Particles of masses 1, 2 and 3kg respectively are arranged at the corners of an equilateral triangle as shown in figure. Obtain the moment of inertia of the system about an axis passing through the centre of mass of the system and perpendicular to its plane. The side of the triangle is 1m.

Question

G1 Physics · Accepted Answer

For point masses about their common centre of mass, I = [Σ over all pairs (mᵢmⱼrᵢⱼ²)]/(total mass). Every pair is separated by 1 m, total mass = 1+2+3 = 6 kg, and the pair-product sum is 1·2+1·3+2·3 = 11. Therefore I = 11/6 kg·m². Hence option C.