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I've derived equations for 2d polygon's moment of inertia using Green's Theorem (constant density \rho)

$$I_y = \frac{\rho}{12}\sum_{i=0}^{i=N-1} ( x_i^2 + x_i x_{i+1} + x_{i+1}^2 ) ( x_i y_{i+1} - x_{i+1} y_i )$$

$$I_x = \frac{\rho}{12}\sum_{i=0}^{i=N-1} ( y_i^2 + y_i y_{i+1} + y_{i+1}^2 ) ( x_{i+1} y_i - x_i y_{i+1} )$$

And I'm trying to add them up for calculating $I_0 = I_x + I_y$.

$$I_0 = \frac{\rho}{12}\sum_{i=0}^{i=N-1} ( x_i^2 - y_i^2 + x_i x_{i+1} - y_i y_{i+1} + x_{i+1}^2 - y_{i+1}^2 ) ( x_i y_{i+1} - x_{i+1} y_i )$$

But I found different(?) equation for $I_0$ on the internet. and many people says below equation is correct.

$$I_0 = \frac{\rho}{6} \frac{ \sum_{i=0}^{i=N-1} ( x_i^2 + y_i^2 + x_i x_{i+1} + y_i y_{i+1} + x_{i+1}^2 + y_{i+1}^2 ) ( x_i y_{i+1} - x_{i+1} y_i ) }{ \sum_{i=0}^{i=N-1} ( x_i y_{i+1} - x_{i+1} y_i ) }$$

So I'm confusing now. I think my equations for $I_x$ and $I_y$ is correct. But how am I gonna calculate for $I_0$ (moment of inertia with respect to origin axis). I couldn't prove both equations are equal.

Could you help me out please ?

(This post has been cross-posted at http://math.stackexchange.com/questions/59470/calculating-moment-of-inertia-in-2d-planar-polygon)

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1  
It's good style to tell people that you're cross-posting (math.stackexchange.com/questions/59470/…); else efforts will be unnecessarily duplicated. –  joriki Aug 24 '11 at 15:21
    
Thanks for pointing out that. I've fix my post. –  juhlnet Aug 24 '11 at 15:57
2  
I think you forgot to normalize by the area. Also note that $I_0$ should be rotation invariant, in particular, invariant by the rotation $(x,y)\mapsto(y,−x)$ (yours is not, it changes sign) –  Pietro Majer Aug 24 '11 at 18:30

2 Answers 2

up vote 1 down vote accepted

Sorry for my mistake. both equations was slightly incorrect. Let me write correct equations

$$I_y = \frac{\rho}{12}\sum_{i=0}^{i=N-1} ( x_i^2 + x_i x_{i+1} + x_{i+1}^2 ) ( x_i y_{i+1} - x_{i+1} y_i )$$

$$I_x = \frac{\rho}{12}\sum_{i=0}^{i=N-1} ( y_i^2 + y_i y_{i+1} + y_{i+1}^2 ) ( x_i y_{i+1} - x_{i+1} y_i )$$

$$I_0 = \frac{\rho}{12}\sum_{i=0}^{i=N-1} ( x_i^2 + y_i^2 + x_i x_{i+1} + y_i y_{i+1} + x_{i+1}^2 + y_{i+1}^2 ) ( x_i y_{i+1} - x_{i+1} y_i )$$

and

$$I_0 = \frac{m}{6} \frac{ \sum_{i=0}^{i=N-1} ( x_i^2 + y_i^2 + x_i x_{i+1} + y_i y_{i+1} + x_{i+1}^2 + y_{i+1}^2 ) ( x_i y_{i+1} - x_{i+1} y_i ) }{ \sum_{i=0}^{i=N-1} ( x_i y_{i+1} - x_{i+1} y_i ) }$$

Note that latter equation changed mass density term($\rho$) to mass(m).

Both equations are equal.

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Hi guys, it might be worth mentioning w.r.t. which axis this is expressed and the direction of the axis. Let me also share this link from wikipedia that reports a similar expression: http://en.wikipedia.org/wiki/List_of_moments_of_inertia

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