Angular momentum conservation and energy considerations

[LeeB]

I have a question about a classical physics problem. The original problem appeared as a homework problem in a physics book and it is really an extension of the problem that causes me an issue. The original problem went something like this:

A frictionless puck travels, with linear velocity, v, in circular motion around an level air table, and the puck is held in circular motion by means of a string attached to its center. The string passes down through a frictionless hole in the center of the table and is held by a student so the length of the string above the table is R. While the puck is rotating, the student pulls in some string so the new length of the string above the table becomes R/2. What happens to the (a) linear velocity of the puck? (b) kinetic energy of the puck, and (c) where does the additional kinetic energy come from?

The answers to these questions are simple: The linear velocity becomes 2v (because of conservation of angular momentum… M*V(1)*R(1) = M*V(2)*R(2) and thus the kinetic energy becomes 4 times as great. The additional energy is produced by the work done by the student as he pulls in the string. So conceptually, the additional energy comes from food eaten by the student and there are no difficulties with energy conservation.

Now consider an extension that came to mind when I did this problem. Suppose we replace the hole in the table and the student with a thin post at the center of the table to which the string is firmly attached. We strike the puck perpendicularly to the string and it begins to rotate around the center pin. At the moment the string length is R, the linear velocity is v (just as before.) The string winds around the thin post so that eventually, the new radius is R/2. Again, the linear velocity needs to be 2v and again, the kinetic energy is four times as great as the original kinetic energy, but this time, I cannot attribute the increase in energy to a particular source. I’m sure the pole did not eat breakfast, and there is no engine or obvious source of energy. Can someone help me here?

 

[Doc Al]

Now consider an extension that came to mind when I did this problem. Suppose we replace the hole in the table and the student with a thin post at the center of the table to which the string is firmly attached. We strike the puck perpendicularly to the string and it begins to rotate around the center pin. At the moment the string length is R, the linear velocity is v (just as before.) The string winds around the thin post so that eventually, the new radius is R/2. Again, the linear velocity needs to be 2v and again, the kinetic energy is four times as great as the original kinetic energy, but this time, I cannot attribute the increase in energy to a particular source. I’m sure the pole did not eat breakfast, and there is no engine or obvious source of energy. Can someone help me here?

This case is quite different than the other scenario. In this case, no work is done on the puck--the string just wraps around, it's not being pulled in. KE is conserved, but angular momentum is not; the string is not perfectly radial--it exerts a torque on the puck.

Good thing too--the last thing we need is another mouth to feed!

 

[LeeB]

Additional question, then...

So if there is no work done, then you are suggesting that the new velocity WON'T be twice the old velocity under these circumstances? When we actually did this on an air table, the results were virtually identical to the first case. The puck DID speed up and although it isn't possible to say the result was EXACTLY the same as the first case, to the degree we could measure it, there was no difference. The puck was definitely traveling at NEARLY twice the original velocity.

 

[Doc Al]

So if there is no work done, then you are suggesting that the new velocity WON'T be twice the old velocity under these circumstances?

That's right.

When we actually did this on an air table, the results were virtually identical to the first case. The puck DID speed up and although it isn't possible to say the result was EXACTLY the same as the first case, to the degree we could measure it, there was no difference. The puck was definitely traveling at NEARLY twice the original velocity.

How did you determine the speed of the puck?

 

[LeeB]

We used a PASCO motion sensor. (And furthermore, it was totally obvious that the puck was moving MUCH faster when the string was shorter.) A torque can't provide more energy because the force is perpendicular to the motion, right? So where did the extra energy come from?

 

[Doc Al]

I'm curious as to how you set it up to make those measurements.

Why was it "totally obvious" that it was moving faster? Certainly the angular speed would be faster.

Didn't you ask this same question here over 3 years ago?

 

[LeeB]

I don't know if it was three years ago, but yes. I did. I never got an answer then and I still don't know if I've gotten one yet here. As for how we set it up, we did this: Instead of going DOWN through the center of the table for the first part, we arranged a bracket with an eye at the center of the table (ABOVE the table) The student pulled the cord UP (not down) and the puck sped up just as in the problem. We measured the speed with a motion sensor and, indeed, when the radius became 1/2 of the old radius, the speed was approximately twice as great. (We need to allow for reasonable error.) Then, for the second part, we replaced the bracket with the eye in it with post that was lowered right to the table top. We got the puck moving in a circle and had pencil marks on the table for R and R/2. When the puck reached the mark for R/2 the measured velocity also changed by a factor of 2 just like it did in the original experiment. I have never found anyone who could explain where the extra energy came from and I'm just trying again.

