leonhard euler (1707 - september...

LEONHARD EULER (1707 - SEPTEMBER 1783)

BEYOND EQUATIONS

Leonhard Euler (1707 - September 1783)

Leonhard Euler was born in Basle, Switzerland;he was in fact a born mathematician, who went onto become the most prolific mathematician of alltime. His contemporaries called him “analysis in-carnate”, who “calculated without apparent effort,as men breathe, or as eagles sustain themselvesin the wind.” He easily constructed algorithmsto solve particular problems. He could work any-where, under any conditions, sometimes writingseveral mathematics papers a day.

Euler was appointed to the Academy in St. Pe-tersburg, where among other things he wrote his 1736 treatise Mechanica, inwhich he used calculus throughout, in contrast to Newton’s Principia, whichused no calculus at all. In 1740 he was invited to join the Berlin Academy, wherehe worked for many years, after which he accepted an invitation to return toSt. Petersburg. As noted in Chapter 3, Euler made extensive contributions tothe calculus of variations; he also contributed strongly to fluid dynamics andrigid-body rotation. In fact, if the reader comes across an important anonymousequation in mechanics, a good guess would be to call it “Euler’s equation.” Hehas not one but two numbers named after him, Euler’s number in calculus, e =2.71828..., and the Euler-Mascheroni constant, γ = 0.57721 ...

Euler had a prodigious memory and ability to calculate in his head. He oncesummed seventeen terms of a complicated convergent series for a particular valueof the variable to fifty decimal places, entirely in his head. This ability wouldbecome particularly useful during the last seventeen years of his life, when hewas blind. He would reportedly dictate mathematical papers to his wife whilegrandchildren played on his knee.

LEONHARD EULER (1707 - SEPTEMBER 1783)

Chapter 12

Rigid Body Dynamics

Watching a frisbee tumble around erratically is certainly entertaining, but– to a physicist without background in rigid body dynamics – it looks verytroubling. There is no net torque acting on the frisbee, yet the rotationalmotion looks rather complicated. With the powerful tools provided by theLagrangian and Hamiltonian formalisms, we are well-equipped to tackle thissubject and go beyond to more complicated examples. We start by definingthe concept of a rigid body in a precise way; we then proceed by introducingthe Euler angles that can be used to describe a snapshot of the orientationof an object in three dimensional space. With the scaffolding established, wecan then describe torque-free dynamics first, then full rotational evolutionwith non-zero torque. Throughout this chapter, we restrict our discussion tonon-relativistic dynamics for simplicity.

12.1 Rigid Bodies

A rigid body is defined as an object of finite size within which the distancebetween any two of its constituent bits is fixed. A metal rod, a glass hoop,a frisbee, all qualify as rigid bodies as long as we can ignore their tendencyto bend slightly or even break apart. Flexible rubber or honey would notbe considered rigid bodies. So far, we have treated a rigid body as a pointparticle located at its center of mass. In this chapter, we want to go beyondthis and describe the dynamics of the orientation of the object in three-dimensional space.

Figure ?? shows an arbitrarily shaped rigid body. The first task is to

447

CHAPTER 12. RIGID BODY DYNAMICS

quantify the object’s position and orientation. Pick any point within thebody, and imagine you tag it with a colored dot. As the object tumblesaround and moves, we can describe the trajectory of the colored dot as apoint particle. But this is not enough to describe the state of the rigidbody. In addition, we need to describe the orientation of the object aboutthe colored dot. Hence, if we provide, for any instant in time, the location ofthe colored dot and the orientation of the object about the colored dot, wecan reconstruct the state of the object at that instant in time. Note that thechoice of the colored dot within the body of the object is arbitrary. We maythen say that the state of a rigid body can be described by a combination ofa translation and a rotation: a translation of a fixed point of our choosingwithin the body that we henceforth call the tagged point ; and a rotation ofthe object about this fixed point. This rather intuitive statement arises fromEuler’s theorem: all spatial transformations that leave distances unchangedmust be a combination of a translation and a rotation.

While it is possible to use any point within a rigid body as the taggedpoint, there are a couple of particularly convenient choices. If the rigid bodyhas a fixed pivot point – like a pendulum swinging from a pivot – then ajudicious choice for the tagged point is the pivot point itself, as shown inFigure ??. On the other hand, if the object does not have an obvious pivot,the best choice for a tagged point is often the center of mass

Rcm =∑i

∆miriM

=1

M

∫r dm =

1

M

∫ρ r dV . (12.1)

Here, we have divided the rigid body into small bits labeled by the index ilocated at positions ri, where the total mass is M =

∑imi (see Figure ??).

The sum is over all bits of the rigid body, which, in the continuum limit, wecan write as an integral of bits of mass dm = ρ dV , where ρ is the volumedensity of the object. Note that a rigid body may not necessarily haveconstant volume density; i.e., we have in general ρ(r). All this is fine andcute, but how do we describe the orientation of the object about the taggedpoint, whatever it may be?

12.2 Rotations

A rotation is a linear transformation of coordinates that leaves distancesunchanged. To describe a rotation in three dimensions, we need three angles:

12.2. ROTATIONS

think for example of the orientation of an airplane, as illustrated in Figure ??.Hence, three numbers are needed to fully describe the orientation of a rigidbody. Another way to see this is by realizing that the orientation of an objectcan be fully described by specifying an axis of rotation and a single angle ofrotation about this axis (see Figure ??). Identifying the axis requires twonumbers – say the two angles in spherical coordinates that define a ray; plusthe single angle of rotation about this axis. This gives again three angles intotal to describe the orientation of the object.

