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Calculations for the moments of inertia of various shapes when rotating around their respective axes. The shapes considered are a rod, a circular disc, a sphere, and a paraboloidal solid. The moments of inertia are determined using the integral of mass times radius squared, with respect to the axis of rotation.
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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 x 3 -direction, so that the rod is in the (x 1 , x 2 )-plane. The angular ve- locity is thus Ω = (0~ , 0 , Ω). The kinetic energy is evi- dently T = 12 IabΩaΩb = 12 I 33 Ω^2. We can then write
y
x
z
Ω
ρ(ξ~ · ξ~ − ξ 3 ξ 3 ) =
ρ
ξ^21 + ξ 22
dξ 1 dξ 2 dξ 3 ρ ξ^21 +
dξ 1 dξ 2 dξ 3 ρ ξ 22
= ρ
(^2) + ρ 2 3
≈ (ρL∆^2 ) L
2 3
2 3
The range of integration is from zero to L for ξ 1 and −∆/ 2 to ∆/ 2 for ξ 2 and ξ 3. In the last line we have used the fact that ∆ L to discard the term proportional to ∆^4. Also M = ρL∆^2 is the mass of the rod. Thus the relevant moment of inertia is
I 33 = M L
2 3
a) A circular disc rotating around its axis
In this case, we can again take the axis of the disc to be along the x 3 -axis, with Ω = (0~ , 0 , Ω). Once again the only relevant component for the moment of in- ertia is I 33. Let ∆ denote the thickness of the disc and let R be its radius. In this case, it is simplest to use cylindrical coordinates because of the obvious symmetry. Thus
x
y
z
R
d^3 ξ ρ (ξ 12 + ξ^22 ) =
0
r^2 dr dϕdz ρ
= ρ R
4 4 2 π ∆ = ρ (πR^2 ∆) R
2 2 = 12 M R^2 (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.
1
a) A sphere rotating around any axis
We can think of the sphere as a stack of discs of ra- dius r and thickness dz. Let R be the radius of the sphere. We also have z^2 + r^2 = R^2. We take the for- mula for I 33 for a disc and integrate over z, from −R to R, to get
I 33 = π 2
−R
dz ρ r^4 = π 2
−R
dz ρ (R^2 − z^2 )^2
= πρ 2
dz R^4 − 2 R^2
dz z^2 +
dz z^4
= πρ 2
5 5
= 8 πρ 15
4 πR^3 3
ρ 2 5
Ω
r
a) A paraboloidal solid rotating around its axis of symmetry
Again, we can consider this as a stack of discs of ra- dius r. The relation between r and z is z = λr^2. This gives the paraboloidal shape for some λ > 0. For radius R, the height is h = λR^2. Thus
I 33 = πρ 2
∫ (^) h
0
dz r^4 = πρ 2 λ^2
∫ (^) h
0
dz z^2 = πρ 2 λ^2
h^3 3
The mass of the solid is given by
M =
dz πr^2 ρ = πρλ
∫ (^) h
0
dz z = πρ 2 λ h^2 (6)
Using this relation
I 33 = M h 2 λ =
x
z
y