Chapter 5

Rotational motion

A rigid body is defined as a body (or collection of particles) where all mass points stay atthe same relative distances at all times. This can be a continuous body, or a collectionof discrete particles: the same equations of motion hold for both cases. A rigid bodywill move as a single entity, but it can change its orientation, and this motion can behighly nontrivial as we shall see.

The notion of a rigid body is an idealisation, since no real completely rigid bodies exist inthe real world. Firstly, real bodies consist of atoms and molecules which always undergovibrations (and electrons, as quantum particles, are never at rest). These complicationscan be ignored for macroscopic bodies. Secondly, it is always possible to deform anactual body, and this can happen even in the absence of external forces if for examplethe internal forces keeping the constituents apart are not strong enough to balance theattractive forces — or vice-versa!

Still, the idealised description is reasonable for many macroscopic solids, and provide agood description of for example tops, gyroscopes, bicycle wheels, falling sandwiches andtumbling cats.

Learning outcomes

At the end of this section, you should be able to

• identify appropriate degrees of freedom and coordinates for a rigid body;

• describe rotations using rotation matrices, and explain the general properties ofrotation matrices;

• define the inertia tensor and explain the relation between the inertia tensor, rota-tional kinetic energy and angular momentum;

• calculate the inertia tensor for simple objects;

• explain what is meant by principal axes of inertia and how they may be found;

• use the equations of motion for rotating bodies (Euler equations) to analyse themotion of rotating systems.

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5.1 How many degrees of freedom do we have?

Let us imagine a rigid body consisting of N discrete particles. Altogether this gives us3N coordinates. The requirement that the body is rigid, ie all the internal distancesare fixed, imposes constraints on these coordinates. We denote the distance betweenparticles i and j by rij. We can work out how many constraints and hence degrees offreedom we have:

N = 2 Here we have a single constraint r12 = r, and hence we have a total of 3 · 2− 1 = 5degrees of freedom.

N = 3 We now have 3 internal distances to be fixed: r12, r13 and r23, giving 3 constraintsand 3N − 3 = 6 degrees of freedom.

N = 4 Here there are 6 internal distances to be fixed, so we have 3 · 4− 6 = 6 degrees offreedom. Strictly speaking, there are 2 different configurations which both satisfythese 6 constraints, corresponding to rigid bodies that are mirror images of eachother. This ambiguity does not correspond to any physical degree of freedom,however.

N = 5 Now it appears there would be 10 constraints, corresponding to the 10 pairs ofparticles we have. However, if you construct a 5-particle body from a 4-particleone, you will see that once the fifth particle has been positioned relative to 3 ofthe others, the position relative to the fourth one is also given. (You may try thisfor yourself!) We therefore only have 9 constraints, and 3 · 5 − 9 = 6 degrees offreedom.

N ≥ 6 As was the case for N = 5, we will need to specify 3 relative distances to positionthe 6th particle relative to the other 5, 3 more for the 7th particle, etc. Thiscancels out the 3 coordinates that each new particle comes with, so we end upwith 6 degrees of freedom in every case.

In summary, we find that a rigid body has 6 degrees of freedom, except for N = 2, whichis a special case, and has only 5 degrees of freedom. A more careful analysis will revealthat there are only 5 degrees of freedom whenever the rigid body is 1-dimensional (allthe constituent particles are located on a single line), and 6 otherwise.

Three of these degrees of freedom can be taken to represent the position of the body,and it is natural to use the centre of mass coordinates of the body as corresponding gen-eralised coordinates. The three remaining degrees of freedom represent the orientationof the body, and it is natural to choose three angles as coordinates. We will come backto how these angles may be chosen later on.

Dynamically, the 3+3 degrees of freedom correspond to two different kinds of motion:the linear motion of the centre of mass, and the rotation of the body about its centreof mass. We can now see why there are 3 such degrees of freedom: they correspond torotations (changes in orientation) about 3 axes going through the centre of mass. Inthe case of a 1-dimensional body, there are only 2 rotational degrees of freedom, sincerotation about the line the body is located on does not correspond to any real motion.

This discussion can be summarised in Chasles’ theorem, which states

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Any motion of a rigid body is the sum of a translation and a rotation.

From now on we will focus on how we can describe and study the rotational motion anddegrees of freedom.

5.1.1 Relative motion as rotation

Assume first that one point in the body is fixed. We can choose this point to be theorigin of our coordinate system. If we now move a point P relative to this origin, itfollows from the definition of a rigid body that although the position ~r of P changes,the distance r of P from the origin does not change, and the displacement d~r ⊥ ~r. Thisdefines a rotation of an angle dφ about some axis passing through the origin. Since allthe points in the body remain at a fixed distance relative to the origin, they all rotateabout an axis through the origin, and since the body retains its shape, they all rotateabout the same axis.

This can be summarised in Euler’s theorem:

Any movement of a rigid body with one point fixed is a rotation about someaxis.

If we say that the vector d~φ points along the axis of rotation, and this vector forms andangle θ with the position vector ~r, we have

|d~r| = |~r| sin θ|d~φ| or d~r = d~φ× ~r . (5.1)

The velocity of the point P is then

~v =d~r

dt=d~φ

dt× ~r = ~ω × ~r . (5.2)

If in addition, the whole body moves with some linear velocity ~V , the total velocity ofthe point P is

~v = ~V + ~vrel = ~V + ~ω × ~r (5.3)

You have in the past studied the case where the axis of rotation, ie the direction of ~ω isfixed. But in the general case, ~ω = ~ω(t) can change both magnitude and direction, andwe need to describe this general situation.

5.2 Rotated coordinate systems and rotation matri-

ces

Before we go on to discuss the kinematics and dynamics of rotational motion, let uslook at the coordinates we can use to describe the rigid body. It is clearly convenient todescribe the shape of the body, or the relative positions of the constituent parts of thebody, using a coordinate system that is sitting in the body and moving with it. It willalso turn out to be convenient to describe the rotational motion of the body in such a

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system, since it is natural to describe this as the body rotating about its own axes. Wecall such a coordinate system the body coordinate system.

But to describe the motion of the body in space we need a coordinate system fixed inspace, not moving with the body. We therefore need to know the relation between thebody coordinate system and this fixed coordinate system.

