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# PHYS 200: Fundamentals of Physics I

## Lecture 11

## - Torque

### Overview

This lecture is a continuation of an analogue to Newton’s law: τ= lα. While previous problems examined situations in which τ is not zero, this time the focus is on extreme cases in which there is no torque at all. If there is no torque, α is zero and the angular velocity is constant. The lecture starts with a simple example of a seesaw and moves on to discuss a collection of objects that are somehow subject to a variety of forces but remain in static equilibrium.

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html## Fundamentals of Physics I## PHYS 200 - Lecture 11 - Torque
So, today, I’m going to consider an extreme case where there is no torque at all. If there’s no torque, we know So, let me start with a simple problem and take some horizontal rod like this. And I say, “What does it take to keep the rod from moving around?” And the first thing, you know, is the forces on it should add up to zero because otherwise So, I’m going to write down now the conditions for a body, a planar body lying in the plane of the blackboard, to stay still and not move. They are that all the forces in the So, simplest problem from the family is a seesaw. Let’s say that’s a point where the seesaw is supported and let’s say we put one kid here applying some force F is the _{1}mg for the kid. And let’s say this end is a distance x from the support. The question is, “How much of a kid should I put here to keep this in balance?” It should not move under the combined weight of these two kids. So, I take the body and I write down all the forces on it. So, I have _{1}F coming down here. I have _{1}F there. Then there is–Of course, that cannot be the whole story because the seesaw is not going into the ground. That is the support and the normal force from the support. And these three forces together should conspire to keep the body in equilibrium. So let’s see. So, here is what I want you to understand. _{2}F is known. Let’s give it some name, 100 Newtons. This distance _{1}x is also known. That distance _{1}x is also known. I’m trying to find out what force _{2}F is needed. So, _{2}F is what I’m trying to solve for. So, I write all the equations I know. So, horizontally, there’s nothing going on. So, it’s zero equals zero. In the vertical direction, I have _{2}F pushing down, _{1}F pushing down equals _{2}N. And that’s not going to be enough to solve this problem because I have two unknowns. I don’t know N and I don’t know F. I just know this guy. So, that’s where the torque equation’s going to come in. Now, here is the subtlety when you do the torque in this problem. So, how do you find the torque due to any force? Can you tell me what you’re supposed to do? Yeah? Yes, you, yes._{2}
N. Two unknowns can only satisfy two equations, not 2,000 equations. So, we’ve got to hope that the extra equations I get by varying the pivot point all say the same thing. And that’s what I will show you.I will show you, first, that if you have a bunch of forces that add up to zero, then it doesn’t matter around which point you compute the torque. If it was zero around one point, then it will be zero around any other point. Suppose I pick some point here and I take each force, multiply the distance of that force from that pivot point and they added up to zero. Now you come along and say, “No, I want to take the torque around that point.” What you will be computing will be the same force, but to every a where a is the distance by which you moved your axis. Now, let’s open this out. You’ll find it is F times sum of all the _{i}x_{i} + aFs. This is zero because for me the torque around that point vanished and this is zero because all the forces add up to zero. In other words, if you find any one point and make sure the torques around that point don’t cause–that they all add up to zero, then the torque around any other point will also add up to zero, provided the sum of the forces adds up to zero. And of course, that’s the problem we’re looking at. So, you’ve got to realize that when you take the torque, you may pick any point you like._{i}Now, that’s where a strategic issue comes up. If you were asked simply to find what F,” I don’t care what _{2}N is” then, there’s a particular choice of pivot point that is optimal. That choice is the one in which this normal force doesn’t get to enter the torque equation and I think you all know what I mean. Pick that point as the point around which you take the torques; then N drops out. So, this is the usual model when you take torques. Always take the torque through a point where an unknown force is acting because the unknown force doesn’t contribute to the torque equation. So in principle, you can take the torque around any point and you can satisfy yourself in your spare time; it doesn’t matter which point you take. It will be the more complicated way to solve the problem because N will come into the picture and no one asked you what N is. So, we take the point of support and then, of course here is the point of support, and what did I have, F here a distance _{1}x, I had _{1}F at a distance _{2}x. This is the point around which I’m doing the torque, so _{2}N doesn’t count. Then, you get a condition F. From that you can solve for _{1}x_{1} = F_{2}x_{2}F._{2}For example, if this was one meter and this was six meters, and this was say 10 Newtons here, then 10 times one has to be six times F and _{2}F will be 10/6 Newtons. I don’t want to use _{2}N because N is also standing for the force here. That’s how we find F. Once you’ve found _{2}F, in case you want to know what is the support going through and how much load is it carrying, you can come back now and add _{2}F, which was given to you, and the _{1}F you just solved for. Then if you want to, you can find _{2}N. Even if you were asked to find N, it’s best to first take torque around a point N so N doesn’t enter, find F, then it’s very trivial to add _{2}F and _{1}F to get _{2}N. This is the easiest prototype. So, you guys have to know how this is done. Everything I do will be more and more bells and whistles on the simple notion of how to take the torque. All right.So now, I’m going to take a slightly more complicated problem. So, here is a wall and there is a rod here of length
