A thin uniform rod of mass M and length L has moment of inertia \tfrac{1}{12}ML^2 about its center. What is its moment of inertia about an axis perpendicular to the rod through one end?

\tfrac{1}{3}ML^2 Parallel-axis: I = I{cm} + Md^2 = \tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac{1}{3}ML^2.

Rotational Inertia

AP Physics 1· difficulty 4/5

M

Explanation

I = I_{cm} + Md^2 = \tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac{1}{3}ML^2

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