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Topic: Abel's Wire Problem (Read 16203 times) 

william wu
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Abel's Wire Problem
« on: Aug 23^{rd}, 2003, 3:17am » 
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Consider a wire resting in the Cartesian plane as shown in the figure above. A small bead of mass m with zero initial velocity is released from rest at the point P_{0}. The bead then proceeds to slide down the wire, eventually passing through the origin, where the potential energy is assumed to be zero. If the bead is released from the point P_{0} = (x_{0},y_{0}), its time of descent to the point 0 can be measured as a function of y_{0}, say, T = [phi](y_{0}). Find the shape of the wire in order that the time of descent be a specified function of y_{0}. Source: I read about this in Engineering Mathematics by Kenneth Miller. Dover, New York: 1956.

« Last Edit: Sep 1^{st}, 2003, 1:57am by william wu » 
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Icarus
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Re: Abel's Wire Problem
« Reply #1 on: Aug 23^{rd}, 2003, 6:28pm » 
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Assuming that the downward force is constant, the specified function [phi] must be increasing, or else the problem has no solution.


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SWF
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Re: Abel's Wire Problem
« Reply #2 on: Aug 24^{th}, 2003, 12:32pm » 
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The function [phi](y) does not have be monotonically increasing. It can be a constant. In that case the equation of the curve is a cycloid. Although with [phi]=constant, for a sufficiently large value of y_{0} it would be impossible to reach the origin in the required time. Also [phi](y) can decrease with y or have any number of local minima. However, if y[to][infty], [phi] must approach [infty] also. It is impossible to take less time to reach the origin than the time it takes for a bead on a vertical wire to slide from a given value of y to y=0 and that goes to [infty] as y[to][infty].


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Icarus
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Re: Abel's Wire Problem
« Reply #3 on: Aug 24^{th}, 2003, 7:03pm » 
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True  I should have thought about it a little more. There may be some other restrictions on [phi], but they are certainly looser than "increasing".


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James Fingas
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Re: Abel's Wire Problem
« Reply #4 on: Aug 26^{th}, 2003, 11:09am » 
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This problem is nasty. No real progress towards a solution, especially as I don't remember how to take derivatives properly. I think the problem will be soluble for any smooth continuous function of time defined over a finite domain. Each function of time will have a minimum value of gravity under which it can be solved. For functions that go off to infinity, they will have to have certain properties to keep the minimum value of gravity bounded away from zero (I think this is what SWF is saying). Given a function x(y), define a function s(y)=integral_{u=0..y}[sqrt](1+x^{2}(u))du Then the time to the origin from a height y_{0} is the following integral: T(y_{0}) = integral_{y=0..y0}(ds/dy)/[sqrt](2g*(y_{0}y))dy where g is the gravity constant. To solve this, just take the derivative of both sides w.r.t. y_{0}. This should give you a formula for (ds/dy) and the integral of the same in terms of T and its derivatives. Solve this integral equation and you'll find s in terms of T. I can't even figure out how to take the derivative! I know it will have two terms: one (positive) for the little bit of time added at the top of the track due to a new piece of wire, and one (negative) due to the higher speed at all other points on the track.

