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 ?
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asked Jan 15, 2016 at 11:46
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3$\begingroup$ your assymtion that neighbouring aprticles only contribute with tangental force is not true. There will be a small componenet of transverse force are not all lined up linearly. $\endgroup$– Mikael FremlingJan 15, 2016 at 11:52
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$\begingroup$ @MikaelFremling's is basically already the answer. The tension in the ring will adjust itself exacly such as to provide the corresponding component towards the center. You only need to do an actual calculation if you want to determine the magnitude of that force. $\endgroup$– Norbert SchuchJan 15, 2016 at 12:22
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$\begingroup$ @MikaelFremling : You mean that the tangential forces on the particle from both the neighbouring particles will have some radial components too ? $\endgroup$– Varun ChandraJan 15, 2016 at 12:28
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$\begingroup$ The ring is curved, so yes. (Try to draw a diagram for a segment with angle $d\vartheta$.) $\endgroup$– Norbert SchuchJan 15, 2016 at 12:46
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$\begingroup$ @VarunChandra Precicely. Added an answer where i elaborate on this. $\endgroup$– Mikael FremlingJan 15, 2016 at 14:11
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2 Answers
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When there is curvature, tensile forces (in a ring, in a string, ...) will give rise to a net force as shown in the following sketch:
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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.
