Talk:Carroll's paradox

This idea can be approximated experimentally. Arrange a greased hook in the ceiling to be cut at the 6:00 position. Screw an eye into the top of your rod. Catch the hook with the eye, and pull back to (nearly) horizontal. Release. Observe the rod as it flies off the hook. Measure the whirling and infer the angular momentum.

At the instant of release, the axis of rotation suddenly changes from the end of the rod to its center of gravity. At that moment, there will be a sudden change in the motion, like a bad manual gear shift. Nonethess, my casual experiments with a pencil seem to indicate that spinning continues.

Thus, one of the assumptions behind the "no angular momentum" assertion must be wrong, as it is empirically falsified.

Another experimental setup: Only a bottom guide. Find a large bowl, grease it, and align your pencil horizontally against the inside. Now, slowly release your finger from the eraser until friction with your finger approaches zero. Thud. Not much angular momentum, but it's not following the other rules either.

Which assumption is false? Probably the idea that there is no force perpendicular to the rod. The inside guide MUST exert such a force, or else the rod cannot "move like the hands of a clock". This force need not be supplied by friction, as demonstrated by the "greased hook" approach. It is, rather, supplied by supporting the inner end of the rod against gravity.

Hi

Well, something pretty odd is going on in the physical situation. The resolution of the paradox is that the constraints on the rod must supply an infinite force to maintain the rod in a radial position (more precisely: as the deviation from radiality tends to zero, the force exterted by the constraints tends to infinity).

There are two papers in the American Journal of Physics that discuss different resolutions. When I get a minute I'll type them up.

best wishes

Robinh 19:01, 28 Mar 2005 (UTC)

Yes, it would be great if someone added information on the resolutions of this "paradox". "The resolution of this paradox is not clear." is at best unsatisfying, and at worse, false. dbenbenn | talk 19:33, 13 August 2005 (UTC)

To be honest, I fail to see the "paradox" in all of it. The problem with looking at the situation apparently lies in the frictionless guidance from the disks (which in turn would require infinite forces to keep the assembly from breaking apart.)

Consider a slightly altered arrangement: You have two concentric disks of different radii which can freely and independently from each other rotate around their centers. At the edges of the disks, there are two pivots. When the pivots are aligned, their distance is exactly the length of the rod in question. Now, attach the rod to the pivots.

You will notice the following:

• ) Although the individual disks can rotate independently, and although the rod could rotate around each single pivot at will, the current assembly makes the whole arrangement effectively rigid: Both disks and the rod must move in unison.

• ) From this point on, considering the disks to be massless, the whole situation is just that of the rod being attached by means of a massless lever to the common center of the disks.

Am I missing something? -- Syzygy 09:22, 27 March 2006 (UTC)

Hi. Sorry, to delete good-faith edits, but I removed the following paragraph:

However, even this resolution does not avoid the main cause of the seeming paradox. The reason why this system does not conserve
angular momentum is because it simply does not have to. All the conservation laws are ony applicable to closed system, those
which contain all the objects that exert forces on the system. By not including the Earth, which is tacitly assumed to produce a
pull on the rod, we are not considering the angular momentum that will be gained as a result of the rod pulling on the Earth.
This 'additional' angular momentum will add with that of the rod to result in sero gain.

the "system" under consideration is the rod, which (according to the second analysis) cannot change its angular momentum, because the constraints cannot exert a force lateral to the rod. I can't see how including the Earth is relevant. What additional angular momentum will be gained as a result of the rod pulling on the Earth? please explain this using Newton's Laws.

Robinh 21:51, 30 April 2006 (UTC)

Gravity

What has been neglected in the above is the ubiquitous force of gravity. The force of gravity is always directed downwards and provides a torque that imparts angular momentum to the rod. If you were to ignore gravity, the rod wouldn't slide in the first place.

I removed this, because it is not quite right. Gravity acts uniformly, effectively on the center of the rod, therefore imparts no torque. Or, if you like: if you ignore the walls, the rod wouldn't rotate in the first place. -Dan 17:42, 11 June 2006 (UTC)

Glaring omission

Who is the Carroll who noted this paradox? Jefferson Anderson 18:43, 19 December 2006 (UTC)

Yikes! Added. 192.75.48.150 19:39, 19 December 2006 (UTC)

Illustration needed

I am having a hard time visualizing the concept. Please provide an illustration--71.245.164.83 (talk) 02:57, 16 February 2011 (UTC)

reversion

Sorry to revert a good-faith addition. The deleted material did not add to the article: the issue is the angular momentum of the rod per se. Robinh 08:14, 14 March 2007 (UTC)

reversion of a reversion

Dear Robinh, I kindly suggest that you read any physics book and try to understand the laws of conservation of angular and translational momentum properly. Momenta conserve only in systems that do contain all the bodies exerting forces on the system, i.e. if you consider a car on the surface of the Earth, that was initially stationary, then after it has accelerated it will have angular momentum. But the Earth will have precisely on opposite angular momentum because the car had to act by a friction force on the Earth and hence gave it a little additional spin. Think about this.

