Re: computing the moment of inertia
- From: Duncan Muirhead <dmuir@xxxxxxxxx>
- Date: Fri, 30 Mar 2007 11:31:25 +0100
On Thu, 29 Mar 2007 10:06:27 -0700, bob wrote:
I was just wondering if anyone knows the best way to compute theYou might want to take a look at the Wikipedia article about the moment of
moment of inertia of a cube about a given axis.
Thank you.
inertia.
For a cube, it's all pretty simple:
First off suppose your cube is {x,y,z | -s/2<=x,y,z<=s/2}, (ie its
sides have length s) and is of constant density and mass M
The moment of inertia tensor J is then (1/6)*M*s*s times the identity
matrix.
If your axis goes through (0,0,0) and has direction u (a vector if
length 1) the moment of inertia about that axis is then u'*J*u,
(' denotes transpose)
so the moment of inertia is (1/6)*M*s*s.
If your axis goes through a point p and has direction u then the
moment of inertia is (1/6)*M*s*s + M*d*d, where d is the distance from
the centre of the cube to the axis, ie the length of the vector
p - (p.u)*u.
Finally if your cube has been rotated by a rotation matrix R and
then translated by a vector t, then we can change coordinates by undoing
the translation and then the rotation; in these coordinates the axis is
described by the point q = p-t, and the vector v = R'*u,
and the moment of inertia can be computed from the formula above, using
q instead of p and v instead of u.
Duncan
.
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