Three light rods, each of length 2e, are joined together to form a triangle. Three particles A, B, C of masses m, 2m, 3m are fixed to the vertices of the triangle. Find the moment of inertia of the resulting body about
an axis passing through A and the midpoint of BC.

Explanation

`I_A` about X' Y' ` = m(0)^2`
`I_B` about X' Y' = `2m (l)^2`
`I_C` about X' Y' `= 3m (l)^2`
Therefore the moment of inertia of the body about X' Y' is
`m(0)^2 + 2m(l)^2 + 3m(l)^2 =5 ml^2`