Q138 / 180 · JIPMER 2016 · Physics
A solid sphere of mass $M$ and radius $2 R$ rolls down an inclined plane of height $h$ without slipping. The speed of its centre of mass when it reaches the bottom is
- A$\sqrt{\frac{6}{7} g h}$
- B$\sqrt{3 g h}$
- Correct answerC$\sqrt{\frac{10}{7} g h}$
- D$\sqrt{\frac{4}{3} g h}$
Solution
When solid sphere rolls down on an inclined plane, then it has both rotational and translational kinetic energy
or $\quad \begin{array}{ll} & K=K_{\text {rot }}+K_{\text {trans }} \\ & K=\frac{1}{2} l \omega^2+\frac{1}{2} M v^2\end{array}$
where, $I=$ moment of inertia of solid sphere $=\frac{2}{5} M R^2$
$\begin{aligned} \therefore \quad K & =\frac{1}{2}\left(\frac{2}{5} M(2 R)^2\right) \omega^2+\frac{1}{2} M v^2 & {[\because R=2 R] } \\ & =\frac{4}{5} M R^2\left(\frac{v}{2 R}\right)^2+\frac{7}{2} M v^2 & {[\because v=r \omega] }\end{aligned}$
$=\frac{1}{5} M v^2+\frac{1}{2} M v^2=\frac{1}{10} M v^2$
Now, gain in $\mathrm{KE}=$ loss in PE
$\begin{aligned} \frac{7}{10} M v^2 & =M g h \\ v & =\sqrt{\frac{10}{7} g h}\end{aligned}$
Hence, the correct answer is option: 'C'.