Ignoring any linear displacements, these coordinate systems will be rotated relative toeach other. We therefore need to know how coordinates change when the coordinatesystem is rotated.

Let us call the original coordinates of a point (eg in the fixed coordinate system)~r = (x, y, z) and the coordinates in the rotated (eg the body) coordinate system~r′ = (x′, y′, z′). We call the angles between the axes in the original and the rotatedcoordinate systems θij, i = x, y, z: for example, θxz is the angle between the x′-axis andthe z-axis. We can then write the position vector (or indeed any vector) as

~r = xx+ yy + zz = x′x′ + y′y′ + z′z′ . (5.4)

Since x′, y′, z′ are orthogonal, we can find the rotated coordinate x′ by

x′ = ~r · x′ = xx · x′ + yy · x′ + zz · x′ = x cos θxx + y cos θxy + z cos θxz . (5.5)

Similarly we find

y′ == x cos θyx + y cos θyy + z cos θyz , z′ == x cos θzx + y cos θzy + z cos θzz , (5.6)

This can be written as a matrix equation,x′y′z′

=

cos θxx cos θxy cos θxzcos θyx cos θyy cos θyzcos θzx cos θzy cos θzz

xyz

⇐⇒ ~r′ = A~r (5.7)

The 3× 3 matrix A is called the rotation matrix.

5.2.1 Active and passive transformations

There are two equivalent ways of thinking about rotations:

1. You rotate the coordinate system, as described above. This is called a passiverotation, since it is not doing anything to the world, only to the mathematicaldescription of it. The rotation matrix then gives the relation between the old andthe new coordinates.

2. You leave the coordinate system in place, but rotate your points (for example thebody you want to describe) in the opposite direction. This is called an activerotation, since you are now doing something to the world. The rotation matrixthen gives the positions of the body after it has been rotated, given the positionbefore.

The two points of view are mathematically equivalent: the relation between old andnew coordinates are exactly the same.

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The two pictures: active and passive, and the equivalence between them, can be extendedto other transformations, such as translations, where it is the same whether you movean object from a position x to a position x + a or you shift the coordinate system by−a.

5.2.2 Elementary rotation matrices

For a rotation about the z-axis, it is clear that the z-coordinates are unchanged. If werotate an angle θ, then we have that θxx = θyy = θ (the angles between the old and newx- and y-axes respectively). The angle θxy between the old y-axis and the new x-axis is90 − θ, while the angle θyx between the old x-axis and the new y-axis is 90 + θ. Therotation matrix is therefore

Az(θ) =

cos θ cos(90 − θ) 0cos(90 + θ) cos θ 0

0 0 1

=

cos θ sin θ 0− sin θ cos θ 0

0 0 1

(5.8)

Similarly we find for rotations about the x- and y-axes,

Ax(θ) =

1 0 00 cos θ sin θ0 − sin θ cos θ

, Ay(θ) =

cos θ 0 − sin θ0 1 0

sin θ 0 cos θ

. (5.9)

5.2.3 General properties of rotation matrices

1. Any combination of two successive rotations about the same point (albeit aboutdifferent axes through that point) is a rotation about that point.

This is ‘obvious’, since the relative distance of each point from the fixed point(origin) remains unchanged throughout. We can therefore describe the combinedrotation by a rotation matrix which is the product of the two rotation matrices.If we call the coordinates after the first rotation ~r′ and after the second rotation~r′′, and the two rotation matrices A (first rotation) and A′ (second rotation), wehave

~r′′ = A′~r′ = A′(A~r) = (A′A)~r = B~r with B = A′A . (5.10)

Note that the first rotation matrix is on the right and the second rotation matrixis on the left.

Example 5.1

Find the rotation matrix corresponding to a 30 rotation about the x-axisfollowed by a 45 rotation about the z-axis.

Solution: A rotation of an angle 30 about the x-axis is, according to (5.9),

Ax(30) =

1 0 00 cos 30 sin 30

0 − sin 30 cos 30

=

1 0 0

0 12

√3 1

2

0 −12

12

√3

. (5.11)

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The matrix for a 45 rotation about the z-axis is

Az(45) =

cos 45 sin 45 0− sin 45 cos 45 0

0 0 1

=

12

√2 1

2

√2 0

−12

√2 1

2

√2 0

0 0 1

. (5.12)

The combined rotation is

A = AzAx =

12

√2 1

2

√2 0

−12

√2 1

2

√2 0

0 0 1

1 0 0

0 12

√3 1

2

0 −12

12

√3

=

12

√2 1

4

√6 1

4

√2

−12

√2 1

4

√6 1

4

√2

0 −12

12

√3

(5.13)

2. Rotations (about different axes) are not commutative: the order in which you dothem matters.

We know that matrix multiplication is not commutative: AB 6= BA if A and Bare matrices. You can also show for yourself that if you for example rotate a book90 about the x-axis followed by 90 about the y-axis, you get something differentfrom doing the two in reverse order.

Example 5.2

The rotation matrix for a 45 rotation about the z-axis followed by a 30

rotation about the x-axis is

B′ = AxAz =

1 0 0

0 12

√3 1

2

0 −12

12

√3

12

√2 1

2

√2 0

−12

√2 1

2

√2 0

0 0 1

=

12

√2 1

2

√2 0

−14

√6 1

4

√6 1

214

√2 −1

4

√2 1

2

√3

6= B (5.14)

3. All rotation matrices are orthogonal, ATA = 11.

Proof: We know that the distance of any point from the origin is the same beforeand after the rotation. Therefore we have that

~r′2 =(x′ y′ z′

)x′y′z′

= r′T r′ = (Ar)T (Ar) = rTATAr = ~r2 = rT r (5.15)

⇐⇒ ATA = 11 . (5.16)

We can check explicitly that all the matrices in the examples above are orthogonal.We can also show that if A and B are orthogonal, then their product AB is also

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orthogonal:

(AB)T (AB) = (BTAT )(AB) = BT (ATA)B = BTB = 11 . (5.17)

4. For every rotation A there is an inverse rotation given by AT which brings us backto our starting point.

It is physically obvious that you can always get back to your starting point byreversing all your rotations in reverse order. It follows from point 3. that theinverse rotation matrix is given by AT .

