8.01x - Lect 24 - Rolling Motion, Gyroscopes, VERY NON-INTUITIVE

With our knowledge of torque... calm down. With our knowledge of torque and angular momentum, we can now attack

rolling

objects

rolling

downhill. For example, the following... I have a cylinder here or it could be a sphere, in fact, and this angle is beta. I prefer not to use alpha because it is angular acceleration and there is a coefficient of friction with the surface, mu, and this object will roll down and get an acceleration in this direction, a. And I'll evaluate the situation when we have a good time. That means the object does not skid or slide.

What is pure roll? If there is an object here, here is the cylinder of radius R, and I am going to rotate it like this and roll it in this direction, the center is called point Q. Once it has made a complete rotation, if then the point Q has moved a distance 2pi R, so we call that pure throw. When we have pure roll, the velocity of this Q point and the velocity of the circle, if you can read that, I'll just put a c in there, they are the same. In other words, vQ is exactly the same as v circumference, and v circumference is always omega R.

This part always holds, but for the pure roll, this holds. You can easily imagine that if there is no friction here, the object could be still, spinning like crazy, but Q wouldn't go anywhere. So we have skidding and slipping and then we don't have the pure rolling situation. If the object skids or slides, then friction must always be maximum. If the object is in pure

motion

, the friction could be substantially less than the maximum possible friction. Now I would like to calculate with you the acceleration that a cylinder would obtain. When it rolls purely down that slope, it has mass M, length l, and radius R.

And I would like you to use your intuition and not be afraid that it is wrong. I'm going to go down two cylinders down this slope. They are both solid, they have the same mass, they have the same length but they have

very

different radii and I'm going to have a race between these two. Which one will reach the bottom first? Then I repeat the problem. Two cylinders, both solid, of the same length, same mass, but one has a larger radius than the other. There's going to be a race. We are going to roll them down, pure nonsense.

Which one will do it? Win win? Will he win? Who do you think the one with the largest radius will win? Who do you think will win the big one with the smallest radius? Who thinks there will be no winner or loser? Wow, your intuition is better than mine. We'll see how it goes. Keep in mind what your vote was and you will see it come out

very

soon. Well, let's put all the forces we know on this object. This is Mg. And we're going to decompose that into a longer slope, which is Mg sine beta, and another perpendicular to the slope.

We've done it millions... millions of times. And this is equal to Mg cosine beta. So here there is a normal force, and the magnitude of that normal force is Mg cos beta, so there is no acceleration in this direction and then we have a friction force here for which I will write F of f. There is an angular velocity at any moment, omega, that will change over time, no doubt. And then this center point Q, which is the center of mass, will get a velocity v and that v will also change over time. And the v of the point Q that changes with time is v of the circumference, because that is the condition of the pure roll.

That equals omega R. This is always true, but it's only true when it's pure bullshit. I take the derivative of time. The derivative of the velocity of that point Q is, by definition, its acceleration, so I get an omega point equal to R, and that is equal to alpha R, with alpha being the angular acceleration. So this is the condition for the pure roll. Now I'm going to take the torque around point Q. When I take the torque around point Q, N has no effect because it goes through Q, and g has no effect because it goes through Q, so there is only one force adding to the torque. .

If this radius is R, the magnitude is RF and the direction is on the board. But for now I'm only interested in the magnitude, so I get R multiplied by the friction force. This should be I alpha, with I being the rotational moment of inertia about this axis through point Q, multiplied by alpha, but I can replace alpha with a/R. Then I get the moment of inertia approximately Q multiplied by a/R. And this is my first equation, and I have the friction force as an unknown, and I have a as an unknown, so I can't solve both.

I need another equation. The next equation I have is obvious, it is Newton's second law: f = MA. For the center of mass, I can consider all the mass here at Q. We must have f = MA. And then M multiplied by the acceleration of that point Q, which is our goal, by the way, is equal to this component, is equal to Mg sine beta. That's the downhill component. And minus Ff, the friction force, which is the uphill component, and this is my equation number two. Now I have two equations with two unknowns. So I can solve it now.

I can eliminate Ff and I'll replace Ff here with this quantity divided by R. And so I get that Ma is equal to Mg sine beta minus the moment of inertia around the point Q, multiplied by a divided by R squared. And now notice that I removed F from f, so now I can solve for a. So I'm going to take out... I take the a to the side. So I get a multiplied by M plus the moment of inertia divided by R squared equals Mg sine beta, and now we have. I multiply both sides by R squared.

