11. Torque
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Transcript: English(auto-generated)
00:01
So, so far, here's what we have done. We've been studying the whole time this analog of Newton's law, τ equals Iα. So, after teaching you what τ is and what I is and what α is,
00:21
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 response to the torques, the body will accelerate. Then, we went to another class of problems where there are no
00:41
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. There's no torque.
01:05
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
01:22
has no angular velocity. Common example is you are trying to do something in your house. You get up on a ladder, you're somewhere there. This ladder should have no angular velocity, right? That is not a trivial statement to a person who is going up there.
01:41
So, you want to ask, what does it take to keep the ladder from falling over? Okay, so 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
02:02
around? Well, the first thing you know is the forces on it should add up to zero, because otherwise F equals 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
02:21
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 force adds 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
02:42
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
03:00
plane of the blackboard, to stay still and not move. So, the conditions for a body, they are that all the forces in the x direction, I is some label, force number one, force number two, 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
03:21
not going anywhere. And finally, to keep the body from moving or spinning, all the torques should add up to zero. So, the 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,
03:41
but still in equilibrium, in static equilibrium. That means they are not moving. So, the simplest problem from the family is a seesaw. Let's say there's a point where the seesaw is supported, and let's say we put one kid here applying some force,
04:01
and the 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
04:21
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
04:42
into the ground. There is a support and a normal force from the support. And these three forces together should conspire to keep the body in equilibrium. So, let's see. So, here's what I want you to understand. F1 is known. Let's give it some name, 100 Newtons.
05:02
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 0 equals 0. In the vertical direction,
05:24
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 have two. I just know this guy.
05:42
So, that's where the torque equation is 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 we're supposed to do? Yeah.
06:01
Yes, you. Yes. Right. What r should I use for this force? Pardon me? x1? Okay, why did you choose that?
06:24
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, it 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 torque to that axis.
06:42
When a body is not rotating, it's not rotating about 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.
07:02
If you change your mind on where you're taking the torque, then the torques are all changing because in the force time 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.
07:23
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
07:43
the same thing, and that's what I will show you now. 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
08:01
point, it would 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 add it up to zero. Now, you come along and say, no, I want to take the torque around that point.
08:21
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 will find it as xi xi plus a times sum of all the fis.
08:47
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
09:00
around that point don't cause, if 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 are 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.
09:25
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, I'll just find F2, I don't care what N is, then that's a particular choice of pivot point that is
09:42
optimal. That choice is the one in which this normal force doesn't get to enter the torque equation. 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.
10:01
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.
10:20
It'll 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 pick the point of support and then, of course, here is the point of support. And what did I have? F1 here, a distance x1, and F2 at a distance x2.
10:40
This is the point around which I'm doing the torques. So, N doesn't count, then you get the condition F1x1 equals F2x2. From that, you can solve for F2. For example, if this was 1 meter and this was 6 meters, and this was say 10 newtons here, then 10x1 has to be
11:04
6xF2. So, 10 equals 6F2, and F2 will be 10 over 6 newtons. I don't want to use N because N is also standing for the force here. That's how we find F2.
11:21
Once you've found F2, in case you want to know what is the support going through, 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.
11:41
Even if you were asked to find N, it's best to first take torque around the 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 this simple notion of how to take the torque. All right, so now I'm going to
12:04
take a slightly more complicated problem. So, here's 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
12:22
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,
12:44
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 at anything or anything, but the pivot is frictionless.
13:00
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 the torque due to gravity is what I want to discuss now. First of all,
13:20
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.
13:41
No matter what's going on, it 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 it's F times
14:03
r times the sine of the angle between them, and sine of the angle is sine of 90, which is 1. How about the torque due to the pull of gravity? Anybody know what we're supposed to do here? Yep? Okay.
14:23
Now, why is it okay to replace the torque by the center of mass? Yep, either of you can answer yes. Right, but why am I allowed to pretend? See, we can pretend a lot,
14:43
but it occasionally has to correspond to reality so that it works. So, why is it okay to pretend? That's the correct answer, 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.
