# Cm acceleration

1. Feb 24, 2010

### enippeas

hello and sorry for my english
Let us consider a rolling whithout slipping cylinder. We can work in two different ways. First considering the motion as a pure translation of the cm plus a rotation about the cm. Second, considering the motion as a pure rotation about the instantaneous axis from the contact point with the ground.
The cylinder is moving with constant velocity. In the first way, the cm has no acceleration. But in the second way, the cm has a centripetal acceleration vcm^2/R.
What is going wrong here?

2. Feb 24, 2010

### torquil

Well, in the second picture, the cylinder really doesn't have a rotational motion about the axis at the contact point. This is only a lowest order approximation to the actual motion. The approximation is so bad that it will only give you the correct result for the velocity of the motion, not the acceleration.

Torquil

3. Feb 24, 2010

### Staff: Mentor

While using an instantaneous axis of rotation at the contact point is OK for some purposes, realize that the contact point is itself accelerating so it is not an inertial frame.

4. Feb 24, 2010

### Bob S

Although all points on the rim of the cylinder have centripetal acceleration relative to the cm, the force is perpendicular to the velocity, so no work is done.

Bob S