= Solution
Let $N$ be circumferential <tension>. A small arc subtending $d\theta$ has inward resultant $N\,d\theta$ and <mass> $\rho A r\,d\theta$. Its <centripetal acceleration> gives $N=\rho A\omega^2r^2$. Circumferential <stress> is $N/A$ and the <strain> is $\Delta(2\pi r)/(2\pi r)=\Delta r/r$. Thus the <centrifugal strain of a thin elastic hoop> is
$$
\boxed{\frac{\Delta r}{r}\simeq\frac{\rho\omega^2r^2}{E}.}
$$
Use the unstrained radius and cross-section at this order. The approximation requires $\rho\omega^2r^2/E\ll1$, a slender hoop and linear elastic response; the area cancels.
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