A practical moves from rest at A on the surface of a smooth circular cylinder of radius `r` as shown . At `B` it leaves the cylinder. The equation relating `alpha and beta` is

Options

A. `3 sin alpha = 2 sin beta`
B. `2 sin alpha = 3 cos beta`
C. `3 sin beta = 2 cos alpha`
D. `2 sin beta = 3 cos alpha`

Answer

C

Explanation

By law of conservation of mechanical energy
`(K + U)_(A) = (K + U)_(B)`
`0 + mgR cos alpha = (1)/(2) mv_(B)^(2) + mgR sin beta`
`v_(B)^(2) = 2gR (cos alpha - sin beta)`
At `B` since partical leavel constant eith the surface hence normal reaction equals zero
`mg sin beta = (mv_(B)^(2))/( R)`
`mg sin beta =(m)/(R) [2 gR (cos alpha - sin beta)]`