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

Speed of each particle at angle `theta` is,
`v=sqrt(2gh)` , (from energy conservation)
where, `h=R(1-costheta)`
:. `v=sqrt(2gR(1-costheta))`
`N+mgcostheta=(mv^(2))/(R)`
or `N+mgcostheta=2mg(1-costheta)`
or `N=2mg-3mgcostheta` ...(i)
The tube breaks its contact with ground when Substituting, `2Ncostheta=Mg`
or `4mgcostheta-6mgcos^(2)theta=Mg`
Substituting `theta=60^(@)`
`2mg-(3mg)/(2)=Mg`
or `(M)/(m)=(1)/(2)`
Intially bnormal reaction on each vball will be radially outward and later it will be radailly inward, so that normal reactions on tube is radially outward to break it off from the groung .