flow rate of about 300,000 L/s. At the gorge, the river narrows to 20 m wide and averages 20 m deep. (a) What is the average speed of the river in the gorge? (b) What is the average speed of the water in the river downstream of the falls when it widens to 60 m and its depth increases to an average of 40 m? The Huka Falls in Taupo, New Zealand, demonstrate flow rate. (credit: RaviGogna, Flickr Solution (a) 0.75 m/s (b) 0.13 m/s 39. A major artery with a cross-sectional area of $\displaystyle 1.00cm^2$ branches into 18 smaller arteries, each with an average cross-sectional area of $\displaystyle 0.400cm^2$. By what factor is the average velocity of the blood reduced when it passes into these branches? 40. (a) As blood passes through the capillary bed in an organ, the capillaries join to form venules (small veins). If the blood speed increases by a factor of 4.00 and the total cross-sectional area of the venules is 10.0cm2, what is the total cross-sectional area of the capillaries feeding these venules? (b) How many capillaries are involved if their average diameter is $\displaystyle 10.0μm$? Solution (a) $\displaystyle 40.0cm^2$ (b) $\displaystyle 5.09×10^7$ 41. The human circulation system has approximately $\displaystyle 1×10^9$ capillary vessels. Each vessel has a diameter of about 8μm. Assuming cardiac output is 5 L/min, determine the average velocity of blood flow through each capillary vessel. 42. (a) Estimate the time it would take to fill a private swimming pool with a capacity of 80,000 L using a garden hose delivering 60 L/min. (b) How long would it take to fill if you could divert a moderate size river, flowing at $\displaystyle 5000m^3/s$, into it? Solution (a) 22 h (b) 0.016 s 43. The flow rate of blood through a $\displaystyle 2.00×10^{–6}-m$ -radius capillary is $\displaystyle 3.80×10^{-9}cm^3/s$. (a) What is the speed of the blood flow? (This small speed allows time for diffusion of materials to and from the blood.) (b) Assuming all the blood in the body passes through capillaries, how many of them must there be to carry a total flow of 90.0cm3/s? (The large number obtained is an overestimate, but it is still reasonable.) 44. (a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rate in cubic meters per second? (c) Would your answers be different if salt water replaced the fresh water in the fire hose? Solution (a) 12.6 m/s (b) $\displaystyle 0.0800m^3/s$ (c) No, independent of density. 45. The main uptake air duct of a forced air gas heater is 0.300 m in diameter. What is the average speed of air in the duct if it carries a volume equal to that of the house’s interior every 15 min? The inside volume of the house is equivalent to a rectangular solid 13.0 m wide by 20.0 m long by 2.75 m high. 46. Water is moving at a velocity of 2.00 m/s through a hose with an internal diameter of 1.60 cm. (a) What is the flow rate in liters per second? (b) The fluid velocity in this hose’s nozzle is 15.0 m/s. What is the nozzle’s inside diameter? Solution (a) 0.402 L/s (b) 0.584 cm 47. Prove that the speed of an incompressible fluid through a constriction, such as in a Venturi tube, increases by a factor equal to the square of the factor by which the diameter decreases. (The converse applies for flow out of a constriction into a larger-diameter region.) 48. Water emerges straight down from a faucet with a 1.80-cm diameter at a speed of 0.500 m/s. (Because of the construction of the faucet, there is no variation in speed across the stream.) (a) What is the flow rate in $\displaystyle cm^3/s$? (b) What is the diameter of the stream 0.200 m below the faucet? Neglect any effects due to surface tension. Solution (a) $\displaystyle 127cm^3/s$ (b) 0.890 cm 49. Unreasonable Results A mountain stream is 10.0 m wide and averages 2.00 m in depth. During the spring runoff, the flow in the stream reaches $\displaystyle 100,000m^3/s$. (a) What is the average velocity of the stream under these conditions? (b) What is unreasonable about this velocity? (c) What is unreasonable or inconsistent about the premises? 12.2: Bernoulli’s Equation 50. Verify that pressure has units of energy per unit volume. Solution $\displaystyle P=\frac{Force}{Area}$, $\displaystyle (P)_{units}=N/m^2=N⋅m/m^3=J/m^3$ $\displaystyle =energy/volume$ 51. Suppose you have a wind speed gauge like the pitot tube shown in [link](b). By what factor must wind speed increase to double the value of $\displaystyle h$ in the manometer? Is this independent of the moving fluid and the fluid in the manometer? 52. If the pressure reading of your pitot tube is 15.0 mm Hg at a speed of 200 km/h, what will it be at 700 km/h at the same altitude? Solution 184 mm Hg 53. Calculate the maximum height to which water could be squirted with the hose in [link] example if it: (a) Emerges from the nozzle. (b) Emerges with the nozzle removed, assuming the same flow rate. 54. Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. Approximately what is the force due to the Bernoulli effect on a roof having an area of $\displaystyle 220m^2$? Typical air density in Boulder is $\displaystyle 1.14kg/m^3$, and the corresponding atmospheric pressure is $\displaystyle 8.89×10^4N/m^2$. (Bernoulli’s principle as stated in the text assumes laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.) Solution $\displaystyle 2.54×10^5N$ 55. a) Calculate the approximate force on a square meter of sail, given the horizontal velocity of the wind is 6.00 m/s parallel to its front surface and 3.50 m/s along its back surface. Take the density of air to be $\displaystyle 1.29 kg/m^3$. (The calculation, based on Bernoulli’s principle, is approximate due to the effects of turbulence.) (b) Discuss whether this force is great enough to be effective for propelling a sailboat. 56. (a) What is the pressure drop due to the Bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire hose while carrying a flow of 40.0 L/s? (b) To what maximum height above the nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.) Solution (a) $\displaystyle 1.58×10^6N/m^2$ (b) 163 m 57. (a) Using Bernoulli’s equation, show that the measured fluid speed v for a pitot tube, like the one in Figure(b), is given by $\displaystyle v=(\frac{2ρ′gh}{ρ})^{1/2}$, where $\displaystyle h$ is the height of the manometer fluid, $\displaystyle ρ′$ is the density of the manometer fluid, $\displaystyle ρ\(\displaystyle is the density of the moving fluid, and \(\displaystyle g\(\displaystyle is the acceleration due to gravity. (Note that v is indeed proportional to the square root of \(\displaystyle h$, as stated in the text.) (b) Calculate $\displaystyle v$ for moving air if a mercury manometer’s $\displaystyle h$ is 0.200 m. 12.3: The Most General Applications of Bernoulli’s Equation 58. Hoover Dam on the Colorado River is the highest dam in the United States at 221 m, with an output of 1300 MW. The dam generates electricity with water taken from a depth of 150 m and an average flow rate of $\displaystyle 650m^3/s$. (a) Calculate the power in this flow. (b) What is the ratio of this power to the facility’s average of 680 MW? Solution (a) $\displaystyle 9.56×10^8W$ (b) 1.4 59. A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.) (a) At takeoff, an aircraft travels at 60.0 m/s, so that the air speed relative to the bottom of the wing is 60.0 m/s. Given the sea level density of air to be $\displaystyle 1.29kg/m^3$, how fast must it move over the upper surface to create the ideal lift? (b) How fast must air move over the upper surface at a cruising speed of 245 m/s and at an altitude where air density is one-fourth that at sea level? (Note that this is not all of the aircraft’s lift—some comes from the body of the plane, some from engine thrust, and so on. Furthermore, Bernoulli’s principle gives an approximate answer because flow over the wing creates turbulence.) 60. The left ventricle of a resting adult’s heart pumps blood at a flow rate of $\displaystyle 83.0cm^3/s$, increasing its pressure by 110 mm Hg, its speed from zero to 30.0 cm/s, and its height by 5.00 cm. (All numbers are averaged over the entire heartbeat.) Calculate the total power output of the left ventricle. Note that most of the power is used to increase blood pressure. Solution 1.26 W 61. A sump pump (used to drain water from the basement of houses built below the water table) is draining a flooded basement at the rate of 0.750 L/s, with an output pressure of $\displaystyle 3.00×10^5N/m^2$. (a) The water enters a hose with a 3.00-cm inside diameter and rises 2.50 m above the pump. What is its pressure at this point? (b) The hose goes over the foundation wall, losing 0.500 m in height, and widens to 4.00 cm in diameter. What is the pressure now? You may neglect frictional losses in both parts of the problem. 12.4: Viscosity and Laminar Flow; Poiseuille’s Law 62. (a) Calculate the retarding force due to the viscosity of the air layer between a cart and a level air track given the following information—air temperature is 20º C , the cart is moving at 0.400 m/s, its surface area is $\displaystyle 2.50×10^{−2}m^2$, and the thickness of the air layer is $\displaystyle 6.00×10^{−5}m$. (b) What is the ratio of this force to the weight of the 0.300-kg cart? Solution (a) $\displaystyle 3.02×10^{−3}N$ (b) $\displaystyle 1.03×10^{−3}$ 63. What force is needed to pull one microscope slide over another at a speed of 1.00 cm/s, if there is a 0.500-mm-thick layer of 20º C water between them and the contact area is $\displaystyle 8.00cm^2$? 64. A glucose solution being administered with an IV has a flow rate of $\displaystyle 4.00cm^3/min$. What will the new flow rate be if the glucose is replaced by whole blood having the same density but a viscosity 2.50 times that of the glucose? All other factors remain constant. Solution $\displaystyle 1.60 cm^3/min$ 65. The pressure drop along a length of artery is 100 Pa, the radius is 10 mm, and the flow is laminar. The average speed of the blood is 15 mm/s. (a) What is the net force on the blood in this section of artery? (b) What is the power expended maintaining the flow? 66. A small artery has a length of $\displaystyle 1.1×10^{−3}m$ and a radius of $\displaystyle 2.5×10^{−5}m$. If the pressure drop across the artery is 1.3 kPa, what is the flow rate through the artery? (Assume that the temperature is 37º C.) Solution $\displaystyle 8.7×10^{−11}m^3/s$ 67. Fluid originally flows through a tube at a rate of $\displaystyle 100cm^3/s$. To illustrate the sensitivity of flow rate to various factors, calculate the new flow rate for the following changes with all other factors remaining the same as in the original conditions. (a) Pressure difference increases by a factor of 1.50. (b) A new fluid with 3.00 times greater viscosity is substituted. (c) The tube is replaced by one having 4.00 times the length. (d) Another tube is used with a radius 0.100 times the original. (e) Yet another tube is substituted with a radius 0.100 times the original and half the length, and the pressure difference is increased by a factor of 1.50. 68. The arterioles (small arteries) leading to an organ, constrict in order to decrease flow to the organ. To shut down an organ, blood flow is reduced naturally to 1.00% of its original value. By what factor did the radii of the arterioles constrict? Penguins do this when they stand on ice to reduce the blood flow to their feet. Solution 0.316 69. Angioplasty is a technique in which arteries partially blocked with plaque are dilated to increase blood flow. By what factor must the radius of an artery be increased in order to increase blood flow by a factor of 10? 70. (a) Suppose a blood vessel’s radius is decreased to 90.0% of its original value by plaque deposits and the body compensates by increasing the pressure difference along the vessel to keep the flow rate constant. By what factor must the pressure difference increase? (b) If turbulence is created by the obstruction, what additional effect would it have on the flow rate? Solution (a) 1.52 (b) Turbulence will decrease the flow rate of the blood, which would require an even larger increase in the pressure difference, leading to higher blood pressure. 71. A spherical particle falling at a terminal speed in a liquid must have the gravitational force balanced by the drag force and the buoyant force. The buoyant force is equal to the weight of the displaced fluid, while the drag force is assumed to be given by Stokes Law, $\displaystyle F_s=6πrηv$. Show that the terminal speed is given by $\displaystyle v=\frac{2R^2g}{9η}(ρ_s−ρ_1),$ where $\displaystyle R$ is the radius of the sphere, $\displaystyle ρ_s$ is its density, and $\displaystyle ρ_1$ is the density of the fluid and $\displaystyle η$ the coefficient of viscosity. 72. Using the equation of the previous problem, find the viscosity of motor oil in which a steel ball of radius 0.8 mm falls with a terminal speed of 4.32 cm/s. The densities of the ball and the oil are 7.86 and 0.88 g/mL, respectively. Solution $\displaystyle 225mPa⋅s$ 73. A skydiver will reach a terminal velocity when the air drag equals their weight. For a skydiver with high speed and a large body, turbulence is a factor. The drag force then is approximately proportional to the square of the velocity. Taking the drag force to be $\displaystyle F_D=\frac{1}{2}ρAv^2$ and setting this equal to the person’s weight, find the terminal speed for a person falling “spread eagle.” Find both a formula and a number for $\displaystyle v_t$, with assumptions as to size. 74. A layer of oil 1.50 mm thick is placed between two microscope slides. Researchers find that a force of $\displaystyle 5.50×10^{−4}N$ is required to glide one over the other at a speed of 1.00 cm/s when their contact area is $\displaystyle 6.00cm^2$. What is the oil’s viscosity? What type of oil might it be? Solution $\displaystyle 0.138 Pa⋅s$ or Olive oil. 75. (a) Verify that a 19.0% decrease in laminar flow through a tube is caused by a 5.00% decrease in radius, assuming that all other factors remain constant, as stated in the text. (b) What increase in flow is obtained from a 5.00% increase in radius, again assuming all other factors remain constant? 76. Example dealt with the flow of saline solution in an IV system. (a) Verify that a pressure of $\displaystyle 1.62×10^4N/m^2$ is created at a depth of 1.61 m in a saline solution, assuming its density to be that of sea water. (b) Calculate the new flow rate if the height of the saline solution is decreased to 1.50 m. (c) At what height would the direction of flow be reversed? (This reversal can be a problem when patients stand up.) Solution (a) $\displaystyle 1.62×10^4N/m^2$ (b) $\displaystyle 0.111 cm^3/s$ (c)10.6 cm 77. When physicians diagnose arterial blockages, they quote the reduction in flow rate. If the flow rate in an artery has been reduced to 10.0% of its normal value by a blood clot and the average pressure difference has increased by 20.0%, by what factor has the clot reduced the radius of the artery? 78. During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.0% of its normal value, and the blood pressure difference across the circulatory system has increased by 50.0%. By what factor has the average radii of her blood vessels increased? Solution 1.59 79. Water supplied to a house by a water main has a pressure of $\displaystyle 3.00×10^5N/m^2$ early on a summer day when neighborhood use is low. This pressure produces a flow of 20.0 L/min through a garden hose. Later in the day, pressure at the exit of the water main and entrance to the house drops, and a flow of only 8.00 L/min is obtained through the same hose. (a) What pressure is now being supplied to the house, assuming resistance is constant? (b) By what factor did the flow rate in the water main increase in order to cause this decrease in delivered pressure? The pressure at the entrance of the water main is $\displaystyle 5.00×10^5N/m^2$, and the original flow rate was 200 L/min. (c) How many more users are there, assuming each would consume 20.0 L/min in the morning? 80. An oil gusher shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter. Neglecting air resistance but not the resistance of the pipe, and assuming laminar flow, calculate the gauge pressure at the entrance of the 50.0-m-long vertical pipe. Take the density of the oil to be $\displaystyle 900kg/m^3$ and its viscosity to be $\displaystyle 1.00(N/m^2)⋅s$ (or $\displaystyle 1.00Pa⋅s$). Note that you must take into account the pressure due to the 50.0-m column of oil in the pipe. Solution $\displaystyle 2.95×10^6N/m^2$(gauge pressure) 81. Concrete is pumped from a cement mixer to the place it is being laid, instead of being carried in wheelbarrows. The flow rate is 200.0 L/min through a 50.0-m-long, 8.00-cm-diameter hose, and the pressure at the pump is $\displaystyle 8.00×10^6N/m^2$. (a) Calculate the resistance of the hose. (b) What is the viscosity of the concrete, assuming the flow is laminar? (c) How much power is being supplied, assuming the point of use is at the same level as the pump? You may neglect the power supplied to increase the concrete’s velocity. 82. Construct Your Own Problem Consider a coronary artery constricted by arteriosclerosis. Construct a problem in which you calculate the amount by which the diameter of the artery is decreased, based on an assessment of the decrease in flow rate. 83. Consider a river that spreads out in a delta region on its way to the sea. Construct a problem in which you calculate the average speed at which water moves in the delta region, based on the speed at which it was moving up river. Among the things to consider are the size and flow rate of the river before it spreads out and its size once it has spread out. You can construct the problem for the river spreading out into one large river or into multiple smaller rivers. 12.5: The Onset of Turbulence 84. Verify that the flow of oil is laminar (barely) for an oil gusher that shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter. The vertical pipe is 50 m long. Take the density of the oil to be $\displaystyle 900 kg/m^3$ and its viscosity to be $\displaystyle 1.00(N/m^2)⋅s$ (or $\displaystyle 1.00 Pa⋅s$). Solution $\displaystyle N_R=1.99×10^2< 2000$ 85. Show that the Reynolds number $\displaystyle N_R$ is unitless by substituting units for all the quantities in its definition and cancelling. 86. Calculate the Reynolds numbers for the flow of water through (a) a nozzle with a radius of 0.250 cm and (b) a garden hose with a radius of 0.900 cm, when the nozzle is attached to the hose. The flow rate through hose and nozzle is 0.500 L/s. Can the flow in either possibly be laminar? Solution (a) nozzle: $\displaystyle 1.27×10^5$, not laminar (b) hose: $\displaystyle 3.51×10^4$, not laminar. 87. A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of $\displaystyle 1.62×10^6N/m^2$. The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Calculate the Reynolds numbers for flow in the fire hose and nozzle to show that the flow in each must be turbulent. 88. Concrete is pumped from a cement mixer to the place it is being laid, instead of being carried in wheelbarrows. The flow rate is 200.0 L/min through a 50.0-m-long, 8.00-cm-diameter hose, and the pressure at the pump is $\displaystyle 8.00×10^6N/m^2$. Verify that the flow of concrete is laminar taking concrete’s viscosity to be $\displaystyle 48.0(N/m^2)⋅s$, and given its density is $\displaystyle 2300 kg/m^3$. Solution 2.54 << 2000, laminar. 89. At what flow rate might turbulence begin to develop in a water main with a 0.200-m diameter? Assume a $\displaystyle 20º C$ temperature. 90. What is the greatest average speed of blood flow at $\displaystyle 37º C$ in an artery of radius 2.00 mm if the flow is to remain laminar? What is the corresponding flow rate? Take the density of blood to be $\displaystyle 1025 kg/m^3$. Solution 1.02 m/s $\displaystyle 1.28×10^{–2}L/s$ 91. In Take-Home Experiment: Inhalation, we measured the average flow rate $\displaystyle Q$ of air traveling through the trachea during each inhalation. Now calculate the average air speed in meters per second through your trachea during each inhalation. The radius of the trachea in adult humans is approximately $\displaystyle 10^{−2}m$. From the data above, calculate the Reynolds number for the air flow in the trachea during inhalation. Do you expect the air flow to be laminar or turbulent? 92. Gasoline is piped underground from refineries to major users. The flow rate is $\displaystyle 3.00×10^{–2}m^3/s$ (about 500 gal/min), the viscosity of gasoline is $\displaystyle 1.00×10^{–3}(N/m^2)⋅s$, and its density is $\displaystyle 680kg/m^3$. (a) What minimum diameter must the pipe have if the Reynolds number is to be less than 2000? (b) What pressure difference must be maintained along each kilometer of the pipe to maintain this flow rate? Solution (a) $\displaystyle ≥ 13.0 m$ (b) $\displaystyle 2.68×10^{−6}N/m^2$ 93. Assuming that blood is an ideal fluid, calculate the critical flow rate at which turbulence is a certainty in the aorta. Take the diameter of the aorta to be 2.50 cm. (Turbulence will actually occur at lower average flow rates, because blood is not an ideal fluid. Furthermore, since blood flow pulses, turbulence may occur during only the high-velocity part of each heartbeat.) 94. Unreasonable Results A fairly large garden hose has an internal radius of 0.600 cm and a length of 23.0 m. The nozzleless horizontal hose is attached to a faucet, and it delivers 50.0 L/s. (a) What water pressure is supplied by the faucet? (b) What is unreasonable about this pressure? (c) What is unreasonable about the premise? (d) What is the Reynolds number for the given flow? (Take the viscosity of water as $\displaystyle 1.005×10^{–3}(N/m^2)⋅s$.) Solution (a) 23.7 atm or $\displaystyle 344 lb/in^2$ (b) The pressure is much too high. (c) The assumed flow rate is very high for a garden hose. (d) $\displaystyle 5.27×10^6 > > 3000$, turbulent, contrary to the assumption of laminar flow when using this equation. 12.7: Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes 95. You can smell perfume very shortly after opening the bottle. To show that it is not reaching your nose by diffusion, calculate the average distance a perfume molecule moves in one second in air, given its diffusion constant $\displaystyle D$ to be $\displaystyle 1.00×10^{–6}m^2/s$. Solution $\displaystyle 1.41×10^{−3}m$ 96. What is the ratio of the average distances that oxygen will diffuse in a given time in air and water? Why is this distance less in water (equivalently, why is $\displaystyle D$ less in water)? 97. Oxygen reaches the veinless cornea of the eye by diffusing through its tear layer, which is 0.500-mm thick. How long does it take the average oxygen molecule to do this? Solution $\displaystyle 1.3×10^2s$ 98. (a) Find the average time required for an oxygen molecule to diffuse through a 0.200-mm-thick tear layer on the cornea. (b) How much time is required to diffuse $\displaystyle 0.500cm^3$ of oxygen to the cornea if its surface area is $\displaystyle 1.00cm^2$? 99. Suppose hydrogen and oxygen are diffusing through air. A small amount of each is released simultaneously. How much time passes before the hydrogen is 1.00 s ahead of the oxygen? Such differences in arrival times are used as an analytical tool in gas chromatography. Solution 0.391 s Contributors and Attributions - Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .	libretexts	2025-03-17T19:53:31.420302	2017-04-20T14:44:08	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/12%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications/12.E%3A_Fluid_Dynamics_and_Its_Biological_and_Medical_Applications_(Exercises)", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "12.E: Fluid Dynamics and Its Biological and Medical Applications (Exercises)", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws	13: Temperature, Kinetic Theory, and the Gas Laws What is heat? How do we define it? How is it related to temperature? What are heat’s effects? How is it related to other forms of energy and to work? We will find that, in spite of the richness of the phenomena, there is a small set of underlying physical principles that unite the subjects and tie them to other fields. Thumbnail: High heat resulted in thermal expansion of these rails, buckling this section of railway. Imaged used with permission (Public domain; U.S. Department of Transportation)	libretexts	2025-03-17T19:53:31.478395	2015-10-27T19:35:21	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "13: Temperature, Kinetic Theory, and the Gas Laws", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.00%3A_Prelude_to_Temperature_Kinetic_Theory_and_the_Gas_Laws	13.0: Prelude to Temperature, Kinetic Theory, and the Gas Laws Heat is something familiar to each of us. We feel the warmth of the summer Sun, the chill of a clear summer night, the heat of coffee after a winter stroll, and the cooling effect of our sweat. Heat transfer is maintained by temperature differences. Manifestations of heat transfer —the movement of heat energy from one place or material to another—are apparent throughout the universe. Heat from beneath Earth’s surface is brought to the surface in flows of incandescent lava. The Sun warms Earth’s surface and is the source of much of the energy we find on it. Rising levels of atmospheric carbon dioxide threaten to trap more of the Sun’s energy, perhaps fundamentally altering the ecosphere. In space, supernovas explode, briefly radiating more heat than an entire galaxy does. What is heat? How do we define it? How is it related to temperature? What are heat’s effects? How is it related to other forms of energy and to work? We will find that, in spite of the richness of the phenomena, there is a small set of underlying physical principles that unite the subjects and tie them to other fields.	libretexts	2025-03-17T19:53:31.535357	2015-11-01T05:37:27	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.00%3A_Prelude_to_Temperature_Kinetic_Theory_and_the_Gas_Laws", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "13.0: Prelude to Temperature, Kinetic Theory, and the Gas Laws", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.01%3A_Temperature	13.1: Temperature Learning Objectives By the end of this section, you will be able to: - Define temperature. - Convert temperatures between the Celsius, Fahrenheit, and Kelvin scales. - Define thermal equilibrium. - State the zeroth law of thermodynamics. The concept of temperature has evolved from the common concepts of hot and cold. Human perception of what feels hot or cold is a relative one. For example, if you place one hand in hot water and the other in cold water, and then place both hands in tepid water, the tepid water will feel cool to the hand that was in hot water, and warm to the one that was in cold water. The scientific definition of temperature is less ambiguous than your senses of hot and cold. Temperature is operationally defined to be what we measure with a thermometer. (Many physical quantities are defined solely in terms of how they are measured. We shall see later how temperature is related to the kinetic energies of atoms and molecules, a more physical explanation.) Two accurate thermometers, one placed in hot water and the other in cold water, will show the hot water to have a higher temperature. If they are then placed in the tepid water, both will give identical readings (within measurement uncertainties). In this section, we discuss temperature, its measurement by thermometers, and its relationship to thermal equilibrium. Again, temperature is the quantity measured by a thermometer. Misconception Alert: Human Perception vs. Reality On a cold winter morning, the wood on a porch feels warmer than the metal of your bike. The wood and bicycle are in thermal equilibrium with the outside air, and are thus the same temperature. They feel different because of the difference in the way that they conduct heat away from your skin. The metal conducts heat away from your body faster than the wood does (see more about conductivity in Conduction ). This is just one example demonstrating that the human sense of hot and cold is not determined by temperature alone. Another factor that affects our perception of temperature is humidity. Most people feel much hotter on hot, humid days than on hot, dry days. This is because on humid days, sweat does not evaporate from the skin as efficiently as it does on dry days. It is the evaporation of sweat (or water from a sprinkler or pool) that cools us off. Any physical property that depends on temperature, and whose response to temperature is reproducible, can be used as the basis of a thermometer. Because many physical properties depend on temperature, the variety of thermometers is remarkable. For example, volume increases with temperature for most substances. This property is the basis for the common alcohol thermometer, the old mercury thermometer, and the bimetallic strip ( Figure ). Other properties used to measure temperature include electrical resistance and color, as shown in Figure , and the emission of infrared radiation, as shown in Figure . Temperature Scales Thermometers are used to measure temperature according to well-defined scales of measurement, which use pre-defined reference points to help compare quantities. The three most common temperature scales are the Fahrenheit, Celsius, and Kelvin scales. A temperature scale can be created by identifying two easily reproducible temperatures. The freezing and boiling temperatures of water at standard atmospheric pressure are commonly used. The Celsius scale (which replaced the slightly different centigrade scale) has the freezing point of water at $0^o \, C$ and the boiling point at $100^o \, C$. Its unit is the degree Celsius $(^oC)$. On the Fahrenheit scale (still the most frequently used in the United States), the freezing point of water is at $37^o \, F$ and the boiling point is at $212^o \, F$. The unit of temperature on this scale is the degree Fahrenheit $(^oF)$. Note that a temperature difference of one degree Celsius is greater than a temperature difference of one degree Fahrenheit. Only 100 Celsius degrees span the same range as 180 Fahrenheit degrees, thus one degree on the Celsius scale is 1.8 times larger than one degree on the Fahrenheit scale $180/100 = 9/5$. The Kelvin scale is the temperature scale that is commonly used in science. It is an absolute temperature scale defined to have 0 K at the lowest possible temperature, called absolute zero . The official temperature unit on this scale is the kelvin , which is abbreviated K, and is not accompanied by a degree sign. The freezing and boiling points of water are 273.15 K and 373.15 K, respectively. Thus, the magnitude of temperature differences is the same in units of kelvins and degrees Celsius. Unlike other temperature scales, the Kelvin scale is an absolute scale. It is used extensively in scientific work because a number of physical quantities, such as the volume of an ideal gas, are directly related to absolute temperature. The kelvin is the SI unit used in scientific work. The relationships between the three common temperature scales is shown in Figure . Temperatures on these scales can be converted using the equations in Table . \| To convert from . . . \| Use this equation . . . \| Also written as . . . \| \|---\|---\|---\| \| Celsius to Fahrenheit \| $T(^oF) = \frac{9}{5}T(C^o) +32$ \| $T_{^oF} = \frac{9}{5}T_{^oC} +32$ \| \| Fahrenheit to Celsius \| $T(^oC) = \frac{9}{5}(T(F^o) -32)$ \| $T_{^oC} = \frac{5}{9}(T_{^oF} - 32)$ \| \| Celsius to Kelvin \| $T(K) = T(^oC) + 273.15$ \| $T_K = T_{^oC} + 273.15$ \| \| Kelvin to Celsius \| $T^oC) = T(K) - 273.15$ \| $T_{^oC} = T_K - 273.15$ \| \| Fahrenheit to Kelvin \| $T(K) = \frac{5}{9}(T(^oF) - 32) + 273.15$ \| $T_K = \frac{5}{9}(T_{^oF} - 32) + 273.15$ \| \| Kelvin to Fahrenheit \| $T(^oF) = \frac{9}{5}(T(K) - 273.15) + 32$ \| $T_{^oF} = \frac{9}{5}(T_K - 273.15) + 32$ \| Notice that the conversions between Fahrenheit and Kelvin look quite complicated. In fact, they are simple combinations of the conversions between Fahrenheit and Celsius, and the conversions between Celsius and Kelvin. Example $\PageIndex{1}$: Converting between Temperature Scales: Room Temperature “Room temperature” is generally defined to be $25^oC$ (a) What is room temperature in $^oF$? (b) What is it in K? Strategy To answer these questions, all we need to do is choose the correct conversion equations and plug in the known values. Solution for (a) 1. Choose the right equation. To convert from $^oC$ to $^oF$, use this equation \[T_{^oF} = \dfrac{9}{5}T_{^oC} + 32.\] 2. Plug the known value into the equation and solve: \[T_{^oF} = \dfrac{9}{5}25^oC + 32 = 77^oF.\] Solution for (b) 1. Choose the right equation. To convert from $^oC$ to $K$, use this equation \[T_K = T_{^oC} + 273.15.\] 2. Plug the known value into the equation and solve: \[T_K = 25^oC + 273.15 = 298 \, K.\] Example $\PageIndex{2}$: Plug the known value into the equation and solve: The Reaumur scale is a temperature scale that was used widely in Europe in the 18th and 19th centuries. On the Reaumur temperature scale, the freezing point of water is $0^oR$ and the boiling temperature is $80^oR$. If “room temperature” is $25^oC$ on the Celsius scale, what is it on the Reaumur scale? Strategy To answer this question, we must compare the Reaumur scale to the Celsius scale. The difference between the freezing point and boiling point of water on the Reaumur scale is $80^oR.$ On the Celsius scale it is $100^oC$. Therefore $100^oC = 80^oR$. Both scales start at $0^o$ for freezing, so we can derive a simple formula to convert between temperatures on the two scales. Solution 1. Derive a formula to convert from one scale to the other: \[T_{^oR} = \dfrac{0.8^oR}{^oC} \times T_{^oC}.\] 2. Plug the known value into the equation and solve: \[T_{^oR} = \dfrac{0.8^oR}{^oC} \times 25^oC = 20^oR.\] Temperature Ranges in the Universe Figure shows the wide range of temperatures found in the universe. Human beings have been known to survive with body temperatures within a small range, from $24^oC$ to $44^oC \, (75^oF$ to $111^oF)$. The average normal body temperature is usually given as $37.0^oC \, (98.6^oF)$, and variations in this temperature can indicate a medical condition: a fever, an infection, a tumor, or circulatory problems (see Figure ). The lowest temperatures ever recorded have been measured during laboratory experiments: $4.5 \times 10^{-10} K$ at the Massachusetts Institute of Technology (USA), and $1.0 \times 10^{-10} K$ at Helsinki University of Technology (Finland). In comparison, the coldest recorded place on Earth’s surface is Vostok, Antarctica at 183 K $(-89^oC)$ and the coldest place (outside the lab) known in the universe is the Boomerang Nebula, with a temperature of 1 K. Making Connections: Absolute Zero What is absolute zero? Absolute zero is the temperature at which all molecular motion has ceased. The concept of absolute zero arises from the behavior of gases. Figure shows how the pressure of gases at a constant volume decreases as temperature decreases. Various scientists have noted that the pressures of gases extrapolate to zero at the same temperature, $-273.15^oC.$ This extrapolation implies that there is a lowest temperature. This temperature is called absolute zero . Today we know that most gases first liquefy and then freeze, and it is not actually possible to reach absolute zero. The numerical value of absolute zero temperature is $-273.15^oC$ or 0K. Thermal Equilibrium and the Zeroth Law of Thermodynamics Thermometers actually take their own temperature, not the temperature of the object they are measuring. This raises the question of how we can be certain that a thermometer measures the temperature of the object with which it is in contact. It is based on the fact that any two systems placed in thermal contact (meaning heat transfer can occur between them) will reach the same temperature. That is, heat will flow from the hotter object to the cooler one until they have exactly the same temperature. The objects are then in thermal equilibrium , and no further changes will occur. The systems interact and change because their temperatures differ, and the changes stop once their temperatures are the same. Thus, if enough time is allowed for this transfer of heat to run its course, the temperature a thermometer registers does represent the system with which it is in thermal equilibrium. Thermal equilibrium is established when two bodies are in contact with each other and can freely exchange energy. Furthermore, experimentation has shown that if two systems, A and B, are in thermal equilibrium with each another, and B is in thermal equilibrium with a third system C, then A is also in thermal equilibrium with C. This conclusion may seem obvious, because all three have the same temperature, but it is basic to thermodynamics. It is called the zeroth law of thermodynamics . The Zeroth Law of Thermodynamics If two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C. This law was postulated in the 1930s, after the first and second laws of thermodynamics had been developed and named. It is called the zeroth law because it comes logically before the first and second laws (discussed in Thermodynamics ). An example of this law in action is seen in babies in incubators: babies in incubators normally have very few clothes on, so to an observer they look as if they may not be warm enough. However, the temperature of the air, the cot, and the baby is the same, because they are in thermal equilibrium, which is accomplished by maintaining air temperature to keep the baby comfortable. Check Your Understanding Does the temperature of a body depend on its size [Hide Solution] No, the system can be divided into smaller parts each of which is at the same temperature. We say that the temperature is an intensive quantity. Intensive quantities are independent of size. Section Summary - Temperature is the quantity measured by a thermometer. - Temperature is related to the average kinetic energy of atoms and molecules in a system. - Absolute zero is the temperature at which there is no molecular motion. - There are three main temperature scales: Celsius, Fahrenheit, and Kelvin. - Temperatures on one scale can be converted to temperatures on another scale using the following equations: \[ T_{^oF} = \frac{9}{5}T_{^oC} +32\] \[ T_{^oC} = \frac{5}{9}T_{^oF} -32)\] \[T_K = T_{^oC} + 273.15\] \[ T_{^oC} = T_K - 273.15\] - Systems are in thermal equilibrium when they have the same temperature. - Thermal equilibrium occurs when two bodies are in contact with each other and can freely exchange energy. - The zeroth law of thermodynamics states that when two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system, C, then A is also in thermal equilibrium with C. Glossary - temperature - the quantity measured by a thermometer - Celsius scale - temperature scale in which the freezing point of water is 0ºC and the boiling point of water is 100ºC - degree Celsius - unit on the Celsius temperature scale - Fahrenheit scale - temperature scale in which the freezing point of water is 32ºF and the boiling point of water is 212ºF - degree Fahrenheit - unit on the Fahrenheit temperature scale - Kelvin scale - temperature scale in which 0 K is the lowest possible temperature, representing absolute zero - absolute zero - the lowest possible temperature; the temperature at which all molecular motion ceases - thermal equilibrium - the condition in which heat no longer flows between two objects that are in contact; the two objects have the same temperature - zeroth law of thermodynamics - law that states that if two objects are in thermal equilibrium, and a third object is in thermal equilibrium with one of those objects, it is also in thermal equilibrium with the other object	libretexts	2025-03-17T19:53:31.630394	2015-11-01T05:37:44	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.01%3A_Temperature", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "13.1: Temperature", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.02%3A_Thermal_Expansion_of_Solids_and_Liquids	13.2: Thermal Expansion of Solids and Liquids Learning Objectives By the end of this section, you will be able to: - Define and describe thermal expansion. - Calculate the linear expansion of an object given its initial length, change in temperature, and coefficient of linear expansion. - Calculate the volume expansion of an object given its initial volume, change in temperature, and coefficient of volume expansion. - Calculate thermal stress on an object given its original volume, temperature change, volume change, and bulk modulus. The expansion of alcohol in a thermometer is one of many commonly encountered examples of thermal expansion , the change in size or volume of a given mass with temperature. Hot air rises because its volume increases, which causes the hot air’s density to be smaller than the density of surrounding air, causing a buoyant (upward) force on the hot air. The same happens in all liquids and gases, driving natural heat transfer upwards in homes, oceans, and weather systems. Solids also undergo thermal expansion. Railroad tracks and bridges, for example, have expansion joints to allow them to freely expand and contract with temperature changes. What are the basic properties of thermal expansion? First, thermal expansion is clearly related to temperature change. The greater the temperature change, the more a bimetallic strip will bend. Second, it depends on the material. In a thermometer, for example, the expansion of alcohol is much greater than the expansion of the glass containing it. What is the underlying cause of thermal expansion? As is discussed in Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature , an increase in temperature implies an increase in the kinetic energy of the individual atoms. In a solid, unlike in a gas, the atoms or molecules are closely packed together, but their kinetic energy (in the form of small, rapid vibrations) pushes neighboring atoms or molecules apart from each other. This neighbor-to-neighbor pushing results in a slightly greater distance, on average, between neighbors, and adds up to a larger size for the whole body. For most substances under ordinary conditions, there is no preferred direction, and an increase in temperature will increase the solid’s size by a certain fraction in each dimension. Linear Thermal Expansion—Thermal Expansion in One Dimension The change in length $\Delta L$ is proportional to length $L$. The dependence of thermal expansion on temperature, substance, and length is summarized in the equation \[\Delta L = \alpha L \Delta T, \label{linear}\] where $\Delta L$ is the change in length $L$, $\Delta T$ is the change in temperature, and $\alpha$ is the coefficient of linear expansion , which varies slightly with temperature. Table $\PageIndex{1}$ lists representative values of the coefficient of linear expansion, which may have units of $1/^oC$ or 1/K. Because the size of a kelvin and a degree Celsius are the same, both $\alpha$ and $\Delta T$ can be expressed in units of kelvins or degrees Celsius. The equation $\Delta L = \alpha L \Delta T$ is accurate for small changes in temperature and can be used for large changes in temperature if an average value of $\alpha$ is used. \| Material \| Coefficient of linear expansion ($\alpha (1/^oC)$) \| Coefficient of volume expansion ($\beta (1/^oC)$) \| \|---\|---\|---\| \| Solids \| \|\| \| Aluminum \| $25 \times 10^{-6}$ \| $75 \times 10^{-6}$ \| \| Brass \| $19 \times 10^{-6}$ \| $56 \times 10^{-6}$ \| \| Copper \| $17 \times 10^{-6}$ \| $51 \times 10^{-6}$ \| \| Gold \| $14 \times 10^{-6}$ \| $42 \times 10^{-6}$ \| \| Iron or Steel \| $12 \times 10^{-6}$ \| $35 \times 10^{-6}$ \| \| Invar (Nickel-Iron alloy) \| $0.9 \times 10^{-6}$ \| $2.7 \times 10^{-6}$ \| \| Lead \| $29 \times 10^{-6}$ \| $87 \times 10^{-6}$ \| \| Silver \| $18 \times 10^{-6}$ \| $54 \times 10^{-6}$ \| \| Glass (ordinary) \| $9 \times 10^{-6}$ \| $27 \times 10^{-6}$ \| \| Glass (Pyrex) \| $3 \times 10^{-6}$ \| $9 \times 10^{-6}$ \| \| Quartz \| $0.4 \times 10^{-6}$ \| $1 \times 10^{-6}$ \| \| Concrete, Brick \| $\approx 12 \times 10^{-6}$ \| $\approx 36 \times 10^{-6}$ \| \| Marble (average) \| $7 \times 10^{-6}$ \| \| \| Liquids \| \|\| \| Ether \| N/A \| $1650 \times 10^{-6}$ \| \| Ethyl alcohol \| N/A \| $1100 \times 10^{-6}$ \| \| Petrol \| N/A \| $950 \times 10^{-6}$ \| \| Glycerin \| N/A \| $500 \times 10^{-6}$ \| \| Mercury \| N/A \| $180 \times 10^{-6}$ \| \| Water \| N/A \| $210 \times 10^{-6}$ \| \| Gases \| \|\| \| Air and most other gases at atm \| N/A \| $3400 \times 10^{-6}$ \| Example $\PageIndex{1}$: Calculating Linear Thermal Expansion of The Golden Gate Bridge The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest. The bridge is exposed to temperatures ranging from $-15^oC$ to $40^oC$. What is its change in length between these temperatures? Assume that the bridge is made entirely of steel. Strategy Use the equation for linear thermal expansion (Equation \ref{linear}) to calculate the change in length, $\Delta L$. Use the coefficient of linear expansion, $\alpha$, for steel from Table $\PageIndex{1}$, and note that the change in temperature, $\Delta T$, is $55^oC$. Solution Plug all of the known values into the equation to solve for $\Delta L$. \[ \begin{align} \Delta L &= \alpha L \Delta T \\[5pt] &= \left(\dfrac{12 \times 10^{-6}}{^oC}\right) (1275 \, m)(55^oC) \\[5pt] &= 0.84 \, m \end{align} \] Discussion Although not large compared with the length of the bridge, this change in length is observable. It is generally spread over many expansion joints so that the expansion at each joint is small. Thermal Expansion in Two and Three Dimensions Objects expand in all dimensions, as illustrated in Figure $\PageIndex{2}$. That is, their areas and volumes, as well as their lengths, increase with temperature. Holes also get larger with temperature. If you cut a hole in a metal plate, the remaining material will expand exactly as it would if the plug was still in place. The plug would get bigger, and so the hole must get bigger too. (Think of the ring of neighboring atoms or molecules on the wall of the hole as pushing each other farther apart as temperature increases. Obviously, the ring of neighbors must get slightly larger, so the hole gets slightly larger). Thermal Expansion in Two Dimensions For small temperature changes, the change in area $\Delta A$ is given by \[\Delta A = 2 \alpha A \Delta T,\] where $\Delta A$ is the change in area $A$, $\Delta T$ is the change in temperature, and $\alpha$ is the coefficient of linear expansion, which varies slightly with temperature. Thermal Expansion in Three Dimensions The change in volume $\Delta V$ is very nearly $\Delta V = 3\alpha V \Delta T$. This equation is usually written as \[\Delta V = \beta V \Delta T, \label{volume}\] where $\beta$ is the coefficient of volume expansion and $\beta \approx 3\alpha$. Note that the values of $\beta$ in Table $\PageIndex{1}$ are almost exactly equal to $3\alpha$. In general, objects will expand with increasing temperature. Water is the most important exception to this rule. Water expands with increasing temperature (its density decreases ) when it is at temperatures greater than $4^oC(40^oF)$. However, it expands with decreasing temperature when it is between $+4^oC$ and $0^oC(40^oF$ to $32^oF)$. Water is densest at $+4^oC$ (Figure $\PageIndex{1}$). Perhaps the most striking effect of this phenomenon is the freezing of water in a pond. When water near the surface cools down to $4^oC$ it is denser than the remaining water and thus will sink to the bottom. This “turnover” results in a layer of warmer water near the surface, which is then cooled. Eventually the pond has a uniform temperature of $4^oC$. If the temperature in the surface layer drops below $4^oC$, the water is less dense than the water below, and thus stays near the top. As a result, the pond surface can completely freeze over. The ice on top of liquid water provides an insulating layer from winter’s harsh exterior air temperatures. Fish and other aquatic life can survive in $4^oC$ water beneath ice, due to this unusual characteristic of water. It also produces circulation of water in the pond that is necessary for a healthy ecosystem of the body of water. Real-World Connections - Filling the Tank Differences in the thermal expansion of materials can lead to interesting effects at the gas station. One example is the dripping of gasoline from a freshly filled tank on a hot day. Gasoline starts out at the temperature of the ground under the gas station, which is cooler than the air temperature above. The gasoline cools the steel tank when it is filled. Both gasoline and steel tank expand as they warm to air temperature, but gasoline expands much more than steel, and so it may overflow. This difference in expansion can also cause problems when interpreting the gasoline gauge. The actual amount (mass) of gasoline left in the tank when the gauge hits “empty” is a lot less in the summer than in the winter. The gasoline has the same volume as it does in the winter when the “add fuel” light goes on, but because the gasoline has expanded, there is less mass. If you are used to getting another 40 miles on “empty” in the winter, beware—you will probably run out much more quickly in the summer. Example $\PageIndex{2}$: Calculating Thermal Expansion: Gas vs. Gas Tank Suppose your 60.0-L (15.9-gal) steel gasoline tank is full of gas, so both the tank and the gasoline have a temperature of $15^oC$. How much gasoline has spilled by the time they warm to $35.0^oC$? Strategy The tank and gasoline increase in volume, but the gasoline increases more, so the amount spilled is the difference in their volume changes. (The gasoline tank can be treated as solid steel.) We can use the equation for volume expansion to calculate the change in volume of the gasoline and of the tank. Solution 1. Use the equation for volume expansion (Equation \ref{volume}) to calculate the increase in volume of the steel tank: \[\Delta V_s = \beta_s V_s \Delta T. \nonumber\] 2. The increase in volume of the gasoline is given by this equation: \[\Delta V_{gas} = \beta_{gas} V_{gas} \Delta T. \nonumber\] 3. Find the difference in volume to determine the amount spilled as \[V_{spill} = \Delta V_{gas} - \Delta V_S. \nonumber\] Alternatively, we can combine these three equations into a single equation. (Note that the original volumes are equal.) \[ \begin{align} V_{spill} &= (\beta_{gas} - \beta_s)V \Delta T \\[5pt] &= [(950 - 35) \times 10^{-6} /^oC] (60.0 \, L)(20.0^oC) \\[5pt] &= 1.10 \, L. \end{align} \] Discussion This amount is significant, particularly for a 60.0-L tank. The effect is so striking because the gasoline and steel expand quickly. The rate of change in thermal properties is discussed in Heat and Heat Transfer Methods . If you try to cap the tank tightly to prevent overflow, you will find that it leaks anyway, either around the cap or by bursting the tank. Tightly constricting the expanding gas is equivalent to compressing it, and both liquids and solids resist being compressed with extremely large forces. To avoid rupturing rigid containers, these containers have air gaps, which allow them to expand and contract without stressing them. Thermal Stress Thermal stress is created by thermal expansion or contraction (see Elasticity: Stress and Strain for a discussion of stress and strain). Thermal stress can be destructive, such as when expanding gasoline ruptures a tank. It can also be useful, for example, when two parts are joined together by heating one in manufacturing, then slipping it over the other and allowing the combination to cool. Thermal stress can explain many phenomena, such as the weathering of rocks and pavement by the expansion of ice when it freezes. Example $\PageIndex{3}$: Calculating Thermal Stress - Gas Pressure What pressure would be created in the gasoline tank considered in Example $\PageIndex{2}$, if the gasoline increases in temperature from $15.0^oC$ to $35.0^oC$ without being allowed to expand? Assume that the bulk modulus $B$ for gasoline is $1.00 \times 10^9 \, N/m^2$. (For more on bulk modulus, see Elasticity: Stress and Strain .) Strategy To solve this problem, we must use the following equation, which relates a change in volume $\Delta V$ to pressure: \[\Delta V = \dfrac{1}{B}\dfrac{F}{A}V_0, \nonumber \] where $F/A$ is pressure, $V_0$ is the original volume, and $B$ is the bulk modulus of the material involved. We will use the amount spilled in Example $\PageIndex{2}$ as the change in volume, $\Delta V$. Solution 1. Rearrange the equation for calculating pressure: \[P = \dfrac{F}{A} = \dfrac{\Delta V}{V_0}B. \nonumber\] 2. Insert the known values. The bulk modulus for gasoline is $B = 1.00 \times 10^9 \, N/m^2$. In the previous example, the change in volume $\Delta V = 1.10 \, L$ is the amount that would spill. Here, $V_0 = 60.0 \, L$ is the original volume of the gasoline. Substituting these values into the equation, we obtain \[P = \dfrac{1.10 \, L}{60.0 \, L}(1.00 \times 10^9 Pa) = 1.83 \times 10^7 \, Pa. \nonumber\] Discussion This pressure is about $2500 \, lb/in^2$, much more than a gasoline tank can handle. Forces and pressures created by thermal stress are typically as great as that in the example above. Railroad tracks and roadways can buckle on hot days if they lack sufficient expansion joints. (Figure $\PageIndex{1}$). Power lines sag more in the summer than in the winter, and will snap in cold weather if there is insufficient slack. Cracks open and close in plaster walls as a house warms and cools. Glass cooking pans will crack if cooled rapidly or unevenly, because of differential contraction and the stresses it creates. (Pyrex is less susceptible because of its small coefficient of thermal expansion.) Nuclear reactor pressure vessels are threatened by overly rapid cooling, and although none have failed, several have been cooled faster than considered desirable. Biological cells are ruptured when foods are frozen, detracting from their taste. Repeated thawing and freezing accentuate the damage. Even the oceans can be affected. A significant portion of the rise in sea level that is resulting from global warming is due to the thermal expansion of sea water. Metal is regularly used in the human body for hip and knee implants. Most implants need to be replaced over time because, among other things, metal does not bond with bone. Researchers are trying to find better metal coatings that would allow metal-to-bone bonding. One challenge is to find a coating that has an expansion coefficient similar to that of metal. If the expansion coefficients are too different, the thermal stresses during the manufacturing process lead to cracks at the coating-metal interface. Another example of thermal stress is found in the mouth. Dental fillings can expand differently from tooth enamel. It can give pain when eating ice cream or having a hot drink. Cracks might occur in the filling. Metal fillings (gold, silver, etc.) are being replaced by composite fillings (porcelain), which have smaller coefficients of expansion, and are closer to those of teeth. Exercise $\PageIndex{1}$ Two blocks, A and B, are made of the same material. Block A has dimensions $l \times w \times h = L \times 2L \times L$ and Block B has dimensions $2L \times 2L \times 2L$. If the temperature changes, what is - the change in the volume of the two blocks, - the change in the cross-sectional area $l \times w$ and - the change in the height $h$ of the two blocks? - Answer a - The change in volume is proportional to the original volume. Block A has a volume of $L \times 2L \times L = 2L^3$. Block B has a volume of $2L \times 2L \times 2L = 8L^3$, which is 4 times that of Block A. Thus the change in volume of Block B should be 4 times the change in volume of Block A. - Answer b - The change in area is proportional to the area. The cross-sectional area of Block A is $L \times 2L = 2L^2$, while that of Block B is $2L \times 2L = 4L^2$. Because cross-sectional area of Block B is twice that of Block A, the change in the cross-sectional area of Block B is twice that of Block A. - Answer c - The change in height is proportional to the original height. Because the original height of Block B is twice that of A, the change in the height of Block B is twice that of Block A. Summary - Thermal expansion is the increase, or decrease, of the size (length, area, or volume) of a body due to a change in temperature. - Thermal expansion is large for gases, and relatively small, but not negligible, for liquids and solids. - Linear thermal expansion is \[\Delta L = \alpha L \Delta T, \nonumber\] where $\Delta L $ is the change in length $L$, $\Delta T$ is the change in temperature, and $\alpha$ is the coefficient of linear expansion, which varies slightly with temperature. - The change in area due to thermal expansion is \[\Delta A = 2\alpha A \delta T, \nonumber\] where $\Delta A$ is the change in area. - The change in volume due to thermal expansion is \[\Delta V = \beta V \delta T, \nonumber\] where $\beta$ is the coefficient of volume expansion and $\beta \approx 3\alpha$. Thermal stress is created when thermal expansion is constrained. Glossary - thermal expansion - the change in size or volume of an object with change in temperature - coefficient of linear expansion - the change in length, per unit length, per 1 degree Celsius, change in temperature; a constant used in the calculation of linear expansion; the coefficient of linear expansion depends on the material and to some degree on the temperature of the material - coefficient of volume expansion - the change in volume, per unit volume, per 1 degree Celsius,change in temperature - thermal stress - stress caused by thermal expansion or contraction	libretexts	2025-03-17T19:53:31.732180	2015-11-01T05:38:01	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.02%3A_Thermal_Expansion_of_Solids_and_Liquids", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "13.2: Thermal Expansion of Solids and Liquids", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.03%3A_The_Ideal_Gas_Law	13.3: The Ideal Gas Law Learning Objectives By the end of this section, you will be able to: - State the ideal gas law in terms of molecules and in terms of moles. - Use the ideal gas law to calculate pressure change, temperature change, volume change, or the number of molecules or moles in a given volume. - Use Avogadro’s number to convert between number of molecules and number of moles. In this section, we continue to explore the thermal behavior of gases. In particular, we examine the characteristics of atoms and molecules that compose gases. (Most gases, for example nitrogen, $N_2$ and oxygen, $O_2$ are composed of two or more atoms. We will primarily use the term “molecule” in discussing a gas because the term can also be applied to monatomic gases, such as helium.) Gases are easily compressed. We can see evidence of this elsewhere , where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same $ \beta$. This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates. The answer lies in the large separation of atoms and molecules in gases, compared to their sizes, as illustrated in Figure $\PageIndex{1}$. Because atoms and molecules have large separations, forces between them can be ignored, except when they collide with each other during collisions. The motion of atoms and molecules (at temperatures well above the boiling temperature) is fast, such that the gas occupies all of the accessible volume and the expansion of gases is rapid. In contrast, in liquids and solids, atoms and molecules are closer together and are quite sensitive to the forces between them. To get some idea of how pressure, temperature, and volume of a gas are related to one another, consider what happens when you pump air into an initially deflated tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the walls limit volume expansion. If we continue to pump air into it, the pressure increases. The pressure will further increase when the car is driven and the tires move. Most manufacturers specify optimal tire pressure for cold tires (Figure $\PageIndex{2}$). At room temperatures, collisions between atoms and molecules can be ignored. In this case, the gas is called an ideal gas, in which case the relationship between the pressure, volume, and temperature is given by the equation of state called the ideal gas law. Definition: Ideal Gas Law The ideal gas law states that \[PV = NkT, \label{igl}\] where $P$ is the absolute pressure of a gas, $V$ is the volume it occupies, $N$ is the number of atoms and molecules in the gas, and and $T$ is its absolute temperature. The constant $k$ is called the Boltzmann constant discussed below. Definition: Boltzmann constant The Boltzmann constant $k$ is named in honor of Austrian physicist Ludwig Boltzmann (1844–1906) and has the value \[k = 1.38 \times 10^{-23} \, J/K.