Question

A sphere of mass M and radius R is moving on a rough fixed surface, having coefficient of friction $\mu$ as shown in the figure above. It will attain a minimum linear velocity after time : $(v_0>{\omega }_{0}R)$

• A. $\frac{v_0}{\mu g}$
• B. $\frac{{\omega }_{0}R}{\mu g}$
• C. $\frac{2(v_0-{\omega }_{0}R)}{5\mu g}$
• D. $\frac{2(v_0-{\omega }_{0}R)}{7\mu g}$

Answer 1

Answer: (D)

$\frac{2(v_0-{\omega }_{0}R)}{7\mu g}$

The friction force will be $\mu mg$ backwards (opposite to direction of velocity)

It will attain a minimum linear velocity when friction force vanishes, i.e. $v^{\prime }={\omega }^{\prime }R$ where $v^{\prime }\&{\omega }^{\prime }$ are attained after time t.

Using equation of motion

$F=\mu mg\therefore a=-\mu g{v}^{\prime }=v_o+at=v_o-\mu gt$

Using Moment of Inertia:

$I\alpha =T\Rightarrow \frac{2}{5}m{R}^2\alpha =\mu mgR\Rightarrow \alpha =\frac{5\mu g}{2R}{\omega }^{\prime }={\omega }_o+\alpha t\Rightarrow {\omega }^{\prime }={\omega }_o+\frac{5\mu g}{2R}t\Rightarrow {\omega }^{\prime }R={\omega }_{o}R+\frac{5\mu g}{2}t\Rightarrow v^{\prime }=v_o-\mu gt={\omega }^{\prime }R={\omega }_{o}R+\frac{5\mu g}{2}t\Rightarrow v_o-{\omega }_{o}R=\frac{7\mu g}{2}t\Rightarrow t=\frac{2(v_o-{\omega }_{o}R)}{7\mu g}$