A small block of mass `m` is placed on top of a smooth hemisphere also of mass `m` which is placed on a smooth horizontal surface. If the block begins to slide down due to a negligible small impulse, show that it will loose contact with the hemisphere when the radial line through vertical makes an angle `theta` given by the equaition `cos^3theta-6costheta+4=0`.

Explanation

Let `v_r` be the relative velocity of block as it leaves contact with the sphere `(N=0)` and v the horizontal velocity of sphere at this instant.

Applying conservation of linear momentum in horizontal direction, we get
`mv=m(v_rcostheta-v)`
or `2v=v_rcostheta` ...(i)
Conservation of mechanincal energy gives,
`mgr(1-costheta)=1/2mv^2+1/2m(v_r^2+v^2-2v v_rcostheta)`
or `gr(1-costheta)=v^2+(v_r^2)/(2)-v v_rcostheta` ...(ii)
Equation of laws of motion gives,
`mgcostheta=(mv_r^2)/(r)` or `gr=(v_r^2)/(costheta)` ...(iii)
Solving Eqs. (i), (ii) and (ii), we get
`cos^3theta-6costheta+4=0`