PHYS 200: Fundamentals of Physics I
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Fundamentals of Physics I
PHYS 200 - Lecture 11 - Torque
Professor Ramamurti Shankar: So, so far here is what we have done. We’ve been studying the whole time this analogue of Newton’s law: τ = Iα. So, after teaching you what τ is and what I is and what α is, in the early days I concentrated on problems where τ is not zero. There is some object on which some torques are acting and in responds to the torques–in response to the torques, the body will accelerate. Then, we went to another class of problems where there were no external torques, there are only torques from inside. Excuse me, will you people who come late use the other door? Yeah, you’re blocking actually the taping right now. Yes. All right.
So, today, I’m going to consider an extreme case where there is no torque at all. If there’s no torque, we know α is zero and the angular velocity is constant. But I’m going to take a case where even angular velocity is zero. There is no motion; there is no torque. So you can say, “You know, what’s there to study?” Well, sometimes it’s of great interest to us that the object has no angular velocity. A common example is, you are trying to do something in your house, you get up on a ladder, that’s somewhere there. This ladder should have no angular velocity. Right? That’s not a trivial statement to a person who is going up there. So you want to ask, “What does it take to keep the ladder from falling over?” Okay, that’s what we’re going to discuss. All right.
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 F = ma tells you this rod’s going to be flying off somewhere. So, I’ve taken a body, a rod of length L, and I don’t want it to move at all. So, total forces adding up to zero we know is not good enough because here is the problem where I apply two forces equal and opposite, add up to zero, but that doesn’t do it. This rod will still rotate. It is still true that when the total forces add up to zero, the center of mass will not accelerate. But that doesn’t mean the body will not move because keeping the center of mass fixed, the body can rotate. All right. So, you have to ask yourself, “What does it take to keep it from rotating?” Then we know that the cause of rotation is the torque.
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 x direction. i is some label, force number 1, force number 2. That should add up to zero. All the forces in the y direction; that should add up to zero. This will make sure the center of mass as a whole is not going anywhere. And finally, to keep the body from moving or spinning, all the torques should add up to zero. So, bulk of the lecture today will just be dealing with this condition. So, what we’re going to study today is a collection of objects that are somehow subject to a variety of forces but still in equilibrium, in static equilibrium. That means they are not moving.
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 F1. So F1 is the mg for the kid. And let’s say this end is a distance x1 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 F1 coming down here. I have F2 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. F1 is known. Let’s give it some name, 100 Newtons. This distance x1 is also known. That distance x2 is also known. I’m trying to find out what force F2 is needed. So, F2 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 F1 pushing down, F2 pushing down equals 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 F2. 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.
Professor Ramamurti Shankar: Right. What r should I use for this force?
Professor Ramamurti Shankar: Pardon me? x1? Okay, why did you choose that?
Professor Ramamurti Shankar: Okay. Her answer was, “That’s the pivot so I chose that distance.” The point is, when we say the body has no net torque, which means it’s not rotating around some axis. If there is an axis around which a body is rotating, then we know how to take the torques through that axis. When a body is not rotating, it’s not rotating around any axis whatsoever. So, you can say it’s not rotating through an axis I draw through here. So, I will take all the torque with respect to that point. Someone else can say, “No, it’s not rotating through that either. I will take all torques from that point.” If you change your mind on where you’re taking the torque, then the torques are all changing because in the force times distance the distances are going to change. So, now we have a predicament where for every possible point around which I compute a torque, I will get a condition involving the different forces and the distances from that point. So, I can write down an infinite number of equations for each different choice of pivot point. But I only have two unknowns, F2 and 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 xi you will add a distance a where a is the distance by which you moved your axis. Now, let’s open this out. You’ll find it is Fixi + a times sum of all the Fis. 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.
Now, that’s where a strategic issue comes up. If you were asked simply to find what F2 is, not to completely solve the problem, which would be also finding what the normal force is–if someone said, “Just find F2,” I don’t care what 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, F1 here a distance x1, I had F2 at a distance x2. This is the point around which I’m doing the torque, so N doesn’t count. Then, you get a condition F1x1 = F2x2. From that you can solve for F2.
