JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Laws of Motion

Using energy conservation: $mgR(1-\cos \theta )=\frac{1}{2}m{v}^2$, so $\frac{m{v}^2}{R}=2mg(1-\cos \theta )$.

The normal force is $N=mg\cos \theta -\frac{m{v}^2}{R}=mg\cos \theta -2mg(1-\cos \theta )=3mg\cos \theta -2mg=3mg(\cos \theta -\frac{2}{3})$.

When $\cos \theta >\frac{2}{3}$ (initially, near the top), $N>0$, meaning the wire pushes the bead radially outward, so the bead pushes the wire radially inward.

When $\cos \theta <\frac{2}{3}$ (later, near the bottom), $N<0$, meaning the wire pulls the bead radially inward, so the bead pushes the wire radially outward.

Therefore, the force is radially inwards initially and radially outwards later.