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Edit: Many of the assertions in this post and the next one are false due to a problem in geometry that Jeff points out. I'd like to thank him dearly for helping me iron them out. I give an "approximate" solution assuming energy is conserved. It turns out (when you use the right geometry) energy IS conserved, so this solution is exact. I think the OP threw me on conservation of energy, and I've been unsure ever since so I threw away my intuition and stuck to equations... often wrong equations.

Before I launch into my argument I should make a few general comments. Tabchouri's diagram looks correct to me, as does most of his algebra.

Jeff: I agree that the puck traces the involute of a circle. Although I haven't showed that my/Tabchouri's equations are consistent with it I suspect that they are - we have given the position of the puck in polar coordinates, but this is not a parameterisation of the motion - to find that you'd have to solve the equations of motion. (That is if we found position r as a function of time t we

There is almost certainly work done. Think of it this way: freeze the puck's spiral at some time. At that time it is moving around a bit and inwards a bit. The string is pulling back a bit and inwards a bit. Unless the tangential contribution exactly cancels with the radial contribution work will be done.

It's a pity too, problems like this are so much easier in the Lagrangian formulation, but if work is done it's back to Newton

[tex]dW=T\cos(\gamma)(\cos(\gamma)-1)dR[/tex]

(I'm not sure if it's equivalent to tabochouri's answer.) However it is clear that the work is

The reason work is done here is that the string is not pulling perpendicular to the trajectory of the puck, and consequently the reaction force of the pole is not perpendicular to the string.

However this disagrees with what the OP observed - the speed can not increase, it must decrease. An important question is what speed did you measure? Was it the total speed of the puck, or the speed radially inwards? If it's the latter I think I have a solution.

Let us assume that the radius of the pole is vanishingly small in comparison to the length of the string, i.e. that [tex]\gamma[/tex] is small. Then [tex]dW=T\gamma dR + O(\gamma^2)[/tex], which we'll take to be negligible.

At small angles [tex]L \approx R[/tex], so from

[tex]\vector{v}=\dot{R}\hat{r}+R\dot{\theta}\hat{t}[/tex]

(where a dot signifies a time derivative) and

[tex]d\theta=-\frac{dL}{r}\approx-\frac{dR}{r}[/tex]

we get kinetic energy

[tex]K\approx\dot{R}^2+(\frac{R\dot{R}}{r})^2[/tex]

So

[tex]\dot{R}^2(r^2+R^2)\approx\dot{R}^2R^2[/tex]

is constant.

Consequently if R is halved, [tex]\dot{R}[/tex] is doubled, so the radial velocity doubles although the tangential velocity remains constant (in this approximation).

So if you were wrapping a really long piece of string around a really small pole, and you didn't let it wrap around too much, the radial speed will be inversely proportional to the distance from the pole (the length of the string).

(I should throw in a derivation of my first result, but it's fairly similar to tabchouri's, and I've already spent a lot of time on this post - if anyone's interested or wants to debate it I'll put it up).

I'm really interested in getting this problem sorted out!

Before I launch into my argument I should make a few general comments. Tabchouri's diagram looks correct to me, as does most of his algebra.

Jeff: I agree that the puck traces the involute of a circle. Although I haven't showed that my/Tabchouri's equations are consistent with it I suspect that they are - we have given the position of the puck in polar coordinates, but this is not a parameterisation of the motion - to find that you'd have to solve the equations of motion. (That is if we found position r as a function of time t we

**should**get an involute, I haven't checked whether we do though...There is almost certainly work done. Think of it this way: freeze the puck's spiral at some time. At that time it is moving around a bit and inwards a bit. The string is pulling back a bit and inwards a bit. Unless the tangential contribution exactly cancels with the radial contribution work will be done.

It's a pity too, problems like this are so much easier in the Lagrangian formulation, but if work is done it's back to Newton

Looks like you dropped a negative sign to me. Your diagram looks convincing (cleared up much of my confusion) and running through the algebra with tabchouri's nomenclature I got[tex]d\gamma = d(arctan(\frac{r}{R})) = \frac{d\frac{r}{R}}{1+(\frac{r}{R})^2} = \frac{r.dR.\frac{-1}{R^2}}{1+(\frac{r}{R})^2} = \frac{r.dR}{R^2+r^2} = \frac{r.dR}{L^2}[/tex]

[tex]dW=T\cos(\gamma)(\cos(\gamma)-1)dR[/tex]

(I'm not sure if it's equivalent to tabochouri's answer.) However it is clear that the work is

**negative**since [tex]\cos(\gamma)\leq 1[/tex]. I would assume that the energy is transferred via the string into heating of the pole (in the same way friction causes heating).The reason work is done here is that the string is not pulling perpendicular to the trajectory of the puck, and consequently the reaction force of the pole is not perpendicular to the string.

However this disagrees with what the OP observed - the speed can not increase, it must decrease. An important question is what speed did you measure? Was it the total speed of the puck, or the speed radially inwards? If it's the latter I think I have a solution.

Let us assume that the radius of the pole is vanishingly small in comparison to the length of the string, i.e. that [tex]\gamma[/tex] is small. Then [tex]dW=T\gamma dR + O(\gamma^2)[/tex], which we'll take to be negligible.

At small angles [tex]L \approx R[/tex], so from

[tex]\vector{v}=\dot{R}\hat{r}+R\dot{\theta}\hat{t}[/tex]

(where a dot signifies a time derivative) and

[tex]d\theta=-\frac{dL}{r}\approx-\frac{dR}{r}[/tex]

we get kinetic energy

[tex]K\approx\dot{R}^2+(\frac{R\dot{R}}{r})^2[/tex]

So

**if**this quantity is conserved (which is approximately true if R>>r) then[tex]\dot{R}^2(r^2+R^2)\approx\dot{R}^2R^2[/tex]

is constant.

Consequently if R is halved, [tex]\dot{R}[/tex] is doubled, so the radial velocity doubles although the tangential velocity remains constant (in this approximation).

So if you were wrapping a really long piece of string around a really small pole, and you didn't let it wrap around too much, the radial speed will be inversely proportional to the distance from the pole (the length of the string).

(I should throw in a derivation of my first result, but it's fairly similar to tabchouri's, and I've already spent a lot of time on this post - if anyone's interested or wants to debate it I'll put it up).

I'm really interested in getting this problem sorted out!

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