• Angular momenta conserve when all bodies exerting torque are considered. But see above: gravity acts on the centre of mass, therefore no torque. --192.75.48.150 18:01, 28 March 2007 (UTC)

Why isn't it...

I think it would act like a gyroscope. It will stay at the 3-o-clock position because it resists turning. So why doesn't it fall straight and pop off the track? It's "constrained". How can it be so constrained and not have the inner track exert a force with a component along the rod? The setup is self-contradictory. —Długosz

Resolution

Am I missing something, because the resolution to this paradox seems to be quite trivial doesn't it? It has nothing to do with gravity or friction. Also, some of the reasoning behind point 2 is pretty flawed.

Angular motion should always be stated with respect to a frame of reference, and it is because the poser of the problem has not been clear in this that the appears to be a paradox.

1. The rod will fall due to gravity regardless of the presence of any friction or not (how strong the frictional force is determines how far it will go. If strong enough, the rod may not even reach the 6 o'clock position). Thus, its angular momentum about the common centre of the circles does indeed change from this perspective.
2. There is no frictional force, so there is indeed no frictional reaction force perpendicular to the rod. Note, however, that there is an overall component perpendicular to the rod due to the reaction force from gravity on the outer circle. There is indeed no overall moment on the rod about the centre of the rod because the components of this reaction force (the ones that are inducing its angular motion about the centre of the circles) are equal in magnitude.

The moment about the centre of the rod is not the same as the angular motion of the rod about the centre of the circle: the rod can undergo two different angular motions. If you haven't quite got the distinction yet, imagine (for the sake of argument) looking down on the earth as it orbits the sun. The earth rotates about its centre as it rotates about the sun: the two angular motions are quite distinct. It is erroneous to compare the two and deduce there is a contradiction.

Thus, there is no contradiction: the rod indeed has no moment acting on it, but this does not mean that it is not able to slide about in between the circles: only that it cannot rotate about its centre when it does this.

I wonder whether this even deserves to be called a paradox: the "paradox" is based on a complete misunderstanding of the physics involved. Krea 02:49, 25 August 2007 (UTC)

Hi. Have you read the two American Journal of Physics papers cited in the article? Robinh 21:57, 25 August 2007 (UTC)

Actually, no I haven't. Yes, I know that I really should have done that before I posted above, but I don't have access to that (or any) journal right now. I've never used that particular journal, and (from what wikipedia says) it doesn't appear to be particularly technical; i.e. it is geared towards students and teachers. I'm trying to be tactful here, but to me it seems that the "paradox" is trivial: I wonder (based on the apparent nature of the journal) whether it was meant to be interpreted as a challenge to students of the subject. I mean, going back to my earth-sun analogy, the problem essentially is saying that the earth should not be able to orbit the sun because some mechanism is preventing the earth from rotating on its axis: this is poor physics. Or am I just going crazy? Krea 03:18, 26 August 2007 (UTC)

Hello again. Thanks for your comments. The paradox is far from trivial. The AJP papers make that very clear, and I think the wiki article is a fair summary of these papers. The rod has no moment acting on it about its centre, but somehow accumulates angular momentum. Think about the angular version of Newton's second law to see the paradox. The two papers cited give different resolutions to the paradox. Your Earth/Sun analogy is way off the mark: sorry. Best wishes, Robinh 10:01, 26 August 2007 (UTC)

Maybe I am going crazy then! When you say, "...but somehow accumulates angular momentum," angular momentum about what? N2L for angular motion is exactly the same form as for linear motion (except with torque and angular momentum): all that says is that changes in angular momentum require a torque (i.e. a moment). Can you walk me through this because I still don't get it? Where don't you (or the paper) agree?

1. The rod can undergo two different rotations: it can spin about its centre, and it can also move around the inner circle. These two motions are distinct and clearly angular.
2. There is no torque that will induce a "spinning", so given that its angular momentum in this regard is zero, it must remain zero.
3. However, there is an overall torque that will cause the the rod to move about in between the circles.
4. I contest that the "paradox" occurs because one is erroneously comparing the two angular motions.

