Homework 24 - sanchez (ds28677) homework 24 Turner (58220)...

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Unformatted text preview: sanchez (ds28677) homework 24 Turner (58220) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A solid sphere of radius 28 cm is positioned at the top of an incline that makes 29 angle with the horizontal. This initial position of the sphere is a vertical distance 2 . 9 m above its position when at the bottom of the incline. The sphere is released and moves down the incline. 28 cm M 29 2 . 9m 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 . Correct answer: 6 . 37181 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 . 9 m) = 6 . 37181 m / s . 002 (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. Correct answer: 7 . 53923 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 . 9 m) = 7 . 53923 m / s . keywords: 003 10.0 points A solid sphere has a radius of 0 . 7 m and a mass of 180 kg. How much work is required to get the sphere rolling with an angular speed of 67 rad / s on a horizontal surface? Assume the sphere starts from rest and rolls without slipping. Correct answer: 2 . 77151 10 5 J. Explanation: Let : r = 0 . 7 m , m = 180 kg , and f = 67 rad / s . I = 2 5 mr 2 and v = r , so the work is W = K = 1 2 mv 2 + 1 2 I 2 = 1 2 m ( r ) 2 + 1 2 parenleftbigg 2 5 mr 2 parenrightbigg 2 = 7 10 mr 2 2 = 7 10 (180 kg) (0 . 7 m) 2 (67 rad / s) 2 = 2 . 77151 10 5 J . sanchez (ds28677) homework 24 Turner (58220) 2 keywords: 004 10.0 points A solid steel sphere of density 7 . 6 g / cm 3 and mass 1 kg spin on an axis through its center with a period of 1 . 3 s. Given V sphere = 4 3 R 3 , what is its angular momentum? Correct answer: 0 . 00192469 kg m 2 / s. Explanation: The definition of density is M V = M 4 3 R 3 , Therefore R = bracketleftbigg 3 M 4 bracketrightbigg 1 / 3 = bracketleftbigg 3 (1 kg) 4 (7600 kg / m 3 ) bracketrightbigg 1 / 3 = 0 . 0315524 m . Using = 2 T = 2 (1 . 3 s) = 4 . 83322 s 1 and I = 2 5 M R 2 = 2 5 (1 kg)(0 . 0315524 m) 2 = 0 . 000398222 kg m 2 , we have L I = 4 M R 2 5 T = 4 (1 kg)(0 . 0315524 m) 2 5 (1 . 3 s) = 0 . 00192469 kg m 2 / s ....
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Homework 24 - sanchez (ds28677) homework 24 Turner (58220)...

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