Question

A block of mass $m$ is placed on a smooth sphere of radius $R$. It slides when pushed slightly. At what vertical 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)

Let the block leave the sphere at point $B$, which is at a distance
$h$ from the top of the sphere.
$\therefore \frac{m{v}^2}{R}=mg\cos \theta -N$
where $N$ is the normal reaction and $v$ is the velocity of block at point $B$.
When the block leaves the sphere at point $B$, the normal reaction $N$ becomes zero. $\therefore \frac{m{v}^2}{R}=mg\cos \theta$ or $\cos \theta =\frac{v^2}{Rg}$
From figure, $\cos \theta =\frac{R-h}{R}$
$\therefore \frac{R-h}{R}=\frac{2gh}{Rg}[\because v=\sqrt{2gh}]$
or $R-h=2h$ or $3h=R$ or $h=\frac{R}{3}$