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Unformatted text preview: salazar (ggs332) – HW09 – TSOI – (58160) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A circularshaped object of mass 11 kg has an inner radius of 15 cm and an outer radius of 21 cm. Three forces (acting perpendicular to the axis of rotation) of magnitudes 12 N, 27 N, and 14 N act on the object, as shown. The force of magnitude 27 N acts 21 ◦ below the horizontal. 12 N 14 N 27 N 21 ◦ ω Find the magnitude of the net torque on the wheel about the axle through the center of the object. Correct answer: 1 . 41 N · m. Explanation: Let : a = 15 cm = 0 . 15 m , b = 21 cm = 0 . 21 m , F 1 = 12 N , F 2 = 27 N , F 3 = 14 N , and θ = 21 ◦ . F 1 F 3 F 2 θ ω The total torque is τ = a F 2 b F 1 b F 3 = (0 . 15 m) (27 N) (0 . 21 m) (12 N + 14 N) = 1 . 41 N · m , with a magnitude of 1 . 41 N · m . 002 (part 1 of 2) 10.0 points A particle is located at the vector position vectorr = (1 . 7 m)ˆ ı + (3 . 6 m)ˆ and the force acting on it is vector F = (2 . 9 N)ˆ ı + (2 . 3 N)ˆ . What is the magnitude of the torque about the origin? Correct answer: 6 . 53 N m. Explanation: Basic Concept: vector τ = vectorr × vector F Solution: Since neither position of the par ticle, nor the force acting on the particle have the zcomponents, the torque acting on the particle has only zcomponent: vector τ = [ x F y y F x ] ˆ k = [(1 . 7 m) (2 . 3 N) (3 . 6 m) (2 . 9 N)] ˆ k = [ 6 . 53 N m] ˆ k . 003 (part 2 of 2) 10.0 points What is the magnitude of the torque about the point having coordinates [ a, b ] = [(0 . 8 m) , (7 m)]? Correct answer: 11 . 93 N m. Explanation: Reasoning similarly as we did in the pre vious section, but with the difference that relative to the point [(0 . 8 m) , (7 m)] the y component of the particle is now [ y (7 m)], we have vector τ = { [ x a ] F y [ y b ] F x } ˆ k = { [(1 . 7 m) (0 . 8 m)] [2 . 3 N] [(3 . 6 m) (7 m)] [2 . 9 N] } ˆ k = { 11 . 93 N m } ˆ k . salazar (ggs332) – HW09 – TSOI – (58160) 2 004 10.0 points A uniform 1 kg rod with length 4 m has a frictionless pivot at one end. The rod is released from rest at an angle of 21 ◦ beneath the horizontal. 2 m 4 m 1 k g 21 ◦ What is the angular acceleration of the rod immediately after it is released? The moment of inertia of a rod about the center of mass is 1 12 mL 2 , where m is the mass of the rod and L is the length of the rod. The moment of inertia of a rod about either end is 1 3 mL 2 , and the acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 3 . 43091 rad / s 2 . Explanation: Let : m = 1 kg , L = 4 m , and θ = 21 ◦ ....
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This note was uploaded on 11/03/2010 for the course PHYSICS 58160 taught by Professor Tsoi during the Spring '10 term at University of Texas.
 Spring '10
 TSOI
 Physics, Mass

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