oldhw 21 - alexander (jra2623) oldhomework 21 Turner...

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Unformatted text preview: alexander (jra2623) oldhomework 21 Turner (92510) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A uniform rod of mass 1 . 4 kg is 17 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 17 m from the center of mass of the rod. Initially the rod makes an angle of 62 with the horizontal. The rod is released from rest at an angle of 62 with the horizontal, as shown in the figure below 17 m 17 m 1 . 4 kg O 62 What is the angular acceleration of the rod at the instant the rod is in a horizontal posi- tion? The acceleration of gravity is 9 . 8 m / s 2 and the moment of inertia of the rod about its center of mass is 1 12 m 2 . Correct answer: 0 . 532127 rad / s 2 . Explanation: Let : = 17 m , d = = 17 m , = 62 , and m = 1 . 4 kg . I = I CM + md 2 = 1 12 m 2 + m 2 = 13 12 m 2 . Since the rod is uniform, its center of mass is located at . Recalling that the weight mg acts at the center of mass, the magnitude of the torque at the horizontal position is = I r F = parenleftbigg 13 12 m 2 parenrightbigg ( mg ) = 13 12 m 2 = 12 g 13 = 12 (9 . 8 m / s 2 ) 13 (17 m) = . 532127 rad / s 2 . 002 10.0 points Three spherical masses are located in a plane at the positions shown in the figure below. A has mass 50 . 7 kg, B has mass 42 . 9 kg, and C has mass 35 . 9 kg. 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 A B C y Distance(m) x Distance (m) Calculate the moment of inertia (of the three masses) with respect to an axis perpen- dicular to the xy plane and passing through x = 4 m and y = 4 . 5 m . Assume the masses are point particles; e.g. , neglect the contri- bution due to moments of inertia about their center of mass. Correct answer: 1753 . 17 kg m 2 . Explanation: alexander (jra2623) oldhomework 21 Turner (92510) 2 Using (4 m , 4 . 5 m) as the coordinate origin (see figure below), we have x a = (1 . 5 m)- (4 m) =- 2 . 5 m , y a = (4 . 5 m)- (4 . 5 m) = 0 m , x b = (7 . 5 m)- (4 m) = 3 . 5 m , y b = (0...
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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 at Austin.

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oldhw 21 - alexander (jra2623) oldhomework 21 Turner...

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