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12homework10

# 12homework10 - Husain Zeena Homework 10 Due Nov 4 2003 4:00...

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Husain, Zeena – Homework 10 – Due: Nov 4 2003, 4:00 am – Inst: H L Berk 1 This print-out should have 19 questions, check that it is complete. Multiple-choice 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 . A bicycle wheel of mass m rotating at an angular velocity ω has its shaft supported on one side, as shown in the figure. When viewing from the left, one sees that the wheel is rotating in a counterclockwise manner. The distance from the center of the wheel to the pivot point is b . We assume the wheel is a hoop of radius R , and the shaft is horizontal. b τ W=mg ϖ The magnitude of the angular momentum of the wheel is given by 1. 1 4 m R 2 ω 2. 1 2 m R 2 ω 3. m R 2 ω 2 4. m R 2 ω correct 5. 1 2 m R 2 ω 2 6. 1 4 m R 2 ω 2 Explanation: Solution: Basic Concepts: ~ τ = d ~ L dt Top view L τ L + L L ∆φ The magnitude of the angular momentum of the wheel, L , is L = I ω = m R 2 ω, since the moment of inertia of the wheel, I , is m R 2 . 002 (part 2 of 3) 10 points Given: the mass 3 kg, the angular velocity 10 rad / s, the axil length b = 0 . 5 m, and the radius of the wheel R = 0 . 47 m. Find the precession angle in the time interval t = 1 . 8 s. Correct answer: 228 . 768 . Explanation: From the figure below, we get ∆ φ = L L . Using the relation, ∆ L = τ t , where τ is the magnitude of the torque, mg · b , we get φ = L L = τ t L = m g b t m R 2 ω = g b t R 2 ω = (9 . 8 m / s 2 )(0 . 5 m)(1 . 8 s) (0 . 47 m) 2 (10 rad / s) = 3 . 99276 rad = 228 . 768 . 003 (part 3 of 3) 10 points The direction of precession as viewed from the top is:

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Husain, Zeena – Homework 10 – Due: Nov 4 2003, 4:00 am – Inst: H L Berk 2 1. along the direction of rotation of the wheel 2. clockwise 3. opposite to the direction of rotation of the wheel 4. counterclockwise correct Explanation: From the figure, we can see the direction of precession is counterclockwise. 004 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . Two weights attached to a uniform beam of mass 26 kg are supported in a horizontal position by a pin and cable as shown in the figure. 41 kg 20 kg 3 . 3 m 7 m 28 26 kg What is the tension in the cable which sup- ports the beam? Correct answer: 1 . 09233 kN. Explanation: Let : m = 26 kg , M 1 = 41 kg , M 2 = 20 kg , L 1 = 3 . 3 m , L 2 = 7 m , and θ = 28 . Basic Concepts: X ~ F = 0 X ~ τ = 0 . Solution: The sum of the torques about the pivot is T L 2 sin θ - mg L 2 2 - M 1 gL 1 - M 2 gL 2 = 0 Solving for the tension T T = m g L 2 2 + M 1 g L 1 + M 2 g L 2 L 2 sin θ = (26 kg)(9 . 8 m / s 2 ) 7 m 2 (7 m) sin θ + (41 kg)(9 . 8 m / s 2 )(3 . 3 m) (7 m) sin θ + (20 kg)(9 . 8 m / s 2 )(7 m) (7 m) sin θ = 1 . 09233 kN .
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