Dynamics

Mathematical Tripos IA:Dynamics

John StewartCentre for Mathematical Sciences

Wilberforce Road Cambridge CB3 [email protected]

http://www.damtp.cam.ac.uk/user/john/c8

Constructive comments or amendments should beset to the email address above. Printed materialis available at the web site in “gzipped Postscript”format to users from “.cam.ac.uk” sites.

I have only one book recommendation for thecourse as a whole.

• M. Lunn, A First Course in Mechanics, Oxford(1991).

I will be recommending other sources for specifictopics.

There are four problem sheets, intended forsupervisions in weeks 3, 5, 7 and 9. Week 1 isnext week; week 9 is next term.

University of Cambridge: Mathematical Tripos IA: Dynamics: c©J.M. Stewart 2002 1

0. Introduction

0.1 Mathematical Preliminaries

R is the set of all real numbers. Rn is a n-dimensional vector space, where n = 1, 2, . . .. If,in addition, we have a scalar or “dot” product, asdefined in course C1/2 the norm or magnitude ofx ∈ Rn is

|x| =√

x.x . (1)

Rn with this norm is called Euclidean space En.

If U and V are vector spaces, U × V is the setof ordered pairs (u, v), where u ∈ U , v ∈ V . U × Vis also a vector space.

Let (i, j) denote the standard basis for E2, i.e.,|i| = |j| = 1, i.j = 0. Then any vector x ∈ E2 canbe written

x = xi + yj or x = (x, y). (2)

Since we usually regard x as a “column vector” itwould be more accurate (and more cluttered) towrite x =

(xy

)or x = (x, y)T .


0.2 Trajectories

A function x : R → E2, t → x(t) = x(t)i+ y(t)j(where y(t) is a function R → R) is a trajectory.

x

yt1

t2

Example 1.

x(t) = (ut, vt − 12gt2),

where u, v and g are constants.

Functions R → E2 can be differentiated in anobvious manner:

dx(t)dt

≡ x(t) =dx

dti +

dy

dtj ≡ (x, y).

Note that dj/dt = 0 etc. Here we regard the basisas being fixed.


x(t1) is called the velocity of the trajectoryx(t) at t = t1.

Further

d2x(t)dt2

≡ x(t) =d2x

dt2i +

d2y

dt2j

is the acceleration of the trajectory x(t) at t.

Exercise 1. Verify that for example 1

x(t) = (u, v − gt), x(t) = (0,−g).

Exercise 2. Define trajectories in E3 and theirvelocities and accelerations. (A typical point inE3 is x = xi + yj + zk.)


0.3 The Gradient Operator

Consider a function f : R2 → R, e.g.,

f(x, y) = log x + sin y + ex cos y.

The partial derivative of f with respect to yis defined by the following prescription: regard x asfixed, so that f is a function of the single variable y,and differentiate it

∂f

∂y= cos y − ex sin y.

We can do the same for x

∂f

∂x=

1x

+ ex cos y,

and we can repeat the operation

∂2f

∂x2≡ ∂

∂x

(∂f

∂x

)= − 1

x2+ ex cos y,


∂2f

∂x∂y≡ ∂

∂x

(∂f

∂y

)= −ex sin y,

∂2f

∂y∂x≡ ∂

∂y

(∂f

∂x

)= −ex sin y,

∂2f

∂y2≡ ∂

∂y

(∂f

∂y

)= − sin y − ex cos y.

It can be shown that for “smooth” functions f ,partial derivatives commute, i.e., ∂2f/∂y∂x =∂2f/∂x∂y.

Each function f : R2 → R defines a vectorwritten as ∇f , pronounced “grad f”, via

∇f =

(∂f

∂x,∂f

∂y

),

and for our example

∇f =(1x

+ ex cos y, cos y − ex sin y).


Exercise 3. [Polar coordinates] Consider the functionsr =

√x2 + y2 and θ = tan−1(y/x), and show

∇r =(x

r,y

r

)= (cos θ, sin θ),

∇θ =(− y

r2,

x

r2

)=

1r(− sin θ, cos θ),

(3)

which will be needed later.

Exercise 4. How would you define partialderivatives and the gradient operator for functionsf : Rn → R?

If for some vector f , there exists a scalar functionU such that f = −∇U , then U is said to be apotential for f . The minus sign is conventional.


1. Introduction to Mechanics

1.1 Mechanics: experimental facts

The following approximately true experimentalfacts are the basis for mechanics.

(A) Our space is three-dimensional, homogeneous,isotropic and Euclidean, and time is one-dimensional.

(B) Galileo’s principle of relativity holds: there existspecial coordinate systems called inertial withthe properties

(1) all the laws of nature at all moments of timeare the same in all such coordinate systems,

(2) all coordinate systems in uniform rectilinearmotion with respect to an inertial oneare themselves inertial, as are those whichdiffer from an inertial one by a constanttranslation, rotation or reflection.

(C) Newton’s principle holds: the initial state of amechanical system (the totality of positions and


velocities of its points at some initial moment intime) uniquely determines all of its motion.

We now examine them in detail.


1.2 Space and time

It is a mistake to say our space is the vectorspace E3, for every vector space contains a privilegedvector, 0, which destroys the homogeneity. Howeverrelative positions behave like vectors.

It doesn’t make much sense (until we agree anorigin of time) to say “the lecture started at 1005”but the assertion “the duration of the lecture was 55minutes” is readily verifiable. The position of King’sCollege Chapel is not a vector, but the displacementfrom there to the Cockroft Lecture Theatre is one.

In fact our space and time constitute a 4-dimensional affine space AR4, whose points arecalled events. For the record1 we give a precisedefinition of an affine space, although you will neverbe asked to use it.

1For further details see e.g., P. Bamberg & S. Sternberg, A course inmathematics for students of physics, vol 1, CUP 1990.


Affine spaces

Let V be a vector space. The affine space AVconsists of a set of points p, q, . . . and an operation⊕ which assigns to each p ∈ AV and each v ∈ Vanother point in AV denoted p ⊕ v. The operationsatisfies the axioms:

1. associativity: (p ⊕ u) ⊕ v = p ⊕ (u + v) for anyp ∈ AV , u,v ∈ V ,

2. identity: p ⊕ 0 = p for any p ∈ AV ,

3. transitivity: given p, q ∈ AV there is a v ∈ Vsuch that q = p ⊕ v,

4. faithfulness: if for all p ∈ AV the equality p⊕u =p ⊕ v holds then u = v.

These imply that given o, p ∈ AV there is aunique u ∈ V such that p = o⊕u and it is convenientto write p � o = u. Thus if we choose a preferred,fixed o, then any point p ∈ AV can be identifiedwith u ∈ V .


From remark 3 above q = o ⊕ (u + v) and soq � p = v. To see that this is independent of thechoice of o consider another choice o′ = o⊕w. Thenp = o′ ⊕ (−w + u) and q = o′ ⊕ (−w + u + v) sothat q � p = v as before.

This construction is implicitly assumed intextbooks.


1.2 Space and time (continued)

For Newtonian dynamics (but not specialrelativity) we assume that our AR4 is actuallyAR×AE3. The first affine space represents time, thesecond physical space. We now choose a particularo ∈ AR × AE3 and measure displacements relativeto o so that our space becomes R × E3. Finally wechoose a standard basis (i, j,k) for E3, see exercise 2.The choice of o and basis is a choice of referenceframe or frame. Because of the arbitrariness wemust permit two types of transformation. The firstreflects the choice of o, the second the choice ofbasis:

1. Translations (t,x) → (t′,x′) = (t + τ,x + ξ),where τ ∈ R, ξ ∈ E3,

2. Rotations/reflections (t,x) → (t′,x′) =(t, U(t)x), where U is a unitary2 3-matrix whichcan depend on time.

2U is unitary if | det(U)| = 1, i.e., U represents a rotation or areflection


1.3 Galilean Relativity

Suppose that we have chosen our frame to bean inertial one in the sense of §1.1. Condition B(2)implies that the change

(t,x) → (t′,x′) = (t,x + ut), (4)

where u ∈ E3 is a fixed relative velocity,defines a new inertial frame. Galileo’s principle(based on empirical evidence) requires that constanttranslations

(t,x) → (t′,x′) = (t + τ,x + ξ), (5)

where τ ∈ R and ξ ∈ E3, and constantrotations/reflections

(t,x) → (t′,x′) = (t, Ux), (6)

(where U is a constant unitary matrix) generatenew inertial frames. These three transformationscan be combined, each has an inverse and includesthe identity. They form the Galilean group oftransformations from one inertial frame to another.


1.4 Special Relativity∗3

The discussion of the previous two sectionsreflects the special status allocated to time in classicalmechanics. Let us, for simplicity, suppress two spacedimensions and consider our spacetime to be anAR2, subject to axiom (B) of §1.1. We now makethe more general assumptions:

A1 We are considering linear transformationsbetween inertial frames S and S′, whosecoefficients depend continuously on the relativevelocity of S′ with respect to S.

A2 The transformations form a group. This is oftencalled the relativity principle. An alternativeversion is that if a physical phenomenon holds inone inertial frame, it must hold in every other one,i.e., physics should be frame-independent.

A3 There is no preferred direction; any physicalsituation remains invariant under t → t, x → −x,provided of course, all velocities are reversed.

3An asterisk denotes material which is non-examinable for this course.


A4 If the velocity of S′ relative to S is positive, thenthe relative velocity of S with respect to S′ isnegative.

A5 If the velocity of S′ relative to S is positive, andthe velocity of a third inertial frame S′′ relative toS′ is positive, then the velocity of S′′ relative to S

is also positive.

The motivation for the first two assumptionshas already been given. A3 is an isotropyassumption based on empirical evidence. Inmany situations there is no preferred spatialdirection. There are situations where this isnot the case—they involve gravitation or moregenerally general relativity. The fourth assumptionasserts simply that if I see you moving awayfrom me, then you see me moving away fromyou. (Note that this is a tautology for Galileantransformations, but might not hold for moreexotic transformation laws.) Similarly assumptionA5 reflects common experience, and is a tautologyfor Galilean transformations.

Now, almost miraculously, there are only twogroups of transformations which satisfy these


assumptions/axioms. Obviously the Galileangroup is one of them. The other turnsout to be the Lorentz group in whichthere is a finite upper limit to the relativespeed between two inertial frames. This isthe basis for special relativity, invented byAlbert Einstein in 1905 from (almost) totallydifferent premises. The justification for theseassertions is not available in most of thestandard textbooks4 but is discussed in detail inhttp://www.damtp.cam.ac.uk/user/john/srib/.

4The exception is W. Rindler, Essential Relativity, Springer, � 1977.


1.5 Newton’s principle

We now examine Newton’s principle, (C) ofsection 1.1. For a system consisting of a singleparticle it asserts that there is a function A suchthat

x = A(t,x, x). (7)

Specification of A is the definition of the system.

However Galilean relativity places someconstraints on the form of A. Recall that timetranslations preserve inertial frames. Thus ifx = ϕ(t) is a solution of (7), then so also isx = ϕ(t + s) for any s ∈ R. It follows that in aninertial frame the right hand side of (7) does notdepend on time

x = Φ(x, x). (8)

Next consider a system of n particles. Fixedspatial translations preserve inertial frames and soif xi = ϕi(t) (i = 1, 2, . . . , n) are solutions of(8), then so are xi = ϕi(t) + r where r ∈ E3


is fixed. It follows that Φ can only depend onrelative positions

xi = Φi({xj − xk}, xl). (9)

By a similar argument, since inertial frames arepreserved by uniform rectilinear motion, Φ canonly depend on relative velocities:

xi = Φi({xj − xk}, {xl − xm}), (10)

where i, j, k, l, m = 1, 2, . . . , n.

Finally let U be a constant 3 × 3 unitary matrix.Because constant reflections/rotations preserveinertial frames we must require that if xi = ϕi(t)is a solution of (10) then so is Uϕi(t), or

Φi(U{xj − xk}, U{xl − xm}) =

UΦi({xj − xk}, {xl − xm}). (11)


Example 2. Consider a system consisting of asingle particle. Then the acceleration vectorA cannot depend on t, x or x and so mustbe a constant vector, which in addition mustbe invariant under rotations. Therefore theacceleration vanishes; the particle moves withuniform rectilinear motion. This is usually calledNewton’s first law.

Exercise 5. You may have encounteredNewton’s first law before. (If not then consulta conventional textbook.) Check for, and ifnecessary, resolve any inconsistencies.

Example 3. Consider a system consisting of twoparticles, which are initially at rest. This systemmust remain invariant under rotations about theline which joins the two particles initially. Thusthe particles remain on this line.

Exercise 6. Consider a system consisting of twoparticles. Show that there exists an inertialcoordinate system in which the two particlesremain in a fixed plane.


Example 4. [Gravitation] Experiments, firstconducted by Galileo, show that for a particle closeto the earth’s surface, the height z is governed by

z = −g, (12)

where g ≈ 9.8m/s2 is a universal constant. If weintroduce the potential U = gz then clearly

z = −dU

dz.

Exercise 7. (The same as example 4 but invectorial notation.)

x

g

The vector g has magnitude g and points verticallydownwards, and x is a general “position vector”.


Show that if U = g.x then

x = −∇U.

Example 5. [Non-localized gravitation] Experimentsshow that Galileo’s equation (12) is invalid forvery large heights and has to be replaced by ageneralized Galileo’s equation

z = r = −gr20

r2, (13)

where r = r0 + z.

zr r0

��

��

Note that if z is small, |z|/r0 1, (13) reducesto Galileo’s equation (12).


This acceleration can also be expressed as (minus)the gradient of a potential, z = −∇U where

U = −gr20

r.


1.6 Mass and Force

Most textbooks assume that you already know,intuitively, what mass and force are. We look alittle more closely at these concepts.

x

Example 6. [Inertial mass, Newton’s second law]Suppose we have a spring with one end fixed anda particle attached to the other end, constrainedto move along the direction of the spring. If thedisplacement of the particle is x, then experimentsshow that the acceleration is

x = −α2x,

where α is a constant, which too can be derivedfrom a potential U = 1

2α2x2. Experiments

show that if the particle is replaced by two


identical ones, then for the same extension x, theacceleration is only half as large. It is establishedexperimentally that for any two bodies the ratio ofthe accelerations x1/x2 under the same extensionof a spring does not depend on the type of springbut only on the bodies themselves. We thereforeassociate a property inertial mass with everybody. We define the ratio of the inertialmasses to be the inverse of the ratio of theiraccelerations

x1

x2=

m2

m1. (14)

Although we have defined this ratio foraccelerations produced by springs, it is anexperimental fact that equation (14) holds forany acceleration mechanism.

Eq.(14) does not define the unit of inertialmass. By convention we take the unit to bea kilogramme, the mass of a litre ( = 10−3 m3)of water5 at a fixed temperature. Then the massm2 of a body can be determined by comparingits acceleration with that of a litre of water.

5It is assumed you are familiar with SI units for length and time.


Experiments show that all masses are positive.It follows from equation (14) that the productof inertial mass and acceleration mx does notdepend on the body and is a characteristic of theextension of the spring. This product is called theforce of the spring acting on the body.

This is an example of a more general,experimentally verified, property

force = inertial mass × acceleration, (15)

usually called Newton’s second law. (We metNewton’s first law in example 2.) As unit of forcewe take the newton with abbreviation N . If onelitre of water is suspended on a vertical spring atthe surface of the earth, the spring acts with aforce of 9.8 newtons.

Newton’s second law in the form (15) is usuallytaken as an axiom of mechanics. Note howeverthat in the experiments used to substantiate it,the mass of the particle is constant. Thus anentirely equivalent formula is

force =d

dt

(inertial mass × velocity

). (16)


We call the product inertial mass × velocity themomentum of the particle, usually denoted p.Thus (16) can be rewritten as

force =d

dtmomentum. (17)


1.7 Gravitational Force

All bodies exert gravitational attraction, andwe can quantify this by assigning every bodya gravitational mass M . Suppose wehave two particles of inertial masses m1, m2,gravitational masses M1, M2 at positions x1 andx2 respectively. It is observed experimentally, andcodified as Newton’s law of gravitation thatparticle 1 is accelerated towards the other, and theacceleration is proportional to M2 and inverselyproportional to the square of their separation

x1 = GM2x2 − x1

|x2 − x1|3 , (18)

where G ≈ 6.67 × 10−11Nm2kg−2 is thegravitational constant. (Note that thisformula is consistent with Newton’s principle(11).) Thus the gravitational force exerted byparticle 2 on particle 1 is

F1←2 = Gm1M2x2 − x1

|x2 − x1|3 . (19)


If we denote by F2←1 the gravitational forceexerted by particle 1 on particle 2 then it isobvious that F2←1 = −(m2/M2)(M1/m1)F1←2.

It is a remarkable experimental fact (the Eotvosexperiment, to be discussed in a later example)that the ratio of masses m/M is the same forall bodies, and by choosing units appropriatelywe can arrange M = m. Thus there is no needto distinguish between inertial and gravitationalmasses despite their very different origins. Infuture we just refer to mass.

One corollary of this result is that F2←1 =−F1←2. This is an example of Newton’s thirdlaw: action and reaction are equal and opposite.The third law appears to hold for all interactionforces, not just gravity.

Exercise 8. We shall prove later that thegravitational acceleration produced by anextended spherically symmetric body at pointsoutside it is the same as that produced by a pointof equal mass situated at the centre of the body.Use this fact to deduce Galileo’s law (13) fromNewton’s law (18).


Example 7. [Freely falling particle] Suppose thatm2 � m1 so that the acceleration of particle 2is negligible. Then setting x2 = 0, x1 = x andr = |x|, equation (18) reduces to

x = −Gm2xr3

.

Suppose that at t = 0, |x| = R and x = 0. Howlong will it take for the particles to meet?

This is clearly a one-dimensional problem

r = −Gm2

r2= −dU

dr, (20)

where U = −Gm2/r, see example 5. The easiestway to integrate this is to note that

d

dt

(12r2)

= rr = −dU

drr = −d

dtU(r(t)),

where the chain rule has been used, so that 12r

2 =Gm2/r + const. The initial conditions give r2 =2Gm2(1/r − 1/R).


Thus the time to the collision is

T = −∫ 0

R

dr

|r|

=∫ R

0

( Rr

2Gm2(R − r)

)1/2

dr

=∫ π/2

0

( R2 sin2 θ

2Gm2R cos2 θ

)1/2

2R sin θ cos θ dθ

=π

4

( 2R3

Gm2

)1/2

,

where the substitution r = R sin2 θ was used.


1.8 Newton’s 3rd Law & Closed

Systems

Consider a system of n particles; the ith particlehas momentum pi, and the kth particle exertsa force Fi←k on it. Such forces are internalto the system, while forces produced by outsideinfluences are external. A closed system hasno external forces. The total momentum ofthe system is P =

∑ni=1 pi.

Recall that for a single particle subject to no forcesthe momentum is conserved, see (17).

Theorem 1. Consider a closed system ofparticles where the internal forces obey Newton’sthird law. Then the total momentum is conserved.

Proof. Using the notation above, eq.(17)generalizes to

dpi

dt=∑k �=i

Fi←k.


ThusdPdt

=∑

i

∑k �=i

Fi←k.

The double sum is a sum over pairs of particles.The contribution from the pair (2, 5) is F2←5 +F5←2 = 0 by Newton’s third law. Thus the sumvanishes and so P is constant. �

Of course the question then arises as to whichsystems are closed.


1.9 Electrical Forces

While all bodies have non-negative mass, theypossess another property, electrical charge q whichcan have either sign or be zero. For time-independent situations the Coulomb law ofelectrostatics holds. In the notation of §1.7the force exerted by particle 2 on particle 1 is

F1←2 = − q1q2

4πε0

x2 − x1

|x2 − x1|3, (21)

where ε0 is a new universal constant. Thuselectrostatics is effectively gravity in disguise. Theelectric field produced at point 1 by particle 2is

E1←2 = − q2

4πε0

x2 − x1

|x2 − x1|3, (22)

so that F1←2 = q1E1←2.

