Question

A solid sphere is released from rest from the top of an inclined plane of inclination $\theta$ and length $l$. If the sphere rolls without slipping, what will be its speed when it reaches the bottom?

• A. $\sqrt{\frac{10}{7}gl\sin \theta }$
• B. $\sqrt{\frac{7}{10}gl\sin \theta }$
• C. $\sqrt{\frac{3}{7}gl\sin \theta }$
• D. $\sqrt{\frac{7}{3}gl\sin \theta }$

Answer 1

The correct option is A $\sqrt{\frac{10}{7}gl\sin \theta }$

Let the mass of the sphere be $m$ and its radius be $r$.

Suppose the final linear speed is $v$.

Change in kinetic energy,

$\Delta KE=K{E}_f-K{E}_i=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2-0=\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\left(\frac{v}{r}\right)}^2+\frac{1}{2}m{v}^2$

$=\frac{7}{10}m{v}^2$

According to work energy theorem,

$\Delta KE=W_{\text{external force}}$

Here, $W_{\text{external force}}=$ Work Done by gravity $=mgl\sin \theta$

$\Rightarrow mgl\sin \theta =\frac{7}{10}m{v}^2$

$\therefore v=\sqrt{\frac{10}{7}gl\sin \theta }$

Hence, option $\left(A\right)$ is the correct answer.