Question

A solid sphere of mass $m$ is released from rest from the rim of a hemspherical cup so that it slides along the surface. If the rim of the hemisphere is kept horizontal, find the normal force exerted by the cup on the ball when the ball reaches the bottom of the cup. Neglect the radius of the ball as compared to the radius of the cup.

• A. $0$
• B. $mg$
• C. $3mg$
• D. $1.2mg$

Answer 1

The correct option is C $3mg$

Let radius of solid sphere $=r$ and radius of hemispherical cup $=R$.
When sphere is at bottom, applying force equation radially,
$N-mg=\frac{m{v}^2}{(R-r)}$
$\Rightarrow$ Total normal force, $N=mg+\frac{m{v}^2}{(R-r)}$

But $r<<R$ (given)
$\Rightarrow N=mg+\frac{m{v}^2}{R}$ ......{i}


Applying energy conservation at release point (say $A$) and bottommost point (say $B$).
${\text{(P.E)}}_A+{\text{(K.E)}}_A={\text{(K. E)}}_B$
(taking $B$ as datum for P.E)
$\Rightarrow mgR+0=\frac{1}{2}m{v}^2$
(sphere is released from rest at point $A$)
$\Rightarrow m{v}^2=2mgR$
or $v^2=2gR$

Putting in equation {i}
$N=mg+\frac{m\left(2gR\right)}{R}$
$\Rightarrow N=3mg$ is the normal force exerted on the ball at the bottom of the cup.