HW 011 - hanna (brh687) hk 11 Opyrchal (11105) 1 This...

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: hanna (brh687) hk 11 Opyrchal (11105) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A bowling ball has a mass of 4 . 7 kg, a moment of inertia of 0 . 127088 kg m 2 , and a radius of . 26 m. It rolls along the lane without slipping at a linear speed of 3 . 7 m / s. What is the kinetic energy of the rolling ball? Correct answer: 45 . 0401 J. Explanation: Let : v = 3 . 7 m / s , r = 0 . 26 m , m = 4 . 7 kg , and I = 0 . 127088 kg m 2 . E = E rot + E kin = 1 2 I 2 + 1 2 mv 2 = 1 2 I v 2 r 2 + 1 2 mv 2 = 1 2 (0 . 127088 kg m 2 ) (3 . 7 m / s) 2 (0 . 26 m) 2 + 1 2 (4 . 7 kg)(3 . 7 m / s) 2 = 45 . 0401 J keywords: 002 10.0 points A solid 101 N ball with a radius of 0.490 m rolls 8.9 m down a ramp that is inclined at 25 with the horizontal. If the ball starts from rest at the top of the ramp, what is the angular speed of the ball at the bottom of the ramp? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 14 . 7054 rad / s. Explanation: Let: F g = 101 N r = 0 . 490 m d = 8 . 9 m = 25 g = 9 . 81 m / s 2 By conservation of energy, U i = K trans,f + K rot,f = 1 2 mv 2 f + 1 2 I 2 mg d (sin ) = 1 2 m ( r f ) 2 + 1 2 parenleftbigg 2 5 mr 2 parenrightbigg 2 f = 1 2 mr 2 2 f + 1 5 mr 2 2 f = 7 10 mr 2 2 f 2 f = 10 7 gd (sin ) r 2 f = radicalbigg 10 gd (sin ) 7 r 2 = radicalBigg 10(9 . 81 m / s 2 )(8 . 9 m)(sin 24 . 6 ) 7(0 . 49 m) 2 = 14 . 7054 rad / s . keywords: /* If you use any of these, fix the comment symbols. 003 (part 1 of 2) 10.0 points A particle is located at the vector position vectorr = (2 . 5 m) + (3 . 7 m) and the force acting on it is vector F = (5 . 5 N) + (4 N) . What is the magnitude of the torque about the origin? Correct answer: 10 . 35 N m. Explanation: hanna (brh687) hk 11 Opyrchal (11105) 2 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 . 5 m) (4 N) (3 . 7 m) (5 . 5 N)] k = [ 10 . 35 N m] k . 004 (part 2 of 2) 10.0 points What is the magnitude of the torque about the point having coordinates [ a, b ] = [(0 . 5 m) , (6 m)]?...
View Full Document

This note was uploaded on 04/07/2012 for the course PHYSICS 121 taught by Professor James during the Spring '12 term at University of Texas at Austin.

Page1 / 6

HW 011 - hanna (brh687) hk 11 Opyrchal (11105) 1 This...

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