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 F2. So, 10 = 6F2 and F2 will be 10/6 Newtons. I don’t want to use N because N is also standing for the force here. That’s how we find F2. Once you’ve found F2, 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 F1, which was given to you, and the F2 you just solved for. Then if you want to, you can find 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 F2, then it’s very trivial to add F1 and F2 to get 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 L and mass M. And it’s supported by some kind of pivot on the wall. And I want to hold it up by supporting it here with a force F. And the question is, “What force should I apply?” Okay. So, we know that if I draw the free body diagram of the rod, I have my force F; I have gravity acting down everywhere on the rod, then the pivot, namely the wall, it can exert a certain vertical force. It can also exert a horizontal force. No one tells you the wall is frictionless or anything or the pivot is frictionless. So, the wall is exerting in general a force, which is at some oblique angle, which I’ve broken up conveniently into horizontal and vertical parts. What about gravity? And torque due to gravity is what I want to discuss now. First of all, I am only asking you to find the force I have to apply to support the rod. So, just like the other problem, I’m not asking you to find v and h. So the trick, once more, is to jump straight to the torque equation. Forget about the force equation. Go to the torque equation and take the torque around this point, then this guy is out. No matter what’s going on – doesn’t matter – the torque due to the pivot is out. So, we have to equate the torque due to the force I apply to the torque due to gravity pulling the rod down. Torque due to force I apply; that torque, is equal to F times the length of the rod because there’s F times r times the sine of the angle between them, and the sine of the angle is 9, the sine of 90, which is one. How about the torque due to the pull of gravity? Anybody know what we’re supposed to do here? Yeah?
Professor Ramamurti Shankar: Okay. Now, why is that? Why is it okay to replace the torque by the center of mass? Yeah? Either of you can answer, yes.
Professor Ramamurti Shankar: Right. But why am I allowed to pretend? See, we can pretend a lot but it occasionally has to correspond to reality so that it works. So, why is it okay to pretend? That’s a correct answer.
Professor Ramamurti Shankar: Right. But if I really wanted to prove it, somebody has a way to really nail it down and show that it’s okay? Yes?
Professor Ramamurti Shankar: Thank you. Yes. Your answer is right. I know where you’re going. That’s the right answer. Look, what you want to do is you always have to go back to first principles. You cannot take anything on faith because you picked it up on the street or someone whispered this to you. That’s not good enough. One lesson I want to show you throughout the lectures is, I cannot invoke new laws or new principles. The only principle we know is F = ma. You’ve got to go back to that all the time. Of course, there is the other principle that gravity is an extra force on the body. But this is an external–This is an extended object. It’s got a center of mass but gravity is not necessarily acting there. It’s pulling each part of it down. So, I want to show you that in the end of course the answer is as if all the mass were concentrated here for the purposes of torques but I want to show you why.
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 xi. The torque, due to gravity acting on that little fellow–It’s the mass of that little guy times xi times g. If mg is the force, then mg times xi is the torque. You want to sum it over all the little pieces. Sum over 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 mixi summed over 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 mi need not be the same. But it’s very important that gravity be constant. We demand that the force of gravity have a constant value over the length of the rod. And it’s not obvious that’s always true. I mean, if I took a tiny rod, force of gravity is constant–took a very big rod, comparable to the size of the Earth, of course the pull of gravity is varying on the length of the rod. But for all rods on the laboratory scale, that’s not an issue. And that’s where this result comes from.
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 Fx. That’s equal to Mg times x, where x is the location of the center of mass. That is going to be half the length of the rod. I’m sorry– this Fx is FL. We cancel the L, then you find the force I apply is Mg divided by two. Yeah?
Professor Ramamurti Shankar: If the pivot is exerting what force?
Professor Ramamurti Shankar: Yeah. But it won’t cause a torque.
Professor Ramamurti Shankar: I am finding–no. Here, what I’m calculating is not the net force. I’m finding the torque around this point due to the pull of gravity on the different segments that make up the rod.
Professor Ramamurti Shankar: Oh. But the pull of gravity is straight down and the separation is horizontal. Is that the question now?
Student: I want to know about the force you’re applying, not the gravity.
Professor Ramamurti Shankar: Well, I’m–oh, the force I am applying. Oh, I am applying the force here, straight up. Oh, you want to know–
Professor Ramamurti Shankar: Ah. Well, you’re quite right. You’re saying we don’t know a priori whether the force I’m applying is vertical or horizontal. Correct? Maybe that’s your question. That’s actually a valid question. It’s clear that I have to apply vertical force and possibly horizontal force to compensate. This one, if there is a horizontal force in the wall. You’re quite right. Absolutely correct. So, this F–I should also divide the force that I apply into a vertical part I call F and a horizontal part that I call H prime. Then, it’ll turn that in this problem an H prime is not needed. You can sort of tell intuitively if you’re holding up a plank, you know. One end is anchored and you’re trying to support it; it’s enough to apply a vertical force. But I don’t have to presume that. So, if you want, you can think of applying horizontal force. But it’s not necessary to hold this in equilibrium.