Consider a rotation described by a three by three matrix R acting on avector r

R · r = r′ . (12.2)

Since the defining property of a rotation is that it does not change distance,we then need

r · r = r′ · r′ . (12.3)

This implies

r · r = r′ · Rt · R · r′ (12.4)

where t stands for matrix transposition. We then need

Rt · R = 1 ; (12.5)

that is, the matrix R must be orthogonal. We note however that R may alsoimplement a reflection of the coordinates while preserving distances and thestatement of orthogonality. We want to separate reflections from rotations.To do this, we define a rotation matrix as one satisfying both of the followingconditions

Rt · R = 1 , Det(R)

= +1 . (12.6)

The condition on the determinant rules out reflections. Equations (12.6)define a rotation in general – in fact in any dimension of space.

We can then ask a mathematics question: what are all three by threematrices R that satisfy the conditions (12.6)? This exercise leads to thefollowing conclusion: any such matrix is parameterized by three independent


angular parameters. For example, a rotation about the z axis would looklike

Rz =

cosαz sinαz 0− sinαz cosαz 0

0 0 1

(12.7)

where αz is the angle of rotation. One about the x and y axes would looklike

Rx =

1 0 00 cosαx sinαx0 − sinαx cosαx

, Ry =

cosαy 0 sinαy0 1 0

− sinαy 0 cosαy

.(12.8)

The total rotation can be written as a product of three such matrices withthree independent parameters αx, αy, and αz

R(αx, αy, αz) = Rx · Ry · Rz . (12.9)

As required, a product of orthogonal matrices with determinants equal tounity is orthogonal and has determinant equal to unity as well. Note thatrotations matrices about different axes are non-commuting; that is, we have

Rx · Ry 6= Ry · Rx , Rx · Rz 6= Rz · Rx etc... . (12.10)

Thus, Euler’s suggestion of decomposing rigid body dynamics into a trans-lation motion of a tagged point and a rotation about it can be summarizedas follows: for every bit of mass ∆mi of the rigid body, one can write itsposition in the laboratory frame as

ri(t) = a(t) + R(αx(t), αy(t), αz(t)) · ri(0) (12.11)

where a(t) is the position of the tagged point as a function of time, with theconvention a(0) = 0.

12.3 Infinitesimal Rotations

Consider the time evolution of a rigid body – decomposed as prescribedinto a translation and a rotation. Let the position of the tagged point bedenoted by r. In a small time increment ∆t, the tagged point translatesby a small amount ∆r. More interestingly, how can we describe the small

12.3. INFINITESIMAL ROTATIONS

angular rotation of the object during this time interval? At the given instantin time, the rotation of the object can be quantified by prescribing an axis ofrotation as shown in Figure ??; and a small rotation about it ∆α. The axisgives a direction in three dimensional space, and the angle a scalar number.We can represent this set of three numbers as a vector ω: the direction ofω is along the instantaneous axis of rotation; and its magnitude is definedas ∆α/∆t. Naturally, this vector is called the angular velocity vector. Asthe system evolves, the direction of the axis of rotation and the infinitesimalrotation change; hence, in general the vector ω evolves in both directionand magnitude. Thus, describing the evolution of ω(t) is tantamount todescribing the rotational evolution of the object about the tagged point.Note that the direction of the spin of the body can be obtained from ωeasily using the right hand rule, with the thumb pointing in the direction ofω.

We can easily relate ω to the change in the orientation of the rigid body.Figure ?? shows how a fixed point within the rigid body whose position isdenoted by r′i evolves in time – given ω at an instant in time. We have

|∆r′i| = ∆α (r′i sin θ) (12.12)

or equivalently

dr′idt

= ω × r′i (12.13)

a relation that should look familiar from our discussion of rotating coordinatesystems in Chapter 9. Note that, since the velocity dr′i/dt (the left hand sideof (12.13)) can be decomposed into a vectorial sum of component velocities –i.e., the Galilean velocity addition rule applies – the angular velocity ω (theright hand side of (12.13)) can also be decomposed into component angularvelocities. For example, we may have

ω = ω1 + ω2 (12.14)

where ω1 and ω2 describe a decomposition of the full infinitesimal rotationinto two alternative ones with different axes and magnitudes, as shown inFigure ??. Since addition is a commutative operation, this means that in-finitesimal rotations are commutative

ω1 + ω2 = ω2 + ω1 . (12.15)


However, once a large rotation is built out of infinitesimal ones, large ro-tations are no more commutative in general since they involve product ofnon-commuting matrices.

