# CW11 - 713 m and moment oF inertia 1 2 M r 2 is used to...

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boland (kbb544) – CW11 – mackie – (10611) 1 This print-out should have 2 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 6 kg has an inner radius oF 6 cm and an outer radius oF 23 cm. Three Forces (acting perpendicular to the axis oF rotation) oF magnitudes 12 N, 23 N, and 15 N act on the object, as shown. The Force oF magnitude 23 N acts 29 below the horizontal. 12 N 15 N 23 N 29 ω ±ind the magnitude oF the net torque on the wheel about the axle through the center oF the object. Correct answer: 4 . 83 N · m. Explanation: Let : a = 6 cm = 0 . 06 m , b = 23 cm = 0 . 23 m , F 1 = 12 N , F 2 = 23 N , F 3 = 15 N , and θ = 29 . F 1 F 3 F 2 θ ω The total torque is τ = a F 2 - b F 1 - b F 3 = (0 . 06 m) (23 N) - (0 . 23 m) (12 N + 15 N) = - 4 . 83 N · m , with a magnitude oF 4 . 83 N · m . 002 10.0 points A cylindrical pulley with a mass oF 5 . 6 kg, ra- dius oF 0

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Unformatted text preview: . 713 m and moment oF inertia 1 2 M r 2 is used to lower a bucket with a mass oF 1 . 8 kg into a well. The bucket starts From rest and Falls For 3 . 4 s. r M m What is the linear acceleration oF the Falling bucket? The acceleration oF gravity is 9 . 8 m / s 2 . Correct answer: 3 . 83478 m / s 2 . Explanation: Let : M = 5 . 6 kg , r = 0 . 713 m , m = 1 . 8 kg , and g = 9 . 8 m / s 2 . Let T be the tension in the cord and α the angular acceleration oF the wheel. Newton’s equation For the mass m is m g-T = m a T = m ( g-a ) and For the disk T r = I α ( m g-m a ) r = p 1 2 M r 2 P ± a r ² 2 m g-2 m a = M a 2 m g = ( M + 2 m ) a boland (kbb544) – CW11 – mackie – (10611) 2 a = 2 m g M + 2 m = 2 (1 . 8 kg) (9 . 8 m / s 2 ) 5 . 6 kg + 2 (1 . 8 kg) = 3 . 83478 m / s 2 ....
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## This note was uploaded on 02/20/2011 for the course PHYS 1062 taught by Professor Mackie during the Spring '11 term at Temple.

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CW11 - 713 m and moment oF inertia 1 2 M r 2 is used to...

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