1. Oh, it’s your standard “boy meets girl, boy loses girl,
boy invents a new deposition technique for ultra-thin
film semiconductors, boy gets girl back” story!

2. Ch 5: Rigid Body Eqtns of Motion
Ch 4: Introduced all kinematical tools to analyze rigid body motion.
Euler angles: ,θ,ψ: A useful set of 3 generalized coordinates for describing the body orientation.
Orthogonal transformation A & associated matrix properties & algebra: A powerful, elegant math formalism for the description of rigid body motion.
Ch 5: Uses these tools & techniques to get general eqtns of motion for rigid body in convenient form  The Euler Dynamical Equations.
Also: Applications of Euler Equations.

3. Sect 5.1: Angular Momentum & KE
Chasles’ Theorem: The most general displacement of a rigid body is a rotation about some axis plus a translation.
 It ought to be possible to divide the problem into 2 separate problems: 1. Translation, 2. Rotation.
Obviously if one point is fixed, this separation is trivially true.
Also possible for general problem in some cases.
Ch. 4 discussion: 6 generalized coordinates are needed to describe rigid body motion: 3 Cartesian coordinates of a fixed point in the body + 3 Euler angles describing the body rotation about an axis through the fixed point.
In general: Equations of motion for all 6 coordinates do not separate into 3 eqtns for translation + 3 for rotation!!

4. Most common special case where fixed point in rigid body  Center of Mass (CM):
Can use some Ch. 1 results for angular momentum & kinetic energy which simplify the problem & separate equations of motion for translational & rotational coordinates into 2 sets of 3, so translation & rotation can be treated separately!

5. A Ch. 1 result for the angular momentum of a system of particles:
O = arbitrary origin (fixed axes), R = CM position with respect to O. V = CM velocity, M = CM mass.
ri´ = position of particle i with respect to CM (body axes)
pi´ = mivi´ = momentum of same particle in this system.
A Ch. 1 result for the angular
momentum of a system of
particles:
L = R  MV + ∑i[ri´  pi´] (1)
The total angular momentum about point O = the angular momentum of the motion of the CM about O + the angular momentum of the motion of particles about the CM

6. L = R  MV + ∑i[ri´  pi´] (1)
Total angular momentum about O = angular momentum of the CM about O + angular momentum of particles about the CM
(1)  In general, L depends on the origin
O, through vector R. Only if the CM is at rest
with respect to O, will first term in (1)
vanish. Then & only then will L be indep
of the point of reference. Then & only then
L = angular momentum about the CM. If we choose the origin of
the body axes to be the CM, then clearly, as far as L is
concerned, we can treat the translation & rotation of the body
separately. Then, the 1st term in (1) depends only on (X,Y,Z) &
the 2nd term depends on (,θ,ψ) : L = R  MV + L´(,θ,ψ)

7. Ch. 1 result for KE of a system of particles:
O = arbitrary origin (fixed axes), R = CM position with respect to O. V = CM velocity, M = CM mass.
ri´ = position of particle i with respect to CM (body axes)
vi´ = velocity of same particle in this system.
Ch. 1 result for KE of a system of
particles:
T = (½)MV2 + (½)∑imi(vi´)2 (2)
The total Kinetic Energy of a many particle system is equal to the Kinetic Energy of the CM plus the Kinetic Energy of motion about the CM.

8. T = (½)MV2 + (½)∑i mi(vi´)2 (2)
Total KE with respect to O = KE of the CM+ KE of motion about the CM.
(2)  In general, T depends on origin
O, through vector V. Only if the CM is at
rest with respect to O, will first term in (2)
vanish. Then & only then will T be indep
of point of reference. Then & only then
T = KE about the CM. If choose origin of body axes to be the
CM, then clearly, as far as T is concerned, we can treat
the translation of body & rotation of body the separately. Then,
the 1st term in (2) depends only on (X,Y,Z), 2nd term depends on
(,θ,ψ): T = (½)MV2 + T´(,θ,ψ)

9. We can make this separation for the angular momentum L & the KE T
We can make this separation for the angular momentum L & the KE T. To do dynamics, using say, the Lagrangian method, we would like to make this same kind of separation in Lagrangian L = T -V
 Obviously, we need to be able to separate the PE V in the same way.
We don’t have a theorem which does this in general for V, like we do for L & T! However, from experience, in a large number of problems (special cases) of practical interest, such a separation is possible.
PE in a uniform gravitational field, PE of a magnetic dipole in uniform magnetic field, others,...
In such cases, we can write: L = L (X,Y,Z) + L´(,θ,ψ)

10. In cases where: L = L (X,Y,Z) + L´(,θ,ψ)
 Lagrange’s Eqtns of motion for translation (X,Y,Z) obviously separate from Lagrange’s Eqtns of motion for rotation (,θ,ψ) & we can treat the translation & rotation as independent.
Now, work on expressions for angular momentum L & KE T for motion about some fixed point in a rigid body. Work on 2nd terms in previous expressions for these. Called L´ & T´ in Eqtns (1) & (2) above. Drop prime notation.
Make repeated use of the Ch. 4 result relating the time derivatives in space & body axes (angular velocity ω):
(d/dt)s = (d/dt)r + ω  (I)