I don't want to drag this out, but you haven't told me where/how I am mistaken, you've just repeated the assertion in the article. Hope to hear from you soon! Krea 15:26, 26 August 2007 (UTC)

Hello Krea. thanks for your comments. Rereading the discussion above, I'll try to summarize the essence of the paper, although I must confess I don't have it to hand. Consider the angular momentum of the rod, taken about the center of mass of the rod. This is zero at t=0, when we hold the rod by hand, at the three o'clock position. Then let go of the rod. There is no torque acting on the rod because there are only three forces acting on it and each of these three exert zero torque on the rod: gravity acts through the centre of mass, and the two reaction forces from the inner and outer rings also act through the cetre of the rod. So there is no force that can exert a torque on the rod, with moments defined about the centre of the rod. NOW... consider what one "expects" to happen. The rod at three o'clock slides down the rings, and at six o'clock the rod is rotating about its centre. Clockwise. So the rod has somehow acquired angular momentum. BUT none of the three forces acting on the rod exert a torque on the rod, when taking moments about the centre of the rod. Is this any clearer or am I mudding the waters further with my late-night editing? :-) Robinh 21:15, 26 August 2007 (UTC)

Ah, wait a minute. I think I may just have gotten what you and the paper are tying to say (I hope!). Let me see if I have this right. Can we imagine a grid centred at the centre of the rod which does not rotate with respect to the circles; i.e., the grid axes are always orientated in the same direction although they may be translated. As the rod falls, it "drags" the grid with it, but, just to be clear, it does not rotate with the rod. Thus, with respect to that grid, the rod, as it falls, has changed angular position and must therefore have gained an angular velocity. Is this what you mean when you say it has gained an angular momentum as it falls? Krea 22:30, 26 August 2007 (UTC)

Hello Krea. That is right. Do you see the paradox now? Robinh 18:48, 27 August 2007 (UTC)

Right, I see. Let me clarify a few things that I said before, and a few things that you have said. There is only one external force that is driving this system: gravity. Splitting that force into components tangential and normal to the point at which the rod contacts the outer circle will explain why the rod is behaving as it does. The normal gravitational component will produce a reaction force from the outer circle that is always centrally directed: it is this force which is producing the circular motion. Before I consider the tangential gravitational component, let me first say that we cannot make statements like, "there is no torque acting on the rod," because such statements are too vague, and in this example incorrect. As I'm sure you know, a torque is r x F (vector product), and hence the existence of a torque is dependent on the frame of reference. In other words, from the point of view of the circles (i.e. from their reference frame) the tangential gravitational component is exerting a torque on the rod. With respect to this reference frame, the rod is changing angular position. But, of course, this is not a problem because there is a torque with respect to this reference frame.

Now, what you and the paper are pointing out is that, from the point of view of the rod, it is also appears to be rotating anomalously (i.e. without the influence of any force). However, this is a mistake since that frame of reference is not the frame of reference of the rod. The frame of reference of the rod is the one which also rotates with the rod. I'll explain why I think this is so soon, but do you follow me so far? Krea 20:41, 28 August 2007 (UTC)

Yes, nothing here requires any frame that is rotating with respect to any other. Robinh 14:16, 29 August 2007 (UTC)

Great! We can get straight to it then. As I see it, it's quite simple: the reference frame of the rod, by definition, is the rod itself. That reference frame that I talked about before (the one that moved with the rod, but did not orientate with it) does not correspond to anything: it is not physical, which is why it is giving non-physical results.

The best argument I can think of to back up my claim is this: the essence of the problem is actually quite general, and can be applied to many systems. For example, consider the rod attached to an inextensible string such that the rod acts like a pendulum. The physics of this system is identical to the physics of the system Carroll proposed (all extant forces are the same in both systems). Thus, the two systems are equivalent. The presence of frictionless circles does not affect the paradox.

Since the essence of the paradox is the consideration of a reference frame that moves, but does not orientate with a body that is undergoing angular motion, it applies to all systems where one body undergoes angular motion wrt to another body. This is, therefore, not a specific problem of this particular setup, but a very general problem which lies at the heart of all angular motion.

I hope I have not made an embarrassingly simple mistake, and I look forward to your reply, Krea 01:39, 30 August 2007 (UTC).

