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Unformatted text preview: This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A woman whose mass is 59 . 9 kg stands at the rim of a horizontal turntable which has a moment of inertia of 341 kg m 2 about the axis of rotation and a radius of 2 . 95 m. The system is initially at rest and the turntable is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim in a clockwise direction (viewed from above) at a constant speed of 0 . 725 m / s relative to the ground. What will the motion of the turntable be, relative to the ground? 1. rotating counterclockwise correct 2. at rest, nonrotating 3. rotating clockwise Explanation: Because angular momentum is conserved, the turntable must rotate in the direction op posite to the woman. This is reflected in the fact that our ω is positive. That is, the direc tion of rotation is counterclockwise. 002 (part 2 of 3) 10.0 points What is its angular speed? Correct answer: 0 . 375692 rad / s. Explanation: The net external torque on the woman turntable system is zero, so angular momen tum is conserved. The angular momentum of the turntable is L t = Iω . The angular momentum of a particle (the woman) is vector L w = vector r × vector p and here the position vector from the center is perpendicular to the velocity, so the mag nitude is simply L w = mvr . We choose coun terclockwise to be positive, so the woman’s angular momentum will be negative. The ini tial and final total angular momenta are L i = 0 + 0 L f = − mvr + Iω and by conservation of angular momentum, L i = L f . So the angular velocity of the turntable is ω = mvr I = 59 . 9 kg(0 . 725 m / s)2 . 95 m 341 kg m...
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This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Mass, Work

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