Could my teacher be wrong?

  • Thread starter charlesc56
  • Start date
  • #1
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
 
Last edited:

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
I don't follow the middle of your argument. It seems simpler to me.
The torque required to follow the linear acceleration is
T = I*α
Ff*R = ½mR²*a/R
μmgR = ½mR*a
μg = a/2
μ = a/(2g) or greater.
Curious that it falls in between your answer and the "right" one!
 
  • #3
Hi Delphi,
Thanks for your reply. Your deduction is right, but we have to express the coefficient in terms of a0 (the acceleration of the plank) and not just a (the acceleration of the cylinder ON the plank - which is due to rotation)
 
  • #4
Any ideas then? ;)
 
  • #5
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
683
Your teacher is right. Here's your mistake:
I = (1/2)MR2 (moment of inertia for a homogeneous cylinder)
That's the moment of inertia for a cylinder rotating about the central axis. That is not the axis of rotation if you want to treat this as pure rotation. If you want to use that formula for the moment of inertia you also need to consider the translational acceleration of the cylinder. Either way (pure rotation about some non-central axis (which one?) or rotation about the central axis + translation of the center of mass), you will arrive at your teacher's answer if you do the math right.
 
  • #6
Thanks a lot D H. I appreciate it =)
 

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