Rotational kinetic energy of rigid systems
About this video
Three round objects with the same mass and radius race down an incline, and the winner depends on rotational kinetic energy. Because rotational KE is (1/2)I*omega^2 and depends on how mass is distributed, objects with more rotational inertia keep more energy tied up in spinning, leaving less for forward motion, so the solid sphere (least rotational inertia) reaches the bottom first.
You'll be able to
- write rotational kinetic energy as (1/2)I*omega^2, analogous to translational (1/2)m*v^2.
- explain that a rolling object's total kinetic energy is the sum of translational and rotational parts.
- use rotational inertia to rank which rolling object reaches the bottom of an incline first.
Key words
- translational kinetic energy
- The energy of straight-line motion, (1/2)m*v^2.
- rotational kinetic energy
- The energy of spinning, (1/2)I*omega^2.
- rotational work-energy theorem
- The total work done by torques equals the change in rotational kinetic energy.
Transcript
- [Lecturer] A solid sphere, a hollow sphere, and a ring or a hoop are placed at the top of an incline. They have the same mass and they have the same radius. The question is if we release them from the same height and at the same time, which of them will reach the bottom first?
What do you think? Well, when I first look at this problem, here's how I was thinking. Since they're released from the same height and they have the same mass, they have the same gravitational potential energy to begin with, and when they reach the bottom, all of that potential energy gets connected to kinetic energy, which means they should all have the same speed as they're moving down.
So, they should all reach the bottom at exactly the same time, but that is wrong. Why? Because they're not just moving forward. It's not just a translatory motion. They're also spinning as they go forward, and that makes things a little bit more interesting. So, let's see how to tackle this problem and which of the three actually reaches the ground first.
Suppose a basketball moves such that at any given moment all the points on that basketball has exactly the same velocity, then we call such a motion, translational motion. And if the basketball has mass m, then the kinetic energy of the ball would be just half mv square.
We call this the translational kinetic energy. In contrast, if you have a basketball that is spinning like this, then look, all particles are going in a circular path about a common axis, we call such a motion rotational motion or just rotation. Now, in this particular case, different particles don't have the same speed at any given moment.
Particles that are far away from the center are covering a lot of distance per second. So, they are traveling with a very high speed compared to the particles that are closer. And the particles which are near the axis or on the axis, they're not moving at all.
So, they have very different speeds, but they're all covering equal angles per second. So, they have the same angular speed. And just like in translational motion, the mass represents inertia. It tells how hard it is to linearly accelerate this object. When it comes to rotation, we have an analogous quantity called rotational inertia.
That is a quantity that tells us how hard it is to angularly accelerate the object, which means how hard is it to get an object to start spinning or to stop spinning, for example. So, rotational inertia here is analogous to mass. And by the way, earlier we used to call this quantity moment of inertia, but it's now more typically referred to as just rotational inertia, because that sounds much more intuitive.
So, can you now guess what will be the expression for rotational kinetic energy? Pause and think about this. All right, it's gonna be pretty analogous to this. It's gonna be half times not M, but rotational inertia, so I times, not we, but angular speed, omega squared.
So, half I omega squared. So, an object is spinning. This is how you calculate kinetic energy. So look, rotational kinetic energy depends on rotational inertia, and that's what makes rotation so interesting. Lemme take an example. Imagine you had two objects, one disc and another one like a disc, but it is hollow in the center, okay?
But more importantly, they both have exactly the same mass. Now, if these two objects were undergoing translational motion with exactly the same speed, would they have the same kinetic energy? The answer is yes. Now they're undergoing translation, so they have the same mass.
They have the same speed. They should have the exact same kinetic energy. So look in translational motion, I don't care about, you know, whether I'm dealing with a solid disc or a hollow disc. I don't care about all of that. Mass is all that matters to me. And the speed, of course.
But now, let's say they're undergoing rotational motion. In fact, let's assume they're spinning with the exact same angular speeds, okay? So, they have the same mass. They have the same angular speed. Would they have the same kinetic energy now? No, because when it comes to rotation, it's not the mass that matters.
It's the rotational inertia that matters. And rotational inertia depends on the distribution of the particle, about the center. If particles are, if there are more particles away from the center of rotation, you get a higher rotational inertia. So over here, because there's a hole punched out, a lot of particles are away from the center, the farther away from the center compared to over here.
So, this one has a higher rotational inertia compared to this one, but they have the same omega, and therefore this one ends up having more kinetic energy now. So when it comes to rotation, it's not just the mass that matters, you also have to think about the distribution of the particles, and that's what makes rotation so interesting.
All right, what if you had an object that's doing both translation and rotation at the same time? For example, a ball that's rolling on the floor. Look, it's moving forward. It's translating, and it's spinning. It's rotating as well. How do you calibrate kinetic energy of such an object?
Well, now you have to add both its translational kinetic energy and the rotational kinetic energy. That'll be the total kinetic energy of such motion. All right, now before we can tackle our original problem, we can solve our original question. We need one more concept, one more idea.
