Consider a long solid, rigid hexagonal prism like a common type of pencil. The mass of the prism is $M$ and is uniformly distributed. The length of each side of cross-sectional hexagon is $a$ . The moment of inertia $I$ of the hexagonal prism about its central axis is $I=\frac { 5 }{ 12 } M{ a }^{ 2 }$ Find the moment of inertia $I$ of the hexagonal prism about an edge of the prism.

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Details and Assumptions
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- $M=1 \text{ kg}$ , $a=1 m$

The answer is 1.416.

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$I'=I_{cm}+Md^{2}$ $I'=\frac{5}{12}Ma^{2} + Ma^{2}$ $I'=\frac{17}{12}Ma^2$ $I'=17/12$

PS. This is a very tiny part of that IPhO problem; if anybody wants to solve the full problem, you can find it at IPhO 1998 Problem 1 (Iceland).