Some Surprising Conservative andNonconservative Moments in the Dynamics of

Rods and Rigid Bodies

Evan G. Hemingway and Oliver M. O’Reilly

Department of Mechanical Engineering, University of California,

Berkeley, CA 94720-1740, USA

Abstract Representations for conservative and nonconservative mo-

ments in classical mechanics are discussed in this expository article.

When the rotation is parameterized by a set of Euler angles, a par-

ticularly transparent representation can be found which has ties to

classic works in mechanics dating to Lagrange in 1780 and joint

coordinate systems that are commonly used in orthopaedic biome-

chanics. The article also surveys connections between Lagrange’s

equations of motion and the Newton-Euler equations of motion. A

variant on the Lagrange top and a satellite dynamics problem are

presented to illustrate some of the key concepts discussed in the

paper.

1 Introduction

In a remarkable paper, Lagrange (1780) presented a dynamic model to ex-plain the oscillations (librations) in the attitude of the Moon as seen by anEarth-based observer. The starting point for his model featured for the firsttime his celebrated equations of motion (cf. Lagrange (1780, Section 11)):

d

dt

(∂T

∂qK

)

−∂T

∂qK= −

∂V

∂qK, (K = 1, . . . , 6) , (1)

where V and T are the respective potential and kinetic energies of the Moon.Later in this work (cf. Lagrange (1780, Section 21)), Lagrange used a setof 3-1-3 Euler angles to parametrize the rotation of the Moon.

A few years prior to Lagrange’s work, Euler published a series of seminalworks on the dynamics of rigid bodies (cf. Euler (1752, 1775)). Among thecontributions from Euler’s works that have had a lasting influence are the

1

Euler angle parameterization of rotations and the Newton-Euler equationsof motion for a rigid body of mass m:

F = m¨x, M = H, H = Jω, (2)

where F is the resultant force acting on the rigid body, M is the resultantmoment relative to the center of mass of the rigid body, J is the momentof inertia tensor of the rigid body relative to its center of mass, x is theposition vector of the center of mass, H is the angular momentum relativeto the center of mass of the rigid body, and ω is the angular velocity vectorof the rigid body.

Lagrange’s treatment of the libration problem is surprising. First, al-though he employs Euler angles and is completely comfortable with massmoments of inertia, he uses an entirely different formulation of the problemthan the Newton-Euler form (2). While he calculates gravitational forcesand approximations to the gravitational potential energy, it is not clear whatthe corresponding gravitational moments are from his work. Indeed the mo-ment in question, known as a gravity-gradient torque (cf. Eqn. (83)2), iscredited to James Mac Cullagh (1809–1847) following the posthumous pub-lication of his lecture notes by Allman (1855).

Assuming that the dynamic equations of motion formulated using La-grange’s equations (1) and the Newton-Euler equations (2) are equivalent,it is natural to ask if some of the partial derivatives − ∂V

∂qKcan be inter-

preted as moment components. Using results on dual Euler basis vectorsfrom O’Reilly (2007) we are able to show how these partial derivatives arespecific components of force and moment vectors and thus facilitate physi-cal interpretations of the partial derivatives − ∂V

∂qK. Our discussion is illumi-

nated with examples from rigid body dynamics and orthopaedic biomechan-ics and also highlights the simplest known representation for a conservativemoment.

An outline of this expository article is as follows. Background on theEuler angle parameterization of a rotation is collected in Section 2. The no-tation we use for the three angles follows Lagrange (1780). We supplementhis treatment with a discussion of the Euler basis and dual Euler basis vec-tors and the representation of vectors using these distinct bases. In Section3, the relationship between Lagrange’s equations of motion and the Newton-Euler equations of motion for a single rigid body are examined. Particularattention is paid to the incorporation of ideal integrable constraints andpotential energies. A simple representation for a conservative moment thatfeatures the dual Euler basis vectors is discussed in Section 4. To dispelpossible confusion, the case where the motion of a rigid body is constrainedto have a fixed axis of rotation is presented in Section 5. We close the pa-

2

per with two examples. First, a derivation of the equations of motion ofa Lagrange top subject to applied forces and moments is discussed. Then,in Section 7, we return to the problem of a satellite in a gravitational fieldthat was the subject of Lagrange (1780). Among other matters, we areable to demonstrate how the partial derivatives − ∂V

∂qKin Eqn. (1) can be

considered as components of a moment vector.The notation and terminology employed in this paper closely follows the

textbook by O’Reilly (2008). We appeal to a recent expository article byO’Reilly and Srinivasa (2014) when discussing constraint forces and con-straint moments. Additional complementary background on rotations canbe found in the exceptional survey article by Shuster (1993) and the onlineresource http://rotations.berkeley.edu/.

2 Background on Euler Angles and Bases

Central to our discussion is the method of Euler angles to parameterize arotation tensor. Of the twelve available sets of Euler angles, we focus ourattention here on the 3-1-3 set. We define a fixed right-handed orthonor-mal basis {E1,E2,E3} and use the rotation tensor Q to define a basis{e1, e2, e3}:

ei = QEi. (3)

It is straightforward to show using the facts that QQT = I and det (Q) = 1that the basis {e1, e2, e3} is right-handed and orthonormal. In addition,the rotation tensor Q has the representations

Q =

3∑

k=1

ek ⊗Ek =

3∑

i=1

3∑

k=1

QikEi ⊗Ek =

3∑

i=1

3∑

k=1

Qikei ⊗ ek. (4)

That is, the components Qik = ek · Ei of Q can be considered as directioncosines.

