In the figure shown, the plank is being pulled to the right with a constant velocity $v$. If the cylinder does not slip, then

If $v_{cm}$ is the velocity of $C.M$, then

From ground frame , velocity of the topmost point of the cylindre is V (leftwards) and velocity of the lowermost point of the cylinder is V(rightwards ).

Therefore $V_{cm}=\frac{-V+V}{2}=0$.

So $\omega =\frac{V-\left(-V\right)}{2R}=\frac{V}{R}$