A solid sphere of radius 0.24 m has a dense spherical core of half the radius (i

Question

A solid sphere of radius 0.24 m has a dense spherical core of half the radius (i.e with radius 0.12 m). The core density is 23700 kg/m^3, and the density in the rest of the sphere is one-third that, i.e. 7900 kg/m^3.
If the core density had been exactly equal to the density of the rest of the sphere, i.e. 7900 kg/m^3, what would the moment of inertia of the sphere have been?

What is the moment of inertia of the actual sphere (i.e. with the dense core)?

Hint: Think of the sphere as a sum of a big sphere of uniform density, and a small sphere whose density is the excess core density.

I'm unsure as to how to get the mass to find I. The picture included is a smaller dark circle (core) within a circle

Explanation / Answer

Mass of the sphere = volume x density
   = ( 4 pi R3 / 3 ) (d)
R is radius and d is density of object.
Mass of sphere is uniform density of 7900 kg/m3 = 4 pi (0.24)3 7900 /3 = 457.2 Kg
Moment of inertia of sphere = 2 M R2 /5

Hence moment of intertia of sphere with unifor density of 7900 kg/m3
   = 2* 457.2*0.242 / 5 = 10.5 kg m2

Moment of inertia of core part with excess density, that is ( 23700 - 7900) = 15800 kg/m3
= 2* mass of core * r2 /5
= 2 ( 4 pi 0.123* 15800/3)* 0.122 / 5
= 0.7 kg m2

Moment of inertia of sphere with dense core =

moment of inertia of sphere with density 7900 + moment of inertia of core part with density 15800
= 10.5 + 0.7 = 11.2 kgm2

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