In the 3-1-3 set of Euler angles, the tensor Q is decomposed into theproduct of three simple rotations:

Q = QE (ϕ,g3)QE (ω,g2)QE (ψ,g1) , (5)

where the function QE (θ, i) describes a rotation about an axis described bya unit vector i through a counterclockwise angle θ:

QE (θ, i) = cos(θ) (I− i⊗ i) + sin(θ)skew (i) + i⊗ i. (6)

In the above representation for a rotation tensor, the operator skew(i) trans-forms i into a skew-symmetric tensor such that i × b = skew(i)b for any

3

vector b. The associated operator ax transforms a skew-symmetric tensorinto a vector: ax (A) × b = Ab where A = −AT is a skew-symmetrictensor.

The basis {g1,g2,g3} is known as the Euler basis. This basis is notorthogonal, and, for the 3-1-3 set of interest here,

g1 = E3

= cos (ω) e3 + sin (ω) (cos (ϕ) e2 + sin (ϕ) e1) ,

g2 = cos (ψ)E1 + sin (ψ)E2

= cos (ϕ) e1 − sin (ϕ) e2,

g3 = cos (ω)E3 + sin (ω) (sin (ψ)E1 − cos (ψ)E2)

= e3. (7)

The angles ψ and ϕ range from 0 to 2π. Because

(g1 × g2) · g3 = − sin (ω) , (8)

in order to ensure that the Euler basis is a basis for E3, we restrict the

second angle ω ∈ (0, π). Other perspectives on the singularity when ω = 0, πinclude noting that ψ and ω are polar coordinates for the axis of rotationg3 = e3. Thus, the singularity arises when multiple values of ψ + ϕ (whenω = π) and ψ − ϕ (when ω = 0) are possible for a given rotation tensor Q.

E1

E2 E3

e∗1e∗2

e∗1e∗2 e∗∗2

e∗∗3

e∗∗2

e1e2

ψψ

ωω

ϕϕ

Figure 1: Schematic of a set of 3-1-3 Euler angles that are used to parame-terize a rotation from the basis {E1,E2,E3} to the basis {e1, e2, e3}. Thethree axes of rotation are g1 = E3 = e∗3, g2 = e∗1 = e∗∗1 , and g3 = e∗∗3 = e3and the inset image is Handmann’s portrait of Leonhard Euler (1707–1783)from 1753. For details on the intermediate bases used to construct thefigure, see Eqn. (9).

Referring to Figure 1, it is straightforward to transform from ei to Ei

and vice versa with the help of two pairs of intermediate bases {e∗1, e∗

2, e∗

3}

4

and {e∗∗1 , e∗∗

2 , e∗∗

3 }:

e∗1e∗2e∗3

=

cos (ψ) sin (ψ) 0− sin (ψ) cos (ψ) 0

0 0 1

E1

E2

E3

,

e∗∗1e∗∗2e∗∗3

=

1 0 00 cos (ω) sin (ω)0 − sin (ω) cos (ω)

e∗1e∗2e∗3

,

e1e2e3

=

cos (ϕ) sin (ϕ) 0− sin (ϕ) cos (ϕ) 0

0 0 1

e∗∗1e∗∗2e∗∗3

. (9)

We note that the three matrices in Eqn. (9) can be combined to provideexpressions for the components Qik = (QEk) · Ei of Q:

Q11 Q12 Q13

Q21 Q22 Q23

Q31 Q32 Q33

=

−s1c2s3 + c1c3 −c1s3 − c3c2s1 s1s2c2c1s3 + c3s1 −s1s3 + c2c1c3 −c1s2

s3s2 s2c3 c2

.

(10)In writing expressions for the components Qik, we have used the helpfulabbreviations c1 = cos (ψ), s1 = sin (ψ), c2 = cos (ω), s2 = sin (ω), c3 =cos (ϕ), and s3 = sin (ϕ).

The Euler basis vectors feature in the representation of the angular ve-locity vector ω associated with the rotation tensor Q. In particular,

ω = ax(

QQT)

= ϕg3 + ωg2 + ψg1

=(

ψ sin (ω) sin (ϕ) + ω cos (ϕ))

e1

+(

ψ sin (ω) cos (ϕ)− ω sin (ϕ))

e2 +(

ψ cos (ω) + ϕ)

e3.

(11)

This representation can be established by direct, but lengthy computationof Q using Eqn. (10) or, more rapidly, using two relative angular velocityvectors as in Casey and Lam (1986).

2.1 The Euler and Dual Euler Bases

In addition to the Euler basis, we also have a companion dual Eulerbasis. Given an Euler basis, the corresponding dual Euler basis vectors aredefined by the nine relations

gk · gi = δki , (i = 1, 2, 3, andk = 1, 2, 3) , (12)

5

(a) (b)g1

g3

g2 = g2

E3 = g1

E3 = g1

e3 = g3e3 = g3

g2

E2E2

E1E1

ϕϕ

ωω

ψψ

Figure 2: Schematic of the Euler and dual Euler basis vectors associatedwith a 3-1-3 set of Euler angles. (a) The Euler basis vectors, g1, g2, andg3, and their relation to the Euler angles (cf. Eqn. (7)) and (b) the corre-sponding dual Euler basis vectors, g1, g2, and g3 (cf. Eqn. (14)).

