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Unformatted text preview: Husain, Zeena – Homework 9 – Due: Oct 28 2003, 4:00 am – Inst: H L Berk 1 This printout should have 22 questions, check that it is complete. Multiplechoice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 3) 10 points Given: g = 9 . 8 m / s 2 . Two pulley wheels, one of radius 0 . 015 m, the other of radius 0 . 03 m, are mounted rigidly on a common axle, as in the figure. The rotational inertia of the two pulleys, which are clamped together, is 1 . 7 kg m 2 . A(n) 1 . 3 kg mass is attached on the left and a(n) 1 . 8 kg mass on the right, as shown. 1 2 m R 1 m R 2 Find the angular acceleration of the system. Take clockwise direction as positive. Correct answer: 0 . 198659 rad / s 2 . Explanation: 1 2 m g R 1 m g R 2 T 1 2 T 2 T T 1 Let’s assume m 2 moves downward and take motion downward as positive for m 2 and mo tion upward as positive for m 1 , take clockwise as positive for the pulley wheels. Since the two masses are connected to the pulleys, their accelerations are a 1 = α R 1 and a 2 = α R 2 . Applying Newton’s second law to m 1 , m 2 , and the pulleys separately, we obtain T 1 m 1 g = m 1 a 1 = m 1 α R 1 (1) m 2 g T 2 = m 2 a 2 = m 2 α R 2 (2) τ net = T 2 R 2 T 1 R 1 = I α . (3) Solving these equations, we find α = m 2 g R 2 m 1 g R 1 I + m 1 R 2 1 + m 2 R 2 2 = 0 . 198659 rad / s 2 . The fact that α > 0 indicates that our pre vious assumption that m 2 moves downward is right. If α < 0, our assumption was not correct and m 2 would move upward, but the equations obtained to calculate T 1 and T 2 would be still correct and we need not reas sume the direction of the motion and do it again. 002 (part 2 of 3) 10 points Find the tension in the cord attached to the 1 . 3 kg mass. Correct answer: 12 . 7439 N. Explanation: T 1 = m 1 ( g + α R 1 ) = 12 . 7439 N . 003 (part 3 of 3) 10 points Find the tension in the cord attached to the 1 . 8 kg mass. Correct answer: 17 . 6293 N. Explanation: T 2 = m 2 ( g α R 2 ) = 17 . 6293 N . 004 (part 1 of 3) 10 points Given: g = 9 . 8 m / s 2 . A thin cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, without slipping, from the top of an inclined plane that is 3 . 4 m above the ground. Husain, Zeena – Homework 9 – Due: Oct 28 2003, 4:00 am – Inst: H L Berk 2 Find the final linear velocity of the thin cylindrical shell. Correct answer: 5 . 77235 m / s. Explanation: Basic Concept: The mechanical energy is conserved K rot + K tran = U i . H S h 1 2 S Solution: Because there is no slipping, v 1 = ω 1 R The rotational inertia of the cylindrical shell is I 1 = m R 2 Thus, from conservation of energy 1 2 I 1 ω 2 1 + 1 2 m ( ω 1 R ) 2 = m g H (1) 1 2 ( m R 2 ) ‡ v 1 R · 2 + 1 2 m v 2 1 = m g H ....
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 Fall '08
 Turner
 Angular Momentum, Kinetic Energy, Work, Moment Of Inertia, Correct Answer, Husain

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