Re: What is the Fastest spinning man made Object?

mmeron_at_cars3.uchicago.edu
Date: 07/12/04 

Date: Mon, 12 Jul 2004 22:47:04 GMT

In article <eca320d0.0407121351.7effd2dc@posting.google.com>, spamspamspam3@netzero.com (Edward Green) writes:
>mmeron@cars3.uchicago.edu wrote in message news:<qlKHc.67$25.17147@news.uchicago.edu>...
>> In article <40EF5274.9AAD48C0@hate.spam.net>, Uncle Al <UncleAl0@hate.spam.net> writes:
>> >mmeron@cars3.uchicago.edu wrote:
>
>Aha. I hadn't noticed when I took my second baby steps with a rim and
>a spoke, you are already posted the entire argument!
>
I was about to answer your previous post but, since we're already here
...

>> >> Note that for an infinitsimally thin surface element the only stresses
>> >> acting are tangential since radial stresses go to zero on the surface.
>> >> But, said tangential stresses still yield a radial force component
>> >> for a curved surface. It is a general result from the theory of
>> >> elesticity (established first by Laplace, I think, but easy to
>> >> rrederive on the back of aq small envelope, that given a surface
>> >> element with area dA and thickness dt, with tangential stresses
>> >> present in the surface, the normal (to the surface) force acting on
>> >> this element is
>> >>
>> >> dF_s = (S1/R1 + S2/R2)*dAdt
>> >>
>> >> where R1, R2 are the main radii of curvature of the surface (at the
>> >> point of evaluation) and S1, S2 are the stresses along the directions
>> >> of the corresponding axes of curvature).
>
>Yes, that relation is a favorite of mine; though I normally think of
>it in terms of surface tensions rather than stresses -- which in your
>case comes out considering a thickness dt.
>
Laplace indeed first introduced the above in the context of surface
tension. If you take the limit of dt -> 0 while S1, S2 go to infinity
in such way that the product Sdt remains finite (that's the tension)
you'll see the connection of the two.

>Would you care to sketch the back of envelope sketch?
>
Well, ASCII is a very cumbersome tool for sketching so I'll limit it
to a 1D problem (you can generalize from there). So, lets do it for a
cylinder (curvature in just one direction). Assum a piece of
cylindrical arc as

                   _______
                  / ____ \
                 / / \ \
             dt |/ \|

(yes, I know this is crude, hard to do better here). The end points
are the one marked dt and its symmetric opposite (dt is the thickness).
you can also assume that it extends some distance h into the page.
the length of the arc is dl and it subtends an angle dtheta (meaning,
each of the ends points downwards at an angle dtheta/2. Note that dl
and dtheta are related through

        dl = R*dtheta

Now, assume a stress S pulling at both edges of the arc. The total
force on each edge is the stress multiplied by area i.e

        dF = S*h*dt

The horizantal components of both forces cancel, being in opposed
directions. The verical components are in the same direction
(downward) though. Each vertical component is dF*sin(dtheta/2)
now, taking into account that we're dealing with infinitsimal
elements, so sin(dtheta/2) -> dtheta/2, we get for the normal force

        dF_n = 2*dF*dtheta/2 = df*dtheta = S*h*dt*dtheta = S*h*dl*dt/R

But h*dl is just the surface area dA of this cylindrical element so we
end with

        dF_n = S*dA*dt/R

You can generalize from here.

>> >> So, lets take our ball, rotating with an angular velocity w, and evaluate
>> >> the forces acting on an equatorial surface of element. For convenience
>> >> lets do the evaluation in the rotating reference frame. In this frame,
>> >> the surface element is acted upon by two forces. First, there is the
>> >> elastic force, described above, which simplifies to
>> >>
>> >> dF_s = (S1 + S2)*dAdt/R
>> >>
>> >> since R1 = R2 = R, the radius of the sphere. This force is pulling
>> >> the surface element inwards, towards the rotation axis. The second
>> >> force is the centrifugal one
>> >>
>> >> dF_c = R*w^2*dm
>> >>
>> >> where dm is the mass of the suface element, given by
>> >>
>> >> dm = dAdt*rho
>> >>
>> >> where rho is the density. So, we get
>> >>
>> >> dF_c = R*w^2*rho*dAdt
>> >>
>> >> And this force acts ouwards, away from the rotation axis. Since the
>> >> surface element remains stationary in the rotating frame (at least,
>> >> remains stationary till the sphere disintegrates:-)), the two forces
>> >> dF_s and dF_c must be equal. So, we've
>> >>
>> >> R*w^2*rho*dAdt = (S1 + S2)*dAdt/R
>> >>
>> >> Cancelling the common factors and reorganizing we get
>> >>
>> >> (R*w)^2 = (S1 + S2)/rho
>> >>
>> >> Now, R*w is simply v, the velocity of the equatorial point. As for S1
>> >> and S2, neither of them can be larger than the tensile strength S.
>> >> So, you get
>> >>
>> >> v^2 <= 2*S/rho
>> >>
>> >> with the = sign obtaining at the limit, before the material gives.
>> >> This is the relationship I used above to find the limiting velocity
>> >> and nte that the radius cancells out of it.
>
>Well, that was essentially my derivation for the rim, with a more
>professional touch, so that's reassuring! ;-) [There really is
>something to be said for this "professional" stuff, isn't there? I
>noticed this effect when I gave the same counter-example as Timo
>Nieman in regards to a signal not having a long term average; except
>he stated as a apothegm what I rendered as an Aesop's fable. Curse
>him. :-)]

