In this case we need to divide the disc up into elements parallel to the axis in

rotation. The inertia element is R squared dM.

The mass element is, of course, row the density times the area element, dA.

Now, using the fact that our distance to the axis is, in fact, x we have an

integrand of X squared times row, times twice the square root of capital R

squared, minus X squared dx. When we integrate this to get the moment

of inertia, we can simplify things a little bit by integrating from 0 to

capital R, and then multiplying by 2, taking advantage of symmetry.

Even so, this integral looks involved. We're going to need to try, at least, a

trygonemetric substitution. Substituting x equals capital R sine

theta will allow us to clear out that square root and obtain the integral of x

squared, and that's R squared sine squared theta, times what happens to the

square root that gives us an R cosine theta * dx, which is going to give

another r cosine theta. The resulting integral has even powers of

sine and cosine and would take a little bit of thinking in order to work through.

Instead of doing that explicitly, let me claim that the answer is not so bad.

It works out to One fourth row times pie times capital r to the fourth.

We can simplify that a little bit. Taking advantage of the fact that the

mass is pie times r squared times row. And that gives us a moment of inertia of

one fourth m r squared. Well to prove that result that we didn't

justify, let's change perspectives a little bit, and compute the moment of

inertia when this disc is rotated about the center.

About the z axis, if you will. In this case, we need to divide into area

elements parallel to where at a fixed distance from the axis of rotation.

Let's call that radial coordinate, R. Then, the inertia element is R squared.

Time to row times the area element, in this case 2 pi r dr help this integral.

It's not going to be so bad, we have to integrate r cubed dr as r goes from zero

to capital 'R' And that of course gives us an answer of pi row over 2 times

capital r to the fourth, and substituting in the mass we obtain one half m, r

squared. That was much simpler but not unrelated

to the problem of rotating about a vertical axis.

Think of it this way: The moment of inertia is the integral of R squared d M

in the x y plane. As we have set thing s up, R squared is

equal to x squared. Plus y squared.

If we distribute that integral over addition and consider each piece, well we

recognize something. The integral of x squared d M really gave

us the moment of inertia of that disc rotated about the vertical Access.

What would the integral of Y squared D M give us?

Well you shouldn't be surprised to see that that's really the moment of inertia

rotated about a horizontal access. But because this is a symmetric domain

these 2 of course equal to one and another.

Therefore we have Given the fact that the moment of inertia about the center is

about 1/2MR squared. We can conclude that the moment of

inertia bout the vertical or horizontal axis is half of that, as we saw in our

previous slide. Let's consider a different, simpler sort

of object. A rectangle.

Let's say, of length, l, and height, h. We're going to rotate that, about,uh,

vertical or horizontal axis. Which do you think, would have, the

greater moment of inertia? Well, let's compute both.

And find out. In this case we need to compute the

integral of r squared, row dA, whereas before row is the density.

In the case of rotation, about a vertical axis, our area element Is that obtained

at some distance X to the Y axis. And it is a vertical strip.

We need to integrate R squared row D A, that is X squared times row Times h dx.

The limits are x goes from negative l over 2 to l over 2.

That's a simple integral. X squared integrates to x cubed over 3,

and evaluating at the limits we get one twelfth row hl cubed...

It's sensible to substitute in the mass, that is rho times l times h, yielding an

inertia of 1 12th m l squared. I'm going to leave it to you to do the

same in the case of rotation about a horizontal axis, where we wind up

following the exact same procedure. But exchanging L for H, and vice versa.

This yields an inertia of 1 twelfth and H squared.

We're going to take a moment and consider a related physical concept.

That of the radius of gyration. The radius of gyration is the answer to

the question. If all of the mass of the object were

focused at a single point, how far from the access would it need to be?

In order to give the same moment of inertia.

Now, you've felt the radius of gyration before if you've ever hit an object with

a bat. If you get it at just the right spot, it

feels Right. That is the radius of gyration.

We denote it capital R sub g, and it satisfies the equation that I equals M,

the entire mass, times R sub g squared. Taking that equation and solving for R

sub g. You obtain a formula of the square root

of I over M. That's simple enough but think about what

we're doing. I is really the integral of little r

squared dM. M is really the integral of 1 dM.

And so, in this ratio. We're really looking at the average of

little r squared, taking the square root of that, gives us the root mean square of

the distance little r to the axis. That is really what this radius of

gyration means it a root mean square of distance to the axis.

Now it's perhaps worth taking a, moment of two and thinking about what this

radius of gyration means and what different examples have.

If you look at a tennis racket or varies kinds of sports objects In fact, if you

go through your house and just pull out random items that you have lying around,

you can physically measure. The radius of gyration and start to get a

feel for how different objects have their masses distributed in different.

Weighs. This entire lesson concerns the question

of how mass is distributed over an object.

If we consider a slightly simpler setting where the object is one dimensional, We

split it up into mass elements based on this coordinate, X.

Then, for irregular-shaped objects, the mass might be concentrated at different

places. The moment of inertia is one way to get

at that. Integrating X squared dM leads us to the

moment of inertia. Or, after normalizing and taking a square

root, the radius of gyration. These are both measures or ways to

characterize. How the mass is distributed.

Sometimes an interval of this form is called the second mass moment.

Well anything that has the name second mass moment would lead you to wonder.

What is the first mass moment? It is, as you might guess, not the

integral of x squared dM but the integral of x dM.

This, too, tells you something about how mass is distributed across the object,

but in this case, it leads To the centroid when properly normalized.

And so both of these integrals are giving you different physical properties about

how mass is distributed. There are other moments as well, all of

which answer that same question. The higher mass moments are integrals of

the form x to the n D m and you may well wonder what sorts of physical properties

do these integrals give. Well, we're not going to have time to

answer that now. But I will say that there is one more

mass moment that you know. That is the zero mass moment.

With the integral of x to the zero dm. That of course is just the integral of

dm, which is m, the mass. This concludes our treatment of solid

bodies through integrals, of masses, middles and moments.

In our next lesson, we'll start learning about probabilities.

It won't be as discontinuous a jump as you might think.

So stay tuned.