There are four solid balls with their centres at the four corners 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 one of the sides 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)]`