Question

A ball of mass `0.6 kg` attached to a light inextensible string rotates in a vertical circle of radius `0.75 m` such that it has speed of `5 ms^(-1)` when the string is horizontal. Tension in the string when it is horizontal on other side is `(g = 10 ms^(-2))`.

Answer 1

Tension in the string when it makes angle `theta` with the vertical
`T=(mv^(2))/(r)+"mg cos"theta`
When the string is horizontal, `theta=90^(@)`
`therefore" "T=(mv^(2))/(r)+mgxx0=(mv^(2))/(r)`
`=(0.6xx(5)^(2))/(0.75)=20N`