#2015_14 From an old Physics Olympiad!

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

Details and Assumptions :

  • M = 1 kg M=1 \text{ kg} , a = 1 m a=1 m
More questions .

The answer is 1.416.

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2 solutions

Potsawee Manakul
Aug 18, 2015

I = I c m + M d 2 I'=I_{cm}+Md^{2} I = 5 12 M a 2 + M a 2 I'=\frac{5}{12}Ma^{2} + Ma^{2} I = 17 12 M a 2 I'=\frac{17}{12}Ma^2 I = 17 / 12 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).

Utkarsh Tiwari
Jun 30, 2015

why have you named the problem as ipho problem it is 100 times diluted then the original. anandhu raj

Well,the root of problem is there.

Anandhu Raj - 5 years, 11 months ago

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