Lesson Video: Angular Momentum in Terms of Moment of Inertia Physics

Video Transcript

In this video, our topic is angular momentum in terms of moment of inertia. We’re going to learn how these two terms relate as well as get practice calculating angular momentum this way. Now, since we’re talking about angular momentum in terms of moment of inertia, we can start out by reminding ourselves about moment of inertia. If we have a block of mass 𝑚 and we say that that mass is stationary, then Newton’s first law of motion tells us that this mass is likely to remain this way unless it’s acted on by an unbalanced force. That is, this mass has inertia, which makes it tend to keep doing what it’s doing, in this case, sitting in place.

So if we wanted to translate this mass, say, by moving it left to right, we would need to exert some force on it. We could also imagine, though, that one way of moving this mass is not by translation, but through rotation about some axis, in this case, through its center. If we think about how this object will resist starting to rotate, rather than talking about its mass and the inertia there, we talk about its moment of inertia, symbolized 𝐼. What an object’s mass is to its resistance to linear motion, an object’s moment of inertia is to its resistance to angular motion. In general, the units of an object’s moment of inertia are kilograms times meters squared. This shows us that we’re talking about some amount of mass some distance away from an axis of rotation squared.

We can think then of moment of inertia as the rotational analog of an object’s mass. In a similar way, say that earlier on, we had indeed given this mass a push in some linear direction, say, to the right, and then as a result, our mass acquired a velocity 𝑣. In that case, we would say that since this mass has a velocity, it has a linear momentum. And linear momentum for an object moving in a straight line is analogous to angular momentum for a rotating object. We represent angular momentum using a capital 𝐿 and its units are kilograms meter squared per second. We see that we can write those units though as kilograms times meters per second times meters because this makes it more clear that we’re talking about some amount of mass moving with some speed some distance away from an axis of rotation.

So, for example, say that our axis of rotation was this blue line here and that we had a mass of value 𝑚 moving at a speed 𝑣 out of the screen a distance 𝑟 away from that axis of rotation. Then we can see that by increasing anyone of these three factors — the mass, the speed, or the distance from the axis of rotation — we could increase angular momentum. As a quick example, if we kept our object’s mass and its distance from the axis of rotation the same but increased its speed, then it would have a greater angular momentum than before. Now, we’ve written out the units of angular momentum this way to make it clear physically what’s going on. But typically, these units are written this way, kilograms meter squared per second.

Expressed that way, we can see a close similarity to the units of moment of inertia, kilograms meters squared. All we need to connect the two is this one-over-seconds factor. And this tells us that we’re looking for some kind of rate to bridge the gap between angular momentum and moment of inertia. We can obtain that rate by considering two of the factors in our sketch here, our mass’s speed and its distance from its axis of rotation. In general, if we have an object with some linear speed, we’ll call it 𝑣, and that linear speed contributes to the object rotating around some axis, then the radius of the circle in which that object moves multiplied by its angular speed 𝜔 is equal to its linear speed 𝑣.

So, here, our object has a linear speed, we’ve called it 𝑣 sub f. And the object is rotating around an axis a distance 𝑟 away, which means that our mass will have some angular speed. And it’s equal to the ratio of the object’s linear speed to its distance from the axis of rotation. The units of angular speed are radians per second. And since a radian is a dimensionless quantity, we can say that the dimensions of angular speed are simply inverse time or one over time.

Knowing this, we’re now ready to bring together angular speed, angular momentum, and moment of inertia in one equation. We can write that an object’s angular momentum 𝐿 is equal to its moment of inertia multiplied by its angular speed. And if we write out all the units involved in these factors, we can see that indeed they equal out since, as we saw, radians are a dimensionless quantity. So here we’ve described angular momentum in terms of moment of inertia. Note, though, that this isn’t the only way we can write this equation. The reason for this is that we can write our angular speed 𝜔, also sometimes called an angular frequency, in terms of linear frequency 𝑓. We’re able to do this by multiplying two 𝜋 times that frequency 𝑓.

But then, as we think further, we recall that frequency 𝑓 can also be expressed as one over the period of rotation of some rotating object. If we call that period capital 𝑇, then 𝑓 is equal to one divided by capital 𝑇. And this means we now have these two equivalent ways of writing the angular momentum of some object. It’s equal to 𝐼 times 𝜔 but also 𝐼 times two 𝜋 times 𝑓 and also 𝐼 times two 𝜋 divided by 𝑇. Whenever we work through exercises involving these equations, it’s important to keep track of the units involved in each term. Doing so gives us a better understanding of physically what’s going on, and it also helps to check our progress as we go. Knowing all this, let’s now get some practice with these ideas through an example exercise.

A solid metal sphere is rotating with an angular speed of 12.6 radians per second. It has a moment of inertia of 0.0200 kilograms meters squared. What is the angular momentum of the sphere?

