Physics: Work and Power in Rotational Motion

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We've been talking about the kinematics and the dynamics of rotational motion. It's a little bit different than motion of something moving back and forth or through space, what you call linear motion, or I like to call it translational motion, because something is translating around through space. Rotational motion is different, but it's closely related. All the concepts that we talk about in rotational motion are closely connected to concepts we've talked about before.
Let's look at some variables that we use. Here are the linear variables. When you want to describe where a particle is, you use x. What do you use to describe where a solid object - I'm thinking about a solid rigid object that has a fixed axis here in my mind for pure rotational motion. Instead of talking about position, we talk about angle. They're analogous. They're not the same things, but it tells you where the object is. Instead of the velocity in translational motion, we think about the angular velocity, . Both tell you how fast something is moving in a certain sense. a, the linear acceleration, becomes , the angular acceleration. These aren't equal signs, the quantities are physically different, but they're connected conceptually to one another.
When we have linear motion and we think about Newton's Laws, we think about mass, it's the important physical quantity, the inertia. In rotational motion, it's the moment of inertia, I, which is the analogous concept. I is not equal to m, in fact, it's the sum of masses of individual chunks times the radius of that chunk squared, but it carries the same physical role. It plays the same role in the equations. Indeed, once you've got these basic concepts, we start writing down equations.
For instance, we wrote down Newton's Second Law, F equals ma. That's the familiar old one, and then when we've got rotational motion, we write it as, instead of force we've got torque. Torque is the angular analog, it's the twistiness or twisty force. And instead of mass, we use I, instead of a, we use . The formula is just what you would think it should be. Torque is the angular analog of force, it's r F. And the same thing with kinetic energy. If you have kinetic energy as mv^2, that's the correct formula for linear motion, it becomes not mass, but I, not v but ^2.
So you start to realize that all the concepts that we introduced - there were more when we talked about linear motion besides just force and kinetic energy. They're going to have a direct analog. For instance, we defined and talked about work, it's a very important idea and it's useful in a lot of problems. A differential amount of work, if you apply a force over a small distance dr, that's the correct formula, force times distance. And remember these are vectors, so you have to worry about the relative orientation of force and dr. It's going to be the exact same story with rotational motion. The amount of work done on some rotating object is going to be given by not force, but torque, not dr, but d. And if you've got torques in some funny directions, not just a fixed position, fixed axis problem, you might have to worry about the vector nature of these quantities. For a simple fixed axis, all you have to worry about is signs. Here in one dimension it was our force and dr in the same direction or opposite plus work or minus work. It's the same story here. If torque and change in angle are both clockwise, it's going to be positive work, if one's clockwise and one's counterclockwise, negative work. I'll show you an example in just a second. This formula is correct, it really works.
Let me show you one more formula that we've talked about before, power. Power was defined as work per time or energy transferred per second. It's force times velocity. And power is still going to be defined in that same generic way for rotational motion. It's still work done per time. And the formula is going to be times .
So you can start guessing these formulas and you can check them, you really need to. Sometimes when you get to rotational motion, the story is a little bit more complicated and there might be some subtle caveats on exactly when you can use them. Mass is fixed. Moment of inertia depends on the distribution, so moment of inertia can change. So you can imagine that there are some subtle issues here. But, fundamentally, these formulas are correct. They're certainly correct in the simple cases of a fixed rigid body rotating around an axis that we've been talking about. And they're very powerful. They're just as powerful in the rotational case, powerful in the English language sense there, as useful for problem solving as these original ideas and formulas.
