Three particles, each of mass m are situated at the vertices of an equilateral triangle ABC of side L as shown in the figure. Find the moment of inertia of the system about the line AX perpendicular to AB in the plane of ABC

Explanation

Perpendicular distance of A from AX=0
Perpendicular distance B from AX=L
Perpendicular distance of DC from AX=L/2
Thus, the moment of inertia of the particle at A=0, of the particle at `B=mL^2` and of the particle at `C=m(L/2)^2`. The moment of inertia of the three particle system about AX is
`0+mL^2+m(L/2)^2=(5mL^2)/4`
Note that the particle on the axis do not contribute to the moment of inertia.