Finding the Period of a Physical Pendulum Attached to a Solid Sphere

Thread starter jboone88
Start date Feb 19, 2006
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Hard Pendulum

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To determine the period of a physical pendulum consisting of a solid sphere attached to a pivot, one must consider the moment of inertia and the center of mass. The period of a simple pendulum with the same period can be expressed in terms of the radius R of the sphere. The formula for the period T of a physical pendulum is T = 2π√(I/mgh), where I is the moment of inertia, m is mass, g is gravitational acceleration, and h is the distance from the pivot to the center of mass. For a solid sphere, the effective length of the simple pendulum can be derived from these parameters. The final answer will relate the period of the physical pendulum to its radius R.

Feb 19, 2006

jboone88

I can't figure this one out:

A point of the surface of a solid sphere (radius = R) is attached directly to a pivot on the ceiling. The sphere swings back and forth as a physical pendulum with a small amplitude. What is the length of a simple pendulum that has the same period as this physical pendulum? Give answer in terms of R.

Thanks for the help

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Feb 19, 2006

Hootenanny

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Well, how far have you got?

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