Question

A particle slides on a surface of fixed smooth sphere of radius r starting from the topmost point the height from the ground where it leaves contact with the sphere is

Answer 1

Let velocity of the particle be ‘v’ when it leaves contact with surface.

see diagram,

at equilibrium,

centripetal force = component of weight

or, mv²/R = mgcos$\theta$

or, v² = Rgcos$\theta$.....(1)

from work - energy theorem,

change in K.E = workdone

or, final K.E - initial K.E = force × displacement

or, 1/2 mv² - 0 = mg(R - Rcos$\theta$)

or, v² = 2gR(1 - cos$\theta$)....(2)

from equations (1) and (2),

Rgcos$\theta$ = 2gR(1 - cos$\theta$)

or, 3gRcos$\theta$ = 2gR

or, cos$\theta$ = 2/3

or, $\theta$ = cos^-1(2/3)

hence, $\theta=cos^{-1}(2/3)$