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[228.91s -> 242.58s] Depending on the scenario, it may or may not be easier to model rotational motion with these equations rather than with the linear ones, where we have to use tangential velocities that vary according to the radius. |
[243.02s -> 249.73s] How can we set a system into rotational motion? We can do this by generating something called a torque. |
[249.73s -> 264.05s] Torque is defined as the ability of a force to rotate an object around some axis, whether we are looking at a wheel, a seesaw, or anything else that exhibits this kind of motion, and for now we can just look at a top-down view |
[264.05s -> 268.80s] a panel attached to a rod, allowing for free rotation. |
[268.80s -> 283.12s] Torque, represented by the Greek letter tau, will be equal to the applied force times the distance over which the force is applied, times sine theta, where theta is the angle between the force vector and the plane of the rotating force vector. |
[283.12s -> 297.57s] object. Torque will therefore have units of Newton meters, and torque will be at a maximum when the applied force is perpendicular to the plane containing the object, because the sine of 90 degrees is one. |
[297.57s -> 311.89s] Any angle less than 90 in either direction will give a sine value that is less than one, and will therefore diminish the torque, while still producing some motion, until the angle is zero, at which point the whole expression will be equal to zero. |
[311.89s -> 324.74s] and no motion can result. By convention we will say that torque is positive if it results in counterclockwise rotation and negative if it results in clockwise rotation, as we are by now familiar. |
[324.74s -> 339.06s] This must be the case, because if two different forces act upon this object to produce opposite torques that are equal in magnitude, the object will not move and the sum of the individual torques must be zero. |
[339.06s -> 351.42s] to reflect this fact. We must also note that the magnitude of torque will be equal to the magnitude of the applied force times the length of the lever arm, expressed in Newton meters. |
[351.42s -> 365.71s] the lever arm, or the |
[365.71s -> 376.62s] from the axis of rotation to the point where the force is applied. This sheds light on the quote by Archimedes which is generally reported as some variation of the following. |
[376.62s -> 390.21s] Give me a lever long enough, and a fulcrum to place it on, and I will move the earth. This implies that with a long enough lever arm, one could generate such a massive torque so as to move the world. |
[390.21s -> 401.36s] Wise words from a wise man, so let's learn about some other things Archimedes said next, but first let's check comprehension. |
[427.54s -> 438.32s] Thanks for watching, guys. Subscribe to my channel for more tutorials, support me on patreon so I can keep making content, and as always feel free to email me, |
[0.75s -> 15.09s] Hi everyone, this is Dr. Manu Krishnan K. Welcome to another session on Let's Hack an X-ray. And today we will be discussing about the Smith's fracture. In the previous session, we have seen how the Collis fracture happens and we have seen the X-ray. |
[15.31s -> 29.62s] Similarly, we will discuss what is Smith's fracture today. So what is Smith's fracture? The fracture of distal end of radius with the ventral or volat displacement of the fragment is termed as the Smith's fracture. |
[29.62s -> 41.09s] Here we have an X-ray which shows a Smith's fracture. And here you can see this is the lower end of radius. You can clearly see the clean margins without any discontinuity here. |
[41.09s -> 54.46s] While on the other side where there is a smith's fracture, you can see the axis. Here the lower end of the radius has been broken and it is displaced forwards or ventrally towards the palmar side. |
[54.46s -> 68.88s] And that's why that is called as a Smith's fracture. So let's compare it with that of the Collis fracture. Here we have both hand in hand where you can see the Smith's fracture where the displaced fracture fragment is. |
[68.88s -> 82.61s] displays forwards and while in case of the college fracture it is displaced backwards so that is a major difference between the smith's fracture and college fracture so let's see the causes of smith's fracture |
[82.61s -> 96.40s] a fall onto the flexed wrist here you can see the diagram which represents the flexed wrist and if you fall on that flexed wrist the chances of the lower end of radius to get fractured and the fragment to get |
