Moment of inertia of a cylinder of mass M, length L and radius R … - Vidyadip

MCQ

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 $\frac LR$ for it to have minimum possible $I$ is

A
$\sqrt{\frac{2}{3}}$
B
$\frac{3}{2}$
✓
$\sqrt{\frac{3}{2}}$
D
$\frac{2}{3}$

✓

Answer

Correct option: C.

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

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 $(1)$

$\frac{ dI }{ clR }= M \left(\frac{2 R }{4}+\frac{2 L }{12} \times \frac{ dI }{ dr }\right)=0$

$\frac{ R }{2}+\frac{ L }{6} \frac{ dI }{ dR }=0$

Substituting value of $\frac{ dI }{ d R }$ from eqution $(2)$

$\frac{ R }{2}+\frac{ L }{6}\left(\frac{-2 L }{ R }\right)=0$

$\frac{ R }{2}=\frac{ L^{2}}{3 R } \Rightarrow \frac{ L }{ R }=\sqrt{\frac{3}{2}}$

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