USAPhO 1999, Полуфинал — Задача 4. A4

• a. Swinging in the plane of the bob.
• b. Swinging perpendicular to the plane of the bob with the rod horizontal.
• c. $\frac{T_{\max}}{T_{\min}} = \sqrt{4/3}$

The only torque acting on the system is due to the gravitational force.

For small angles

We will use the parallel axis theorem to find the moment of inertia of the system. Swinging in the plane of the bob, pivot on rim

Swinging perpendicular to the bob, we expect the extreme cases to be rod horizontal and rod vertical

vertical: $I_{rod} = \tfrac{4}{3}mR^2$ horizontal: $I_{rod} = mR^2$

For the ring:

vertical:

horizontal:

The maximum period occurs when ω is smallest or $I$ is largest. Since $I_{tot} \ge I_{vtot} \ge I_{htot}$ the maximum period occurs when it swings in the plane of the bob. The minimum period occurs when it swings perpendicular to the plane of the bob and the wire is horizontal.