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Unformatted text preview: sousse (res2468) HW 9 Kleinman (58225) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. This HW covers Chs. 13 and 9. 001 (part 1 of 2) 10.0 points The motor driving a grinding wheel with a rotational inertia of 0 . 82 kg m 2 is switched off when the wheel has a rotational speed of 21 rad / s. After 7 . 1 s, the wheel has slowed down to 16 . 8 rad / s. What is the absolute value of the constant torque exerted by friction to slow the wheel down? Correct answer: 0 . 485071 N m. Explanation: We have t = L = ( I ) , so that   = I  1  t 1 = (0 . 82 kg m 2 ) (21 rad / s 16 . 8 rad / s) 7 . 1 s = 0 . 485071 N m . 002 (part 2 of 2) 10.0 points If this torque remains constant, how long after the motor is switched off will the wheel come to rest? Correct answer: 35 . 5 s. Explanation: When the wheel comes to rest, its angular speed is 2 = 0; hence t 2 = vextendsingle vextendsingle vextendsingle vextendsingle I ( 2 ) vextendsingle vextendsingle vextendsingle vextendsingle = I   = (0 . 82 kg m 2 ) (21 rad / s) (0 . 485071 N m) = 35 . 5 s . 003 10.0 points A wooden bucket filled with water has a mass of 48 kg and is attached to a rope that is wound around a cylinder with a radius of 0.067 m. A crank with a turning radius of 0.33 m is attached to the end of the cylinder. What minimum force directed perpendicu larly to the crank handle is required to raise the bucket? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 95 . 6029 N. Explanation: Let : m = 48 kg , r = 0 . 067 m , d = 0 . 33 m and g = 9 . 81 m / s 2 . Since = 90 , sin = 1 and = F d sin = F d min = ( mg ) r F min d = mg r F min = mg r d = (48 kg)(9 . 81 m / s 2 )(0 . 067 m) . 33 m = 95 . 6029 N . 004 (part 1 of 2) 10.0 points A string is wound around a uniform disc of radius 0 . 62 m and mass 2 . 7 kg . 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 . 62 m 2 . 7 kg As the disc descends, calculate the tension in the string. Correct answer: 8 . 82 N. Explanation: sousse (res2468) HW 9 Kleinman (58225) 2 Let : R = 0 . 62 m , M = 2 . 7 kg , and g = 9 . 8 m / s 2 . Basic Concepts summationdisplay vector F = mvectora summationdisplay vector = I vector U + K rot + K trans = 0 Solution summationdisplay F = T M g = M a and (1) summationdisplay = T R = I = 1 2 M R 2 parenleftBig a R parenrightBig . (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 parenleftbigg g 2 T M parenrightbigg = M g 2 T 3 T = M g T = M g 3 (4) = (2 . 7 kg) (9 . 8 m / s 2 ) 3 = 8 . 82 N ....
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This note was uploaded on 04/08/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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