A thin rod of length $L$ and mass $M$ is bent at its midpoint into two halves so that the angle between them is $90^{\circ}$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

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Aicharya · Accepted Answer

Since rod is bent at the middle, so each part of it will have same length $\left(\frac{L}{2}\right)$ and mass $\left(\frac{M}{2}\right)$ as shown.

Moment of inertia of each part through its one end
$=\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2$
Hence, net moment of inertia through its middle point $O$ is
$\begin{aligned}
I & =\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2+\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2 \\
& =\frac{1}{3}\left[\frac{M L^2}{8}+\frac{M L^2}{8}\right]=\frac{M L^2}{12}
\end{aligned}$ .

A thin rod of length $L$ and mass $M$ is bent at its midpoint into two halves so that the angle between them…

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