15:05
Thank you. Yes. Your answer is right. I'm just going to, I'm going to tell you this. 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,
15:21
and not good enough. That's the 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 equals ma. You've got to go back to that all the time. Of course, that is the other principle, that gravity is an extra force in the body. But this is an extended object.
15:42
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
16:01
tiny pieces, each of which is small enough to say it has a definite location xi. Now, the torque due to gravity acting on that little fellow is the mass of the little guy times xi times g, because mg is the force and mg times xi is the torque. You want to sum it over all
16:22
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 stare at this combination, mi xi summed over i divided by m, that is the center of mass. So, it does look like mgx,
16:42
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 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
17:03
over the length of the rod. And it's not obvious that's constant. I mean, if I took a tiny rod, force of gravity is constant. If I 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 in the laboratory scale, that's not an issue, and that's where this result
17:23
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, and that's equal to mg times x, where x is the location of the center of mass now. That is going to be half the
17:43
length of the rod. I'm sorry, this Fx is FL. You cancel the L, then you find the force I apply is mg divided by 2. Yep, if the pivot is exerting what force?
18:08
Student asks a question. Professor Ramamurti
18:20
Shankar asks a question. I'm 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.
18:40
Oh, but the pull of gravity is straight down and the separation is horizontal. Is that the question now? Oh, the force I'm applying. Oh, I'm applying a force here, straight up. Oh, you want to know, right? Ah, well, you're quite right.
19:08
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 a vertical force and possibly a horizontal force to
19:22
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 horizontal part that I call F, and a horizontal part that I call H prime. Then, it'll turn out that in this problem an H prime is not needed.
19:42
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. It's not necessary to hold this in equilibrium. Maybe the main point is a purely vertical force will
20:01
suffice to keep this in equilibrium, is the point that I'm making. In fact, if that force is equal to mg over 2, it won't rotate. Now, we can go back. Now that I find the force, I have to apply in this direction, what are the forces at the pivot? Well, the vertical force on the pivot is pointing up. My force that I calculated is
20:23
pointing up, and together they should add up to mg. Now, F is equal to mg over 2, so the vertical force will be equal to mg over 2. Now, horizontal force in this problem, there is no need for you to apply the horizontal
20:42
force, and there's no need for the pivot to fight you back with the horizontal force. It's not needed for equilibrium, but suppose you insisted, not only are you lifting it, but you're 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, the walls got to push you out, and h prime will be equal to h.
21:04
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 this section, and it goes like this. Here is a rod,
21:23
and it's supported by a wire, and the wire has a certain tension T, it's anchored to the wall, the length of the rod is L, the mass of the rod is m. And at this end is some pivot which is exerting some unknown force, which can have a
21:42
horizontal and vertical part. It'll have a horizontal part h and a vertical part b. Now, I know for sure that's a horizontal part. I think you can tell, can you guys tell in your mind whether it's a horizontal part? This force that's supporting it is pushing the rod into the wall and also pulling it up.
22:01
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's 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.
22:24
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 a formula in F times R times sin θ,
22:42
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 over 2, turning it this way. So, this force exerts a torque, which is equal to the value of
23:05
the tension, the length of the rod, and 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 over 2 sin θ.
23:28
That's why θ cannot be too small. If θ's too small, at some point t will become so large that the string will break. So, when you want to anchor a sine, you've got to try to take a big angle like that so that sine θ is not too small.
23:43
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 1000 Newtons? Well, then you put 1000 Newtons here, you know the m and the g, you solve for the θ at which it crosses 1000 Newtons, and your θ should be bigger
24:00
than that. Yep, that was the problem. 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
24:24
how that happened. I'm trying to find the analog of F equals ma, analog 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,
24:44
okay? Look, you've got to understand that the right 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
25:00
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.
