Obtain An Expression For The Moment Of Inertia Of A Hollow Cylinder (i) About Its Own Axis, (ii) About An Axis Passing Through The Centre Of Mass Of The Cylinder And Perpendicular To Its Length.

Let us consider a hollow cylinder of inner radius \( r_1 \), outer radius \( r_2 \) and length \( l \). So the volume of this hollow cylinder is \( \pi(r_2^2-r_1^2)l \). The cylinder is rotating about its own AB.
Let \( M \) be the mass of this hollow cylinder, so the mass density i.e., mass per unit volume of this cylinder is \( \frac{M}{\pi(r_2^2-r_1^2)l} \).

To calculate the moment of inertia of this cylinder about the axis of rotation AB, let us consider an elementary coaxial cylindrical shell of radius \( x \) and thickness \( dx \). The volume of this cylindrical shell is \( 2\pi{x}dx{l} \). So mass of this cylindrical shell is \( \frac{2\pi{xdxl}M}{\pi(r_2^2-r_1^2)l} \) = \( \frac{2Mxdx}{r_2^2-r_1^2} \).

Moment of inertia of this cylindrical shell about the axis of rotation AB is \( \frac{2Mx^3dx}{r_2^2-r_1^2} \).
So the moment of inertia of this whole hollow cylinder is given by,

\( \displaystyle{I=\int_{r_1}^{r_2}{\frac{2Mx^3dx}{r_2^2-r_1^2}}\\=\frac{2M}{r_2^2-r_1^2}\int_{r_1}^{r_2}x^3{dx}\\=\frac{2M}{4(r_2^2-r_1^2)}{[x^4]}_{r_1}^{r_2}\\=\frac{M}{2(r_2^2-r_1^2)}(r_2^4-r_1^4)} \)

\( or,\ \displaystyle{I=\frac{1}{2}M(r_1^2+r_2^2)} \)

CASE – II :
In case of a solid cylinder, \( r_1=0 \) and \( r_2=r \)(say), which is the radius of the solid cylinder,
Moment of inertia of this solid cylinder about the axis of rotation AB is
\( \displaystyle{I'=\frac{1}{2}Mr^2} \)

(ii) Moment Of Inertia Of A Hollow Cylinder About An Axis Passing Through its centre of mass and perpendicular to its length:

Let us consider a hollow cylinder of inner radius \( r_1 \), outer radius \( r_2 \), and length \( l \). This hollow cylinder is rotating about the axis CD passes through the centre of mass of the hollow cylinder and perpendicular to its length. To calculate the moment of inertia of this hollow cylinder about the axis \( CD \), let us consider an elementary ring at a distance \( x \) from the axis CD, of thickness \( dx \), as shown in the above Fig.2. Now the volume of this elementary ring is \( \pi(r_2^2-r_1^2)dx \) and the mass is \( \frac{M}{\pi(r_2^2-r_1^1)l}\times{\pi(r_2^2-r_1^2)dx} \) = \( \frac{M}{l}dx \).

So the moment of inertia of this elementary ring about its diameter EF is

\( \frac{1}{4}\frac{M}{l}dx(r_1^2+r_2^2) \)

[ To know the derivation of moment of inertia of a circular annular disc about its own diameter Click Here ]

Since the diameter EF is parallel to the axis of rotation CD and the perpendicular distance between these two axes is \( x \), so by applying the theorem of parallel axis we can write that the moment of inertia of this elementary disc is about the axis of rotation CD is

\( \frac{M(r_1^2+r_2^2)dx}{4l}+(\frac{M}{l}dx\times{x^2}) \)

Now the moment of inertia of the whole hollow cylinder about the axis of rotation CD is given by,

\( I_1=\displaystyle{\int_{-\frac{l}{2}}^{\frac{l}{2}}} (\frac{M(r_1^2+r_2^2)dx}{4l}+\frac{M}{l}x^2{dx}) \)

or, \( I_1=\frac{M(r_1^2+r_2^2)}{4l}\displaystyle{\int_{-\frac{l}{2}}^{\frac{l}{2}}}dx\\+\frac{M}{l}\displaystyle{\int_{-\frac{l}{2}}^{\frac{l}{2}}}x^2{dx} \)

or, \( I_1=\frac{M(r_1^2+r_2^2)}{4l}[\frac{l}{2}-\frac{-l}{2}]\\+\frac{M}{3l}[{(\frac{l}{2})}^3-{(\frac{-l}{2})}^3] \)

or, \( I_1=\frac{M(r_1^2+r_2^2)}{4l}l+\frac{M}{3l}\frac{2l^3}{8} \)

or, \( I_1=\frac{M(r_1^2+r_2^2)}{4}+\frac{Ml^2}{12} \)

or, \( \displaystyle{I_1=M\left(\frac{(r_1^2+r_2^2)}{4}+\frac{l^2}{12}\right)} \)

CASE – I :
In case of a solid cylinder, \( r_1=0 \) and \( r_2=r \)(say), which is the radius of the solid cylinder,
Moment of inertia of this solid cylinder about an axis passing through the centre of mass of this solid cylinder and perpendicular to its length is
\( \displaystyle{I'_1=M\left( \frac{r^2}{4}+\frac{l^2}{12}\right)} \)