11. Follow text & write using summation convention: L = mi[ri  vi] (1)
Goldstein proves that angular velocity ω is indep of choice of the origin of body axes. See pages for details. Intuitively obvious by definition of rigid body?
Angular momentum L of rigid body about a fixed point in the body. Dropping the primes of before on positions, momenta, & velocities: L = ∑i[ri  pi], pi = mivi
Follow text & write using summation convention:
L = mi[ri  vi] (1)
ri = fixed with respect to body axes (definition of rigid!)
 The only contribution to vi is from the body rotation.
 Use (I) vi = (dri/dt)s = (dri/dt)r + ω  ri = 0 + ω  ri
 (1) becomes: L = mi [ri  (ω  ri)] (2)

12. Use the vector identity for double cross product:
 L = mi[ω(ri)2 - ri(riω)] (3)
Look at the x, y, & z components:
 Lx = ωxmi[(ri)2 - (xi)2] - ωymixiyi - ωzmixizi
Ly = ωymi[(ri)2 - (yi)2] - ωxmiyixi - ωzmiyizi
Lx = ωzmi[(ri)2 - (zi)2] - ωymiziyi - ωxmizixi (4)
(4)  Each component of L = Linear combination of components of angular velocity ω. Or: The angular momentum vector is related to the angular velocity vector by a linear transformation!
 Define inertia tensor Ijk (summation convention!) :
Lj  Ijkωk (5)
Ijk also called the moment of inertia coefficients

13. Inertia tensor Ijk (summation convention!) : Lj  Ijkωk
Or: Lx = Ixxωx + Ixyωy + Ixzωz
Ly = Iyxωx + Iyyωy + Iyzωz
Lz = Izxωx + Izyωy + Izzωz
Diagonal elements  Moment of inertia coefficients:
Ixx = mi[(ri)2 - (xi)2], Iyy = mi[(ri)2 - (yi)2], Izz = mi[(ri)2 - (zi)2]
Off-diagonal elements  Products of inertia:
Ixy = Iyx = - mixiyi , Ixz = Izx = - mixizi , Iyz = Izy = - miyizi

14. Compact form: Ijk  mi[(ri)2δjk - (ri)j(ri)k]
Inertia tensor Ijk (summation convention, i = particle label, j, k = Cartesian x,y,z = 1,2,3 labels!) :
Lj  Ijkωk
Compact form: Ijk  mi[(ri)2δjk - (ri)j(ri)k]
All of this is in notation for discrete systems of particles.
Continuous bodies: Sum over particles  integral over volume V. Define: ρ(r)  Mass density at position r
 Ijk  ∫Vρ(r)[r2δjk - xjxk]dV
Or: Ixx = ∫Vρ(r)[r2 - x2]dV, Iyy = ∫Vρ(r)[r2 - y2]dV
Izz = ∫Vρ(r)[r2 - z2]dV
Ixy = Iyx = ∫Vρ(r)[r2 - xy]dV, Ixz = Izx = ∫Vρ(r)[r2 - xz]dV
Iyz = Izy = ∫Vρ(r)[r2 - yz]dV

15. Recall: The coordinate system is the BODY axis system (the primes from the last chapter are dropped).
Since the body is rigid, all elements Ijk are constants in time. Clearly, they DO depend on the origin of coordinates (different if taken about the CM or elsewhere!).
We can summarize Lj = Ijkωk in an even more compact form: L  Iω where I  Inertia Matrix
This is in the form of the orthogonal transformation of Ch. 4. Clearly we need to take the interpretation that I acts on the vector ω to produce L, rather than the interpretation of I acting on the coordinate system.

16. Sect 5.2: Tensors
More pure math discussion. Brief! Hopefully, a review!
I  Tensor of 2nd rank. Now, a general discussion.
In Cartesian 3d-space, DEFINE a tensor of the Nth rank T  a quantity with 3N elements or components Tijk.. (N indices) which transform under an orthogonal transformation of coords A (from Ch. 4) as (summation convention!):
T´ijk..(x´) = ail ajm akn … Tlmn..(x)
(N factors of matrix elements aij of A)
Following the discussion of pseudo-vectors in Ch. 4, we also need to define a pseudo-tensor, which transforms as a tensor except under inversion. See footnote, p 189.

17. T´ij(x´) = aik ajl Tkl(x)
Tensor of 0th rank: 1 element  scalar. Invariant under an orthogonal transformation A.
Tensor of 1st rank: 3 elements  ordinary vector. Transforms under orthogonal transformation A as:
T´i(x´) = aij Tj(x)
Tensor of 2nd rank: 9 elements. Transforms under orthogonal transformation A as:
T´ij(x´) = aik ajl Tkl(x)
Rigorously, we have a distinction between the 2nd rank tensor T & the square matrix formed from its elements: Tensor: Defined only by its transform properties under orthogonal transforms. Matrix: A representation of a tensor in a particular coordinate system. Every tensor equation  In some coordinate system, a corresponding matrix equation. Further discussion, p. 189.
Various math properties of tensors, p : Read on your own!

