## 10. Rotations, Part II: Parallel Axis Theorem

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 Title 10. Rotations, Part II: Parallel Axis Theorem Title of Series Fundamentals of Physics I Part Number 10 Number of Parts 24 Author License CC Attribution - NonCommercial - ShareAlike 3.0 Unported:You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal and non-commercial purpose as long as the work is attributed to the author in the manner specified by the author or licensor and the work or content is shared also in adapted form only under the conditions of this license. Identifiers 10.5446/46675 (DOI) Publisher Release Date 2006 Language English

 Subject Area Physics Abstract Part II of Rotations. The lecture begins with an explanation of the Parallel Axis Theorem and how it is applied in problems concerning rotation of rigid bodies. The moment of inertia of a disk is discussed as a demonstration of the theorem. Angular momentum and angular velocity are examined in a variety of problems. 00:00 - Chapter 1. Review and Derive the Parallel Axis Theorem 16:27 - Chapter 2. For System of Masses: Derive KEtotal = ½ MV2 + ½ ICM2 27:55 - Chapter 3. Derive KEtotal in Terms of Equivalent Rotation about Stationary Point 38:40 - Chapter 4. Effect of Rotational Kinetic Energy on Translational Motion for No Skid 43:41 - Chapter 5. Example Problem: Torque on a Disk 49:30 - Chapter 6. Advanced Example Problem: Pulley Rotating and Translating 01:02:14 - Chapter 7. Example Problem: Systems with Angular Moment Conserved 01:09:09 - Chapter 8. Application: Angular Momentum Changes for Spinning Ballerina

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Professor Ramamurti Shankar: All right, so let's begin Part II of Rotations. I thought I should summarize for you the main points, because you learned a whole lot of stuff, but you don't have to carry all that in your head. What I'm going to write down is the essential subject you should know. So, we were going to do rotation of rigid bodies, which are forced to lie on the xy plane. They have some shape. And here are the main ideas we should know. You first pick a point around which you want to rotate the body by driving a little skewer through that. That's the point of rotation. Then, here are the things we learned. First of all, the body has a mass M, which is a simple thing. You put it on the weighing machine, that's the mass. Then, it has a moment of inertia, which, if the body were made of a discreet set of points, would be the mass of each point, m i times the square of its distance from the point of rotation. For a continuous distribution of matter like this one, the sum is replaced by an integral. For example, for a rod it was ML^(2)/12 around one of the ends, around the center of mass. And for a disk, it was MR^(2)/2. So, every body has a moment of inertia. Here is the important caveat, with respect to a particular point. There is nothing called "the moment of inertia," because depending on the point, these distances will vary. Okay. Then, if you want the rigid body to start doing interesting things, you wanted it to move around, it will have an angle of velocity ?, which is rate at--which goes round and round. It's the usual frequency times 2? because 2? radians are full circle. Then, it has an angular momentum, which is the analog of the mass, times analog of the velocity, as is universally noted by the symbol L. Then, if you want to change this angular momentum, it's like asking, "How do you change the ordinary momentum of a particle?" You apply a force. So, now we apply what's called a torque. And the torque is the reason the angular momentum changes. That's--The analog of force is the reason the usual momentum changes. I remind you, P = mv, so it is really mdv/dt = ma. So, you can also write analogously here, ? = I? because if you take the d/dt of this guy, that's a constant, and ? is a rate of change of ?. The last thing I have to tell you is what is the torque of the total torque on a body. Well, if there is a force, for example, acting in that direction, let me just take one force, and it's separated by distance r, or vector r from the point of rotation. Define that angle ? as the angle between this point of separation distance, and separation r, and the applied force. Then, the torque is F times r times sin ?. If you took F to be perpendicular to this separation vector, then sin ? = 1. If you took F to be parallel to the separation vector, torque vanishes because it's no use applying a force in the line joining you to the center, because you don't produce rotations then. The sin ? tells you how much of the force is good at producing rotations. If you got many forces and many torques, you got to add all the torques. There's one thing to be careful about. For example, you can have a body where one force is trying to do that, and here maybe a second force acting that way, trying to do the opposite. So, we keep track of the torque by saying either it's clockwise or counter-clockwise. Each force you can tell intuitively by looking at it. This guy's making it go counter-clockwise. This is rotating clockwise. Counter-clockwise will be taken