In time dependent problems there is anothervector, the magnetic field to contend with.Suppose we have a “test particle”, ie., one whosecharge (or mass) is so small that it does not affectthe electrical fields E(t,x) and B(t,x). (This isthe same approximation as that made in example


7.) Then if the particle is at x(t) with velocityx(t), its acceleration is given by the Lorentzforce

mx(t) = FLorentz

= q(E(t,x(t)) + x(t) × B(t,x(t))

).(23)

We shall study simplified versions of this later in§3.4.


1.10 Frictional Forces

The world is a complicated place and we oftenmake gross but useful simplifications. It’s difficultcycling against the wind, but rather than discussthe effect of all of the 1023 air molecules hittingus, we model it by a simple phenomenologicalfrictional force.

Example 8. Linear friction models the forceas proportional to, but opposing the velocity. E.g.,consider a particle of mass m falling under gravity,with position z measured downwards

mz = mg − kz, (24)

or

z = g − αz, (25)

where α = k/m.


t

z

V0 =g

α

It is clear that z = V0 ≡ g/α is a solution of (25).In fact all solutions z(t) → V0 as t → ∞. V0 isthe terminal velocity.

Exercise 9. Suppose that z(0) = U . Show thatthe solution of equation (25) is

z(t) = V0 − (V0 − U)e−αt,

and hence confirm the statement made above.

Example 9. A slab of armour plate of mass Mand length a lies on a smooth horizontal benchand a bullet of mass m is fired at it with speedu. The resistive force on the bullet is known to


be kmM/(M + m) times the relative speed ofthe bullet with respect to the plate. Show thatthe bullet passes through if u > ka, and if sodetermine the displacement of the plate and thefinal speed of the bullet.

Let x denote the displacement of the plate.Because of the form of the frictional force it isconvenient to describe the position of the bullet byits displacement y relative to the plate, so thatthe absolute displacement is x + y. The equationof motion of the plate is Mx = kmMy/(M +m)or

x =kmy

M + m,

and by a similar argument the equation of motionof the bullet is

x + y = − kMy

M + m.

We can deduce y = −ky so that y + ky = const.Initially x = y = 0, x = 0, y = u and so y +ky =u. It is straightforward to obtain y = u exp(−kt)and y = (u/k)(1 − exp(−kt)). The bullet passesthrough iff y > 0 at y = a ie., u > ka, the firstresult required. If the bullet reaches the far side


y = a at t = t1 then exp(−kt1) = 1 − ka/u. Wenow have x = kmu exp(−kt)/(M + m) leadingto x = mu(1 − exp(−kt))/(M + m) and

x =mut

M + m− mu

(M + m)k(1 − e−kt).

At t = t1 the displacement is

x =mu

(M + m)k

[log( u

u − ka

)− ka

u

],

and the speed of the bullet is

x + y = u − kMa

M + m.


Example 10. Quadratic friction is sometimesmore realistic. Equation (24) is replaced by

mz = mg − Kz2sign(z), (26)

or, setting v = z, β2 = K/m, γ2 = g andassuming v = U > 0 at t = 0 and v � 0

v = γ2 − β2v2. (27)

Note that there is a terminal velocity V0 = γ/βand that the figure of example 8 is valid here.Thus

t =∫ v

U

dv

γ2 − β2v2=

12γ

∫ v

U

(1

γ − βv+

1γ + βv

)dv

=1

2βγlog{

γ + βv

γ + βU

γ − βU

γ − βv

},

and so

v = V0

{(V0 + U)e2βγt − (V0 − U)(V0 + U)e2βγt + (V0 − U)

}.

As before z = v → V0 as t → ∞ independent ofthe initial conditions. See also example 24.


Example 11. In other situations the Coulombor dry friction model may be the mostappropriate one. Here the force is

F = −Ksign(z). (28)

Example 12. A variant on this idea is theconcept of constraint forces or constraints.Imagine a rigid straight wire along the x-axis,y = z = 0, with a bead sliding along it. The beadis constrained to stay at y = z = 0 becauseany attempt to move away is opposed by the(extremely strong) reaction force of the wire.


2 Dimensional Analysis

We have three fundamental quantities, and witheach of them we can associate an abstractdimension and possible units

• Mass: [M ]: kg, ounce, ton, stone, hundredweight,carat,

• Length: [L]: m, cm, inch, mile, perch, verst,

• Time: [T ]: second, day, month, year, leapyear,millennium.

Now it makes sense to say “length1 = length2”,for if it’s true in metres then it’s true in miles.However a statement “length1 = time2” couldbe a numerical coincidence only in one set ofunits. Schematically we allow [L] = [L] but not[L] = [T ]. Besides the fundamental quantities,we can construct derived ones with their abstractdimensions, e.g.,

[area] = [L2], [velocity] = [LT−1],

[density] = [mass/volume] = [ML−3],


and in an equation a = b + c, a, b and c musthave the same abstract dimension, a useful checkon the algebra.

Dimensioned quantities occur only in power laws6,e.g., volume = length3, but never in otherfunctions; what does sin(2 inches) mean?

Only quantities which are dimensionless can occurin general functions, e.g., in sinα = sin(length ×length/area) where [α] = [L][L]/[L2] =[L2]/[L2] = [1]. This idea, often calleddimensional analysis, although superficiallysimple, has powerful consequences because onecan obtain significant results from (apparently)insignificant information. We study someexamples of dimensional analysis.

6This key result is proved in the appendix to this chapter.


Example 13. [Pythagoras] We obtain Pythagoras’theorem from “elementary” premises.

A

B Ca

bc

θ

We can specify a right-angled triangle by givingits hypotenuse (longest side) a and smallest angleθ. Its area A must therefore be a function of aand θ, A = A(a, θ). Notice that θ and A/a2 aredimensionless and so A/a2 = f(θ), or A(a, θ) =a2f(θ).

A

B CDa

bc

θ

θ

Similarly the areas ABD and ADC are c2f(θ)


and b2f(θ) and their sum is the original area, i.e.,

c2f(θ) + b2f(θ) = a2f(θ),

ora2 = b2 + c2.

You should examine, very carefully, what hasbeen assumed here (Euclidean geometry leadingto similarity of triangles, addition of areas), andwhat has not been assumed, an explicit formulafor f(θ), which (trigonometric identity) impliesthe Pythagoras theorem.

Example 14. [Simple pendulum] As a morephysical example, consider a simple pendulum,consisting of a point mass m connected to a fixedpoint by a massless thread of length . Thependulum performs oscillations in a vertical plane(under the influence of gravity) of frequency ω.What can be said using dimensional analysis?

We first establish the dimensions of m, , g andω

[m] = [M ], [] = [L], [g] = [LT−2], [ω] = [T−1].


Next we construct all of the dimensionlessquantities that can be formed from these. Thereis essentially one, τ = g−1ω−2. Of course 2τ andτ2 are also dimensionless, but they are not reallydifferent. As was stated above, in any physicalequation the dimensions must match. It followsthat they should be expressible in dimensionlessform, and so the physics of pendulums impliesthat τ will satisfy a relation of the form f(τ) = 0.Now apart from exceptional cases this fixes τ asa (positive) zero of f , i.e., τ = C−2 a constant.Thus g−1ω−2 = C−2 whence

ω = C

√g

.

Those of you who have seen the simple pendulumbefore can verify that (for a particular value of C)this is the correct formula.

Exercise 10. There is of course a systematicway to obtain the form of τ . Posit a relation

τ = gαlβωγ,

where α, β and γ are constants and determine


their values by requiring that τ be dimensionless.


Example 15. [Flow in a pipe] Consider a longstraight pipe of length and circular cross sectionof radius a. Suppose that a liquid with viscositycoefficient η flows through the tube. ([η] =[ML−1T−1].) We are interested in the flow Qper unit time as a function of the pressure dropP per unit length. Now

[P ] = [force][area]−1[L−1] = [ML−2T−2],

and [Q] = [L3T−1]. Note that the physicssuggests that is irrelevant here. Thusessentially the only dimensionless parameter isθ = Pa4/(Qη) and so by our usual argument

Q =Ca4P

η,

for some constant C. Thus if an oil refinerydoubles the radius of a pipe it can send 16times as much oil down it. We all have twolarge important pipes in our bodies, the left andright coronary arteries. When fatty material isdeposited on their walls the tubes narrow. There


is a substantial safety factor built in, but thefourth power law means that once the dangerlevel is reached, quite small further deposits canhave catastrophic results.

Example 16. [Surface waves] Consider surfacewaves in an ocean of depth H These havewavelength λ and travel with speed c. Howdoes c depend on H, λ, g and ρ the density ofwater?

H

λ

��

��

We have already written down the dimensions ofall of the quantities involved and it is easy toverify that there are essentially two independentdimensionless quantities τ1 = λ/H and τ2 =c2/(gH). The physics of water waves presumably


implies a dimensionless equation of the form

f(τ1, τ2) = 0,

and in general this is all that we can say. Subjectto smoothness requirements though, the implicitfunction theorem implies τ1 = τ1(τ2). Howeverwithout extra information we cannot proceedfurther

Suppose though that the ocean is very deep, i.e.,λ H. Then one might expect the physics to beindependent of H, or in our notation f(τ1, τ2) =g(τ2/τ1) = g(c2/(gλ)). Thus c2/(gλ) = A2, a(constant) positive zero of g, and we deduce thatfor deep waves.

c = A√

gλ.

The other extreme is a very shallow ocean,H λ. In this case we expect c to be almostλ-independent, or in dimensionless notation,f(τ1, τ2) = h(τ2). By a similar argument wecan deduce

c = B√

gH,


where B is a constant. Both types of behaviourare verified experimentally.

Example 17. [Atomic bombs] Our final examplerelates to atomic bomb explosions. In the period1944–48 the US government regarded the amountof energy E released as top secret. They didhowever release movies of explosions, and fromthese the Cambridge mathematician G.I. Taylordeduced and published the result E ≈ 1021erg,causing great embarrassment all round. How didhe do this?

Taylor modelled the explosion as the release ofenergy E in a small volume, essentially a point, att = 0. There is a spherical blast wave of radius r.He argued that r can depend only on t, E and ρthe density of air. We know the dimensions of allquantities, viz. [r] = [L], [t] = [T ], [ρ] = [ML−3]and you should know [E] = [ML2T−2]. Positinga relation r = KtαρβEγ where K, α, β andγ are dimensionless constants, Taylor used thetechnique of exercise 10 to obtain the exponents

α = 25, β = −1

5, γ = 15.


Taylor knew from theoretical reasoning that K ≈1 and so

r ≈(

t2E

ρ

)1/5

.

The value of ρ is no secret, and so plotting ras a function of t using the movie film, Taylordeduced, correctly, the value of E.


Appendix A: Composite dimensions

are always power laws∗

Earlier in this chapter it was claimed that allphysical quantities have dimensions given bypower laws e.g., [velocity] = LT−1. We nowprove this result. Suppose the quantity a hasdimension given by

[a] = f(L, M, T ),

and that in some standard unit system it takesthe value a0. Suppose now we move to a newsystem in which the units are decreased by factorsL1, M1 and T1. Then the new value will be a1 =a0f(L1,M1, T1). Suppose that a2 is similarlydefined. Then

a2

a1=

f(L2, M2, T2)f(L1, M1, T1)

.

However we could regard system 1 as the standard,and system 2 having been obtained by decreasingthe units by factors L2/L1, M2/M1 and T2/T1,i.e.

a2

a1= f(L2/L1, M2/M1, T2/T1).


Equating the two forms for a2/a1 we obtain afunctional equation for f

f(L2, M2, T2)f(L1, M1, T1)

= f(L2/L1,M2/M1, T2/T1).

To solve it we first take the partial derivative withrespect to L2 keeping all other quantities fixed.

∂f∂L(L2, M2, T2)f(L1, M1, T1)

=1L1

∂f

∂L

(L2

L1,M2

M1,T2

T1

).

Next set L2 = L1 = L etc.

∂f∂L(L, M, T )f(L,M, T )

=1L

∂f

∂L(1, 1, 1) =

α

L,

say. Now the solution of y′(x)/y(x) = α/x isy = cxα, where c is a constant. Here we aredifferentiating with respect to L holding M andT fixed, and so the “constant of integration” candepend on M and T . Thus

f(L, M, T ) = g(M, T )Lα.


Substituting this back into the original functionalequation, we obtain a new one of identical type

g(M2, T2)g(M1, T1)

= g(M2/M1, T2/T1).

We solve this by the same technique findingg(M, T ) = h(T )Mβ and a further applicationof the technique gives h(T ) = cT γ, where cis a constant “constant of integration”. Thusf(L,M, T ) = cLαMβT γ, a power law as claimed.


3 Simple Examples

3.1 Oscillations

The extension of an elastic spring of unstretchedlength 0 and actual length > 0 is ( − 0)/0.It is observed empirically, Hooke’s law, that thespring exerts a force or tension T given by

T = λ

( − 0

0

), (29)

where λ is the modulus of elasticity of thespring (see example 6). An elastic string behavessimilarly whenever the extension is positive.

Example 18. [Longitudinal oscillations] An unstretchespring PO has one end P fixed at x = −0 andthe other end O free at x = 0. A particle of massm is attached at O and is given a displacementin the x-direction. What happens?

If the particle is at x(t) then the length of thespring is 0 + x(t) and the tension is λx(t)/0.


If x > 0 the tension acts on the particle in thenegative x-direction and so

mx = −T = −λx/0,

or

x + ω2x = 0, ω2 = λ/(m0). (30)

This is the equation of simple harmonicmotion (SHM). To integrate equation (30)multiply it by 2x

2xx + ω2(2xx) = 0,

ord

dt(x2 + ω2x2) = 0,

which implies

x2 + ω2x2 = a2, const. (31)


a

ω

a

x

x

Clearly x = (a/ω) sin θ(t), x = a cos θ(t) satisfiesequation (31) for any choice of θ(t). Imposingthe condition x = dx/dt gives θ = ω, i.e., x =(a/ω) sin(ωt + θ0). The motion is periodic withperiod τ = 2π/ω. Note that the physics requiresa non-negative length for the spring. Thus a/ω <0.


Example 19. [Transverse oscillations] A particleof mass m is attached to the mid point P ofa tightly stretched string AB of length 2b andtension T . It is displaced slightly through yperpendicular to the direction of the string. Findthe period of the resulting oscillations.

A BO

P

T T

y

b

Note that AP =√

b2 + y2 = b(1 + (y/b)2)1/2 ≈b(1 + 1

2(y/b)2) ≈ b if terms quadratic in y/b areneglected. Thus the tension remains T and

my = −2Ty√

b2 + y2≈ −2T

y

b

by the same argument. This is SHM of period2π√

mb/(2T ).


Example 20. [Large oscillations] A simple pendulumconsists of a rigid massless rod OP of length with the end O fixed. P carries a mass m and thesystem is released from rest at an angle α to thevertical. Find the period of the oscillations.

Q

O

P

T

mg

θ

θ

Because we are not interested in the tensionT we resolve vectors along the direction PQperpendicular to OP . P has an instantaneousvelocity θ and acceleration θ along PQ and thisis opposed by the component of gravity mg sin θalong PQ. Thus mlθ = −mg sin θ or

θ + ω2 sin θ = 0, ω2 =g

. (32)


We try the usual trick, multiplication by 2θ,finding

d

dt(θ2 − 2ω2 cos θ) = 0

so that θ2 − 2ω2 cos θ = C, a constant. Whenθ = α, θ = 0 and so

θ2 = 2ω2(cos θ − cosα). (33)

Thus the time taken for θ to change from 0 to αis

τ =1

ω√

2

∫ α

0

dθ√cos θ − cosα

, (34)

and the period is 4τ . The integral (34)cannot be evaluated in terms of simple functions;the indefinite integral is a Jacobean ellipticfunction. If α is small we can find τeither by approximating the integral (34) or byapproximating the equation (33). In the lattercase we may set cosα ≈ 1 − 1

2α2 and, since

|θ| � α, cos θ ≈ 1−12θ

2, finding θ2+ω2θ2 = ω2α2

which is SHM with period 2π/ω.


3.2 Projectiles

We look at a selection of problems involvingmotion under gravity.

Example 21. A ball is thrown from the originO with speed V at an angle α to the horizontal.Show that if V is sufficiently large, there willbe two trajectories passing through the arbitrarypoint P , (x = X, y = Y ), and that the timestaken satisfy t1t2 = 2

√X2 + Y 2/g.

Neglecting air resistance we have x = 0, x =V cosα and x = V t cosα. Similarly y = −g,y = V sinα − gt and y = V t sinα − 1

2gt2. Nowset x = X , y = Y . The answer appears not toinvolve α and so we eliminate it:

X2+(Y +12gt2)2 = V 2t2 cos2 α+V 2t2 sin2 α = V 2t2,

or

14g2t4 + (gY − V 2)t2 + X2 + Y 2 = 0.

This is a quadratic for t2. Provided gY − V 2 <0 and (gY − V 2)2 > g2(X2 + Y 2) (which will


certainly be the case if V is sufficiently large)there will be two positive roots t1

2 and t22 with

t12t2

2 = 4(X2 + Y 2)/g2.

Example 22. [Transition to vector equations]Recall the dynamics content of the previousexample. We had

x = 0, x = V cosα, x = V t cosα,

y = −g, y = V sinα − gt, y = V t sin α − 12gt2.

We can combine these two sets of equations,setting x = x(t)i + y(t)j, V = V (cosαi + sinαj)and g = −gj. Then

x = g, x = V + gt, x = Vt + 12gt2.

Notice that we could start from the vector form ofthe equation of motion and integrate it directly.This is a very useful technique, and it works forany linear problem.


The next example involves both the use of vectorsand linear friction.

Example 23. A ball is thrown from the originwith initial velocity V in a crosswind of velocityU. Assuming that the air resistance produces aforce of −k times the relative velocity, determinethe orbit.

Let the ball have position x(t), and let thegravitational force be mg and the frictional forcebe F = −k(x−U). Thus mx = mg−k(x−U),or

x + βx = g + βU, (35)

where β = k/m. The complementary function iseasily seen to be x = c+de−βt where c and d areconstant vectors. We need a particular integral,and so we guess x = at finding a = U + g/β.The general solution of equation (35) will be thesum of these two solutions. We fix c and d byrequiring x(0) = 0 and x(0) = V so that

x(t) =1β

(V − 1βg − U)(1 − e−βt) + (U +

1βg)t.

It should be noted that this is planar motion.


The next example extends example 10on quadratic friction and illustrates someapproximation techniques.

Example 24. A particle is fired verticallyupwards with speed V . Besides gravity thereis an acceleration due to air resistance opposingthe motion with magnitude kv2 where v is thespeed. Assuming air resistance is small, find themaximum height and the time taken to reach it.

What does “small” mean? Clearly kV 2/g isdimensionless and we assume ε ≡ kV 2/g 1.We can proceed as in example 10 or as follows. Letv = y. On the upward path v = −g − kv2. Nowdv/dt = (dv/dy)(dy/dt) = v(dv/dy) = −g−kv2

and sodv2

dy= −2g − 2kv2.

Separation of variables yields log(2kv2 + 2g) =−2ky + C. Since v = V at y = 0

log(

kv2 + g

kV 2 + g

)= −2ky.

The highest point occurs when v = 0 and 2ky =log((kV 2 + g)/g) = log(1 + ε). Now if |x| < 1,


log(1+x) = x− 12x

2 + . . . and so 2ky ≈ ε(1− 12ε)

or

y ≈ V 2

2g

(1 − kV 2

2g

).

In order to find the time we start from v =−g − kv2 again. Separating variables directly(dv/(g + kv2) = −dt) we find (recall v = V att = 0)

1√kg

tan−1

(√k

gv

)= −t+

1√kg

tan−1

(√k

gV

)

The highest point occurs when v = 0, t =(tan−1

√ε)/

√kg. Now Gregory’s series is

tan−1 x = x − 13x

3 + . . . when |x| < 1, and thisleads, after some simple algebra, to

t ≈√

ε√kg

(1 − ε

3

)≈ V

g

(1 − kV 2

3g

).

