Explanation Section 1: To determine: The initial moment of inertia. Solution: The initial moment of inertia is 5.97 × 10 8 kg ⋅ m 2 . Given information: The radius of station is 100 m , the moment of inertia of station is 5 × 10 8 kg ⋅ m 2 , the average mass of a crew of 150 people is 65 kg , the free fall acceleration of the crew is g and the time period of ball fallen is 15 min . The normal force from the rim provides a centripetal force on any person equals to the weight of person. Then, m v 2 r = m g v 2 r = g v is the speed. r is the radius of station. The formula for initial angular speed is, ω i = v r v = ω i r Substitute ω i r for v in above equation. ( ω i r ) 2 r = g ω i 2 = g r The space station is isolated, so its angular momentum is conserved. When the people move to the centre the station’s moment of inertia decreases. Its angular speed increases and the effective value of gravity decreases. From law of conservation of angular momentum, I i ω i = I f ω f ω f ω i = I i I f ω i is the initial angular speed. ω f is the final angular speed The initial moment of the inertia is calculated as, I i = I station + I ( 150 ) i = I station + 150 ( m r 2 ) I station is the moment of inertia of the station. I ( 150 ) i is the moment of inertia due to 150 people. m is the mass of people. Substitute 5 × 10 8 kg ⋅ m 2 for I station , 100 m for r and 65 kg for m in above equation to find I i . I i = ( 5 × 10 8 kg ⋅ m 2 ) + 150 ( 65 kg ) ( 100 m ) 2 = 5.97 × 10 8 kg ⋅ m 2 Section 2: To determine: The final moment of inertia. Solution: The final moment of inertia is 5.32 × 10 8 kg ⋅ m 2 . Given information: The radius of station is 100 m , the moment of inertia of station is 5 × 10 8 kg ⋅ m 2 , the average mass of a crew of 150 people is 65 kg , the free fall acceleration of the crew is g and the time period of ball fallen is 15 min

Physics for Scientists and Engineers with Modern Physics

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ISBN: 9781337671729

Author: SERWAY

Publisher: Cengage

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Chapter 11, Problem 28P

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