A particle of mass m is kept on the top of a smooth sphere of radius R. It is given a sharp impulse which imparts it a horizontal speed v. a. find the normal force between the sphere and the particle just after the impulse. B. What should be the minimum value of v for which the particle does not slip on the sphere? c. Assuming the velocity v to be half the minimum calculated in part, d. find the angle made by the radius through the particle with the vertical when it leaves the sphere.

Answer

A, B, C, D

Explanation

a. Radius =R
Horizontal speed=v
From the free body diagram ure
N=normal force
`=mg-(mv^2)/R`
b. When the particle is givenmaximum velocity so that the centrifugal force balances the weight, the particle does not slip on the sphere,
`rarr (mv^2)/R=mg`
`rarr v=sqrt(gR)`
c if the body is given velocity `v_1` at the top such that
`v_1=sqrt(gR)/2`
`v_1^2=(gR)/4` Let the velocity be `v_2` when it leaves contact with the surface, ure

So, (mv_2^2)/R=mgcostheta`
`rarrv_2^2=Rgcostheta`........i
`Again `(1/2)mv_2^2-(1/2)mv_1^2`
`=mgR(1-costheta)`
`rarr v_2^2=v_1^2+2gR(1-costheta)`............ii
From equatioon i and equation ii
`Rgcostheta=((Rg)/4)+2gR(1-costheta)`
`rarr costheta=(1/4+2-2costheta)`
`rarr 3costheta=(9/4)`
`rarr theta=cos^-1(3/4)`