When an object is rolling without slipping, the rolling

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

Problem 101CP Chapter 10

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 101CP

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that ?x and ?z are approximately zero and vx and ?z are approximately constant. Rolling without slipping means vx = r?z and ax = r?z. If an object is set in motion on a surface ?without? these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass M and radius R, rotating with angular speed ?0 about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is ┬Ák. (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations ax of the center of mass and ?z of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially ?z = ?0 but vx = 0. Rolling without slipping sets in when vx = r?z. Calculate the ?distance? the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

Step-by-Step Solution:

Solution 101CP Step 1 of 3: a) f = n = Mg, so a = g k k k The magnitude of the angular acceleration is fR = kMgR I (2)MR2 2kg = R Step 2 of 3: b)v = at = R = ( t0R R 0 t = a+R = R0 k+2 k R0 = 3 k d = at 2 2 2 1 R 0 2 = (2 g)k 3 kg) R 2 = 0 18 kg

Step 3 of 3

Chapter 10, Problem 101CP is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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