 

[rcgldr]

Well there's the obvious factor that when a person pulls on the string, there is work done, corresponding to the integral of force on the string times distance the string is pulled, and this work is going to increase the kinetic energy of the puck.

While the puck is being pulled in, it has a inwards component of velocity as well as it's tangental component of velocity, following a spiral path. Again, the work done will be equal to the integral of the force times the inwards distance moved, and the kinetic energy will increase by the amount of work done.

For those that are doubting this, note that if the tension in the string was modulated to be relative to 1/R^2, the puck would follow an elliptical path, same as a planet orbiting the sun, and it's speed would vary by this formula:

$V = \sqrt{U(2/R-1/A)}$

where R is distance from the "central body", A is length of major axis, and U is gravitational parameter:

http://en.wikipedia.org/wiki/Elliptic_orbit

Using the elliptical path, to double velocity at R/2:

(2/(R/2)-1/A) = 4
4/R - 1/A = 4
1/A = 4/R - 4 = (4-4R)/R
A = R/(4-4R)


I won't go through all the math, but here are the initial and final results of your experiment, where T is tension, M is mass of puck, and S = speed of puck:

$T_0\ =\ M\ (S_0)^2/(R_0)$

$T_1\ =\ M\ (2\ S_0)^2/(R_0/2)\ =\ 8\ M\ (S_0)^2/R_0\ =\ 8\ T_0$

Based on these two values

$T(R)\ =\ M\ (S_0)^2\ (R_0)^2\ /\ R^3$

 

[LeeB]

Jeff...

Okay, you say there's work being done because there's an inward component of velocity. My question is... "What's the source of the new energy?" In the first case, it ultimately came from the energy in food eaten by the student. There is no obvious source of new energy here, yet the puck increases in speed (and thus kinetic energy!)

 

[rcgldr]

Okay, you say there's work being done because there's an inward component of velocity.

An inward component of movment times an inwards component of force. Work = force x distance = change in energy. In the elliptical orbit case, the total energy remains constant, but the kinetic and potential energy vary, and the work done is related to the change in these energies. Net work done per obit is zero. However the puck case is different, because there the poential energy of gravity is replaced by the energy of the person pulling on the string, allowing for a large variation in puck paths and energy changes.

update - No net work only occurs when the net force is always perpendicular to the velocity (direction of travel).

 

[Doc Al]

Well there's the obvious factor that when a person pulls on the string, there is work done, corresponding to the integral of force on the string times distance the string is pulled, and this work is going to increase the kinetic energy of the puck.

While the puck is being pulled in, it has a inwards component of velocity as well as it's tangental component of velocity, following a spiral path. Again, the work done will be equal to the integral of the force times the inwards distance moved, and the kinetic energy will increase by the amount of work done.

When the puck is being pulled in because the string is being pulled in by the person (effectively shortening the string), then work is being done on the puck, increasing its energy. But when the string just wraps around the fixed pole, no work is being done. The motion of the puck is tangential to the string tension. Note that the string is at an angle due to the radius of the pole. That's what allows it to exert a torque, changing the angular momentum.

The only way the KE could increase is if the string were elastic and you exchanged elastic PE for KE. I am assuming an unstretchable string here.

I just don't buy that the puck increases speed. (I wish I had such a lab set up to do the experiment myself.)

 

[Doc Al]

However the puck case is different, because there the poential energy of gravity is replaced by the energy of the person pulling on the string, allowing for a large variation in puck paths and energy changes.

The case where the person pulls the string is not the one in question: in that case energy clearly is added by the person. The situation that LeeB is concerned with is the one where there is no person, just a string wrapping around a fixed pole.

 

[rcgldr]

Just a string wrapping around a fixed pole.

Sorry, I was asleep at the screen, I sit corrected. Glad I didn't waste my time integrating T(R)dR to figure out work done in the person pulling case.

update - a better explanation

The pole situation is a no work situation. In the case of any spiral, the "normal", a line perpendicular to the path of the spiral, doesn't cross the center. For most spirals, the distance from the center to the normal varies, but for one special case, the "involute of circle", the normal is always a fixed distance from the center. This situation is exactly what occurs in the case of the pole. The path of the puck is a "involute of circle", and the string coincides with the normal, it's alway perpendicular to the path of the puck, so the tension force is always perpendicular to the path of the puck and no work is peformed. I made pictures of both the string wrapping around a pole, and a string pulled through a hole:

The pole case where no work is done, the string is perpendicular to the path at all points in the path:

pole.jpg

The hole case path could be just about any path, but any path other than a circle will result in periods of time where the tension force is not perpendicular to the path of the puck, so work is done. For this example spiral, I used an equal angular spiral, the normal is always at a fixed angle compared to the line from the center point to a point on the path.

hole.jpg

In a later post, I include links to more info, including some animated gif's.

