A circular tube of mass `M` placed veticlly on a horizontal surface as shown in the figure. Two small spheres, each of mass `m` , just fit in the tube, are released from the top. If `theta` gives the angle between radius vector of eirther ball with the vertical, obtain the value of the ratio `M//m` if the tube breacks its contact with ground when `theta=60^(@)` . Neglect any friction.

Explanation

Using conservation of energy principle, if `v` be the speed of either ball when its radius vector makes angle `theta` with vertically upward direction.
`mgR[1-costheta]=(1)/(2)mv^(2)`
`rArr (mv^(2))/(R )=2mg[1-costheta]`

From F.B.D `(i)`
`N=mgcostheta-(mv^(2))/(R )`
`=mgcostheta-2mg[1-costheta]`
From F.B.D. `(ii)`
`N'=2Ncostheta+Mg`
At the instant tube breaks its contact with ground
`N'=0`
`rArr Mg+[mgcostheta-2mg(1-costheta)]2costheta=0`
for `theta=60^(@)`, we get `mM=2`.