Question

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 ?

• A. $\frac{3}{\sqrt{2}}$
• B. $\sqrt{\frac{3}{2}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $1$

Answer 1

The correct option is B $\sqrt{\frac{3}{2}}$

Given,

The length of the cylinder $=l$,

Radius of cylinder$=R$

And $I=$moment of inertia.

Now,

$I=\frac{m{R}^2}{4}+\frac{m{l}^2}{12}$

$I=\frac{m}{4}(R^2+\frac{l^2}{3})$

$I=\frac{m}{4}(\frac{V}{\pi l}+\frac{l^2}{3})(\because V=\pi R^{2}l)$

Now differentiate $I$with respect to $l$

So, $\frac{dI}{dl}=\frac{m}{4}(\frac{-V}{\pi l^2}+\frac{2l}{3})$

For maxima and minima,$\frac{dI}{dl}=0$

So,$\frac{m}{4}(\frac{-V}{\pi l^2}+\frac{2l}{3})=0$

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

$\frac{R^2}{l}=\frac{2l}{3}(\because V=\pi R^{2}l)$

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

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

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