HW9 Solution - salazar(ggs332 HW09 TSOI(58160 This...

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salazar (ggs332) – HW09 – TSOI – (58160) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A circular-shaped 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 ±ind 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 vr = (1 . 7 m)ˆ ı + (3 . 6 m)ˆ and the Force acting on it is v 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: v τ = × v F Solution: Since neither position oF the par- ticle, nor the Force acting on the particle have the z -components, the torque acting on the particle has only z -component: v 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 di²erence that relative to the point [(0 . 8 m) , (7 m)] the y - component oF the particle is now [ y - (7 m)], we have v 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 .
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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 kg 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 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 . The rod’s moment of inertia about its end- point is I = 1 3 2 , so the angular accelera- tion of the rod is α = τ I = 1 2 mg L cos θ 1 3 2 = 3 2 g L cos θ = 3 2 9 . 8 m / s 2 4 m cos 21 = 3 . 43091 rad / s 2 . 005 10.0 points Consider the setup shown, where the inclined plane has a frictionless surface. The blocks have masses m 2 and m 1 . The pulley has mass m 3 , and is a uniform disc with radius R . Assume the pulley to be frictionless. m 1 T 1 R m 3 m 2 μ = 0 T a θ What is the acceleration of the blocks? As- sume the mass m 1 is more massive and is descending with acceleration a . The moment of inertia of a disk is 1 2 M R 2 . The accelera- tion of gravity is 9 . 8 m / s 2 .
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HW9 Solution - salazar(ggs332 HW09 TSOI(58160 This...

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