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Unformatted text preview: alexander (jra2623) – oldhomework 22 – Turner – (92510) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A giant yoyo of mass 430 kg measuring about . 9 m in radius was dropped from a platform 60 m high. One end of the string was tied to the platform, so the yoyo unwinds as it descended. . 9 m mg . 2 m T Assuming that the axle of the yoyo has a radius of 0 . 2 m, find the velocity of descent at the end of the fall. The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 10 . 2867 m / s. Explanation: Let : m = 430 kg , R = 0 . 9 m , h = 60 m , r = 0 . 9 m , and g = 9 . 81 m / s 2 . Analyzing the torque, summationdisplay τ = T r = I α = parenleftbigg 1 2 mR 2 parenrightbigg parenleftBig a r parenrightBig T = 1 2 ma R 2 r 2 . Using Newton’s second law, summationdisplay F x = mg T = ma mg maR 2 2 r 2 = ma 2 r 2 mg = mR 2 a + 2 r 2 ma a = 2 r 2 g R 2 + 2 r 2 = 2 (0 . 2 m) 2 (9 . 81 m / s 2 ) (0 . 9 m) 2 + 2 (0 . 2 m) 2 = 0 . 881798 m / s 2 . Since v = 0 , v 2 = v 2 + 2 a Δ h = 2 a Δ h v = √ 2 a Δ h = radicalBig 2 (0 . 881798 m / s 2 ) (60 m) = 10 . 2867 m / s . keywords: 002 (part 1 of 2) 10.0 points A spool of wire of mass 4 . 7 kg and radius . 31 m is unwound under a constant wire tension 7 N. Assume the spool is a uniform solid cylinder that rolls without slipping. R F M Find the friction force on the bottom of the spool. Let right be positive. Correct answer: 2 . 33333 N. Explanation: Let : M = 4 . 7 kg , R = 0 . 31 m , and F = 7 N . alexander (jra2623) – oldhomework 22 – Turner – (92510) 2 The spool’s angular acceleration (from the net torque in the centerofmass frame) is Iα = F R f R where I ≡ I cm = M R 2 2 . Since the spool rolls without slipping, α = a R . For a wire tension F and friction force f act ing on the bottom of the spool, by Newton’s Second Law, the spool’s center of mass(which is in the middle of the spool’s axis) accelerates according to Ma = F + f M Ra 2 = F R f R M a = 2 F 2 f Thus 2 F 2 f = F + f f = F 3 = 7 N 3 = 2 . 33333 N ....
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This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas.
 Summer '08
 Kaplunovsky
 Mass, Work

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