There are four solid balls with their centres at the four comers of a square of side `a`. the mass of each sphere is `m` and radius is `r`. Find the moment of inertia of the system about (i) one of the sides of the square (ii) one of the diagonals of the square.

Explanation

a. Moment of inertia of the arrangement about the diagonal `AC`:
The moment of inertia of each of spheres `A` and `C` about their common diameter `AC=(2/5)ma^(2).` The moment of inerta of each of spheres `B` and `D` about an axis passing through their centres and parallel to `AC` is `(2/5)ma^(2)`. The distance between the axis and `AC` is `b/(sqrt(2))`, `b` being side of the square. therefore the moment of inertia of each spheres `B` and `D` about `AC` by the theorem of parallel axis is
`2/5 ma^(2)+m(b/sqrt(2))^(2)=2/5 ma^(2)-(mb^(2))/2`
Therefore, the moment of inertias of all the four spheres about diagonal `AC` is
`I=2(2/5 ma^(2))+2[2/5ma^(2)+(mb^(2))/2]`
b. The moment of inertia of each of spheres `A` and `D` about side `AD` is `(2/5) ma^(2)`.
The moment of inertia of each of sphere `B` and `C` about `AD` is `(2/5)ma^(2)+mb^(2)`.
Therefore the moment of inertia of the whole arrangement about side `AD` is
`2[2/5ma^(2)+(2/5ma^(2)+mb^(2))]=(2m)/5[4^(2)+5b^(2)]`