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A thin wire of length ' $L$ ' and uniform linear mass density ' $\mathrm{m}$ ' is bent into a circular coil. The moment of inertia of this coil about tangential axis and in plane of the coil is
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Verified Answer
The correct answer is:
$\frac{3 m L^3}{8 \pi^2}$
$\therefore \quad$ Moment of inertia of thin wire:
$$
\begin{aligned}
\mathrm{I} & =\frac{\mathrm{MR}^2}{2} \\
\mathrm{M} & =\mathrm{V} \times \mathrm{m} \text { and } \mathrm{L}=2 \pi \mathrm{R} \\
\mathrm{R} & =\frac{\mathrm{L}}{2 \pi} \\
\therefore \quad \mathrm{I} & =\frac{\mathrm{Lm}}{2}\left(\frac{\mathrm{L}}{2 \pi}\right)^2 \\
\mathrm{I} & =\frac{\mathrm{mL}^3}{8 \pi^2}
\end{aligned}
$$
$\therefore \quad$ Using Parallel axis theorem:
$$
\begin{aligned}
& I^{\prime}=I+M R^2 \\
& I=\frac{m L^3}{8 \pi^2}+L m\left(\frac{L}{2 \pi}\right)^2 \\
& I=\frac{3 m L^3}{8 \pi^2}
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{I} & =\frac{\mathrm{MR}^2}{2} \\
\mathrm{M} & =\mathrm{V} \times \mathrm{m} \text { and } \mathrm{L}=2 \pi \mathrm{R} \\
\mathrm{R} & =\frac{\mathrm{L}}{2 \pi} \\
\therefore \quad \mathrm{I} & =\frac{\mathrm{Lm}}{2}\left(\frac{\mathrm{L}}{2 \pi}\right)^2 \\
\mathrm{I} & =\frac{\mathrm{mL}^3}{8 \pi^2}
\end{aligned}
$$
$\therefore \quad$ Using Parallel axis theorem:
$$
\begin{aligned}
& I^{\prime}=I+M R^2 \\
& I=\frac{m L^3}{8 \pi^2}+L m\left(\frac{L}{2 \pi}\right)^2 \\
& I=\frac{3 m L^3}{8 \pi^2}
\end{aligned}
$$
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