18. Sect 5.3: Inertia Tensor & Moment of Inertia
Inertia tensor I = 2nd rank tensor, transforms as such under orthogonal transformation A.
The angular momentum of a rigid body is: L = Iω. Should more properly be written L = Iω
Where   matrix or tensor multiplication.
Now, we’ll derive a form for the KE T for a rigid body in terms of I and ω.
Start with KE of motion about a point in form (summation convention, dropped prime from earlier) :
T  (½)mi(vi)2

19. T  (½)mi(vi)2 = (½)mivivi
Rigid body  The only contribution to vi is from body rotation.  As before use
vi = (dri/dt)s = (dri/dt)r + ω  ri = 0 + ω  ri
Convenient to use with only one of the vi in the dot product:
 T = (½)mivi(ω  ri )
Permuting vectors in triple product:
 T = (½)ω[mi(ri  vi )]  (½)ωL
Using L = Iω this becomes:
T = (½)ωIω
Note: Order matters & product with ω when  is to left of I & to right of I is clearly different. T is clearly a scalar, as it should be!
Often, I may write it as simply T = (½)ωIω (leaving out )

20. I  nIn  Moment of Inertia (about rotation axis)
n  unit vector along rotation axis, so that ω = ωn
 Can write (1) as:
T = (½)ω2nIn  (½)Iω (2)
I  nIn  Moment of Inertia (about rotation axis)
Using forms of I from before, can write:
I = nIn = mi[(ri)2 - (rin)2] (3)
Text proves (p. 192, figure) that
definition, (3), of I is consistent with
elementary definition as sum over all
particles of products of particle masses
times square of  distances from
rotation axis. I = mi[(ri)2] . Read!

21. I  nIn  Moment of Inertia (about rotation axis)
n  unit vector along rotation axis:
 T = (½)ω2nIn  (½)Iω (2)
I  nIn  Moment of Inertia (about rotation axis)
Clearly, moment of inertia I depends on the direction n of the rotation axis. Also, for a rigid body in motion, that direction & thus the direction of ω can be time dependent.  In general, I = I(t)
In the (important!) special case where the body is constrained to rotate about a fixed axis, clearly then, I = constant.
When this is the case, T in the form of (1) can be used in the Lagrange formalism if we can write ω as the time derivative of some angle.

22. I  nIn  Moment of Inertia (about rotation axis)
Inertia tensor I & moment of inertia I depend on the choice of the body axes origin. However, the moment of inertia about a given axis is simply related to the moment of inertia about a parallel axis through the CM ( Parallel Axis Theorem).
Figure: 2 || rotation axes a & b,
direction n. Axis b passes through
CM. R, ri = positions of CM &
particle i with respect to origin O.
ri´ = position of particle i with
respect to CM. ri = R + ri´.
Moment of inertia about axis a:
Ia = ∑imi(ri  n)2 = ∑imi[(R + ri´)  n]2
= ∑imi(R  n)2 + ∑imi(ri´  n)2 + 2∑imi(R n)(ri´  n)

23. Moment of inertia about axis a:
Ia = ∑imi(R  n)2 + ∑imi(ri´ n)2
+ 2 ∑imi(R n)(ri´  n)
Moment of inertia about axis b:
Ib = ∑imi(ri´ n)2
Note that: ∑ imi(R  n)2  M(R  n)2
 Ia = Ib + M(R  n)2 + 2∑imi(R n)(ri´ n)
Rewrite 3rd term as: 2(R n)(∑imiri´  n)
Note that, by definition of CM, ∑imiri´ = 0
 2∑imi(R n)(ri´ n) = 0
 Ia = Ib + M(R  n)2 For θ = angle between R & rotation axis n,  distance between axes a & b = |R  n| = Rsinθ = r.  Ia = Ib + MR2sin2θ
Or: Ia = Ib + M(r)2  Parallel Axis Theorem

24. Ia = Ib + M(r)2  Parallel Axis Theorem
In words: The moment of inertia about an arbitrary axis is equal to the moment of inertia about a parallel axis passing through the center of mass plus the moment of inertia of the body about the arbitrary axis, taken as if all of the mass M of the body were at the center of mass.

25. Ijk  mi[(ri)2δjk - (ri)j(ri)k]
Summary: Rotational KE in terms of inertia tensor:
T = (½)ωIω (1)
Can rewrite in terms of tensor elements as (summation convention):
T = (½)ωjIjkωk (2)
Again (summation convention, i = particle label, j, k = Cartesian x,y,z = 1,2,3 labels!) :
Ijk  mi[(ri)2δjk - (ri)j(ri)k]
Continuous bodies: Sum over particles  integral over volume V. Define: ρ(r)  Mass density at position r
 Ijk  ∫Vρ(r)[r2δjk - xjxk]dV