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Unformatted text preview: Overton, Mays Homework 25 Due: Apr 1 2005, 4:00 am Inst: Turner 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A space station shaped like a giant wheel has a radius of 193 m and a moment of inertia of 5 . 4 10 8 kg m 2 (when it is unmanned). A crew of 150 live on the rim, and the station is rotating so that the crew experience an apparent acceleration of 1 g . When 100 people move to the center of the station, the angular speed changes. The acceleration of gravity is 9 . 8 m / s 2 . What apparent acceleration is experienced by those remaining at the rim? Assume: The average mass of for each in habitant is 68 . 7 kg. Correct answer: 18 . 7474 m / s 2 . Explanation: Given : r = 193 m , I = 5 . 4 10 8 kg m 2 , and m = 150 68 . 7 kg . For an apparent acceleration of g, X F c = m a c = m g v 2 r = g r 2 = g = r g r Initially, L = I + ( 150 m r 2 ) and after 100 people move to the center, L f = I + 50 m r 2 Angular momentum is conserved, so L f = L i ( I + 150 m r 2 ) = ( I + 50 m r 2 ) = I + 150 m r 2 I + 50 m r 2 r g r and the apparent gravity is g n = r 2 = I + 150 m r 2 I + 50 m r 2 2 g Since I + 150 m r 2 = 5 . 4 10 8 kg m 2 + 150(68 . 7 kg)(193 m) 2 = 9 . 23851 10 8 m / s 2 and I + 50 m r 2 = 5 . 4 10 8 kg m 2 + 50(68 . 7 kg)(193 m) 2 = 6 . 6795 10 8 m / s 2 , then g n = 9 . 23851 10 8 m / s 2 6 . 6795 10 8 m / s 2 2 ( 9 . 8 m / s 2 ) = 18 . 7474 m / s 2 . 002 (part 1 of 1) 10 points Consider a thin 44 m rod pivoted at one end. A uniform density spherical object (whose mass is 4 kg and radius is 12 m) is attached to the free end of the rod. The moment of inertia of the rod about an end is I rod = 1 3 m L 2 . Overton, Mays Homework 25 Due: Apr 1 2005, 4:00 am Inst: Turner 2 The moment of inertia of the sphere about its centerofmass is I sphere = 2 5 m r 2 . Note: The length C mass in the figure rep resents the location of the centerofmass of the rod plus mass system. The rod is ini tially at rest at 30 below the horizontal. The acceleration of gravity g = 9 . 8 m / s 2 . 4 4 m C m a s s 4 k g 4 kg radius 12 m 30 What is the angular acceleration of the rod immediately after it is released? Correct answer: 0 . 172441 rad / s 2 . Explanation: Basic Concepts: The moment of inertia, the center of mass and the energy conserva tion. The rotational kinetic energy is K rot = 1 2 I 2 . Let : m = 4 kg , L = 44 m , r = 12 m , and = 30 . Solution: The moment of inertia of the rod, I rod , with respect to the pivot point is I rod = 1 3 m L 2 , and the moment of inertia I sphere of the mass m with respect to the pivot point is I sphere = 2 5 m r 2 + m ( L + r ) 2 , Then, the moment of inertia of the system I is I = I rod + I m = 1 3 m L 2 + m ( L + r ) 2 + 2 5 m r 2 = 1 3 (4 kg) (44 m)...
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 Spring '08
 Turner
 Physics, Work

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