I was also interested in the question how large the tension force on any point in a ring is due to its spinning. The logical answer is already given by Mark Eichenlaub; I just want to add some geometry and mathematical considerations.
We can try to find the correct value of the tension force in the ring at any given point by assuming that the ring of mass $m_R$ resembles a chain of $N$ finite point masses $M_N$ connected by tethers. The tethers then experience the tension force and have to produce the centripetal force which keeps the point masses on their circular path (i.e. in the ring).
Here's a picture of two such chains, once for $N = 4$ and once for $N = 8$:
If $N$ goes to infinity we will find out the tension force in a real ring. So that's our goal.
First, have a look at the diagrams. There I drew accelerations instead of forces. This seemed appropriate because the point mass and thus the centripetal force change with $N$ while the centripetal acceleration stays constant. Since $F = m \cdot a$ and since $M_N = \frac{m_R}{N}$ is constant for each $N$, this is simply proportional and will give us the tension force eventually.
$a_z$ is this constant centripetal acceleration, $a_1$ and $a_2$ are the accelerations due to the pulling of the tethers which we know result in this centripetal acceleration. $a_{t1}$ and $a_{t2}$ are the accelerations (which cancel each other out unless the ring breaks) of the pure tension forces pulling on the point masses to both sides.
We can see that with growing $N$ the accelerations along the tension $a_{t1}$ and $a_{t2}$ will grow to infinity as well (notice the increment from the left to the right diagram). But since $M_N$ will shrink to 0 in the same process, their product (the force) might be finite, so let's go on.
Some basic geometry as formulas:
$a_{t1} \cdot \tan\alpha = \frac{a_z}{2}$
$\implies a_{t1} = \frac{a_z}{2 \cdot \tan\alpha}$
$\alpha = \frac{2\pi}{2 \cdot N} = \frac{\pi}{N}$ (It is half the angle between the point masses. And concerning your original question: Here you can see again, no matter how large a finite $N$ is, the angle $\alpha$ will never be zero.)
$\implies a_{t1} = \frac{a_z}{2 \cdot \tan{\frac{\pi}{N}}}$
For very small arguments to $\tan$ (which we have here if N goes to infinity), $\tan x \approx x$, so we can simplify this to:
$a_{t1} = \frac{a_z}{2 \cdot \frac{\pi}{N}} = \frac{N}{2\pi} a_z$
If we now multiply with the mass $M_N$ we get our forces $F_T$ (tension force) and $F_z$ (centripetal force):
$F_T = a_{t1} M_N = \frac{N}{2\pi} a_z M_N = \frac{N}{2\pi} a_z \frac{m_R}{N} = \frac{1}{2\pi} a_z m_R = \frac{1}{2\pi} F_z$
So $F_T = \frac{F_z}{2\pi}$.
Be aware that the centripetal force $F_z = a_z m_R$ is given in relation to the mass of the whole ring. What does that mean?
Well, consider "splitting" the ring into just one single point mass. So $F_z$ is the force we need to to apply on an object of the mass of the ring to rotate that object around the center in a distance equal to the radius of the ring.
Example:
We have a metal ring with a radius $r = 1\mbox{m}$ and a mass $m_R = 1\mbox{kg}$. We rotate the ring with a velocity $v = 1\frac{\mbox{m}}{\mbox{s}}$.
The formula for the centripetal force to rotate a point mass of 1kg in a radius of 1m with this speed is $F_z = m\frac{v^2}{r} = 1\frac{1}{1} \mbox{kg}\frac{\mbox{m}}{\mbox{s}^2} = 1\mbox{N}$. According to our computation above the tension force $F_T$ in any point of the ring is then $\frac{1}{2\pi} \mbox{N}$.