 

[tabchouri]

When the puck is being pulled in because the string is being pulled in by the person (effectively shortening the string), then work is being done on the puck, increasing its energy. But when the string just wraps around the fixed pole, no work is being done. The motion of the puck is tangential to the string tension. Note that the string is at an angle due to the radius of the pole. That's what allows it to exert a torque, changing the angular momentum.

The only way the KE could increase is if the string were elastic and you exchanged elastic PE for KE. I am assuming an unstretchable string here.

I just don't buy that the puck increases speed. (I wish I had such a lab set up to do the experiment myself.)

Both case are in some extent exactly the same: in both cases there is something shortening the string : a human, or the string warping around the pole.

And Yes, there is work in both cases because the tension of the string is not perpendicular to velocity of the object when the string is pulled (or shortened for that matter) : the velocity has a tangential component (tangent to the circle of center the fixed pole and radius of "distance between the pole and the object), but it also has a radial component ! (remember that the trajectory is ellipsoid not circle !

so there is work, there is an energy increase.
That increase is the result of the work done of course, the string did not eat breakfast in any cases :)
The additional force that created the work is shortening of the string, which creates an attractive force (besides the tension of the string holding the object when there is no shortening).

 

[Doc Al]

Both case are in some extent exactly the same: in both cases there is something shortening the string : a human, or the string warping around the pole.

The cases are quite different. In the human-pulling case, the puck moves in the direction of the string tension; not so in the wrapping around the pole case.

And Yes, there is work in both cases because the tension of the string is not perpendicular to velocity of the object when the string is pulled (or shortened for that matter) : the velocity has a tangential component (tangent to the circle of center the fixed pole and radius of "distance between the pole and the object), but it also has a radial component ! (remember that the trajectory is ellipsoid not circle !

No, there is no work done in the wrapping around case. Realize that the string is not radially oriented, but lies at an angle. Sure, the puck gets closer to the pole, but it always moves tangential to the string so the string tension does no work.

so there is work, there is an energy increase.
That increase is the result of the work done of course, the string did not eat breakfast in any cases :)

If you think that, then tell us the source of the energy.

 

[tabchouri]

The cases are quite different. In the human-pulling case, the puck moves in the direction of the string tension; not so in the wrapping around the pole case.

No, there is no work done in the wrapping around case. Realize that the string is not radially oriented, but lies at an angle. Sure, the puck gets closer to the pole, but it always moves tangential to the string so the string tension does no work.


If you think that, then tell us the source of the energy.

I maintain my position on that,
The ebergy came form the work, the work came from the additional tension of the string, which came from shortening the string (warping arount the pole).

now here is an illustration of the case : http://ghazi.bousselmi.free.fr/illustrate.GIF

assume:
$$\theta$$ : the angle of the object,
$$d\theta$$, the instantaneous angular velocity
$$dL = -r.d\theta$$, is the rate at which the string shortens

as seen in the picture :
$$\gamma = arctan(\frac{r}{R})$$
$$R = L.cos(\gamma)$$
$$dR = dL.cos\gamma-L.sin\gamma = -r.d\theta.cos\gamma-L.sin\gamma.d\gamma$$

or
$$d\gamma = d(arctan(\frac{r}{R})) = \frac{d\frac{r}{R}}{1+(\frac{r}{R})^2} = \frac{r.dR.\frac{-1}{R^2}}{1+(\frac{r}{R})^2} = \frac{r.dR}{R^2+r^2} = \frac{r.dR}{L^2}$$

thus
$$dR=-r.d\theta.cos\gamma - \frac{sin\gamma.r.dR}{L}$$

so
$$dR=\frac{-L.r.d\theta.cos\gamma}{L+sin\gamma.r}$$

or:
$$\stackrel{\rightarrow}{v} = R.d\theta. \stackrel{\rightarrow}{t} + dR. \stackrel{\rightarrow}{r} = R.d\theta. \stackrel{\rightarrow}{t} - \frac{L.r.d\theta.cos\gamma}{L+sin\gamma.r} . \stackrel{\rightarrow}{r} =$$ (the linear velocity of the object)

and
$$\stackrel{\rightarrow}{T} = ||\stackrel{\rightarrow}{T}|| . (-sin\gamma.\stackrel{\rightarrow}{t} - cos\gamma.\stackrel{\rightarrow}{r})$$ the tension of the string

