Rolling cylinder on a conveyor Suppose we have a rolling cylinder on a moving conveyor. The velocity of the conveyor is such that the cylinder remains at rest against the ground.
In this picture, the green dot remains in repaus (the blue cylinder it is at rest against the ground).

The question is:What happens to the cylinder after the conveyor suddenly stops?
 A: 
We assume that the cylinder was rotating (but not translating) without slipping. In that case the angular velocity $\omega$ and conveyor belt speed $v$ relate as:
$$v=\omega R$$
Where $R$ is the radius of the cylinder.
We also assume the conveyor belt stops suddenly ($\Delta t_\textrm{stop}=0$).
A friction force $F_f$, depicted in red, now acts on the cylinder and in the absence of drag, rolling resistance or other forces in the horizontal plane the cylinder will now start accelerating in the horizontal plane and to the left, acc. Newton's second:
$$F_f=ma$$
Where $m$ is the cylinder's mass and $a$ its acceleration.
$F_f$ is usually modelled as:
$$F_f=\mu mg$$
Where $\mu$ is a friction coefficient.
So:
$$\mu mg=ma \implies a=\mu g$$
A: The conveyor has a (linear) speed of $v$. In order for the cylinder to remain at the same position is has to rotate with a certain angular velocity $\omega$. If you have the velocity of the conveyor $v$ and know that the cylinder does not slip on the belt, you can compute $\omega$.
Then you know that the cylinder rotates with a certain angular velocity. Then use Newton's first axiom, namely that bodies in motion (or at rest) stay in motion (or at rest) as long as no force acts on them. What forces act on the cylinder when the conveyor stopped? Is the motion of the cylinder guided in any way?
If you have thought about those questions, leave me a comment and I can expand this answer.
A: Assuming that the speed of the centre of mass of the cylinder relative to the ground is zero when the conveyor belt stops, there will be a frictional force $\mu_k m g$ to the left, where $\mu_k$ is the coefficient of kinetic friction and $m$ is the mass of the cylinder.  
The frictional force on the cylinder will do two things to the cylinder.


*

*The frictional force will accelerate the centre of mass of the cylinder to the left and give it an increasing linear speed to the left $v_{\rm CM}$

*The clockwise torque about the centre of mass produced by the frictional force will produce a clockwise angular acceleration of the cylinder which will produce a decreasing anti-clockwise angular speed of the cylinder $\omega$.
This will continue until the no slip condition $v_{\rm CM}=r\omega$, where $r$ is the radius of the cylinder,  is reached and then there is no frictional force acting on the cylinder.