In translational motion, if you have a force that's acting on an object, and let's say it displaces the object about some direction, the discussion of the object need not be in the direction of this force if there are multiple forces acting on it, okay? And let's say it does so in such a way that it's only translating, it's not rotating at all.
Then we can define work done by this force as the product of the component of the force in the direction of the displacement. That's what this F parallel stands for. You don't just multiply the force, but you take the component of the force in the direction of the displacement times the displacement, right?
So, you take this product and we call that as the work done. Its units becomes joules or Newton meters, okay? Why do we care about work done? Well, because if there's more than one force acting on an object, then the total work done on the object by all the forces, that equals the change in the kinetic energy.
So, it'll be equal to the change in the translational kinetic energy. We call this the work-energy theorem. And this is super useful in making certain predictions as we will see. Now, the same thing can be applied over here when it comes to rotation. If you have an object on where there's a force acting, which puts a torque on this object and makes it spin through some angle, delta theta for example, then we can define the work done by the torque in a very analogous manner.
We would say the work done equals the torque acting on that object times the angular displacement. Again, that makes sense, right? Torque is like the rotational version of the force, and angular displacement is like the rotational version of the linear displacement.
So you get very, very analogous expression for the work done. And again, we could ask, "Why do we care about work done over here?" Same idea. If there are multiple forces acting on an object producing multiple torques, then the total work done by all the torque equals the change in the rotational kinetic energy of that object.
Of course, one small caveat is that we can apply this expression when the net torque is 0. So there is an unbalanced force, but there are no unbalanced torques. So that's why you know there's only translational motion. And we apply this expression when the net force is 0, but there is some unbalanced torque.
That's why it's only affecting its spin, for example. But what if you have both, you know, unbalanced force and an unbalanced torque? Well then again, we have to treat them separately. We have to calculate the work done by the unbalanced forces separately equate it to change in translational kinetic energy.
And we have to calculate the work done by, the total work done by the unbalanced torque separately and equate that to change in rotational kinetic energy. Okay, one final thing, when it comes to translational motion, work done equals a product of force and displacement, right?
So, we can calculate it as the signed area under the graph of the force as a function of position, right? Similarly, when it comes to rotational work done, we can calculate it, you can calculate the, this particular work done as the signed area under the graph of torque as a function of angle theta.
Again, the situation is super duper analogous. So if we come back to this example, let's say initially both these objects are at rest, they're not spinning, and then we put a torque on them and we make them spin, and we get them to the same angular speed. In which case, would we have to do more work?
Can you pause and think about this? All right, well, we already know that this one ends up having more kinetic energy compared to this one because it has more rotational inertia. So when we made it spin, this one ended up with more change in kinetic energy compared to this one.
Therefore, more work has to be done on this one compared to this one. So if we rotated them to the same angle, for example, then the net torque on this one must have been greater compared to this one. So, these are the kind of analysis that we can do using the work-energy theorem.
All right, now let's go back to our original problem. We know initially they all have the exact same gravitational potential energy to begin with because they have the same mass, they're the same height, all right? And when they reach the bottom, all of that potential energy would've been converted to kinetic energy.
Now, if these were not spinning at all, okay, assume that they were just sliding only translator motion, then all of that potential energy would get converted to translational kinetic energy. And so, all of them would end up having the same amount of translational kinetic energy.
And since they have the same mass, they would have the same speed. And then our in, you know, in my instinct would be right, they would all reach the ground, they would all reach bottom at the same time. But they're not just gonna go in translating motion, they're also spinning, which means when they reach the bottom, they not only have translational kinetic energy, but along with that they also have rotational kinetic energy.
So, some of that potential energy is converted into rotational kinetic energy and only the remaining will be the translational kinetic energy. And now, the question is how does it get distributed for different objects? Well, we already know objects that have higher rotational inertia, well, they end up having higher rotational kinetic energy, you know, for a given speed.
Now, of the three which has the highest rotational inertia? Well, they all have the same radius but when it comes to this hoop, all the particles, pretty much all the particles are at the maximum distance from the center. So they have the highest, this hoop has the highest rotational inertia.
Then comes the whole sphere. It has lesser rotational inertia, because there are certain particles closer to the axis of rotation, because it's a sphere. But then comes the solid sphere because it's solid on the inside, a lot of particles are closer to the center, and so it has the least rotational inertia.
So if I were to only look at their rotational kinetic energy, I would expect this one to have the maximum rotational kinetic energy and this one to have the minimum rotational kinetic energy. But since the total kinetic energy must be the same for all three of them when they reach the bottom, that means this will end up having the maximum translational kinetic energy and this will end up having the minimum translational kinetic energy.
Does that make sense? Therefore, this one will have the least speed when it reaches the bottom, and this one will have the maximum speed when it reaches the bottom. That's why we can now predict that this one should reach the bottom first, because it speeds up.
It's linear speed increases much more quickly compared to this one. It achieves a higher linear speed by the time it reaches the bottom. So, this one must have reached first. Then, comes this one. And this one reaches the bottom the last. So if you were to look at the animation, this is kind of what it would look like.
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