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A disc of moment of inertia $4 \mathrm{~kg}-\mathrm{m}^2$ revolving with $16 \mathrm{rad} / \mathrm{s}$ is placed on another disc of moment of inertia $8 \mathrm{Kg}-\mathrm{m}^2$ revolving $4 \mathrm{rad} / \mathrm{s}$. The angular frequency of composite disc
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Verified Answer
The correct answer is:
$8 \mathrm{rad} / \mathrm{s}$
Given that, moment of inertia of disc, $I_1=4 \mathrm{~kg}-\mathrm{m}^2$
$$
I_2=8 \mathrm{~kg} \cdot \mathrm{m}^2
$$
Angular velocities of discs,
$$
\begin{aligned}
& \omega_1=16 \mathrm{rad} / \mathrm{s} \\
& \omega_2=4 \mathrm{rad} / \mathrm{s}
\end{aligned} .
$$
From angular momentum conservation principle,
$$
\begin{array}{rlrl}
& & I_1 \omega_1+I_2 \omega_2 & =\left(I_1+I_2\right) \omega \\
\Rightarrow & 4 \times 16+8 \times 4 & =(4+8) \omega \\
\Rightarrow & & 64+32 & =12 \omega \\
\Rightarrow & & \omega & =\frac{96}{12}=8 \mathrm{rad} / \mathrm{s}
\end{array}
$$
$$
I_2=8 \mathrm{~kg} \cdot \mathrm{m}^2
$$
Angular velocities of discs,
$$
\begin{aligned}
& \omega_1=16 \mathrm{rad} / \mathrm{s} \\
& \omega_2=4 \mathrm{rad} / \mathrm{s}
\end{aligned} .
$$
From angular momentum conservation principle,
$$
\begin{array}{rlrl}
& & I_1 \omega_1+I_2 \omega_2 & =\left(I_1+I_2\right) \omega \\
\Rightarrow & 4 \times 16+8 \times 4 & =(4+8) \omega \\
\Rightarrow & & 64+32 & =12 \omega \\
\Rightarrow & & \omega & =\frac{96}{12}=8 \mathrm{rad} / \mathrm{s}
\end{array}
$$
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