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 + f_s = ½ Ma

The force equation is:

F - f_s = 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.