5. (a) Any (proper) rotation can be expressed as a combination of elementary ro-tations about coordinate axes.

(b) No combination of such rotations can produce a reflection ~r → −~r.(c) All proper rotations have detA = 1. Improper rotations (involving an odd

number of reflections) have detA = −1.

The proof of (a) is complicated, but we will use this fact later on when we willdefine coordinates corresponding to the the rotational degrees of freedom.

Statement (c) follows from the properties of determinants. It is straightforwardto show that the elementary rotation matrices all have detA = 1. We also have

det(AB) = (detA)(detB) (5.18)

so any combination of elementary rotations must have determinant 1. Moreover,any orthogonal matrix A must have detA = ±1:

det 11 = 1 = det(ATA) = detAT detA = (detA)2 . (5.19)

Finally, the reflection matrix, which takes (x, y, z)→ (−x,−y,−z) is

R =

−1 0 00 −1 00 0 −1

, detR = −1 . (5.20)

It is clearly impossible to construct this from a product of matrices with determi-nant 1. Physically, this means that it is impossible (in 3 dimensions) to rotate abody into its mirror image.

5.2.4 The rotation group [optional]

A group is defined as a set of elements, together with a composition (multiplication)operation, with the following properties:

1. There exists an identity element (called 1 or e) which is a member of the group.

2. The combination a · b of two elements a and b is also a member of the group.

3. For every member a of the group there is an inverse a−1 which is also a memberof the group, such that a · a−1 = a−1 · a = 1.

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We see that rotations form a group according to this definition: properties 1 and 4 ofSec. 5.2.3 correspond to properties 2 and 3 above. The identiy element is the operationof doing nothing, corresponding to the identity matrix.

We can use the rotation matrices to define this group, and this gives the rotation groupits name: O(3), or “the group of real orthogonal 3×3 matrices. The set of all proper rota-tions also form a (smaller) group, called SO(3), or “the group of all special [determinant1] real orthogonal 3×3 matrices”. However, this is merely a particular representation ofthe general operation that we call rotations. We could equally well represent the groupelements by actual rotations in space, or by 3 angles (with a suitably defined multiplica-tion operation), or in many other ways. All these different representations would sharethe same multiplication table, which is what ultimately defines the group.

This theory of different representations of the same group is mathematically extremelypowerful, and the group-theory properties of rotations are extremely important in mod-ern physics. For example, particles and bodies in general may be classified according tohow they behave when rotated, and this turns out to be a fundamental classification.

It is possible to express any rotation matrix formally as

A = e~L·~T , with T1 =

0 0 00 0 10 −1 0

, T2 =

0 0 −10 0 01 0 0

, T3 =

0 1 0−1 0 00 0 0

.

(5.21)This expression is very useful theoretically, but useless if you want to find an explicitform for A. It does however give a direct connection between rotations and angularmomentum. The T matrices obey the commutation relations

[T1, T2] = −2T3 , [T2, T3] = −2T1 , [T3, T1] = −2T2 . (5.22)

These are essentially the commutation relations of the angular momentum operatorsin quantum mechanics, and the vector ~L is proportional to the angular momentumof the particle in question. But it turns out that there is another group with thesame multiplication table, namely the group of 2 × 2 complex unitary matrices withdeterminant 1, called SU(2). The operations of this group actually describe the rotationsof fermions, while bosons such as photons are described by the usual 3 × 3 rotationmatrices. The SU(2) matrices can be written in a similar form to (5.21),

A = ei~S·~σ , (5.23)

where

σ1 =

(0 11 0

), σ2 =

(0 −i−i 0

), σ3 =

(1 00 −1

), (5.24)

are the three Pauli matrices, which obey the same commutation relations as the T -matrices in (5.21) (up to a factor i). The vector ~S is a new form of angular momentumcalled spin, which corresponds to an ‘internal rotation’ degree of freedom. A curiousresult is that we must rotate a fermion by 4π (2 full rotations) to get back to where westarted!

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Figure 5.1: The Euler angles

5.3 Euler angles

We now go on to discuss which coordinates we can use to describe the orientation of arigid body, or alternatively, which three parameters can be used to uniquely obtain arotated coordinate system from an original one. There are several possibilities:

• The most natural would be to use rotation angles about the x-, y- and z-axes. Thisleads to the three Tait–Bryan angles, which are widely used for aircraft. However,since rotations do not commute, unless supplemented with a prescription for theorder of the three rotations, these angles are not unique, except for small rotations.

• The next most natural parameter would be to find the axis and angle of rotation,ie the vector ~φ above. This is called the Cayley–Klein or Euler parameters. Theseare mathematically very nice, and can be related to the vector ~L given above, butare not very practical.

• The third, most widely used parametrisation in mechanics is in terms of the threeEuler angles. Here, a general rotation is constructed from 3 successive elementaryrotations:

1. The body is rotated an angle φ ∈ [0, 2π〉 about the z-axis. The x- and y-axesmove, while the z-axis is unchanged.

2. The body is rotated an angle θ ∈ [0, π] about its x-axis. The z- and y-axesmove, but the x′-axis is unchanged.

3. The body is rotated an angle ψ ∈ [0, 2π〉 about its z-axis. As in the first step,the x- and y-axes move, while the z′-axis is unchanged.

Note that there is no rotation about any y-axis here; instead there are two rotationsabout (different) z-axes. These three rotations are summarised in figure 5.1. Itcan be proved that any rotation can be expressed in this way.

5.3.1 Rotation matrix for Euler angles

We can now construct the general rotation matrix explitly as a function of the Eulerangles. Calling the vectors in the intermediate coordinate systems ~ρ, ~ρ′ respectively, the

77

first rotation gives us ~ρ = Az(φ)~r. The second rotation gives us ~ρ′ = Ax(θ)~ρ, and thefinal rotation ~r′ = Az(ψ)~ρ′. Putting all this together, we find

A = Az(ψ)Ax(θ)Az(φ)

=

cosψ sinψ 0− sinψ cosψ 0

0 0 1

1 0 00 cos θ sin θ0 − sin θ cos θ

cosφ sinφ 0− sinφ cosφ 0

0 0 1

=

cosψ cosφ− sinψ cos θ sinφ cosψ sinφ+ sinψ cos θ cosφ sinψ sin θ− sinψ cosφ− cosψ cos θ sinφ − sinψ sinφ+ cosψ cos θ cosφ cosψ sin θ

sin θ sinφ − sin θ cosφ cos θ

.

