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Two identical particles each of mass ' $\mathrm{m}$ ' are separated by a distance ' $d$ '. The axis of rotation passes through the midpoint of ' $\mathrm{d}$ ' and is perpendicular to the length $\mathrm{d}$. If ' $\mathrm{K}$ ' is the average rotational kinetic energy of the system, then the angular frequency is
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The correct answer is:
$\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{K}}{\mathrm{m}}}$
Moment of inertia $\mathrm{I}=2 \mathrm{~m}\left(\frac{\mathrm{d}}{2}\right)^2=\frac{\mathrm{md}^2}{2}$
Kinetic energy $K=\frac{1}{2} I \omega^2$
$$
\begin{aligned}
& \therefore \omega^2=\frac{2 \mathrm{~K}}{\mathrm{I}}=2 \mathrm{~K} \cdot \frac{2}{\mathrm{md}^2}=\frac{4 \mathrm{~K}}{\mathrm{md}^2} \\
& \therefore \omega=\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}
\end{aligned}
$$
Kinetic energy $K=\frac{1}{2} I \omega^2$
$$
\begin{aligned}
& \therefore \omega^2=\frac{2 \mathrm{~K}}{\mathrm{I}}=2 \mathrm{~K} \cdot \frac{2}{\mathrm{md}^2}=\frac{4 \mathrm{~K}}{\mathrm{md}^2} \\
& \therefore \omega=\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}
\end{aligned}
$$
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