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Q.
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?
Laws of Motion
Solution:
Let the block leave the sphere at point $B$, which is at a distance
$h$ from the top of the sphere.
$\therefore \frac{m v^{2}}{R}=m g \cos \theta-N$
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. $\therefore \frac{m v^{2}}{R}=m g \cos \theta$ or $\cos \theta=\frac{v^{2}}{R g}$
From figure, $\cos \theta=\frac{R-h}{R}$
$\therefore \frac{R-h}{R}=\frac{2 g h}{R g} \quad[\because v=\sqrt{2 g h}]$
or $ R-h=2h$ or $3 h=R \quad$ or $\quad h=\frac{R}{3}$
