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

Posted 1 month ago
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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[[1]] + (xx[[2]] - xx[[1]]) u + (xx[[3]] - xx[[1]]) 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]}]

@Lavanya Reddimasi was notified of this answer by email. Thank you.

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