Question

A horizontal plank of mass m is lying on a smooth horizontal surface. A sphere of same mass and radius r is spinned to an angular frequency ${\omega }_0$ and gently placed on the plank with its axis horizontal as shown in the figure. If coefficient of friction between the plank and sphere is $\mu$. The distance moved by the plank till the sphere starts pure rolling on the plank. The plank is long enough.

• A. $-\frac{2{\omega }_0^2{r}^2}{81\mu g}$
• B. $\frac{2{\omega }_0^2{r}^2}{91\mu g}$
• C. $\frac{{\omega }_0^2{r}^2}{9\mu g}$
• D. $-\frac{{\omega }_0^2{r}^2}{\mu g}$

Answer 1

The correct option is A $-\frac{2{\omega }_0^2{r}^2}{81\mu g}$

Normal reaction between plank and sphere is: $N=mg$

Friction acting is: $F_r=\mu N=\mu mg$

Acceleration of sphere is: $a=\frac{F_r}{m}=\mu g$

In plank's frame of reference, acceleration of sphere is $a^{\prime }=2a=2\mu g$

From first equation of motion,

$v=u+at$

$v=2\mu gt.................\left(i\right)$

Torque acting on the sphere is:

$\tau =F_{r}r=\mu mgr$

$I\alpha =\mu mgr$

$\frac{2}{5}m{r}^2\alpha =\mu mgr$

$\alpha =\frac{5\mu g}{2r}$

From first equation of motion for rotational motion,

$\omega ={\omega }_o-\alpha t$

$\omega r={\omega }_{o}r-\frac{5\mu g}{2r}rt$

At pure rolling, $v=r\omega$

$v={\omega }_{o}r-5\mu gt/2...........\left(ii\right)$

From (i) & (ii),

$t=\frac{2{\omega }_{o}r}{9\mu g}$

Distance moved by plank is:

$s=ut+.5a{t}^2$

$s=.5\times \mu g\times \frac{4{\omega }_o^2{r}^2}{81{\mu }^2{g}^2}$

$s=\frac{2{\omega }_o^2{r}^2}{81\mu g}$