Question

The track shown in the diagram ends in a circular track of radius $r$ with centre at $O$. A small solid sphere of mass $m$ rolls from rest without slipping from a point $A$, which is at a height $h=6r$ from the ground level. What is the speed of the sphere, when it reaches point $B$, which is at height $r$ above the ground level?

• A. $\sqrt{10gr}$
• B. $\sqrt{\frac{50}{7}gr}$
• C. $\sqrt{\frac{22}{7}gr}$
• D. zero

Answer 1

The correct option is B $\sqrt{\frac{50}{7}gr}$

Let $v$ be the speed of the sphere when it reaches $B$.
Then, loss in $P.E=$ gain in translational $K.E+$ gain in rotational $K.E$,
i.e,
$mg(6r-r)=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

$5mgr=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{r}^2\times \frac{v^2}{r^2}$ $(\because \text{for pure rolling}\omega =\frac{v}{r})$

$\Rightarrow 5mgr=\frac{7}{10}m{v}^2$

$\therefore v=\sqrt{\frac{50gr}{7}}$

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