2
$\begingroup$

In this article about Ehrenfest's paradox, an introductory remark on classical rigidity is made:

Any rigid object made from real materials that is rotating with a transverse velocity close to the speed of sound in the material must exceed the point of rupture due to centrifugal force, because centrifugal pressure can not exceed the shear modulus of material.

$${\frac {F}{S}}=\frac {mv^{2}}{rS}<\frac {mc_{s}^{2}}{rS}\approx \frac {mG}{rS\rho }\approx G$$

where $ c_{s}$ is speed of sound, $\rho$ is density and $G$ is shear modulus. Therefore, when considering velocities close to the speed of light, it is only a thought experiment.

I'm not sure I understand the relation between the rupture point and the speed of sound. What happens if the material rotates with the speed of sound? Why is that a special velocity? Because $G$, $c_s$, and $\rho$ are interdependent? What happens physically at a transverse velocity equal to the speed of sound in the material in question.

$\endgroup$

1 Answer 1

2
$\begingroup$

[LATER EDIT, 9 hours after submitting the answer]
I did some googling, to see if I could find any corroboration of the claim there is a connection between the speed-of-sound-in-the-material and the point where the material ruptures due to the tensile stress. I have not found such corroboration.

I am very suspicious that the assertion is unfounded. The section with the shear stress statement was added in june 2011, by a contributor who had not created a wikipedia account. My recommendation is that you do not trust that statement.
[END LATER EDIT]


We have that the centrifugal effect of rapid rotation gives rise to tensile stress in the material. Failure occurs when the amount of required centripetal force exceeds the amount of elastic force that a material can produce upon being subjected to tensile stress.

Generally a stiffer material will have more structural integrity.


Longitudinal propagation of sound in a material is highly correlated with the response of that material to the compressive/tensile stress of the sound vibration.

Generally a stiffer material will feature faster propagation of sound vibration.


Let's compare copper and a high strength steel. Wrap copper wire around the rim of a flywheel. Spin the flywheel, steadily increasing angular velocity. At some spin rate the copper wire will snap. The high strength steel will snap at a much higher spin rate.


Put the copper wire under tension, and generate longitudinal sound vibrations at one end. Those sound vibrations will travel at a particular velocity.

Put the high strength steel wire under tension, and generate longitudinal sound vibrations at one end. The vibration will propagate faster than in the case of the copper wire.


I'm not sure whether there will be a 1-on-1 correlation between the amount of tensile stress where the material will fail, and the speed of propagation of sound in that material, but I do expect a strong correlation.

$\endgroup$
3
  • $\begingroup$ Thanks for the effort! True physical curiosity! :) $\endgroup$ Feb 19, 2022 at 10:33
  • $\begingroup$ By the way, your Hamiltonian stationary action seems interesting. Could QFT be described by it (instead of Lagrangian)? I post a new question maybe. $\endgroup$ Feb 19, 2022 at 10:41
  • $\begingroup$ @Felicia Please note: the name I use is: Hamilton's stationary action. Historically there was another action concept, proposed by Maupertuis. To disambiguate I use the expression "Hamilton's stationary action" to refer to the action concept used in classical mechanics. (There is the expression 'Hamiltonian' too, also introduced by Hamilton; but it refers to something else in mechanics.) A concept of stationary action is used in multiple areas of physics. Generally, whenever some physics taking place can be described with differential equations it can be restated in a stationary action form. $\endgroup$
    – Cleonis
    Feb 19, 2022 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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