Moment of Inertia of a Solid Cylinder

Moment of Inertia of a Solid cylinder:

1) About its own axis of cylindrical symmetry:

A solid cylinder is just a thick circular disc or a no. of discs( all of the same radius) piled up one over the other, so that its axis of cylindrical symmetry is the same as the axis passing through the centre of the disc and perpendicular to its plane.

Therefore, M.I of the solid cylinder about its principal axis is equal to the M.I of the thick disc of the same mass and radius, about the axis through its centre and perpendicular to its plane.

\begin{align*}\frac{MR^{2}}{2}\end{align*}

Where M is the mass of the solid cylinder and R is its radius.

2) About an axis perpendicular to the axis of cylinder symmetry and passing through its centre (CG) [-l/2 to l/2]:

Consider a solid cylinder of mass M and radius R. consider an axis XY perpendicular to the principal axis of the cylinder and passing through the CG, about which M.I of the cylinder is to be determined.

Let us divide the cylinder into large no. of elementary discs along the principal axis. Then let us take any one of the elementary disc having thickness ‘dx’ and radius R at a distance x from the axis of rotation.

Here, mass per unit length of the cylinder= \(\begin{align*}\frac{M}{l}\end{align*}\)

Where, l is the length of the cylinder.

Mass of the elementary disc, m= \(\begin{align*}\frac{Mdx}{l}\end{align*}\)

But we know that the M.I of the disc about the diameter is given by,

\begin{align*}dI_{d}=\frac{MR^{2}}{4}\end{align*}

So, M.I of the disc about the axis of rotation is given by the parallel axis theorem. i.e.

\begin{align*}
dI=I_{c}+m\times (distance)^{2}\end{align*}\begin{align*}=\frac{Mdx}{4l}R^{2}+\frac{Mx^{2}}{l}dx\end{align*}

Since, these rings are varying from x= -l/2 to x= l/2 then the total M.I of the cylinder is given by,

\begin{align*}I=\int dI\end{align*}\begin{align*}=\int_{-l/2}^{l/2}(\frac{Mdx}{l})\frac{R^{2}}{4}+\int_{-l/2}^{l/2}\frac{Mx^{2}}{l}dx\end{align*}\begin{align*}=\frac{MR^{2}}{4}+\frac{Ml^{2}}{12}\end{align*}\begin{align*}\therefore I=\frac{MR^{2}}{4}+\frac{Ml^{2}}{12}\end{align*}

References:

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014