Three rods each of mass $1 \mathrm{~kg}$ and length $2 \mathrm{~m}$ are joined together end-to-end to form an equilateral triangle $A B C$. Find the moment of inertia of this system about an axis passing through its centre of mass and perpendicular to the plane of the triangle.

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Aicharya · Accepted Answer

Given, mass of rods, $m=1 \mathrm{~kg}$ Length of rod, $l=2 \mathrm{~m}$ Let moment of inertia of rod about centre of rod, $ I_{\mathrm{CM}}=\frac{m l^2}{12} $

$D, E$ and $F$ are mid points of side $A C, A B$ and $B C, O$ is centroid. $ \begin{array}{ll} \because & \tan 30^{\circ}=\frac{O E}{E B} \Rightarrow \frac{1}{\sqrt{3}}=\frac{O E}{l / 2} \\ \Rightarrow & O E=\frac{l}{2 \sqrt{3}}=O D=O F=d \end{array} $ By using parallel axis theorem, $ I=I_{\mathrm{CM}}+m d^2 $ and for three rods, $ \begin{aligned} I & =3\left[\frac{m l^2}{12}+m\left(\frac{l}{2 \sqrt{3}}\right)^2\right] \\ & =3\left[\frac{m l^2}{12}+\frac{m l^2}{12}\right] \\ & =\frac{m l^2}{2}=\frac{1 \times 2^2}{2}=2 \mathrm{~kg}-\mathrm{m}^2 \end{aligned} $

Three rods each of mass $1 \mathrm{~kg}$ and length $2 \mathrm{~m}$ are joined together end-to-end to form…

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