[96.40s -> 109.79s] displaced forwards is more and that's why this particular fracture happens or it can also be caused by a direct blow to the back of the wrist so here we have a clear representation you can see the lower end |
[109.79s -> 123.93s] and the displaced fragment is moved forwards and that is smith's fracture so if you have any further queries regarding the radiological anatomy do post them as the comments below thank you |
[0.30s -> 12.61s] Let's get into monosaccharides. So first, let's expand on the definition of a monosaccharide that you put in your 20.1 notes. |
[12.61s -> 24.94s] In addition to everything that I've already said about monosaccharides, let's expand our definition to say that it is either a ketone or an aldehyde. |
[27.60s -> 35.95s] with three to six carbon atoms. And I told you that six is the most common. |
[38.51s -> 50.19s] And in this monosaccharide, every one of the carbon atoms is going to have either |
[51.38s -> 60.94s] a double bond to an oxygen, so carbon-oxygen double bond, and that will be the ketone part or the aldehyde part. Or if it doesn't have that, |
[60.94s -> 74.99s] it has a bond to an OH. So every single carbon atom has an oxygen on it. Monosaccharides can be called, they can be classified as either an aldose, |
[76.43s -> 79.31s] Or a ketose. |
[85.55s -> 96.78s] In the aldose molecules, the aldose monosaccharides, the carbon-oxygen double bond is on the first carbon, carbon number one. |
[96.78s -> 101.39s] of our chain and so it is an aldehyde. |
[103.92s -> 117.20s] And we'll draw a picture of that. In the ketose molecule, the carbon-oxygen double bond is on carbon number two. And so it is going to be a ketone. |
[120.50s -> 135.47s] So let's draw a couple of pictures. Now remember when we drew one of these before, when we draw monosaccharides, we draw them up and down vertically, which takes up a lot of space. |
[136.91s -> 147.92s] So this is going to be a four carbon chain. The carbon on the top, carbon number one, is going to have the carbon-oxygen double bond. |
[149.14s -> 163.73s] And even though in section 20.1 I told you that down here at the end of the chain we normally draw this part of the monosaccharide in condensed notation, I'm expanding it out in this particular section. |
[163.73s -> 178.72s] drawing. So this would be an aldose with our carbons. When we number them, we start at the top and we number down to the bottom like this. So our carbon oxygen double bond is on number one. It's an aldehyde and it's an aldose. |
[178.72s -> 191.79s] let's draw a ketose. So again, we're going to have a carbon chain. This one's going to be a five carbon chain. And since we're drawing a ketose, we want to put the double bond on carbon number two. |
[191.79s -> 200.30s] We'll fill those numbers in in a second. And every other carbon atom has to have an OH. So we'll fill those in. |
[200.69s -> 207.98s] And then we'll add hydrogens as we need to to make sure that every carbon has four bonds. |
[208.78s -> 223.02s] And let's put numbers on our carbons. There's one, two, three, four, five. I left a hydrogen off of carbon number five. So there's our ketose with the carbon oxygen double bond on carbon number two. |
[223.92s -> 238.70s] So in addition to classifying a monosaccharide as an aldose or a ketose, we can also classify it based on how many carbon atoms it has. So if it has three carbon atoms, |
[240.56s -> 253.04s] Whether it's an aldose or a ketose, three carbon atoms, we are also going to call it a triose. Ose is the suffix for sugar. |
[253.04s -> 265.36s] or carbohydrate. So triose, tri meaning three, and the ose telling us that we're looking at a sugar molecule in general. If we have four carbon atoms |
[266.61s -> 281.10s] we will call that a tetros, where T-E-T-R is going to be our prefix meaning four, and os is our suffix telling us that it's a saccharide. If we have five carbon atoms, |
[282.29s -> 291.09s] We can also call it a pentose, pent for five. And if we have six carbon atoms, |
[291.79s -> 305.25s] We can call it a hexose. And these names, triose, tetrose, pentose, hexose, they apply to both the aldoses and the ketoses. So for our two... |
[305.25s -> 319.57s] um molecules like in looking at this guy right here if we wanted to describe this we could call it an aldose and that would be correct and it would be kind of a generic name because when we say aldose |
[319.57s -> 333.17s] The only thing that we're communicating in that name is the position of the carbon oxygen double bond. If we wanted to be more specific, we could call it an aldo. |
[334.16s -> 335.92s] Tetros. |
