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Moment of inertia of a (homogenous) triangle in R3

Posted 2 years ago

In a forgoing post

https://community.wolfram.com/groups/-/m/t/2332921

the moment of inertia of a triangle in R2 rotating around a point was calculated.

Here a method is given to do this in R3, the triangle rotating around an axis thru x4 and with direction b

(* square of distance of a point p to an axis thru x and direction b *)

Clear[dist]
dist2[p_, x_, b_] := Module[{aa, fL, vB},
  aa = p - x;
  fL = b.aa/(b.b);
  vB = p - x - b fL;
  vB.vB]

(* Moment of Inertia of a triangle given by x1, x2, x3 with respect to axis thru x4 and direction b  *)

Clear[fMI]
fMI[x1_, x2_, x3_, x4_, b_] := Module[{d2, d3, fe},
  d2 = x2 - x1;
  d3 = x3 - x1;
  fe = Cross[d2, d3]; 
  fe = Sqrt[fe.fe];  (*surface-element of Triangle*)
  Integrate[ fe  dist2[x1 + u d2 + v d3, x4, b], {v, 0, 1}, {u, 0, 1 - v}]
  ]

Check with the result of forgoing post

fMI[{4, 1, 0}, {1, 3, 0}, {2, 7, 0}, {.5, 5, 9}, {0, 0, 3}]
POSTED BY: Hans Dolhaine
Posted 2 years ago

Hans showed very nice how to calculate the moment of inertia. If one is only interested in the result one can use the function provided by Mathematica which has a similiar interface compared to the function of Hans. Main difference is that not a list of coordinates has to be provided but rather a object in form of a region (here: triangle).

With[{pointOfRotation = {0.5, 5, 9}, direction = {0, 0, 3}}, 
 MomentOfInertia[Triangle[{{4, 1, 0}, {1, 3, 0}, {2, 7, 0}}], 
  pointOfRotation, direction]]
POSTED BY: Michael Helmle
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