A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 revolutions per minute in a horizontal plane. What is the tension in the spring?

Explanation

Given m = 0.25 kg, R=1.5m, `omega = 40 rev.min^(-1)`
i.e `omega= (40 xx 2 pi)/(60) rad s^(-1) " " i.e omega=(40 xx 2 pi)/(60)`
`omega = (4 xx 3.14)/(3)= 4.19 rad// s^(-1)`
Linear speed `=v=r omega = 1.5 xx 4.19 =6.285 ms^(-1)` Tension in the spring = centripetal force `= (mv^2)/(R)= (0.25 xx 6.285 xx 6.285)/(1.5)= 6.584 N` `therefore T=6.584 N`