\] The ideal gas law can be derived from basic principles, but was originally deduced from experimental measurements of Charles’ law (that volume occupied by a gas is proportional to temperature at a fixed pressure) and from Boyle’s law (that for a fixed temperature, the product $PV$ is a constant). In the ideal gas model, the volume occupied by its atoms and molecules is a negligible fraction of $V$. The ideal gas law describes the behavior of real gases under most conditions. (Note, for example, that $N$ is the total number of atoms and molecules, independent of the type of gas.) Let us see how the ideal gas law is consistent with the behavior of filling the tire when it is pumped slowly and the temperature is constant. At first, the pressure $P$ is essentially equal to atmospheric pressure, and the volume $V$ increases in direct proportion to the number of atoms and molecules $N$ put into the tire. Once the volume of the tire is constant, Equation \ref{igl} predicts that the pressure should increase in proportion to the number N of atoms and molecules . Example $\PageIndex{1}$: Calculating Pressure Changes Due to Temperature Changes - Tire Pressure Suppose your bicycle tire is fully inflated, with an absolute pressure of $7.00 \times 10^5 \, Pa$ (a gauge pressure of just under $90.0 \, lb/in^2$) at a temperature of $18.0^oC$. What is the pressure after its temperature has risen to $35.0^oC$? Assume that there are no appreciable leaks or changes in volume. Strategy The pressure in the tire is changing only because of changes in temperature. First we need to identify what we know and what we want to know, and then identify an equation to solve for the unknown. We know the initial pressure $P_0 = 7.00 \times 10^5 \, Pa$, the initial temperature $T_0 = 18.0^oC$, and the final temperature $T_F = 35.0^oC$. We must find the final pressure $P_f$. How can we use Equation \ref{igl}? At first, it may seem that not enough information is given, because the volume $V$ and number of atoms $N$ are not specified. What we can do is use the equation twice: \[P_0V_0 = NkT_0 \nonumber\] and \[P_fV_f = NkT_f \nonumber.\] If we divide $P_fV_f$ by $P_0V_0$, we can come up with an equation that allows us to solve for $P_f$ \[\dfrac{P_fV_f}{P_0V_0} = \dfrac{N_fkT_f}{N_0kT_0} \nonumber\] Since the volume is constant, $V_f$ and $V_0$ are the same and they cancel out. The same is true for $N_f$ and $N_0$, and $k$, which is a constant. Therefore, \[\dfrac{P_f}{P_0} = \dfrac{T_f}{T_0}. \nonumber\] We can then rearrange this to solve for $P_f$: \[P_f = P_0\dfrac{T_f}{T_0},\label{ex1a}\] where the temperature must be in units of kelvins, because $T_0$ and $T_f$ are absolute temperatures. Solution 1. Convert temperatures from Celsius to Kelvin. \[T_0 = (18.0 +273)K = 291 \, K \nonumber\] \[T_f = (35.0 + 273)K = 308 \, K \nonumber\] 2. Substitute the known values into Equation \ref{ex1a}. \[\begin{align} P_f &= P_0\dfrac{T_f}{T_0} \\[5pt] &= 7.00 \times 10^5 \, Pa \left(\dfrac{308 \, K)}{291 \, K}\right) \\[5pt] &= 7.41 \times 10^5 \, Pa \end{align} \] Discussion The final temperature is about 6% greater than the original temperature, so the final pressure is about 6% greater as well. Note that absolute pressure and absolute temperature must be used in the ideal gas law. Take-Home Experiment - Refrigerating a Balloon The Inflate a balloon at room temperature. Leave the inflated balloon in the refrigerator overnight. What happens to the balloon, and why? We left a balloon in a the freezer for a bit then I pulled it out to see what would happen How many molecules are in a typical object, such as gas in a tire or water in a drink? We can use the ideal gas law (Equation \ref{igl}) to give us an idea of how large $N$ typically is. Example $\PageIndex{2}$: Calculating the Number of Molecules in a Cubic Meter of Gas Calculate the number of molecules in a cubic meter of gas at standard temperature and pressure (STP), which is defined to be $0^oC$ and atmospheric pressure. Strategy Because pressure, volume, and temperature are all specified, we can use the ideal gas law (Equation \ref{igl}) to find $N$. Solution 1. Identify the knowns. - $T = 0^oC = 273 \, K$ - $P = 1.01 \times 10^5 \, Pa$ - $V = 1.00 \, m^3$ - $k = 1.38 \times 10^{-23} \, J/K$ 2. Identify the unknown: number of molecules, $N$. 3. Rearrange the ideal gas law to solve for $N$. \[\begin{align} N &= \dfrac{PV}{kT} \\[5pt] &= \dfrac{(1.01 \times 10^5\, Pa)(1.00 \, m^3)}{(1.38 \times 10^{-23} \, J/K)(273 \, K)} \\[5pt] &= 2.68 \times 10^{25} \, molecules \end{align}\] Discussion This number is undeniably large, considering that a gas is mostly empty space. $N$ is huge, even in small volumes. For example, $1\, cm^3$ of a gas at STP has $2.68 \times 10^{10}$ molecules in it. Once again, note that $N$ is the same for all types or mixtures of gases. Moles and Avogadro’s Number It is sometimes convenient to work with a unit other than molecules when measuring the amount of substance. A mole (abbreviated mol) is defined to be the amount of a substance that contains as many atoms or molecules as there are atoms in exactly 12 grams (0.012 kg) of carbon-12. The actual number of atoms or molecules in one mole is called Avogadro’s number $(N_A)$, in recognition of Italian scientist Amedeo Avogadro (1776–1856). He developed the concept of the mole, based on the hypothesis that equal volumes of gas, at the same pressure and temperature, contain equal numbers of molecules. That is, the number is independent of the type of gas. This hypothesis has been confirmed with an estimate of the value of Avogadro’s number. Definition: Avogadro’s Number One mole always contains $6.02 \times 10^{23}$ particles (atoms or molecules), independent of the element or substance. A mole of any substance has a mass in grams equal to its molecular mass, which can be calculated from the atomic masses given in the periodic table of elements. \[N_A = 6.02 \times 10^{23} \, mol^{-1}\] Exercise $\PageIndex{2}$ - The active ingredient in a Tylenol pill is 325 mg of acetaminophen $(C_8H_9NO_2)$. Find the number of active molecules of acetaminophen in a single pill. - Answer - We first need to calculate the molar mass (the mass of one mole) of acetaminophen. To do this, we need to multiply the number of atoms of each element by the element’s atomic mass. (8 moles of Carbon)(12 grams/mole) + (9 moles hydrogen)(1 gram/mole) + (1 mole of nitrogen)(14 grams/mole) + (2 moles oxygen)(16 grams/mole) = 151 grams. Then we need to calculate the number of moles in 325 mg. \[ \left ( \frac{325 \, mg}{151 \, grams/mole} \right) \left( \frac{1 \, gram}{1000 \, mg}\right) = 2.15 \times 10^{-3} \, moles \nonumber\] Then use Avogadro’s number to calculate the number of molecules. \[N = (2.15 \times 10^{-3} \, moles)(6.02 \times 10^{23} \, molecules/mole)\]\[ = 1.30 \times 10^{21} \, molecules \nonumber\] Example $\PageIndex{3}$: Calculating Moles per Cubic Meter and Liters per Mole Calculate: - the number of moles in $1.00 \, m^3$ of gas at STP, and - the number of liters of gas per mole. Strategy and Solution (a) We are asked to find the number of moles per cubic meter, and we know from Example that the number of molecules per cubic meter at STP is $2.68 \times 10^{25}$. The number of moles can be found by dividing the number of molecules by Avogadro’s number. We let $n$ stand for the number of moles, \[n \, mole/m^3 = \dfrac{N \, molecules/m^3}{6.02 \times 10^{23} \, molecules/mol} = \dfrac{2.68 \times 10^{25} \, molecules/m^3}{6.02 \times 10^{23} \, molecules/mol} = 44.5 \, mol/m^3\] (b) Using the value obtained for the number of moles in a cubic meter, and converting cubic meters to liters, we obtain \[\dfrac{(10^3 \, L/m^3)}{44.5 \, mol/m^3} = 22.5 \, L/mol.\] Discussion This value is very close to the accepted value of 22.4 L/mol. The slight difference is due to rounding errors caused by using three-digit input. Again this number is the same for all gases. In other words, it is independent of the gas. The (average) molar weight of air (approximately 80% $N_2$ and 20% $O_2$ is $M = 28.8 \, g$. Thus the mass of one cubic meter of air is 1.28 kg. If a living room has dimensions $5 \, m \times 5 \, m \times 3 \, m$, the mass of air inside the room is 96 kg, which is the typical mass of a human. Exercise $\PageIndex{3}$ The density of air at standard conditions $(P = 1 \, atm$ and $T = 20^oC)$ is $1.28 \, kg/m^3$. At what pressure is the density $0.64 \, kg/m^3$ if the temperature and number of molecules are kept constant? - Answer - The best way to approach this question is to think about what is happening. If the density drops to half its original value and no molecules are lost, then the volume must double. If we look at the equation $PV = NkT$, we see that when the temperature is constant, the pressure is inversely proportional to volume. Therefore, if the volume doubles, the pressure must drop to half its original value, and $P_f = 0.50 \, atm$. The Ideal Gas Law Restated Using Moles A very common expression of the ideal gas law uses the number of moles, $n$, rather than the number of atoms and molecules ($N$). We start from the ideal gas law, \[PV = NkT,\] and multiply and divide the equation by Avogadro’s number $N_A$. This gives \[PV = \dfrac{N}{N_A}N_AkT.\] Note that $n = N/N_A$ is the number of moles. We define the universal gas constant $R = N_Ak$, and obtain the ideal gas law in terms of moles. Definition: Ideal Gas Law (in terms of moles) The ideal gas law (in terms of moles) is \[PV = nRT. \label{iglmoles}\] where $P$ is the absolute pressure of a gas, $V$ is the volume it occupies, $n$ is the number of moles of atoms and molecules in the gas, and and $T$ is its absolute temperature. The constant $R$ is called the gas constant and varies depending on the units of the pressure and volume used. Definition: Gas Constant The numerical value of $R$ in SI units is \[ \begin{align} R &= N_Ak \\[5pt] &= (6.02 \times 10^{23} \, mol^{-1}) (1.38 \times 10^{-23} \, J/K) \\[5pt] &= 8.31 \, J/mol \cdot K. \end{align}\] In other units, - $R = 1.99 \, cal/mol \cdot K$ - $R = 0.0821 \, L \cdot atm/mol \cdot K$ You can use whichever value of $R$ is most convenient for a particular problem. Example $\PageIndex{4}$: Calculating Number of Moles - Gas in a Bike Tire How many moles of gas are in a bike tire with a volume of $2.00 \times 10^{-3} \,m^3(2.00 \, L)$, a pressure of $7.00 \times 10^5 \, Pa$ (a gauge pressure of just under $90.0 \, lb/in^2$), and at a temperature of $18.0^oC$? Strategy Identify the knowns and unknowns, and choose an equation to solve for the unknown. In this case, we solve the ideal gas law (Equation \ref{iglmoles}) for the number of moles $n$. Solution 1. Identify the knowns. - $P = 7.00 \times 10^5 \, Pa$ - $V = 2.00 \times 10^{-3} m^3$ - $T = 18.0^oC = 291 \, K$ - $R = 8.31 \, J/mol \cdot K$ 2. Rearrange Equation \ref{iglmoles} to solve for $n$ and substitute known values. \[\begin{align} n &= \dfrac{PV}{RT} \\[5pt] &= \dfrac{(7.00 \times 10^5 Pa)(2.00 \times 10^{-3} m^3)}{(8.31 \, J/mol\cdot K)(291 \, K)} \\[5pt] &= 0.579 \, mol \end{align}\] Discussion The most convenient choice for $R$ in this case is $8.31 \, J/mol \cdot K$, because our known quantities are in SI units. The pressure and temperature are obtained from the initial conditions in Example $\PageIndex{1}$, but we would get the same answer if we used the final values. The ideal gas law can be considered to be another manifestation of the Law of Conservation of Energy . Work done on a gas results in an increase in its energy, increasing pressure and/or temperature, or decreasing volume. This increased energy can also be viewed as increased internal kinetic energy, given the gas’s atoms and molecules. The Ideal Gas Law and Energy Let us now examine the role of energy in the behavior of gases. When you inflate a bike tire by hand, you do work by repeatedly exerting a force through a distance. This energy goes into increasing the pressure of air inside the tire and increasing the temperature of the pump and the air. The ideal gas law is closely related to energy: the units on both sides are joules. The right-hand side of the ideal gas law in $PV = NkT$ is $NkT$. This term is roughly the amount of translational kinetic energy of $N$ atoms or molecules at an absolute temperature $T$, as we shall see formally in Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature . The left-hand side of the ideal gas law is $PV$ which also has the units of joules. We know from our study of fluids that pressure is one type of potential energy per unit volume, so pressure multiplied by volume is energy. The important point is that there is energy in a gas related to both its pressure and its volume. The energy can be changed when the gas is doing work as it expands—something we explore in Heat and Heat Transfer Methods —similar to what occurs in gasoline or steam engines and turbines. Problem-Solving Strategy: The Ideal Gas Law - Step 1 Examine the situation to determine that an ideal gas is involved. Most gases are nearly ideal. - Step 2 Make a list of what quantities are given, or can be inferred from the problem as stated (identify the known quantities). Convert known values into proper SI units (K for temperature, Pa for pressure, $m^3$ for volume, molecules for $N$, and moles for $n$. - Step 3 Identify exactly what needs to be determined in the problem (identify the unknown quantities). A written list is useful. - Step 4 Determine whether the number of molecules or the number of moles is known, in order to decide which form of the ideal gas law to use. The first form is $PV = NkT$ and involves $N$, the number of atoms or molecules. The second form is $PV = nRT$ and involves $n$, the number of moles. - Step 5 Solve the ideal gas law for the quantity to be determined (the unknown quantity). You may need to take a ratio of final states to initial states to eliminate the unknown quantities that are kept fixed. - Step 6 Substitute the known quantities, along with their units, into the appropriate equation, and obtain numerical solutions complete with units. Be certain to use absolute temperature and absolute pressure. - Step 7 Check the answer to see if it is reasonable: Does it make sense? Exercise $\PageIndex{4}$ Liquids and solids have densities about 1000 times greater than gases. Explain how this implies that the distances between atoms and molecules in gases are about 10 times greater than the size of their atoms and molecules. - Answer - Atoms and molecules are close together in solids and liquids. In gases they are separated by empty space. Thus gases have lower densities than liquids and solids. Density is mass per unit volume, and volume is related to the size of a body (such as a sphere) cubed. So if the distance between atoms and molecules increases by a factor of 10, then the volume occupied increases by a factor of 1000, and the density decreases by a factor of 1000. Summary - The ideal gas law relates the pressure and volume of a gas to the number of gas molecules and the temperature of the gas. - The ideal gas law can be written in terms of the number of molecules of gas: \[PV = NkT,\] where $P$ is pressure, $V$ is volume, $T$ is temperature, $N$ is number of molecules, and $k$ is the Boltzmann constant \[ k = 1.38 \times 10^{-23} J/K.\] - A mole is the number of atomes in a 12-g sample of carbon-12. - The number of molecules in a mole is called Avogadro’s number $N_A$, \[N_A = 6.02 \times 10^{-23} mol^{-1}.\] - A mole of any substance has a mass in grams equal to its molecular weight, which can be determined from the periodic table of elements. - The ideal gas law can also be written and solved in terms of the number of moles of gas: \[PV = nRT,\] where $n$ is number of moles and $R$ is the universal gas constant, \[R = 8.31 \, J/mol\cdot K.\] - The ideal gas law is generally valid at temperatures well above the boiling temperature. Glossary - ideal gas law - the physical law that relates the pressure and volume of a gas to the number of gas molecules or number of moles of gas and the temperature of the gas - Boltzmann constant - $k$, a physical constant that relates energy to temperature; $k=1.38×10^{–23}J/K$ - Avogadro’s number - $N_A$, the number of molecules or atoms in one mole of a substance; $N_A=6.02×10^{23}$ particles/mole - mole - the quantity of a substance whose mass (in grams) is equal to its molecular mass	libretexts	2025-03-17T19:53:31.837269	2015-11-01T05:38:36	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.03%3A_The_Ideal_Gas_Law", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "13.3: The Ideal Gas Law", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.04%3A_Kinetic_Theory-_Atomic_and_Molecular_Explanation_of_Pressure_and_Temperature	13.4: Kinetic Theory- Atomic and Molecular Explanation of Pressure and Temperature Learning Objectives By the end of this section, you will be able to: - Express the ideal gas law in terms of molecular mass and velocity. - Define thermal energy. - Calculate the kinetic energy of a gas molecule, given its temperature. - Describe the relationship between the temperature of a gas and the kinetic energy of atoms and molecules. - Describe the distribution of speeds of molecules in a gas. We have developed macroscopic definitions of pressure and temperature. Pressure is the force divided by the area on which the force is exerted, and temperature is measured with a thermometer. We gain a better understanding of pressure and temperature from the kinetic theory of gases, which assumes that atoms and molecules are in continuous random motion. Figure shows an elastic collision of a gas molecule with the wall of a container, so that it exerts a force on the wall (by Newton’s third law). Because a huge number of molecules will collide with the wall in a short time, we observe an average force per unit area. These collisions are the source of pressure in a gas. As the number of molecules increases, the number of collisions and thus the pressure increase. Similarly, the gas pressure is higher if the average velocity of molecules is higher. The actual relationship is derived in the Things Great and Small feature below. The following relationship is found: \[PV = \dfrac{1}{3} Nm\overline{v^2},\] where $P$ is the pressure (average force per unit area), $V$ is the volume of gas in the container, $N$ is the number of molecules in the container, $m$ is the mass of a molecule, and $\overline{v^2}$ is the average of the molecular speed squared. What can we learn from this atomic and molecular version of the ideal gas law? We can derive a relationship between temperature and the average translational kinetic energy of molecules in a gas. Recall the previous expression of the ideal gas law: \[PV = NkT.\] Equating the right-hand side of this equation with the right-hand side of $PV = \frac{1}{3}Nmv^2$ gives \[\dfrac{1}{3}Nm\overline{v^2} = NkT.\] Making Connections: Things Great and Small—Atomic and Molecular Origin of Pressure in a Gas Figure shows a box filled with a gas. We know from our previous discussions that putting more gas into the box produces greater pressure, and that increasing the temperature of the gas also produces a greater pressure. But why should increasing the temperature of the gas increase the pressure in the box? A look at the atomic and molecular scale gives us some answers, and an alternative expression for the ideal gas law. The figure shows an expanded view of an elastic collision of a gas molecule with the wall of a container. Calculating the average force exerted by such molecules will lead us to the ideal gas law, and to the connection between temperature and molecular kinetic energy. We assume that a molecule is small compared with the separation of molecules in the gas, and that its interaction with other molecules can be ignored. We also assume the wall is rigid and that the molecule’s direction changes, but that its speed remains constant (and hence its kinetic energy and the magnitude of its momentum remain constant as well). This assumption is not always valid, but the same result is obtained with a more detailed description of the molecule’s exchange of energy and momentum with the wall. If the molecule’s velocity changes in the x-direction, its momentum changes from $-mv_x$ to $+mv_x$. Thus, its change in momentum is $\Delta mv = +mv_x -(-mv_x) = 2mv_x$. The force exerted on the molecule is given by \[F = \dfrac{\Delta p}{\Delta t} = \dfrac{2mv_x}{\Delta t}.\] There is no force between the wall and the molecule until the molecule hits the wall. During the short time of the collision, the force between the molecule and wall is relatively large. We are looking for an average force; we take $\Delta t$ to be the average time between collisions of the molecule with this wall. It is the time it would take the molecule to go across the box and back (a distance $2l$) at a speed of $v_x.$ Thus $\Delta t = 2l/v_x$, and the expression for the force becomes \[F = \dfrac{2mv_x}{2l/v_x} = \dfrac{mv_x^2}{l}.\] This force is due to one molecule. We multiply by the number of molecules $N$ and use their average squared velocity to find the force \[F = N\dfrac{m\overline{v_x^2}}{l},\] where the bar over a quantity means its average value. We would like to have the force in terms of the speed $v$, rather than the x-component of the velocity. We note that the total velocity squared is the sum of the squares of its x-components, so that \[\overline{v^2} = \overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2}. \] Because the velocities are random, their average components in all directions are the same: \[ \overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} \] Thus, \[ \overline{v^2} = 3 \overline{v_x^2}, \] or \[ \overline{v_x^2} = \dfrac{1}{3} \overline{v^2}. \] Substituting $\frac{1}{3} \overline{v^2}$ into the expression for $F$ gives \[F = N\dfrac{m\overline{v^2}}{33l}.\] The pressure is $F/A$, so that we obtain \[P = \dfrac{F}{A} = N\dfrac{m\overline{v^2}}{3Al} = \dfrac{1}{3} \dfrac{Nm\overline{v^2}}{V},\] where we used $V = Al$ for the volume. This gives the important result. \[PV = \dfrac{1}{3}Nm\overline{v^2}.\] This equation is another expression of the ideal gas law. We can get the average kinetic energy of a molecule, $\dfrac{1}{2} mv^2$, from the right-hand side of the equation by canceling $N$ and multiplying by 3/2. This calculation produces the result that the average kinetic energy of a molecule is directly related to absolute temperature. \[\overline{KE} = \dfrac{1}{2}m\overline{v^2} = \dfrac{3}{2}kT\] The average translational kinetic energy of a molecule, $\overline{KE}$, is called thermal energy . The equation $\overline{KE} = \frac{1}{2}m\overline{v^2} = \frac{3}{2}kT$ is a molecular interpretation of temperature, and it has been found to be valid for gases and reasonably accurate in liquids and solids. It is another definition of temperature based on an expression of the molecular energy. It is sometimes useful to rearrange $\overline{KE} = \frac{1}{2}m\overline{v^2} = \frac{3}{2}kT$ and solve for the average speed of molecules in a gas in terms of temperature, \[\sqrt{\overline{v^2}} = v_{rms} = \sqrt{\dfrac{3kT}{m}},\] where $v_{rms}$ stands for root-mean-square (rms) speed. Example $\PageIndex{1}$: Calculating Kinetic Energy and Speed of a Gas Molecule (a) What is the average kinetic energy of a gas molecule at $20^oC$ (room temperature)? (b) Find the rms speed of a nitrogen molecule $(N_2)$ at this temperature. Strategy for (a) The known in the equation for the average kinetic energy is the temperature. \[\overline{KE} = \dfrac{1}{2}m\overline{v^2} = \dfrac{3}{2}kT\] Before substituting values into this equation, we must convert the given temperature to kelvins. This conversion gives $T = (20.0 + 273) k = 293 \, K$. Solution for (a) The temperature alone is sufficient to find the average translational kinetic energy. Substituting the temperature into the translational kinetic energy equation gives \[\overline{KE} = \dfrac{3}{2}kT = \dfrac{3}{2}(1.38 \times 10^{-23} \, J/K)(293 \, K) = 6.07 \times 10^{-21} \, J.\] Strategy for (b) Finding the rms speed of a nitrogen molecule involves a straightforward calculation using the equation \[\sqrt{\overline{v^2}} = v_{rms} = \sqrt{\dfrac{3kT}{m}},\] but we must first find the mass of a nitrogen molecule. Using the molecular mass of nitrogen $N_2$ from the periodic table, \[m = \dfrac{2(14.0067) \times 10^{-3} \, kg/mol}{6.02 \times 10^{23} \, mol^{-1}} = 4.65 \times 10^{-26} \, kg.\] Solution for (b) Substituting this mass and the value for $k$ into the equation for $v_{rms}$ yields \[v_{rms} = \sqrt{\dfrac{3(1.38 \times 10^{-23} \, J/K)(293)}{4.65 \times 10^{-26} \, kg}} = 511 \, m/s.\] Discussion Note that the average kinetic energy of the molecule is independent of the type of molecule. The average translational kinetic energy depends only on absolute temperature. The kinetic energy is very small compared to macroscopic energies, so that we do not feel when an air molecule is hitting our skin. The rms velocity of the nitrogen molecule is surprisingly large. These large molecular velocities do not yield macroscopic movement of air, since the molecules move in all directions with equal likelihood. The mean free path (the distance a molecule can move on average between collisions) of molecules in air is very small, and so the molecules move rapidly but do not get very far in a second. The high value for rms speed is reflected in the speed of sound, however, which is about 340 m/s at room temperature. The faster the rms speed of air molecules, the faster that sound vibrations can be transferred through the air. The speed of sound increases with temperature and is greater in gases with small molecular masses, such as helium. (See Figure .) Making Connections: Historical Note—Kinetic Theory of Gases - The kinetic theory of gases was developed by Daniel Bernoulli (1700–1782), who is best known in physics for his work on fluid flow (hydrodynamics). Bernoulli’s work predates the atomistic view of matter established by Dalton. Distribution of Molecular Speeds The motion of molecules in a gas is random in magnitude and direction for individual molecules, but a gas of many molecules has a predictable distribution of molecular speeds. This distribution is called the Maxwell-Boltzmann distribution , after its originators, who calculated it based on kinetic theory, and has since been confirmed experimentally. (See Figure .) The distribution has a long tail, because a few molecules may go several times the rms speed. The most probable speed is less than the rms speed $v_{rms}$. Figure shows that the curve is shifted to higher speeds at higher temperatures, with a broader range of speeds. The distribution of thermal speeds depends strongly on temperature. As temperature increases, the speeds are shifted to higher values and the distribution is broadened. What is the implication of the change in distribution with temperature shown in Figure for humans? All other things being equal, if a person has a fever, he or she is likely to lose more water molecules, particularly from linings along moist cavities such as the lungs and mouth, creating a dry sensation in the mouth. Example $\PageIndex{2}$: Calculating Temperature: Escape Velocity of Helium Atoms In order to escape Earth’s gravity, an object near the top of the atmosphere (at an altitude of 100 km) must travel away from Earth at 11.1 km/s. This speed is called the escape velocity . At what temperature would helium atoms have an rms speed equal to the escape velocity? Strategy Identify the knowns and unknowns and determine which equations to use to solve the problem. Solution 1. Identify the knowns: $v$ is the escape velocity, 11.1 km/s. 2. Identify the unknowns: We need to solve for temperature, $T$. We also need to solve for the mass $m$ of the helium atom. 3. Determine which equations are needed. - To solve for mass $m$ of the helium atom, we can use information from the periodic table: \[m = \dfrac{molar \, mass}{number \, of \, atoms \, per \, mole}.\] - To solve for temperature $T$, we can rearrange either \[\overline{KE} = \dfrac{1}{2}m\overline{v^2} = \dfrac{3}{2}kT\] or \[\sqrt{\overline{v^2}} = v_{rms} = \sqrt{\dfrac{3kT}{m}}\] - \[T = \dfrac{\overline{mv^2}}{3k},\] where $k$ is the Boltzmann constant and $m$ is the mass of a helium atom. 4. Plug the known values into the equations and solve for the unknowns. \[m = \dfrac{molar \, mass}{number \, of \, atoms \, per \, mole} = \dfrac{4.0026 \times 10^{-3} kg/mole}{6.02 \times 10^{23} mol} = 6.65 \times 10^{-27} \, kg\] \[T = \dfrac{(6.65 \times 10^{-27} \, kg)(11.1 \times 10^3 \, m/s)^2}{3(1.38 \times 10^{-23} \, J/K)} = 1.98 \times 10^4 \, K\] Discussion This temperature is much higher than atmospheric temperature, which is approximately 250 K $(-25^oC $ or $-10^oF)$ at high altitude. Very few helium atoms are left in the atmosphere, but there were many when the atmosphere was formed. The reason for the loss of helium atoms is that there are a small number of helium atoms with speeds higher than Earth’s escape velocity even at normal temperatures. The speed of a helium atom changes from one instant to the next, so that at any instant, there is a small, but nonzero chance that the speed is greater than the escape speed and the molecule escapes from Earth’s gravitational pull. Heavier molecules, such as oxygen, nitrogen, and water (very little of which reach a very high altitude), have smaller rms speeds, and so it is much less likely that any of them will have speeds greater than the escape velocity. In fact, so few have speeds above the escape velocity that billions of years are required to lose significant amounts of the atmosphere. Figure shows the impact of a lack of an atmosphere on the Moon. Because the gravitational pull of the Moon is much weaker, it has lost almost its entire atmosphere. The comparison between Earth and the Moon is discussed in this chapter’s Problems and Exercises. Check Your Understanding If you consider a very small object such as a grain of pollen, in a gas, then the number of atoms and molecules striking its surface would also be relatively small. Would the grain of pollen experience any fluctuations in pressure due to statistical fluctuations in the number of gas atoms and molecules striking it in a given amount of time? [Hide Solution] Yes. Such fluctuations actually occur for a body of any size in a gas, but since the numbers of atoms and molecules are immense for macroscopic bodies, the fluctuations are a tiny percentage of the number of collisions, and the averages spoken of in this section vary imperceptibly. Roughly speaking the fluctuations are proportional to the inverse square root of the number of collisions, so for small bodies they can become significant. This was actually observed in the 19th century for pollen grains in water, and is known as the Brownian effect. PhET Explorations: Gas Properties Pump gas molecules into a box and see what happens as you change the volume, add or remove heat, change gravity, and more. Measure the temperature and pressure, and discover how the properties of the gas vary in relation to each other. Summary - Kinetic theory is the atomistic description of gases as well as liquids and solids. - Kinetic theory models the properties of matter in terms of continuous random motion of atoms and molecules. - The ideal gas law can also be expressed as \[PV = \dfrac{1}{3}Nm\overline{v^2},\] where $P$ is the pressure (average force per unit area), $V$ is the volume of gas in the container, $N$ is the number of molecules in the container, is the mass of a molecule, and $\overline{v^2}$ is the average of the molecular speed squared. - The temperature of gases is proportional to the average translational kinetic energy of atoms and molecules. \[\overline{KE} = \dfrac{1}{2}m\overline{v^2} = \dfrac{3}{2}kT\] or \[\sqrt{\overline{v^2}} = v_{rms} = \sqrt{\dfrac{3kT}{m}}.\] - The motion of individual molecules in a gas is random in magnitude and direction. However, a gas of many molecules has a predictable distribution of molecular speeds, known as the Maxwell-Boltzmann distribution .	libretexts	2025-03-17T19:53:31.923378	2015-11-01T05:38:52	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.04%3A_Kinetic_Theory-_Atomic_and_Molecular_Explanation_of_Pressure_and_Temperature", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "13.4: Kinetic Theory- Atomic and Molecular Explanation of Pressure and Temperature", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.05%3A_Phase_Changes	13.5: Phase Changes Learning Objectives By the end of this section, you will be able to: - Interpret a phase diagram. - State Dalton’s law. - Identify and describe the triple point of a gas from its phase diagram. - Describe the state of equilibrium between a liquid and a gas, a liquid and a solid, and a gas and a solid. Up to now, we have considered the behavior of ideal gases. Real gases are like ideal gases at high temperatures. At lower temperatures, however, the interactions between the molecules and their volumes cannot be ignored. The molecules are very close (condensation occurs) and there is a dramatic decrease in volume, as seen in Figure $\PageIndex{1}$. The substance changes from a gas to a liquid. When a liquid is cooled to even lower temperatures, it becomes a solid. The volume never reaches zero because of the finite volume of the molecules. High pressure may also cause a gas to change phase to a liquid. Carbon dioxide, for example, is a gas at room temperature and atmospheric pressure, but becomes a liquid under sufficiently high pressure. If the pressure is reduced, the temperature drops and the liquid carbon dioxide solidifies into a snow-like substance at the temperature $-78^oC$. Solid $CO_2$ is called “dry ice.” Another example of a gas that can be in a liquid phase is liquid nitrogen $(LN_2)$. $LN_2$ is made by liquefaction of atmospheric air (through compression and cooling). It boils at 77 K $(-196^oC)$ at atmospheric pressure. $LN_2$ is useful as a refrigerant and allows for the preservation of blood, sperm, and other biological materials. It is also used to reduce noise in electronic sensors and equipment, and to help cool down their current-carrying wires. In dermatology, $LN_2$ is used to freeze and painlessly remove warts and other growths from the skin. PV Diagrams We can examine aspects of the behavior of a substance by plotting a graph of pressure versus volume, called a PV diagram . When the substance behaves like an ideal gas, the ideal gas law describes the relationship between its pressure and volume. That is, \[PV = NkT \, (ideal \, gas).\] Now, assuming the number of molecules and the temperature are fixed, \[PV - constant \, (ideal \, gas, \, constant \, temperature).\] For example, the volume of the gas will decrease as the pressure increases. If you plot the relationship $PV$ = constant on a $PV$ diagram, you find a hyperbola. Figure $\PageIndex{2}$ shows a graph of pressure versus volume. The hyperbolas represent ideal-gas behavior at various fixed temperatures, and are called isotherms . At lower temperatures, the curves begin to look less like hyperbolas—the gas is not behaving ideally and may even contain liquid. There is a critical point —that is, a critical temperature —above which liquid cannot exist. At sufficiently high pressure above the critical point, the gas will have the density of a liquid but will not condense. Carbon dioxide, for example, cannot be liquefied at a temperature above $31.0^oC$. Critical pressure is the minimum pressure needed for liquid to exist at the critical temperature. Table $\PageIndex{1}$ lists representative critical temperatures and pressures. \| Substance \| Critical Temperature \| Critical pressure \| \|\| \|---\|---\|---\|---\|---\| \| $K$ \| $^oC$ \| $Pa$ \| $atm$ \| \| \| Water \| 647.4 \| 374.3 \| $22.12 \times 10^6$ \| 219.0 \| \| Sulfur dioxide \| 430.7 \| 157.6 \| $7.88 \times 10^6$ \| 78.0 \| \| Ammonia \| 405.5 \| 132.4 \| $11.28 \times 10^6$ \| 111.7 \| \| Carbon dioxide \| 304.2 \| 31.1 \| $7.39 \times 10^6$ \| 73.2 \| \| Oxygen \| 154.8 \| −118.4 \| $5.08 \times 10^6$ \| 50.3 \| \| Nitrogen \| 126.2 \| −146.9 \| $3.39 \times 10^6$ \| 33.6 \| \| Hydrogen \| 33.3 \| −239.9 \| $1.30 \times 10^6$ \| 12.9 \| \| Helium \| 5.3 \| −267.9 \| $0.229 \times 10^6$ \| 2.27 \| Phase Diagrams The plots of pressure versus temperatures provide considerable insight into thermal properties of substances. There are well-defined regions on these graphs that correspond to various phases of matter, so $PT$ graphs are called phase diagrams . Figure $\PageIndex{3}$ shows the phase diagram for water. Using the graph, if you know the pressure and temperature you can determine the phase of water. The solid lines—boundaries between phases—indicate temperatures and pressures at which the phases coexist (that is, they exist together in ratios, depending on pressure and temperature). For example, the boiling point of water is $100^oC$ at 1.00 atm. As the pressure increases, the boiling temperature rises steadily to $374^oC$ at a pressure of 218 atm. A pressure cooker (or even a covered pot) will cook food faster because the water can exist as a liquid at temperatures greater than $100^oC$ without all boiling away. The curve ends at a point called the critical point , because at higher temperatures the liquid phase does not exist at any pressure. The critical point occurs at the critical temperature, as you can see for water from Table . The critical temperature for oxygen is $-118^oC$, so oxygen cannot be liquefied above this temperature. Similarly, the curve between the solid and liquid regions in Figure gives the melting temperature at various pressures. For example, the melting point is $0^oC$ at 1.00 atm, as expected. Note that, at a fixed temperature, you can change the phase from solid (ice) to liquid (water) by increasing the pressure. Ice melts from pressure in the hands of a snowball maker. From the phase diagram, we can also say that the melting temperature of ice rises with increased pressure. When a car is driven over snow, the increased pressure from the tires melts the snowflakes; afterwards the water refreezes and forms an ice layer. At sufficiently low pressures there is no liquid phase, but the substance can exist as either gas or solid. For water, there is no liquid phase at pressures below 0.00600 atm. The phase change from solid to gas is called sublimation . It accounts for large losses of snow pack that never make it into a river, the routine automatic defrosting of a freezer, and the freeze-drying process applied to many foods. Carbon dioxide, on the other hand, sublimates at standard atmospheric pressure of 1 atm. (The solid form of $CO_2$ is known as dry ice because it does not melt. Instead, it moves directly from the solid to the gas state.) All three curves on the phase diagram meet at a single point, the triple point , where all three phases exist in equilibrium. For water, the triple point occurs at 273.16 K $(0.01^oC)$ and is a more accurate calibration temperature than the melting point of water at 1.00 atm, or 273.15 K $(0.0^oC)$. See Table for the triple point values of other substances. Equilibrium Liquid and gas phases are in equilibrium at the boiling temperature. (See Figure .) If a substance is in a closed container at the boiling point, then the liquid is boiling and the gas is condensing at the same rate without net change in their relative amount. Molecules in the liquid escape as a gas at the same rate at which gas molecules stick to the liquid, or form droplets and become part of the liquid phase. The combination of temperature and pressure has to be “just right”; if the temperature and pressure are increased, equilibrium is maintained by the same increase of boiling and condensation rates. \| Substance \| Temperature \| Pressure \| \|\| \|---\|---\|---\|---\|---\| \| $K$ \| $^oC$ \| Pa \| atm \| \| \| Water \| 273.16 \| 0.01 \| $6.10 \times 10^2$ \| 0.00600 \| \| Carbon dioxide \| 216.55 \| −56.60 \| $5.16 \times 10^5$ \| 5.11 \| \| Sulfur dioxide \| 197.68 \| −75.47 \| $1.67 \times 10^3$ \| 0.0167 \| \| Ammonia \| 195.40 \| −77.75 \| $6.06 \times 10^3$ \| 0.0600 \| \| Nitrogen \| 63.18 \| −210.0 \| $1.25 \times 10^4$ \| 0.124 \| \| Oxygen \| 54.36 \| −218.8 \| $1.52 \times 10^2$ \| 0.00151 \| \| Hydrogen \| 13.84 \| −259.3 \| $7.04 \times 10^3$ \| 0.0697 \| One example of equilibrium between liquid and gas is that of water and steam at $100^oC$ and 1.00 atm. This temperature is the boiling point at that pressure, so they should exist in equilibrium. Why does an open pot of water at $100^oC$ boil completely away? The gas surrounding an open pot is not pure water: it is mixed with air. If pure water and steam are in a closed container at $100^oC$ and 1.00 atm, they would coexist—but with air over the pot, there are fewer water molecules to condense, and water boils. What about water at $20.0^oC$ and 1.00 atm? This temperature and pressure correspond to the liquid region, yet an open glass of water at this temperature will completely evaporate. Again, the gas around it is air and not pure water vapor, so that the reduced evaporation rate is greater than the condensation rate of water from dry air. If the glass is sealed, then the liquid phase remains. We call the gas phase a vapor when it exists, as it does for water at $20.0^oC$, at a temperature below the boiling temperature. Exercise $\PageIndex{1}$: Check your Understanding Explain why a cup of water (or soda) with ice cubes stays at $0^oC$, even on a hot summer day. - Answer - The ice and liquid water are in thermal equilibrium, so that the temperature stays at the freezing temperature as long as ice remains in the liquid. (Once all of the ice melts, the water temperature will start to rise.) Vapor Pressure, Partial Pressure, and Dalton’s Law Vapor pressure is defined as the pressure at which a gas coexists with its solid or liquid phase. Vapor pressure is created by faster molecules that break away from the liquid or solid and enter the gas phase. The vapor pressure of a substance depends on both the substance and its temperature—an increase in temperature increases the vapor pressure. Partial pressure is defined as the pressure a gas would create if it occupied the total volume available. In a mixture of gases, the total pressure is the sum of partial pressures of the component gases , assuming ideal gas behavior and no chemical reactions between the components. This law is known as Dalton’s law of partial pressures , after the English scientist John Dalton (1766–1844), who proposed it. Dalton’s law is based on kinetic theory, where each gas creates its pressure by molecular collisions, independent of other gases present. It is consistent with the fact that pressures add according to Pascal’s Principle . Thus water evaporates and ice sublimates when their vapor pressures exceed the partial pressure of water vapor in the surrounding mixture of gases. If their vapor pressures are less than the partial pressure of water vapor in the surrounding gas, liquid droplets or ice crystals (frost) form. Exercise $\PageIndex{1}$ Is energy transfer involved in a phase change? If so, will energy have to be supplied to change phase from solid to liquid and liquid to gas? What about gas to liquid and liquid to solid? Why do they spray the orange trees with water in Florida when the temperatures are near or just below freezing? - Answer - Yes, energy transfer is involved in a phase change. We know that atoms and molecules in solids and liquids are bound to each other because we know that force is required to separate them. So in a phase change from solid to liquid and liquid to gas, a force must be exerted, perhaps by collision, to separate atoms and molecules. Force exerted through a distance is work, and energy is needed to do work to go from solid to liquid and liquid to gas. This is intuitively consistent with the need for energy to melt ice or boil water. The converse is also true. Going from gas to liquid or liquid to solid involves atoms and molecules pushing together, doing work and releasing energy. PHET EXPLORATIONS: STATES OF MATTER - BASICS Heat, cool, and compress atoms and molecules and watch as they change between solid, liquid, and gas phases. Section Summary - Most substances have three distinct phases: gas, liquid, and solid. - Phase changes among the various phases of matter depend on temperature and pressure. - The existence of the three phases with respect to pressure and temperature can be described in a phase diagram. - Two phases coexist (i.e., they are in thermal equilibrium) at a set of pressures and temperatures. These are described as a line on a phase diagram. - The three phases coexist at a single pressure and temperature. This is known as the triple point and is described by a single point on a phase diagram. - A gas at a temperature below its boiling point is called a vapor. - Vapor pressure is the pressure at which a gas coexists with its solid or liquid phase. - Partial pressure is the pressure a gas would create if it existed alone. - Dalton’s law states that the total pressure is the sum of the partial pressures of all of the gases present. Glossary - PV diagram - a graph of pressure vs. volume - critical point - the temperature above which a liquid cannot exist - critical temperature - the temperature above which a liquid cannot exist - critical pressure - the minimum pressure needed for a liquid to exist at the critical temperature - vapor - a gas at a temperature below the boiling temperature - vapor pressure - the pressure at which a gas coexists with its solid or liquid phase - phase diagram - a graph of pressure vs. temperature of a particular substance, showing at which pressures and temperatures the three phases of the substance occur - triple point - the pressure and temperature at which a substance exists in equilibrium as a solid, liquid, and gas - sublimation - the phase change from solid to gas - partial pressure - the pressure a gas would create if it occupied the total volume of space available - Dalton’s law of partial pressures - the physical law that states that the total pressure of a gas is the sum of partial pressures of the component gases	libretexts	2025-03-17T19:53:32.025365	2015-11-01T05:39:10	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.05%3A_Phase_Changes", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "13.5: Phase Changes", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.06%3A_Humidity_Evaporation_and_Boiling	13.6: Humidity, Evaporation, and Boiling Learning Objectives By the end of this section, you will be able to: - Explain the relationship between vapor pressure of water and the capacity of air to hold water vapor. - Explain the relationship between relative humidity and partial pressure of water vapor in the air. - Calculate vapor density using vapor pressure. - Calculate humidity and dew point. The expression “it’s not the heat, it’s the humidity” makes a valid point. We keep cool in hot weather by evaporating sweat from our skin and water from our breathing passages. Because evaporation is inhibited by high humidity, we feel hotter at a given temperature when the humidity is high. Low humidity, on the other hand, can cause discomfort from excessive drying of mucous membranes and can lead to an increased risk of respiratory infections. Figure $\PageIndex{1}$: Dew drops like these, on a banana leaf photographed just after sunrise, form when the air temperature drops to or below the dew point. At the dew point, the rate at which water molecules join together is greater than the rate at which they separate, and some of the water condenses to form droplets. (credit: Aaron Escobar, Flickr) When we say humidity, we really mean relative humidity . Relative humidity tells us how much water vapor is in the air compared with the maximum possible. At its maximum, denoted as saturation , the relative humidity is 100%, and evaporation is inhibited. The amount of water vapor in the air depends on temperature. For example, relative humidity rises in the evening, as air temperature declines, sometimes reaching the dew point . At the dew point temperature, relative humidity is 100%, and fog may result from the condensation of water droplets if they are small enough to stay in suspension. Conversely, if you wish to dry something (perhaps your hair), it is more effective to blow hot air over it rather than cold air, because, among other things, the increase in temperature increases the energy of the molecules, so the rate of evaporation increases. The amount of water vapor in the air depends on the vapor pressure of water. The liquid and solid phases are continuously giving off vapor because some of the molecules have high enough speeds to enter the gas phase; see Figure (a). If a lid is placed over the container, as in Figure (b), evaporation continues, increasing the pressure, until sufficient vapor has built up for condensation to balance evaporation. Then equilibrium has been achieved, and the vapor pressure is equal to the partial pressure of water in the container. Vapor pressure increases with temperature because molecular speeds are higher as temperature increases. Table gives representative values of water vapor pressure over a range of temperatures. Relative humidity is related to the partial pressure of water vapor in the air. At 100% humidity, the partial pressure is equal to the vapor pressure, and no more water can enter the vapor phase. If the partial pressure is less than the vapor pressure, then evaporation will take place, as humidity is less than 100%. If the partial pressure is greater than the vapor pressure, condensation takes place. In everyday language, people sometimes refer to the capacity of air to “hold” water vapor, but this is not actually what happens. The water vapor is not held by the air. The amount of water in air is determined by the vapor pressure of water and has nothing to do with the properties of air. \| Temperature $^oC$ \| Vapor Pressure $(Pa)$ \| Saturation vapor density $g/m^3$ \| \| −50 \| 4.0 \| 0.039 \| \| -20 \| $1.04 \times 10^2$ \| 0.89 \| \| -10 \| $2.60 \times 10^2$ \| 2.36 \| \| 0 \| $6.10 \times 10^2$ \| 4.84 \| \| 5 \| $8.68 \times 10^2$ \| 6.80 \| \| 10 \| $1.19 \times 10^3$ \| 9.40 \| \| 15 \| $1.69 \times 10^3$ \| 12.8 \| \| 20 \| $2.33 \times 10^3$ \| 17.2 \| \| 25 \| $3.17 \times 10^3$ \| 23.0 \| \| 30 \| $4.24 \times 10^3$ \| 30.4 \| \| 37 \| $6.31 \times 10^3$ \| 44.0 \| \| 40 \| $7.34 \times 10^3$ \| 51.1 \| \| 50 \| $1.23 \times 10^4$ \| 82.4 \| \| 60 \| $1.99 \times 10^4$ \| 130 \| \| 70 \| $3.12 \times 10^4$ \| 197 \| \| 80 \| $4.73 \times 10^4$ \| 294 \| \| 90 \| $7.01 \times 10^4$ \| 418 \| \| 95 \| $8.59 \times 10^4$ \| 505 \| \| 100 \| $1.01 \times 10^5$ \| 598 \| \| 120 \| $1.99 \times 10^5$ \| 1095 \| \| 150 \| $4.76 \times 10^5$ \| 2430 \| \| 200 \| $1.55 \times 10^6$ \| 7090 \| \| 220 \| $2.32 \times 10^6$ \| 10,200 \| Saturation Vapor Density of Water Example $\PageIndex{1}$: Calculating Density Using Vapor Pressure Table gives the vapor pressure of water at .$20.0^oC$ as $2.33 \times 10^3 \, Pa$. Use the ideal gas law to calculate the density of water vapor in $g/m^3$ that would create a partial pressure equal to this vapor pressure. Compare the result with the saturation vapor density given in the table. Strategy To solve this problem, we need to break it down into a two steps. The partial pressure follows the ideal gas law, \[PV = nRT,\] where $n$ is the number of moles. If we solve this equation for $n/V$ to calculate the number of moles per cubic meter, we can then convert this quantity to grams per cubic meter as requested. To do this, we need to use the molecular mass of water, which is given in the periodic table. Solution 1. Identify the knowns and convert them to the proper units: - temperature $T = 20^oC = 203 \, K$ - vapor pressure $P$ of water at $20^oC$ is $2.33 \times 10^3 \, Pa$ - molecular mass of water is $18.0 \, g/m$ 2. Solve the ideal gas law for $n?V$. \[\dfrac{n}{V} = \dfrac{P}{RT}\] 3. Substitute known values into the equation and solve for $n/V$. \[\dfrac{n}{V} = \dfrac{P}{RT} = \dfrac{2.33 \times 10^3 \, Pa}{(8.31 \, J/mol \cdot K)(293 \, K)} = 0.957 \, mol/m^3\] 4. Convert the density in moles per cubic meter to grams per cubic meter. \[\rho = \left(0.957\dfrac{mol}{m^3}\right)\left(\dfrac{18.0 \, g}{mol}\right) = 17.2 \, g/m^3\] Discussion The density is obtained by assuming a pressure equal to the vapor pressure of water at $20.0^oC$. The density found is identical to the value in Table , which means that a vapor density of $17.2 \, g/m^3$ at $20.9^oC$ creates a partial pressure of $2.33 \times 10^3 \, Pa$, equal to the vapor pressure of water at that temperature. If the partial pressure is equal to the vapor pressure, then the liquid and vapor phases are in equilibrium, and the relative humidity is 100%. Thus, there can be no more than 17.2 g of water vapor per $m^3$ at $20.0^oC$, so that this value is the saturation vapor density at that temperature. This example illustrates how water vapor behaves like an ideal gas: the pressure and density are consistent with the ideal gas law (assuming the density in the table is correct). The saturation vapor densities listed in Table are the maximum amounts of water vapor that air can hold at various temperatures. Percent Relative Humidity We define percent relative humidity as the ratio of vapor density to saturation vapor density, or \[percent \, relative \, humidity = \dfrac{vapor \, density}{saturation \, vapor \, density} \times 100\] We can use this and the data in Table to do a variety of interesting calculations, keeping in mind that relative humidity is based on the comparison of the partial pressure of water vapor in air and ice. We can use this and the data in Table to do a variety of interesting calculations, keeping in mind that relative humidity is based on the comparison of the partial pressure of water vapor in air and ice. Exampl $\PageIndex{2}$: Calculating Humidity and Dew Point (a) Calculate the percent relative humidity on a day when the temperature is $25.0^oC$ and the air contains 9.40 g of vapor per $m^3$. (b) At what temperature will this air reach 100% relative humidity (the saturation density)? This temperature is the dew point. (c) What is the humidity when the air temperature is $25.0^oC$ and the dew point is $-10.0^oC$? Strategy and Solution (a) Percent relative humidity is defined as the ratio of vapor density to saturation vapor density. \[percent \, relative \, humidity = \dfrac{vapor \, density}{saturation \, vapor \, density} \times 100\] The first is given to be $9.40 \, g/m^3$, and the second is found in Table to be $23.0 \, g/m^3$. Thus, \[percent \, relative \, humidity = \dfrac{9.40 \, g/m^3}{23.0 \, g/m^3} \times 100 = 40.9 \%\] (b) The air contains $9.40 \, g/m^3$ of water vapor. The relative humidity will be 100% at a temperature where $9.40 \, g/m^3$ is the saturation density. Inspection of Table reveals this to be the case at $10.0^oC$, where the relative humidity will be 100%. That temperature is called the dew point for air with this concentration of water vapor. (c) Here, the dew point temperature is given to be $10.0^oC$. Using Table , we see that the vapor density is $2.36 \, g/m^3$, because this value is the saturation vapor density at $-10.0^oC$. The saturation vapor density at $25.0^oC$ is seen to be $23.0 \, g/m^3$. Thus, the relative humidity at $25.0^oC$ is \[percent \, relative \, humidity = \dfrac{2.36 \, g/m^3}{23.0 \, g/m^3} \times 100 = 10.3 \%\] Discussion The importance of dew point is that air temperature cannot drop below $10.0^oC$ in part (b), or $-10.0^oC$ in part (c), without water vapor condensing out of the air. If condensation occurs, considerable transfer of heat occurs (discussed in Heat and Heat Transfer Methods ), which prevents the temperature from further dropping. When dew points are below $0^oC$, freezing temperatures are a greater possibility, which explains why farmers keep track of the dew point. Low humidity in deserts means low dew-point temperatures. Thus condensation is unlikely. If the temperature drops, vapor does not condense in liquid drops. Because no heat is released into the air, the air temperature drops more rapidly compared to air with higher humidity. Likewise, at high temperatures, liquid droplets do not evaporate, so that no heat is removed from the gas to the liquid phase. This explains the large range of temperature in arid regions. Why does water boil at $100^oC$? You will note from Table that the vapor pressure of water at $100^oC$ is $1.01 \times 10^5 \, Pa$, or 1.00 atm. Thus, it can evaporate without limit at this temperature and pressure. But why does it form bubbles when it boils? This is because water ordinarily contains significant amounts of dissolved air and other impurities, which are observed as small bubbles of air in a glass of water. If a bubble starts out at the bottom of the container at $20^oC$, it contains water vapor (about 2.30%). The pressure inside the bubble is fixed at 1.00 atm (we ignore the slight pressure exerted by the water around it). As the temperature rises, the amount of air in the bubble stays the same, but the water vapor increases; the bubble expands to keep the pressure at 1.00 atm. At $100^oC$, water vapor enters the bubble continuously since the partial pressure of water is equal to 1.00 atm in equilibrium. It cannot reach this pressure, however, since the bubble also contains air and total pressure is 1.00 atm. The bubble grows in size and thereby increases the buoyant force. The bubble breaks away and rises rapidly to the surface—we call this boiling! (See Figure .) Check Your Understanding - Freeze drying is a process in which substances, such as foods, are dried by placing them in a vacuum chamber and lowering the atmospheric pressure around them. How does the lowered atmospheric pressure speed the drying process, and why does it cause the temperature of the food to drop? [Hide Solution] Decreased the atmospheric pressure results in decreased partial pressure of water, hence a lower humidity. So evaporation of water from food, for example, will be enhanced. The molecules of water most likely to break away from the food will be those with the greatest velocities. Those remaining thus have a lower average velocity and a lower temperature. This can (and does) result in the freezing and drying of the food; hence the process is aptly named freeze drying. PHET EXPLORATIONS: STATES OF MATTER Watch different types of molecules form a solid, liquid, or gas. Add or remove heat and watch the phase change. Change the temperature or volume of a container and see a pressure-temperature diagram respond in real time. Relate the interaction potential to the forces between molecules. Summary - Relative humidity is the fraction of water vapor in a gas compared to the saturation value. - The saturation vapor density can be determined from the vapor pressure for a given temperature. - Percent relative humidity is defined to be \[percent \, relative \, humidity = \dfrac{vapor \, density}{saturation \, vapor \, density} \times 100\%\] - The dew point is the temperature at which air reaches 100% relative humidity. Glossary - dew point - the temperature at which relative humidity is 100%; the temperature at which water starts to condense out of the air - saturation - the condition of 100% relative humidity - percent relative humidity - the ratio of vapor density to saturation vapor density - relative humidity - the amount of water in the air relative to the maximum amount the air can hold	libretexts	2025-03-17T19:53:32.119079	2015-11-01T05:39:26	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.06%3A_Humidity_Evaporation_and_Boiling", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "13.6: Humidity, Evaporation, and Boiling", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.E%3A_Work_Energy_and_Energy_Resources_(Exercise)	13.E: Work, Energy, and Energy Resources (Exercise) - - Last updated - Save as PDF Conceptual Questions 13.1: Temperature 1. What does it mean to say that two systems are in thermal equilibrium? 2. Give an example of a physical property that varies with temperature and describe how it is used to measure temperature. 3. When a cold alcohol thermometer is placed in a hot liquid, the column of alcohol goes down slightly before going up. Explain why. 4. If you add boiling water to a cup at room temperature, what would you expect the final equilibrium temperature of the unit to be? You will need to include the surroundings as part of the system. Consider the zeroth law of thermodynamics. 13.2: Thermal Expansion of Solids and Liquids 5. Thermal stresses caused by uneven cooling can easily break glass cookware. Explain why Pyrex®, a glass with a small coefficient of linear expansion, is less susceptible. 6. Water expands significantly when it freezes: a volume increase of about 9% occurs. As a result of this expansion and because of the formation and growth of crystals as water freezes, anywhere from 10% to 30% of biological cells are burst when animal or plant material is frozen. Discuss the implications of this cell damage for the prospect of preserving human bodies by freezing so that they can be thawed at some future date when it is hoped that all diseases are curable. 7. One method of getting a tight fit, say of a metal peg in a hole in a metal block, is to manufacture the peg slightly larger than the hole. The peg is then inserted when at a different temperature than the block. Should the block be hotter or colder than the peg during insertion? Explain your answer. 8. Does it really help to run hot water over a tight metal lid on a glass jar before trying to open it? Explain your answer. 9. Liquids and solids expand with increasing temperature, because the kinetic energy of a body’s atoms and molecules increases. Explain why some materials shrink with increasing temperature. 13.3: The Ideal Gas Law 10. Find out the human population of Earth. Is there a mole of people inhabiting Earth? If the average mass of a person is 60 kg, calculate the mass of a mole of people. How does the mass of a mole of people compare with the mass of Earth? 11. Under what circumstances would you expect a gas to behave significantly differently than predicted by the ideal gas law? 12. A constant-volume gas thermometer contains a fixed amount of gas. What property of the gas is measured to indicate its temperature? 13.4: Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature 13. How is momentum related to the pressure exerted by a gas? Explain on the atomic and molecular level, considering the behavior of atoms and molecules. 13.5: Phase Changes 14. A pressure cooker contains water and steam in equilibrium at a pressure greater than atmospheric pressure. How does this greater pressure increase cooking speed? 15. Why does condensation form most rapidly on the coldest object in a room—for example, on a glass of ice water? 