where δki is the Kronecker delta: δki = 1 if i = k and is otherwise 0. Thesolution to these nine equations is known in differential geometry:1

g1 =g2 × g3

(g1 × g2) · g3

,

g2 = g2,

g3 =g1 × g2

(g1 × g2) · g3

. (13)

The expression for g2 is greatly simplified because the Euler basis vector g2

is perpendicular to the other two: g1 ⊥ g2 and g3 ⊥ g2. With the help ofEqn. (13), we now compute the dual Euler basis vectors for the 3-1-3 set ofEuler angles:

g1 = cosec (ω) (sin (ϕ) e1 + cos (ϕ) e2) ,

g2 = cos (ϕ) e1 − sin (ϕ) e2,

g3 = cot (ω) (− cos (ϕ) e2 − sin (ϕ) e1) + e3. (14)

The Euler and dual Euler bases are sketched in Figure 2.We observe from Eqn. (14) that the dual Euler basis vectors are not

defined when ω = 0, π. The dual Euler basis vectors were first introduced

1We are exploiting the correspondence between the Euler and dual Euler basis sets and

covariant and contravariant sets of basis vectors in differential geometry and are able to

use a well-known result. See, e.g., Green and Zerna (1968, Eqn. (1.9.13)) or Simmonds

(1982, Exercise 2.11)).

6

in O’Reilly and Srinivasa (2002) and O’Reilly (2007). They are also relatedto the dual vectors described by Howard et al. (1998) and Zefran and Kumar(2002) in their discussion of screw motions for rigid bodies.

2.2 Vector Representations

As discussed in Nichols and O’Reilly (2017) and O’Reilly et al. (2013),the dual Euler basis feature in representations for the joint moment vectorthat is commonly used in orthopaedic biomechanics. To elaborate, a vectorb has multiple representations including

b = b1g1 + b2g2 + b3g3

= b1g1 + b2g

2 + b3g3, (15)

where

bk = b · gk, bk = b · gk. (16)

Referring to Figure 3, in biomechanics of anatomical joints, the first andthird Euler basis vectors are identified with landmarks on the respectivebones and moment components M · gk are computed.2 Consequently, inorder to reconstruct the moment vector, the dual Euler basis vectors areneeded: M =

∑3

k=1 (M · gk)gk.

3 Lagrange’s Equations of Motion and the

Newton-Euler Equations of Motion

Consider a rigid body of mass m which has an inertia tensor J relative toits center of mass. We assume that a set of six coordinates are used tocharacterize the kinematics of the rigid body:

x = x(q1, . . . , q6

),

Q = Q(q1, . . . , q6

). (17)

These coordinates are usually chosen to readily accommodate the integrableconstraints on the system. Referring to Figure 4, for the Lagrange topq1, q2, and q3 are usually chosen to be a set of Euler angles while q4, q5, andq6 are chosen to be the Cartesian coordinates of the fixed point O of thetop. In satellite dynamics problems where a steady motion of the satelliteinvolves a 1-1 locking of the orbital angular speed and the angular velocity,

2See, for example, Desroches et al. (2010), Grood and Suntay (1983), and Schache and

Baker (2007).

7

j1

j1

j3

j3j2 ‖ j1 × j3

C1C2α

β

γ

Femur

Tibia

Figure 3: Schematic of a joint coordinate system for the human knee joint.The axis j1 corotates with the femur and j3 corotates with the tibia. Theangles α, β, γ and the axes {j1, j2, j3} can be identified with a set of 3-2-1(or 1-2-3) Euler angles and their associated Euler basis vectors.

a set of cylindrical polar coordinates are chosen for q1, q2 = ϑ, and q3 andthe rotation is parameterized by a set of Euler angles and the polar angle:Q = Q

(ϑ, q4 = ψ, q5 = ω, q6 = ϕ

).

It is straightforward to show that

v =6∑

K=1

qK∂x

∂qK,

∂x

∂qK=

∂v

∂qK. (18)

Further,

ω =

6∑

K=1

qKwK , wK =∂ω

∂qK= ax

(∂Q

∂qKQT

)

. (19)

If a set of Euler angles are used to parameterize Q,

Q = Q(q6 = ϕ, q5 = ω, q4 = ψ

), (20)

8

(a) (b)

e1e1

e2e2

e3

e3

E1E1

E2

E2

E3

E3

er

X

X

x

g O

O

m

M

ϑ

Figure 4: (a) Schematic of a rigid body freely rotating about a fixed pointO. This rigid body is commonly known as a Lagrange top and the insetimage is a portrait of Joseph-Louis Lagrange (1736–1813). (b) Schematicof a rigid body of mass m in motion about a fixed rigid body of mass M .

then w3+k = gk and w1,2,3 = 0.The kinetic energy of the rigid body has the representation

T =m

2v · v +

1

2ω · Jω. (21)

Typically the corotational (or body-fixed) basis vectors ei are chosen to bethe principal axis of the body. In this case, J can be expressed as

J = λ1e1 ⊗ e1 + λ2e2 ⊗ e2 + λ3e3 ⊗ e3, (22)

where λk are the principal mass moments of inertia.It can be shown that Lagrange’s equations of motion are equivalent to a

linear combination of the Newton-Euler balance laws:3

d

dt

(∂T

∂qK

)

−∂T

∂qK= F ·

∂v

∂qK+M ·

∂ω

∂qK. (23)

That is,

∂T

∂qK= mv ·

∂v

∂qK+H ·

∂ω

∂qK,

∂T

∂qK= mv ·

d

dt

(∂v

∂qK

)

+H ·d

dt

(∂ω

∂qK

)

. (24)

3A proof of this correspondence can be found in Casey (1995). Casey’s proof is discussed

in the textbook O’Reilly (2008) where additional examples are presented.