The professional touch may be reassuring but the similarity doesn't go
unnoticed. So, yes, that's essentially your derivation for the rim,
indeed. And your observation that for a rotating spoke that maximum
stress point is on the hub, is also true.
>
>> >> The ratio S/rho, which has the dimensions of energy/mass can be simply
>> >> interpreted as the binding energy per unit mass of the material. With
>> >> this, the result above can be simply interpreted as saying that the
>> >> ball will hold together as long as the kinetic energy per unit mass
>> >> (in any locality) is no larger than the binding energy per unit mass.
>> >> This also provides you with a "sanity check" for any exaggerated
>> >> claims as you know that at absolute strength limit, assuming that all
>> >> the material is held together by molecular bonds, with no faults and
>> >> flaws whatsoever, the binding energy per unit mass still doesn exceed
>> >> something of the order of eV/amu. That was the source of the number I
>> >> gave in the previous post (the "10 miles/sec or so") before I had any
>> >> actual numbers regarding tensile strength of diamonds available.
>> >>
>> >> > I seem to remember a paper published in
>> >> >"Sceince" or "Nature" about opticaly levitated fused silica balls
>> >> >a decade or so back that did around that. Silica is nowhere near
>> >> >as strong as diamond.
>> >> >
>> >> Indeed, it isn't. So, the numbers were exaggerated or presented in a
>> >> misleading form. Remember, even when dealing with papers in
>> >> prestigious journals, the attitude should be "trust, but verify".
>> >
>> >We thus have the curious case that equatorial surface velocity is
>> >a scale-free absolute limit based upon lattice binding energy
>> >(less flaws) and all one can do is manipulate rpm vs. radius.
>> >Thus there is no circumstance in which a spinning test mass can
>> >go outside the Newtonian regime. Pity.
>
>That was my first thought! So an Ehrenfest disk using ordinary matter
>is out of the question. Damn. One might have thought that by
>building a huge ring one could get the velocity relativistic.
>
With puppeteer hull metal, maybe:-)

>> Well, that's true for small masses. However ...
>>
>> >If one wishes to really roll it, other binding forces must come
>> >into play
>>
>> Yes
>>
>> >like surface gravitation of a neutron star, abetted by
>> >the stiffness of nuclear matter. Even here, the surface is a few
>> >meters' depth of iron edging into degenerate matter from
>> >overlying pressure - which can slough.
>> >
>> It can, but it'll take a hell of a lot to do it.
>>
>> Basically, we can view the issue as "escape velocity". For the case
>> discussed (way) above, the limit of stability is reached when the
>> surface velocity is high enough to make the total energy (binding plus
>> kinetic) of a surface element positive. So, this limit velocity is just
>> the escape velocity, required to escape from the binding potential.
>> Now, adding gravity to the mix, the velocity becomes the escape
>> velocity from the combined, binding and gravitational potential. And
>> here you do get size dependence. This is so because unlike the
>> chemical binding energy where you've a constant per unit mass value,
>> regardless what other mass is there, the gravitational one does depend
>> on all the other mas that's there. So, for the neutron star you
>> mention the limit velocity will be tremendous even if the surface is
>> just iron, not neutronium. But, heck, even for a mundane planet like
>> Earth, the gravitational binding energy of a surface element exceeds
>> the chemical one (and that even if the Earth would've been a single
>> crystal).
>>
>> Steve Harris wrote once about the distinction between small,
>> chemically bound bodies and large, gravitationally bound ones. This
>> is relevant here.
>
>I've long had a puzzlement about escape velocity: given a spherically
>symmetric potential, I have no problem imagining a body flying
>straight up (as determined by the radial direction of the potential),
>having a certain critical velocity, will just escape to infinity.
>What I'm uncertain about is what happens to body's flying obliquely at
>this altitude, but having the same velocity. It seems to me a body
>whose orbit doesn't pass through the center of the potential may have
>a non-zero tangential velocity at perigee, hence store some KE which
>can't be used to escape, hence fail to escape even if it had escape
>velocity.

Ahh, but the further you go the less of a tangential component is
there, since the direction of "tangential" changes. To convince
yourself, draw a circle and a straigh line (that's just for
simplicity) tangent to it. At the contact point the direction of the
line is purely tangential. As you go further, the tangential part
keeps going down and the radial up.
>
>Yet I also feel I've heard the assertion that a body with escape speed
>will escape, no matter which way it's pointed.
>
>> >>We see something like this in CHI calculations of crystal
>> >>lattices. In the most restrictive case of DSI=0 and COR=1 with
>> >>CHI rapidly building with radius, every added surface atom exerts
>> >>a substantial effect at scale. CHI is bound to moments of
>> >>inertia that vary as the square of the radius. Adding few atoms
>> >>to the surface of a centimeter alpha-quartz sphere won't affect
>> >>the bulk CHI. However, despite the insignificant mass increment,
>> >>it will make the tail end of log(1-CHI) = (-2/3)log(atoms) + K go
>> >>absolutely wild. Nature sems to have a thing about surface area,
>> >>geometric parity divergence or black hole information.
>>
>> Yes, very much so.
>>
>> >> Surface area is deeply embedded in General Relativity.
>>
>> So it appears.
>> >
>> >Thanks for posting the analysis.
>> >
>> You're welcome, I was glad to have the opportunity.
>>

Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"

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