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 Mg/2, it won’t rotate. Now, we can go back–now that I find the force I have to apply in this direction, what force is at the pivot? Well, the vertical force on the pivot is pointing up. My force that I calculated is pointing up. And together they should add up to Mg. Now, F = Mg/2 so the vertical force will be Mg/2. Now, horizontal force–In this problem, there is no need for you to apply the horizontal force and there’s no need for the pivot to fight you back with the horizontal force. It’s not needed for equilibrium, but supposed you insisted. Not only are we lifting it but we are leaning into the rod. Yes, then, you will apply horizontal force. It’s not going to contribute to the torque equation, but if you push it in, wall’s got to push you out, and H prime will be equal to H.
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 T. It is anchored to the wall; the length of the rod is L, the mass of the rod is M. And at the–This end is some pivot, which is exerting some unknown force, which can have a horizontal and vertical part. We’ll have a horizontal part H and a vertical part V. Now, I know for sure there’s a horizontal part. I think you can tell–can you guys tell in your mind where there’s a horizontal part? This force that’s supporting it is pushing the rod into the wall and also pulling it up. The part that’s pushing it to the left has to be compensated by H.
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 V and H. I hope that’s clear to all of you. A force going through the pivot point in any direction is incapable of producing any torque. That’s so intuitively clear, we need not belabor that. If you want the formula in F times r times sin θ, r is zero. So, we don’t care what the angle is. So, once again, we’ll take torques around this point and see what it will tell us. So, the torque due to gravity, we all agreed, we can now replace by MgL/2, turning it this way. So, this force exerts a torque which is equal to the value of the tension, the length of the rod and the sine of the angle between the two, which is this angle here. So, the nice thing is, you can solve for the tension directly in one shot, which is just Mg/2 sin θ. That’s why θ cannot be too small. If θ is too small, at some point T will become so large that the string will break. So, if you want to anchor a sign, you’ve got to try to take a big angle like that so that sin θ is not too small. So, one of the problems you could be asked is, “What’s the minimum angle so that the rope doesn’t break, given that it can support a maximum tension of 1,000 Newtons? Well, then you put a 1,000 Newtons here, you know the M and the g, you solve for the θ at which it crosses a thousand Newtons. And your θ should be bigger than that. Yeah?
Professor Ramamurti Shankar: That was a definition of the torque. If you go back, then you’ve got to go back to what we did earlier and you’ll find I deduced the other day–I said–Maybe I’ll remind you how that happened. I’m trying to the find the analogue of F = ma. Analogue of ma is Iα. And I said, “Who is the guy who can play the role on the left-hand side?” And I found the quantity that plays the role of Iα; that’s equal to Iα – is the force times distance times sin θ. Okay? So, that’s– you– look, you’ve got to understand why torque has a sin θ in it. If you want to rotate a body – take any body that you want to rotate – you know, here is a rod and you want to rotate it, you pick a point, you want to apply a force. We know intuitively that any force applied along this direction is no use, in terms of rotating it. The best way to apply a force of a given magnitude is to be perpendicular to it. But if, for some reason, you can only apply it at some angle, then this part of the force is good for rotation and that part of the force is not good for rotation. And the sin θ projects out this part. If that angle is θ, then F sin θ is really the perpendicular part of F. So, that’s how the torque is defined. Okay. So, we found the tension here. If that’s all you want to know, that’s the end of the problem. You’re got the tension. But you may be asked to do more.
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 x direction, the total forces have to add to zero. So, H should be equal to F, which should be equal to tension times cos θ because this T has a T sin θ that way and T cos θ that way. And this is T cos θ. So, T cos θ is pushing in, the wall should push out. And now I know the value of T. I put that in. Mg/2 cos θ because that’s cosine over sine. All right. So, that’s the horizontal part. How about the vertical part? In the vertical direction, upward vertical force plus T sin θ has to be the downward force, which is Mg. Well, we know what T sin θ is. We just did that. T sin θ, if you look up here, T sin θ is Mg/2 = Mg. That means V = Mg/2. That means, the pivot provides half the support, half the mass. And the tension’s vertical part provides the other part.
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 x component of the force adds to zero, the y component adds to zero, the torque adds to zero. If you’ve got three equations, you can plug the numbers into any of them in any order you like. In the end, you will all get the same answer. But the way, to get the answer most effectively is to take the torque equation, because the beauty of the torque equation is that neither the vertical part nor the horizontal part have anything to do with the pivot enters that. So, bear that in mind.
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 m, on top of the mass of the rod. I am actually hanging a weight here. Well, all you have to do when you go to the torque equation, is to remember that this weight going this way will provide an extra torque. That will be little m times g times the whole length of the rod, because I’m taking the torque around that point. You put that in and you solve for the tension T; then go back and, again, do the force part. Any questions about this so far? Because I’m going to do now the last of these equilibrium problems before I do something else.
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 θ, you can ask yourself, “Is there a limit to the θ?” Do you think there’s going to be some kind of limit, and if so, what kind of limit will it be? A lower limit or an upper limit? Would you like to guess?