9

It is an interesting exercise to establish these results first for a single particleof mass m - where the angular momentum terms can be ignored. Indeedthis case is discussed in the classic textbook Synge and Griffith (1959).

3.1 A Force FA Acting at a Point A

The right hand side of Lagrange’s equations (23) have several simplifi-cations. First, suppose that a force FA acts at a point XA which has aposition vector xA and velocity vector vA:

vA = v + ω× (xA − x) . (25)

The position vectors xA and x can be expressed as functions of the sixcoordinates q1, . . . , q6. Unlike v, ω, and vA, this pair of position vectors donot depend on the velocities q1, . . . , q6.

For the force FA, a simple differentiation of Eqn. (25) with respect toqK can be used to show that

FA ·∂v

∂qK+ ((xA − x)× FA) ·

∂ω

∂qK= FA ·

∂vA

∂qK. (26)

This identity is helpful in several respects. First, it enables a direct compar-ison of treatments of Lagrange’s equations where a virtual work argument isused to prescribe nonconservative generalized forces on the right-hand sideof Lagrange’s equations.4 In addition, if FA is a conservative force with apotential energy function

U = U (xA) , (27)

then

FA = −∂U

∂xA

. (28)

With the help of the identity

vA = xA =

6∑

K=1

qK∂xA

∂qK, (29)

it is straightforward to show using the chain rule that

FA ·∂vA

∂qK= FA ·

∂xA

∂qK

= −∂U

∂xA

·∂xA

∂qK

= −∂U

∂qK. (30)

4See, for example, the lucid discussion in Baruh (1999, Chapter 4, Section 9).

10

Finally, Eqn. (26) allows one to easily write down Lagrange’s equations incases where a nonconservative follower force or a dynamic Coulomb frictionforce acts on the mechanical system. The most celebrated instance of theformer case is Ziegler’s pendulum.

3.2 Ideal Integrable Constraints

For the second simplification to the right-hand side of Lagrange’s equa-tions of motion, suppose that an integrable constraint is imposed on thesystem:

q6 − f(t) = 0. (31)

Then, if the constraint force Fc and constraint moment Mc satisfy La-grange’s prescription, it can be shown that5

Fc ·∂v

∂qK+Mc ·

∂ω

∂qK= µδK6 , (32)

where µ is a scalar function (Lagrange multiplier) and δK6 is the Kroneckerdelta. This remarkable result enables Lagrange’s equations of motion to de-couple into two sets: one for the unconstrained (or generalized) coordinatesand the other for the function µ.

In most treatments of Lagrange’s equations of motion interest is re-stricted to the former set and the equations of motion for the generalizedcoordinates are formulated. For multibody systems involving rigid bodiesconnected by pin joints the savings in algebraic computations when atten-tion is restricted to the generalized coordinates can be considerable. Forinstance, for the planar double pendulum, the number of equations of mo-tion for the generalized coordinates is two while the remaining ten equationsgive expressions for the reaction forces and reaction moments at the two pinjoints.

3.3 Potential Energies

The third and final simplification we wish to mention occurs if a conser-vative force Fcon acting at the center of mass and a conservative moment(relative to the center of mass) Mcon act on the rigid body. The combinedmechanical power of these quantities is assumed to be equal to the negativerate of change of a potential energy function:

Fcon · v +Mcon · ω = −U , (33)

5This constraint in this case is sometimes known as ideal because Fc and Mc have no

frictional components. For further details on constraint forces and constraint moments,

see the expository paper by O’Reilly and Srinivasa (2014).

11

where

U = U (x,Q) , U =6∑

K=1

∂U

∂qKqK . (34)

Assuming that Fcon and Mcon are independent of the rates q1, . . . , q6, itfollows that

Fcon ·∂v

∂qK+Mcon ·

∂ω

∂qK= −

∂U

∂qK. (35)

Thus, we find that the partial derivatives of U with respect to the coor-dinates qK are linear combinations of the components of the conservativeforce and conservative moment. Expressions for Fcon and Mcon can be es-tablished as gradients of U with respect to x and Q, respectively. We shallexamine one such representation for Mcon shortly.

3.4 A Canonical Form, Equilibria, and Linearization

Lagrange’s equations of motion reveals a canonical form of the equationsof motion. To elaborate, consider a system withN degrees-of-freedomwhosekinetic energy can be expressed as a quadratic form and whose potentialenergy function depends on the N generalized coordinates:

T =1

2

N∑

I=1

N∑

K=1

aIK qI qK , U = U

(q1, . . . , qN

), (36)

where aIK = aIK(q1, . . . , qN

). Such systems are pervasive in mechanics.

For such a system, it is known that6

d

dt

(∂T

∂qK

)

−∂T

∂qK=

N∑

I=1

aKI qI +

N∑

S=1

N∑

J=1

[SJ,K] qS qJ . (37)

Here, we have used the Christoffel symbols of the first kind [SJ,K] to collectthe quadratic velocity terms:

[SJ,K] =1

2

(∂aKJ

∂qS+∂aKS

∂qJ−∂aSJ

∂qK

)

, (J,K, S = 1, . . . , N) . (38)

If we assume that the only generalized forces acting on the system areconservative, then the equations of motion can be expressed as

N∑

I=1

aKI qI +

N∑

S=1

N∑

J=1

[SJ,K] qS qJ = −∂U

∂qK, (K = 1, . . . , N) . (39)

6See, for example, the classic text by McConnell (1947).

