Two identical solid spheres each of mass 2 kg and radii 10 cm are fixed at the ends of a light rod. The separation between the centres of the spheres is 40 cm. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is __________ × 10⁻³ kg m²

Solution

Physics sir · 4 steps

Each sphere spins about its own centre and also swings around the middle, so add both parts.

1. Distance to the axis. The axis passes through the middle of the rod. Each centre is half the separation away: d = 20 cm = 0.2 m.
2. Own moment of inertia. About its own centre, a solid sphere has I_(cm) = 2/5mr² = 2/5 × 2 × (0.1)² = 8 × 10⁻³ kg m².

   I = 2(2/5mr² + md²)

3. Parallel axis theorem. Shift each sphere to the middle axis by adding md² = 2 × (0.2)² = 80 × 10⁻³ kg m². One sphere then gives 88 × 10⁻³ kg m². The light rod adds nothing.
4. So. For both spheres, I = 2 × 88 × 10⁻³ = 176 × 10⁻³ kg m², so the answer is 176.

The common slip: 160. Treating the spheres as point masses and dropping 2/5mr² gives 2 × 80 = 160, not 176.

The idea to keep. For a body whose centre is away from the axis, use the parallel axis theorem: its own I_(cm) plus md². Measure d from the axis, not between the two bodies.

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