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Determine the moments of inertia of the shaded area shown with respect to the $x$ and $y$ axes.

Physics 101 Mechanics

Chapter 9

Distributed Forces: Moments of Inertia

Section 2

Parallel-Axis Theorem and Composite Areas

Moment, Impulse, and Collisions

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

University of Winnipeg

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.

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Determine the moments of i…

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Determine the moment of in…

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Determine the moment of i…

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in this problem were asked to determine the polar area moment of the shaded region That is a square with some half circular holes cut in it. Um, and we want to find their polar area moment about the X and Y axes, which are, or have an origin down here. So in this problem, what we're gonna do is we're going to, um So it's easy to find the polar area moment or the, um, the area moment, um, of the square about this point. So we'll do that. And then we'll find the area moments of thes to cut outs about this point and subtract those area moments from the area moment here. So the area moment of this guy since it is hazard width and height of four a from in its corner is 1/3 for a to the fourth power or 256 over three a square. And that's the same for X and Y. Um, obviously their symmetry about that, um, access. And we can now get to the more difficult part finding the area moments of these cut outs about this point. So we found in the previous problem we found the area moment of 1/2 circle over a semicircle about it centrally. And that was pi over eight, minus eight over nine pi times a to the fourth thes have a radius of a And why about the centrally is just pi over 88 of the fourth. Now we can figure out the area moment with respect to this point by using the parallel access Theo. And so we have, um, the area moment about the central eight of this run times It's area. Oh, plus, it's area times the distance from this ax access to the x axis of the central square. And that is its three halves A up to here and then minus 4/3 pie down to the centrally. So that's the distance squared there to this century. And then, for why? It's a little simpler. Um oh, we can, um, plug this in and then do some biology operetta and simplified this and we get for a square eight of the fourth over four times eight plus five pi is the, um, area moment about this x axis of this region here. No, why is a little easier? Because, um, the central little moments is simpler, and then the distance here is his easier. And that is just to a so that I'll become a 17 pi over a to the fourth. And now, if we look at the upper hemisphere, um, semicircle, we get again. It's it's area moment about the X axis is its area about moment, about this and title X axis. Um and it's that that's the same as for this one here, since these are have the same shape just relocated 180 degrees and then the distance than the area times the distance squared. I think I forgot a square here. The distance squared that distance. It was five. Five Have a up to here and then we add another 4/3 pi A to that. And so we get the, um, area moment about the EC, the X axis that we're looking for. Aziz this and then we can see that the area moment about the Y axis is the same as for this one, because these guys, these two are just rotated and flipped. I mean, flipped and translated about the Y axis. So they're gonna be have the same area moment about Why, No, We can figure out the area moments about the net shape by taking the square minus the semi circles. And if we do that, that becomes a to the fourth over to times a quantity 160 minus nine pie. Or about UM, 65.868 of the fourth, and likewise for why we can take the area moment with his rectal y of square and minus that for the semi circles. And those two are equal, so we can just want to play that by two. And that gives us eight of the fourth over 12 times 1024 minus 51 pie or 71.988 to the fourth.

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