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Three identical thin rods, each of mass $m$ and length $\ell$, are joined to form an equilateral triangular frame. The moment of inertia of the frame about an axis parallel to its one side and passing through the opposite vertex is
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$\frac{5}{4} m \ell^2$
Moment of Inertia of each of rod $\mathrm{AC}$ and $\mathrm{BC}$ about the given axis $00^{\prime}$ is
$\left.I_{A C}=I_{B C}=\frac{m \ell^2}{3} \sin ^2 60^{\circ}=\frac{m \ell^2}{4} \right\rvert\,$ and $\mathrm{M}$.I. of rod $\mathrm{AB}$ about the given axis $0 0^{\prime}$ is
$I_{A B}=m\left(\frac{\ell \sqrt{3}}{2}\right)^2=\frac{3}{4} m \ell^2$
Hence, $\left.I=I_{A C}+I_{B C}+I_{A B}=\frac{m \ell^2}{4}+\frac{m \ell^2}{4}+\frac{3}{4} m \ell^2=\frac{5}{4} m \ell^2 \right\rvert\,$\n
$\left.I_{A C}=I_{B C}=\frac{m \ell^2}{3} \sin ^2 60^{\circ}=\frac{m \ell^2}{4} \right\rvert\,$ and $\mathrm{M}$.I. of rod $\mathrm{AB}$ about the given axis $0 0^{\prime}$ is
$I_{A B}=m\left(\frac{\ell \sqrt{3}}{2}\right)^2=\frac{3}{4} m \ell^2$
Hence, $\left.I=I_{A C}+I_{B C}+I_{A B}=\frac{m \ell^2}{4}+\frac{m \ell^2}{4}+\frac{3}{4} m \ell^2=\frac{5}{4} m \ell^2 \right\rvert\,$\n
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