A uniform thin rod of $120 \mathrm{~cm}$ length and $1600 \mathrm{~g}$ mass is bent as shown in the figure. The moment of inertia of the bent rod about an axis passing through the point ' $O$ ' and perpendicular to the plane of the paper is $\mathrm{kg}-\mathrm{m}^2$.

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Aicharya · Accepted Answer

There are four rods each of mass $\mathrm{m}=400 \mathrm{~g}=0.4 \mathrm{~kg}$ and each having a length of $\ell=30 \mathrm{~cm}=0.3 \mathrm{~m}$ So, we can calculate moment of inertia as follows

$\begin{aligned} & h^2=l^2-\left(\frac{l}{2}\right)^2=\frac{3}{4} l^2 \\ & \text { M.I. system }=(\text { MI of I }) \times 2+(\text { MI of II }) \times 2 \\ & =\frac{m l^2}{3} \times 2+2\left(\frac{m l^2}{12}+m h^2\right) \\ & =\frac{2}{3} m l^2+2\left(\frac{m l^2}{12}+\frac{3}{4} m l^2\right) \\ & =\frac{7}{3} m l^2=\frac{7}{3} \times 0.4 \times 0.3 \times 0.3=0.084 \mathrm{~kg}-\mathrm{m}^2\end{aligned}$ ,

A uniform thin rod of $120 \mathrm{~cm}$ length and $1600 \mathrm{~g}$ mass is bent as shown in the figure.…

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