hw9 - cerutti (cpc566) HW09 TSOI (58160) 1 This print-out...

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: cerutti (cpc566) HW09 TSOI (58160) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A circular-shaped object of mass 7 kg has an inner radius of 14 cm and an outer radius of 25 cm. Three forces (acting perpendicular to the axis of rotation) of magnitudes 11 N, 22 N, and 13 N act on the object, as shown. The force of magnitude 22 N acts 26 below the horizontal. 11 N 13 N 22 N 26 Find the magnitude of the net torque on the wheel about the axle through the center of the object. Correct answer: 2 . 92 N m. Explanation: Let : a = 14 cm = 0 . 14 m , b = 25 cm = 0 . 25 m , F 1 = 11 N , F 2 = 22 N , F 3 = 13 N , and = 26 . F 1 F 3 F 2 The total torque is = a F 2- b F 1- b F 3 = (0 . 14 m) (22 N)- (0 . 25 m) (11 N + 13 N) =- 2 . 92 N m , with a magnitude of 2 . 92 N m . 002 (part 1 of 2) 10.0 points A particle is located at the vector position vectorr = (2 . 2 m) + (3 . 9 m) and the force acting on it is vector F = (3 . 5 N) + (4 . 3 N) . What is the magnitude of the torque about the origin? Correct answer: 4 . 19 N m. Explanation: Basic Concept: vector = vectorr vector F Solution: Since neither position of the par- ticle, nor the force acting on the particle have the z-components, the torque acting on the particle has only z-component: vector = [ x F y- y F x ] k = [(2 . 2 m) (4 . 3 N)- (3 . 9 m) (3 . 5 N)] k = [- 4 . 19 N m] k . 003 (part 2 of 2) 10.0 points What is the magnitude of the torque about the point having coordinates [ a, b ] = [(2 m) , (7 m)]? Correct answer: 11 . 71 N m. Explanation: Reasoning similarly as we did in the pre- vious section, but with the difference that relative to the point [(2 m) , (7 m)] the y- component of the particle is now [ y- (7 m)], we have vector = { [ x- a ] F y- [ y- b ] F x } k = { [(2 . 2 m)- (2 m)] [4 . 3 N]- [(3 . 9 m)- (7 m)] [3 . 5 N] } k = { 11 . 71 N m } k . cerutti (cpc566) HW09 TSOI (58160) 2 004 10.0 points A uniform 2 kg rod with length 40 m has a frictionless pivot at one end. The rod is released from rest at an angle of 30 beneath the horizontal. 2 m 4 m 2 k g 30 What is the angular acceleration of the rod immediately after it is released? The moment of inertia of a rod about the center of mass is 1 12 mL 2 , where m is the mass of the rod and L is the length of the rod. The moment of inertia of a rod about either end is 1 3 mL 2 , and the acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 318264 rad / s 2 . Explanation: Let : m = 2 kg , L = 40 m , and = 30 ....
View Full Document

Page1 / 14

hw9 - cerutti (cpc566) HW09 TSOI (58160) 1 This print-out...

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