Moment of Inertia of a Cylinder – Formula, Derivation

To calculate the moment of inertia of a cylinder, let’s consider two situations. First, when the cylinder is hollow, and second when the cylinder is solid.

Moment of Inertia of a Hollow Cylinder about its Axis

The moment of inertia of a hollow cylinder is given by

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle I=M{{R}^{2}}}}$

Derivation

Suppose $M$ is the mass of a hollow cylinder of radius $R$.

As every element of this cylinder is at the same perpendicular distance ($r = R$) from the axis of the cylinder, therefore, the moment of inertia of the hollow cylinder about its axis is

$\displaystyle I=\int{{{{r}^{2}}}}dm={{R}^{2}}\int{}dm$

$\displaystyle I=M{{R}^{2}}$

This expression is the same as the moment of inertia of a uniform (hollow) circular ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring. This is because a hollow cylinder is made of a large number of identical circular rings placed one over the other.

Moment of Inertia of a Uniform Solid Cylinder about its Axis

The moment of inertia of a uniform solid cylinder about its axis is given by

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle I=\frac{1}{2}M{{R}^{2}}}}$

Derivation

Let $M$ be the mass, $l$ be the length, and $R$ be the radius of a solid cylinder as shown in figure. We have to calculate the moment of inertia of this solid cylinder about its axis.

Volume of the cylinder = $\pi R^2 l$ $\tag{i}$

Mass per unit volume of the cylinder,

\[
\rho = \frac{M}{\pi R^2 l} \tag{ii}
\]

Draw two cylindrical surfaces of radii $x$ and $(x + dx)$ coaxial with the given cylinder as shown in the figure. This part of the cylinder may be considered as a hollow cylinder of radius $x$ and thickness $dx$.

Area of cross section of the wall of this hollow cylinder $= 2 \pi x \, dx$.

Volume of material in this elementary hollow cylinder $= (2 \pi x \, dx) l$.

Mass of the elementary hollow cylinder

$\displaystyle m=(2\pi xdx)l\times \rho $

$\displaystyle m=(2\pi xdx)l\times \frac{M}{{\pi {{R}^{2}}l}}$

\[
m = \frac{2M}{R^2} x \, dx
\]

As the radius of this cylinder is $x$, the moment of inertia of the elementary cylinder about the given axis is

$  \displaystyle dI=m{{x}^{2}}=\frac{{2M}}{{{{R}^{2}}}}xdx\times {{x}^{2}}$

$ \displaystyle dl=\frac{{2M}}{{{{R}^{2}}}}{{x}^{3}}dx$

Therefore, the moment of inertia of the solid cylinder about the given axis is

$ \displaystyle \begin{array}{l}I=\int_{0}^{R}{{\dfrac{{2M}}{{{{R}^{2}}}}}}{{x}^{3}}dx\\I=\dfrac{{2M}}{{{{R}^{2}}}}\int_{0}^{R}{{{{x}^{3}}}}dx\\I=\dfrac{{2M}}{{{{R}^{2}}}}\left[ {\dfrac{{{{x}^{4}}}}{4}} \right]_{0}^{R}\end{array}$

\[
I = \frac{2M}{R^2} \times \frac{R^4}{4} = \frac{MR^2}{2}
\]

$\displaystyle I=\frac{1}{2}M{{R}^{2}}$

Note that this formula for $I$ does not depend upon the length of the cylinder.

Further, the expression for the moment of inertia of a solid cylinder about its axis is the same as the moment of inertia of a uniform circular disc about an axis passing through the center of the disc and perpendicular to its plane. It is due the fact that a solid cylinder is made up of a number of uniform circular discs placed one over the other.