Lesson Video: The Conservation of Angular Momentum Physics

Video Transcript

In this video, we’re talking about the conservation of angular momentum. As we learn this principle, we’ll see how it applies to small systems like this top spinning on a table, all the way up to very large systems, such as planets orbiting around stars. As we get started, we can imagine having some mass, we’ve called it 𝑚, in motion at a speed we’ll call 𝑣. Now, just by virtue of the fact that we have a mass 𝑚 moving at a speed 𝑣, we can say that this mass has some amount of linear momentum 𝑝. That’s equal to 𝑚 multiplied by 𝑣.

But then, what if this moment in time, where the mass is moving upward with a speed 𝑣, is just one snapshot of its overall motion? We could say that our object is actually following a circular path where the radius of that circle we’ll call 𝑟. When this is the case, we have a mass in motion around some axis, then we say that this mass possesses angular momentum, capital 𝐿. This is equal to 𝑚 multiplied by 𝑣, just like linear momentum, but then also multiplied by the radius of the circular path in which this mass moves. Moving this way, not only does our object have angular momentum, but we can also say that in addition to its linear speed 𝑣, it has an angular speed 𝜔. And in general, that angular speed, when it’s multiplied by the radius of the circle an object moves in, is equal to its instantaneous linear speed.

Realizing that 𝑟 times 𝜔 is equal to 𝑣, we could substitute 𝑟 times 𝜔 in for 𝑣 in our equation for angular momentum. And when we do that, this can be written as 𝑚 times 𝑟 squared times 𝜔. Now, since we have some mass of object rotating about some center, that means that our mass has a moment of inertia. Moment of inertia 𝐼 quantifies just how easy or difficult it is to get a particular mass moving around a certain axis. The higher 𝐼 is, the more difficult we could say it is to get that particular shape rotating in that way. Now, if we suppose that our mass here is a point mass, that is, it occupies an infinitely small amount of space, then the moment of inertia of an object like that is defined as its mass times the radial distance between the mass and its rotation axis squared.

And this is very interesting because we see 𝑚𝑟 squared in our definition for angular momentum. Since that’s equal to the moment of inertia of a point mass, when we consider specifically a point mass, like we are here, we can replace 𝑚𝑟 squared with 𝐼. And now we have this new expression for angular momentum. It’s important to note that this expression is true in general, whether or not our object has a moment of inertia equal to 𝑚𝑟 squared. So then, for a given system, the angular momentum of that system is equal to its moment of inertia multiplied by its angular speed. And as we saw earlier, this also equals 𝑚 times 𝑣 times 𝑟.

Our focus here is on the fact that this quantity, angular momentum, is conserved for a closed system. Here’s an example of what that means practically. Say that we have a stool, and while its legs are fixed in place on the ground, the stool is free to rotate and that it does so frictionlessly. So if we were to step on the stool and then started rotating, that would mean that we turn along with the stool at the same rate. But then imagine this. Say that we stretch our arms out. When we do this, we’re effectively increasing our moment of inertia. We’re doing this by moving mass, the mass in our arms, farther away from the axis of rotation that runs straight down through our center.

So, considering again our equation for angular momentum, what we’re saying is that we’re raising our moment of inertia by moving our arms out. But then, 𝐿, the angular momentum of the system, which consists of us and this rotating stool, must stay the same. That’s what it means for it to be conserved. Therefore, the angular speed at which we’re spinning must decrease. And indeed, if we put our arms out, we’ll find that that happens. We spin at a slower rate. And then, if we were to bring our arms in, that would decrease our moment of inertia and therefore raise our angular speed. We would spin more quickly.

All of this can be experienced by finding the right kind of stool that can spin this way. And by the way, to really have some fun with angular momentum conservation, it’s great to pick up a wheel that can be rotated, hold on to an axle through its center, get the wheel spinning so that it itself has angular momentum, and then start to move it around and hold it in different positions.

If we think about the total angular momentum of this system, including us, the stool, and the wheel, we can call that total 𝐿. We can say that that total is equal to the angular momentum of the person and the stool, we’ll call it 𝐿 sub ps, plus the angular momentum of the wheel, 𝐿 sub w. If this system is closed, that is, no angular momentum is put into it from the outside, and none leaves, so to speak, then this means that no matter how we change 𝐿 sub w, that is, no matter how we move the wheel in our hands, our angular momentum and that of the stool will respond to any changes in 𝐿 sub w so that 𝐿 overall remains the same.

This means that if some sort of torque was exerted on the wheel, and say the effect of this was to decrease 𝐿 sub w, then 𝐿 sub ps would correspondingly increase by that same amount. In a similar way, if 𝐿 sub ps were somehow to decrease, then 𝐿 sub w would increase again by that same amount. All throughout, the overall angular momentum of the system remains constant.

