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Solid Cube Problem
(a) Inertia tensor in COM frame
The cube is located with a vertex at the origin of a Cartesian coordinate
system whose
x
,
y
and
z
axes lie along edges of the cube. The edges of the cube
are of length
a
.
x
z
y
For this coordinate system, the inertia tensor has the structure (given in
class notes)
I
=
b
12
⎛
⎝
8
−
3
−
3
−
38
−
3
−
3
−
⎞
⎠
,
(1)
where
b
=
Ma
2
and
M
is the mass of the cube.
The next step is to
f
nd the principal moments with respect to the center of
mass. In the given coordinate system, the center of mass is at
(
a/
2
,a/
2
2)
.
We use the Parallel Axis Theorem in the tensor form
I
=
I
0
+
I
R
(2)
and construct the inertia tensor
I
R
for the center of mass vector
R
in the
original coordinate system with unit basis vectors denoted by
x
,
y
,and
z
. Here,
obviously,
R
is
R
=(
a/
2)(
x
+
y
+
z
)
.
(3)
From the de
f
nition,
I
R
=
M
[
R
2
1
−
RR
]
,
(4)
we obtain the explicit form
I
R
=(
2
/
4)[3
1
−
(
x
+
y
+
z
)(
x
+
y
+
z
)]
,
(5)
after noting that
R
2
=3
a
2
/
4
, and substituting for
R
with Eq.(3) and for the
unit tensor
1
=
xx
+
yy
+
zz
.
(6)
1
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View Full DocumentWe can write Eq.(5) in a matrix form similar to Eq.(1) as
I
R
=
b
4
⎛
⎝
2
−
1
−
1
−
12
−
1
−
1
−
⎞
⎠
.
(7)
By the parallel axis theorem, the inertia tensor
I
0
we require is
I
0
=
I
−
I
R
,
(8)
and with Eqs.(1) and (7), we see that in the center of mass coordinate system
the inertia tensor is automatically diagonalized
I
0
=
b
6
⎛
⎝
100
010
001
⎞
⎠
.
(9)
This tells us that when the original coordinate axes are translated to the center
of mass, while maintaining their original orientation, they serve as principal
axes.
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 Fall '10
 Wilemski
 mechanics, Inertia

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