# Moment of inertia of a triangle in R2 with respect to a point

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 A couple of hours ago someone asked how to calculate the moment of inertia of a (homogenous) triangle with respect to a point of rotation. I think it could be done like this xx = {{4, 1}, {1, 3}, {2, 7}}; (*start*) rr = {.5, 5}; (*point of rotation*) a = .522; (* angle of rotation , to move the triangle around*) (*rotated triangle*) xx = # + rr & /@ ({{Cos[a], -Sin[a]}, {Sin[a], Cos[a]}}.(# - rr) & /@ xx); (* point ( element R2 ) in triangle-cooridantes *) \[Xi]1 = xx[] + (xx[] - xx[]) u + (xx[] - xx[]) v; (*cofactor for volume-element in triangle-coordinates*) jj = Abs[Det[D[\[Xi]1, {{u, v}}]]] (*moment of inertia of homogenous triangle with respect to rr *) theta = Integrate[(\[Xi]1 - rr).(\[Xi]1 - rr) jj, {v, 0, 1}, {u, 0,1 - v}] ParametricPlot[\[Xi]1, {v, 0, 1}, {u, 0, 1 - v}, AxesOrigin -> {0, 0}, Epilog -> {Red, PointSize[.025], Point /@ xx, Blue, Point[rr]}] Answer Answer