Problem  Inertia ellipsoid of a right circular cone
The height of the cone is
h
and its base diameter is 2
a
. The cone is oriented
so that its symmetry axis lies along the
z
axis of a Cartesian frame whose origin
is at the vertex of the cone. In this orientation, there are only three nonzero
elements of the inertia tensor, the diagonal elements.
All of the o
f
diagonal
elements are zero. In other words, these
xyz
axes are the principal axes. We
will verify this below.
For the inertia ellipsoid of the right circular cone to
be that of a sphere, all of the principal moments must be equal. This means
that our task becomes one simply of
f
nding general expressions for the three
principal moments in terms of
h
and
a
andtheneva
luat
ingtherat
io
a/h
that
makes them all equal.
Start with the general expression for the
αβ
component of the inertia tensor
(
α
,
β
=
x
,
y
,
z
)
I
αβ
=
Z
ρ
(
r
)(
r
2
δ
αβ
−
x
α
x
β
)
d
3
r.
(1)
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 Fall '10
 Wilemski
 mechanics, Cartesian Coordinate System, Angular Momentum, Inertia, Moment Of Inertia, Coordinate system, Poinsot's ellipsoid

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