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Anderson Video - Rotational Energy

Professor Anderson
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>> Hello, class. Professor Anderson here. Let's talk a little bit about rotational energy and how it applies to angular momentum, and again, let's go back to the ball on the string problem. So the ball on the string looks like this. We're spinning around at some radius r. We're heading around like this. It goes in a circle, and we're moving at some particular v. What we saw last time was if we're going to shrink r to half its length, then the ball speeds up, okay. So r going to r over 2. We figured out very quickly what the new speed was. And how did we do that? We just satisfied angular momentum, okay. Conservation of angular momentum says initial has to equal final. We know exactly what initial is, we know exactly what final is, and so you can solve this for v final. What do you get? V final is equal to r initial over r final times v initial, okay. If you go to half for r final, you get a factor of 2. So what about the energy, right? If you're swinging this ball around -- and this is the top view, right, so we're swinging it in a horizontal circle. If you swing the ball around, and now you pull down on the string, the ball starts spinning faster and faster. What happens to the energy? Well, we know what energy is, right. Energy is just one-half mv squared. Specifically, kinetic energy is just one-half mv squared. All right. Does that mean that the ball somehow has more energy when it's at half the distance and spinning faster? Yeah, it seems that way, right? We said our v final goes up, so kinetic energy must have gone up. How does this play into rotational energy? How do we see this, right? One-half mv squared is certainly the kinetic energy of the particle, but we can rewrite this, because v is equal to omega times r, right. And look. We've got one-half m. I'm going to pull the r squared out, and then I have an omega squared left over, but if it's a ball on the end of a string, mr squared is the moment of inertia, and so this thing is exactly equal to one-half I omega square. So rotational energy is really just the same idea as kinetic energy: How fast is that thing moving? We're just going to write it in terms of these other variables, moment of inertia and omega, okay. So does omega go up as you pull it in? Absolutely, just like v went up. Let's figure out exactly what the increase in energy is. Okay, the rotational energy that we just talked about, we're going to write with an r, and we said it's one-half I omega square. So when the ball is out at its original position, we had one-half I initial times omega initial quantity square, but I initial is mr initial quantity squared. Omega initial is what? It is v initial divided by r initial quantity squared. And look what happens. We just get back to our good old kinetic energy, one-half mvi squared. It all sort of makes sense. But what about r final? When we pull this thing in, we've got one-half I final omega final squared, and that is, of course, by the same argument going to become one-half times mv final square, and we know exactly what v final is. V final is r initial over r final times v initial, and then we have to remember that we're squaring the whole thing. So in our example, where r went to r by 2, let's see what we get for energy. So we've got rotational final energy is one-half m times r initial, which we said was r, r final, which we said was r by two, and then we're multiplying by v initial, and that whole thing is squared. The r's, of course, cancel out. We get a 2 up there, and then we're going to square the whole thing, so we get 2vi squared. Two squared is 4. When I take that 4 and I multiply it by a half, I get 2mvi squared. How does that compare to the initial? The initial is right there. It was just one-half mvi square, and so our final energy has increased by a factor of 4. Somehow just pulling on that string, half the length of the string, that ball is now rotating with four times the energy that we had before. And so let's ask you a follow-up question. Where did this energy come from? In other words, what did the work? Well, energy is not created or destroyed. Energy has to go somewhere, so the ball just can't gain energy without something or somebody doing work on it. What did the work? Was it the string as it went around? Probably not, because the tension in the string is always at a right angle to the velocity, and we know if tension and velocity are right angles, you can't do any work, right, because you have that cosine of the angle between them. The angle between them is 90 degrees. Cosine of 90 degrees is zero. So the string didn't do the work. Something else must have done the work. Eric, what do you think? What did the work? >> (student speaking) Something else. >> Something else. Correct. What did the work? You did the work. By pulling on that string, it required force, okay. Force over a distance equals work, and so the amount of work that you had to do pulling on that string is exactly equal to the change in the kinetic energy that the particle experienced. You don't pull on the string, the ball just goes around at the exact same speed, okay. So you did the work by pulling on that string. It's kind of neat. All right, hopefully that one's clear. If not, come see me in office hours. Cheers.