Circular Cross-Section

For a circular cross-section of radius $a$, Eq.$\,$(E.7) tells us that the squared radius of gyration about any line passing through the center of the cross-section is given by

$$\begin{eqnarray*} r_0^2 &=& \frac{1}{\pi a^2} \int_{-a}^a dx \int_{-\sqrt{a^2-x^... ...frac{4a^2}{3\pi} \int_{0}^{\frac{\pi}{2}} \cos^4(\theta)d\theta. \end{eqnarray*}$$

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

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

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

$$r_0^2 = \frac{4a^2}{3\pi} \frac{3\pi}{16} = \frac{a^2}{4}.$$

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

$$\zbox {r_0 = \frac{a}{2}.}$$

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

$$\zbox {r_0 = \frac{\sqrt{a^2-b^2}}{2}.} \protect$$ (E.8)