[337.58s -> 349.62s] The aldo tetros name, the aldo part is telling us that it is an aldose, and the tetros part is telling us that it is a total of four carbons. |
[350.48s -> 363.17s] So this name is a little bit more specific, but it's still not really all that specific. It's still kind of a generic name. Looking at our other molecule over here. |
[363.17s -> 369.95s] If we wanted to classify this molecule, we could say that it is a ketose. |
[369.95s -> 383.44s] When we say that it's a ketose, all that we're really saying is that we have a carbon-oxygen double bond on carbon number two. We're not saying anything about the entire molecule. We could also call it a keto. |
[383.82s -> 393.10s] pentose. This name is a little bit more specific. Keto means that it's a ketose. |
[393.84s -> 402.56s] And the pentose part is telling us that it has a total of five carbons. So that name is even more specific. However... |
[402.56s -> 416.85s] These names, these four names that we've come up with, the ketose, the aldose, the aldotetrose, the ketopentose, those names are also still pretty generic because there are multiple ways in which we could draw a ketopentose. |
[416.85s -> 431.06s] by changing the position of the OH groups in this molecule, whether we're drawing the OH group on the right-hand side like we did here, or we could also draw it over on the left-hand side. So as we get more specific... |
[431.06s -> 441.44s] with our molecules, we'll come up with even more names to be able to describe them more accurately. But these are the generic classifications of monosaccharides. |
[0.34s -> 6.10s] Hi guys, I've put this presentation together for solving problems in a transformer circuit |
[6.10s -> 20.37s] It's not fixing or maintaining a transformer, it's more about calculating the relationship between what's going on on the primary windings and what's happening on the secondary windings. So it's a relationship between the two different circuits, the primary circuit and the secondary circuit. |
[20.37s -> 33.30s] not getting your hands dirty fixing the transformer okay so the objective of this presentation is using the supplied formula we're going to solve problems in the transformer circuit |
[33.62s -> 38.77s] so um what have we got okay so in your |
[38.77s -> 51.23s] Test or exam you're going to receive sort of like a question that's going to look something a little bit like this But the values are going to be be different, but the principle for solving or completing the problem |
[51.23s -> 59.89s] is the same so uh what you're gonna get so you're gonna get the um the transformer equation okay and you're gonna have your um |
[60.34s -> 75.02s] Formula will here. Okay, so just a recap on this. So what you want to find is in the center Okay, so if you're looking for watts, okay, you go into this quarter here power and once you go in this quarter here volts this quarter here |
[75.02s -> 82.45s] Ohms and resistance in this quarter here and amps for current which is in this quarter here |
[82.86s -> 93.36s] so again this is this presentation is all about finding out solving the transformer sort of like looking at what we have okay and trying to work out |
[95.34s -> 104.05s] what we don't know effectively okay so we're looking for the relationship between what's on the primary and what's on the secondary winding okay |
[104.66s -> 115.06s] So let's have a just quick recap of the terminology and some of the abbreviations that we're using. So what are we looking for and what do we have? |
[115.76s -> 130.03s] When you start your question the ideally what you need to do is sort of write down what you have Sort of break it down into sort of primary and secondary winding so you could say like the one sorry VP IP |
[130.10s -> 143.78s] RP, NP, and PP. So V is voltage, I is current, R is resistance. N is the number of turns on the coil. So NP would be the number of turns on the primary coil. |
[143.78s -> 154.32s] NS would be the number of turn on turns on the secondary coil P is the power in watts and the way we sort of |
[154.83s -> 168.94s] identify between the difference between primary and the secondary winding is we have a subscript okay so you'll have a capital N and then down here you'll have a little tiny P or a little tiny s okay that is just to indicate that there's you know |
[168.94s -> 174.96s] What part of the transformer we are working to or working with? Okay |
[176.05s -> 184.50s] So the problem okay, we've been given a problem like this So what we need to do is you'll probably end up having to calculate so we're going to calculate the number of turns |
