Moment of Inertia of a Hollow cylinder

Moment of Inertia of a Hollow cylinder:

1) About the principal axis:

A hollow cylinder may be considered to be a thick annular disc or a combination of thin annular discs each of external and internal radii, placed adjacent to each other, the axis of the cylinder (i.e. its axis of cylinder symmetry) being the same as the axis passing through the centre of the thick annular disc (or the combination of thin annular discs) and perpendicular to its plane.

The moment of inertia of the hollow cylinder about its principal axis is the same as that of a thick annular disc of the same mass and external and internal radii R and r respectively about the axis passing through its center and perpendicular to its plane. i.e.

\begin{align*}I=\frac{M}{2}\times (R^{2}+r^{2})\end{align*}

Where M is the mass of the hollow cylinder.

2) About an axis passing through its center and perpendicular to its own axis:

Consider a hollow cylinder of mass ‘M’ external radius ‘R’ and internal radius ‘r’. let us take an axis YOY’ perpendicular to the principal axis of the cylinder and passing through the CG of the cylinder about which the M.I of the cylinder is to be found.

Let us divide the hollow cylinder into large no. of elementary annular discs along the principal axis. Consider any one of such discs having thickness ‘dx’ and diameter PQ which is at a distance ‘x’ from the axis of rotation.

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

Where, the l= length of the cylinder.

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

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

\begin{align*}I_{d}=\frac{M}{4}(R^{2}+r^{2})\end{align*}

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

\begin{align*}
dI=I_{d}+m(distance)^{2}\end{align*}\begin{align*}=(\frac{Mdx}{4l})(\frac{R^{2}+r^{2}}{4})+(\frac{Mdx}{l})x^{2}\end{align*}
hence, M.I of the entire hollow cylinder about YOY’ axis is given by, integrating dI with limit from x= -l/2 to x= l/2 i.e.

\begin{align*}I=\int dI\end{align*}\begin{align*}=\int_{-l/2}^{l/2}[(\frac{Mdx}{l})(\frac{R^{2}+r^{2}}{4})+(\frac{Mdx}{l})x^{2}]\end{align*}\begin{align*}=M(\frac{R^{2}+r^{2}}{4}+\frac{l^{2}}{12})\end{align*}\begin{align*}\therefore I=M(\frac{R^{2}+r^{2}}{4}+\frac{l^{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