A particle of mass `m` is released on a fixed. Smooth sphere of radius `R` at a position, where the radius through the particle makes an angle of `alpha` with the verticle. What is the angle made by radius through the particle when the particle loses contact with sphere? `(cos alpha=3//4)`

Explanation

`h=R(cos alpha-cosbeta)`
`v^(2)=u^(2)+2gh=0+2gR(cos alpha-cos beta)`
`mg cos beta-N=(mv^(2))/(R )`
when the particle loses contact
`N=0`
`v^(2)=gR cos beta=2gR(cos alpha-cos beta)`
`3 cos beta=2 cos alpha=2xx(3)/(4)`
`cos beta=(1)/(2)impliesbeta=60^(@)`