Question
The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length $L(R < L)$ about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as M ) :
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Solution
$$\begin{matrix} & I_{net} = 2\left( I_{1} + I_{2} \right) \\ & \ = 2\left( \frac{M'R^{2}}{4} + \frac{M'\mathcal{l}^{2}}{12} \right) + 2\left( \frac{M'R^{2}}{2} + M'\left( \frac{\mathcal{l}}{2} \right)^{2} \right) \\ & \ = \frac{M'R^{2}}{2} + \frac{M'R^{2}}{6} + M'R^{2} + \frac{M'\mathcal{l}^{2}}{2} \\ & \ = \frac{3M'R^{2}}{2} + \frac{2M'\mathcal{l}^{2}}{3} \end{matrix}$$
Given masses $M' = \frac{M}{4}$
So, $I = \frac{3(M/4)R^{2}}{2} + 2\frac{(M/4)\mathcal{l}^{2}}{3}$
$$I = \frac{3}{8}{MR}^{2} + \frac{M\mathcal{l}^{2}}{6}$$
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