« Last Edit: Aug 26^{th}, 2003, 11:11am by James Fingas » 
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Icarus
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Re: Abel's Wire Problem
« Reply #5 on: Aug 30^{th}, 2003, 7:17am » 
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AAARRRRGGGGGHHHH! Can someone point out to me what I am doing wrong here? I keep going over it and I cannot find the problem: Let ([smiley=x.gif]([smiley=s.gif]), [smiley=y.gif]([smiley=s.gif])) be a parametric formula for the wire with respect to arclength [smiley=s.gif]. Denoting differentiation with respect to [smiley=s.gif] with a prime ', we have [smiley=x.gif]'[sup2] + [smiley=y.gif]'[sup2] = 1 I will denote differentiation with respect to time [smiley=t.gif] by using ^{[cdot]}. Then [smiley=x.gif]^{[cdot]} = [smiley=x.gif]'s^{[cdot]} [smiley=x.gif]^{[cdot][cdot]} = [smiley=x.gif]'s^{[cdot][cdot]} + [smiley=x.gif]''s^{[cdot]}[sup2] Similarly,[smiley=y.gif]^{[cdot][cdot]} = [smiley=y.gif]'s^{[cdot][cdot]} + [smiley=y.gif]''s^{[cdot]}[sup2] Now the force acting on the bead can broken into two components: one acting along the wire, and one normal to it. The second component is counteracted by the stiffness of the wire, so only the force acting along the wire causes movement in the bead. The force of gravity is represented by the vector [smiley=cf.gif] = (0, [smiley=m.gif][smiley=g.gif]). The unit tangent vector to the wire at position ([smiley=x.gif]([smiley=s.gif]), [smiley=y.gif]([smiley=s.gif])) is given by ([smiley=x.gif]', [smiley=y.gif]'). The component of an arbitrary vector [smiley=cv.gif] in the direction of an arbitrary unit vector [smiley=n.gif] is ([smiley=n.gif][cdot][smiley=cv.gif])[smiley=n.gif] Applying this, the force acting on the bead along the curve is (([smiley=x.gif]', [smiley=y.gif]')[cdot](0, [smiley=m.gif][smiley=g.gif])) ([smiley=x.gif]', [smiley=y.gif]') = ([smiley=m.gif][smiley=g.gif][smiley=y.gif]'[smiley=x.gif]', [smiley=m.gif][smiley=g.gif][smiley=y.gif]'[sup2]) By Newton's law, this is equal to [smiley=m.gif][smiley=a.gif] = ([smiley=m.gif][smiley=x.gif]^{[cdot][cdot]}, [smiley=m.gif][smiley=y.gif]^{[cdot][cdot]}). Combining, dividing out the [smiley=m.gif], and applying the equations for 2nd derivative with respect to time gives:[smiley=x.gif]'s^{[cdot][cdot]} + [smiley=x.gif]''s^{[cdot]}[sup2] = [smiley=g.gif][smiley=y.gif]'[smiley=x.gif]' [smiley=y.gif]'s^{[cdot][cdot]} + [smiley=y.gif]''s^{[cdot]}[sup2] = [smiley=g.gif][smiley=y.gif]'[sup2] Multiply the first line by [smiley=y.gif]', the second line by [smiley=x.gif]', and take the difference. This leaves ([smiley=y.gif]'[smiley=x.gif]''  [smiley=x.gif]'[smiley=y.gif]'')[smiley=s.gif]^{[cdot]}[sup2] = 0. [smiley=s.gif]'[sup2] is the square of the speed of the bead. If the bead does not stop on its slide, this is never zero. Thus [smiley=y.gif]'[smiley=x.gif]''  [smiley=x.gif]'[smiley=y.gif]'' = 0 or,[smiley=x.gif]''/[smiley=x.gif]' = [smiley=y.gif]''/[smiley=y.gif]' Integrate to get [smiley=l.gif][smiley=n.gif] [smiley=x.gif]' = [smiley=l.gif][smiley=n.gif] [smiley=y.gif]' + [smiley=cc.gif] and so, [smiley=x.gif]' = [smiley=ck.gif][smiley=y.gif]' Combining this with [smiley=x.gif]'[sup2] + [smiley=y.gif]'[sup2] = 1 implies that both [smiley=x.gif]' and [smiley=y.gif]' are constant. And so the wire must be straight. That is, the only wire down which a bead will slide is a perfectly straight one !!! . Obviously somewhere I have done something that is only true for straight wires, but I can't see it. James  if you are willing to take the advice of someone who can't figure out his own stuff, you can differentiate your formula by replacing each instance of y0 with different variable, which themselves are all equal to y0. Then apply the chain rule for partial derivatives: dT/dy0 = ([partial]T/[partial]u)(du/dy0) + ([partial]T/[partial]v)(dv/dy0)

