# Moment of inertia of regular polygonal frames

The moment of inertia with respect to the orthogonal barycentral axis on the plane where it lies:

• a homogeneous triangular (equilateral) frame is equal $I_C = (1)\,\frac{M\,L^2}{6}$;
• a homogeneous quadrangular frame is equal $I_C = (2)\,\frac{M\,L^2}{6}$;
• a homogeneous pentagonal (regular) frame is equal $I_C = \left(2+\frac{3}{\sqrt{5}}\right)\frac{M\,L^2}{6}$;
• a homogeneous hexagonal (regular) frame is equal $I_C = (5)\,\frac{M\,L^2}{6}$;
• a homogeneous heptagon (regular) frame is equal $I_C = \dots \,$ boh;
• a homogeneous octagon (regular) frame is equal $I_C = \left(5+3\,\sqrt{2}\right) \frac{M\,L^2}{6}$;
• $\dots$

Is there a formula to write that moment of inertia for any regular polygon frame?

I emphasize the fact that these are frames, not laminates.

Thank you.

Thanks to the answers we received, for any regular $n$-sided polygon of side $L$:

• $\text{perimeter} = n\, L$;
• $\text{fixed_number} = \frac{1}{2\,\tan\left(\frac{\pi}{n}\right)}$;
• $\text{apothem} = \text{fixed_number}\cdot L$;
• $\text{area} = \, \frac{\text{perimeter}\cdot \text{aphotem}}{2}$;
• $\text{moment_inertia}_C = \frac{M\,L^2}{12} + M\,\text{apothem}^2$ (homogeneous frame of mass $M$);

and for completeness (and also for curiosity) I also add:

• $\text{moment_inertia}_C = \frac{\frac{M}{2}\,L^2}{12} + \frac{M}{2}\,\text{apothem}^2$ (homogeneous lamina of mass $M$).
• So what is $L$? – David G. Stork Sep 29 '17 at 22:53
• With frames, I clearly assuming the boundary of the respective polygonal lamina. L is the length of the side. Thank you. – TeM Sep 29 '17 at 23:11
• So the total circumference is $nL$? Please be more careful and thorough when posing questions. I'm not going to take time now to go back and fix my solution. – David G. Stork Sep 29 '17 at 23:14
• I tried to illustrate the problem. If anyone was able to give me a hand I would be grateful to him! – TeM Sep 29 '17 at 23:37
• And why is $C$ not the origin? What difference does the location make? None of your answers depend upon the center location. – David G. Stork Sep 29 '17 at 23:49

The moment of inertia of a bar around the perpendicular axis through the center is $\frac{mL^2}{12}$. In your case $m=\frac{M}{N}$. Using the parallel axis theorem, the moment of inertia of a bar with respect to an axis at distance d from the center is $$\frac{mL^2}{12}+md^2$$ so the total moment of inertia for your system is $$I_N= \frac{ML^2}{12}+Md_N^2$$ Now all you need to do is calculate what is the distance $d_N$ from the center of a regular polygon with $N$ sides to the side. In the case $N=3$, $d_3=\frac{L}{2\sqrt{3}}$, for $N=4$ $d_4=\frac{L}{2}$ and so on. You have $$\tan\frac{2\pi}{2N}=\frac{L/2}{d}$$

• Excellent, the formula collimation with the data calculated above, thank you! – TeM Sep 30 '17 at 13:17

To calculate these moments of inertia, we must simply integrate one side and multiply the result by n.

If we imagine C to be the origin, a side can be expressed as $x = \frac{L}{2\tan(\frac\pi n)}, -\frac L2 < y<\frac L2$. We can easily integrate to find moment of inertia of this segment:

$$I_C\cdot L = \int^{\frac L2}_{-\frac L2}\frac{M}n\cdot D(x, y)^2dy = \frac{M}n\int^{\frac L2}_{-\frac L2}\left(\sqrt[]{x^2+y^2}\right)^2dy = \frac{M}n\int^{\frac L2}_{-\frac L2}\left(\frac{L^2}{4\tan^2(\frac\pi n)} + y^2\right)dy\\ =\frac{M}n\left(\frac{L^2}{4\tan^2(\frac\pi n)}y + \frac13y^3\right)\biggr|^{\frac L2}_{-\frac L2} = \frac{M}n\cdot\left(\frac{L^3}{4\tan^2(\frac\pi n)} + \frac{L^3}{12}\right).$$

Simply multiplying by $n$ yields the desired moments:

$$I_C=M\left(\frac{L^2}{4\tan^2(\frac\pi n)} + \frac{L^2}{12}\right)$$

• Thank you too! I think there is an error: mass density is not M/L, but M/(nL); in this way the result collides with that of Andrei and in turn all the above particular cases (verifiably in literature) are embedded! – TeM Sep 30 '17 at 12:25