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Unformatted text preview: Le (tl8426) HW10 gentle (56245) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A bowling ball is given an initial speed v on an alley such that it initially slides without rolling. The coefficient of friction between ball and alley is . Find the speed of the balls center of mass v CM at the time pure rolling motion occurs. 1. v CM = 1 5 v 2. v CM = 1 3 v 3. v CM = 3 7 v 4. v CM = 5 7 v correct 5. v CM = 3 5 v Explanation: Basic Concepts: F fr = N L = I . Solution: The frictional force F = M g acts to slow the ball v = v F M t. (1) It also exerts a torque RF and starts the ball rolling I = RF t. The ball starts rolling without slipping at time t = , when = v R = v R = v F M R = RF I . Solve for with I = 2 5 M R 2 and F = M g = v R RF I + F M R = 2 7 v g . Substituting the expression for into (1) yields v ( ) = v M g M = 5 7 v . 002 (part 2 of 2) 10.0 points Since the frictional force provides the decel eration, from Newtons second law it follows that a CM = g . Find the distance x it has traveled. When pure rolling motion occurs, v CM = R . 1. x = 12 v 2 25 g 2. x = 2 v 2 7 g 3. x = 3 v 2 14 g 4. x = 12 v 2 49 g correct 5. x = 5 v 2 81 g Explanation: The distance traveled by the ball from t = 0 to t = is x = integraldisplay t = t =0 v dt = integraldisplay t = t =0 ( v g t ) dt = 12 49 v 2 g keywords: 003 (part 1 of 2) 10.0 points A string is wound around a uniform disc of radius 0 . 37 m and mass 2 . 8 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 . Le (tl8426) HW10 gentle (56245) 2 h . 37 m 2 . 8 kg As the disc descends, calculate the tension in the string. Correct answer: 9 . 14667 N. Explanation: Let : R = 0 . 37 m , M = 2 . 8 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 . 8 kg) (9 . 8 m / s 2 ) 3 = 9 . 14667 N . 004 (part 2 of 2) 10.0 points Calculate the speed of the center of mass when, after starting from rest, the center of mass has fallen 1 . 5 m. Correct answer: 4 . 42719 m / s. Explanation: From conservation of mechanical energy we have U + K rot + K trans = 0 1 2 parenleftbigg 1 2 M R 2 parenrightbigg 2 + 1 2 M v 2 = M g h....
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 Fall '07
 Swinney
 mechanics

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