### Video Transcript

In this video, we’re going to learn about moment of inertia. We’ll learn what it is, why it matters, and how to calculate it. To get started, imagine you’re holding a race called “The Shapes Race.” At the top of an inclined plane, you’ve placed a hollow ring, a solid ball, a long solid cylinder, and a hollow spherical shell. On cue, all four of these shapes are released at the same time and allowed to start rolling down the incline. Certain of these shapes roll downhill more readily than others. And they pull ahead in the race.

If you wanted to predict ahead of time which of the shapes will win the race, it will be helpful to know something about moment of inertia. We can begin to understand the notion of moment of inertia by realizing that certain linear variables have rotational analogs, and vice versa. For example, consider the linear variable 𝑠, which stands for distance.

In the rotational world, the variable that corresponds to 𝑠 we can call 𝜃, for an angular distance. These two variables correspond to one another. They both describe a distance traveled. It might be surprising to learn that mass is also a linear variable. Here’s what we mean by that. Whenever we draw the free body diagram of forces acting on a mass, we know that those forces move through the center of the mass, that is, its center of mass.

Forces that act this way tend to translate rather than rotate our mass 𝑚. So the mass in response to forces such as these or any forces we might draw in using a free body diagram tends to move in a line. But we’re now going to imagine a different scenario. Instead of our forces always originating or passing through the center of mass of our object, we’ll now consider scenarios where the forces act along some different axis than the line through the center of mass. Forces such as these tend to make our mass rotate. And depending on the shape and distribution of our mass, that rotation will be more or less difficult.

This is the notion behind moment of inertia, often symbolized with a capital 𝐼. Instead of our forces moving through the center of mass of an object, which will lead to its linear translation, our forces now act outside of that center. Or even if the forces do move through the mass’s center, we might be considering the motion of the mass relative to an axis of rotation some distance away from its center.

We can see that moment of inertia always has to deal with rotation. And it’s in that sense that the moment of inertia is like a rotational mass. Say that we have a randomly shaped mass, subject to a force that moves through the center of mass of the object. And we’ll say that this force causes the mass to move in a circular path around an axis of rotation. Given that we have a force on a mass that is moving in a circular path, by Newton’s second law of motion, we can write a linear expression of this as 𝐹 equals 𝑚 times 𝑎.

Interestingly, there’s a rotational version of Newton’s second law. And we can find that version by considering that our mass is at a distance 𝑟 from the center of rotation. And if the vector 𝐹 and the vector 𝑟 are perpendicular to one another, we can write that 𝐹 times 𝑟 is equal to the moment of inertia, rotational mass remember, multiplied by angular acceleration 𝛼. So not only do individual variables have linear and rotational pairs, but so does Newton’s second law. There’s a linear and rotational version. It’s this rotational expression of Newton’s second law of motion that gives us a little bit of a better feel for just what moment of inertia 𝐼 is.

As we consider this connection between mass and moment of inertia, the question may come up: measuring mass is straightforward enough. As long as we know the acceleration due to gravity, we can put a body on a scale and solve for its mass that way. But what about 𝐼? What about an object’s moment of inertia?

It turns out that an object’s moment of inertia 𝐼 can be measured by using a simple pendulum. If we attach the shape whose moment of inertia we want to measure to the arm of a pendulum and let it swing freely under the influence of gravity, we can see that the back and forth motion the pendulum makes is a rotation. We can explore just how easy or difficult it is for the mass to make this rotation by varying the length of the pendulum’s arm as well as the amplitude from which it’s released.

For more common shapes, an example of which is the hollow cylinder we have here, the moment of inertia of those shapes about particular axes has already been calculated. And we can look them up in a table. As examples of various shapes and their corresponding moments of inertia, the moment of inertia of a point mass is equal to the mass of that point multiplied by the square of its distance from the axis of rotation. A solid sphere, on the other hand, rotating about its center, has a moment of inertia of two-fifths its mass times its radius squared. And a solid rod rotating about one end of the rod has a moment of inertia of one-third its mass times the length of the rod squared.

This brings up an important point about moment of inertia. We’ve been talking about how moment of inertia 𝐼 and mass are similar to one another. An object’s moment of inertia 𝐼 depends not only on the object’s shape and density, like mass does, but 𝐼 also depends on the axis of rotation. For example, if we measure the moment of inertia of our hollow cylinder in this case, then that measured value would depend not just on the mass of our cylinder, but also on its position or its orientation.

If we move the cylinder so that now, instead of its center, one of its ends is attached to the pendulum arm or if we rotated it so it was in line with the pendulum arm, in each of these three cases, we will probably measure a different moment of inertia, same shape different moment. So when we do go to a table to look up the moment of inertia of a particular shape, we’ll want to be careful to ensure that the axis of rotation of our shape matches our scenario. Otherwise, we’ll write down an incorrect or unmatching moment of inertia.

In addition to looking up an object’s moment of inertia, it’s also possible to calculate it. To consider how we might do this, let’s consider again our moment of inertia of a point mass, that is, an infinitely small mass with mass value 𝑚 a distance 𝑟 away from an axis of rotation. Under these conditions, that point mass’s moment of inertia has been determined to be its mass times 𝑟 squared.

Now let’s imagine that we envelop or surround our point mass with a much larger extended mass. The point mass 𝑚 is an infinitesimally small component of this larger mass. And we could even say that this larger mass is simply composed of infinitely many infinitesimally small point masses. If, for each infinitesimally small mass element that makes up this larger mass, we calculate that mass element times the square of its distance from the axis of rotation, then adding up all those parts through integration, we would solve for the overall moment of inertia of this extended mass with random shape.

