Physics: The Kinetic Energy of Rotation

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Consider some rigid body that's rotating. Think about a symmetrical one like a bicycle wheel. It's spinning around its central axis. It's spinning around its center of mass. And now let me pose the following simple question. Does this thing have kinetic energy? So you stop and think and you say kinetic energy, it's a concept that we defined, we developed in terms of point objects. So this is a large object, but let me consider this big object, should all of it's mass be located at its center of mass? That's usually what we do when we have complex objects. So if all of its mass is located at its center of mass, then think about it. The center of mass is right in the middle of the axis. It's not going anywhere. It's sitting still. The velocity of the center of mass is zero. So as this thing is spinning, is zero.
So that says there's no kinetic energy here. But you look at this object and you say, "Well this is ridiculous. I mean there's a bunch of little pieces of wheel. And they all have little masses. And they're all moving like crazy." Each little individual chunk certainly has plenty of kinetic energy. And kinetic energy is always positive. So it just adds up. This thing must have a huge amount of kinetic energy. So what's going on? Is it completely wrong to argue that this object can be considered to be at its center of mass? Well the answer is yes. When you are worried about energy arguments, you have to be careful about that.
Let me give you another example, and then we'll try to figure out what to do about this. Here's a yo-yo. And I'll show you sort of a standard yo-yo trick. It's called sleeping. The yo-yo just sits. And then at a certain point it climbs up the rope. So think about conservation of energy. It was sitting down here. It had zero potential energy, because it's right next to the ground, zero gravitational potential energy. And how about kinetic energy? Well its center of mass was just sitting right here . Again, if you think of that as the kinetic energy, it's got no energy. And then all of a sudden it climbs up the little string. And it gained in potential energy. Where did that energy come from? It's a physical demonstration that a rotating object does contain kinetic energy. And when you think about the wheel example, and you think about the yo-yo, it certainly makes physical sense. All of the little chunks are moving. There's plenty of kinetic energy in there.
How can we quantify this? How much kinetic energy does an object have? So suppose the object has a mass m, and let's just think of a wheel with radius R. And let's think of a wheel that's spinning about its central axis. And what am I going to write down for kinetic energy? is the formula that you would write down when you have an object that's moving through space. So you're making some wild guess. You know this thing isn't moving through space, but it is rotating. I have an angular velocity. How about , that's replacing v with its angular partner? And that's a nice idea, but it can't be right, because the units aren't right. This is kilograms per second squared. It's not kilogram-meters squared per second squared, so that's not right.
Rather than guessing, let me start to calculate. Suppose that my wheel was really not a big solid wheel, but just a single point mass m here and it's rotating about the central axis. This is the world's simplest complicated object. It's not a point, but it really is just one mass running around in a circle. And it's running around with some angular velocity, omega. And let's suppose that omega is constant just for the moment. Later we'll just talk about instantaneous kinetic energy anyway. So at this moment, it's got omega. So what's the physical velocity? What's the linear velocity of this object? This little mass has velocity equals to, well, that's the connection between angular and linear velocities, it's R omega. That's how fast it's going around the circle. So now I can plug in, because now I've got the linear velocity. And now I say oh that's just a point mass moving with velocity v. It's got . So it's R^2 . And that's looking better. It has got units of kilogram-meters squared per second squared. So it's got the right units. And it's the correct formula for this little simple case.
But what about when you have a complex object? What do you do when you have an object that looks something like this? And you've got it fixed on some axle and it's rotating. What's the kinetic energy? One half m, what radius am I going to use? This doesn't have a definite radius. Here's the trick. Break it up into a bunch of little pieces. It's the usual story. Break it up into a bunch of little dms. And that little dm has a radius R[I]. And I'll label it dm[I]. So I again is just counting. Chunk number, chunk number two, chunk number three, and I'm going to add up all the chunks that make up this body. And each chunk is going to have m[I]R[I]^2 . Each little chunk is moving in a circle. Even though the shape of this thing is funny, this little chunk is a fixed radius. Remember this is a rigid body. It's a fixed radius. It's just going around in a circle with angular velocity omega. All chunks have the same omega. And so this has velocity R omega. And so this is .
