Question

A uniform cylinder of mass $M$ and radius $R$ is placed on a rough horizontal board, which in turn is placed on a smooth surface. The coefficient of friction between the board and the cylinder is $\mu$. If the board starts accelerating with constant acceleration $a$, as shown in the figure, then

• A. For pure rolling motion of the cylinder, the direction of frictional force is forward and its magnitude is $\frac{Ma}{3}$
• B. The maximum value of $a$, so that the cylinder performs pure rolling is $3\mu g$
• C. The acceleration of the center of mass of the cylinder under pure rolling condition for the given $a$ is $\frac{2a}{3}$
• D. None of the above

Answer 1

Answer: (A), (B), (C)

The correct options are
A For pure rolling motion of the cylinder, the direction of frictional force is forward and its magnitude is $\frac{Ma}{3}$
B The maximum value of $a$, so that the cylinder performs pure rolling is $3\mu g$
C The acceleration of the center of mass of the cylinder under pure rolling condition for the given $a$ is $\frac{2a}{3}$

From the frame of reference, attached to the horizontal board, there is a pseudo force $Ma$ on the cylinder in the backward direction.

So to prevent slipping, the frictional force acts in the forward direction.

Let this force be $f$.

So the linear acceleration of the cylinder is, $A=(Ma-f)/M$ in the backward direction.

Torque due to this frictional force is $fR$ and corresponding angular acceleration is given as, $\alpha =fR/I=fR/0.5M{R}^2=2f/MR$

For pure rolling,

$A=\alpha R⟹(Ma-f)/M=2f/M⟹f=Ma/3$

Now the maximum value of frictional force is, $\mu Mg$

So the maximum value of $a$ is given by, $\mu Mg=M{a}_{max}/3⟹a_{max}=3\mu g$, for pure rolling to happen.

Now acceleration of the CoM is, $A=(Ma-f)/M=2a/3$