Moment of Inertia of a Spherical shell (hollow sphere)

Moment of Inertia of a Spherical shell (hollow sphere):

1) About a diameter:

Let us consider a spherical shell of mass M and radius R. Let us take a diameter XOX’ of the spherical shell about which moment of inertia of the shell is to be determined.

Now, let us divide the spherical shell into large no. of elementary rings by a plane, perpendicular to the diameter XOX’. Let us take any one of these elementary rings having radius ‘y’ and thickness ‘dx’. If dθ be the angle made by the thickness dx of the ring at the centre of the shell, then we can write

From figure,

\begin{align*}dx=Rd\theta \;and\; y=Rsin\theta\end{align*}

Mass per unit surface area of the shell= \(\begin{align*}\frac{M}{4\pi R^{2}}\end{align*}\)

Surface area of the elementary ring=\(\begin{align*}2\pi ydx\end{align*}\)

Mass of the ring, m=\(\begin{align*}\frac{M}{4\pi R^{2}}\times 2\pi ydx\end{align*}\)

\begin{align*}m=\frac{msin\theta d\theta }{2}\end{align*}

Now, moment of inertia of the elementary ring is given by,

\begin{align*}
dI=mass\times (radius)^{2}(\because for a ring I=MR^{2})\end{align*}\begin{align*}=(\frac{Msin\theta d\theta }{2})y^{2}\end{align*}\begin{align*}=\frac{MR^{2}sin^{3}\theta d\theta }{2}\end{align*}

Since, these rings are varying from θ= 0 at X to θ= π at X’, the moment of inertia of the shell is given by,

\begin{align*}I=\int dI\end{align*}\begin{align*}=\int_{0}^{\pi}\frac{MR^{2}sin^{3}\theta d\theta }{2}\end{align*}\begin{align*}=\frac{MR^{2}}{2}[-\int_{0}^{\pi}dcos\theta +\int_{0}^{\pi}cos^{2}\theta dcos\theta ]\end{align*}\begin{align*}I=\frac{MR^{2}}{2}\times \frac{4}{3}\end{align*}\begin{align*}I=\frac{2MR^{2}}{3}\end{align*}

2) About a tangent:

Consider TXT’ is a tangent to the spherical shell at point X and parallel to the diameter of the shell AOA’. We need to find the M.I of the spherical shell about the tangent TXT’. For this we make the use of parallel axis theorem. i.e. M.I of the spherical shell is given by the algebraic sum of M.I of the shell about AOA’ and the product of mass of the shell and square of the distance between the two axes i.e.

\begin{align*}I_{TXT'}=I_{AOA'}+MR^{2}\end{align*}\begin{align*}I_{T}=I_{d}+MR^{2}(where, I_{T}=I_{TXT'})\end{align*}\begin{align*}I_{T}=\frac{5MR^{2}}{3}\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