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(A) The moment of inertia of a uniform solid cylinder of mass $m$,length $l$,and radius $R$ about its perpendicular bisector is given by:$I = \frac{mR

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

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(A) The moment of inertia of a uniform solid cylinder of mass $m$,length $l$,and radius $R$ about its perpendicular bisector is given by:$I = \frac{mR^2}{4} + \frac{ml^2}{12}$Assuming the density $\rho$ is constant,the mass $m = \rho V = \rho (\pi R^2 l)$.Substituting $m$ in the expression for $I$:$I = \frac{\rho \pi R^2 l R^2}{4} + \frac{\rho \pi R^2 l^3}{12} = \frac{\rho \pi R^4 l}{4} + \frac{\rho \pi R^2 l^3}{12}$Since the volume $V = \pi R^2 l$ is constant,we can write $R^2 = \frac{V}{\pi l}$. Substituting this into the expression for $I$:$I = \frac{m}{4} \left( \frac{V}{\pi l} + \frac{l^2}{3} \right)$To find the minimum moment of inertia,we differentiate $I$ with respect to $l$ and set it to zero:$\frac{dI}{dl} = \frac{m}{4} \left( -\frac{V}{\pi l^2} + \frac{2l}{3} \right) = 0$$\frac{V}{\pi l^2} = \frac{2l}{3} \implies V = \frac{2\pi l^3}{3}$Since $V = \pi R^2 l$,we have:$\pi R^2 l = \frac{2\pi l^3}{3} \implies R^2 = \frac{2l^2}{3} \implies \frac{l^2}{R^2} = \frac{3}{2}$Therefore,the ratio is $\frac{l}{R} = \sqrt{\frac{3}{2}}$.

The moment of inertia of a uniform cylinder of length $l$ and radius $R$ about its perpendicular bisector is $I$. What is the ratio $l/R$ such that the moment of inertia is minimum?

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