EXAMPLE 12-1: Rotations in higher dimensions

Consider D dimensional space, where D ≥ 2. What would a rotation be in such aspace of arbitrary dimensions? The defining property of rotations still comes from requiringlinear transformations that preserve distances. Hence, we still would use equation (12.6) asa condition on a D by D rotation matrix R. In the boring world of three dimensions, wecould have three independent rotations: one about each of the x, y, and z axes. Lookingat the form of the rotation matrix, we see that the orthogonality condition gets satisfied bya cosine-sine mixing of two directions at a time. For example, for rotation about the z axis,we mix the x and y axes using the familiar trigonometric functions; and orthogonality followsfrom cos2 + sin2 = 1. In higher dimensions, we can still satisfy the orthogonality condition bymixing any two of the D directions of the coordinates with the same cosine and sine pattern.Hence, a general rotation in D dimensions is a product of a number of rotations, each mixingtwo of the D space dimensions. The number of different rotations in D dimensions is thensimply the number of independent ways we can pair up D axes. That is

Number of independent angles in D dimensions =D(D − 1)

2. (12.16)

For example, for D = 2, we have naturally one angle. D = 3 gives three. But for D = 9,

we would have 9× 8/2 = 36 different independent rotations – and not nine as you may have

naively guessed...

12.4 The Euler Angles

We know that an arbitrary orientation of a rigid body involves a three bythree rotation matrix decomposable into a product of three independent rota-tions. Hence, the orientation state of the object is described by three angularparameters. This decomposition is not unique. For example, we may write

R(αx, αy, αz) = Rx(αx) · Ry(αy) · Rz(αz) . (12.17)

Or even

R(α′x, α′y, α

′z) = Ry(α

′y) · Rx(α

′x) · Rz(α

′z) = R(αx, αy, αz) . (12.18)

with a different set of angles α′x, α′y, and α′z. Hence, we need to establish

a convention. If your name happens to be Euler, then the convention you

12.4. THE EULER ANGLES

introduce has a greater chance to stick around. Hence, we will adhere toEuler’s convention, and the corresponding set of angles knowns as the Eulerangles.

A general rotation is henceforth defined through three rotations

R(ϕ, θ, ψ) = R3(ψ) · R2(θ) · R1(ϕ) (12.19)

where we have:

• A rotation about z axis by ϕ

R1(ϕ) =

cosϕ sinϕ 0− sinϕ cosϕ 0

0 0 1

(12.20)

• A rotation about the x axis by θ

R2(θ) =

1 0 00 cos θ sin θ0 − sin θ cos θ

(12.21)

• A rotation about the z axis by ψ

R3(ψ) =

cosψ sinψ 0− sinψ cosψ 0

0 0 1

(12.22)

Figure ?? depicts these three angles ϕ, θ, and ψ known as the Euler angles.Say we are given a vector a in the coordinate system of the lab; we denote

the components by (ax, ay, az). A rigid body is oriented with respect to thiscoordinate system in such a way that its state is described by three Eulerangles ϕ, θ, and ψ. We want the coordinates of the vector a as seen from theperspective of the rigid body. Denoting these coordinates by (ax′ , ay′ , az′),we then write wx

wywz

= R(ϕ, θ, ψ).

wx′wy′wz′

⇒

wx′wy′wz′

=[R(ϕ, θ, ψ)

]t.

wxwywz

(12.23)


where we transpose the rotation matrix to invert it, given that it is an or-thogonal matrix. We also note that[

R(ϕ, θ, ψ)]t

=[R1(ϕ)

]t.[R2(θ)

]t.[R3(ψ)

]t= R1(−ϕ).R2(−θ).R3(−ψ) . (12.24)

In this manner, we can transform back and forth between the laboratory andrigid body frames. For example, the vector w may be chosen as the positionvector of a fixed point within the body of the object; this implies that itscomponents would be constant in time in the rigid body frame. From thelaboratory perspective, the components of this vector would then evolve intime as the object tumbles away. This evolution would then be quantifiedthrough three functions of time, ϕ(t), θ(t), and ψ(t). Before looking into howone finds these functions of time from the laws of dynamics, let us look at aparticularly important example.

EXAMPLE 12-2: Angular velocity transformation

Consider the angular momentum vector ω introduced earlier that describes the instanta-neous rate of rotation of a rigid body. Remember that, given ω at a snapshot in time, weknow the object of interest is spinning about the direction of this vector at a rate ω at thatinstant in time. Hence, if we can find out the time evolution of the components of the angularvelocity vector from the lab perspective, we can reconstruct the rotational dynamics of therigid body.

As before, we label the components of ω in the laboratory frame by (ωx, ωy, ωz); andthose in the rigid body frame by (ωx′ , ωy′ , ωz′). And we now know how to relate these givenϕ(t), θ(t), and ψ(t).

Since the orientation of the object is tracked by the Euler angles, it should be possible towrite the angular velocity vector in terms of ϕ, θ, and ψ. To do this, we divide ω into threeparts, as shown in Figure ??

ω = ω(I) + ω(II) + ω(III) . (12.25)

We can read off these individual spin rates from the Figure as

ω(I) = ϕ , ω(II) = θ , ω(III) = ψ , (12.26)

each aligned as shown in the Figure. And we can then write the components of ω(I), ω(II),ω(III) in the body frame, and transform back to the laboratory frame. For example, we have

ω(I) = R(ϕ, θ, ψ).

00ϕ

=

ϕ sin θ sinψϕ sin θ cosψϕ cos θ

. (12.27)

12.5. ROTATIONAL KINETIC ENERGY

For ω(II), we need to be more careful since its components need to be rotated back to thelaboratory only partially. We have

ω(II) = R3(ψ).

