Question

The moment of inertia of a uniform cylinder of length $l$ and radius $R$ about its perpendicular bisector is $l$. 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. 1
• D. $\frac{3}{\sqrt{2}}$

Answer 1

Answer: (A)

$I=\frac{m{R}^2}{4}+\frac{m{l}^2}{12}$
$I=\frac{m}{4}\left[R^2+\frac{l^2}{3}\right]$
$=\frac{m}{4}\left[\frac{v}{\pi l}+\frac{l^2}{3}\right]$
$\frac{dl}{dl}=\frac{m}{4}\left[\frac{-v}{\pi l^2}+\frac{2l}{3}\right]=0$
$\frac{v}{\pi l^2}=\frac{2l}{3}$
$v=\frac{2\pi l^3}{3}$
$\pi R^{2}l=\frac{2\pi l^3}{3}$
$\frac{l^2}{R^2}=\frac{3}{2}$
$\frac{l}{R}=\sqrt{\frac{3}{2}}$