PlanetPhysics/Rotational Inertia of a Solid Sphere

The [[../MomentOfInertia/|Rotational Inertia]] or moment of inertia of a solid sphere rotating about a diameter is

$I=\frac{2}{5}M{R}^2$

This can be shown in many different ways, but here we have chosen integration in spherical coordinates to give the reader practice in this coordinate [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|system]]. If we choose an axis such as the z axis, then we just have one moment of inertia given by

$I=\int z^{2}dm$

It is important to understand this distinction and the more general case about an arbitrary axis is handled by the [[../InertiaTensor/|inertia tensor]]. Since we have chosen z as our axis of rotation, then z in [[../Formula/|formula]] (2) is the distance from dm (dV) to the z axis. In the figure below this is shown as the purple line.

\begin{figure} \includegraphics[scale=.6]{InertiaSphere.eps} \caption{Rotational inertia of a solid sphere rotating about a diameter, z} \end{figure}

Then from spherical coordiantes we obtain z through

$z=r\sin \theta$

leaving us with the integral

$I=\int r^2{\sin }^2\theta dm$

Assuming a constant density throughout the sphere converts the infinitesimal [[../Mass/|mass]] dm to

$dm=\rho dV$

and in spherical coordinates the infinitesmal [[../Volume/|volume]] dV is given by

$dV=r^2\sin \theta d\theta d\varphi$

giving the final [[../Bijective/|function]] to integrate as

$I=\rho \underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\pi }{\int }}\underset{0}{\overset{R}{\int }}{r}^4{\sin }^3\theta drd\theta d\varphi$

Integrating the r term is simply

$I=\frac{R^5\rho }{5}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\pi }{\int }}{\sin }^3\theta d\theta d\varphi$

The $\theta$ term is a little more involved and we substitude in the trigonometric [[../Bijective/|relation]] ${\sin }^2\theta =(1-{\cos }^2\theta )$

and the integrand for the $\theta$ term becomes

$\underset{0}{\overset{\pi }{\int }}(\sin \theta -{\cos }^2\theta \sin \theta d\theta )$

Using the technique of u substitution to solve this

$u=\cos \theta$ $du=-\sin \theta d\theta$ $d\theta =\frac{-du}{\sin \theta }$

so

$\underset{0}{\overset{\pi }{\int }}\sin \theta d\theta +u^{2}du$

completing the integration yields

$I=\frac{4R^5\rho }{15}\underset{0}{\overset{2\pi }{\int }}d\varphi$

Finally, the $\varphi$ term integrates to $2\pi$ so

$I=\frac{8\pi R^5\rho }{15}$

Using the simple formula for density

$\rho =M/V$

and we know that the volume of a sphere is

$V=\frac{4}{3}\pi R^3$

plugging these into (3) gives us our original equation in (1)

$I=\frac{2}{5}M{R}^2$

References

[1] Halliday, D., Resnick, R., Walker, J.: "[[../CosmologicalConstant/|fundamentals of physics".\,]] 5th Edition, John Wiley \& Sons, New York, 1997.