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Homework 15 Solutions

Homework 15 Solutions - lim(kl9356 Homework 15...

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lim (kl9356) – Homework 15 – Weathers – (17104) 1 This print-out should have 12 questions. Multiple-choice 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 . 65 m and 4 . 47 m long and have masses of 57 . 6 kg and 95 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 . 10432 kg · m 2 / s. Explanation: Basic Concepts: L = I ω 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 = (95 kg)(4 . 47 m) 2 3 = 632 . 728 kg · m 2 The angular momentum is L min = I min ω min = ( 632 . 728 kg · m 2 ) (0 . 00174533 1 / s) = 1 . 10432 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 . 0196105 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 = (57 . 6 kg)(2 . 65 m) 2 3 = 134 . 832 kg · m 2 The angular momentum is L hr = I hr ω hr = (134 . 832 kg · m 2 )(0 . 000145444 1 / s) = 0 . 0196105 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 1 . 5 m 9 kg 31 Calculate the magnitude of the angular momentum of the bob about the support- ing point. Correct answer: 14 . 8299 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 θ . Let : = 1 . 5 m , θ = 31 , g = 9 . 8 m / s 2 , and m = 9 kg .

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lim (kl9356) – Homework 15 – Weathers – (17104) 2 Use the free body diagram below. T m g θ Solution: The second Newton’s law in the vertical and horizontal projections, respec- tively, in our case reads T cos θ - m g = 0 T sin θ - m ω 2 sin θ = 0 , where T is the force with which the wire acts on the bob and the radius of the orbit is R = sin θ . From this system of equations we find ω = radicalbigg g cos θ = radicalBigg (9 . 8 m / s 2 ) (1 . 5 m) cos(31 ) = 2 . 7608 rad / s .
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Homework 15 Solutions - lim(kl9356 Homework 15...

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