$$\stackrel{\rightarrow}{v}.\stackrel{\rightarrow}{T} = ||\stackrel{\rightarrow}{T}||.(-R.d\theta.sin\gamma +\frac{L.r.d\theta.cos^2\gamma}{L+sin\gamma.r}) = ||\stackrel{\rightarrow}{T}||.\frac{d\theta(r(L.cos^2\gamma-R.sin^2\gamma)-L.R.sin\gamma)}{L+sin\gamma.r}$$
$$= ||\stackrel{\rightarrow}{T}|| .\frac{d\theta(r(cos^2\gamma-cos\gamma.sin^2\gamma)-R.sin\gamma)}{1+sin^2\gamma}$$
$$= ||\stackrel{\rightarrow}{T}|| .\frac{r.d\theta(cos^2\gamma-cos\gamma.sin^2\gamma-cotan\gamma.sin\gamma)}{1+sin^2\gamma}$$
$$= ||\stackrel{\rightarrow}{T}|| .\frac{r.d\theta.cos\gamma(cos\gamma-sin^2\gamma-1)}{1+sin^2\gamma}$$
$$= ||\stackrel{\rightarrow}{T}|| r.d\theta.cos\gamma(\frac{cos\gamma}{1+sin^2\gamma}-1)$$


This scalar product is not null (continually), as a matter of fact it is never null (becoz $$\gamma \neq 0 [\pi]$$, i.e. $$\gamma \neq k.\pi,\ k\ interger$$)
so THERE is a work beeing done.


As for where is it coming from ? as i said: is it the result of shortening the string, giving an extra pull force on the object. You can attribute it to the physical coehence of the material fabric of the string and to the force of attachment of the pole to the table. If the object is too massive of the velocity is too quick, either the string would break or the pole would be detached from the table, or the table would break ...

existance of force = energy (potential or kinetic ...)
It's like you ask : where the kinetic energy comes from when a proton attracts an electron !

 

[rcgldr]

Link to picture of puck winding clockwise around a pole (I used the same radius for both).

pole.jpg

From this picture, it should be clear that the tension in the line is pependicular to the path of the puck. The name of the spiral is "involute of circle". The distance from the side of the pole to the center of the puck equals the radius of the pole times the angle, which is the chord (cord? bad pun here) length around the circumference of the pole.

Defining "r" to be the radius of the pole, the paremetric equations for this curve are:

$x\ =\ r\ (\ cos(\theta) \ +\ \theta \ sin(\theta)\ )$

$y\ =\ r\ (\ sin(\theta) \ -\ \theta \ cos(\theta)\ )$


Link to picture of puck being drawn in towards a hole while moving clockwise.

hole.jpg

From this picture, it should be clear that the tension in the line is not quite pependicular to the path of the puck, causing an increase in energy as the puck is pulled inwards. In this case, with the line being drawn into (or out of) the hole, the path could be anything, like an ellipse, but in all cases if the line is moved inwards or outwards there's a change in energy. For this example, I used an "equal angular spiral" since it looks nice. The (polar) equation for this is:

$$r\ =\ e^{a \theta}$$

 

[tabchouri]

so, you telling me that the velocity is perpendicular to the tension of the string ? so I've spend more than half an our writing those equations in LaTex in vain :) ?

"From this picture, it should be clear that the tension in the line is pependicular to the path of the puck."
By the way, are you basing your conslusions sololy on the picture ?

 

[rcgldr]

Note that I updated my previous post to include the string though the hole example as well.

So, you telling me that the velocity is perpendicular to the tension of the string ?

Yes.

So I've spend more than half an our writing those equations in LaTex in vain?

Yes. You needed to start with the right equation for the path of the puck, which is "involute of circle".

Are you basing your conslusions solely on the picture?

No, it's based on the characteristics of that curve. As I mentioned, in the case of "involute of circle" the distance from the side of the circle to the spiral path is equal to the angle times the radius of the circle. If you were to wrap one end of a string around a post, say a flashlight, so the string couldn't slip, and the other end around a pencil, then drew a line with the pencil as you wound (or unwound) the string around the post, then "involute of circle" is the graph you would be drawing.

You can redo your equations, now that you know the equation for the path of the puck ("involute of circle"). Calculate the tangent and then the vector perpendicular (normal) to this path, and you'll find that it directly points to the edge of the post.

 

[tabchouri]

now the point is, you have fixed the path and the curve in advance, without any practcal considerations (btw, what are those equation ? are they meant to be the coordinates of the puck ? if so, they are Wrong, just compute the radius = sqrt(x²+y²).

I've based my calculation considering the system, its configuration, and as only input : the angular velocity of the puck (on which all other equation are based) !

try to re-see the process i undertook :)

 

[rcgldr]

what are those equation ? are they meant to be the coordinates of the puck?

Yes.

Now the point is, you have fixed the path and the curve in advance, without any practical considerations

Once again, if you were to wrap one end of a string around a flashlight so the string couldn't slip, and the other end around a pencil, then drew a line with the pencil as you wound (or unwound) the string around the post, then "involute of circle" is the graph you would be drawing.

