Suppose a ring is rotating in space with an angular velocity $\omega .$ Then each element of the ring is having an acceleration of $m\omega^2 r$ ($r$ is the radius of the ring) but what force is causing this acceleration in each element? The acceleration is inward so the force should also be inward but there is nothing to provide this force . The neighbouring elements can only provide a tangential force on a particular element, this force will not have any component along the radius so this can not be the reason for the centripetal acceleration. So how are the elements accelerating and moving with a constant angular velocity ?
Expanding on the comment, here is an answer.
Getting all the microcosmic force correct is a bit outside of the scope of this answer, so let me answer the question with a toy model instead.
Assume $N$ particles on a ring, attached to each other with perfect springs. The force of the springs is then $F=kx$ where $x$ is the displacement from equilibrium. We assume we start in equilibrium, by the way.
The angle in a corner of or circle will be $\theta=\pi\cdot(N-2)/N=\pi-2\pi/N$. This means that the force from the springs to the left and right of a particle will not cancel completely.
Assume now a small displacement of a particle (for whatever reason) by a distance $x$ along the direction of one of the springs. The compression of the other spring will then be almost $x$ (not precisely though since the two springs are not parallel). The spring force response is therefore $F=kx$. This force will have a and a parallel component $$F_p=kx\sin(\theta)=kx\cos(2\pi/N)\approx kx $$ but also a transverse component $$F_t=kx\cos(\theta)=kx\sin(2\pi/N)\approx 2\pi kx/N $$
It is this transverse component that gives the centripetal acceleration needed to keep the ring spinning.