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A thin uniform rod of mass M and length L is bent at its center so that the two segments are perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through the point where the two segments meet.

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MechEngineer

Answer:

[tex]\frac{1}{12}ML^2[/tex]

Explanation:

The moments of the whole object is the sum of the moments of the 2 segments of rod at their ends of which length is L/2 and mass M/2:

[tex]I = 2I_{end} = 2\frac{1}{3}\frac{M}{2}\left(\frac{L}{2}\right)^2[/tex]

[tex]I = \frac{1}{3}M\frac{L^2}{4}[/tex]

[tex]I = \frac{1}{12}ML^2[/tex]

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