Or you can just look at the pretty animated graphic at Wiki:

Animated_involute_of_circle.gif

It's from this article:

http://en.wikipedia.org/wiki/Involute

Another animation from Mathworld:

http://mathworld.wolfram.com/Involute.html

Or instead of a pencil and a flashlight, a goat and a post: "The simplest kind of spiral to draw and understand. It is, for example, the path that a goat, tethered to a post, would follow if it walked around and around in the same direction, keeping its tether taught until it wound its way to the center"

http://www.daviddarling.info/encyclopedia/C/circle_involute.html


There's even a pratical purpose for "involute of circle", gears:

http://en.wikipedia.org/wiki/Involute_gear

http://www.abbeyclock.com/gearing4.html

For the rest of you that have lost interest by now, there are at least some pretty animations and pictures here.

 

[tabchouri]

you are right on one thing: there is a little flow in my calculation, and the funny part is the first ones before I edit the post was right, i changed it thinking it was wrong :)

but also with the old calculation, the tension is of the string was not perpendicular to the velocity.

PS: i can repost the right formulas for the sake of good reading

 

[rcgldr]

You can redo your equations, now that you know the equation for the path of the puck ("involute of circle"). Calculate the tangent and then the vector perpendicular (normal) to this path, and you'll find that it directly points to the edge of the post.

This is already done in the articles. The tangent at angle $\theta \ =\ \theta$, and the normal is $\theta \ +\ \pi / 2$, pointing at the edge of the involuted circle.

 

[Doc Al]

I maintain my position on that,
The ebergy came form the work, the work came from the additional tension of the string, which came from shortening the string (warping arount the pole).

The string starts with some length L and after wrapping around the pole still has length L--it has not shortened. (Unless you are assuming a stretchy, elastic string--which is not the case here.)

existance of force = energy (potential or kinetic ...)
It's like you ask : where the kinetic energy comes from when a proton attracts an electron !

A force is not energy! :yuck:

It's perfectly clear why the charged particles speed up as they get closer--work has been done: a force has acted through a distance along the line of the force. Is there an energy source? You bet! Electric potential energy has been "consumed".

If you really think work is being done when the string wraps around, despite its length remaining unchanged and its tension always remaining perpendicular to the velocity of the puck, then specify the energy source. Additional tension does not mean additional energy.

(It looks like Jeff might have straightened things out for you. )

 

[tabchouri]

The string starts with some length L and after wrapping around the pole still has length L--it has not shortened. (Unless you are assuming a stretchy, elastic string--which is not the case here.)


A force is not energy! :yuck:

It's perfectly clear why the charged particles speed up as they get closer--work has been done: a force has acted through a distance along the line of the force. Is there an energy source? You bet! Electric potential energy has been "consumed".

If you really think work is being done when the string wraps around, despite its length remaining unchanged and its tension always remaining perpendicular to the velocity of the puck, then specify the energy source. Additional tension does not mean additional energy.

(It looks like Jeff might have straightened things out for you. )

the length of the string shrinks : i guess you have guessed I'm speaking of the string that's straight and not already rolled around the pole.
and by force = energy, i suppose you have also guessed that i'am paraphrasing this "when a net force is applyed to an object, that object has a potential energy" :)

 

[fantispug]

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 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

$$d\gamma = d(arctan(\frac{r}{R})) = \frac{d\frac{r}{R}}{1+(\frac{r}{R})^2} = \frac{r.dR.\frac{-1}{R^2}}{1+(\frac{r}{R})^2} = \frac{r.dR}{R^2+r^2} = \frac{r.dR}{L^2}$$

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
$$dW=T\cos(\gamma)(\cos(\gamma)-1)dR$$

(I'm not sure if it's equivalent to tabochouri's answer.) However it is clear that the work is negative since $$\cos(\gamma)\leq 1$$. 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 $$\gamma$$ is small. Then $$dW=T\gamma dR + O(\gamma^2)$$, which we'll take to be negligible.
At small angles $$L \approx R$$, so from
$$\vector{v}=\dot{R}\hat{r}+R\dot{\theta}\hat{t}$$
(where a dot signifies a time derivative) and
$$d\theta=-\frac{dL}{r}\approx-\frac{dR}{r}$$
we get kinetic energy
$$K\approx\dot{R}^2+(\frac{R\dot{R}}{r})^2$$
So if this quantity is conserved (which is approximately true if R>>r) then
$$\dot{R}^2(r^2+R^2)\approx\dot{R}^2R^2$$
is constant.
Consequently if R is halved, $$\dot{R}$$ 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!