16. What is the vapor pressure of solid carbon dioxide (dry ice) at $\displaystyle –78.5ºC$? The phase diagram for carbon dioxide. The axes are nonlinear, and the graph is not to scale. Dry ice is solid carbon dioxide and has a sublimation temperature of $\displaystyle –78.5ºC$. 17. Can carbon dioxide be liquefied at room temperature ($\displaystyle 20ºC$)? If so, how? If not, why not? (See Figure.) 18. Oxygen cannot be liquefied at room temperature by placing it under a large enough pressure to force its molecules together. Explain why this is. 19. What is the distinction between gas and vapor? 13.6: Humidity, Evaporation, and Boiling 20. Because humidity depends only on water’s vapor pressure and temperature, are the saturation vapor densities listed in Table valid in an atmosphere of helium at a pressure of $\displaystyle 1.01×10^5N/m^2$, rather than air? Are those values affected by altitude on Earth? 21. Why does a beaker of $\displaystyle 40.0ºC$ water placed in a vacuum chamber start to boil as the chamber is evacuated (air is pumped out of the chamber)? At what pressure does the boiling begin? Would food cook any faster in such a beaker? 22. Why does rubbing alcohol evaporate much more rapidly than water at STP (standard temperature and pressure)? Problems & Exercises 13.1: Temperature 23. What is the Fahrenheit temperature of a person with a $\displaystyle 39.0ºC$ fever? Solution $\displaystyle 102ºF$ 24. Frost damage to most plants occurs at temperatures of $\displaystyle 28.0ºF$ or lower. What is this temperature on the Kelvin scale? 25. To conserve energy, room temperatures are kept at $\displaystyle 68.0ºF$ in the winter and $\displaystyle 78.0ºF$ in the summer. What are these temperatures on the Celsius scale? Solution $\displaystyle 20.0ºC$ and $\displaystyle 25.6ºC$ 26. A tungsten light bulb filament may operate at 2900 K. What is its Fahrenheit temperature? What is this on the Celsius scale? 27. The surface temperature of the Sun is about 5750 K. What is this temperature on the Fahrenheit scale? Solution $\displaystyle 9890ºF$ 28. One of the hottest temperatures ever recorded on the surface of Earth was $\displaystyle 134ºF$ in Death Valley, CA. What is this temperature in Celsius degrees? What is this temperature in Kelvin? 29. (a) Suppose a cold front blows into your locale and drops the temperature by 40.0 Fahrenheit degrees. How many degrees Celsius does the temperature decrease when there is a $\displaystyle 40.0ºF$ decrease in temperature? (b) Show that any change in temperature in Fahrenheit degrees is nine-fifths the change in Celsius degrees. Solution (a) $\displaystyle 22.2ºC$ (b) $\displaystyle ΔT(ºF)=T_2(ºF)−T_1(ºF)$ $\displaystyle =\frac{9}{5}T_2(ºC)+32.0º−(\frac{9}{5}T_1(ºC)+32.0º)$ $\displaystyle =\frac{9}{5}(T_2(ºC)−T_1(ºC))=\frac{9}{5}ΔT(ºC)$ 30. (a) At what temperature do the Fahrenheit and Celsius scales have the same numerical value? (b) At what temperature do the Fahrenheit and Kelvin scales have the same numerical value? 13.2: Thermal Expansion of Solids and Liquids 31. The height of the Washington Monument is measured to be 170 m on a day when the temperature is $\displaystyle 35.0ºC$. What will its height be on a day when the temperature falls to $\displaystyle -10.0ºC$? Although the monument is made of limestone, assume that its thermal coefficient of expansion is the same as marble’s. Solution 169.95 m 32. How much taller does the Eiffel Tower become at the end of a day when the temperature has increased by $\displaystyle 15ºC$? Its original height is 321 m and you can assume it is made of steel. 33. What is the change in length of a 3.00-cm-long column of mercury if its temperature changes from $\displaystyle 37.0ºC$ to $\displaystyle 40.0ºC$, assuming the mercury is unconstrained? Solution $\displaystyle 5.4×10^{−6}m$ 34. How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature $\displaystyle 35.0ºC$ greater than when they were laid? Their original length is 10.0 m. 35. You are looking to purchase a small piece of land in Hong Kong. The price is “only” $60,000 per square meter! The land title says the dimensions are $\displaystyle 20m×30m$. By how much would the total price change if you measured the parcel with a steel tape measure on a day when the temperature was $\displaystyle 20ºC$ above normal? Solution Because the area gets smaller, the price of the land DECREASES by $\displaystyle ~$17,000.$ 36. Global warming will produce rising sea levels partly due to melting ice caps but also due to the expansion of water as average ocean temperatures rise. To get some idea of the size of this effect, calculate the change in length of a column of water 1.00 km high for a temperature increase of $\displaystyle 1.00ºC$. Note that this calculation is only approximate because ocean warming is not uniform with depth. 37. Show that 60.0 L of gasoline originally at 15.0ºC size 12{"15" "." 0°C} {} will expand to 61.1 L when it warms to 35.0ºC, size 12{"35" "." 0°"C,"} {} as claimed in Example. Solution $\displaystyle V=V_0+ΔV=V_0(1+βΔT)$ $\displaystyle =(60.00 L)[1+(950×10^{−6}/ºC)(35.0ºC−15.0ºC)]$ $\displaystyle =61.1L$ 38. (a) Suppose a meter stick made of steel and one made of invar (an alloy of iron and nickel) are the same length at $\displaystyle 0ºC$. What is their difference in length at $\displaystyle 22.0ºC$? (b) Repeat the calculation for two 30.0-m-long surveyor’s tapes. 39. (a) If a 500-mL glass beaker is filled to the brim with ethyl alcohol at a temperature of $\displaystyle 5.00ºC$, how much will overflow when its temperature reaches $\displaystyle 22.0ºC$? (b) How much less water would overflow under the same conditions? Solution (a) 9.35 mL (b) 7.56 mL 40. Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0-L capacity when at $\displaystyle 10.0ºC$. What volume of radiator fluid will overflow when the radiator and fluid reach their $\displaystyle 95.0ºC$ operating temperature, given that the fluid’s volume coefficient of expansion is $\displaystyle β=400×10^{–6}/ºC$? Note that this coefficient is approximate, because most car radiators have operating temperatures of greater than $\displaystyle 95.0ºC$. 41. A physicist makes a cup of instant coffee and notices that, as the coffee cools, its level drops 3.00 mm in the glass cup. Show that this decrease cannot be due to thermal contraction by calculating the decrease in level if the $\displaystyle 350cm^3$ of coffee is in a 7.00-cm-diameter cup and decreases in temperature from $\displaystyle 95.0ºC$ to $\displaystyle 45.0ºC$. (Most of the drop in level is actually due to escaping bubbles of air.) Solution 0.832 mm 42. (a) The density of water at $\displaystyle 0ºC$ is very nearly $\displaystyle 1000kg/m^3$ (it is actually $\displaystyle 999.84 kg/m^3$), whereas the density of ice at $\displaystyle 0ºC$ is $\displaystyle 917 kg/m^3$. Calculate the pressure necessary to keep ice from expanding when it freezes, neglecting the effect such a large pressure would have on the freezing temperature. (This problem gives you only an indication of how large the forces associated with freezing water might be.) (b) What are the implications of this result for biological cells that are frozen? 43. Show that $\displaystyle β≈3α$, by calculating the change in volume $\displaystyle ΔV$ of a cube with sides of length $\displaystyle L$. Solution We know how the length changes with temperature: $\displaystyle ΔL=αL_0ΔT$. Also we know that the volume of a cube is related to its length by $\displaystyle V=L^3$ , so the final volume is then $\displaystyle V=V_0+ΔV=(L_0+ΔL)^3$. Substituting for $\displaystyle ΔL$ gives $\displaystyle V=(L_0+αL_0ΔT)^3=L^3_0(1+αΔT)^3$. Now, because $\displaystyle αΔT$ is small, we can use the binomial expansion: $\displaystyle V≈L^3_0(1+3αΔT)=L^3_0+3αL^3_0ΔT.$ So writing the length terms in terms of volumes gives $\displaystyle V=V_0+ΔV≈V_0+3αV_0ΔT$, and so $\displaystyle ΔV=βV_0ΔT≈3αV_0ΔT,$ or $\displaystyle β≈3α$. 13.3: The Ideal Gas Law 44. The gauge pressure in your car tires is $\displaystyle 2.50×10^5N/m^2$ at a temperature of $\displaystyle 35.0ºC$ when you drive it onto a ferry boat to Alaska. What is their gauge pressure later, when their temperature has dropped to $\displaystyle –40.0ºC$? Solution 1.623 atm 45. Convert an absolute pressure of $\displaystyle 7.00×10^5N/m^2$ to gauge pressure in $\displaystyle lb/in^2$. (This value was stated to be just less than $\displaystyle 90.0 lb/in^2$ in Example. Is it?) 46. Suppose a gas-filled incandescent light bulb is manufactured so that the gas inside the bulb is at atmospheric pressure when the bulb has a temperature of $\displaystyle 20.0ºC$. (a) Find the gauge pressure inside such a bulb when it is hot, assuming its average temperature is $\displaystyle 60.0ºC$ (an approximation) and neglecting any change in volume due to thermal expansion or gas leaks. (b) The actual final pressure for the light bulb will be less than calculated in part (a) because the glass bulb will expand. What will the actual final pressure be, taking this into account? Is this a negligible difference? Solution (a) 0.136 atm (b) 0.135 atm. The difference between this value and the value from part (a) is negligible. 47. Large helium-filled balloons are used to lift scientific equipment to high altitudes. (a) What is the pressure inside such a balloon if it starts out at sea level with a temperature of $\displaystyle 10.0ºC$ and rises to an altitude where its volume is twenty times the original volume and its temperature is $\displaystyle –50.0ºC$? (b) What is the gauge pressure? (Assume atmospheric pressure is constant.) 48. Confirm that the units of $\displaystyle nRT$ are those of energy for each value of $\displaystyle R$: (a) $\displaystyle 8.31J/mol⋅K$, (b) $\displaystyle 1.99 cal/mol⋅K$, and (c) $\displaystyle 0.0821 L⋅atm/mol⋅K$. Solution (a) $\displaystyle nRT=(mol)(J/mol⋅K)(K)=J$ (b) $\displaystyle nRT=(mol)(cal/mol⋅K)(K)=cal$ (c) $\displaystyle nRT=(mol)(L⋅atm/mol⋅K)(K)$ $\displaystyle =L⋅atm=(m^3)(N/m^2)$ $\displaystyle =N⋅m=J$ 49. In the text, it was shown that $\displaystyle N/V=2.68×10^{25}m^{−3}$ for gas at STP. (a) Show that this quantity is equivalent to $\displaystyle N/V=2.68×10^{19}cm^{−3}$, as stated. (b) About how many atoms are there in one $\displaystyle μm^3$ (a cubic micrometer) at STP? (c) What does your answer to part (b) imply about the separation of atoms and molecules? 50. Calculate the number of moles in the 2.00-L volume of air in the lungs of the average person. Note that the air is at $\displaystyle 37.0ºC$ (body temperature). Solution $\displaystyle 7.86×10^{−2}mol$ 51. An airplane passenger has $\displaystyle 100cm^3$ of air in his stomach just before the plane takes off from a sea-level airport. What volume will the air have at cruising altitude if cabin pressure drops to $\displaystyle 7.50×10^4N/m^2$? 52. (a) What is the volume (in $\displaystyle km^3$) of Avogadro’s number of sand grains if each grain is a cube and has sides that are 1.0 mm long? (b) How many kilometers of beaches in length would this cover if the beach averages 100 m in width and 10.0 m in depth? Neglect air spaces between grains. Solution (a) $\displaystyle 6.02×10^5km^3$ (b) $\displaystyle 6.02×10^8km$ 53. An expensive vacuum system can achieve a pressure as low as $\displaystyle 1.00×10^{–7}N/m^2$ at $\displaystyle 20ºC$. How many atoms are there in a cubic centimeter at this pressure and temperature? 54. The number density of gas atoms at a certain location in the space above our planet is about $\displaystyle 1.00×10^{11}m^{−3}$, and the pressure is $\displaystyle 2.75×10^{–10}N/m^2$ in this space. What is the temperature there? Solution $\displaystyle −73.9ºC$ 55. A bicycle tire has a pressure of $\displaystyle 7.00×10^5N/m^2$ at a temperature of $\displaystyle 18.0ºC$ and contains 2.00 L of gas. What will its pressure be if you let out an amount of air that has a volume of $\displaystyle 100cm^3$ at atmospheric pressure? Assume tire temperature and volume remain constant. 56. A high-pressure gas cylinder contains 50.0 L of toxic gas at a pressure of $\displaystyle 1.40×10^7N/m^2$ and a temperature of $\displaystyle 25.0ºC$. Its valve leaks after the cylinder is dropped. The cylinder is cooled to dry ice temperature ($\displaystyle –78.5ºC$) to reduce the leak rate and pressure so that it can be safely repaired. (a) What is the final pressure in the tank, assuming a negligible amount of gas leaks while being cooled and that there is no phase change? (b) What is the final pressure if one-tenth of the gas escapes? (c) To what temperature must the tank be cooled to reduce the pressure to 1.00 atm (assuming the gas does not change phase and that there is no leakage during cooling)? (d) Does cooling the tank appear to be a practical solution? Solution (a) $\displaystyle 9.14×10^6N/m^2$ (b) $\displaystyle 8.23×10^6N/m^2$ (c) 2.16 K (d) No. The final temperature needed is much too low to be easily achieved for a large object. 57. Find the number of moles in 2.00 L of gas at $\displaystyle 35.0ºC$ and under $\displaystyle 7.41×10^7N/m^2$ of pressure. 58. Calculate the depth to which Avogadro’s number of table tennis balls would cover Earth. Each ball has a diameter of 3.75 cm. Assume the space between balls adds an extra 25.0% to their volume and assume they are not crushed by their own weight. Solution 41 km 59. (a) What is the gauge pressure in a $\displaystyle 25.0ºC$ car tire containing 3.60 mol of gas in a 30.0 L volume? (b) What will its gauge pressure be if you add 1.00 L of gas originally at atmospheric pressure and $\displaystyle 25.0ºC$? Assume the temperature returns to $\displaystyle 25.0ºC$ and the volume remains constant. 60. (a) In the deep space between galaxies, the density of atoms is as low as $\displaystyle 10^6atoms/m^3$, and the temperature is a frigid 2.7 K. What is the pressure? (b) What volume (in $\displaystyle m^3$) is occupied by 1 mol of gas? (c) If this volume is a cube, what is the length of its sides in kilometers? Solution (a) $\displaystyle 3.7×10^{−17}Pa$ (b) $\displaystyle 6.0×10^{17}m^3$ (c) $\displaystyle 8.4×10^2km$ 13.4: Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature 61. Some incandescent light bulbs are filled with argon gas. What is $\displaystyle v_{rms}$ for argon atoms near the filament, assuming their temperature is 2500 K? Solution $\displaystyle 1.25×10^3m/s$ 62. Average atomic and molecular speeds ($\displaystyle v_{rms}$) are large, even at low temperatures. What is $\displaystyle v_{rms}$ for helium atoms at 5.00 K, just one degree above helium’s liquefaction temperature? 63. (a) What is the average kinetic energy in joules of hydrogen atoms on the $\displaystyle 5500ºC$ surface of the Sun? (b) What is the average kinetic energy of helium atoms in a region of the solar corona where the temperature is $\displaystyle 6.00×10^5K$? Solution (a) $\displaystyle 1.20×10^{−19}J$ (b) $\displaystyle 1.24×10^{−17}J$ 64. The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would oxygen molecules (molecular mass is equal to 32.0 g/mol) have an average velocity $\displaystyle v_{rms}$ equal to Earth’s escape velocity of 11.1 km/s? 65. The escape velocity from the Moon is much smaller than from Earth and is only 2.38 km/s. At what temperature would hydrogen molecules (molecular mass is equal to 2.016 g/mol) have an average velocity $\displaystyle v_{rms}$ equal to the Moon’s escape velocity? Solution $\displaystyle 458K$ 66. Nuclear fusion, the energy source of the Sun, hydrogen bombs, and fusion reactors, occurs much more readily when the average kinetic energy of the atoms is high—that is, at high temperatures. Suppose you want the atoms in your fusion experiment to have average kinetic energies of $\displaystyle 6.40×10^{–14}J$. What temperature is needed? 67. Suppose that the average velocity ($\displaystyle v_{rms}$) of carbon dioxide molecules (molecular mass is equal to 44.0 g/mol) in a flame is found to be $\displaystyle 1.05×10^5m/s$. What temperature does this represent? Solution $\displaystyle 1.95×10^7K$ 68. Hydrogen molecules (molecular mass is equal to 2.016 g/mol) have an average velocity $\displaystyle v_{rms}$ equal to 193 m/s. What is the temperature? 69. Much of the gas near the Sun is atomic hydrogen. Its temperature would have to be $\displaystyle 1.5×10^7K$ for the average velocity $\displaystyle v_{rms}$ to equal the escape velocity from the Sun. What is that velocity? Solution $\displaystyle 6.09×10^5m/s$ 70. There are two important isotopes of uranium— $\displaystyle ^{235}U$ and $\displaystyle ^{238}U$; these isotopes are nearly identical chemically but have different atomic masses. Only $\displaystyle ^{235}U$ is very useful in nuclear reactors. One of the techniques for separating them (gas diffusion) is based on the different average velocities $\displaystyle v_{rms}$ of uranium hexafluoride gas, $\displaystyle UF_6$. (a) The molecular masses for $\displaystyle ^{235}U$ $\displaystyle UF_6$ and $\displaystyle ^{238}U$ $\displaystyle UF_6$ are 349.0 g/mol and 352.0 g/mol, respectively. What is the ratio of their average velocities? (b) At what temperature would their average velocities differ by 1.00 m/s? (c) Do your answers in this problem imply that this technique may be difficult? 13.6: Humidity, Evaporation, and Boiling 71. Dry air is 78.1% nitrogen. What is the partial pressure of nitrogen when the atmospheric pressure is $\displaystyle 1.01×10^5N/m^2$? Solution $\displaystyle 7.89×10^4Pa$ 72. (a) What is the vapor pressure of water at $\displaystyle 20.0ºC$? (b) What percentage of atmospheric pressure does this correspond to? (c) What percent of $\displaystyle 20.0ºC$ air is water vapor if it has 100% relative humidity? (The density of dry air at $\displaystyle 20.0ºC$ is $\displaystyle 1.20kg/m^3$.) 73. Pressure cookers increase cooking speed by raising the boiling temperature of water above its value at atmospheric pressure. (a) What pressure is necessary to raise the boiling point to $\displaystyle 120.0ºC$? (b) What gauge pressure does this correspond to? Solution (a) $\displaystyle 1.99×10^5Pa$ (b) 0.97 atm 74. (a) At what temperature does water boil at an altitude of 1500 m (about 5000 ft) on a day when atmospheric pressure is $\displaystyle 8.59×10^4N/m^2$? (b) What about at an altitude of 3000 m (about 10,000 ft) when atmospheric pressure is $\displaystyle 7.00×10^4N/m^2$? 75. What is the atmospheric pressure on top of Mt. Everest on a day when water boils there at a temperature of $\displaystyle 70.0ºC$? Solution $\displaystyle 3.12×10^4Pa$ 76. At a spot in the high Andes, water boils at $\displaystyle 80.0ºC$, greatly reducing the cooking speed of potatoes, for example. What is atmospheric pressure at this location? 77. What is the relative humidity on a $\displaystyle 25.0ºC$ day when the air contains $\displaystyle 18.0g/m^3$ of water vapor? Solution 78.3% 78. What is the density of water vapor in $\displaystyle g/m^3$ on a hot dry day in the desert when the temperature is $\displaystyle 40.0ºC$ and the relative humidity is 6.00%? 79. A deep-sea diver should breathe a gas mixture that has the same oxygen partial pressure as at sea level, where dry air contains 20.9% oxygen and has a total pressure of $\displaystyle 1.01×10^5N/m^2$. (a) What is the partial pressure of oxygen at sea level? (b) If the diver breathes a gas mixture at a pressure of $\displaystyle 2.00×10^6N/m^2$, what percent oxygen should it be to have the same oxygen partial pressure as at sea level? Solution (a) $\displaystyle 2.12×10^4Pa$ (b) $\displaystyle 1.06%$ 80. The vapor pressure of water at $\displaystyle 40.0ºC$ is $\displaystyle 7.34×10^3N/m^2$. Using the ideal gas law, calculate the density of water vapor in $\displaystyle g/m^3$ that creates a partial pressure equal to this vapor pressure. The result should be the same as the saturation vapor density at that temperature ($\displaystyle 51.1 g/m^3$). 81. Air in human lungs has a temperature of $\displaystyle 37.0ºC$ and a saturation vapor density of $\displaystyle 44.0 g/m^3$. (a) If 2.00 L of air is exhaled and very dry air inhaled, what is the maximum loss of water vapor by the person? (b) Calculate the partial pressure of water vapor having this density, and compare it with the vapor pressure of $\displaystyle 6.31×10^3N/m^2$. Solution (a) $\displaystyle 8.80×10^{−2}g$ (b) $\displaystyle 6.30×10^3Pa$; the two values are nearly identical. 82. If the relative humidity is 90.0% on a muggy summer morning when the temperature is $\displaystyle 20.0ºC$, what will it be later in the day when the temperature is $\displaystyle 30.0ºC$, assuming the water vapor density remains constant? 83. Late on an autumn day, the relative humidity is 45.0% and the temperature is $\displaystyle 20.0ºC$. What will the relative humidity be that evening when the temperature has dropped to $\displaystyle 10.0ºC$, assuming constant water vapor density? Solution 82.3% 84. Atmospheric pressure atop Mt. Everest is $\displaystyle .30×10^4N/m^2$. (a) What is the partial pressure of oxygen there if it is 20.9% of the air? (b) What percent oxygen should a mountain climber breathe so that its partial pressure is the same as at sea level, where atmospheric pressure is $\displaystyle 1.01×10^5N/m^2$? (c) One of the most severe problems for those climbing very high mountains is the extreme drying of breathing passages. Why does this drying occur? 85. What is the dew point (the temperature at which 100% relative humidity would occur) on a day when relative humidity is 39.0% at a temperature of $\displaystyle 20.0ºC$ ? Solution $\displaystyle 4.77ºC$ 86. On a certain day, the temperature is $\displaystyle 25.0ºC $ and the relative humidity is 90.0%. How many grams of water must condense out of each cubic meter of air if the temperature falls to $\displaystyle 15.0ºC$? Such a drop in temperature can, thus, produce heavy dew or fog. 87. Integrated Concepts The boiling point of water increases with depth because pressure increases with depth. At what depth will fresh water have a boiling point of $\displaystyle 150ºC$, if the surface of the water is at sea level? Solution $\displaystyle 38.3m$ 88. Integrated Concepts (a) At what depth in fresh water is the critical pressure of water reached, given that the surface is at sea level? (b) At what temperature will this water boil? (c) Is a significantly higher temperature needed to boil water at a greater depth? 89. Integrated Concepts To get an idea of the small effect that temperature has on Archimedes’ principle, calculate the fraction of a copper block’s weight that is supported by the buoyant force in $\displaystyle 0ºC$ water and compare this fraction with the fraction supported in $\displaystyle 95.0ºC$ water. Solution $\displaystyle \frac{(F_B/w_{Cu})}{(F_B/w_{Cu})′}=1.02$. The buoyant force supports nearly the exact same amount of force on the copper block in both circumstances. 90. Integrated Concepts If you want to cook in water at $\displaystyle 150ºC$, you need a pressure cooker that can withstand the necessary pressure. (a) What pressure is required for the boiling point of water to be this high? (b) If the lid of the pressure cooker is a disk 25.0 cm in diameter, what force must it be able to withstand at this pressure? 91. Unreasonable Results (a) How many moles per cubic meter of an ideal gas are there at a pressure of $\displaystyle 1.00×10^{14}N/m^2$ and at $\displaystyle 0ºC$? (b) What is unreasonable about this result? (c) Which premise or assumption is responsible? Solution (a) $\displaystyle 4.41×10^{10}mol/m^3$ (b) It’s unreasonably large. (c) At high pressures such as these, the ideal gas law can no longer be applied. As a result, unreasonable answers come up when it is used. 92. Unreasonable Results (a) An automobile mechanic claims that an aluminum rod fits loosely into its hole on an aluminum engine block because the engine is hot and the rod is cold. If the hole is 10.0% bigger in diameter than the $\displaystyle 22.0ºC$ rod, at what temperature will the rod be the same size as the hole? (b) What is unreasonable about this temperature? (c) Which premise is responsible? 93. Unreasonable Results The temperature inside a supernova explosion is said to be $\displaystyle 2.00×10^{13}K$. (a) What would the average velocity $\displaystyle v_{rms}$ of hydrogen atoms be? (b) What is unreasonable about this velocity? (c) Which premise or assumption is responsible? Solution (a) $\displaystyle 7.03×10^8m/s$ size 12{ size 11{7 "." "03" times "10" rSup { size 8{8} } `"m/s"}} {} (b) The velocity is too high—it’s greater than the speed of light (c) The assumption that hydrogen inside a supernova behaves as an idea gas is responsible, because of the great temperature and density in the core of a star. Furthermore, when a velocity greater than the speed of light is obtained, classical physics must be replaced by relativity, a subject not yet covered. 94. Unreasonable Results Suppose the relative humidity is 80% on a day when the temperature is $\displaystyle 30.0ºC$. (a) What will the relative humidity be if the air cools to $\displaystyle 25.0ºC$ and the vapor density remains constant? (b) What is unreasonable about this result? (c) Which premise is responsible? Contributors and Attributions - Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .	libretexts	2025-03-17T19:53:32.245129	2018-05-04T03:01:47	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/13%3A_Temperature_Kinetic_Theory_and_the_Gas_Laws/13.E%3A_Work_Energy_and_Energy_Resources_(Exercise)", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "13.E: Work, Energy, and Energy Resources (Exercise)", "author": null }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods	14: Heat and Heat Transfer Methods Energy can exist in many forms and heat is one of the most intriguing. Heat is often hidden, as it only exists when in transit, and is transferred by a number of distinctly different methods. Heat transfer touches every aspect of our lives and helps us understand how the universe functions. It explains the chill we feel on a clear breezy night, or why Earth’s core has yet to cool. This chapter defines and explores heat transfer, its effects, and the methods by which heat is transferred. These topics are fundamental, as well as practical, and will often be referred to in the chapters ahead. - - 14.1: Heat - If two objects at different temperatures are brought in contact with each other, energy is transferred from the hotter to the colder object until equilibrium is reached and the bodies reach thermal equilibrium (i.e., they are at the same temperature). No work is done by either object, because no force acts through a distance. The transfer of energy is caused by the temperature difference, and ceases once the temperatures are equal. Heat is the spontaneous transfer of energy due to a temperature - - 14.2: Temperature Change and Heat Capacity - One of the major effects of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase change and that no work is done on or by the system. Experiments show that the transferred heat depends on three factors—the change in temperature, the mass of the system, and the substance and phase of the substance. - - 14.3: Phase Change and Latent Heat - Energy is required to melt a solid because the cohesive bonds between the molecules in the solid must be broken apart such that, in the liquid, the molecules can move around at comparable kinetic energies; thus, there is no rise in temperature. Similarly, energy is needed to vaporize a liquid, because molecules in a liquid interact with each other via attractive forces. There is no temperature change until a phase change is complete. - - 14.4: Heat Transfer Methods - Whenever there is a temperature difference, heat transfer occurs. Heat transfer may occur rapidly, such as through a cooking pan, or slowly, such as through the walls of a picnic ice chest. Every process involving heat transfer takes place by only three methods: Heat is transferred by three different methods: conduction, convection, and radiation. - - 14.5: Conduction - Heat conduction is the transfer of heat between two objects in direct contact with each other. The rate of heat transfer $Q/t$ (energy per unit time) is proportional to the temperature difference $T_2 - T_1$ and the contact area $A$ and inversely proportional to the distance between the objects: \[\dfrac{Q}{t} = \dfrac{kA(T_2 - T_1)}{d}.\] - - 14.6: Convection - Convection is heat transfer by the macroscopic movement of mass. Convection can be natural or forced and generally transfers thermal energy faster than conduction. Table gives wind-chill factors, indicating that moving air has the same chilling effect of much colder stationary air. Convection that occurs along with a phase change can transfer energy from cold regions to warm ones. - - 14.7: Radiation - Heat is transferred by radiation. That is, the hot body emits electromagnetic waves that are absorbed by our skin: no medium is required for electromagnetic waves to propagate. Different names are used for electromagnetic waves of different wavelengths: radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. Thumbnail: Different flame types of a Bunsen burner depend on oxygen supply. On the left a rich fuel with no premixed oxygen produces a yellow sooty diffusion flame; on the right a lean fully oxygen premixed flame produces no soot. (GNU Free Documentation License, Version 1.2; Jan Fijałkowski).	libretexts	2025-03-17T19:53:32.311942	2015-11-01T04:12:19	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "14: Heat and Heat Transfer Methods", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.00%3A_Prelude_to_Heat_and_Heat_Transfer_Methods	Energy can exist in many forms and heat is one of the most intriguing. Heat is often hidden, as it only exists when in transit, and is transferred by a number of distinctly different methods. Heat transfer touches every aspect of our lives and helps us understand how the universe functions. It explains the chill we feel on a clear breezy night, or why Earth’s core has yet to cool. This chapter defines and explores heat transfer, its effects, and the methods by which heat is transferred. These topics are fundamental, as well as practical, and will often be referred to in the chapters ahead.	libretexts	2025-03-17T19:53:32.458019	2015-11-01T05:40:12	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.00%3A_Prelude_to_Heat_and_Heat_Transfer_Methods", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "14.0: Prelude to Heat and Heat Transfer Methods", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.01%3A_Heat	14.1: Heat Learning Objectives By the end of this section, you will be able to: - Define heat as transfer of energy. In Work, Energy, and Energy Resources , we defined work as force times distance and learned that work done on an object changes its kinetic energy. We also saw in Temperature, Kinetic Theory, and the Gas Laws that temperature is proportional to the (average) kinetic energy of atoms and molecules. We say that a thermal system has a certain internal energy: its internal energy is higher if the temperature is higher. If two objects at different temperatures are brought in contact with each other, energy is transferred from the hotter to the colder object until equilibrium is reached and the bodies reach thermal equilibrium (i.e., they are at the same temperature). No work is done by either object, because no force acts through a distance. The transfer of energy is caused by the temperature difference, and ceases once the temperatures are equal. These observations lead to the following definition of heat : Heat is the spontaneous transfer of energy due to a temperature difference. As noted in Temperature, Kinetic Theory, and the Gas Laws , heat is often confused with temperature. For example, we may say the heat was unbearable, when we actually mean that the temperature was high. Heat is a form of energy, whereas temperature is not. The misconception arises because we are sensitive to the flow of heat, rather than the temperature. Owing to the fact that heat is a form of energy, it has the SI unit of joule (J). The calorie (cal) is a common unit of energy, defined as the energy needed to change the temperature of 1.00 g of water by $1.00^oC$ —specifically, between $14.5^oC$ and $15.5^oC$, since there is a slight temperature dependence. Perhaps the most common unit of heat is the kilocalorie (kcal), which is the energy needed to change the temperature of 1.00 kg of water by $1.00^oC$. Since mass is most often specified in kilograms, kilocalorie is commonly used. Food calories (given the notation Cal, and sometimes called “big calorie”) are actually kilocalories $(1 \, kilocalorie = 1000 \, calories)$, a fact not easily determined from package labeling. Mechanical Equivalent of Heat It is also possible to change the temperature of a substance by doing work. Work can transfer energy into or out of a system. This realization helped establish the fact that heat is a form of energy. James Prescott Joule (1818–1889) performed many experiments to establish the mechanical equivalent of heat — the work needed to produce the same effects as heat transfer . In terms of the units used for these two terms, the best modern value for this equivalence is \[ 1.000 \, kcal = 4186 \, J.\] We consider this equation as the conversion between two different units of energy. Figure $\PageIndex{2}$ shows one of Joule’s most famous experimental setups for demonstrating the mechanical equivalent of heat. It demonstrated that work and heat can produce the same effects, and helped establish the principle of conservation of energy. Gravitational potential energy (PE) (work done by the gravitational force) is converted into kinetic energy (KE), and then randomized by viscosity and turbulence into increased average kinetic energy of atoms and molecules in the system, producing a temperature increase. His contributions to the field of thermodynamics were so significant that the SI unit of energy was named after him. Heat added or removed from a system changes its internal energy and thus its temperature. Such a temperature increase is observed while cooking. However, adding heat does not necessarily increase the temperature. An example is melting of ice; that is, when a substance changes from one phase to another. Work done on the system or by the system can also change the internal energy of the system. Joule demonstrated that the temperature of a system can be increased by stirring. If an ice cube is rubbed against a rough surface, work is done by the frictional force. A system has a well-defined internal energy, but we cannot say that it has a certain “heat content” or “work content”. We use the phrase “heat transfer” to emphasize its nature. Exercise $\PageIndex{1}$ Two samples (A and B) of the same substance are kept in a lab. Someone adds 10 kilojoules (kJ) of heat to one sample, while 10 kJ of work is done on the other sample. How can you tell to which sample the heat was added? - Answer - Heat and work both change the internal energy of the substance. However, the properties of the sample only depend on the internal energy so that it is impossible to tell whether heat was added to sample A or B. Summary - Heat and work are the two distinct methods of energy transfer. - Heat is energy transferred solely due to a temperature difference. - Any energy unit can be used for heat transfer, and the most common are kilocalorie (kcal) and joule (J). - Kilocalorie is defined to be the energy needed to change the temperature of 1.00 kg of water between $14.5^oC$ and $15.5^oC$. - The mechanical equivalent of this heat transfer is $1.00 \, kcal = 4186 \, J.$ Glossary - heat - the spontaneous transfer of energy due to a temperature difference - kilocalorie - 1kilocalorie=1000caloriessize 12{1`"kilocalorie=1000"`"calories"} {} - mechanical equivalent of heat - the work needed to produce the same effects as heat transfer	libretexts	2025-03-17T19:53:32.525775	2015-11-01T05:40:26	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.01%3A_Heat", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "14.1: Heat", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.02%3A_Temperature_Change_and_Heat_Capacity	14.2: Temperature Change and Heat Capacity Learning Objectives By the end of this section, you will be able to: - Observe heat transfer and change in temperature and mass. - Calculate final temperature after heat transfer between two objects. One of the major effects of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase change and that no work is done on or by the system. Experiments show that the transferred heat depends on three factors—the change in temperature, the mass of the system, and the substance and phase of the substance. The dependence on temperature change and mass are easily understood. Owing to the fact that the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the absolute temperature and the number of atoms or molecules. Owing to the fact that the transferred heat is equal to the change in the internal energy, the heat is proportional to the mass of the substance and the temperature change. The transferred heat also depends on the substance so that, for example, the heat necessary to raise the temperature is less for alcohol than for water. For the same substance, the transferred heat also depends on the phase (gas, liquid, or solid). Heat Transfer and Temperature Change The quantitative relationship between heat transfer and temperature change contains all three factors: \[Q = mc\Delta T,\] where $Q$ is the symbol for heat transfer, $m$ is the mass of the substance, and $\Delta T$ is the change in temperature. The symbol $c$ stands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by $1.00^oC$. The specific heat $c$ is a property of the substance; its SI unit is $J/(kg \cdot K)$ or $J/(kg \cdot ^oC)$. Recall that the temperature change $(\Delta T)$ is the same in units of kelvin and degrees Celsius. If heat transfer is measured in kilocalories, then the unit of specific heat is $kcal/(kg \cdot ^oC)$. Values of specific heat must generally be looked up in tables, because there is no simple way to calculate them. In general, the specific heat also depends on the temperature. Table $\PageIndex{1}$ lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak. We see from this table that the specific heat of water is five times that of glass and ten times that of iron, which means that it takes five times as much heat to raise the temperature of water the same amount as for glass and ten times as much heat to raise the temperature of water as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth. Example $\PageIndex{1}$: Calculating the Required Heat: Heating Water in an Aluminum Pan A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from $20.0^oC$ to $80.0^oC$. (a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water? Strategy The pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and the pan is increased by the same amount. We use the equation for the heat transfer for the given temperature change and mass of water and aluminum. The specific heat values for water and aluminum are given in Table $\PageIndex{1}$. Solution Because water is in thermal contact with the aluminum, the pan and the water are at the same temperature. - Calculate the temperature difference: \[\Delta T = T_f - T_i = 60.0^oC.\] - Calculate the mass of water. Because the density of water is $1000 \, kg/m^3$, one liter of water has a mass of 1 kg, and the mass of 0.250 liters of water is $m_w = 0.250 \, kg$. - Calculate the heat transferred to the water. Use the specific heat of water in Table $\PageIndex{1}$ \[Q_w = m_wc_w\Delta T = (0.250 \, kg)(4186 \, J/kg ^oC)(60.0^oC) = 62.8 \, kJ.\] - Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table $\PageIndex{1}\: \[Q_{Al} = m_{Al}c_{Al}\Delta T = (0.500 \, kg)(900 \, J/kg^oC)(60.0^oC) = 27.0 kJ.\] - Compare the percentage of heat going into the pan versus that going into the water. First, find the total transferred heat: \[Q_{Total} = Q_W + Q_{Al} = 62.8 \, kJ + 27.0 \, kJ = 89.8 \, kJ.\] - Thus, the amount of heat going into heating the pan is \[\dfrac{62.8 \, kJ}{89.8 \, kJ} \times 100\% = 69.9\%\] Discussion In this example, the heat transferred to the container is a significant fraction of the total transferred heat. Although the mass of the pan is twice that of the water, the specific heat of water is over four times greater than that of aluminum. Therefore, it takes a bit more than twice the heat to achieve the given temperature change for the water as compared to the aluminum pan. Example \(\PageIndex{2}$: Calculating the Temperature Increase from the Work Done on a Substance: Truck Brakes Overheat on Downhill Runs Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into increased internal energy (higher temperature) of the brake material. This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck. The problem is that the mass of the truck is large compared with that of the brake material absorbing the energy, and the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment. Calculate the temperature increase of 100 kg of brake material with an average specific heat of $800.0 \, J/kg \cdot ^oC$ if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at a constant speed. Strategy If the brakes are not applied, gravitational potential energy is converted into kinetic energy. When brakes are applied, gravitational potential energy is converted into internal energy of the brake material. We first calculate the gravitational potential energy $(Mgh)$ that the entire truck loses in its descent and then find the temperature increase produced in the brake material alone. Solution - Calculate the change in gravitational potential energy as the truck goes downhill \[Mgh = (10,000 \, kg)(9.80 \, m/s^2)(75.0 \, m) = 7.35 \times 10^6 \, J.\] - Calculate the temperature from the heat transferred using $Q = Mgh$ and \[\Delta T = \dfrac{Q}{mc},\] where $m$ is the mass of the brake material. Insert the values $m = 100 \, kg$ and $c = 800 \, J/kg \cdot ^oC$ to find \[\Delta T = \dfrac{(7.35v \times 10^6 \, J)}{(100 \, kg)(800 \, J/kg^oC)} = 92^oC.\] Discussion This temperature is close to the boiling point of water. If the truck had been traveling for some time, then just before the descent, the brake temperature would likely be higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of the brake material above the boiling point of water, so this technique is not practical. However, the same idea underlies the recent hybrid technology of cars, where mechanical energy (gravitational potential energy) is converted by the brakes into electrical energy (battery). \| Substances \| Specific heat (c) \| \| \|---\|---\|---\| \| Solids \| $J/kg\cdot^oC$ \| $kcal/kg\cdot^oC$ \| \| Aluminum \| 900 \| 0.215 \| \| Asbestos \| 800 \| 0.19 \| \| Concrete, granite (average) \| 840 \| 0.20 \| \| Copper \| 387 \| 0.0924 \| \| Glass \| 840 \| 0.20 \| \| Gold \| 129 \| 0.0308 \| \| Human body (average at 37 °C) \| 3500 \| 0.83 \| \| Ice (average, -50°C to 0°C) \| 2090 \| 0.50 \| \| Iron, steel \| 452 \| 0.108 \| \| Lead \| 128 \| 0.0305 \| \| Silver \| 235 \| 0.0562 \| \| Wood \| 1700 \| 0.4 \| \| Liquids \| \|\| \| Benzene \| 1740 \| 0.415 \| \| Ethanol \| 2450 \| 0.586 \| \| Glycerin \| 2410 \| 0.576 \| \| Mercury \| 139 \| 0.0333 \| \| Water (15.0 °C) \| 4186 \| 1.000 \| \| Gases 3 \| \|\| \| Air (dry) \| 721 (1015) \| 0.172 (0.242) \| \| Ammonia \| 1670 (2190) \| 0.399 (0.523) \| \| Carbon dioxide \| 638 (833) \| 0.152 (0.199) \| \| Nitrogen \| 739 (1040) \| 0.177 (0.248) \| \| Oxygen \| 651 (913) \| 0.156 (0.218) \| \| Steam (100°C) \| 1520 (2020) \| 0.363 (0.482) \| Note that Example $\PageIndex{2}$ is an illustration of the mechanical equivalent of heat. Alternatively, the temperature increase could be produced by a blow torch instead of mechanically. Example $\PageIndex{3}$: Calculating the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan Suppose you pour 0.250 kg of $20.0^oC$ water (about a cup) into a 0.500-kg aluminum pan off the stove with a temperature of $150^oC$. Assume that the pan is placed on an insulated pad and that a negligible amount of water boils off. What is the temperature when the water and pan reach thermal equilibrium a short time later? Strategy The pan is placed on an insulated pad so that little heat transfer occurs with the surroundings. Originally the pan and water are not in thermal equilibrium: the pan is at a higher temperature than the water. Heat transfer then restores thermal equilibrium once the water and pan are in contact. Because heat transfer between the pan and water takes place rapidly, the mass of evaporated water is negligible and the magnitude of the heat lost by the pan is equal to the heat gained by the water. The exchange of heat stops once a thermal equilibrium between the pan and the water is achieved. The heat exchange can be written as $\|Q_{hot}\| = Q_{cold}.$ Solution - Use the equation for heat transfer $Q = mc\Delta T$ to express the heat lost by the aluminum pan in terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final temperature: \[Q_{hot} = m_{Al}c_{Al}(T_f - 150^oC).\] - Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water and the final temperature: \[Q_{cold} = m_wc_w(T_f - 20.0^oC).\] - Note that $Q_{hot}<0$ and $Q_{cold} >0$ and that they must sum to zero because the heat lost by the hot pan must be the same as the heat gained by the cold water: \[Q_{cold} + Q_{hot} = 0,\] \[Q_{cold} = -Q_{hot},\] \[m_wc_w(T_f - 20.0^oC) = -m_{Al}c_{Al}(T_f - 150.0^oC).\] - Bring all terms involving $T_f$ on the left hand side and all other terms on the right hand side. Solve for $T_f$, $T_f=\dfrac{m_{Al}c_{Al}(150ºC)+m_Wc_W(20.0ºC)}{m_{Al}c_{Al}+m_Wc_W}$, and insert the numerical values: $T_f=\dfrac{(0.500 kg)(900 J/kgºC)(150ºC)+(0.250 kg)(4186 J/kgºC)(20.0ºC)}{(0.500 kg)(900 J/kgºC)+(0.250 kg)(4186 J/kgºC)}=\dfrac{88430 J}{1496.5 J/ºC}=59.1ºC.$ Discussion This is a typical calorimetry problem—two bodies at different temperatures are brought in contact with each other and exchange heat until a common temperature is reached. Why is the final temperature so much closer to 20.0ºC than 150ºC ? The reason is that water has a greater specific heat than most common substances and thus undergoes a small temperature change for a given heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase its temperature appreciably. This explains why the temperature of a lake stays relatively constant during a day even when the temperature change of the air is large. However, the water temperature does change over longer times (e.g., summer to winter). TAKE-HOME EXPERIMENT: TEMPERATURE CHANGE OF LAND AND WATER What heats faster, land or water? To study differences in heat capacity: - Place equal masses of dry sand (or soil) and water at the same temperature into two small jars. (The average density of soil or sand is about 1.6 times that of water, so you can achieve approximately equal masses by using $50%$ more water by volume.) - Heat both (using an oven or a heat lamp) for the same amount of time. - Record the final temperature of the two masses. - Now bring both jars to the same temperature by heating for a longer period of time. - Remove the jars from the heat source and measure their temperature every 5 minutes for about 30 minutes. Which sample cools off the fastest? This activity replicates the phenomena responsible for land breezes and sea breezes. Exercise $\PageIndex{1}$ If 25 kJ is necessary to raise the temperature of a block from 25ºC to 30ºC , how much heat is necessary to heat the block from 45ºC to 50ºC ? - Answer - The heat transfer depends only on the temperature difference. Since the temperature differences are the same in both cases, the same 25 kJ is necessary in the second case. Summary - The transfer of heat $Q$ that leads to a change $ΔT$ in the temperature of a body with mass $m$ is $Q=mcΔT$, where $c$ is the specific heat of the material. This relationship can also be considered as the definition of specific heat. Footnotes 1 The values for solids and liquids are at constant volume and at 25ºC , except as noted. 2 These values are identical in units of cal/g⋅ºC . 3 cv at constant volume and at 20.0ºC , except as noted, and at 1.00 atm average pressure. Values in parentheses are $c_p$ at a constant pressure of 1.00 atm. Glossary specific heat the amount of heat necessary to change the temperature of 1.00 kg of a substance by 1.00 ºC	libretexts	2025-03-17T19:53:32.618357	2015-11-01T05:40:43	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.02%3A_Temperature_Change_and_Heat_Capacity", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "14.2: Temperature Change and Heat Capacity", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.03%3A_Phase_Change_and_Latent_Heat	14.3: Phase Change and Latent Heat Learning Objectives By the end of this section, you will be able to: - Examine heat transfer. - Calculate final temperature from heat transfer. So far we have discussed temperature change due to heat transfer. No temperature change occurs from heat transfer if ice melts and becomes liquid water (i.e., during a phase change). For example, consider water dripping from icicles melting on a roof warmed by the Sun. Conversely, water freezes in an ice tray cooled by lower-temperature surroundings. Energy is required to melt a solid because the cohesive bonds between the molecules in the solid must be broken apart such that, in the liquid, the molecules can move around at comparable kinetic energies; thus, there is no rise in temperature. Similarly, energy is needed to vaporize a liquid, because molecules in a liquid interact with each other via attractive forces. There is no temperature change until a phase change is complete. The temperature of a cup of soda initially at $0^oC$ stays at $0^oC$ until all the ice has melted. Conversely, energy is released during freezing and condensation, usually in the form of thermal energy. Work is done by cohesive forces when molecules are brought together. The corresponding energy must be given off (dissipated) to allow them to stay together ( see Figure $\PageIndex{2}$) The energy involved in a phase change depends on two major factors: the number and strength of bonds or force pairs. The number of bonds is proportional to the number of molecules and thus to the mass of the sample. The strength of forces depends on the type of molecules. The heat $Q$ required to change the phase of a sample of mass $m$ is given by \[ Q = mL_f (melting/freezing),\] \[Q = mL_v (vaporization/condensation),\] where the latent heat of fusion, $L_f$, and latent heat of vaporization, $L_v$, are material constants that are determined experimentally. See (Table $\PageIndex{1}$). Latent heat is measured in units of J/kg. Both $L_f$ and $L_v$ depend on the substance, particularly on the strength of its molecular forces as noted earlier. $L_f$ and $L_v$ are collectively called latent heat coefficients . They are latent , or hidden, because in phase changes, energy enters or leaves a system without causing a temperature change in the system; so, in effect, the energy is hidden. Table $\PageIndex{1}$ lists representative values of $L_f$ and $L_v$, together with melting and boiling points. The table shows that significant amounts of energy are involved in phase changes. Let us look, for example, at how much energy is needed to melt a kilogram of ice at $0^oC$ to produce a kilogram of water at $0^oC$. Using the equation for a change in temperature and the value for water from Table $\PageIndex{1}$, we find that \[Q = mL_f = (1.0 \, kg)(334 \, kJ/kg) = 334 \, kJ\] is the energy to melt a kilogram of ice. This is a lot of energy as it represents the same amount of energy needed to raise the temperature of 1 kg of liquid water from $0^oC$ to $79.8^oC$. Even more energy is required to vaporize water; it would take 2256 kJ to change 1 kg of liquid water at the normal boiling point $(100^oC$ at atmospheric pressure) to steam (water vapor). This example shows that the energy for a phase change is enormous compared to energy associated with temperature changes without a phase change. \| $L_f$ \| $L_v$ \| \|\|\|\|\| \|---\|---\|---\|---\|---\|---\|---\| \| Substance \| Melting point (ºC) \| kJ/kg \| kcal/kg \| Boiling point (°C) \| kJ/kg \| kcal/kg \| \| Helium \| −269.7 \| 5.23 \| 1.25 \| −268.9 \| 20.9 \| 4.99 \| \| Hydrogen \| −259.3 \| 58.6 \| 14.0 \| −252.9 \| 452 \| 108 \| \| Nitrogen \| −210.0 \| 25.5 \| 6.09 \| −195.8 \| 201 \| 48.0 \| \| Oxygen \| −218.8 \| 13.8 \| 3.30 \| −183.0 \| 213 \| 50.9 \| \| Ethanol \| −114 \| 104 \| 24.9 \| 78.3 \| 854 \| 204 \| \| Ammonia \| −75 \| 108 \| −33.4 \| 1370 \| 327 \| \| \| Mercury \| −38.9 \| 11.8 \| 2.82 \| 357 \| 272 \| 65.0 \| \| Water \| 0.00 \| 334 \| 79.8 \| 100.0 \| 2256 2 \| 539 3 \| \| Sulfur \| 119 \| 38.1 \| 9.10 \| 444.6 \| 326 \| 77.9 \| \| Lead \| 327 \| 24.5 \| 5.85 \| 1750 \| 871 \| 208 \| \| Antimony \| 631 \| 165 \| 39.4 \| 1440 \| 561 \| 134 \| \| Aluminum \| 660 \| 380 \| 90 \| 2450 \| 11400 \| 2720 \| \| Silver \| 961 \| 88.3 \| 21.1 \| 2193 \| 2336 \| 558 \| \| Gold \| 1063 \| 64.5 \| 15.4 \| 2660 \| 1578 \| 377 \| \| Copper \| 1083 \| 134 \| 32.0 \| 2595 \| 5069 \| 1211 \| \| Uranium \| 1133 \| 84 \| 20 \| 3900 \| 1900 \| 454 \| \| Tungsten \| 3410 \| 184 \| 44 \| 5900 \| 4810 \| 1150 \| Phase changes can have a tremendous stabilizing effect even on temperatures that are not near the melting and boiling points, because evaporation and condensation (conversion of a gas into a liquid state) occur even at temperatures below the boiling point.Take, for example, the fact that air temperatures in humid climates rarely go above $35.0^oC$, which is because most heat transfer goes into evaporating water into the air. Similarly, temperatures in humid weather rarely fall below the dew point because enormous heat is released when water vapor condenses. We examine the effects of phase change more precisely by considering adding heat into a sample of ice at $-20^oC$ (Figure $\PageIndex{3}$). The temperature of the ice rises linearly, absorbing heat at a constant rate of $0.50 \, cal/g \cdot ^oC$ until it reaches $0^oC$. Once at this temperature, the ice begins to melt until all the ice has melted, absorbing 79.8 cal/g of heat. The temperature remains constant at $0^oC$ during this phase change. Once all the ice has melted, the temperature of the liquid water rises, absorbing heat at a new constant rate of $1.00 \, cal/g \cdot ^oC$. At $100^oC$, the water begins to boil and the temperature again remains constant while the water absorbs 539 cal/g of heat during this phase change. When all the liquid has become steam vapor, the temperature rises again, absorbing heat at a rate of $0.482 \, cal/g \cdot ^oC$. Water can evaporate at temperatures below the boiling point. More energy is required than at the boiling point, because the kinetic energy of water molecules at temperatures below $100^oC$ is less than that at $100^oC$, hence less energy is available from random thermal motions. Take, for example, the fact that, at body temperature, perspiration from the skin requires a heat input of 2428 kJ/kg, which is about 10 percent higher than the latent heat of vaporization at $100^oC$. This heat comes from the skin, and thus provides an effective cooling mechanism in hot weather. High humidity inhibits evaporation, so that body temperature might rise, leaving unevaporated sweat on your brow. Example $\PageIndex{1}$: Calculate Final Temperature from Phase Change: Cooling Soda with Ice Cubes Three ice cubes are used to chill a soda at $20^oC$ with mass $m_{soda} = 0.25 \, kg$. The ice is at $0^oC$ and each ice cube has a mass of 6.0 g. Assume that the soda is kept in a foam container so that heat loss can be ignored. Assume the soda has the same heat capacity as water. Find the final temperature when all ice has melted. Strategy The ice cubes are at the melting temperature of $0^oC$. Heat is transferred from the soda to the ice for melting. Melting of ice occurs in two steps: first the phase change occurs and solid (ice) transforms into liquid water at the melting temperature, then the temperature of this water rises. Melting yields water at $0^oC$, so more heat is transferred from the soda to this water until the water plus soda system reaches thermal equilibrium, \[Q_{ice} = -Q_{soda}.\] The heat transferred to the ice is $Q_{ice} = m_{ice}L_f + m_{ice} c_w (T_f - 0^oC).$. The heat given off by the soda is $Q_{soda} = m_{soda}c_w (T_f - 20^oC).$ Since no heat is lost, $Q_{ice} = -Q_{soda}$, so that \[m_{ice}Lf + m_{ice} c_w (T_f - 0^oC) = -m_{soda} c_w (T_f - 20^oC).\] Bring all terms involving $T_f$ on the left-hand-side and all other terms on the right-hand-side. Solve for the unknown quantity $T_f$: \[T_f = \dfrac{m_{soda} c_w (20^oC) - m_{ice} L_f}{(m_{soda} + m_{ice}) c_w}.\] Solution - Identify the known quantities. The mass of ice is $m_{ice} = 3 \times 6.0 \, g = 0.018 \, kg $ and the mass of soda is $m_{soda} = 0.25 \, kg$. - Calculate the terms in the numerator: \[m_{soda} c_w (20^oC) = (0.25 \, kg) (4186 \, J/kg \cdot ^oC) (20^oC) = 20,930 \, J\] and \[m_{ice}L_f = (0.018 \, kg)(334,000 \, J/kg) = 6012 \, J.\] - Calculate the denominator: \[(m_{soda} + m_{ice})c_w = (0.25 \, kg + 0.018 \, kg)(4186 \, J/kg \cdot ^oC)= 1122 \, J/^oC\] - Calculate the final temperature: \[T_f = \dfrac{20,930 \, J - 6012 \, J}{1122 \, J/^oC} = 13^oC.\] Discussion This example illustrates the enormous energies involved during a phase change. The mass of ice is about 7 percent the mass of water but leads to a noticeable change in the temperature of soda. Although we assumed that the ice was at the freezing temperature, this is incorrect: the typical temperature is $-6^oC$. However, this correction gives a final temperature that is essentially identical to the result we found. Can you explain why? We have seen that vaporization requires heat transfer to a liquid from the surroundings, so that energy is released by the surroundings. Condensation is the reverse process, increasing the temperature of the surroundings. This increase may seem surprising, since we associate condensation with cold objects—the glass in the figure, for example. However, energy must be removed from the condensing molecules to make a vapor condense. The energy is exactly the same as that required to make the phase change in the other direction, from liquid to vapor, and so it can be calculated from $Q = mL_v$. Real-World Application Energy is also released when a liquid freezes. This phenomenon is used by fruit growers in Florida to protect oranges when the temperature is close to the freezing point $(0^oC)$. Growers spray water on the plants in orchards so that the water freezes and heat is released to the growing oranges on the trees. This prevents the temperature inside the orange from dropping below freezing, which would damage the fruit. Sublimation is the transition from solid to vapor phase. You may have noticed that snow can disappear into thin air without a trace of liquid water, or the disappearance of ice cubes in a freezer. The reverse is also true: Frost can form on very cold windows without going through the liquid stage. A popular effect is the making of “smoke” from dry ice, which is solid carbon dioxide. Sublimation occurs because the equilibrium vapor pressure of solids is not zero. Certain air fresheners use the sublimation of a solid to inject a perfume into the room. Moth balls are a slightly toxic example of a phenol (an organic compound) that sublimates, while some solids, such as osmium tetroxide, are so toxic that they must be kept in sealed containers to prevent human exposure to their sublimation-produced vapors. All phase transitions involve heat. In the case of direct solid-vapor transitions, the energy required is given by the equation $Q = mL_s$, where $L_s$ is the heat of sublimation , which is the energy required to change 1.00 kg of a substance from the solid phase to the vapor phase. $L_s$ is analogous to $L_f$ and $L_v$, and its value depends on the substance. Sublimation requires energy input, so that dry ice is an effective coolant, whereas the reverse process (i.e., frosting) releases energy. The amount of energy required for sublimation is of the same order of magnitude as that for other phase transitions. The material presented in this section and the preceding section allows us to calculate any number of effects related to temperature and phase change. In each case, it is necessary to identify which temperature and phase changes are taking place and then to apply the appropriate equation. Keep in mind that heat transfer and work can cause both temperature and phase changes. Problem-Solving Strategies for the Effects of Heat Transfer - Examine the situation to determine that there is a change in the temperature or phase. Is there heat transfer into or out of the system? When the presence or absence of a phase change is not obvious, you may wish to first solve the problem as if there were no phase changes, and examine the temperature change obtained. If it is sufficient to take you past a boiling or melting point, you should then go back and do the problem in steps—temperature change, phase change, subsequent temperature change, and so on. - Identify and list all objects that change temperature and phase. - Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. - Make a list of what is given or what can be inferred from the problem as stated (identify the knowns). - Solve the appropriate equation for the quantity to be determined (the unknown). If there is a temperature change, the transferred heat depends on the specific heat (see Temperature Change and Heat Capacity ) whereas, for a phase change, the transferred heat depends on the latent heat. See Table $\PageIndex{1}$. - Substitute the knowns along with their units into the appropriate equation and obtain numerical solutions complete with units. You will need to do this in steps if there is more than one stage to the process (such as a temperature change followed by a phase change). - Check the answer to see if it is reasonable: Does it make sense? As an example, be certain that the temperature change does not also cause a phase change that you have not taken into account. Exercise $\PageIndex{1}$ Why does snow remain on mountain slopes even when daytime temperatures are higher than the freezing temperature? - Answer - Snow is formed from ice crystals and thus is the solid phase of water. Because enormous heat is necessary for phase changes, it takes a certain amount of time for this heat to be accumulated from the air, even if the air is above $0^oC$. The warmer the air is, the faster this heat exchange occurs and the faster the snow melts. Summary - Most substances can exist either in solid, liquid, and gas forms, which are referred to as “phases.” - Phase changes occur at fixed temperatures for a given substance at a given pressure, and these temperatures are called boiling and freezing (or melting) points. - During phase changes, heat absorbed or released is given by $Q = mL,$ where $L$ is the latent heat coefficient. Footnotes 1 Values quoted at the normal melting and boiling temperatures at standard atmospheric pressure (1 atm). 