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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.
Options:
mass and perpendicular to the plane of the triangle.
Solution:
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Verified Answer
The correct answer is:
$2 \mathrm{~kg}-\mathrm{m}^2$
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,
$$
\begin{aligned}
I & =3\left[\frac{m l^2}{12}+m\left(\frac{l}{2 \sqrt{3}}\right)^2\right] \\
& =3\left[\frac{m l^2}{12}+\frac{m l^2}{12}\right] \\
& =\frac{m l^2}{2}=\frac{1 \times 2^2}{2}=2 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}
$$
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,
$$
\begin{aligned}
I & =3\left[\frac{m l^2}{12}+m\left(\frac{l}{2 \sqrt{3}}\right)^2\right] \\
& =3\left[\frac{m l^2}{12}+\frac{m l^2}{12}\right] \\
& =\frac{m l^2}{2}=\frac{1 \times 2^2}{2}=2 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}
$$
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