θ00

=

θ cosψ

−θ sinψ0

. (12.28)

Finally, for ω(III), we have

ω(III) = R3(ψ).

00

ψ

=

00

ψ

. (12.29)

The total angular velocity vector is a sum of the three component vectors; in the laboratoryframe, this becomes

ω =(ϕ sin θ sinψ + θ cosψ, ϕ sin θ cosψ − θ sinψ, ϕ cos θ + ψ

). (12.30)

Given ϕ(t),θ(t), and ψ(t), we can then determine the angular velocity vector at any instant

in time. Conversely, if laws of physical dynamics were to tell us the components of ω in the

laboratory frame, we would have three first order differential equations for ϕ(t),θ(t), and ψ(t).

12.5 Rotational kinetic energy

In order to write the Lagrangian for a rigid body, we need to write its kineticand potential energies. We start with kinetic energy. We set up our labora-tory coordinate system as shown in Figure ?? and divide up the rigid bodyinto bits of small mass ∆mi. We denote the tagged point’s location by theposition vector R. We then have

ri = R+ r′i (12.31)

for every bit of the rigid body. The velocity of a bit is then given by

vi = R+ ω × r′i = V + ω × r′i (12.32)

using (12.13), where V is the translational velocity of the tagged point. Thekinetic energy of the rigid body is simply the sum of the kinetic energies ofall the bits; we write

T =1

2

∑i

∆miv2i =

1

2

∑i

∆mi (V + ω × r′i) · (V + ω × r′i) . (12.33)


Expanding this expression, we identify three different parts

T =1

2MV 2 +

∑i

∆miV · (ω × r′i) +1

2

∑i

∆mi (ω × r′i) · (ω × r′i)

= T1 + T2 + T3 (12.34)

To simplify things further, it helps to choose a strategic tagged point in thedecomposition of the dynamics into translation and rotational parts. Asmentioned earlier, we consider two possibly choices:

• We choose R at center of mass of the rigid body

R = Rcm (12.35)

We then have

T1 =1

2MVcm

2 (12.36)

where Vcm is the linear velocity of the center of mass; and

T2 = M Vcm · (ω ×R′cm) = 0 (12.37)

since R′cm = 0: this is the position vector of the center of mass in thecenter of mass reference frame, which is obviously the zero vector; thatis, the center of mass is at the origin of the center of mass referenceframe.

• We choose R at a fixed pivot point

R = Constant⇒ V = 0 . (12.38)

We then immediately have

T1 = T2 = 0 . (12.39)


We are then left with T3 to deal with

T3 = Trot =1

2

∑i

∆mi (ω × r′i) · (ω × r′i) . (12.40)

Suggestively, we call it the rotational kinetic energy. To simplify thisexpression further, we make use of two vector identities

A · (B ×C) = B · (C ×A) = C · (A×B) (12.41)

A×B = −B ×A (12.42)

which allow us to write

(ω × r′i) · (ω × r′i) = −r′i · (ω × (ω × r′i)) . (12.43)

We now use the identity

A× (B ×C) = (A ·C)B − (A ·B)C (12.44)

which allows us to simplify things further

(ω × r′i) · (ω × r′i) = −r′i ·((ω · r′i)ω − ω2r′i

)= − (ω · r′i)

2+ ω2r′i

2(12.45)

Putting things back into the original expression for Trot, we get

Trot =1

2

∑i

∆mi

(ω2r′i

2 − (ω · r′i)2)

=1

2ωa

[∑i

∆mi

(r′i

2δab − (r′i)a (r′i)b

)]ωb

=1

2ωaIabωb =

1

2ωt · I · ω (12.46)

where we define

Iab ≡∑i

∆mi

(r′i

2δab − (r′i)a (r′i)b

)=

∫dm(r′

2δab − (r′)a (r′)b

)(12.47)


In the last step, we take the continuum limit where ∆mi → dm with

dm = ρdV = σdS = λdl (12.48)

where ρ, σ, or λ are respectively the volume, surface, or linear mass densitiesof the rigid body. In essence, we have factored away the angular velocitiesfrom the expression for rotational kinetic energy; and the rest, denoted as amatrix I, depends only on the way mass is distributed within the shape ofthe rigid body. It plays the role of rotational inertia for a rigid body, and iscalled the moment of inertia matrix.

Hence, given a rigid body, we would compute its moment of inertia matrixfrom (12.47); and, given its angular velocity ω, we can compute its rotationalkinetic energy from

Trot =1

2ωt · I · ω . (12.49)

Note that, when this expression is expanded in component form, we need tochoose a coordinate system. Hence, much like the components of the angularvelocity vector depend on the reference frame, so do the components of themoment of inertia matrix Iab. Under an arbitrary rotation, these componentswould transform as

I → R · I · Rt(12.50)

as it follows from (12.47). Furthermore, looking back at (12.47), we noticethat the components also depend on the choice of tagged point in the rigidbody: changing the tagged point will shift the r′i’s, and hence change Iab.

Thus, the components of the moment of inertia matrix depend on thecoordinate system and the location of the tagged point. However, the ro-tational kinetic energy is a scalar under rotations: this means that under arotation of the coordinate system, we would have

Trot =1

2ωt · I ·ω → 1

2ωt · Rt · R · I · Rt · R ·ω =

1

2ωt · I ·ω .(12.51)

Under a rotation of the coordinate system, the rotational kinetic energy isunchanged. Under a translation of the coordinate system however, whilethe components ω remain unchanged, the moment of inertia matrix andconsequently the rotational kinetic energy will change.