2 At 37.0ºC (body temperature), the heat of vaporization $L_v$ for water is 2430 kJ/kg or 580 kcal/kg 3 At 37.0ºC (body temperature), the heat of vaporization $L_v$ for water is 2430 kJ/kg or 580 kcal/kg Glossary - heat of sublimation - the energy required to change a substance from the solid phase to the vapor phase - latent heat coefficient - a physical constant equal to the amount of heat transferred for every 1 kg of a substance during the change in phase of the substance - sublimation - the transition from the solid phase to the vapor phase Contributors and Attributions Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .	libretexts	2025-03-17T19:53:32.719474	2015-11-01T05:40:58	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.03%3A_Phase_Change_and_Latent_Heat", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "14.3: Phase Change and Latent Heat", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.04%3A_Heat_Transfer_Methods	14.4: Heat Transfer Methods Learning Objectives By the end of this section, you will be able to: - Discuss the different methods of heat transfer. Equally as interesting as the effects of heat transfer on a system are the methods by which this occurs. Whenever there is a temperature difference, heat transfer occurs. Heat transfer may occur rapidly, such as through a cooking pan, or slowly, such as through the walls of a picnic ice chest. We can control rates of heat transfer by choosing materials (such as thick wool clothing for the winter), controlling air movement (such as the use of weather stripping around doors), or by choice of color (such as a white roof to reflect summer sunlight). So many processes involve heat transfer, so that it is hard to imagine a situation where no heat transfer occurs. Yet every process involving heat transfer takes place by only three methods: - Conduction is heat transfer through stationary matter by physical contact. (The matter is stationary on a macroscopic scale—we know there is thermal motion of the atoms and molecules at any temperature above absolute zero.) Heat transferred between the electric burner of a stove and the bottom of a pan is transferred by conduction. - Convection is the heat transfer by the macroscopic movement of a fluid. This type of transfer takes place in a forced-air furnace and in weather systems, for example. - Heat transfer by radiation occurs when microwaves, infrared radiation, visible light, or another form of electromagnetic radiation is emitted or absorbed. An obvious example is the warming of the Earth by the Sun. A less obvious example is thermal radiation from the human body. We examine these methods in some detail in the three following modules. Each method has unique and interesting characteristics, but all three do have one thing in common: they transfer heat solely because of a temperature difference Figure $\PageIndex{1}$. Exercise $\PageIndex{1}$ Name an example from daily life (different from the text) for each mechanism of heat transfer. - Answer - - Conduction: Heat transfers into your hands as you hold a hot cup of coffee. - Convection: Heat transfers as the barista “steams” cold milk to make hot cocoa. - Radiation: Reheating a cold cup of coffee in a microwave oven. Summary - Heat is transferred by three different methods: conduction, convection, and radiation. Glossary - conduction - heat transfer through stationary matter by physical contact - convection - heat transfer by the macroscopic movement of fluid - radiation - heat transfer which occurs when microwaves, infrared radiation, visible light, or other electromagnetic radiation is emitted or absorbed	libretexts	2025-03-17T19:53:32.785008	2015-11-01T05:41:14	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.04%3A_Heat_Transfer_Methods", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "14.4: Heat Transfer Methods", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.05%3A_Conduction	14.5: Conduction Learning Objectives By the end of this section, you will be able to: - Calculate thermal conductivity. - Observe conduction of heat in collisions. - Study thermal conductivities of common substances. Your feet feel cold as you walk barefoot across the living room carpet in your cold house and then step onto the kitchen tile floor. This result is intriguing, since the carpet and tile floor are both at the same temperature. The different sensation you feel is explained by the different rates of heat transfer: the heat loss during the same time interval is greater for skin in contact with the tiles than with the carpet, so the temperature drop is greater on the tiles. Some materials conduct thermal energy faster than others. In general, good conductors of electricity (metals like copper, aluminum, gold, and silver) are also good heat conductors, whereas insulators of electricity (wood, plastic, and rubber) are poor heat conductors. Figure $\PageIndex{2}$ shows molecules in two bodies at different temperatures. The (average) kinetic energy of a molecule in the hot body is higher than in the colder body. If two molecules collide, an energy transfer from the molecule with greater kinetic energy to the molecule with less kinetic energy occurs. The cumulative effect from all collisions results in a net flux of heat from the hot body to the colder body. The heat flux thus depends on the temperature difference $\Delta T = T_{hot} - T_{cold}$ Therefore, you will get a more severe burn from boiling water than from hot tap water. Conversely, if the temperatures are the same, the net heat transfer rate falls to zero, and equilibrium is achieved. Owing to the fact that the number of collisions increases with increasing area, heat conduction depends on the cross-sectional area. If you touch a cold wall with your palm, your hand cools faster than if you just touch it with your fingertip. A third factor in the mechanism of conduction is the thickness of the material through which heat transfers. The figure below shows a slab of material with different temperatures on either side. Suppose that $T_2$ is greater than $T_1$ so that heat is transferred from left to right. Heat transfer from the left side to the right side is accomplished by a series of molecular collisions. The thicker the material, the more time it takes to transfer the same amount of heat. This model explains why thick clothing is warmer than thin clothing in winters, and why Arctic mammals protect themselves with thick blubber. Lastly, the heat transfer rate depends on the material properties described by the coefficient of thermal conductivity . All four factors are included in a simple equation that was deduced from and is confirmed by experiments. The rate of conductive heat transfer through a slab of material, such as the one in Figure $\PageIndex{3}$, is given by \[\dfrac{Q}{t} = \dfrac{kA(T_2 - T_!)}{d},\] where $Q/t$ is the rate of heat transfer in watts or kilocalories per second, $k$ s the thermal conductivity of the material, $A$ and $d$ are its surface area and thickness and $(T_2 - T_1)$ is the temperature difference across the slab. Table $\PageIndex{1}$gives representative values of thermal conductivity. Example $\PageIndex{1}$: Calculating Heat Transfer Through Conduction: Conduction Rate Through an Ice Box A Styrofoam ice box has a total area of $0.950 \, m^2$ and walls with an average thickness of 2.50 cm. The box contains ice, water, and canned beverages at $0^oC$. The inside of the box is kept cold by melting ice. How much ice melts in one day if the ice box is kept in the trunk of a car at $35.0^oC$? Strategy This question involves both heat for a phase change (melting of ice) and the transfer of heat by conduction. To find the amount of ice melted, we must find the net heat transferred. This value can be obtained by calculating the rate of heat transfer by conduction and multiplying by time. Solution - Identify the knowns. \[A = 0.950 \, m^2; \, d = 2.50 \, cm = 0.0250 \, m; \, T_1 = 35.0^oC, \, t = 1 \, day = 24 \, hours = 86,400 \, sec.\] - Identify the unknowns. We need to solve for the mass of the ice $m$. We will also need to solve for the net heat transferred to melt the ice, $Q$. - Determine which equations to use. The rate of heat transfer by conduction is given by \[\dfrac{Q}{t} = \dfrac{kA(T_2 - T_1)}{d}.\] - The heat is used to melt the ice: $Q = mL_f$. - Insert the known values: \[\dfrac{Q}{t} = \dfrac{(0.010 \, J/s \cdot m \cdot ^oC)(0.950 \, m^2)(35.0^oC - 0^oC)}{0.0250 \, m} = 13.3 \, J/s.\] - Multiply the rate of heat transfer by the time $(1 \, day = 86,400 s)$: \[Q = (Q/t)t = (13.3 \, J/s)(86,400 \, s) = 1.15 \times 10^6 \, J.\] - Set this equal to the heat transferred to melt the ice: $Q = mL_f$. Solve for the mass $m$: \[m = \dfrac {Q}{L_f} = \dfrac{1.15 \times 10^6 \, J}{334 \times 10^3 \, J/kg} = 3.44 \, kg.\] Discussion The result of 3.44 kg, or about 7.6 lbs, seems about right, based on experience. You might expect to use about a 4 kg (7–10 lb) bag of ice per day. A little extra ice is required if you add any warm food or beverages. Inspecting the conductivities in Table $\PageIndex{1}$ shows that Styrofoam is a very poor conductor and thus a good insulator. Other good insulators include fiberglass, wool, and goose-down feathers. Like Styrofoam, these all incorporate many small pockets of air, taking advantage of air’s poor thermal conductivity. \| Substance \| Thermal conductivity \| \|---\|---\| \| Air \| 0.023 \| \| Aluminum \| 220 \| \| Asbestos \| 0.16 \| \| Concrete brick \| 0.84 \| \| Copper \| 390 \| \| Cork \| 0.042 \| \| Down feathers \| 0.025 \| \| Fatty tissue (without blood) \| 0.2 \| \| Glass (average) \| 0.84 \| \| Glass wool \| 0.042 \| \| Gold \| 318 \| \| Ice \| 2.2 \| \| Plasterboard \| 0.16 \| \| Silver \| 420 \| \| Snow (dry) \| 0.10 \| \| Steel (stainless) \| 14 \| \| Steel iron \| 80 \| \| Styrofoam \| 0.010 \| \| Water \| 0.6 \| \| Wood \| 0.08–0.16 \| \| Wool \| 0.04 \| Thermal Conductivities of Common Substances 1 A combination of material and thickness is often manipulated to develop good insulators—the smaller the conductivity $k$ and the larger the thickness $d$, the better. The ratio of $d/k$ will thus be large for a good insulator. The ratio $d/k$ is called the $R$ factor . The rate of conductive heat transfer is inversely proportional to $R$. The larger the value of $R$, the better the insulation. $R$ factors are most commonly quoted for household insulation, refrigerators, and the like—unfortunately, it is still in non-metric units of $ft^2 \cdot ^oF \cdot h/Btu$, although the unit usually goes unstated (1 British thermal unit [Btu] is the amount of energy needed to change the temperature of 1.0 lb of water by 1.0 °F). A couple of representative values are an $R$ factor of 11 for 3.5-in-thick fiberglass batts (pieces) of insulation and an $R$ factor of 19 for 6.5-in-thick fiberglass batts. Walls are usually insulated with 3.5-in batts, while ceilings are usually insulated with 6.5-in batts. In cold climates, thicker batts may be used in ceilings and walls. Note that in Table $\PageIndex{1}$, the best thermal conductors—silver, copper, gold, and aluminum—are also the best electrical conductors, again related to the density of free electrons in them. Cooking utensils are typically made from good conductors. Example $\PageIndex{1}$: Calculating the Temperature Difference Maintained by a Heat Transfer: Conduction Through an Aluminum Pan Water is boiling in an aluminum pan placed on an electrical element on a stovetop. The sauce pan has a bottom that is 0.800 cm thick and 14.0 cm in diameter. The boiling water is evaporating at the rate of 1.00 g/s. What is the temperature difference across (through) the bottom of the pan? Strategy Conduction through the aluminum is the primary method of heat transfer here, and so we use the equation for the rate of heat transfer and solve for the temperature difference . \[T_2 - T_1 = \dfrac{Q}{t} \left(\dfrac{d}{kA}\right).\] Solution - Identify the knowns and convert them to the SI units. The thickness of the pan, $d = 0.800 \, cm = 8.0 \times 10^{-3} m$, the area of the pan, $A = \pi(0.14/2)^2 \, m^2 = 1.54 \times 10^{-2} \, m^2$, and the thermal conductivity, $k = 220 \, J/s \cdot m \cdot ^oC$. - Calculate the necessary heat of vaporization of 1 g of water: \[Q = mL_v = (1.00 \times 10^{-3} \, kg)(2256 \times 10^3 \, J/kg) = 2256 \, J.\] - Calculate the rate of heat transfer given that 1 g of water melts in one second: \[ Q/t = 2256 \, J/s \, or 2.26 kW.\] - Insert the knowns into the equation and solve for the temperature difference: \[T_2 - T_1 = \dfrac{Q}{t}\left(\dfrac{d}{kA} \right) = (2256 \, J/s) \dfrac{8.00 \times 10^{-3} m}{(220 \, J/s \cdot m \cdot ^oC)(1.54 \times 10^{-2} \, m^2)} = 5.33 ^oC.\] Discussion The value for the heat transfer $Q/t = 2.26 \, kW \, or 2256 \, J/s$ is typical for an electric stove. This value gives a remarkably small temperature difference between the stove and the pan. Consider that the stove burner is red hot while the inside of the pan is nearly $100^oC$ because of its contact with boiling water. This contact effectively cools the bottom of the pan in spite of its proximity to the very hot stove burner. Aluminum is such a good conductor that it only takes this small temperature difference to produce a heat transfer of 2.26 kW into the pan. Conduction is caused by the random motion of atoms and molecules. As such, it is an ineffective mechanism for heat transport over macroscopic distances and short time distances. Take, for example, the temperature on the Earth, which would be unbearably cold during the night and extremely hot during the day if heat transport in the atmosphere was to be only through conduction. In another example, car engines would overheat unless there was a more efficient way to remove excess heat from the pistons. Exercise $\PageIndex{1}$: Check your understanding How does the rate of heat transfer by conduction change when all spatial dimensions are doubled? - Answer - Because area is the product of two spatial dimensions, it increases by a factor of four when each dimension is doubled $(A_{final} = (2d)^2 = 4d^2 = 4A_{initial})$. The distance, however, simply doubles. Because the temperature difference and the coefficient of thermal conductivity are independent of the spatial dimensions, the rate of heat transfer by conduction increases by a factor of four divided by two, or two: \[\left(\dfrac{Q}{t} \right)_{final} = \dfrac{kA_{final}(T_2 - T_1)}{d_{final}} = \dfrac{k(4A_{initial})(T_2 - T_1)}{2d_{initial}} = 2 \dfrac{kA_{initial}(T_2 - T_1)}{d_{initial}} = 2\left(\dfrac{Q}{t}\right)_{initial}\] Summary Heat conduction is the transfer of heat between two objects in direct contact with each other. The rate of heat transfer $Q/t$ (energy per unit time) is proportional to the temperature difference $T_2 - T_1$ and the contact area $A$ and inversely proportional to the distance between the objects: \[\dfrac{Q}{t} = \dfrac{kA(T_2 - T_1)}{d}.\] Footnotes - At temperatures near 0ºC. Glossary - R factor - the ratio of thickness to the conductivity of a material - rate of conductive heat transfer - rate of heat transfer from one material to another - thermal conductivity - the property of a material’s ability to conduct heat	libretexts	2025-03-17T19:53:32.869276	2015-11-01T05:41:31	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.05%3A_Conduction", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "14.5: Conduction", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.06%3A_Convection	14.6: Convection Learning Objectives By the end of this section, you will be able to: - Discuss the method of heat transfer by convection. Convection is driven by large-scale flow of matter. In the case of Earth, the atmospheric circulation is caused by the flow of hot air from the tropics to the poles, and the flow of cold air from the poles toward the tropics. (Note that Earth’s rotation causes the observed easterly flow of air in the northern hemisphere). Car engines are kept cool by the flow of water in the cooling system, with the water pump maintaining a flow of cool water to the pistons. The circulatory system is used the body: when the body overheats, the blood vessels in the skin expand (dilate), which increases the blood flow to the skin where it can be cooled by sweating. These vessels become smaller when it is cold outside and larger when it is hot (so more fluid flows, and more energy is transferred). The body also loses a significant fraction of its heat through the breathing process. While convection is usually more complicated than conduction, we can describe convection and do some straightforward, realistic calculations of its effects. Natural convection is driven by buoyant forces: hot air rises because density decreases as temperature increases. The house in Figure $\PageIndex{1}$ is kept warm in this manner, as is the pot of water on the stove in Figure $\PageIndex{2}$. Ocean currents and large-scale atmospheric circulation transfer energy from one part of the globe to another. Both are examples of natural convection. Take-Home Experiment: Convection Rolls in a Heated Pan Take two small pots of water and use an eye dropper to place a drop of food coloring near the bottom of each. Leave one on a bench top and heat the other over a stovetop. Watch how the color spreads and how long it takes the color to reach the top. Watch how convective loops form. Example $\PageIndex{1}$: Calculating Heat Transfer by Convection: Convection of Air Through the Walls of a House Most houses are not airtight: air goes in and out around doors and windows, through cracks and crevices, following wiring to switches and outlets, and so on. The air in a typical house is completely replaced in less than an hour. Suppose that a moderately-sized house has inside dimensions $12.0m \times 18.0 m \times 3.00 m$ high, and that all air is replaced in 30.0 min. Calculate the heat transfer per unit time in watts needed to warm the incoming cold air by $10.0^oC$ thus replacing the heat transferred by convection alone. Strategy Heat is used to raise the temperature of air so that $Q = mc\Delta T$. The rate of heat transfer is then $Q/t$, where $t$ is the time for air turnover. We are given that $\Delta T$ is $10.0^oC$, but we must still find values for the mass of air and its specific heat before we can calculate $Q$. The specific heat of air is a weighted average of the specific heats of nitrogen and oxygen, which gives $c = c_p \approx 1000 \, J/kg \cdot ^oC$ from the table (note that the specific heat at constant pressure must be used for this process). Solution - Determine the mass of air from its density and the given volume of the house. The density is given from the density $\rho$ and the volume \[m = \rho V = (1.29 \, kg/m^3)912.0 \, m \times 18.0 \, m \times 3.00 \, m) = 836 \, kg\] - Calculate the heat transferred from the change in air temperature $Q = mc \Delta T$ so that \[Q = (836 \, kg)(1000 \, J/kg \cdot^oC)(10.0^oC) = 8.36 \times 10^6 \, J.\] - Calculate the heat transfer from the heat $Q$ and the turnover time $t$. Since air is turned over in $t = 0.500 \, h = 1800 \, s$, the heat transferred per unit time is \[\dfrac{Q}{t} = \dfrac{8.36 \times 10^6 \, J}{1800 \, s} = 4.64 \, kW.\] Discussion This rate of heat transfer is equal to the power consumed by about forty-six 100-W light bulbs. Newly constructed homes are designed for a turnover time of 2 hours or more, rather than 30 minutes for the house of this example. Weather stripping, caulking, and improved window seals are commonly employed. More extreme measures are sometimes taken in very cold (or hot) climates to achieve a tight standard of more than 6 hours for one air turnover. Still longer turnover times are unhealthy, because a minimum amount of fresh air is necessary to supply oxygen for breathing and to dilute household pollutants. The term used for the process by which outside air leaks into the house from cracks around windows, doors, and the foundation is called “air infiltration.” A cold wind is much more chilling than still cold air, because convection combines with conduction in the body to increase the rate at which energy is transferred away from the body. The table below gives approximate wind-chill factors, which are the temperatures of still air that produce the same rate of cooling as air of a given temperature and speed. Wind-chill factors are a dramatic reminder of convection’s ability to transfer heat faster than conduction. For example, a 15.0 m/s wind at $0^oC$ has the chilling equivalent of still air at about $-18^oC$. Moving air temperature Wind speed (m/s) \| $^oC$ \| 2 \| 5 \| 10 \| 15 \| 20 \| \|---\|---\|---\|---\|---\|---\| \| 5 \| 3 \| -1 \| -8 \| -10 \| -12 \| \| 2 \| 0 \| -7 \| -12 \| -16 \| -18 \| \| 0 \| -2 \| -9 \| -15 \| -18 \| -20 \| \| -5 \| -7 \| -15 \| -22 \| -26 \| -29 \| \| -10 \| -12 \| -21 \| -29 \| -34 \| -36 \| \| -20 \| -23 \| -34 \| -44 \| -50 \| -52 \| \| -40 \| -44 \| -59 \| -73 \| -82 \| -84 \| Although air can transfer heat rapidly by convection, it is a poor conductor and thus a good insulator. The amount of available space for airflow determines whether air acts as an insulator or conductor. The space between the inside and outside walls of a house, for example, is about 9 cm (3.5 in) —large enough for convection to work effectively. The addition of wall insulation prevents airflow, so heat loss (or gain) is decreased. Similarly, the gap between the two panes of a double-paned window is about 1 cm, which prevents convection and takes advantage of air’s low conductivity to prevent greater loss. Fur, fiber, and fiberglass also take advantage of the low conductivity of air by trapping it in spaces too small to support convection, as shown in the figure. Fur and feathers are lightweight and thus ideal for the protection of animals. Some interesting phenomena happen when convection is accompanied by a phase change . It allows us to cool off by sweating, even if the temperature of the surrounding air exceeds body temperature. Heat from the skin is required for sweat to evaporate from the skin, but without air flow, the air becomes saturated and evaporation stops. Air flow caused by convection replaces the saturated air by dry air and evaporation continues. Example $\PageIndex{2}$: Calculate the Flow of Mass during Convection: Sweat-Heat Transfer away from the Body The average person produces heat at the rate of about 120 W when at rest. At what rate must water evaporate from the body to get rid of all this energy? (This evaporation might occur when a person is sitting in the shade and surrounding temperatures are the same as skin temperature, eliminating heat transfer by other methods.) Strategy Energy is needed for a phase change $(Q = mL_v)$. Thus, the energy loss per unit time is \[\dfrac{Q}{t} = \dfrac{mL_v}{t} = 120 \, W = 120 \, J/s.\] We divide both sides of the equation by $L_v$ to find that the mass evaporated per unit time is \[\dfrac{m}{t} = \dfrac{120 \, J/s}{L_v}.\] Solution (1) Insert the value of the latent heat from [link] , $L_v = 2430 \, kJ/kg = 2430 \, J/g$. This yields \[\dfrac{m}{t} = \dfrac{120 \, J/s}{2430 \, J/g} = 0.0494 \, g/s = 2.96 \, g/min.\] Discussion Evaporating about 3 g/min seems reasonable. This would be about 180 g (about 7 oz) per hour. If the air is very dry, the sweat may evaporate without even being noticed. A significant amount of evaporation also takes place in the lungs and breathing passages. Another important example of the combination of phase change and convection occurs when water evaporates from the oceans. Heat is removed from the ocean when water evaporates. If the water vapor condenses in liquid droplets as clouds form, heat is released in the atmosphere. Thus, there is an overall transfer of heat from the ocean to the atmosphere. This process is the driving power behind thunderheads, those great cumulus clouds that rise as much as 20.0 km into the stratosphere. Water vapor carried in by convection condenses, releasing tremendous amounts of energy. This energy causes the air to expand and rise, where it is colder. More condensation occurs in these colder regions, which in turn drives the cloud even higher. Such a mechanism is called positive feedback, since the process reinforces and accelerates itself. These systems sometimes produce violent storms, with lightning and hail, and constitute the mechanism driving hurricanes. The movement of icebergs is another example of convection accompanied by a phase change. Suppose an iceberg drifts from Greenland into warmer Atlantic waters. Heat is removed from the warm ocean water when the ice melts and heat is released to the land mass when the iceberg forms on Greenland. Exercise $\PageIndex{1}$ Explain why using a fan in the summer feels refreshing! - Answer - Using a fan increases the flow of air: warm air near your body is replaced by cooler air from elsewhere. Convection increases the rate of heat transfer so that moving air “feels” cooler than still air. Summary - Convection is heat transfer by the macroscopic movement of mass. Convection can be natural or forced and generally transfers thermal energy faster than conduction. The table gives wind-chill factors, indicating that moving air has the same chilling effect of much colder stationary air. Convection that occurs along with a phase change can transfer energy from cold regions to warm ones.	libretexts	2025-03-17T19:53:32.948775	2015-11-01T05:41:46	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.06%3A_Convection", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "14.6: Convection", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.07%3A_Radiation	14.7: Radiation Learning Objectives By the end of this section, you will be able to: - Discuss heat transfer by radiation. - Explain the power of different materials. You can feel the heat transfer from a fire and from the Sun. Similarly, you can sometimes tell that the oven is hot without touching its door or looking inside—it may just warm you as you walk by. The space between the Earth and the Sun is largely empty, without any possibility of heat transfer by convection or conduction. In these examples, heat is transferred by radiation. That is, the hot body emits electromagnetic waves that are absorbed by our skin: no medium is required for electromagnetic waves to propagate. Different names are used for electromagnetic waves of different wavelengths: radio waves, microwaves, infrared radiation , visible light, ultraviolet radiation, X-rays, and gamma rays. The energy of electromagnetic radiation depends on the wavelength (color) and varies over a wide range: a smaller wavelength (or higher frequency) corresponds to a higher energy. Because more heat is radiated at higher temperatures, a temperature change is accompanied by a color change. Take, for example, an electrical element on a stove, which glows from red to orange, while the higher-temperature steel in a blast furnace glows from yellow to white. The radiation you feel is mostly infrared, which corresponds to a lower temperature than that of the electrical element and the steel. The radiated energy depends on its intensity, which is represented in the figure below by the height of the distribution. Electromagnetic Waves explains more about the electromagnetic spectrum and Introduction to Quantum Physics discusses how the decrease in wavelength corresponds to an increase in energy. All objects absorb and emit electromagnetic radiation. The rate of heat transfer by radiation is largely determined by the color of the object. Black is the most effective, and white is the least effective. People living in hot climates generally avoid wearing black clothing, for instance. Similarly, black asphalt in a parking lot will be hotter than adjacent gray sidewalk on a summer day, because black absorbs better than gray. The reverse is also true—black radiates better than gray. Thus, on a clear summer night, the asphalt will be colder than the gray sidewalk, because black radiates the energy more rapidly than gray. An ideal radiator is the same color as an ideal absorber , and captures all the radiation that falls on it. In contrast, white is a poor absorber and is also a poor radiator. A white object reflects all radiation, like a mirror. (A perfect, polished white surface is mirror-like in appearance, and a crushed mirror looks white.) Gray objects have a uniform ability to absorb all parts of the electromagnetic spectrum. Colored objects behave in similar but more complex ways, which gives them a particular color in the visible range and may make them special in other ranges of the nonvisible spectrum. Take, for example, the strong absorption of infrared radiation by the skin, which allows us to be very sensitive to it. The rate of heat transfer by emitted radiation is determined by the Stefan-Boltzmann law of radiation : \[\dfrac{Q}{t} = \sigma e AT^4,\] where $\sigma = 5.67 \times 10^{-8} \, J/s \cdot m^2 \cdot k^4$ is the Stefan-Boltzmann constant, $A$ is the surface area of the object, and $T$ is its absolute temperature in kelvin. The symbol $e$ stands for the emissivity of the object, which is a measure of how well it radiates. An ideal jet-black (or black body) radiator has $e = 1$, whereas a perfect reflector has $e = 0$. Real objects fall between these two values. Take, for example, tungsten light bulb filaments which have an $e$ of about 0.5, and carbon black (a material used in printer toner), which has the (greatest known) emissivity of about 0.99. The radiation rate is directly proportional to the fourth power of the absolute temperature—a remarkably strong temperature dependence. Furthermore, the radiated heat is proportional to the surface area of the object. If you knock apart the coals of a fire, there is a noticeable increase in radiation due to an increase in radiating surface area. Skin is a remarkably good absorber and emitter of infrared radiation, having an emissivity of 0.97 in the infrared spectrum. Thus, we are all nearly (jet) black in the infrared, in spite of the obvious variations in skin color. This high infrared emissivity is why we can so easily feel radiation on our skin. It is also the basis for the use of night scopes used by law enforcement and the military to detect human beings. Even small temperature variations can be detected because of the $T^4$ dependence. Images, called thermographs , can be used medically to detect regions of abnormally high temperature in the body, perhaps indicative of disease. Similar techniques can be used to detect heat leaks in homes Figure $\PageIndex{5}$, optimize performance of blast furnaces, improve comfort levels in work environments, and even remotely map the Earth’s temperature profile. All objects emit and absorb radiation. The net rate of heat transfer by radiation (absorption minus emission) is related to both the temperature of the object and the temperature of its surroundings. Assuming that an object with a temperature $T_1$ is surrounded by an environment with uniform temperature $T_2$ the net rate of heat transfer by radiation is \[\dfrac{Q_{net}}{t} = \sigma e A(T_2^4 - T_1^4),\] where $e$ is the emissivity of the object alone. In other words, it does not matter whether the surroundings are white, gray, or black; the balance of radiation into and out of the object depends on how well it emits and absorbs radiation. When $T_2 > T_1$, the quantity $Q_{net}/t$ is positive; that is, the net heat transfer is from hot to cold. Take-Home Experiment: Temperature in the Sun Place a thermometer out in the sunshine and shield it from direct sunlight using an aluminum foil. What is the reading? Now remove the shield, and note what the thermometer reads. Take a handkerchief soaked in nail polish remover, wrap it around the thermometer and place it in the sunshine. What does the thermometer read? Example $\PageIndex{1}$: Calculate the Net Heat Transfer of a Person: Heat Transfer by Radiation What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is $22.0^oC$. The person has a normal skin temperature of $33.0^oC$ and a surface area of $1.50 \, m^2$. The emissivity of skin is 0.97 in the infrared, where the radiation takes place. Strategy We can solve this by using the equation for the rate of radiative heat transfer. Solution Insert the temperatures values $T_2 = 295 \, K$ and $T_1 = 306 \, K$, so that \[\dfrac{Q}{t} = \sigma e A(T_2^4 - T_1^4)\] \[= (5.67 \times 10^{-8} \, J/s \cdot m^2 \cdot k^4)(0.97)(1.50 \, m^2) [(295 \, K)^4 - (306 \, K)^4]\] \[= -99 \, J/s = -99 \, W.\] Discussion This value is a significant rate of heat transfer to the environment (note the minus sign), considering that a person at rest may produce energy at the rate of 125 W and that conduction and convection will also be transferring energy to the environment. Indeed, we would probably expect this person to feel cold. Clothing significantly reduces heat transfer to the environment by many methods, because clothing slows down both conduction and convection, and has a lower emissivity (especially if it is white) than skin. The Earth receives almost all its energy from radiation of the Sun and reflects some of it back into outer space. Because the Sun is hotter than the Earth, the net energy flux is from the Sun to the Earth. However, the rate of energy transfer is less than the equation for the radiative heat transfer would predict because the Sun does not fill the sky. The average emissivity (e) of the Earth is about 0.65, but the calculation of this value is complicated by the fact that the highly reflective cloud coverage varies greatly from day to day. There is a negative feedback (one in which a change produces an effect that opposes that change) between clouds and heat transfer; greater temperatures evaporate more water to form more clouds, which reflect more radiation back into space, reducing the temperature. The often mentioned greenhouse effect is directly related to the variation of the Earth’s emissivity with radiation type (see the figure given below). The greenhouse effect is a natural phenomenon responsible for providing temperatures suitable for life on Earth. The Earth’s relatively constant temperature is a result of the energy balance between the incoming solar radiation and the energy radiated from the Earth. Most of the infrared radiation emitted from the Earth is absorbed by carbon dioxide $(CO_2)$ and water $(H_2O)$ in the atmosphere and then re-radiated back to the Earth or into outer space. Re-radiation back to the Earth maintains its surface temperature about $40^oC$ higher than it would be if there was no atmosphere, similar to the way glass increases temperatures in a greenhouse. The greenhouse effect is also central to the discussion of global warming due to emission of carbon dioxide and methane (and other so-called greenhouse gases) into the Earth’s atmosphere from industrial production and farming. Changes in global climate could lead to more intense storms, precipitation changes (affecting agriculture), reduction in rain forest biodiversity, and rising sea levels. Heating and cooling are often significant contributors to energy use in individual homes. Current research efforts into developing environmentally friendly homes quite often focus on reducing conventional heating and cooling through better building materials, strategically positioning windows to optimize radiation gain from the Sun, and opening spaces to allow convection. It is possible to build a zero-energy house that allows for comfortable living in most parts of the United States with hot and humid summers and cold winters. Conversely, dark space is very cold, about $3K (-454^oF)$, so that the Earth radiates energy into the dark sky. Owing to the fact that clouds have lower emissivity than either oceans or land masses, they reflect some of the radiation back to the surface, greatly reducing heat transfer into dark space, just as they greatly reduce heat transfer into the atmosphere during the day. The rate of heat transfer from soil and grasses can be so rapid that frost may occur on clear summer evenings, even in warm latitudes. Exercise $\PageIndex{1}$ What is the change in the rate of the radiated heat by a body at the temperature $T_1 = 20^oC$ compared to when the body is at the temperature $T_2 = 40^oC$? - Answer - The radiated heat is proportional to the fourth power of the absolute temperature. Because $T_1 = 293 \, K$ and $T_2 = 313 \, K$, the rate of heat transfer increases by about 30 percent of the original rate. Career Connection: Energy Conservation Consultation The cost of energy is generally believed to remain very high for the foreseeable future. Thus, passive control of heat loss in both commercial and domestic housing will become increasingly important. Energy consultants measure and analyze the flow of energy into and out of houses and ensure that a healthy exchange of air is maintained inside the house. The job prospects for an energy consultant are strong. Problem Solving Strategies for the Methods of Heat Transfer - Examine the situation to determine what type of heat transfer is involved. - Identify the type(s) of heat transfer—conduction, convection, or radiation. - Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is very useful. - Make a list of what is given or can be inferred from the problem as stated (identify the knowns). - Solve the appropriate equation for the quantity to be determined (the unknown). - For conduction, equation $\dfrac{Q}{t} = \dfrac{kA(T_2 - T_1)}{d} $ is appropriate. [link] lists thermal conductivities. For convection, determine the amount of matter moved and use equation $Q = mc\Delta T$, to calculate the heat transfer involved in the temperature change of the fluid. If a phase change accompanies convection, equation $Q = mL_f$ or $Q = mL_v$ is appropriate to find the heat transfer involved in the phase change. [link] lists information relevant to phase change. For radiation, equation $\dfrac{Q_{net}}{t} = \sigma e A (T_2^4 - T_1^4)$ gives the net heat transfer rate. - Insert the knowns along with their units into the appropriate equation and obtain numerical solutions complete with units. - Check the answer to see if it is reasonable. Does it make sense? Summary - Radiation is the rate of heat transfer through the emission or absorption of electromagnetic waves. - The rate of heat transfer depends on the surface area and the fourth power of the absolute temperature: \[\dfrac{Q}{t} = \sigma eAT^4,\] where $\sigma = 5.67 \times 10^{-8} \, J/s \cdot m^2 \cdot K^4$ is the Stefan-Boltzmann constant and $e$ is the emissivity of the body. For a black body, $e = 1$ whereas a shiny white or perfect reflector has $e = 0$, with real objects having values of e between 1 and 0. The net rate of heat transfer by radiation is \[\dfrac{Q_{net}}{t} = \sigma eA(T_2^4 - T_1^4)\] where $T_1$ is the temperature of an object surrounded by an environment with uniform temperature $T_2$ and $e$ is the emissivity of the object . Glossary - emissivity - measure of how well an object radiates - greenhouse effect - warming of the Earth that is due to gases such as carbon dioxide and methane that absorb infrared radiation from the Earth’s surface and reradiate it in all directions, thus sending a fraction of it back toward the surface of the Earth - net rate of heat transfer by radiation - is $\frac{Q_{net}}{t}=σeA(T^4_2−T^4_1)$ - radiation - energy transferred by electromagnetic waves directly as a result of a temperature difference - Stefan-Boltzmann law of radiation - $\frac{Q}{t}=σeAT^4$, where σ is the Stefan-Boltzmann constant, $A$ is the surface area of the object, $T$ is the absolute temperature, and $e$ is the emissivity	libretexts	2025-03-17T19:53:33.027044	2015-11-01T05:42:02	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.07%3A_Radiation", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "14.7: Radiation", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.E%3A_Heat_and_Heat_Transfer_Methods_(Exercise)	14.E: Heat and Heat Transfer Methods (Exercise) - - Last updated - Save as PDF Conceptual Questions 14.1: Heat 1. How is heat transfer related to temperature? 2. Describe a situation in which heat transfer occurs. What are the resulting forms of energy? 3. When heat transfers into a system, is the energy stored as heat? Explain briefly. 14.2: Temperature Change and Heat Capacity 4. What three factors affect the heat transfer that is necessary to change an object’s temperature? 5. The brakes in a car increase in temperature by $\displaystyle ΔT$ when bringing the car to rest from a speed $\displaystyle v$. How much greater would $\displaystyle ΔT$ be if the car initially had twice the speed? You may assume the car to stop sufficiently fast so that no heat transfers out of the brakes. 14.3: Phase Change and Latent Heat 6. Heat transfer can cause temperature and phase changes. What else can cause these changes? 7. How does the latent heat of fusion of water help slow the decrease of air temperatures, perhaps preventing temperatures from falling significantly below $\displaystyle 0ºC$, in the vicinity of large bodies of water? 8. What is the temperature of ice right after it is formed by freezing water? 9. If you place $\displaystyle 0ºC$ ice into $\displaystyle 0ºC$ water in an insulated container, what will happen? Will some ice melt, will more water freeze, or will neither take place? 10. What effect does condensation on a glass of ice water have on the rate at which the ice melts? Will the condensation speed up the melting process or slow it down? 11. In very humid climates where there are numerous bodies of water, such as in Florida, it is unusual for temperatures to rise above about 35ºC(95ºF). In deserts, however, temperatures can rise far above this. Explain how the evaporation of water helps limit high temperatures in humid climates. 12. In winters, it is often warmer in San Francisco than in nearby Sacramento, 150 km inland. In summers, it is nearly always hotter in Sacramento. Explain how the bodies of water surrounding San Francisco moderate its extreme temperatures. 13. Putting a lid on a boiling pot greatly reduces the heat transfer necessary to keep it boiling. Explain why. 14. Freeze-dried foods have been dehydrated in a vacuum. During the process, the food freezes and must be heated to facilitate dehydration. Explain both how the vacuum speeds up dehydration and why the food freezes as a result. 15. When still air cools by radiating at night, it is unusual for temperatures to fall below the dew point. Explain why. 16. In a physics classroom demonstration, an instructor inflates a balloon by mouth and then cools it in liquid nitrogen. When cold, the shrunken balloon has a small amount of light blue liquid in it, as well as some snow-like crystals. As it warms up, the liquid boils, and part of the crystals sublimate, with some crystals lingering for awhile and then producing a liquid. Identify the blue liquid and the two solids in the cold balloon. Justify your identifications using data from Table. 14.4: Heat Transfer Methods 17. What are the main methods of heat transfer from the hot core of Earth to its surface? From Earth’s surface to outer space? 14.5: Conduction 18. Some electric stoves have a flat ceramic surface with heating elements hidden beneath. A pot placed over a heating element will be heated, while it is safe to touch the surface only a few centimeters away. Why is ceramic, with a conductivity less than that of a metal but greater than that of a good insulator, an ideal choice for the stove top? 19. Loose-fitting white clothing covering most of the body is ideal for desert dwellers, both in the hot Sun and during cold evenings. Explain how such clothing is advantageous during both day and night. A jellabiya is worn by many men in Egypt. (credit: Zerida) 14.6: Convection 20. One way to make a fireplace more energy efficient is to have an external air supply for the combustion of its fuel. Another is to have room air circulate around the outside of the fire box and back into the room. Detail the methods of heat transfer involved in each. 21. On cold, clear nights horses will sleep under the cover of large trees. How does this help them keep warm? 14.7 Radiation 22. When watching a daytime circus in a large, dark-colored tent, you sense significant heat transfer from the tent. Explain why this occurs. 23. Satellites designed to observe the radiation from cold (3 K) dark space have sensors that are shaded from the Sun, Earth, and Moon and that are cooled to very low temperatures. Why must the sensors be at low temperature? 24. Why are cloudy nights generally warmer than clear ones? 25. Why are thermometers that are used in weather stations shielded from the sunshine? What does a thermometer measure if it is shielded from the sunshine and also if it is not? 26. On average, would Earth be warmer or cooler without the atmosphere? Explain your answer. Problems & Exercises 14.2: Temperature Change and Heat Capacity 27. On a hot day, the temperature of an 80,000-L swimming pool increases by $\displaystyle 1.50ºC$. What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation. Solution $\displaystyle 5.02×10^8J$ 28. Show that $\displaystyle 1cal/g⋅ºC=1kcal/kg⋅ºC$. 29. To sterilize a 50.0-g glass baby bottle, we must raise its temperature from $\displaystyle 22.0ºC$ to $\displaystyle 95.0ºC$. How much heat transfer is required? Solution $\displaystyle 3.07×10^3J$ 30. The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at $\displaystyle 20.0ºC$: (a) water; (b) concrete; (c) steel; and (d) mercury. 31. Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm per rub, and with an average frictional force of 40.0 N, what is the temperature increase? The mass of tissues warmed is only 0.100 kg, mostly in the palms and fingers. Solution $\displaystyle 0.171ºC$ 32. A 0.250-kg block of a pure material is heated from $\displaystyle 20.0ºC$ to $\displaystyle 65.0ºC$ by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed. 33. Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water? Solution 10.8 34. (a) The number of kilocalories in food is determined by calorimetry techniques in which the food is burned and the amount of heat transfer is measured. How many kilocalories per gram are there in a 5.00-g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a $\displaystyle 54.9ºC$ temperature increase? (b) Compare your answer to labeling information found on a package of peanuts and comment on whether the values are consistent. 35. Following vigorous exercise, the body temperature of an 80.0-kg person is $\displaystyle 40.0ºC$. At what rate in watts must the person transfer thermal energy to reduce the the body temperature to $\displaystyle 37.0ºC$ in 30.0 min, assuming the body continues to produce energy at the rate of 150 W? ( 1 watt = 1 joule/second or 1 W = 1 J/s ). Solution 617 W 36. Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails ( 1 watt = 1 joule/second or 1 W = 1 J/s and 1 MW = 1 megawatt ). (a) Calculate the rate of temperature increase in degrees Celsius per second ($\displaystyle ºC/s$) if the mass of the reactor core is $\displaystyle 1.60×10^5kg$ and it has an average specific heat of $\displaystyle 0.3349 kJ/kgº⋅C$. (b) How long would it take to obtain a temperature increase of $\displaystyle 2000ºC$, which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the $\displaystyle 5×10^5-kg$ steel containment vessel would also begin to heat up.) 14.3: Phase Change and Latent Heat 37. How much heat transfer (in kilocalories) is required to thaw a 0.450-kg package of frozen vegetables originally at $\displaystyle 0ºC$ if their heat of fusion is the same as that of water? Solution 35.9 kcal 38. A bag containing $\displaystyle 0ºC$ ice is much more effective in absorbing energy than one containing the same amount of 0ºC water. a. How much heat transfer is necessary to raise the temperature of 0.800 kg of water from $\displaystyle 0ºC$ to $\displaystyle 30.0ºC$? b. How much heat transfer is required to first melt 0.800 kg of $\displaystyle 0ºC$ ice and then raise its temperature? c. Explain how your answer supports the contention that the ice is more effective. 39. (a) How much heat transfer is required to raise the temperature of a 0.750-kg aluminum pot containing 2.50 kg of water from $\displaystyle 30.0ºC$ to the boiling point and then boil away 0.750 kg of water? (b) How long does this take if the rate of heat transfer is 500 W 1 watt = 1 joule/second ( 1 W = 1 J/s )? Solution (a) 591 kcal (b) $\displaystyle 4.94×10^3s$ 40. The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.00 g of condensation forms on a glass containing both water and 200 g of ice, how many grams of the ice will melt as a result? Assume no other heat transfer occurs. 41. On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at $\displaystyle 0ºC$ and completely melts to $\displaystyle 0ºC$ water in exactly one day 1 watt = 1 joule/second ( 1 W = 1 J/s )? Solution 13.5 W 42. On a certain dry sunny day, a swimming pool’s temperature would rise by $\displaystyle 1.50ºC$ if not for evaporation. What fraction of the water must evaporate to carry away precisely enough energy to keep the temperature constant? 43. (a) How much heat transfer is necessary to raise the temperature of a 0.200-kg piece of ice from $\displaystyle −20.0ºC$ to $\displaystyle 130ºC$, including the energy needed for phase changes? (b) How much time is required for each stage, assuming a constant 20.0 kJ/s rate of heat transfer? (c) Make a graph of temperature versus time for this process. Solution (a) 148 kcal (b) 0.418 s, 3.34 s, 4.19 s, 22.6 s, 0.456 s 44. In 1986, a gargantuan iceberg broke away from the Ross Ice Shelf in Antarctica. It was approximately a rectangle 160 km long, 40.0 km wide, and 250 m thick. (a) What is the mass of this iceberg, given that the density of ice is $\displaystyle 917 kg/m^3$? (b) How much heat transfer (in joules) is needed to melt it? (c) How many years would it take sunlight alone to melt ice this thick, if the ice absorbs an average of $\displaystyle 100 W/m^2$, 12.00 h per day? 45. How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to cool the coffee from $\displaystyle 95.0ºC$ to $\displaystyle 45.0ºC$? You may assume the coffee has the same thermal properties as water and that the average heat of vaporization is 2340 kJ/kg (560 cal/g). (You may neglect the change in mass of the coffee as it cools, which will give you an answer that is slightly larger than correct.) Solution 33.0 g 46. (a) It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases $\displaystyle 2.80×10^7J$ of energy when burned. To illustrate this difficulty, calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water has its temperature raised from $\displaystyle 20.0ºC$ to $\displaystyle 100ºC$, it boils, and the resulting steam is raised to $\displaystyle 300ºC$. (b) Discuss additional complications caused by the fact that crude oil has a smaller density than water. Solution (a) 9.67 L (b) Crude oil is less dense than water, so it floats on top of the water, thereby exposing it to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less efficient in absorbing the heat generated by the oil. 47. The energy released from condensation in thunderstorms can be very large. Calculate the energy released into the atmosphere for a small storm of radius 1 km, assuming that 1.0 cm of rain is precipitated uniformly over this area. 48. To help prevent frost damage, 4.00 kg of $\displaystyle 0ºC$ water is sprayed onto a fruit tree. (a) How much heat transfer occurs as the water freezes? (b) How much would the temperature of the 200-kg tree decrease if this amount of heat transferred from the tree? Take the specific heat to be $\displaystyle 3.35 kJ/kg⋅ºC$, and assume that no phase change occurs. Solution a) 319 kcal b) $\displaystyle 2.00ºC$ 49. A 0.250-kg aluminum bowl holding 0.800 kg of soup at $\displaystyle 25.0ºC$ is placed in a freezer. What is the final temperature if 377 kJ of energy is transferred from the bowl and soup, assuming the soup’s thermal properties are the same as that of water? Explicitly show how you follow the steps in Problem-Solving Strategies for the Effects of Heat Transfer. 50. A 0.0500-kg ice cube at $\displaystyle −30.0ºC$ is placed in 0.400 kg of $\displaystyle 35.0ºC$ water in a very well-insulated container. What is the final temperature? Solution $\displaystyle 20.6ºC$ 51. If you pour 0.0100 kg of $\displaystyle 20.0ºC$ water onto a 1.20-kg block of ice (which is initially at $\displaystyle −15.0ºC$), what is the final temperature? You may assume that the water cools so rapidly that effects of the surroundings are negligible. 52. Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of $\displaystyle 500ºC$ rock must be placed in 4.00 kg of $\displaystyle 15.0ºC$ water to bring its temperature to $\displaystyle 100ºC$, if 0.0250 kg of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings and take the average specific heat of the rocks to be that of granite. Solution 4.38 kg 53. What would be the final temperature of the pan and water in Calculating the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan if 0.260 kg of water was placed in the pan and 0.0100 kg of the water evaporated immediately, leaving the remainder to come to a common temperature with the pan? 54. In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3.00-hour delivery trip requires 200 L of liquid nitrogen, which has a density of $\displaystyle 808 kg/m^3$. (a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to $\displaystyle 3.00ºC$. (Use $\displaystyle c_p$ and assume it is constant over the temperature range.) This value is the amount of cooling the liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt-hours? (c) Compare the amount of cooling obtained from melting an identical mass of 0ºC ice with that from evaporating the liquid nitrogen. Solution (a) $\displaystyle 1.57×10^4kcal$ (b) $\displaystyle 18.3 kW⋅h$ (c) $\displaystyle 1.29×10^4kcal$ 55. Some gun fanciers make their own bullets, which involves melting and casting the lead slugs. How much heat transfer is needed to raise the temperature and melt 0.500 kg of lead, starting from 25.0ºC ? 14.5: Conduction 56. (a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The surface area of the walls is $\displaystyle 120m^2$ and their inside surface is at $\displaystyle 18.0ºC$, while their outside surface is at $\displaystyle 5.00ºC$. (b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction? Solution (a) $\displaystyle 1.01×10^3$W (b) One 57. The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly the windows transfer heat by conduction, calculate the rate of conduction in watts through a $\displaystyle 3.00-m^2$ window that is $\displaystyle 0.635 cm$ thick (1/4 in) if the temperatures of the inner and outer surfaces are $\displaystyle 5.00ºC$ and $\displaystyle −10.0ºC$, respectively. This rapid rate will not be maintained—the inner surface will cool, and even result in frost formation. 58. Calculate the rate of heat conduction out of the human body, assuming that the core internal temperature is $\displaystyle 37.0ºC$, the skin temperature is $\displaystyle 34.0ºC$, the thickness of the tissues between averages $\displaystyle 1.00 cm$, and the surface area is $\displaystyle 1.40m^2$. Solution 84.0 W 59. Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet, making contact over an area of $\displaystyle 80.0cm^2$ with each foot. Both the ceramic and the carpet are 2.00 cm thick and are $\displaystyle 10.0ºC$ on their bottom sides. At what rate must heat transfer occur from each foot to keep the top of the ceramic and carpet at $\displaystyle 33.0ºC$? 60. A man consumes 3000 kcal of food in one day, converting most of it to maintain body temperature. If he loses half this energy by evaporating water (through breathing and sweating), how many kilograms of water evaporate? Solution 2.59 kg 61. (a) A firewalker runs across a bed of hot coals without sustaining burns. Calculate the heat transferred by conduction into the sole of one foot of a firewalker given that the bottom of the foot is a 3.00-mm-thick callus with a conductivity at the low end of the range for wood and its density is $\displaystyle 300 kg/m^3$. The area of contact is $\displaystyle 25.0 cm^2$, the temperature of the coals is $\displaystyle 700ºC$, and the time in contact is 1.00 s. (b) What temperature increase is produced in the $\displaystyle 25.0 cm^3$ of tissue affected? (c) What effect do you think this will have on the tissue, keeping in mind that a callus is made of dead cells? 62. (a) What is the rate of heat conduction through the 3.00-cm-thick fur of a large animal having a $\displaystyle 1.40-m^2$ surface area? Assume that the animal’s skin temperature is $\displaystyle 32.0ºC$, that the air temperature is $\displaystyle −5.00ºC$, and that fur has the same thermal conductivity as air. (b) What food intake will the animal need in one day to replace this heat transfer? Solution (a) 39.7 W (b) 820 kcal 63. A walrus transfers energy by conduction through its blubber at the rate of 150 W when immersed in $\displaystyle −1.00ºC$ water. The walrus’s internal core temperature is $\displaystyle 37.0ºC$, and it has a surface area of $\displaystyle 2.00m^2$. What is the average thickness of its blubber, which has the conductivity of fatty tissues without blood? Walrus on ice. (credit: Captain Budd Christman, NOAA Corps) 64. Compare the rate of heat conduction through a 13.0-cm-thick wall that has an area of $\displaystyle 10.0 m^2$ and a thermal conductivity twice that of glass wool with the rate of heat conduction through a window that is 0.750 cm thick and that has an area of $\displaystyle 2.00 m^2$, assuming the same temperature difference across each. Solution 35 to 1, window to wall 65. Suppose a person is covered head to foot by wool clothing with average thickness of 2.00 cm and is transferring energy by conduction through the clothing at the rate of 50.0 W. What is the temperature difference across the clothing, given the surface area is $\displaystyle 1.40 m^2$? 66. Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.600 cm thick and heat conduction occurs through the same area and at the same rate as computed in Example, what is the temperature difference across it? Ceramic has the same thermal conductivity as glass and brick. Solution $\displaystyle 1.05×10^3K$ 67. One easy way to reduce heating (and cooling) costs is to add extra insulation in the attic of a house. Suppose the house already had 15 cm of fiberglass insulation in the attic and in all the exterior surfaces. If you added an extra 8.0 cm of fiberglass to the attic, then by what percentage would the heating cost of the house drop? Take the single story house to be of dimensions 10 m by 15 m by 3.0 m. Ignore air infiltration and heat loss through windows and doors. 68. (a) Calculate the rate of heat conduction through a double-paned window that has a $\displaystyle 1.50-m^2$ area and is made of two panes of 0.800-cm-thick glass separated by a 1.00-cm air gap. The inside surface temperature is $\displaystyle 15.0ºC$, while that on the outside is $\displaystyle −10.0ºC$. (Hint: There are identical temperature drops across the two glass panes. First find these and then the temperature drop across the air gap. This problem ignores the increased heat transfer in the air gap due to convection.) (b) Calculate the rate of heat conduction through a 1.60-cm-thick window of the same area and with the same temperatures. Compare your answer with that for part (a). Solution (a) 83 W (b) 24 times that of a double pane window. 69. Many decisions are made on the basis of the payback period: the time it will take through savings to equal the capital cost of an investment. Acceptable payback times depend upon the business or philosophy one has. (For some industries, a payback period is as small as two years.) Suppose you wish to install the extra insulation in Exercise. If energy cost $1.00 per million joules and the insulation was $4.00 per square meter, then calculate the simple payback time. Take the average $\displaystyle ΔT$ for the 120 day heating season to be $\displaystyle 15.0ºC$. 