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HS000010

# HS000010 - Busto Anthony Homework 10 Due 11:00 pm Inst...

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Busto, Anthony – Homework 10 – Due: Nov 17 2006, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A cylindrical pulley with a mass of M = 4 . 7 kg and a radius of r = 0 . 562 m is used to lower a bucket with a mass of m = 2 kg into a well. The bucket starts from rest and falls for 2 . 7 s. The acceleration of gravity is 9 . 8 m / s 2 . r M m What is the linear acceleration of the falling bucket? Correct answer: 4 . 50575 m / s 2 . Explanation: The disk has moment of inertia, I = 1 2 M r 2 . Let T be the tension in the cord and let α be the angular acceleration of the wheel. Newton’s equation for the disk is T r = I α . (1) Newton’s equation for the mass m is m g - T = m a . (2) The linear acceleration a of the mass is related to the angular acceleration α of the wheel by the formula, a = α r . From Eq. (2) the tension is T = m ( g - a ). Putting this into Eq. (1) and solving for a gives ( m g - m a ) r = 1 2 M r 2 a r · 2 m g - 2 m a = M a 2 m g = ( M + 2 m ) a a = 2 m g M + 2 m . Or a = g 1 1 + M 2 m = ( 9 . 8 m / s 2 ) 1 1 + (4 . 7 kg) 2 (2 kg) = 4 . 50575 m / s 2 . keywords: 002 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . Consider a thin rod of mass m = 1 . 7 kg, length = 5 . 8 m and uniform density. The rod is pivoted at one end on a frictionless horizontal pin. 5 . 8 m 1 . 7 kg O 63 The rod is initially held in horizontal po- sition but eventually allowed to swing down.

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HS000010 - Busto Anthony Homework 10 Due 11:00 pm Inst...

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