25:21
If that angle is θ, then F sin θ is really the perpendicular part of F. So, that's how the torque was defined. Okay, so we found the tension here. If that's all you want to know, that's the end of the problem. You got the tension. But you may be asked to do
25:41
more. You could be asked to now find what's happening at the pivot. Then, you 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,
26:01
because this T has a T sin θ that way and a 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 over 2 cotangent θ, because that's cosine
26:22
over sine. All right, so that's the horizontal part. How about the vertical part? The vertical direction, upward vertical force, plus T sin θ has to be the downward force,
26:41
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 over 2, equal to mg. That means v is mg over 2. That means the pivot provides
27:01
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,
27:21
and maybe 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,
27:41
and gravity, for this purpose, I will 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.
28:01
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, nor anything to do with the pivot, enters that. So, bear that in mind. Now, you can embellish the
28:20
problem. I don't want to do that. You can hang a weight here. I don't think that should cause a new problem. Suppose there's an extra weight, little m, on top of the mass of the rod, I'm 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'll be little m times g
28:43
times the whole length of the rod, because I'm taking the torque around that point. You put that in, 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
29:02
the last of these equilibrium problems before I do something else. The last of the equilibrium problem is very standard, fair 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
29:29
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?
29:40
A lower limit or an upper limit? Would you like to guess? A lower limit on the θ? By that you mean what? If θ is less than some amount, what'll happen? It'll fall. Yeah, very good. So, for equilibrium, there'll be a lower limit on θ.
30:01
I think it's clear to all of us that if you try to climb a building and you put a ladder like this, you're asking for trouble, right? Yep. Look, I agree. You mean 91 degrees ladder like this?
30:21
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. Philosopher's, even more difficult. What is good?
30:40
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 or 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 time this question has come up.
31:02
So, I learn something every day. All right. Anyway, let's say we are 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? Yep?
31:21
Yes. Between the what? Okay. So, in fact, it turns out even the coefficient of friction with this wall may help, but we are going to rely on a problem where the wall is frictionless and the floor has a friction. And it's got a static friction, μ, which I'm just going to give to you. So, here is our challenge now.
31:44
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'll all happen. The first condition is, well, the zeroth condition is draw
32:03
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, center of the rod. The wall is very important. If it's a frictionless wall,
32:21
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,
32:42
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.
33:01
For the x motion, the fact that all the, maybe I should write it, sum of all the x forces is zero, simply says w equals F. Sum of all the y forces equals, tells me,
33:21
n equals mg. 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
33:41
some crazy point, and then n and f will all come into the 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?
34:00
For w, it is w times L times sine theta. 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
34:21
of separation, w is the force, the distance is L, the length of the ladder, and this theta and that theta 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 sine theta you want to use is that. That's the torque. That's the torque due to
34:42
the wall. The torque due to gravity is mgL over 2 cosine theta. Now, why did I write cosine theta? Am I making a mistake instead of sine theta? Why is it cosine theta?
35:05
Yes? So, let me repeat the answer. When you write torque as mgL sine theta, that theta is not some fixed angle that somebody drew in the
35:21
diagram. It varies from force to force. It's the angle that force makes with the line connecting the point of application of 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. You really want the sine of
35:41
this angle, but sine of that angle is cosine of this angle. So, you can cancel that. So, I find W equals mg over 2 cotangent theta. Just the torque equation tells me that's the W I need. So, the wall's going to push
36:04
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 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
36:22
friction times the normal force. 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
36:41
force, you take the torque, you take the force, you join the point of application of the force to the point around which you're doing the torque, and take the angle between those two lines. So, it's really that angle, which is not the theta here. But if I draw the picture here, you can see this is theta, this is 90 minus theta, this angle that I
37:01
want. But then, sine of 90 minus theta is cosine theta, and that's why you use that. Sine of 90 minus theta is cosine theta, because when you change theta at 90 minus theta, what's opposite becomes adjacent and what's adjacent becomes opposite in any right triangle. If that's your angle,
37:21
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 theta goes to 90 minus theta. 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 equals Mg.