Finally, we not that the moment of inertia matrix is necessarily symmetric

Iab = Iba (12.52)


as can be seen from (12.47). If the kinetic energy is to be quadratic in theangular velocity, the momenta of inertia tensor had to be symmetry: anyarbitrary matrix Jab can be written as a sum of a symmetric matrix and ananti-symmetric one; and the expression ωaJabωb would not get a contribu-tion from any anti-symmetric part of Jab because of the commutativity ofmultiplication.

EXAMPLE 12-3: A hoop

Consider a hoop of radius R, mass M , and constant mass density. We want to computethe moment of inertia matrix for this rigid body. First, we need to choose a tagged point anda coordinate system. On possibility is shown in Figure ??, where the tagged point is at thecenter of the hoop, and the coordinate axes are aligned as shown. Given that this is an objectwith some constant linear mass density λ, we start from

Iab =

∫λ(r′

2δab − (r′)a (r′)b

)dl (12.53)

where the integral goes around the hoop with dl = Rdθ, and

λ =M

2π R. (12.54)

The coordinates of the bits in the hoop are then traced by the angle θ

x′ = R cos θ , y′ = R sin θ , z′ = 0 . (12.55)

We then get

Ixx =M

2π R

∫ 2π

0

(R2 −R2 cos2 θ

)dθ (12.56)

Iyy =M

2π R

∫ 2π

0

(R2 −R2 sin2 θ

)dθ (12.57)

Izz =M

2π R

∫ 2π

0

(R2 − 0

)dθ (12.58)

for the diagonal components. For the off-diagonal components, we only need to compute halfof them due to the fact that the matrix is symmetric

Ixy = Iyx =M

2π R

∫ 2π

0

(R2 cos θ sin θ

)dθ (12.59)


Ixz = Izx =M

2π R

∫ 2π

0

(0) dθ (12.60)

Iyz = Izy =M

2π R

∫ 2π

0

(0) dθ (12.61)

Simplifying things, we get

Ixx =M R2

2, Iyy =

M R2

2, Izz = M R2 , Ixy = Ixz = Iyz = 0 . (12.62)

In matrix notation, the moment of inertia matrix of the hoop about its center is then

I =

M R2/2 0 00 M R2/2 00 0 M R2

(12.63)

The reader may want to attempt the same computation with the origin of the coordinate

system shifted to the rim of the hoop. We will come back to this case later with a trick up

our sleeves.

12.6 Potential Energy

We next tackle the potential energy of a rigid body. We focus on the uniformgravitational potential energy, but the procedure for any other potential en-ergy is similar. In general, we divide the rigid body into bits as before andwe write the total potential energy as the sum of the potential energies ofthe small bits

U =∑i

Ui (12.64)

For uniform gravity, we have

U =∑i

∆mig hi =∑i

∆mig ri · z = M gRcm · z = M gH (12.65)

where H is the height of the center of mass. Hence, as far as uniform gravityis concerned, the rigid body behaves as if all its mass is concentrated at itscenter of mass. We say that gravity pulls on the rigid body at its centerof mass. This is partly why splitting the dynamics of a rigid body into thetranslation dynamics of the center of mass plus a rotational one about thecenter of mass can sometimes be very useful.

12.7. ANGULAR MOMENTUM

With other potentials, this exercise may not lead to such a simple ex-pression. For example, even for the full form of Newtonian gravity, we wouldhave

U =∑i

−GM ∆mi

ri(12.66)

which in general does not simplify further.

12.7 Angular Momentum

It will be useful to also develop a simple expression for the angular momentumof a rigid body. To do this, once again we start by dividing up the rigid bodyinto small bits of mass ∆mi, and we write the total angular momentum withrespect to a tagged point as the sum of individual angular momenta of thebits

L =∑i

ri × (∆mivi) =∑i

(R+ r′i)× (V + ω × r′i)

= R× (M V ) +R× (ω ×M R′cm) +∑i

∆mir′i × (ω × r′i)

+ M R′cm × V = L1 +L2 +L3 +L4 (12.67)

where R′cm is the position vector for the center of mass determined withrespect to the tagged point as the origin∑

∆mir′i = M R′cm . (12.68)

Contrast this with∑∆miri = M Rcm . (12.69)

where Rcm is determined from the origin of the laboratory frame. We nowconsider two convenient scenarios as before:

• We choose R at center of mass

R = Rcm ⇒ R′cm = 0⇒ L2 = L4 = 0 (12.70)

We necessarily have R′cm = 0; i.e., the center of mass is at the originin the center of mass coordinate system. We then also have

V = Vcm ⇒ L1 = Rcm × (M Vcm) (12.71)


• We choose R at a fixed pivot point

R = Constant⇒ V = 0 . (12.72)

We then have

L1 = L4 = 0 (12.73)

Often we choose our coordinate system at the pivot point and computethe angular momentum with respect to the pivot point; we then have

R = 0⇒ L2 = 0 (12.74)

The last piece of the puzzle can be written as

L3 = Lrot =∑i

∆mir′i × (ω × r′i)

