2. 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: L = Iω becomes:
L1 = I1ω1, L2 = I2ω2, L3 = I3ω3.
the KE T = (½)ωIω becomes:
T = (½)I1(ω1)2 + (½)I2(ω2)2 + (½)I3(ω3)2

3. 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:
I I1, I2, I3  eigenvalues of I
ID = I I1, I2, I3  principal Components of I
0 0 I3
Directions x, y, z defined by the eigenvectors  principal axes of I

4. The matrix algebra method to diagonalize 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 = (1)
 The Eigenvalue Problem
Values of I which satisfy (1)  Eigenvalues (I1, I2, I3)
Vectors R which satisfy (1)  Eigenvectors x, y, z

5. 3 solutions to (2): Eigenvalues (I1, I2, I3)
(I - I1)R = (1)
 Ixx - I Ixy Ixz
Ixy Iyy - I Iyz = (2)
Ixz Iyz Izz - I
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.

6. n  αi +βj + γk Work out the details of I:
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+ 2Ixzρ1ρ (A)
(A)  I = I(ρ1,ρ2,ρ3)
Equation of a surface in “ρ” space  Inertial Ellipsoid

7. I = I1(ρ1)2 + I2(ρ2)2 + I3(ρ3)2 (B)
ρ = n(I)-½ = ρ1i + ρ2j + ρ3k
I = Ixx(ρ1)2 +Iyy(ρ2)2 +Izz(ρ3)2 + 2Ixyρ1 ρ2 + 2Iyzρ2ρ3 + 2Ixzρ1ρ (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) (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!

8. 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.

9. 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.

10. 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  Mb  I =
Symmetry:

11. 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:

12. 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.

13. Example 3 from Marion 1 0 0  ID = = (1/6)Mb2 0 1 0 0 0 1
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 = = (1/6)Mb
Or: ID = (1/6)Mb2 1

14. 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.

15.  Lagrangian similarly divides: L = Ltrans + Lrot
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

16. L = Ltrans + Lrot = Lc(qc,qc) + Lb(qb,qb)
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!

17.  (dL/dt)s = (dL/dt)b + ω  L (3)
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:

18. (dLi/dt) + ijkωjLk = Ni (4)
(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)

19. 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) = N (5a)
I2(dω2/dt) - ω3ω1(I3 -I1) = N (5b)
I3(dω3/dt) - ω1ω2(I1 -I2) = N (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)