(5.25)

Example 5.3

Find the Euler angles corresponding to a rotation of an angle ϑ about the y-axis.

The rotation matrix is

Ay(ϑ) ==

cosϑ 0 − sinϑ0 1 0

sinϑ 0 cosϑ

. (5.26)

Comparing the bottom right (zz) element of the matrices, we immediately see thatθ = ϑ. Comparing the bottom left elements then, we see that sinφ = 1, so φ = π/2.Finally, from the top right element we see that sinψ = −1, so ψ = −π/2. You maythen confirm that all the other elements come out as desired. You may also checkfor yourself that you may indeed achieve a rotation about the y axis by the followingcombination of rotations:

1. rotate it by 90 about the z-axis;

2. rotate it about the x-axis (by your desired angle);

3. rotate it back by 90 about the z-axis.

5.3.2 Euler angles and angular velocity

The total angular velocity can be constructed as the sum of angular velocities that resultfrom the changes in each of the three Euler angles. Note that simply adding these threevectors together is ok, since these correspond to infinitesimal changes in orientation,and for such changes the order does not matter. Hence we can write

~ω = ~φ+ ~θ + ~ψ . (5.27)

The magnitude of each of these three vectors is the respective angular velocity: φ, θ, ψ.But in which directions are they pointing? We will in the end want to express theangular velocity in the body coordinate system, since it will turn out that the equationsof motion are best expressed in these coordinates.

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~φ This vector points along the original z-axis, which is unchanged by the first rota-tion. The second rotation is about the x-axis, and changes the vector (0, 0, φ) →(0, φ sin θ, φ cos θ). After the final rotation of an angle ψ about the z-axis, we get

~φ = (φ sin θ sinψ, φ sin θ cosψ, φ cos θ) . (5.28)

~θ This vector poins along the intermediate x-axis, so after the second rotation we

have ~θ = (θ, 0, 0). After the final rotation about the z-axis it becomes

~θ = (θ cosψ,−θ sinψ, 0) . (5.29)

~ψ This vector points along the body’s z-axis, so ~ψ = (0, 0, ψ).

Adding up these, we get

~ω = ~φ+ ~θ + ~ψ =

φ sin θ sinψ + θ cosψ

φ sin θ cosψ − θ sinψ

φ cos θ + ψ

(5.30)

We will use this to derive the equations of motion. But first we need to determine thekinetic energy.

5.4 The inertia tensor

5.4.1 Rotational kinetic energy

Let us now work out the total kinetic energy of a rigid body. From (5.3) we get that

T =1

2

∑α

mα(~V + ~ω × ~rα)2 =1

2

∑α

mα

[~V 2 + 2~V · (~ω × ~rα) + (~ω × ~rα)2

]=

1

2M~V 2 + ~V · ~ω ×

(∑α

mα~rα

)+

1

2

∑α

mα(~ω × ~rα)2

(5.31)

• If one point P in the body is fixed, ie the motion is pure rotation then the one canchose the origon of the body coorinate system at P . Then ~V = 0 so the first twoterms vanish, and the third is the rotational energy Trot, where ~rα is the distancefrom particle number α to the point P .

• If we take the origin to be the centre of mass, then we can show that the secondterm vanishes, since then

∑αmα~rα = 0. The kinetic energy can then be written

as the sum of the centre of mass energy and the rotational energy about the centreof mass:

R = TCM + Trot =1

2M~V 2

CM +1

2

∑α

mα(~ω × ~rα)2 , (5.32)

where ~rα is the distance from particle number α in the body to the centre of mass.

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We now use that the angular velocity ~ω is the same for the whole body. We furthermoreuse the identity

( ~A× ~B)2 = A2B2 sin2 θAB = A2B2(1− cos2 θAB) = A2B2 − ( ~A · ~B)2 . (5.33)

With this, (5.32) becomes

Trot =1

2

∑α

mα

[r2αω

2 − (~rα · ω)2]

(5.34)

=1

2

∑α

mα

[r2α

∑i

ω2i −

∑ij

(xα,iωi)(xα,jωj)]

(5.35)

=1

2

∑ij

∑α

mα

(∑k

x2α,kδij − xα,ixα,j

)ωiωj (5.36)

=1

2

∑ij

Iijωiωj =1

2~ω · I · ~ω . (5.37)

We end up with

Trot =1

2

∑ij

Iijωiωj , (5.38)

Iij =∑α

mα(r2αδij − xα,ixα,j

)→

∫ρ(~x)(~x2δij − xixj)d3x (5.39)

The quantity I = Iij is called the inertia tensor of the body. It characterises therotational properties of the body.

If ~ω is directed purely along one of the coordinate axes, for example ~ω = (0, 0, ω), wesee that Trot reduces to the expression we have encountered before,

Trot =1

2

∑ij

Iijωiωj =1

2Izzω

2 , (5.40)

where Izz is the moment of inertia about the z-axis. But the expression (5.38) is muchmore general, and will hold in any (orthogonal) coordinate system, regardless of whichdirection the rotation vector is pointing.

We can non calculate the inertia tensor once and for all in the body coordinate system(indeed, we will later use it to define this coordinate system). Once we know I inone coordinate system, we shall see that we can relatively easily find it in any othercoordinate system.

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Example 5.4

6

-x

y

. ............ ............ ............ ............θ

x

zm1

m2

r2

Figure 5.2: A dumbbell

Find the inertia tensor of the dumbbell pictured infig. 5.2, and find the kinetic energy if it rotates withangular velocity ω

1. about the y-axis,

2. about the z-axis,

3. about its own axis.

Answer:

Particle 1 is located at ~r1 = (r1 sin θ, r1 cos θ, 0).Particle 2 is located at ~r2 = (−r2 sin θ,−r2 cos θ, 0).

