Question

A hollow sphere is projected horizontally along a rough surface with speed $v$ and angular velocity ${\omega }_0$, find out the ratio $\left(\frac{v}{{\omega }_0}\right)$, so that the sphere stops moving after some time.

• A. $\frac{2R}{3}$
• B. $\frac{4R}{3}$
• C. $\frac{7R}{3}$
• D. $R$

Answer 1

The correct option is A $\frac{2R}{3}$

Let the given situation be as shown in the below figure,


Torque about lowest point of a sphere.
$f_k\times R=I\alpha$
or, $\mu mg\times R=\frac{2}{3}m{R}^2\alpha$
$\Rightarrow \alpha =\frac{3\mu g}{2R}$, angular accelaration in opposite direction of angular velocity.
Now we have,
$\omega ={\omega }_0-\alpha t$ (final angular velocity $\omega =0$)
$\Rightarrow {\omega }_0=\frac{3\mu g}{2R}\times t$

$\Rightarrow t=\frac{{\omega }_0\times 2R}{3\mu g}$

Also, acceleration $a=\mu g$
so, we have
$v_f=v-at$ ( final velocity $v_f=0$)
$\Rightarrow v=\mu g\times t$
$\Rightarrow t=\frac{v}{\mu g}$

Now, to stop the sphere, time at which $v\text{\&}\omega$ are zero, should be same.
Thus,
$\frac{v}{\mu g}=\frac{2{\omega }_{0}R}{3\mu g}\Rightarrow \frac{v}{{\omega }_0}=\frac{2R}{3}$
Hence, $\left(B\right)$ is correct answer.