70. For the human body, what is the rate of heat transfer by conduction through the body’s tissue with the following conditions: the tissue thickness is 3.00 cm, the change in temperature is $\displaystyle 2.00ºC$, and the skin area is $\displaystyle 1.50 m^2$. How does this compare with the average heat transfer rate to the body resulting from an energy intake of about 2400 kcal per day? (No exercise is included.) Solution 20.0 W, 17.2% of 2400 kcal per day 14.6: Convection 71. At what wind speed does $\displaystyle −10ºC$ air cause the same chill factor as still air at $\displaystyle −29ºC$? Solution 10 m/s 72. At what temperature does still air cause the same chill factor as $\displaystyle −5ºC$ air moving at 15 m/s? 73. The “steam” above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot coffee. What is the final temperature of 250 g of hot coffee initially at $\displaystyle 90.0ºC$ if 2.00 g evaporates from it? The coffee is in a Styrofoam cup, so other methods of heat transfer can be neglected. Solution $\displaystyle 85.7ºC$ 74. (a) How many kilograms of water must evaporate from a 60.0-kg woman to lower her body temperature by $\displaystyle 0.750ºC$? (b) Is this a reasonable amount of water to evaporate in the form of perspiration, assuming the relative humidity of the surrounding air is low? 75. On a hot dry day, evaporation from a lake has just enough heat transfer to balance the $\displaystyle 1.00 kW/m^2$ of incoming heat from the Sun. What mass of water evaporates in 1.00 h from each square meter? Explicitly show how you follow the steps in the Problem-Solving Strategies for the Effects of Heat Transfer. Solution 1.48 kg 76. One winter day, the climate control system of a large university classroom building malfunctions. As a result, $\displaystyle 500 m^3$ of excess cold air is brought in each minute. At what rate in kilowatts must heat transfer occur to warm this air by $\displaystyle 10.0ºC$ (that is, to bring the air to room temperature)? 77. The Kilauea volcano in Hawaii is the world’s most active, disgorging about $\displaystyle 5×10^5m^3$ of $\displaystyle 1200ºC$ lava per day. What is the rate of heat transfer out of Earth by convection if this lava has a density of $\displaystyle 2700kg/m^3$ and eventually cools to $\displaystyle 30ºC$? Assume that the specific heat of lava is the same as that of granite. Lava flow on Kilauea volcano in Hawaii. (credit: J. P. Eaton, U.S. Geological Survey) Solution $\displaystyle 2×10^4 MW$ 78. During heavy exercise, the body pumps 2.00 L of blood per minute to the surface, where it is cooled by $\displaystyle 2.00ºC$. What is the rate of heat transfer from this forced convection alone, assuming blood has the same specific heat as water and its density is $\displaystyle 1050 kg/m^3$? 79. A person inhales and exhales 2.00 L of $\displaystyle 37.0ºC$ air, evaporating $\displaystyle 4.00×10^{−2}g$ of water from the lungs and breathing passages with each breath. (a) How much heat transfer occurs due to evaporation in each breath? (b) What is the rate of heat transfer in watts if the person is breathing at a moderate rate of 18.0 breaths per minute? (c) If the inhaled air had a temperature of $\displaystyle 20.0ºC$, what is the rate of heat transfer for warming the air? (d) Discuss the total rate of heat transfer as it relates to typical metabolic rates. Will this breathing be a major form of heat transfer for this person? Solution (a) 97.2 J (b) 29.2 W (c) 9.49 W (d) The total rate of heat loss would be $\displaystyle 29.2 W+9.49 W=38.7W$. While sleeping, our body consumes 83 W of power, while sitting it consumes 120 to 210 W. Therefore, the total rate of heat loss from breathing will not be a major form of heat loss for this person. 80. A glass coffee pot has a circular bottom with a 9.00-cm diameter in contact with a heating element that keeps the coffee warm with a continuous heat transfer rate of 50.0 W (a) What is the temperature of the bottom of the pot, if it is 3.00 mm thick and the inside temperature is $\displaystyle 60.0ºC$? (b) If the temperature of the coffee remains constant and all of the heat transfer is removed by evaporation, how many grams per minute evaporate? Take the heat of vaporization to be 2340 kJ/kg. 14.7 Radiation 81. At what net rate does heat radiate from a $\displaystyle 275-m^2$ black roof on a night when the roof’s temperature is $\displaystyle 30.0ºC$ and the surrounding temperature is $\displaystyle 15.0ºC$? The emissivity of the roof is 0.900. Solution $\displaystyle −21.7 kW$ Note that the negative answer implies heat loss to the surroundings. 82. (a) Cherry-red embers in a fireplace are at $\displaystyle 850ºC$ and have an exposed area of $\displaystyle 0.200 m^2$ and an emissivity of 0.980. The surrounding room has a temperature of $\displaystyle 18.0ºC$. If 50% of the radiant energy enters the room, what is the net rate of radiant heat transfer in kilowatts? (b) Does your answer support the contention that most of the heat transfer into a room by a fireplace comes from infrared radiation? 83. Radiation makes it impossible to stand close to a hot lava flow. Calculate the rate of heat transfer by radiation from $\displaystyle 1.00 m^2$ of $\displaystyle 1200ºC$ fresh lava into $\displaystyle 30.0ºC$ surroundings, assuming lava’s emissivity is 1.00. Solution $\displaystyle −266 kW$ 84. (a) Calculate the rate of heat transfer by radiation from a car radiator at $\displaystyle 110°C$ into a $\displaystyle 50.0ºC$ environment, if the radiator has an emissivity of 0.750 and a $\displaystyle 1.20-m^2$ surface area. (b) Is this a significant fraction of the heat transfer by an automobile engine? To answer this, assume a horsepower of $\displaystyle 200hp(1.5kW)$ and the efficiency of automobile engines as 25%. 85. Find the net rate of heat transfer by radiation from a skier standing in the shade, given the following. She is completely clothed in white (head to foot, including a ski mask), the clothes have an emissivity of 0.200 and a surface temperature of $\displaystyle 10.0ºC$, the surroundings are at $\displaystyle −15.0ºC$, and her surface area is $\displaystyle 1.60m^2$. Solution $\displaystyle −36.0 W$ 86. Suppose you walk into a sauna that has an ambient temperature of $\displaystyle 50.0ºC$. (a) Calculate the rate of heat transfer to you by radiation given your skin temperature is $\displaystyle 37.0ºC$, the emissivity of skin is 0.98, and the surface area of your body is $\displaystyle 1.50m^2$. (b) If all other forms of heat transfer are balanced (the net heat transfer is zero), at what rate will your body temperature increase if your mass is 75.0 kg? 87. Thermography is a technique for measuring radiant heat and detecting variations in surface temperatures that may be medically, environmentally, or militarily meaningful. (a) What is the percent increase in the rate of heat transfer by radiation from a given area at a temperature of $\displaystyle 34.0ºC$ compared with that at $\displaystyle 33.0ºC$, such as on a person’s skin? (b) What is the percent increase in the rate of heat transfer by radiation from a given area at a temperature of $\displaystyle 34.0ºC$ compared with that at $\displaystyle 20.0ºC$, such as for warm and cool automobile hoods? Artist’s rendition of a thermograph of a patient’s upper body, showing the distribution of heat represented by different colors. Solution (a) 1.31% (b) 20.5% 88. The Sun radiates like a perfect black body with an emissivity of exactly 1. (a) Calculate the surface temperature of the Sun, given that it is a sphere with a $\displaystyle 7.00×10^8-m$ radius that radiates $\displaystyle 3.80×10^{26} W$ into 3-K space. (b) How much power does the Sun radiate per square meter of its surface? (c) How much power in watts per square meter is that value at the distance of Earth, $\displaystyle 1.50×10^{11} m$ away? (This number is called the solar constant.) 89. A large body of lava from a volcano has stopped flowing and is slowly cooling. The interior of the lava is at $\displaystyle 1200ºC$, its surface is at $\displaystyle 450ºC$, and the surroundings are at $\displaystyle 27.0ºC$ (a) Calculate the rate at which energy is transferred by radiation from $\displaystyle 1.00 m^2$ of surface lava into the surroundings, assuming the emissivity is 1.00. (b) Suppose heat conduction to the surface occurs at the same rate. What is the thickness of the lava between the $\displaystyle 450ºC$ surface and the $\displaystyle 1200ºC$ interior, assuming that the lava’s conductivity is the same as that of brick? Solution (a) $\displaystyle −15.0 kW$ (b) 4.2 cm 90. Calculate the temperature the entire sky would have to be in order to transfer energy by radiation at $\displaystyle 1000W/m^2$—about the rate at which the Sun radiates when it is directly overhead on a clear day. This value is the effective temperature of the sky, a kind of average that takes account of the fact that the Sun occupies only a small part of the sky but is much hotter than the rest. Assume that the body receiving the energy has a temperature of $\displaystyle 27.0ºC$. 91. (a) A shirtless rider under a circus tent feels the heat radiating from the sunlit portion of the tent. Calculate the temperature of the tent canvas based on the following information: The shirtless rider’s skin temperature is $\displaystyle 34.0ºC$ and has an emissivity of 0.970. The exposed area of skin is $\displaystyle 0.400 m^2$. He receives radiation at the rate of 20.0 W—half what you would calculate if the entire region behind him was hot. The rest of the surroundings are at $\displaystyle 34.0ºC$. (b) Discuss how this situation would change if the sunlit side of the tent was nearly pure white and if the rider was covered by a white tunic. Solution (a) $\displaystyle 48.5ºC$ (b) A pure white object reflects more of the radiant energy that hits it, so a white tent would prevent more of the sunlight from heating up the inside of the tent, and the white tunic would prevent that heat which entered the tent from heating the rider. Therefore, with a white tent, the temperature would be lower than $\displaystyle 48.5ºC$, and the rate of radiant heat transferred to the rider would be less than 20.0 W. 92. Integrated Concepts One $\displaystyle 30.0ºC$ day the relative humidity is $\displaystyle 75.0%$, and that evening the temperature drops to $\displaystyle 20.0ºC$, well below the dew point. (a) How many grams of water condense from each cubic meter of air? (b) How much heat transfer occurs by this condensation? (c) What temperature increase could this cause in dry air? 93. Integrated Concepts Large meteors sometimes strike the Earth, converting most of their kinetic energy into thermal energy. (a) What is the kinetic energy of a $\displaystyle 10^9$kg meteor moving at 25.0 km/s? (b) If this meteor lands in a deep ocean and $\displaystyle 80%$ of its kinetic energy goes into heating water, how many kilograms of water could it raise by $\displaystyle 5.0ºC$? (c) Discuss how the energy of the meteor is more likely to be deposited in the ocean and the likely effects of that energy. Solution (a) $\displaystyle 3×10^{17} J$ (b) $\displaystyle 1×10^{13} kg$ (c) When a large meteor hits the ocean, it causes great tidal waves, dissipating large amount of its energy in the form of kinetic energy of the water. 94. Integrated Concepts Frozen waste from airplane toilets has sometimes been accidentally ejected at high altitude. Ordinarily it breaks up and disperses over a large area, but sometimes it holds together and strikes the ground. Calculate the mass of $\displaystyle 0ºC$ ice that can be melted by the conversion of kinetic and gravitational potential energy when a $\displaystyle 20.0$ piece of frozen waste is released at 12.0 km altitude while moving at 250 m/s and strikes the ground at 100 m/s (since less than 20.0 kg melts, a significant mess results). 95. Integrated Concepts (a) A large electrical power facility produces 1600 MW of “waste heat,” which is dissipated to the environment in cooling towers by warming air flowing through the towers by $\displaystyle 5.00ºC$. What is the necessary flow rate of air in $\displaystyle m^3/s$? (b) Is your result consistent with the large cooling towers used by many large electrical power plants? Solution (a) $\displaystyle 3.44×10^5 m^3/s$ (b) This is equivalent to 12 million cubic feet of air per second. That is tremendous. This is too large to be dissipated by heating the air by only $\displaystyle 5ºC$. Many of these cooling towers use the circulation of cooler air over warmer water to increase the rate of evaporation. This would allow much smaller amounts of air necessary to remove such a large amount of heat because evaporation removes larger quantities of heat than was considered in part (a). 96. Integrated Concepts (a) Suppose you start a workout on a Stairmaster, producing power at the same rate as climbing 116 stairs per minute. Assuming your mass is 76.0 kg and your efficiency is $\displaystyle 20.0%$, how long will it take for your body temperature to rise $\displaystyle 1.00ºC$ if all other forms of heat transfer in and out of your body are balanced? (b) Is this consistent with your experience in getting warm while exercising? 97. Integrated Concepts A 76.0-kg person suffering from hypothermia comes indoors and shivers vigorously. How long does it take the heat transfer to increase the person’s body temperature by $\displaystyle 2.00ºC$ if all other forms of heat transfer are balanced? Solution 20.9 min 98. Integrated Concepts In certain large geographic regions, the underlying rock is hot. Wells can be drilled and water circulated through the rock for heat transfer for the generation of electricity. (a) Calculate the heat transfer that can be extracted by cooling $\displaystyle 1.00 km^3$ of granite by $\displaystyle 100ºC$. (b) How long will this take if heat is transferred at a rate of 300 MW, assuming no heat transfers back into the 1.00 km of rock by its surroundings? 99. Integrated Concepts Heat transfers from your lungs and breathing passages by evaporating water. (a) Calculate the maximum number of grams of water that can be evaporated when you inhale 1.50 L of $\displaystyle 37ºC$ air with an original relative humidity of 40.0%. (Assume that body temperature is also $\displaystyle 37ºC$.) (b) How many joules of energy are required to evaporate this amount? (c) What is the rate of heat transfer in watts from this method, if you breathe at a normal resting rate of 10.0 breaths per minute? Solution (a) $\displaystyle 3.96×10^{-2} g$ (b) $\displaystyle 96.2 J$ (c) $\displaystyle 16.0 W$ 100. Integrated Concepts (a) What is the temperature increase of water falling 55.0 m over Niagara Falls? (b) What fraction must evaporate to keep the temperature constant? 101. Integrated Concepts Hot air rises because it has expanded. It then displaces a greater volume of cold air, which increases the buoyant force on it. (a) Calculate the ratio of the buoyant force to the weight of 50.0ºC50.0ºC air surrounded by 20.0ºC20.0ºC air. (b) What energy is needed to cause 1.00m31.00 m3 of air to go from 20.0ºC20.0ºC to 50.0ºC50.0ºC? (c) What gravitational potential energy is gained by this volume of air if it rises 1.00 m? Will this cause a significant cooling of the air? Solution (a) 1.102 (b) $\displaystyle 2.79×10^4J$ (c) 12.6 J. This will not cause a significant cooling of the air because it is much less than the energy found in part (b), which is the energy required to warm the air from 20.0ºC to 50.0ºC 102. Unreasonable Results (a) What is the temperature increase of an 80.0 kg person who consumes 2500 kcal of food in one day with 95.0% of the energy transferred as heat to the body? (b) What is unreasonable about this result? (c) Which premise or assumption is responsible? Solution (a) 36ºC (b) Any temperature increase greater than about $\displaystyle 3ºC$ would be unreasonably large. In this case the final temperature of the person would rise to 73ºC ( 163ºF ). \|(c) The assumption of $\displaystyle 95%$ heat retention is unreasonable. 103. Unreasonable Results A slightly deranged Arctic inventor surrounded by ice thinks it would be much less mechanically complex to cool a car engine by melting ice on it than by having a water-cooled system with a radiator, water pump, antifreeze, and so on. (a) If $\displaystyle 80.0%$ of the energy in 1.00 gal of gasoline is converted into “waste heat” in a car engine, how many kilograms of $\displaystyle 0ºC$ ice could it melt? (b) Is this a reasonable amount of ice to carry around to cool the engine for 1.00 gal of gasoline consumption? (c) What premises or assumptions are unreasonable? 104. Unreasonable Results (a) Calculate the rate of heat transfer by conduction through a window with an area of $\displaystyle 1.00 m^2$ that is 0.750 cm thick, if its inner surface is at $\displaystyle 22.0ºC$ and its outer surface is at $\displaystyle 35.0ºC$. (b) What is unreasonable about this result? (c) Which premise or assumption is responsible? Solution (a) 1.46 kW (b) Very high power loss through a window. An electric heater of this power can keep an entire room warm. (c) The surface temperatures of the window do not differ by as great an amount as assumed. The inner surface will be warmer, and the outer surface will be cooler. 105. Unreasonable Results A meteorite 1.20 cm in diameter is so hot immediately after penetrating the atmosphere that it radiates 20.0 kW of power. (a) What is its temperature, if the surroundings are at $\displaystyle 20.0ºC$ and it has an emissivity of 0.800? (b) What is unreasonable about this result? (c) Which premise or assumption is responsible? 106. Construct Your Own Problem Consider a new model of commercial airplane having its brakes tested as a part of the initial flight permission procedure. The airplane is brought to takeoff speed and then stopped with the brakes alone. Construct a problem in which you calculate the temperature increase of the brakes during this process. You may assume most of the kinetic energy of the airplane is converted to thermal energy in the brakes and surrounding materials, and that little escapes. Note that the brakes are expected to become so hot in this procedure that they ignite and, in order to pass the test, the airplane must be able to withstand the fire for some time without a general conflagration. 107. Construct Your Own Problem Consider a person outdoors on a cold night. Construct a problem in which you calculate the rate of heat transfer from the person by all three heat transfer methods. Make the initial circumstances such that at rest the person will have a net heat transfer and then decide how much physical activity of a chosen type is necessary to balance the rate of heat transfer. Among the things to consider are the size of the person, type of clothing, initial metabolic rate, sky conditions, amount of water evaporated, and volume of air breathed. Of course, there are many other factors to consider and your instructor may wish to guide you in the assumptions made as well as the detail of analysis and method of presenting your results. Contributors and Attributions - Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .	libretexts	2025-03-17T19:53:33.166950	2018-05-04T03:02:23	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods/14.E%3A_Heat_and_Heat_Transfer_Methods_(Exercise)", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "14.E: Heat and Heat Transfer Methods (Exercise)", "author": null }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics	15: Thermodynamics Thermodynamics is the branch of science concerned with heat and temperature and their relation to energy and work. It states that the behavior of these quantities is governed by the four laws of thermodynamics, irrespective of the composition or specific properties of the material or system in question. Thermodynamics applies to a wide variety of topics in science and engineering, especially physical chemistry, chemical engineering, and mechanical engineering. Thumbnail: The Steam engine and gas and oil engines. By John Perry. 1899. This file is in the public domain because its copyright has expired in the United States.	libretexts	2025-03-17T19:53:33.226631	2015-11-01T04:12:52	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "15: Thermodynamics", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.00%3A_Prelude_to_Thermodynamics	15.0: Prelude to Thermodynamics Heat transfer is energy in transit, and it can be used to do work. It can also be converted to any other form of energy. A car engine, for example, burns fuel for heat transfer into a gas. Work is done by the gas as it exerts a force through a distance, converting its energy into a variety of other forms—into the car’s kinetic or gravitational potential energy; into electrical energy to run the spark plugs, radio, and lights; and back into stored energy in the car’s battery. But most of the heat transfer produced from burning fuel in the engine does not do work on the gas. Rather, the energy is released into the environment, implying that the engine is quite inefficient. It is often said that modern gasoline engines cannot be made to be significantly more efficient. We hear the same about heat transfer to electrical energy in large power stations, whether they are coal, oil, natural gas, or nuclear powered. Why is that the case? Is the inefficiency caused by design problems that could be solved with better engineering and superior materials? Is it part of some money-making conspiracy by those who sell energy? Actually, the truth is more interesting, and reveals much about the nature of heat transfer. Basic physical laws govern how heat transfer for doing work takes place and place insurmountable limits onto its efficiency. This chapter will explore these laws as well as many applications and concepts associated with them. These topics are part of thermodynamics —the study of heat transfer and its relationship to doing work.	libretexts	2025-03-17T19:53:33.283397	2015-11-01T05:43:09	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.00%3A_Prelude_to_Thermodynamics", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "15.0: Prelude to Thermodynamics", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.01%3A_The_First_Law_of_Thermodynamics	15.1: The First Law of Thermodynamics Learning Objectives By the end of this section, you will be able to: - Define the first law of thermodynamics. - Describe how conservation of energy relates to the first law of thermodynamics. - Identify instances of the first law of thermodynamics working in everyday situations, including biological metabolism. - Calculate changes in the internal energy of a system, after accounting for heat transfer and work done. If we are interested in how heat transfer is converted into doing work, then the conservation of energy principle is important. The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is \[\Delta U = Q - W. \label{first}\] Here $\Delta U$ is the change in internal energy $U$ of the system. $Q$ is the net heat transferred into the system —that is, $Q$ is the sum of all heat transfer into and out of the system. $W$ is the net work done by the system —that is, $W$ is the sum of all work done on or by the system. We use the following sign conventions: if $Q$ is positive, then there is a net heat transfer into the system; if $W$ is positive, then there is net work done by the system. So positive $Q$ adds energy to the system and positive $W$ takes energy from the system. Thus $\Delta U = Q - W$. Note also that if more heat transfer into the system occurs than work done, the difference is stored as internal energy. Heat engines are a good example of this—heat transfer into them takes place so that they can do work (Figure $\PageIndex{2}$). We will now examine $Q, \, W$ and $\Delta U$ further. LAW OF THERMODYNAMICS AND LAW OF CONSERVATION OF ENERGY The first law of thermodynamics is actually the law of conservation of energy stated in a form most useful in thermodynamics. The first law gives the relationship between heat transfer, work done, and the change in internal energy of a system. Heat Q and Work W Heat transfer $Q$ and doing work $W$ are the two everyday means of bringing energy into or taking energy out of a system. The processes are quite different. Heat transfer, a less organized process, is driven by temperature differences. Work, a quite organized process, involves a macroscopic force exerted through a distance. Nevertheless, heat and work can produce identical results.For example, both can cause a temperature increase. Heat transfer into a system, such as when the Sun warms the air in a bicycle tire, can increase its temperature, and so can work done on the system, as when the bicyclist pumps air into the tire. Once the temperature increase has occurred, it is impossible to tell whether it was caused by heat transfer or by doing work. This uncertainty is an important point. Heat transfer and work are both energy in transit—neither is stored as such in a system. However, both can change the internal energy $U$ of a system. Internal energy is a form of energy completely different from either heat or work. Internal Energy U We can think about the internal energy of a system in two different but consistent ways. The first is the atomic and molecular view, which examines the system on the atomic and molecular scale. The internal energy $U$ of a system is the sum of the kinetic and potential energies of its atoms and molecules. Recall that kinetic plus potential energy is called mechanical energy. Thus internal energy is the sum of atomic and molecular mechanical energy. Because it is impossible to keep track of all individual atoms and molecules, we must deal with averages and distributions. A second way to view the internal energy of a system is in terms of its macroscopic characteristics, which are very similar to atomic and molecular average values. Macroscopically, we define the change in internal energy $\Delta U$ to be that given by the first law of thermodynamics (Equation \ref{first}): \[\Delta U = Q - W \nonumber\] Many detailed experiments have verified that $\Delta U = Q - W$, where $\Delta U$ is the change in total kinetic and potential energy of all atoms and molecules in a system. It has also been determined experimentally that the internal energy $U$ of a system depends only on the state of the system and not how it reached that state . More specifically, $U$ is found to be a function of a few macroscopic quantities (pressure, volume, and temperature, for example), independent of past history such as whether there has been heat transfer or work done. This independence means that if we know the state of a system, we can calculate changes in its internal energy $U$ from a few macroscopic variables. MACROSCOPIC vs. MICROSCOPIC In thermodynamics, we often use the macroscopic picture when making calculations of how a system behaves, while the atomic and molecular picture gives underlying explanations in terms of averages and distributions. We shall see this again in later sections of this chapter. For example, in the topic of entropy, calculations will be made using the atomic and molecular view. To get a better idea of how to think about the internal energy of a system, let us examine a system going from State 1 to State 2. The system has internal energy $U_1$ in State 1, and it has internal energy $U_2$ in State 2, no matter how it got to either state. So the change in internal energy \[\Delta U = U_2 - U_1\] is independent of what caused the change. In other words, $\delta U$ is independent of path . By path, we mean the method of getting from the starting point to the ending point. Why is this independence important? Both $Q$ and $W$ depend on path , but $\Delta U$ does not (Equation \ref{first}). This path independence means that internal energy $U$ is easier to consider than either heat transfer or work done. Example $\PageIndex{1}$: Calculating Change in Internal Energy - The Same Change in $U$ is Produced by Two Different Processes - Suppose there is heat transfer of 40.00 J to a system, while the system does 10.00 J of work. Later, there is heat transfer of 25.00 J out of the system while 4.00 J of work is done on the system. What is the net change in internal energy of the system? - What is the change in internal energy of a system when a total of 150.00 J of heat transfer occurs out of (from) the system and 159.00 J of work is done on the system (Figure $\PageIndex{3}$)? Strategy In part (a), we must first find the net heat transfer and net work done from the given information. Then the first law of thermodynamics (Equation \ref{first}). can be used to find the change in internal energy. In part (b), the net heat transfer and work done are given, so the equation can be used directly. Solution for (a) The net heat transfer is the heat transfer into the system minus the heat transfer out of the system, or \[ \begin{align} Q &= 40.00 \, J - 25.00 \, J \\[5pt] &= 15.00 \, J \end{align}\] Similarly, the total work is the work done by the system minus the work done on the system, or \[ \begin{align} W &= 10.00 \, J - 4.00 \, J \\[5pt] &= 6.00 \, J. \end{align}\] Discussion on (a) No matter whether you look at the overall process or break it into steps, the change in internal energy is the same. Solution for (b) Here the net heat transfer and total work are given directly to be $Q = -150.00 \, J$ and $W = -159.00 \, J$, so that \[ \begin{align} \Delta U &= Q - W = -150.00 - (-159.00) \\[5pt] &= 9.00 \, J. \end{align}\] Discussion on (b) A very different process in part (b) produces the same 9.00-J change in internal energy as in part (a). Note that the change in the system in both parts is related to $\Delta U$ and not to the individual $Q$s or $W$s involved. The system ends up in the same state in both (a) and (b). Parts (a) and (b) present two different paths for the system to follow between the same starting and ending points, and the change in internal energy for each is the same—it is independent of path. Human Metabolism and the First Law of Thermodynamics Human metabolism is the conversion of food into heat transfer, work, and stored fat. Metabolism is an interesting example of the first law of thermodynamics in action. We now take another look at these topics via the first law of thermodynamics. Considering the body as the system of interest, we can use the first law to examine heat transfer, doing work, and internal energy in activities ranging from sleep to heavy exercise. What are some of the major characteristics of heat transfer, doing work, and energy in the body? For one, body temperature is normally kept constant by heat transfer to the surroundings. This means $Q$ is negative. Another fact is that the body usually does work on the outside world. This means $W$ is positive. In such situations, then, the body loses internal energy, since $\Delta U = Q - W$ is negative. Now consider the effects of eating. Eating increases the internal energy of the body by adding chemical potential energy (this is an unromantic view of a good steak). The body metabolizes all the food we consume. Basically, metabolism is an oxidation process in which the chemical potential energy of food is released. This implies that food input is in the form of work. Food energy is reported in a special unit, known as the Calorie. This energy is measured by burning food in a calorimeter, which is how the units are determined. In chemistry and biochemistry, one calorie (spelled with a lowercase c) is defined as the energy (or heat transfer) required to raise the temperature of one gram of pure water by one degree Celsius. Nutritionists and weight-watchers tend to use the dietary calorie, which is frequently called a Calorie (spelled with a capital C). One food Calorie is the energy needed to raise the temperature of one kilogram of water by one degree Celsius. This means that one dietary Calorie is equal to one kilocalorie for the chemist, and one must be careful to avoid confusion between the two. Again, consider the internal energy the body has lost. There are three places this internal energy can go—to heat transfer, to doing work, and to stored fat (a tiny fraction also goes to cell repair and growth). Heat transfer and doing work take internal energy out of the body, and food puts it back. If you eat just the right amount of food, then your average internal energy remains constant. Whatever you lose to heat transfer and doing work is replaced by food, so that, in the long run, $\Delta U = 0$. If you overeat repeatedly, then $\Delta U$ is always positive, and your body stores this extra internal energy as fat. The reverse is true if you eat too little. If $\Delta U$ is negative for a few days, then the body metabolizes its own fat to maintain body temperature and do work that takes energy from the body. This process is how dieting produces weight loss. Life is not always this simple, as any dieter knows. The body stores fat or metabolizes it only if energy intake changes for a period of several days. Once you have been on a major diet, the next one is less successful because your body alters the way it responds to low energy intake. Your basal metabolic rate (BMR) is the rate at which food is converted into heat transfer and work done while the body is at complete rest. The body adjusts its basal metabolic rate to partially compensate for over-eating or under-eating. The body will decrease the metabolic rate rather than eliminate its own fat to replace lost food intake. You will chill more easily and feel less energetic as a result of the lower metabolic rate, and you will not lose weight as fast as before. Exercise helps to lose weight, because it produces both heat transfer from your body and work, and raises your metabolic rate even when you are at rest. Weight loss is also aided by the quite low efficiency of the body in converting internal energy to work, so that the loss of internal energy resulting from doing work is much greater than the work done.It should be noted, however, that living systems are not in thermal equilibrium. The body provides us with an excellent indication that many thermodynamic processes are irreversible . An irreversible process can go in one direction but not the reverse, under a given set of conditions. For example, although body fat can be converted to do work and produce heat transfer, work done on the body and heat transfer into it cannot be converted to body fat. Otherwise, we could skip lunch by sunning ourselves or by walking down stairs. Another example of an irreversible thermodynamic process is photosynthesis. This process is the intake of one form of energy—light—by plants and its conversion to chemical potential energy. Both applications of the first law of thermodynamics are illustrated in Figure $\PageIndex{4}$. One great advantage of conservation laws such as the first law of thermodynamics is that they accurately describe the beginning and ending points of complex processes, such as metabolism and photosynthesis, without regard to the complications in between. Summary The table presents a summary of terms relevant to the first law of thermodynamics. \| Term \| Definition \| \|---\|---\| \| $U$ \| Internal energy—the sum of the kinetic and potential energies of a system’s atoms and molecules. Can be divided into many subcategories, such as thermal and chemical energy. Depends only on the state of a system (such as its $P$, $V$ and $T$, not on how the energy entered the system. Change in internal energy is path independent. \| \| $Q$ \| Heat—energy transferred because of a temperature difference. Characterized by random molecular motion. Highly dependent on path. $Q$ entering a system is positive. \| \| $W$ \| Work—energy transferred by a force moving through a distance. An organized, orderly process. Path dependent. $W$ done by a system (either against an external force or to increase the volume of the system) is positive. \| - The first law of thermodynamics is given as $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy of a system, $Q$ is the net heat transfer (the sum of all heat transfer into and out of the system), and $W$ is the net work done (the sum of all work done on or by the system). - Both $Q$ and $W$ are energy in transit; only $\Delta U$ represents an independent quantity capable of being stored. - The internal energy $U$ of a system depends only on the state of the system and not how it reached that state. - Metabolism of living organisms, and photosynthesis of plants, are specialized types of heat transfer, doing work, and internal energy of systems. Glossary - first law of thermodynamics - states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system - internal energy - the sum of the kinetic and potential energies of a system’s atoms and molecules - human metabolism - conversion of food into heat transfer, work, and stored fat	libretexts	2025-03-17T19:53:33.365140	2015-11-01T05:43:23	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.01%3A_The_First_Law_of_Thermodynamics", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "15.1: The First Law of Thermodynamics", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.02%3A_The_First_Law_of_Thermodynamics_and_Some_Simple_Processes	15.2: The First Law of Thermodynamics and Some Simple Processes Learning Objectives By the end of this section, you will be able to: - Define the first law of thermodynamics. - Describe how conservation of energy relates to the first law of thermodynamics. - Identify instances of the first law of thermodynamics working in everyday situations, including biological metabolism. - Calculate changes in the internal energy of a system, after accounting for heat transfer and work done. One of the most important things we can do with heat transfer is to use it to do work for us. Such a device is called a heat engine . Car engines and steam turbines that generate electricity are examples of heat engines. Figure $\PageIndex{2}$ shows schematically how the first law of thermodynamics applies to the typical heat engine. Figure $\PageIndex{3}$: (a) Heat transfer to the gas in a cylinder increases the internal energy of the gas, creating higher pressure and temperature. (b) The force exerted on the movable cylinder does work as the gas expands. Gas pressure and temperature decrease when it expands, indicating that the gas’s internal energy has been decreased by doing work. (c) Heat transfer to the environment further reduces pressure in the gas so that the piston can be more easily returned to its starting position. The illustrations above show one of the ways in which heat transfer does work. Fuel combustion produces heat transfer to a gas in a cylinder, increasing the pressure of the gas and thereby the force it exerts on a movable piston. The gas does work on the outside world, as this force moves the piston through some distance. Heat transfer to the gas cylinder results in work being done. To repeat this process, the piston needs to be returned to its starting point. Heat transfer now occurs from the gas to the surroundings so that its pressure decreases, and a force is exerted by the surroundings to push the piston back through some distance. Variations of this process are employed daily in hundreds of millions of heat engines. We will examine heat engines in detail in the next section. In this section, we consider some of the simpler underlying processes on which heat engines are based. PV Diagrams and their Relationship to Work Done on or by a Gas A process by which a gas does work on a piston at constant pressure is called an isobaric process . Since the pressure is constant, the force exerted is constant and the work done is given as \[W = Fd\] See the symbols as shown in Figure $\PageIndex{4}$. Now $F = PA$, and so \[W = PAd.\] Because the volume of a cylinder is its cross-sectional area $A$ times its length $d$, we see that $Ad = \Delta V$, the change in volume; thus, \[W = P\Delta V \, (isobaric \, process).\] Note that if $\Delta V$ is positive, then $W$ is positive, meaning that work is done by the gas on the outside world. (Note that the pressure involved in this work that we’ve called $P$ is the pressure of the gas inside the tank. If we call the pressure outside the tank $P_{ext}$, an expanding gas would be working against the external pressure; the work done would therefore be $W = -P_{ext}\delta V$ (isobaric process). Many texts use this definition of work, and not the definition based on internal pressure, as the basis of the First Law of Thermodynamics. This definition reverses the sign conventions for work, and results in a statement of the first law that becomes $\Delta U = Q + W$.) It is not surprising that $W = P\Delta V$, since we have already noted in our treatment of fluids that pressure is a type of potential energy per unit volume and that pressure in fact has units of energy divided by volume. We also noted in our discussion of the ideal gas law that $PV$ has units of energy. In this case, some of the energy associated with pressure becomes work. Figure $\PageIndex{5}$ shows a graph of pressure versus volume (that is, a $PV$ diagram for an isobaric process. You can see in the figure that the work done is the area under the graph. This property of $PV$ diagrams is very useful and broadly applicable: the work done on or by a system in going from one state to another equals the area under the curve on a $PV$ diagram . We can see where this leads by considering Figure $\PageIndex{6}$(a), which shows a more general process in which both pressure and volume change. The area under the curve is closely approximated by dividing it into strips, each having an average constant pressure $P_{i(ave)}$. The work done is $W_i = P_{i(ave)}\Delta V_i$ for each strip, and the total work done is the sum of the $W_i$. Thus the total work done is the total area under the curve. If the path is reversed, as in Figure $\PageIndex{6}$(b), then work is done on the system. The area under the curve in that case is negative, because $\Delta V$ is negative. $PV$ diagrams clearly illustrate that the work done depends on the path taken and not just the endpoints . This path dependence is seen in Figure $\PageIndex{7}$(a), where more work is done in going from A to C by the path via point B than by the path via point D. The vertical paths, where volume is constant, are called isochoric processes. Since volume is constant, $\Delta V = 0$, and no work is done in an isochoric process. Now, if the system follows the cyclical path ABCDA, as in Figure $\PageIndex{7}$(b), then the total work done is the area inside the loop. The negative area below path CD subtracts, leaving only the area inside the rectangle. In fact, the work done in any cyclical process (one that returns to its starting point) is the area inside the loop it forms on a $PV$ diagram, as Figure $\PageIndex{7}$(c) illustrates for a general cyclical process. Note that the loop must be traversed in the clockwise direction for work to be positive—that is, for there to be a net work output. Example $\PageIndex{1}$: Total Work Done in a Cyclical Process Equals the Area Inside the Closed Loop on a PV Diagram Calculate the total work done in the cyclical process ABCDA shown in Figure $\PageIndex{7}$(b) by the following two methods to verify that work equals the area inside the closed loop on the $PV$ diagram. (Take the data in the figure to be precise to three significant figures.) (a) Calculate the work done along each segment of the path and add these values to get the total work. (b) Calculate the area inside the rectangle ABCDA. Strategy To find the work along any path on a $PV$ diagram, you use the fact that work is pressure times change in volume, or $W = P\Delta V$. So in part (a), this value is calculated for each leg of the path around the closed loop. Solution for (a) The work along path AB is \[W_{AB} = P_{AB}\Delta Y_{AB}\] \[= (1.50 \times 10^6 \, N/m^2)(5.00 \times 10^{-4} \, m^3) = 750 \, J.\] Since the path BC is isochoric, $\Delta V_{BC} = 0$, and so $W_{BC} = 0$. The work along path CD is negative, since $\Delta V_{CD} $ is negative (the volume decreases). The work is \[W_{CD} = P_{CD}\Delta V_{CD}\] \[= (2.00 \times 10^5 \, N/m^2)(-5.00 \times 10^{-4} \, m^3) = -100 \, J.\] Again, since the path DA is isochoric, $\delta V_{DA} = 0$, and so $W_{DA} = 0$. Now the total work is \[W = W_{AB} + W_{BC} + W_{CD} + W_{DA}\] \[ = 750 \, J + 0 + (-100 \, J) + 0 = 650 \, J.\] Solution for (b) The area inside the rectangle is its height times its width, or \[area = (P_{AB} - P_{CD})\Delta V\] \[= [(1.50 \times 10^6 \, N/m^2) - (2.00 \times 10^5 \, N/m^2)] (5.00 \times 1-^{-4} m^3)\] \[= 650 \, J.\] Thus, \[area = 650 \, J = W.\] Discussion The result, as anticipated, is that the area inside the closed loop equals the work done. The area is often easier to calculate than is the work done along each path. It is also convenient to visualize the area inside different curves on $PV$ diagrams in order to see which processes might produce the most work. Recall that work can be done to the system, or by the system, depending on the sign of $W$. A positive $W$ is work that is done by the system on the outside environment; a negative $W$ is work that is done by the system on the outside environment; a negative $W$ represents work done by the environment on the system. Figure $\PageIndex{8}$(a) shows two other important processes on a $PV$ diagram. For comparison, both are shown starting from the same point A. The upper curve ending at point B is an isothermal process—that is, one in which temperature is kept constant. If the gas behaves like an ideal gas, as is often the case, and if no phase change occurs, then $PV = nRT$. Since $T$ is constant, $PV$ is a constant for an isothermal process. We ordinarily expect the temperature of a gas to decrease as it expands, and so we correctly suspect that heat transfer must occur from the surroundings to the gas to keep the temperature constant during an isothermal expansion. To show this more rigorously for the special case of a monatomic ideal gas, we note that the average kinetic energy of an atom in such a gas is given by \[\dfrac{1}{2}m\overline v^2 = \dfrac{3}{2}kT.\] The kinetic energy of the atoms in a monatomic ideal gas is its only form of internal energy, and so its total internal energy $U$ is \[U = N\dfrac{1}{2}m\overline v^2 = \dfrac{3}{2}NkT, \, (monatomic \, ideal \, gas),\] where $N$ is the number of atoms in the gas. This relationship means that the internal energy of an ideal monatomic gas is constant during an isothermal process—that is, $\Delta U = 0$. If the internal energy does not change, then the net heat transfer into the gas must equal the net work done by the gas. That is, because $\Delta U = Q - W = 0$ here, $Q = W$. We must have just enough heat transfer to replace the work done. An isothermal process is inherently slow, because heat transfer occurs continuously to keep the gas temperature constant at all times and must be allowed to spread through the gas so that there are no hot or cold regions. Also shown in Figure $\PageIndex{8}$(a) is a curve AC for an adiabatic process, defined to be one in which there is no heat transfer—that is, $Q = 0$. Processes that are nearly adiabatic can be achieved either by using very effective insulation or by performing the process so fast that there is little time for heat transfer. Temperature must decrease during an adiabatic expansion process, since work is done at the expense of internal energy: \[U = \dfrac{3}{2}NkT.\] (You might have noted that a gas released into atmospheric pressure from a pressurized cylinder is substantially colder than the gas in the cylinder.) In fact, because $Q = 0, \, \Delta U = -W$ for an adiabatic process. Lower temperature results in lower pressure along the way, so that curve AC is lower than curve AB, and less work is done. If the path ABCA could be followed by cooling the gas from B to C at constant volume (isochorically), Figure $\PageIndex{8}$(b), there would be a net work output. Reversible Processes Both isothermal and adiabatic processes such as shown in Figure $\PageIndex{8}$ are reversible in principle. A reversible process is one in which both the system and its environment can return to exactly the states they were in by following the reverse path. The reverse isothermal and adiabatic paths are BA and CA, respectively. Real macroscopic processes are never exactly reversible. In the previous examples, our system is a gas (like that in Figure $\PageIndex{4}$), and its environment is the piston, cylinder, and the rest of the universe. If there are any energy-dissipating mechanisms, such as friction or turbulence, then heat transfer to the environment occurs for either direction of the piston. So, for example, if the path BA is followed and there is friction, then the gas will be returned to its original state but the environment will not—it will have been heated in both directions. Reversibility requires the direction of heat transfer to reverse for the reverse path. Since dissipative mechanisms cannot be completely eliminated, real processes cannot be reversible. There must be reasons that real macroscopic processes cannot be reversible. We can imagine them going in reverse. For example, heat transfer occurs spontaneously from hot to cold and never spontaneously the reverse. Yet it would not violate the first law of thermodynamics for this to happen. In fact, all spontaneous processes, such as bubbles bursting, never go in reverse. There is a second thermodynamic law that forbids them from going in reverse. When we study this law, we will learn something about nature and also find that such a law limits the efficiency of heat engines. We will find that heat engines with the greatest possible theoretical efficiency would have to use reversible processes, and even they cannot convert all heat transfer into doing work. The table summarizes the simpler thermodynamic processes and their definitions. \| Isobaric \| Constant pressure \| $W = P\Delta V$ \| \|---\|---\|---\| \| Isochoric \| Constant volume \| $W = 0$ \| \| Isothermal \| Constant temperature \| $Q = W$ \| \| Adiabatic \| No heat transfer \| $Q = 0$ \| PHET EXPLORATIONS: STATES OF MATTER Watch different types of molecules form a solid, liquid, or gas in teh States of Matter simuator. Add or remove heat and watch the phase change. Change the temperature or volume of a container and see a pressure-temperature diagram respond in real time. Relate the interaction potential to the forces between molecules. Summary - One of the important implications of the first law of thermodynamics is that machines can be harnessed to do work that humans previously did by hand or by external energy supplies such as running water or the heat of the Sun. A machine that uses heat transfer to do work is known as a heat engine. - There are several simple processes, used by heat engines, that flow from the first law of thermodynamics. Among them are the isobaric, isochoric, isothermal and adiabatic processes. - These processes differ from one another based on how they affect pressure, volume, temperature, and heat transfer. - If the work done is performed on the outside environment, work $(W)$ will be a positive value. If the work done is done to the heat engine system, work $(W)$ will be a negative value. - Some thermodynamic processes, including isothermal and adiabatic processes, are reversible in theory; that is, both the thermodynamic system and the environment can be returned to their initial states. However, because of loss of energy owing to the second law of thermodynamics, complete reversibility does not work in practice. Glossary - heat engine - a machine that uses heat transfer to do work - isobaric process - constant-pressure process in which a gas does work - isochoric process - a constant-volume process - isothermal process - a constant-temperature process - adiabatic process - a process in which no heat transfer takes place - reversible process - a process in which both the heat engine system and the external environment theoretically can be returned to their original states	libretexts	2025-03-17T19:53:33.446257	2015-11-01T05:43:42	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.02%3A_The_First_Law_of_Thermodynamics_and_Some_Simple_Processes", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "15.2: The First Law of Thermodynamics and Some Simple Processes", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.03%3A_Introduction_to_the_Second_Law_of_Thermodynamics_-_Heat_Engines_and_their_Efficiency	15.3: Introduction to the Second Law of Thermodynamics - Heat Engines and their Efficiency Learning Objectives By the end of this section, you will be able to: - State the expressions of the second law of thermodynamics. - Calculate the efficiency and carbon dioxide emission of a coal-fired electricity plant, using second law characteristics. - Describe and define the Otto cycle. The second law of thermodynamics deals with the direction taken by spontaneous processes. Many processes occur spontaneously in one direction only—that is, they are irreversible, under a given set of conditions. Although irreversibility is seen in day-to-day life—a broken glass does not resume its original state, for instance—complete irreversibility is a statistical statement that cannot be seen during the lifetime of the universe. More precisely, an irreversible process is one that depends on path. If the process can go in only one direction, then the reverse path differs fundamentally and the process cannot be reversible. For example, as noted in the previous section, heat involves the transfer of energy from higher to lower temperature. A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter. Furthermore, mechanical energy, such as kinetic energy, can be completely converted to thermal energy by friction, but the reverse is impossible. A hot stationary object never spontaneously cools off and starts moving. Yet another example is the expansion of a puff of gas introduced into one corner of a vacuum chamber. The gas expands to fill the chamber, but it never regroups in the corner. The random motion of the gas molecules could take them all back to the corner, but this is never observed to happen. (See Figure $\PageIndex{2}$) The fact that certain processes never occur suggests that there is a law forbidding them to occur. The first law of thermodynamics would allow them to occur—none of those processes violate conservation of energy. The law that forbids these processes is called the second law of thermodynamics. We shall see that the second law can be stated in many ways that may seem different, but which in fact are equivalent. Like all natural laws, the second law of thermodynamics gives insights into nature, and its several statements imply that it is broadly applicable, fundamentally affecting many apparently disparate processes. The already familiar direction of heat transfer from hot to cold is the basis of our first version of the second law of thermodynamics . The Second Law of Thermodynamics (first expression) Heat transfer occurs spontaneously from higher- to lower-temperature bodies but never spontaneously in the reverse direction. Another way of stating this: It is impossible for any process to have as its sole result heat transfer from a cooler to a hotter object. Heat Engines Now let us consider a device that uses heat transfer to do work. As noted in the previous section, such a device is called a heat engine, and one is shown schematically in $\PageIndex{3}$(b). Gasoline and diesel engines, jet engines, and steam turbines are all heat engines that do work by using part of the heat transfer from some source. Heat transfer from the hot object (or hot reservoir) is denoted as $Q_f$, while heat transfer into the cold object (or cold reservoir) is $Q_c$, and the work done by the engine is $W$. The temperatures of the hot and cold reservoirs are $T_h$ and $T_c$, respectively. Because the hot reservoir is heated externally, which is energy intensive, it is important that the work is done as efficiently as possible. In fact, we would like $W$ to equal $Q_h$ and for there to be no heat transfer to the environment $(Q_c = 0)$. Unfortunately, this is impossible. The second law of thermodynamics also states, with regard to using heat transfer to do work (the second expression of the second law): The Second Law of Thermodynamics (second expression) It is impossible in any system for heat transfer from a reservoir to completely convert to work in a cyclical process in which the system returns to its initial state. A cyclical process brings a system, such as the gas in a cylinder, back to its original state at the end of every cycle. Most heat engines, such as reciprocating piston engines and rotating turbines, use cyclical processes. The second law, just stated in its second form, clearly states that such engines cannot have perfect conversion of heat transfer into work done. Before going into the underlying reasons for the limits on converting heat transfer into work, we need to explore the relationships among $W, \, Q_h$, and $Q_c$ and to define the efficiency of a cyclical heat engine. As noted, a cyclical process brings the system back to its original condition at the end of every cycle. Such a system’s internal energy $U$ is the same at the beginning and end of every cycle—that is, $\Delta U = 0$. The first law of thermodynamics states that \[\Delta U = Q - W,\] where $Q$ is the net heat transfer during the cycle $(Q = Q_h - Q_c)$ and $W$ is the net work done by the system. Since $\Delta U = 0$ for a complete cycle, we have \[0 = Q - W,\] so that \[W = Q.\] Thus the net work done by the system equals the net heat transfer into the system, or \[W = Q_h - Q_c \, (cyclical \, process),\] just as shown schematically in Figure $\PageIndex{3}$(b). The problem is that in all processes, there is some heat transfer $Q_c$ to the environment—and usually a very significant amount at that. In the conversion of energy to work, we are always faced with the problem of getting less out than we put in. We define conversion efficiency $Eff$ to be the ratio of useful work output to the energy input (or, in other words, the ratio of what we get to what we spend). In that spirit, we define the efficiency of a heat engine to be its net work output $W$ divided by heat transfer to the engine $Q_h$, that is, \[Eff = \dfrac{W}{Q_h}.