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$\bullet$ A thin, light string is wrapped around the rim of a 4.00 kg solid uniform disk that is 30.0 $\mathrm{cm}$ in diameter. A person pulls on the string with a constant force of 100.0 $\mathrm{N}$ tangent to the disk, as shown in Figure $10.49 .$ The disk is not attached to anything and is free tomove and tum. (a) Find the angular acceleration of the disk about its center of mass and the linear acceleration of its center of mass. (b) If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the accelerations in part (a)?

a) 50 $\mathrm{m} / \mathrm{s}^{2}$b) 25 $\mathrm{m} / \mathrm{s}^{2}$

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

Cornell University

Rutgers, The State University of New Jersey

Simon Fraser University

University of Winnipeg

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So here we have a mass of four kilograms. We know that the moment of inertia of a disc is going to be equal to 1/2 Um are squared. We know the diameter of the discus 30 centimeters. Or we can say that the radius is equaling 300.15 meters and we know that the force applied is going to be equal to 100 Newton's and we're going to say that clockwise is positive. So let's draw a diagram. We havethe e x and Y axes and let's draw this circle here. This distance right here is of course, the radius actually lower case R radius and we're going to say that the angular acceleration is clockwise. That's why Riccio's clockwise to be positive and the force applied acts right on the edge of the disc. This will be the force applied and ah, in the positive ex direction is the linear velocity in linear acceleration of the center of mass. So to find the linear acceleration of the center of mass and the angular acceleration of this, we can say that the sum of forces in the ex direction is going to be equal to the mass times the acceleration of the center of mass and this will simply be equal to force applied. So we know that the ace of CM will be equal to force applied, divided by the mass. And so this was going to be equal to 100 Newton's divided by four meters or other four kilograms and the only 25.0 meters per second squared. We know that the sum of torques is going to be equal to the moment of inertia, times the angle, angular acceleration and this will be equal to the force Applied times the radius from the center of mass for the distance from the center of Mass. We can say that the angular acceleration is going to be equal to F. Applied times are divided by the moment of inertia and we can say that this is going to be equal to two times the force applied, divided by the mass and let's solve. So this will be two times 100 divided by oh rather, the moment of inertia is half em are squared. So I forgot. A are in the denominator. The the angular acceleration will be two times the force applied, divided by mass divided by mass times The radius. So it will be two times 100 divided by four kilograms, times 0.15 meters and the angular acceleration is going to be equal to 333.33 radiance per second squared. So this will be your answer for these will be your two answers for party for part B. Now we have a hollow, thin walled cylinder. So this means that the moment of inertia of the center of mass will be simply equal to M. R. Squared again. The linear acceleration of the center of mass has nothing to do with the ah with the dimensions of the object itself. Again, it's only based on the force applied, divided by the mass of the object. So here the linear acceleration of the center of mass is going to stay the same. This will be 25 meters per second squared. However, for the angular acceleration, this will simply be equal to the force applied, divided by M r. Again, that factor of two is eliminated because the moment the moment of inertia is changing from half em are squared to M R squared. And so this is going to be equal to 100 divided by four times 40.15 and this is giving us 166.67 radiance per second squared. So these air be your two answers for part B and again note that the angular acceleration is not equal toothy, linear exploration of the center of mass divided by our and this is because there's a force applied. This relationship only exists for a uniform circular motion. If the object is not in uniform circular motion, this does not apply, so this will not apply for these two cases. Again. Your answer for a part B is 25 meters per second squared, and your angular acceleration is 166.67 radiance per second squared. That is the end of the solution. Thank you for watching

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