# why does rolling without slipping imply no work done by the frictional force?

Suppose we have a cylinder (mass $m$, radius $R$, and moment of inertia $I$) that rolls without slipping straight down an inclined plane which is at an angle $\alpha$ from the horizontal. The quickest way to solve for the motion is to use conservation of energy, assuming that there is no work down by the frictional force.

My question is: why does rolling without slipping imply no work done by the frictional force? One explanation you will sometimes read is that this is because the point of contact of the cylinder is instantaneously at rest relative to the ramp surface, so when we compute the work via $W = F\ dx$ we have $dx = 0$ and hence $dW = 0$.

However, consider a related problem. Suppose the cylinder is held fixed in place (say by a rod drilled down its middle) but is free to rotate around its center. Now suppose we take the ramp up against the cylinder and move the ramp so as to get the cylinder spinning. In this case the frictional force obviously did work on the cylinder, yet once again the contact point of the cylinder is at rest relative to the ramp. Question: what is the difference between the two cases?

• I tried to make it a little more clear what you're asking here (I'm afraid there's a knee-jerk reaction here to mentioning homework problems by number). Feel free to roll back my changes if you don't like them. – Chris May 20 '18 at 2:01

Note that "work done" is a quantity that depends on the reference frame your are working in. Your argument says that there is no work done on the cylinder in the reference frame where $dx=0$. That is only the same as the lab frame if the ramp is not moving.