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Unformatted text preview: hernandez (ejh742) – homework 24 – Turner – (58120) 1 This printout should have 13 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 solid sphere of radius 28 cm is positioned at the top of an incline that makes 21 ◦ angle with the horizontal. This initial position of the sphere is a vertical distance 4 . 3 m above its position when at the bottom of the incline. The sphere is released and moves down the incline. 28 cm M μ ℓ 21 ◦ 4 . 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 . Correct answer: 7 . 75887 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 ω 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 ) (4 . 3 m) = 7 . 75887 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: 9 . 18041 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 ) (4 . 3 m) = 9 . 18041 m / s . keywords: 003 10.0 points A solid sphere has a radius of 0 . 54 m and a mass of 120 kg. How much work is required to get the sphere rolling with an angular speed of 82 rad / s on a horizontal surface? Assume the sphere starts from rest and rolls without slipping. Correct answer: 1 . 647 × 10 5 J. Explanation: Let : r = 0 . 54 m , m = 120 kg , and ω f = 82 rad / s . I = 2 5 mr 2 and v = r ω , so the work is W = Δ K = 1 2 mv 2 + 1 2 I ω 2 = 1 2 m ( r ω ) 2 + 1 2 parenleftbigg 2 5 mr 2 parenrightbigg ω 2 = 7 10 mr 2 ω 2 = 7 10 (120 kg) (0 . 54 m) 2 (82 rad / s) 2 = 1 . 647 × 10 5 J . hernandez (ejh742) – homework 24 – Turner – (58120) 2 keywords: 004 10.0 points A solid steel sphere of density 7 . 79 g / cm 3 and mass 0 . 6 kg spin on an axis through its center with a period of 1 . 4 s. Given V sphere = 4 3 π R 3 , what is its angular momentum? Correct answer: 0 . 000750378 kg m 2 / s. Explanation: The definition of density is ρ ≡ M V = M 4 3 π R 3 , Therefore R = bracketleftbigg 3 M 4 π ρ bracketrightbigg 1 / 3 = bracketleftbigg 3 (0 . 6 kg) 4 π (7790 kg / m 3 ) bracketrightbigg 1 / 3 = 0 . 0263942 m . Using ω = 2 π T = 2 π (1 . 4 s) = 4 . 48799 s − 1 and I = 2 5 M R 2 = 2 5 (0 . 6 kg)(0 . 0263942 m) 2 = 0 . 000167197 kg m 2 , we have L ≡ I ω = 4 π M R 2 5 T = 4 π (0 . 6 kg)(0 . 0263942 m) 2 5 (1 . 4 s) = 0 . 000750378 kg m 2 / s ....
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Work

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