MomentOfInertiaOfACircularDisk 2006-07-11 00:08:58 2006-07-11 00:08:58 Example moment of inertia % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) \usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Here we look at two cases for the moment of inertia of a homogeneous circular disk (a) about its geometrical axis, (b) about one of the elements of its lateral surface. Let \emph{m} be the mass, \emph{a} the radius, \emph{l} the thickness, and $\tau$ the density of the disk. Then choosing a circular ring for the element of mass we have $$dm = \tau \cdot l \cdot 2\pi r \cdot dr$$ where $r$ is the radius of the ring and $dr$ its thickness. \begin{figure} \includegraphics[scale=.6]{Fig87.eps} \vspace{20 pt} \end{figure} Therfore the moment of inertia about the axis of the disk is $$I = 2\pi l \tau \int_0^a r^3 dr$$ $$I = \frac{\tau l \pi a^4}{2}$$ $$I = \frac{ma^2}{2}$$ The moment of inertia about the element is obtained easily by the help of theorem II. Thus $$I' = I + ma^2$$ $$I' = \frac{3}{2}ma^2$$ It will be noticed that the thickness of the disk does not enter into the expressions for $I$ and $I'$ except through the mass of the disk. Therefore these expressions hold good whether the disk is thick enough to be called a cylinder or thin enough to be called a circular lamina.