Question

A solid sphere of radius $R$ is set into motion on a rough horizontal surface with a linear speed $v_0$ in forward direction and an angular velocity ${\omega }_0=v_0/2R$ in counter coefficient of friction is $\mu$, then find the time after which sphere starts pure rolling.

• A. $t_0=\frac{3v_0}{7\mu g}$
• B. $t_0=\frac{2v_0}{7\mu g}$
• C. $t_0=\frac{4v_0}{7\mu g}$
• D. $t_0=\frac{v_0}{7\mu g}$

Answer 1

The correct option is A $t_0=\frac{3v_0}{7\mu g}$

The point of contact of sphere has non-zero velocity w.r.t. ground in the forward direction, frictional force will apply in the backward direction.

This would decrease both the linear velocity and the angular velocity.

frictional force is $f=\mu mg$.

Linear deceleration is $a=f/m=\mu g$

Let the time at which pure rolling starts be $t$.

So the linear velocity at this time gives by,

$v_t=v_o-at=v_o-\mu gt$

Angular acceleration is $\alpha =\frac{fR}{2m{R}^2/5}=5\mu g/2R$

At time $t$, the angular velocity is given by,

${\omega }_t=-{\omega }_o+\alpha t=-v_o/2R+5\mu gt/2R$

We have taken angular velocity as negative here to indicate the reverse spin, and angular acceleration positive as it is in the forward spin direction.

For pure rolling,

$v_t={\omega }_{t}R⟹v_o-\mu gt=-v_o/2+5\mu gt/2⟹3v_0/2=7\mu gt/2⟹t=3v_o/7\mu g$