>> Welcome to Calculus. 
I'm Professor Ghrist. We're about to begin lecture 41 on 
moments and gyrations. >> Centers of mass are wonderfully 
visiral /g. You can just feel it when you balance 
something right at its center. There are other properties of solid 
bodies that integrals help explain. In this lesson, we will consider moments 
of inertia, a measure of resistance to rotation about an axis. 
You may recall from your previous exposure to physics the moment of 
inertia, denoted I. It is a a measure of resistance to 
rotation. About an axis, in this it's not unlike 
mass which is a measure of resistance to translation. 
Thinking of M, as the constant proportionality between force and 
acceleration. Allows us to consider a moment of inertia 
I as the constant of proportionality between torque and angular acceleration. 
Let's begin in the case where our object is a point mass. 
We choose some axis about which to rotate. 
What does that moment of inertia depend on? 
Well, it certainly depends on the mass. The heavier the object, the more 
difficult it is to rotate it. It also depends on the distance, R, to 
the axis. The longer the distance, the greater 
moment of inertia. In this setting of a point mass The 
moment of inertia is equaled to r squared times m. 
Now that's the simple case in the mass concentrated at a point. 
What happens when we have a more interesting object whose mass is 
distributed. Over some domain. 
Well, we're going to think in terms of calculus. 
What would the appropriate technique be? You will be surprised to see that the 
right way to proceed is to divide the object up into differential elements. 
And integrate. If we consider a mass element dM as that 
portion of the object, distance r to the axis then the moment of inertia element 
dI is given as rsquared dM. And so to compute, the moment of inertia, 
we integrate, that element. Integrating r squared dm. 
Now, a good physical understanding of moment of inertia, is crucial for 
interpreting our results. Consider a simple experiment, where we 
take three round objects Whose mass is distributed in different ways, and let 
them roll down an inclined plane. As they roll, they'll be rotating about 
their axis. The way that the mass is distributed 
influences the speed at which the gravitational force acts to cause 
rotation. An object whose mass is all distributed 
at the outside is going to have much greater moment of inertia than an object 
whose mass is distributed evenly over the entire disc. 
Let us compute the moment of inertia in the specific case of a disc, a flat disc, 
of radius. Are rotated about say, the y axis, 
centering this in the xy plane. 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.