The ball is released from the horizontal position (level with O). The string length is $l$.
The nail P is located at a distance $r$ from O. Based on the diagram and solution context, P is located at an angle $\theta$ from the vertical.

The string collides with the nail when the ball reaches angle $\theta$ (when the string aligns with OP). At this collision point (let’s call it C):

• Height of ball below O: $h_C = l \cos \theta$.
• Velocity $v_C$ is derived from energy conservation (Initial height 0): $$ \frac{1}{2}m v_C^2 = mg(l \cos \theta) \implies v_C^2 = 2gl \cos \theta $$

After the string catches on nail P, the ball begins to rotate around P with a reduced radius $R’ = l – r$. For the ball to make a complete circular turn around P, it must reach the highest point of the new circle (vertically above P) with sufficient velocity to keep the string taut.

Geometry of the Top Point (T):
P is at vertical depth $r \cos \theta$ below O.
The top of the loop T is at height $R’ = l – r$ above P.
Therefore, the depth of T below O is: $$ h_T = r \cos \theta – (l – r) $$ (Note: If this value is negative, T is above O, which is allowed).

Energy at release (Level O) = Energy at Top T. $$ 0 = -mgh_T + \frac{1}{2}m v_{top}^2 $$ We can also use depth. Loss in PE = Gain in KE. $$ mg \cdot (\text{Depth of T}) = \frac{1}{2} m v_{top}^2 $$ $$ mg (r \cos \theta – (l – r)) = \frac{1}{2} m v_{top}^2 $$ $$ v_{top}^2 = 2g (r \cos \theta – l + r) $$

To complete the circle, the tension at the top must be $\ge 0$. The critical condition is: $$ \frac{m v_{top}^2}{R’} \ge mg \implies v_{top}^2 \ge g R’ $$ Substitute $R’ = l – r$ and the expression for $v_{top}^2$: $$ 2g (r \cos \theta – l + r) \ge g(l – r) $$ Cancel $g$ and expand: $$ 2r \cos \theta – 2l + 2r \ge l – r $$ Group terms with $r$ on the left and $l$ on the right: $$ 2r \cos \theta + 3r \ge 3l $$ $$ r(2 \cos \theta + 3) \ge 3l $$ $$ r \ge \frac{3l}{2 \cos \theta + 3} $$