2. When Can We Find Principal Axes?
• We can always write down the cubic equation that one
must solve to determine the principal moments
• But if we want to interpret these as physically meaningful
quantities, the roots of that cubic have to be real
– Recall that in general, cubics can have two complex roots
• Fortunately, we’re not in the general case here
• The inertia tensor is both real and symmetric – in
particular, it satisfies:
• Matrices that satisfy this restriction are called Hermitian
• For such matrices, the principal moments can always be
found, and they are always real (see proof in text)
*
ij ij
I I
=
This mathematics will come up again in Quantum Mechanics
Principal Moments ↔ Eigenvalues
Principal Axes ↔ Eigenfunctions
3. Inertia Tensor Under Translation
• We’ve already seen how the inertia tensor transforms
under a rotation of the coordinate system
• Now we’ll see how it behaves under translation – that is,
keeping the direction of the the coordinate axes fixed, but
changing the origin
• In particular, we’ll compare I for a coordinate system at
the center of mass of an object with that for a different
origin:
x1
x2
x3
O
X1
X2
X3
Q
rCM
O is at the center of
mass; Q can be
anywhere
4. • In the X reference frame (the one with origin at Q), the
position of any point in the rigid body can be written as:
which means the inertia tensor becomes:
α
α = +
CM
R r
r Position in center-of-
mass frame
( )
( ) ( )( )
( )
( )
(
)
( )
( )
2
, ,
2
, , ,
2
, , , , , ,
2 2
, , , ,
, , , , , , , ,
,
2
, , , ,
2
2
ij ij k i j
k
ij k CM k i CM i j CM j
k
ij k CM k k CM k
k
i j CM i j i CM j CM i CM j
ij CM k k
ij k i j
k
CM k CM
k
I m X X X
m x x x x x x
m x x x x
x x x x x x x x
m x x x
m x x x
x
α α α α
α
α α α α
α
α α α
α
α α α α
α
α α
α
α α
α
δ
δ
δ
δ
δ
= −
= + − + +
= + +
− − − −
=
+ + −
−
( )
, , , , ,
i j i CM j CM i CM j
x x x x x
α α
α
− −
Inertia tensor about the center of mass
5. • So the inertia tensor about Q is:
• Each term in the second line has a factor that looks like:
• For a general reference frame, this quantity equals the mass of
the system times the position of the ith component of the center
of mass
– But the xi and in the center of mass frame, so all these terms are
zero
• Therefore we have the result (Steiner’s Parallel Axis Theorem):
( )
( )
2
, , , ,
, , , , , ,
2
ij CM ij ij CM k CM i CM j
k
ij CM k k CM i j i CM j
k
I I x x x m
m x x x x x x
α
α
α α α α
α
δ
δ
= + −
+ − −
,i
m x
α α
α
( )
2
, , , ,
ij CM ij ij CM k CM i CM j
k
I I M x x x
δ
= + −
6. Eulerian Angles
• In analyzing the motion of a rigid body, it is most
convenient to use a reference frame that is fixed to the
body
– For example, a symmetry axis of the body could be the z
axis in this “body frame”
– But that’s usually not an inertial frame
• So we need to define a translation between some inertial
“fixed” frame and the body frame
• Three angles are needed to specify how one coordinate
system is rotated with respect to another (in three
dimensions)
– These angles determine the elements of the rotation matrix λ
λ
λ
λ
7. • There are many ways the three angles can be specified
