Angular Momentum Paradox


by cmcraes
Tags: angular, momentum, paradox
cmcraes
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#1
Apr28-13, 01:41 PM
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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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Nugatory
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Apr28-13, 02:06 PM
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Quote Quote by cmcraes View Post
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.
sophiecentaur
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Apr28-13, 05:53 PM
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Quote Quote by Nugatory View Post
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.

AlephZero
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Apr28-13, 08:47 PM
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Angular Momentum Paradox


The paradox is that this is a limiting case that only exists as an idealized theorietical model.

First, note the constraints at the ends of the rod must be able to apply a force both radially inwards or outwards. Otherwise, the rod would just start to fall without rotation, losing contact with the inside circle, and there is no paradox about what would happen after that.

Next, consider a rod whcih is slightly longer than r2-r1, When it is at the "3 o clock" position, it will be at a slight angle to the horizontal. The two constraint forces can apply a moment to the rod becauise they are not exactly aligned with the rod, and again there is no paradox describing the motion.

Now imagine the rod is gradually shortened towards length r2-r1. By considering the KE of the rod compared with the work done by its weight as it falls, the motion of the rod from 3 o clock to 6 o clock will remain constant, to a good approximation. Therefore the moment required to rotate the rod is also unchanged. As the rod becomes more closely aligned to the radial direction, the constraiint forces on the ends will increase to provide the same moment.

So in your limiting situation, you have a finite (non-zero) moment caused by an "infinitely large" force times a "zero" distance.

If you make the model more realistic by assuming the rod is flexible not rigid, again the pardox disappears, because the rod can stretch slightly and twist away from the radial direction, and the constraints can then apply a moment to it.

Of course if you model the system using energy methods (or Lagrangian mechanics, which amounts to the same thing) you can get an answer without bothering about where the moment comes from!

If you tried to make this device in practice, the rod would not move at all. It would stay at 3 o clock, wedged at a very small angle to the radial, because of the inevitable small amount of friction in the system.
cmcraes
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Apr28-13, 08:59 PM
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I agree; An apparent resolution of this paradox is that the physical situation cannot occur. To maintain the rod in a radial position the circles have to exert an infinite force. In real life it would not be possible to construct guides that do not exert a significant reaction force perpendicular to the rod.
sophiecentaur
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Apr29-13, 07:00 AM
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If the rod had mass and everything were ideally rigid then there would presumably be a place where you could exert a force that could allow each end of the rod to accelerate tangentially at the right rate as the point of application accelerated. If that's acceptable then could not the rod have nonuniform mass distribution and achieve the same thing?
AlephZero
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Apr29-13, 02:26 PM
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You could certainly find the angular acceleration required for any position of the rod (using conservation of energy) and hence find where the center of mass of the rod should be give the correct proportion of linear and angular acceleration.

But without doing the math, it's not obvious (to me) whether or not a fixed position for the CM would work for every position of the rod as it moved around the circle.

(Doing the math is straightforward enough, but I'm not sufficiently interested in knowing the answer to do it myself!)


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