Question

A solid sphere of radius $r$ is gently placed on a rough horizontal surface with an initial angular speed ${\omega }_0$ but no linear velocity. If the coefficient of friction is $\mu$, then the time $t$ when the slipping will stop is

• A. $\frac{2}{7}\frac{r{\omega }_0}{\mu g}$
• B. $\frac{3}{7}\frac{r{\omega }_0}{\mu g}$
• C. $\frac{4}{7}\frac{r{\omega }_0}{\mu g}$
• D. $\frac{r{\omega }_0}{\mu g}$

Answer 1

The correct option is A $\frac{2}{7}\frac{r{\omega }_0}{\mu g}$

Let $m$ be the mass of the sphere.
Since, it is a case of backward slipping, force of friction is in forward direction.


Linear acceleration, (by $F_x=ma$)
$a=\frac{f}{m}=\frac{\mu mg}{m}=\mu g$ ...(i)
Angular retardation (by ${\tau }_{com}=I_{com}\alpha$)
$\alpha =\frac{\tau }{I}=\frac{f\times r}{\frac{2}{5}m{r}^2}=\frac{5}{2}\frac{\mu g}{r}$ ...(ii)
[$\because f=\mu mg$]

Slipping will be stopped when $v=r\omega$
i.e $(v_0+at)=r\times ({\omega }_0-\alpha t)$
[Here, -ve because $\alpha$ is in anti-clockwise direction and $\omega$ is in clockwise direction]
But, initially $v_0=0$.
Hence, $at=r({\omega }_0-\alpha t)$
From eqn (i) and (ii),
$\mu gt=r({\omega }_0-\frac{5}{2}\frac{\mu gt}{r})$
$\Rightarrow \frac{7}{2}\mu gt=r{\omega }_0$
$\therefore t=\frac{2}{7}\frac{r{\omega }_0}{\mu g}$