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 the frictional force at point P?

Let us analyze from the frame of reference of the rug which is moving towards the left with an acceleration a. The net horizontal force on the cylinder = Ma (pseudo force) - Ffriction towards the right. The linear acceleration of the center of mass of the cylinder = (Ma - Ffriction)/M towards the right. The net torque on the cylinder = FfrictionR, where R is the radius of the cylinder. The angular acceleration of the cylinder α = FfrictionR/I clockwise, where I is the moment of inertia of the cylinder with respect to the axis passing through the center parallel to it; I = MR²/2. From the above two, the linear acceleration of point P = (Ma - Ffriction)/M - αR. Since the cylinder does not slip, point P is at rest in our frame of reference, ⇒ (Ma - Ffriction)/M = αR ⇒ (Ma - Ffriction)/M = FfrictionR/(MR²/2)R ⇒ Ma - Ffriction = 2Ffriction