# Angular momentum of a two body system

This question comes from the Kleppner and Kolenkow textbook.

A ring of mass M and radius R lies on its side on a frictionless table. It is pivoted to the table at its rim. A bug of mass m walks around the ring with speed v, starting at the pivot. What is the rotational velocity of the ring when the bug is (a) halfway around and (b) back at the pivot.

The answer key states that angular momentum for the system should be conserved but this doesn't make sense to me. Doesn't a gravitational force act on the bug that results in a net external torque about the pivot? Why is angular momentum conserved for this system? Thanks

• please add a sketch of the setup. If the ring lies in a horizontal plane and rotates around a vertical axis, gravity is equilibrated by the reaction of the horizontal surface and thus no net external moment acts on the system Commented Dec 23, 2022 at 19:11
• I surmise the intention of that exercise question is as follows: the plane of the ring is perpendicular to the direction of gravity. K&K introduce a bit of extra complexity by not making the center of the ring the axis of rotation (of the ring). Instead some form of axial bearing is secured to a point along the perimeter. The ring pivots around that point, with the plane of the ring maintained perpendicular to the direction of gravity Commented Dec 23, 2022 at 19:13