A particle of mass m is kept on a fixed, smooth sphere of radius R at a position, where the radius through the particle makes an angle of 30 ∘ with the vertical. The particle is released from this position. (a) What is the force exerted by the sphere on the particle just after the release? (b) Find the distance traveled by the particle before it leaves contact with the sphere.

Explanation

a. At the time of release, the speed of box is zero. It will push the sphere only with the normal component of its weight. Along radial direction, we have
`N=mg cos 30^@impliesN=sqrt3/2mg`
b. The situation is shown in figure.
Let the box lose contact with the sphere at an angle `theta` from the vertical. At this instant, its normal reaction becomes zero.
Thus, we have at this point.

`(mv^2)/(R)=mg cos theta`
`v=sqrt(Rgcos theta)`
Velocity of particle at this point can be given by energy conservation as it has fallen a distance h, we have
`h=R(cos 30^@-costheta)`
and `v=sqrt(2gh)=sqrt(2gR(sqrt3/2-costheta))` (ii)
Equating Eqs. (i) and (ii), we have
`cos theta=sqrt3-2cos theta=1/sqrt3=theta=cos^-1(1/sqrt3)`
Angular displacement of the box before leaving the sphere is
`cos^-1(1/sqrt3)-pi/6`
Distance travelled by the box is
`[cos^-1(1/sqrt3)-pi/6]R`