Rotating Ring Tension

$\angle AOB = \delta\theta$ $\angle AOO' = \frac{\delta\theta}{2}$

Consider an arc element of the ring as shown in the figure:

Its mass is:

$\delta m = \frac{M}{2\pi R}R\delta\theta$

Let the tension in the ring be $T$ . The tension is denoted by the red arrows. Applying Newton's second law along the vertical gives:

$T\sin{\frac{\delta\theta}{2}} + T\sin{\frac{\delta\theta}{2}} = (\delta m) \omega^2 R$ $2T\sin{\frac{\delta\theta}{2}} = \left(\frac{M}{2\pi R}R\delta\theta\right)\omega^2 R$

Since $\delta\theta$ is small, the approximation $\sin{\delta\theta} = \delta\theta$ is made. This leads to the answer:

$\boxed{T = \frac{M\omega^2 R}{2\pi}}$

Let tension in the ring be $F_{T}$ , unit tangent vector at a point on the ring $\hat{T}$ , and unit normal vector $\hat{N}$ : $\begin{aligned} dm\,\omega^{2}R \cdot \hat{N} &= d(F_{T}\cdot \hat{T}) \\ \frac{M}{2\pi R}ds\,\omega^{2}R \cdot \hat{N}&= F_{T}\cdot \frac{d\hat{T}}{ds}ds \\ \frac{M}{2\pi}\omega^{2}\cdot \hat{N} &=F_{T}\cdot \frac{\hat{N}}{R} \\ \implies F_{T} &= \frac{M\omega^{2}R}{2\pi},\end{aligned}$ where $\frac{d\hat{T}}{ds}$ is a curvature vector (curvature of a circle is just the reciprocal of its radius) and we know that tension is uniform throughout the ring. This makes answer $20,000\pi$ .