Why does stress decrease as you move away from the rotational center?

A while ago Real Engineering posted a video "Can we create artificial gravity". During the video he states that the designers of Elysium took note of the centrifugal and centripetal forces acting on Elysium and compensated by having thicker beams as you get closer to the center. Closer?

Centrifugal force increases as you move further away from the rotating center. So why does the stress decrease as you move further away from the rotating center?

Would like to hear thoughts and answers on this. Or a link to a similar question. Wouldn't mind some good old mathematics thrown in the bunch either.

1 Answer

It's true that the centripetal acceleration increases as you move away from a rotating center, by the equation $a = v^2/r = r\omega^2$. And so is the centrifugal force (which is the outward force 'felt' by the object at a point on a rotating frame) and centripetal force (which is the inward force needed to keep the object at that point maintain a circular path) also increase as you move away from the center.

But the tensional force (the stress experienced by the material) at the inner parts of the Elysium is needed to provide for all the centripetal force at the outer parts, otherwise, what provides it? The tensional force at an inner part is actually equal to the sum of the centripetal forces of the outer parts.

Lets say we have parts A, B, C, D, E, of the spinning ship, with A on the innermost, and E on the outermost. part E is rotating, and so needs a force $F_E$ to keep it in circular path. This force is provided by the tension force at the D part and inward (C, B, A). part D is also rotating and needs a centripetal force $F_D$ to keep it on a circular path, which are provided by the tension force at point C and inward. So now, part C provides a tensional force for both D and E (Why does part C provide force for part E when D is already providing it? Consider it like this: D is pulling E inward and C is pulling D inward, so C must also carry the weight of E), while part D provides tensional force for part E only. As you go inward, there in a need to provide for more tensional force to provide for the outer parts circular motion.

The equation for the tensional force provided at r = a is: $$T_a = \sum mr\omega^2 = \omega^2\int_a^{r_{max}}r\lambda(r)dr$$ where $\lambda(r)dr$ is the mass of the strip dr as a function of the distance from the center.