Some sample calculations of the moment of inertia
a) A rod rotating around an axis perpendicular to it at one end
We consider a rod of length L, and thickness in the
two transverse directions as ∆, with ∆L. We take
the axis of rotation to be along the x3-direction, so
that the rod is in the (x1, x2)-plane. The angular ve-
locity is thus ~
Ω = (0,0,Ω). The kinetic energy is evi-
dently T=1
2IabΩaΩb=1
2I33 Ω2. We can then write
y
x
z
Ω
I33 =Zρ(~
ξ·~
ξ−ξ3ξ3) = Zρξ2
1+ξ2
2
=Zdξ1dξ2dξ3ρ ξ2
1+Zdξ1dξ2dξ3ρ ξ2
2
=ρL3
3∆2+ρ2
3∆
23
∆L
≈(ρL∆2)L2
3=ML2
3(1)
The range of integration is from zero to Lfor ξ1and −∆/2to ∆/2for ξ2and ξ3. In the last line
we have used the fact that ∆Lto discard the term proportional to ∆4. Also M=ρL∆2is the
mass of the rod. Thus the relevant moment of inertia is
I33 =ML2
3(2)
a) A circular disc rotating around its axis
In this case, we can again take the axis of the disc to
be along the x3-axis, with ~
Ω = (0,0,Ω). Once again
the only relevant component for the moment of in-
ertia is I33. Let ∆denote the thickness of the disc
and let Rbe its radius. In this case, it is simplest to
use cylindrical coordinates because of the obvious
symmetry. Thus
x
y
z
R
I33 =Zd3ξ ρ (ξ2
1+ξ2
2) = ZR
0
r2dr dϕdz ρ
=ρR4
42π∆ = ρ(πR2∆) R2
2
=1
2MR2(3)
Notice that, in this case, we are not taking ∆to be very small, so the result also applies to a
cylinder of height ∆rotating around its own axis.
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