Question

Four spheres, each of radius $a$ and mass $m$, are placed with their centres on four corners of a square of side $b$. Calculate the moment of inertia of the arrangement about any (a) diagonal of the square and (b) any side of the square

Answer 1

(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 $AC=\left(2/5\right)m{a}^2$. The moment of inertia of each of spheres $B$ and $D$ about an axis passing through their centres and parallel to $AC$ is $\left(2/5\right)m{a}^2$. 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

$\frac{2}{5}m{a}^2+m{\left(\frac{b}{\sqrt{2}}\right)}^2=\frac{2}{5}m{a}^2+\frac{m{b}^2}{2}$

Therefore, the moment of inertia of all the four spheres about diagonal $AC$ is

$I=2\left(\frac{2}{5}m{a}^2\right)+2[\frac{2}{5}m{a}^2+\frac{m{b}^2}{2}]$

(b) The moment of inertia of each of spheres $A$ and $D$ about side $AD$ is $\left(2/5\right)m{a}^2$

The moment of inertia of each of spheres $B$ and $C$ about $AD$ is $\left(2/5\right)m{a}^2+m{b}^2$

Therefore, the moment of inertia of the whole arrangement about side $AD$ is

$2[\frac{2}{5}m{a}^2+(\frac{2}{5}m{a}^2+\frac{m{b}^2}{2})]=\frac{2m}{5}(4a^2+5b^2)$