Question

A solid cylinder is projected on a rough surface having coefficient of friction $\mu$ with velocity $v_0$ and angular velocity ${\omega }_0$$($where ${\omega }_0=\frac{v_0}{2R},R$ is radius$)$. The time $t$ at which rolling without slipping occurs is :

• A. $\frac{v_0}{3\mu g}$
• B. $\frac{v_0}{6\mu g}$
• C. $\frac{v_0}{2\mu g}$
• D. $\frac{v_0}{\mu g}$

Answer 1

The correct option is B $\frac{v_0}{6\mu g}$

The velocity of point of contact is $v_o-{\omega }_{o}R=v_o/2$. So the rigid body is slipping on the ground in forward direction, which means frictional force will act in backward direction.

This will reduce the linear velocity and increase the angular velocity till there is rolling without slipping.

Frictional force is $\mu mg$. Le the time after which rolling-without-slipping starts is $t$.

So the linear velocity at time $t$, is $v_t=v_o-\left(\mu mg/m\right)t=v_o-\mu gt$

the angular velocity after time $t$ is, ${\omega }_t={\omega }_o+\frac{\mu mgR}{m{R}^2/2}t={\omega }_o+2\mu gt/R=v_o/2R+2\mu gt/R$

for rolling-without-slipping,

$v_t={\omega }_{t}R⟹v_o-\mu gt=v_o/2+2\mu gt⟹t=\frac{v_o}{6\mu g}$