PlanetPhysics/Moment of Inertia of a Circular Disk

Here we look at two cases for the [[../MomentOfInertia/|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/|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

$I=2\pi l\tau \underset{0}{\overset{a}{\int }}{r}^{3}dr$ $I=\frac{\tau l\pi a^4}{2}$ $I=\frac{m{a}^2}{2}$

The moment of inertia about the element is obtained easily by the help of [[../Formula/|theorem]] II. Thus

$I^{\prime }=I+m{a}^2$ $I^{\prime }=\frac{3}{2}m{a}^2$

It will be noticed that the thickness of the disk does not enter into the expressions for $I$ and $I^{\prime }$ 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.

References

This article is a derivative of the public [[../Bijective/|domain]] book, "Analytical [[../Mechanics/|mechanics]]" by Haroutune M. Dadourian, 1913. Made available by the internet archive