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Unformatted text preview: Muraj, Hamza – Homework 26 – Due: Apr 3 2006, 4:00 am – Inst: Florin 1 This printout should have 9 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 3) 10 points A string is wound around a uniform disc whose mass is 2 . 8 kg and radius is 0 . 38 m , see figure below. The disc is released from rest with the string vertical and its top end tied to a fixed support. The acceleration of gravity is 9 . 8 m / s 2 . h . 38 m 2 . 8 kg ω As the disc descends, calculate the tension in the string. Correct answer: 9 . 14667 N. Explanation: Let : R = 0 . 38 m , M = 2 . 8 kg , and g = 9 . 8 m / s 2 . Basic Concepts X ~ F = m~a X ~ τ = I ~α Δ U + Δ K rot + Δ K trans = 0 Solution X F = T M g = M a and (1) X τ = T R = I α = 1 2 M R 2 ‡ a R · . (2) Solving for a in (2), a = 2 T M . (3) Using a from Eq. (3) and solving for T in (1), T = M ( g a ) = M µ g 2 T M ¶ = M g 2 T 3 T = M g T = M g 3 (4) = (2 . 8 kg) (9 . 8 m / s 2 ) 3 = 9 . 14667 N . 002 (part 2 of 3) 10 points Calculate the magnitude of the acceleration of its center of mass. 1. a cm = 7 . 252 m / s 2 2. a cm = 5 . 94533 m / s 2 3. a cm = 6 . 53333 m / s 2 correct 4. a cm = 5 . 684 m / s 2 Explanation: From Eq. (3), we have a = 2 T M = 2 M µ M g 3 ¶ = 2 3 g = 2 3 (9 . 8 m / s 2 ) = 6 . 53333 m / s 2 ....
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Turner

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