The moment of inertia of a cylinder of mass M | vedclass

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(C) Given the moment of inertia $I = M \left(\frac{R^2}{4} + \frac{L^2}{12}\right)$.Since the mass $M$ and density $\rho$ are constant,the volume $V =

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$\sqrt{\frac{3}{2}}$

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(C) Given the moment of inertia $I = M \left(\frac{R^2}{4} + \frac{L^2}{12}\right)$.Since the mass $M$ and density $\rho$ are constant,the volume $V = \pi R^2 L$ must be constant.Thus,$R^2 L = K$ (where $K$ is a constant).Differentiating with respect to $R$,we get $2RL + R^2 \frac{dL}{dR} = 0$,which implies $\frac{dL}{dR} = -\frac{2L}{R}$.To minimize $I$,we set $\frac{dI}{dR} = 0$:$\frac{dI}{dR} = M \left(\frac{2R}{4} + \frac{2L}{12} \frac{dL}{dR}\right) = 0$.$\frac{R}{2} + \frac{L}{6} \left(-\frac{2L}{R}\right) = 0$.$\frac{R}{2} - \frac{L^2}{3R} = 0$.$\frac{R}{2} = \frac{L^2}{3R} \Rightarrow \frac{L^2}{R^2} = \frac{3}{2}$.Therefore,$\frac{L}{R} = \sqrt{\frac{3}{2}}$.

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