Example 25. [Example 24 continued] What isits speed U when it reaches the ground again, andwhy is it different from V ?


Here we have to be careful about the sign of theair resistance. On the downward path we have

dv2

dy= −2g + 2kv2,

with first integral log(2g−2kv2) = 2ky+C, whereC is a constant. (The apparently gratuitous factorof −1 is to ensure that we take log of a positiveargument.) When v = 0 we know (see top ofthe previous page) that 2ky = log(1 + ε) and soC = log(2g/(1 + ε)). Finally when y = 0 we find

(1 + ε)(

1 − kU2

g

)= 1,

or

U =√

g

k

√ε

1 + ε≈ V

(1 − kV 2

2g

).

U < V because air resistance has reduced theparticle’s energy.


3.3 Rockets

A rocket has mass m(t), velocity v(t) andis acted upon by a force F(t). It emitsexhaust gas at a rate |m| (where m < 0)with (usually constant) velocity U relative to therocket. The details of this problem involve verycomplicated physics. Fortunately there are somany microscopic particles in the gas that we caninvoke statistical physics, which tells us that wecan regard a packet of identical particles as amacroscopic particle with mass equal to the sumof the microscopic masses and velocity equal tothe average velocity.

Let us compare the system at times t and t + δt.At time t we have a rocket mass m(t), velocityv(t). At time t+ δt the rocket has mass m+ δm,velocity v + δv and in addition there are exhaustgases of mass −δm with (approximate) velocityv−U. Suppose for the moment that F = 0, i.e.,we have a closed system. Then, Theorem 1, thetotal momentum is conserved.


m,v

m + δm,v + δv −δm,v − U

t

t + δt

At time t it is mv and at time t + δt it is(m + δm)(v + δv) + (−δm)(v−U+ O(δt)), seefigure. Thus

mv + mδv + δmv − δmv + δmU− mv ≈ 0,

valid for small δt, where the approximation isignoring terms of order (δt)2. If we include theexternal force F we obtain

mδv + δmU ≈ Fδt.

valid for small δt. Now divide by δt and take thelimit as δt → 0. There is a first order small termwhich vanishes and we are left with the (exact)rocket equation

mv = F − mU. (36)


Example 26. A rocket of initial mass m0 burnsfuel at a constant rate α and lifts off verticallyunder gravity. Compute the speed as a functionof time.

Here m(t) = m0 − αt. Resolve equation (36)vertically, setting F = −mg, and dividing bym(t)

dv

dt= −g +

αU

m0 − αt.

Since v = 0 at t = 0 a simple integration gives

v = −gt − U log(

1 − αt

m0

). (37)

Two points should be noticed.

• The condition for liftoff is (dv/dt) > 0 at t = 0,i.e., Uα/m0 > g. If this condition is satisfied then(dv/dt) > 0 for 0 < t < m0/α.

• Let the “payload” or “final mass” (no more fuel)be βm0. Then the final speed V0 can be computedfrom (37) as V0 = −gt − U log β. If β 1,V0/U can be arbitrarily large, although gravityeventually slows the rocket.


Another approach is to take m as the independentvariable.

Example 27. Consider a rocket, as above,subject to air resistance with a force kv opposingthe motion, and for simplicity ignore gravity.The rocket equation is mv = αU − kv. Now(dv/dt) = (dv/dm)(dm/dt) = −α(dv/dm) sothat after separating the variables in the rocketequation

(k/α)dv

(k/α)v − U=

k

α

dm

m.

The integration of this differential equation isstraightforward. Using the condition v = 0 atm = m0 we find

1 − k

α

v

U=(

m

m0

)k/α

.

Finally m = βm0, v = V0 where V0 = (αU/k)(1−βk/α), and so air resistance imposes an upper limiton V0, a terminal velocity .


The opposite to a rocket,(m > 0) is an avalancheor a condensing raindrop. This is a longer exampleillustrating three different ways of solving the sameproblem, which involves expansions in terms of asmall parameter. Quite often it is hard to decidewhich is the optimal approach, and so one needsto keep a store of applicable methods.

Example 28. [Falling raindrop] A small raindropfalling through a stationary cloud acquiresmoisture by condensation from the cloud. Whenthe mass of the raindrop is m, the rate of increaseof the mass per unit time is km. The raindropstarts from rest. Neglecting resistance due tomotion, find the relation between the velocity vand the distance fallen y, and prove that if k issmall enough the velocity is given approximatelyby

v2 = 2gy

[1 − 2

3k

√2y

g

].

First we establish the dynamics. Obviously wecan use the rocket equation (36), resolving in thedownward vertical direction. The force F is m(t)g


and m = km. Because the cloud is at rest therelative velocity is U = v. Thus my = mg− kmyor, with the obvious change in notation,

v = g − kv.

The naıve, autopilot, approach is to spot thatthis is a linear equation with constant coefficients.Setting y = v = 0 at t = 0 we have

v =g

k

(1 − e−kt

), y =

g

kt − g

k2

(1 − e−kt

).

While the integration is easy, and the resultsare correct, it is hard to obtain what was askedfor, because we next need to eliminate t betweenthese two equations. The first stage is to note thatt = −(1/k) log (1 − kv/g). The details are clearlymessy and are left as an exercise for masochists.

A little more thought might suggest that sincethe answer does not involve t there is little pointin introducing t as the independent variable. Wehave another choice for v = (dv/dy)(dy/dt), so


that if we regard v as a function of y rather thant we have

vdv

dy= g − kv.

This is, admittedly, non-linear, but it is offirst order and is easily solved by separation ofvariables. Using v = 0 at y = 0 we obtain

y = −v

k− g

k2log(

1 − kv

g

),

which is the required relation between v and y.

Next we assume that kv/g 1 and expand thelog term as a Taylor series, finding

y = −v

k− g

k2

(− kv

g− 1

2k2v2

g2− 1

3k3v3

g3− . . .

)

= −v

k+

v

k+

v2

2g+

kv3

3g2+ . . .

=v2

2g

(1 +

23kv

g+ . . .

)

=v2

2g

(1 + O(kv/g)

).


Thus v2 = 2gy(1 + O(kv/g)

)which implies

v =√

2gy(1+O(kv/g)

). Inserting this back into

the series,

y =v2

2g

(1 +

23k

g

√2gy + O((kv/g)2)

),

and this is easily manipulated into

v2 = 2gy

(1 − 2

3k

√2y

g+ O((kv/g)2)

).

Well that was long, but at least it got the answer!

The third method of solution relies on the factthat, provided all functions are smooth, i.e., theyhave Taylor series expansions in the appropriatesmall parameter, then successive approximation(as above) commutes with solving ordinarydifferential equations, even non-linear ones!

We return to our differential equation

vdv

dy= g − kv = g

(1 + O(kv/g)

),


The zero order equation vv′ = g integratesimmediately to give v2 = 2gy (for we required v =0 at y = 0) and we deduce v2 = 2gy+O(kv/g) orv =

√2gy(1 + O(kv/g)). Substituting this back

into the equation

vdv

dy= g − k

√2gy1/2 + O

((kv/g)2

),

which integrates to give

v2 = 2(

gy − 23k√

2gy3/2 + O((kv/g)2

))

= 2gy

[1 − 2

3k

√2y

g+ O

((kv/g)2

)],

the answer!

The purpose of this extended example is toemphasize that there is often more than one wayto solve a problem. To find which method isoptimal for a given problem requires some insightand some experience.


3.4 Electrodynamics

We recall the Lorentz force described by equation(23)

mx(t) = FLorentz = q(E(t,x(t))+x(t)×B(t,x(t))

).

(38)For a first study of this equation we make thesignificant simplification that both E and B areconstant both in time and space. We consider firstthe case of single fields, and then a specializationof the general case. The case B = 0, i.e., x =(q/m)E is easily despatched for this is constantacceleration. The remaining cases make extensiveuse of the following result.

Lemma 1. If vectors x and y depend on time

(x.y) = x.y + x.y.

Proof. We sketch the proof in two dimensionswhere x.y = x1y1 + x2y2. Then (x.y) = x1y1 +x2y2 + x1y1 + x2y2. �


Exercise 11. With the same notation show that

(x × y) = x× y + x × y.

Example 29. Now suppose that E = 0, andchoose axes so that B = Bk where k is a unitvector in the z-direction and B = |B|. Then theLorentz equation (38) can be written as

x = ωx × k, ω = qB/m, (39)

where it is easily seen that [ω] = [T−1] so that ωis a frequency, the gyromagnetic frequency.

There are many ways to solve eq.(39). A simple-minded approach is to write x = (x, y, z), and toresolve the equation in components

x = ωy, y = −ωx, z = 0. (40)

We obtain immediately z = z0+w0t and y = a0−ωx where z0, w0 and a0 are constants. Eliminatingy from the first equation we obtain x+ω2x = ωa0.


It is straightforward to solve this SHM equationfinding x = a0/ω + (V0/ω) cos(ωt − α) where V0

and α are constants of integration. We now havey = a0 − ωx = −V0 cos(ωt − α) and finally,

x = x0+V0

ωcos(ωt−α), y = y0−V0

ωsin(ωt−α),

(41)where x0 = a0/ω and y0 are constants.

A more sophisticated (and generalizable) wayto solve the first two equations (40) is tointroduce the complex variable ζ = x + iyfinding ζ + iωζ = 0. We can solve this asζ = [−iV0 exp(iα)] exp(−iωt), where the termin square brackets is the “constant of integration”and V0 and α are real constants. This is easilyintegrated to give

ζ = ζ0 +V0

ωe−i(ωt−α),

where ζ0 is a complex constant. Setting ζ0 =x0 + iy0 we recover (41).

Another powerful approach is to treat eq.(39) as avector equation, setting x = xi+yj+zk. Dotting


the equation with k gives k.x = 0 or z = 0 whichimplies z = z0 + w0t as before. Dotting theequation with x gives x.x = 0. Using the lemmawe can deduce x2 = V 2

0 + w20 a constant. This

implies x2 + y2 = V 20 and so we set

x = −V0 sin θ(t), y = −V0 cos θ(t),

where θ(t) is to be determined. Now dotting(39) with i gives x = ωy which implies θ = ωand θ = ωt − α where α is a constant. Anotherintegration now gives (41).

Suppose we shift the origin to (x0, y0, z0), and forthe moment set w0 = 0. Then eq.(41) describesmotion in a circle in the (x, y)-plane with constantangular velocity −ω.

i

j


If we allow z = w0t the circle becomes a helixwith axis parallel to k,

x =V0

ωcos(ωt − α)i − V0

ωsin(ωt − α)j + w0tk,

which is a helix with axis in the 3-direction.

Example 30. Now we assume that both E andB are nonzero, but we restrict attention to the,physically important, case E.B = 0. Making useof the results from the previous example we setx = (x, y, z), E = (0, E, 0) and B = (0, 0, B),as well as ω = qB/m, and λ = qE/m. Now theLorentz equation (39) implies

x = ωy, y = λ − ωx, z = 0,

and so z = z0 + w0t as before. In order to solvethe other two equations we set ζ = x+ iy, finding

ζ = iλ − iωζ.

Here the complementary function is ζ = C +ζ0e

−iωt, where C and ζ0 are complex constants,


and a suitable particular integral is ζ = (λ/ω)t.The constant C corresponds to a fixed change ororigin, and so we discard it. Setting ζ0 = R0e

iα

(R0 = V0/ω) we have

x =λ

ωt+R0 cos(ωt−α), y = −R0 sin(ωt−α),

so that the helix is sheared into the x-direction.

Notice that λ/ω = (qE/m)/(m/(qB)) = E/Bdepends only on the fields and is the same forall particles with q = 0. The velocity vL =(E/B, 0, 0) is the Larmor drift velocity

vL =E× B|B|2 .

The easiest way to visualize the motion is tochoose an inertial frame so that x(0) = x(0) = 0.then it is easy to verify that, setting R = E/(Bω),

x = R(ωt − sinωt, 1 − cosωt, 0),

andx = Rω(1 − cos ωt, sinωt, 0).


Note that

(x − Rωt)2 + (y − R)2 = R2.

Consider a bicycle wheel radius R in the z = 0plane rolling without slipping along the x-axis withangular velocity ω. Then the centre has velocity(Rω, 0, 0). The above equation implies that theparticle behaves like a point on the rim and sotraces out a cycloid.


4 Energy and Stability

4.1 One degree of freedom

A system is said to possess one degree offreedom if it can be described by a scalarequation of motion of the form

mx = F (x). (42)

The potential energy is

U(x) = −∫ x

F (ξ) dξ, (43)

defined up to an arbitrary additive constant. Thenmx = −dU/dx which we have seen before. Thekinetic energy is T = 1

2mx2 and the totalenergy is E(x, x) = T +U . The following resultis fundamental.

Theorem 2. For such a system E is a constant.


Proof.

dE

dt=

d

dt

(12mx2 + U(x)

)= mxx + U ′(x)x

= x(mx − F )

= 0 by (42).

�

Note that the more general case F = F (x, x)is excluded. E.g., mx = −mg − kx. SettingU = mgx we have (d/dt)(T + U) = −kx2 < 0.

Surprisingly the general case with more degreesof freedom has not yet been solved. There ishowever an important special case where we canmake progress.


4.2 Conservative systems

Suppose we consider systems described by thevector equation of motion

mx = F(x). (44)

The system is said to be conservative if ∃ U(x)such that F = −∇U . Then the potentialenergy is U(x) and the kinetic energy isT = 1

2mx.x.

Theorem 3. For a conservative system thetotal energy E = T + U is conserved.

Proof.

dE

dt= mx.x + (∇U).x

= x.(mx − F) = 0.

�

Note that the potential and total energies are notdefined for non-conservative systems.


4.3 When is a system conservative?

O

P

x

F

Suppose we have a constant force F whose pointof application is displaced through x from O toP . Then the work done is W = F.x. Moregenerally we want to consider varying forces andcurved paths. We split the path into a largenumber, N , of tiny segments δx(i) and let theforce, restricted to the ith segment be F(i).

O

P

δx(i)

F(i)

Define the work function by

W = limN→∞

∑i

F(i).δx(i) →∫ P

O

F.dx. (45)


In one dimension W =∫ P

OF dx = −(U(P ) −

U(O)) but how do we calculate (45) in higherdimensions? Suppose the path is specified byx = x(s) where s = 0 at O and s = 1 at P .Then dx = (dx/ds) ds = x′ ds and

W =∫ 1

0

[F(x(s)).x′(s)

]ds, (46)

where the integrand is a scalar function of s andthe integral is computable by elementary methods.


Example 31. [Path independence] In two dimensionslet O be (0, 0) and P be (1, 1) and consider thethree paths shown. (b is given by y = x2.)

O

P

M

a b c

Suppose F = (y, x).

For path a, x(s) = (s, s), x′ = (1, 1) and Wa =∫ 1

0(s, s).(1, 1) ds =

∫ 1

02s ds = 1.

For path b, x(s) = (s, s2), x′ = (1, 2s) and

Wb =∫ 1

0(s2, s).(1, 2s) ds =

∫ 1

03s2 ds = 1.

For the path OM x = (s, 0), x′ = (1, 0)and Wci =

∫ 1

0(0, s).(1, 0) ds = 0. For the

path MP x = (1, s), x′ = (0, 1) and Wcii =∫ 1

0(s, 1).(0, 1) ds = 1, and so Wc = 1.

The work integral appears to be path-independent.


Exercise 12. [Path non-independence] Considerthe same setup but with F = (y,−x). ShowWa = 0, Wb = −1

3 and Wc = −1, so that theintegral depends on the path taken.

In the case of example 31, any path from O toP will give the same answer. For fixed O, theanswer depends only on P and not on the pathtaken. Thus setting P at x we can define

U(x) = −∫ x

0F(ξ).dξ. (47)

Then if Q is at x + δx,

δU = U(x + δx) − U(x) = −∫ x+δx

xF(ξ).dξ

≈ −F(x).δx

≈ −F1δx − F2δy,

where we have set F = (F1, F2). It follows thatF = (−∂U/∂x,−∂U/∂y) = −∇U and we have aconservative system. (For example 31 U = −xy.)


In exercise 12 none of this goes through and thereis no potential energy.

So what’s the difference? Most textbooks referyou to other books at this point, but we will tryto be a little more complete. First a notationalpoint. In the next lemma and theorem δx, δy etcwill be small but variable quantities and we shallfrequently take limits as they tend to zero. Aquantity which has magnitude |(δx)m(δy)n| willbe denoted by O(εm+n)

Lemma 2. [Taylor Series] Suppose f : R2 →R is a smooth function. Then

f(x+δx, y+δy) = f(x, y)+∂f

∂xδx+

∂f

∂yδy+O(ε2),

(48)where the partial derivatives are computed at thepoint (x, y).

Proof. Suppose first that δy = 0. Then (48) isequivalent to the assertion

∂f

∂x= lim

δx→0

f(x + δx, y) − f(x, y)δx

.


A similar argument applies for the second (andindeed any further) dimension. �


Theorem 4. The following assertions areequivalent:

1. F = −∇U for some U(x),

2.∫ B

AF.dx is path-independent,

3.∮

CF.dx = 0 for all closed curves C,

4.∮

F.dx = 0 for all closed infinitesimal rectangles,

5. ∇ × F = 0.

Here

∇×F =(∂F3

∂y− ∂F2

∂z,∂F1

∂z− ∂F3

∂x,∂F2

∂x− ∂F1

∂y

).

Proof.

1 ⇒ 2:∫ B

AF.dx = − ∫ B

A∇U.dx = − ∫ B

AdU =

U(A) − U(B) independent of the path chosen.


2 ⇒ 1: Regard A as fixed and B as a variablepoint. Define a function U via U(B) = U(A) -∫ B

AF.dx which is independent of the path chosen.

Then

δU = U(x + δx) − U(x)

= −∫ x+δx

xF.dx = −F(x).δx + O(ε2).

Choosing δx = (δx, 0, 0) gives δU = −F1δxwhich implies F1 = −∂U/∂x. A similar resultholds for the other components.

A

B

C1

C2

2 ⇒ 3: Suppose that A and B lie on a closed

curve C. Then∫ B

AF.dx is the same whether we

use path C1 or C2. Thus

(∫C1

−∫

C2

)F.dx = 0 =

∮C

F.dx.


3 ⇒ 2: Choose any two paths C1 and C2 fromA to B, and let C = C1 − C2. Then

∫C1

F.dx =∫C2

F.dx.

3 ⇒ 4: This step is obvious.

4 ⇒ 3:

Regard a smooth closed curve C and its interior asbeing composed of a large number of infinitesimalrectangles Rn with sides parallel to the coordinateaxes. By condition 4,

∑n

∮Rn

F.dx = 0. Howeverall the interior sides are traversed twice in oppositesenses and so cancel out. Thus

∮CF.dx = 0.

1 ⇒ 5: This step is an (almost) obvious identity.

5 ⇒ 4:


Consider a small rectangular path from (x, y) to(x + δx, y + δy) and back again, and integrateF around it. The contribution from the verticalsides is∫ δy

0

[F2(x + δx, y + s) − F2(x, y + s)

]ds

=∫ δy

0

[F2(x, y) +

∂F2

∂xδx +

∂F2

∂ys − F2(x, y) − ∂F2

∂ys]ds

=∂F2

∂xδxδy + O(ε3),

where the Taylor lemma 2 was used. Including thecontribution from the horizontal sides we have∮

F.dx =[∂F2

∂x− ∂F1

∂y

]δxδy + O(ε3).

Thus if ∇ × F = 0 the integral vanishes. �

Thus a system with many degrees of freedom is


conservative iff ∇ × F = 0.


4.4 Qualitative dynamics with one

degree of freedom

A system with one degree of freedom is necessarilyconservative, theorem 1, and the energy equation12mx2 + U(x) = E constant, can be rewritten as

x2 = f(x) ≡ 2(E − U)/m. (49)

Differentiating with respect to time, and dividingby 2x gives

x =12f ′(x). (50)

This has only been derived for points where x = 0.In general zeros of f will be isolated and so (50)holds, by continuity, at these points too.

