Sample Problem

A cylinder of mass M and radius R has a string wrapped around it, with the string coming off the cylinder above the cylinder. If the string is pulled to the right with a force F, what is the acceleration of the cylinder if the cylinder rolls without slipping?

  1. a = F/M
  2. a < F/M
  3. a > F/M









One key to the problem is realizing that there is a static friction force acting on the cylinder. Let's assume this force acts to the left.

Draw the free-body diagram. Take positive to the right, and clockwise.

The normal force cancels Mg vertically. Apply Newton's Second Law for horizontal forces, and for torques:
Forces | Torques
ΣFx = Ma | Στ = I α
F - fs = Ma | RF + Rfs = ½ MR2α
For rolling without slipping, α =
a
R

The torque equation becomes:
RF + Rfs =
MR2a
2R

All the R's cancel, leaving:

F + fs = ½ Ma

The force equation is:

F - fs = Ma

Subtracting the force equation from the torque equation gives:
2fs = -
Ma
2
        so     fs = -
Ma
4

The minus sign means the force of friction is not to the left, as we assumed, but to the right.
The equations also show that fs = -
F
3
and that a =
4 F
3 M

This is rather surprising, but seeing as we've carefully applied Newton's Laws we can be confident in our answer.