Consider a case of fixed smooth sphere of radius R . A block of mass m is placed at the most poit of the sphere . A sharp impulse is applied on the block to impart it a speed v .
If a speed given at top is `sqrt((gR)/3)` , then what is the angle made by radius through block when it leaves the sphere ?

Options

A. `cos^(-1)(2/3)`
B. `cos^(-1)(3/4)`
C. `cos^(-1)(5/9)`
D. `cos^(-1)(7/9)`

Answer

D

Explanation

First we can apply conservation of mechanical energy between top point and the point where radius through the aprticle makes an angle `theta ` with the vertical .
Gain in K.E = Loss of gravitational potential energy
`rArr " "1/2 mv_(2)^(2) -1/2 mv_(1)^(2) = mgR(1-cos theta)`
Substituting `v_(1) = sqrt((gR)/3)` we get the following :
`rArr 1/2 mv_(2)^(2) -1/2 m(gR)/3 = mgR(1-cos theta )`
`rArr 1/2mv_(2)^(2) - (mgR)/6 = mgR (1-cos theta )`
`rArr " " 1/2 mv_(2)^(2) = mgR (7/6 - cos theta )`
`rArr " " (mv_(2)^(2))/R = 2mg(7/6-cos theta ) " " ...(ii)`
According to circular dynamics we can write the following :
`mg cos theta -N_(2) = (mv_(2)^(2))/R `
Substituting from equation (ii) and `N_(2) = 0 `, we get
`mg cos theta = 2mg (7/6 - cos theta ) rArr cos theta = 2 (7/6 - cos theta) `
`rArr " " cos theta = 7/3 - 2 co theta rArr 3 cos theta = 7/3 rArr cos theta = 7/9 `