Question

A small body of mass $m$ slides without friction from the top of a hemisphere of radius $r$. At what height will the body be detached from the centre of the hemisphere?

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

Answer 1

Answer: (C)

$\frac{2}{3}r$

Answer 2

Body will be detached if $\frac{m \nu^{2}}{r}=m g \cos \theta$ As $\nu^{2}-0^{2}=2 g d$ $\nu=\sqrt{2 g d}$ and $\cos \theta=\frac{h}{r}$ So, from E (i), $\frac{m \times 2 g d}{r}=m g \frac{h}{r}$ or $h=2 d$ or $h=2(r-h)$ ie, $h=(2 / 3) r$