Question

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.

• A. $Mg$
• B. $Ma$
• C. $\frac{Ma}{2}$
• D. $\frac{Ma}{3}$

Answer 1

Answer: (D)

$\frac{Ma}{3}$

Let us analyse from the frame of reference of the rug which is moving towards left with an acceleration $a$
net horizontal force on the cylinder = $Ma$ (pseudo force) $-F_{friction}$ towards right
linear acceleration of the center of mass of the cylinder = $\frac{Ma-F_{friction}}{M}$ towards right
net torque on the cylinder = $F_{friction}R$ where R is the radius of the cylinder
angular acceleration of the cylinder $\alpha =\frac{F_{friction}R}{I}$ clockwise
where I is the moment of inertia of the cylinder with respect to the axis passing through the center parallel to it; $I=M{R}^2/2$
From above two, linear acceleration of point P$=\frac{Ma-F_{friction}}{M}-\alpha R$
Since the cylinder does not slip, point P is at rest in our frame of reference,
$⟹\frac{Ma-F_{friction}}{M}=\alpha R$
$⟹\frac{Ma-F_{friction}}{M}=\frac{F_{friction}R}{\left(M{R}^2/2\right)}R$
$⟹Ma-F_{friction}=2F_{friction}$
Alternate solution: In the ground frame of reference, point P moves towards left with an acceleration $a$ and the pseudo force $Ma$ is absent. Now the equation is $\frac{F_{friction}}{M}+\alpha R=a$ which is basically a modified form of above equation.