Question

A particle rests on the top of a smooth hemisphere of radius $r$. It is imparted a horizontal velocity of $\sqrt{gr}$. Find the normal reaction when it makes an angle $\theta ={60}^{\circ }$ with vertical.

• A. $1.5mg$
• B. $2.5mg$
• C. $-1.5mg$
• D. $0$

Answer 1

The correct option is D $0$


Using energy conservation (taking $U=0$ at the base of hemisphere):

$U_1+K_1=U_2+K_2$
$mgr+\frac{1}{2}m(\sqrt{gr}{)}^2=mgr\cos {60}^{\circ }+K_2$
$\Rightarrow mg(r-r\cos {60}^{\circ })+\frac{1}{2}mgr=\frac{1}{2}m{v}_2^2$
$\Rightarrow \frac{mgr}{2}+\frac{mgr}{2}=\frac{1}{2}m{v}_2^2$
$\Rightarrow 2mgr=m{v}_2^2$
$\Rightarrow \frac{v_2^2}{r}=2g$

Force balance in radial direction at $\theta ={60}^{\circ }$
$mg\cos \theta -N=\frac{m{v}_2^2}{r}$
or $N=mg\cos {60}^{\circ }-\frac{2mgr}{r}=\frac{mg}{2}-2mg$
$N=-1.5mg$
$N$ is negative, it means that particle has lost contact with surface, therefore, $N=0$
(since normal reaction cannot be negative)