Moment of Inertia of a Ring

Moment of Inertia of a Ring:

1) About an axis perpendicular to the plane and passing through CG (0 to 2πR):

Consider a circular ring of mass M and radius r. Let us take an axis AB perpendicular to the plane of the ring and passing through the CG (centre of the ring) about which M.I is to be determined.

Let us divide the ring in to large no. of elementary pieces along the circumference and take any one piece length dl.

Mass of the elementary piece=$\begin{align*}(\frac{M}{2\pi R})dl\end{align*}$

M.I of this elementary piece is dI=$\begin{align*}mass\times (distance)^{2}\end{align*}$ $$\begin{align*}=\frac{MR^{2}}{2\pi R}dl\end{align*}$$

Since such elementary pieces are varying from dl= 0 to dl= 2πR, total M.I of the ring about AB- axis is given by,

$$\begin{align*}I=\int dl\end{align*}$$ $$\begin{align*}=\frac{MR^{2}}{2\pi R}\int_{0}^{2\pi R}dl\end{align*}$$ $$\begin{align*}I=MR^{2}\end{align*}$$

2) About an axis along the plane and passing through CG (i.e. about a diameter):

To determine the M.I of a ring about one of it’s diameter we have to apply the principle of perpendicular axis. For this let us consider two diameters XOX’ and YOY’ perpendicular to each other, about which M.I of the ring is I_d.

Consider an imaginary axis AOB perpendicular to the plane of the ring and passing through its centre then by applying, the theorem of perpendicular axis, we can write

$$\begin{align*}I_{AB}=I_{XOX'}+I_{YOY'}\end{align*}$$ $$\begin{align*}I_{d}=\frac{MR^{2}}{2}\end{align*}$$

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