« Last Edit: Aug 30^{th}, 2003, 10:27am by Icarus » 
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SWF
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Re: Abel's Wire Problem
« Reply #6 on: Aug 30^{th}, 2003, 12:36pm » 
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on Aug 30^{th}, 2003, 7:17am, Icarus wrote:Applying this, the force acting on the bead along the curve is (([smiley=x.gif]', [smiley=y.gif]')[cdot](0, [smiley=m.gif][smiley=g.gif])) ([smiley=x.gif]', [smiley=y.gif]') = ([smiley=m.gif][smiley=g.gif][smiley=y.gif]'[smiley=x.gif]', [smiley=m.gif][smiley=g.gif][smiley=y.gif]'[sup2]) By Newton's law, this is equal to [smiley=m.gif][smiley=a.gif] = ([smiley=m.gif][smiley=x.gif]^{[cdot][cdot]}, [smiley=m.gif][smiley=y.gif]^{[cdot][cdot]}). 
 I haven't looked at the equations carefully, but I see an error. You are saying F=m(x[cdot][cdot]+y[cdot][cdot])=force acting tangent to curve. You need to include force acting normal to the curve too, because by moving on a curved path the direction of acceleration is not tangent to the curve. The same reason something moving in a circle at constant angular velocity has a radial acceleration.


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Icarus
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Re: Abel's Wire Problem
« Reply #7 on: Aug 30^{th}, 2003, 2:20pm » 
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Thanks! I knew that there had to be something like that which was wrong, but my thoughts refused to break out of the path I had worn. I was mixing two different approaches to the problem in the physics. Removing the normal forces like this is a part of solving the problem using intrinsic coordinates, where the normal accelerations are hidden, but [smiley=x.gif] and [smiley=y.gif] are extrinsic (as functions of [smiley=t.gif]). The correct intrinsic equation would be [smiley=g.gif][smiley=y.gif]'([smiley=s.gif]) = [smiley=s.gif]^{[cdot][cdot]}


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SWF
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Re: Abel's Wire Problem
« Reply #8 on: Aug 30^{th}, 2003, 3:25pm » 
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Why are you working on this problem, Icarus? It is not on your unsolved Hard riddles list.


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Icarus
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Re: Abel's Wire Problem
« Reply #9 on: Aug 31^{st}, 2003, 11:11am » 
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I only add them to the list when they've been around awhile without solution. If this is solved soon, I won't have to update my list!


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SWF
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Re: Abel's Wire Problem
« Reply #10 on: Sep 1^{st}, 2003, 1:12pm » 
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The suggestion by James Fingas to solve by taking derivative looks like it would just complicate the problem. The derivative with respect to x of an integral that has x in both the limits and integrand will have a term for each integration limit plus another integral. From conservation of energy, an expression similar to James' is found: To simplfy future expressions I will define h(u)=[sqrt](1+ (dx(u)/du)[sup2]). The goal is to find the unknown function x(u), which requires finding h(u). I guessed the answer would be some integral with [phi] in it, and figured one should multiply both sides by some function and integrate. The hard part was figuring out which function would make the right hand side get more simple when integrated. A trick that turned out to work is: divide both side by [sqrt](yy[sub0]) and integrate with respect to y[sub0] from 0 to y. Note that order of integration was changed, which modifies the integration limits. The key here is that the last integral in the line above equals [pi] (complete the square to get something related to inverse sine), which greatly simplfies everything: Taking derivative of both sides with respect to y and replacing h(u) by its definition leaves a differential equation to be solved for x(y): Doing the integral is not always going to be easy. But if [phi]=constant, I know the solution should be cycloid (that is a well known differential equations problem). Also, if the path is a straight line then [phi](y0)=[sqrt](2*(x0[sup2]+y0[sup2])/(g*y0)). Both of these check out with the above solution.


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