The mass m is released from rest at point A. When the mass reaches vertical position, the rope turns around a nail which is located at a distance of r from the end of the rope. Find the angle α that will cause a complete rotation of the mass around the nail.
We will take the axes so that the x direction is pointing to the center of the rotation and the y direction is always tangential to the circle in this axes system the equations are simpler:
The velocity of the mass at the vertical position when the rope reaches the nail is calculated by energy methods.
The initial potential energy is converted to kinetic energy: 

Falling distance is: 
h = R − R cosα = R (1 − cosα) 

The velocity at point B is: 

(1) 
In order to complete a hole rotation around the nail the kinetic energy at point B (vertical position) should be equal to the potential energy at height 2r (upper position of the mass after swirling around the nail) plus the kinetic energy at top due to the velocity v_{T} .
Energy balance: 

(2) 
From free body diagram 

(3) 
From eq. (3) 
v_{T}^{2} = g r 
(4) 
Substitute eq. (1) and (4) into (2) we get: 
m 2 g h / 2 = m g 2 r + m g r / 2 

h = 5 r / 2 
Substitute the value of h to the previous found value of h we get: 

If point A is horizontally located relative to point O and the ball released from this position, find the maximum value of r that will cause a complete rotation around the nail.
In this case: h = R and from eq. (1) v_{B}^{2} = 2 g R substitute this value into eq. (2) we get:
 
And finally:  r = 2 R / 5 
