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Unformatted text preview: Version 006 TEST03 TSOI (58160) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A cylindrical pulley with a mass of 5 . 6 kg, radius of 0 . 98 m and moment of inertia 1 2 M r 2 is used to lower a bucket with a mass of 3 . 1 kg into a well. The bucket starts from rest and falls for 2 . 8 s. r M m What is the linear acceleration of the falling bucket? The acceleration of gravity is 9 . 8 m / s 2 . 1. 5.14915 2. 4.50973 3. 4.11765 4. 4.9 5. 4.32353 6. 4.61176 7. 5.41333 8. 5.05313 9. 5.84561 10. 4.21037 Correct answer: 5 . 14915 m / s 2 . Explanation: Let : M = 5 . 6 kg , r = 0 . 98 m , m = 3 . 1 kg , and g = 9 . 8 m / s 2 . Let T be the tension in the cord and the angular acceleration of the wheel. Newtons equation for the mass m is mg T = ma T = m ( g a ) and for the disk T r = I ( mg ma ) r = parenleftbigg 1 2 M r 2 parenrightbigg parenleftBig a r parenrightBig 2 mg 2 ma = M a 2 mg = ( M + 2 m ) a a = 2 mg M + 2 m = 2 (3 . 1 kg) (9 . 8 m / s 2 ) 5 . 6 kg + 2 (3 . 1 kg) = 5 . 14915 m / s 2 . 002 (part 1 of 2) 10.0 points A solid sphere of radius 44 cm is positioned at the top of an incline that makes 27 angle with the horizontal. This initial position of the sphere is a vertical distance 2 . 3 m above its position when at the bottom of the incline. The sphere is released and moves down the incline. 44 cm M 27 2 . 3m Calculate the speed of the sphere when it reaches the bottom of the incline if it rolls without slipping. The acceleration of gravity is 9 . 8 m / s 2 . The moment of inertia of a sphere with respect to an axis through its center is 2 5 M R 2 . 1. 5.6745 2. 7.29383 3. 4.73286 4. 5.01996 5. 7.93725 6. 5.54977 7. 7.75887 8. 5.91608 9. 5.79655 10. 6.03324 Version 006 TEST03 TSOI (58160) 2 Correct answer: 5 . 6745 m / s. Explanation: From conservation of energy we have U i = K trans,f + K rot,f M g h = 1 2 M v 2 + 1 2 I 2 = 1 2 M v 2 + 1 2 parenleftbigg 2 5 M R 2 parenrightbigg parenleftbigg v 2 R 2 parenrightbigg = 7 10 M v 2 v 1 = radicalbigg 10 7 g h = radicalbigg 10 7 (9 . 8 m / s 2 ) (2 . 3 m) = 5 . 6745 m / s . 003 (part 2 of 2) 10.0 points Calculate the speed of the sphere if it reaches the bottom of the incline by slipping friction lessly without rolling. 1. 8.51587 2. 7.9196 3. 6.41561 4. 6.10246 5. 9.28655 6. 6.71416 7. 6.85857 8. 6.26099 9. 7.4081 10. 8.85438 Correct answer: 6 . 71416 m / s. Explanation: From conservation of energy we have U i = K trans,f M g h = 1 2 M v 2 v 2 = radicalbig 2 g h = radicalBig 2 (9 . 8 m / s 2 ) (2 . 3 m) = 6 . 71416 m / s . keywords: 004 10.0 points A skater spins with an angular speed of 7 . 3 rad / s with her arms outstretched. She lowers her arms, decreasing her moment of inertia by a factor of 1 . 5....
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This note was uploaded on 05/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Mass

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