### Demos: 1S-01 Ring Pendulum

A large metal ring has a small support attached, the latter being placed over a knife edge so that the ring can oscilate about this edge as a pendulum. A simple pendulum is suspended from the same point and it is shown that the periods match when the length of the simple pendulum is equal to 2R. (The simple pendulum is shown separated in the diagram at the right for clarity.)

The period of the ring (compound pendulum) can be shown to be

If we assume that d R, then I is (using the parallel-axis theorem):

I = MR2 + MR2

Therefore

The period of a simple pendulum is

so that the equivalent length of the simple pendulum is

Directions: The simple pendulum is set up so that the length of the string may be varied. Adjust the length until the two periods match.

Suggestions for Presentation: Students often believe that the length of a pendulum is the distance from the support to the center of gravity. Suggest that since the center of gravity of the ring is at its geometric center, the ball of the simple pendulum should be placed at this point to achieve a common period. Try it and show that the periods are very much different. Then go through the calculations above and show that the length of the SP is actually 2R. Adjust the pendulum to this length and show that the periods are now identical. (Note: Although d R is okay for the I calculation, the length of the pendulum should be set to R itself. Because the support is above the ring, the pendulum ball will be slightly above the bottom of the ring.)

Applications:

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Last Updated: May 9, 2016 11:44 AM