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**arildno** · May18-04, 12:07 PM

The closed form expression of the moment of inertia is:
$LaTeX Code: I=\\frac{M_{P}}{6A_{P}}\\sum_{i=1}^{M}A_{T,i}(\\vec{T }_{i,1}^{2}+\\vec{T}_{i,1}\\cdot\\vec{T}_{i,2}+\\vec{T }_{i,1}\\cdot\\vec{T}_{i,3}+\\vec{T}_{i,2}\\cdot\\vec{T }_{i,3}+\\vec{T}_{i,2}^{2}+\\vec{T}_{i,3}^{2})$

where:
1. The polygon P (of mass $LaTeX Code: M_{P}$ )consists of M triangles $LaTeX Code: T_{i}$
2. $LaTeX Code: \\vec{T}_{i,s}$ is the position vector of the s'th vertex in the i'th triangle (measured from the rotation axis)
3. $LaTeX Code: A_{T,i}=\\frac{1}{2}||(\\vec{T}_{i,2}-\\vec{T}_{i,1})\\times(\\vec{T}_{i,3}-\\vec{T}_{i,1})||$ is the area of triangle $LaTeX Code: T_{i}$
4. $LaTeX Code: A_{P}=\\sum_{i=1}^{M}A_{T,i}$

May18-04, 12:07 PM	Last edited by arildno; May18-04 at 12:40 PM..
arildno arildno is Online: Posts: 10,153 Recognitions: PF Contributor Homework Helper Science Advisor	The closed form expression of the moment of inertia is: $LaTeX Code: I=\\frac{M_{P}}{6A_{P}}\\sum_{i=1}^{M}A_{T,i}(\\vec{T }_{i,1}^{2}+\\vec{T}_{i,1}\\cdot\\vec{T}_{i,2}+\\vec{T }_{i,1}\\cdot\\vec{T}_{i,3}+\\vec{T}_{i,2}\\cdot\\vec{T }_{i,3}+\\vec{T}_{i,2}^{2}+\\vec{T}_{i,3}^{2})$ where: 1. The polygon P (of mass $LaTeX Code: M_{P}$ )consists of M triangles $LaTeX Code: T_{i}$ 2. $LaTeX Code: \\vec{T}_{i,s}$ is the position vector of the s'th vertex in the i'th triangle (measured from the rotation axis) 3. $LaTeX Code: A_{T,i}=\\frac{1}{2}\|\|(\\vec{T}_{i,2}-\\vec{T}_{i,1})\\times(\\vec{T}_{i,3}-\\vec{T}_{i,1})\|\|$ is the area of triangle $LaTeX Code: T_{i}$ 4. $LaTeX Code: A_{P}=\\sum_{i=1}^{M}A_{T,i}$