I get MR squared g sin beta above, and below I get MR squared plus the moment of inertia around that Q point. This is my result, and all I have to enter now is the rotational moment of inertia around that axis . But I want to remind you that this is only true if we find ourselves in a situation of pure nonsense. So now we can substitute in there the values ​​that we have for a solid cylinder. If we have a solid cylinder, then the moment of inertia about this axis through the center of mass, which I have called Q, is equal to 1/2 MR squared.

And if I substitute that in here, notice that all my M... MR squares disappear. I get 1 + 1/2, which is 1 1/2. The other way around it becomes 2/3. So a = 2/3 times g multiplied by the sine of beta. There is no M, there is no L and there is no R. So if I have two cylinders, solid cylinders with totally different mass, totally different radii, totally different length and they have a stroke, neither one wins. Very un

intuitive

. Every time I see it I find it something amazing. Observe that everything disappears. M, R and l disappear. So those who said that if I take two cylinders with the same mass, different radii, those who said that there is no winner or loser, they were right.

But even more surprising is that even the dough can be changed. You can change anything as long as both cylinders are solid. That's what matters. So if we take a hollow cylinder, then the moment of inertia about this axis that goes through the center of mass, by Q, if this... if really most of the mass is actually on the surface, then it's very close to MR squared, and then the acceleration... if I substitute MR squared here, I get a 2 there... it's equal to 1/2 times g times sine beta. So this acceleration is less than this one. Thus, the hollow cylinder will lose in any race against a solid cylinder, regardless of its mass, radius or length.

And I want to show you. We have a setup here and I'll try to show it to you on the screen as well, but for those of you sitting nearby, it's probably much better to watch the demo right here. I have here... ooh. Uh-uh. I have to start with a very heavy cylinder made of brass and this one is made of aluminum. They have very different masses, same radii, same length. There shouldn't be any difference. There should be no winner and no loser. I'm going to start with them at the same time. I hope you can see that there.

This is... this is the starting point. You can lower it a little. I'll count down from three to zero and then you'll see they hit the bottom almost at the same time. Very different in mass. The difference in mass is at least a factor of three. All other dimensions are the same. Three, two, one, zero. Completely in unison. It's not

intuitive

to me. Now I have one that has a very small radius compared to this one. This is a small aluminum rod. Maybe you can see it here, television. This is much heavier, almost 30 times heavier. There shouldn't be any difference.

As long as it's solid, there shouldn't be any difference. Neither winner nor loser. The radii are different, the masses are different. There shouldn't be any difference. Well? Here we start the race. Three, two, one, zero. And they hit rock bottom at the same time. But now here I have a hollow one, and you better believe it, it is hollow. So now all the mass is on the circumference and now it takes longer. Now the acceleration as it was... as you will see, is half of g sine beta; in the other case it was 2/3. And you might want to think about it tonight, why this one requires more.

It has to do, of course, with the moment of inertia, but again, it is independent of mass, radius and length. So it's purely a question of geometry. This one will be the loser and this one, regardless of its mass or length, will be the winner. Then you see them. One is hollow, the other is not. This is very light; this is very heavy. I'll put the hole on your side. Three, two, one, zero. The hollow lost and even fell to the ground. Yes, I always find these things quite surprising, that nature works this way, and I'm impressed that most of you or many of you had the right intuition when you said that there would be no difference for the two solid cylinders.

Now we come to the least intuitive part of all of 8.01 and possibly the most difficult part of all of physics and that has to do with

gyroscopes

. And I really urge you to pay close attention and don't even miss ten seconds, because you're going to see some mind-blowing demos that are so incredibly unintuitive that unless you've followed the steps leading up to them, you won't have any idea. what you are looking at. It will be fun, it will be cute, but it won't do anything for you. Imagine that you and I go to outer space. Without gravity.

We are somewhere in outer space and we have this bicycle wheel there. And I'm going to torque this bicycle wheel in this direction, so I'm going to put my right hand toward you and my left hand away from you. And I'll do it for a short period of time and let it go. It's obvious what's going to happen. This wheel is going to turn like this forever and ever. I gave it a little torque. That means that if there is torque, there is a change in angular momentum. The change in angular momentum should be the torque multiplied by delta t.