12

An equilibrium of these equations satisfies the following 2N conditions:

qK = 0, qK = qK0 . (40)

Examining the equations of motion (39), we observe that at an equilibriumthe potential energy is extremized:

∂U

∂qK

(q10 , . . . , q

N0

)= 0. (41)

To establish the linearized equations of motion in the neighborhood ofan equilibrium, we consider the following asymptotic expansion:

q1 = q10 + ǫη1, . . . , qN = qN0 + ǫηN . (42)

After substituting into (39), performing Taylor series expansions of U , aIK ,and [SJ,K], using the equilibrium conditions, and ignoring terms of or-der ǫ2 and higher, we find the following equations governing the linearizeddynamics in a neighborhood of the equilibrium:

M0η + K0η = 0. (43)

The mass matrix M0 and stiffness matrix K0 are both symmetric:

η =[η1, · · · , ηN

]T, (44)

M0 =

a11(q10 , . . . q

N0

)· · · a1N

(q10 , . . . q

N0

)

.... . .

...a1N

(q10 , . . . q

N0

)· · · aNN

(q10 , . . . q

N0

)

, (45)

K0 =

∂2U∂q1∂q1

(q10 , . . . q

N0

)· · · ∂2U

∂q1∂qN

(q10 , . . . q

N0

)

.... . .

...∂2U

∂q1∂qN

(q10 , . . . q

N0

)· · · ∂2U

∂qN∂qN

(q10 , . . . q

N0

)

. (46)

Thus, for many mechanical systems, Lagrange’s equations of motion allowsus to infer the equilibria of the system and the equations governing thelinearized dynamics by simply computing the kinetic and potential energiesand the derivatives of the latter energy.

4 Simple Conservative Moments

As remarked by Ziegler (1968, Page 30), the simplest nonconservative mo-ment is a constant moment. He showed that a constant moment was non-conservative by examining the work done in rotating a rigid body through

13

180◦. Such a motion can be accomplished in two equivalent manners. Thefirst method is direct while the second involves successively rotating thebody through 180◦ about two perpendicular axis. The work done by themoment in the latter case is zero and in the former case is non-zero. Whence,the constant moment is not conservative.

Ziegler’s conclusion is surprising and involves an ingenious use of theHamilton-Rodrigues theorem on finite rotations. His work also begs thequestion that if a constant moment isn’t conservative, then which momentis? This question was also posed by Simmonds and answered by Antman(1972) and later by Simmonds (1984) himself. Antman’s solution uses theEuler representation for a rotation featuring an axis of rotation and an angleof rotation. A simpler answer can be found using Euler angles.7

4.1 A Simple Representation for a Conservative Moment

Consider a potential energy U that depends solely on the orientation ofthe rigid body. Thus, we can express U as a function of the Euler anglesand

U =∂U

∂ψψ +

∂U

∂ωω +

∂U

∂ϕϕ. (47)

However, using the dual Euler basis vectors,

ω · g1 = ψ, ω · g2 = ω, ω · g3 = ϕ. (48)

Substituting into the expression for U ,

U =

(∂U

∂ψg1 +

∂U

∂ωg2 +

∂U

∂ϕg3

)

· ω. (49)

Paralleling the case of a conservative force, we define a conservative momentMcon by postulating that

U = −Mcon · ω. (50)

Assuming in addition that Mcon is independent of ω, we find the represen-tation

Mcon = −∂U

∂ψg1 −

∂U

∂ωg2 −

∂U

∂ϕg3. (51)

This is the simplest known representation of a conservative moment.8 Ithas evident parallels to a representation of a conservative force acting on aparticle that features a gradient expressed using contravariant basis vectors.

7Our developments and discussion are based on the works O’Reilly and Srinivasa (2002)

and O’Reilly (2007, 2008).8A compilation of representations for the gradient of U for various representations of

the rotation Q can be found in O’Reilly (2008, Section 6.10).

14

4.2 Ziegler’s Example Revisited

Returning to Ziegler’s example, suppose that M0E3 (where M0 is a con-stant) is conservative. Then, the potential energy function associated withthis moment would have to satisfy the following set of partial differentialequations:

−∂U

∂ψ=M0E3 · g1 =M0E3 ·E3 =M0,

−∂U

∂ω=M0E3 · g2 =M0E3 · e

∗

1 = 0,

−∂U

∂ϕ=M0E3 · g3 =M0E3 · e3 =M0 cos (ω) . (52)

However, the statements ∂U∂ω

= 0 and ∂U∂ϕ

= −M0 cos (ω) are contradictoryand we conclude that no such U can exist. Thus, M0E3 is nonconservative.

4.3 Torsional Springs

We can use the representation for a conservative moment to establishan expression for the moment provided by a torsional spring. Suppose thatthe torsional spring’s potential energy function is

Uspring =K

2(ψ − ψ0)

2, (53)

where ψ0 is a constant and K is the spring constant. Invoking Eqn. (51),we find that

Mspring = −K (ψ − ψ0)g1

= −K (ψ − ψ0) (E3 − cot (ω) (− cos (ψ)E2 − sin (ψ)E1)) . (54)

Observe that the spring moment has components orthogonal to the E3 di-rection. The fact that these components become unbounded as ω → 0, π isa manifestation of the singularity in the 3-1-3 set of Euler angles at thesevalues of ω. Unless the rigid body is constrained, it is necessary to switch toa complementary set of Euler angles such as the 3-2-1 set as ω approachesthese singular values.

