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Unformatted text preview: reddy (ar38357) Newtons Second Law applied to Rotational Motion clancy (SCI411-2) 1 This print-out should have 3 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A cable passes over a pulley. Because the cable grips the pulley and the pulley has non- zero mass, the tension in the cable is not the same on opposite sides of the pulley. The force on one side is 124 N, and the force on the other side is 93 N. Assuming that the pulley is a uniform disk of mass 0 . 91 kg and radius 0 . 481 m, determine the magnitude of its angular acceleration. Correct answer: 141 . 646 rad / s 2 . Explanation: Let : f 1 = 124 N f 2 = 93 N m = 0 . 91 kg and r = 0 . 481 m . The resultant torque is given by = f 1 r- f 2 r = ( f 1- f 2 ) r and the moment of inertia is I = 1 2 mr 2 , so = I ( f 1- f 2 ) r = parenleftbigg 1 2 mr 2 parenrightbigg = 2 ( f 1- f 2 ) r mr 2 = 2 ( f 1- f 2 ) mr = 2(124 N- 93 N)...
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This note was uploaded on 11/21/2010 for the course PHYS 12231 taught by Professor Sontar during the Fall '10 term at Ill. Chicago.
- Fall '10