Re: Inertia tensor of tetrahedron

In article
<1165383509.500925.129160@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
,
Jthompson5213@xxxxxxxxx wrote:

Hi there,

I'm in the process of determining the inertia tensor for a
homogeneous tetrahedron with peaks:

A = (-a/2, 0, 0)
B = (0, a/2, 0)
C = (0, 0, 3^(1/2)*a/2)
D = (0, 3^(1/2)*a/2, 0)

With much algebra, I've come to find the inertia tensor to be (sorry
for the format):


2/3 0 0
{I} = 0 7/20 0
0 0 7/20

with the whole matrix multiplied by B, where B = M*a^2 with M being the
mass of the tetrahedron.

My concern is with the elements I12 = I13 = I21 = I31, and I32 = I23
(by symmetry) all being zero. Could anyone tell me if I've made a
mistake in my algebra (which is entirely possible), or is it true that
I11, I22, and I33 are simply the principal moments of inertia? Any
input is appreciated!

I did not check your diagonal elements. In general an
element of the inertia tensor is given by
1/M * int_{Body} x_i * x_j dm. The body specified above
has all mass concentrated on the coordinate axes,
therefore all forms x_i * x_j dm, i <> j are
identically zero.

--
Michael Press
.