5 The Case of a Fixed Axis of Rotation

The singular behavior of the torsional spring moment in the previous sectionbegs the question of how to deal with the case when the axis of rotation isconstrained to be fixed. In this situation, choosing E3, say, to be parallel

15

to the axis of rotation, we find that the angular velocity vector has therepresentation ω = ψE3. In addition, the motion of the rigid body issubject to a pair of constraints that can be expressed as

ω · E1 = 0, ω · E2 = 0. (55)

A constraint moment with two independent components is needed to enforcethese constraints:

Mc = µ1E1 + µ2E2. (56)

The moment Mc does no work and is nonconservative. If a pin joint isused to ensure that the axis of rotation stays constant, then the constraintmoment Mc can be considered as a reaction moment provided by the pinjoint. If the body is sliding on a flat surface, then Mc is the resultantmoment provided by the normal forces acting on the surface of the bodycontacting the plane.

The potential energy of a torsional spring in this case is again given by

Uspring =K

2(ψ − ψ0)

2. (57)

Now, however, we seek solutions Mspring to the equation U = −Mspring · ω

where ω = ψE3. The resulting solution is

Mspring = −K (ψ − ψ0)E3. (58)

Fortunately, this expression has none of the issues associated with its three-dimensional counterpart (54). On a related note, as remarked in Ziegler(1968, Chapter 5), in the dynamics of rods where terminal moments ofthe form M0E3 are applied, the boundary conditions often restrict the endrotation of the rod to be along E3. In this case, M0E3 is a conservativemoment. We refer the reader to O’Reilly (2017, Section 5.15) for furtherdiscussion of this case.

6 The Lagrange Top

To illustrate many of the previous developments, we now consider the classicexample of an axisymmetric rigid body which is free to rotate about a pointO. The rigid body is under the action of a vertical gravitational force.This mechanical system is known as the Lagrange top and its celebrateddynamics have a long and storied history.9

9For additional references and discussion, see Baruh (1999), Lewis et al. (1992), and

Marsden and Ratiu (1999).

16

For the purposes of exposition, we establish the equation of motion forthis system and include the effects of a constant moment MaE3 acting onthe body and a follower force Fae1 acting at the tip of the top. Further,we allow the point O to be given a prescribed vertical motion f(t)E3. Thismotion can be imagined by assuming that the top is freely spinning abouta point O and then the point O is oscillated in a vertical manner.

6.1 Kinematical Considerations

To establish the equations of motion for the top, we assign a set of 3-1-3Euler angles to describe its orientation. The translational motion of therigid body is characterized by a set of coordinates to describe the motionof O:

q1 = ψ, q2 = ω, q3 = ϕ,

q4 = xO · E1, q5 = xO ·E2, q6 = xO · E3. (59)

We assume that the position vector of the center of mass X relative to O is

x− xO = ℓe3. (60)

The mass of the top is denoted by m and its inertia tensor is

J = λt (I− e3 ⊗ e3) + λae3 ⊗ e3. (61)

The velocity vector of the center of mass of the top has the representation

v = q4E1 + q5E2 + q6E3 + ω2ℓe1 − ω1ℓe2. (62)

In this expression, the components ωk = ω · ek can be easily read off fromEqn. (11).

For future purposes, we note that

∂ω

∂qk= gk,

∂ω

∂qk+3= 0,

∂vO

∂qk= 0,

∂vO

∂qk+3= Ek, (63)

where k = 1, 2, 3.

6.2 Constraints and Constraint Forces

The motion of the top is subject to three constraints. We have chosenthe six coordinates in anticipation of these constraints being imposed. Thethree constraints can be expressed as follows:

q4 = 0, q5 = 0, q6 − f(t) = 0. (64)

17

These constraints can also be expressed in terms of a velocity vector:

vO ·E1 = 0, vO · E2 = 0, vO · E3 − f = 0. (65)

Referring to Section 3.2, this representation of the constraints allows us toeasily appeal to Lagrange’s prescription to write down a representation forthe constraint force acting on the the top:

Fc = µ1E1 + µ2E2 + µ3E3 acting at O. (66)

This force is none other than the reaction force at O. With the help of theearlier results (63) we find that

Fc ·∂vO

∂qk= 0, Fc ·

∂vO

∂qk+3= µk. (67)

Thus, as anticipated, the constraint force will be absent from three of thesix Lagrange’s equations of motion.

6.3 Kinetic and Potential Energies

The unconstrained kinetic energy of the top is

T =m

2

(3∑

i=1

q3+iEi + ω2ℓe1 − ω1ℓe2

)

·

(3∑

k=1

q3+kEk + ω2ℓe1 − ω1ℓe2

)

+λa

2

(

ϕ+ ψ cos (ω))2

+λt

2

(

ω2 + ψ2 sin2 (ω))

. (68)

The unconstrained potential energy of the top is

U = mgE3 · (xO + ℓe3) . (69)

This potential energy is a function of q6 and the Euler angle q2 = ω. Weornament T and U with hats · to distinguish them from their constrainedcounterparts.

