A ball of  mass  M  is  attached to a  chord of  length 10 m. The chord is tied at the top to a swivel. The swivel is rotated with an angular velocity of 1.3 rad /s about its own axis. Find the radius of the circle about which the ball rotates . a ) 7.91 m b ) 5.81 m c ) 9.24 m d ) 8.14 m Let w be the angular velocity of the swivel ; l, be the length of the string; T, the tension in the string ; v, the velocity of the ball ; M , the mass of the ball ; R , the radius of the circle of rotation of the ball and q the angle that the chord makes with the vertical. The free - body diagram of the ball is as shown. The equation of motion in the direction parallel to the horizontal at the given instant is ,    { ( M v 2 ) / R } - T sin q = 0         or    T =  ( M v 2 ) / ( R sin q ) The equation of motion in the vertical direction is ,    T cos q - M g = 0 Substituting the value of T from the previous equation and rearranging we get    tan q = v 2 / ( R g ) But v = R w . substituting we get ,    tan q = ( R w )2 / ( R g ) = R w 2 / ( g ) Now, tan q is got from the trigonometric identities of the triangle with l as the hypoteneous and R as one of the sides.    tan q = R / { l 2 - R 2 } 0.5 Substituting this in the above equation we get ,    { l 2 - R 2 } 0.5 = g / w 2 squaring both sides and solving for R we get ,    R 2 = { l 2 - ( g / w 2 )2 }           or ,    R = { l 2 - ( g / w 2 )2 } 0.5   Substituting the respective values in the above equation we get the radius of the circle of the rotation of the ball as, R = 8.14 m