=∑i

∆mi

(r′i

2)ω −∆mi (ω · r′i) r′i (12.75)

which we have written in terms of an expression looking like the moment ofinertia matrix using the same vector identities encountered earlier. We callthis piece of the total angular momentum rotational angular momentum.Since the angular momentum is a vector, its components are then

(Lrot)a =∑i

∆mir′i2ωa −∆mi (r

′i)a (r′i)b ωb

=

[∑i

∆mir′i2δab −∆mi (r

′i)a (r′i)b

]ωb = Iabωb . (12.76)

In short, we find

Lrot = I · ω . (12.77)

Once again, the moment of inertia matrix plays the role of inertia for ro-tational dynamics. Note that this expression depends on the origin of thecoordinate system, and the orientation of the axes. Unlike rotational ki-netic energy, both a rotation of the axes or a translation of the origin wouldchanges the rotational angular momentum. For example, a rotation changethe components of the angular momentum as it does those of any vector

Lrot → R ·Lrot . (12.78)

12.8. TORQUE

12.8 Torque

Torque is defined as the rate of change of angular momentum. Let us remindourselves of its origin. We start by writing Newton’s second law for every bitof mass making up a rigid body,

Fi = ∆miai = ∆midvidt

. (12.79)

We now cross both sides of this equation by the position vector of each bitri and sum over i∑

i

ri × Fi =d

dt

∑i

∆miri × vi =dL

dt. (12.80)

On the left hand side, the quantity is then define as the torque

τ ≡∑i

ri × Fi . (12.81)

Typically, in this sum, forces between the bits making up the rigid bodycancel pairwise due to Newton’s third law. So, we may write instead

τ =∑i

ri × F exti . (12.82)

where F exti account for external forces only acting on the rigid body. We then

have a different form of Newton’s second law useful for studying rotationaldynamics

τ =dL

dt. (12.83)

As for angular momentum, we can divide up torque into two differentparts by splitting the dynamics of the rigid body into a translational and arotational part. Substituting ri = r′i+R into the definition of torque (12.82),we get

τ =∑i

r′i × F exti +R× F ext

i = τrot +R× F ext , (12.84)

where the second term is the torque acting on the rigid body as a whole.


12.9 Summary

Let us take a moment to summarize the conclusions. Given the dynamics ofa rigid body, we decompose the motion into a translational part about whatwe call a tagged point, and a rotation about the tagged point. Usually, wechoose the tagged point as a stationary pivot (if any) or the center of mass.The total kinetic energy of the rigid body is then given by

T = 12ωt · I · ω Stationary pivot decomposition

T = 12M V 2

cm + 12ωt · I · ω Center of mass decomposition .

.(12.85)

The total angular momentum is given by

L = I · ω Stationary pivot decomposition

L = Rcm ×M Vcm + I · ω Center of mass decomposition ..(12.86)

Newton’ second law can be recast into a rotational form. We have

τ = R× F ext + τrot =dL

dt(12.87)

which splits into two pieces: (1) For either the stationary pivot or center ofmass decomposition, we have

τrot =d

dt

(I · ω

). (12.88)

In addition, for the center of mass decomposition, we can also write

R× F ext =d

dt(Rcm ×M Vcm) (12.89)

as a statement about the angular momentum of the center of mass.

12.10 Principal Axes

Since the choice of coordinate system affects the components of the momentof inertia matrix, we would want to choose the coordinate system that ismost strategic in making the computation of the moment of inertia easiest.While it may be tempting to choose the laboratory coordinate system forall computations, a rigid body that is tumbling around with respect to thelaboratory would have the components of its moment of inertia matrix change

12.10. PRINCIPAL AXES

in time from such a perspective! The distribution of mass bits in the bodywould be changing all the time relative to the laboratory. This is generallynot a good thing and we want to confine the time evolution dynamics toquantities like the angular velocity and orientation of the rigid body. Thisimplies that we need to choose the coordinate system fixed with respect to theorientation of the body. Then the mass distribution remains constant fromthis perspective, no matter how the object tumbles around in the laboratory.

Next, we want to fix the location of the origin of this coordinate system.The natural choice is the tagged point with respect to which the motion ofthe rigid body was decomposed into translational and rotational ones. Thisis generally a good choice.

We are not finished, however. Within these criteria, we can still orientour axes in infinitely many ways, as pictured in Figure ??. All these config-urations are related to each other by rotations, which change the moment ofinertia matrix components by

I → R · I · Rt. (12.90)

Since I is a real symmetric matrix, we can use this freedom to orient our axesso that the moment of inertia matrix is diagonal: effectively, we can alwaysfind an orthogonal transformation – a rotation – that diagonalizes any realsymmetric matrix. As shown in the Figure, this typically corresponds toaligning the axes with the symmetry axes of the shape of the rigid body.Hence, in practice, you do not need to find a hideous transformation to dothe job if your rigid body has enough symmetries; you can quickly guessat the appropriate orientation of your axes. This choice of axes for yourcoordinate system is referred to as the choice of the principal axes.

Henceforth, we always would compute the moment of inertia matrix inthe principal axes coordinate system – fixed at the tagged point in the rigidbody. In computing the rotational kinetic energy or the rotational angularmomentum, we need then to make sure that the components of the angularvelocity vector are expressed in this same coordinate system.