The two systems are not equivalent. The "pendulum" rod will move differently because it will not always "point" towards the centre of the circles. Robinh 07:14, 30 August 2007 (UTC)

Sorry, what do you mean when you say, "it will not always 'point' towards the...circles"? If the rod-pendulum is initially positioned identically as the rod in Carroll's system and then allowed to freely move in both systems, I believe it will move identically in both systems because the forces present in both systems are identical: the gravitation forces are clearly the same; and the reaction force from the outer circle is the same as the reaction force on the string (which produces a tensile force acting along the wire pointing to the other end of the string). Again, if the initial conditions are the same, then the only way for the systems to behave differently is if there is at least one force acting in one system that is different to the other system. Have I missed one? Krea 16:25, 30 August 2007 (UTC)

The systems are not identical; the difference between them is (close to) the essence of the paradox. The reaction force from the outer circle is missing in the pendulum case. Try it! You will see that, in the pendulum case, the rod will stay close to horizontal for a long time while falling downward with an acceleration close to g. (to see this, consider the behaviour at very small t). The angle of the rod will "lag behind" that of the string. In this case the torque is supplied by the string, because the rod is non-radial. Best wishes, Robinh 21:29, 30 August 2007 (UTC)

Hi. Well, I'd argue that the reaction force from the circles is present: it is now the reaction force from the string. The reaction force is a consequence of gravity: if the relevant gravitational component is the same, then the reaction will be the same. It doesn't matter of what the reaction force occurs.

About the rod, yes, I was being too vague again (damn it!). Yes, the rod will do that if the string has a mass (which way the rod will go depends on the relative masses of the string and the rod; if the string is lighter than the rod, the rod will actually move the other way around: the rod will go more vertical rather than stay horizontal). I neglected to mention that I was considering the string to be massless. If the string is massless, then the string will not undergo any gravitational motion: the only motion the string will undergo will be induced by a motion of the rod. Since any motion of the string is induced by a motion of the rod, a rotation of the rod will produce exactly the same rotation in the string.

Let me summarize: with a massless string, I still think the forces in both scenarios are equivalent. Yes, the reaction forces are being induced of different media, but they are equal in magnitude and direction. From the point of view of the physics, they are therefore equivalent.

P.S. Thanks for staying patient. I know we're arguing and counter arguing, but at least we're not going around in circles. If that were to happen, then this really would be pointless. Krea 00:05, 31 August 2007 (UTC)

Hi again. The two systems are not equivalent, although teasing out why they are not equivalent is very close to the essence of the paradox. Consider the pendulum (rod plus massless string) starting at three o'clock (I'll call this 90 degrees). Hold the rod by hand and at t=0, release it. What happens to the rod at very small times? Well, it has a bodily acceleration downwards of magnitude g. This is because gravity is the only vertical force acting on it. It also has zero horizontal acceleration if we assume that the string is inextensible. So far so good. Now, what is the initial angular acceleration of the rod? Well, zero because the only forces acting on the rod exert zero torque on it (taking moments about the centre of the rod). So, if we assume that the string stays straight (valid for small times, at the very least), then to first order in t the string's angle will be something like 90.00001 degrees and the rod will remain exactly (well, to first order in t) 90 degrees. So the rod's angle will "lag behind" the angle of the string. You gotta understand this before tackling the next bit. You must understand what "to first order in t" means (if you don't, I will be very happy to explain :-). I seriously don't want to be annoying here!

Now consider a rod that moves while constrained to move between the concentric circles. According to one view, it starts to descend: it moves from the 90.0000 degrees 3 o'clock) position to the 90.0001 degrees position. One corollary of this is that the rod picks up angular momentum about its centre. Where does this angular momentum come from? Well, that is the essence of the paradox because no force exerts a torque about the centre of the rod. Hope this helps, best wishes, Robinh 19:27, 2 September 2007 (UTC)

Partial resolution by non-rigidity

Hello. I think I have a solution, but it might depend on loosening one of the assumptions; I'm not sure.

When the rod starts at the 3-o'-clock position, the endpoints are situated at places where the tracks are vertical. When the rod is released, at time t=0, there is obviously nothing to rotate the rod. But what about at time t=δ small?

Without angular momentum, the rod has fallen a distance, call it Ε, and remained horizontal. The endpoints have also dropped by Ε, and since the tracks r1 and r2 are not entirely vertical, these endpoints have also undergone horizontal displacement. So with the tracks being concentric circles of unequal radius, these displacements ε1 and ε2 are also unequal, and the horizontal rod must have stretched.

(This is where I hope I made an unneeded alteration. Does the rod have infinite rigidity, or just arbitrarily great rigidity? If the latter, we can pick a correspondingly small δ and continue. I am less sure how to deal with the former case.)

The distorted rod is undergoing tension, which of course is along its length. The tracks, meanwhile, are exerting their own forces along their radii. The resolution comes, then, by recognizing that these three forces don't line up at time δ, so the tracks do in fact impart angular momentum. (As a side note, gravity seems not to be involved; just impart an impulse at t=0 and the same situation should emerge.) --Bryce Herdt, 76.25.100.146 (talk) 19:58, 22 January 2011 (UTC)