We start with the obvious remark that the motionis possible only for those points where f(x) � 0,and if f(x) > 0 two motions, right-moving (x >0) and left-moving (x < 0) are possible.


x

f

0

The figure shows a typical configuration. Motionis only possible for x � 0 and the tangent to thegraph of f at x = 0 is shown. If for x > 0the motion is right-moving it remains so, andsince f ′(x) > 0 it accelerates. On the otherhand if it is left-moving it approaches x = 0decelerating as it does so. In order to discuss whathappens as x → 0+ we approximate f locally byits tangent f(x) ≈ c2x where c2 = f ′(0) > 0.Thus x2 = c2x. Initially x < 0 and so x = −c

√x.

It is easy to integrate this equation, finding

√x =

12c(t0 − t),

where t0 is a constant. The interpretation is thatas t → t0− the particle comes to rest at theorigin. Because of equation (50) it has positive


acceleration there and so the velocity becomespositive. Thus for t � t0 we have to integratex = +c

√x, with solution

√x = 1

2c(t − t0). Wecan combine both cases as x2 = 1

4c2(t − t0)2.

Although this behaviour is exact only for linearf the general behaviour is qualitatively correct.The left-moving system reaches rest at x = 0in a finite time (assuming f ′(0) > 0) and thenbecomes a right-mover.

If we reflect the figure in the y-axis this qualitativebehaviour is still correct provided we interchange“left” and “right”. Further there is nothing specialabout x = 0; any zero of f would do.


Example 32. For motion in the earth’sgravitational field, U = −mgr2

0/r, example 5,and so, with a slight change of notation,

r2 = f(r) =2m

(E +

mgr20

r

).

Suppose a rocket is fired vertically upwards withspeed V0. If E < 0 then it is easy to see thatf has a single zero at a finite value rmax of r.Thus the rocket will ascend to r = rmax and thenreturn to earth. However if E > 0, f(r) > 0 forall r, and so the particle can escape the earth’sgravitational field. The critical case is whereE = 0 which corresponds to V0 =

√2gr0, the

so-called escape velocity. (V0 ≈ 11.2km s−1.)

Of course we do not need all this abstractmachinery to treat simple situations like example32. Its real purpose is to reduce the study ofstability of systems with one degree of freedom toa sequence of simple “A/S level” problems.


4.5 The stability of systems with one

degree of freedom

We consider systems described by equations (49)and (50). The system is said to have anequilibrium point at x = x0 if x(t) = x0,constant, is a possible motion of the system—inother words if x does not change. Now supposethat the system has an equilibrium point at x0 andthe system is given a “small perturbation”, i.e.,the functional form of f(x) is changed slightly.If the system remains in a small neighbourhoodof x = x0 then the equilibrium is said to bestable, otherwise it is unstable. (Note that theperturbed system may have no equilibrium pointsat all. Stability is the requirement that the motionis confined to a small neighbourhood.)

Example 33. Empirical experience shows thata simple pendulum consisting of a point massattached to a fixed point via a weightless rigid rodhas two equilibria. That in which the pendulumhangs vertically downwards is stable, while thatwhere the point mass is exactly above the pivotis unstable. (We shall see soon how to treat thisexample quantitatively.)


Theorem 5. [The stability algorithm] The setof equilibria of systems governed by equations(49) and (50) is the set of critical pointsx0 of f (f ′(x0) = 0) with critical valueszero, i.e., f(x0) = 0. Stable critical pointscorrespond to maxima, unstable ones to minima.(The case of an inflection point needs specialtreatment.) Perturbed motion about a stablecritical point is simple harmonic motion with

frequency ω =√−1

2f′′(x0).

Thus the investigation of stability reduces to thedetermination of maxima and minima of functionsf : R → R.

Proof. If x = x0 is an equilibrium point, then x(t)and all derivatives thereof must vanish at x = x0.In particular we must have f(x0) = f ′(x0) = 0from equations (49) and (50). Suppose converselythat f(x0) = f ′(x0) = 0, so that x and x bothvanish when x = x0. Differentiating (50) givesd3x/dt3 = 1

2f′′(x)x, and so the third derivative

of x vanishes at x = x0. By a simple inductionargument we see that all derivatives of x(t) vanish


at x = x0 and so we have an equilibrium. Thisproves the first sentence of the theorem.

Suppose that x = x0 is an equilibrium point, andconsider a motion x(t) = x0 + ε(t) where ε isinitially small. Then, using a Taylor expansion,

x = ε =12f ′(x0 + ε)

=12f ′(x0) +

12f ′′(x0)ε + O(ε2)

=12f ′′(x0)ε + O(ε2),

where Taylor’s theorem was used in the secondline. If f ′′(x0) > 0 then ε(t) grows exponentiallyfast, and the solution passes outside any smallneighbourhood of x = x0. Conversely if f ′′(x0) <0 we have ε(t) + ω2ε(t) = O(ε2) (with ω definedin the statement of the theorem) which proves thesecond sentence. �


It is instructive to examine the two cases usingthe qualitative methods of the previous section.

x

stable

unstable

a b

In the unstable case a minimum (with value 0) isshown as a solid curve, together with two possibleperturbations (dotted curves). In both cases themotion moves out of a small neighbourhood of thecritical point. In the stable case one perturbationis impossible for f(x) < 0. For the other f(x) > 0only for x ∈ (a, b) and so we have boundedmotion.


Example 34. [Simple stability] A bead P ofmass m can slide on a fixed smooth wire shapedas a circle of radius a, centre O. The circle liesin a vertical plane. Determine the equilibria andtheir stability. Let θ be the angle between OPand the downward vertical.

θ

O

P

Clearly T = 12ma2θ2, U = −mga cos θ + const.

and we set E = T + U After some simple algebra

θ2 = f(θ) ≡ 2ma2

E + 2g

acos θ,

and we can deduce that f ′(θ) = −2(g/a) sin θand f ′′(θ) = −2(g/a) cos θ. Equilibrium requires

f(θ) = 0 = f ′(θ).


The second of these fixes θ = 0 or θ = π. Thefirst is boring—it merely fixes E. However E isonly defined up to the arbitrary constant whichoccurs in U , and so its value is usually of nophysical interest.

Next note that f ′′(π) > 0, and so θ = π isunstable.

However f ′′(0) = −2g/a < 0 and so θ = 0 isstable, and the frequency for small oscillations isω =

√g/a.

Example 35. A massless spring of naturallength 0, modulus λ = αmg has the topend fixed, while the bottom end carries amass m. Show that there is a single, stable,equilibrium position and find the frequency ofsmall oscillations about it.

We measure y downwards from the unstretchedposition. Clearly T = 1

2my2, and there is agravitational contribution to the potential energyUg = −mgy + const. Now the tension in thespring is −λ(y−0)/0 = −dUel/dy and so Uel =12λ(y−0)2/0+const. Setting E = T +Ug +Uel


and absorbing all energy constants into E we have

y2 = f(y) ≡ 2E

m+ 2gy − αg

0(y − 0)2.

This is the equation of an upturned parabola.There is a single maximum when 0 = α(y −0), i.e., y = y0 = (1 + α−1)0, and so for anappropriate (but boring) choice of E we have astable equilibrium point. The frequency of smalloscillations is ω where ω2 = −1

2f′′(y0) = αg/0.


Example 36. A light elastic loop of unstretchedlength 4a passes over two small pegs at the samelevel, distance 2a apart. A particle is attachedto the loop, and in equilibrium the loop forms anequilateral triangle. Show that the frequency of

vertical oscillations is ω =√

7g

4√

3a.

Let the particle have mass m and let its depthbelow the pegs be ay where y is dimensionless.

a a

ay

m

The kinetic energy is T = 12ma2y2 and

the gravitational potential energy is (ignoringinessential constants) Ug = −mgay. The

stretched length of the loop is 2a+2√

a2 + a2y2.Since the unstretched length is 0 = 4a the elasticenergy is Uel = 1

2λ(4a)−1(2a + 2√

a2 + a2y2 −4a)2 = 1

2λa(√

1 + y2 − 1)2, where λ is the


modulus of elasticity of the string. The energyequation can be reduced to

y2 = f(y) ≡ 2E

ma2+

2gy

a− λ

ma

(√1 + y2 − 1

)2

.

It is elementary to compute

f ′(y) =2g

a− 2λ

ma

(√1 + y2 − 1

) y√1 + y2

,

f ′′(y) = − 2λ

ma

(1 − 1

(1 + y2)1/2+

y2

(1 + y2)3/2

).

At equilibrium we know that y =√

3 and sof ′(

√3) = 0, which implies λ = 2mg/

√3. Then

f ′′(√

3) = −7g/(2√

3a) and so we have stableequilibrium with the frequency of small oscillationsas advertised.


Example 37. [Planar stability] A particle Pmoves in a plane under the influence of n fixedcentres of attraction A1, A2,. . . , An in the plane.The magnitude of the attractive force to As ismλsrs, where λs is a positive constant and rs isthe distance PAs. If G is the centre of gravity ofmasses λ1, λ2, . . . , λn at A1, A2,. . . , An, provethat G is the unique position of stable equilibrium,and that the frequency ω of small oscillationsabout it satisfies ω2 = Λ ≡∑λs.

Let the position of the particle be (x, y). Itskinetic energy is 1

2m(x2+y2) and, by analogy withelasticity, its potential energy is (up to an additiveconstant) U = 1

2m∑

λs[(x−xs)2 +(y− ys)2]. IfE denotes the total energy, then simple algebragives

x2 + y2 = f(x, y) ≡2E/m −

∑s

λs[(x − xs)2 + (y − ys)2].

For an equilibrium position, the condition of zeroacceleration in the x-direction implies ∂f/∂x = 0,


or∑

λs(x − xs) = 0 or x = xG = Λ−1∑

λsxs,the x-coordinate of the centre of mass. A similarresult holds for the y-direction and so the centroidG is the unique equilibrium point.

Now write (x, y) = (xG + ε(t), yG + δ(t)). Thenusing

∑λs(xG − xs) = 0 etc., it is easy to see

thatε2 + δ2 = 2E′/m − Λ(ε2 + δ2).

This is the equation for SHM with frequency√

Λin each direction.


5 Impulses and collisions

5.1 Impulses

When two billiard balls collide at t = t0 theyalmost instantaneously exchange momentum. Wemodel this by saying that ball 1 suffers a forceF1 = I1δ(t − t0) or an impulse I1 at time t0.If ball 1 has momentum m1u1 before and m1v1

after the impulse then

m1v1 − m1u1 = I1. (51)

Note that we do not try to analyze the detailedphysics of the actual collision. Equation (51)is merely an empirical summary of the dynamicswhich we henceforth treat as an axiom.


5.2 Binary collisions in one dimension

We examine first the collision in one dimension,where m2, u2 and v2 are the correspondingquantities for ball 2. The force that ball 2 exertson ball 1 is F1←2 = I1←2δ(t − t0), and F2←1 issimilarly defined. However by Newton’s third lawF2←1 = −F1←2 and so I2←1 = −I1←2. Clearly

m1v1−m1u1 = I1←2, m2v2−m2u2 = I2←1.

Since we do not have, as yet, any furtherinformation on the impulses we eliminate themfinding

m1v1 + m2v2 = m1u1 + m2u2, (52)

expressing conservation of momentum during thecourse of the impulse. Note that this accords withthe discussion of closed systems in section 1.8.

In order to determine v1 and v2 in termsof u1 and u2 we need a further (vector)equation. Newton observed empirically that for 1-dimensional collisions the relative velocities after


and before impact were linearly related. Moreprecisely

v2 − v1 = −e(u2 − u1), (53)

where e is the coefficient of restitution. Itdepends on the physics of the collision, and(usually) e � 0. This provides the required secondrelation.

Actually ui and vi are not the most usefulvariables. We define the total mass M =m1+m2 and the centre of mass X = (m1x1+m2x2)/M (where xi are the positions of the balls).Now equation (52) asserts that MX is continuousat collisions, MV = MU, in a (hopefully) obviousnotation. Next set u1 = U + u1 etc., so that

m1v1 + m2v2 = 0 = m1u1 + m2u2. (54)

In this notation Newton’s law becomes

v2 − v1 = −e(u2 − u1), (55)


and simple algebra on equations (54) and (55)shows

v1 = em2

M(u2 − u1), v2 = −e

m1

M(u2 − u1),

(56)giving v1 and v2, and hence v1 and v2.

Let us look at the energies involved. Initially

Einit =12m1(U + u1)2 +

12m2(U + u2)2

=12M |U|2 +

12(m1|u1|2 + m2|u2|2)+

U.(m1u1 + m2u2),

but the last term vanishes by virtue of (54).Using this property again we have

Einit =12M |U|2+1

2M

[m1(m1 + m2)|u1|2 + m2(m1 + m2)|u2|2

]−1

2M|m1u1 + m2u2|2

=12M |U|2 +

12m1m2

M|u1 − u2|2,


after some algebra. A similar expression obtainsfor the final energy

Efin =12M |V|2 +

12m1m2

M|v1 − v2|2,

and using V = U and Newton’s law (55) we find

ΔE ≡ Efin −Einit =12m1m2

M(e2 − 1)|u1− u2|2.

(57)We expect to lose energy in a collision, e.g., sound,internal heating etc., and we deduce e2 � 1. Thecase e = 1 ⇔ ΔE = 0 is said to be perfectlyelastic.

Example 38. Three small spheres, A, B and C,whose masses are 8m, m and 7m are at rest inline, and AB = BC = a. The middle sphere B isprojected towards C with velocity U . Assumingall of the spheres to be perfectly elastic, discussthe subsequent motion, and illustrate this by adiagram showing the positions of the spheres atany time after the start.

We first determine the velocities after the variouscollisions; times and positions can then be


calculated. We are making extensive use of (52)and (53). If v and w are the velocities of B andC after the first impact then U = v +7w = w−vgiving

v = −34U, w = 1

4U.

If u and v are now the velocities of A and B afterthe second impact then −3

4U = 8u + v = u − vgiving

u = −16U, v = 7

12U.

If v and w are now the velocities of B and C afterthe third impact then

v + 7w = 712U + 7

4U, w − v = 712U − 1

4U,

giving v = 0, w = 13U , and there are no further

collisions. The velocities in the four stages ofmotion are shown below.


A B C

A B C

A B C

A B C

U

14U

14U

13U

712U

34U

16U

16U

Suppose the initial positions to be (−a, 0, a). Thetime to the first collision is t = a/U after whichthe positions are (−a, a, a).

The time to the second collision is t = 2a/(34U) =

83a/U , and it is straightforward to calculate thenew positions to be (−a,−a, 5

3a).

You should now verify that the third collisiontakes place after a further time 8a/U at positions(−7

3a, 113 a, 11

3 a).


5.2 Newton’s law in two and three

dimensions

When two bodies collide there will be a line/planeof contact Σ with unit normal n.

Σ

n

The one-dimensional law (53) holds for thevelocity components parallel to n while thecomponents of v1−u1 and v2−u2 perpendicularto n (i.e., tangent to Σ) are continuous. Formally

n.(v2 − v1) = −en.(u2 − u1),

n× (v1 − u1) = 0 = n× (v2 − u2),(58)

and there are of course “hatted” versions.

The only point in the previous section where wereally used the assumption of one dimension was


Newton’s law (53). In particular the derivationsof Einit and Efin are unchanged. (You shouldverify this.) Thus

ΔE =12m1m2

M

(|v1 − v2|2 − |u1 − u2|2).

Now for any vector y,

|y|2 = (n.y)2 + |n× y|2,

by Pythagoras’ theorem. Thus, using (58),

ΔE =12m1m2

M(e2 − 1)

(n.(u1 − u2)

)2,

with a similar interpretation to (57).


Example 39. [Bouncing ball] A ball is throwninto the air with initial speed V0 at an angle α0.

A0 A1 A2

V0 V0 V1

α0 α0 α1

Let the ground be object 2 so that u2 = v2 = 0,and drop suffix 1 from the ball. At A1 the initialand final velocities of the ball are

u = (V0 cosα0,−V0 sinα0), v = (V1 cosα1, V1 sinα1).

Setting n = (0, 1) equations (58) imply

V1 sinα1 = eV0 sinα0, V1 cos α1 = V0 cosα0,

and so tanα1 = e tanα0. Now assume 0 < e < 1and let the times and horizontal distances for each


bounce be t0, t1, . . . and d0, d1, . . .. By elementarymechanics

t0 =2gV0 sinα0, d0 =

2g(V0 sinα0)(V0 cosα0)

and it is easy to see that t1 = et0, d1 = ed0 sothat tn = ent0 and dn = end0. Thus the ballbounces an infinite number of times in a durationof t0/(1 − e) covering a distance d0/(1 − e). (Ofcourse air resistance, linear friction, has beenneglected!)


6 Rotation in two dimensions

6.1 Angular velocity and momentum

Up to now we have been using inertial frameswith constant orthonormal bases, eg., (i, j) whichwe now write as (e1, e2). For many purposesit is more convenient to use orthonormal baseswhich are in motion relative to our standardones. Of course an equation like mx = F isbasis-independent, but when we resolve it intocomponents with respect to these new bases wemay expect some unfamiliar terms to arise.

Of particular interest is the orthonormal basis(er, eθ) related to polar coordinates (r, θ).

x

y

θ

ereθ


Note the basic relation

er = cos θe1+sin θe2, eθ = − sin θe1+cos θe2.(59)

Now consider two frames, a fixed inertial onebased on (e1, e2) and a non-inertial one basedon (er, eθ) which have a common origin. Ingeneral θ = θ(t) and θ is the angular velocity.Although e1 and e2 are fixed, eθ is changing intime, precisely because θ is changing. Thus wehave

eθ = −θ cos θe1 − θ sin θe2

= −θ er,

and similarly er = θ eθ, i.e.,

er = θ eθ, eθ = −θ er. (60)

The concepts of position, velocity and accelerationare fundamental to dynamics. What do thevectors look like in the two frames?


The general position vector is x(t) = (x, y) =(r cos θ, r sin θ) in cartesians. This means

x = xe1 + ye2 = r cos θ e1 + r sin θ e2 = r er

using (59). Thus in polar coordinates x = (r, 0).

For velocity we have

x = x e1 + y e2 = (x, y) in cartesians,

= (r er) = r er + rθ eθ

= (r, rθ) in polars.

(61)

For acceleration

x = x e1 + y e2 = (x, y) in cartesians,

= (r er + rθ eθ)

= r er + rθ eθ + rθ eθ + rθeθ − rθ2 er

= (r − rθ2, rθ + 2rθ) in polars,

= (r − rθ2, r−1(r2θ)˙).

(62)


For the trajectory specified by x(t), θ is alsocalled the angular velocity. Recall that fora particle of mass m the linear momentum isp = mx. The angular momentum (withrespect to the origin) of the trajectory isa three-dimensional concept defined by

h = x × p

= m(xy − yx) e3 in cartesians,

= r er × m(r er + rθ eθ)

= mr2θ e3 in polars.

(63)


6.2 Central force problems

For the moment we are working in threedimensions. A central force problem (CFP)is one of the type

x = −f(r)r−1x, (64)

where r =√

x2 + y2 + z2 = |x|. These are bothphysically important and relatively simple.

Theorem 6. All CFPs are planar, i.e., two-dimensional.

Proof. [Quite often one is required to show thata trajectory x(t) lies in a plane. The techniqueis to guess a constant vector k normal to theplane, and to define ψ(t) = k.x(t). Using theequation of motion for the trajectory one tries tofind an ordinary differential equation for ψ(t), andto show ψ(t) ≡ 0. This means that the trajectorylies in the plane perpendicular to k.]

Suppose that at some initial time t0, x(t0) = x0

and x(t0) = v0. Define k = x0 × v0. If k = 0


then x0 and v0 are parallel. The system (64) andthe initial conditions are invariant under rotationsabout the line joining the origin to x0 and so themotion remains on this line, which is definitelyplanar. If k = 0 define ψ(t) = k.x(t). Then (64)implies

ψ(t) = −f(r)r−1ψ(t),and by construction ψ(t0) = 0 = ψ(t0). Now twoinitial conditions of this type are sufficient to fixuniquely the solution of the ordinary differentialequation for ψ. By inspection ψ ≡ 0 satisfies theequation and the initial conditions. We deducethat this is the solution. Thus the orbit lies in theplane perpendicular to k, which we choose to bein the e3 direction.

