alleryMoment of inertia of an annular disc (inner radius r1, outer r2) about perpendicular axis through center:
A(1/2)M(r2² - r1²)
B(1/2)M(r1² + r2²)
CM(r2² - r1²)
DM(r2² + r1²)
Answer & Solution
Correct answer: B. (1/2)M(r1² + r2²)
I_annular = (1/2)M(r1² + r2²).
When r1 → 0, this becomes (1/2)Mr2² (solid disc).
When r1 → r2, becomes Mr2² (thin ring).
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