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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

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

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who asked to determine the second moment of area and the redistrict freight of generation of the shaded area expected the X axis. So the shaded area is bounded by K one X squared and que two squared of axe. We know that at X equals a. They both equal be pan, so you can get K one and K two in terms of A and B. The integral we need to do to get the second moment of area about the X axis is to integrate y squared over the area, and the area goes from y equals y one to y two on X course from zero to a you can crank through that and we get that I set backs equals 3 35th a be cute. That should be a cute and we get that the area. If we instead of putting, why here we just put one so that will give us the area and we get that the area is a B over three, and then the release gyration about the X axis is square root of I Cybex all over a and that gives us three B all over a squared or 35

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