Moment of inertia of a solid circular cylinder about its own axis:

Let us consider a solid circular cylinder of length l and radius r be rotating about its own axis AB. So the volume of the cylinder is \( \pi{r^2}l \). Let M be the mass of the cylinder, so the mass density i.e., mass per unit volume of the cylinder is \( \frac{M}{\pi{r^2}l} \).

To calculate the moment of inertia of the cylinder about the axis AB, let us consider a coaxial cylindrical shell if radius \( x \) and thickness \( dx \). The volume of this cylindrical shell is \( 2\pi{x}dx\ l \), so the mass of this cylindrical shell is given by
\( \frac{M}{\pi{r^2}l}\times{2\pi{x}dx\ l} \) = \( \frac{2M}{r^2}xdx \).

Moment of inertia of this cylindrical shell about the axis AB is given by, \( \frac{2M}{r^2}xdx\times{x^2} \) = \( \frac{2M}{r^2}{x^3}dx \).

So the moment of inertia of the whole cylinder about the axis AB is given by,

\( \displaystyle{I=\frac{2M}{r^2}\int_{0}^{r}x^3{dx}\\=\frac{2M}{r^2}\frac{1}{4}{[x^4]}_0^r\\=\frac{M}{2r^2}r^4} \)

\( or,\ \displaystyle{I=\frac{1}{2}Mr^2} \)

Moment of inertia of a solid circular cylinder about an axis passing through its centre of mass and perpendicular to its length:

Let us consider a solid circular cylinder of radius r and length l be rotating about an axis CD passing through its centre of mass and perpendicular to its length. Let M be the total mass of the cylinder, then the mass per unit volume of the cylinder is \( \frac{M}{2\pi{r}l} \), where \( 2\pi{rl} \) is the volume of this cylinder.

To calculate the moment of inertia of this cylinder about the axis CD, let us consider an elementary disc of thickness \( dx \) at a distance \( x \) from the axis of rotation CD. The volume of this elementary disc is \( 2\pi{r\ dx} \), so the total mass of this elementary disc is \( 2\pi{r\ dx}\times{\frac{M}{2\pi{r}l}} \) = \( \frac{M}{l}dx \). EF is the diameter of this elementary disc, so we can write thet, the moment of inertia of this elementary disc about this diameter is given by,

\( I_0=\frac{1}{4}(\frac{M}{l}dx)r^2 \)

[To know the derivation of the moment of inertia of a circular plane disc about its own diameter (click here) ]

Since the axis EF is parallel to the axis of rotation CD, and the perpendicular distance between these two axes is \( x \). So by using the theorem of parallel axes we can write that the moment of inertia of the circular disc about the axis of rotation CD is

\( \frac{1}{4}(\frac{M}{l}dx)r^2+(\frac{M}{l}dx)\cdot{x} \)

So the moment of inertia of the whole cylinder about the axis of rotation CD is given by

\( I_1=\displaystyle{\int_{-\frac{l}{2}}^{\frac{l}{2}}}(\frac{Mr^2}{4l}dx+\frac{Mx^2}{l}dx) \)

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

= \( \frac{Mr^2}{4l}(\frac{l}{2}+\frac{l}{2})\\+\frac{M}{3l}( \frac{l^3}{8}+\frac{l^3}{8} ) \)

= \( \frac{Mr^2}{4l}\cdot{l}+\frac{M}{3l}\cdot\frac{2l^3}{8} \)

= \( \frac{Mr^2}{4}+\frac{Ml^2}{12} \)

or, \( \displaystyle{I_1=M[\frac{r^2}{4}+\frac{l^2}{12}]} \)

In case of thin rod, \( r\simeq{0} \)
CASE – I :
So the moment of inertia of a thin rod of length l and mass M about an axis passing through its centre of mass and perpendicular to its length is \( I_1=\frac{1}{12}Ml^2 \)
CASE – II :
If this thin rod rotates about an axis passing through one end and perpendicular to its length. By using theorem of parallel axes we can write that the moment of inertia of this thin rod about that axis is
\( I’_1=\frac{Ml^2}{12}+M\cdot{(\frac{l}{2})}^2\\=\frac{Ml^2}{12}+\frac{Ml^2}{4} \)
\( or,\ I’_1=\frac{Ml^2}{3} \)