# Angular momentum conservation and energy considerations

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!

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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.

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rcgldr
Homework Helper
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

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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.

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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.

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rcgldr
Homework Helper
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.

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

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

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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
Homework Helper
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 momemtum 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.

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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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rcgldr
Homework Helper
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.
Are you sure about this? After going through the math, it seems that the puck shouldn't have an increase in energy, but this is based on the assumption that the puck does follow an involute of circle path, and this sure looks like the proper model for anything winding around a post. Assuming the puck is attached at the side, then it's angular momentum (rotation wise) should slow things down a bit.

I do realize that the string is moving inwards with tension, but it's tension is perpendicular to the path of the puck so I'm not sure where the increase in energy would originate from. Relative to the string (which isn't an inertial frame since it's rotating), the string is being wrapped around the pole similar to the pole rotating and pulling in the string. Using this analogy, the only work done is to increase the potential energy (relative to the pole), but not the kinetic energy of the puck, similar to rasing a weight with a string (increase in potential energy, no increase in kinetic energy).

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Measurement error

Straightforward thermodynamics says that the linear velocity cannot increase when you do this. If this were not so, David could have whacked an entire army of Goliaths by simply shortening up his sling. If you have measured otherwise, the problem is in the measurement which is difficult to make for small circles. Measure omega and calculate v.

rcgldr
Homework Helper
Regarding the post case (involute of circle), and taking into account the post experiences a torque and applies this torque to the earth, increasing the earth's angular momentum, then angular momentum for the entire system, earth, post, and puck are conserved. In the hole case, no there no torque applied at the hole or to the earth, so there's no change in angular momentum to the earth, so the puck retains it's angular momentum.

Wow Jeff, I've gotta dig this one up to thank you big time for your explanation. Pretty much every post enlightened me at least a little bit on angular momentum, major kudos to you.