Homework 15 - Gozick Brandon Homework 15 Due Nov 3 2006...

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Gozick, Brandon – Homework 15 – Due: Nov 3 2006, 5:00 pm – Inst: D Weathers 1 This print-out should have 9 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 27 kg mass and a 17 kg mass are suspended by a pulley that has a radius of 6 . 3 cm and a mass of 8 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 1 . 3 m apart. Treat the pulley as a uniform disk. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 3 m 6 . 3 cm 8 kg ω 27 kg 17 kg Determine the speeds of the two masses as they pass each other. Correct answer: 1 . 62916 m / s. Explanation: Let : M = 8 kg , R = 6 . 3 cm , m 1 = 27 kg , m 2 = 17 kg , h = 1 . 3 m , v = ω R , I = 1 2 M R 2 , and K disk = 1 2 I ω 2 = 1 4 M v 2 . From conservation of energy K 1 + K 2 + K disk = - Δ U or m 1 v 2 2 + m 2 v 2 2 + M v 2 4 = ( m 1 - m 2 ) g h 2 m 1 + m 2 + M 2 v 2 = ( m 1 - m 2 ) g h , where h 2 is the height. Taking no slipping into account, we can solve for v v = s 2 ( m 1 - m 2 ) g h 2 ( m 1 + m 2 ) + M (1) = s 127 . 4 J 48 kg = 1 . 62916 m / s . Alternative Solution: The forces in the vertical directions for m 1 , m 2 , and the torque on the pulley give us m 1 g - T 1 = + m 1 a m 2 g - T 2 = - m 2 a , so T 1 = m 1 ( g - a ) T 2 = m 2 ( g + a ) , and I α = [ T 1 - T 2 ] R 1 2 M R 2 a R · = h m 1 ( g - a )] R - [ m 2 ( g + a ) i R 1 2 M a = m 1 g - m 1 a - m 2 a - m 2 g a = 2 ( m 1 - m 2 ) g M + 2 ( m 1 + m 2 ) . The velocity is v 2 f = v 2 i + a Δ y v f = r 2 a h 2 = s 2 ( m 1 - m 2 ) g h 2 ( m 1 + m 2 ) + M , which is the same as Eq. 1. keywords:
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Gozick, Brandon – Homework 15 – Due: Nov 3 2006, 5:00 pm – Inst: D Weathers 2 002 (part 1 of 1) 10 points A uniform disk of radius 1 . 3 m and mass 1 . 3 kg is suspended from a pivot 0 . 13 m above its center of mass. The acceleration of gravity is 9 . 8 m / s 2 . axis Find the angular frequency ω for small os- cillations. Correct answer: 1 . 21578 rad / s. Explanation: Basic Concepts The physical pendulum: τ = I α = - m g d sin θ α = d 2 θ dt 2 so that the angular frequency for small oscil- lations (sin θ θ ) is ω = r m g d I .
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