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Determine the moments of inertia $\bar{I}_{x}$ and $\bar{I}_{y}$ of the area shown with respect to centroidal axes respectively parallel andperpendicular to side $A B .$

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

Simon Fraser University

Hope College

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.

05:14

Determine the moments of i…

04:57

05:42

he was asked to determine the area moments about the central oId um, I x and I y bar of the area shown with respect to the central axis. Parallel the B. Um, so these are the axes that I'm going to use initially to find an area moment. Um, and so these air parallel and parallel and perpendicular to the axes to the side. A B And so what we have here is kind of Ah, hybrid L C shape, beam, cross section. And I've listed off the various dimensions I used throughout the problem here. Um, then if we take, we look at this, uh, remember here. So we're going to divide it into three sections, Three rectangles. And look at this top rectangle. It's area moment. About the X axis that I've defined is 1/12 be one h one cubed. Plus, it's area times H two minus H 1/2. That's the distance from this point up to here and then back down to the central rate of that area and he plug in the numbers and get 18.5 inches to the fourth. Now for why, uh, we have again 1 12 H one B one Cube and then times the area and then plus B to a I plus a one times the quantity B two plus B 1/2 squared, and we get a value of 0.67 inches to the fourth. Which we would expect this to be much smaller than this, because this piece here is much farther away from the X accidents as it is from the y axis. No, If we take a look at our second member here, then that's that's it. Simple one, because we have the area moments. If we have the origin at the corner and that is 1/3 to be to H two tube or 24.8 inches to the fourth, and I why is 1/3 age to be to to open that is 0.22 inches to the fourth and again would expect this to be much bigger than this, because this has much more area away from the X axis than it does from the Y axis. Now we have, um, the third piece, this bottom leg here and about X. It is 1 12 to be three h three cube closets area times H three this distance here, um, over to the distance from the X axis here to the central. And we get zero point, um, 1 to 9 inches to the fourth. And that's again a very small number, as we expect, because there's not much area away from this axis. Now. We need, um, uh, the area moment about the Y axis for that piece. And that is 1 12 83 B three Cube Plus a three times this year we got be to plus B 3/2 to get us to the century there, um, squared. And we get 7.76 interests of the fourth, and that makes sense. It's not huge, but again, there's a lot of area away from the Y axis here. Now we can get the total, um, area moments about X. The extradite defying is 43.44 inches of the fourth, and about why is 8.67667 inches to four. The next thing we need to do is find the centrally and the coordinate system that I defined, and that's the total area times the X centrally, the distance to the centrally in the X direction. And then we got to add up the Pete, all of the pieces. And so we have the first area times its central right in the In the X direction that I defined second area times. It's centrally. The third area comes this sense, right, and we know all those values and we can get that the centrally 0.912 inches in the X direction. So 0.912 b two is about 1/2 B three. So this is it's it's out here somewhere. This is this Total distance is a C 3.9. So it's kind of in here, I guess somewhere in here, which you'd expect, it's pretty. It's not very far from this access because all of this mass here no, we need the why central coordinate and we can do the same thing. Total area area. The first element. Why position of the central of the first element? Very the second element. Why position of the century, The second element third element area. The third time's the Y position of that the century of that element, and that we get 2.25 inches. So H two is 5.3. So we're up, Um, a little over this year. 5.2505 A little over a little over half way up. You kind of expect, because this is actually a one inch and this guy is only half an inch. So it's providing quite a bigger mass up here off the Y axis. And obviously, this one to this one doesn't contribute much. Um, so now we have the central it'll coordinates, and we can then play again. And we know that the, um area moment about the X axis that central it'll X axis is, um, I about my X axis minus the total area times y bar square it, and that becomes 18.13 inches to the fore. And why I y bar is i y minus a Times X r squared. And that becomes 4.51 inches to the fourth

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