wrote in message
ups.com...
Todd wrote:
I think it's possible to arrange for the helix to rotate freely about its
axis without any wobbling by choosing just the right amount of winding
and
by also adding a point mass of just the right magnitude to the helix
halfway
along the helix (bear with me).

Like dynamically balancing a tire with one well-positioned
weight... seems plausible.

yes

This is based on some calculations that I'll
summarize here. I would be grateful to anyone who bothers to check my
results. All the integrations are elementary.

I parameterized the helix as

(x, y, z) = (a*theta, R*cos(theta), R*sin(theta))

where theta is the parameter that runs from -thmax to +thmax. The helix
winds around the x-axis from x = -a*thmax to x = +a*thmax with radius R
and
pitch 2*Pi*a. At x = 0, the helix passes through the positive y axis at
y =
R.

Letting lambda be the mass per unit length, the total mass is found to be

M = 2*lambda*Sqrt(a^2+R^2)*thmax

For a general value of thmax, the center of mass of the helix is located
at

(Xcm, Ycm, Zcm) = (0, R*sin(thmax)/thmax, 0).

For integral number of windings (thmax = 0, Pi, 2Pi, 3Pi, etc.) the
center
of mass is located at the origin of the coordinate system. Otherwise, it
is
located somewhere on the positive or negative y-axis.

In order for no wobbling to occur as the helix rotates about the x-axis,
it
is necessary that the x-axis be one of the principle axes of the moment
of
inertia tensor. This will be the case if all the off-diagonal elements
I_xy, I_yx, and I_xz of the tensor are zero. Recall that these are
defined
as

I_xy = - integral [x*y*dm], etc.

where dm is an element of mass.

For the helix with an arbitrary value of thmax, I find

I_xy = I_yz = 0 (odd integrands)

I_xz = 2*lambda*a*R*Sqrt(a^2+R^2)*( sin(thmax) - thmax*cos(thmax) )

= M*a*R*( sin(thmax)/thmax - cos(thmax) )


I left out the an overall negative sign, but that doesn't change anything.

Thus, we can force I_xz to vanish by choosing thmax to be a positive root
of

tan(thmax) = thmax

Call these roots: th1, th2, th3, th4, etc. (in order of increasing
magnitude). These roots must be found numerically. For these values of
thmax the helix has a non-integral number of windings.

FWIW I get 4.49341, 7.72525, 10.90412 for the first three.
There are fortunately no solutions less than pi, which agrees
with the "obvious" requirement that there should be more than
one complete turn in the helix.


Good, I get the same numerical values.


For these special values of thmax, all off-diagonal elements of the
moment
of inertia tensor vanish. We need one further requirement in order for
the
helix to be able to freely rotate about the x-axis without any wobble.
We
need the center of mass of the helix to be located on the x-axis. From
above, we had Xcm=Zcm=0 and

Ycm = R*sin(thmax)/thmax

If thmax is a root of tan(thmax)=thmax, this may be written as

Ycm = R*cos(thmax).

It is easy to check that for th1, th3, th5, etc., the value of cos(thmax)
is
negative and the center of mass is located on the negative y-axis. For
th2,
th4, th6, etc. the value of cos(thmax) is positive and the center of mass
is
located on the positive y-axis.

Thus, for th1, th3, th5, etc. we can shift the center of mass to the
origin
by just adding a point mass m0 to the helix at (x, y, z) = (0, R, 0).
For
example, if we choose thmax = th1, then m0 = M*cos(th1). For thmax =
th3,
m0 = M*cos(th3), etc.

This looks and feels right to me. Note that as n increases,
thn gets closer and closer to (2n+1)pi/2, i.e. for more and
more turns the required mass gets closer and closer to zero.


Yes


Thus, for thmax = th1, th3, th5, etc., we can arrange for the center of
mass
to be at the origin. It is easy to check that adding the point mass does
not mess up the vanishing of the off diagonal elements of the inertia
tensor. Thus, the system should be able to rotate freely about the
x-axis
without wobbling.

If the pitch of the helix is chosen large enough, then there will exist a
'moving' inertial frame in which the rotating helix appears to be
completely
unwound into a straight line parallel to the x-axis (with the point mass
m0
attached to the midpoint). The system orbits around the x-axis as it
also
moves parallel to the x-axis. And it does this without any externally
applied forces! Strange.

But we really need to do some calculations to see whether
this is physically possible from the standpoint of the
speed of sound in the material. I have speculated in my
earlier post that it is in fact not possible, but the
calculation does not appear easy even to set up. If I were
Feynman, perhaps I'd have worked it all out already, but as
anyone can see, I'm not. So for now, I can only talk about
what I think here, not what I know.

(OTOH the quickness and certainty with which Tom Roberts
answered this, suggests that it is given somewhere in the
literature. Unfortunately Roberts has not responded to my
followup questions, moreover there are some points in his
answer that don't seem quite right to me. Which could of
course be my problem, not his.)

Note that the corkscrew must be almost straight to begin
with (i.e. even 2piR is a pretty long pitch, and anything
close to that would require a relativistic spin) yet we
require it to provide, through its own torsional rigidity,
the centripetal force to accelerate a relativistic mass
through a small-radius arc. I think this requirement is
too great.


Well, as a 'thought experiment', I don't see why we can't consider a helix
of *very* large radius. Then the helix could rotate so that points of the
helix move at relativistic speed and yet have small centripetal
acceleration. So, the rigidity would not have to be great.


Either something's wrong with this analysis or else such a strange motion
is
actually allowed. Linear momentum and angular momentum do not *appear*
to
be conserved for the system as viewed in the moving frame. However, it
could be that they are in fact conserved if the momentum contributions of
the internal stresses in the helix are taken into account.

What I find a bit perplexing is that even if it unwinds
as little as would be required to put thmax at the next-
lower multiple of pi (i.e. not anywhere close to straight)
it seems you are still guaranteed to get a wobble. So, by
making the corkscrew really long, with a lot of turns, it
seems you can make it wobble even if the angular velocity
is quite small. OTOH the longer you make it, the more
demands you put on its rigidity to maintain the helical
shape, so I guess that doesn't necessarily lead to a
contradiction.

Assuming it unwinds only a very little bit, within what I am
calling the allowed range, does it make sense that the two
ends of the helix will gain in their contribution to the
moment of inertia enough to keep the rotation wobble-free?
This should be a much easier calculation but I'm not sure
how to do even that.


Good point. There is still an apparent paradox even if we just look at the
rotating helix in a 'moving frame' in which the helix is only slightly
unwound. And for this, we don't need to assume that the helix is rotating
at relativistic speeds.

For the slightly unwound helix, thmax would change such that the
condition tan(thmax) = thmax would no longer hold. So, it would *appear*
that the rotating helix would have to wobble in this reference frame. But
on the other hand, it is clear that if the rotating helix doesn't wobble in
the original 'rest frame' then it can't wobble when transforming to the
'moving frame' in which the helix is slightly unwound. So, there appears to
be a paradox. The thought experiment in which we arrange for the helix to
be completely unwound is just a more dramatic example of the same paradox.
I still think the solution probably lies in taking into account the momentum
contributions from the stresses in the helix.

Todd