Let’s say that this here is our sphere, and in our problem statement, we’re told that it’s rotating with a certain angular speed. Now, we don’t know where the axis of rotation of the sphere is. It could be, for example, through a point on the sphere or that axis could be somewhere outside the sphere. But wherever it is, we know the sphere’s angular speed as it moves about that axis. We’ll refer to that speed as 𝜔. And we’re also told about the sphere that it has a given moment of inertia. This quantity, which we’ll represent using the letter capital 𝐼, is a rotational analog to mass. In other words, what mass is to linear motion, how it affects that motion through inertia, the moment of inertia 𝐼 is to rotational or angular motion.

We could say that the larger an object’s moment of inertia is, the harder it is to begin rotating or turning that object. Knowing moment of inertia and angular speed for the sphere, we want to calculate its angular momentum. And that is equal to the product of moment of inertia and angular speed. We could think of angular momentum as the product of the degree of difficulty in getting some object to rotate by the actual speed with which it is rotating. And this gives us how much momentum the object has in an angular direction. So when we substitute in the given values for 𝐼 and 𝜔, to three significant figures, we find a result of 0.252 kilograms meters squared per second. Note that we’d leave out of our final answer the units of radians because these are dimensionless. And so the angular momentum of our solid metal sphere is 0.252 kilograms meter squared per second.

Let’s look now at another example exercise.

The wheel of a car has a moment of inertia of 2.25 kilograms meters squared about its axis of rotation. As the car is driven along a road, the wheel has an angular momentum of 70.7 kilograms meter squared per second. With what angular speed is the wheel rotating?

Let’s say that this is the wheel we’re considering and that as the car drives along, the wheel rotates about an axis through its center. This rotation happens at some angular speed, we’ll call it 𝜔, and it’s this value we want to solve for. We can do this by taking into account that we’re given the moment of inertia of the wheel as well as its angular momentum. We can recall that an object’s angular momentum 𝐿 is equal to its moment of inertia multiplied by its angular speed. And so we can equivalently write that angular speed is equal to angular momentum divided by moment of inertia.

In our scenario, the wheel’s angular momentum is 70.7 kilograms meter squared per second and its moment of inertia is 2.25 kilograms meters squared. Looking at the units in this expression, notice that kilograms and meters squared cancel out from top and bottom and that we’ll be left with one over seconds in our final answer. When we calculate our result, which to three significant figures has a numerical value of 31.4, we can recall that 𝜔, an angular speed, has units of radians per second. Now, the reason that radians didn’t appear in our fraction is that the radian is a dimensionless quantity. Nonetheless, it’s a unit that we know to include when we’re talking about angular motion, as in this case with angular speed. So the angular speed with which this car wheel is rotating is 31.4 radians per second.

Let’s look now at one last example exercise.

Earth has a moment of inertia about its axis of rotation of 9.69 times 10 to the 37th kilograms meters squared and an angular speed of 7.29 times 10 to the negative fifth radians per second. What is the angular momentum of Earth due to its rotation?

Let’s say that this is the Earth and the axis that it rotates about. And we’re told that as it does this, it has a particular value for its moment of inertia, we’ll call that 𝐼, and that this rotation happens at a given angular speed, we’ll call it 𝜔. Given this information, we want to solve for the angular momentum of Earth due to its rotation, we can call this capital 𝐿. We can recall that, in general, the angular momentum of an object is equal to its moment of inertia multiplied by the speed with which it rotates. So 𝐿 is equal to 𝐼 times 𝜔. And if we substitute in the given values for these terms and then calculate 𝐿, to three significant figures, we find it’s 7.06 times 10 to the 33rd kilograms meter squared per second.

Note that, in this result, we don’t have the units of radians. That’s because radians are a dimensionless quantity. When we report our final answer then, we say that Earth has an angular momentum of 7.06 times 10 to the 33rd kilograms meter squared per second.

Let’s summarize now what we’ve learned about angular momentum in terms of moment of inertia. In this lesson, we saw that moment of inertia, symbolized with a capital 𝐼, is analogous to mass. While an object’s mass helps us understand its linear motion, moment of inertia helps us understand an object’s rotational motion. Along with this, we saw that an object’s angular speed, represented 𝜔, has units of radians per second where radians, we saw, are a dimensionless quantity.

This led us to the important result that angular momentum 𝐿 is equal to moment of inertia times angular speed. And since angular speed 𝜔 can also be expressed as angular frequency, which is equal to two times 𝜋 times the linear frequency 𝑓, this means that angular momentum can be written as 𝐼 times two times 𝜋 times 𝑓. And since the frequency 𝑓 is equal to the inverse of a period of rotation, where that period is represented with a capital 𝑇, we saw we could also write angular momentum as the moment of inertia times two 𝜋 divided by the period 𝑇. This is a summary of angular momentum in terms of moment of inertia.