Question

A hollow sphere is projected horizontally along a rough surface with speed v and angular velocity ${\omega }_0$ find out the ratio $\frac{v}{{\omega }_0}$. So that the sphere stop moving after some time.

• A. $\frac{2R}{3}$
• B. $\frac{R}{3}$
• C. $\frac{5R}{6}$
• D. $\frac{2R}{7}$

Answer 1

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

Torque about lowest point of sphere.
$f_k\times R=I\alpha$
$\mu mg\times R=\frac{2}{3}m{R}^2\alpha$
$\alpha =\frac{3\mu g}{2R}$ angularacceleration in opposition directionof angularvelocity.
$\omega ={\omega }_0-\alpha t$ (final angular velocity $\omega =0$ )
${\omega }_0=\frac{3\mu g}{2R}\times t$
$t=\frac{{\omega }_0\times 2R}{3\mu g}$
acceleration 'a = $\mu g^{\prime }$
$v_1=v-at(finalvelocity{v}_1=0)$
$v=\mu g\times t$
$t=\frac{v}{\mu g}$
To stop the sphere time at which v & $\alpha$ are zero, should be same.
$\frac{v}{\mu g}=\frac{2{\alpha }_{0}R}{3\mu g}=\frac{v}{{\alpha }_0}=$$\frac{2R}{3}$