Mathematically, we can express the conservation of angular momentum like this. For a given system, whatever objects that system includes, so long as the system is closed, the angular momentum of the system overall at some initial moment is equal to its angular momentum overall at some later time. Now, we mentioned earlier that the conservation of angular momentum applies to systems of any physical size. That includes orbiting bodies, such as moons or planets. To see how this is so, let’s say that our mass here, which until now we’ve been considering a point mass, is actually much larger. Let’s say that it’s a moon orbiting a planet like this. Here we have a circular orbit, which means that 𝑟 in this equation for angular momentum is always the same. And therefore, the conservation of angular momentum as the moon moves through its circular orbit tells us that its speed is also constant all throughout.

At any given instant then, this moon is always the same distance from the center of mass of the planet, and it’s always moving with the same speed. This all changes, though, if instead of a circular orbit, our moon is moving in an ellipse. Moving this way, we can see that the radial distance, from the center of the moon to the center of the planet it orbits, is no longer a constant. So if 𝑟 in this equation is no longer the same for this orbiting body, does that mean that angular momentum is no longer conserved? Well, it doesn’t. And the way this works is, since the mass of our moon is constant, the moon’s speed 𝑣 is adjusted, we could say, for the various radial distances.

When 𝑟 is small, that is, when the moon and the planet are relatively close to one another, the speed 𝑣 goes up. And then when 𝑟 increases, when the moon moves farther away from the planet, its orbital speed 𝑣 decreases. Overall then, the planet is moving relatively slowly out here and relatively quickly in here. In this way, the overall angular momentum of our system, the moon and the planet, is conserved. That is, at any two moments in time, and we could call these initial and final moments, the angular momentum is the same. Now, it is true that in a closed system, the angular momentum of that system is conserved. But if we take the objects in our system and we apply a torque to them, then that will change the overall angular momentum. Our system is no longer closed.

As an example of this, let’s say that our system is planet Earth, which we know rotates about an axis roughly through its poles. Since the Earth has mass and is rotating about some axis, therefore it has angular momentum. Let’s say, though, that we bring in something outside of our Earth system. In particular, we bring in a person and set them standing on the surface of the Earth. Now, if this person started to walk across the surface of the Earth, technically they would be applying a torque to planet Earth. This torque would have some effect on its angular momentum.

Before we get worried, though, about possibly upsetting the balance of the Earth by walking or running about on it, we can recall that its angular momentum is equal to its moment of inertia multiplied by its angular speed. The moment of inertia of the Earth is a huge quantity. Therefore, the torque that we exert on Earth by walking across its surface is negligibly small. It has no measurable effect on Earth’s overall angular momentum. Nonetheless, by bringing in something outside of our system and having that apply a torque to the system, technically, we then could no longer expect the angular momentum of the Earth to be conserved. Knowing all this about angular momentum conservation, let’s get some practice of these ideas now through an example.

The diagram shows three disks that are all the same size and made of the same material, and they can rotate around an axle. If the angular momenta of disks one and three increase by 20 kilograms meter squared per second each, by how much must the angular momentum of disk two change in order to counterbalance the increase in angular momentum of the other two?

In our diagram, we see these three identical disks — one, two, and three — all set up on the same axle. Our question tells us that the angular momenta of disks one and three increases by this specific amount, 20 kilograms meter squared per second. If we recall that the angular momentum of an object is equal to its moment of inertia multiplied by its angular speed, then if we assume that for these two disks their moment of inertia stays constant, that must mean that for their angular momentum to increase their angular speed must be doing that.

Given this change in the angular momentum of disks one and three, the question is, by how much must the angular momentum of disk two change so that it counterbalances the increase in angular momentum of disks one and three? Thinking about the total change in angular momentum of this system of three disks, if we call that change Δ𝐿, we can write that it’s equal to the change in the angular momentum of disk one plus the change in the angular momentum of disk two plus the change in the angular momentum of disk three. And very importantly, we want to enforce the constraint that this overall change is zero. That’s what it means for the change in the angular momentum of disk two to counterbalance that in disks one and three.

In this equation then, we want to solve for Δ𝐿 sub two, the change in the angular momentum of disk two. And to do it, we’ll focus on this equality. The sum of all these changes is equal to zero. And our problem statement tells us what Δ𝐿 sub one and Δ𝐿 sub three are. Since the angular momentum of each of those disks increases by 20 kilograms meter squared per second, then we can substitute that value in for Δ𝐿 sub one and Δ𝐿 sub three. And now to solve for Δ𝐿 sub two, all we need to do is subtract two times 20 kilograms meter squared per second from both sides. When we do this, we find that Δ𝐿 sub two is negative 40 kilograms meter squared per second. This is how much the angular momentum of disk two needs to change in order to counterbalance the increase in angular momentum of disks one and three.