 

[fantispug]

It works! I had to redraw the diagram to get there though... I thought the right angle in the diagram was between R and a (which makes all my conclusions in my previous post suspect), but it's actually between L and a (since the string is tangent to the circle).
If a is the radius of the inner circle, and L is the length of the string
$$dL=-ad\theta$$
And from tabchouri's diagram
$$r^2=L^2+a^2$$
Where r is the distance of the puck from the centre of the circle. Differentiating yields:
$$rdr=LdL$$
Substituting:
$$dr=-\frac{La}{r}d\theta$$
Or in other words:
$$\frac{dr}{d\theta}=-\frac{a\sqrt{r^2-a^2}}{r}$$.
Going the other way, $$r=a\sqrt{\theta^2+1}$$ is the polar parametric equation of a circle involution, differentiating yields
$$\frac{dr}{d\theta}=\frac{a\theta}{\sqrt{\theta^2+1}}=\frac{a^2\theta}{r}$$

Now rearranging the defining equation for theta yields
$$\theta=\pm\frac{\sqrt{r^2-a^2}}{a}$$
so
$$\frac{dr}{d\theta}=\pm\frac{a\sqrt{r^2-a^2}}{r}$$

As expected! (Take the negative sign since r decreases as theta increases)

I think my brain has worn out on this problem, I might have to come back to it after a break.

I tried to make an analogous experiment by tethering a ball to something and swinging it around, but it kept flinging off. I tried nailing it to the shoelace I was using to tether, but the lace wasn't thick enough so the ball just flew out nail and all.

 

[rcgldr]

The $\theta$ is relative to the circle, not the polar coordinate one.

Using $\phi$ to represent the inner circle angle, I get:

$r\ = \ R\ \sqrt{\phi ^2\ +\ 1}$

Then to get the polar coordinate $\theta$, I get:

$\theta \ = \ \phi\ -\ tan^{-1}(\phi)$

Here is a picture: invcir.jpg

 

[fantispug]

Graphing it up, Jeff, you appear to be right (very odd coincidence though). My solution gives something that looks like, but isn't quite, the involution of a circle.

I've attached an image comparing what I got (inside in green) to the involution of the circle (outside in blue). Qualitatively both solutions look roughly like what I would expect, but you convincingly argued that it should be an involution.

I'm most frustrated. In case anyone is interested I got the solution
$$\ddot{r}=-r\dot{r}^2\frac{r^2-3a^2}{r^4-a^4}=-\frac{\dot{r}^2}{r}+\frac{3a^2\dot{r}^2}{r^3}+O(a^4)$$

However I don't know how to solve these differential equations, so I can't get an exact solution. All I need is one integration so I could get $$\dot{r}$$ in terms of r and t, and analyse the effect of decreasing r on the motion is, assuming I haven't made any mistakes, which isn't beyond question seeing as the solution is not the involution of a circle.
Though all the assumptions that went into it seem physically plausible to me.

It feels like I'm using a sledgehammer to crack a rather soft nut. Maybe a different approach would be simpler. It's just hard if the string does work on the puck, since then the Lagrangian approach won't work. And the motion isn't obvious, though after thinking about it and throwing some balls around I strongly doubt the speed increases significantly as the puck moves inwards, and I doubt the kinetic energy of the puck increases.

Frustrating.

Edit: I think I can see what happened. I said as the string moves through a small angle $$d\theta$$ the string's length decreases by $$Rd\theta$$, but as Jeff Reid pointed out, this angle is not the same as the polar angle. I'm far too bad at picturing situations.
Again I have come up with equations that are pretty much worthless... not sure how many more times I'll have to go through it.

 

[fantispug]

Ok, you're probably sick of me now, but I'll post what I've got thus far, sans details. From my failed previous attempts you're probably not too confident in my result - I'm not either, but each time I'm getting closer to the solution.
I got, using something like Jeff's geometry (l is instantaneous length of string, r is radial distance from the centre, theta and phi are as in Jeff's diagrams)
$$d\phi=d\theta+\frac{adr}{rl}$$
Which using
$$dl=ad\phi$$
and geometry gives
$$rdr=ldr$$
All together we get:
$$\frac{dr}{d\theta}=\frac{ar}{\sqrt{r^2-a^2}}$$
Now use a bit of geometry to solve for the forces, crank the algebraic handle, and I got
$$\ddot{r}=-\frac{\dot{r}^2(r^2-2a^2)}{r^3}$$
(My exact solution is rather close to my approximate solution from my last run through basically because the differential in theta is close to the differential in phi for small angles - and I used the wrong angle last time).

