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A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity $ω$ . Another disc of same dimensions but of mass $\frac{1}{4} M$ is placed gently on the first disc co-axially. The angular velocity of the system is

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$\frac{2}{3} ω$

$\frac{4}{5} ω$

$\frac{3}{4} ω$

$\frac{1}{3} ω$

answer is B.

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Detailed Solution

$∴ I_{1} = \frac{M R^{2}}{2} and I_{2} = \frac{M R^{2}}{2} + \frac{1}{2} (\frac{M}{4}) R^{2} = \frac{5}{8} M R^{2}$

By conservation of angular momentum,

$I ω_{1} = I_{2} ω_{2} \Rightarrow ω_{2} = \frac{I_{1}}{I_{2}} ω_{1}$

$= \frac{(1 / 2) M R^{2}}{(5 / 8) M R^{2}} ω \Rightarrow ω_{2} = \frac{4}{5} ω$

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