Question

A small body of mass $m$ slides without friction released from the top of a hemisphere of radius $R$. At what height below the body be detached from the surface of hemisphere ?

• A. $h=R$
• B. $h=\frac{R}{3}$
• C. $h=\frac{R}{2}$
• D. $h=\frac{2R}{3}$

Answer 1

The correct option is D $h=\frac{2R}{3}$

$\begin{array}{c}\text{Let the Block leaves the sphere for an angle}\theta \\ \text{as shown in the figure, and let the velocity at}\\ \text{the instant be}v\text{-}\\ \Rightarrow \text{By conservation of Energy}\end{array}$

$\begin{array}{c}\frac{1}{2}m{v}^2=mgh\\ \Rightarrow v=\sqrt{2gh}\\ \text{For the block to leave the sphere-}\\ \text{Fout}⩾mg\cos \theta \\ \Rightarrow \frac{m{v}^2}{R}⩾mg\cos \theta \end{array}$

$\begin{array}{c}\text{at that instant -}\\ \begin{array}{cc}\cos \theta ⩽\frac{2h}{R}& \text{we have}h=R\left(1\cos \theta \right)\\ \text{from figure}& \end{array}\end{array}$

$\begin{array}{cc}\Rightarrow & \cos \theta =2(1-\cos \theta )\\ & \cos \theta =\frac{2}{3}\\ \Rightarrow & h=R(1-\frac{2}{3})=\frac{R}{3}\end{array}$