37:40
So, bring this equation here. So, that's equal to μs times Mg. So, I have the restriction that Mg over two cotangent theta. So, let's compare that to that, and we find the restriction that Mg over two
38:00
cotangent theta less than or equal to μs times Mg. You cancel the Mg and you find cotangent theta has to be bigger than or equal to 2 times μs.
38:21
I like to write it as tan theta has to be bigger than or equal to 1 over 2 times μs. Do you know how I 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 ten,
38:42
one third is bigger than one tenth. Same thing. So, I'm somehow used to changing theta to bigger than rather than with cotangent, so I flip this over and I flip that over, and I also change less than to bigger than. This way, we can understand tan theta has to be bigger than some number. That means theta has to be bigger than some number, because tan theta increases with
39:02
theta. So, you tell me what your coefficient of friction is. Maybe it's 0.5, then this is 1. Tan theta should be bigger than 1, theta should be bigger than 45 degrees, okay, or π over 4. Maybe I should keep using radians so you guys get used to
39:24
radians as a way to think about it. So, what's the famous show in CNN? Anderson, Cooper, what? Anderson, Cooper, 2π. That's what it is for this class. Think radians. Are you going to screw up somewhere?
39:41
Okay, angle is what we use every day, but radians is what we use here. When I write a tan theta and I write a tan theta and I don't tell you what it is. Or, if I tell you tangential velocity is ω times R, that angular velocity ω is radians per second. A lot of people get tripped over this, okay?
40:00
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
40:21
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 forces 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
40:40
top or a potato. So, it's 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
41:04
limited context. A limited context I've set for myself right now is to just apply it to a single point of mass. What does the analog of τ equals Iα when the mass is not moving in the xy plane
41:20
but just running around all over three dimensions? How do we define torque? How do we define angular momentum in 3D? You'll find that 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
41:42
middle 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 it in a
42:03
counterclockwise direction. If the forces are opposite, it'll rotate in a clockwise direction. So, whenever I did a torque calculation, the torque was either plus something or minus something. Plus meant counterclockwise, minus meant clockwise. If I got ten different torques, I just add up all of them with proper attention to plus and minus signs.
42:23
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
42:41
force F. So, what does that torque do? The torque will certainly provide a counter, let me see, a counterclockwise rotation, but around this axis, piercing the sample. So, we know it's counterclockwise, but it's not enough to say it's counterclockwise because
43:02
in this problem, it is counterclockwise 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 counterclockwise, but the axis itself has to be specified. As long as your body was
43:22
stuck to the xy plane, it had to rotate around the z axis. I'm just telling you, if you liberate the condition and throw it in free space like a 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
43:43
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
44:02
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,
44:22
but the torque we suddenly learn is actually a vector. So, if it's a vector, it needs a magnitude and it needs a direction. The magnitude of the vector is the whole thing, FR sin θ.
44:42
I hope you people understand that in three dimensions, even in three dimensions, if R and F are two vectors, 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 need. So, here is a vector R.
45:02
If you want, extend the vector R like that. That's the θ I'm talking about. Now, once you have two vectors defining a plane, here is one vector, here is another vector, that is perpendicular to that plane. Every plane has a perpendicular.
45:21
There's only one ambiguity, 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 tau. And that's where you use the right-hand rule. The rule says there are lots
45:43
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 x F. Its magnitude is this, length of R,
46:00
length of F, sine of the angle. I'm trying to pick a direction. Either the 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.
46:24
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
46:41
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 and that's the torque. But I find it much easier to do another right-hand rule which is like this.
47:02
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, R cross F is minus F cross
47:20
R. So, you've got to be very careful when you take this cross product. The order in which you multiply the vector is very important. Okay, so in this example, R cross F, you guys ready now? I'm not going to tell you the answer, but think for a second. R cross F, in this example,
47:41
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 up 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,
48:00
doing whatever you want. That's the direction of the torque. So, torque is a vector. By the way, that's the 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
48:23
to is that the length of c is the length of a, the length of b, and the sign 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
48:40
dimensions, if I give you two vectors like these two fingers, well, that's a 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.
