 So now that you've seen how to calculate the center of a body to approximate a continuum body, a rigid body by a single point, as well as calculating the second moment of area, also known as the area moment of inertia, to capture the bending resistance of a beam when we reduce it down to a line, we still need to cover one more property. And that's the mass moment of inertia which relates to the rotational resistance of a body. And in order to do that, we're going to do a small experiment with two cylinders here, a solid cylinder and a hollow cylinder that are the same length and have the overall same mass. As you see here, the mass is off by a little over a gram, but in terms of the overall mass of the two cylinders, that's practically negligible. So let's look at what would happen if we take these two cylinders and we approximate them as points, which is one of the things we do often in statics, we reduce down a body to a point or a line. So if I take these two cylinders that have the overall length here L and reduce it down to a point, what we would see is that the two would have the exact same center of mass, right? Because it's a constant cross-section of an overall length L for both of these, it's going to be right in the middle at L over 2 from the one edge which was located A from our axis. So the location of the center of mass for both of these would be A plus L over 2. So they would have the same center of mass, same centroid and center of gravity because they're all equivalent with it being the same material. But are these two cylinders, are they actually going to behave the same and specifically in terms of rotational resistance? So in order to do this, in order to look at this, let's do a small experiment. We'll put these two cylinders on an inclined plane and see what happens. So here you can see we set up the two cylinders on an inclined plane with a bar preventing them from rolling down. And when I lift that bar, the two cylinders will roll down the same inclined plane. Now remember they effectively have the same mass, but as you can see there's a very big difference in how they roll down. The solid cylinder seems to have less resistance to rolling and rolls down that inclined plane quite a bit faster than the hollow cylinder. So what's going on here? In order to explain what's going on here it's good to think about this in terms of energy. So when the cylinders were at the top of the inclined plane they both had the same potential energy. Because they were both at the same height, they both have the same mass. And then that potential energy is converted into kinetic energy. And for a sliding mass, for just a mass that's translating we can calculate the kinetic energy using the familiar formula one-half mass times velocity squared. So that's our equation for kinetic energy for a translating body. But our two cylinders are also rolling and that rolling has a different motion to it. So if we consider a cylinder here, and in this case it is a hollow cylinder it will be rotating with an angular velocity omega. And that rotating body also has kinetic energy. So we can look at that by taking a small infinitesimal slice of this cylinder material and it will have a mass dm. And if you think about the instantaneous velocity of that mass it is going to be translating tangent to the cylinder with a velocity that is equal to omega times r. The radius times the angular velocity will give us the instantaneous tangential velocity. So that small element dm will have kinetic energy associated with it. And then if we integrate all the way around the cylinder we will get all of those dms times their velocity squared which is omega r. So omega r all squared in brackets as shown here times one half to have the exact same sort of format of the formula there. So it's still one half mass times velocity squared but the velocity is omega r squared, all squared. Or the velocity is omega r sorry and then the velocity is squared. For this the angular velocity at any given point is the same So the fact that we took the angular velocity out of the integral doesn't mean the angular velocity doesn't change over time. It just means that at this snapshot in time the angular velocity is the same for all elements around the cylinder. So we can take it out of the integral and we get one half the integral of r squared dm times omega squared. And if you look at this the integral of r squared dm that's actually similar to the integral of y squared dA. It's looking at the distribution of area, well in this case mass which would be area times density or volume times density. So they're kind of similar and in fact we give them a similar name. We call it the mass moment of inertia rather than the area moment of inertia. And this is the mass moment of inertia about the axis A coming in and out of the page. Now the confusing thing is that we denote the mass moment of inertia exactly the same way we denote moment of inertia for a beam. It's I with a subscript of its axis. I know that's confusing, it's just what we do. But you can really tell the difference when it comes down to the units. And again always give you units in your results and values because mass moment of inertia will have distance squared times mass. So it's going to have the units of kilogram meter squared where the moment of inertia of a beam will have distance to the power of 4 because it's y squared dA so it will have meters to the power of 4. Now this mass moment of inertia shows up a lot in dynamics of rotational bodies. So it's a very important property. It's the resistance to rotational motion just like moment of inertia is a resistance to bending in a beam. You might sometimes hear it referred to as something else and that is as rotational mass. And the reason it is often referred to as rotational mass is if you look at, if I reduce this equation I get one half mass moment of inertia times omega squared. Well that looks very similar to one half mv squared where omega is the angular velocity. So the mass moment of inertia takes the place of mass in that kinetic energy, rotational kinetic energy formula. So let's summarize this. The mass moment of inertia about axis in this case for our generalized engineering potato will have axis O. So the mass moment of inertia about axis O is the integral over the entire mass domain of r squared dm. So not too different than the moment of inertia we've seen. It's just with respect to mass and its radial position. So it goes all the way around the axis. It's not because it's a volume, it's not an area. So there is a slight difference there. It has the units of kilogram times meter squared as I said. And the nice thing is that actually that parallel axis theorem we've been using for moments of inertia of beams still works for a mass moment of inertia. So if we have a parallel axis P and that original axis O is our centroidal axis. So an axis passing through the center of gravity because the parallel axis theorem only helps you translate from an axis going through the centroid to a parallel axis. So we would get the mass moment of inertia about axis P is the mass moment of inertia about the centroidal axis plus mass times distance squared. So the only difference here is instead of it being area for the area moment of inertia, it's mass for the mass moment of inertia. So we can use this to conveniently transform a mass moment of inertia to a parallel axis.