
ABCD is square of sides 'b' and four spheres each of mass 'M' and radius 'a' are placed at the four corners.
Let us calculate the moment of inertia of the system about side BC as axis.
M.I. of the sphere placed at corner A of the square ABCD. M.I. of the sphere about its diameter parallel to the axis BC = 2/2Ma2.
Applying parallel axis theorem, the M.I of the sphere at A about axis of BC = 2/5 Ma2 +Mb2.........(i)
Similarly the M.I. of the sphere placed at the corner D of the square about the axis of BC = 2/5 Ma2 +Mb2.........(ii)
M.I. of the sphere placed at the corner B of the square about the axis of BC
=2/5 Ma2........(iii)
Similarly the M.l. of the sphere placed at the corner C of the square about the axis of BC = 2/5 Ma2...........(iv)
= M.I. of the system as a whole = (i) + (ii) + (iii)+ (iv)