Cutting to the chase I don't know how to solve analytically. I hectically threw something together in matlab, toyed with the initial conditions until the rounding errors disappeared (it doesn't like it when r gets too close to a, as you'd expect). The solution looked something like a involution, but as I've already learned looks can be deceiving, and I haven't substituted the equations for an involution back in for theta (though they look close to me, again looks aren't everything).
Cut to the chase. Over time $$\dot{r}$$ increased as r decreased. In fact a plot of r versus $$\frac{1}{\dot{r}}$$ was a straight line. This agrees with my approximate solution from post 26 (again my bad geometry becomes irrelevant at small angles, thankfully).
Though I'm sure you're not completely convinced by my analysis, and neither am I.

In any case, unless someone else has evidence otherwise, I propose:
The kinetic energy is always conserved. It drops out beautifully of these new equations that the angular work just cancels the radial work. A Lagrangian approach is hence fine. (My apologies to more physically minded people like Jeff that saw this right off the bat - I was having trouble with it).

Angular momentum is not conserved - the string exerts a torque on the puck.

The total velocity of the puck in the thin pole approximation is constant, but the radial velocity is inversely proportional to radius. The tangential velocity remains constant.

I suggest thus that the OP was measuring the radial speed. Alternatively he could have been measuring the angular velocity.
If there are any experimentalists out there that can give this simple experiment a go, I'd much appreciate the reality check.

 

[rcgldr]

I suggest thus that the OP was measuring the radial speed.

The OP mentions an increase in speed when the pole is replaced by a very small hole and someone pulls the string up through the hole, in this case there is work done, (math already done in this case, see earlier posts). The OP was wondering why a pole doesn't have the same effect as a hole, and it's because the curve for the pole case is an involute of circle, whilethe hole case could be anything, even an ellipse if the tension is varied to be 1/r^2 (like gravity). I did the string through the hole case first, then the pole case.

 

[rcgldr]

Cutting to the chase I don't know how to solve analytically.

Use cartesian coordiantes instead.

Equation of involute of circle:
$x\ =\ R\ (\ cos(\phi) \ +\ \phi \ sin(\phi)\ )$
$y\ =\ R\ (\ sin(\phi) \ -\ \phi \ cos(\phi)\ )$

Derivative of involute of circle:
$d(y(\phi))/d(x(\phi)=(R(cos(\phi)\ + \phi \ sin(\phi) \ - cos(\phi)))\ / \ (R(-sin(\phi)\ + \phi \ cos(\phi) \ + sin(\phi)))$
$d(y(\phi))/d(x(\phi)=(R\ \phi \ sin(\phi))\ /\ (R\ \phi \ cos(\phi)) \ = \ tan(\phi)$

Equation of normal line of involute of circle:
$-(dy_0/dx_0)(y - y_0)=(x - x_0)$
$-tan(\phi _0)(y - y_0)=(x - x_0)$

Equation of circle:
$x\ = \ R\ cos(\phi)$
$y\ = \ R\ sin(\phi)$

Derivative of circle:
$d(y(\phi))/d(x(\phi)=-R sin(\phi) \ /\ R cos(\phi) = -cot(\phi)$

Equation of tangent line of circle:
$(y - y_0)=(dy_0/dx_0)(x - x_0)$
$(y - y_0)=(-cot(\phi _0))(x - x_0)$
$-tan(\phi _0)(y - y_0)=(x - x_0)$

For both equations:
$-tan(\phi _0)(y - y_0)=(x - x_0)$
$y - y_0=-cot(\phi _0)(x - x_0)$
$y + cot(\phi _0)x = y_0 + cot(\phi _0)x_0$

For involute of circle:
$y + cot(\phi _0)x = R(sin(\phi_0)-\phi_0 cos(\phi_0) + cot(\phi _0)R(cos(\phi_0)+\phi_0 sin(\phi_0))$
$y + cot(\phi _0)x = R(sin(\phi_0)-\phi_0 cos(\phi_0) + cos(\phi_0)^2/sin(\phi_0) + \phi_0 cos(\phi_0))$
$y + cot(\phi _0)x = R(sin(\phi_0) + cos(\phi_0)^2/sin(\phi_0))$

For circle:
$y + cot(\phi _0)x = R(sin(\phi_0)) + cot(\phi _0)R(cos(\phi_0))$
$y + cot(\phi _0)x = R(sin(\phi_0)) + cos(\phi_0)^2/sin(\phi_0))$

So the the equations for the normal line of the involute of circle and tangent line of the circle are the same with respect to $\phi$ , and no work is done since the tangent line of the circle, which is the string and the line of force, is normal to the path of the puck. (and now my head hurts). Examples of the normal+tangent line equations for specific values of $\phi$ .

