by cmcraes
 P: 90 Consider two concentric circles of radius r1 and r2 as might be drawn on the face of a wall clock. Suppose a uniform rigid heavy rod of length | r2-r1 | is somehow constrained between these two circles so that one end of the rod remains on the inner circle and the other remains on the outer circle. Motion of the rod along these circles, acting as guides, is frictionless. The rod is held in the three o'clock position so that it is horizontal, then released. Now consider the angular momentum about the centre of the rod: After release, the rod falls. Being constrained, it must rotate as it moves. When it gets to a vertical six o'clock position, it has lost potential energy and, because the motion is frictionless, will have gained kinetic energy. It therefore possesses angular momentum. The reaction force on the rod from either circular guide is frictionless, so it must be directed along the rod; there can be no component of the reaction force perpendicular to the rod. Taking moments about the center of the rod, there can be no moment acting on the rod, so its angular momentum remains constant. Because the rod starts with zero angular momentum, it must continue to have zero angular momentum for all time. Is there any solution to this paradox?
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P: 3,660
 Quote by cmcraes Is there any solution to this paradox?
Yes.
Although the interaction of the rod ends and the two rings is frictionless, it does not follow that the reaction forces at the two ends have to be directed along the length of the rod. The constraints are holding the rod against the rings so (for example) when the rod is still in its initial position will resist rotating the rod by lifting the left end while the right end stays put. This can only happen if the constraints are pivots or other hinge-like devices, and these are can exert sidewise forces as well as longitudinal ones.
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PF Gold
P: 12,132
 Quote by Nugatory Yes. Although the interaction of the rod ends and the two rings is frictionless, it does not follow that the reaction forces at the two ends have to be directed along the length of the rod. The constraints are holding the rod against the rings so (for example) when the rod is still in its initial position will resist rotating the rod by lifting the left end while the right end stays put. This can only happen if the constraints are pivots or other hinge-like devices, and these are can exert sidewise forces as well as longitudinal ones.
Doesn't the OP imply that the contact points on the rod are a fixed length apart? If the rod starts of parallel with a radius of the circles then it will always point radially.

I drew a diagram and I think there is no paradox because the radial forces from the two circular tracks will not be equal. They will be ωr1 and ωr2, where r1 and r2 are the two radii. So you will have three forces on the rod, the weight force and the two radial forces - giving you a net tangential acceleration force. I think the solution should just 'fall out' as they say.