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Prove that the centroidal polar moment of inertia of a given area $A$ cannot be smaller than $A^{2} / 2 \pi$. (Hint: Compare the moment of inertia of the given area with the moment of inertia of a circle that has the same area and the same centroid.)

Physics 101 Mechanics

Chapter 9

Distributed Forces: Moments of Inertia

Section 1

Moments of Inertia of Areas

Moment, Impulse, and Collisions

University of Michigan - Ann Arbor

University of Washington

Hope College

Lectures

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

03:30

In physics, impulse is the integral of a force, F, over the time interval, t, for which it acts. Given a force, F, applied for a time, t, the resulting change in momentum, p, is equal to the impulse, I. Impulse applied to a mass, m, is also equal to the change in the object's kinetic energy, T, as a result of the force acting on it.

03:05

Determine the polar moment…

who asked to prove that the central area is insert Central Polar area Moment of inertia, of one out of inertia of a given area A cannot be smaller than a squared all over two pi. This is actually a really difficult problem. And if we depending on what you mean, what they mean by prove, um, you probably don't have the math background to actually prove it, But, um and the answer is that I've seen in this problem all basically use a lot of hand waving. So I will, um I will try toe, show you one way, convince a couple ways to convince yourself and then actually the way toe solve it properly. Um, So what we have here is we have the different the change in the polar area moment of, um, this This blob here is unknown, um, area and that is given by this expression here. If our is a function of data which actually is an assumption because we could have a blob where are is not a function of data. If it came back in on itself like that, then we have to go to some other methods. But this isn't a completely general case. But if are as a function of theater, that's basically that for any ray coming out of here. We only crossed the across the surface border once. Then we have this express expression for the difference. Okay, so then we can say, Well, the, um the areas have to be the same. So this is the area of, uh, this unknown area are unknown region, and this is the area of the circle. Is pi r squared? So we can say that. Okay, r squared. Then we can get our to the fourth, which we need up here. So are the fourth is one over four pi squared times this integral here squared again. An integral of a function square. Um, over some finite region is going to be positive. That square this is gonna be positive. So I really get a positive value for our here. So that would be the the Radius if we wanted to calculate what the radius is of the circle. We could use this formula here if we knew what our they'd awas. But what we can do is we can put, um, used this in here and what we see What we get is 1/4 the integral from 0 to 2 pi over our data to the fourth power D theta minus 1/8 pi times this in ago 0 to 2 pi our data squared B thing squared and there's actually a a proof and there's actually approving mathematics that you can show that this is in fact, bigger than this. So if you integrate the fourth power of a function, um, it's gonna be bigger than, um, again 1/2 pi times the integral of this function and then squaring it, okay. And I obviously one thing that you don't can't do we can't just pull this square into here and say that, um, one over a pie is less than 1/4 with these things being equal because obviously wouldn't take a function and integrate, take the square of a function and integrated and then take that Value Square is not the same as taking the fourth power of the function and integrating it. So that's kind of a hand wavy way too to do it. And you can check for yourself, plug in some various functions in the here and see that this is always true. Another way of doing it a little bit different is to assume that, um that our data is the mean radius. Um, plus some small thing times. Um, some traded up difference function dealt the data. If we do that, what are area constraint Says if we say that the area of this read that's founded by this region must be, um pi r m squared. We get this formula, which is that the in agro rm times the in agro of Delta? I think so. Um, plus 1/2 absalon times in ago of dealt that they squared equals zero. Or we can get a relationship between the inner girls of what these functions are. If we're given, um, forgiven our end, then we have a relationship between what the integral of these functions can be. Note that this thing is always positive. If Absalon is positive that this thing must be negative because this is positive. So we can. What I did here is I, instead of writing these inner grows out all the time. I called this one I one So I want is the integral of delta over from 0 to 2 pi I, too was integral. Felt a square from 0 to 2 pi. So we get this relationship here that I won must be equal to minus Epsilon over to r m. I to now if we expand out the if we plug this into our formula for the polar area moment and expanded out in powers of Epsilon. Okay, so we just have this thing to the fourth, and we can do the math to expand that we get. This is two pi pi over to R M squared over to the fourth. This is also equal to, um, a square all over two pi if you use the relationship between A and R M. And so then we have this term here where we have this integral. And then we have this term where we have the integral of delta squared. Then we have an integral of Delta Cube and an