[math-fun] brachistochrone for rolling ball
The underlying formula for a sliding frictionless bead was V = sqrt(2*h*g) where g=gravity acceleration, h=vertical height dropped so far, V=speed. For a ball (or other rotationally symmetric) rolling object with zero slip, a constant fraction of the energy is in rotational, and a constant fraction in translational, energy. Hence V=const*sqrt(2*h*g). This is the same formula, just with an altered value of g. Therefore, I conclude the brachistochrone, isochrone, etc will have unaltered (still cycloid) shape, although of course you want for the center of mass of the ball to move along the cycloid, not the contact point of ball with track -- hence track would need to be a "1-radius-expanded" version of cycloid. It is unnecessary to make balls have foamy outside and dense inside to reduce their moment of inertia. Any MoI is fine. Galileo was not aware of brachist. or iso property of cycloid, but he was aware of, or at least conjectured, cycloid as best possible shape for thin beam arch (constant beam width) -- which I have discussed here, the cycloid has uniform compressive stress but not uniform bending stress. Apparently Galileo invented the cycloid. The notion of "uniform stress" is something that can be defined without needing calculus -- probably for somebody as ingenious as Galileo, pure Euclidean geometry would suffice. (Archimedes similarly computed area & volume of sphere without needing calculus.)
This reminds me of a question I've wondered about. The cycloid is both the tautochrone and the brachistochrone. I've seen proofs of both of these facts, but it seems a remarkable coincidence that the same curve satisfies both these properties. Is there a proof that the tautochrone and the brachistochrone are the same curve that is simpler than just finding the equations of both, and proving that they are in fact the same? (Brachistochrone = Given points x and y, find the curve so that a ball rolling from x to y gets there fastest. Tautochrone = Find a curve so that the time to roll from an arbitrary starting point on the curve to a fixed ending point is the same, regardless of the choice of starting point). Andy (I suppose I should say "sliding frictionlessly", rather than "rolling", but Warren's argument below shows that the curve will be the same). On Thu, Nov 22, 2012 at 10:11 AM, Warren Smith <warren.wds@gmail.com> wrote:
The underlying formula for a sliding frictionless bead was V = sqrt(2*h*g) where g=gravity acceleration, h=vertical height dropped so far, V=speed.
For a ball (or other rotationally symmetric) rolling object with zero slip, a constant fraction of the energy is in rotational, and a constant fraction in translational, energy. Hence V=const*sqrt(2*h*g).
This is the same formula, just with an altered value of g.
Therefore, I conclude the brachistochrone, isochrone, etc will have unaltered (still cycloid) shape, although of course you want for the center of mass of the ball to move along the cycloid, not the contact point of ball with track -- hence track would need to be a "1-radius-expanded" version of cycloid.
It is unnecessary to make balls have foamy outside and dense inside to reduce their moment of inertia. Any MoI is fine.
Galileo was not aware of brachist. or iso property of cycloid, but he was aware of, or at least conjectured, cycloid as best possible shape for thin beam arch (constant beam width) -- which I have discussed here, the cycloid has uniform compressive stress but not uniform bending stress. Apparently Galileo invented the cycloid. The notion of "uniform stress" is something that can be defined without needing calculus -- probably for somebody as ingenious as Galileo, pure Euclidean geometry would suffice. (Archimedes similarly computed area & volume of sphere without needing calculus.)
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-- Andy.Latto@pobox.com
There's some useful information here mainly about the brachistochrone property, but also about the tautochrone property. If you can get through the writing style: < http://www.maa.org/pubs/Calc_articles/ma060.pdf >. Maybe there's a way to use this to show the two properties are equivalent. --Dan On 2012-11-22, at 9:07 AM, Andy Latto wrote:
This reminds me of a question I've wondered about. The cycloid is both the tautochrone and the brachistochrone. I've seen proofs of both of these facts, but it seems a remarkable coincidence that the same curve satisfies both these properties. Is there a proof that the tautochrone and the brachistochrone are the same curve that is simpler than just finding the equations of both, and proving that they are in fact the same?
(Brachistochrone = Given points x and y, find the curve so that a ball rolling from x to y gets there fastest. Tautochrone = Find a curve so that the time to roll from an arbitrary starting point on the curve to a fixed ending point is the same, regardless of the choice of starting point).
I don't see how they can be equivalent since in the limit of y directly below x there is no tautochrone. Brent On 11/22/2012 11:55 AM, Dan Asimov wrote:
There's some useful information here mainly about the brachistochrone property, but also about the tautochrone property. If you can get through the writing style: < http://www.maa.org/pubs/Calc_articles/ma060.pdf>.
Maybe there's a way to use this to show the two properties are equivalent.
--Dan
On 2012-11-22, at 9:07 AM, Andy Latto wrote:
This reminds me of a question I've wondered about. The cycloid is both the tautochrone and the brachistochrone. I've seen proofs of both of these facts, but it seems a remarkable coincidence that the same curve satisfies both these properties. Is there a proof that the tautochrone and the brachistochrone are the same curve that is simpler than just finding the equations of both, and proving that they are in fact the same?
(Brachistochrone = Given points x and y, find the curve so that a ball rolling from x to y gets there fastest. Tautochrone = Find a curve so that the time to roll from an arbitrary starting point on the curve to a fixed ending point is the same, regardless of the choice of starting point).
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Thanks for the link. I thought that this was a very nice paper. At 11:55 AM 11/22/2012, Dan Asimov wrote:
There's some useful information here mainly about the brachistochrone property, but also about the tautochrone property. If you can get through the writing style: < http://www.maa.org/pubs/Calc_articles/ma060.pdf >.
participants (5)
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Andy Latto -
Dan Asimov -
Henry Baker -
meekerdb -
Warren Smith