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Determine the moment of inertia and the radius of gyration of the shaded 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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were asked to find the second moment of area in the radius of gyration of the shaded area shown with respect to the X axis. And that area is bounded by why want equal see, signed K X and Y two equals M X plus B. We know we have four conditions on these functions. That why one at X equals A S H by two at X equals A is 24. Why one and why two at two. A. Or both. Zero. So that tells us we can get these for Constance here. Um, in terms of a terms of h. So we get that M equals minus two h over a B equals for h C equals eight and K equals pi over two A. To get the second moment of area about the X axis we need to integrate. Why squared over the area of interest? And that area goes from why was why one toe y two and X equals a to a and we can crank through these ugly in a girls. And when we do that, we get that ice of X equals to a cube all over nine pi times three pi minus two. If we calculate that numerically we get 0.525 a cube, we can find the area by integrating one over the area of interest. And that area is a H times one minus two over pie. And then the radius situation is just the square root of ice of X all over a and that is K X, and that gives us a value of zero of 1.202 times h.
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