We divide the computation of I as I = Idiag − Iprod

where Idiagi,j = δi,j

∑α r

2α and Iprod

i,j =∑

αmαxα,ixα,j.The two terms are computed as

Idiag = (m1r21 +m2r

22)1

=

m1r21 +m2r

22 0 0

0 m1r21 +m2r

22 0

0 0 m1r21 +m2r

22

and

Iprod = m1

r21 sin2 θ r2

1 sin θ cos θ 0r2

1 sin θ cos θ r21 cos2 θ 0

0 0 0

+m2

r22 sin2 θ r2

2 sin θ cos θ 0r2

2 sin θ cos θ r22 cos2 θ 0

0 0 0

= (m1r

21 +m2r

22)

sin2 θ sin θ cos θ 0sin θ cos θ cos2 θ 0

0 0 0

Adding them together we get the full intertial tensor I as

I = (m1r21 +m2r

22)

cos2 θ︷ ︸︸ ︷

1− sin2 θ − sin θ cos θ 0

− sin θ cos θ

sin2 θ︷ ︸︸ ︷1− cos2 θ 0

0 0 1

1. For rotation about the y-axis, ~ω = (0, ω, 0), so

Trot =1

2Iyyω

2 =1

2(m1r

21 +m2r

22)ω2 sin2 θ

2. For rotation about the z-axis, ~ω = (0, 0, ω), so

Trot =1

2Izzω

2 =1

2(m1r

21 +m2r

22)ω2 .

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3. For rotation about the body axis, ~ω = (ω sin θ, ω cos θ, 0), so

Trot =1

2Ixxω

2x +

1

2Ixyωxωy +

1

2Iyxωyωx +

1

2Iyyω

2y

=1

2(m1r

21 +m2r

22)ω2[cos2 θ sin2 θ − 2(sin θ cos θ)(sin θ cos θ)− sin2 θ cos2 θ]

= 0 .

It should not be a surprise that we get Trot = 0 here, since nothing is actuallymoving in this case!

Example 5.5

Find the inertia tensor for a homogeneous cube with mass M and length L with theorigin at one corner and edges along coordinate axes.

Answer:

The density of the cube is ρ = M/L3. We find

Ixx =

L∫0

L∫0

L∫0

M

L3(y2 + z2)dxdydz =

M

L2

L∫0

L∫0

(y2 + z2)dydz

=M

L2

L∫0

(L3

3+ z2L

)dz =

M

L2

(L4

3+L3

3L)

=2

3ML2 ,

(5.41)

Ixy =

L∫0

L∫0

L∫0

M

L3(−xy)dxdydz

= −ML2

L∫0

1

2L2ydy = −M

2

(1

2L2)

= −1

4ML2 .

(5.42)

Since the x, y and z axes are completely symmetric, it is clear that Ixx = Iyy = Izzand Ixy = Ixz = Iyz, and the inertia tensor is

I =

23ML2 −1

4ML2 −1

4ML2

−14ML2 2

3ML2 −1

4ML2

−14ML2 −1

4ML2 2

3ML2

= ML2

23−1

4−1

4

−14

23−1

4

−14−1

423

. (5.43)

5.4.2 What is a tensor? Scalars, vectors and tensors.

In practice, you can think of a tensor as a kind of matrix, where the rows and columnscorrespond to directions in space. You may then treat the expression for Trot as avector–matrix–vector multiplication.

82

In principle, a tensor is a generalisation of a vector, ie a physical (or mathematical)quantity with more than one direction. Tensors (and vectors) are defined by their trans-formation properties, and in particular how they change when they, or the coordinatesystem, are rotated. We shall see that in a rotated coordinate system, the inertia tensoris given by

I ′ij =∑kl

AikAjlIkl . (5.44)

This is the defining property of a tensor (to be precise, a rank-2 tensor, ie a tensor with2 indices).

This mirrors the definition of scalars and vectors, which are defined as follows:

• A scalar is a quantity that does not change when you rotate your coordinatesystem. Examples of this is the length of a vector, or the kinetic energy.

• A vector is a quantity v that transforms in the same way as the position vector, ie

v′i =∑j

Aijvj . (5.45)

There are however some vector and scalar type quantities that transform differentlyunder reflections :

• An ordinary vector will change its sign when seen in a mirror, but for example theangular momentum vector, ~L = ~r × ~p, will keep the same sign (since both ~r and~p change sign). Such vectors are called pseudovectors or axial vectors.

• Finally, ordinary scalars will be unchanged under reflections, but some quantitieschange sign. These are called pseudoscalars. For example, ~v ·~L (the scalar productof the velocity and angular momentum) would be a pseudoscalar.

Other examples of tensors

6

-

y

x

-----

The stress tensor. The picture on the left depicts a fluid flowingwith a velocity ~u in the x-direction. The fluid flows faster at largery, so that ∂yux 6= 0. In a viscous fluid this will create stress (shear)forces in the y-direction. This can be expressed through the stresstensor σxy. The diagonal components of this tensor represent thepressure of the fluid, eg σxx is the pressure in the x-direction.

The same picture also governs stress forces in solid materials, if spatially varying forcesare applied.

The (outer) product of two vectors. If we define

T = ~a⊗~b ⇐⇒ Tij = aibj , (5.46)

then we can easily see that this satisfies the rotation transformation property T ′ij =∑klAikAjlTkl. The vector product ~c = ~a×~b can be obtained as the antisymmetric part

of this tensor,

cx = Tyz − Tzy , cy = Txz − Tzx , cz = Txy − Tyx . (5.47)

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The electromagnetic field. In 4-dimensional spacetime, the electric field ~E andmagnetic field ~B form the (antisymmetric) field tensor Fµν = −Fνµ,

Fµν =

0 Ex Ey Ey−Ex 0 Bz −By

−Ex −Bz 0 Bx

−Ez By −Bx 0

. (5.48)

(Note that in 4-dimensional space-time, ‘rotations’ include Lorentz boosts.)