Professor Ramamurti Shankar: A lower limit on the θ? By that you mean what, the θ is less than some amount? What’ll happen?
Professor Ramamurti Shankar: It’ll fall. Very good. So, for equilibrium, there’ll be a lower limit on θ. I think it’s clear to all of us that if you’re trying to climb a building and you put a ladder like this, you are asking for trouble. Right? Yeah?
Student: So, there’s also an upper limit of θ[inaudible] you know, 90 degrees [inaudible]
Professor Ramamurti Shankar: Look, I agree. You mean 91 degrees, ladder like this. Yes. I agree with you. I don’t recommend that either. So, between–Are you a mathematician? Because they always find out something wrong in anything I say. I say, “Good morning,” they say, “Well, not quite you know, let’s look into this.” Philosophers, even more difficult, “What is good? What is morning? It’s morning here, it’s nighttime in California.” But look, every time I make up an exam for a question I run it by a lot of students and a lot of TAs. Because when I’m talking to you, I have a certain notion of what I’m talking about, but the same words can mean something else. And quite often, I am tripped up by this, though this is the first that this question has come up. So, I learn something every day.
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?
Professor Ramamurti Shankar: The [inaudible] between the what?
Student: The bottom of the ladder and the [inaudible]
Professor Ramamurti Shankar: Right. Okay. So, in fact, it turns out even the coefficient of the friction with this wall may help, but we are going to rely on a problem where the wall is frictionless and the floor has the friction. And it’s got a static friction μ, which I’m just going to give to you. So, here is our challenge now. We want to see why this rod is not falling. So, let’s just write down all the equations. So, whenever you get any rotation problem, rule number one is “don’t panic.” Write down the three great conditions and it will all happen. The first condition is, well, zero condition is, draw all the forces on this guy. So, what forces do you want to draw? I think we have learned over and over Mg can be assumed to be acting in the middle of the rod, the center of the rod. The wall is very important. If it’s a frictionless wall, then the only force it can exert is perpendicular to itself. That’s the meaning of frictionless. You cannot apply a force along the length of the rod. So, let’s give that name W for wall. I come to the floor now. The floor is exerting a force N, standard name for normal or perpendicular. Then, it’s exerting a frictional force f, which will be inward. Because the ladder is trying to slide out, you will apply a static frictional force inward to keep it from moving. That’s it. These are the forces, and subject to these forces we have to write down all the conditions. This angle is θ here. For the x motion, the fact that all the – maybe I should write it – sum of all the x forces is zero, simply it says W = F. Sum of all the y forces equals–tells me N = Mg. Right.
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 N and f will all come into problem and you will have to work that much harder. So, the trick again, take the torque here. Then N and f are gone, and what do I have? I have Mg trying to rotate it this way; I have W trying to rotate it that way. I’m just going to balance the magnitudes of the two torques. So, what equations will I get? For W, it is W times L times sin θ. Now, here is where you shouldn’t simply write it down. You’ve got to make sure that you would have written this down. Going back to the definition of torque, force times distance from the pivot point times the sine of the angle between the force and the line of separation. W is the force, the distance is L – the length of the ladder – and this θ and that θ are equal by some theorem we learned in high school that if you’ve got two parallel lines the angle is the same. So, the sin θ you want to use is that. That’s the torque. That’s the torque due to the wall. The torque due to gravity is MgL/2 cos θ. Now, why did I write cos θ? Am I making a mistake? Instead of sin θ, why is it cos θ? Yes?
Student: Because that angle it makes with the ladder is the complimentary of θ.
Professor Ramamurti Shankar: Yeah. So, let me repeat the answer. When you write torque as MgL sin θ, the θ is not some fixed angle that somebody drew in the diagram. It varies from force to force. It’s the angle that force makes with the line connecting the point of application with the force to the point of rotation. So, Mg is acting here. You want the angle between the direction of Mg and the direction of the line of separation, which is this angle. We really want the sine of this angle, but sine of that angle is cosine of this angle. So, you can cancel that.
So, I find W equals Mg/2 cos θ. Just the torque equation tells me that’s the W I need. The wall’s going to push out with that W, and you better be able to fight that. To fight that, that W should be equal to f, which is what we’ve got here. That’s fine too. But now we have a restriction. f cannot be arbitrarily big. f is less than or equal to the coefficient of static friction times the normal force. Yeah?
Student: Could you explain why you used the cos θ?
