A particle of mass m begins to slide down on a fixed smooth sphere from the top, when it breaks off the sphere, the tangential accleration will be

Options

A. `2g//3`
B. g
C. `sqrt(5)/(3)g`
D. `(g)/(2)`

Explanation

`mg cos theta=(mv^(2))/(R)`
`mg(R-R cos theta)`
`=(1)/(2)mv^(2)`
`rArr cos theta=(2)/(3)`
`a_(1)=g sin theta=gxxsqrt(5)/(3)`