Question

A ball of mass $1$ Kg moves inside a fixed smooth spherical shell of radius $1$ m with an initial velocity $v=5$ m/s from the bottom. If the total force acting on the particle at point $B$ is $F$ Newton, find the value of $\frac{F^2}{25}$.

Answer 1

Since it is a smooth surface, energy is conserved.

$\frac{m{v}_i^2}{2}=mgh+\frac{m{v}_f^2}{2}$

$\frac{1\times 5^2}{2}=10+\frac{v_f^2}{2}$

$v_f=\sqrt{5}$

Net force towards the centre $=\frac{m{v}_f^2}{R}=5$

Net vertical force $=mg=10$

Net force $=\sqrt{{10}^2+5^2}=\sqrt{125}$

$\therefore$ $\frac{F^2}{25}=\frac{125}{25}=5$