5.4.3 Rotations and the inertia tensor

How does the inertia tensor change if we rotate our coordinate system? To answer thisquestion, we can use the fact that the kinetic energy must be the same regardless ofhow we choose our coordinates (it is a scalar):

Trot =1

2

∑ij

ωiIijωj =1

2

∑ij

ω′iI′ijω′j . (5.49)

But ~ω is a vector, and changes in the same way as the position vector when rotated1,

ω′i =∑k

Aikωk . (5.50)

Substituting this into (5.49), we get

Trot =1

2

∑ij

∑k

AikωkI′ij

∑l

Ajlωl =1

2

∑kl

(∑ij

AikI′ijAjl

)ωl =

1

2

∑kl

ωkIklωl . (5.51)

So we find that Ikl =∑

ij AikI′ijAjl or I = AT I′A.

5.4.4 Angular momentum and the inertia tensor

We want to find the angular momentum ~L of a rigid body about some point O. Thispoint can be

• the centre of mass (for a body tumbling freely in space, for example), or

• some point fixed in space (for example a spinning top).

The momentum of some particle α is ~pα = mα~vα = mα~ω×~rα. Using the vector identity

~A× ( ~B × ~A) = A2 ~B − ( ~A · ~B) ~A (5.52)

1Strictly speaking, this is not quite the case: under a reflection, ~r → −~r, but ~ω is unchanged. Wecall a vector which behaves this way a pseudovector or axial-vector. The angular momentum is anotherexample of such a pseudovector.

84

we can write the total angular momentum as

~L =∑α

~rα × ~pα =∑α

~rα × (~ω × ~rα) =∑α

mα

(r2α~ω − (~rα · ~ω)~rα

)(5.53)

Li =∑α

mα(r2αωi −

∑j

xα,jωjxα,i) =∑α

mα

∑j

(r2αδij − xα,jxα,i)ωj (5.54)

=∑j

[∑α

mα(r2αδij −−xα,jxα,i)

]ωj =

∑j

Iijωj . (5.55)

So we find

Li =∑j

Iijωj or ~L = I · ~ω (5.56)

5.5 Principal axes of inertia

Life would be a lot simpler if, in some coordinate system, the inertia were diagonal,

I =

I1 0 00 I2 00 0 I3

(Iij = Iiδij) . (5.57)

We would then have

Li = Iiωi ; Trot =1

2

∑i

Iiω2i . (5.58)

In particular, if the body rotates about one of the axes of such a coordinate system, wehave ~L = I~ω, Trot = 1

2Iω2.

The good news is that it is always possible to find such a set of (body) coordinateaxes. These axes are called principal axes of inertia, and the corresponding I1, I2, I3 arethe principal moments of inertia.

The question then is how we find these axes. There are two methods:

• Find a rotation matrix A such at AIAT = I′ is diagonal. The coordinate axes inthe rotated system are then the principal axes.

• Find vectors (directions) ~ω such that ~L = I · ~ω = I~ω. These vectors form theprincipal axes of inertia, and the numbers I are the principal moments.

In fact, both these methods are mathematically identical. Let us first look at method2. If ~ω points along a principal axis of inertia, we have

Li =∑j

Iijωj = Iωi , (5.59)

85

where I is the corresponding moment of inertia, or

L1 = Iω1 = I11ω1 + I12ω2 + I13ω3

L2 = Iω2 = I21ω1 + I22ω2 + I23ω3

L3 = Iω3 = I31ω1 + I32ω2 + I33ω3

(5.60)

This is an eigenvalue problem. The condition for a nontrivial solution is det(I−I11) = 0.This gives us a (third-degree) equation for I, called the characteristic equation. Thesolutions are the principal moments of inertia (or eigenvalues of I). Once we havefound one of the solutions (eigenvalues), we can find the corresponding principal axis

by substituting the values of I into the equation for ~L = I~ω = I · ~ω. This gives us thedirection of ~ω, or equivalently, the ratios ω1 : ω2 : ω3. The vectors ~ω are the eigenvectorsof I.

Example 5.6

Show that the homogeneous cube with the origin at one corner has a principalmoment of inertia I = 1

6ML2, and find the corresponding principal axis of inertia.

Answer:

We found that the inertia tensor was

I = ML2

23−1

4−1

4

−14

23−1

4

−14−1

423

. (5.61)

I is a principal moment of inertia if det(I− I11) = 0. Setting I = 16ML2, we find∣∣∣∣∣∣

23ML2 − I −1

4ML2 −1

4ML2

−14ML2 2

3ML2 − I −1

4ML2

−14ML2 −1

4ML2 2

3ML2 − I

∣∣∣∣∣∣ =

∣∣∣∣∣∣12ML2 −1

4ML2 −1

4ML2

−14ML2 1

2ML2 −1

4ML2

−14ML2 −1

4ML2 1

2ML2

∣∣∣∣∣∣=(1

4ML2

)3

∣∣∣∣∣∣2 −1 −1−1 2 −1−1 −1 2

∣∣∣∣∣∣ =(1

4ML2

)3[2(2 · 2− 1) + 1(−1 · 2− 1)− 1(1− 2 · (−1)

]= 0 . (5.62)

The corresponding axis is given by

Iω1 = I11ω1 + I12ω2 + I13ω3 =⇒ 1

6ω1 =

2

3ω1 −

1

4ω2 −

1

4ω3 (5.63)

Iω2 = I21ω1 + I22ω2 + I23ω3 =⇒ 1

6ω2 = −1

4ω1 +

2

3ω2 −

1

4ω3 (5.64)

Iω3 = I31ω1 + I32ω2 + I33ω3 =⇒ 1

6ω3 = −1

4ω1 −

1

4ω2 +

2

3ω3 (5.65)

Only two of these equations are independent, since (5.65) is equivalent to

86

(5.63)+(5.64). We thus have

1

2ω1 −

1

4ω2 −

1

4ω3 = 0 (5.63)

−1

4ω1 −

1

2ω2 −

1

4ω3 = 0 (5.64)

(5.63)− (5.64) :1

2ω1 −

1

2ω2 = 0 ⇐⇒ ω2 = ω1 (5.66)

(5.63) =⇒ 1

2ω1 −

1

4ω1 −

1

4ω3 =

1

4ω1 −

1

4ω3 = 0 ⇐⇒ ω3 = ω1 (5.67)

We thus have ω1 = ω2 = ω3, or the principal axis is along the diagonal (1,1,1).

