# hw25 - Overton Mays – Homework 25 – Due Apr 1 2005 4:00...

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Unformatted text preview: Overton, Mays – Homework 25 – Due: Apr 1 2005, 4:00 am – Inst: Turner 1 This print-out should have 13 questions. Multiple-choice 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 ω 2 = g ω = r g r Initially, L = I ω + ( 150 m r 2 ) ω and after 100 people move to the center, L f = I ω + 50 m r 2 ω Angular momentum is conserved, so L f = L i ( I + 150 m r 2 ) ω = ( I + 50 m r 2 ) ω ω = I + 150 m r 2 I + 50 m r 2 r g r and the apparent gravity is g n = r ω 2 = µ 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 = µ 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 center-of-mass is I sphere = 2 5 m r 2 . Note: The length C mass in the figure rep- resents the location of the center-of-mass 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 ω 2 . Let : m = 4 kg , L = 44 m , r = 12 m , and θ = 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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## This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw25 - Overton Mays – Homework 25 – Due Apr 1 2005 4:00...

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