Talk:PlanetPhysics/Rotational Inertia of a Solid Sphere

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: rotational inertia of a solid sphere %%% Primary Category Code: 45.40.-f %%% Filename: RotationalInertiaOfASolidSphere.tex %%% Version: 6 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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The \htmladdnormallink{Rotational Inertia}{http://planetphysics.us/encyclopedia/MomentOfInertia.html} or moment of inertia of a solid sphere rotating about a diameter is

\begin{equation} I = \frac{2}{5}M R^2 \end{equation}

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 \htmladdnormallink{system}{http://planetphysics.us/encyclopedia/SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence.html}. If we choose an axis such as the z axis, then we just have one moment of inertia given by

\begin{equation} I = \int {z^2 dm} \end{equation}

It is important to understand this distinction and the more general case about an arbitrary axis is handled by the \htmladdnormallink{inertia tensor}{http://planetphysics.us/encyclopedia/InertiaTensor.html}. Since we have chosen z as our axis of rotation, then z in \htmladdnormallink{formula}{http://planetphysics.us/encyclopedia/Formula.html} (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 \htmladdnormallink{mass}{http://planetphysics.us/encyclopedia/Mass.html} dm to

$$ dm = \rho dV $$

and in spherical coordinates the infinitesmal \htmladdnormallink{volume}{http://planetphysics.us/encyclopedia/Volume.html} dV is given by

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

giving the final \htmladdnormallink{function}{http://planetphysics.us/encyclopedia/Bijective.html} to integrate as

$$ I = \rho \int_{0}^{2\pi} \int_{0}^{\pi} \int_0^R { r^4 \sin^3 \theta \, \, dr d\theta d\phi} $$

Integrating the r term is simply

$$ I = \frac{R^5 \rho}{5} \int_{0}^{2\pi} \int_{0}^{\pi} { \sin^3 \theta \, \, d\theta d\phi} $$

The $\theta$ term is a little more involved and we substitude in the trigonometric \htmladdnormallink{relation}{http://planetphysics.us/encyclopedia/Bijective.html} $$ \sin^2 \theta = (1 - \cos^2 \theta) $$

and the integrand for the $\theta$ term becomes

$$ \int_{0}^{\pi} { (\sin \theta - \cos^2 \theta \sin \theta \, \, d\theta )} $$

Using the technique of u substitution to solve this

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

so

$$ \int_{0}^{\pi} { \sin \theta \, d\theta + u^2 \, du } $$

completing the integration yields

$$ I = \frac{4 R^5 \rho}{15} \int_{0}^{2\pi} { d\phi} $$

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

\begin{equation} I = \frac{8 \pi R^5 \rho}{15} \end{equation}

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)

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

\subsection{References}

[1] Halliday, D., Resnick, R., Walker, J.: "\htmladdnormallink{fundamentals of physics".\,}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html} 5th Edition, John Wiley \& Sons, New York, 1997.

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