A uniform rod of mass M and length L is pivoted at one end. What is its moment of inertia about an axis perpendicular to the rod and passing through the pivot?

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TestHub · Accepted Answer

The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is $I_{CM} = \frac{1}{12} M L^{2}$ .

Using the parallel axis theorem, the moment of inertia about an axis passing through one end (at a distance $d = L /2$ from the center of mass) is $I = I_{CM} + M d^{2}$ . Therefore, $I = \frac{1}{12} M L^{2} + M (\frac{L}{2})^{2} = \frac{1}{12} M L^{2} + \frac{1}{4} M L^{2} = \frac{1}{12} M L^{2} + \frac{3}{12} M L^{2} = \frac{4}{12} M L^{2} = \frac{1}{3} M L^{2}$ .

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