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