And now I just have to sum over all the little chunks. And I can put this dm in there. And if I do that, notice that omega is constant. It's the same for everybody. So when I pull it out, I have kinetic energy is one half times something. That something was the sum on I R^2dm. It's the sum on I R[I]^2dm[I], which if you took the limit that these chunks are really tiny, would be the integral of R^2dm integrated over the whole body. And it's such an ugly looking expression people just give it a name. They call it capital I, it's got nothing to do with impulse. It's a totally different variable. We've just run out of letters. It's kind of got something to do with inertia. We're going to talk more about how this quantity is related to rotational inertia. But for now it's just called the moment of inertia or sometimes the rotational inertia of the object. So is the formula for the kinetic energy of any arbitrarily shaped object. If I tell you its moment of inertia, it's a quick and easy calculation. It's as quick and easy as the old formula for linear motion.
Let's do an example. Let's think about that spinning yo-yo. The spinning yo-yo has some moment of inertia. It's got a bunch of little chunks of mass. And each one is at a different radius. We're going to talk in the future about how to calculate the moment of inertia. A decent estimate of the moment of inertia for a yo-yo of that size, it's about a 60-gram yo-yo, its moment of inertia is about 2 x 10^-5 kilogram-meters squared. Look at the units. It's mass times R^2, which is correct. It's a tiny number and that's right. It's a small object. Those radii, the R^2 in meters are really small numbers. And that's why this is such a small number.
So the kinetic energy, , you have to know how fast a spinning yo-yo is spinning. What's its angular velocity? I took a guess at what seems like a reasonable number. I plugged in 250 radians per second. That's about 40 revolutions per second. It's spinning like 40 times each second. It seems like a reasonable order of magnitude number. I plugged in 2 x 10^-5 x 250 radians per second quantity squared. I get kilogram-meters squared per second squared. That's joules, 0.6 joules, a small but reasonable amount of energy.
How much energy is that? Suppose that I converted all of that kinetic energy into gravitational potential energy, which is exactly what the yo-yo ends up doing. How high would it go? Conservation of energy says the initial kinetic energy plus the initial potential energy, the total initial mechanical energy, which is just this because it's down at ground level, should equal the total final energy. And when it gets up to the top into my hand it's at rest. It's no longer spinning. It's got mgh of gravitational potential energy. And mass is 60 grams, and g is 9.8. So I plug in and I get h is about a meter, which is about what it climbs. So all of these numbers seem to make some sense. It's a physical demonstration of the reality of this fancy looking formula, , which tells you the kinetic energy of a rotating object rotating about a fixed axis.
The moment of inertia, I, for simple objects is just the sum of radius squared times the mass of the chunk. So if you have an object like this, it's easy to compute the moment of inertia, at least calculating the moment of inertia about the center. You always have to pick a center when you calculate this, because where are you measuring your radius from? You get to pick. So I've got a mass over here, and a mass over here. The moment of inertia is mR^2 + mR^2. It's easy to compute. It's some number. And if I spin this thing around, it's got a certain amount of rotational kinetic energy. If I take these two masses and I slide them further out toward the edge, I'm not changing the mass of the system. And yet the moment of inertia changes. The formula says, as you slide these masses out, the R[I] is getting bigger. And it's getting a lot bigger, because you're squaring it. So this has much bigger moment of inertia. If I spin it with that same omega as before, the same angular rotational speed, this has more kinetic energy. And it makes sense. These masses, if I'm going at the same omega, they're going a much larger distance, because they're going through a bigger circle. Bigger distance in the same period, they're moving faster. So they do have more kinetic energy. So there's a lot of physical sense to this formula.
When you're working with moment of inertia, you use for the kinetic energy. And what's important to realize is that in rotational motion, it's not just the mass that's important like it was in . It's also the distribution of mass. Is the mass located near the middle or is it located toward the outside? Once you know the moment of inertia, kinetic energy is a snap.
The Physics of Extended Objects
Angular Momentum
The Kinetic Energy of Rotation Page [3 of 3]