Sect 5.4: Eigenvalues of I & Principal
Axis Transformation
• Definition of inertia tensor (continuous body):
Ijk  ∫Vρ(r)[r2δjk - xjxk]dV
– Clearly, Ijk is symmetric: Ijk = Ikj
 Out of the 9 elements Ijk only 6 are independent.
Ijk depend on the location of the origin of the body axes set &
on the orientation of the body axes with respect to the body.
• Symmetry  There exists a set of coordinates
 Principal Axes in which the tensor Ijk is diagonal with
3 Principal Values  I1, I2, I3. In this system, the angular
momentum:
the KE
L = Iω
becomes:
L1 = I1ω1, L2 = I2ω2, L3 = I3ω3.
T = (½)ωIω
becomes:
T = (½)I1(ω1)2 + (½)I2(ω2)2 + (½)I3(ω3)2
• Get principal axes set and principal values of tensor I by
diagonalizing I. That is, by finding eigenvalues (principal
values I1, I2, I3) & eigenvectors (defining principal axes).
• From Ch. 4, do this by solving determinant eigenvalue
problem or by a similarity transformation on I.
• Given inertia matrix I, principal axes & principal values can be
found by finding a suitable rotation matrix (finding a proper set of
Euler angles ,θ,ψ) A and performing the similarity
transformation: ID = AIA-1 = AIÃ such that ID is a
diagonal matrix. That is, want ID to have form:
I1 0 0
I1, I2, I3  eigenvalues of I
I D = 0 I2 0
I1, I2, I3  principal Components of I
0 0 I3
Directions x, y, z defined by the eigenvectors 
principal axes of I
• Once I is diagonalized, Principal Components (I1,
I2, I3) & principal Axes (x, y, z) are known.
– Then, can get I relative to any other axes set by another
similarity transformation: I = AI(A)-1 = A IÃ
– Also Parallel Axis Theorem might be used to shift rotation
axis.
• The matrix algebra method to diagonalize I
 Solve secular eqtn: (I - I1)R = 0
(1)
 The Eigenvalue Problem
Values of I which satisfy (1)  Eigenvalues (I1, I2, I3)
Vectors R which satisfy (1)  Eigenvectors x, y, z
(I - I1)R = 0

Ixx - I
Ixy
Ixz
Ixy Iyy - I
Iyz
Ixz
Iyz Izz - I
= 0
(1)
(2)
– Have used Ijk = Ikj .
• 3 solutions to (2): Eigenvalues (I1, I2, I3)
• Put these into (1) & get: Eigenvectors
Ri = x, y, z = Principal Axes.
• Often, can know principal axes by the object symmetry.
• Some Properties of the Eigenvalues (I1, I2, I3)
1. Can’t be < 0!
Ii > 0,
(i = 1,2,3)
2. If one Ii = 0, the body is vanishingly small in the
direction given by the corresponding eigenvector.
• Principal axes from geometry:
• Moment of inertia about the rotation axis n: I  nIn
• Body Cartesian axes unit vectors: i,j,k
Define:
n  αi +βj + γk
Work out the details of I:
 I = Ixxα2 +Iyyβ2 +Izzγ 2 +2Ixyαβ + 2Iyzβγ + 2Ixzαγ
Define the vector ρ  n/(I)½ . |ρ| = (I)-½
Write:
ρ  ρ1i + ρ2j + ρ3k

I = Ixx(ρ1)2 + Iyy(ρ2)2 + Izz(ρ3)2
+ 2Ixyρ1ρ2 + 2Iyzρ2ρ3+ 2Ixzρ1ρ3
(A)
(A)

I = I(ρ1,ρ2,ρ3)
Equation of a surface in “ρ” space  Inertial Ellipsoid
ρ = n(I)-½ = ρ1i + ρ2j + ρ3k
I = Ixx(ρ1)2 +Iyy(ρ2)2 +Izz(ρ3)2 + 2Ixyρ1 ρ2 + 2Iyzρ2ρ3 + 2Ixzρ1ρ3 (A)
Equation of a surface in “ρ” space  Inertial Ellipsoid
• Diagonalizing the inertia tensor I  Transforms to principal
axes where the moment of inertia I has the form:
I = I1(ρ1)2 + I2(ρ2)2 + I3(ρ3)2
(B)
The normal form for an ellipsoid!
 A geometric interpretation of the principal moments
of inertia (I1,I2,I3): They are exactly the lengths of the
axes of the inertial ellipsoid. If 2 Ij’s are equal, this
ellipsoid has 2 equal axes & this is an ellipsoid of
revolution. If all 3 are equal, this is a sphere!
• Define: The Radius of Gyration R0 in terms of the
total mass M & the moment of inertia I: I  M(R0)2
I is written as if all mass M were a distance R0
from the rotation axis.
• From the definition of the vector ρ = n(I)-½, we can write:
ρ = n(R0)-1(M)-½
 A radius vector to a point on the inertia ellipsoid is
inversely proportional to the radius of gyration about
the direction of that same vector.
Emphasize:
• The inertia tensor I & all associated quantities
(principal axes, principal moments, inertia ellipsoid, moment
of inertia, radius of gyration, etc.) are defined relative to
some fixed point in the body.
• If we shift this point to somewhere else, all of these in general
are changed. The parallel axis theorem can be used.
• The principal axis transformation which diagonalizes
I about an axis through CM will not necessarily
diagonalize it about another axis! It will be diagonal with
respect to both axes only if the shift is along a vector parallel
to one of the original principal axes & only if that axis passes
through the CM.
Example 1 from Marion
• Calculate the inertia tensor of a homogeneous cube of density
ρ, mass M, and side length b. Let one corner be at the origin,
and let the 3 adjacent edges lie along the coordinate axes (see
figure). (For this choice of coord axes, it should be obvious that
the origin does NOT lie at the CM!)
Ijk  ∫Vρ(r)[r2δjk - xjxk]dV