So I do this and I let it go and it will spin around this axis forever and ever. Simple, right? Well. Now I'm going to squeeze it in this direction. So we're in outer space, the wheel is stopped and all I do is do this and I let it go. Then it will turn forever and ever and ever and ever in this direction. That's clear. Now comes the very unintuitive part. Now I'm going to turn it in your direction and now I'm going to press like this again. What will happen now? You will say... or you could say.

I'm not accusing you of anything. You might say, well, you give the wheel a spin so this... the wheel will probably continue to spin and you do this, then maybe what you're going to see is that it will spin just like it did before and at the same time the wheel will be spinning. But that can't be because if the wheel were spinning like this, then the angular momentum of the spinning wheel is in this direction. And if I were to give it a twist and it continued to spin and it rotated like this, then this spin angular momentum would spin like this and that can't be because there's no torque in the system, because once I let go, there's no torque anymore.

So it is not possible for the wheel to continue spinning, and as a result of this torque that I give it, it just spins. That is not possible. How is nature going to deal with that? I'll show you on a visualization graph. It is very unintuitive what will happen. And then also... I will also prove it to you. So here is the situation exactly as I described it to you. You, as an observer, are in this direction. This is the address of 26,100. So you're looking at the wheels like this. They are turning towards you. That's what I'll do shortly and that's what I hinted at now.

This is my right hand and this is my left hand. The separation between my height... right and left hand is the small "b". So the torque that I apply is bF, it's this arm, so to speak, multiplied by this force. And the force is perpendicular to the arm. I apply a torque for a certain period of time: delta t. When I do that, I apply, I add angular momentum in this direction. But the wheel was turning in... in... in this direction. You see it. And then the angular momentum of rotation of the wheel is in this direction.

I add angular momentum in this direction and you see it here. So this was originally the turning angular momentum of the wheel. I tighten the time delta t, so I add angular momentum like this. And then I stop. I just squeeze for a short period of time and stop. That means that after I have stopped, the angular momentum of the system as a whole can no longer change because there is no torque in the system, and the only way nature can now solve that problem is to tilt this wheel in the way in which I have done it. indicated here, and to make it turn in this direction and it will stop.

In other words, I hold it in my hand, the wheel... I'll pick it up... I'll give it a spin. I have it in my hand, I turn it towards you... and I'm going toput my right hand towards you and my left hand away from you. The spin angular momentum is now in this direction, so I'll give it a torque like this. That means up. And what will the wheel do? The wheel will do this. Very unintuitive. Look at it. Isn't that strange? You wouldn't expect it. I'll do it again. I go to turn pushing my hand towards you and the steering wheel does something completely unexpected.

He just leans in. If I tighten the other way around, then the torque will of course make it turn like this. I'm going to give it a little bit more turning angular momentum, so now I move my left hand towards you and my right hand towards me, and then I wait for the wheel to do this. And that's what he does. Extremely unintuitive. These torques applied to a spinning wheel always do something unexpected. However, there is one thing that always helps me in terms of guiding me and that is that you can always predict that the spin angular momentum will always move in the direction of the torque, which is this external torque that I applied.

Let's go over that again. Here we have the spin angular momentum that you saw. I was pointing in this direction. And I applied the torque in this direction, the vector. And what does spin angular momentum do? It goes in the direction of the pair. And then when I stop with my torque, then of course nothing changes anymore, and what happens is this wheel tilts. But notice that L, the spin angular momentum, has moved from here to here. And it was tightening in this direction, so it moved towards torque. If you have digested this, then you can test yourself now.

Now we have the same wheel. I'm going to turn it in exactly the same direction, but now I'm not going to press like this, in the Z direction, or like this, in the minus Z direction. Now I'm going to do this. Or I'm going to do this. Try now to really concentrate on what I just taught you. And try to answer the following question. The wheel is turning. I'm holding it in my hand and I'm going to tighten it like this so that the torque factor is in your direction. The angular momentum is like this, the torque is like this.

What will the angular momentum vector do? Move in the direction of... of the pair. What will the spin angular momentum vector do then if the torque is in this direction? Will do this. It will move in the horizontal plane. Very unintuitive, but that's what it will do. And I'll show you. I'm going to spin this wheel. I'm going to spin it with high angular momentum, high spin angular momentum. And then I'm going to sit on this stool and apply torque exactly like you see in the image on the right. I'm going to press like this. And as long as you tighten it like this, the spin angular momentum wants to spin in the horizontal plane.

