A solid sphere of radius 0.13 m has a dense spherical core of half the radius (i

Question

A solid sphere of radius 0.13 m has a dense spherical core of half the radius (i.e with radius 0.065 m). The core density is 25200 kg/ms, and the density in the rest of the sphere is one-third that, i.e. 8400 kg/m, If the core density had been exactly equal to the density of the rest of the sphere, i.e. 8400 kg/m, what would the moment of inertia of the sphere have been? kg m2 What is the moment of inertia of the actual sphere (i.e. with the dense core)? kg m 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 Submit

Explanation / Answer

a) Total mass of the sphere with uniform density,

M = density*volume

= 8400*(4/3)*pi*0.13^3

= 77.3 kg

so, I = (2/5)*M*R^2

= (2/5)*77.3*0.13^2

= 0.522 kg.m^2

b) mass of core, m1 = density*volume

= 25200*(4/3)*pi*0.065^3

= 28.99 kg

mass of the core uniform density, m1' = density*volume

= 8400*(4/3)*pi*0.065^3

= 9.66 kg

Total moment of inertia = (2/5)*m1*R'^2 + (2/5)*M*R^2 - (2/5)*m1'*R'^2

= (2/5)*28.99*0.065^2 + 0.522 - (2/5)*9.66*0.065^2

= 0.555 kg.m^2

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