Question

A uniform cylinder has a radius $R$ and length $L$. If the moment of inertia of this cylinder about an axis passing through its centre and normal to its circular face is equal to the moment of inertia of the same cylinder about an axis passing through its centre and perpendicular to its length, then:

• A. $L=\sqrt{3}R$
• B. $L=\sqrt{6}R$
• C. $L=\frac{R}{\sqrt{6}}$
• D. $L=\frac{R}{\sqrt{3}}$

Answer 1

The correct option is A $L=\sqrt{3}R$

Consider the diagram shown below.

For axis perpendicular to circular face, $I_1=\frac{M{R}^2}{2}$

For axis perpendicular to its length, consider a disc of width $dx$ at a distance $x$ from the axis. Let its mass is $dM$. Then

$dM=\frac{M}{L}dx$

Moment of inertia of this about an axis parallel to $l_2$.

$d{I}_1=\frac{dM{R}^2}{4}$

Its moment of inertia about axis of $l_{2x}$.

$dI=\frac{dM{R}_2}{4}+dM{x}^2$

Therefore,

$I_2=\frac{M{R}^2}{4}+\frac{M{L}^2}{\left(3\right)\left(4\right)}$

According to the question,

$\frac{M{R}^2}{2}=\frac{M{R}^2}{4}+\frac{M{L}^2}{12}$

$\frac{L^2}{12}=\frac{R^2}{4}$

$\Rightarrow L=\sqrt{3}R$

Hence, this is the required result.