Four particles with a mass of 1 kg, 2 kg, 3 kg and 4 kg are situated at the corners of a square with side 1 m (as shown in the figure). The moment of inertia of the system, about an axis passing through the point O and perpendicular to the plane of the square, is ______________ kg m².

Solution

Physics sir · 3 steps

O is the centre, so all four masses sit at the same distance: add the masses, then multiply by r².

1. Find the distance. O is where the diagonals cross. Each diagonal is √2 m long, so every corner is half a diagonal from O: r = √2/2 = 1/√2 m. That gives r² = 1/2 m².
2. Sum over particles. For point masses, I = Σ mᵢ rᵢ². All four rᵢ are equal, so r² comes out as a common factor.

   I = (1 + 2 + 3 + 4)×1/2 = 10 × 1/2 = 5 kg m²

3. So. The moment of inertia about the axis through O, perpendicular to the square, is 5 kg m².

The common slip: 10. Taking r as the side length, 1 m, instead of half the diagonal gives 10 × 1 = 10, not 5.

The idea to keep. For point masses, I = Σ m r², where r is the perpendicular distance from the axis. Find that distance from the geometry first. The side length is not the distance unless the axis lies on a corner.

Answer from the NTA final answer key. Solution by PrepWiser.