A ball of mass M is hanging from a pivot with the help of a massless chord of length 10 m .Find the minimum horizontal velocity that has to be applied to the stationary ball such that the chord is taut at the highest point of rotation . a ) 19.90 m / s b ) 22.15 m / s c ) 19.81 m / s d ) 20.90 m / s Let l be the length of the string; T, the tension in the string ; v f , the final velocity of the ball ; v 0 , the initial velocity of the ball ; M , the mass of the ball and R , the radius of the circle of rotation of the ball . At the highest point of rotation the equation of motion in the radial direction is,      { ( M vf2 ) / R }  - M g - T = 0           or        T = { ( M vf2 ) / R }  - M g For the chord to remain taut, the tension in it should be greater than zero . Hence the minimum velocity occurs when T just becomes zero . Rewriting the above equation to suit the condition ,     { ( M vf2 ) / R } - M g = 0   or   ( vf2 ) = g R Now consider the motion of the ball at some instant when the chord makes an angle of q initial position. Let the tangential deceleration of the ball be a t here .The equation of motion in the tangential direction at this instant is ,     M g sin q = M ( - a t ) But we have ,     a t = v { dv / ( R dq ) } Substituting a t in the previous equation and rearranging we get ,    - v dv = R g sin q dq Integrating both sides, Substituting the values of g and R in the above , the minimum initial horizontal velocity that should be applied to the ball for the chord to be taut at the highest point is 22.15 m / s.