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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Husain, Zeena Homework 10 Due: Nov 4 2003, 4:00 am Inst: H L Berk 1 Thisprint-outshouldhave19questions, 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 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 ....
View Full Document

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.

Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online