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Unformatted text preview: Thevalingam, Donald Homework 32 Due: May 24 2007, midnight Inst: Eslami 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points A bicycle wheel of radius 0 . 32 m rolls down a hill without slipping. Its linear velocity increases constantly from 0 to 5 m / s in 2 . 5 s. What is its angular acceleration? Correct answer: 6 . 25 rad / s 2 . Explanation: v B Consider point B at the bottom of the wheel. It is at rest since there is no slip ping. Its velocity v B = v R = 0 Therefore, = v R = t = 1 R v t = 1 . 32 m 5 m / s 2 . 5 s = 6 . 25 rad / s 2 002 (part 2 of 3) 10 points Through what angle does the wheel turn in the 2 . 5 s? Correct answer: 19 . 5312 rad. Explanation: = 1 2 t 2 = 1 2 (6 . 25 rad / s 2 )(2 . 5 s) 2 = 19 . 5312 rad 003 (part 3 of 3) 10 points How many revolutions does it make? Correct answer: 3 . 10849 rev. Explanation: We know 1 rev = 2 rad, the revolutions made is: 19 . 5312 rad 2 = 3 . 10849 rev keywords: 004 (part 1 of 1) 10 points A coin with a diameter of 3 . 39 cm is dropped onto a horizontal surface. The coin starts out with an initial angular speed of 8 . 66 rad / s and rolls in a straight line without slipping. If the rotation slows with an angular deceleration of magnitude 0 . 763 rad / s 2 , how far does the coin roll before coming to rest? Correct answer: 0 . 833011 m. Explanation: Basic Concept 2 f = 2 + 2 There is a deceleration, so = 2 2 = (8 . 66 rad / s) 2 2( . 763 rad / s 2 ) = 49 . 1452 rad . s = r = d 2 so s = 3 . 39 cm 2 (49 . 1452 rad) = 83 . 3011 cm = 0 . 833011 m . keywords: 005 (part 1 of 3) 10 points A 4 . 17 kg hollow cylinder with inner radius . 12 m and outer radius 0 . 52 m is pulled by a horizontal string with a force of 33 . 7 N, as shown in the figure. Thevalingam, Donald Homework 32 Due: May 24 2007, midnight Inst: Eslami 2 33 . 7 N 4 . 17 kg . 12 m . 52 m The frictional force exerted on the wheel by the floor 1. is to the left. 2. is to the right. correct 3. is zero. Explanation: If there were no friction the bottom of the wheel would spin to the left; the frictional force will oppose this spinning motion. Therefore, the frictional force is directed to the right. 006 (part 2 of 3) 10 points What must be the magnitude of the force of friction if the cylinder is to roll without slipping? Correct answer: 10 . 4496 . Explanation: Let : R in = 0 . 12 m , R out = 0 . 52 m , F = 33 . 7 N , m = 4 . 17 kg , and f = force of friction . Suppose the cylinder has a length of ; then the density of the cylinder is = m ( R 2 out R 2 in ) ....
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This note was uploaded on 03/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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