Angular momentum conservation and energy considerations

1. Sep 17, 2007

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?

2. Sep 17, 2007

Staff: Mentor

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!

3. Sep 17, 2007

LeeB

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.

4. Sep 17, 2007

Staff: Mentor

That's right.
How did you determine the speed of the puck?

5. Sep 17, 2007

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?

6. Sep 17, 2007

Staff: Mentor

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?

7. Sep 17, 2007

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.

8. Sep 17, 2007

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

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

Last edited: Sep 17, 2007
9. Sep 17, 2007

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!)

10. Sep 17, 2007

rcgldr

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).

Last edited: Sep 17, 2007
11. Sep 17, 2007

Staff: Mentor

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.)

12. Sep 17, 2007

Staff: Mentor

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.

13. Sep 17, 2007

rcgldr

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.

Last edited: Sep 18, 2007
14. Sep 17, 2007

tabchouri

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).

Last edited: Sep 17, 2007
15. Sep 17, 2007

Staff: Mentor

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.

Last edited: Sep 17, 2007
16. Sep 17, 2007

tabchouri

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 wich 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 !!!

Last edited: Sep 17, 2007
17. Sep 17, 2007

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:

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

Last edited: Sep 18, 2007
18. Sep 18, 2007

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 ?

19. Sep 18, 2007

rcgldr

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

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

Last edited: Sep 18, 2007
20. Sep 18, 2007

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 :)

Last edited: Sep 18, 2007