Question

A small block slides down from the top of hemisphere of radius $R=3\text{m}$ as shown in the figure.The height ${}^{\prime }{h}^{\prime }$ at which the block will lose contact with the surface of the sphere is $\text{(in m)}$.
(Assume there is no friction between the block and the hemisphere)

Answer 1

The condition can be represented as
Also, FBD of the block can be shown as

So, the force equation along the radial direction, at point $A$ can be written as
$mg\cos \theta =N+F_c$
At point $A$ block will lose contact i.e. $N=0$
$\therefore mg\cos \theta =F_c=\frac{m{v}^2}{R}...\left(1\right)$ where, $v$ is the tangential velocity of block at point $A$.
Also, $\cos \theta =\frac{h}{R}...\left(2\right)$

Now, from energy conservation, we have
$mg(R-h)=\frac{1}{2}m{v}^2...\left(3\right)$
From $\left(1\right),\left(2\right)$ and $\left(3\right)$ we have
$\Rightarrow mg\left(\frac{h}{R}\right)=\frac{2mg(R-h)}{R}$
$\Rightarrow h=\frac{2R}{3}=2\text{m}$