# Tension in a rotating ring [duplicate]

To calculcate tension in a rotating ring (about its own axis) the usual approach is to consider an element of angle $$d\theta$$ and consider the tension forces acting on it but isn't an element of circle taken to be a straight line? But in that case the tension forces would cancel out and the ring wouln't be able to rotate. So are we taking elements such that forces do not cancel? That seems to be a biased approach to me.

I understand your uneasiness, but all is well if you think carefully about the powers of $$\theta$$ involved...
If the arc subtends angle $$\theta$$, then the component of tension force, T, acting on the arc, in a direction antiparallel to the radius meeting the midpoint of the arc is $$2T \sin {(\theta/2)} = 2T \left(\frac{\theta}{2}-\frac{\theta ^3}{48} + ...\right)$$
Now the centripetal force needed by the arc of the ring, assumed to be thin and to have a mass $$\mu$$ per unit length, is $$(\mu r \theta) r\omega ^2$$.
So we have $$2T \left(\frac{\theta}{2}-\frac{\theta ^3}{48} + ...\right) = (\mu r \theta) r\omega ^2.$$
In the limit as $$\theta$$ goes to zero, the term in $$\theta^3$$ disappears and the equation simplifies nicely.
If you're still not happy you might prefer a virtual work approach: mathematically trivial, but physically less transparent. Suppose the ring's radius were to expand by $$\Delta r$$ due to the rotation. Then the circumference would expand by $$2\pi \Delta r$$, so work $$T 2\pi \Delta r$$ would be done against the tension. This work is equal to the centripetal force times the increase in radius, that is $$(\mu 2\pi r) (r \omega ^2) \Delta r$$. Equating the two gives you the same result as by considering forces on the arc.