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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 $0 .$

Physics 101 Mechanics

Chapter 9

Distributed Forces: Moments of Inertia

Section 1

Moments of Inertia of Areas

Moment, Impulse, and Collisions

Cornell University

Rutgers, The State University of New Jersey

University of Washington

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.

02:22

Determine the polar moment…

01:26

01:39

05:06

in this problem were asked to calculate the polar second moment of area and the polar radius of gyration of the shaded region that shone. And it's outlined by a spiral shape. Where are is a function of data and peoples to a plus K data. We know our at pie is to a so that kay must be a over pie. Then we can set up our integral and we use polar coordinates. So we have r squared D A and our goes from zero to our fada. This expression here that that it goes from zero to pi and we can calculate, uh, crank through our inter girls here and we get that this is 31/20 pie 1/4 or roughly 8.8, 4.87 times eight in the fourth to check our answers to see if they seem reasonable. We can look at this is the, um, polar area moment of 1/2 disc with the radius of a So it's pi over 48 in the fourth, and we can see that this is clearly greater than that. As we would expect. This is the polar area moment of 1/2 disk of radius to a and it is for pie 84th. And it is clearly this is clearly bigger than this, as we would expect. And this is the polar area moment of 1/2 disc with Radius three have a and so kind of halfway in between these two with radius, and we can see that it's polar area. Moment is about 48 of the fourth, which is roughly in the ballpark of this guy, so we can have some pretty. We can have some confidence that this seems like a reasonable answer now, where asked to calculate the polar radius of innovation and so we can plug in, substitute one infinite R squared here and calculate the area that is 7/6 pie a squared or 3.67 a squared, and we can take the square root of this divided by this to get the polar radius of innovation and that square root of 93/70 a or about 1.15 a

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