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Unformatted text preview: Husain, Zeena Homework 10 Due: Nov 4 2003, 4:00 am Inst: H L Berk 1 Thisprintoutshouldhave19questions, 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 . A bicycle wheel of mass m rotating at an angular velocity 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 W=mg The magnitude of the angular momentum of the wheel is given by 1. 1 4 mR 2 2. 1 2 mR 2 3. mR 2 2 4. mR 2 correct 5. 1 2 mR 2 2 6. 1 4 mR 2 2 Explanation: Solution: Basic Concepts: ~ = d ~ L dt Top view L L + L L The magnitude of the angular momentum of the wheel, L , is L = I = mR 2 , since the moment of inertia of the wheel, I , is mR 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 precessionangleinthetimeinterval t = 1 . 8s. Correct answer: 228 . 768 . Explanation: From the figure below, we get = L L . Using the relation, L = t , where is the magnitude of the torque, mg b , we get = L L = t L = mg b t mR 2 = g b t R 2 = (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: 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 = 28 . Basic Concepts: X ~ F = 0 X ~ = 0 . Solution: The sum of the torques about the pivot is T L 2 sin  mg L 2 2 M 1 gL 1 M 2 gL 2 = 0 Solving for the tension T T = mg L 2 2 + M 1 g L 1 + M 2 g L 2 L 2 sin = (26 kg)(9 . 8 m / s 2 ) 7 m 2 (7 m)sin + (41 kg)(9 . 8 m / s 2 )(3 . 3 m) (7 m)sin + (20 kg)(9 . 8 m / s 2 )(7 m) (7 m)sin = 1 . 09233 kN ....
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This note was uploaded on 01/27/2009 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
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