And when you press the other way around, it will return to the horizontal plane. I'm going to tighten exactly as you see in the image. Are you ready? I stop the torque; no problem. I squeeze back; I keep pushing. I keep pushing. I feel it in my hand. I really have to push. I keep pushing. And I stop squeezing and it stops. The vector of angular momentum chases, so to speak, the torque. Isn't that intuitive? Very unintuitive? Sometimes it is also dangerous. We call this movement the stool and, in this case, the movement of the spinning wheel, we call it precession.

Then you apply a torque to a spinning wheel. So what you get is a precession. I can show you precession in another way that is actually very intriguing. Suppose I have a rope here, a rope, like the one we have there. And I get into that rope, I tie this wheel to the rope, just like that. And I let it go. Well, we all know what will happen. Pfff, thud. It is empty. Alright. But now I'm going to think about it before I let it go. So here at the bottom, at this point P, there is a loop.

And here's the... the bicycle wheel rotation shaft, which is solid brass, is one solid piece, and I give it a small length "r", not to be confused with the capital "R", What is the radius? of the bicycle wheel. So this is capital R. And you can rotate here with reasonable freedom. I call that center point Q and let this be the part of the wheel that is on your side. I'm trying to make you see it a little three-dimensional. Suppose I now turn it in this direction. omega s. "S" means "spin." In which direction is the spin angular momentum now?

Use your hands, your thumbs. Turning in this direction. Yes. When rotating in this direction, the angular momentum is in this direction. That's spin angular momentum. L-turn. Well, there is a force on this system, Mg, and that force is in this direction. It has a mass M, the bicycle wheel, and it has a radius, uppercase "R", and this part is lowercase "r". So, relative to point P, there is a torque and the torque is R multiplied by Mg. This is 90 degrees, so the cross product is good. The sine of the angle is one. So the torque relative to point P is r times M times g.

In what direction is that torque? R crosses F. In which direction is that torque? Use your hands, thumbs, whatever you want. Do you think in this direction? I disagree. I disagree. R cross F is... you've got to be kidding me. On the blackboard. It's not off the board; It's on the board. R cross F is on the board. There are a couple at this address. Nature, gravity provides that torque. What will spin angular momentum do? It will move in the direction of the torque. It's going to chase torque. So what will it do if the angular momentum is here?

What will he do? Will do this. And as it moves, the twist will always be perpendicular to the plane through the string and r. You can see for yourself why this is so. At this very moment, when the angular momentum is like this, the torque is on the board because it is r transverse to F. But when I am here, this r has changed position and always remains perpendicular to the wheel. Then the torque will also change direction and therefore this angular momentum, the spin angular momentum, will continue to chase the torque and start spinning freely.

That's exactly what I was doing when I was sitting on the stool, except I had to apply that torque to my hands this way. It's exactly the same address. I had to apply it all the time and when I stopped, the precession stopped. Here, however, the torque will never stop because this Mg will always be there, and I will show it to you shortly. You may say, "You must be crazy" because you are violating Newton's second law, f = MA. "This object had to fall. "There is only one force on that object. "f = MA." How not? The center of mass must fall with acceleration g." Aha.

There is not a single force on that object. What do you think is here? The tension in this cable, T, will be exactly Mg. And then the net, the sum of all the forces on that object is zero. There is no net force on that wheel, but there is a net torque, and that is why it is going to precede. If there had been a net force, then it would also decrease, if this were the case. greater than this. So nature is very clever, the way it approaches these quite difficult problems. Before I show you this demonstration of spinning this wheel and then hanging it on that rope, I want to mention that the angular frequency of precession, which.

It should never be confused with the angular frequency of the spin, it is derived for you... it's just a three or four minute job, on page 344 of your book. Now, I won't derive it here but what comes out of it, that. is the torque which is what we have here in this case, divided by the spin angular momentum. That gives you the frequency of the precession. In our case, for our bicycle wheel, it is rMg and the spinning angular momentum of this wheel, if it rotates with an angular velocity omega of s, would be I multiplied by omega.

Remember, he is I times omega of a spinning wheel. So here I have I rotating around point Q, this is the axis of rotation, multiplied by omega of the spin. This is the spin and this is the precession. And then the precession period would be 2pi divided by the omega precession. Let's take a look at that equation and see if it intuitively makes sense. First of all, if you increase the torque, above, then it says that the precession frequency will increase. That makes sense to me because the torque is persuading the angular momentum to follow it.

