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A ball of mass m and radius r is rolling without slipping inside a hemispherical shell of radius R. It is released from rest from point A as shown in figure. The angular velocity of centre of the ball in position B about the centre of the shell is

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$\sqrt[]{\frac{10 g}{7 (R - r)}}$

$\sqrt{\frac{2 g}{7 (R - r)}}$

$\sqrt{\frac{5 g}{2 (R - r)}}$

$2 \sqrt[]{\frac{g}{5 (R - r)}}$

answer is B.

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Detailed Solution

K.E. of ball in position B = $mg (R - r)$
Here m = mass of ball.
Since it rolls without slipping the ratio of rotational to translational kinetic energy will be $\frac{2}{5}$ .

$\frac{K_{R}}{K_{T}} = \frac{2}{5}$

$K_{T} = \frac{5}{7} mg (R - r)$

$\frac{1}{2} {mv}^{2} = \frac{5}{7} mg (R - r)$

$v = \sqrt{\frac{10 g (R - r)}{7}}$

$ω = \frac{v}{R - r} = \sqrt[]{\frac{10 g}{7 (R - r)}}$

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