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Determine the moment of inertia and the radius of gyration of theshaded area shown with respect to the $x$ axis.
Physics 101 Mechanics
Chapter 9
Distributed Forces: Moments of Inertia
Section 1
Moments of Inertia of Areas
Moment, Impulse, and Collisions
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this problem has to calculate the second moment of area and the readers of gyration of the shaded area expect to the X or the Y axis. And our area is bounded by why equal see, signed K X and Y equals MX plus B Now where we know a few things about, um, thes two functions that, uh a, um why one is h and why to is 24 and at two A they are both zero. So that gives this kind of four equations before knowns and we can crank through the algebra and find out that a R M equals minus two h over a b equals for h c equals H and K equals pi all over two a. Now we could do the inter girls. So our area for isil why we need to integrate X squared over that area and that area goes from white was why one toe white to these two expressions here and X goes from a to to a Now we can plug into everything and start cranking through the calculus and it gets pretty ugly but weaken Do it theoretically. And what we get for our answer is this ugly expression. You get a cube H all over six times a quantity. 11 plus I'm 48 times two plus pi minus pi squared all over Pie Cube. And obviously we can calculate this all numerically. And if we do that, we get that I supply equals 0.613 a cube age now were asked to find the radius of gyration. We could go through and calculate find the area by integrating one over the area. And we can crank through that calculus and we get that A equals the area equals any times each times one minus two over pie. And if we take, find the radius information, which is the square root of this divided by this, we find that that is 1.299 a.
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