Professor Ramamurti Shankar: Yes. So, I want to find the torque due to the force of gravity acting there with respect to this point. By the definition of the torque, you take the force, you join the point of application of the force to the point around which you are doing the torque and take the angle between those two lines. So, it’s really that angle, which is not the θ here. But if I draw the picture here, you can see this is θ, this is 90 – θ, this angle that I want. But then, sine of 90 - θ is cos θ. And that’s why you use that. Sine of 90 - θ is cos θ because when you change θ and 90 - θ, what’s opposite becomes adjacent and what’s adjacent becomes opposite. In any right triangle, if that’s your angle, that’s called the adjacent side. But if that’s your angle, that’s called the adjacent side. So, sine and cosine exchange places when θ goes to 90 - θ. All right.
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 N = Mg. So, bring this equation here, so that’s equal to μS times 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 μS times Mg. You cancel the Mg, and you find cos θ has to be less than or equal to 2 times μS. I like to write it as tan θ has to be bigger than or equal to 1 over 2 times μS. 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 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 π /4. Maybe I should keep using radians so you guys get used to radians as a way to think about it. So, what’s the famous show on CNN, Anderson Cooper–what? Anderson Cooper 2π. That’s what it is for this class. Think radians or you’re going to screw up somewhere. Okay? Angle is what we use every day, but radians is what we use here. When I write a tan θ and I don’t tell you what it is, or if I tell you tangential velocity is ω times r, that velocity–angular velocity ω is radians per second. A lot of people get tripped over this. Okay? It’s a very natural way to measure angles, and you guys get used to that.
Okay. So, this is a collection of problems all in the xy plane. Rigid body dynamics is quite manageable as long as everything lies in the xy plane. If the plane of the blackboard is the xy plane, notice that everything is happening in the xy plane. All the mass is concentrated on the plane, all the forces are in the plane; all the vectors separating the point of the torque to the point of the force are in the plane. But now, you have to deal with the fact that in real life objects are not just planar. They are chunks like a top or a potato. So, living in three dimensions, and rigid body dynamics in three dimensions is one of the most complicated fields. So, we’re not going to do that. We’re going to take a rather simpler version of the rigid body problem, basically the problem of torques and forces in 3D, but apply it in a very limited context.
The limited context I have set for myself right now is to just apply it to single point mass. What does the analogue of τ = Iα when the mass is not moving in the xy plane but just running around all over three dimensions. How do you define torque? How do you define angular momentum in 3D? You will find this a–it’s not a straightforward extension. You have to think a little harder. So, let me motivate the problem we have. Suppose I have a body in the xy plane. Right? It’s a sheet of metal or something. And it is spinning and a torque is applied here. This is r, and a force F is applied here. This is the kind of problem we’ve been studying. We know that the torque in this case is going to rotate in a counter-clockwise direction. If the forces are opposite it will rotate in a clockwise direction. So, whenever I did a torque calculation, the torque was either plus something or minus something. Plus meant counter-clockwise, minus meant clockwise. If I got ten different torques, I just add up all of them with proper attention to plus and minus signs.
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 xy plane but floating somewhere in space, and somewhere there on its surface is a force pulling it with a separation r. What does that torque do? The torque will certainly provide a counter – let me see – a counter-clockwise rotation, but around this axis, piercing the sample. So, we know it’s counter-clockwise, but it’s not enough to say it’s counter-clockwise because in this problem it is counter-clockwise around that axis. So, when a body is rotated in more than two dimensions, the body will tend to rotate around an axis either clockwise or counter-clockwise, but the axis itself has to be specified. As long as your body was stuck to the xy plane, it had to rotate around az axis. I’m just telling you, if you liberate the condition and throw it in free space like Frisbee–A Frisbee can rotate at any instant. Here’s a flying Frisbee; it’s rotating around some axis. So, in order to say that the torque is producing rotations, we want to characterize not only the magnitude of the torque but also tell you the direction in which it will produce rotations.
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 τ; τ, which was so far a scalar is going to be now promoted to a vector. And the vector will be denoted by this symbol, called a “cross product.” But I will now tell you how to get τ as a vector. We know r and we know F. But the torque, we suddenly learn, is actually a vector. So, well, if it’s a vector it needs a magnitude and it needs a direction. The magnitude of the vector is the old thing, Fr sin θ. I hope you people understand that in three dimensions, even in three dimensions, if r and F are two vectors, you know, they can move around like these two fingers; they define a plane. And in that plane, there is a clear angle between these two vectors. That’s the angle I mean. So, here is the vector of r. If you want, extend the vector r like that, that’s the θ that I’m talking about. Now, once you have two vectors defining a plane – here is one vector, here is another vector – there is a perpendicular to that plane. Every plane has a perpendicular. There’s only one ambiguity; it is that if I give you something planar like this, I can draw two perpendiculars, one going up and one going down. Both are perpendicular to the plane. So, we have to make a choice on what direction I want to associate with this τ. And that’s where you use the right-hand rule. The rule says–There are lots of names for this rule.