Imposing the constraints (65), the constrained kinetic and potential en-ergies can be found:

T =λa

2

(

ϕ+ ψ cos (ω))2

+λt +mℓ2

2

(

ω2 + ψ2 sin2 (ω))

+m

2f2 −mωfℓ sin (ω) ,

U = mgf +mgℓ cos (ω) . (70)

Observe that λOt = λt +mℓ2 is a mass moment of inertia relative to O.

18

6.4 The Equations of Motion

The rigid body is subject to an applied moment MaE3 and a force Fae1which follows e1 and acts at the tip of the top. The latter point, which wedenote by Xt is assumed to have a position vector ℓ1e3 relative to O. Theequations governing the generalized coordinates ψ, ω, and ϕ are obtainedfrom Lagrange’s equation of motion:

d

dt

(∂T

∂qk

)

−∂T

∂qk= −

∂U

∂qk+MaE3 ·

∂ω

∂qk+ Fae1 ·

∂vt

∂qk. (71)

As vt = vO + ω× ℓ1e3,∂vt

∂qk= ℓ1gk × e3, (72)

where k = 1, 2, 3. Whence, the right-hand side of Lagrange’s equations canbe simplified:

d

dt

(∂T

∂ψ

)

−∂T

∂ψ= −

∂U

∂ψ+ (Ma + ℓ1Fa sin (ω) cos (ϕ)) ,

d

dt

(∂T

∂ω

)

−∂T

∂ω= −

∂U

∂ω− ℓ1Fa sin (ϕ) ,

d

dt

(∂T

∂ϕ

)

−∂T

∂ϕ= −

∂U

∂ϕ+Ma cos (ω) . (73)

Notice that the nonconservative follower force and nonconservative momentboth contribute terms that are coordinate dependent to the right-hand sidesof Lagrange’s equations of motion.

Evaluating the derivatives of T and U , we find that

λa cos2(ω) + λOt sin2(ω) 0 λa cos(ω)

0 λOt 0λa cos(ω) 0 λa

︸ ︷︷ ︸

M

ψ

ω

ϕ

+

α1

α2

α3

=

β1β2β3

,

(74)where

α1

α2

α3

=

2(λOt − λa

)ωψ cos (ω) sin (ω)− λaϕω sin (ω)

λaϕψ sin (ω)−(λOt − λa

)ψ2 cos (ω) sin (ω)

−λaψω sin (ω)

,

β1β2β3

=

Ma + ℓ1Fa sin (ω) cos (ϕ)

m(

g + f)

ℓ sin(ω)− ℓ1Fa sin (ϕ)

Ma cos (ω)

. (75)

19

The mass matrix M on the left-hand side of Eqn. (74) can be inferredfrom the expression for the kinetic energy function (70)1. The α1,2,3 termsare related to sums of Christoffel symbols featuring the derivatives of thecomponents of M with respect to the coordinates.

We observe from the equations of motion that the effect of the vertical,time-varying motion of the point O is equivalent to changing the gravita-tional constant from g to g + f .

6.5 Equilibria and Linearized Equations of Motion

Suppose that the point O is fixed. A static equilibrium of the top cor-responds to stationary values of ψ, ω, and ϕ. The static values, which aredistinguished by a subscript 0, are

Ma + ℓ1Fa sin (ω0) cos (ϕ0) = 0,

mgℓ sin (ω0)− ℓ1Fa sin (ϕ0) = 0,

Ma cos (ω0) = 0. (76)

Whence, ψ0 is arbitrary, and Fa and Ma must satisfy the latter pair of thefollowing relations for a static equilibrium to exist:

ω0 =π

2, sin (ϕ0) =

mgℓ

Faℓ1, cos (ϕ0) = −

Ma

Faℓ1. (77)

Thus, an infinite number of equilibrium states exist. As can be seen inFigure 5, the top is tilted at 90◦ to the vertical, ψ0 is arbitrary, andtan (ϕ0) = −mgℓ

Ma.

We now consider perturbations from an equilibrium position:

ψ = ψ0 + ǫη1, ω = ω0 + ǫη2, ϕ = ϕ0 + ǫη3. (78)

Inserting these expressions into the equations of motion (74), using the equi-librium conditions (77), performing Taylor series expansions, and ignoringterms of order ǫ2, we find the linearized equations

λOt 0 00 λOt 00 0 λa

︸ ︷︷ ︸

M0

η1η2η3

+

0 0 mgℓ

0 0 −Ma

0 Ma 0

︸ ︷︷ ︸

K0

η1η2η3

=

000

. (79)

The linearized system has six eigenvalues:

0, 0, ±

√√√√±

√

−M2

a

λaλOt

. (80)

20

Fae1

e2

e3

E1

E2

E3

ψ0 +π2

X

g

O

MaE3

Figure 5: Equilibrium configuration of the Lagrange top subject to a followerforce Fae1 and a nonconservative moment MaE3.

The pair of zero eigenvalues are a reflection of the arbitrariness of ψ0. Fornon-zero values of Ma, the remaining four eigenvalues form a quartet. Astwo of the quartet have positive real parts, we conclude that the equilibriumis unstable.

Observe that the mass matrix M0 is symmetric and can be readily de-duced from the mass matrix M associated with the nonlinear equations ofmotion (74). In contrast to the conservative case presented in Eqn. (43),the stiffness matrix K0 is asymmetric. This asymmetry can be attributed tothe follower force load Fae1 and the nonconservative moment MaE3 actingon the top.

