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Moments of Inertia Calculations for Different Shapes, Study notes of Physics

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.

What you will learn

  • How does the moment of inertia of a circular disc differ from that of a rod?
  • What is the moment of inertia of a rod rotating around an axis perpendicular to it at one end?
  • What is the moment of inertia of a sphere rotating around any axis?

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2021/2022

Uploaded on 09/12/2022

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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 x3-direction, so
that the rod is in the (x1, x2)-plane. The angular ve-
locity is thus ~
= (0,0,Ω). The kinetic energy is evi-
dently T=1
2Iabab=1
2I33 2. We can then write
y
x
z
I33 =Zρ(~
ξ·~
ξξ3ξ3) = Zρξ2
1+ξ2
2
=Z123ρ ξ2
1+Z123ρ ξ2
2
=ρL3
32+ρ2
3
23
L
(ρL2)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=ρL2is 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 ~
= (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.
1
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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

I 33 =

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

L^3

(^2) + ρ 2 3

∆ L

≈ (ρL∆^2 ) L

2 3

= M L

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

I 33 =

d^3 ξ ρ (ξ 12 + ξ^22 ) =

∫ R

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

−R

dz ρ r^4 = π 2

∫ R

−R

dz ρ (R^2 − z^2 )^2

= πρ 2

[∫

dz R^4 − 2 R^2

dz z^2 +

dz z^4

]

= πρ 2

[

2 R^5 − 2 R^2

R^3 + 2 R

5 5

]

= 8 πρ 15

R^5 =

4 πR^3 3

ρ 2 5

R^2

M R^2 (4)

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 =

3 M R

x

z

y