We might have guessed this from the symmetry of the cube, and rotated our systemso that the new x-axis is along the diagonal. This can be obtained by a 45 rotationabout the x-axis, followed by a rotation of − cos−1

√2/3 about the y-axis. The

matrix for this rotation is

A =

√

23

0 1√3

0 1 0

− 1√3

0√

23

1√

21√2

0

− 1√2

1√2

0

0 0 1

=1√6

√2√

2√

2

−√

3√

3 0−1 −1 2

. (5.68)

The rotated inertia tensor is

I′ = AIAT =ML2

6

√2√

2√

2

−√

3√

3 0−1 −1 2

23−1

4−1

4

−14

23−1

4

−14−1

423

√2 −√

3 −1√2√

3 −1√2 0 2

=ML2

12

2 0 00 11 00 0 11

. (5.69)

This also gives us the other two principal moments of inertia, which we find to beequal.

5.5.1 Comments

1. Finding the principal moments and axes of inertia (diagonalising the inertia tensor)by hand can be a very tedious process. Sometimes it can be simplified by symmetryconsiderations (see below), but in most cases it is better left to computers.

2. For the cube with the origin at one corner, we found that two of the principalmoments of inertia were equal, I2 = I3 = 11

12ML2. This means that the two

corresponding principal axes can be any orthogonal pair of axes perpendicular tothe diagonal — ie, the moment of inertia is the same about any axis orthogonalto the diagonal.

3. For a body where all three principal moments of inertia are equal, all directions oraxes are equivalent. We can see this by noting that in this case the inertia tensor in

87

the coordinate system defined by the principal axes is proportional to the identity(matrix), I = I11, which commutes with all rotation matrices, so I′ = IA11AT = I11for all A ∈ O(3).

4. Any body which is rotationally symmetric about some axis has one principal axisof inertia along that axis. The two other axes can be arbitrarily chosen in theplane perpendicular to the first principal axis, ie I2 = I3.

Definitions

• A body with I1 = I2 = I3 is called a spherical top.

• A body with I1 = I2 6= I3 is called a symmetric top.

• A body where all principal moments are different is called an asymmetric top.

• A body with I1 = 0, I2 = I3 is called a rotor.

For example, the cube with the origin at the centre is a spherical top; the cube with theorigin at one corner is a symmetric top, and the dumbbell (or a diatomic molecule) is arotor.

5.6 Equations of motion

Having found the principal axes of inertia, we can now use them to define the bodycoordinate system. This will greatly simplify the equations of motion. The kineticenergy is given by

T =1

2

∑i

Iiω2i . (5.70)

Using (5.30) we can write this as

T =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 . (5.71)

The equations of motion for generic values of the moments of inertia and a genericpotential energy will become very complicated. We will instead focus on two specificcases: a symmetric top (I1 = I2) with one point fixed and under the influence of constantgravity, and force-free motion of an asymmetric top (all I1, I2, I3 are different).

5.6.1 The symmetric heavy top

For a symmetric top, we can take I1 = I2, and (5.71) becomes

T =1

2I1(θ2 + φ2 sin2 θ) +

1

2I3(φ cos θ + ψ)2 . (5.72)

We now take the top to be rotating about a fixed point at the bottom of the symmetryaxis, under the influence of a constant gravitational field g. We take θ to be the angle

88

the symmetry axis forms with the vertical. In that case, the potential energy of the topis V = Mgh cos θ, where M is the mass of the top, and h is the distance from the baseof the top to the centre of mass along the symmetry axis. The lagrangian is therefore

L =1

2I1(θ2 + φ2 sin2 θ) +

1

2I3(φ cos θ + ψ)2 −Mgh cos θ . (5.73)

The meaning of the three Euler angles in this case is:

• θ denotes the angle the axis of the top makes with the vertical.

• φ denotes the orientation of the (tilted) axis relative to the fixed reference coordi-nate system.

• ψ denotes the orientation of the top relative to its own axis.

We will find that in general, θ will oscillate between a minimum and a maximum, whileboth φ and ψ will be monotonically increasing (or decreasing). The motion in ψ reflectsthe top spinning around its own axis. The motion in φ corresponds to the orientationof the axis precessing around the vertical axis. Finally, the motion in θ corresponds toperiodic “wobbles” in the tilt of the top, called nutation.

We may now derive the Euler–Lagrange equations for the top and use them to study thismotion in detail. We will not do this here, but instead perform a qualitative analysis ofthe possible motion. First we determine the canonical momenta,

pφ =∂L

∂φ= I1φ sin2 θ + I3ω3

∂ω3

∂φ= I1φ sin2 θ + I3 cos θ(φ cos θ + ψ) , (5.74)

pθ =∂L

∂θ= I1θ (5.75)

pψ =∂L

∂ψ= I3ω3

∂ω3

∂ψ= I3(φ cos θ + ψ) . (5.76)

Since L does not depend explicitly on either φ or ψ, ∂L/∂φ = ∂L/∂ψ = 0, and we seestraightaway that the canonical momenta pφ, pψ are conserved. In particular, pψ = I3ω3,so the top spins with a constant angular velocity ω3 about its own axis. However, theprecession rate, governed by the constant pφ is more complicated.

We now proceed to derive the hamiltonian of the system. From (5.76) we see thatω3 = φ cos θ + ψ = pψ/I3. We can use this to rewrite (5.74),

pφ = I1φ sin2 θ + I3 cos θω3 = I1φ sin2 θ + pψ cos θ (5.77)

⇐⇒ pφ − pψ cos θ = I1φ sin2 θ ⇐⇒ φ =pφ − pψ cos θ

I1 sin2 θ. (5.78)

Since the kinetic energy is quadratic in the generalised velocities, and the potentialenergy does not depend on velocities, the hamiltonian is equal to the total energy,

H = T + V

=1

2I1

[(pφ − pψ cos θ

I1 sin2 θ

)2

sin2 θ +(pθI1

)2]

+1

2I3

(pψI3

)2

+Mgh cos θ

=p2θ

2I1

+p2ψ

2I3

+(pφ − pψ cos θ)2

2I1 sin2 θ+Mgh cos θ .