If you are unhappy with this argument youmay prefer the following one. The displayedequation implies ψ(t0) = 0. Now differentiateit, remembering that r depends on t. The righthand side is the sum of terms each containingψ or ψ as a factor. Thus the third derivativeof ψ vanishes at t = t0. By a simple inductionargument one can conclude that ψ(t) and all ofits derivatives vanish at t = t0. Thus ψ(t) has atrivial Taylor series. �


Theorem 7. CFPs have conservative forcefields.

Proof. Because of the previous theorem we canwork in a plane with cartesian coordinates (x, y).Recall that r2 = x2+y2 implies r dr = x dx+y dy.Suppose the particle has mass m. Then from (64)the work integral is

∫F.dx = m

∫fr−1 (x, y).(dx, dy)

= m

∫fr−1(x dx + y dy)

= m

∫f(r)r−1r dr by the remark above

= m

∫f(r) dr = −U(r) say.

�


Theorem 8. CFPs have one degree of freedom.

Proof. Using equation (62) we can resolve thecomponents of (64) with respect to our polarbasis

r − rθ2 = −f(r), r−1(r2θ)˙= 0. (65)

The second equation implies that

r2θ = h, a constant. (66)

This equation, plus (63) imply that the angularmomentum h is constant (hence the notation usedabove). It is instructive to verify this directly from(63):

h = m(x × x)˙

= m(x × x + x × x)

= mx × (−fr−1)x = 0 using (64).

Finally the first equation (65) and (66) imply

r = −f(r) + h2r−3, (67)


which is of the form (42). �


6.3 Ellipses and hyperbolae

In this section we review the equations ofellipses and hyperbolae. These curves dependon two real parameters a > 0 and e � 0. Incartesian coordinates the focus is (ae, 0), andthe directrix is the line x = a/e. In the figurewe assume e < 1.

x

y

F

PD

ae a a/e

The curves are the loci of points P such thatPF = e × PD.

Setting P = (x, y) we obtain (x − ae)2 + y2 =e2(a/e − x)2 or x2(1 − e2) + y2 = a2(1 − e2).Assuming e < 1 set b2 = a2(1 − e2). Then thecurve, an ellipse, has the cartesian equation

x2

a2+

y2

b2= 1.

University of Cambridge: Mathematical Tripos IA: Dynamics: c©J.M. Stewart 2002 133 University of Cambridge: Mathematical Tripos IA: Dynamics: c©J.M. Stewart 2002 134

x

y

a

b

Another useful parametric form for the ellipseequation in cartesian coordinates is

(x, y) = (a cosψ, b sinψ). (68)

We shall also need the form of the ellipse in polarcoordinates. Note that the origin of polarcoordinates is chosen to be F , r = FP andFP subtends an angle θ with the x-axis. Thenthe defining equation implies

r = e(a/e − ae − r cos θ),

or

r =a(1 − e2)1 + e cos θ

(69)

University of Cambridge: Mathematical Tripos IA: Dynamics: c©J.M. Stewart 2002 135 University of Cambridge: Mathematical Tripos IA: Dynamics: c©J.M. Stewart 2002 136

Exercise 13. [Hyperbolae] Verify all of thestatements (and figures) for ellipses. Construct7

an equivalent theory for hyperbolae, where e > 1.What about parabolae, where e = 1?

7

ThedetailscanbefoundinappendixoneofLunn


6.4 Kepler’s laws of planetary motion

After many years of patient observation of theorbits of the planets, Kepler announced threelaws. The motion of an individual planet arounda central Sun is a CFP, and these laws werediscovered before, and motivated, Newton’s threelaws and also his law of gravitation.

If we consider the motion of planets around theSun there is an important simplification. Themass of the Sun is M ≈ 2 × 1030kg, while themass of the earth is m ≈ 6×1024kg. Since actionand reaction forces are equal and opposite, theacceleration of the Sun due to the earth is aboutone millionth of the acceleration of the earth dueto the Sun. Thus we may regard the Sun as fixedat the (polar coordinate) origin, around which theearth orbits.


Example 40. [Kepler’s second law] A straightline joining the Sun and a planet sweeps out equalareas in equal times.

S

e

r

r + δr

t

t + δt

δθ

The area of the triangle is δA = 12r(r +

δr) sin δθ = 12r

2δθ to first order. Therefore

A =12r2θ =

12h,

from (66). Thus Kepler’s second law is anobservational confirmation of the conservation ofangular momentum of solitary planets orbiting thesun. In fact Newton used it as evidence that the


gravitational field of the sun is a central forceproblem.


We now want to apply the CFP conceptto planetary motion. The central force isgravitational acceleration and so we set f(r) =GM/r2 in equation (64), where M is the mass ofthe Sun. Then (67) becomes

r = −GM

r2+

h2

r3. (70)

The standard way to handle this is to multiplyfirst by 2r and then to integrate with respect totime obtaining the energy equation

r2 = 2(E − V ) where V (r) = −GM

r+

12h2

r2,

(71)and E is a constant. Formally we can solve thisby

t =∫

dr√2(E − V (r))

, (72)

but not in terms of simple functions.


It is more instructive to return to equation (70)and to look for a solution r(θ) rather thanr(t), θ(t). In fact it pays to study u(θ) whereu = 1/r. Note that (66) implies θ = hu2 and so

r = (u−1) = −u−2(du/dθ)θ = −u−2u′hu2 = −hu′,

where u′ ≡ du/dθ. Further r = −hu′′θ =−h2u2u′′. Now equation (70) reduces to

u′′ + u =GM

h2, (73)

whose general solution is

u(θ) = GM/h2 + β cos(θ − θ0), (74)

where β and θ0 are constants, and without lossof generality θ0 = 0. The energy equation (71)becomes

h2(u′2 + u2) = 2E + 2GMu, (75)

and inserting (74) in this gives, after a littlealgebra,

β2 = (2E + G2M2/h2)/h2. (76)


Now recall the energy equation (71)

r2 = 2(

E +GM

r− 1

2h2

r2

).

The right hand side is negative for small r and forlarge r it approaches E. We need real solutions,i.e., a positive right hand side for some range ofr. This requires that the quadratic expression

Q = 2Er2 + 2GMr − h2

has two positive zeros if E < 0 and one if E > 0.The condition for real zeros is

(GM)2 + 2Eh2 > 0,

which guarantees that β2 defined by (76) ispositive, i.e., β is real. Notice, from the expressionfor Q above, that if E < 0 the sum of the zerosof Q is positive as is the product. Thus if theroots are real they are both positive, as required.Conversely if E > 0 the sum and product ofthe zeros are both negative. For real zeros we


must have one positive and one negative zero, asrequired.

Now set p = h2/(GM) and e = pβ =√1 + 2Eh2/(GM)2 so that (74) becomes pu =

1 + e cos θ or

r =p

1 + e cos θ. (77)

Comparing this with equation (69) we see thatwe have an ellipse if e < 1 ⇔ E < 0. (Thecase E > 0 leads to unbound hyperbolic orbits.)Thus we have verified Kepler’s first law: thepath of each planet is an ellipse with thesun at its focus. Newton used the observedform (77) to compute the f(u) of (64) and thusto verify that the gravitational force obeyed aninverse square law.

We next investigate Kepler’s third law:the square of each planet’s period isproportional to the cube of the semi-major axis of the elliptic orbit. The semi-axes are, from (69) and (77) a = p/(1 − e2)and b ≡ a

√1 − e2 = p/

√1 − e2. We can

parametrize the equation of the ellipse by (68),


i.e., x = a cosψ, y = b sinψ The area is

A = 4∫ a

0

y dx

= 4ab

∫ π/2

0

sin2 ψ dψ

= 4ab(π/2)12

= πab

= πa2√

1 − e2

= πa3/2p1/2.

Now from example 40, A = 12h. Thus the period

is

T =A

A= 2πa3/2p1/2h−1 =

2π√GM

a3/2, (78)

which verifies the third law, for M is the samefor all planets. Newton used this to deduce thatgravity was an acceleration, i.e. the force actingon a planet was proportional to the mass of theplanet.


6.5 Miscellaneous examples

We now look at some consequences of planetarymotion. First we ask the question why doesour world have three spatial dimensions? Letthe number be N . Clearly N > 2, for N = 2places severe topological restrictions on our bodyfunctions, but why not N = 11, N = 26? Notethat we exist. This means that there must existplanetary orbits which are close to circular and arestable. (This kind of argument is an anthropicprinciple.) We first investigate what gravitylooks like in N dimensions, and then we findwhich values of N permit stable planetary orbits.

Example 41. [Gravity in N dimensions] If Uis the gravitational potential energy, then g =−∇U . What scalar equation does U satisfy?The simplest is ∇2U ≡ ∇.∇U = 0. In 2dimensions ∇U = (∂U/∂x, ∂U/∂y), ∇2U =∂2U/∂x2 + ∂2U/∂y2, and in N dimensions

(∂2

∂x12

+ . . . +∂2

∂xN2

)U = 0. (79)


We look for a solution U(r) where r = (x12 +

. . . + xN2)1/2. Now

∂U

∂x1=

dU

dr

∂r

∂x1= U ′(r)

x1

r

∂2U

∂x12

= U ′′ x12

r2+

U ′

r− U ′x1

2

r3.

Thus

∇2U = U ′′∑i

xi2

r2+ N

U ′

r− U ′∑

i

xi2

r3

= U ′′ + (N − 1)U ′/r.

Solving ∇2U = 0 gives

U ′ = C/rN−1, U = D/rN−2, (80)

where we are assuming N > 2, and C > 0,D < 0 are constants. (The signs are determinedby requiring gravity to be attractive.) Note thatfor N = 3 we obtain the “correct” results.


Example 42. [Circular orbits in a central force]Recall equation (67)

r = −f(r) + h2r−3,

which has the first integral

r2 = H(r) ≡ 2E − 2∫ r

f(s) ds − h2/r2. (81)

We now use the stability algorithm, theorem 5 of§4.5: a circular orbit at r = a requires H(a) =0 = H ′(a) and will be stable if H ′′(a) < 0.We can, by appropriate choice of E always setH(a) = 0. Notice

H ′(r) = −2f(r)+2h2/r3, H ′′(r) = −2f ′(r)−6h2/r4.

Thus H ′(a) = 0 fixes h2/a3 = f(a) and stabilityrequires H ′′(a) = −2f ′(a) − 6h2/a4 < 0 or

af ′(a) + 3f(a) > 0. (82)


We can now resolve the question as to the numberof spatial dimensions. Recall from equation (80)that the gravitational force is proportional to1/rN−1 with a positive proportionality constant.Thus we substitute f(r) = 1/rN−1 in the stabilitycriterion (82) finding N < 4 for stability. It is noaccident that we live in three spatial dimensions.

We insert here the proof of a result used in §1.6.

Example 43. [Gravitational field of a sphere]In vacuum the gravitational potential U satisfies∇2U = 0 and in three dimensions, assumingspherical symmetry, i.e., U = U(r), U ′′ +2r−1U ′ ≡ r−2(r2U ′)′ = 0. Thus in vacuum

r2U ′(r) = C, C a constant. (83)

Now suppose we have a sphere mass M , radius Rmade of material with mass density ρ(r). Insidethe sphere Laplace’s equation ∇2U = 0 has tobe generalized to Poisson’s equation ∇2U =4πGρ, and for spherical symmetry

(r2U ′)′ = 4πGρr2.


Assuming that U ′ is finite at r = 0,

r2U ′(r) = 4πG

∫ r

0

ρ(s)s2 ds.

In particular, at r = R, R2U ′(R) = GM . Outsidethe sphere (83) implies r2U ′(r) = C, and bycontinuity C = GM . Thus outside the body thegravitational field is −U ′(r) = −GM/r2. Thisis as if all of the mass was concentrated at thecentre. (This result was used in example 7.)

Example 44. [Startrek] A spaceship S has shutdown its engine and is coasting near a strangelooking object O. The commander notices

• his angular momentum with respect to the objectis constant,

• the orbit is a part of a circle, radius a, into theobject.

What can he deduce about the object?


O

S

θ P

��

��

��

The commander deduces he has a central forceproblem r2θ = h constant, governed by equation(67), r = −f(r) + h2/r3. Recall also that ifu(θ) = 1/r then r = −h2u2u′′ and so

u′′ + u =f(1/u)h2u2

.

Clearly r = OS = OP cos θ = 2a cos θ and sou = 1/(2a cos θ). Thus u′′ + u = 8a2u3 andf = 8h2a2r−5. It was a strange object!

Example 45. A particle is acted upon by anattractive central force, and satisfies (67) with

f(r) = k(4(a/r)2 − 3(a/r)3

). Initially the

particle is distance a from the centre with boththe radial component of the velocity as well asthat in a perpendicular direction both equal to√

ak. Show that the maximum and minimum


distances of the particle from the centre are 2aand 2a/3, and find the angle turned between amaximum and the next minimum.

From the initial conditions we have (setting themass m = 1) angular momentum h = a

√ak.

Using the displayed equation from example 44, wehave u′′ + u = 4/a − 3u, or

u′′ + 4u =4a,

for which the solution is

u =1a

(1 + α cos(2θ) + β sin(2θ)

),

where α and β are constants. (Thecomplementary function is obvious and theparticular integral is easily guessed.) Initially weset θ = 0. Initially rθ =

√ak = u−1hu2 =

hu = a√

aka−1(1 + α) and so α = 0. Furtherat θ = 0, r =

√ak = −hu′ =

√ak(−2β) and so

β = −12. Clearly u achieves a minimum value of

1/(2a) at 2θ = π/2 and a maximum of 3/(2a)at 2θ = 3π/2. Thus r achieves a maximum of


2a followed by a minimum of 2a/3 after turningthrough a further Δθ = π/2.

Example 46. A particle of unit mass moves on asmooth horizontal table and is attached to a fixedpoint of the table by an elastic string of naturallength and modulus of elasticity λ. The particleis held with the string just taut, and projected atright angles to the string with velocity 2

√λ/3.

Show that the greatest length of the string in theresulting motion is 2.

This is clearly a CFP, and so angular momentumis conserved. We obviously use polar coordinateswith centre at the fixed point, and θ = 0 alongthe initial direction of the string. Then it is easyto compute h = × 2

√λ/3 from the initial

conditions. One approach is to start from r =−f(r)+h2/r3, where (from Hooke’s law) f(r) =λ(r − )/. However when the string achievesits greatest length, r = 0; we are interested in rrather than r.

Obviously we can multiply the above equationby r and integrate, obtaining the energyequation. Alternatively we can write down the


energy equation directly! The kinetic energyis 1

2(r2 + r2θ2) = 1

2(r2 + h2/r2) and is equal

to 12

(2√

λ/3)2

initially. The elastic energy is12λ(r − )2/ and is initially zero. Writing downthe (conserved) energy and equating it to its initialvalue gives (after trivial algebra)

r2 = 43λ − λ(r − )2/ − 4

33λ/r2

or, setting r = x, where x is dimensionless,x2 = (λ/)f(x) where

f(x) =43

(1 − 1

x2

)− (x − 1)2.

Oh dear, they want us to examine a quarticequation! (And we have to be careful becauseif f(x) < 0, or if x < 1, then the physicsused becomes invalid!) However, from the initialconditions we know that r = 0 initially, and sox = 1 is a root. Guessing the answer, we canverify that x = 2 is a root. It is easy to compute

f ′(x) = 2(1 − x) +8

3x3, f ′′(x) = −2 − 8

x4.


Because f ′′ < 0 we can deduce that f ′(x) has,at most, one zero. It is easy to verify f ′(1) > 0and f ′(2) < 0. Thus f(x) increases from zeroat x = 1. It reaches a positive maximum inx ∈ (1, 2) and returns to a zero value at x = 2.The assertions made in the question have beenverified.


Finally in this chapter we consider an exampleof unbound orbits produced by a repulsiveelectrostatic force.

Example 47. [Rutherford scattering] A particleof unit mass, charge q initial speed v isapproaching a stationary mass of charge Q withimpact parameter b.

Q

q Vb

We choose Qq > 0 so that the charges repel. Bywhat angle Θ is the orbit bent?


Q θ = 0

Θ/2

Θ/2

This is an electrostatics problem with an inversesquare central force, so that we can use Kepleriantheory of §6.4 provided we replace GM byk ≡ −Qq/(4πε0), see §1.9 (and recall that weare treating a particle of unit mass). Noticethat initially r is large, so that the electrostaticpotential energy is negligible. Thus E = 1

2v2 and

h = bv = b√

2E. Since E > 0 the eccentricity ofthe orbit is

e =

√1 +

2Eh2

k2=

√1 +(2Eb

k

)2

> 1, (84)


and so we have a hyperbola, see equation (77)

1r

=1p(1 + e cos θ), (85)

where p = h2/k = 2Eb2/k. Note that θ = 0corresponds to the minimum value of r. Asr → ∞, θ → ±θ0 = cos−1(−1/e) where θ0 =12π + 1

2Θ. Thus sin(12Θ) = − cos θ0 = 1/e and

finally Θ = 2 sin−1(1/e).

We can also attempt this problem from firstprinciples, rather than rely on established results,but now we configure the geometry differently,and we allow the particle to have mass m.

QθΘ r

Because this is a central force problem weobviously have r2θ = h/m = bv. The initialhorizontal component of the momentum is mv,and the initial energy is 1

2mv2. At late times


the electrostatic potential energy is again zero,and so the final energy, all kinetic, is 1

2mv2.It follows that the final speed is again v, andso the final horizontal momentum componentis mv cosΘ. The horizontal component ofthe repulsive force is F = −kr−2 cos θ wherek = −Qq/(4πε0) as above. Thus the reductionin the horizontal component of the momentumshould be

∫ t=∞t=−∞ Fdt.

In order to perform this integral we change thevariable from t to θ using the expression for θgiven above. Thus the two expressions for thechange in horizontal momentum give

mv(cos Θ − 1) = −∫ π−Θ

θ=0

(− k cos θ

r2

)(r2

bv

)dθ

or

mv(1 − cos Θ) = − k

bvsinΘ,

which implies

tan 12Θ =

Qq

4πε0mv2b.


The first approach established sin 12Θ = 1/e which

would imply tan 12Θ = 1/

√e2 − 1. Using the

displayed formula (84) for e it is easy to verify theconsistency of the two approaches.

Of course both of these approaches will also workwhen the force is attractive, provided the orbit isunbound, i.e., a hyperbola.


7 Rotation in three dimensions

7.1 Introduction

For many purposes we can treat a frame centredon the Sun, at rest with respect to the fixedstars as inertial. However the centre of the earthrotates with respect to the sun, and the surfaceof the earth rotates relative to the centre. Framesattached to the earth, although convenient, arenot inertial. How then are we to proceed?

We shall discuss two frames S and S′, with acommon origin. S will always be inertial, and S′

is in general rotating with respect to S. By thiswe mean that if v is any vector then |v|2 = v.vis the same in both frames, but v = v/|v| is not.Rotations preserve the lengths of vectors, but nottheir directions.

Lemma 3. Rotations preserve angles betweentwo vectors u and v, i.e., u.v is the same in bothframes.

Proof.

|u + v|2 = |u|2 + |v|2 + 2u.v.


�


Thus rotations preserve shapes but notorientations. However as we soon discover, ratesof change of vectors in the two frames will bedifferent.

Let {e1, e2, e3} be an orthonormal basis, fixed inS, and let u(t) be an arbitrary vector, so thatu =

∑i ui(t)ei. The we define the rate of

change of u in S as

(dudt

)S

=∑

i

dui

dtei, (86)

i.e., we demand

(dei

dt

)S

= 0. (87)

Now let {e′i} be an orthonormal basis fixed in S′,

i.e. (de′i/dt)S′ = 0 for each i = 1, 2, 3. Each e′kcan of course be expressed as a linear combinationof the {ei}. We express this as

e′i = Uei, i = 1, 2, 3,


where U is a 3 × 3 matrix. If we choose asa basis of S, e1 = (1, 0, 0)T , e2 = (0, 1, 0)T ,e3 = (0, 0, 1)T , then it is easy to see that the

columns of U are the {e′i} ,

U =(e′1|e

′2|e

′3

).