So, the torque is coaxing the change in spin angular momentum. Well, if the torque is stronger, then it is more powerful, so the precession frequency is expected to be higher. However, if spin angular momentum is very powerful, then spin angular momentum says, "Sorry, torque, I'm not going to go as fast as you want me to go." So when the spin angular momentum on the wheel increases, it is also intuitive that the precession frequency will decrease. As the wheel spins, it has spin angular momentum, but as it goes this way, there will also be angular momentum in this direction because it spins like this.

Therefore there is a total angular momentum that is the vector sum of the two. This equation will only hold as long as the spin angular momentum actually dominates the total angular momentum and you can see that right away, because let's say you make the spin angular momentum zero, which is not spinning at all. Do you really think the precession frequency will be infinitely high? Of course not. So this is only valid in situations where the spin angular momentum is much larger than the angular momentum you get due to precession. So there are restrictions. When... when the wheel stops, when it doesn't spin anymore, you better believe it, then the thing will make noise.

The precession mode no longer exists. For our bicycle wheel, to get an idea of ​​how long the precession will take, we can substitute the numbers in there, our bicycle wheel, the... the... the rod, the brass rod, the little r has a length 17 centimeters, and the... the radius of the bicycle wheel is approximately 29 centimeters. And let me assume that all the mass of the bicycle wheel is in the circumference, which is not very accurate, but close to that. I mean, there are some radii here, but let's assume everything is here, then the moment of inertia is MR squared.

Well, if you now take a frequency of five hertz, the spin frequency, you can now calculate the omega of the spin frequency. Omega is equivalent to 2pi times the rotation frequency. And now I know I can substitute that in there, so I get an omega precession which is now equal to rMg times the moment of inertia. I assume all the mass is in the circumference, an approximation, so we get MR squared and then we get omega S, which we have here. We lose the M, so we get rg divided by omega s times R squared. That is the angular frequency of the precession, and the period of the precession is 2pi divided by omega and the period of the precession is then found to be about ten seconds.

So if you gave it a spin frequency of five hertz with these dimensions and with this approximation that all the mass is in the circumference, you would expect it to precede very smoothly in about ten seconds. But I have very little control over that frequency, so I may have given it seven hertz, I may have given it three hertz. But I'll do what I can. In fact, I'll give you the most I can. That is always guaranteed success. Where is the wheel? The wheel is here. So we'll speed it up and then put it here. Notice the way I'm turning it.

I'm moving it away from me now and I'll change it and do it differently next. And there it goes. About ten seconds. Isn't it amazing? And it rotates, seen from below, clockwise. Now it goes this way and I'm going to redo the experiment, changing the direction of rotation, and then it will go the other way. And now the angular momentum rotates like this, points this way. Spin angular momentum is pointing like this, torque is like that, so spin angular momentum is changing that... chasing that torque. I am the spin angular momentum. I am the pair. This is the pair.

He's chasing him. Alright. So I have this in my right hand. That's ok. And now I'm going to do it... so when I spin it, that's it. So now let me change direction. I turn it over and I'm going to turn it over again. The angular momentum is now in this direction. Look, it's changing backwards. The angular momentum is in this direction. The torque is now towards me. The angular momentum chases the pair. I changed the direction of the spin angular momentum. I have not changed the direction of the torque and it is now rotating, seen from below, counterclockwise.

Before it rotated clockwise. If I can increase the torque by putting some weight here on the shaft, I have this... this actually extends, in our case, and I can put some weight here, then I add torque and then you'll see. which is... it goes faster. The frequency of precession increases. So I'll put some weight there. So let it spin first, which was about ten seconds, roughly calculated, and now I'm going to put two kilograms here at the end. And now you will see an instant increase in the precession frequency. You will see that it goes much faster now.

I take it off and then it goes back to about ten seconds. So what I did was I increased this torque but not at the expense of M, because the reason M cancels out is because the moment of inertia has an M, but if I just hang this object on it, that doesn't change the moment. of inertia of the spinning wheel. None of this is intuitive. None of this is intuitive. You can do all this with a $5 toy gyroscope. And I want to show you this. This is my toy gyroscope. I have it in my office.

It is fun. And this toy gyro is doing exactly the same thing as this one. Let me first show you the gyroscope.toy. Toy gyroscope. Oh yeah. Here is a toy gyroscope. You can see it? Maybe I should make it a little darker here. Can you see my toy gyroscope? Yeah? I'm going to rotate it and then hang it exactly the same way it was hanging. I'm going to rotate it, for those sitting nearby... whoosh... and then I'll turn it horizontally and hang it from a rope, and you'll see the exact same thing happening. And now, of course, due to friction, all this fun finally comes to an end.

