 Solution

Let $N$ be circumferential tension. A small arc subtending $d\theta$ has inward resultant $Nd\theta$ and mass $\rho Ard\theta$. Its centripetal acceleration gives $N=\rho A{\omega }^2{r}^2$. 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 isUse the unstrained radius and cross-section at this order. The approximation requires $\rho {\omega }^2{r}^2/E\ll 1$, a slender hoop and linear elastic response; the area cancels.