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

[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

 

[Hootenanny]

Well, how far have you got?

 

[kyle8921]

!

The period of a physical pendulum is given by the equation T = 2π√(I/mgh), where T is the period, I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and h is the distance from the pivot point to the center of mass.

In this case, the solid sphere acts as a point mass at its center of mass, so the moment of inertia can be calculated as I = 2/5mR^2, where m is the mass of the sphere and R is the radius.

The distance h is equal to R, since the pivot point is directly attached to a point on the surface of the sphere. Therefore, the period of this physical pendulum is T = 2π√(2/5Rg).

To find the length of a simple pendulum with the same period, we can use the equation T = 2π√(L/g), where L is the length of the pendulum. Setting this equal to the period of the physical pendulum, we get 2π√(2/5Rg) = 2π√(L/g).

Solving for L, we get L = (2/5)R, which is the length of the simple pendulum that has the same period as the physical pendulum attached to the solid sphere.

Therefore, the length of the simple pendulum is directly proportional to the radius of the sphere. As the sphere's radius increases, so does the length of the simple pendulum needed to have the same period.