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Rigid rod problem



 
 
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  #71  
Old September 13th 05 posted to sci.physics.relativity
russell@mdli.com
external usenet poster
 
Posts: 256
Default Rotating corkscrew (Was: Rigid rod problem)

Todd wrote:
wrote in message
oups.com...
Todd wrote:
wrote in message
ups.com...
Todd wrote:


[snip]

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.


I guess I should not have referred to the radius as
necessarily small.



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.


Yes it would, because we would have miles and miles more
material in each turn. For a given thickness of material,
and constant angular velocity, the rigidity must increase
as R increases. It would of course be different if we were
dealing here with a closed hoop, but we're not.

[snip]

Yes, I goofed in thinking that increasing R would reduce the stress.
Instead I believe that the stress would be independent of R just as for a
closed hoop.

For a closed hoop spinning such that the material has a speed u, the tension
in the hoop will equal lambda*u^2 which is independent of R (at least for
nonrelativistic speeds).

For the helix I suspect that the tension will still be lambda*u^2 except for
an extra numerical factor that would depend on the ratio of the pitch to the
radius.


But this ignores an important difference: the helix's
two ends have nothing to pull on, so the tension stress
must be zero there. The only thing that keeps the
material curved at all is tension on the inside of the
curve and compression on the outside, similar to the
case of a straight bar subject to a bending stress.
This is a huge demand.

A one-mile ring of steel can easily be spun so that it
keeps its circular shape. Not so a one-mile bar of steel
bent in a helix of one or two turns.

As you scale up the helix with larger radius, you would also scale
up the pitch so that the ratio of pitch to radius would stay constant. (For
example, we might choose the pitch to equal 2*Pi*R) So, I think the tension
would be independent of R for a given rotational speed u and a given mass
per unit length lambda. I also think that the tension stress would be
essentially independent of the length (number of wraps) of the helix.


But the tension-compression stress that is *really*
responsible for the shape is dependent on the total
length of the bar; you can't just keep adding turns
and length to the helix, keeping other things constant,
and expect it not to distort under the centrifugal
force. At least this is what I think, without doing
any calculations...


Anyway, I was wrong in believing that increasing R would help.
Nevertheless, I don't believe that the resolution of the paradox lies in
establishing that there are no real materials that are rigid enough to
withstand the rotations imagined in the thought experiment. For some
reason, that doesn't seem to me to get to the heart of the paradox.


It does seem unsatisfying, yes. Btw it bothered me that
we hadn't considered the density of the material at all,
and then I remembered that speed of sound of course depends
on density as well as stiffness, so maybe I'm not completely
full of crap after all.

I could
be wrong of course. As in David's original post, we can consider a thought
experiment in which the pitch is astronomically long and then we can get by
with a slow rotation speed. I think the stress would then be small and we
wouldn't have to worry about the strength of the material.


The stress would *not* be small.

Btw before you posted your nifty helix-balancing calc, I
thought of some other ways to rectify the principal axis,
e.g. by using a double-helix with the two strands attached
only by a single cross-bar at one end. The strands could
then unwind into perfectly straight rods attached at one
end, like a very long tuning fork (and of course we would
not have the paradox we did before). But just imagine what
that would mean for the stress in the two rods. It would
be enormous. I think your single helix would be subject to
exactly the same stresses from the end to the middle, because
for all that length it wouldn't really "know" that it wasn't
attached by a crossbar to some partner helix. And I think
this probably also gets to Tom Roberts' notion of "driving"
the rotation even though as I said, I didn't really understand
everything he wrote.

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  #73  
Old September 13th 05 posted to sci.physics.relativity
Todd
external usenet poster
 
Posts: 8
Default Rotating corkscrew (Was: Rigid rod problem)


wrote in message
oups.com...
Todd wrote:
wrote in message
oups.com...
Todd wrote:
wrote in message
ups.com...
Todd wrote:

[snip]

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.

I guess I should not have referred to the radius as
necessarily small.



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.

Yes it would, because we would have miles and miles more
material in each turn. For a given thickness of material,
and constant angular velocity, the rigidity must increase
as R increases. It would of course be different if we were
dealing here with a closed hoop, but we're not.

[snip]

Yes, I goofed in thinking that increasing R would reduce the stress.
Instead I believe that the stress would be independent of R just as for a
closed hoop.

For a closed hoop spinning such that the material has a speed u, the
tension
in the hoop will equal lambda*u^2 which is independent of R (at least for
nonrelativistic speeds).

For the helix I suspect that the tension will still be lambda*u^2 except
for
an extra numerical factor that would depend on the ratio of the pitch to
the
radius.


But this ignores an important difference: the helix's
two ends have nothing to pull on, so the tension stress
must be zero there. The only thing that keeps the
material curved at all is tension on the inside of the
curve and compression on the outside, similar to the
case of a straight bar subject to a bending stress.
This is a huge demand.

A one-mile ring of steel can easily be spun so that it
keeps its circular shape. Not so a one-mile bar of steel
bent in a helix of one or two turns.


Yes indeed, thanks. It helped me to think about a rotating circular ring in
which you make a cross sectional slice at one point so that the surfaces of
the slice are free of stress. Then, as you say, in the rest of the rotating
ring you get inner tension and outer compression that increase as you move
around to the diametrically opposite point. These stresses can be huge and
they would increase with an increase in radius of the ring. I hope I've got
it right now.

As you scale up the helix with larger radius, you would also scale
up the pitch so that the ratio of pitch to radius would stay constant.
(For
example, we might choose the pitch to equal 2*Pi*R) So, I think the
tension
would be independent of R for a given rotational speed u and a given mass
per unit length lambda. I also think that the tension stress would be
essentially independent of the length (number of wraps) of the helix.


But the tension-compression stress that is *really*
responsible for the shape is dependent on the total
length of the bar; you can't just keep adding turns
and length to the helix, keeping other things constant,
and expect it not to distort under the centrifugal
force. At least this is what I think, without doing
any calculations...


Anyway, I was wrong in believing that increasing R would help.
Nevertheless, I don't believe that the resolution of the paradox lies in
establishing that there are no real materials that are rigid enough to
withstand the rotations imagined in the thought experiment. For some
reason, that doesn't seem to me to get to the heart of the paradox.


It does seem unsatisfying, yes. Btw it bothered me that
we hadn't considered the density of the material at all,
and then I remembered that speed of sound of course depends
on density as well as stiffness, so maybe I'm not completely
full of crap after all.

I could
be wrong of course. As in David's original post, we can consider a
thought
experiment in which the pitch is astronomically long and then we can get
by
with a slow rotation speed. I think the stress would then be small and
we
wouldn't have to worry about the strength of the material.


The stress would *not* be small.

Agreed

Btw before you posted your nifty helix-balancing calc, I
thought of some other ways to rectify the principal axis,
e.g. by using a double-helix with the two strands attached
only by a single cross-bar at one end. The strands could
then unwind into perfectly straight rods attached at one
end, like a very long tuning fork (and of course we would
not have the paradox we did before). But just imagine what
that would mean for the stress in the two rods. It would
be enormous. I think your single helix would be subject to
exactly the same stresses from the end to the middle, because
for all that length it wouldn't really "know" that it wasn't
attached by a crossbar to some partner helix.


Interesting

And I think
this probably also gets to Tom Roberts' notion of "driving"
the rotation even though as I said, I didn't really understand
everything he wrote.


I didn't understand what Tom was getting at either.

Todd

 




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