A particle of mass m is kept on a fixed , smooth sphere of radius R at a position, where the rdius thrugh the particle makes an angle of `30^0` with the vertical. The particle is releases from this positon. a. What is the force exerted by the sphere on teh particle just after the release? b. Find the distance travelled 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`