# Moment of Inertia

Four small spheres, each of which you can regard as a point of mass $$0.2$$Kg are arranged in a square $$0.4$$m on a side and connected by extremely light rods. Find the moment of inertia of the system about an axis through the centre of the square.
Discussion:

The length of each side of a square is $$0.4$$m. Thus, the perpendicular bisector is at $$0.2$$m. The length of the bisector dropped from the point of intersection of the diagonals is $$0.2$$m. Therefore, the distance of the vertices from the point of intersection of diagonals is $$r=\sqrt{(0.2)^2+(0.2^2)}=0.2828m$$
The moment of inertia $$I=MR^2= 4(0.2)(0.2828)=0.0640kg m^2$$