Verification of moment of inertia calculationby Yaridovich Tags: calculation, inertia, moment, verification 

#1
Jun1910, 03:59 PM

P: 6

I was looking at the moment of inertia list that they have on Wikipedia and noticed that the moment of inertia for a regular polygon was rather complicated. I did the calculation myself and found a significantly simpler result of
I = (m/6)(3+tan(pi/n)^2)*R^2: m is the mass of the polygon, n is the number of edges of the polygon, R is the length of the line segment from the center of the polygon to one of its edges, where the line segment is perpendicular to that edge. I just wanted to verify that this is correct; I can submit a proof of how I calculated this if need be. 



#2
Jun1910, 04:18 PM

PF Gold
P: 1,999

How about testing it out first for a few known polygons.If n is infinity you will have a thin solid disc for which the MI=(mR^2)/4.I think your equation gives (mR^2)/2
Whoops,we need to clarify this.If the axis of rotation is through the centre and at 90 degrees to the plane of the disc then the MI=(mR^2)/2.If you are considering the same axis then your equation seems to give the right answer.Try some other polygons. 



#3
Jun2010, 08:45 AM

P: 83

moment of inertia of square:
[tex]I_c=\frac{m(h^2+w^2)}{12}=\frac{m(2h^2)}{12}=\frac{m(\sqrt{2}R)^2}{6}=\ frac{mR^2}{3}[/tex] but your formula shows: [tex]I_c=\frac{2mR^2}{3}[/tex] 


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