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A thin uniform circular disc of mass ' $\mathrm{M}$ ' and radius ' $R$ ' is rotating with angular velocity ' $\omega$ ', in a horizontal plane about an axis passing through its centre and perpendicular to its plane. Another disc of same radius but of mass $\left(\frac{M}{2}\right)$ is placed gently on the first disc co-axially. The new angular velocity will be
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The correct answer is:
$\frac{2}{3} \omega$
Angular momentum $=\mathrm{I} \omega$ By conservation of angular momentum, $\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2$
Here, $I_1=\frac{M^2}{2}, I_2=\frac{(M+M / 2)}{2} R^2=\frac{3 M^2}{4}$
$$
\begin{aligned}
& \therefore \quad \frac{\mathrm{MR}^2}{2} \omega_1=\frac{3 \mathrm{MR}^2}{4} \omega_2 \\
& \therefore \quad \omega_2=\frac{2}{3} \omega_1
\end{aligned}
$$
Here, $I_1=\frac{M^2}{2}, I_2=\frac{(M+M / 2)}{2} R^2=\frac{3 M^2}{4}$
$$
\begin{aligned}
& \therefore \quad \frac{\mathrm{MR}^2}{2} \omega_1=\frac{3 \mathrm{MR}^2}{4} \omega_2 \\
& \therefore \quad \omega_2=\frac{2}{3} \omega_1
\end{aligned}
$$
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