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Unformatted text preview: zhou (cz3574) HW9 mackie (10611) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A can of soup has a mass 570 g, height 14 cm, and diameter 6 . 6 cm. It is placed at rest on the top of an incline that is 4 . 4 m long at 24 to the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . Calculate the moment of inertia of the can if it takes 1 . 7 s to reach the bottom of the incline. Correct answer: 0 . 000191834 kg m 2 . Explanation: The speed of the center of mass of the can at the bottom of the incline is v f = 2 v = 2 L t . Thus the angular speed at the bottom of the incline is f = v f R . Using energy conservation, we have m g L sin = 1 2 m v 2 f + 1 2 I 2 f . Solving for I , I = 2 m 2 f bracketleftbigg g L sin  1 2 v 2 f bracketrightbigg = 2 (570 g) (156 . 863 rad / s) 2 bracketleftbigg (9 . 8 m / s 2 )(4 . 4 m) sin(24 ) 1 2 (5 . 17647 m / s) 2 bracketrightbigg = 0 . 000191834 kg m 2 . 002 10.0 points A solid steel sphere of density 7 . 68 g / cm 3 and mass 0 . 9 kg spin on an axis through its center with a period of 3 . 8 s. Given V sphere = 4 3 R 3 , what is its angular momentum? Correct answer: 0 . 000548564 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 (0 . 9 kg) 4 (7680 kg / m 3 ) bracketrightbigg 1 / 3 = 0 . 0303574 m . Using = 2 T = 2 (3 . 8 s) = 1 . 65347 s 1 and I = 2 5 M R 2 = 2 5 (0 . 9 kg)(0 . 0303574 m) 2 = 0 . 000331765 kg m 2 , we have L I = 4 M R 2 5 T = 4 (0 . 9 kg)(0 . 0303574 m) 2 5 (3 . 8 s) = 0 . 000548564 kg m 2 / s . 003 10.0 points A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. zhou (cz3574) HW9 mackie (10611) 2 v 9 . 8 m / s 2 1 . 7 m 7 kg 3 3 Calculate the magnitude of the angular mo mentum of the bob about a vertical axis through the supporting point. The acceler ation of gravity is 9 . 8 m / s 2 . Correct answer: 15 . 7328 kg m 2 / s. Explanation: Let : = 1 . 7 m , = 33 , g = 9 . 8 m / s 2 , and m = 7 kg ....
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This note was uploaded on 03/25/2011 for the course PHYS 122 taught by Professor Hiyishi during the Spring '10 term at Temple College.
 Spring '10
 Hiyishi
 Mass

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