Question: 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 \({a_x}\) and \({a_z}\) are approximately zero and \({v_x}\) and\({\omega _z}\) are approximately constant. Rolling without slipping means\({v_x} = r{\omega _z}\;{\rm{and}}\;{a_x} = r{\alpha _z}\). If an object is set in motion on a surface withoutthese 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 \({\omega _0}\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \({\mu _k}\). (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 \({a_x}\) of the center of mass and \({a_z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \({\omega _z} = {\omega _0}\) but \({v_x} = 0\). Rolling without slipping sets in when \({v_x} = r{\omega _z}\). 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

01

Given Data

Kinetic friction coefficient is \({\mu _K}\)

Angular speed about an axis is\({\omega _0}\)

02

Concept

The rolling friction force is defined as the force that resists the movement of the rolling body.

03

Step 3(a)(i): Draw a free-body diagram

The force of kinetic friction in the electron of motion is

\(\omega \)is the supported by the surface due to normal force

04

Step 3(a)(i): Calculate the accelerations

The frictional force,

\(\begin{array}{c}{F_K} = {\mu _K}Mg\\\tau = I{a_z}\\I = \frac{1}{2}M{R^2}\end{array}\)

To find the accelerations,

\(\begin{array}{c}{F_K} = {\mu _K}Mg\\M{a_x} = {\mu _K}Mg\\{a_x} = {\mu _K}g\end{array}\)

\(\begin{array}{c}\tau = I{a_z}\\{F_K} = I{a_z}\\{a_z} = \frac{{{F_K}}}{I}\\ = \frac{{{\mu _K}MgR}}{{\frac{1}{2}M{R^2}}}\\{a_z} = \frac{{2{\mu _K}g}}{R}\end{array}\)

Hence the accelerations are \({a_x} = {\mu _K}g\), \({a_z} = \frac{{2{\mu _K}g}}{R}\)

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