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.