test3 - Version 006 TEST03 TSOI(58160 1 This print-out...

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Unformatted text preview: Version 006 TEST03 TSOI (58160) 1 This print-out should have 15 questions. Multiple-choice 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.

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test3 - Version 006 TEST03 TSOI(58160 1 This print-out...

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