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.

Answer

B, C, D

Explanation

a. When the particle is released from rest ure the centrifugal force is zero.
N force =`mgcostheta=mgcos30^0`
`=sqrt3/2`mg
b. When the particle leaves contact with the surface ure `N=theta`

`So, (mv^2)/R=mgcostheta`
`rarr v^2=Rcostheta` ........i
`Again
(1/2)mv^2=mgR(cos30^0-costheta)`
`rarr v^2=2rg(sqrt3/2-costheta)`........ii
`from equation i and equation ii
`Rgcostheta=2Rg[sqrt3/2-costheta]`
`rarr 3costheta=sqrt(3)`
`rarr costheta=1/sqrt3`
`or theta=cos^-1 1/sqrt3`
so, the distance travelld by the particle before leaving contact
`L=R(theta-pi/6)`
[because `30^0=(pi/6)`]
`putting the value of theta we get`
`L=0.43R`