Question

A particle of mass $m$ is projected with speed $\sqrt{\frac{Rg}{4}}$ from top of a smooth hemisphere as shown in figure. If the particle starts slipping from the highest point, then the horizontal distance between the point where it leaves contact with sphere and the point at which the body was placed is:

Answer 1

The equation of motion:

$N-mg\cos \theta =\frac{m{V}^2}{R}$

When the body loses contact, $N=0$:

$\frac{m{V}^2}{R}=-mg\cos \theta$

$⟹\frac{Rg}{4R}=-g\cos \theta$

$⟹\frac{1}{4}=-\cos \theta$

$⟹\theta ={\cos }^{-1}\left(\frac{1}{4}\right)={75.5}^o$

$\therefore \varphi ={90}^o-{75.5}^o={14.4}^o$

$⟹\cos \varphi =0.96$ $R$

$\therefore$ Horizontal distance $=0.96$ $R$