Talk:Carroll's paradox

message: 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)
replies: text: 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)

message: :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)
thread_title:  Gravity 

message: Who is the Carroll who noted this paradox? Jefferson Anderson 18:43, 19 December 2006 (UTC)
replies: text: Yikes! Added. 192.75.48.150 19:39, 19 December 2006 (UTC)
thread_title:  Glaring omission 

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

message: 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)
thread_title: reversion

message: 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)
thread_title: reversion of a reversion

message: 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)
thread_title:  Partial resolution by non-rigidity