So, the trick proposed by this gentleman here–Imagine dividing the rod into tiny pieces, each of which is small enough to say it has a definite location x times _{i}g. If mg is the force, then mg times x is the torque. You want to sum it over all the little pieces. Sum over _{i}i is summing over all the little pieces. So, let me make my life easy by dividing by the total mass and multiplying by the total mass. If I started this combination m summed over _{i}x_{i}I divided by M, that is the center of mass. So, it does look like Mgx, where x is the center of mass of the rod.Now, the rod really doesn’t have to be uniform for me to say this. It doesn’t have to be a uniform rod. The mass can be distributed any way it likes. Each Okay. So, now that we know where the result comes from, we will now apply it in the torque equation. This is the torque due to the force I apply
Maybe the main point is, a purely vertical force will suffice to keep this in equilibrium, is the point that I’m making. In fact, if that force is equal to In fact, let’s turn to a problem where the support, in fact, exerts a horizontal and vertical force. So, that’s the next level of difficulty. This is one of the standard problems in the section and it goes like this. Here is a rod and it’s supported by a wire. And the wire has a certain tension But suppose the question they ask you is, “What is the tension on that rope?” And suppose that’s the crucial issue because if the tension is too much the rope is going to break. If you want only the tension, once again the trick is, forget the force equation and go to the torque equation because, if you go to the torque equation, you banish both
You could be asked to now find what’s happening at the pivot. Then, you’ve got to go back to the forces. So, in the horizontal direction, in the So, I will give you a couple of seconds to think about this. This is the kind of problem you guys should be easily able to do. You isolate the rod, you write all the forces on it. If you want, maybe this–If this picture is not clear to you, what I have in mind is, I pull out the rod and say, “Why is this rod not turning and moving under all the forces acting on it?” Then I write the forces on it. The tension is acting in that direction; the pivot has a vertical force, the pivot has a horizontal force, and gravity for this purpose I’ll replace as acting there. Under the action of all these forces, the rod is neither translating nor rotating. That means the Now, you can embellish the problem. I don’t want to do that. You can hang a weight here. I don’t think it should cause a new problem. Suppose there’s an extra with, a little The last of the equilibrium problems is very standard fare for this course, which is the problem I mentioned in the beginning to motivate all of this. And that is, if I have a ladder leaning on a wall at some angle
All right. Anyway, let’s say we’re going to limit ourselves 0 to 90. And what angle can it take? What’s the minimum angle? What do you think is going to control that angle? Yeah?
All that is left is the torque, and that’s where you have the choice. You can take the torque around any point you like. You can take that point; in fact, you can take a point here. Okay. If you want to really punish yourself, you can take a torque around some crazy point and then
So, I find W equals
So, coming back again, I did this cancellation. I got that to be equal to the force of friction but that’s bounded by this. But I know Mg. So, I have the restriction that Mg/2 cos θ–;So, let’s compare that to that and we find the restriction that Mg/2 cos θ less than or equal to μ times _{S}Mg. You cancel the Mg, and you find cos θ has to be less than or equal to 2 times μ. I like to write it as _{S}tan θ has to be bigger than or equal to 1 over 2 times μ. Do you know how we went from here to here? If this number is smaller than this number, the reciprocal of this number is bigger than the reciprocal of this number. Three is smaller than 10; 1/3 is bigger than 1/10. Same thing. So, I’m somehow used to tangent better than the cos, so I flip this over and I flip that over and I also changed less than to bigger than. This way we can understand _{S}tan θ has to be bigger than some number. That means θ has to be bigger than some number, because tan θ increases with θ. So, you tell me what your coefficient of friction is. Maybe it’s .5, then this is 1. Tan θ should be bigger than 1, θ should be bigger than 45 degrees. Okay.Or Okay. So, this is a collection of problems all in the The limited context I have set for myself right now is to just apply it to single point mass. What does the analogue of But now, imagine that this body is actually–You don’t even need a potato for this. Just take the same flat sheet, but it’s now not living in the So, there is a trick we use. The trick is, we are going to somehow indicate that the effect of this torque is to produce a rotation around that axis. So, we are going to introduce a new vector Let’s first take a simple case. Here is So, in this example So, torque is a vector. By the way, that’s an accidental property of living in three dimensions. If you take two vectors Now, the whole cross product is something one can go on and on. I don’t want to talk for too long. Let me just point out a couple of properties of the cross product. One is that By the way, you will find with your hands one turn would be easy. The other would require all kinds of contortions. So, what I suggest you do–do it for the easier one and then flip the sign because I saw you going through some unnecessary torment in the front row. I have the same problem. I’ve seen people doing– you know, at my age I shouldn’t even be doing the cross product because you can end up in the hospital. So, anyway, it’s a game for the young. You can do the cross product your way, but one day you will learn to follow my