A string of length `50cm` is fixed at one end and carries a mass of `200g` at the other end. The string makes `5//2pi` revolution per second around the verticle axis through the fixed end. Calculate (a) the tension in the string (b) the angle of incilination of the string with verticle and (c) linear velocity of the mass.

Explanation

Given: `m=200g=0.2kg,L=50cm=0.5m`
`omega=2piN=2pi((5)/(2pi))=5rad//s`
`T sin theta=m omega^(2)r`
`T(r )/(L)=momega^(2)r`
`implies T=momega^(2)L=0.2xx(5)^(2)xx0.5=2.5N`
`T cos theta=mg`
`cos theta=(mg)/(T)=(0.2xx10)/(2.5)=(2)/(2.5)=(4)/(5)impliestheta=37^(@)`
`v=romega=(L sin theta) omega=(0.5 sin 37^(@))xx5`
`=(0.5xx(3)/(5))xx5=1.5 m//s`