Four particles of masses `4 kg, 2 kg, 3kg and 5 kg` are respectively located at the four corners `A, B, C, D` of a square of side `1 m`. Calculate the moment of inertia of the system about
(i) the axis passing through point of intersection of the diagonals and perpendicualr to the plane of the square. (ii) side `AB` (ii) diagonal `BD`.

Explanation

In Fig.
`AC = BD = sqrt(1^(2) + 1^(2)) = sqrt(2)m`
`:. AO = BO = CO = DO = (sqrt(2))/(2) = (1)/(sqrt(2))m`
(i) About the axis passing through `O and _|_` to plane of square,
`I = 4 ((1)/(sqrt(2)))^(2) + 2((1)/(sqrt(2)))^(2) +3 ((1)/(sqrt(2)))^(2) + 5((1)/(sqrt(2)))^(2)`
`= 14 xx (1)/(2) = 7 kgm^(2)`
(ii) About `AB`
`I = 4 xx (0)^(2) + 2(0)^(2) + 3(1)^(2) + 5(1)^(2) = 8 kg m^(2)`
About diagonal `BD`
`I = 4 ((1)/(sqrt(2)))^(2) + 2(0)^(2) + 3((1)/(sqrt(2)))^(2) + 5(0)^(2)`
`= 3.5 kg m^(2)`