A pendulum, comprising a light string of length Z and a small sphere, swings in the vertical plane. The string hits a peg located a distance d below the point of suspension (Fig). If the pendulum is released from rest at the horizontal position (0: 90') and is to swing in a complete circle centered on the peg, the minimum value of dis.

Energy is conserved in the swing of the pendulum, and the stationary peg does no work. So the ball${\mathrm{U}}_{\mathrm{i}} = \mathrm{mg}2(\mathrm{L}-\mathrm{d})$'s speed does not change when the string hits or leaves the peg, and the ball swings equally high on both sides.

The ball will swing in a circle of radius R = (L - d) about the peg. If the ball is to travel in the circle, the minimum centripetal acceleration at the top of the circle must be that of gravity:
 

$\frac{{\mathrm{mv}}^2}{\mathrm{R}} = \mathrm{mg} \Rightarrow {\mathrm{v}}^2 = \mathrm{g}(\mathrm{L}-\mathrm{d})$

When the ball is released from rest, ${\mathrm{U}}_{\mathrm{i}} = \mathrm{mgL},$ and when it is at the top of the circle, ${\mathrm{U}}_{\mathrm{i}} = \mathrm{mg}2(\mathrm{L}-\mathrm{d})$,  where height is measured from the bottom of the swing. By energy conservation,

$\mathrm{mgL} = \mathrm{mg}2(\mathrm{L}-\mathrm{d}) + \frac{1}{2}{\mathrm{mv}}^2$

From this and the condition on ${\mathrm{v}}^2$ we find $\mathrm{d} = \frac{3\mathrm{L}}{5}$