Let’s first take a simple case. Here is r; and let me bring the F vector and draw it like this. I want to define the cross product, r cross F. Its magnitude is this, length of r, length of F, sine of the angle. I’m trying to pick a direction. Either direction, in this case, is going to be out of the blackboard or into the blackboard. And the convention is stated in many different ways. One convention is, you take this r and you turn a screwdriver from r to F. Okay? You guys imagine in your mind; take a screwdriver, turn the screwdriver from r to F. Which way will this–That’s right. You see. That’s what all the rap guys are doing. They’re trying to use the right-hand rule, because the right-hand rule says that if you have an r and if you have an F that’s the torque. I’ve not followed the lyrics very closely, but I believe they’re muttering something about forces and torques. Okay. So, here is the right-hand rule. If this is r and that is F, then that’s the torque. But I find it much easier to do another right-hand rule, which is like this. You can see that in books too. Curl your fingers. The rule says, from r to F, and whichever way your thumb points, that’s the direction of r cross F. I think it’s very clear if you write it that way; that r cross F is minus F cross r. So, you’ve got to be very careful when you take this cross product. The order in which you multiply the vectors is very important. Okay.
So, in this example r cross F–Are you guys ready now. I’m not going to tell you the answer, but think for a second. r cross F in this example, if you turn a screwdriver or do the right-hand rule, it’s coming out of the blackboard. That is the direction of the torque. You can imagine these two vectors, which are in the xy plane; lift them upon and turn them into free space. It doesn’t matter. They still define a perpendicular direction uniquely, by turning the screwdriver from r to F or curling your fingers, or doing whatever you want; that’s the direction of the torque.
So, torque is a vector. By the way, that’s an accidental property of living in three dimensions. If you take two vectors A and B, we can define a cross product C equals A cross B. The rule for what C is equal to, is that the length of C is a length of A, the length of B and the sine of the angle between them. And the direction of C is found by taking the plane in which A and B lie and drawing a perpendicular to the plane in the sense that you rotate a screwdriver from A to B. If you’re living in four dimensions, if I give you two vectors like these two fingers, well, that’s perpendicular to it, you can see, but there’s one other whole fourth dimension you don’t see. So, in four dimensions, there is not one perpendicular direction to a plane. There are two perpendicular directions. So, you cannot take the cross product and turn it into a vector in four dimensions. It’s a beautiful property of three dimensions that one is able to take two vectors and manufacture from them a third vector. So, that’s an accident of 3D, but we live in 3D so we exploit this a lot, and it’s a natural quantity to bring in when you talk about torques.
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 A cross B is equal to minus B cross A, because when you turn the screwdriver from A to B it advances one way. If it’s B to A, it goes the other way. One particular consequence of this is that A cross A is zero because A cross A is minus A cross A; that means A cross A is zero. It also follows from the fact that the angle between A and A is zero degrees, so the sin θ vanishes. If you want to get practice, you can practice some simple cross products. Here is a unit vector I, here is a unit vector J, here is a unit vector K. So let’s try I cross J. I cross J is a vector perpendicular to I and J, namely, pointing in the Z direction, whose length is equal to 1 times 1 times sine of 90, which is 1. Therefore, this is none other than K. I cross J is K, but that’s equal to minus of J cross I. If you went from J to I, you turned the screwdriver from J to I, the screw will go down the z axis. So, I cross J and J cross I are opposite of each other, and I cross I is zero.
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 A dot B, is length of A length of B cosine of the angle between them. But I told you another way to do A dot B, AxBx + AyBy. You get that by writing A as I times Ax plus J times Ay, likewise for B and taking all the different dot products when you open all the brackets, that’s where you get AxBx + AyBy. There, because 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 Ax + J times Jy + K times Az, cross product with similar things for B. I times Bx + J times By, etc. K times Bz. There are, in principle, nine terms you can get in the cross product. Three of them will vanish, because when you take 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 AxBy - AyBx [correction: should have said AyBz - AzBy] plus 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 r cross F. I hope you know what that means. If a force is acting at the point with vector r measured from some origin, the torque due to that force is r cross F. Now, I’m going to introduce, for a single point mass somewhere here, an angular momentum, which is also going to be a vector. Previously, angular momentum Iω was not a vector. We just say it’s spinning clockwise or counter-clockwise. Now, for a single point mass, moving in some direction, I’m going to introduce the angular momentum to be r cross p. This is an inspired definition and you will find that it’s a good definition because things will work out. What’s the first thing this definition to satisfy? Anybody have an idea what we expect? What relation do I expect between torque and L?