Typically, when calculating the integral to solve for an object’s moment of inertia, we express the differential mass element 𝑑𝑚 in terms of 𝑑𝑟 in some way, depending on the shape of the mass we want to solve for. Beyond calculating an object’s moment of inertia from scratch, there’s one more moment of inertia tool we’ll want to learn before getting some practice with these ideas.

This tool is known as the parallel axis theorem. Here’s one way to think about this theorem. Say that we have a solid sphere that’s rotating about an axis that runs through its center. We saw earlier that the moment of inertia of this shape rotating about its center of mass is two-fifths its mass times its radius squared. And we noted that this result depends on the position of the axis of rotation. If that axis were to move, then so would the overall moment of inertia of this mass.

Let’s say, for instance, that we move our axis of rotation from the center of the sphere to some other point. And we’ll say that that point is separated from the center of the sphere by a distance 𝑑. Here’s what the parallel axis theorem tells us about this situation. This theorem says that if we move our rotation axis to another position but keep the axis parallel to its original orientation. Then the new moment of inertia of a rotating object is equal to the sum of the moment of inertia of the object rotating about its center of mass plus its mass times the distance squared between the original and final axes of rotation.

That’s a mouthful. But what it means is that we can take advantage of the moment of inertia of an object rotating about its center of mass when solving for its moment of inertia rotating about some other axis. This tool is very helpful in solving for moment of inertia when our rotation axis shifts. And it’s important to know that, to use it, our two axes must be parallel to one another. Knowing all this, let’s get some practice with moment of inertia through an example.

A rod and a sphere are combined to form a system. The rod’s length 𝐿 is 0.50 meters. And its mass is 2.0 kilograms. The sphere’s radius 𝑅 is 20.0 centimeters. And its mass is 1.0 kilograms. The system can either rotate about the point 𝐴 at the opposite end of the rod to the sphere or about the point 𝐵 where the rod and the sphere connect, as shown in the diagram. Find the moment of inertia of the system about the point 𝐴. Find the moment of inertia of the system about the point 𝐵.

We can call these two values 𝐼 sub 𝐴 and 𝐼 sub 𝐵. Considering the information we’ve been given that we’re told the length of the rod, its mass, the radius of the sphere, and its mass, we can begin solving for 𝐼 sub 𝐴 by writing out this moment of inertia as the sum of the components of this system of the rod and the sphere. 𝐼 sub 𝐴 is equal to the moment of inertia of the rod rotating about point 𝐴 plus the moment of inertia of the sphere rotating about the same point.

When we consider looking up the moments of inertia of the two parts of our system, we know we’ll be able to find the moment of inertia of a rod rotating about its end as we have in this case. But for the moment of inertia of a sphere, we’ll be able to find that value for a sphere rotating about its center. But in this case, the sphere does not rotate about its center, but about the point 𝐴.

To help us out, we can recall the parallel axis theorem. This theorem says that if we have a mass 𝑚 with a rotation axis shifted a distance 𝑑 from the center of mass of 𝑚, then the overall moment of inertia of this object is equal to the moment of inertia of the object about its center of mass plus its mass times the distance 𝑑 squared between the two parallel axes. This means that when it comes to the moment of inertia of the sphere rotating about point 𝐴, that moment of inertia is equal to the moment of inertia of the sphere about its center of mass plus its mass times the distance from the axis of rotation, in our case 𝐿 plus 𝑅, squared.

If we then go and look up in a table the moment of inertia of a rod rotating about one of its ends, we see that it’s one-third the mass of the rod times the length of the rod squared. Moreover, when we look up the moment of inertia of a sphere rotating about its center, we see that value is equal to two-fifths the sphere’s mass times its radius squared. All this means that we can rewrite the moment of inertia of our system rotating about the point 𝐴 as one-third the mass of the rod times its length squared plus two-fifths the mass of the sphere times its radius squared plus, because of the parallel axis theorem, the mass of the sphere multiplied by the length of the rod plus the radius of the sphere quantity squared.

Since we’re given the values for all four of these variables in our problem statement, we’re ready to plug in and solve for 𝐼 sub 𝐴. When we plug in for all these values, we’re careful to convert the radius of our sphere from units of centimeters to units of meters to agree with the units in the rest of our expression. Adding these three terms together, we find a result, to two significant figures, of 0.67 kilograms meter squared. That’s the moment of inertia of this system rotating about point 𝐴.

Next, we want to consider the same system, but a different rotation axis. Now our rotation axis is at the place where the rod and the sphere join. Once again, the moment of inertia of our system is equal to that of the rod plus that of the sphere for this particular rotation axis. Since the rod is, once again, rotating about one of its ends, we can again use the relationship for its moment of inertia about such an axis. Likewise, the sphere is not rotating about its center of mass, whose moment of inertia we know, but is rotating about an axis a distance of its radius 𝑅 away from that center.

Our equation for 𝐼 sub 𝐵 is the same as our equation for 𝐼 sub 𝐴 was, except for one single term. Instead of our distance 𝑑 being 𝐿 plus 𝑅, now it’s simply 𝑅, the radius of our sphere. When we plug in for these values, again converting the radius of our sphere to units of meters, we find that 𝐼 sub 𝐵, to two significant figures, is 0.22 kilograms meter squared. That’s the moment of inertia of this system rotating about an axis through point 𝐵.

Let’s summarize what we’ve learned about moment of inertia. We’ve seen that an object’s moment of inertia is equivalent to its rotational mass. Moment of inertia, symbolized by capital 𝐼, depends on an object’s mass, shape, and the axis that the object rotates around. Common moments of inertia can be looked up in a table. Or they can be calculated using the relationship the moment of inertia is equal to the integral of each infinitesimal mass element that makes up a larger mass multiplied by each element’s distance from the axis of rotation squared.