Consider a cylinder of mass M resting on a rough horizontal rug that is pulled out from under it with acceleration 'a' perpendicular to the axis of the cylinder. What is $$F_{friction}$$ at point P? It is assumed that the cylinder does not slip.

For translational motion of the center of mass $$O$$: $$f = M a_0 \quad \dots (1)$$, where $$a_0$$ is the linear acceleration of the center of mass.

For rotational motion about the center of mass $$O$$: $$\tau = I \alpha \implies f \cdot R = \left( \frac{1}{2} M R^2 \right) \alpha$$

$$f = \frac{1}{2} M R \alpha \quad \dots (2)$$

The condition for no slipping at point $$P$$ is: $$a = a_0 + R \alpha$$

$$a = \frac{f}{M} + \frac{2f}{M}$$

$$a = \frac{3f}{M}$$

$$f = \frac{Ma}{3}$$