positive; clockwise will be taken negative. Similarly, angular acceleration will be positive if it's counter-clockwise, increasing. So, this is a summary of everything we did yesterday, on Monday. You guys follow that? So, there's the analogy to F = ma. Everything has a rotational counterpart, but life is more difficult because in the case of F = ma, m was simply given to you. Here, the moment of inertia has to be calculated from scratch, given the mass distribution that forms the body. Similarly, you may have been simply given the forces in F = ma. But here, even after I give you the forces, you got to do some work, calculating the torque. But every force you got to find how far it is from the point of rotation, watch the angle between the force and the vector separating where the force is to where the rotation is taking place and calculating the sin ?. Okay, that's all we did. And what I did towards the end of the class was to calculate the moment of inertia for a couple of objects. I showed you for a disk, it's MR^(2)/2. For a rod, I did two calculations: one was around this end, and the moment of inertia around the end was ML^(2)/3, and the moment of inertia around the center was ML^(2)/12. And at the end of the lecture I said, "Notice the following fact. The moment of inertia through the end and the moment of inertia through the center of mass are related in a very simple way, which is the moment of inertia with respect to the new point, the moment of inertia with respect to the center, plus the mass of the entire rod times the square of the distance, (L/2)^(2) separating the axis going through the center of mass, and the new axis. And that's called the Parallel Axis Theorem. And I'm going to show you in general why it is true. The Parallel Axis Theorem says, "If you know the moment of inertia through the center of mass, you are done," because then the moment of inertia through any other point does not require you to go back and do the whole integral one more time. I'll tell you the answer once and for all. The answer is: take the moment of inertia at the center of mass, add to it, in general the mass of the whole body times the square of the distance between the center of mass and your new axis. For example, you want to rotate it around this point. That's the distance d; add that Md^(2).
So, I said I don't want to just give you a result; I want to tell you where everything comes from. So, I will now prove this. So, let's start with that proof. So, this is the proof of the Parallel Axis Theorem. Before I do the Parallel Axis Theorem, I'm going to use one result in vectors, so you might as well be reminded of that, because I use it freely, and I don't want you to get caught in that. If A is a vector, then the length of A^(2) is found by taking A dot A. A.A is the length of A times the cosine of the angle between A and A. Well, the angle between A and A is zero, and cosine is just 1; therefore, the way to find the length of a vector is to take the dot product of the vector with itself. Another way to say this is, the dot product of A and B, you remember is AxBx + AyBy, so A with itself is Ax^(2) + Ay^(2) which is the length squared of A. The result I want to use today is, if you take vector A and add to it vector B, and you want to find the length squared of A + B, well, you take the dot product of that with itself, and when you do the dot product you can open out the brackets as if it is an ordinary product. So, you can do this in your head, you'll get the length of A^(2) from doing the A.A, length of B^(2) from doing the B.B; then, you'll get an A.B + B.A, and these are both equal; then, you will get twice A.B. That's the result I want to use. So, here is the calculation. Let's take some arbitrary object. And you want to find the center, the moment of inertia around that point. This is the center of mass. This is a part of the body that I'm calling i. Actually, you should integrate over it, but let's do it as a sum. Mentally, divide this body into a hundred thousand million tiny little squares, which when tiled together form this object because it's easier for me to work with a sum than
with an integral. So, this fellow has some mass m i. It has a location, which is r i and I'm going to call it r i prime, because the prime means it's the distance measured from this new axis. I'm going to reserve the symbol r i for the separation from the center of mass. Because the center of mass is a privileged point, all distance measured from it won't carry off prime. If you pick an arbitrary point, I want to put a prime for that one. Then, d is the vector that tells you how to go from your chosen axis to the center of mass. So, what is the moment of inertia I'm trying to calculate with respect to this point? It's the sum over i, the mass of the little fellow here, m i times r i prime squared. But r i prime squared, you can see, is the length of r i + d dot, product with r i + d, using the result I established above. Okay, now, let's start opening out the brackets and look at the terms you will get. First, you'll take dot product of this guy or this guy, because r i^(2), that's going to be m ir i^(2), sum over I. I think you guys can recognize that as the moment of inertia with respect to the center of mass. That's what you would do to find the moment of inertia with respect to the center of mass. You'll take the square of that distance. So, the first time I get is just the I with respect to center of mass. Now, I purposely want to skip the middle term and go the last term. In the last term, there is no sum over i because d is a fixed vector, it doesn't have anything to do with the particle i. I get d.d, which is just d^(2) times sum over all the masses, which is just the total mass, and that's part of the theorem because I got the two parts I like. I = I CM + Md^(2). The trouble is I got what I want, but I got more stuff. I got to hope that the stuff I don't