For $\phi = \pi / 2$, $y = R$
For $\phi = \pi / 4$, $y = -x + \sqrt{2} \ R$

 

[fantispug]

Then, for the second part, we replaced the bracket with the eye in it with post that was lowered right to the table top. We got the puck moving in a circle and had pencil marks on the table for R and R/2. When the puck reached the mark for R/2 the measured velocity also changed by a factor of 2 just like it did in the original experiment. I have never found anyone who could explain where the extra energy came from and I'm just trying again.

This seems to imply to me that LeeB is claiming that when the string wraps around the pole the puck speeds up in an inversely proportional manner. I argue with this claim on theoretical and physical grounds (I wish I had the apparatus to carry out a reasonable experiment).

Anyway Jeff I checked and my equations are consistent with the path being involute of a circle. I don't think that 2nd order DE I presented is right.

Anyway I ended up trying again, getting kinetic energy
$$T=\frac{m\dot{r}^2r^2}{2a^2}$$
Which leads to a DE in terms of R:
$$\ddot{r}{r}+\dot{r}^2$$
Which has solution
$$r=\sqrt{2r_0\dor{r}_0t+r_0^2}$$
where the puck has initial position $$r_o$$ and velocity $$\dot{r}_0$$. The precise motion can be worked out using the relation between r and theta
$$\frac{dr}{d\theta}=\frac{ar}{\sqrt{r^2-a^2}}$$
Which is the differential form of the involute of a circle, which has solutions
$$r=R\sqrt{\phi^2+1}$$
where phi is given implicitly by
$$\theta=\phi-\arctan(\phi)$$
where theta is the polar coordinate.

The explicit solution isn't too helpful though. Conservation of energy, since there are no sources of potential energy, implies the speed is constant throughout the motion. As Jeff pointed out this is very different to the case where someone is pulling on the string supplying energy.

 

[rcgldr]

The string through the hole case, assuming that angular momentum is conserved when the string is pulled through a hole. For example, angular momentum is m x r x s, so if r is halfed, then to conserve momentum, it becomes m x (r/2) x (2s). Based on this assumption, then the relation ship between tension and r is:

$$t(r)\ =\ -m\ (s_0)^2\ (r_0)^2\ /\ r^3$$

So if string is pulled from $r_0$ to $r_0/2$, then work done is:

$$\int _{r_0} ^{r_0/2} (-m\ (s_0)^2\ (r_0)^2\ /\ r^3)\ dr$$

$$\left[ \frac{1}{2} m\ (s_0)^2\ (r_0)^2\ /\ r^2 \right]_{r_0}^{r_0/2}$$

$$\frac{1}{2} m\ (s_0)^2\ (r_0)^2\ /\ (\frac{r_0}{2})^2 - \frac{1}{2} m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2$$

$$\frac{4}{2} m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2 - \frac{1}{2}\ m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2$$

$$\frac{3}{2} m\ (s_0)^2\ (r_0)^2\ /\ (r_0)^2$$

$$\frac{3}{2} m\ (s_0)^2$$

Original KE:

$${KE}_0 = \frac{1}{2} m (s_0)^2$$

KE after work done:

$${KE}_1= \frac{1}{2} m(s_0)^2 + \frac{3}{2} m\ (s_0)^2 = \frac{4}{2} m\ (s_0)^2$$

$${KE}_1= \frac{1}{2} m (2\ s_0)^2$$

So the speed is doubled if the radius is decreased by 1/2. In the case of the post, relative to a circular path, the post applies a backwards and inwards tension, which may account for the reason that angular momentum is not conserved in that case. Also in the case of the post, the tension is normal to the actual path of the puck, so no work is done and the puck's speed remains constant.

 

[tabchouri]

Use cartesian coordiantes instead.

Equation of involute of circle:
$x\ =\ R\ (\ cos(\phi) \ +\ \phi \ sin(\phi)\ )$
$y\ =\ R\ (\ sin(\phi) \ -\ \phi \ cos(\phi)\ )$

Derivative of involute of circle:
$d(y(\phi))/d(x(\phi)=(R(cos(\phi)\ + \phi \ sin(\phi) \ - cos(\phi)))\ / \ (R(-sin(\phi)\ + \phi \ cos(\phi) \ + sin(\phi)))$
$d(y(\phi))/d(x(\phi)=(R\ \phi \ sin(\phi))\ /\ (R\ \phi \ cos(\phi)) \ = \ tan(\phi)$

Equation of normal line of involute of circle:
$-(dy_0/dx_0)(y - y_0)=(x - x_0)$
$-tan(\phi _0)(y - y_0)=(x - x_0)$

You are right, now i see the error of ly ways :)
Error in my algebra led to my erronous conclusions.
I opologies for any trouble.

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