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