Question

Moment of inertia of a cylinder of mass $M$ length $L$ and radius $R$ about an axis passing through its centre and perpendicular to the axis of the cylinder is $I = M \left(\frac{ R ^{2}}{4}+\frac{ L ^{2}}{12}\right)$. If such a cylinder is to be made for a given mass of material, the ratio $L / R$ for it to have minimum possible $I$ is :

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

Answer 1

Answer: (C)

$I=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}\right)$....(1)
as mass is constant $\Rightarrow m =\rho V = constant$
$V =$ constant
$\pi^{2} R l=$ constant
$\Rightarrow R ^{2} L =$ constant
$2 RL + R ^{2} \frac{ dL }{ dR }=0$....(2)
From equation (1)
$\frac{ dI }{ dR }= M \left(\frac{2 R }{4}+\frac{2 L }{12} \times \frac{ dL }{ dr }\right)=0$
$\frac{ R }{2}+\frac{ L }{6} \frac{ dL }{ dR }=0$
Substituting value of $\frac{ dL }{ d R }$ from eqution (2)
$\frac{R}{2}+\frac{L}{6}\left(\frac{-2 L}{R}\right)=0$
$\frac{R}{2}=\frac{L^{2}}{3 R}$
$\Rightarrow \frac{L}{R}=\sqrt{\frac{3}{2}}$