\] Since $W = Q_h - Q_c$ in a cyclical process, we can also express this as \[Eff = \dfrac{Q_h - Q_c}{Q_h} = 1 - \dfrac{Q_c}{Q_h} \, (cyclical \, process),\] making it clear that an efficiency of 1, or 100%, is possible only if there is no heat transfer to the environment $(Q_c = 0)$. Note that all $Q$s are positive. The direction of heat transfer is indicated by a plus or minus sign. For example, $Q-c$ is out of the system and so is preceded by a minus sign. Example $\PageIndex{1}$: Daily Work Done by a Coal-Fired Power Station, Its Efficiency and Carbon Dioxide Emissions A coal-fired power station is a huge heat engine. It uses heat transfer from burning coal to do work to turn turbines, which are used to generate electricity. In a single day, a large coal power station has $2.50 \times 10^{14} \, J$ of heat transfer from coal and $1.48 \times 10^14 \, J$ of heat transfer into the environment. (a) What is the work done by the power station? (b) What is the efficiency of the power station? (c) In the combustion process, the following chemical reaction occurs: $C + O_2 \rightarrow CO_2$ This implies that every 12 kg of coal puts 12 kg + 16 kg + 16 kg = 44 kg of carbon dioxide into the atmosphere. Assuming that 1 kg of coal can provide $2.5 \times 10^6 \, J$ of heat transfer upon combustion, how much $CO_2$ is emitted per day by this power plant? Strategy for (a) We can use $W = Q_h - Q_c$ to find the work output $W$, assuming a cyclical process is used in the power station. In this process, water is boiled under pressure to form high-temperature steam, which is used to run steam turbine-generators, and then condensed back to water to start the cycle again. Solution for (a) Work output is given by: \[W = Q_h - Q_c. \nonumber\] Substituting the given values: \[\begin{align} W &= 2.50 \times 10^14 \, J - 1.48 \times 10^{14} \, J \\[4pt] &= 1.02 \times 10^{14} \, J. \end{align}\] Strategy for (b) The efficiency can be calculated with $Eff = \dfrac{W}{Q_h}$ since $Q_H$ is given and work $W$ was found in the first part of this example. Solution for (b) Efficiency is given by $Eff = \dfrac{W}{Q_h}$. The work $W$ was just found to be $1.01 \times 10^{14}$, and $Q_h$ is given, so the efficiency is \[Eff = \dfrac{1.01 \times 10^{14} \, J}{2.50 \times 10^{14} \, J} = 0.408, \, or \, 40.8\%\] Strategy for (c) The daily consumption of coal is calculated using the information that each day there is \[\dfrac{2.50 \times 10^{14} \, J}{2.50 \times 10^6 \, J/kg} = 1.0 \times 10^8 \, kg. \nonumber\] Assuming that the coal is pure and that all the coal goes toward producing carbon dioxide, the carbon dioxide produced per day is \[1.0 \times 10^8 \, kg \, coal \times \dfrac{44 \, kg \, CO_2}{12 \, kg \, coal} = 3.7 \times 10^8 \, kg \, CO_2. \nonumber\] This is 370,000 metric tons of $CO2$ produced every day. Discussion If all the work output is converted to electricity in a period of one day, the average power output is 1180 MW (this is left to you as an end-of-chapter problem). This value is about the size of a large-scale conventional power plant. The efficiency found is acceptably close to the value of 42% given for coal power stations. It means that fully 59.2% of the energy is heat transfer to the environment, which usually results in warming lakes, rivers, or the ocean near the power station, and is implicated in a warming planet generally. While the laws of thermodynamics limit the efficiency of such plants—including plants fired by nuclear fuel, oil, and natural gas—the heat transfer to the environment could be, and sometimes is, used for heating homes or for industrial processes. The generally low cost of energy has not made it economical to make better use of the waste heat transfer from most heat engines. Coal-fired power plants produce the greatest amount of $\ce{CO2}$ per unit energy output (compared to natural gas or oil), making coal the least efficient fossil fuel. With the information given in Example $\PageIndex{1}$, we can find characteristics such as the efficiency of a heat engine without any knowledge of how the heat engine operates, but looking further into the mechanism of the engine will give us greater insight. Figure $\PageIndex{4}$ illustrates the operation of the common four-stroke gasoline engine. The four steps shown complete this heat engine’s cycle, bringing the gasoline-air mixture back to its original condition. The Otto cycle shown in Figure $\PageIndex{5a}$ is used in four-stroke internal combustion engines, although in fact the true Otto cycle paths do not correspond exactly to the strokes of the engine. The adiabatic process AB corresponds to the nearly adiabatic compression stroke of the gasoline engine. In both cases, work is done on the system (the gas mixture in the cylinder), increasing its temperature and pressure. Along path BC of the Otto cycle, heat transfer $Q_h$ into the gas occurs at constant volume, causing a further increase in pressure and temperature. This process corresponds to burning fuel in an internal combustion engine, and takes place so rapidly that the volume is nearly constant. Path CD in the Otto cycle is an adiabatic expansion that does work on the outside world, just as the power stroke of an internal combustion engine does in its nearly adiabatic expansion. The work done by the system along path CD is greater than the work done on the system along path AB, because the pressure is greater, and so there is a net work output. Along path DA in the Otto cycle, heat transfer $Q_c$ from the gas at constant volume reduces its temperature and pressure, returning it to its original state. In an internal combustion engine, this process corresponds to the exhaust of hot gases and the intake of an air-gasoline mixture at a considerably lower temperature. In both cases, heat transfer into the environment occurs along this final path. The net work done by a cyclical process is the area inside the closed path on a $PV$ diagram, such as that inside path ABCDA in Figure $\PageIndex{5}$. Note that in every imaginable cyclical process, it is absolutely necessary for heat transfer from the system to occur in order to get a net work output. In the Otto cycle, heat transfer occurs along path DA. If no heat transfer occurs, then the return path is the same, and the net work output is zero. The lower the temperature on the path AB, the less work has to be done to compress the gas. The area inside the closed path is then greater, and so the engine does more work and is thus more efficient. Similarly, the higher the temperature along path CD, the more work output there is (Figure $\PageIndex{6}$. So efficiency is related to the temperatures of the hot and cold reservoirs. In the next section, we shall see what the absolute limit to the efficiency of a heat engine is, and how it is related to temperature. Summary - The two expressions of the second law of thermodynamics are: (i) Heat transfer occurs spontaneously from higher- to lower-temperature bodies but never spontaneously in the reverse direction; and (ii) It is impossible in any system for heat transfer from a reservoir to completely convert to work in a cyclical process in which the system returns to its initial state. - Irreversible processes depend on path and do not return to their original state. Cyclical processes are processes that return to their original state at the end of every cycle. - In a cyclical process, such as a heat engine, the net work done by the system equals the net heat transfer into the system, or $W = Q_h - Q_c$, where $Q_h$ is the heat transfer from the hot object (hot reservoir), and $Q_c$ is the heat transfer into the cold object (cold reservoir). - Efficiency can be expressed as $Eff = \dfrac{W}{Q_h}$, the ratio of work output divided by the amount of energy input. - The four-stroke gasoline engine is often explained in terms of the Otto cycle, which is a repeating sequence of processes that convert heat into work. Glossary - irreversible process - any process that depends on path direction - second law of thermodynamics - heat transfer flows from a hotter to a cooler object, never the reverse, and some heat energy in any process is lost to available work in a cyclical process - cyclical process - a process in which the path returns to its original state at the end of every cycle - Otto cycle - a thermodynamic cycle, consisting of a pair of adiabatic processes and a pair of isochoric processes, that converts heat into work, e.g., the four-stroke engine cycle of intake, compression, ignition, and exhaust	libretexts	2025-03-17T19:53:33.527722	2015-11-01T05:43:58	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.03%3A_Introduction_to_the_Second_Law_of_Thermodynamics_-_Heat_Engines_and_their_Efficiency", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "15.3: Introduction to the Second Law of Thermodynamics - Heat Engines and their Efficiency", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.04%3A_Carnots_Perfect_Heat_Engine-_The_Second_Law_of_Thermodynamics_Restated	15.4: Carnot’s Perfect Heat Engine- The Second Law of Thermodynamics Restated Learning Objectives By the end of this section, you will be able to: - Identify a Carnot cycle. - Calculate maximum theoretical efficiency of a nuclear reactor. - Explain how dissipative processes affect the ideal Carnot engine. We know from the second law of thermodynamics that a heat engine cannot be 100% efficient, since there must always be some heat transfer $Q_c$ to the environment, which is often called waste heat. How efficient, then, can a heat engine be? This question was answered at a theoretical level in 1824 by a young French engineer, Sadi Carnot (1796–1832), in his study of the then-emerging heat engine technology crucial to the Industrial Revolution. He devised a theoretical cycle, now called the Carnot cycle , which is the most efficient cyclical process possible. The second law of thermodynamics can be restated in terms of the Carnot cycle, and so what Carnot actually discovered was this fundamental law. Any heat engine employing the Carnot cycle is called a Carnot engine . What is crucial to the Carnot cycle—and, in fact, defines it—is that only reversible processes are used. Irreversible processes involve dissipative factors, such as friction and turbulence. This increases heat transfer $Q_c$ to the environment and reduces the efficiency of the engine. Obviously, then, reversible processes are superior. Carnot Engine Stated in terms of reversible processes, the second law of thermodynamics has a third form: A Carnot engine operating between two given temperatures has the greatest possible efficiency of any heat engine operating between these two temperatures. Furthermore, all engines employing only reversible processes have this same maximum efficiency when operating between the same given temperatures. Figure $\PageIndex{2}$ shows the $PV$ diagram for a Carnot cycle. The cycle comprises two isothermal and two adiabatic processes. Recall that both isothermal and adiabatic processes are, in principle, reversible. Carnot also determined the efficiency of a perfect heat engine—that is, a Carnot engine. It is always true that the efficiency of a cyclical heat engine is given by: \[Eff = \dfrac{Q_h - Q_c}{Q_h} = 1 - \dfrac{Q_c}{Q_h}.\] What Carnot found was that for a perfect heat engine, the ratio $Q_c/Q_h$ equals the ratio of the absolute temperatures of the heat reservoirs. That is, $Q_c/Q_h = T_c/T_h$ for a Carnot engine, so that the maximum or Carnot efficiency $Eff_c$ is given by \[Eff_c = 1 - \dfrac{T_c}{T_h},\] where $T_h$ and $T_c$ are in kelvins (or any other absolute temperature scale). No real heat engine can do as well as the Carnot efficiency—an actual efficiency of about 0.7 of this maximum is usually the best that can be accomplished. But the ideal Carnot engine, like the drinking bird above, while a fascinating novelty, has zero power. This makes it unrealistic for any applications. Carnot’s interesting result implies that 100% efficiency would be possible only if $T_c = 0$ - that is, only if the cold reservoir were at absolute zero, a practical and theoretical impossibility. But the physical implication is this—the only way to have all heat transfer go into doing work is to remove all thermal energy, and this requires a cold reservoir at absolute zero. It is also apparent that the greatest efficiencies are obtained when the ratio $T_c/T_h$ is as small as possible. Just as discussed for the Otto cycle in the previous section, this means that efficiency is greatest for the highest possible temperature of the hot reservoir and lowest possible temperature of the cold reservoir. (This setup increases the area inside the closed loop on the $PV$ diagram; also, it seems reasonable that the greater the temperature difference, the easier it is to divert the heat transfer to work.) The actual reservoir temperatures of a heat engine are usually related to the type of heat source and the temperature of the environment into which heat transfer occurs. Consider the following example. Example $\PageIndex{1}$: Maximum Theoretical Efficiency for a Nuclear Reactor A nuclear power reactor has pressurized water at $300^oC$. (Higher temperatures are theoretically possible but practically not, due to limitations with materials used in the reactor.) Heat transfer from this water is a complex process (see Figure $\PageIndex{3}$). Steam, produced in the steam generator, is used to drive the turbine-generators. Eventually the steam is condensed to water at $27^oC$ and then heated again to start the cycle over. Calculate the maximum theoretical efficiency for a heat engine operating between these two temperatures. Strategy Since temperatures are given for the hot and cold reservoirs of this heat engine, $Eff_c = 1 - \dfrac{T_c}{T_h}$ can be used to calculate the Carnot (maximum theoretical) efficiency. Those temperatures must first be converted to kelvins. Solution The hot and cold reservoir temperatures are given as $300^oC$ and $27^oC$, respectively. In kelvins, then, $T_h = 573 \, K$ and $T_c = 300 \, K$, so that the maximum efficiency is \[Eff_c = 1 - \dfrac{T_c}{T_h}.\] Thus, \[Eff_c = 1 - \dfrac{300 \, K}{573 \, K}\]\[= 0.476, \, or \, 47.6 \%.\] Discussion A typical nuclear power station’s actual efficiency is about 35%, a little better than 0.7 times the maximum possible value, a tribute to superior engineering. Electrical power stations fired by coal, oil, and natural gas have greater actual efficiencies (about 42%), because their boilers can reach higher temperatures and pressures. The cold reservoir temperature in any of these power stations is limited by the local environment. Figure $\PageIndex{4}$ shows (a) the exterior of a nuclear power station and (b) the exterior of a coal-fired power station. Both have cooling towers into which water from the condenser enters the tower near the top and is sprayed downward, cooled by evaporation. Since all real processes are irreversible, the actual efficiency of a heat engine can never be as great as that of a Carnot engine, as illustrated in Figure $\PageIndex{5}$(a). Even with the best heat engine possible, there are always dissipative processes in peripheral equipment, such as electrical transformers or car transmissions. These further reduce the overall efficiency by converting some of the engine’s work output back into heat transfer, as shown in Figure $\PageIndex{5}$(b). Summary - The Carnot cycle is a theoretical cycle that is the most efficient cyclical process possible. Any engine using the Carnot cycle, which uses only reversible processes (adiabatic and isothermal), is known as a Carnot engine. - Any engine that uses the Carnot cycle enjoys the maximum theoretical efficiency. - While Carnot engines are ideal engines, in reality, no engine achieves Carnot’s theoretical maximum efficiency, since dissipative processes, such as friction, play a role. Carnot cycles without heat loss may be possible at absolute zero, but this has never been seen in nature. Glossary - Carnot cycle - a cyclical process that uses only reversible processes, the adiabatic and isothermal processes - Carnot engine - a heat engine that uses a Carnot cycle - Carnot efficiency - the maximum theoretical efficiency for a heat engine	libretexts	2025-03-17T19:53:33.673341	2015-11-01T05:44:18	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.04%3A_Carnots_Perfect_Heat_Engine-_The_Second_Law_of_Thermodynamics_Restated", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "15.4: Carnot’s Perfect Heat Engine- The Second Law of Thermodynamics Restated", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.05%3A_Applications_of_Thermodynamics-_Heat_Pumps_and_Refrigerators	15.5: Applications of Thermodynamics- Heat Pumps and Refrigerators Learning Objectives By the end of this section, you will be able to: - Describe the use of heat engines in heat pumps and refrigerators. - Demonstrate how a heat pump works to warm an interior space. - Explain the differences between heat pumps and refrigerators. - Calculate a heat pump’s coefficient of performance. Heat pumps, air conditioners, and refrigerators utilize heat transfer from cold to hot. They are heat engines run backward. We say backward, rather than reverse, because except for Carnot engines, all heat engines, though they can be run backward, cannot truly be reversed. Heat transfer occurs from a cold reservoir $Q_c$ and into a hot one. This requires work input $W$, which is also converted to heat transfer. Thus the heat transfer to the hot reservoir is $Q_h = Q_c + W$. (Note that $Q_h$, $Q_c$ and $W$ are positive, with their directions indicated on schematics rather than by sign.) A heat pump’s mission is for heat transfer $Q-h$ to occur into a warm environment, such as a home in the winter. The mission of air conditioners and refrigerators is for heat transfer $Q_c$ to occur from a cool environment, such as chilling a room or keeping food at lower temperatures than the environment. (Actually, a heat pump can be used both to heat and cool a space. It is essentially an air conditioner and a heating unit all in one. In this section we will concentrate on its heating mode.) Heat Pumps The great advantage of using a heat pump to keep your home warm, rather than just burning fuel, is that a heat pump supplies $Q_h = Q_c + W$. Heat transfer is from the outside air, even at a temperature below freezing, to the indoor space. You only pay for $W$, and you get an additional heat transfer of $Q_c$ from the outside at no cost; in many cases, at least twice as much energy is transferred to the heated space as is used to run the heat pump. When you burn fuel to keep warm, you pay for all of it. The disadvantage is that the work input (required by the second law of thermodynamics) is sometimes more expensive than simply burning fuel, especially if the work is done by electrical energy. The basic components of a heat pump in its heating mode are shown in Figure $\PageIndex{3}$. A working fluid such as a non-CFC refrigerant is used. In the outdoor coils (the evaporator), heat transfer $Q_c$ occurs to the working fluid from the cold outdoor air, turning it into a gas. The electrically driven compressor (work input $W$) raises the temperature and pressure of the gas and forces it into the condenser coils that are inside the heated space. Because the temperature of the gas is higher than the temperature inside the room, heat transfer to the room occurs and the gas condenses to a liquid. The liquid then flows back through a pressure-reducing valve to the outdoor evaporator coils, being cooled through expansion. (In a cooling cycle, the evaporator and condenser coils exchange roles and the flow direction of the fluid is reversed.) The quality of a heat pump is judged by how much heat transfer $Q_h$ occurs into the warm space compared with how much work input $W$ is required. In the spirit of taking the ratio of what you get to what you spend, we define a heat pump’s coefficient of performance $(COP_{hp})$ to be \[COP_{hp} = \dfrac{Q_h}{W}.\] Since the efficiency of a heat engine is $Eff = W/Q_h$, wee see that $COP_{hp} = 1/Eff$, an important and interesting fact. First, since the efficiency of any heat engine is less than 1, it means that $COP_{hp}$ is always greater than 1—that is, a heat pump always has more heat transfer $Q_h$ than work put into it. Second, it means that heat pumps work best when temperature differences are small. The efficiency of a perfect, or Carnot, engine is $Eff_c = 1 - (T_c?T_h)$ thus, the smaller the temperature difference, the smaller the efficiency and the greater the $COP_{hp}$ (because $COP_{hp} = 1/Eff$). In other words, heat pumps do not work as well in very cold climates as they do in more moderate climates. Friction and other irreversible processes reduce heat engine efficiency, but they do not benefit the operation of a heat pump—instead, they reduce the work input by converting part of it to heat transfer back into the cold reservoir before it gets into the heat pump. Example $\PageIndex{1}$: The Best $COP_{hp}$ of a Heat Pump for Home Use A heat pump used to warm a home must employ a cycle that produces a working fluid at temperatures greater than typical indoor temperature so that heat transfer to the inside can take place. Similarly, it must produce a working fluid at temperatures that are colder than the outdoor temperature so that heat transfer occurs from outside. Its hot and cold reservoir temperatures therefore cannot be too close, placing a limit on its $COP{hp}$. (See Figure $\PageIndex{5}$.) What is the best coefficient of performance possible for such a heat pump, if it has a hot reservoir temperature of $45.0^oC$ and a cold reservoir temperature of $-15.0^oC$? Strategy A Carnot engine reversed will give the best possible performance as a heat pump. As noted above, $COP_{hp} = 1/Eff$, so that we need to first calculate the Carnot efficiency to solve this problem. Solution Carnot efficiency in terms of absolute temperature is given by : \[Eff_c = 1 - \dfrac{T_c}{T_h}.\] The temperatures in kelvins are $T_h = 318 \, K$ and $T_c = 258 \, K$, so that \[Eff_c = 1 - \dfrac{258}{318} K = 0.1887.\] Thus, from the discussion above, \[COP_{hp} = \dfrac{1}{0.1887} = 5.30,\] or \[COP_{hp} = \dfrac{Q_h}{W} = 5.30,\] so that \[Q_h = 5.30 \, W.\] Discussion This result means that the heat transfer by the heat pump is 5.30 times as much as the work put into it. It would cost 5.30 times as much for the same heat transfer by an electric room heater as it does for that produced by this heat pump. This is not a violation of conservation of energy. Cold ambient air provides 4.3 J per 1 J of work from the electrical outlet. Real heat pumps do not perform quite as well as the ideal one in the previous example; their values of $COP_{hp}$ range from about 2 to 4. This range means that the heat transfer $Q_h$ from the heat pumps is 2 to 4 times as great as the work $W$ put into them. Their economical feasibility is still limited, however, since $W$ is usually supplied by electrical energy that costs more per joule than heat transfer by burning fuels like natural gas. Furthermore, the initial cost of a heat pump is greater than that of many furnaces, so that a heat pump must last longer for its cost to be recovered. Heat pumps are most likely to be economically superior where winter temperatures are mild, electricity is relatively cheap, and other fuels are relatively expensive. Also, since they can cool as well as heat a space, they have advantages where cooling in summer months is also desired. Thus some of the best locations for heat pumps are in warm summer climates with cool winters. Figure $\PageIndex{6}$ shows a heat pump, called a “ reverse cycle” or “ split-system cooler” in some countries. Air Conditioners and Refrigerators Air conditioners and refrigerators are designed to cool something down in a warm environment. As with heat pumps, work input is required for heat transfer from cold to hot, and this is expensive. The quality of air conditioners and refrigerators is judged by how much heat transfer $Q_c$ occurs from a cold environment compared with how much work input $W$ is required. What is considered the benefit in a heat pump is considered waste heat in a refrigerator. We thus define the coefficient of performance $COP_{ref}$ of an air conditioner or refrigerator to be \[COP_{ref} = \dfrac{Q_c}{W}.\] Noting again that $Q_h = Q_c + W$, we can see that an air conditioner will have a lower coefficient of performance than a heat pump, because $COP_{hp} = Q_h/W$ and $Q_h$ is greater than $Q_c$. In this module’s Problems and Exercises, you will show that \[COP_{ref} = COP_{hp} - 1\] for a heat engine used as either an air conditioner or a heat pump operating between the same two temperatures. Real air conditioners and refrigerators typically do remarkably well, having values of $COP_{ref}$ ranging from 2 to 6. These numbers are better than the $COP_{hp}$ values for the heat pumps mentioned above, because the temperature differences are smaller, but they are less than those for Carnot engines operating between the same two temperatures. A type of $COP$ rating system called the “energy efficiency rating” ($EER$) has been developed. This rating is an example where non-SI units are still used and relevant to consumers. To make it easier for the consumer, Australia, Canada, New Zealand, and the U.S. use an Energy Star Rating out of 5 stars—the more stars, the more energy efficient the appliance. $EER$s are expressed in mixed units of British thermal units (Btu) per hour of heating or cooling divided by the power input in watts. Room air conditioners are readily available with $EER$s ranging from 6 to 12. Although not the same as the $COP$ just described, these $EER$s are good for comparison purposes—the greater the $EER$, the cheaper an air conditioner is to operate (but the higher its purchase price is likely to be). The $EER$ of an air conditioner or refrigerator can be expressed as \[EER = \dfrac{Q_c/t_1}{W/t_2},\] where $Q_c$ is the amount of heat transfer from a cold environment in British thermal units, $t_1$ is time in hours, $W$ is the work input in joules, and $t_2$ is time in seconds. PROBLEM SOLVING STRATEGIES FOR THERMODYNAMICS - Examine the situation to determine whether heat, work, or internal energy are involved . Look for any system where the primary methods of transferring energy are heat and work. Heat engines, heat pumps, refrigerators, and air conditioners are examples of such systems. - Identify the system of interest and draw a labeled diagram of the system showing energy flow. - Identify exactly what needs to be determined in the problem (identify the unknowns) . A written list is useful. Maximum efficiency means a Carnot engine is involved. Efficiency is not the same as the coefficient of performance. - Make a list of what is given or can be inferred from the problem as stated (identify the knowns) . Be sure to distinguish heat transfer into a system from heat transfer out of the system, as well as work input from work output. In many situations, it is useful to determine the type of process, such as isothermal or adiabatic. - Solve the appropriate equation for the quantity to be determined (the unknown). - Substitute the known quantities along with their units into the appropriate equation and obtain numerical solutions complete with units. - Check the answer to see if it is reasonable: Does it make sense? For example, efficiency is always less than 1, whereas coefficients of performance are greater than 1. Summary - An artifact of the second law of thermodynamics is the ability to heat an interior space using a heat pump. Heat pumps compress cold ambient air and, in so doing, heat it to room temperature without violation of conservation principles. - To calculate the heat pump’s coefficient of performance, use the equation $COP_{hp} = \dfrac{Q_h}{W}$. - A refrigerator is a heat pump; it takes warm ambient air and expands it to chill it. Glossary - heat pump - a machine that generates heat transfer from cold to hot - coefficient of performance - for a heat pump, it is the ratio of heat transfer at the output (the hot reservoir) to the work supplied; for a refrigerator or air conditioner, it is the ratio of heat transfer from the cold reservoir to the work supplied	libretexts	2025-03-17T19:53:33.748473	2015-11-01T05:44:37	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.05%3A_Applications_of_Thermodynamics-_Heat_Pumps_and_Refrigerators", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "15.5: Applications of Thermodynamics- Heat Pumps and Refrigerators", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.06%3A_Entropy_and_the_Second_Law_of_Thermodynamics-_Disorder_and_the_Unavailability_of_Energy	15.6: Entropy and the Second Law of Thermodynamics- Disorder and the Unavailability of Energy Learning Objectives By the end of this section, you will be able to: - Define entropy. - Calculate the increase of entropy in a system with reversible and irreversible processes. - Explain the expected fate of the universe in entropic terms. - Calculate the increasing disorder of a system. There is yet another way of expressing the second law of thermodynamics. This version relates to a concept called entropy . By examining it, we shall see that the directions associated with the second law—heat transfer from hot to cold, for example—are related to the tendency in nature for systems to become disordered and for less energy to be available for use as work. The entropy of a system can in fact be shown to be a measure of its disorder and of the unavailability of energy to do work. MAKING CONNECTIONS: ENTROPY, ENERGY, AND WORK Recall that the simple definition of energy is the ability to do work. Entropy is a measure of how much energy is not available to do work. Although all forms of energy are interconvertible, and all can be used to do work, it is not always possible, even in principle, to convert the entire available energy into work. That unavailable energy is of interest in thermodynamics, because the field of thermodynamics arose from efforts to convert heat to work. We can see how entropy is defined by recalling our discussion of the Carnot engine. We noted that for a Carnot cycle, and hence for any reversible processes, $Q_c/Q_h = T_c/T_h$. Rearranging terms yields \[\dfrac{Q_c}{T_c} = \dfrac{Q_h}{T_h}\] for any reversible process. $Q_c$ and $Q_h$ are absolute values of the heat transfer at temperatures $T_c$ and $T_h$, respectively. This ratio of $Q/T$ is defined to be the change in entropy $\Delta S$ for a reversible process, \[\Delta S = \left(\dfrac{Q}{T} \right)_{rev},\] where $Q$ is the heat transfer, which is positive for heat transfer into and negative for heat transfer out of, and $T$ is the absolute temperature at which the reversible process takes place. The SI unit for entropy is joules per kelvin (J/K). If temperature changes during the process, then it is usually a good approximation (for small changes in temperature) to take $T$ to be the average temperature, avoiding the need to use integral calculus to find $\Delta S$. The definition of $\Delta S$ is strictly valid only for reversible processes, such as used in a Carnot engine. However, we can find $\Delta S$ precisely even for real, irreversible processes. The reason is that the entropy $S$ of a system, like internal energy $U$ depends only on the state of the system and not how it reached that condition. Entropy is a property of state. Thus the change in entropy $\Delta S$ of a system between state 1 and state 2 is the same no matter how the change occurs. We just need to find or imagine a reversible process that takes us from state 1 to state 2 and calculate $\Delta S$ for that process. That will be the change in entropy for any process going from state 1 to state 2. (See Figure $\PageIndex{2}$.) Now let us take a look at the change in entropy of a Carnot engine and its heat reservoirs for one full cycle. The hot reservoir has a loss of entropy $\Delta S_h = -Q_h/T_h$, because heat transfer occurs out of it (remember that when heat transfers out, then $Q$ has a negative sign). The cold reservoir has a gain of entropy $\Delta S_c = Q_c/T_c$, because heat transfer occurs into it. (We assume the reservoirs are sufficiently large that their temperatures are constant.) So the total change in entropy is \[\Delta S_{tot} = \Delta S_h + \Delta S_c.\] Thus, since we know that $Q_h/T_h = Q_c/T_c$ for a Carnot engine, \[\Delta S_{tot} = -\dfrac{Q_h}{T_h} + \dfrac{Q_c}{T_c} = 0.\] This result, which has general validity, means that the total change in entropy for a system in any reversible process is zero. The entropy of various parts of the system may change, but the total change is zero. Furthermore, the system does not affect the entropy of its surroundings, since heat transfer between them does not occur. Thus the reversible process changes neither the total entropy of the system nor the entropy of its surroundings. Sometimes this is stated as follows: Reversible processes do not affect the total entropy of the universe. Real processes are not reversible, though, and they do change total entropy. We can, however, use hypothetical reversible processes to determine the value of entropy in real, irreversible processes. The following example illustrates this point. Example $\PageIndex{1}$: Entropy Increases in an Irreversible (Real) Process Spontaneous heat transfer from hot to cold is an irreversible process. Calculate the total change in entropy if 4000 J of heat transfer occurs from a hot reservoir at $T_h = 600 \, K \, (327^oC) $ to a cold reservoir at $T_c = 250 \, K \, (-23^oC)$, assuming there is no temperature change in either reservoir. (See Figure $\PageIndex{3}$.) Strategy How can we calculate the change in entropy for an irreversible process when $\Delta S_{tot} = \Delta S_h + \Delta S_c$ is valid only for reversible processes? Remember that the total change in entropy of the hot and cold reservoirs will be the same whether a reversible or irreversible process is involved in heat transfer from hot to cold. So we can calculate the change in entropy of the hot reservoir for a hypothetical reversible process in which 4000 J of heat transfer occurs from it; then we do the same for a hypothetical reversible process in which 4000 J of heat transfer occurs to the cold reservoir. This produces the same changes in the hot and cold reservoirs that would occur if the heat transfer were allowed to occur irreversibly between them, and so it also produces the same changes in entropy. Solution We now calculate the two changes in entropy using $\Delta S_{tot} = \Delta S_h + \Delta S_c$. First, for the heat transfer from the hot reservoir, \[\Delta S_h = \dfrac{-Q_h}{T_h} = \dfrac{-4000 \, J}{600 \, K} = -6.67 \, J/K.\] And for the cold reservoir, \[\Delta S_c = \dfrac{Q_c}{T_c} = \dfrac{4000 \, J}{250 \, K} = 16.0 \, J/K.\] Thus the total is \[\Delta S_{tot} = \Delta S_h + \Delta S_c\] \[= (-6.67 + 16.0) \, J/K\] \[= 9.33 \, J/K.\] Discussion There is an increase in entropy for the system of two heat reservoirs undergoing this irreversible heat transfer. We will see that this means there is a loss of ability to do work with this transferred energy. Entropy has increased, and energy has become unavailable to do work. It is reasonable that entropy increases for heat transfer from hot to cold. Since the change in entropy is $Q/T$ there is a larger change at lower temperatures. The decrease in entropy of the hot object is therefore less than the increase in entropy of the cold object, producing an overall increase, just as in the previous example. This result is very general: There is an increase in entropy for any system undergoing an irreversible process. With respect to entropy, there are only two possibilities: entropy is constant for a reversible process, and it increases for an irreversible process. There is a fourth version of the second law of thermodynamics stated in terms of entropy : The total entropy of a system either increases or remains constant in any process; it never decreases. For example, heat transfer cannot occur spontaneously from cold to hot, because entropy would decrease. Entropy is very different from energy. Entropy is not conserved but increases in all real processes. Reversible processes (such as in Carnot engines) are the processes in which the most heat transfer to work takes place and are also the ones that keep entropy constant. Thus we are led to make a connection between entropy and the availability of energy to do work. Entropy and the Unavailability of Energy to Do Work What does a change in entropy mean, and why should we be interested in it? One reason is that entropy is directly related to the fact that not all heat transfer can be converted into work. The next example gives some indication of how an increase in entropy results in less heat transfer into work. Example $\PageIndex{2}$: Less Work is Produced by a Given Heat Transfer When Entropy Change is Greater (a) Calculate the work output of a Carnot engine operating between temperatures of 600 K and 100 K for 4000 J of heat transfer to the engine. (b) Now suppose that the 4000 J of heat transfer occurs first from the 600 K reservoir to a 250 K reservoir (without doing any work, and this produces the increase in entropy calculated above) before transferring into a Carnot engine operating between 250 K and 100 K. What work output is produced? (See Figure $\PageIndex{4}$.) Strategy In both parts, we must first calculate the Carnot efficiency and then the work output. Solution (a) The Carnot efficiency is given by \[Eff_c = 1 - \dfrac{T_c}{T_h}.\] Substituting the given temperatures yields \[Eff_c = 1 - \dfrac{100 \, K}{600 \, K} = 0.833.\] Now the work output can be calculated using the definition of efficiency for any heat engine as given by \[Eff = \dfrac{W}{Q_h}.\] Solving for $W$ and substituting known terms gives \[W = Eff_cQ_h\]\[= (0.833)(4000 \, J) = 3333 \, J.\] Solution (b) Similarly, \[Eff_c = 1 - \dfrac{T_c}{T'_c} = 1 - \dfrac{100 \, K}{250 \, K} = 0.600,\] so that \[W = Eff_cQ_h\] \[ = (0.600)(4000 \, J) = 2400 \, J\] Discussion There is 933 J less work from the same heat transfer in the second process. This result is important. The same heat transfer into two perfect engines produces different work outputs, because the entropy change differs in the two cases. In the second case, entropy is greater and less work is produced. Entropy is associated with the un availability of energy to do work. When entropy increases, a certain amount of energy becomes permanently unavailable to do work. The energy is not lost, but its character is changed, so that some of it can never be converted to doing work—that is, to an organized force acting through a distance. For instance, in the previous example, 933 J less work was done after an increase in entropy of 9.33 J/K occurred in the 4000 J heat transfer from the 600 K reservoir to the 250 K reservoir. It can be shown that the amount of energy that becomes unavailable for work is \[W_{unavail} = \Delta S \cdot T_0,\] where $T_0$ is the lowest temperature utilized. In the previous example, \[W_{unavail} = (9.33 \, J/K)(100 \, K) = 933 \, J\] Heat Death of the Universe: An Overdose of Entropy In the early, energetic universe, all matter and energy were easily interchangeable and identical in nature. Gravity played a vital role in the young universe. Although it may have seemed disorderly, and therefore, superficially entropic, in fact, there was enormous potential energy available to do work—all the future energy in the universe. As the universe matured, temperature differences arose, which created more opportunity for work. Stars are hotter than planets, for example, which are warmer than icy asteroids, which are warmer still than the vacuum of the space between them. Most of these are cooling down from their usually violent births, at which time they were provided with energy of their own—nuclear energy in the case of stars, volcanic energy on Earth and other planets, and so on. Without additional energy input, however, their days are numbered. As entropy increases, less and less energy in the universe is available to do work. On Earth, we still have great stores of energy such as fossil and nuclear fuels; large-scale temperature differences, which can provide wind energy; geothermal energies due to differences in temperature in Earth’s layers; and tidal energies owing to our abundance of liquid water. As these are used, a certain fraction of the energy they contain can never be converted into doing work. Eventually, all fuels will be exhausted, all temperatures will equalize, and it will be impossible for heat engines to function, or for work to be done. Entropy increases in a closed system, such as the universe. But in parts of the universe, for instance, in the Solar system, it is not a locally closed system. Energy flows from the Sun to the planets, replenishing Earth’s stores of energy. The Sun will continue to supply us with energy for about another five billion years. We will enjoy direct solar energy, as well as side effects of solar energy, such as wind power and biomass energy from photosynthetic plants. The energy from the Sun will keep our water at the liquid state, and the Moon’s gravitational pull will continue to provide tidal energy. But Earth’s geothermal energy will slowly run down and won’t be replenished. But in terms of the universe, and the very long-term, very large-scale picture, the entropy of the universe is increasing, and so the availability of energy to do work is constantly decreasing. Eventually, when all stars have died, all forms of potential energy have been utilized, and all temperatures have equalized (depending on the mass of the universe, either at a very high temperature following a universal contraction, or a very low one, just before all activity ceases) there will be no possibility of doing work. Either way, the universe is destined for thermodynamic equilibrium—maximum entropy. This is often called the heat death of the universe , and will mean the end of all activity. However, whether the universe contracts and heats up, or continues to expand and cools down, the end is not near. Calculations of black holes suggest that entropy can easily continue for at least $10^{100}$ years. Order to Disorder Entropy is related not only to the unavailability of energy to do work—it is also a measure of disorder. This notion was initially postulated by Ludwig Boltzmann in the 1800s. For example, melting a block of ice means taking a highly structured and orderly system of water molecules and converting it into a disorderly liquid in which molecules have no fixed positions. (See Figure $\PageIndex{5}$.) There is a large increase in entropy in the process, as seen in the following example. Example $\PageIndex{3}$: Entropy Associated with Disorder Find the increase in entropy of 1.00 kg of ice originally at $0^oC$, that is melted to form water at $0^oC$. Strategy As before, the change in entropy can be calculated from the definition of $\Delta S$ once we find the energy $Q$ needed to melt the ice. Solution The change in entropy is defined as: \[\Delta S = \dfrac{Q}{T}.\] Here $Q$ is the heat transfer necessary to melt 1.00 kg of ice and is given by \[Q = mL_f,\] where $m$ is the mass and $L_f$ is the latent heat of fusion. $L_f = 334 \, kJ/kg$ for water, so that \[Q = 1.00 \, kg)(334 \, kJ/kg) = 3.34 \times 10^5 \, J.\] Now the change in entropy is positive, since heat transfer occurs into the ice to cause the phase change; thus, \[\Delta S = \dfrac{Q}{T} = \dfrac{3.34 \times 10^5 \, J}{T}.\] $T$ is the melting temperature of ice. That is $T = 0^oC = 273 \, K$. So the change in entropy is \[\Delta S = \dfrac{3.34 \times 10^5 \, J}{273 \, K}\]\[ = 1.22 \times 10^3 \, J/K.\] Discussion This is a significant increase in entropy accompanying an increase in disorder. In another easily imagined example, suppose we mix equal masses of water originally at two different temperatures, say $20.0^oC$ and $40.0^oC$. The result is water at an intermediate temperature of $30.0^oC$. Three outcomes have resulted: entropy has increased, some energy has become unavailable to do work, and the system has become less orderly. Let us think about each of these results. First, entropy has increased for the same reason that it did in the example above. Mixing the two bodies of water has the same effect as heat transfer from the hot one and the same heat transfer into the cold one. The mixing decreases the entropy of the hot water but increases the entropy of the cold water by a greater amount, producing an overall increase in entropy. Second, once the two masses of water are mixed, there is only one temperature—you cannot run a heat engine with them. The energy that could have been used to run a heat engine is now unavailable to do work. Third, the mixture is less orderly, or to use another term, less structured. Rather than having two masses at different temperatures and with different distributions of molecular speeds, we now have a single mass with a uniform temperature. These three results—entropy, unavailability of energy, and disorder—are not only related but are in fact essentially equivalent. Life, Evolution, and the Second Law of Thermodynamics Some people misunderstand the second law of thermodynamics, stated in terms of entropy, to say that the process of the evolution of life violates this law. Over time, complex organisms evolved from much simpler ancestors, representing a large decrease in entropy of the Earth’s biosphere. It is a fact that living organisms have evolved to be highly structured, and much lower in entropy than the substances from which they grow. But it is always possible for the entropy of one part of the universe to decrease, provided the total change in entropy of the universe increases. In equation form, we can write this as \[\Delta S_{tot} = \Delta S_{syst} + \Delta S_{envir} > 0.\] Thus $\Delta S_{yst}$ can be negative as long as $\Delta S_{envir}$ is positive and greater in magnitude. How is it possible for a system to decrease its entropy? Energy transfer is necessary. If I pick up marbles that are scattered about the room and put them into a cup, my work has decreased the entropy of that system. If I gather iron ore from the ground and convert it into steel and build a bridge, my work has decreased the entropy of that system. Energy coming from the Sun can decrease the entropy of local systems on Earth—that is, $\Delta S_{syst}$ is negative. But the overall entropy of the rest of the universe increases by a greater amount—that is, $\Delta S_{envir}$ is positive and greater in magnitude. Thus, $\Delta S_{tot} = \Delta S_{syst} + \Delta S_{envir} > 0 $, and the second law of thermodynamics is not violated. Every time a plant stores some solar energy in the form of chemical potential energy, or an updraft of warm air lifts a soaring bird, the Earth can be viewed as a heat engine operating between a hot reservoir supplied by the Sun and a cold reservoir supplied by dark outer space—a heat engine of high complexity, causing local decreases in entropy as it uses part of the heat transfer from the Sun into deep space. There is a large total increase in entropy resulting from this massive heat transfer. A small part of this heat transfer is stored in structured systems on Earth, producing much smaller local decreases in entropy. (See Figure $\PageIndex{6}$.) PHET EXPLORATIONS: REVERSIBLE REACTIONS Watch a reaction proceed over time. How does total energy affect a reaction rate? Vary temperature, barrier height, and potential energies. Record concentrations and time in order to extract rate coefficients. Do temperature dependent studies to extract Arrhenius parameters. This simulation is best used with teacher guidance because it presents an analogy of chemical reactions. $\PageIndex{7}$: Reversible Reaction Summary - Entropy is the loss of energy available to do work. - Another form of the second law of thermodynamics states that the total entropy of a system either increases or remains constant; it never decreases. - Entropy is zero in a reversible process; it increases in an irreversible process. - The ultimate fate of the universe is likely to be thermodynamic equilibrium, where the universal temperature is constant and no energy is available to do work. - Entropy is also associated with the tendency toward disorder in a closed system. Glossary - entropy - a measurement of a system's disorder and its inability to do work in a system - change in entropy - the ratio of heat transfer to temperature $Q/T$ - second law of thermodynamics stated in terms of entropy - the total entropy of a system either increases or remains constant; it never decrease	libretexts	2025-03-17T19:53:33.838118	2015-11-01T05:44:56	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.06%3A_Entropy_and_the_Second_Law_of_Thermodynamics-_Disorder_and_the_Unavailability_of_Energy", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "15.6: Entropy and the Second Law of Thermodynamics- Disorder and the Unavailability of Energy", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.07%3A_Statistical_Interpretation_of_Entropy_and_the_Second_Law_of_Thermodynamics-_The_Underlying_Explanation	15.7: Statistical Interpretation of Entropy and the Second Law of Thermodynamics- The Underlying Explanation Learning Objectives By the end of this section, you will be able to: - Identify probabilities in entropy. - Analyze statistical probabilities in entropic systems. The various ways of formulating the second law of thermodynamics tell what happens rather than why it happens. Why should heat transfer occur only from hot to cold? Why should energy become ever less available to do work? Why should the universe become increasingly disorderly? The answer is that it is a matter of overwhelming probability. Disorder is simply vastly more likely than order. When you watch an emerging rain storm begin to wet the ground, you will notice that the drops fall in a disorganized manner both in time and in space. Some fall close together, some far apart, but they never fall in straight, orderly rows. It is not impossible for rain to fall in an orderly pattern, just highly unlikely, because there are many more disorderly ways than orderly ones. To illustrate this fact, we will examine some random processes, starting with coin tosses. Coin Tosses What are the possible outcomes of tossing 5 coins? Each coin can land either heads or tails. On the large scale, we are concerned only with the total heads and tails and not with the order in which heads and tails appear. The following possibilities exist: \[5 \, heads, \, 0 \, tails\] \[4 \, heads, \, 1 \, tail\] \[3 \, heads, \, 2 \, tails\] \[2 \, heads, \, 3 \, tails\] \[1 \, head, \, 4 \, tails\] \[0 \, heads, \, 5 \, tails\] These are what we call macrostates . A macrostate is an overall property of a system. It does not specify the details of the system, such as the order in which heads and tails occur or which coins are heads or tails. Using this nomenclature, a system of 5 coins has the 6 possible macrostates just listed. Some macrostates are more likely to occur than others. For instance, there is only one way to get 5 heads, but there are several ways to get 3 heads and 2 tails, making the latter macrostate more probable. Table $\PageIndex{1}$ lists of all the ways in which 5 coins can be tossed, taking into account the order in which heads and tails occur. Each sequence is called a microstate —a detailed description of every element of a system. \| Individual microstates \| Number of microstates \| \| \|---\|---\|---\| \| 5 heads, 0 tails \| HHHHH \| 1 \| \| 4 heads, 1 tail \| HHHHT, HHHTH, HHTHH, HTHHH, THHHH \| 5 \| \| 3 heads, 2 tails \| HTHTH, THTHH, HTHHT, THHTH, THHHT HTHTH, THTHH, HTHHT, THHTH, THHHT \| 10 \| \| 2 heads, 3 tails \| TTTHH, TTHHT, THHTT, HHTTT, TTHTH, THTHT, HTHTT, THTTH, HTTHT, HTTTH \| 10 \| \| 1 head, 4 tails \| TTTTH, TTTHT, TTHTT, THTTT, HTTTT \| 5 \| \| 0 heads, 5 tails \| TTTTT \| 1 \| \| Total: 32 \| The macrostate of 3 heads and 2 tails can be achieved in 10 ways and is thus 10 times more probable than the one having 5 heads. Not surprisingly, it is equally probable to have the reverse, 2 heads and 3 tails. Similarly, it is equally probable to get 5 tails as it is to get 5 heads. Note that all of these conclusions are based on the crucial assumption that each microstate is equally probable. With coin tosses, this requires that the coins not be asymmetric in a way that favors one side over the other, as with loaded dice. With any system, the assumption that all microstates are equally probable must be valid, or the analysis will be erroneous. The two most orderly possibilities are 5 heads or 5 tails. (They are more structured than the others.) They are also the least likely, only 2 out of 32 possibilities. The most disorderly possibilities are 3 heads and 2 tails and its reverse. (They are the least structured.) The most disorderly possibilities are also the most likely, with 20 out of 32 possibilities for the 3 heads and 2 tails and its reverse. If we start with an orderly array like 5 heads and toss the coins, it is very likely that we will get a less orderly array as a result, since 30 out of the 32 possibilities are less orderly. So even if you start with an orderly state, there is a strong tendency to go from order to disorder, from low entropy to high entropy. The reverse can happen, but it is unlikely. \| Macrostate \| Number of Microstates \| \| \|---\|---\|---\| \| Heads \| Tails \| ( W ) \| \| 100 \| 0 \| 1 \| \| 99 \| 1 \| 100 \| \| 95 \| 5 \| $7.5 \times 10^7$ \| \| 90 \| 10 \| $1.7 \times 10^{13}$ \| \| 75 \| 25 \| $2.4 \times 10^{23}$ \| \| 60 \| 40 \| $1.4 \times 10^{28}$ \| \| 55 \| 45 \| $6.1 \times 10^{28}$ \| \| 51 \| 49 \| $9.9 \times 10^{28}$ \| \| 50 \| 50 \| $1.0 \times 10^{29}$ \| \| 49 \| 51 \| $9.9 \times 10^{28}$ \| \| 45 \| 55 \| $6.1 \times 10^{28}$ \| \| 40 \| 60 \| $1.4 \times 10^{28}$ \| \| 25 \| 75 \| $2.4 \times 10^{23}$ \| \| 10 \| 90 \| $1.7 \times 10^{13}$ \| \| 5 \| 95 \| $7.5 \times 10^7$ \| \| 1 \| 99 \| 100 \| \| 0 \| 100 \| 1 \| \| Total: \| $1.27 \times 10^{30}$ \| This result becomes dramatic for larger systems. Consider what happens if you have 100 coins instead of just 5. The most orderly arrangements (most structured) are 100 heads or 100 tails. The least orderly (least structured) is that of 50 heads and 50 tails. There is only 1 way (1 microstate) to get the most orderly arrangement of 100 heads. There are 100 ways (100 microstates) to get the next most orderly arrangement of 99 heads and 1 tail (also 100 to get its reverse). And there are $1 \times 10^{29} ways to get 50 heads and 50 tails, the least orderly arrangement. Table \(\PageIndex{2}$ is an abbreviated list of the various macrostates and the number of microstates for each macrostate. The total number of microstates—the total number of different ways 100 coins can be tossed—is an impressively large $1.27 \times 10^{30}$. Now, if we start with an orderly macrostate like 100 heads and toss the coins, there is a virtual certainty that we will get a less orderly macrostate. If we keep tossing the coins, it is possible, but exceedingly unlikely, that we will ever get back to the most orderly macrostate. If you tossed the coins once each second, you could expect to get either 100 heads or 100 tails once in $2 \times 10^{2}$ years! This period is 1 trillion ($10^{12}$ times longer than the age of the universe, and so the chances are essentially zero. In contrast, there is an 8% chance of getting 50 heads, a 73% chance of getting from 45 to 55 heads, and a 96% chance of getting from 40 to 60 heads. Disorder is highly likely. Disorder in a Gas The fantastic growth in the odds favoring disorder that we see in going from 5 to 100 coins continues as the number of entities in the system increases. Let us now imagine applying this approach to perhaps a small sample of gas. Because counting microstates and macrostates involves statistics, this is called statistical analysis . The macrostates of a gas correspond to its macroscopic properties, such as volume, temperature, and pressure; and its microstates correspond to the detailed description of the positions and velocities of its atoms. Even a small amount of gas has a huge number of atoms: $1.0 \, cm^3$ of an ideal gas at 1.0 atm and $0^oC$ has $2.7 \times 10^{19}$ atoms. So each macrostate has an immense number of microstates. In plain language, this means that there are an immense number of ways in which the atoms in a gas can be arranged, while still having the same pressure, temperature, and so on. The most likely conditions (or macrostates) for a gas are those we see all the time—a random distribution of atoms in space with a Maxwell-Boltzmann distribution of speeds in random directions, as predicted by kinetic theory. This is the most disorderly and least structured condition we can imagine. In contrast, one type of very orderly and structured macrostate has all of the atoms in one corner of a container with identical velocities. There are very few ways to accomplish this (very few microstates corresponding to it), and so it is exceedingly unlikely ever to occur. (See Figure $\PageIndex{2}$(b).) Indeed, it is so unlikely that we have a law saying that it is impossible, which has never been observed to be violated—the second law of thermodynamics. The disordered condition is one of high entropy, and the ordered one has low entropy. With a transfer of energy from another system, we could force all of the atoms into one corner and have a local decrease in entropy, but at the cost of an overall increase in entropy of the universe. If the atoms start out in one corner, they will quickly disperse and become uniformly distributed and will never return to the orderly original state (Figure $\PageIndex{2}$(b)). Entropy will increase. With such a large sample of atoms, it is possible—but unimaginably unlikely—for entropy to decrease. Disorder is vastly more likely than order. The arguments that disorder and high entropy are the most probable states are quite convincing. The great Austrian physicist Ludwig Boltzmann (1844–1906)—who, along with Maxwell, made so many contributions to kinetic theory—proved that the entropy of a system in a given state (a macrostate) can be written as \[S = klnW,\] where $k = 1.38 \times 10^{-23} \, J/K$ is Boltzmann’s constant, and $lnW$ is the natural logarithm of the number of microstates $W$ corresponding to the given macrostate. $W$ is proportional to the probability that the macrostate will occur. Thus entropy is directly related to the probability of a state—the more likely the state, the greater its entropy. Boltzmann proved that this expression for $S$ is equivalent to the definition $\Delta S = Q/T$ which we have used extensively. Thus the second law of thermodynamics is explained on a very basic level: entropy either remains the same or increases in every process. This phenomenon is due to the extraordinarily small probability of a decrease, based on the extraordinarily larger number of microstates in systems with greater entropy. Entropy can decrease, but for any macroscopic system, this outcome is so unlikely that it will never be observed. Example $\PageIndex{1}$: Entropy Increases in a Coin Toss Suppose you toss 100 coins starting with 60 heads and 40 tails, and you get the most likely result, 50 heads and 50 tails. What is the change in entropy? Strategy Noting that the number of microstates is labeled $W$ in Table $\PageIndex{2}$ for the 100-coin toss, we can use $\Delta S = S_f - S_i = klnW_f - klnW_i$ to calculate the change in entropy. Solution The change in entropy is \[\Delta S = S_f - S_i = klnW_f - klnW_i,\] where the subscript i stands for the initial 60 heads and 40 tails state, and the subscript f for the final 50 heads and 50 tails state. Substituting the values for $W$ from Table $\PageIndex{2}$ gives \[\Delta S = (1.38 \times 10^{-23} \, J/K)[ln(1.0 \times 10^{29}) - ln(1.4 \times 10^{28})]\]\[= 2.7 \times 10^{-23} \, J/K\] Discussion This increase in entropy means we have moved to a less orderly situation. It is not impossible for further tosses to produce the initial state of 60 heads and 40 tails, but it is less likely. There is about a 1 in 90 chance for that decrease in entropy $(-2.7 \times 10^{-23} \, J/K)$ to occur. If we calculate the decrease in entropy to move to the most orderly state, we get $\Delta S = -92 \times 10^{-23} \, J/K$. There is about 1 in $10^{30}$ chance of this change occurring. So while very small decreases in entropy are unlikely, slightly greater decreases are impossibly unlikely. These probabilities imply, again, that for a macroscopic system, a decrease in entropy is impossible. For example, for heat transfer to occur spontaneously from 1.00 kg of $0^oC$ ice to its $0^oC$ environment, there would be a decrease in entropy of $1.22 \times 10^3 \, J/K$. Given that a $\Delta S$ of $10^{-21} \, J/K$ corresponds to about a 1 in $10^{30}$ chance, a decrease of this size $(10^3 \, J/K)$ is an utter impossibility. Even for a milligram of melted ice to spontaneously refreeze is impossible. PROBLEM-SOLVING STRATEGIES FOR ENTROPY - Examine the situation to determine if entropy is involved. - Identify the system of interest and draw a labeled diagram of the system showing energy flow. - Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. - Make a list of what is given or can be inferred from the problem as stated (identify the knowns). You must carefully identify the heat transfer, if any, and the temperature at which the process takes place. It is also important to identify the initial and final states. - Solve the appropriate equation for the quantity to be determined (the unknown). Note that the change in entropy can be determined between any states by calculating it for a reversible process. - Substitute the known value along with their units into the appropriate equation, and obtain numerical solutions complete with units. - To see if it is reasonable: Does it make sense? For example, total entropy should increase for any real process or be constant for a reversible process. Disordered states should be more probable and have greater entropy than ordered states. Summary - Disorder is far more likely than order, which can be seen statistically. - The entropy of a system in a given state (a macrostate) can be written as $s = KLNw,$ where $k = 1.38 \times 10^{-23} \, J/K$ is Boltzmann’s constant, and $lnW$ is the natural logarithm of the number of microstates $W$ corresponding to the given macrostate. Glossary - macrostate - an overall property of a system - microstate - each sequence within a larger macrostate - statistical analysis - using statistics to examine data, such as counting microstates and macrostates	libretexts	2025-03-17T19:53:33.929801	2015-11-01T05:45:10	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.07%3A_Statistical_Interpretation_of_Entropy_and_the_Second_Law_of_Thermodynamics-_The_Underlying_Explanation", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "15.7: Statistical Interpretation of Entropy and the Second Law of Thermodynamics- The Underlying Explanation", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.E%3A_Heat_and_Heat_Transfer_Methods_(Exercise)	15.E: Heat and Heat Transfer Methods (Exercise) - - Last updated - Save as PDF Conceptual Questions 15.1: The First Law of Thermodynamics 1. Describe the photo of the tea kettle at the beginning of this section in terms of heat transfer, work done, and internal energy. How is heat being transferred? What is the work done and what is doing it? How does the kettle maintain its internal energy? 2. The first law of thermodynamics and the conservation of energy, as discussed in Conservation of Energy, are clearly related. How do they differ in the types of energy considered? 3. Heat transfer $\displaystyle Q$ and work done $\displaystyle W$ are always energy in transit, whereas internal energy $\displaystyle U$ is energy stored in a system. Give an example of each type of energy, and state specifically how it is either in transit or resides in a system. 4. How do heat transfer and internal energy differ? In particular, which can be stored as such in a system and which cannot? 5. If you run down some stairs and stop, what happens to your kinetic energy and your initial gravitational potential energy? 6. Give an explanation of how food energy (calories) can be viewed as molecular potential energy (consistent with the atomic and molecular definition of internal energy). 7. Identify the type of energy transferred to your body in each of the following as either internal energy, heat transfer, or doing work: (a) basking in sunlight; (b) eating food; (c) riding an elevator to a higher floor. 15.2: The First Law of Thermodynamics and Some Simple Processes 8. A great deal of effort, time, and money has been spent in the quest for the so-called perpetual-motion machine, which is defined as a hypothetical machine that operates or produces useful work indefinitely and/or a hypothetical machine that produces more work or energy than it consumes. Explain, in terms of heat engines and the first law of thermodynamics, why or why not such a machine is likely to be constructed. 9. One method of converting heat transfer into doing work is for heat transfer into a gas to take place, which expands, doing work on a piston, as shown in the figure below. (a) Is the heat transfer converted directly to work in an isobaric process, or does it go through another form first? Explain your answer. (b) What about in an isothermal process? (c) What about in an adiabatic process (where heat transfer occurred prior to the adiabatic process)? 10. Would the previous question make any sense for an isochoric process? Explain your answer. 11. We ordinarily say that $\displaystyle ΔU=0$ for an isothermal process. Does this assume no phase change takes place? Explain your answer. 12. The temperature of a rapidly expanding gas decreases. Explain why in terms of the first law of thermodynamics. (Hint: Consider whether the gas does work and whether heat transfer occurs rapidly into the gas through conduction.) 