want will go away. Okay, it's like putting a washing machine or something, and when you're done you find three or four parts on the ground. You don't know why they're there. In that case, there's no solution to that problem. But in this problem, there's a very nice solution. There's stuff I don't want that is going to vanish. It's going to vanish. And let's see why it vanishes. Let's take the cross terms. The cross terms are like the 2A.B terms. So, here is the 2. One of the vectors A is the vector d. And it's going to be taking the dot product with m ir i summed over i because there's going to be a d.r i twice. But d is not dependent on the summation index. It's common to all the terms in the sum and you pull it out. Now, you can think about what this is. This is every mass multiplied by its position. So, take this guy, sum over i, m ir i, divide it by the total mass, and multiply it by the total mass. But what is this? Yep, that's the center of mass, but that's a center of mass in a coordinate system with the center of mass as the origin. So, you will get zero. If I use this as the origin, take m ir i, that's the center of mass. It will have some coordinates from that origin. But these r i's are all measured from the center of mass itself. To compute the center of mass with the center of mass as your origin of coordinates, then it's a zero. One way to say that is if you're sitting at the center of mass, then all the masses are surrounding the center of mass in such a way that the way of some of the positions will be the center of mass, but which will be then zero, if that's your origin of coordinates. That's the reason this will vanish. This is a very useful result. The center of mass coordinates, in a coordinate system with the center of mass as the origin, will therefore be zero, and that's why this sum vanishes; then, I get the result I mentioned to you. I is I center of mass + Md^(2). And the rod was a special case of this where I put d = L/2 to do that example. But it's generally true. Okay. One place where you might find the result useful is the following. Let's think about the moment of inertia of a coin, or a disk. If I run the moment of inertia through the center, I truly have to calculate it. You divide it into concentric rings. The answer for each ring is simple, you do an integral, you get MR^(2)/2. But suppose I say I want the moment of inertia through that point. Do you know what I mean? Stick the pin through that and rotate it. With some wobbly motion, it will perform with that at the center. The moment of inertia through that point, if you just went back to basics, is very hard to calculate because you can try your little trick with concentric circles, that'll be nice. That'll be nice. Then, you'll run to the edge, and after that, the circles are not really full circles. And depending on how far you go, you take different fractions of a circle, it's not easy to do that summation. Luckily, the moment of inertia was through the center, the whole object is divided into concentric rings; then, you can do the sum easily. But here is where this theorem comes into play. You don't have to do any more work through this point. It is MR^(2)/2, that's the answer through the center plus Md^(2), where d is simply the distance from the center of mass to where you set your axis. Okay? All right, so that's one demonstration of the Parallel Axis Theorem, of the proof of the Parallel Axis Theorem. Now I'm going to show you a similar result, because that's going to be also useful in the study of rigid bodies. It has the very same flavor, so I want to do that side-by-side. And that's the following question. Suppose you have, in two dimensions, a whole bunch of masses: m 1 with the location r 1, m 2 with a location r 2, and they're all moving around. You know that given all the masses that it's something, some right here, that's the center of mass, R. As the bodies move, this is not a fixed situation. The whole--It could be a bunch of planets traveling as seen by another galaxy. The whole Solar System is drifting. As the whole thing is drifting, the center of mass, which is the weighted average of the location, will also drift with that. Now, I want to write a formula for the kinetic energy of this complex, which is ?, sum over i, m i v i^(2). But now, I want to look at the velocity of this object, of every object v i; I want to see it as the velocity of the center of mass, plus the velocity relative to the center of mass, which I will denote maybe with the symbol r for relative. Do you see that? Every little speck is moving. But let's ride in a frame moving with the center of mass. Wherever it goes, we go with that. We latch onto it. As seen by us, the body will have a velocity v i relative to us, but from the point of view of the ground, we should add to that velocity, the velocity of the center of mass to get the velocity with respect to the ground, or some fixed coordinate system. So let's do that same trick with the expansion, ? m i. Now, I think it will let me do this a little faster, because we've done the thing similar to it before. Now, I'm squaring velocities and not squaring positions. So, one term will be this capital V, I remind you is the velocity of the center of mass. I just don't want to show it subscript every time, but capital V is center of mass. So, when I squared this v i^(2), I'll get a V^(2) term; then, I'll get a v i relative squared term. Then, I'll get a cross-term twice V, v i relative, summed over all the particles. So, let's expand the three terms.