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Unformatted text preview: khounvivongsy (sk27799) Homework 16 Weathers (17104) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points The hour and minute hands of Big Ben in London are 2 . 79 m and 4 . 4 m long and have masses of 62 . 2 kg and 97 kg respectively. Calculate the total angular momentum of the minute hand about the center point. Treat the hand as long, thin rod. Treat into the clock as the positive direction. Correct answer: 1 . 09253 kg m 2 / s. Explanation: Basic Concepts: L = I Moment of inertia of a thin rod pivoted through one of its ends is I = m l 2 3 , where m is its mass and l is its length. Solution: Applying this formula to the minute hand we obtain its moment of inertia: I min = m min l 2 min 3 = (97 kg)(4 . 4 m) 2 3 = 625 . 973 kg m 2 The angular momentum is L min = I min min = ( 625 . 973 kg m 2 ) (0 . 00174533 1 / s) = 1 . 09253 kg m 2 / s 002 (part 2 of 2) 10.0 points Calculate the total angular momentum of the hour hand about the center point. Treat the hand as long, thin rod, and into the clock as the positive direction. Correct answer: 0 . 0234733 kg m 2 / s. Explanation: This part is to be solved similarly to the previ ous part. We obtain for the moment of inertia of the hour hand: I hr = m hr l 2 hr 3 = (62 . 2 kg)(2 . 79 m) 2 3 = 161 . 39 kg m 2 The angular momentum is L hr = I hr hr = (161 . 39 kg m 2 )(0 . 000145444 1 / s) = 0 . 0234733 kg m 2 / s 003 10.0 points A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9 . 8 m / s 2 . v 9 . 8 m / s 2 3 . 1 m 6 kg 3 Calculate the magnitude of the angular mo mentum of the bob about the supporting point. Correct answer: 27 . 5411 kg m 2 / s. Explanation: Basic Concepts: Angular Momentum L = m vectorr vectorv . Note: In this problem vectorr and vectorv are perpendic ular, where r = sin . Let : = 3 . 1 m , = 30 , g = 9 . 8 m / s 2 , and m = 6 kg . khounvivongsy (sk27799) Homework 16 Weathers (17104) 2 Use the free body diagram below. T m g Solution: The second Newtons law in the vertical and horizontal projections, respec tively, in our case reads T cos  m g = 0 T sin  m 2 sin = 0 , where T is the force with which the wire acts on the bob and the radius of the orbit is R = sin . From this system of equations we find = radicalbigg g...
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This note was uploaded on 05/09/2010 for the course PHYS 1710 taught by Professor Weathers during the Spring '09 term at University of North Carolina Wilmington.
 Spring '09
 weathers
 mechanics, Work

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