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>> Welcome to Calculus. I'm Professor Ghrist.

Â We're about to begin lecture 41 on moments and gyrations.

Â >> Centers of mass are wonderfully visiral /g.

Â You can just feel it when you balance something right at its center.

Â There are other properties of solid bodies that integrals help explain.

Â In this lesson, we will consider moments of inertia, a measure of resistance to

Â rotation about an axis. You may recall from your previous

Â exposure to physics the moment of inertia, denoted I.

Â It is a a measure of resistance to rotation.

Â About an axis, in this it's not unlike mass which is a measure of resistance to

Â translation. Thinking of M, as the constant

Â proportionality between force and acceleration.

Â Allows us to consider a moment of inertia I as the constant of proportionality

Â between torque and angular acceleration. Let's begin in the case where our object

Â is a point mass. We choose some axis about which to

Â rotate. What does that moment of inertia depend

Â on? Well, it certainly depends on the mass.

Â The heavier the object, the more difficult it is to rotate it.

Â It also depends on the distance, R, to the axis.

Â The longer the distance, the greater moment of inertia.

Â In this setting of a point mass The moment of inertia is equaled to r squared

Â times m. Now that's the simple case in the mass

Â concentrated at a point. What happens when we have a more

Â interesting object whose mass is distributed.

Â Over some domain. Well, we're going to think in terms of

Â calculus. What would the appropriate technique be?

Â You will be surprised to see that the right way to proceed is to divide the

Â object up into differential elements. And integrate.

Â If we consider a mass element dM as that portion of the object, distance r to the

Â axis then the moment of inertia element dI is given as rsquared dM.

Â And so to compute, the moment of inertia, we integrate, that element.

Â Integrating r squared dm. Now, a good physical understanding of

Â moment of inertia, is crucial for interpreting our results.

Â Consider a simple experiment, where we take three round objects Whose mass is

Â distributed in different ways, and let them roll down an inclined plane.

Â As they roll, they'll be rotating about their axis.

Â The way that the mass is distributed influences the speed at which the

Â gravitational force acts to cause rotation.

Â An object whose mass is all distributed at the outside is going to have much

Â greater moment of inertia than an object whose mass is distributed evenly over the

Â entire disc. Let us compute the moment of inertia in

Â the specific case of a disc, a flat disc, of radius.

Â Are rotated about say, the y axis, centering this in the xy plane.

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

Â rotation. The inertia element is R squared dM.

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

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

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

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

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

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

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

Â trygonemetric substitution. Substituting x equals capital R sine

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

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

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

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

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

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

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

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

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

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

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

Â inertia when this disc is rotated about the center.

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

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

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

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

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

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

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

Â squared. That was much simpler but not unrelated

Â to the problem of rotating about a vertical axis.

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

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

Â equal to x squared. Plus y squared.

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

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

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

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

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

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

Â these 2 of course equal to one and another.

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

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

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

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

Â of object. A rectangle.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â This yields an inertia of 1 twelfth and H squared.

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

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

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

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

Â In order to give the same moment of inertia.

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

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

Â feels Right. That is the radius of gyration.

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

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

Â sub g. You obtain a formula of the square root

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

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

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

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

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

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

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

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

Â radius of gyration means and what different examples have.

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

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

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

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

Â Weighs. This entire lesson concerns the question

Â of how mass is distributed over an object.

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

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

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

Â places. The moment of inertia is one way to get

Â at that. Integrating X squared dM leads us to the

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

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

Â characterize. How the mass is distributed.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â bodies through integrals, of masses, middles and moments.

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

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

Â So stay tuned.

Â