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An round disc of moment of inertia $I_2$ about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia $\mathrm{I}_1$ rotating with an angular velocity $\omega$ about the same axis. The final angular velocity of the combination of discs is:
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Verified Answer
The correct answer is:
$\frac{I_1 \omega}{I_1+I_2}$
From law of conservation of angular momentum we have:
$$
\begin{aligned}
& I_1 \omega_1=\left(I_1+I_2\right) \omega_2 \\
& \therefore \quad \omega_2=\frac{I_1 \omega_1}{\left(I_1+I_2\right)}=\frac{I_1 \omega}{I_1+I_2} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& I_1 \omega_1=\left(I_1+I_2\right) \omega_2 \\
& \therefore \quad \omega_2=\frac{I_1 \omega_1}{\left(I_1+I_2\right)}=\frac{I_1 \omega}{I_1+I_2} \\
&
\end{aligned}
$$
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