integral of guilt to the fourth. I. So what we can do here is, um, is take this. Um, let me actually make this a little cleaner here. This is our epsilon to the third power. But we know from our area constraint that that has to be true. So we plugged that into here. And what we see is that the polar area moment is the area Moment of circle with me and radius with me, Green ladies, RM plus sometimes here and notice that this term has to be positive because the integral of a function squared over some finite rain is positive because this is always positive. So if you integrate it, you're going to sum up a bunch of positive numbers. So this is positive. This is positive. This is positive. So to leaving order when Epsilon is small, um, we get the radius, um, the polar area moment of the circle with the radius R M plus some positive number. So that means that this this has to be bigger than this now, as absolute gets bigger and figure that we could say, Well, what happens, this could be negative. Here. This has to be positive. Well, this will always, um even if Absalon is very is large bigger than one. This thing is always gonna wind up being positive also. So we have that we have our base polar area moment, plus some positive number Now to actually prove it, you need to use a thing called calculus of variations, which you probably are not aware of, and the calculus of variations. Basically, it takes what's called functional that's actually do a little thing here. So a functional a lot of times will be noted. Um, but I say a script I and it is a functional is a is a function of a function. So we say we put a function in here so it takes a function in a spits out of value. And you've seen this ever since you started doing calculus. Because ever any definite integral is a functional because it takes you, give me a function and I would give you a value. And you give me another function. I'll give you another value, another function on other value. So a definite integral is just a functional. It takes a function and spits out of value. The difference between a function and a functional is a function takes value in, spits out another value a functional, takes a function and spits out another put spits out of value. So what we have here is the area. The polar area moment is a functional off. Are you? Give me a function are and I can tell you is a function of data. And I can tell you what um, the polar area moment is Give me another function. I'll tell you what it is. So give me your phone. Give me your function. I'll give you a number. And so what you want to do is we want to find our are here that minimizes what number like it. So you keep giving me functions. I keep giving you numbers and you keep giving me functions. And finally, you find out over this one's got to be minimum here. Okay. So many moments here. And what function is that? What function minimizes this integral now way, Obviously. What? One thing we could say. Well, if we don't have any constraints that well, um, this thing well could be zero R zero. Well, that that minimizes everything because that gives us a zero here. But what we need is a constraint. And that constraint is is that the area has to be for a fixed area. So we have a constraint that the area is a not and that equals then ago, from of RDR data. So that's our constraints. We have tohave give me an area. Then I confined. Give me an area and then let me find an R that has that a region surrounded by that pie are that that has that area that then minimizes to pull the area and the calculus of variations But we can do is when we have a constraint here we can put on the ground. You multiplier here, um, and add the constraint. Okay, so we know that, um, now, now we have an unconstrained minimization problem since we added our constraint and did the minimization problem. Um, and we know this. So now we need to find a function that minimizes our Q minus. Lambda are over it. The air of the region, um, surrounded by our and Lambda isn't unknown. So what? What we can do, though, is the calculus of variations basically tells us how to minimize this thing, and it turns out that it looks if you have a simple functions like this, it looks a lot like calculus. So we want to minimize this functional I got called. The f of our overall functions are and we do that uses a little delta simple here to say the variation of the functional R was zero s. So it's like taking the setting the derivative of a function zero to find an extreme. And the variation of dysfunctional actually is is zero only when, um, the derivative of this with respect are is zero. So that says that we need that are equals three R squared, minus dealt by. The slander must be zero. Okay, so this whole step here is all about calculus of variations. Then you could take a class of math class on that, probably in grad school. Um, but what this tells us is that r squared must Atlanta's a constant so r squared must be a constant, So our must be a constant. So if we have given the area okay, we can find the radius. Okay, The circle enclosing that area. And then we know that they're polar area moment. Is that so? We know that basically given an area we confined in our and then that are that we confined our that have a circle that encloses that so are is a constant and our is a constant here, so we know that it's actually given an area that we surrounded by a circle given up on it. Value of our area area. If we surrounded by a circle than that has the minimum polar radius of gyration polar area Moment, um, of any of any region with that area.

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