13. Which cyclical process represented by the two closed loops, ABCFA and ABDEA, on the $\displaystyle PV$ diagram in the figure below produces the greatest net work? Is that process also the one with the smallest work input required to return it to point A? Explain your responses. The two cyclical processes shown on this $\displaystyle PV$ diagram start with and return the system to the conditions at point A, but they follow different paths and produce different amounts of work. 14. A real process may be nearly adiabatic if it occurs over a very short time. How does the short time span help the process to be adiabatic? 15. It is unlikely that a process can be isothermal unless it is a very slow process. Explain why. Is the same true for isobaric and isochoric processes? Explain your answer. 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency 16. Imagine you are driving a car up Pike’s Peak in Colorado. To raise a car weighing 1000 kilograms a distance of 100 meters would require about a million joules. You could raise a car 12.5 kilometers with the energy in a gallon of gas. Driving up Pike's Peak (a mere 3000-meter climb) should consume a little less than a quart of gas. But other considerations have to be taken into account. Explain, in terms of efficiency, what factors may keep you from realizing your ideal energy use on this trip. 17. Is a temperature difference necessary to operate a heat engine? State why or why not. 18. Definitions of efficiency vary depending on how energy is being converted. Compare the definitions of efficiency for the human body and heat engines. How does the definition of efficiency in each relate to the type of energy being converted into doing work? 19. Why—other than the fact that the second law of thermodynamics says reversible engines are the most efficient—should heat engines employing reversible processes be more efficient than those employing irreversible processes? Consider that dissipative mechanisms are one cause of irreversibility. 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated 20. Think about the drinking bird at the beginning of this section (Figure). Although the bird enjoys the theoretical maximum efficiency possible, if left to its own devices over time, the bird will cease “drinking.” What are some of the dissipative processes that might cause the bird’s motion to cease? 21. Can improved engineering and materials be employed in heat engines to reduce heat transfer into the environment? Can they eliminate heat transfer into the environment entirely? 22. Does the second law of thermodynamics alter the conservation of energy principle? 15.5: Applications of Thermodynamics: Heat Pumps and Refrigerators 23. Explain why heat pumps do not work as well in very cold climates as they do in milder ones. Is the same true of refrigerators? 24. In some Northern European nations, homes are being built without heating systems of any type. They are very well insulated and are kept warm by the body heat of the residents. However, when the residents are not at home, it is still warm in these houses. What is a possible explanation? 25. Why do refrigerators, air conditioners, and heat pumps operate most cost-effectively for cycles with a small difference between $\displaystyle T_h$ and $\displaystyle T_c$? (Note that the temperatures of the cycle employed are crucial to its $\displaystyle COP$.) 26. Grocery store managers contend that there is less total energy consumption in the summer if the store is kept at a low temperature. Make arguments to support or refute this claim, taking into account that there are numerous refrigerators and freezers in the store. 27. Can you cool a kitchen by leaving the refrigerator door open? 15.6: Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy 28. A woman shuts her summer cottage up in September and returns in June. No one has entered the cottage in the meantime. Explain what she is likely to find, in terms of the second law of thermodynamics. 29. Consider a system with a certain energy content, from which we wish to extract as much work as possible. Should the system’s entropy be high or low? Is this orderly or disorderly? Structured or uniform? Explain briefly. 30. Does a gas become more orderly when it liquefies? Does its entropy change? If so, does the entropy increase or decrease? Explain your answer. 31. Explain how water’s entropy can decrease when it freezes without violating the second law of thermodynamics. Specifically, explain what happens to the entropy of its surroundings. 32. Is a uniform-temperature gas more or less orderly than one with several different temperatures? Which is more structured? In which can heat transfer result in work done without heat transfer from another system? 33. Give an example of a spontaneous process in which a system becomes less ordered and energy becomes less available to do work. What happens to the system’s entropy in this process? 34. What is the change in entropy in an adiabatic process? Does this imply that adiabatic processes are reversible? Can a process be precisely adiabatic for a macroscopic system? 35. Does the entropy of a star increase or decrease as it radiates? Does the entropy of the space into which it radiates (which has a temperature of about 3 K) increase or decrease? What does this do to the entropy of the universe? 36. Explain why a building made of bricks has smaller entropy than the same bricks in a disorganized pile. Do this by considering the number of ways that each could be formed (the number of microstates in each macrostate). 15.7: Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation 37. Explain why a building made of bricks has smaller entropy than the same bricks in a disorganized pile. Do this by considering the number of ways that each could be formed (the number of microstates in each macrostate). Problems & Exercises 15.1: The First Law of Thermodynamics 38. What is the change in internal energy of a car if you put 12.0 gal of gasoline into its tank? The energy content of gasoline is $\displaystyle 1.3×10^8J/gal$. All other factors, such as the car’s temperature, are constant. Solution $\displaystyle 1.6×10^9J$ 39. How much heat transfer occurs from a system, if its internal energy decreased by 150 J while it was doing 30.0 J of work? 40. A system does $\displaystyle 1.80×10^8J$ of work while $\displaystyle 7.50×10^8J$ of heat transfer occurs to the environment. What is the change in internal energy of the system assuming no other changes (such as in temperature or by the addition of fuel)? Solution $\displaystyle −9.30×10^8J$ 41. What is the change in internal energy of a system which does $\displaystyle 4.50×10^5J$ of work while $\displaystyle 3.00×10^6J$ of heat transfer occurs into the system, and $\displaystyle 8.00×10^6J$ of heat transfer occurs to the environment? 42. Suppose a woman does 500 J of work and 9500 J of heat transfer occurs into the environment in the process. (a) What is the decrease in her internal energy, assuming no change in temperature or consumption of food? (That is, there is no other energy transfer.) (b) What is her efficiency? Solution (a) $\displaystyle −1.0×10^4J$, or $\displaystyle −2.39 kcal$ (b) 5.00% 43. (a) How much food energy will a man metabolize in the process of doing 35.0 kJ of work with an efficiency of 5.00%? (b) How much heat transfer occurs to the environment to keep his temperature constant? Explicitly show how you follow the steps in the Problem-Solving Strategy for thermodynamics found in Problem-Solving Strategies for Thermodynamics. 44. (a) What is the average metabolic rate in watts of a man who metabolizes 10,500 kJ of food energy in one day? (b) What is the maximum amount of work in joules he can do without breaking down fat, assuming a maximum efficiency of 20.0%? (c) Compare his work output with the daily output of a 187-W (0.250-horsepower) motor. Solution (a) 122 W (b) $\displaystyle 2.10×10^6J$ (c) Work done by the motor is $\displaystyle 1.61×10^7$; thus the motor produces 7.67 times the work done by the man 45. (a) How long will the energy in a 1470-kJ (350-kcal) cup of yogurt last in a woman doing work at the rate of 150 W with an efficiency of 20.0% (such as in leisurely climbing stairs)? (b) Does the time found in part (a) imply that it is easy to consume more food energy than you can reasonably expect to work off with exercise? 46. (a) A woman climbing the Washington Monument metabolizes $\displaystyle 6.00×10^2kJ$ of food energy. If her efficiency is 18.0%, how much heat transfer occurs to the environment to keep her temperature constant? (b) Discuss the amount of heat transfer found in (a). Is it consistent with the fact that you quickly warm up when exercising? Solution (a) 492 kJ (b) This amount of heat is consistent with the fact that you warm quickly when exercising. Since the body is inefficient, the excess heat produced must be dissipated through sweating, breathing, etc. 15.2: The First Law of Thermodynamics and Some Simple Processes 47. A car tire contains $\displaystyle 0.0380m^3$ of air at a pressure of $\displaystyle 2.20×10^5N/m^2$ (about 32 psi). How much more internal energy does this gas have than the same volume has at zero gauge pressure (which is equivalent to normal atmospheric pressure)? Solution $\displaystyle 6.77×10^3J$ 48. A helium-filled toy balloon has a gauge pressure of 0.200 atm and a volume of 10.0 L. How much greater is the internal energy of the helium in the balloon than it would be at zero gauge pressure? 49. Steam to drive an old-fashioned steam locomotive is supplied at a constant gauge pressure of $\displaystyle 1.75×10^6N/m^2$ (about 250 psi) to a piston with a 0.200-m radius. (a) By calculating $\displaystyle PΔV$, find the work done by the steam when the piston moves 0.800 m. Note that this is the net work output, since gauge pressure is used. (b) Now find the amount of work by calculating the force exerted times the distance traveled. Is the answer the same as in part (a)? Solution (a) $\displaystyle W=PΔV=1.76×10^5J$ (b) $\displaystyle W=Fd=1.76×10^5J$. Yes, the answer is the same. 50. A hand-driven tire pump has a piston with a 2.50-cm diameter and a maximum stroke of 30.0 cm. (a) How much work do you do in one stroke if the average gauge pressure is $\displaystyle 2.40×10^5N/m^2$ (about 35 psi)? (b) What average force do you exert on the piston, neglecting friction and gravitational force? 51. Calculate the net work output of a heat engine following path ABCDA in the figure below. Solution $\displaystyle W=4.5×10^3J$ 52. What is the net work output of a heat engine that follows path ABDA in the figure above, with a straight line from B to D? Why is the work output less than for path ABCDA? Explicitly show how you follow the steps in the Problem-Solving Strategies for Thermodynamics. 53. Unreasonable Results What is wrong with the claim that a cyclical heat engine does 4.00 kJ of work on an input of 24.0 kJ of heat transfer while 16.0 kJ of heat transfers to the environment? Solution $\displaystyle W$ is not equal to the difference between the heat input and the heat output. 54. (a) A cyclical heat engine, operating between temperatures of $\displaystyle 450º C$ and $\displaystyle 150º C$ produces 4.00 MJ of work on a heat transfer of 5.00 MJ into the engine. How much heat transfer occurs to the environment? (b) What is unreasonable about the engine? (c) Which premise is unreasonable? 55. Construct Your Own Problem Consider a car’s gasoline engine. Construct a problem in which you calculate the maximum efficiency this engine can have. Among the things to consider are the effective hot and cold reservoir temperatures. Compare your calculated efficiency with the actual efficiency of car engines. 56. Construct Your Own Problem Consider a car trip into the mountains. Construct a problem in which you calculate the overall efficiency of the car for the trip as a ratio of kinetic and potential energy gained to fuel consumed. Compare this efficiency to the thermodynamic efficiency quoted for gasoline engines and discuss why the thermodynamic efficiency is so much greater. Among the factors to be considered are the gain in altitude and speed, the mass of the car, the distance traveled, and typical fuel economy. 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency 57. A certain heat engine does 10.0 kJ of work and 8.50 kJ of heat transfer occurs to the environment in a cyclical process. (a) What was the heat transfer into this engine? (b) What was the engine’s efficiency? Solution (a) $\displaystyle 18.5kJ$ (b) $\displaystyle 54.1%$ 58. With $\displaystyle 2.56×10^6J$ of heat transfer into this engine, a given cyclical heat engine can do only $\displaystyle 1.50×10^5J$ of work. (a) What is the engine’s efficiency? (b) How much heat transfer to the environment takes place? 59. (a) What is the work output of a cyclical heat engine having a 22.0% efficiency and $\displaystyle 6.00×10^9J$ of heat transfer into the engine? (b) How much heat transfer occurs to the environment? Solution (a) $\displaystyle 1.32 × 10^9J$ (b) $\displaystyle 4.68 × 10^9J$ 60. (a) What is the efficiency of a cyclical heat engine in which 75.0 kJ of heat transfer occurs to the environment for every 95.0 kJ of heat transfer into the engine? (b) How much work does it produce for 100 kJ of heat transfer into the engine? 61. The engine of a large ship does $\displaystyle 2.00×10^8J$ of work with an efficiency of 5.00%. (a) How much heat transfer occurs to the environment? (b) How many barrels of fuel are consumed, if each barrel produces $\displaystyle 6.00×10^9J$ of heat transfer when burned? Solution (a) $\displaystyle 3.80 × 10^9J$ (b) 0.667 barrels 62. (a) How much heat transfer occurs to the environment by an electrical power station that uses $\displaystyle 1.25×10^{14}J$ of heat transfer into the engine with an efficiency of 42.0%? (b) What is the ratio of heat transfer to the environment to work output? (c) How much work is done? 63. Assume that the turbines at a coal-powered power plant were upgraded, resulting in an improvement in efficiency of 3.32%. Assume that prior to the upgrade the power station had an efficiency of 36% and that the heat transfer into the engine in one day is still the same at $\displaystyle 2.50×10^{14}J$. (a) How much more electrical energy is produced due to the upgrade? (b) How much less heat transfer occurs to the environment due to the upgrade? Solution (a) $\displaystyle 8.30 × 10^{12}J$, which is 3.32% of $\displaystyle 2.50 × 10^{14}J$. (b) $\displaystyle –8.30 × 10^{12}J$, where the negative sign indicates a reduction in heat transfer to the environment. 64. This problem compares the energy output and heat transfer to the environment by two different types of nuclear power stations—one with the normal efficiency of 34.0%, and another with an improved efficiency of 40.0%. Suppose both have the same heat transfer into the engine in one day, $\displaystyle 2.50×10^{14}J$. (a) How much more electrical energy is produced by the more efficient power station? (b) How much less heat transfer occurs to the environment by the more efficient power station? (One type of more efficient nuclear power station, the gas-cooled reactor, has not been reliable enough to be economically feasible in spite of its greater efficiency.) 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated 65. A certain gasoline engine has an efficiency of 30.0%. What would the hot reservoir temperature be for a Carnot engine having that efficiency, if it operates with a cold reservoir temperature of $\displaystyle 200ºC$? Solution $\displaystyle 403ºC$ 66. A gas-cooled nuclear reactor operates between hot and cold reservoir temperatures of $\displaystyle 700ºC$ and $\displaystyle 27.0ºC$. (a) What is the maximum efficiency of a heat engine operating between these temperatures? (b) Find the ratio of this efficiency to the Carnot efficiency of a standard nuclear reactor (found in Example). 67. (a) What is the hot reservoir temperature of a Carnot engine that has an efficiency of 42.0% and a cold reservoir temperature of $\displaystyle 27.0ºC$? (b) What must the hot reservoir temperature be for a real heat engine that achieves 0.700 of the maximum efficiency, but still has an efficiency of 42.0% (and a cold reservoir at $\displaystyle 27.0ºC$? (c) Does your answer imply practical limits to the efficiency of car gasoline engines? Solution (a) $\displaystyle 244ºC$ (b) $\displaystyle 477ºC$ (c) Yes, since automobiles engines cannot get too hot without overheating, their efficiency is limited. 68. Steam locomotives have an efficiency of 17.0% and operate with a hot steam temperature of $\displaystyle 425ºC$. (a) What would the cold reservoir temperature be if this were a Carnot engine? (b) What would the maximum efficiency of this steam engine be if its cold reservoir temperature were $\displaystyle 150ºC$? 69. Practical steam engines utilize $\displaystyle 450ºC$ steam, which is later exhausted at $\displaystyle 270ºC$. (a) What is the maximum efficiency that such a heat engine can have? (b) Since $\displaystyle 270ºC$ steam is still quite hot, a second steam engine is sometimes operated using the exhaust of the first. What is the maximum efficiency of the second engine if its exhaust has a temperature of $\displaystyle 150ºC$? (c) What is the overall efficiency of the two engines? (d) Show that this is the same efficiency as a single Carnot engine operating between $\displaystyle 450ºC$ and $\displaystyle 150ºC$. Explicitly show how you follow the steps in the Problem-Solving Strategies for Thermodynamics. Solution (a) $\displaystyle Eff_1=1−\frac{T_{c,1}}{T_{h,1}}=1−\frac{543K}{723K}=0.249$ or $\displaystyle 24.9%$ (b) $\displaystyle Eff_2=1−\frac{423K}{543K}=0.221$ or $\displaystyle 22.1%$ (c) $\displaystyle Eff_1=1−\frac{T_{c,1}}{T_{h,1}}⇒T_{c,1}=T_{h,1}(1,−,eff_1)$ similarly, $\displaystyle T_{c,2}=T_{h,2}(1−Eff_2)$ using $\displaystyle T_{h,2}=T_{c,1} \text{in above equation gives} T_{c,2}=T_{h,1}(1−Eff_1)(1−Eff_2)≡T_{h,1}(1−Eff_{overall})$ $\displaystyle ∴(1−Eff_{overall})=(1−Eff_1)(1−Eff_2)$ $\displaystyle Eff_{overall}=1−(1−0.249)(1−0.221)=41.5%$ (d) $\displaystyle Eff_{overall}=1−\frac{423K}{723K}=0.415$ or $\displaystyle 41.5%$ 70. A coal-fired electrical power station has an efficiency of 38%. The temperature of the steam leaving the boiler is $\displaystyle 550ºC $. What percentage of the maximum efficiency does this station obtain? (Assume the temperature of the environment is $\displaystyle 20ºC$.) 71. Would you be willing to financially back an inventor who is marketing a device that she claims has 25 kJ of heat transfer at 600 K, has heat transfer to the environment at 300 K, and does 12 kJ of work? Explain your answer. Solution The heat transfer to the cold reservoir is $\displaystyle Q_c=Q_h−W=25kJ−12kJ=13kJ$, so the efficiency is $\displaystyle Eff=1−\frac{Q_c}{Q_h}=1−\frac{13kJ}{25kJ}=0.48$. The Carnot efficiency is $\displaystyle Eff_C=1−\frac{T_c}{T_h}=1−\frac{300K}{600K}=0.50$. The actual efficiency is 96% of the Carnot efficiency, which is much higher than the best-ever achieved of about 70%, so her scheme is likely to be fraudulent. 72. Unreasonable Results (a) Suppose you want to design a steam engine that has heat transfer to the environment at 270ºC and has a Carnot efficiency of 0.800. What temperature of hot steam must you use? (b) What is unreasonable about the temperature? (c) Which premise is unreasonable? 73. Unreasonable Results (a) Calculate the cold reservoir temperature of a steam engine that uses hot steam at $\displaystyle 450ºC$ and has a Carnot efficiency of 0.700. (b) What is unreasonable about the temperature? (c) Which premise is unreasonable? Solution (a) –56.3ºC (b) The temperature is too cold for the output of a steam engine (the local environment). It is below the freezing point of water. (c) The assumed efficiency is too high. 15.5: Applications of Thermodynamics: Heat Pumps and Refrigerators 74. What is the coefficient of performance of an ideal heat pump that has heat transfer from a cold temperature of $\displaystyle −25.0ºC$ to a hot temperature of $\displaystyle 40.0ºC$? Solution 4.82 75. Suppose you have an ideal refrigerator that cools an environment at $\displaystyle −20.0ºC$ and has heat transfer to another environment at $\displaystyle 50.0ºC$. What is its coefficient of performance? 76. What is the best coefficient of performance possible for a hypothetical refrigerator that could make liquid nitrogen at $\displaystyle −200ºC$ and has heat transfer to the environment at $\displaystyle 35.0ºC$? Solution 0.311 77. In a very mild winter climate, a heat pump has heat transfer from an environment at $\displaystyle 5.00ºC$ to one at $\displaystyle 35.0ºC$. What is the best possible coefficient of performance for these temperatures? Explicitly show how you follow the steps in the Problem-Solving Strategies for Thermodynamics. 78. (a) What is the best coefficient of performance for a heat pump that has a hot reservoir temperature of $\displaystyle 50.0ºC$ and a cold reservoir temperature of $\displaystyle −20.0ºC$? (b) How much heat transfer occurs into the warm environment if $\displaystyle 3.60×10^7J$ of work (10.0kW⋅h) is put into it? (c) If the cost of this work input is $\displaystyle 10.0 cents/kW⋅h$, how does its cost compare with the direct heat transfer achieved by burning natural gas at a cost of 85.0 cents per therm. (A therm is a common unit of energy for natural gas and equals $\displaystyle 1.055×10^8J$.) Solution (a) 4.61 (b) $\displaystyle 1.66×10^8J$ or $\displaystyle 3.97×10^4kcal$ (c) To transfer $\displaystyle 1.66×10^8J$, heat pump costs $1.00, natural gas costs $1.34. 79. (a) What is the best coefficient of performance for a refrigerator that cools an environment at $\displaystyle −30.0ºC$ and has heat transfer to another environment at $\displaystyle 45.0ºC$? (b) How much work in joules must be done for a heat transfer of 4186 kJ from the cold environment? (c) What is the cost of doing this if the work costs 10.0 cents per $\displaystyle 3.60×10^6J$? (d) How many kJ of heat transfer occurs into the warm environment? (e) Discuss what type of refrigerator might operate between these temperatures. 80. Suppose you want to operate an ideal refrigerator with a cold temperature of $\displaystyle −10.0ºC$, and you would like it to have a coefficient of performance of 7.00. What is the hot reservoir temperature for such a refrigerator? Solution 27.6ºC 81. An ideal heat pump is being considered for use in heating an environment with a temperature of $\displaystyle 22.0ºC$. What is the cold reservoir temperature if the pump is to have a coefficient of performance of 12.0? 82. A 4-ton air conditioner removes $\displaystyle 5.06×10^7J$ (48,000 British thermal units) from a cold environment in 1.00 h. (a) What energy input in joules is necessary to do this if the air conditioner has an energy efficiency rating ($\displaystyle EER$) of 12.0? (b) What is the cost of doing this if the work costs 10.0 cents per $\displaystyle 3.60×10^6J $ (one kilowatt-hour)? (c) Discuss whether this cost seems realistic. Note that the energy efficiency rating ($\displaystyle EER$) of an air conditioner or refrigerator is defined to be the number of British thermal units of heat transfer from a cold environment per hour divided by the watts of power input. Solution (a) $\displaystyle 1.44×10^7J$ (b) 40 cents (c) This cost seems quite realistic; it says that running an air conditioner all day would cost $9.59 (if it ran continuously). 83. Show that the coefficients of performance of refrigerators and heat pumps are related by $\displaystyle COP_{ref}=COP_{hp}−1$. Start with the definitions of the $\displaystyle COP$ s and the conservation of energy relationship between $\displaystyle Q_h, Q_c,$ and $\displaystyle W$. 15.6: Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy 84. (a) On a winter day, a certain house loses $\displaystyle 5.00×10^8J$ of heat to the outside (about 500,000 Btu). What is the total change in entropy due to this heat transfer alone, assuming an average indoor temperature of $\displaystyle 21.0º C$ and an average outdoor temperature of $\displaystyle 5.00º C$? (b) This large change in entropy implies a large amount of energy has become unavailable to do work. Where do we find more energy when such energy is lost to us? Solution (a) $\displaystyle 9.78×10^4J/K$ (b) In order to gain more energy, we must generate it from things within the house, like a heat pump, human bodies, and other appliances. As you know, we use a lot of energy to keep our houses warm in the winter because of the loss of heat to the outside. 85. On a hot summer day, $\displaystyle 4.00×10^6J$ of heat transfer into a parked car takes place, increasing its temperature from $\displaystyle 35.0º C$ to $\displaystyle 45.0º C$. What is the increase in entropy of the car due to this heat transfer alone? 86. A hot rock ejected from a volcano’s lava fountain cools from $\displaystyle 1100º C$ to $\displaystyle 40.0º C$, and its entropy decreases by 950 J/K. How much heat transfer occurs from the rock? Solution $\displaystyle 8.01×10^5J$ 87. When $\displaystyle 1.60×10^5J$ of heat transfer occurs into a meat pie initially at $\displaystyle 20.0º C$, its entropy increases by 480 J/K. What is its final temperature? 88. The Sun radiates energy at the rate of $\displaystyle 3.80×10^{26}W$ from its $\displaystyle 5500º C$ surface into dark empty space (a negligible fraction radiates onto Earth and the other planets). The effective temperature of deep space is $\displaystyle −270º C$. (a) What is the increase in entropy in one day due to this heat transfer? (b) How much work is made unavailable? Solution (a) $\displaystyle 1.04×10^{31}J/K$ (b) $\displaystyle 3.28×10^{31}J$ 89. (a) In reaching equilibrium, how much heat transfer occurs from 1.00 kg of water at $\displaystyle 40.0º C$ when it is placed in contact with 1.00 kg of $\displaystyle 20.0º C$ water in reaching equilibrium? (b) What is the change in entropy due to this heat transfer? (c) How much work is made unavailable, taking the lowest temperature to be $\displaystyle 20.0º C$? Explicitly show how you follow the steps in the Problem-Solving Strategies for Entropy. 90. What is the decrease in entropy of 25.0 g of water that condenses on a bathroom mirror at a temperature of $\displaystyle 35.0º C$, assuming no change in temperature and given the latent heat of vaporization to be 2450 kJ/kg? Solution 199 J/K 91. Find the increase in entropy of 1.00 kg of liquid nitrogen that starts at its boiling temperature, boils, and warms to $\displaystyle 20.0º C$ at constant pressure. 92. A large electrical power station generates 1000 MW of electricity with an efficiency of 35.0%. (a) Calculate the heat transfer to the power station, $\displaystyle Q_h$, in one day. (b) How much heat transfer $\displaystyle Q_c$ occurs to the environment in one day? (c) If the heat transfer in the cooling towers is from $\displaystyle 35.0º C$ water into the local air mass, which increases in temperature from $\displaystyle 18.0º C$ to $\displaystyle 20.0º C$, what is the total increase in entropy due to this heat transfer? (d) How much energy becomes unavailable to do work because of this increase in entropy, assuming an $\displaystyle 18.0º C$ lowest temperature? (Part of $\displaystyle Q_c$ could be utilized to operate heat engines or for simply heating the surroundings, but it rarely is.) Solution (a) $\displaystyle 2.47×10^{14}J$ (b) $\displaystyle 1.60×10^{14}J$ (c) $\displaystyle 2.85×10^{10}J/K$ (d) $\displaystyle 8.29×10^{12}J$ 93. (a) How much heat transfer occurs from 20.0 kg of $\displaystyle 90.0º C$ water placed in contact with 20.0 kg of $\displaystyle 10.0º C$ water, producing a final temperature of $\displaystyle 50.0º C$? (b) How much work could a Carnot engine do with this heat transfer, assuming it operates between two reservoirs at constant temperatures of $\displaystyle 90.0º C$ and $\displaystyle 10.0º C$? (c) What increase in entropy is produced by mixing 20.0 kg of $\displaystyle 90.0º C$ water with 20.0 kg of $\displaystyle 10.0º C$ water? (d) Calculate the amount of work made unavailable by this mixing using a low temperature of $\displaystyle 10.0º C$, and compare it with the work done by the Carnot engine. Explicitly show how you follow the steps in the Problem-Solving Strategies for Entropy. (e) Discuss how everyday processes make increasingly more energy unavailable to do work, as implied by this problem. 15.7: Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation 94. Using Table, verify the contention that if you toss 100 coins each second, you can expect to get 100 heads or 100 tails once in $\displaystyle 2×10^{22}$ calculate the time to two-digit accuracy. Solution It should happen twice in every $\displaystyle 1.27×10^{30}s$ or once in every $\displaystyle 6.35×10^{29}s$ $\displaystyle (6.35×10^{29}s)(\frac{1 h}{3600 s})=(\frac{1 d}{24 h})(\frac{1 y}{365.25 d})=2.0×10^{22}y$ 95. What percent of the time will you get something in the range from 60 heads and 40 tails through 40 heads and 60 tails when tossing 100 coins? The total number of microstates in that range is $\displaystyle 1.22×10^{30}$. (Consult Table.) 96. (a) If tossing 100 coins, how many ways (microstates) are there to get the three most likely macrostates of 49 heads and 51 tails, 50 heads and 50 tails, and 51 heads and 49 tails? (b) What percent of the total possibilities is this? (Consult Table.) Solution (a) $\displaystyle 3.0×10^{29}$ (b) 24% 97. (a) What is the change in entropy if you start with 100 coins in the 45 heads and 55 tails macrostate, toss them, and get 51 heads and 49 tails? (b) What if you get 75 heads and 25 tails? (c) How much more likely is 51 heads and 49 tails than 75 heads and 25 tails? (d) Does either outcome violate the second law of thermodynamics? 98. (a) What is the change in entropy if you start with 10 coins in the 5 heads and 5 tails macrostate, toss them, and get 2 heads and 8 tails? (b) How much more likely is 5 heads and 5 tails than 2 heads and 8 tails? (Take the ratio of the number of microstates to find out.) (c) If you were betting on 2 heads and 8 tails would you accept odds of 252 to 45? Explain why or why not. Solution (a) $\displaystyle −2.38×10^{–23}J/K$ (b) 5.6 times more likely (c) If you were betting on two heads and 8 tails, the odds of breaking even are 252 to 45, so on average you would break even. So, no, you wouldn’t bet on odds of 252 to 45. 99. (a) If you toss 10 coins, what percent of the time will you get the three most likely macrostates (6 heads and 4 tails, 5 heads and 5 tails, 4 heads and 6 tails)? (b) You can realistically toss 10 coins and count the number of heads and tails about twice a minute. At that rate, how long will it take on average to get either 10 heads and 0 tails or 0 heads and 10 tails? 100. (a) Construct a table showing the macrostates and all of the individual microstates for tossing 6 coins. (Use Table as a guide.) (b) How many macrostates are there? (c) What is the total number of microstates? (d) What percent chance is there of tossing 5 heads and 1 tail? (e) How much more likely are you to toss 3 heads and 3 tails than 5 heads and 1 tail? (Take the ratio of the number of microstates to find out.) Solution (b) 7 (c) 64 (d) 9.38% (e) 3.33 times more likely (20 to 6) 101. In an air conditioner, 12.65 MJ of heat transfer occurs from a cold environment in 1.00 h. (a) What mass of ice melting would involve the same heat transfer? (b) How many hours of operation would be equivalent to melting 900 kg of ice? (c) If ice costs 20 cents per kg, do you think the air conditioner could be operated more cheaply than by simply using ice? Describe in detail how you evaluate the relative costs. Contributors and Attributions - Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .	libretexts	2025-03-17T19:53:34.077614	2018-05-04T03:03:11	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/15%3A_Thermodynamics/15.E%3A_Heat_and_Heat_Transfer_Methods_(Exercise)", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "15.E: Heat and Heat Transfer Methods (Exercise)", "author": null }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves	16: Oscillatory Motion and Waves - - 16.1: Hooke’s Law - Stress and Strain Revisited - An oscillation is a back and forth motion of an object between two points of deformation. An oscillation may create a wave, which is a disturbance that propagates from where it was created. The simplest type of oscillations and waves are related to systems that can be described by Hooke’s law. - - 16.2: Period and Frequency in Oscillations - We define periodic motion to be a motion that repeats itself at regular time intervals, such as exhibited by the guitar string or by an object on a spring moving up and down. The time to complete one oscillation remains constant and is called the period. Its units are usually seconds, but may be any convenient unit of time. - - 16.3: Simple Harmonic Motion- A Special Periodic Motion - Simple Harmonic Motion (SHM) is the name given to oscillatory motion for a system where the net force can be described by Hooke’s law, and such a system is called a simple harmonic oscillator. If the net force can be described by Hooke’s law and there is no damping (by friction or other non-conservative forces), then a simple harmonic oscillator will oscillate with equal displacement on either side of the equilibrium position. - - 16.4: The Simple Pendulum - Pendulums are in common usage. Some have crucial uses, such as in clocks; some are for fun, such as a child’s swing; and some are just there, such as the sinker on a fishing line. For small displacements, a pendulum is a simple harmonic oscillator. A simple pendulum is defined to have an object that has a small mass, also known as the pendulum bob, which is suspended from a light wire or string. - - 16.6: Uniform Circular Motion and Simple Harmonic Motion - If studied in sufficient depth, simple harmonic motion produced in this manner can give considerable insight into many aspects of oscillations and waves and is very useful mathematically. In our brief treatment, we shall indicate some of the major features of this relationship and how they might be useful. A projection of uniform circular motion undergoes simple harmonic oscillation. - - 16.7: Damped Harmonic Motion - Although we can often make friction and other non-conservative forces negligibly small, completely undamped motion is rare. In fact, we may even want to damp oscillations, such as with car shock absorbers. For a system that has a small amount of damping, the period and frequency are nearly the same as for simple harmonic motion, but the amplitude gradually decreases. This occurs because the non-conservative damping force removes energy from the system, usually in the form of thermal energy. - - 16.8: Forced Oscillations and Resonance - In this section, we shall briefly explore applying a periodic driving force acting on a simple harmonic oscillator. The driving force puts energy into the system at a certain frequency, not necessarily the same as the natural frequency of the system. The natural frequency is the frequency at which a system would oscillate if there were no driving and no damping force. - - 16.9: Waves - a wave is a disturbance that propagates, or moves from the place it was created. For water waves, the disturbance is in the surface of the water, perhaps created by a rock thrown into a pond or by a swimmer splashing the surface repeatedly. For sound waves, the disturbance is a change in air pressure, perhaps created by the oscillating cone inside a speaker. For earthquakes, there are several types of disturbances, including disturbance of Earth’s surface and pressure under the surface. Thumbnail: Mavericks Surf Contest 2010. (CC-SA-BY; Shalom Jacobovitz ).	libretexts	2025-03-17T19:53:34.149717	2015-11-01T04:13:23	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16: Oscillatory Motion and Waves", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.00%3A_Prelude_to_Oscillatory_Motion_and_Waves	16.0: Prelude to Oscillatory Motion and Waves What do an ocean buoy, a child in a swing, the cone inside a speaker, a guitar, atoms in a crystal, the motion of chest cavities, and the beating of hearts all have in common? They all oscillate —-that is, they move back and forth between two points. Many systems oscillate, and they have certain characteristics in common. All oscillations involve force and energy. You push a child in a swing to get the motion started. The energy of atoms vibrating in a crystal can be increased with heat. You put energy into a guitar string when you pluck it. Some oscillations create waves . A guitar creates sound waves. You can make water waves in a swimming pool by slapping the water with your hand. You can no doubt think of other types of waves. Some, such as water waves, are visible. Some, such as sound waves, are not. But every wave is a disturbance that moves from its source and carries energy . Other examples of waves include earthquakes and visible light. Even subatomic particles, such as electrons, can behave like waves. By studying oscillatory motion and waves, we shall find that a small number of underlying principles describe all of them and that wave phenomena are more common than you have ever imagined. We begin by studying the type of force that underlies the simplest oscillations and waves. We will then expand our exploration of oscillatory motion and waves to include concepts such as simple harmonic motion, uniform circular motion, and damped harmonic motion. Finally, we will explore what happens when two or more waves share the same space, in the phenomena known as superposition and interference. Glossary - oscillate - moving back and forth regularly between two points - wave - a disturbance that moves from its source and carries energy	libretexts	2025-03-17T19:53:34.209183	2015-11-01T05:48:16	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.00%3A_Prelude_to_Oscillatory_Motion_and_Waves", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.0: Prelude to Oscillatory Motion and Waves", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.01%3A_Hookes_Law_-_Stress_and_Strain_Revisited	16.1: Hooke’s Law - Stress and Strain Revisited Learning Objectives By the end of this section, you will be able to: - Explain Newton’s third law of motion with respect to stress and deformation. - Describe the restoration of force and displacement. - Calculate the energy in Hook’s Law of deformation, and the stored energy in a string. Newton’s first law implies that an object oscillating back and forth is experiencing forces. Without force, the object would move in a straight line at a constant speed rather than oscillate. Consider, for example, plucking a plastic ruler to the left as shown in Figure $\PageIndex{1}$. The deformation of the ruler creates a force in the opposite direction, known as a restoring force . Once released, the restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. However, by the time the ruler gets there, it gains momentum and continues to move to the right, producing the opposite deformation. It is then forced to the left, back through equilibrium, and the process is repeated until dissipative forces dampen the motion. These forces remove mechanical energy from the system, gradually reducing the motion until the ruler comes to rest. The simplest oscillations occur when the restoring force is directly proportional to displacement. When stress and strain were covered in Newton’s Third Law of Motion , the name was given to this relationship between force and displacement was Hooke’s law: \[F = -kx\] Here, $F$ is the restoring force, $x$ is the displacement from equilibrium or deformation , and $k$ is a constant related to the difficulty in deforming the system. The minus sign indicates the restoring force is in the direction opposite to the displacement. The force constant $k$ is related to the rigidity (or stiffness) of a system—the larger the force constant, the greater the restoring force, and the stiffer the system. The units of $k$ are newtons per meter (N/m). For example, $k$ is directly related to Young’s modulus when we stretch a string. Figure $\PageIndex{3}$ shows a graph of the absolute value of the restoring force versus the displacement for a system that can be described by Hooke’s law—a simple spring in this case. The slope of the graph equals the force constant $k$ in newtons per meter. A common physics laboratory exercise is to measure restoring forces created by springs, determine if they follow Hooke’s law, and calculate their force constants if they do. Example $\PageIndex{1}$: How Stiff Are Car Springs? What is the force constant for the suspension system of a car that settles 1.20 cm when an 80.0-kg person gets in? Strategy Consider the car to be in its equilibrium position $x = 0$ before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position $x = -1.20 \times 10^{-2} m$. At that point, the springs supply a restoring force $F$ equal to the person’s weight $w = mg = (80.0 \, kg)(9.80 \, m/s^2) = 784 \, N.$ We take this force to be $F$ in Hooke's law. Knowing $F$ and $x$, we can then solve the force constant $k$. Solution 1. Solve Hooke’s law, $F = -kx$, for $k$: \[k = -\dfrac{F}{x}. \nonumber\] 2. Substitute known values and solve $k$: \[\begin{align} k &= -\dfrac{784 \, N}{-1.20 \times 10^{-2}m} \\[5pt] &=6.53 \times 10^4 \, N/m.\end{align}\] Discussion Note that $F$ and $x$ have opposite signs because they are in opposite directions—the restoring force is up, and the displacement is down. Also, note that the car would oscillate up and down when the person got in if it were not for damping (due to frictional forces) provided by shock absorbers. Bouncing cars are a sure sign of bad shock absorbers. Energy in Hooke’s Law of Deformation In order to produce a deformation, work must be done. That is, a force must be exerted through a distance, whether you pluck a guitar string or compress a car spring. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy. The potential energy stored in a spring is $PE_{el} = \dfrac{1}{2}kx^2$. Here, we generalize the idea to elastic potential energy for a deformation of any system that can be described by Hooke’s law. Hence, \[PE_{el} = \dfrac{1}{2}kx^2,\] where $PR_{el}$ is the elastic potential energy stored in any deformed system that obeys Hooke’s law and has a displacement $x$ from equilibrium and a force constant $k$. It is possible to find the work done in deforming a system in order to find the energy stored. This work is performed by an applied force $F_{app}$. The applied force is exactly opposite to the restoring force (action-reaction), and so $F_{app} = kx$. Figure $\PageIndex{5}$ shows a graph of the applied force versus deformation $x$ for a system that can be described by Hooke’s law. Work done on the system is force multiplied by distance, which equals the area under the curve or $\dfrac{1}{2} kx^2$ (Method A in the figure). Another way to determine the work is to note that the force increases linearly from 0 to $kx$, so that the average force is $(1/2)kx$, the distance moved is $x$, and thus $W = F_{app} d = [(1/2)kx](x) = (1/2)kx^2$ (Method B in the figure). Example $\PageIndex{2}$: Calculating Stored Energy: A Tranquilizer Gun Spring We can use a toy gun’s spring mechanism to ask and answer two simple questions: - How much energy is stored in the spring of a tranquilizer gun that has a force constant of 50.0 N/m and is compressed 0.150 m? - If you neglect friction and the mass of the spring, at what speed will a 2.00-g projectile be ejected from the gun? Strategy for a (a): The energy stored in the spring can be found directly from elastic potential energy equation, because $k$ and $x$ are given. Solution for a Entering the given values for $k$ and $x$ yields \[PE_{el} = \dfrac{1}{2}kx^2 = \dfrac{1}{2}(50.0 \, N/m)(0.150 \, m)^2 = 0.563 \, N \cdot m = 0.563 \, J\] Strategy for b Because there is no friction, the potential energy is converted entirely into kinetic energy. The expression for kinetic energy can be solved for the projectile’s speed. Solution for b - Identify known quantities: \[ KE_f = PE_{el} \, or \, 1/2 mv^2 = (1/2)kx^2 = PE_{el} = 0.563 \, J\] 2. Solve for $v$: \[ v = \left[\dfrac{2PE_{el}}{m}\right] = \left[\dfrac{2(0.563 \, J)}{0.002 \, kg}\right] = 23.7(J/K)^{1/2}\] 3. Convert units: $23.7 \, m/s$ Discussion (a) and (b): This projectile speed is impressive for a tranquilizer gun (more than 80 km/h). The numbers in this problem seem reasonable. The force needed to compress the spring is small enough for an adult to manage, and the energy imparted to the dart is small enough to limit the damage it might do. Yet, the speed of the dart is great enough for it to travel an acceptable distance. Exercise $\PageIndex{1}$: Check your Understanding Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations of the ruler. In what way could you modify this simple experiment to increase the rigidity of the system? - Answer - You could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment. Exercise $\PageIndex{2}$: Check your Understanding If you apply a deforming force on an object and let it come to equilibrium, what happened to the work you did on the system? - Answer - It was stored in the object as potential energy. Summary - An oscillation is a back and forth motion of an object between two points of deformation. - An oscillation may create a wave, which is a disturbance that propagates from where it was created. - The simplest type of oscillations and waves are related to systems that can be described by Hooke’s law: $ F = -kx$ where $F$ is the $x$ restoring force, $PE_{el}$ is the displacement from equilibrium or deformation, and $PE_{el} = (1/2)kx^2.$ Glossary - deformation - displacement from equilibrium - elastic potential energy - potential energy stored as a result of deformation of an elastic object, such as the stretching of a spring - force constant - a constant related to the rigidity of a system: the larger the force constant, the more rigid the system; the force constant is represented by k - restoring force - force acting in opposition to the force caused by a deformation	libretexts	2025-03-17T19:53:34.289080	2015-11-01T05:48:36	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.01%3A_Hookes_Law_-_Stress_and_Strain_Revisited", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.1: Hooke’s Law - Stress and Strain Revisited", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.02%3A_Period_and_Frequency_in_Oscillations	16.2: Period and Frequency in Oscillations Learning Objectives By the end of this section, you will be able to: - Observe the vibrations of a guitar string. - Determine the frequency of oscillations. When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time. Each successive vibration of the string takes the same time as the previous one. We define periodic motion to be a motion that repeats itself at regular time intervals, such as exhibited by the guitar string or by an object on a spring moving up and down. The time to complete one oscillation remains constant and is called the period $T$. Its units are usually seconds, but may be any convenient unit of time. The word period refers to the time for some event whether repetitive or not; but we shall be primarily interested in periodic motion, which is by definition repetitive. A concept closely related to period is the frequency of an event. For example, if you get a paycheck twice a month, the frequency of payment is two per month and the period between checks is half a month. Frequency $f$ is defined to be the number of events per unit time. For periodic motion, frequency is the number of oscillations per unit time. The relationship between frequency and period is \[f = \dfrac{1}{T},\] The SI unit for frequency is the cycle per second , which is defined to be a hertz (Hz): \[1 \, Hz = 1 \dfrac{cycle}{sec} \, or 1 \, Hz = \dfrac{1}{s}\] A cycle is one complete oscillation. Note that a vibration can be a single or multiple event, whereas oscillations are usually repetitive for a significant number of cycles. Example $\PageIndex{1}$: Determine the Frequency of Two Oscillations, Medical Ultrasound and the Period of Middle C We can use the formulas presented in this module to determine both the frequency based on known oscillations and the oscillation based on a known frequency. Let’s try one example of each. - A medical imaging device produces ultrasound by oscillating with a period of 0.400 µs. What is the frequency of this oscillation? - The frequency of middle C on a typical musical instrument is 264 Hz. What is the time for one complete oscillation? Strategy Both questions (a) and (b) can be answered using the relationship between period and frequency. In question (a), the period $T$ is given and we are asked to find frequency $f$. In question (b), the frequency is given and we are asked to find the period $T$. Solution a Substitute 0.400 $\mu s$ for $T$ in $f = \dfrac{1}{T}$: \[f = \dfrac{1}{T} = \dfrac{1}{0.400 \times 10^{-6} s}.\] Solve to find \[f = 2.50 \times 10^6 \, Hz.\] Discussion a The frequency of sound found in (a) is much higher than the highest frequency that humans can hear and, therefore, is called ultrasound. Appropriate oscillations at this frequency generate ultrasound used for noninvasive medical diagnoses, such as observations of a fetus in the womb. Solution b - Identify the known values: The time for one complete oscillation is the period $T$: \[f = \dfrac{1}{T}.\] - Solve for $T$: \[T = \dfrac{1}{f}.\] - Substitute the given value for the frequency into the resulting expression: \[T = \dfrac{1}{f} = \dfrac{1}{264 \, Hz} = \dfrac{1}{264 \, cycles/s} = 3.79 \times 10^{-3} s = 3.79 \, ms.\] Discussion The period found in (b) is the time per cycle, but this value is often quoted as simply the time in convenient units (ms or milliseconds in this case). Exercise $\PageIndex{1}$ Identify an event in your life (such as receiving a paycheck) that occurs regularly. Identify both the period and frequency of this event. - Answer - I visit my parents for dinner every other Sunday. The frequency of my visits is 26 per calendar year. The period is two weeks. Summary - Periodic motion is a repetitious oscillation. - The time for one oscillation is the period $T$. - The number of oscillations per unit time is the frequency $f$. - These quantities are related by $f = \dfrac{1}{T}.$ Glossary - period - time it takes to complete one oscillation - periodic motion - motion that repeats itself at regular time intervals - frequency - number of events per unit of time	libretexts	2025-03-17T19:53:34.359857	2015-11-03T20:07:23	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.02%3A_Period_and_Frequency_in_Oscillations", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.2: Period and Frequency in Oscillations", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.03%3A_Simple_Harmonic_Motion-_A_Special_Periodic_Motion	16.3: Simple Harmonic Motion- A Special Periodic Motion Learning Objectives By the end of this section, you will be able to: - Describe a simple harmonic oscillator. - Explain the link between simple harmonic motion and waves. The oscillations of a system in which the net force can be described by Hooke’s law are of special importance, because they are very common. They are also the simplest oscillatory systems. Simple Harmonic Motion (SHM) is the name given to oscillatory motion for a system where the net force can be described by Hooke’s law, and such a system is called a simple harmonic oscillator . If the net force can be described by Hooke’s law and there is no damping (by friction or other non-conservative forces), then a simple harmonic oscillator will oscillate with equal displacement on either side of the equilibrium position, as shown for an object on a spring in Figure $\PageIndex{1}$. The maximum displacement from equilibrium is called the amplitude $X$. The units for amplitude and displacement are the same, but depend on the type of oscillation. For the object on the spring, the units of amplitude and displacement are meters; whereas for sound oscillations, they have units of pressure (and other types of oscillations have yet other units). Because amplitude is the maximum displacement, it is related to the energy in the oscillation. TAKE-HOME EXPERIMENT: SHM AND THE MARBLE Find a bowl or basin that is shaped like a hemisphere on the inside. Place a marble inside the bowl and tilt the bowl periodically so the marble rolls from the bottom of the bowl to equally high points on the sides of the bowl. Get a feel for the force required to maintain this periodic motion. What is the restoring force and what role does the force you apply play in the simple harmonic motion (SHM) of the marble? What is so significant about simple harmonic motion? One special thing is that the period $T$ and frequency $f$ of a simple harmonic oscillator are independent of amplitude. The string of a guitar, for example, will oscillate with the same frequency whether plucked gently or hard. Because the period is constant, a simple harmonic oscillator can be used as a clock. Two important factors do affect the period of a simple harmonic oscillator. The period is related to how stiff the system is. A very stiff object has a large force constant $k$, which causes the system to have a smaller period. For example, you can adjust a diving board’s stiffness—the stiffer it is, the faster it vibrates, and the shorter its period. Period also depends on the mass of the oscillating system. The more massive the system is, the longer the period. For example, a heavy person on a diving board bounces up and down more slowly than a light one. In fact, the mass $m$ and the force constant $k$ are the only factors that affect the period and frequency of simple harmonic motion. Period of Simple Harmonic Oscillator The period of a simple harmonic oscillator is given by \[T = 2\pi \sqrt{\dfrac{m}{k}}\] and, because $f = 1/T$, the frequency of a simple harmonic oscillator is \[f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}.\] Note that neither $T$ nor $f$ has any dependence on amplitude. TAKE-HOME EXPERIMENT: MASS AND RULER OSCILLATIONS Find two identical wooden or plastic rulers. Tape one end of each ruler firmly to the edge of a table so that the length of each ruler that protrudes from the table is the same. On the free end of one ruler tape a heavy object such as a few large coins. Pluck the ends of the rulers at the same time and observe which one undergoes more cycles in a time period, and measure the period of oscillation of each of the rulers. Example $\PageIndex{1}$: Calculate the Frequency and Period of Oscillations: Bad Shock Absorbers in a Car If the shock absorbers in a car go bad, then the car will oscillate at the least provocation, such as when going over bumps in the road and after stopping (See Figure $\PageIndex{2}$). Calculate the frequency and period of these oscillations for such a car if the car’s mass (including its load) is 900 kg and the force constant $k$ of the suspension system is $6.53 \times 10^4 \, N/m$. Strategy The frequency of the car’s oscillations will be that of a simple harmonic oscillator as given in the equation $f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}$. The mass and the force constant are both given. Solution - Enter the known values of k and m : \[f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}} = \dfrac{1}{2\pi} \sqrt{\dfrac{6.53 \times 10^4 \, N/m}{900 \, kg}}. \nonumber\] - Calculate the frequency: \[\dfrac{1}{2\pi} \sqrt{72.6/s^{-2}} = 1.3656/s^{-1} \approx 1.36/s^{-1} = 1.36 \, Hz\nonumber\] - You could use $T = 2\pi \sqrt{\dfrac{m}{k}}$ to calculate the period, but it is simpler to use the relationship $T = 1/f$ and substitute the value just found for $f$: \[T = \dfrac{1}{f} = \dfrac{1}{1.356 \, Hz} = 0.738 \, s.\nonumber\] Discussion The values of $T$ and $f$ both seem about right for a bouncing car. You can observe these oscillations if you push down hard on the end of a car and let go. The Link between Simple Harmonic Motion and Waves If a time-exposure photograph of the bouncing car were taken as it drove by, the headlight would make a wavelike streak, as shown in Figure $\PageIndex{2}$. Similarly, Figure $\PageIndex{3}$ shows an object bouncing on a spring as it leaves a wavelike "trace of its position on a moving strip of paper. Both waves are sine functions. All simple harmonic motion is intimately related to sine and cosine waves. The displacement as a function of time $t$ in any simple harmonic motion—that is, one in which the net restoring force can be described by Hooke’s law, is given by \[x(t) = X \, cos \dfrac{2\pi t}{T},\] where $X$ is amplitude. At $T = 0$, the initial position is $x_0 = X$, and the displacement oscillates back and forth with a period $T$. (When $t = T$, we get $x = X$ again because $cos \, 2\pi = 1$). Furthermore, from this expression for $x$, the velocity $v$ as a function of time is given by: \[v(t) = -v_{max}sin \left(\dfrac{2\pi t}{T}\right),\] where $v_{max} = 2\pi X/T = X \sqrt{k/m}$. The object has zero velocity at maximum displacement—for example, $v = 0$ when $t = 0$, and at that time $x = X$. The minus sign in the first equation for $v(t)$ gives the correct direction for the velocity. Just after the start of the motion, for instance, the velocity is negative because the system is moving back toward the equilibrium point. Finally, we can get an expression for acceleration using Newton’s second law. [Then we have $x(t), \, v(t), \, t,$ and $a(t)$, the quantities needed for kinematics and a description of simple harmonic motion.] According to Newton’s second law, the acceleration is $a = F/m = kx/m$. So, $a(t)$ is also a cosine function: \[a(t) = -\dfrac{kX}{m}cos \dfrac{2\pi t}{T}.\] Hence, $a(t)$ is directly proportional to and in the opposite direction to $x(t)$. Figure $\PageIndex{4}$ shows the simple harmonic motion of an object on a spring and presents graphs of $x(t)$, $v(t)$, and $a(t)$ versus time. The most important point here is that these equations are mathematically straightforward and are valid for all simple harmonic motion. They are very useful in visualizing waves associated with simple harmonic motion, including visalizing how waves add with one another. Exercise $\PageIndex{1}$ Suppose you pluck a banjo string. You hear a single note that starts out loud and slowly quiets over time. Describe what happens to the sound waves in terms of period, frequency and amplitude as the sound decreases in volume. - Answer - Frequency and period remain essentially unchanged. Only amplitude decreases as volume decreases. Exercise $\PageIndex{2}$ A babysitter is pushing a child on a swing. At the point where the swing reaches $x$, where would the corresponding point on a wave of this motion be located? - Answer - $x$ is the maximum deformation, which corresponds to the amplitude of the wave. The point on the wave would either be at the very top or the very bottom of the curve. PHET EXPLORATIONS: MASSES AND STRINGS A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can even slow time. Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energy for each spring. Summary - Simple harmonic motion is oscillatory motion for a system that can be described only by Hooke’s law. Such a system is also called a simple harmonic oscillator. - Maximum displacement is the amplitude $X$. The period $T$ and frequency $f$ of a simple harmonic oscillator are given by $T = 2\pi \sqrt{\frac{m}{k}}$ and $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$, where $m$ is the mass of the system. - Displacement in simple harmonic motion as a function of time is given by $x = X \, cos \, \frac{2\pi t}{T}$. - The velocity is given by $v(t) = -v_{max}sin \frac{2\pi t}{T}$, where $v_{max} = \sqrt{k/m}X$. - The acceleration is found to be $a = -\frac{kX}{m} \, cos \frac{2\pi t}{T}$. Glossary - amplitude - the maximum displacement from the equilibrium position of an object oscillating around the equilibrium position - simple harmonic motion - the oscillatory motion in a system where the net force can be described by Hooke’s law - simple harmonic oscillator - a device that implements Hooke’s law, such as a mass that is attached to a spring, with the other end of the spring being connected to a rigid support such as a wall	libretexts	2025-03-17T19:53:34.437719	2015-11-03T20:07:52	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.03%3A_Simple_Harmonic_Motion-_A_Special_Periodic_Motion", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.3: Simple Harmonic Motion- A Special Periodic Motion", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.04%3A_The_Simple_Pendulum	16.4: The Simple Pendulum Learning Objectives By the end of this section, you will be able to: - Measure acceleration due to gravity. Pendulums are in common usage. Some have crucial uses, such as in clocks; some are for fun, such as a child’s swing; and some are just there, such as the sinker on a fishing line. For small displacements, a pendulum is a simple harmonic oscillator. A simple pendulum is defined to have an object that has a small mass, also known as the pendulum bob, which is suspended from a light wire or string, such as shown in Figure $\PageIndex{1}$. Exploring the simple pendulum a bit further, we can discover the conditions under which it performs simple harmonic motion, and we can derive an interesting expression for its period. We begin by defining the displacement to be the arc length $s$. We see from Figure $\PageIndex{1}$ that the net force on the bob is tangent to the arc and equals $mg \, sin \, \theta$. (The weight $mg$ has components $mg \, cos \, \theta$ along the string and $mg \, sin \, \theta$ tangent to the arc.) Tension in the string exactly cancels the component $mg \, cos \theta$ parallel to the string. This leaves a net restoring force back toward the equilibrium position at $\theta = 0$. Now, if we can show that the restoring force is directly proportional to the displacement, then we have a simple harmonic oscillator. In trying to determine if we have a simple harmonic oscillator, we should note that for small angles (less than about $15^o$), $sin \, \theta \approx \theta \, (sin \, \theta$ and $\theta$ differ by about 1% or less at smaller angles). Thus, for angles less than about $15^o$, the restoring force $F$ is \[F \approx -mg\theta.