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In the diagram shown below all three rods are of equal length \(L\) and equal mass \(M\). The system is rotated such that \(\operatorname{rod} B\) is the axis. What is the moment of inertia of the system?

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The correct answer is:
\(\frac{M L^2}{6}\)
Since the width is negligible, the moment of inertia of \(B\) is zero along its length.
\(I_A+I_C=\frac{m L^2}{12}+\frac{m L^2}{12}=\frac{m L^2}{6}\)

\(I_A+I_C=\frac{m L^2}{12}+\frac{m L^2}{12}=\frac{m L^2}{6}\)

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