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 $\frac{L}{R}$ such that the moment of inertia is minimum?

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

Answer 1

The correct option is A

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

Step 1: Given Data:

Length of cylinder$=L$

Radius of cylinder$=R$

Step 2: Calculation:

$\begin{array}{rcl}I& =& \frac{m{R}^2}{4}+\frac{m{L}^2}{12}\\ & =& \frac{m}{4}\left(\frac{R^2}{1}+\frac{L^2}{3}\right)\\ & =& \frac{m}{4}\left(\frac{V}{\mathrm{\pi L}}+\frac{L^2}{3}\right)\dots \dots \dots \dots \dots \left(V={\mathrm{\pi R}}^{2}L\right)\end{array}$

Step 3: Differentiating $I$ with respect to $L$

From the above formula

$\frac{dI}{dL}=\frac{m}{4}\left(\frac{-V}{\pi L^2}+\frac{2L}{3}\right)$

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

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

Now,

$\begin{array}{rcl}\frac{V}{{\mathrm{\pi L}}^2}& =& \frac{2L}{3}\\ \frac{R^2}{L^2}& =& \frac{2}{3}\dots \dots \dots \dots \left(V=\mathrm{\pi }{R}^{2}L\right)\\ \frac{R^2}{L^2}& =& \frac{2}{3}\\ \sqrt{\frac{R^2}{L^2}}& =& \sqrt{\frac{2}{3}}\\ \frac{R}{L}& =& \sqrt{\frac{2}{3}}\\ & & or\\ \frac{L}{R}& =& \sqrt{\frac{3}{2}}\end{array}$

Hence option A is correct.