A uniform solid sphere of radius `r = ( R)/(5)` is placed on the inside surface of a hemisherical bowl with radius `R (= 5 r)`. The sphere is released from rest at an angle `theta = 37^@` to the vertical and rolls without slipping (Fig.) The angular speed of the sphere when it reaches the bottom of the bowl is.
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Answer

B

Explanation

(b) For the isolated sphere-Earth system, energy conserved,
`Delta U + Delta K_(rat) + Delta K_(trans) = 0`
`mg(R - r)(cos theta - 1)+[(1)/(2) mv^2 - 0]+ (1)/(2)[(2)/(5) mr^2]omega^2 = 0`
Substituting `v = r omega` we obtain
`mg(R - r)(cos theta - 1) + [(1)/(2)m(r omega)^2 - 0]`
`+(1)/(2)[(2)/(5) mr^2] omega^2 = 0`
`mg(R - r)(cos theta - 1)+[(1)/(2) +(1)/(5)]mr^2 omega^2 = 0`
`omega = sqrt(((10)/(7))((R - r)(1 - cos theta)g)/(r^2))` ...(i)
Here we have `r = (R )/(5) and theta = 37^@`
Substituting this values in equation (i), we get
`omega = sqrt((40)/(7)Rg)`.
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