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(adsbygoogle = window.adsbygoogle || []).push({});

| 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?

**Physics Forums - The Fusion of Science and Community**

# Angular Momentum Paradox

Have something to add?

- Similar discussions for: Angular Momentum Paradox

Loading...

**Physics Forums - The Fusion of Science and Community**