Since we know that the columns of U areorthonormal, it follows that U is an orthogonalmatrix. Although this has been deduced in aparticular basis, it is of course a basis-independentproperty. Let eiβ be the β-component withrespect to the given basis of ei. In other wordseiβ = δiβ. Let e′jα be the α-component of e′j withrespect to the same basis.

Then the equation e′j = Uej means

e′jα =∑

β

Uαβejβ.

Indeed substituting ejβ = δjβ we have e′jα = Uαj,confirming the displayed equation above.

To obtain another useful form for the components


of U note that

ei . e′j =∑α

eiαe′jα

=∑α

∑β

eiαUαβejβ

=∑α

∑β

δiαUαβδjβ

= Uij.

We now obtain some useful results aboutorthogonal matrices.

Lemma 4. If U is a real orthogonal matrix, thenits eigenvalues are of the form λ = exp(±iθ)where θ is real, and are either real or occur incomplex conjugate pairs.

Proof. Because U is real, if e is an eigenvectorwith eigenvalue λ then e∗ is an eigenvector witheigenvalue λ∗. Thus

λ∗λe∗.e = (λ∗e∗)T (λe)

= (Ue∗)T (Ue) = (e∗)TUTUe = e∗.e,


which implies λ∗λ = 1. �

Theorem 9. [Euler] Let U be an orthogonal 3×3 matrix which is proper, i.e., U = I. Then 1is an eigenvalue of U of multiplicity 1, i.e., thereexists a vector e, unique up to multiplication, suchthat

Ue = e. (88)

Proof. The eigenvalues are the roots of thesecular equation

det(U − λI) = 0. (89)

Because U is orthogonal UUT = I. This meansthat UT (U − I) = I − UT , and we may deducedet(UT ) det(U − I) = det(I − UT ) = det((I −U)T ).

Now det(AT ) = det(A) for any square matrix Aand det(U) = 1 for rotations. Thus

det(U − I) = det(I − U) = det(−(U − I))

= (−1)3 det(U − I).


We therefore deduce that det(U − I) = 0.Comparison with the secular equation (89) shows8

that 1 is indeed an eigenvalue of U . Nowlet the eigenvalues be λ1, λ2, λ3 = 1. Sincedet(U) = λ1λ2λ3 = 1 we deduce that λ1λ2 = 1.Thus the eigenvalues are (1, 1, 1) (which is ruledout since U is proper) or (eiθ, e−iθ, 1) where0 < θ < 2π. Thus λ3 = 1 has multiplicity1. �

Corollary. Suppose we choose coordinates sothat the eigenvector e lies in the z-direction.Then elementary algebra shows that

U =

⎛⎝ cos θ sin θ 0− sin θ cos θ 0

0 0 1

⎞⎠

Hopefully some of these results are familiar, butit is convenient to collect them all in one place.We now return to dynamics.

8There is nothing special about 3 here; any odd dimension would do.


7.2 Kinematical theorems

Theorem 10. [Euler] The general rotation ofan orthonormal frame with fixed origin leaves, ateach instant, points in one direction, the axis ofrotation at rest.

Proof. This is an immediate consequence oftheorem 9. Since Ue = e, where e is the uniqueeigenvector (up to a factor) with eigenvalue 1,points in the direction of e are not moved. Notehowever that U and hence e depend on time.Thus the axis can, and usually does, change withtime. �

Theorem 11. [Angular velocity] If {ei} and

{e′i} are orthonormal bases fixed in frames S

and S′ respectively, then there exists an angularvelocity vector ω such that

(de

′i

dt

)S

= ω × e′i, i = 1, 2, 3. (90)


Proof. We have e′i = Uei which implies ei =

UTe′i. Thus

(de

′i

dt

)S

=(

dU

dt

)S

ei + U

(dei

dt

)S

.

The last term vanishes by definition. Thecomponents of U are Uij = ei.e

′j and are the

same in either frame by lemma 3. We write(dU/dt)S = (dU/dt)S′ = U . Thus

(de

′i

dt

)S

= Uei = UUTe′i.

Now UUT = I implies UUT + UUT = 0 orUUT = −(UUT )T so that UUT is antisymmetric,and we can write

UUT =

⎛⎝ 0 −ω3 ω2

ω3 0 −ω1

−ω2 ω1 0

⎞⎠ .


Then for any vector a,

UUTa =

⎛⎝ω2a3 − ω3a2

ω3a1 − ω1a3

ω1a2 − ω2a1

⎞⎠ = ω × a,

where ω = (ω1, ω2, ω3)T . In particularequation(90) holds. �

Corollary. Note that if e′i is parallel to ω

then (de′i/dt)S = 0. Thus the direction ω is

instantaneously constant. It is the instantaneousaxis of rotation implied by Euler’s theorem.


Theorem 12. [Rotating axes] For any vectora(t), (

dadt

)S

=(

dadt

)S′

+ ω × a. (91)

Proof. Set a =∑

a′ie

′i. Then

(dadt

)S

=∑(

da′i

dt

)S

e′i +∑

a′i

(de

′i

dt

)S

.

Since a′i = a.e

′i, lemma 3 implies ((da

′i)/dt)S =

((da′i)/dt)S′. We note also that ((de

′i)/dt)S′ = 0

by definition. Thus

(dadt

)S

=(∑ d

dta′ie

′i

)S′

+∑

a′iω × e

′i

=(

dadt

)S′

+ ω ×∑

a′ie

′i

=(

dadt

)S′

+ ω × a.

�


In the angular velocity theorem we implicitlyasserted that ω was a vector without provingit. The next lemma shows that angular velocitiesadd, which is what we require.

Lemma 5. If frame S′′ rotates with angularvelocity ω′ with respect to S′, it rotates withangular velocity ω + ω′ with respect to S.

Proof. Recall that for an arbitrary vector a(t),

(dadt

)S′

=(

dadt

)S′′

+ ω′ × a.

Then (dadt

)S

=(

dadt

)S′

+ ω × a

=(

dadt

)S′′

+ ω′ × a + ω × a

=(

dadt

)S′′

+ (ω + ω′) × a.

�


Theorem 13. [Acceleration] For an arbitraryvector a(t),

(d2adt2

)S

=(

d2adt2

)S′

+ 2ω ×(

dadt

)S′+

ω × (ω × a) + ω × a.

(92)

Proof. We have(d2adt2

)S

=(

d

dt

[dadt

]S

)S

=(

d

dt

[dadt

]S

)S′

+ ω ×(

dadt

)S

using (91),

=(

d

dt

[(dadt

)S′

+ ω × a])

S′+ ω ×

(dadt

)S

=(

d

dt

[dadt

]S′

)S′+(

dω

dt

)S′×a + ω ×

(dadt

)S′+

ω ×(

dadt

)S

where (91) has been used again and again.


Now (dω/dt)S = (dω/dt)S′ + ω ×ω = (dω/dt)S′is unambiguous and we write it as ω. Expandingthe last term using (91)

(d2adt2

)S

=(

d2adt2

)S′

+ ω × a + ω ×(

dadt

)S′+

ω ×(

dadt

)S′

+ ω × (ω × a),

which is equation (92). �

Note that in almost all applications ω = 0 andso we usually discard this term, often withoutexplicitly saying so.

The term 2ω × (da/dt)S′ is the Coriolisacceleration and ω×(ω×a) is the centripetalacceleration.


7.3 Applications to dynamics

In an inertial frame S a particle of mass m at x(t)subject to a force F satisfies Newton’s second law

m

(d2xdt2

)S

= F.

Thus, from (92)

m

(d2xdt2

)S′

+2mω×(

dxdt

)S′

+mω×(ω×x) = F.

(93)The extra terms reflect the non-inertial nature ofS′. Physicists love to confuse by “simplifying”,and rewrite (93) as

m

(d2xdt2

)S′

= F−2mω×(

dxdt

)S′−mω×(ω×x).

(94)They call the extra terms on the right handside the Coriolis force and centrifugal forcerespectively. (Note that these are −mass ×acceleration!)


Example 48. [Fairground wall] This is a largecircular drum of radius a which can rotate aboutits vertical axis. Visitors pay to stand insideagainst the wall, the drum rotates, and when itreaches full speed the floor slides down, leavingthe visitors stuck to the wall.

Let S be an inertial frame attached to thefairground, and let S′ be attached to the drum,with the vertical axes e3 and e

′3 coinciding. In

S′ the visitor is at rest, x = ae′1, (dx/dt)S′ = 0,

(d2x/dt2)S′ = 0. Then, setting ω = ωe′3, (93)

gives

m

(d2xdt2

)S

= mωe′3×(ωe

′3×ae

′1) = −mge3+R+F,

where R is the horizontal reaction force and F isthe vertical frictional force. Resolving horizontallywe obtain −maω2e

′1 = R, so the reaction is

directed inwards, and resolving vertically 0 =−mge3 + F, so that the frictional force isupwards. Empirical friction laws suggest thatslipping will not occur if |F| � μ|R|, whereμ is a dimensionless friction coefficient, i.e., ifmg � μmaω2. Thus the visitor “sticks” if aω2/g


is sufficiently large.

Example 49. [Newton bucket experiment] A bucketof water is rotated with vertical angular velocityΩ about its axis. The shape of the surface adjustsso that the normal to the surface is antiparallelto the net force acting on the water. Using polarcoordinates (r, θ, z) determine the shape of thesurface.

As usual S is inertial, fixed in space, and S′ isfixed in the bucket.

Ω

n

t

F′

Because of friction, the water is at rest relative tothe bucket. Using (94) the net force as seen in S′

isF′ = mg − m(Ω.x)Ω + m|Ω|2x.


Setting Ω = Ωe′3, g = (0, 0,−g) and x =

(r, 0, z), we find F′ = m(Ω2r, 0,−g). Thisis antiparallel to the normal n to the surface,as shown. An element t = (dr, 0, dz) of thesurface,i.e., perpendicular to n is shown. Thismust be perpendicular to F′ and so Ω2r dr −g dz = 0, or dz/dr = Ω2r/g so that the surfaceis a parabola z = z0 + 1

2Ω2r2/g, which is what

Newton observed.

Example 50. [Eotvos experiment] Eotvos constructedsimple pendulums with different materials for thebobs.

Ω

T

Fg

Fc

He argued that the bob is in equilibrium underthree forces:

• Fg, the gravitational force proportional to M the


gravitational mass of the bob,

• Fc, the centrifugal force proportional to m theinertial mass,

• T, the tension in the string, required to balancethe other two forces.

Thus the pendulum will be deflected slightlyfrom the vertical (except at the equator) and,resolving perpendicular to T, we see that the smalldeflection depends on the dimensionless ratio|Fc|/|Fg|. He observed that whatever materialwas used for the bob, the deflection remained thesame, and hence he deduced that m/M was auniversal constant, and equal to 1 by a suitablechoice of units. This justifies the assertion madein §1.6.

All of these examples have involved onlycentripetal accelerations. We now turn toexamples where the Coriolis acceleration plays arole. There are easier relevant examples in thebook by Lunn, and you may wish to study thosefirst, before looking at the next two examples.


Example 51. A bead of mass m slides on a rigidsmooth circular loop of radius a which is forcedto rotate with uniform angular speed ω about afixed vertical diameter. Determine the points ofequilibrium, and their stability.

e′3

e′1

θm�

��

��

��

��

We choose axes e′1, e

′2 and e

′3 fixed in the loop

with e′3 vertical and the loop lying in the 13-plane.

We may set

x = a(sin θ, 0,− cos θ),

x ≡ (dx/dt)S′ = aθ(cos θ, 0, sin θ),

x ≡ (d2x/dt2)S′ = aθ(cos θ, 0, sin θ) − xθ2.


We now want to use equation (93) in the form

m

(d2xdt2

)S′+2mω ×

(dxdt

)S′

+ mω × (ω × x)

= F = mg + R,

where R is the reaction of the wire, and isperpendicular to it since the wire is smooth. If,as here, we are not interested in R we can “dot”the above equation with x which is parallel to thewire. Incidentally this also removes the Coriolisterm. This give

a2θ − a2ω2 sin θ cos θ = −ag sin θ.

Equilibrium requires θ = θ = 0 and so eithersin θ = 0 or cos θ = ν ≡ g/(aω2). Therefore θ =0 and θ = π are always equilibria. If 0 < ν < 1there is a third equilibrium point at θ = θ0 =cos−1 ν.

The stability algorithm of theorem 5 does applyhere, even though energy is not conserved(because work is done maintaining the constancyof ω). However the displayed equation


corresponds to one degree of freedom. It canbe integrated (in the usual way) to give

θ2 = const. − ω2(12

cos 2θ − 2ν cos θ).

It is now straightforward to verify that if ν > 1then θ = 0 is stable, while θ = π is unstable, butif ν < 1 then θ = θ0 is stable and both θ = 0 andθ = π are unstable.


Example 52. A smooth plane revolves withuniform angular speed ω about a fixed verticalaxis which intersects the plane at O. The normalto the plane is at a fixed angle θ to the vertical.A particle of mass m is placed at rest at O andis released. Show that in the subsequent motion,the particle’s speed v relative to the plane, thedistance r of the particle from the axis and thedepth h of the particle below O are related by

v2 = r2ω2 + 2gh.

It is exceedingly difficult to describe the motionof the plane and particle with respect to a fixedinertial frame S and so we won’t try it! We use anon-inertial frame S′ attached to the plane, withe′3 normal to the plane, and we choose cartesian

coordinates (x, y, z) adapted to the {e′i} with

origin at O. The rotation axis is fixed in S andit is a consequence of the rotating axes theoremthat it is also fixed in S′. We can therefore chooseit to lie in the xz-plane. We now want to use


equation (93) in the form

m

(d2xdt2

)S′+2mω ×

(dxdt

)S′

+ mω × (ω × x)

= F = mg + R,

where R is the reaction of the plane. Herex = (x, y, 0), (dx/dt)S′ = (x, y, 0), (d2x/dt2)S′ =(x, y, 0), and R = (0, 0, R).

ω

θ

O

xz

Further ω = (ω sin θ, 0, ω cos θ) and g =(−g sin θ, 0,−g cos θ). Resolving in the x- andy-directions gives

x − 2ωy cos θ − ω2x cos2 θ = −g sin θ,

y + 2ωx cos θ − ω2y = 0.

(We don’t need to resolve in the z-direction,because we are not interested in the reactionforce.)


These equations imply

xx + yy − ω2(xx cos2 θ + yy) = −gx sin θ,

with first integral

x2 + y2 − ω2(x2 cos2 θ + y2) = −2gx sin θ + C,

where C is a constant. But initially x = y = x =y = 0 and so C = 0. Now identify v2 = x2 + y2,r2 = x2 cos2 θ + y2 and h = −x sin θ to obtainthe required result.


7.4 Motion on the surface of the

earth

ω

λ

Rx

O

O′

We would like to use a frame with origin at O′, apoint fixed on the surface of the earth. Howeverin our development we have always assumed thatframes S and S′ have a common origin. Thus wemust replace x in equation (93) by R + x whereR is fixed in S′. For a particle subject only togravity (93) becomes

(d2xdt2

)S′

+2ω×(

dxdt

)S′

+ω×(ω×(R+x)) = g.

(95)In almost all applications |x| |R| and so wedrop the x-contribution to the centripetal term.


We move the R-contribution to the right handside, defining

g′ = g − ω × (ω × R) (96)

as the effective gravity. Since |R| ≈ 6.4 ×106m, it is easy to verify that the relativedifference between g′ and g is tiny and we usuallyignore the difference. With these approximationswe have (

d2xdt2

)S′

+ 2ω ×(

dxdt

)S′

= g, (97)

and this allows us to study the Coriolis terms in aparticularly clean way. A convenient form of thisequation, suitable for physicists, is

(d2xdt2

)S′

= g + C, C = −2ω ×(

dxdt

)S′.

(98)

A basic question is in which direction is a particle,moving relative to the surface of the earth,accelerated? We choose a standard right handed


set of axes, centred at O′ on the surface of theearth as follows: O′1 is east and horizontal, O′2is north and horizontal while O′3 is vertical, (seefigure).


ω

λ

2 3

O

O′

We set (dxdt

)S′

= (U cos θ, U sin θ, V ),

ω = (0, ω cos λ, ω sinλ),

so that

C = 2ω(U sin θ sinλ − V cosλ,−U cos θ sinλ,

U cos θ cosλ).

We can disentangle two important cases here.If U = 0, so that the particle is movingvertically, then we have C = 2ωV cosλ(−1, 0, 0).We deduce: a rising(falling) particle isdeflected to the west(east).


U

1(east)

2(north)

O′θ

horiz. compt. of C

If V = 0, so that the particle ismoving horizontally, then we have C =2ωU sinλ(sin θ,− cos θ, .). (We have notcomputed the 3-component because it willbe dwarfed by gravity.) We deduce: aparticle with a horizontal velocity suffersa horizontal deflection to the right in thenorthern hemisphere and to the left in thesouthern one.


Example 53. [Projectiles in S′] The projectilehas initial position x0, velocity V0 at t = 0.Because we are ignoring O(ω2) terms (see thecomment just after equation (95)) we can use thetechnique of “successive approximations” to solve(98). First we write it as(

d2xdt2

)S′

= g − 2ω ×(

dxdt

)S′

+ O(ω2) (99)

= g + O(ω). (100)

Integrating (100) is easy:(dxdt

)S′

= V0 + gt + O(ω).

We now insert this into the right hand side of(99):(

d2xdt2

)S′

= g − 2ω × V0 − 2ω × gt + O(ω2),

which integrates to give

x = x0+V0t+12(g−2ω×V0)t2−1

3ω×gt3+O(ω2).


As an example consider dropping a particle fromrest down a lift shaft. We have, to this order ofapproximation,

z = −12gt2, x =

13gt3ω cosλ,

a deflection to the east.


Example 54. [Foucault pendulum] This is asimple pendulum consisting of a bob of massm attached by a string of length to a fixedpoint. We set the origin O′ at (0, 0, ) so thatin equilibrium (ignoring rotation) the bob is atx = (0, 0, 0). In the motion we assume x/ andy/ are both small. It follows that z, z and z aresecond order small and we ignore them. Let T bethe tension in the string, and write (dx/dt)S′ = xetc. Then (97) becomes

x + 2ω × x = g +Tm

,

with components

x − 2ω sinλy = −(T/m)(x/),

y + 2ω sinλx = −(T/m)(y/),

−2ω cosλx = −g + T/m.

Thus T/m = g plus first order quantities.However in the other equations T/m is multipliedby a small quantity, e.g., x/. Thus this


approximation is good enough, and, settingn2 = g/ we have, correct to first order,

x−2ω sinλy+n2x = 0, y+2ω sinλx+n2y = 0.

Setting ζ = x + iy we combine these as

ζ + 2iω sinλζ + n2ζ = 0,

with solution

ζeiωt sin λ = Aeiμt + Be−iμt,

where μ2 = n2 + ω2 sin2 λ, and A and B arecomplex constants. Now choose new axes Ox′′,Oy′′ which are rotating around Oz with angularvelocity −ω sinλ. Then

x′′ + iy′′ = (x + iy)eiωt sin λ = Aeiμt + Be−iμt.

For simplicity we consider the case where A andB are real and set C = A + B, D = A − B so


that

x′′ = C cosμt, y′′ = D sinμt,

Clearly x′′ and y′′ trace out an ellipse withfrequency μ, and since μ = n + O(ω2) this isexactly what one would expect if the earth werenot rotating. However these coordinate axes arerotating with angular velocity −ω sinλ relativeto the surface of the earth, and so the ellipserotates with this angular velocity in the clockwisedirection. Most science museums have a Foucaultpendulum.


