A thin uniform rod of mass $M$ and length $L$ rotates about an axis perpendicular to the rod and passing through one of its ends. Its moment of inertia is:
A$\dfrac{ML^2}{12}$
B$\dfrac{ML^2}{3}$
C$\dfrac{ML^2}{4}$
D$\dfrac{ML^2}{2}$
Answer & Solution
Correct answer: B. $\dfrac{ML^2}{3}$
Using the parallel-axis theorem: $I_{\text{end}} = I_{\text{cm}} + M d^2 = \dfrac{ML^2}{12} + M\left(\dfrac{L}{2}\right)^2 = \dfrac{ML^2}{12} + \dfrac{ML^2}{4} = \dfrac{ML^2}{3}$.
Option A is the centre-axis value — the classic mistake of forgetting to shift the axis.
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