Welcome to Calculus. I'm Professor Greist, and we're about to begin Lecture 41, bonus material. One of the properties of moments of inertia that you're going to find very helpful when doing problems is that of additivity. This comes from the fact that moments are integrals. And you can break them up into pieces. For concreteness, let's consider a, an example, a simple H shape with the dimensions being given as shown, in terms of constants, a and b. Now, compare what the moment of inertia of that shape is going to be to the same shape, rotated by 90 degrees. Let's consider a horizontal axis that passes through the center of each shape, an H shape, and an I shape. Well, let's begin with the H. What is the moment of inertia? We can decompose this shape or region into rectangles. And we know the moment of inertia of a rectangle about a horizontal axis by computing the dimensions of this rectangle. The height is going to be a. The width is going to be a minus b over 2. Then we can get the moment of inertia of this rectangle as 1 12th a cubed times a minus b over 2. And of course there's a corresponding rectangle on the right that is going to have the same moment of inertia. We add it to what we computed before and then we have one more term to add the small square in the middle of dimensions b by b. That contributes 1/12th b cubed times b. When we add all of these together and do a little bit of algebraic simplification, we obtain moment of inertia of 1/12th quantity a to the 4th plus b to the 4th minus a cubed. Now, on the right, we can't do exactly the same thing with rectangles about that axis. But, look what we can do, we can compute the moment of inertia of the outer square that bounds this I. That's 1/12th, a cubed times a, and then we can subtract off the moments of inertia of the holes that we would cut through on the left and on the right. And again we know the dimensions of these rectangles to be b in height and a minus b over 2 in width. When we combine these terms, we obtain a moment of inertia that is similar in form, but slightly different. I is equal to 1 twelfth, a to the fourth plus b to the fourth, minus ab cubed. Now, how might this be useful? Well, remember we once talked about the differential equation associated with the deflection of a beam. And one of the coefficients in that equation was I, the moment of inertia of the cross section of the beam about a horizontal axis. Now, if you had two different choices for your beams, an H beam or an I beam, well, which one is best? Perhaps you should look around you and see which choices get used in practice. But from our computation we can see which one has the more preferable moment of inertia because these two terms differ only at the very end. Minus ab cubed b for the h beam, and minus ab cubed for the I beam. We know that a is bigger than b. There is one more result that is going to be extremely helpful in doing computations,this is called the parallel axis theorem. We begin with the domain and consider the moment of inertia about an axis that passes through the center of mass. This moment i knot is the integral of r squared dm, where r is the distance to this center axis. Now consider what happens when you choose an axis that is parallel to this center of mass axis but some distance D, away. What is the moment of inertia about this axis? Well, it's clear that it's more, but by how much? Well, if we compute this moment, call it I sub d, this is the integral of the distance to this axis squared dM. What is that distance? Well, we could write this as R plus d where r is again the distance from a center of mass axis. Now let's take that integrand and expand it out. What do we get? We get three integrals with respect dm. The first is the integral of r squared dm... That is, of course, our moment of inertia about the center of mass. What about that second term? We're integrating Rdm. Now remember our definition of the center of mass. This is precisely where that comes from, and in our coordinates system the center of mass has r value equal to 0. Therefore this second integral vanishes. R balances because we're passing through that center of mass the last term... Is, simply, the constant D squared times the integral of DM. The integral of DM is M, and so we obtain that I sub D is I not plus D squared M. This is the content of the Parallel Axis Theorem, and allows for a simple computation of some moments. For example, let's consider the moment of a flat, round disc of radius R, about axis, located some distance D, away from the center. What would that be? Well, we know the moment of inertia. Passing through the center of the disk. That's one quarter MR squared. By the parallel axis theorem, the moment of inertia about this parallel axis a distance D away is one quarter MR squared plus MD squared. You could find this result useful in many different contexts. For example, you could go back to the I beam result and compute the moment of inertia based on translated rectangles. I'll leave it to you to practice through some of the homework problems and see where you might make use of the parallel axis theorem and of additivity.