Example 55. [Introduction to meteorology] Considerthe large scale horizontal wind circulation. Massesof air tend to move from regions of high pressureto regions of low pressure. In the vertical directionair close to the earth is warmer than higherair, and the resulting pressure gradient roughlybalances gravitational forces. In the horizontaldirection there are persistent long-range motionsof air masses, usually called winds. Here thepressure gradients are comparable to the Coriolisforces, and one can see both effects, particularlyin cyclones, regions where the pressure achievesa minimum.

low pressure

high pressure

no rotation

with rotation

If the earth did not rotate wind would flow fromhigh pressure to low pressure as shown. In


the northern hemisphere Coriolis forces deflectit to the right as shown in the figure. If thedeflection causes the wind to move nearly parallelto the isobars (contours of equal pressure) asshown, then the Coriolis force becomes almostanti-parallel to the pressure gradient, and canmatch it. One can then obtain geostrophicwinds which follow the isobars (shown red inthe figure). Once friction is taken into accountthese winds are slowed and directed towards thecyclone (green arrows). Thus in cyclonic weatherthe weather maps show direct evidence of Corioliseffects.

low


8 Systems of particles

8.1 Centre of Mass

Up to now we have concentrated on systems whichcan be formulated in terms of a few or even a singlepoint particle. We now want to extend this tosystems of many particles, labelled by A,B, . . . =1, 2, . . . , N . (A continuum approach where sumsare replaced by integrals is also possible.) TheAth particle has mass mA, position xA, and isacted on by forces FA +

∑B �=A FA←B. The

internal forces are required to satisfy Newton’sthird law, FB←A = −FA←B. If, in addition, theyare directed along xB − xA,

FA←B = KAB(xA − xB), (101)

they are said to be torque-free. (Some authorsassume, implicitly, that all internal forces aretorque-free.)

We now generalize two definitions originally givenin §5.2. The total mass is M =

∑A mA, and

the centre of mass or centroid is at X(t)


where MX =∑

A mAxA. In this section all timederivatives take place within an inertial frame S

and so we replace (d/dt)S by .

Theorem 14. The centroid behaves like aparticle of mass M subject to the net externalforce F =

∑A FA, i.e.,

MX = F. (102)

Proof.

MX =∑A

mAxA =∑A

(FA +

∑B �=A

FA←B

)

= F +∑A

∑B �=A

FA←B.

Consider the contribution to the double sumfrom a pair of particles, e.g., 2 and 7. Wehave F2←7 + F7←2 = 0. Thus the double sumvanishes. �

Corollary. If F = 0, then the centre of massframe (CoM) defined by X = 0 is also inertial.


Theorem 15. Let TCoM denote the kineticenergy, as measured in the centre of mass frame.Then the total kinetic energy is given by

T =12M |X|2 + TCoM . (103)

Proof. Write xA = X + xA, where xA is theposition of the particle in the CoM frame, and setxA = uA, X = U and ˙xA = uA. Then

T =12

∑A

mAuA.uA =12

∑A

mA(U + uA).(U + uA)

=12MU.U + U.

∑A

mAuA +12

∑A

mAuA.uA

=12M |X|2 + TCoM .

(The middle term in the middle line vanishes bythe obvious generalization of equation (54).) �


Example 56. [Reduced mass] Things are simplestin the CoM frame when N = 2. Suppose we havetwo stars moving under their mutual gravity, andset r = x2 − x1. Then Newton’s law of gravityimplies

m1x1 = Gm1m2r−3r, m2x2 = −Gm1m2r

−3r,

and MX = 0 since there are no external forces.Also

r = x2− x1 = −G(m1 + m2)r−3r = −GMr−3r,

so that the separation r evolves like the positionof a planet about a sun of mass M , see chapter6.

In the CoM frame m1x1+m2x2 = 0 and x2−x1 =r imply x1 = −(m2/M)r and x2 = (m1/M)r sothat

TCoM =12m1

(m2

M

)2

r.r +12m2

(m1

M

)2

r.r

=12μ|r|2


where μ = m1m2/M is the reduced mass.

The reduced mass also plays a role in the equationof motion, which can be written as

μr = − Gm1m2

r3r.


8.2 Angular momentum

We defined angular momentum with respect tothe origin O of a single particle by (63),

hO = x × mx.

Note that if mx = F then hO = x× F.

x y

F

O

The couple or moment or torque of F aboutO is

gO = x × F. (104)

Note that gO doesn’t depend on the choice of x.For if y − x is parallel to F then gO = y × F aswell. Thus we have

h0 = g0. (105)

We now generalize this to systems.


Now let O be an arbitrary point, not necessarilythe origin. The total angular momentumabout O is

HO =∑A

(xA − xO) × mA(xA − xO), (106)

and the total torque or net couple about Ois

GO =∑A

(xA − xO) × FA. (107)

Theorem 16. Suppose the internal forces aretorque-free, i.e., (101) holds. Then

HO = GO + M(xO − X) × xO. (108)

In particular if either O = CoM or xO = 0 then

HO = GO, (109)

the obvious generalization of (105).


Proof. The deduction of (109) from (108) istrivial. From (106)

HO =∑A

(xA − xO) × mA(xA − xO)

=∑A

(xA − xO) × FA +∑A

(xA − xO) ×∑B �=A

FA←B

−∑A

mAxA × xO +∑A

mAxO × xO.

The first term on the right hand side is GO. Thethird and fourth are −MX × x0 and MxO × xO

respectively. The contribution of particles 2 and7 to the double sum is

(x2 − xO) × F2←7 + (x7 − xO) × F7←2

= (x2 − xO − x7 + xO) × F2←7 by Newton 3

= (x2 − x7) × K27(x2 − x7) by (101)

= 0.

�


Example 57. A bead of mass m1 can slide on asmooth straight horizontal wire, and a particle ofmass m2 is attached to the bead by a light stringof length . The system moves in the verticalplane through the wire, and the string remainstaut. The horizontal displacement of the centreof gravity, G, at time t is denoted by x and theangle the string makes with the downward verticalby θ. Prove that during the motion x and

12(m1 + m2 sin2 θ)θ2 − (m1 + m2)(g/l) cos θ

remain constant. Find the period of smalloscillations about the vertical.

Gθ

m1

m2

The coordinates of G are X = (x,−(m2/M) cos θ),where M = m1+m2. Because the external forces(gravity and reaction at m1) are vertical, theorem


14 implies that the horizontal component of X isconstant, i.e., x is constant. The potential energyof the system is obviously U = −m2g cos θ plusa constant. To compute the kinetic energy weuse theorem 13. The kinetic energy of thecentre of mass is clearly 1

2M |X|2 = 12M(x2 +

((m2/M)θ sin θ)2). Now particle 1 has relativecoordinates x1 = (m2/M)(sin θ,− cos θ) and soparticle 1 contributes 1

2m1(m2/M)22θ2 towardsTCoM and there is a similar contribution fromparticle 2. Now theorem 15 implies that the totalkinetic energy is

T =12Mx2 +

12(m2/M)(m1 + m2 sin2 θ)2θ2

Conservation of energy implies

12(m1 + m2 sin2 θ)2θ2 − Mg cos θ

is constant, which is equivalent to the requiredresult.

For small values of θ we have correct to secondorder, 1

2m12θ2−Mg(1− 1

2θ2) = constant, which


is simple harmonic motion with frequency ω whereω2 = (M/m1)(g/l).

We now look at two examples involving angularmomentum.

Example 58. A horizontal wheel of radius r withbuckets on its circumference revolves about africtionless vertical axis. When the wheel hasangular velocity ω, the angular momentum ofthe wheel and buckets is Iω. Water falls into thebuckets at a uniform rate of mass m per unit time.Treating the buckets as small compared with thewheel, find the angular velocity ω(t) of the wheelafter time t if Ω be its initial value. Show thatafter time t the wheel has turned through an angle

IΩmr2

log(1 +

mr2t

I

).

The motion of and forces on the water and bucketsform a complicated problem, but a moment’sthought reveals that all of the forces in thisproblem are vertical. Now let O be the centre


of the wheel. From equation (107) we can deducethat the total torque of the forces has a vanishingcomponent in the vertical direction. Now theorem16, equation (109), implies that the verticalcomponent of the total angular momentum isconstant. Initially this is IΩ. At time t thewheel and buckets contribute Iω(t). However thebuckets now contain a mass mt of water all at adistance r from O. This water has a velocity ofmagnitude rω(t) directed circumferentially, andso it contributes mtr2ω(t) to the total angularmomentum. Thus we have

(I + mr2t)ω = IΩ or ω =IΩ

I + mr2t.

The angle turned is obviously∫

ω dt which givesthe result displayed above.


Example 59. A particle of mass m is free tomove in a thin smooth uniform straight tube ofmass 3m and length a. The tube can turn freelyin a horizontal plane about one end O which isfixed. Initially the tube has angular velocity Ωand the particle is at rest relative to the tube atits midpoint. Find the velocity with which theparticle leaves the tube.

[A result, proved in chapter 9, is that when thetube has angular velocity ω its angular momentumis ma2ω and its kinetic energy is 1

2ma2ω2.]

r

θ

O

We describe the position of the particle by polarcoordinates (r, θ). There are internal forcesbetween the tube and the particle and an externalreaction force at O, but there are no externaltorques about O. Thus we have conservation ofangular momentum about O, theorem 16 , and of


course conservation of total kinetic energy. Thefirst implies

ma2θ + mr2θ = ma2Ω + m(a/2)2Ω = 54ma2Ω,

and the second

12ma2θ2 +

12m(r2 + r2θ2) =

12ma2Ω2 +

12m(a/2)2Ω2

=58ma2Ω2.

It is simple algebra to conclude that at r = a,θ = 5

8Ω and r =√

15/32aΩ, so that the ejectionvelocity is (√15

32aΩ,

58aΩ).


9 Rigid bodies

9.1 Introduction

Collections of particles are all very well, but we areactually interested in macroscopic rigid bodies. Arigid body is a collection of particles labelled byA,B, . . . = 1, 2, . . . , N such that

1. |xA(t) − xB(t)| = const., ∀A,B,

2. the internal forces are torque-free; condition (101)holds.

We can specify its position by giving the situationof three points (not on a straight line) at eachtime t. Clearly x1(t) requires 3 numbers, butx2(t) requires only 2 (for |x2 − x1| is fixed) andx3(t) needs only one.

According to Newton’s principle, §1.1, we shallalso need 6 parameters to describe the velocity.An evident choice is the velocity of one point andin addition the angular velocity about it. In totalsix parameters are needed.


9.2 Linear momentum

We learned in chapter 8, theorem 14, that theCoM behaves like a particle of mass M subject tothe net external force F =

∑A FA,

MX = F, (110)

where X = M−1∑

A mAxA is the position of thecentroid.


9.3 Angular velocity

We need here to revert to the notation andconventions of chapter 7. Let S be an inertialframe fixed in space, and let S’ be a (possiblynon-inertial) frame fixed in the body.

If A and B are two points fixed in the body, thenxA − xB is fixed in S′. It follows from theorem119 of chapter 7 that there exists a unique angularvelocity vector ω, independent of A and B suchthat

xA − xB = ω × (xA − xB), (111)

where, as throughout this chapter, we write(da/dt)S as a.


If at time t there is a line of points instantaneouslyat rest then this is the (instantaneous) axis ofrotation.

Theorem 17. An axis of rotation exists iffω.X = 0, and it is parallel to ω.

Proof. The axis would have to satisfy

0 = x = x +(ω × (x − X)

),

where we have set xA = x and xB = X in (111).By dotting this equation with ω we see that it isinconsistent unless ω.x = 0. If this condition issatisfied then the solution is

x = X + |ω|−2ω × x + λω,

where λ is arbitrary, and this defines a line parallelto ω. �


9.4 Angular momentum

Let O be an arbitrary point of the body. Wedefine the angular momentum about O bythe usual formula, see (106),

HO =∑A

(xA − xO) × mA(xA − xO), (112)

It follows from theorem 16 that

HO = GO + M(xO − X) × xO. (113)

where GO =∑

A(xA − xO) × FA is the nettorque of the external forces about O. Note thesimplification if

1. xO = X, i.e., O is the centre of mass,

2. xO = 0, in particular if O is at rest.


Example 60. [Moment of inertia] A bicyclewheel, centre O, radius a, is mounted on anaxle which is initially in the e1 direction. The rimis lined with lead, so that essentially all of themass is in the rim. The wheel spins with angularspeed ω. Compute ω and HCoM .

Obviously ω = ωe1. For any point A on the rim,xA − X lies in the 23-plane. Clearly x = 0. Now(111) implies xA = ω × (xA − X) and then thedefinition (106) implies

HCoM =∑A

mA(xA − X) ×(ω × (xA − X)

)(114)

=∑A

mA(xA − X).(xA − X)ω (115)

= Ma2ω. (116)

Here H and ω are proportional, and the constantof proportionality, Ma2, is the moment ofinertia about the axle.


Example 61. [Continuation of example 60] Supposenow that someone applies a torque G = Ge2 atO.

1

2

3

O

G

H

H + Gδt

δϕ

Now (113) implies H = G. In time δt, H →H + Gδt. Note that H.G = 0 initially so thatH = |H| is unchanged to first order, but His rotated through a small angle δϕ = Gδt/Hin the 12-plane. Thus the wheel acquires anangular velocity component dϕ/dt = G/H in the3-direction.

Clearly the relationship between ω, H and G isnon-trivial. We now explore it further, lookingfirst at energy, which is a scalar concept.


9.5 Energy in the CoM frame

Recall that for a single particle with velocity x,linear momentum p, the kinetic energy is T =12x.p. We might conjecture that 1

2ω.H has asimilar interpretation. Suppose we look in theCoM frame. Equation (111) implies that for eachparticle A

xA − x = ω × (xA − X). (117)

Setting O to be the centre of mass (112) asserts

HCoM =∑A

mA(xA − X) × (xA − x),

and so9

12ω.HCoM =

12

∑A

mA(xA − x).ω × (xA − X)

=12

∑A

mA|xA − x|2

= TCoM ,9We use the cyclic property of the scalar triple product in the first

equation and (117) in the second.


where TCoM is the kinetic energy in the CoMframe.


Now recall theorem 15. The total kinetic energyis

T =12M |x|2 + TCoM ,

which can be written as

T =12M |x|2 +

12ω.HCoM . (118)

The first term on the right is called thetranslational kinetic energy and the secondis called the rotational kinetic energy.Another way to write this is

T =12X.P +

12ω.HCoM , (119)

where P = M x is the linear momentum of thecentre of mass.

We know that P and x are proportional. Wehave seen from example 61 that the relationbetween ω and H is likely to be considerably morecomplicated, but we might reasonably expectlinearity.


9.6 Inertia tensor and moments of

inertia

In this section let O be any preferred point of thebody (and not necessarily the centre of mass).Equation (111) implies that for each particle A

xA − xO = ω × (xA − xO),

while (112) asserts

HO =∑A

mA(xA − xO) × (xA − xO).

Combining these we have

HO =∑A

mA(xA − xO) ×(ω × (xA − xO)

).

(120)Clearly we can write this as HO = IO(ω) where IO

is a linear map R3 → R3, the inertia operator.This is the linear relationship alluded to at the endof the last section. A linear map from one vectorspace (here “angular velocity space”) to another(here “angular momentum space”) is called a


tensor relationship, and will be explored in detailin the Methods course in term 4. However weknow from the course C1/2 that we can interpretI as a 3 × 3 matrix Iik. Invoking summationconvention (in this section only) we write (120)as

HOi = IOikωk, (121)

where the IOik are the components of theinertia tensor at O.

We now want to obtain an explicit formula forthe inertia tensor at an arbitrary point O. Itis very convenient to introduce the abbreviationrA ≡ xA − xO. Then (120) shortens to

HO =∑A

mArA × (ω × rA). (122)


Converting to suffix notation we have

HOi =∑A

mAεimnrAmεnkpωkrAp

= εnimεnkpωk

∑A

mArAmrAp

= (δikδmp − δipδmk)ωk

∑A

mArAmrAp.

Comparing this with (121) we have

IOik = (δikδmp − δipδmk)∑A

mArAmrAp,

or

IOik =∑A

mA

[(rAmrAm)δik − rAirAk

], (123)

where rAi = xAi − xOi.


If we wish to regard the rigid body as continuousthen we need to change the sum to an integral.We set mA = ρ(x) dV = ρ d3x, where ρ(x) is thedensity, and dV = d3x is a volume element. Then(123) becomes

IOik =∫

Body

ρ(x)[(rmrm)δik−rirk

]d3x, (124)

where ri = xi − xOi.

Notice that IO is symmetric, i.e., IOki = IOik.This means that we can choose coordinate axesalong the eigenvectors of IOik, so that IOik =diag(I1, I2, I3), where the Ik are the eigenvaluesof IOik. It is obvious from the definition thatthese axes are fixed in the body; they are calledthe principal directions at O and the Ik arecalled the (principal) moments of inertia atO.

Masochists will note, with relish, that thecomputation of the principal moments requires6 integrals (to form IOik) followed by aneigenvalue/eigenvector determination. However,with some finesse, we can often avoid this.


Consider first the evaluation of these quantities atthe centroid. (The general point O will be treatedshortly.) If we are lucky the body will possessa symmetry axis through the centroid. Becausethe body is invariant under rotations about thisaxis, the principal directions (an intrinsic propertyof the body) must also share this invariance.Thus one principal axis at the centroid must bealong the symmetry axis, and the other two areperpendicular to it. Further the principal momentsabout the other two directions must be equal, sothat we only need two integrals to compute theeigenvalues directly.

The next lemma shows how to compute principalmoments about the centroid in this special case.


Lemma 6. Let e1 be a principal directionthrough the CoM, and let RA be the perpendiculardistance of particle A from e1. Then

I1 =∑A

mARA2. (125)

e1

A

RACoM

Proof. One way to verify the result is to seti = k = 1 in (123). However it is constructive toproceed as follows. Suppose the body was rotatingwith angular velocity ω = ωe1. Clearly |rA| =RAω and so the kinetic energy relative to thecentroid is TCoM = 1

2

∑A mARA

2ω2. But from(118), TCoM = 1

2HCoM .ω. However because e1

is an eigenvector of ICom ik, HCoM = I1ω and soTCoM = 1

2I1ω2 which proves the result. �

If we know the inertia tensor at the centroid wecan compute it at an arbitrary point O.


Theorem 18. [Parallel axes] For any point O,

HO = HCoM + M(xO − X) × [ω × (xO − X)],

(126)and

IOik = ICoM ik + M [(xOm − Xm)(xOm − Xm)δik

− (xOi − Xi)(xOk − Xk)]. (127)

Proof. Recall equation (122)

HO =∑A

mArA × (ω × rA),

where rA = xA − xO = (xA −X) − (xO −X) =RA − RO say. Then

HO =∑A

mARA × (ω × RA) −∑A

mARA × (ω × RO)

−∑A

mARO × (ω × RA) +∑A

mARO × (ω × R


The middle two terms vanish since∑

A mARA =0 and the first and the last give (126). Equation(127) follows on writing this out in suffixnotation. �

Another useful result is

Theorem 19. [Perpendicular axes] Consider alamina, a 2-dimensional object with principaldirections e1 and e2 in the plane, and e3

perpendicular to the plane. Then the principalmoments at the centroid satisfy

I3 = I1 + I2. (128)

Proof. Notice that the laminar is invariant underreflections in its normal. The principal axes mustshare this invariance and so can be taken in thestated directions. An algebraic version of thisargument follows if we choose coordinates so that


the plane is x3 = 0. Then equation (123) implies

IOik =

⎛⎝ . . 0

. . 00 0 .

⎞⎠

so that O3 is a principal axis, and the other twolie in the plane.

Choosing coordinates adapted to the principalaxes, lemma 6 implies

I1 =∑A

mA(xA22 + xA3

2),

I2 =∑A

mA(xA32 + xA1

2),

I3 =∑A

mA(xA12 + xA2

2),

and since xA3 ≡ 0 the result follows. �


9.7 Computation of moments of

inertia

Example 62. Consider a uniform rod of massM , length 2a, centre O, one end E.

e1

e3

O

E

By symmetry the principal axes are e1 along therod and any two perpendicular directions. Thedensity is ρ = M/(2a). Using the continuumform

IO3 =∫ a

−a

M

2ax2 dx =

13Ma2.