Let’s look now at a second example exercise.

The diagram shows three disks, which can all rotate around an axle. Disks one and three have the same moment of inertia as each other. Disk two has a moment of inertia four times that of disk one. If the angular velocity of disk two increases by two radians per second, by how much must the angular velocity of disks one and three change in order to counterbalance the change in the angular momentum of disk two? Assume that disks one and three must have the same change in angular velocity as each other.

In our diagram, we see these three disks — one, two, and three — set up on this axle so that they can rotate around it. Our problem statement tells us that disks one and three, which we see in our diagram look identical, have the same moment of inertia as each other. We can write that like this. 𝐼 one, the moment of inertia of disk one, is equal to 𝐼 three, that of disk three. Disk two, our statement goes on to tell us, has a moment of inertia four times that of disk one. We can put that this way. 𝐼 two is equal to four times 𝐼 one.

We then imagine that the angular velocity, that is, the rate at which disk two spins around this axis, increases by two radians per second. The question then becomes, by how much must the angular velocity of disks one and three change in order to counterbalance this change in angular momentum of disk two? And then we’re to assume that disks one and three experience the same change in angular velocity. To begin working this out, let’s clear some space on screen. And having done that, let’s now write out what we know has happened to the angular velocity of disk two. If we describe that using 𝜔 sub two, we know that, whatever its value, eventually the angular speed of disk two became that value plus two radians per second. In other words, the angular speed of disk two increased by two radians per second.

Now, if we recall that the angular momentum of some system is equal to its moment of inertia times its angular speed, then we can see that in the case of disk two, with its angular speed increasing and its moment of inertia staying the same, that means its angular momentum must go up overall. Our question, though, asks us to think about the angular momentum of our entire system, all three disks together. In particular, it asks us to consider the condition that the change in angular momentum for this whole system is zero. That is, its overall angular momentum before the angular speed of disk two increased is the same as its angular momentum after.

If we allow that the moment of inertia of disk one and disk three do not change through this process, and that’s a reasonable assumption to make, then the only way that this condition can be satisfied, that the change in angular momentum of our system overall is zero, is for the angular speeds of disks one and three to somehow change in order to counterbalance the change brought about by disk two’s angular speed change.

So here’s what we’re saying. In this process overall, the system’s change in angular momentum is zero and that the moment of inertia of these three disks also doesn’t change, but that the angular speeds of these disks, that is, 𝜔 one, 𝜔 two, and 𝜔 three, do change. Based on this, we can write that Δ𝐿, the change in angular momentum of the system, is equal to 𝐼 one times the change in angular speed of disk one plus 𝐼 two times two radians per second, that’s the change in disk two’s angular speed, plus 𝐼 three times Δ𝜔 three, the change in disk three’s angular speed. And then we saw in the line above that Δ𝐿 is zero. So then, this is the equation we want to solve. And in it we’re told that this change, Δ𝜔 one, and this one, Δ𝜔 three, are the same.

To recognize that fact mathematically, let’s change this symbol so it’s now Δ𝜔 one three to represent that this change in angular speed applies to both disks. It’s Δ𝜔 one three that we want to solve for. And to do it, we can first subtract two radians per second times 𝐼 two from both sides of the equation. And then next, we factor out Δ𝜔 sub one three from both terms on the right. Lastly, if we divide both sides by 𝐼 one plus 𝐼 three, we get this equation. To solve for Δ𝜔 one three, all we need to do now is figure out what this ratio is. Looking down to the bottom right of our screen, we see that 𝐼 two is equal to four times 𝐼 one and also that 𝐼 one is equal to 𝐼 three. Therefore, we could write this entire fraction in terms of 𝐼 one, four 𝐼 one divided by 𝐼 one plus 𝐼 one or four 𝐼 one over two 𝐼 one, which is equal simply to two.

When we replace the fraction in our equation with this number, we find that Δ𝜔 one three is equal to negative two radians per second times two or equally negative four radians per second. This is how much the angular velocity of disks one and three would need to change so that the change in system angular momentum caused by the change in angular speed of disk two is counterbalanced.

Let’s now summarize what we’ve learned about the conservation of angular momentum. In this lesson, we recalled that for an object of mass 𝑚 moving with a speed 𝑣 in a circular path of radius 𝑟, its angular momentum can be written as 𝑚 times 𝑣 times 𝑟. And this is also equal to its moment of inertia multiplied by its angular speed. Building on this, we saw that within a closed system, that is, a system that experiences no external torque, angular momentum is conserved. The mathematical expression of this is the system’s initial angular momentum equals its final angular momentum. And lastly, we saw that for both circular and elliptical orbits, angular momentum is conserved.