The fastest rotational speed for a disc shaped object was achieved in 1985 . The flywheel was spun up to a tip speed of 1405 m/s, at which point it failed, due to the stresses in the materials involved.

We consider a tower of flywheels, rigidly mounted on top of each other in such a way that the rotational motion of the lower ones is automatically transferred to the ones above, but each flywheel has its own independent source of additional rotational energy, that can give it an extra "kick". 

We assume that these flywheels are put in motion in sequence, from bottom to top. Each flywheel is given enough energy in order to put it in rotational motion (same sense of rotation, let's say counterclockwise) with the same angular velocity, on top of what it "inherited"  from the lower ones.

For an external observer, assuming a large number of flywheels, is it possible (in principle only because from an engineering perspective it is clearly not feasible) that points on the rim of the topmost flywheel could reach relativistic tangential velocities?

And yes, I must confess, I'm also thinking about Tipler cylinders,  but "below"  and "above" would translate to "inner" and "outer", and this is not part of the question. It would become a problem of rotational dynamics.

Edit. Added question. Could these systems of connected flywheels store energy more efficiently,?

  • $\begingroup$ To me your question doesn't look as a kinematical one, since you're talking about "rotational energy". $\endgroup$
    – Elio Fabri
    Apr 6, 2019 at 12:46
  • $\begingroup$ We are not assuming an engineering perspective, so we are not worried about structural failures, right? Does this mean we are asking if we can imagine a disk spinning at relativistic speeds? Yes, I think so. $\endgroup$ Apr 6, 2019 at 15:38
  • $\begingroup$ The question (in principle) is whether we can design a system that would lead to a disc spinning at relativistic speeds, when we only have access to limited sources of energy , not simply imagine a fast spinning disc. $\endgroup$ Apr 6, 2019 at 15:53

1 Answer 1


(Thank you for clarifying.) We imagine the topmost disk will go from rest to spinning at relativistic speeds. The work done by the net force acting on the disk will equal the change in kinetic energy:

$$W = \Delta K = \frac{1}{2}I \omega ^2.$$

For a thin disk of mass $m$ and radius $r$, spinning with tangential speed $v$,

$$W = \Delta K = \frac{1}{4}mv^2.$$

A U.S. quarter has a mass of about 0.006 kg. If the tangential speed $v$ is 0.1$c$, $2.99 \times 10^7$ m/s, the work done by the applied force is

$$W = \Delta K = \frac{1}{4}mv^2 = \frac{1}{4}(0.006 \textrm{ kg})(2.99 \times 10^7 \textrm{ m/s})^2 = 1.34 \times 10^{12} \textrm{ J}.$$

This amount of energy is comparable to a small atomic bomb, so we are not entirely limited by the sheer quantity of energy needed. But note that this is the energy transferred to the top-most quarter by work, and doesn't include the energy required for all the other quarters in the stack plus the other parts of the mechanism. And if a mere 0.000001% of the energy goes into heating the quarter--made mostly of copper-- it will very quickly melt.

And beyond all that, there is the question of how we design a mechanism that can apply a force to a quarter that will transfer that quantity of energy via work. There's nothing special about a stack of quarters in this regard because the topmost quarter is not 'aware' in any physics sense of any of the other quarters in the stack except for the one it sits upon. All that matters for the topmost quarter the net force acting on it.

  • $\begingroup$ The discs are not independent. The semi rigid connection between them allows a disc to rotate at least as fast as the disc below (and maybe faster). Each disc receives an additional rotational energy E, over the energy "inherited " from the lower discs through the rigid connection. If the bottom disc rotates with velocity v_0 the one above will rotate with velocity v_1 given by 1/4* m* v_1^2 = 1/4*mv_0^2 + E. The second disc will rotate with velocity v_2 given by 1/4 m* v_2^2 = 1/4*m*v_1^2 + E, and so on. The question is whether this scheme has an advantage.. $\endgroup$ Apr 7, 2019 at 5:45
  • $\begingroup$ Thank you @RodneyDunning for your answer. As I said in my comment above, the question is whether this scheme has any advantage ( in terms of engineerical feasibility), as compared to having just one disc and transfering to it the total rotational energy necessary. Probably not, but I'm not sure. And the relations above should be modified. The mass m in the relations above is not the same for every disc, because the energy received by one disc is transferred (through the rigid connection) to all the discs above it. $\endgroup$ Apr 7, 2019 at 5:57
  • $\begingroup$ Let's assume we have N discs of mass m, and each receives rotational energy E. Then for the bottom disc we have E = 1/4*Nmv_0^2, from where we find v_0. For the disc above we have E = 1/4*(N-1)*m*(v_1^2-v_0^2) , from where we find v_1. For the next disc we have E = 1/4*(N-2)*m*(v_2^2-v_1^2), from where we find v_2, and so on. You get the pucture. Basically, I'm trying to avoid transferring a lot of energy to just one disc in a short period of time , but increase the rotation speed gradually, incrementally, in order to avoid structural problems, among many others. $\endgroup$ Apr 7, 2019 at 6:35
  • $\begingroup$ Let's forget Bout relativistic speeds. Maybe these systems of connected flywheels could store e ergy more e $\endgroup$ Apr 7, 2019 at 6:49
  • $\begingroup$ Maybe these systems of connected flywheels could store energy efficiently, and in a more stable way. $\endgroup$ Apr 7, 2019 at 6:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.