1
$\begingroup$

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?

enter image description here

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 $M$).
$\endgroup$
  • $\begingroup$ So what is $L$? $\endgroup$ – David G. Stork Sep 29 '17 at 22:53
  • $\begingroup$ With frames, I clearly assuming the boundary of the respective polygonal lamina. L is the length of the side. Thank you. $\endgroup$ – TeM Sep 29 '17 at 23:11
  • 1
    $\begingroup$ 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. $\endgroup$ – David G. Stork Sep 29 '17 at 23:14
  • $\begingroup$ I tried to illustrate the problem. If anyone was able to give me a hand I would be grateful to him! $\endgroup$ – TeM Sep 29 '17 at 23:37
  • $\begingroup$ And why is $C$ not the origin? What difference does the location make? None of your answers depend upon the center location. $\endgroup$ – David G. Stork Sep 29 '17 at 23:49
1
$\begingroup$

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 $$\tan\frac{2\pi}{2N}=\frac{L/2}{d}$$

$\endgroup$
  • $\begingroup$ Excellent, the formula collimation with the data calculated above, thank you! $\endgroup$ – TeM Sep 30 '17 at 13:17
1
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – TeM Sep 30 '17 at 12:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.