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Unformatted text preview: hiarker (srh959) – homework 24 – Turner – (58220) 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 29 cm is positioned at the top of an incline that makes 24 ◦ angle with the horizontal. This initial position of the sphere is a vertical distance 2 . 2 m above its position when at the bottom of the incline. The sphere is released and moves down the incline. 29 cm M μ ℓ 24 ◦ 2 . 2m 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: 5 . 54977 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 ) (2 . 2 m) = 5 . 54977 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: 6 . 56658 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 . 2 m) = 6 . 56658 m / s . keywords: 003 10.0 points A solid sphere has a radius of 0 . 74 m and a mass of 190 kg. How much work is required to get the sphere rolling with an angular speed of 77 rad / s on a horizontal surface? Assume the sphere starts from rest and rolls without slipping. Correct answer: 4 . 31814 × 10 5 J. Explanation: Let : r = 0 . 74 m , m = 190 kg , and ω f = 77 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 (190 kg) (0 . 74 m) 2 (77 rad / s) 2 = 4 . 31814 × 10 5 J . hiarker (srh959) – homework 24 – Turner – (58220) 2 keywords: 004 10.0 points A solid steel sphere of density 7 . 62 g / cm 3 and mass 0 . 5 kg spin on an axis through its center with a period of 2 s. Given V sphere = 4 3 π R 3 , what is its angular momentum? Correct answer: 0 . 000393367 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 . 5 kg) 4 π (7620 kg / m 3 ) bracketrightbigg 1 / 3 = 0 . 0250212 m . Using ω = 2 π T = 2 π (2 s) = 3 . 14159 s − 1 and I = 2 5 M R 2 = 2 5 (0 . 5 kg)(0 . 0250212 m) 2 = 0 . 000125212 kg m 2 , we have L ≡ I ω = 4 π M R 2 5 T = 4 π (0 . 5 kg)(0 . 0250212 m) 2 5 (2 s) = 0 . 000393367 kg m 2 / s ....
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 Spring '09
 NUI
 Physics, Angular Momentum, Kinetic Energy, Work, Moment Of Inertia, Rotation, ωf, L2 Lω

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