A ball of 200 g is at one end of a string of length 20 cm. It is revolved in a horizontal circle at an angular frequency of 6 rpm. Find (i) the angular velocity, (ii) the linear velocity, (iii) the centeripetal acceleration, (iv) the centeripetal force and(v) the tension in the string.

A ball of 200 g is at one end of a string of length 20 cm. It is revol

1.	A ball of 200 g is at one end of a string of length 20 cm. It is revolved in a horizontal circle at an angular frequency of 6 rpm. Find (i) the angular velocity, (ii) the linear velocity, (iii) the centeripetal acceleration, (iv) the centeripetal force and(v) the tension in the string.
Answer» Solution :(i) The angular velocity, `omega = (2pi N)/(t)=(2pi xx 6)/(60)=(pi)/(5)=0.6284 rad s^(-1)` (ii) The linear velocity, `v = r omega` `= 0.20xx0.6284` `= 0.1257 ms^(-1)` (iii) The centripetal ACCELERATION, `a_(c )=r omega^(2)` `= 0.20xx(0.6284)^(2)` `= 0.0790 ms^(-2)`. (iv) The centripetal force, `F_(c )=mr omega^(2)` `= 0.200xx0.20xx(0.6284)^(2)` `= 0.0158 N`. (v) The tension in the string `= 0.0158 N`.