The asymmetry of the stiffness matrix in the linearized equations (79) is ageneric feature of equilibria of nonconservative mechanical systems includingmodels for brake squeal.10 The asymmetry of the stiffness matrix also makesthe equilibrium susceptible to dissipation-induced destabilization.11

10For additional details on brake squeal, the reader is referred to the review article

Kinkaid et al. (2003).11For a pseudospectral perspective on this topic see Kessler et al. (2007).

21

6.6 Solving for the Reaction Force

The constraint force Fc acting on the top can be determined from theequations of motion

d

dt

(

∂T

∂qk+3

)

−∂T

∂qk+3= −

∂U

∂qk+3+MaE3 ·

∂ω

∂qk+3

+ Fae1 ·∂vt

∂qk+3+ Fc ·

∂vO

∂qk+3. (81)

These equations are computed using the unconstrained kinetic and potentialenergy functions. Using the identity vt = vO + ω × ℓ1e3 and Eqns. (63)and (67), we find the following expressions for the components of Fc =∑3

k=1 µkEk:

µ1 =d

dt

(

∂T

∂q4

)

−∂T

∂q4+∂U

∂q4− Fae1 · E1,

µ2 =d

dt

(

∂T

∂q5

)

−∂T

∂q5+∂U

∂q5− Fae1 · E2,

µ3 =d

dt

(

∂T

∂q6

)

−∂T

∂q6+∂U

∂q6− Fae1 · E3. (82)

In other words, these three Lagrange’s equations of motion are simply theprojections of F = m ˙v onto the Cartesian basis vectors.

7 The Satellite Dynamics Problem

We now return to the problem considered by Lagrange. Suppose a rigidbody of mass m is orbiting a fixed spherically symmetric rigid body of massM (cf. Figure 4(b)). We locate the origin of our coordinate system at thecenter of the body of mass M . Mac Cullagh’s approximate expressions forthe gravitational force Fn, moment Mn, and potential energy Un on therigid body of mass m are

Fn ≈ mg,

Mn ≈

(3GM

R3

)

eR × (JeR) ,

Un ≈ −GMm

R−

(GM

2R3

)

tr(J) +

(3GM

2R3

)

(eR · (JeR)) , (83)

22

where

mg = −GMm

R2eR −

3GM

2R4(2J+ (tr(J) − 5eR · JeR) I) eR, (84)

and

R = ||x|| , eR =x

||x||. (85)

Notice that the unit vector eR points from the center of mass of the spher-ically symmetric body to the center of mass of the body of mass m. Inaddition, if the moment of rotational inertia of the rigid body is ignored,then the moment vanishes and these expressions reduce to the familiar ex-pression for a gravitational force on a particle of mass m. The moment Mn

is often known as a gravity gradient torque in the satellite dynamics litera-ture and features in studies of the precession of the equinoxes and librationsof the Moon.12

Suppose a set of cylindrical polar coordinates (r, ϑ, z) are used to pa-rameterize the position vector x and a set of 3-1-3 Euler angles are used toparameterize Q:

q1 = r, q2 = ϑ, q3 = z, q4 = ψ, q5 = ω, q6 = ϕ. (86)

That isx = rer + zE3, v = rer + rϑeϑ + zE3, (87)

where

er = cos (ϑ)E1 + sin (ϑ)E2, eϑ = cos (ϑ)E2 − sin (ϑ)E1. (88)

Then, with the help of Eqn. (35), we can readily identify the terms on theright-hand side of Lagrange’s equations of motion (1) with the conservativeforce and conservative moment acting on the rigid body of mass m:

−∂Un

∂r= Fn · er, −

∂Un

∂ϑ= Fn · reϑ, −

∂Un

∂z= Fn ·E3,

−∂Un

∂ψ= Mn · g1, −

∂Un

∂ω= Mn · g2, −

∂Un

∂ϕ= Mn · g3. (89)

Explicit expressions for Un, Fn, and Mn in terms of the Euler angles andthe cylindrical polar coordinates can be readily obtained but they are verylengthy and disguise the important relations (89). We note that

Fn = −∂Un

∂rer −

1

r

∂Un

∂ϑeϑ −

∂Un

∂zE3,

Mn = −∂Un

∂ψg1 −

∂Un

∂ωg2 −

∂Un

∂ϕg3. (90)

12See, e.g., Goldstein (1980, Section 5-8), Hughes (1986), or Kane et al. (1983).

23

These relations follow readily from Eqns. (15) and (89). Thus, we havebeen able to present transparent representations for the forces and momentsfeaturing in Lagrange’s equations (1).

After substituting for the Euler angles and the cylindrical polar coordi-nates into the expression for Un, one finds the well-known result that Un

can be expressed as a function of ψ + ϑ. With the help of Eqn. (89)2,4, wecan conclude that Fn · reϑ = Mn · E3. As is often assumed in examiningthe dynamics of artificial and natural satellites, if the body is axisymmetricwith λ1 = λ2 then one finds that Un is independent of ϕ. Using Eqn. (89)6,we deduce that Mn has no component along the axis of symmetry e3 andlies entirely in the plane spanned by e1 and e2. Additionally, we can thenconclude from Lagrange’s equations of motion that the angular momentumcomponent H · e3 is conserved.

Acknowledgement

The work of Evan Hemingway was supported by a Berkeley Fellowship fromthe University of California at Berkeley and a U.S. National Science Foun-dation Graduate Research Fellowship.

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