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