Professor Ramamurti Shankar: Torque should be the derivative of L. Right. So, were you going to say that? So, I expect if things are working out τ should be dL/dt. So, let’s see what dL/dt is. The rate of change of angular momentum is going to be dr/dt cross p plus r cross dp/dt. I just used a rule of calculus. When you’ve got a product, you take the derivative of this and then the derivative of that. But dr/dt is the velocity cross momentum plus r cross the force, if the rate of change of momentum is the force. Now, this guy we like. That’s the torque. This guy we don’t like. But that’ll be zero. So, you know why that is zero, v cross p?
Professor Ramamurti Shankar: No. Yes?
Student: Because p = mv where v is the same as velocity.
Professor Ramamurti Shankar: Right. So, velocity and momentum are parallel. They’re just connected by a number m. But if you take two parallel vectors, their cross product will vanish because sin θ = 0. So luckily, that drops out and we do have this very nice result that if I define angular momentum this way, then the rate of change of angular momentum is the torque where torque was defined as r cross F. So, let me convince you that these definitions in three dimensions are fully compatible with what we have been doing all this time in two dimensions. Let me convince you this is not a brand new definition.
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 r cross p. This is r. Right? If it’s a part of the rigid body, it can only move that way. I’m first taking the case where the mass I’m considering is embedded on a rigid body. Then when you rotate it, it has to go along a circle with that as the radius. So, the momentum p will look like that, and the r measured from any point will look like this. r cross p, which is what I’m calling L; the length of L will be r cross p, the lengthy of r cross p, which will be r. For p, let me write mv. v is tangential velocity here. But we both know that that is equal to m times ω times r because the tangential velocity is ωr. That something we did long back. But then this becomes mω2r. I’m sorry–mr2ω. Good. And mr2 is the moment of inertia of this point mass, and 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 r never changed and we just went round. At any instant there is some ω, it has some angular momentum. But I want you to think about the following fact. Whenever you take a rigid body that’s made up of many bodies, each one of them has angular momentum Iω. You added them all up. But as a vector, each one has an angular momentum coming out of the blackboard or going into the blackboard. Because if r and p lie in the xy plane, then r cross p lies along the z axis, either out or in. That is why both torque and angular momentum were treated as scalars and not as vectors. Because if a vector is constrained to lie up or down the z axis, we can forget about the fact it’s a vector. We just say it’s plus if it’s up and minus if it’s down. That’s why in all the early calculations I did today with torques it was treated as a scalar, but actually the torque was a vector coming out of the blackboard or going into the blackboard. Next, I’ve shown you that the angular momentum you get by this definition, when applied to a rigid body – to a tiny speck on a rigid body – is the same as the angular momentum we’ve been using.
But now, I want you to think about the following notion. Here is the xy plane seen from the top. There is some body which is rotating. That’s the speck on the body, for which the angular momentum written as r cross p was Iω. But the speck now flies off. It breaks loose and takes off. Does it still have angular momentum? If you’re doing rigid body dynamics, you may not be so sure whether or not it has angular momentum. Because angular momentum we think of as coming from going round and round something. Well, this guy’s not going around anything. Right? Because it is liberated, it’s taking off like this. But if you take my definition that angular momentum as r cross p, you’ll find that it has angular momentum even when it’s here or there or anywhere else. And in fact, it has the same angular momentum. I want to tell you why.
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; r was in this direction, and the angular momentum was r times this momentum p. What when it’s come here? When it’s come there, r is this bigger vector r, p is the same vector p. It’s going at a constant speed. There are no forces on it. Has the angular momentum changed because the r has become bigger? But we don’t want just r times p. We want r times p times sin θ. But r sin θ is in fact–I’m sorry, θ is the angle between p and r. r sin θ is that perpendicular distance. That is not changing. So, maybe I’ll draw a better figure so you understand what I mean.
Take a body in the xy plane. It is just moving like this, at constant momentum. What’s its angular momentum, is what we are asking. The easiest time to compute angular momentum is when it’s here. Then, the angular momentum is that distance r times the momentum p times sine of 90, which is 1. If you are over here, then what happens is, p is still the same vector p, r looks like a longer vector, but the θ here, if you put it in, r sin θ is still the perpendicular distance. In other words, the angular momentum of a particle with respect to some origin is the product of the momentum times the perpendicular distance you get by extrapolating its velocity so you can drop a perpendicular on it. Draw a line through the momentum and drop a perpendicular on it from the origin. That distance times its momentum is the angular momentum. So, every particle, unless it’s going through the center for which I’m finding the angular momentum, will have an angular momentum. It doesn’t have to rotate, is the main point. Even a particle in a straight line has angular momentum, but I’m trying to show you the angular momentum here is not changing with time. It’s not changing because even as it moves, this component of r, the r -per p, is not changing. We can see that. No matter where you are, this distance is not changing. That’s the key to the angular momentum being a constant. All right.
