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

Option 1)
$\sqrt{\frac{3}{2}}$

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

Option 3)
$1$

Option 4)
$\frac{3}{\sqrt{2}}$

Answers (1)

best_answer

As we learnt in

Moment of inertia of uniform rod of length(l) -

$I= \frac{Ml^{2}}{12}$

- wherein

About axis passing through its centre & perpendicular to its length.

We know that solid cylinder about an axis which is perpendicular bisector.

$I=\frac{mR^{2}}{4}+\frac{mR^{2}}{12}=\frac{m}{4}\left [ R^{2}+\frac{l^{2}}{3} \right ]$

$I=\frac{m}{4}\left [ \frac{\pi lR^{2}}{\pi l}+\frac{R^{2}}{3} \right ]=\frac{m}{4}\left [ \frac{v}{\pi l}+\frac{l^{2}}{3} \right ]$

$\frac{dx}{dl}=0$ $\Rightarrow\ \;\frac{m}{4}\left [ \frac{-v}{\pi l^{2}}+\frac{2l}{3} \right ]=0\ \;\Rightarrow\ \;\frac{-v}{\pi l^{2}}+\frac{2l}{3}=0$

$\frac{v}{\pi l^{2}}=\frac{2l}{3}\ \Rightarrow \; v=\frac{2\pi l^{3}}{3}$

$\pi R^{2}l=2\frac{\pi l^{3}}{3}\ \; \Rightarrow\ \; R^{2}=\frac{2l^{2}}{3}$

$\frac{l^{2}}{R^{2}}=\frac{3}{2}$

$\frac{l}{R}=\sqrt{}\frac{3}{2}$

Option 1)

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

This is the correct option.

Option 2)

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

This is an incorrect option.

Option 3)

$1$

This is an incorrect option.

Option 4)

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

This is an incorrect option.

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