8.3 Volume Integrals, Gauss 135
Fig. 8.15Brick stone with
sidesa,b,c. The coordinate
axes coincide with the
principal axes of the moment
of inertia tensor
a biaxial, uniaxial and spherical symmetry, respectively, of the mass density. For a
constant density inside a solid body, its shape determines the symmetry.
As a simple example, a brick stone with constant mass densityρ 0 and with edges
a,b,c, is considered, cf. Fig.8.15. The center of mass is in the middle, the principal
axes go through it and they are perpendicular to the side planes of the brick. The
volume element is dxdydz, the integration goes over the intervals[−a/ 2 ,a/ 2 ],
[−b/ 2 ,b/ 2 ],[−c/ 2 ,c/ 2 ].ThemassisM=ρ 0 abc. The moment of inertiaΘ(^3 )is
associated with the rotation about thez-axis. Then the square of the distance from
the axis isr⊥^2 =x^2 +y^2. The integral (8.67) leads to
Θ(^3 )=ρ 01
12
(a^3 b+ab^3 )c=1
12
M(a^2 +b^2 ),which is the mass times the square of the length of the diagonal of the side perpen-
dicular to the principal axis. The two other principal moments are
Θ(^1 )=
1
12
M(b^2 +c^2 ), Θ(^2 )=1
12
M(a^2 +c^2 ).In general, a brick stone is an asymmetric top. For, e.g.a=b, it becomes a symmetric
top. Then the moment of inertia tensor is, as in (5.20),
Θμν=Θ‖uμuν+Θ⊥(δμν−uμuν), (8.68)withΘ‖=Θ(^3 ),Θ⊥=Θ(^1 )=Θ(^2 ), anduis a unit vector parallel to the symmetry
axis.
A cube corresponds toa=b=cwhich implies three equal moments of inertia.
So the cube is a spherical top with an isotropic moment of inertia tensor
Θμν=Θδμν,Θ=1
6
Ma^2.