β  Mb2
Symmetry:
I=
Example 2 from Marion
Find the principal moments of inertia &
the principal axes for the same cube:
 Solve secular eqtn: (I - I1)R = 0

So:
Row manipulation:
• This results in:
• Or:
• Giving:
 Diagonal I =
To get the principal axes, substitute I1, I2, I3 into the
secular equation (I - Ij1)Rj = 0 (j = 1,2,3) & solve for Rj.
Find: R1 along cube diagonal. R2, R3  each other & R1.
Example 3 from Marion
• Calculate the inertia tensor of the
same cube in a coord system with
origin at the CM. (figure).
a = (½)b(1,1,1)
Again: Ijk  ∫Vρ(r)[r2δjk - xjxk]dV
Student exercise to show:
I11 = I22 = I33 = (1/6)Mb2, I12 = I21 = I13 = I31 = I23 = I32 = 0
 ID =
Or:
= (1/6)Mb2
ID = (1/6)Mb2 1
1 0 0
0 1 0
0 0 1
Sect 5.5: Solving Rigid Body Problems;
The Euler Equations
• We now have the tools to solve rigid body dynamics
problems. Usually assume holonomic constraints. If
not (like rolling friction) need to use special methods.
• Usually start by seeking a reference point in the body
such that the problem can be split into pure
translational + pure rotational parts.
• If one point in the body is fixed, then obviously all
that is needed is to treat the dynamics of the rotation
about that point.
• If one point is not fixed, it is most useful to choose
the reference point in the body to be the CM.
• We have already seen that, for case where the reference
point is the CM, the KE splits into KE of translation of
CM + KE of rotation about an axis through the CM:
• We had: T = Ttrans + Trot
–
–
–
–
1st term = Ttrans = (½)Mv2 Translational KE of the CM
2nd term = Trot = (½)∑imi(vi)2. Rotational KE about CM
We’ve written second term as: Trot = (½)ωIω
For n  unit vector along the rotation axis:
Trot = (½)ω2nIn  (½) Iω2
• So:
T = (½)Mv2 + (½)Iω2
• For all problems considered here, we can make a similar
division for the PE:
V = Vtrans + Vrot
 Lagrangian similarly divides: L = Ltrans + Lrot
• More generally (Goldstein notation):
L = Ltrans + Lrot = Lc(qc,qc) + Lb(qb,qb)
Lc = CM Lagrangian, qc,qc = generalized coordinates &
velocities of the CM.
Lb = Body Lagrangian, qb,qb = generalized coordinates &
velocities of body (rotation).
• Can use either Newtonian or Lagrangian methods, of course.
In either case, often convenient to work in the principal axes
system so that the rotational KE takes the simple form:
Trot = (½)I1(ω1)2 + (½)I2(ω2)2 + (½)I3(ω3)2
• The most convenient generalized coordinates to use are the
Euler angles: ,θ,ψ. They are cumbersome, but useable.
• If the motion is confined to 2 dimensions (the rotation axis is
fixed in direction), then only 1 angle describes motion!
• Follow Goldstein & start with the Newtonian approach to
describe the rotational motion about an axis through a fixed
point (like the CM):
• Consider either an inertial frame with the origin at a fixed
point in body or a space axes system with CM as the origin.
• In this case, a Ch. 1 result is: (dL/dt)s = N
(1)
 Newton’s 2nd Law (rotational motion): Time derivative of
the total angular momentum L (taken with respect to the space
axes) is equal to the total external torque N.
• Make use of the Ch. 4 result relating the time derivatives in the
space & body axes (angular velocity ω):
(d/dt)s = (d/dt)b + ω 
(2)

(dL/dt)s = (dL/dt)b + ω  L
(3)
• Equating (1) & (3) & dropping the “body” subscript (which is
understood in what follows) gives:
(dL/dt) + ω  L = N
(4)
• (4): Newtonian eqtn of motion relative to body axes.
• In terms of inertia tensor, I, in general, L = Iω needs to be
substituted into (4).
• For ith component, (4) becomes:
(dLi/dt) + ijkωjLk = Ni
(4)
– Levi-Civita density ijk  0, any 2 indices equal
ijk  1, if i, j, k even permutation of 1,2,3
ijk  -1, if i, j, k odd permutation of 1,2,3
122 = 313 = 211 = 0 , etc. 123= 231= 312=1, 132= 213= 321= -1
• Take the body axes to be the principal axes of I, (relative to
the reference point)  Li = Iiωi
(i = 1,2,3)
• (4) becomes (no sum on i)
Ii(dωi/dt) + ijkωjωkIk = Ni
(5)
• Newtonian eqtns of motion relative to the body axes are:
Ii(dωi/dt) + ijkωjωkIk = Ni
(5)
• Write component by component:
I1(dω1/dt) - ω2ω3(I2 -I3) = N1
(5a)
I2(dω2/dt) - ω3ω1(I3 -I1) = N2
(5b)
I3(dω3/dt) - ω1ω2(I1 -I2) = N3
(5c)
 Euler’s Equations of Motion for a Rigid
Body with one point fixed.
• Could derive these from Lagrange’s Eqtns, treating torques as
generalized forces corresponding to Euler angles as generalized
coords.
• Special Case: I1 = I2  I3: In this case, a torque in the “1-2”
plane (N3 = 0) causes a change in ω1 & ω2 while leaving ω3 =
constant. A very important special case! (Sect. 5.7)