A body weighing 0.4 kg is whirled in a vertical circle making 2 revolutions per second. If the radius of the circle is 1.2 m, find the tension in the string, when the body is (a) at the top of the circle (b) at the bottom of the circle. Given : g = 9.8 `ms^(-2)` and `pi` = 1.2 m

Explanation

Mass m = 0.4 kg time period = `(1)/(2)` second and
radius, r =1.2 m
Angular velocity, ` omega =(2pi)/(1//2) = 4pi " rad " s^(-1)`
= 12.56 rad `s^(-1)`
(a) At the top of the circle,
`T = (mv^(2))/(r ) - mg = mr omega^(2) - mg = m (r omega^(2) - g)`
= 0.4 (`1.2 xx 12.56 xx 12.56 - 9.8`) N = 71.8 N
(b) At the lowest point , T = m`(r omega^(2) + g)` = 79.64 m