# Will variable but symmetrical density spheres which are otherwise identical swing at the same rate?

If you have a pendulum with the same length and different spheres on the end will the period be the same if the only variation is the distribution of mass from the center of the sphere? This is to say the radius and mass of each sphere is identical but one may be solid, one may be hollow, and another may be composed of many heavier masses suspended around the center in a symmetrical manner (e.g. in a dodecahedral or icosahedral configuration.)

• So the center of mass remains fixed, hence the torque from gravity is the same (at a given tilt) for all configurations--but is the moment of inertia the same for each configuration? – JEB Feb 13 '18 at 4:32
• @JEB If I'm understanding it correctly the rotational inertia would be different between a solid and a hollow sphere with the same radius and mass, but I'm not sure how to calculate what impact this would have on the swing of a pendulum (for say a 10mm diameter sphere on a foot-long string.) I'm guessing the effect would be negligible, but would like an exact formula to follow for calculating it because my objective is to build a device which has a single optical sensor at the bottom of the pendulum which counts the period and spits out the exact inertia of the mass. – CoryG Feb 13 '18 at 13:50
• @JEB I like to experiment but suck with the math so would like to get this down to an algorithm that has to be written once and never looked at again (the sphere will all be more or less symmetrical with dodecahedral or icosahedral arrangements of weights suspended around a fixed radius, which I'm pretty sure can be treated as shells of the average mass for the volume since they're platonic solids.) At most there would be 5 layers of shells (e.g. filler, dodecahedral+filler, dodecahedral+icosahedral+filler, icosahedral+filler, filler - based on the geometry they can be packed most densely.) – CoryG Feb 13 '18 at 13:54
• @JEB On the really simple side they would be a solid weight in the middle surrounded by a filler material (e.g. lead for the former and plastic for the latter) and there may also be arrangements that are purely dodecahedral or purely icosahedral, resulting in a filler-mass-filler arrangement around the centerpoint. – CoryG Feb 13 '18 at 13:56
• for small angle displacement, the torque is $\tau = mgL\theta$, which causes motion via: $\tau = I \ddot{\theta}$. (Hint: set $\theta(t) = A\cos{\omega t}$ and apply the equations). – JEB Feb 13 '18 at 19:13