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Unformatted text preview: practice 10 – GADHIA, TEJAS – Due: Mar 28 2007, 5:00 pm 1 Question 1 part 1 of 1 10 points A box, with its centerofmass offcenter as indicated by the dot, is placed on an inclined plane. In which of the four orientations shown, if any, does the box tip over? 1. b 2. b correct 3. b 4. b 5. None of the orientations will cause the box to top over. Explanation: In order to tip over, the box must pivot about its bottom left corner. The only orien tation where gravity creates a torque sufficient to tip the block is b Question 2 part 1 of 1 10 points A cable passes over a pulley. Because the cable grips the pulley and the pulley has non zero mass, the tension in the cable is not the same on opposite sides of the pulley. The force on one side is 118 N, and the force on the other side is 96 . 4 N. Assuming that the pulley is a uniform disk of mass 3 . 35 kg and radius 1 . 27 m, determine the magnitude of its angular acceleration. Correct answer: 10 . 154 rad / s 2 (tolerance ± 1 %). Explanation: The resultant torque is given by (118 N)(1 . 27 m) − (96 . 4 N)(1 . 27 m) = 27 . 432 N m The moment of inertia is: I = 1 2 mr 2 = 1 2 (3 . 35 kg)(1 . 27 m) 2 = 2 . 70161 kg m 2 . Then, τ = I α gives α = τ I = 27 . 432 N m 2 . 70161 kg m 2 = 10 . 154 rad / s 2 . Question 3 part 1 of 2 10 points A constant horizontal force of 220 N is ap plied to a lawn roller in the form of a uniform solid cylinder of radius 0 . 59 m and mass 14 kg. The acceleration of gravity is 9 . 8 m / s 2 . M R F If the roller rolls without slipping, calculate the acceleration of the center of mass. Correct answer: 10 . 4762 m / s 2 (tolerance ± 1 %). Explanation: practice 10 – GADHIA, TEJAS – Due: Mar 28 2007, 5:00 pm 2 Basic Concepts: summationdisplay vector F = mvector a summationdisplay vector τ = Ivectorα summationdisplay F = F − f = M a τ = fR = Iα Using I = 1 2 M R 2 + M R 2 = 3 2 M R 2 , we find 3 2 M R 2 a R = F R . Therefore a = 2 F 3 M (1) = 2 (220 N) 3 (14 kg) = 10 . 4762 m / s 2 . Question 4 part 2 of 2 10 points Find the minimum coefficient of friction necessary to prevent slipping. Correct answer: 0 . 5345 (tolerance ± 1 %). Explanation: When there is no slipping, f = μM g . Sub stituting this into the torque equation of the previous part, we have μM g R = 1 2 M Ra, solving for μ and substituting a from Eq. 1, we have μ = 1 2 a g = F 3 M g = (220 N) 3 (14 kg) (9 . 8 m / s 2 ) = 0 . 5345 Question 5 part 1 of 1 10 points A long rod is pivoted (without friction) at one end. It is released from rest at A in a horizontal position and swings down, passing positions B and C . A B C b The angular acceleration at B compared to C is 1. α B > α C . correct 2. There is not enough information....
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 Spring '08
 Turner
 Angular Momentum, Kinetic Energy, Mass, Moment Of Inertia, Rotation, li

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