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1. Homework Statement

This is a relatively simple problem in classical mechanics, but my attempt at a solution seems to conflict with the result my teacher announced.

The problem is as follows:

A homogeneous cylinder with mass M and radius R lies at rest on a plank. The plank is then moved horizontally to one side with acceleration A0. What is the condition of the coefficient of friction to allow rolling without slipping?

My teacher announced the solution to be [itex]\mu[/itex]s [itex]\geq[/itex] a0/(3g)

2. Homework Equations

[itex]\alpha[/itex] * r = a (rolling without slipping)

[itex]\sum[/itex] [itex]\tau[/itex] = I * [itex]\alpha[/itex]

fstatic [itex]\leq[/itex] [itex]\mu[/itex]s Mg

a = fs / m - a0 (in the non-inertial frame of reference following the plank)

I = (1/2)MR2 (moment of inertia for a homogeneous cylinder)

3. The Attempt at a Solution

The only torque on the cylinder must be the friction force, so:

R*fs = I [itex]\alpha[/itex]

By applying the rolling without slipping condition (with a being the acceleration due to rolling) and the moment of inertia we get:

R*fs = (1/2)MR^2 * (a/R) [itex]\Rightarrow[/itex] fs = (1/2)*m*a.

Now the acceleration in the non-inertial frame of reference (if I'm not mistaking) is described by a = fs / m - a0 , so that means:

fs = (1/2)*m*a [itex]\Rightarrow[/itex] fs = (1/2)*(fs / m - a0 ) * M [itex]\Rightarrow[/itex] fs = -a0 * M

Finally if the motion has to be possible this condition has to be true: fstatic [itex]\leq[/itex] [itex]\mu[/itex]s Mg , so :

-a0 * M [itex]\leq[/itex] [itex]\mu[/itex]s Mg (ops, there is a sign error, but this isn't that important) [itex] \Rightarrow[/itex] [itex]\mu[/itex]s [itex]\geq[/itex] (a0/g).

This is not what my teacher reported. What's wrong? :P

This is a relatively simple problem in classical mechanics, but my attempt at a solution seems to conflict with the result my teacher announced.

The problem is as follows:

A homogeneous cylinder with mass M and radius R lies at rest on a plank. The plank is then moved horizontally to one side with acceleration A0. What is the condition of the coefficient of friction to allow rolling without slipping?

My teacher announced the solution to be [itex]\mu[/itex]s [itex]\geq[/itex] a0/(3g)

2. Homework Equations

[itex]\alpha[/itex] * r = a (rolling without slipping)

[itex]\sum[/itex] [itex]\tau[/itex] = I * [itex]\alpha[/itex]

fstatic [itex]\leq[/itex] [itex]\mu[/itex]s Mg

a = fs / m - a0 (in the non-inertial frame of reference following the plank)

I = (1/2)MR2 (moment of inertia for a homogeneous cylinder)

3. The Attempt at a Solution

The only torque on the cylinder must be the friction force, so:

R*fs = I [itex]\alpha[/itex]

By applying the rolling without slipping condition (with a being the acceleration due to rolling) and the moment of inertia we get:

R*fs = (1/2)MR^2 * (a/R) [itex]\Rightarrow[/itex] fs = (1/2)*m*a.

Now the acceleration in the non-inertial frame of reference (if I'm not mistaking) is described by a = fs / m - a0 , so that means:

fs = (1/2)*m*a [itex]\Rightarrow[/itex] fs = (1/2)*(fs / m - a0 ) * M [itex]\Rightarrow[/itex] fs = -a0 * M

Finally if the motion has to be possible this condition has to be true: fstatic [itex]\leq[/itex] [itex]\mu[/itex]s Mg , so :

-a0 * M [itex]\leq[/itex] [itex]\mu[/itex]s Mg (ops, there is a sign error, but this isn't that important) [itex] \Rightarrow[/itex] [itex]\mu[/itex]s [itex]\geq[/itex] (a0/g).

This is not what my teacher reported. What's wrong? :P

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