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A disc of moment of inertia $2 \mathrm{~kg}-\mathrm{m}^2$ revolving with $8 \mathrm{rad} / \mathrm{s}$ is placed on another disc of moment of inertia $4 \mathrm{~kg}-\mathrm{m}^2$ revolving $4 \mathrm{rad} / \mathrm{s}$. What is the angular frequency of composite disc?
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The correct answer is:
$\frac{16}{3} \mathrm{rad} / \mathrm{s}$
Given that, moment of inertia of discs,
$\mathrm{I}_1=2 \mathrm{~kg}-\mathrm{m}^2, \mathrm{I}_2=4 \mathrm{~kg}-\mathrm{m}^2$
Angular velocity of discs, $\omega_1=8 \mathrm{rad} / \mathrm{s}$, $\omega_2=4 \mathrm{rad} / \mathrm{s}$
From angular momentum conservation principle,
$\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2=\left(\mathrm{I}_1+\mathrm{I}_2\right) \omega$
or $\quad 2 \times 8+4 \times 4=(2+4) \omega$
or $\quad \omega=\frac{16+16}{6}=\frac{32}{6}=\frac{16}{3} \mathrm{rad} / \mathrm{s}$
$\mathrm{I}_1=2 \mathrm{~kg}-\mathrm{m}^2, \mathrm{I}_2=4 \mathrm{~kg}-\mathrm{m}^2$
Angular velocity of discs, $\omega_1=8 \mathrm{rad} / \mathrm{s}$, $\omega_2=4 \mathrm{rad} / \mathrm{s}$
From angular momentum conservation principle,
$\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2=\left(\mathrm{I}_1+\mathrm{I}_2\right) \omega$
or $\quad 2 \times 8+4 \times 4=(2+4) \omega$
or $\quad \omega=\frac{16+16}{6}=\frac{32}{6}=\frac{16}{3} \mathrm{rad} / \mathrm{s}$
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