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 τ = r cross F or dLi/dt–dLi/dt = τi. Angular momentum of body 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 A exerts a force on B in the line joining them.
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, τ = Iα. Well, the torque on this gyro is because the mass of the – the rod has no mass – the disk which is sitting there has a mass mg. So, torque is equal to mg times L where L is the length of this rod. That will produce an Iα. But let’s do the torque as a vector. What is the vector torque? I want you guys to think about that. Here is r. The vector r looks like this. Right? Horizontal. The vector mg looks like this, treated as a vector. So, r cross F, which is really not mgL, but r cross F is a general formula which reduces to mgL. Now, do the screwdriver rule. From r to mg, it goes into the board. So, we say the angular momentum of this rod gets an increment pointing into the board. We should be very, very clear that when angular momentum points into the board doesn’t mean the gyro goes into the board. It just means the axis around which the gyro rotates is pointing into the board. It’s another way of saying it rotates like that. So, the gyro will pick up an α, which we also assign as going into the board, it will just fall down. So, if you take a gyro and it doesn’t occur to you to spin the gyro, you might as well get a piece of stick with an orange at the end. What makes a gyro interesting is the following fact. You can spin the gyro now. Now it’s a different ballgame. Suppose the gyro now has an initial angular momentum. Which way is it pointing? That’s the first thing you’ve got to understand. This is a spinning disk. Every part of it is spinning, and the angular momentum of every part – if you do r cross P – will all point outside. So, one way to find the angular momentum of the spinning disk is to have your fingers curl along the direction of rotation and the thumb points in the direction of angular momentum. So, the gyro has an angular momentum L pointing that way.
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 L in this direction. Okay, this is a pretty dicey thing. So, you may have to go home and read it again. I don’t know how much I can do right now. But now let us look at the impact of the gravitational torque. The angular momentum change ΔL is equal to τ times the time Δt we agreed was going into the blackboard. That means this L is getting a little contribution ΔL in that direction and that just means the gyro is going to come to that location. So, what happens is, if you have an L to begin with, you act with the ΔL like this, you get the new L + ΔL, which is just a rotated L. That’s why the gyro doesn’t fall. See, for a static rod, got an angular momentum going into the board, it really means it’s falling down like this. But if a rotating gyro got an angular momentum on top of an existing angular momentum, then the new angular momentum is now just a rotated angular momentum. So, when seen from the top, the gyro will be doing what’s called a “precession.” If you support it, it will just slowly go round and round with this tip here. Now, does this remind you of anything you have seen before? Yeah?
Student: The Earth rotating around the Sun.
Professor Ramamurti Shankar: Right. When we did the Earth rotating around the Sun or any satellite going around, it was the same thing. Here is an apple. If it’s not moving, the force of gravity acts on it, causes an acceleration toward the center. That means what you and I think it means, falling toward the center. But if you take an apple with a huge velocity to begin with, add to it an acceleration toward the center or a tiny increment of velocity in this direction – we will just give it that velocity – and that’s the velocity it has when it has come here. Same thing with the gyro. The gyro does get the same angular momentum ΔL that it would whether or not it was rotating. But in one case, you add it to zero and conclude it’s going to fall down; in the other case, you add it to L and conclude that it will rotate. So, the gyro is constantly trying to fall down but it’s chasing itself. Just like this is constantly trying to fall down and chasing itself.
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 L. A little later L + ΔL looks like that. This is ΔL. And it rotates in angle Δθ in some time Δt. Then, just from the radians business you have learned, I think you guys know that ΔL is L times Δθ if θ is in radians. That’s why I’m cautioning you to always think in radians. So, if you rotate the vector by an angle Δθ, that segment for small inner segments is LΔθ. So, divide both sides by the time over which it happens. ΔL/Δt on the left-hand side is the torque, and this is L times what’s called the precessional frequency, ωP. That’s what we are trying to calculate.
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 mg times L. The angular momentum of the gyro is some ½ Iω squared–I’m sorry, Iω. Okay, so this is–Unfortunately, we don’t have the ability to demonstrate anything in this room. But if you looked at the stuff we have in Sloane [Physics building], if you take a big gyro and you spin it, you suspend one end on the tip of a tower and it’s spinning like this, you attach some weight to it, you will find it starts going around. If the gyro is not spinning, it will just fall down. If it’s spinning, it’s got angular momentum in that direction to which it gets an increment in that direction. So, it goes on rotating. And if you add more weights to the gyro, it will rotate faster because if you add more weights to the gyro besides its own mass you will increase this quantity. All right. So, we are done with the nasty part. So, next week we start relativity so that you guys can get something more contemporary. Before you go, any final questions about–yes? Mid-term? Yeah?
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