• One convenient choice is the Eulerian Angles, in which the
3D rotation is the result of three 1D rotations:
1. Rotate CCW by φ around axis
2. Rotate CCW by θ around axis
3. Rotate CCW by ψ around axis
3 3
x x
′ ′′
=
1
x′
2
x′
φ
θ
1
x
ψ
1 2 3
( , , )
x x x
′ ′ ′
1 2 3
( , , )
x x x
′′ ′′ ′′
1 2 3
( , , )
x x x
′′′ ′′′ ′′′
2
x′′
1
x′′
1
x′
2
x′
3 3
x x
′ ′′
=
2
x′′′
3
x′′′
1 1
x x
′′ ′′′
=
3 3
x x
′′′=
1
x′
3
x′
2
x′
2
x
Line of
nodes
1 2 3
( , , )
x x x
3
x′
1
x′′
3
x′′′
8. • Note that in each rotation, one axis (from the previous
rotation) is held fixed. That means the rotation matrices
have the form:
• The complete rotation is given by:
• i.e., the elements of λ
λ
λ
λ are obtained by multiplying the three
individual rotation matrices (see Equation 11.99 in the text
for their values)
cos sin 0 1 0 0
sin cos 0 ; 0 cos sin
0 0 1 0 sin cos
cos sin 0
sin cos 0
0 0 1
φ φ
φ φ θ θ
θ θ
ψ ψ
ψ ψ
= − =
−
= −
′ ′
= ≡
x x x
9. Angular Velocity in the Body Frame
• It’s going to be easiest to find the equations of motion for
the object in the body frame
• In general, we can write:
• To write the components of this vector in the body frame
we need to do some geometry:
= + +
1
x
ψ
3
x
1
x′
3
x′
2
x′
2
x
φ
θ
• is along the x3 direction:
• is in the x1-x2 plane:
• has components in all three
directions:
ψ
= 3
e
2
cos sin
θ ψ θ ψ
= −
1
e e
2 3
sin sin sin cos cos
φ θ ψ φ θ ψ φ θ
= + +
1
e e e
10. • Adding up these vectors, we find that the components of ω
ω
ω
ω
in the body frame are:
1
2
3
sin sin cos
sin cos sin
cos
ω φ θ ψ θ ψ
ω φ θ ψ θ ψ
ω φ θ ψ
= +
= −
= +
11. Euler’s Equations for Rigid Body
Motion
• Now that we’ve defined the geometry, we can set up the
Lagrangian for rigid body motion
• To make things easy, we’ll first consider the case where no
external forces act on the body, so:
• In that case, we can put the origin of both the fixed
(meaning inertial) and body frames at the center of mass
• Furthermore, we can cleverly choose the body frame to
coincide with the principal axes of the body, so that:
L T
=
2
1
2
i i
i
I
T ω
=
Principal moment of inertia
associated with the ith axis
12. • Since the Eulerian angles fully specify the orientation of
the rigid body in the inertial frame, we can take them to be
our generalized coordinates
• The equation of motion for ψ is:
which becomes (using the chain rule):
• One piece that enters everywhere is which is simply:
0
T d T
dt
ψ ψ
∂ ∂
− =
∂ ∂
0
i i
i i
i i
T d T
dt
ω ω
ω ψ ω ψ
∂ ∂
∂ ∂
− =
∂ ∂ ∂ ∂
,
i
T
ω
∂
∂
i i
i
T
I ω
ω
∂
=
∂
13. • The other relations we need are:
• Putting these back into the Lagrange equation gives:
1
2
2
1
3
sin cos sin
sin sin cos
0
ω
φ θ ψ θ ψ ω
ψ
ω
φ θ ψ θ ψ ω
ψ
ω
ψ
∂
= − =
∂
∂
= − = −
∂
∂
=
∂
1
2
3
0
0
1
ω
ψ
ω
ψ
ω
ψ
∂
=
∂
∂
=
∂
∂
=
∂
( )
( )
1 1 2 2 2 1 3 3
1 2 1 2 3 3
0
0
d
I I I
dt
I I I
ω ω ω ω ω
ω ω ω
+ − − =
− − =
14. • The choice of which axis to call x3 was completely
arbitrary
– That means we can find similar relations for the other
components of the angular velocity
( )
( )
( )
1 2 1 2 3 3
3 1 3 1 2 2
2 3 2 3 1 1
0
0
0
I I I
I I I
I I I
ω ω ω
ω ω ω
ω ω ω
− − =
− − =
− − =
Euler’s Equations
for force-free motion