Referring to the moment of inertia matrix in this coordinate system withthe zero subscript, we then have

I0 =

I1 0 00 I2 00 0 I3

. (12.91)


In this coordinate system, we then need to compute only three moments ofinertia, one along each of the three principal axes. Hence, the physical contentof the moment of inertia matrix is quantified by three independent numbers,out of the six possible ones for a symmetric three by three matrix. Therest are related to these three by orthogonal transformations. The rotationalkinetic energy is then

Trot =1

2I1ω

21 +

1

2I2ω

22 +

1

2I3ω

23 (12.92)

while the rotational angular momentum is

Lrot = (I1ω1, I2ω2, I3ω3) (12.93)

given in the principal axes coordinate system. We may then also write

Trot =L2

1

2 I1

+L2

2

2 I2

+L2

3

2 I3

. (12.94)

Hence, once I1, I2, and I3 are computed, one can construct simple expres-sions for rotational kinetic energy and angular momentum.

EXAMPLE 12-4: Fixed Axis Rotation

Consider a rigid body restricted to rotate about a fixed axis which is also a principal axisof the object. This could be a barrel rolling down an incline or a two dimensional pendulum, asshown in Figure ??. In these scenarios, rotational dynamics simplifies significantly. We needto track the evolution of only one Euler angle. We may align the principal axes such that thebody z′ axis is along the fixed axis of rotation z, and describe the rotation of the body throughthe angle ϕ. The rotational angular velocity is then always aligned along the fixed axis andhas a magnitude ϕ. The angular momentum is also aligned with the same axis

L1 = L2 = 0 , L3 = I3ω3 = I3ϕ (12.95)

where the z or z′ direction is labeled 3. The rotational kinetic energy is then simply

Trot =1

2I3ω

23 =

1

2I3ϕ

2 =L2

3

2 I3. (12.96)

Throughout, the only component of the moment of inertia matrix that matters is I3, theinertia ‘along the fixed axes’. Dynamics then changes the rotational angular momentum inmagnitude but not in direction. Torque – defined as the rate of change of angular momentum

τ =dL

dt(12.97)

12.10. PRINCIPAL AXES

lies then along the same fixes axis. For the rotational part, we have

τrot =dL3

dt= I3ω3 = I3α (12.98)

where α = ϕ is called angular acceleration. The torque is still to be determined by the laws

of physics describing the environment of the rigid body, as usual.

EXAMPLE 12-5: Principal Axis Shifts

Consider a rigid body whose moments of inertia matrix is known in terms of the principalaxis coordinate system with the center of mass at the origin; that is, we already know I1, I2,and I3. The moment of inertia matrix however depends on the choice of the origin of thebody coordinate system used. Consider the scenario depicted in Figure ??: the origin of thebody coordinate system is shifted by a constant vector D. In this new coordinate system,the moment of inertia matrix is still diagonal since a constant shift is not an orthogonaltransformation that realigns the axes of the coordinate system. However, the values of I1, I2,and I3 will change. We want to find out the new eigenvalues of the moment of inertia matrix.

Going back to the definition of the definition of the moment of inertia matrix givenin (12.47) and shift

r′i → r′i +D . (12.99)

We then have for the relevant diagonal components

Iaa =∑i

∆mi

(r′i

2δaa − (r′i)a (r′i)a

)→

∑i

∆mi

((r′i +D)2 − (r′i +D)a (r′i +D)a

)(12.100)

with no sum over a implied. Noting that the center of mass is located at the origin of a centerof mass coordinate system, we then have all cross terms dropping

Iaa → Iaa −∑i

∆miDaDa = Iaa −MDaDa . (12.101)

That is, we shift the eigenvalues in the direction of D as if the entire mass is located at thecenter of mass. For example, going back to the moment of inertia hoop encountered earlier,if we were to shift to the rim of the hoop along say the 1 direction, we get

I =

M R2/2 0 00 M R2/2 00 0 M R2

+

M R2 0 00 0 00 0 0

=

3M R2/2 0 00 M R2/2 00 0 M R2

(12.102)


12.11 Torque Free Dynamics

We start with a deceptively simple scenario: torque free motion. In somecultures, throwing a shoe at someone is a expression of deep discontent –a fact that a U.S. president became particularly aware of. Imagine youthrow your shoe at a friend as a gesture of discontent. Decomposing themotion about the center of mass, the center of mass dynamics of the shoeis a simple problem in trajectory physics: it follows a classic parabolic path.The interesting part is the rotational motion about the center of mass. Wefocus on the rotational dynamics only. Since our discussion of uniform gravityearlier indicated that the gravitational pull acts at the center of mass, thereis no torque acting on the shoe

τrot ≡ R′cm × F ext = 0 (12.103)

since R′cm = 0. So, the rotational dynamics is torque-free. Yet, the shoe willtumble in a complex pattern as it flies towards your friend.

