Tensors for Physics

402 Appendix: Exercises...

I≡

∮

v·dr=R^2 w

∫ 2 π

0

dφ= 2 πR^2 w.

The equalityS=Iis in accord with the Stokes law.

8.4 Moment of Inertia Tensor of a Half-Sphere(p.136)
A half-sphere with radiusaand constant mass density is considered. The orientation
is specified by the unit vectoru, pointing from the center of the sphere to the center
of mass of the half-sphere. For the present geometry, the moment of inertia tensor is
uniaxial and of the form

Θ‖uμuν+Θ⊥(δμν−uμuν).

Thez-axis is chosen parallel tou, just as in examples discussed in Sects.8.3.2and
8.3.3. First, moments of inertia are computed with respect to the geometric center of
the sphere. The pertaining moment of inertiaΘ‖effis

Θ‖eff=ρ 02 π 2

∫a

0

r^4 dr

∫ 1

0

ζ^2 dζ=

4 π

15

ρ 0 a^5 =

2

5

Ma^2 ,

whereM = ( 2 π/ 3 )ρ 0 a^3 is the mass of the half-sphere. As before,ζ =cosθ
is used. The momentΘeff⊥ is inferred from the mean moment of inertiaΘ ̄eff,via
Θ⊥eff=( 3 Θ ̄eff−Θeff‖ )/2. Equation (8.69) yields

3 Θ ̄eff= 2

∫

V

ρ(r)r^2 d^3 r=

4 π

5

ρ 0 a^5 =

6

5

Ma^2.

This impliesΘ⊥eff=Θ‖eff, i.e. the moment of inertia tensor is isotropic, when evalu-
ated with respect to the geometric center. According to the law of Steiner (8.66), the
moment of inertia tensor with respect to the center of mass is

Θμν≡Θcmμν=Θμνeff−M(R^2 δμν−RμRν)=

2

5

Ma^2 δμνMR^2 (δμν−uμuν),

withR=^38 ais the distance between the center of mass and the geometric center,
cf. (8.63). Thus one finds

Θ‖=

2

5

Ma^2 ,Θ⊥=Θ‖−

9

64

Ma^2 =

83

128

Θ‖.

As expected,Θ⊥is smaller thanΘ‖.