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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 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 = I . 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 womans angular momentum will be negative. The ini tial and final total angular momenta are L i = 0 + 0 L f = mvr + I and by conservation of angular momentum, L i = L f . So the angular velocity of the turntable is = mvr I = 59 . 9 kg(0 . 725 m / s)2 . 95 m 341 kg m...
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 Spring '08
 Turner
 Mass, Work

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