Q132 / 160 · TS EAMCET 2018 (05 May Shift 2) · Physics
A solid sphere is rolling without slipping on a semi-circular track of radius $10 \mathrm{~m}$ as shown in the figure. The radius of solid sphere is much smaller than the radius of semi-circular track. At the lowest point, it has a velocity $10 \mathrm{~m} / \mathrm{s}$. To what maximum angle $\theta$ from the vertical will the sphere travel before it comes back down? Neglect the rolling friction between the sphere and the track.
(Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
- A$\sin ^{-1}\left(\frac{3}{5}\right)$
- B$\sin ^{-1}\left(\frac{3}{7}\right)$
- Correct answerC$\cos ^{-1}\left(\frac{3}{10}\right)$
- D$\cos ^{-1}\left(\frac{1}{3}\right)$
Solution
As there is no slip, energy is conserved.
So, mechanical energy of ball at bottom $=$ mechanical energy of ball at top
$\Rightarrow \quad m g r+\frac{1}{2} m v_{\mathrm{COM}}^2+\frac{1}{2} I_{\mathrm{COM}} \omega^2=m g h$
$\Rightarrow \quad \frac{3}{4} v^2=g(h-r) \quad\left[\therefore I=\frac{m r^2}{2}\right.$ and $\left.v=r \omega\right]$
and $r$ can be neglected.
So,
$$
h=\frac{3 v^2}{4 g}
$$
Also, $h=R(1-\cos \theta) \Rightarrow \frac{3 v^2}{4 g}=R(1-\cos \theta)$
With values, we get
$$
\cos \theta=\frac{3}{10} \Rightarrow \theta=\cos ^{-1}\left(\frac{3}{10}\right)
$$
Hence, the correct answer is option: 'C'.