### Video Transcript

In this video, we’re going to learn about the conservation of angular momentum. We’ll review what angular momentum is. We’ll see why it’s conserved. And we’ll learn why the conservation of angular momentum is important.

To get started, imagine that, as a hobbyist who enjoys balancing spinning plates, one day as you’re reading the Guinness Book of World Records, you discover that the world record for the most spinning plates balanced at one time by one person is six. That doesn’t sound like a lot to you. And you wonder if you could beat that record.

With the help of a friend who gets the plates spinning before you start to balance them, you work your way up to the record level. But unfortunately, before you can get there, some of the plates put on earlier start to slow their spin. And eventually they fall. You wonder if there’s anything you can do regarding the plate rotation rate to help them stay stable for longer.

To understand this better, it’s helpful to know about the conservation of angular momentum. As a reminder, angular momentum, which we symbolize with a capital letter 𝐿, is a property of massive systems that rotate. That means that, any time we have a mass of object that’s rotating about some axis, then that object has angular momentum. And we’ve seen that that angular momentum is equal to the cross product of the position vector of the object and its linear momentum 𝑝.

Based on this relationship, we could say that angular momentum is momentum of rotational motion. We’ve also seen that there’s another way to express angular momentum mathematically. Looking back at our diagram, if our object is rotating with an angular velocity 𝜔 and if the object itself has a moment of inertia capital 𝐼, then we can say that the object’s angular momentum is equal to the product of those two variables. Written this way, we’ve expressed angular momentum in terms of rotational variables only, moment of inertia and angular velocity.

Speaking of rotational variables, we can see why angular momentum is conserved by looking at a different relationship and converting that into rotational variables too. The version of Newton’s second law of motion that we may be most familiar with says that the net force acting on an object is equal to the object’s mass times its linear acceleration 𝑎.

If we were to write a rotational version of the second law, we would replace force with torque 𝜏, we would replace mass with moment of inertia 𝐼, and we would replace linear acceleration 𝑎 with angular acceleration 𝛼. And just like in the linear version, we would be referring to the net torque acting on some object.

Looking at this equation, we realize we can substitute for the angular acceleration 𝛼, the change in angular velocity 𝜔 per unit time. That’s because the definition of acceleration is a change in velocity per unit time.

Considering this rewritten expression for Newton’s second law, the rotational version, we see that the right side actually looks a lot like the right side of our equation for angular momentum. Both equations have just the moment of inertia and the angular velocity 𝜔 in them. But our net torque equation has a time derivative.

If we assume that our moment of inertia 𝐼 doesn’t change in time, that means we can factor out the time derivative. And what’s inside our parentheses is simply the object’s angular momentum 𝐿. One thing this resulting equation tells us is that if the net torque on our object is zero, then that means the time rate of change of angular momentum is also zero. Or, in other words, momentum doesn’t change in time. It’s conserved.

What we’ve done then is we found the condition needed in order for angular momentum to be conserved. The requirement is that the net torque acting on our object is zero. We can write this out by saying that angular momentum is conserved in a closed system where there’s no net torque.

You might recognize this as similar to our statement of linear conservation momentum, where we said that linear conservation is conserved in a closed system with no net force. One way we can understand conservation of angular momentum better is to see a few practical examples of this happening.

A classic example of angular momentum conservation is an ice skater standing in place but rotating quickly. When the skater holds his or her arms out to full extension, they’ll rotate with a particular speed. We can call it 𝜔 sub 𝑖. But then if they tuck their arms in as close as possible to their torso, they’ll rotate with a new speed 𝜔 sub 𝑓, which is greater than the original angular speed.

We can explain this effect from a conservation of angular momentum perspective. One way of writing the magnitude of angular momentum is that it’s equal to an object’s mass times its distance from the axis of rotation squared times its angular speed.

In the case of our ice skater, when they move their arms in closer to their chest, that radial distance decreased. And as 𝑟 went down in order for 𝐿 to be conserved, 𝜔 went up. They spun faster. The effect is actually pretty dramatic, as you might guess from the fact that 𝑟 is squared. So cutting that value in half will require 𝜔 to go up by a factor of four in order for angular momentum to stay the same.

Other examples of angular momentum conservation are related to objects that move in a stable way because of the angular momentum they have. For example, if you throw a frisbee, in order to do so, you give it a spin. Because the frisbee is spinning, it’s able to fly in a stable path. In a sense, the frisbee tries to conserve its angular momentum. And the best way to do that is to keep moving as it has been.

Another example of this is a bicycle rider who becomes more stable in their motion as they move faster. This is because the spinning wheels of the bicycle want to conserve their angular momentum as they rotate and therefore resist change to their motion. This is why it’s easier to balance on a bicycle that’s in motion than one that’s moving very slowly or stopped.

Just like was the case with the conservation of linear momentum, the conservation of angular momentum is a very useful concept because it lets us solve problems we might not otherwise be able to. Let’s see how that works practically through an example.

An ice skater is preparing for a jump in which he will rotate while in the air. When he is on the ground with his arms extended, his moment of inertia is 2.2 kilograms meter squared. And he is spinning at 0.31 revolutions per second. He launches himself into the air at a speed of 12.6 meters per second at an angle of 45 degrees above the horizontal.

At the moment that he leaves the ground, he contracts his arms, taking negligible time, and changes his moment of inertia to 0.62 kilograms meter squared. How many revolutions can he complete while airborne?

