Sol. The ball first of all describes a quadrant of a circle of radius equal to the length λ of the thread. Then the thread touches the nail O, which was driven into the wall, and the ball describes an arc of a circle of radius twice as small as the previous one. Finally, when the weight of the ball gives it the centripetal force necessary to motion in a circle, the tension in the thread will be reduced to zero. Let this happen at point M (fig.) We can find this point’s position in the following manner. The component of the ball’s weight acting along the line of the radius equals P cos α , where α is the angle made by the thread at that moment. Further, at point M, v 2 equals 2gh, where H=AB=AO– BO = λ /2 – λ cos α /2. Therefore the centripetal force at point M equals

= = 2P (1–cos α ).

Thus at point M we have the equation P cos α = 2P (1 – cos α ), from which cos α = 2/3. The ball continues its path just as a body thrown at an angle of α to the horizontal with an initial velocity v = . In this case, the highest point of the parabola lies above the point of throwing at a distance of = .

The vertical line which passes through the point of suspension lies at a distance from M of MB = = .

To travel this distance horizontally the ball will require time t = = . .

During this time the ball will travel, in height, a distance of v sin α t – = ,

i.e. it will cut the vertical line AO at a point lying 5 λ /96 below point B.