Q121 / 180 · JIPMER 2015 · Physics
A particle slides on surface of a fixed smooth sphere starting from topmost point. The angle rotated by the radius through the particle, when it leaves contact with the sphere, is
- A$\theta=\cos ^{-1}\left(\frac{1}{3}\right)$
- Correct answerB$\theta=\cos ^{-1}\left(\frac{2}{3}\right)$
- C$\theta=\tan ^{-1}\left(\frac{1}{3}\right)$
- D$\theta=\sin ^{-1}\left(\frac{4}{3}\right)$
Solution
See the diagram
Let the velocity be $v$ when the body leaves the surface. From free body diagram,
$\frac{m v^2}{R}=m g \cos \theta$.
$\Rightarrow \quad v^2=R g \cos \theta$ ...(i)
Again from work energy principle, change in $\mathrm{KE}=$ work done
$\Rightarrow \frac{1}{2} m v^2-0=m g[R-R \cos \theta]$
$v^2=2 g R(1-\cos \theta)$ ...(ii)
From Eqs. (i) and (ii)
$\begin{aligned} R g \cos \theta & =2 g R(1-\cos \theta) \\ 3 g R \cos \theta & =2 g R \\ \Rightarrow \quad \cos \theta & =\frac{2}{3} \Rightarrow \theta=\cos ^{-1}\left(\frac{2}{3}\right)\end{aligned}$
Let the velocity be $v$ when the body leaves the surface. From free body diagram,
$\frac{m v^2}{R}=m g \cos \theta$.
$\Rightarrow \quad v^2=R g \cos \theta$ ...(i)
Again from work energy principle, change in $\mathrm{KE}=$ work done
$\Rightarrow \frac{1}{2} m v^2-0=m g[R-R \cos \theta]$
$v^2=2 g R(1-\cos \theta)$ ...(ii)
From Eqs. (i) and (ii)
$\begin{aligned} R g \cos \theta & =2 g R(1-\cos \theta) \\ 3 g R \cos \theta & =2 g R \\ \Rightarrow \quad \cos \theta & =\frac{2}{3} \Rightarrow \theta=\cos ^{-1}\left(\frac{2}{3}\right)\end{aligned}$Hence, the correct answer is option: 'B'.