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 cylinder. What is $F_{friction}$ at point $P$ ? It is assumed that the cylinder does not slip.

Question

Find external force F so that block can move on inclined plane with constant velocity.

Answer 1

To determine the frictional force $F_{friction}$ at point $P$, consider a cylinder of mass $M$ placed on a rough horizontal rug. The rug is pulled with acceleration $a'$ perpendicular to the axis of the cylinder. We assume that the cylinder does not slip on the rug.

Since the cylinder does not slip, it means that both the translational and rotational effects are maintained such that:

The net translational force on the cylinder due to friction must produce an acceleration equal to $a'$. Thus,

F_{friction} = Ma' \ ... (1)

Now, consider the rotational motion of the cylinder. The frictional force also contributes to a torque about the center of mass, thereby inducing angular acceleration \alpha \.

From the torque equation, since the radius is $R$,

\tau = I \alpha = F_{friction} \cdot R \ ... (2)

Where I is the moment of inertia of the cylinder: I = \frac{1}{2}MR^2 for the central axis of a cylinder.

Upon combining equation (1) and (2) and substituting \alpha = \frac{a'}{R} since a = R\alpha for rolling without slipping:

F_{friction} = I \cdot \frac{a'}{R^2} \Rightarrow \frac{1}{2}MR^2 \cdot \frac{a'}{R^2} = F_{friction}

Now substitute:

F_{friction} = \frac{Ma'}{3}

Thus, the correct answer is \frac{Ma}{3}

Answer 2