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A round disc of moment of inertia $\mathrm{I}_{2}$ 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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The correct answer is:
$\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}$
Total angular momentum of the system initially \(\mathrm{L}_{\mathrm{i}}=\mathrm{I}_{1} \mathrm{w}+\mathrm{I}_{2}(0)=\mathrm{I}_{1} \mathrm{w}\)
Total angular momentum of the system finally \(L_{f}=\left(I_{1}+I_{2}\right) w_{2}\)
According to conservation of angular momentum i.e. \(\mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}}\)
\(\begin{aligned}
&\Rightarrow \mathrm{I}_{1} \omega=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega_{2} \\
&\Rightarrow \omega_{2}=\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}
\end{aligned}\)
Total angular momentum of the system finally \(L_{f}=\left(I_{1}+I_{2}\right) w_{2}\)
According to conservation of angular momentum i.e. \(\mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}}\)
\(\begin{aligned}
&\Rightarrow \mathrm{I}_{1} \omega=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega_{2} \\
&\Rightarrow \omega_{2}=\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{I}_{2}}
\end{aligned}\)
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