\] The displacement $s$ is directly proportional to $\theta$. When $\theta$ is expressed in radians, the arc length in a circle is related to its radius ($L$ in this instance) by: \[s = L\theta,\] so that \[\theta = \dfrac{s}{L}.\] For small angles, then, the expression for the restoring force is: \[F \approx -\dfrac{mg}{L}s.\] This expression is of the form: \[F = -kx,\] where the force constant is given by $k = mg/L$ and the displacement is given by $x = s$. For angles less than about $15^o$ the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator. Using this equation, we can find the period of a pendulum for amplitudes less than about $15^o$. For the simple pendulum: \[T = 2\pi \sqrt{\dfrac{m}{k}} = 2\pi \sqrt{\dfrac{m}{mg/L}}.\] for the period of a simple pendulum. This result is interesting because of its simplicity. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass. As with simple harmonic oscillators, the period $T$ for a pendulum is nearly independent of amplitude, especially if $\theta$ is less than about $15^o$. Even simple pendulum clocks can be finely adjusted and accurate. Note the dependence of $T$ on $g$. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity. Consider the following example. Example $\PageIndex{1}$: Measuring Acceleration due to Gravity: The Period of a Pendulum What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s? Strategy We are asked to find $g$ given the period $T$ and the length $L$ of a pendulum. We can solve $T = 2\pi \sqrt{\frac{L}{g}}$ for $g$, assuming only that the angle of deflection is less than $15^o$. Solution - Square $T = 2\pi \sqrt{\frac{L}{g}}$ and solve for $g$: \[g = 4\pi^2 \dfrac{L}{T^2}.\] - Substitute known values into the new equation: \[g = 4\pi^2 \dfrac{0.75000 \, m}{(1.7357 \, s)^2}.\] - Calculate to find $g$: \[g = 9.8281 \, m/s^2.\] Discussion This method for determining $g$ can be very accurate. This is why length and period are given to five digits in this example. For the precision of the approximation $sin \, \theta \approx \theta$ to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about $0.5^o$. MAKING CAREER CONNECTIONS Knowing $g$ can be important in geological exploration; for example, a map of $g$ over large geographical regions aids the study of plate tectonics and helps in the search for oil fields and large mineral deposits. TAKE-HOME EXPERIMENT: DETERMINING $g$ Use a simple pendulum to determine the acceleration due to gravity $g$ in your own locale. Cut a piece of a string or dental floss so that it is about 1 m long. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). Starting at an angle of less than $10^o$, allow the pendulum to swing and measure the pendulum’s period for 10 oscillations using a stopwatch. Calculate $g$. How accurate is this measurement? How might it be improved? Exercise $\PageIndex{1}$ An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of $10 \, kg$. Pendulum 2 has a bob with a mass of $100 \, kg$. Describe how the motion of the pendula will differ if the bobs are both displaced by $12^o$. - Answer - The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum’s length) and by the acceleration due to gravity. PHET EXPLORATIONS: PENDELUM LAB Play with one or two pendulums and discover how the period of a simple pendulum depends on the length of the string, the mass of the pendulum bob, and the amplitude of the swing. It’s easy to measure the period using the photogate timer. You can vary friction and the strength of gravity. Use the pendulum to find the value of $g$ on planet X. Notice the anharmonic behavior at large amplitude. Glossary - simple pendulum - an object with a small mass suspended from a light wire or string	libretexts	2025-03-17T19:53:34.506411	2015-11-03T20:08:18	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.04%3A_The_Simple_Pendulum", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.4: The Simple Pendulum", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.05%3A_Energy_and_the_Simple_Harmonic_Oscillator	16.5: Energy and the Simple Harmonic Oscillator Learning Objectives By the end of this section, you will be able to: - Determine the maximum speed of an oscillating system. To study the energy of a simple harmonic oscillator, we first consider all the forms of energy it can have. We know from Hooke’s Law: Stress and Strain Revisited that the energy stored in the deformation of a simple harmonic oscillator is a form of potential energy given by: \[PR_{el} = \dfrac{1}{2}kx^2.\] Because a simple harmonic oscillator has no dissipative forces, the other important form of energy is kinetic energy $KE$. Conservation of energy for these two forms is: \[KE + PE_{el} = constant\] or \[\dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2 = constant.\] This statement of conservation of energy is valid for all simple harmonic oscillators, including ones where the gravitational force plays a role. Namely, for a simple pendulum we replace the velocity with $v = L\omega$, the spring constant with $k = mg/L$ and the displacement term with $x = L\theta$. Thus \[\dfrac{1}{2}mL^2\omega^2 + \dfrac{1}{2}mgL\theta^2 = constant.\] In the case of undamped simple harmonic motion, the energy oscillates back and forth between kinetic and potential, going completely from one to the other as the system oscillates. So for the simple example of an object on a frictionless surface attached to a spring, as shown again in Figure $\PageIndex{1}$, the motion starts with all of the energy stored in the spring. As the object starts to move, the elastic potential energy is converted to kinetic energy, becoming entirely kinetic energy at the equilibrium position. It is then converted back into elastic potential energy by the spring, the velocity becomes zero when the kinetic energy is completely converted, and so on. This concept provides extra insight here and in later applications of simple harmonic motion, such as alternating current circuits. The conservation of energy principle can be used to derive an expression for velocity $v$. If we start our simple harmonic motion with zero velocity and maximum displacement ($x = X$), then the total energy is \[\dfrac{1}{2}kX^2.\] This total energy is constant and is shifted back and forth between kinetic energy and potential energy, at most times being shared by each. The conservation of energy for this system in equation form is thus: \[\dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2 = \dfrac{1}{2}kX^2.\] Solving this equation for $v$ yields: \[v = \pm \sqrt{\dfrac{k}{m}(X^2 - x^2)}.\] Manipulating this expression algebraically gives: \[v = \pm\sqrt{\dfrac{k}{m}}X\sqrt{1 - \dfrac{x^2}{X^2}}\] and so \[v = \pm v_{max} \sqrt{1 - \dfrac{x^2}{X^2}},\] where \[v_{max} = \sqrt{\dfrac{k}{m}}X.\] From this expression, we see that the velocity is a maximum ($v_{max}$) at $x = 0$, as stated earlier in $v(t) = - v_{max} \, sin \, \frac{2\pi t}{T}$. Notice that the maximum velocity depends on three factors. Maximum velocity is directly proportional to amplitude. As you might guess, the greater the maximum displacement the greater the maximum velocity. Maximum velocity is also greater for stiffer systems, because they exert greater force for the same displacement. This observation is seen in the expression for $v_{max}$ it is proportional to the square root of the force constant $k$ Finally, the maximum velocity is smaller for objects that have larger masses, because the maximum velocity is inversely proportional to the square root of $m$. For a given force, objects that have large masses accelerate more slowly. A similar calculation for the simple pendulum produces a similar result, namely: \[\omega_{max} = \sqrt{\dfrac{g}{L}}\theta_{max}.\] Example $\PageIndex{1}$: Determine the Maximum Speed of an Oscillating System: A Bumpy Road Suppose that a car is 900 kg and has a suspension system that has a force constant $k = 6.53 \times 10^4 \, N/m$. The car hits a bump and bounces with an amplitude of 0.100 m. What is its maximum vertical velocity if you assume no damping occurs? Strategy We can use the expression for $v_{max}$ given in $v_{max} = \sqrt{\frac{k}{m}}X$ to determine the maximum vertical velocity. The variables $m$ and $k$ are given in the problem statement, and the maximum displacement $X$ is 0.100 m. Solution - Identify known. - Substitute known values into $v_{max} = \sqrt{\frac{k}{m}}X$. \[v_{max} = \sqrt{\dfrac{6.53 \times 10^4 \, N/m}{900 \, kg}}(0.100 \, m).\] - Calculate to find $v_{max} = 0.852 \, m/s$. Discussion This answer seems reasonable for a bouncing car. There are other ways to use conservation of energy to find $v_{max}$. We could use it directly, as was done in the example featured in Hooke’s Law: Stress and Strain Revisited . The small vertical displacement $y$ of an oscillating simple pendulum, starting from its equilibrium position, is given as \[y(t) = a \, sin \, \omega t,\] where $a$ is the amplitude, $\omega$ is the angular velocity and $t$ is the time taken. Substituting $\omega = \frac{2\pi}{T},$ we have \[y(t) = a \, sin \left(\dfrac{2\pi t}{T} \right).\] Thus, the displacement of pendulum is a function of time as shown above. Also the velocity of the pendulum is given by \[v(t) = \dfrac{2a\pi}{T} \, cos \left(\dfrac{2\pi t}{T}\right),\] so the motion of the pendulum is a function of time. Exercise $\PageIndex{1}$ Why does it hurt more if your hand is snapped with a ruler than with a loose spring, even if the displacement of each system is equal? Solution The ruler is a stiffer system, which carries greater force for the same amount of displacement. The ruler snaps your hand with greater force, which hurts more. Exercise $\PageIndex{2}$:Check Your Understanding You are observing a simple harmonic oscillator. Identify one way you could decrease the maximum velocity of the system. Solution You could increase the mass of the object that is oscillating. Summary - Energy in the simple harmonic oscillator is shared between elastic potential energy and kinetic energy, with the total being constant: \[\dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2 = constant. \nonumber\] - Maximum velocity depends on three factors: it is directly proportional to amplitude, it is greater for stiffer systems, and it is smaller for objects that have larger masses: \[v_{max} = \sqrt{\dfrac{k}{m}}X. \nonumber\]	libretexts	2025-03-17T19:53:34.577317	2015-11-03T20:08:53	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.05%3A_Energy_and_the_Simple_Harmonic_Oscillator", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.5: Energy and the Simple Harmonic Oscillator", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.06%3A__Uniform_Circular_Motion_and_Simple_Harmonic_Motion	16.6: Uniform Circular Motion and Simple Harmonic Motion Learning Objectives By the end of this section, you will be able to: - Compare simple harmonic motion with uniform circular motion. There is an easy way to produce simple harmonic motion by using uniform circular motion. Figure $\PageIndex{2}$ shows one way of using this method. A ball is attached to a uniformly rotating vertical turntable, and its shadow is projected on the floor as shown. The shadow undergoes simple harmonic motion. Hooke’s law usually describes uniform circular motions ($\omega$ constant) rather than systems that have large visible displacements. So observing the projection of uniform circular motion, as in Figure $\PageIndex{2}$, is often easier than observing a precise large-scale simple harmonic oscillator. If studied in sufficient depth, simple harmonic motion produced in this manner can give considerable insight into many aspects of oscillations and waves and is very useful mathematically. In our brief treatment, we shall indicate some of the major features of this relationship and how they might be useful. Figure $\PageIndex{3}$ shows the basic relationship between uniform circular motion and simple harmonic motion. The point P travels around the circle at constant angular velocity $\omega$. The point P is analogous to an object on the merry-go-round. The projection of the position of P onto a fixed axis undergoes simple harmonic motion and is analogous to the shadow of the object. At the time shown in the figure, the projection has position $x$ and moves to the left with velocity $v$. The velocity of the point P around the circle equals $\bar{v}_{max}$. The projection of $\bar{v}_{max}$ on the $x$-axis is the velocity $v$ of the simple harmonic motion along the $x$-axis. To see that the projection undergoes simple harmonic motion, note that its position $x$ is given by: \[x = X \cos{\theta},\] is the constant angular velocity, and $X$ is the radius of the circular path. Thus, \[x = X \cos{\omega t}.\] The angular velocity $\omega$ is in radians per unit time; in this case $2 \pi$ radians is the time for one revolution $T$. That is, $\omega = 2\pi/T$. Substituting this expression for $\omega$, we see that the position $x$ is given by: \[x(t) = \cos{\left(\frac{2\pi t}{T}\right)}.\] This expression is the same one we had for the position of a simple harmonic oscillator in Simple Harmonic Motion: A Special Periodic Motion . If we make a graph of position versus time as in Figure $\PageIndex{4}$ we see again the wavelike character (typical of simple harmonic motion) of the projection of uniform circular motion onto the $x$-axis. Now let us use Figure $\PageIndex{3}$ to do some further analysis of uniform circular motion as it relates to simple harmonic motion. The triangle formed by the velocities in the figure and the triangle formed by the displacements ($X$, $x$, and $\sqrt{X^{2} - x^{2}}$) are similar right triangles. Taking ratios of similar sides, we see that \[\frac{v}{v_{max}} = \frac{\sqrt{X^{X} - x^{2}}}{X} = \sqrt{1-\frac{x^{2}}{X^{2}}}.\] We can solve this equation for the speed $v$ or \[v = v_{max}\sqrt{1 - \frac{x^2}{X^{2}}}.\] This expression for the speed of a simple harmonic oscillator is exactly the same as the equation obtained from conservation of energy considerations in Energy and the Simple Harmonic Oscillator. You can begin to see that it is possible to get all of the characteristics of simple harmonic motion from an analysis of the projection of uniform circular motion. Finally, let us consider the period $T$ of the motion of the projection. This period is the time it takes the point P to complete one revolution. That time is the circumference of the circle $2 \pi X$ divided by the velocity around the circle, $v_{max}$. Thus, the period $T$ is \[T = \frac{2\pi X}{v_{max}}.\] We know from conservation of energy considerations that \[v_{max} = \sqrt{\frac{k}{m}}X.\] Solving this equation for $X/v_{max}$ gives \[\frac{X}{v_max} = \sqrt{\frac{m}{k}}.\] Substituting this expression into the equation for $T$ yields \[tT = 2\pi \sqrt{\frac{m}{k}}.\] Thus, the period of the motion is the same as for a simple harmonic oscillator. We have determined the period for any simple harmonic oscillator using the relationship between uniform circular motion and simple harmonic motion. Some modules occasionally refer to the connection between uniform circular motion and simple harmonic motion. Moreover, if you carry your study of physics and its applications to greater depths, you will find this relationship useful. It can, for example, help to analyze how waves add when they are superimposed. Exercise $\PageIndex{1}$ Identify an object that undergoes uniform circular motion. Describe how you could trace the simple harmonic motion of this object as a wave. - Answer - A record player undergoes uniform circular motion. You could attach dowel rod to one point on the outside edge of the turntable and attach a pen to the other end of the dowel. As the record player turns, the pen will move. You can drag a long piece of paper under the pen, capturing its motion as a wave. Summary - A projection of uniform circular motion undergoes simple harmonic oscillation.	libretexts	2025-03-17T19:53:34.644262	2017-01-25T17:53:54	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.06%3A__Uniform_Circular_Motion_and_Simple_Harmonic_Motion", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.6: Uniform Circular Motion and Simple Harmonic Motion", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.07%3A_Damped_Harmonic_Motion	16.7: Damped Harmonic Motion Learning Objectives By the end of this section, you will be able to: - Compare and discuss underdamped and overdamped oscillating systems. - Explain critically damped system. A guitar string stops oscillating a few seconds after being plucked. To keep a child happy on a swing, you must keep pushing. Although we can often make friction and other non-conservative forces negligibly small, completely undamped motion is rare. In fact, we may even want to damp oscillations, such as with car shock absorbers. For a system that has a small amount of damping, the period and frequency are nearly the same as for simple harmonic motion, but the amplitude gradually decreases as shown in Figure $\PageIndex{2}$. This occurs because the non-conservative damping force removes energy from the system, usually in the form of thermal energy. In general, energy removal by non-conservative forces is described as \[W_{nc} = \Delta (KE + PE),\] where $W_{nc}$ is work done by a non-conservative force (here the damping force). For a damped harmonic oscillator, $W_{nc}$ is negative because it removes mechanical energy (KE + PE) from the system. If you gradually increase the amount of damping in a system, the period and frequency begin to be affected, because damping opposes and hence slows the back and forth motion. (The net force is smaller in both directions.) If there is very large damping, the system does not even oscillate—it slowly moves toward equilibrium. Figure $\PageIndex{3}$shows the displacement of a harmonic oscillator for different amounts of damping. When we want to damp out oscillations, such as in the suspension of a car, we may want the system to return to equilibrium as quickly as possible Critical damping is defined as the condition in which the damping of an oscillator results in it returning as quickly as possible to its equilibrium position The critically damped system may overshoot the equilibrium position, but if it does, it will do so only once. Critical damping is represented by Curve A in Figure $\PageIndex{3}$. With less-than critical damping, the system will return to equilibrium faster but will overshoot and cross over one or more times. Such a system is underdamped ; its displacement is represented by the curve in Figure $\PageIndex{2}$. Curve B in Figure $\PageIndex{3}$ represents an overdamped system. As with critical damping, it too may overshoot the equilibrium position, but will reach equilibrium over a longer period of time. Critical damping is often desired, because such a system returns to equilibrium rapidly and remains at equilibrium as well. In addition, a constant force applied to a critically damped system moves the system to a new equilibrium position in the shortest time possible without overshooting or oscillating about the new position. For example, when you stand on bathroom scales that have a needle gauge, the needle moves to its equilibrium position without oscillating. It would be quite inconvenient if the needle oscillated about the new equilibrium position for a long time before settling. Damping forces can vary greatly in character. Friction, for example, is sometimes independent of velocity (as assumed in most places in this text). But many damping forces depend on velocity—sometimes in complex ways, sometimes simply being proportional to velocity. Example $\PageIndex{1}$: Damping an Oscillatory Motion: Friction on an Object Connected to a Spring Damping oscillatory motion is important in many systems, and the ability to control the damping is even more so. This is generally attained using non-conservative forces such as the friction between surfaces, and viscosity for objects moving through fluids. The following example considers friction. Suppose a 0.200-kg object is connected to a spring as shown in Figure $\PageIndex{4}$, but there is simple friction between the object and the surface, and the coefficient of friction $\mu_k$ is equal to 0.0800. (a) What is the frictional force between the surfaces? (b) What total distance does the object travel if it is released 0.100 m from equilibrium, starting at $v = 0$? The force constant of the spring is $k = 50.0 \, N/m$. Strategy This problem requires you to integrate your knowledge of various concepts regarding waves, oscillations, and damping. To solve an integrated concept problem, you must first identify the physical principles involved. Part (a) is about the frictional force. This is a topic involving the application of Newton’s Laws. Part (b) requires an understanding of work and conservation of energy, as well as some understanding of horizontal oscillatory systems. Now that we have identified the principles we must apply in order to solve the problems, we need to identify the knowns and unknowns for each part of the question, as well as the quantity that is constant in Part (a) and Part (b) of the question. Solution a - Choose the proper equation: Friction is $f = \mu_kmg$. - Identify the known values. - Enter the known values into the equation: \[f = (0.0800)(0.200 \, kg)(9.80 \, m/s^2).\] - Calculate and convert units: $f = 0.157 \, N$. Discussion a The force here is small because the system and the coefficients are small. Solution b Identify the known: - The system involves elastic potential energy as the spring compresses and expands, friction that is related to the work done, and the kinetic energy as the body speeds up and slows down. - Energy is not conserved as the mass oscillates because friction is a non-conservative force. - The motion is horizontal, so gravitational potential energy does not need to be considered. - Because the motion starts from rest, the energy in the system is initially $PE_{el,i} = (1/2)kX^2$. This energy is removed by work done by friction $W_{nc} = -fd$, where $d$ is the total distance traveled and $f = \mu_kmg$ is the force of friction. When the system stops moving, the friction force will balance the force exerted by the spring, so $PE_{el,f} = (1/2)kx^2$, where $x$ is the final position and is given by \[F_{el} = f\]\[kx = \mu_kmg.\]\[x = \dfrac{\mu_kmg}{k}\] - By equating the work done to the energy removed, solve for the distance $d$. - The work done by the non-conservative forces equals the initial, stored elastic potential energy. Identify the correct equation to use: \[W_{nc} = \Delta (KE + PE) = PE_{el,f} - PE_{el,i} = \dfrac{1}{2}k \left(\left(\dfrac{\mu_kmg}{k}\right)^2 - X^2\right).\] - Recall that $W_{nc} = -fd$. - Enter the friction as $f = \mu_kmg$ into $W_{nc} = - fd$, thus \[W_{nc} = \mu_kmgd.\] - Combine these two equations to find \[\dfrac{1}{2}k \left( \left(\dfrac{\mu_kmg}{k}\right)^2 - X^2\right) = - \mu_kmgd.\] - Solve the equation for $d$:\[d = \dfrac{k}{2\mu_kmg} \left(X^2 - \left(\dfrac{\mu_kmg}{k}\right)^2\right).\] - Enter the known values into the resulting equation: \[d = \dfrac{50.0 \, N/m}{2(0.0800)(0.200 \, kg)(9.80 \, m/s^2)}\left((0.100 \, m)^2 - \left(\dfrac{(0.0800)(0.200 \, kg)(9.80 \, m/s^2)}{50.0 \, N/m} \right)^2 \right).\] - Calculate $d$ and convert units: \[d = 1.59 \, m\] Discussion b This is the total distance traveled back and forth across $x = 0$, which is the undamped equilibrium position. The number of oscillations about the equilibrium position will be more than $d/X = (1.59 \, m)(0.100 \, m) = 15.9$ because the amplitude of the oscillations is decreasing with time. At the end of the motion, this system will not return to $x = 0$ for this type of damping force, because static friction will exceed the restoring force. This system is underdamped. In contrast, an overdamped system with a simple constant damping force would not cross the equilibrium position $x = 0$ a single time. For example, if this system had a damping force 20 times greater, it would only move 0.0484 m toward the equilibrium position from its original 0.100-m position. This worked example illustrates how to apply problem-solving strategies to situations that integrate the different concepts you have learned. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknowns using familiar problem-solving strategies. These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example, you can see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course, such as in your profession, in other science disciplines, and in everyday life. Exercise $\PageIndex{1}$:Check Your Understanding Why are completely undamped harmonic oscillators so rare? - Answer - Friction often comes into play whenever an object is moving. Friction causes damping in a harmonic oscillator. Exercise $\PageIndex{2}$:Check Your Understanding Describe the difference between overdamping, underdamping, and critical damping. - Answer - An overdamped system moves slowly toward equilibrium. An underdamped system moves quickly to equilibrium, but will oscillate about the equilibrium point as it does so. A critically damped system moves as quickly as possible toward equilibrium without oscillating about the equilibrium. Summary - Damped harmonic oscillators have non-conservative forces that dissipate their energy. - Critical damping returns the system to equilibrium as fast as possible without overshooting. - An underdamped system will oscillate through the equilibrium position. - An overdamped system moves more slowly toward equilibrium than one that is critically damped. Glossary - critical damping - the condition in which the damping of an oscillator causes it to return as quickly as possible to its equilibrium position without oscillating back and forth about this position - over damping - the condition in which damping of an oscillator causes it to return to equilibrium without oscillating; oscillator moves more slowly toward equilibrium than in the critically damped system - under damping - the condition in which damping of an oscillator causes it to return to equilibrium with the amplitude gradually decreasing to zero; system returns to equilibrium faster but overshoots and crosses the equilibrium position one or more times	libretexts	2025-03-17T19:53:34.718390	2015-11-03T20:09:13	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.07%3A_Damped_Harmonic_Motion", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.7: Damped Harmonic Motion", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.08%3A_Forced_Oscillations_and_Resonance	16.8: Forced Oscillations and Resonance Learning Objectives By the end of this section, you will be able to: - Observe resonance of a paddle ball on a string. - Observe amplitude of a damped harmonic oscillator. Sit in front of a piano sometime and sing a loud brief note at it with the dampers off its strings. It will sing the same note back at you—the strings, having the same frequencies as your voice, are resonating in response to the forces from the sound waves that you sent to them. Your voice and a piano’s strings is a good example of the fact that objects—in this case, piano strings—can be forced to oscillate but oscillate best at their natural frequency. In this section, we shall briefly explore applying a periodic driving force acting on a simple harmonic oscillator. The driving force puts energy into the system at a certain frequency, not necessarily the same as the natural frequency of the system. The natural frequency is the frequency at which a system would oscillate if there were no driving and no damping force. Most of us have played with toys involving an object supported on an elastic band, something like the paddle ball suspended from a finger in Figure $\PageIndex{2}$. Imagine the finger in the figure is your finger. At first you hold your finger steady, and the ball bounces up and down with a small amount of damping. If you move your finger up and down slowly, the ball will follow along without bouncing much on its own. As you increase the frequency at which you move your finger up and down, the ball will respond by oscillating with increasing amplitude. When you drive the ball at its natural frequency, the ball’s oscillations increase in amplitude with each oscillation for as long as you drive it. The phenomenon of driving a system with a frequency equal to its natural frequency is called resonance . A system being driven at its natural frequency is said to resonate . As the driving frequency gets progressively higher than the resonant or natural frequency, the amplitude of the oscillations becomes smaller, until the oscillations nearly disappear and your finger simply moves up and down with little effect on the ball. Figure $\PageIndex{3}$ shows a graph of the amplitude of a damped harmonic oscillator as a function of the frequency of the periodic force driving it. There are three curves on the graph, each representing a different amount of damping. All three curves peak at the point where the frequency of the driving force equals the natural frequency of the harmonic oscillator. The highest peak, or greatest response, is for the least amount of damping, because less energy is removed by the damping force. It is interesting that the widths of the resonance curves shown in Figure $\PageIndex{3}$ depend on damping: the less the damping, the narrower the resonance. The message is that if you want a driven oscillator to resonate at a very specific frequency, you need as little damping as possible. Little damping is the case for piano strings and many other musical instruments. Conversely, if you want small-amplitude oscillations, such as in a car’s suspension system, then you want heavy damping. Heavy damping reduces the amplitude, but the tradeoff is that the system responds at more frequencies. These features of driven harmonic oscillators apply to a huge variety of systems. When you tune a radio, for example, you are adjusting its resonant frequency so that it only oscillates to the desired station’s broadcast (driving) frequency. The more selective the radio is in discriminating between stations, the smaller its damping. Magnetic resonance imaging (MRI) is a widely used medical diagnostic tool in which atomic nuclei (mostly hydrogen nuclei) are made to resonate by incoming radio waves (on the order of 100 MHz). A child on a swing is driven by a parent at the swing’s natural frequency to achieve maximum amplitude. In all of these cases, the efficiency of energy transfer from the driving force into the oscillator is best at resonance. Speed bumps and gravel roads prove that even a car’s suspension system is not immune to resonance. In spite of finely engineered shock absorbers, which ordinarily convert mechanical energy to thermal energy almost as fast as it comes in, speed bumps still cause a large-amplitude oscillation. On gravel roads that are corrugated, you may have noticed that if you travel at the “wrong” speed, the bumps are very noticeable whereas at other speeds you may hardly feel the bumps at all. Figure $\PageIndex{4}$ shows a photograph of a famous example (the Tacoma Narrows Bridge) of the destructive effects of a driven harmonic oscillation. The Millennium Bridge in London was closed for a short period of time for the same reason while inspections were carried out. In our bodies, the chest cavity is a clear example of a system at resonance. The diaphragm and chest wall drive the oscillations of the chest cavity which result in the lungs inflating and deflating. The system is critically damped and the muscular diaphragm oscillates at the resonant value for the system, making it highly efficient. Exercise $\PageIndex{1}$ A famous magic trick involves a performer singing a note toward a crystal glass until the glass shatters. Explain why the trick works in terms of resonance and natural frequency. - Answer - The performer must be singing a note that corresponds to the natural frequency of the glass. As the sound wave is directed at the glass, the glass responds by resonating at the same frequency as the sound wave. With enough energy introduced into the system, the glass begins to vibrate and eventually shatters. Summary - A system’s natural frequency is the frequency at which the system will oscillate if not affected by driving or damping forces. - A periodic force driving a harmonic oscillator at its natural frequency produces resonance. The system is said to resonate. - The less damping a system has, the higher the amplitude of the forced oscillations near resonance. The more damping a system has, the broader response it has to varying driving frequencies. Glossary - natural frequency - the frequency at which a system would oscillate if there were no driving and no damping forces - resonance - the phenomenon of driving a system with a frequency equal to the system's natural frequency - resonate - a system being driven at its natural frequency	libretexts	2025-03-17T19:53:34.785842	2015-11-03T20:09:54	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.08%3A_Forced_Oscillations_and_Resonance", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.8: Forced Oscillations and Resonance", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.09%3A_Waves	16.9: Waves Learning Objectives By the end of this section, you will be able to: - State the characteristics of a wave. - Calculate the velocity of wave propagation. What do we mean when we say something is a wave? The most intuitive and easiest wave to imagine is the familiar water wave. More precisely, a wave is a disturbance that propagates, or moves from the place it was created. For water waves, the disturbance is in the surface of the water, perhaps created by a rock thrown into a pond or by a swimmer splashing the surface repeatedly. For sound waves, the disturbance is a change in air pressure, perhaps created by the oscillating cone inside a speaker. For earthquakes, there are several types of disturbances, including disturbance of Earth’s surface and pressure disturbances under the surface. Even radio waves are most easily understood using an analogy with water waves. Visualizing water waves is useful because there is more to it than just a mental image. Water waves exhibit characteristics common to all waves, such as amplitude, period, frequency and energy. All wave characteristics can be described by a small set of underlying principles. A wave is a disturbance that propagates, or moves from the place it was created. The simplest waves repeat themselves for several cycles and are associated with simple harmonic motion. Let us start by considering the simplified water wave in Figure $\PageIndex{2}$ . The wave is an up and down disturbance of the water surface. It causes a sea gull to move up and down in simple harmonic motion as the wave crests and troughs (peaks and valleys) pass under the bird. The time for one complete up and down motion is the wave’s period $T$. The wave’s frequency is $f = 1/T$, as usual. The wave itself moves to the right in the figure. This movement of the wave is actually the disturbance moving to the right, not the water itself (or the bird would move to the right). We define wave velocity $v_w$ to be the speed at which the disturbance moves. Wave velocity is sometimes also called the propagation velocity or propagation speed , because the disturbance propagates from one location to another. MISCONCEPTION ALERT Many people think that water waves push water from one direction to another. In fact, the particles of water tend to stay in one location, save for moving up and down due to the energy in the wave. The energy moves forward through the water, but the water stays in one place. If you feel yourself pushed in an ocean, what you feel is the energy of the wave, not a rush of water. The water wave in Figure $\PageIndex{2}$ also has a length associated with it, called its wavelength $\lambda$ the distance between adjacent identical parts of a wave. ($\lambda$ is the distance parallel to the direction of propagation.) The speed of propagation $v_w$ is the distance the wave travels in a given time, which is one wavelength in the time of one period. In equation form, that is \[v_w = \dfrac{\lambda}{T} \label{eq1}\] or \[v_w = f\lambda. \label{eq2}\] This fundamental relationship holds for all types of waves. For water waves, $v_w$ is the speed of a surface wave; for sound, $v_w$ is the speed of sound; and for visible light, $v_w$ is the speed of light, for example. TAKE-HOME EXPERIMENT: WAVES IN A BOWL Fill a large bowl or basin with water and wait for the water to settle so there are no ripples. Gently drop a cork into the middle of the bowl. Estimate the wavelength and period of oscillation of the water wave that propagates away from the cork. Remove the cork from the bowl and wait for the water to settle again. Gently drop the cork at a height that is different from the first drop. Does the wavelength depend upon how high above the water the cork is dropped? Example $\PageIndex{1}$: Calculate the Velocity of Wave Propagation: Gull in the Ocean Calculate the wave velocity of the ocean wave in Figure $\PageIndex{2}$ if the distance between wave crests is 10.0 m and the time for a sea gull to bob up and down is 5.00 s. Strategy We are asked to find $v_w$. The given information tells us that $\lambda = 10.0 \, m$ and $T = 5.00 \, s$. Therefore, we can use Equation \ref{eq1} to find the wave velocity. Solution - Enter the known values into Equation \ref{eq1}: \[\begin{align} v_w &= \frac{\lambda}{T} \\[5pt] &= \dfrac{10.0 \, m}{5.00 \, s}. \end{align}\] - Solve for $v_w$ to find $v_w = 2.00 \, m/s.$ Discussion This slow speed seems reasonable for an ocean wave. Note that the wave moves to the right in the figure at this speed, not the varying speed at which the sea gull moves up and down. Transverse and Longitudinal Waves A simple wave consists of a periodic disturbance that propagates from one place to another. The wave in Figure $\PageIndex{3}$ propagates in the horizontal direction while the surface is disturbed in the vertical direction. Such a wave is called a transverse wave or shear wave; in such a wave, the disturbance is perpendicular to the direction of propagation. In contrast, in a longitudinal wave or compressional wave, the disturbance is parallel to the direction of propagation. Figure $\PageIndex{4}$ shows an example of a longitudinal wave. The size of the disturbance is its amplitude X and is completely independent of the speed of propagation $v_w$. Waves may be transverse, longitudinal, or a combination of the two . (Water waves are actually a combination of transverse and longitudinal. The simplified water wave illustrated in Figure $\PageIndex{2}$shows no longitudinal motion of the bird.) The waves on the strings of musical instruments are transverse—so are electromagnetic waves, such as visible light. Sound waves in air and water are longitudinal. Their disturbances are periodic variations in pressure that are transmitted in fluids. Fluids do not have appreciable shear strength, and thus the sound waves in them must be longitudinal or compressional. Sound in solids can be both longitudinal and transverse. Earthquake waves under Earth’s surface also have both longitudinal and transverse components (called compressional or P-waves and shear or S-waves, respectively). These components have important individual characteristics—they propagate at different speeds, for example. Earthquakes also have surface waves that are similar to surface waves on water. Exercise $\PageIndex{1}$:Check Your Understanding Why is it important to differentiate between longitudinal and transverse waves? - Answer - In the different types of waves, energy can propagate in a different direction relative to the motion of the wave. This is important to understand how different types of waves affect the materials around them. PHET EXPLORATIONS: WAVE ON A STRING Watch a string vibrate in slow motion with this PhET simulation. Wiggle the end of the string and make waves, or adjust the frequency and amplitude of an oscillator. Adjust the damping and tension. The end can be fixed, loose, or open. Summary - A wave is a disturbance that moves from the point of creation with a wave velocity $v_w$. - A wave has a wavelength $\lambda$ which is the distance between adjacent identical parts of the wave. - Wave velocity and wavelength are related to the wave’s frequency and period by $v_w = \frac{\lambda}{T}$ or $v_w = f\lambda$. - A transverse wave has a disturbance perpendicular to its direction of propagation, whereas a longitudinal wave has a disturbance parallel to its direction of propagation. Glossary - longitudinal wave - a wave in which the disturbance is parallel to the direction of propagation - transverse wave - a wave in which the disturbance is perpendicular to the direction of propagation - wave velocity - the speed at which the disturbance moves. Also called the propagation velocity or propagation speed - wavelength - the distance between adjacent identical parts of a wave	libretexts	2025-03-17T19:53:34.936995	2015-11-03T20:10:16	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.09%3A_Waves", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.9: Waves", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.10%3A_Superposition_and_Interference	16.10: Superposition and Interference Learning Objectives By the end of this section, you will be able to: - Explain standing waves. - Describe the mathematical representation of overtones and beat frequency. Most waves do not look very simple. They look more like the waves in Figure $\PageIndex{1}$ than like the simple water wave considered in Waves . (Simple waves may be created by a simple harmonic oscillation, and thus have a sinusoidal shape). Complex waves are more interesting, even beautiful, but they look formidable. Most waves appear complex because they result from several simple waves adding together. Luckily, the rules for adding waves are quite simple. When two or more waves arrive at the same point, they superimpose themselves on one another. More specifically, the disturbances of waves are superimposed when they come together—a phenomenon called superposition . Each disturbance corresponds to a force, and forces add. If the disturbances are along the same line, then the resulting wave is a simple addition of the disturbances of the individual waves—that is, their amplitudes add. Figure $\PageIndex{2}$ and Figure $\PageIndex{3}$ illustrate superposition in two special cases, both of which produce simple results. Figure $\PageIndex{2}$ shows two identical waves that arrive at the same point exactly in phase. The crests of the two waves are precisely aligned, as are the troughs. This superposition produces pure constructive interference . Because the disturbances add, pure constructive interference produces a wave that has twice the amplitude of the individual waves, but has the same wavelength. Figure $\PageIndex{3}$ shows two identical waves that arrive exactly out of phase—that is, precisely aligned crest to trough—producing pure destructive interference . Because the disturbances are in the opposite direction for this superposition, the resulting amplitude is zero for pure destructive interference—the waves completely cancel. While pure constructive and pure destructive interference do occur, they require precisely aligned identical waves. The superposition of most waves produces a combination of constructive and destructive interference and can vary from place to place and time to time. Sound from a stereo, for example, can be loud in one spot and quiet in another. Varying loudness means the sound waves add partially constructively and partially destructively at different locations. A stereo has at least two speakers creating sound waves, and waves can reflect from walls. All these waves superimpose. An example of sounds that vary over time from constructive to destructive is found in the combined whine of airplane jets heard by a stationary passenger. The combined sound can fluctuate up and down in volume as the sound from the two engines varies in time from constructive to destructive. These examples are of waves that are similar. An example of the superposition of two dissimilar waves is shown in Figure $\PageIndex{4}$. Here again, the disturbances add and subtract, producing a more complicated looking wave. Standing Waves Sometimes waves do not seem to move; rather, they just vibrate in place. Unmoving waves can be seen on the surface of a glass of milk in a refrigerator, for example. Vibrations from the refrigerator motor create waves on the milk that oscillate up and down but do not seem to move across the surface. These waves are formed by the superposition of two or more moving waves, such as illustrated in Figure $\PageIndex{5}$ for two identical waves moving in opposite directions. The waves move through each other with their disturbances adding as they go by. If the two waves have the same amplitude and wavelength, then they alternate between constructive and destructive interference. The resultant looks like a wave standing in place and, thus, is called a standing wave . Waves on the glass of milk are one example of standing waves. There are other standing waves, such as on guitar strings and in organ pipes. With the glass of milk, the two waves that produce standing waves may come from reflections from the side of the glass. A closer look at earthquakes provides evidence for conditions appropriate for resonance, standing waves, and constructive and destructive interference. A building may be vibrated for several seconds with a driving frequency matching that of the natural frequency of vibration of the building—producing a resonance resulting in one building collapsing while neighboring buildings do not. Often buildings of a certain height are devastated while other taller buildings remain intact. The building height matches the condition for setting up a standing wave for that particular height. As the earthquake waves travel along the surface of Earth and reflect off denser rocks, constructive interference occurs at certain points. Often areas closer to the epicenter are not damaged while areas farther away are damaged. Standing waves are also found on the strings of musical instruments and are due to reflections of waves from the ends of the string. Figure $\PageIndex{6}$ and Figure $\PageIndex{7}$ show three standing waves that can be created on a string that is fixed at both ends. Nodes are the points where the string does not move; more generally, nodes are where the wave disturbance is zero in a standing wave. The fixed ends of strings must be nodes, too, because the string cannot move there. The word antinode is used to denote the location of maximum amplitude in standing waves. Standing waves on strings have a frequency that is related to the propagation speed $v_w$ of the disturbance on the string. The wavelength $\lambda$ is determined by the distance between the points where the string is fixed in place. The lowest frequency, called the fundamental frequency , is thus for the longest wavelength, which is seen to be $\lambda_1 = 2L$. Therefore, the fundamental frequency is $f_1 = v_w/\lambda_1 = v_w/2L$. In this case, the overtones or harmonics are multiples of the fundamental frequency. As seen in Figure $\PageIndex{7}$, the first harmonic can easily be calculated since $\lambda_2 = L$. Thus, $f_2 = v_w/\lambda_2 = v_w/2L = 2f_1$. Similarly, $f_3 = 3f_1$, and so on. All of these frequencies can be changed by adjusting the tension in the string. The greater the tension, the greater $v_w$ is and the higher the frequencies. This observation is familiar to anyone who has ever observed a string instrument being tuned. We will see in later chapters that standing waves are crucial to many resonance phenomena, such as in sounding boxes on string instruments. Beats Striking two adjacent keys on a piano produces a warbling combination usually considered to be unpleasant. The superposition of two waves of similar but not identical frequencies is the culprit. Another example is often noticeable in jet aircraft, particularly the two-engine variety, while taxiing. The combined sound of the engines goes up and down in loudness. This varying loudness happens because the sound waves have similar but not identical frequencies. The discordant warbling of the piano and the fluctuating loudness of the jet engine noise are both due to alternately constructive and destructive interference as the two waves go in and out of phase. Figure $\PageIndex{8}$ illustrates this graphically. The wave resulting from the superposition of two similar-frequency waves has a frequency that is the average of the two. This wave fluctuates in amplitude, or beats , with a frequency called the beat frequency . We can determine the beat frequency by adding two waves together mathematically. Note that a wave can be represented at one point in space as \[x = X \, cos \left(\dfrac{2\pi t}{T} \right) = X \, cos \, (2\pi \, ft),\] where $f = 1/T$ is the frequency of the wave. Adding two waves that have different frequencies but identical amplitudes produces a resultant \[x = x_1 + x_2.\] More specifically, \[x = X \, cos (2\pi \, f_1 t) + X \, cos (2\pi \, f_2 t).\] Using a trigonometric identity, it can be shown that \[x = 2X \, cos (\pi \, f_Bt) cos (2\pi \, f_{ave}t),\] where \[f_B = \|f_1 - f_2\|\] is the beat frequency, and $f_{ave}$ is the average of $f_1$ and $f_2$. These results mean that the resultant wave has twice the amplitude and the average frequency of the two superimposed waves, but it also fluctuates in overall amplitude at the beat frequency $f_B$. The first cosine term in the expression effectively causes the amplitude to go up and down. The second cosine term is the wave with frequency $f_{ave}$. This result is valid for all types of waves. However, if it is a sound wave, providing the two frequencies are similar, then what we hear is an average frequency that gets louder and softer (or warbles) at the beat frequency. MAKING CAREER CONNECTIONS Piano tuners use beats routinely in their work. When comparing a note with a tuning fork, they listen for beats and adjust the string until the beats go away (to zero frequency). For example, if the tuning fork has a $256 hZ$ frequency and two beats per second are heard, then the other frequency is either $254$ or $258 \, Hz$. Most keys hit multiple strings, and these strings are actually adjusted until they have nearly the same frequency and give a slow beat for richness. Twelve-string guitars and mandolins are also tuned using beats. While beats may sometimes be annoying in audible sounds, we will find that beats have many applications. Observing beats is a very useful way to compare similar frequencies. There are applications of beats as apparently disparate as in ultrasonic imaging and radar speed traps. Exercise $\PageIndex{1}$ Imagine you are holding one end of a jump rope, and your friend holds the other. If your friend holds her end still, you can move your end up and down, creating a transverse wave. If your friend then begins to move her end up and down, generating a wave in the opposite direction, what resultant wave forms would you expect to see in the jump rope? - Answer - The rope would alternate between having waves with amplitudes two times the original amplitude and reaching equilibrium with no amplitude at all. The wavelengths will result in both constructive and destructive interference. Exercise $\PageIndex{2}$ Add text here. For the automatic number to work, you need to add the "AutoNum" template (preferably at the end) to the page. - Answer - Nodes are areas of wave interference where there is no motion. Antinodes are areas of wave interference where the motion is at its maximum point. Exercise $\PageIndex{3}$ You hook up a stereo system. When you test the system, you notice that in one corner of the room, the sounds seem dull. In another area, the sounds seem excessively loud. Describe how the sound moving about the room could result in these effects. - Answer - With multiple speakers putting out sounds into the room, and these sounds bouncing off walls, there is bound to be some wave interference. In the dull areas, the interference is probably mostly destructive. In the louder areas, the interference is probably mostly constructive. PHET EXPLORATIONS: WAVE INTERFERENCE Make waves with a dripping faucet, audio speaker, or laser! Add a second source or a pair of slits to create an interference pattern. Summary - Superposition is the combination of two waves at the same location. - Constructive interference occurs when two identical waves are superimposed in phase. - Destructive interference occurs when two identical waves are superimposed exactly out of phase. - A standing wave is one in which two waves superimpose to produce a wave that varies in amplitude but does not propagate. - Nodes are points of no motion in standing waves. - An antinode is the location of maximum amplitude of a standing wave. - Waves on a string are resonant standing waves with a fundamental frequency and can occur at higher multiples of the fundamental, called overtones or harmonics. - Beats occur when waves of similar frequencies $f_1$ and $f_2$ are superimposed. The resulting amplitude oscillates with a beat frequency given by $f_B = \|f_1 - f_2\|.$ Glossary - antinode - the location of maximum amplitude in standing waves - beat frequency - the frequency of the amplitude fluctuations of a wave - constructive interference - when two waves arrive at the same point exactly in phase; that is, the crests of the two waves are precisely aligned, as are the troughs - destructive interference - when two identical waves arrive at the same point exactly out of phase; that is, precisely aligned crest to trough - fundamental frequency - the lowest frequency of a periodic waveform - nodes - the points where the string does not move; more generally, nodes are where the wave disturbance is zero in a standing wave - overtones - multiples of the fundamental frequency of a sound - superposition - the phenomenon that occurs when two or more waves arrive at the same point	libretexts	2025-03-17T19:53:35.019292	2015-11-03T20:10:38	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.10%3A_Superposition_and_Interference", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.10: Superposition and Interference", "author": "OpenStax" }
https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.11%3A_Energy_in_Waves-_Intensity	16.11: Energy in Waves- Intensity Learning Objectives By the end of this section, you will be able to: - Calculate the intensity and the power of rays and waves. All waves carry energy. The energy of some waves can be directly observed. Earthquakes can shake whole cities to the ground, performing the work of thousands of wrecking balls. Loud sounds pulverize nerve cells in the inner ear, causing permanent hearing loss. Ultrasound is used for deep-heat treatment of muscle strains. A laser beam can burn away a malignancy. Water waves chew up beaches. The amount of energy in a wave is related to its amplitude. Large-amplitude earthquakes produce large ground displacements. Loud sounds have higher pressure amplitudes and come from larger-amplitude source vibrations than soft sounds. Large ocean breakers churn up the shore more than small ones. More quantitatively, a wave is a displacement that is resisted by a restoring force. The larger the displacement $x$ the larger the force $F = kx$ needed to create it. Because work $W$ is related to force multiplied by distance ($F_x$) and energy is put into the wave by the work done to create it, the energy in a wave is related to amplitude. In fact, a wave’s energy is directly proportional to its amplitude squared because \[W \propto F_x = kx^2.\] The energy effects of a wave depend on time as well as amplitude. For example, the longer deep-heat ultrasound is applied, the more energy it transfers. Waves can also be concentrated or spread out. Sunlight, for example, can be focused to burn wood. Earthquakes spread out, so they do less damage the farther they get from the source. In both cases, changing the area the waves cover has important effects. All these pertinent factors are included in the definition of intensity $I$ as power per unit area: \[I = \frac{P}{A}\] where $P$ is the power carried by the wave through area $A$. The definition of intensity is valid for any energy in transit, including that carried by waves. The SI unit for intensity is watts per square meter $(W/m^2)$. For example, infrared and visible energy from the Sun impinge on Earth at an intensity of $1300 \, W/m^2$ just above the atmosphere. There are other intensity-related units in use, too. The most common is the decibel. For example, a 90 decibel sound level corresponds to an intensity of $10^{-3} \, W/m^2$ (This quantity is not much power per unit area considering that 90 decibels is a relatively high sound level. Decibels will be discussed in some detail in a later chapter. Example $\PageIndex{1}$: Calculating intensity and power: How much energy is in a ray of sunlight? The average intensity of sunlight on Earth’s surface is about $700 \, W/m^2$. - Calculate the amount of energy that falls on a solar collector having an area of $0.500 \, m^2$ in $4.00 \, h$. - What intensity would such sunlight have if concentrated by a magnifying glass onto an area 200 times smaller than its own? Strategy a Because power is energy per unit time or $P = \frac{E}{t}$, the definition of intensity can be written as $I = \frac{P}{A} = \frac{E/t}{A}$, and this equation can be solved for E with the given information. Solution a - Begin with the equation that states the definition of intensity: \[I = \dfrac{P}{A}. \nonumber\] - Replace $P$ with its equivalent $E/t$: \[I = \dfrac{E/t}{A}. \nonumber\] - Solve for $E$: \[E = IAt. \nonumber\] - Substitute known values into the equation: \[E = (700 \, W/m^2)(0.500 \, m^2)[(4.00 \, h)(3600 \, s/h)]. \nonumber\] - Calculate to find $E$ and convert units: \[5.04 \times 10^6 \, J. \nonumber\] Discussion a The energy falling on the solar collector in 4 h in part is enough to be useful—for example, for heating a significant amount of water. Strategy b Taking a ratio of new intensity to old intensity and using primes for the new quantities, we will find that it depends on the ratio of the areas. All other quantities will cancel. Solution b - Take the ratio of intensities, which yields: \[\dfrac{I'}{I} = \dfrac{P'A'}{P/A} = \dfrac{A}{A'} \nonumber\]The powers cancel because $P' = P$. - Identify the knowns: \[A = 200 A',\] \[\dfrac{I'}{I} = 200. \nonumber\] - Substitute known quantities: \[ I' = 200 I = 200(700 \, W/m^2). \nonumber\] - Calculate to find $I'$: \[I' = 1.40 \times 10^5 \, W/m^2. \nonumber\] Discussion b Decreasing the area increases the intensity considerably. The intensity of the concentrated sunlight could even start a fire. Example $\PageIndex{2}$: Determine the combined intensity of two waves: Perfect constructive interference If two identical waves, each having an intensity of $1.00 \, W/m^2$, interfere perfectly constructively, what is the intensity of the resulting wave? Strategy We know from Superposition and Interference that when two identical waves, which have equal amplitudes $X$ interfere perfectly constructively, the resulting wave has an amplitude of $2X$. Because a wave’s intensity is proportional to amplitude squared, the intensity of the resulting wave is four times as great as in the individual waves. Solution - Recall that intensity is proportional to amplitude squared. - Calculate the new amplitude: \[I' \propto (X')^2 = (2X)^2 = 4X^2.\] - Recall that the intensity of the old amplitude was: \[I \propto X^2.\] - Take the ratio of new intensity to the old intensity. This gives: \[\dfrac{I'}{I} = 4.\] - Calculate to find $I'$: \[I' = 4I = 4.00 \, W/m^2.\] Discussion The intensity goes up by a factor of 4 when the amplitude doubles. This answer is a little disquieting. The two individual waves each have intensities of $1.00 \, W/m^2$, yet their sum has an intensity of $4.00 \, W/m^2$, which may appear to violate conservation of energy. This violation, of course, cannot happen. What does happen is intriguing. The area over which the intensity is $4.00 \, W/m^2$ is much less than the area covered by the two waves before they interfered. There are other areas where the intensity is zero. The addition of waves is not as simple as our first look in Superposition and Interference suggested. We actually get a pattern of both constructive interference and destructive interference whenever two waves are added. For example, if we have two stereo speakers putting out $1.00 \, W/m^2$ each, there will be places in the room where the intensity is $4.00 \, W/m^2$, other places where the intensity is zero, and others in between. Figure $\PageIndex{2}$ shows what this interference might look like. We will pursue interference patterns elsewhere in this text. Exercise $\PageIndex{1}$ Which measurement of a wave is most important when determining the wave's intensity? - Answer - Amplitude, because a wave’s energy is directly proportional to its amplitude squared. Summary - Intensity is defined to be the power per unit area: $I = \frac{P}{A}$ and has units of $W/m^2$. Glossary - intensity - power per unit area	libretexts	2025-03-17T19:53:35.091639	2015-11-03T20:11:00	{ "license": "Creative Commons - Attribution - https://creativecommons.org/licenses/by/4.0/", "url": "https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.11%3A_Energy_in_Waves-_Intensity", "book_url": "https://commons.libretexts.org/book/phys-1419", "title": "16.11: Energy in Waves- Intensity", "author": "OpenStax" }

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