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Circular Cross-Section

For a circular cross-section of radius $ a$ , Eq.(B.11) tells us that the squared radius of gyration about any line passing through the center of the cross-section is given by

\begin{eqnarray*} R_g^2 &=& \frac{1}{\pi a^2} \int_{-a}^a dx \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} y^2 dy \quad=\quad \frac{4}{\pi a^2} \int_0^a dx \int_0^{\sqrt{a^2-x^2}} y^2 dy\\ [10pt] &=& \frac{4}{\pi a^2} \int_0^a \left[\frac{1}{3}(a^2-x^2)^{\frac{3}{2}}\right]dx \quad=\quad \frac{4a}{3\pi} \int_0^a \left[1-\left(\frac{x}{a}\right)^2\right]^{\frac{3}{2}}dx\\ [10pt] && \qquad\hbox{[let $\sin(\theta)=x/a\,\,\Rightarrow\,\,dx=a\cos(\theta)d\theta$]}\\ [10pt] &=& \frac{4a^2}{3\pi} \int_{0}^{\frac{\pi}{2}} \cos^4(\theta)d\theta. \end{eqnarray*}

Using the elementrary trig identity $ \cos(2\theta)=2\cos^2(\theta)-1$ , we readily derive

$\displaystyle \cos^4(\theta) = \frac{1}{8}\cos(4\theta) + \frac{1}{2}\cos(2\theta) + \frac{3}{8}. $

The first two terms of this expression contribute zero to the integral from 0 to $ \pi /2$ , while the last term contributes $ 3\pi/16$ , yielding

$\displaystyle R_g^2 = \frac{4a^2}{3\pi} \frac{3\pi}{16} = \frac{a^2}{4}. $

Thus, the radius of gyration about any midline of a circular cross-section of radius $ a$ is

$\displaystyle R_g = \frac{a}{2}. $

For a circular tube in which the mass of the cross-section lies within a circular annulus having inner radius $ b$ and outer radius $ a$ , the radius of gyration is given by

$\displaystyle R_g = \frac{\sqrt{a^2-b^2}}{2}. \protect$ (B.12)

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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4
Copyright © 2024-06-28 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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