Re: Rigid rod problem

Todd wrote:

[snip my rotating corkscrew paradox]

> I think the solution to this 'paradox' is that we need to include the
> momentum contributed by the _stresses_ in the corkscrew. These stresses
> arise because the corkscrew is rotating. In the frame in which the COM is
> screaming along at relativistic speeds, these contributions are large.

I guess your word "scream" is more appropriate than my "zip".
But in principle the issue exists at any speed.

>
> It must turn out that the total momentum (including the contributions from
> the stresses) is indeed conserved if no external forces are applied to the
> system. To actually check this, you would have to first work out the
> stresses inside the rotating corkscrew in the frame in which the COM of the
> corkscrew is at rest (yuck), and then transform these components to the
> moving frame (pretty easy).
>
> A nice treatment of stress contributions to momentum is given in Tolman's
> _Relativity, Thermodynamics, and Cosmology_ which is cheaply available from
> Dover Pub. A famous example of a paradox that arises if you don't include
> the stress momentum is the so-called Right-Angled Lever Paradox (see page 79
> of Tolman ).

Thank you, Todd, that is just the sort of expert contribution
that I was hoping for. Similarity to the Right-Angled Lever
had actually occurred to me separately, but I've never seen
that one worked out. I've heard about Tolman's book, and now
I guess I'll have to get a copy.

>
> Aside 1: A helix with an integral number of windings will have its COM on
> the symmetry axis of the helix. By 'symmetry axis', I mean the axis of the
> cylinder on which the helix may be imagined to be wound. However, I don't
> believe the symmetry axis of the helix will be one of the _principle axes_
> of the moment of inertia tensor of the helix. This means that if you start
> the helix rotating about its symmetry axis, it will wobble in the absence of
> any external forces. So, this is an added complication.

No doubt you're right about that. But actually I was only
trying for something easy to visualize; I don't think it's
in any way a crucial detail.

>
> Aside 2: If I'm not mistaken: In order for there to exist an inertial
> frame in which the rotating helix appears straight, the helix must be wound
> such that in the helix's own rest frame the 'pitch' of the helix (distance
> between consecutive windings) is greater than 2*Pi*R where R is the radius
> of the helix. I think that's kind of interesting.

Doesn't it depend on omega? But I guess you mean that at
the theoretically limiting omega (c/R) this is the minimum
pitch, and therefore it is minimum overall. I think you
are right.

This injects a further dose of healthy reality into my
scenario of the ordinary steel corkscrew: unless the spin
is some ungodly rate (too much for steel) there won't be
much of an effect to measure.

>
> Todd

.

Relevant Pages

  • Rotating corkscrew (Was: Rigid rod problem)
    ... and I made the claim that the corkscrew must be straight ... still a problem in the moving frame: ... > helix unwound as a straight line. ... >>> the symmetry axis of the helix. ...
    (sci.physics.relativity)
  • Re: Rigid rod problem
    ... momentum contributed by the _stresses_ in the corkscrew. ... between the rest frame of the rotating helix and the frame that sees the helix unwound as a straight line. ...
    (sci.physics.relativity)