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Determine the moment of inertia and the radius of gyration of the shaded area with respect to the $x$ axis.

Physics 101 Mechanics

Chapter 9

Distributed Forces: Moments of Inertia

Section 2

Parallel-Axis Theorem and Composite Areas

Moment, Impulse, and Collisions

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Washington

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who asked to determine the moment of the second area moment of the and the radius of gyration of the shaded area. With respect to the X axis, Mr are shaded area looks something like like what I've sketched here and what we're gonna do is break it up into into three parts. So the first part, it will be the center beam here or the center section. So this part here and then to would be this upper flan and three will be this lower flan so we can base, weaken, take, find the area moments for each section and then add them up to find for the total. And we want to do it about, um, this X axis here, which is in the center of this vertical member. So the, um, formula for the area moment of a rectangular shape about its central aid, which this is that, um, is 1/3 be one h one cube. So be one. Is this thickness And H is this distance here, and we're giving those in the problem. And those there they are. And now for two, we can take the area moment about its center of mass. All right. It's, um, sense. Right? And then we need to shift it using the parallel axis deer. I'm from here to here. And so that distance ties. So then we need the area of this, which is B to H two and then this distance here. Um, and, um, that's deep. Well, call that D to so we know, we know a, um that's totally And so here's the values for, um, for here. And then we have the third part, which is this part here. And so we have the center. We have the center, the area moment about the central right, and then plus the area times this distance from the central to our X axis. I call that the three. And these were all given in the problem. Calculate the total area of this cross section and that we get 8.16 centimeters squared. Um, and then we can calculate the area moment about X, and we just need to add all these things up. So this is I x one I x two and I x three. And they seem reasonable because, you know, this is gonna pretty high area moment about excuse. It's tall this one because it's far away from the X axis and this one is bigger than this one and because it's much wider. So that looks reasonable. And if we had everything up, we get 39.0 centimeters to the fourth. And then if we want to find the radius of gyration, we just need to take the square root of this divided by this and we get 2.19 centimeters.

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