While we can solve this problem by setting up a Lagrangian for it, itwill be better to follow a more traditional strategy to help us develop betterintuition about rotational dynamics. The Lagrangian version of this problemwill be left as an exercise to the reader. We start from Newton’s second lawin rotational form

dLrot

dt

∣∣∣∣lab

= τrot = 0 (12.104)

noting that we have written this statement in the inertial laboratory frame.We henceforth drop the ‘rot’ subscripts from expressions. It is more conve-nient to describe the dynamics first from the perspective of the body referenceframe. But we can shift to the body reference frame by

dL

dt

∣∣∣∣lab

=dL

dt

∣∣∣∣body

+ ω ×L = 0 (12.105)

demonstrating already that the angular momentum will be doing interestingthings as see from the body perspective even when there are no torques.Choosing the principal axes for the body coordinate system, we have

L = (I1ω1, I2ω2, I3ω3) (12.106)

and

Ia = 0 . (12.107)

12.11. TORQUE FREE DYNAMICS

Equation (12.105) then becomes a set of first order coupled differential equa-tions of ω1, ω2, and ω3

I1ω1 = (I2 − I3) ω2 ω3 (12.108)

I2ω2 = (I3 − I1) ω1 ω3 (12.109)

I3ω3 = (I1 − I2) ω1 ω2 (12.110)

While this system can readily be solved in terms of elliptic integrals, let usassume for the sake of simplicity that the rigid body is a surface of revolution

I1 = I2 = I (12.111)

as depicted in Figure ??. We then immediately have

I3ω3 = 0⇒ ω3 = ω0 = constant (12.112)

which implies

ω1 =I − I3

Iω2 ω0 , ω2 = −I − I3

Iω1 ω0 . (12.113)

These two equations can now be solved easily by differentiating with respectto time

ω1 = −ω20

(I − I3)2

I2ω1 = −Ω2ω1 . (12.114)

Using the boundary conditions

ω1(0) = 0 . ω2(0) = A (12.115)

leads to the solution

ω = (A sin (Ω t) , A cos (Ω t) , ω0) (12.116)

Note that these are the coordinates of the angular velocity vector in thebody principal axes reference frame. Figure ?? depicts the implied dynamicsfrom this perspective. Remember that ω is along the instantaneous axisof rotation; hence, as the vector spins around the symmetry axis of therigid body, this implies the body is tumbling around. To see this better,remember that the angular momentum vector is constant in time as seenfrom the laboratory; constant both in direction and magnitude. From the


laboratory perspective, the tumbling would look like as shown in Figure ??:the

EXAMPLE 12-6: Adding Angular Momenta

Ω0 = ω −Ω (12.117)∑i

∆ri =∑i

ω∆t× r (12.118)

Ω20 = Ω2 + ω2 − 2 Ωω cosα (12.119)

cosα =ω0

ω(12.120)

Ω0 =L

I(12.121)

More

Ω = ω0I − I3

I(12.122)

Sign

Ω ' ω0

300(12.123)

T = 300 days, lack of rigidity gives T = 400 days.

I1 6= I2 6= I3 (12.124)

Two conserved quantities

L2 = L21 + L2

2 + L23 (12.125)

Sphere

T =L2

1

2 I1

+L2

2

2 I2

+L2

3

2 I3

(12.126)

Ellipsoid with 2T I1, 2T I2, 2T I3

Choose

I1 > I2 > I3 (12.127)√2T I3 ≤ L ≤

√2T I1 (12.128)

Direction 2 is unstable

12.12. GYROSCOPES

12.12 Gyroscopes

We are now ready to tackle the full problem of a rigid body under the influ-ence of non-zero torque. This more complicated scenario lends itself to thepowerful technology of Lagrangian mechanics. The premise is to write theLagrangian of the rigid body as

L = T − U (12.129)

T = Trot =1

2I1ω

21 +

1

2I2ω

22 +

1

2I3ω

23

=1

2I1

(ϕ sin θ sinψ + θ cosψ

)2

+1

2I2

(ϕ sin θ cosψ − θ sinψ

)2

+1

2I3

(ϕ cos θ + ψ

)2

(12.130)

V = M gH −M g l cos θ (12.131)

I1 = I2 = I 6= I3 (12.132)

T =1

2I(θ2 + ϕ2 sin2 θ

)+

1

2I3

(ϕ cos θ + ψ

)2

(12.133)

pψ =∂L

∂ψ= I3

(ψ + ϕ cos θ

)(12.134)

pϕ =∂L

∂ϕ= ϕ

(I sin2 θ + I3 cos2 θ

)+ ψ I3 cos θ (12.135)

H = T + U = E =I

2

(θ2 + ϕ2 sin2 θ

)+

I

2 I3

p2ψ +M g l cos θ (12.136)

θ2 +M g l

Icos θ +

(pϕ − pψ)2

I2 sin2 θ−

2E I3 − p2ψ

I I3

= 0 (12.137)

u = cos θ (12.138)

u2 +(pϕ − pψ)2

I2+

(2E I3 − p2

ψ

I I3

− M g l

Iu

)(u2 − 1

)= 0 (12.139)

LYAPUNOV

BEYOND EQUATIONS

Lyapunov

Life of ?

LYAPUNOV

Chapter 13

Complex systems

13.1 Chaos

475

CHAPTER 13. COMPLEX SYSTEMS

HOOKE

BEYOND EQUATIONS

Hooke

Life of ?

Chapter 14

Small oscillations

479

SEEDS OF QUANTIZATION

Beyond The Basics III:

Chapter 15

Seeds of Quantization

15.1 Beyond classical phase space

leonhard euler (1707 - september...

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