PlanetPhysics/Moment of Inertia of a Circular Disk

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 m be the mass, a  the radius, 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.