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Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to point $P$.

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

University of Michigan - Ann Arbor

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.

02:22

Determine the polar moment…

01:26

01:39

02:32

01:20

and this problem, we're told to calculate the polar second moment of area in the polar radius of gyration of the shaded region. That's basically a a disk with the bottom cut off of it. So there are various ways of doing this. But what I did is I broken into two regions may so that we can integrate over each of those regions and some of them up to get our total poor area more polar area moment. So one way is to note that Okay, let's do Let's do our an integral in polar coordinates around heater. So basically, take this wedge here. And so this angle here is minus pi over six. Because we know this is this distance here is our over two and this distances are we can figure out what that that angle is here. And this angle then is seven pi over six. Okay, so we can do you set up are integral and in polar coordinates. We just need to integrate r squared over a And the differential element in polar coordinates is our DRD data. And here we have our goes from zero to capital are I've called the radius capital arms that little are so we don't get confused with this that are of integration. Coordinate our and then our angle goes from data one here to there to here. And if we do that, we just get that the polar area moment of this upper like part of, ah, circle with the disk with the wedge cut out of it is pi over three Art of the fourth. Now we need to calculate the pool area moment about this point for the second region here and that one because we could try to figure out what our is is a function of thinking that, but that would be pretty ugly. So I just used um Oh, we can also calculate the area here of this guy up here, and that's just 2/3 pi r squared. And so what I did is for this second region. I used Cartesian coordinates, and so we need to calculate X squared than ago X squared plus y squared over, um region of this region here, too, and we can figure out what these line the equations for these lines are. And I did it that X in terms of why and so we need to go from X equals square of three y two. X equals minus square 23 Why? So that's these distances here. Yeah. And then why goes from, um minus are over too. So our origin is here. Some minus are over. Two up 20 So we crank through these in a girls and we get that The polar second area moment it is square to 3/16 are to the fourth for this region. And our area is, um, square to three for over four r squared. So now we can start Something's up and so are total polar. Second moment of area is this square to 3/16 plus pi over three. All times are to the fore in our area is quoted a 4 3/4 plus two pi over three. All times are square numerically, these are 1.155 are to the fourth and 2.5 to 7 are to the r squared and are polar radius of gyration is obviously this divided by the square root of this divided by this and that gives us 0.676 are now You can kind of take a look at, um if we have a full circle, Um, I sabar is just given by this expression here on, we can see that we would expect that this would be less then it is for a full disk, and I'll let you calculate what it is for. A. This is for half. This gets pi over four are to the fourth, and we can see that that is clearly less than this. So this is in between what it would be for 1/2 of a disc and a full disk. So we have some pretty decent confidence that that seems to be reasonable answer.

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