A solid uniform disc of mass m and radius R is pivoted about a horizontal axis through its centre and a small body of mass m is attached to the rim of the disc. If the disc is released from rest with the small body, initially at the same level as the centre, the angular velocity when the small body reaches the lowest point of the disc is

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$\sqrt{\frac{12 g}{13 R}}$

$\sqrt{\frac{11 g}{12 R}}$

$\sqrt{\frac{4 g}{3 R}}$

$\sqrt{\frac{7 g}{11 R}}$

answer is C.

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Detailed Solution

The conservation of energy yields

$∆ K E + ∆ P E = 0 \frac{1}{2} (\frac{1}{2} m R^{2}) ω^{2} + \frac{1}{2} m {(ω R)}^{2} + (- m g R) = 0 ω = \sqrt{\frac{4 g}{3 R}}$ ;

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