49:01
So, this is 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
49:21
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
49:42
zero, because a cross a is minus a cross a, but 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 sine theta vanishes. So, if you want to get practice, you can practice some simple cross products. Here is the unit vector
50:02
j, here's the unit vector i, here's the unit vector j, here's the 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.
50:24
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
50:41
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're
51:01
going to 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 it's a homework problem. Remember the dot product I said is a dot b,
51:21
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 plus 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 you get when you open all the brackets. That's where you get axbx plus ayby.
51:42
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 plus j times ay plus k times az, cross product with similar things for b,
52:01
i times bx plus 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 0, when you take i cross j you will get something, when you take j cross i you'll get minus something, you put them all together. I won't do this now.
52:21
I'm not going to dwell on this. It will look like i times axby minus aybx plus j times something and k times something. I don't need this, 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
52:42
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've introduced torque as r cross F.
53:00
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, and angular momentum, which is also going to be a vector. Previously, angular momentum,
53:22
iω, is not a vector. You just say it's spinning clockwise or counterclockwise. 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 it's a good definition because things will work out.
53:43
What's the first thing I want this definition to satisfy? Anybody have an idea what we expect? What relation do I expect between torque and L? Torque should be the derivative of L, right? So, were you going to say
54:01
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
54:21
dt. Just use the 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. The rate of change of momentum is the force. Now, this guy we like.
54:45
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?
55:10
Yes? So, velocity and momentum are parallel. They're just connected by a number m. But if you take two parallel vectors, the cross product will vanish because sine theta
55:22
is zero. 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, but torque was defined as r cross F. So, let me convince you that
55:45
these definitions in three dimensions are fully compatible with what we've 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.
56:04
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.
56:21
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
56:41
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 length of r cross P, which will be r. For P, let me write mv. mv, v is the tangential velocity here. But we both know that that
57:03
is equal to m times ω times r, because tangential velocity is ωr. That's something we did long back. But then, this becomes m ω square r. I'm sorry, mr square ω. Good.
57:22
And mr square 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 have, 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.
57:42
So, this is why this r never changed, and it just went round. At any instant, if 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 as made up of many bodies, each one of them has angular momentum Iω,
58:01
you added them all up. But as a vector, each one has angular momentum coming out of the blackboard, or going into the blackboard. So, I want you to think about this, because if r and p lie in the xy plane, then r x 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.
58:24
Because if a vector is constrained to lie up or down the z axis, we can forget about the fact that 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
58:40
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. There is a speck on the
59:02
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
59:20
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's been deliberated, it's taking off like this. But if you take my definition, that angular momentum is r cross p, you'll find that it has angular momentum even when it's here or there or
59:41
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 an epsilon before it took off from this rigid body. It was still stuck to the rigid body. Then the momentum was in this direction, r was in this direction, and the angular momentum was
01:00:00
R times is 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
01:00:23
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. Maybe I'll draw a better figure so you understand what I
01:00:40
mean. Take a body in the xy plane. It is just moving like this, at constant momentum. What is angular momentum is what we are asking. The easiest time to compute the angular momentum is when it's here. Then, the angular momentum is that distance R times the momentum P times sin of 90,
01:01:03
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 this θ here, if you put it in, R sin θ is still the perpendicular distance. In other words,
01:01:21
the angular momentum of a particle with respect to some origin is the product of the momentum times the perpendicular momentum. So, the angular 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
01:01:42
origin. That distance times its momentum is the angular momentum. So, every particle, unless it's going through the center, about which I'm finding that angular momentum, will have an angular momentum. It doesn't have to rotate.
01:02:01
It's 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 perp, is not changing. You can see that. No matter where you are, this distance is not
01:02:20
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 interatomic forces and they form a rigid body, then you can use tau equals R cross F, or dL dt.