Let me work out a quick example for you. It may or may not be quick. It's an object, mass m, fixed by a rigid rod to a pivot. So this is the world's simplest complicated object. It's capable of rotating around in a circle and that's it, it can swing around. Let me apply an external force to it. And this circle has radius R and I'm going to refer to the position vector, pointing at any moment in time from the center to the object, as R vector. F is also a vector. And I would like to ask, "How much work is done as this object moves over a small distance, r. Now, this small distance is itself a vector, it's pointing tangents to the circle. And it's not that the radius itself is changing, it's that the direction here is changing. The object is certainly moving. It has moved a distance, and let me complete the triangle here and call this angle that it's moved through . So what I'd like to really do is show that my little formula work is F times r. That's our old familiar formula and that's a vector formula. I want to show that the new angular formula times is in agreement with this.
So let's just think about what this formula says. It's really the component of force in this direction. So what shall I call that? I'd rather not call it F parallel, because if I decompose this force vector into this piece and that piece, this piece is really tangent to the circle. So let me call this F[Tan]. That's a nice name for that component of force and what I want is F[Tan] times the length of the displacement. And how far did it go? If it's a circle of radius R and you move through an angle - that's the definition of radiants, is that is distance divided by a radius. So the distance traveled is just R. That's geometry.
So that's the old formula for work. What was I saying the new formula was? It's supposed to be times . Is this really torque? Yes, indeed, look at the picture. You've got a force off at some funny angle. Remember the definition of torque, it's r vector F vector or, if you prefer, it's the magnitude of r times the tangential component of force. And that's exactly what I've got, magnitude of r times F[Tan]. So it really works. It's equal to times . So that's nice and you can often use this idea in problems. Let me work sort of a more practical problem. That one was really kind of verifying that the formula works.
Here I've got a spinning wheel. And supposing I know what its moment of inertia is and I know what [i] is and I want to stop it. I want to stop it fairly rapidly. I would like to stop it in one rotation. I'm going to apply an external friction force, so I'm going to put my hand there and stop it in one rotation. And I'd like to know what external friction force was required. So what's the direction of the external friction force? The friction force is kinetic friction, it's sliding friction. It's nice and perpendicular to the vector going from the center out towards the force. So here I've got my wheel and here I've got my external force. The circle has radius R. What's the torque? Well, is equal to R times F. Torque has a sign, let's just think about our sign conventions. Let me pick this sense, clockwise, to be positive. So what's the direction of torque? Well, r F is out of the page or, if you want to think about it one-dimensionally, the torque is trying to slow down the motion, so it's a negative torque. That minus sign is physically significant. So what's the work? The work is times , so it's -RF, and what's ? The change in angle is 2 radiants. And the change in angle is indeed in the positive sense here. I'm increasing the angle by 2. And so the work done was negative, which makes sense. It was a frictional force. Look at the units; meters times Newtons, that's the units of torque. You multiply them by radiants. You don't change the units, so that's Newton meters, that is indeed the corrects units, joules, for work.
This is the work done. Why did I want the work? Well, it's not what I was asking for. I was asking to figure out what the force is. And here I'm going to go to the work energy theorem. It's a familiar old formula. Work net is supposed to equal to change in kinetic energy. It's still true for rotational motion. And it's a really very useful theorem in lots of problems. Here I've just worked out what the net work is and what's the change in kinetic energy. Well, -RF times 2, that's the work done by the external force, is equal to final kinetic energy minus - that's zero, minus initial kinetic energy, I[i]^2. And here's my formula, I can solve it for the frictional force required to bring this thing to a halt. The minus signs cancel, as they should, and I'll get a magnitude of the friction force.
Problems with rotational motion, they often look intimidating, all these formulas, they look Greek. They're filled with Greek letters. Don't let them intimidate you. Every formula, every quantity used in rotational motion does have an analogy with linear motion, translational motion formulas. And it's really nice, once you kind of notice that all these formulas are really analogous to the old ones, it helps you understand them. And its' going to help us continue on to finish this story of rotational motion. For instance, we talked about momentum. It was a very important idea when we talked about translational motion. We're going to talk about angular momentum and you can almost guess what the formula is going to be, just in the same way, by analogy.
The Physics of Extended Objects
The Dynamics of Rotational Motion
Work and Power in Rotational Motion Page [2 of 2]