Question

The moment of inertia of a uniform cylinder of length $l$ and radius $R$ about it's perpendicular bisector is $I$. What is the ratio $\frac{l}{R}$ 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 $\sqrt{\frac{3}{2}}$

$I=\frac{m{l}^2}{12}+\frac{m{R}^2}{4}$
$I=\frac{m}{4}(\frac{l^2}{3}+R^2).......\left(1\right)$
Also $m=\rho \pi R^{2}l$
$R^2=\frac{m}{\pi \rho l}$
For minimum moment of inertia $\frac{dI}{dl}=0$
$\frac{dI}{dl}=\frac{m}{4}(\frac{2l}{3}-\frac{m}{\pi l^2\rho })=0$
$\frac{2l}{3}=\frac{m}{\pi l^2\rho }\Rightarrow \frac{2l}{3}=\frac{\pi R^{2}l\rho }{\pi l^2\rho }$
$\frac{2l}{3}=\frac{R^2}{l}$
$\frac{l^2}{R^2}=\frac{3}{2}$
$\frac{l}{R}=\sqrt{\frac{3}{2}}$