01:02:41
dLi dt is tau I. 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 the rigid body made up of many
01:03:03
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
01:03:22
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. Remember, the internal force is cancelled in pairs, and therefore we could drop that, and therefore rate of change of total momentum was
01:03:42
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,
01:04:00
you multiply the force you exert on me, cross product with the distance of me from the origin, whereas with you is the distance of you from the origin, so they don't necessarily cancel. And you can show that only if the force one body exerted on the other was in the line joining them, then that would cancel and angular momentum would be
01:04:21
conserved. So, angular momentum conservation requires more than just usual momentum conservation. Only if at a microscopic level, forces that bodies exert on other bodies is in the line joining them, this angular momentum conserved. Fortunately, that turns out to be true for every force we know, gravitation, Coulomb's law, electrostatic forces.
01:04:42
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,
01:05:03
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. Maybe we can draw it like this. It's a disk, in this case,
01:05:22
which is spinning like that. Everybody with me now? There is a little tower 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'm going to, I'm holding it here, and I'm going to let go.
01:05:43
What's going to happen to the gyro, is what we're asking. Gyro is not spinning. Everybody should know, you don't need physics 200 or 100 or anything to know what's going to happen. You let it go, it's going to fall. You want to make sure that it agrees with the equation that we have, τ equals Iα. Well, the torque on this
01:06:04
gyro is because the mass of 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 the rod. That will produce an Iα. But let's do the torque as a
01:06:22
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,
01:06:44
but R cross F is the 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,
01:07:00
very clear that when angular momentum points into the board, it 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, gyro will pick up an α, which we also assign as going into the board, it'll just fall down. So, if you take a gyro and it
01:07:21
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 ball game. Suppose the gyro now has an angular momentum, the initial angular momentum. Which way is it pointing? It's the first thing you've
01:07:41
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 x P, will all point outside. So, one way to find the angular momentum of a spinning disk is to have your fingers curl along the direction of rotation and the thumb points in the direction of angular momentum.
01:08:00
So, 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.
01:08:23
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, L times ΔL is equal to τ times the time Δt.
01:08:40
We agreed it 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 add with a ΔL like this,
01:09:01
you get the new L plus ΔL, which is just the rotated L. That's why the gyro doesn't fall. It does get an angle. See, if 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
01:09:23
existing angular momentum, 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, you'll just slowly go round and round with this tip here. Now, does this remind you of anything you've seen before?
01:09:44
Yep, 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,
01:10:01
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, you just give it that velocity and that's the velocity it has when it has come here. The same thing with the gyro. The gyro does get the same
01:10:22
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 another 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 changing itself.
01:10:42
So, the last thing I want to calculate is the rate at which it goes round and round. It's a pretty easy calculation. I will do that and I will let you go. So, let's ask the following question. Here is a gyro whose angular momentum is L. A little later, L plus ΔL looks like that.
01:11:00
This is ΔL and it rotates an angle Δθ in some time Δt. Then, just from the radiance business you've learned, I think you guys know that ΔL is L times Δθ, if θ isn't radiant. That's why I'm cautioning you, always think in radiance.
01:11:23
So, if you rotate a vector by an angle Δθ, that segment for small inner segments is L Δθ. So, divide both sides by the time over which it happens. ΔL over Δt, on the left-hand side, is the torque and this is L times what's called the
01:11:44
precessional frequency Ωp. That's what we're trying to calculate. So, the precessional frequency in this problem is equal to the torque divided by angular momentum of the gyro. So, the torque, now we can put the numbers in,
01:12:04
is mg times L. Angular momentum of the gyro is some ½ iω square, I'm sorry, times iω. Okay, so this is,
01:12:24
unfortunately, we don't have the ability to demonstrate anything in this room. But if you looked at the stuff we have in Sloan, 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.
01:12:42
The gyro is not spinning, it'll just fall down. If it's spinning, it's got an 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'll 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
01:13:04
with the nasty part. So, next week we start talking about relativity, so that you guys can get something more contemporary.