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_(B) ` about X' Y' = 2 m `(l)^(2)`
`I_(C)` about X' Y' = 3 m ` (l)^(2)`
Therefore the moment of inertia of the body about X'Y' is
` m(0)^(2) + 2 m(l)^(2) + 3 m (l)^(2) = 5 ml^(2)`