alleryThe moment of inertia of a uniform thin rod of mass M and length L about an axis perpendicular to the rod through its centre is:
AM L² / 12
BM L² / 3
CM L² / 6
DM L²
Answer & Solution
Correct answer: A. M L² / 12
1. Integrate I = ∫ r² dm with linear density λ = M/L, from −L/2 to L/2.
2. I = ∫_(−L/2)^(L/2) x² · (M/L) dx = (M/L) · [x³/3] from −L/2 to L/2.
3. = (M/L) · 2 · (L/2)³ / 3 = M L² / 12.
4. About one end the value is M L² / 3 (by parallel-axis theorem).
_Source: syllabus Physics Ch 6 §6.9 (rod about CM)_
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