Question

A block of mass $m$ is placed on a smooth sphere of radius $R$. It slides when pushed slightly. At what distance $h$, from the top, will it leave the sphere?

• A. $\frac{R}{4}$
• B. $\frac{R}{3}$
• C. $\frac{R}{2}$
• D. $R$

Answer 1

Answer: (B)

$\frac{R}{3}$

Answer 2

Suppose the block will leave the sphere at point $B$,
which is at a distance $h$ from the sphere
$\frac{m v^{2}}{R}=m g \cos \theta N$
When block leaves the sphere at point $B$, the normal reaction $N$ becomes zero
$\therefore \frac{m v^{2}}{R} =m g \cos \theta$
$\cos \theta =\frac{v^{2}}{R g}$
From figure
$\cos \theta=\frac{R-h}{R}$
$\therefore \frac{R-h}{R}=\frac{v^{2}}{R g}$
$\frac{R-h}{R} =\frac{2 g h}{R g}$
$\left[\therefore v^{2}=2 g h\right]$
$h =\frac{R}{3}$