Question

A small block slides down from the top of a inverted hemisphere of radius $r$. It is assumed that there is no friction between the block and the hemisphere. At what height $h$ from ground, will the block lose contact with the surface of sphere ?

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

Answer 1

The correct option is B $\frac{2r}{3}$

The block will lose contact with the surface of hemisphere when the centripetal acceleration becomes equal to the component of accleration due to gravity along the radius. Suppose it happens the point S as shown in the adjoining figure. The velocity at the point S is given by $v=\left[2g\right(r-h){]}^{1/2}$]

The centripetal acceleration should be equal to the component of $g$ along SO.
$i.e.,\frac{v^2}{r}=gcos\theta \Rightarrow \frac{2g(r-h)}{r}=g\times \frac{h}{r}$
$\Rightarrow 2(r-h)=h\therefore h=\frac{2r}{3}$
Hence, the correct answer is option (b).