Question

A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle subtended by the motion of particle from top to the point when it leaves contact with the sphere.

• A. ${\cos }^{-1}\left(\frac{2}{3}\right)$
• B. ${\tan }^{-1}\left(\frac{2}{3}\right)$
• C. ${\sin }^{-1}\left(\frac{2}{3}\right)$
• D. ${\cos }^{-1}\left(\frac{9}{2}\right)$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{2}{3}\right)$


(Assume at an angle $\theta$, particle leaves contact)
F.B.D. of particle,


Here, $N=$ normal reaction force
So, $mg\cos \theta -N=\frac{m{v}^2}{R}$
But, at the instant of leaving contact $N=0$
$\therefore mg\cos \theta =\frac{m{v}^2}{R}$
$v^2=gR\cos \theta ....\left(1\right)$
From conservation of mechanical energy,
$P{E}_i+K{E}_i=P{E}_f+K{E}_f$
$mgR+0=mgR\cos \theta +\frac{1}{2}m{v}^2$
$2gR(1-\cos \theta )=v^2....\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$2gR(1-\cos \theta )=gR\cos \theta$
$\Rightarrow 2=3\cos \theta$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{2}{3}\right)$