We’ll call this number of revolutions capital 𝑁. And we’ll start by drawing a sketch of the situation. When the ice skater jumps, he has an initial moment of inertia we’ve called 𝐼 sub 𝑖 and an initial angular speed we’ve called 𝜔 sub 𝑖, both given to us in the problem statement. He launches himself at an angle we’ve called 𝜃 above the horizontal, with an initial take-off speed we’ve called 𝑣.

Just as he launches off the ice, he pulls his arms into his chest and achieves a final moment of inertia we’ve called 𝐼 sub 𝑓, also given to us. While the skater is in the air, he’s rotating at an angular speed we’ve called 𝜔 sub 𝑓 and, after rotating at this speed while he’s airborne, lands back on the ice. Over that time and spinning at that rate, he completes some number of revolutions we’ve called capital 𝑁. And that’s what we wanna solve for.

We’ll begin by solving for 𝜔 sub 𝑓, the skater’s angular speed while he’s in the air. And to do that, we’ll rely on the principle of the conservation of angular momentum. In a closed system that experiences no net torques, we can say that this system’s angular momentum is equal at any one time to the angular momentum of this system at any other time. That is, the system’s initial angular momentum, wherever we define initial to be, is equal to its final angular momentum.

In our case, we define the initial moment to be a time before the skater has left the ice and the final time to be after the skater has left the ice and changed his moment of inertia. So we can write that 𝐿 sub 𝑖 is equal to 𝐿 sub 𝑓. And recalling that the magnitude of angular momentum is equal to moment of inertia multiplied by an object’s angular speed, we can expand this statement to say that 𝐼 sub 𝑖 times 𝜔 sub 𝑖 is equal to 𝐼 sub 𝑓 times 𝜔 sub 𝑓.

In all this, it’s 𝜔 sub 𝑓 we want to solve for. We see it’s equal to the ratio of the initial moment of inertia to the final moment of inertia multiplied by the initial angular speed. We’re given all these values in the problem statement. So we can plug in and solve for 𝜔 sub 𝑓.

Plugging these values in and entering them on our calculator, we find 𝜔 sub 𝑓 is 1.1 revolutions per second. That’s the angular speed of the skater when he’s in the air. Knowing that, if we knew how long the skater was in the air, then we could figure out how many revolutions he performed and solve for capital 𝑁.

While the skater is airborne, he’s essentially a projectile, under the rules of projectile motion. If we call the total horizontal range that the skater travels capital 𝑅, there’s a relationship we can recall which helps us solve for 𝑅 based on the initial launch speed 𝑣 and the angle 𝜃.

Sometimes called the range equation, this relationship says that the horizontal range of an object which is launched and lands at the same elevation is equal to the square of its initial speed, 𝑣 sub zero, times the sin of two times its launch angle 𝜃, all divided by the acceleration due to gravity 𝑔.

If we let 𝑔 be exactly 9.8 meters per second squared, then we can say that the range our ice skater travels while in the air is equal to 𝑣 squared sin two 𝜃 over 𝑔, or 12.6 meters per second squared times the sin of two times 45 degrees, or the sin of 90 degrees, all divided by 9.8 meters per second squared.

We can simplify this last equation a bit because we know that the sin of 90 degrees is equal to one. Before we calculate this result though, let’s consider how it fits into the bigger picture of solving for 𝑁. We can recall that the average speed of an object is equal to the distance it travels divided by the time it takes to travel that distance.

This range value 𝑅 that we’re calculating is the distance the skater travels in the horizontal direction. That means that if we were to apply this relationship to our scenario, we would use the ice skater’s horizontal speed 𝑣 times the cos of 𝜃 and set that equal to the horizontal distance traveled 𝑅 divided by the time the skater is in the air 𝑡.

Really, it’s not 𝑅 we wanna solve for, but 𝑡, because if we have 𝑡 and we multiply it by 𝜔 sub 𝑓, we’ll get a total number of revolutions the skater has traveled. In other words, we’ll have solved for capital 𝑁.

We can rearrange and say that 𝑡 is equal to 𝑅 divided by 𝑣 times the cos of 𝜃. And now rather than calculating the range 𝑅, we can calculate the time 𝑡 by dividing the range 𝑅 expression by 𝑣 times the cos of 𝜃. So now we’re solving for 𝑡, which is equal to 𝑣 squared over 𝑔 times 𝑣 times the cos of 45 degrees. And we see that a factor of the speed 𝑣 cancels out from numerator and denominator.

Entering this simplified fraction on our calculator, we find that 𝑡 is approximately 1.818 seconds. That’s the amount of time that the skater is in the air. We’re now ready to solve for capital 𝑁, which is 𝑡 times 𝜔𝑓, or 1.818 seconds times 1.1 revolutions per second. To two significant figures, this is equal to 2.0 revolutions. That’s how many turns the skater would make while airborne.

Let’s summarize what we’ve learned about conservation of angular momentum. In this video, we’ve seen that angular momentum, typically symbolized with a capital 𝐿, is a measure of an object’s tendency to continue its rotational motion. We’ve also seen that the angular momentum of a system that experiences no net torques over time stays the same. That is to say, it’s conserved.

And finally, we saw that angular momentum conservation makes this quantity so useful because it helps us solve problems, as it did in the case of our spinning ice skater. It’s that problem-solving utility that makes angular momentum such an important point of focus and so worthwhile to understand.