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Verification of moment of inertia calculation

by Yaridovich
Tags: calculation, inertia, moment, verification
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Yaridovich
#1
Jun19-10, 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.
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Dadface
#2
Jun19-10, 04:18 PM
PF Gold
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P: 2,009
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.
kntsy
#3
Jun20-10, 08:45 AM
P: 82
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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