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

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

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

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

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

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

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

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