(5.79)

89

Since pφ and pψ are both constant, there is effectively only one degree of freedom θ, andthe second and third term in H can be combined with the final term V = Mgh cos θ toform an effective potential Veff(θ). The first term is the usual kinetic energy term.

The first term in Veff is just a constant, so it has no effect on the dynamics of the system,other than increasing the minimum energy. Unless pφ and pψ are both precisely zeroor pφ = ±pψ, the second term is nonzero and diverges to +∞ as θ → 0 and θ → π.2

Therefore, the motion in θ is bounded for all but very special values of pφ, pψ, with θmax

in general being smaller the larger the value of pψ. If θmax < π/2 this ensures the topdoes not fall over.

5.6.2 Euler’s equations for rigid bodies

Consider now force-free motion of a rigid body. In this case, we have3

L = T =∑i

Iiω2i . (5.80)

The Euler–Lagrange equations are given by

d

dt

∂L

∂φ=∂L

∂φ;

d

dt

∂L

∂θ=∂L

∂θ;

d

dt

∂L

∂ψ=∂L

∂ψ. (5.81)

We will however only consider the third of those, and then derive two additional equa-tions of motion from symmetry considerations.

We can write the Euler–Lagrange equation for ψ as

d

dt

∂L

∂ψ=

d

dt

(I3(φ cos θ + ψ)

)=

d

dt(I3ω3) =

∂L

∂ψ=

3∑i=1

Iiωi∂ωi∂ψ

. (5.82)

We first note that

∂ω1

∂ψ= φ sin θ cosψ − θ sinψ = ω2 , (5.83)

∂ω2

∂ψ= −φ sin θ sinψ − θ cosψ = −ω1 . (5.84)

Since ω3 does not depend on ψ the last expression in (5.82) becomes

∂L

∂ψ= I1ω1

∂ω1

∂ψ+ I2ω2

∂ω2

∂ψ= I1ω1ω2 + I3ω2(−ω1) = (I1 − I2)ω1ω2 , (5.85)

so (5.82) becomes

I3dω3

dt= (I1 − I2)ω1ω2 . (5.86)

However, the labels 1, 2 and 3 for the three axes is arbitrary, and we may just aswell rename them, as long as we ensure the coordinate system remains right-handed(corresponding to cyclic permutations. This gives us the three equations

2If pφ = pψ the second term in Veff diverges as θ → π but goes to 0 as θ → 0, and conversely ifpφ = −pψ. This, however, requires finely balanced initial conditions.

3Note that the translational energy can be ignored since ~R is cyclic.

90

I1dω1

dt= (I2 − I3)ω2ω3 , (5.87)

I2dω2

dt= (I3 − I1)ω3ω1 , (5.88)

I3dω3

dt= (I1 − I2)ω1ω2 . (5.89)

These are Euler’s equations for force-free motion.

5.6.3 Stability of rigid-body rotations

Let us now look at what happens when we set a body rotating about one of the threeprincipal axes. For example, if it rotates purely about the first axis, we have ω1 6=0, ω2 = ω3 = 0. In practice, it is not possible to have exactly zero rotation about theother two axes, so what we have is that ω2 and ω3 are both much smaller (in magnitude)than ω1. We now want to find out how ω1, ω2 and ω3 each evolve with time. If ω2 and ω3

remain small and are either damped to zero or fluctuate around zero, the rotation aboutthe first axis is said to be stable. On the other hand, if the magnitude of ω2 and/or ω3

grows with time, they will eventually become as large as ω1 and the body is no longerrotating about its original axis. In that case, the rotation is unstable.

Without any loss of generality, we can take I1 > I2 > I3 since we are allowed to labelour axes as we wish. We then have three different cases to deal with:

1. If ω2 ∼ ω3 ω1 the three equations become

I1ω1 = (I2 − I1)ω2ω3 ≈ 0 =⇒ ω1 = constant , (5.90)

I2dω2

dt= (I3 − I1)ω3ω1 =⇒ ω2 =

(I3 − I1

I2

ω1

)ω3 , (5.91)

I3dω3

dt= (I1 − I2)ω1ω2 =⇒ ω3 =

(I1 − I2

I3

ω1

)ω2 , (5.92)

Differentiating (5.91) and using (5.92) we get

ω2 =I3 − I1

I2

ω1ω3 =(I3 − I1)(I1 − I2)

I2I3

ω21ω2 (5.93)

⇐⇒ ω2 + Ω21ω2 = 0 with Ω2

1 =(I3 − I1)(I2 − I1)

I2I3

ω21 > 0 . (5.94)

This has the solution

ω2(t) = A cos Ω1t+B sin Ω1t (5.95)

ω3(t) ∝ ω2(t) = A′ cos Ω1t+B′ sin Ω1t . (5.96)

So we see that ω2, ω3 will oscillate about equilibrium values ω2 = ω3 = 0.

91

2. If ω1 ∼ ω2 ω3, by the same procedure as in the first case, we obtain

ω3 = const , ω1 ∝ ω3 , ω2 + Ω3ω2 = 0 , Ω23 =

(I3 − I2)(I3 − I1)

I1I2

ω23 > 0 ,

which again has solutions

ω1(t) = A cos Ω3t+B sin Ω3t , ω2(t) = A′ cos Ω3t+B′ sin Ω3t . (5.97)

3. If ω1 ∼ ω3 ω2, then, using the same procedure again, we now find the equations

ω2 = constant , (5.98)

ω3 =I1

(I2 − I3)ω2

ω1 , (5.99)

ω1 =(I3 − I2)(I2 − I1)

I1I3

ω22ω1 = Ω2

2ω1 . (5.100)

The general solution to (5.100) is

ω1(t) = AeΩ2t +Be−Ω2t , (5.101)

so ω1 (and ω3) will increase exponentially with time, at least until the approxima-tions ω1 ω2 and ω3 ω2 are no longer valid.

The upshot of this is that rotations about the “long” and “short” axes are stable, whilethose about the intermediate axis are unstable. You can verify this for yourself by tossinga rectangular block (for example a a book held together by an elastic band) up in theair, rotating it about each of its three axes.

92