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(II) A uniform circular plate of radius 2$R$ has a circular hole of radius $R$ cut out of it. The center $C^{\prime}$ of the smaller circle is a distance 0.80$R$ from the center $C$ of the larger circle, Fig. $45 .$ What is the position of the center of mass of the plate? [Hint: Try subtraction.]

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0.27$R$

Physics 101 Mechanics

Chapter 9

Linear Momentum

Motion Along a Straight Line

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

University of Winnipeg

Lectures

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In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.

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In physics, the kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. The kinetic energy of a rotating object is the sum of the kinetic energies of the object's parts.

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we have it. Uh, we have a shaded circle. Um, and that is and then cut out from it. Cut out from the bigger circle with radius to our, um is a is a white circle. So the shaded circle is, uh uh, the white circle his radius one hour. And so the answer them the shaded circles denoted by a sage white circles demoted by W on. And so the center mass of the whole system is at zero, of course. And relative to that, the center of mass of the white part is that point plus 0.0.8 arms. So what is the center of last coordinates off the shaded circle, right? Eh? So the idea is is actually pretty simple. You first used the center of mass equation. Um, so center mass of the system is just Ah, the is a massive the shaded region Time center mass of the shaded region, uh, plus mass of the white beach in times center of mass of the white region, over the mass of the shaded region, plus mass of the white Megyn. Now we know, uh, everything but this senator. Massive shaded regions, so center mass of the shaded fusion will just be, um, malu inches equation. So what we'll have mobile get is, uh, uh, let's call this called the combined masticated Master. Shady region. Massive white region last as big. So this is big M times X time center of massive system minus massive white region time center, Mass of white region divided by, um, of course masses a shade of region. Right. So since we already know that massive the center mass, the whole system is zero the same cops out. And so we're left with, uh, massive negative, massive white pigeon time center mass of the shaded vision over, uh, master, the shaded region, which is just the mass over the total mess minus the mass of the white region. This is our equation on, and so and sense And okay, so let's, um so work out with these masses are the center of mass. The mass of the white region will be, um, the density will have the same density is, by the way, density row times the volume pie, um, R squared pi r squared times h just some dummy height. This variable. Cancel out, by the way. So we don't need toe really worry about it, but just so that just so we're complete with everything. And so the mass of the total the total mass will just be, um, grow times, uh, grow pi times to our quietly square times a Senate be rope. I four r squared h. Okay, uh, and so the mass of the shaded region therefore will be this minus that. All right, so let's, um go ahead and write that down center of us. Coordinates of the shade region will be recalled to negative times. Ah, will be negative. Row times pi r squared. Age. Um, again, the actual cancel. It's just dummy variable for height. Uh, times. Oops. I made an error here. This will be Ah, that's X. Coordinate our center mass of the y of the white region. So that is times a 0.8 are over there. Divide that by four. Um, so take pi rho pi r squared h common. So you get Hai. Rill are square times four minus one. Since the stree uh, noticed that the are squares, Cancel the pies. Cancel the rose. Cancel h is also there, cancels. So what we're left with is negative 0.8 are over three. That comes out to be a negative 30.27 times the radiance of the circle. Um, and that's it.

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