[184.82s -> 196.85s] the primary okay and now the current on the primary the resistance on the primary and the power on the primary coil so that's all relating to this part of the transformer here |
[198.54s -> 212.56s] On the secondary coil, we've already got the watts, the ohms, and the number of turns. So what we're going to be looking for on the secondary coil is the voltage and the current. So we're looking for voltage and current on the secondary. |
[212.56s -> 215.70s] okay so you know |
[216.69s -> 230.86s] and seven items that we need to find okay so how are we going to do that so firstly I would write down okay everything you have or everything you need to find and everything that you have and lay it out in a logical order so like you've got |
[230.86s -> 244.14s] your primary on one side and your secondary on another side and then start putting in the information that you've got so you can populate the details and just use that as a master reference so every time you work something out |
[244.14s -> 255.02s] pop the new value in there okay so what do we have and what can we find out okay so |
[256.37s -> 268.75s] We've got our formula wheel up here. Okay, and we've got our transformer Equation over here. Okay. So at the moment we know the number of turns on the primary. That's why not number turns the voltage on the primary |
[269.04s -> 271.76s] Okay, we know the number of turns on the secondary |
[272.82s -> 287.60s] But that's about it. Okay, so we can't really use that one because we haven't got enough detail Okay, so what we need to do is now need to look at our formula wheel Okay, and we can look at what we've got and what we need to find. Okay with this formula wheel |
[287.98s -> 297.26s] With the formula wheel you can't mix them up so you can't use the voltage on the primary and the resistance on the secondary. |
[297.90s -> 311.82s] okay to work out for example the current okay so they must if you're using this wheel they must be related on each side you can only use what the information you have on each side of the coil you can't mix |
[313.10s -> 327.57s] Primary and secondary information. Okay, so it must must remain, you know So say for example if you want to work out the current on this one, for example We've got the voltage and we've got the resistance. We've got the voltage on the primary and the resistance on the secondary |
[327.57s -> 333.97s] We can't mix those because one's on the primary, one's on the secondary. So we need to look at what other information we have. |
[336.46s -> 350.24s] So therefore, so looking at what we've got, so we've got number of turns, okay? We can't use number of turns with this formula wheel, so let's scratch that out. But we do have power on the secondary in watts and resistance on the secondary in ohms. |
[350.24s -> 363.07s] okay so with that we can actually use that to find out either our current or our voltage okay so if we look at the power and the resistance okay so we look for if we're looking for voltage |
[363.07s -> 374.82s] we could use so this is the voltage so we've got the power and resistance so we could square root power times the resistance okay and that would give us our voltage or we could use the |
[374.82s -> 387.63s] Power divided by the resistance square root that a square root power divided by the resistance to give us our current Okay, so we've got two options there. So there's two different pathways you can go Okay, so |
[387.82s -> 401.34s] What we have there, so we've got We're going to use this one here So we're going to use this to find out our current so we're going to work out current on the secondary first So that's our formula. Okay, so when you do this, don't forget to do the square root. Okay, so do this in the |
[401.34s -> 415.34s] inside the square root first or in the root first okay so do p divided by r and then square root the answer okay otherwise if you do this part so 50 divided by 8 i think you end up with like 6.25 |
[415.34s -> 429.42s] okay so what we need is to make sure that you square root the answer and then you end up with our 2.25 amps okay so that's our current on the secondary so once you've worked that out then just pop that in there okay |
[430.58s -> 441.68s] like so and then we can start looking at our next stage as well okay so we can now we can use we could use our current |
[442.00s -> 450.54s] And our resistance okay to work out our voltage so we can do I times R or R times I Okay, so we can do |
[452.27s -> 466.48s] Our resistance times our current equals our voltage, which would be 8 ohms times 2.5 amps. And that gives us 20 volts. So that's our voltage sorted on our secondary coil. |
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