Moment of Inertia of Solid sphere

Moment of Inertia of Solid sphere:

1) About a diameter:

Consider a solid sphere of mass ‘M’ and radius R. Let us take a diameter AOA’ of the sphere about which M.I is to be found.

Let us divide the sphere into large no. of elementary discs by a plane perpendicular to the diameter AOA’. In this case plane of each disc is perpendicular to the axis of rotation. Consider one of such elementary disc of radius ‘y’ and thickness ‘dx’ at a distance x from the centre of the sphere.

Here, Mass per unit volume of the sphere=\(\begin{align*}\frac{M}{4/3\pi R^{3}}\end{align*}\)

Volume of the elementary disc=\(\begin{align*}\pi y^{2}dx\end{align*}\)

Mass of the elementary disc, m=\(\begin{align*}\frac{M}{4/3\pi R^{3}}\times \pi y^{2}dx\end{align*}\)\begin{align*}m=\frac{3M}{4R^{3}}y^{2}dx\end{align*}

Now, M.I of the elementary disc about AOA’ axis is given by,

\begin{align*}dI=\frac{mass\times (radius)^{2}}{2}(\because for a disc I=\frac{MR^{2}}{2})\end{align*}\begin{align*}dI=\frac{3M}{8R^{3}}y^{4}dx\end{align*}

Since, these disc are varying from x= -R at A to x= R at A, total M.I of the sphere about AOA’ is given by,

\begin{align*}I=\int dI\end{align*}\begin{align*}=\frac{3M}{8R^{3}}\int_{-R}^{R}y^{4}dx--------(1)\end{align*}

From figure,

\begin{align*}R^{2}=x^{2}+y^{2}\end{align*}\begin{align*}y^{4}=(R^{2}-x^{2})^{2}=R^{4}-2R^{2}x^{2}+x^{4}\end{align*}

Substituting y⁴ in equation (1) , we get,

\begin{align*}I=\frac{3M}{8R^{3}}\int_{-R}^{R}(R^{4}-2R^{2}x^{2}+x^{4})dx\end{align*}\begin{align*}=\frac{3M}{8R^{3}}[R^{4}\int_{-R}^{R}dx-2R^{2}\int_{-R}^{R}x^{2}dx+\int_{-R}^{R}x^{4}dx]\end{align*}\begin{align*}=\frac{3M}{8R^{3}}(2R^{5}-\frac{4R^{5}}{3}+\frac{2R^{5}}{5})\end{align*}\begin{align*}=\frac{3M}{8R^{3}}\times \frac{16R^{5}}{15}\end{align*}\begin{align*}I=\frac{2MR^{2}}{5}\end{align*}

2) About a tangent:

Consider TXT’ is a tangent to the sphere at X, about which M.I of the sphere is to be determined. Consider a diameter AOA’ of the sphere, which is parallel to the given tangent TXT’. Then M.I of the sphere about the tangent can be determined by using the parallel axis theorem.

i.e.\begin{align*}I_{TXT'}=I_{d}+MR^{2}\end{align*}

Where, I_d= M.I of the sphere about its diameter AOA’

\begin{align*}I_{T}=\frac{2MR^{2}}{5}+MR^{2}\end{align*}\begin{align*}I_{T}=\frac{7MR^{2}}{5}\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