way, which is do the easy product, then turn the sign around. All right. So, I don’t want to spend too much time, but let me just mention one thing because of the homework problem. Remember the dot product I said is _{x} + A. You get that by writing _{y}B_{y}A as I times A plus _{x}J times A, likewise for _{y}B and taking all the different dot products when you open all the brackets, that’s where you get A. There, because _{x}B_{x} + A_{y}B_{y}I dot I is 1 and J dot J is 1 and I dot J vanishes for the dot product. Cross product is exactly the opposite. If you write I times A + _{x}J times J + _{y}K times A, cross product with similar things for _{z}B. I times B + _{x}J times B, etc. _{y}K times B. There are, in principle, nine terms you can get in the cross product. Three of them will vanish, because when you take _{z}I cross I you will get zero. When you take I cross J, you will get something. When you take J cross I, you will get minus something. You put them all together – I won’t do this now – it will look like I times A - _{x}B_{y}A [correction: should have said _{y}B_{x}A] plus _{y}B_{z} - A_{z}B_{y}J times something and K times something. I don’t need this, so I’m not going to dwell on this. It’s good enough for our purposes to know the cross product can be visualized as the vector perpendicular to the two vectors forming the product and of length equal to length of A length of B sine of the angle between them. Okay.So, why did I introduce a cross product? Well, here is the point. I have introduced torque as
Let’s go back to 2D and ask, “What does all of this look like when I apply it to 2D?” So, go to two dimensions now. Let’s put that in the–Let’s take some object. Take a tiny piece of gum stuck to some rotating rigid body first. What do we say is the angular momentum according to this formula? Angular momentum is mr. Good. And ^{2}ωmr is the moment of inertia of this point mass, and ^{2}ω is the angular velocity of the point mass. That’s what we’ve been using all this time.So, it’s fully compatible with the definition of angular momentum we had but it’s broader than this one. This was applicable only to bodies which are circling some point because they’re stuck on a rigid body. That’s why this But now, I want you to think about the following notion. Here is the When you are here–Let’s take it that epsilon before it took off from this rigid body, it was still stuck to the rigid body, and the momentum was in this direction; Take a body in the Now, if you’ve got many many bodies, you glue them together or they have inter-atomic forces and they form a rigid body; then, you can use dL. Angular momentum of body _{i}/dt = τ_{i}i is changing due to the torque on body i. You add everything. Left-hand side will be the total rate of change of angular momentum. Right-hand side will be the total torque. Now, the total torque on that rigid body made up of many point particles, even if it’s not a rigid body–Just take a bunch of bodies moving around and find the rate of change of total angular momentum. That’s the sum of the total torques. You can divide the torque into the external torque and the torque internally due to the forces of one part of the system on another part of the system. I’m going in the same way as I did with momentum. The rate of change of total momentum of a system is the external force plus the internal forces. You remember the internal forces cancelled in pairs. And therefore we could drop that, and therefore rate of change of total momentum was the external force.For torque, you may like to, again, cancel the internal forces – internal torques – because the forces are opposite. But it’s not so obvious you can cancel them because even though the force that I exert on you and you exert on me are equal and opposite, to find the torque, you multiply the force you exert on me cross product with the distance of me from the origin; whereas, with you it’s the distance of you from the origin. So, they don’t necessarily cancel. And you can show that only if the force the one body exerted on the other was in the line joining them; then that would cancel and angular momentum would be conserved. So, angular momentum conservation requires more than just usual momentum conservation. Only for the microscopic level, forces that bodies exert on other bodies is in the line joining them is angular momentum conserved. Fortunately, that turns out to be true for every force we know – gravitation, Coulomb’s Law, electro-static forces. All of them have the property that So, let’s take that for granted and then do the last problem I want to do for you guys, which is the gyroscope. So, here is the gyro. Now, again, I have to draw this gyro. It’s another humiliating experience. I’m going to try. So, here is a little tower on which the gyro is supported. Now, maybe we can draw it like this. It’s a disk in this case, which is spinning like that. Everybody with me now? There is a little tower and on the apex of that you take a massless rod, if you like, put a gyro here. And first of all suppose it’s not spinning. I am going to–I’m holding it here and I’m going to let it go. What’s going to happen to the gyro is what we’re asking. The gyro is not spinning. Everybody should know. You don’t need physics 200 or 100 or anything to know what’s going to happen. If you let it go it’s going to fall. You want to make sure that agrees with the equation that we have, Let me look at the top view of this gyro. The top view of the gyro is like this. This is the top of the tower. Here is my gyro. It is spinning, as seen from the top, like this. And it’s got an angular momentum
So, the last thing I wanted to calculate is the rate at which it goes round and round. That’s a pretty easy calculation. I will do that and I will let you go. So, let’s ask the following question. Here is the gyro whose angular momentum was So, the precessional frequency in this problem is equal to the torque divided by angular momentum of the gyro. The torque, and now we can put the numbers in, is [end of transcript] Back to Top |
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