I have something very special for you, or I may have something very special for you. That depends on my assistant who is here behind the scenes. I hear it. He is there. Excellent. Did you arrive at full speed? I go to the airport and I get a little tired and I ask one of my friends to help me. Could you help me and... just take this suitcase? Pick it up, walk a little, do some laps. Turn inside out, please. (laughter) STUDENT: What the hell? LEWIN: Hell yeah. Exactly. What are you doing man? You are behaving so strange.

Do some more laps, man. We have to go... we have to take the plane. He doesn't do exactly what you think he'll do, does he? So here you guessed it. STUDENT: A big spinning wheel or something like that. LEWIN: Spinning wheel. And when you do this, you apply torque to it and it does exactly what you least expect: it lifts up. Isn't it fun? Yes. You might get arrested when you go to Logan Airport with this suitcase. Thank you so much. It's great. Spinning objects have a stabilizing effect. If you take a bicycle wheel, and we have one, and I put it here and do nothing, it will fall off.

Nobody is surprised. However, if I give it a little twist, it doesn't fall off. Because? Because it has angular momentum. It has spin angular momentum. And so it doesn't fall. And it's not just a bicycle wheel. Look. Very stable. Not just with a bicycle wheel. Here... take a coin and put a coin like this on your desk. You can bet your life it will fall. Roll it; becomes stable. You give it spin angular momentum and it becomes stable. Take a blouse. You put a board on the table and you fall. You turn it, you turn it and the top is stable.

So spin angular momentum has the property of stabilizing things. And you'll see that addressed in one of your assignments, when I want you to address it quantitatively. This is the basic idea behind inertial guidance systems. In inertial guidance systems, the wheel rotates, at least in the days when guidance systems had mechanical wheels. but that spinning wheel is mounted in such a way that no torque can be applied to the axis of rotation of the spinning wheel. This is how it is set up. We call that three-axis

gyroscopes

. So the moment you apply torque to it, the casings (in this case, the yellow one and the black one) will start spinning, and you never managed to get that torque on the spinning wheel.

You never get it... on this axis. And therefore, if you now put it on your ship or on a plane, or on a missile, if you can never put a torque on the spinning wheel and if the angular momentum for turning is in this direction, it will stay there forever and ever. , assuming we have no friction losses. And if the plane then spins, the direction of the spin angular momentum won't change, but what will happen, of course, is that this yellow frame will spin or this black frame will spin. And in these bearings there are shaft encoders, and they sense the rotation that the outer casing makes to keep this pointing in the same direction.

And that signal is sent to the autopilot and that keeps the plane flying in the direction you want. So you use, as a reference all the time, the spin angular momentum of your gyroscope, which is now mounted in such a way that you cannot apply torque to it, even when the plane changes direction, and I want to show that to you. Okay, this is the direction of my spin angular momentum and I am the plane and I'm going to fly. Look at that spin angular momentum. He has no respect for me. It stays in the same direction no matter how I fly.

And the arrow signals that come from the yellow case bearings and the black case bearings, those arrow signals are returned to the autopilot and so the airplane will maintain its course. Now, what I can do for you to come up with a final proof of his thinking is that this wheel is suspended in such a way that there is no gravitational torque on it like there was here. But I can torque it by simply putting some weights on the shaft. And what do you think will happen now if I put some weight here on the axle?

So the wheel is spinning, but now I'm going to torque it here. It's turning in this direction. The angular momentum points directly towards me, away from you. I'm going to torque it like this, put a little bit of weight in there. The torque will be in this direction. What will spin angular momentum do? The pair is in this direction; the spin angular momentum is in this direction. The spin angular momentum will begin to chase the torque. Look at it. There it goes. The spin angular momentum chases the torque. You see exactly the same thing that I have shown you before.

And if I increase the torque, then the precession frequency will increase. See, now it stops immediately when I take it off. Put it back on. Keep going. Put more. It goes much faster. What happens now if I put the weight on this side? Then I change the direction of the torque. If I put it on this side, the torque is now in this direction, the spin angular momentum is in this direction. It's going to reverse direction. Here we go. And you see that it is so. Surprisingly unintuitive. If you're struggling with this, you're not alone. See you on Wednesday.