Exercise 14. Show that IO1 = 0. What doesthe perpendicular axes theorem say here?


Example 63. [Continuation of example 62] Whatis the moment of inertia about an axis through Eperpendicular to the rod?

Suppose the rod has angular velocity ω = ωe3.Then HCoM = 1

3Ma2ω. Now use the parallelaxes theorem, replacing O by E.

HE = HCoM + M(xE − X) × [ω × (xE − X)]

= HCoM + M |xE − X|2ω

=13Ma2ω + Ma2ω

= 43Ma2ω

where we used the result ω.(xE − X) = 0. Thisimplies IE3 = 4

3Ma2.

This is a typical use of the parallel axes theorem.Note however that here the geometry is so simplethat we could compute

IE3 =∫ 2a

0

M

2ax2 dx = 4

3Ma2

directly.


Example 64. Consider a circular ring of massM , radius a lying in the 12-plane.

e1

e2

e3

O

Clearly IO3 = Ma2. By symmetry IO1 = IO2,and so the perpendicular axes theorem impliesthat IO1 = IO2 = 1

2Ma2.

Example 65. The same setup, but now auniform disc of radius a, mass M , densityρ = M/(πa2). Then if O is the centroid,

IO3 =M

πa2

∫∫(x2 + y2) dx dy

=M

πa2

∫ a

r=0

∫ 2π

θ=0

r2 dr rdθ

=M

πa22π

∫ a

0

r3 dr =12Ma2


and IO1 = IO2 = 14Ma2.

Example 66. Next consider a uniform cylinderof mass M , radius a and height h. Choosingthe 3-axis along the axis of symmetry, obviouslya principal axis, we have, taking O to be thecentroid,

IO3 =M

πa2h

∫∫∫(x2 + y2) dV

=M

πa2h

∫ h/2

z=−h/2

∫ a

r=0

∫ 2π

θ=0

r2dr rdθ dz,

but there is no need to evaluate this integral!We can collapse the 3-axis, turning this into theintegral of the previous example, with the sameresult, I03 = 1

2Ma2. Clearly IO2 = IO1 where

I01 =M

πa2h

∫∫∫(x2 + z2) dV

=M

πa2h

∫ h/2

z=−h/2

∫ a

r=0

∫ 2π

θ=0

(r2 cos2 θ + z2)dr rdθ dz

=M

12(3a2 + h2),


by conventional methods.


Example 67. Next consider a uniform sphere,centre O, of mass M , radius a, density ρ =3M/(4πa3). Any diameter is a principal axis, theprincipal moments are all equal and, e.g.,

IO3 = ρ

∫∫∫(x2 + y2) dV.

Here it is easiest to decompose the sphere intodiscs z = const., thickness dz, radius

√a2 − z2

giving

IO3 = ρ

∫ a

z=−a

(∫ 2π

θ=0

∫ √a2−z2

r=0

r2 dr rdθ

)dz

= 2πρ

∫ a

z=−a

14(a2 − z2)2 dz

= 25Ma2.

Exercise 15. What is the relation between theheight h and radius a of a uniform cylinder forit to behave dynamically like a sphere, i.e., for


equality among the principal moments of inertiaat its centroid?


Example 68. Finally consider the problem ofdetermining the inertia tensor at the centroid Oof a uniform cube of mass M , side 2a.

We choose coordinates so that O is (0, 0, 0) andthe vertices are (±a,±a,±a). By symmetry Ozis a principal axis and

IO3 =M

8a3

∫ a

−a

∫ a

−a

∫ a

−a

(x2 + y2) dx dy dz

= 23Ma2.

By symmetry IO1 = IO2 = IO3 and so

IO = 23Ma2

⎛⎝1 0 0

0 1 00 0 1

⎞⎠

Note that for a sphere

IO = 25Ma2

⎛⎝1 0 0

0 1 00 0 1

⎞⎠

so that a cube behaves dynamically, pretty muchlike a sphere.


9.8 Miscellaneous examples

Example 69. [Yet another pendulum] A uniformrod, mass M , length 2a swings in a vertical planeabout one end which is fixed. Determine thefrequency of small oscillations.

C

D

Mg

θ

O

��

��

The component of HO perpendicular to the planeis 4

3Ma2θ, see example 63. The only externalforces are gravity and the reaction at O. Thereaction has zero couple at O and we can regardthe gravitational force as acting through thecentroid C. Thus the net couple is −Mga sin θ. Itfollows from (113), specialized to the case where


O is at rest, that

43Ma2θ = −Mga sin θ,

orθ + (3g/4a) sin θ = 0,

corresponding to a simple pendulum with ω2 =(3g/4a). We can regard this as being a simplependulum of equivalent length = 4a/3.

Example 70. [Example 69 continued] Now adisc of mass 12M , radius 1

3a can be clamped sothat its centre D is on the rod at a distance xfrom O. Show that the length of the equivalentsimple pendulum satisfies 2

3a � � 2a.

The disc has moment of inertia 12(12M)(a/3)2

about a horizontal axis through D, example 65,and moment 2

3Ma2 + 12Mx2 about a horizontalaxis through O, parallel axes theorem. Thus thetotal moment about O is M(12x2 + 2a2). Thenet couple is −(12Mgx + Mga) sin θ and so

=12x2 + 2a2

12x + a,


which has a minimum of 2a/3 at x = a/3 andachieves its greatest value at both x = 0 andx = 2a.

Example 71. A uniform wire of length 3a isbent to form a sector of a circle, with the tworadii OA, OB and the arc AB each of lengtha. The point O is fixed and the wire swings ina vertical plane. Show that the period of smalloscillations is

2π

√5a

3g(cos 12 + 2 sin 1

2).

O

B

θ0

A

Simple geometry shows that the angle AOB is 1(in radians). Thus θ0 = 1

2. Let the total massbe 3M . Then the moment of inertia about a


horizontal axis through O is 13Ma2+1

3Ma2+Ma2,where the contributions come from OA, OB andAB respectively. (Examples 63 and 64 applyhere.)

The main computational problem here would seembe to determine the position of the centroid, andhence the gravitational torque. It is actuallysimpler to compute the gravitational potentialenergy directly. Further we need only computeit in the rest, equilibrium, position. For whenthe system is deflected through an an angle θ,the potential energy picks up a factor cos θ. Atrest the potential energy of OA (and OB) is−Mg(a/2) cos 1

2. The potential energy of AB is

−Mga∫ θ0

−θ0cos θ dθ = −2Mga sin 1

2. Thus thetotal potential energy, at equilibrium, is

−Mga

(cos

12

+ 2 sin12

).

Therefore the total energy, in general position, is

E =12

(53Ma2

)θ2−Mga

(cos

12+2 sin

12

)cos θ.


Comparison with the standard theory for thesimple pendulum now produces the requiredanswer.


Example 72. A uniform rod AB of length 2aslides with its ends fixed on a smooth circular wireloop whose centre is O. If b denotes the distanceof the centre C of AB from O, and θ is the anglethat OC makes with the vertical, prove that

θ2 =6bg

a2 + 3b2(cos θ − cosα),

where α is the maximum value of θ. Find thefrequency of small oscillations when α 1.

O B

θ

A

C

Example 62 shows that IC = 13Ma2 and the

parallel axes theorem shows that IO = 13Ma2 +

Mb2. Thus the total energy is

12M

(13a2 + b2

)θ2 − Mgb cos θ,


which has to be a constant, and this impliesthe stated form for θ2 = f(θ). Now thestability algorithm, theorem 5, implies that atequilibrium f ′(θ) ∝ sin θ has to vanish, and thestable equilibrium is obviously at θ = 0. Furtherthe frequency of small oscillations is given byω2 = −1

2f′′(0) or

ω2 =bg

a2 + 3b2.


Example 73. A lamina of mass M is at rest ona smooth horizontal table. A horizontal impulsiveforce of magnitude F is applied at a point Adistant a from the centroid G, and the direction ofthe force is perpendicular to GA. What happens?

G

F

A

Let IG be the moment of inertia about a verticalaxis through G. After the impact G will havea linear velocity v in the direction of F and thelamina will have a vertical angular velocity ω aboutG.

By (110) Mv = F and so v = F/M . We shallassume that the internal forces are torque-free.The vertical axis through G is a principal axisand so HG = IGω. By (113) IGω = aF and soω = aF/IG.


Example 74. [Rolling Disc] A uniform disc ofmass M and radius a rolls without slipping down aline of greatest slope of a plane inclined at an angleα to the horizontal. Determine its accelerationand the reaction forces.

α

D

G

Mg

NFMx

12Ma2θ

The centre of gravity moves with acceleration xparallel to the plane. Because there is no slippingthe angle turned is given by θ = x/a. The massaccelerations and forces are shown in the figure.

One way to obtain F and N is by resolving forcesparallel to and perpendicular to the plane:

Mg sinα − F = Mx, Mg cosα − N = 0.


Further we can find F directly by taking momentsabout G using (113) finding

aF =12Ma2θ =

12Max,

and sox = 2

3g sin α,

a constant value, but smaller than that of aparticle sliding down the plane. Also,

F = 13Mg sinα, N = Mg cosα.

Notice that F/N = 13 tan α and so the friction

coefficient must exceed this value if no slipping isto occur.


Example 75. [Rolling cylinder] A uniform solidcircular cylinder, of mass M and radius a, rollswithout slipping on the inside of a fixed hollowcircular cylinder of radius a + b. The axes of thecylinders are horizontal and parallel. Investigatethe motion, and find the least coefficient of frictionif the motion starts from rest with the point ofcontact at an angular distance α < 1

2π from thelowest point of the fixed cylinder.

θ

θ

D

G

O

Cϕ

Denote by θ the angle OG makes with the vertical.Let C be a fixed point on the small cylinder, such


that GC is vertical and downwards when thecylinder is in its lowest position, and let ϕ be theangle between GC and the downward vertical ina general position. To establish the no-slippingcondition notice that the distance between Cand D is both a(θ + ϕ) and (a + b)θ, whenceaϕ = bθ. The forces on the cylinder are gravityand reaction forces N and F as in the previousexample. The cylinder has a clockwise angularacceleration ϕ about G, and the centre of masshas a tangential acceleration bθ and a radialacceleration bθ2 towards O.

Resolving forces in the tangential and radialdirections we obtain

F−Mg sin θ = Mbθ, N−Mg cos θ = Mbθ2,

and taking moments about G gives

aF = −12Ma2ϕ = −1

2Mabθ,

so that

θ +2g

3bsin θ = 0,


with first integral

θ2 =4g

3b(cos θ − cosα).

It is now easy to show that

F = 13Mg sin θ, N = 1

3Mg(7 cos θ−4 cosα).

NowF

N=

sin θ

7 cos θ − 4 cosαis an increasing function of θ and so achieves itsgreatest value, 1

3 tan α at θ = α.

There is another way to view this problem.The constraint forces act at point D which isinstantaneously at rest. Therefore they do nowork and energy is conserved. The translationalkinetic energy is 1

2M(bθ)2 and the rotationalkinetic energy is 1

2(12Ma2)ϕ2, so that the total

kinetic energy is 34Mb2θ2. The potential energy

is U = −Mgb cos θ, and we easily obtain theequation for θ2 given above.


Example 76. [Backspin] A sphere of mass M ,radius a, is launched with initial speed V and zeroangular velocity on a rough horizontal surfacewith friction coefficient μ. Show that slippingstops after time t = 2V/(7μg), and that the finalkinetic energy is 5

14MV 2.

θ

MgRF

There is no vertical motion and so the reaction ofthe surface R = Mg. Thus while slipping occursthere is a frictional force F = μR = Mμg. If xis the horizontal displacement, Newton’s secondlaw implies Mx = −F , and an easy integral givesx = V − μgt. Using example 66, and equatingthe rate of change of angular momentum to thefrictional torque, 2

5Ma2θ = aF , and integration

gives aθ = 52μgt. Clearly x is decreasing while

aθ is increasing and they become equal at timet = 2V/(7μg). Thereafter slipping stops, F → 0


and we get pure rolling with x = aθ = 57V . Thus

the final kinetic energy is

T =12Mx2 +

12

25Ma2θ2 =

514

MV 2.

Example 77. [Backspin continued] As above,but the sphere has an initial angular velocity −Ω.Show that the ball returns, rolling, to its initialposition if 5V < 2aΩ < 12V and determine itsfinal speed.

Again we have x = −μg, aθ = 52μg while slipping

occurs and so

x = V −μgt, x = V t−12μgt2, aθ = −aΩ+5

2μgt.

So if the ball is slipping it reverses velocity att1 = V/μg) and returns to x = 0 at time t3 =2V/(μg). If it is to return rolling, then thetransition from slipping to rolling (when x = aθ)at t2 = 2

7(V +aΩ)/(μg) must satisfy t1 < t2 < t3which implies the inequality above. The finalspeed is of course the transition speed x(t2) =−(2aΩ − 5V )/7.


10. Phase plane techniques

10.1 Introduction

In chapter 4 we looked at systems with one degreeof freedom, described by

x2 = f(x), (129)

and we showed that equilibrium at x = a requiresf(a) = f ′(a) = 0. Further for small perturbationsx = a + ε(t),

ε − 12f

′′(a)ε = O(ε2). (130)

This equation has the first integral

ε2 − 12f

′′(a)ε2 = ΔE + O(ε3), (131)

where ΔE is a constant. If the equilibrium isstable then f ′′(a) < 0. The solution curves of(131) are small ellipses around x = a, x = 0in the xx-plane. This is a centre in standardordinary differential equation (ODE) notation.


x

x

a

If the equilibrium is unstable then f ′′(a) > 0, andthe solution curves are hyperbolae.

x

x

In ODE notation this is a saddle.


10.2 Plane autonomous systems

Equation (129) in the form x =√

f(x), x =12f

′(x) is a special case of a plane autonomoussystem (PAS)

x = F (x, y), y = G(x, y). (132)

A fixed point is a point (x0, y0) such thatF (x0, y0) = 0 = G(x0, y0).

Theorem 20. A solution curve of a PAS canstart, end or self-intersect only at a fixed point.

Proof. Equation (132) implies

dy

dx=

y

x=

G(x, y)F (x, y)

.

Thus if at most one of F or G vanishes at a point,there is a well-defined tangent direction (whichmay be horizontal or vertical) and the curve canbe continued. This fails at a fixed point. �

It is an immediate corollary that a solution curveof (129) can self-intersect only at a saddle. A


solution curve through a saddle point is called aseparatrix. The set of separatrixes separatesclasses of solution curves with different globalbehaviour.

Example 78. [The simple pendulum] With astandard notation we have

E = 12m2θ2 + U(θ),

where U(θ) = −mg cos θ. For simplicity wechoose units to set m = g = = 1. We caninterpret this equation as saying E = E(θ, θ) anddrawing this a surface above the θθ-plane, thephase plane.


θ

θ

E

-2 -1 0 1 2 3 4 5 -2-1.5

-1-0.5

00.5

11.5

2-1

-0.50

0.51

1.52

2.53

However along the solution curves E = const.Thus it is more informative to draw the contourlines E = const. of the surface, for these are thesolution curves.


θ

θ

−6 −4 −2 0 2 4 6−3

−2

−1

0

1

2

3

Notice that the points θ = 2nπ, θ = 0 correspondto stable equilibrium with E = −mg, and appearas centres on the phase plane. However the pointsθ = (2n + 1)π, θ = 0 corresponding to E = mgare unstable equilibria and appear as saddles inthe phase plane.

The curves E = mg define the separatrixes.These curves are defined by

12m2θ2 − mg cos θ = mg,

or

θ = ±2√

g

cos(1

2θ).


θ

θ

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

-6 -4 -2 0 2 4 6

Curves with −1 < E/(mg) < 1 lie inside theseparatrixes and are bound, i.e. ∃ θ1, θ2 such thatθ1 < θ < θ2, and the motion librates, i.e., θchanges sign regularly. Curves with E > mglie outside the separatrixes and are unbound.(They correspond to a rotating pendulum.) Thusthe separatrix curves separate solution curvescorresponding to radically different behaviour.Note that one can even describe a dampedpendulum in the phase plane diagram of anundamped one. This will be a curve on whichE decreases and so it crosses the contour lines,ending up at a stable centre.

Example 79. [Anharmonic oscillator] We consider


a one-dimensional system with energy given by

E = 12x

2 + U(x),

where

U(x) = 12ω

2x2 +

{14A

2x4 x < 0,

−14A

2x4 x > 0,

where A > 0. (The case A = 0 is the harmonicoscillator.)

0 x0x1

E0

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

-1 -0.5 0 0.5 1 1.5

We have stable equilibrium with a centre at x =0 = x. There is an unstable equilibrium at x =x0 = ω/A, x = 0, which corresponds to an energy


E = E0 = 14ω

4/A2. This separatrix intersects the

x-axis at x = 0, x = ±ω2/(√

2A) and intersects

the x-axis again at x = x1 =√√

2 − 1(ω/A),x = 0.

Here is the phase portrait.

−1 −0.5 0 0.5 1 1.5−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1Phase portait for anharmonic oscillator

x

xdot

Notice that orbits outside the separatrix areunbounded and as x → ∞, x ∼ ±Ax2/

√2.

Note once more that stable equilibria correspondto centres in the phase portrait, while unstable


equilibria correspond to saddles which may bejoined by separatrixes.

In order to discuss the general case we need tointroduce the idea of structural stability. Aprecise definition involves serious mathematics,and so we give a heuristic version, together withexamples. Basically if a system has a featurewhich remains invariant under small continuousperturbations of the system, then the feature isstructurally stable (SS).

1. If a smooth function f(x) has an isolated simplezero then this property is SS. There is an intervalsuch that the value of f changes sign at thetwo endpoints, and this is preserved by smallperturbations.

2. The property that f has an isolated double rootis not SS. Consider e.g., f(x) = x2. Adding asmall constant to f produces 0 or 2 real zeros.

3. The property that f has an isolated stationarypoint is SS. Simply apply item 1 to f ′(x).

4. If a smooth function F : R2 → R2, F (x, y) =


(f(x, y), g(x, y)) has an isolated zero, then it isSS. We can construct a neighbourhood of the zeroin which the curves f(x, y) = 0 and g(x, y) = 0intersect transversally, and this is obviously SS.

5. If a smooth function f(x, y) has an isolatedstationary point then it is SS. (Use item 4 appliedto ∇f .)

6. A zero eigenvalue of a matrix is not SS. Consider

a =(

1 11 1 + ε

)

with eigenvalues

λ = 1 − 12ε ±√

1 + 14ε

2.

Now we are interested in the stationary points ofthe surface E = E(x, x). These are of course SSby item 5. Further their type is determined bythe signs of the eigenvalues of the hessian matrix


of E. For a SS system they must be nonzeroby item 6. If they both have the same sign wehave a maximum or a minimum, and becauseof the nature of the kinetic energy term, only aminimum can occur. If they have oppposite signsthen we have a saddle. Thus the analysis we havegiven applies to any SS system with one degreeof freedom.

In particular we can analyze any smooth systemwith one degree of freedom, e.g. the system withthe potential energy as shown.

x

U

A

B

C

D

Motion will be possible only for small |x| whenx < 0. There will be centres at A and C, andsaddles at B and D and the D-separatrix enclosesthe B-one. Sketching the phase portrait is left asan exercise.


10.3 Limit cycles

In the case of general systems, a closed periodiccurve is a limit cycle.

Example 80. Consider the system defined by

r = cr(r − 1), θ = 1,

where c is a constant. Clearly r(t) = 1, θ(t) = tdefines a limit cycle. If c > 0 then it is unstablefor r > 0 if r > 1 and r < 0 if r < 1. By a similarargument it is stable if c < 0.

Another, almost trivial, example are the simplependulum orbits which are bound. They areneutrally stable.


10.4 The not-so-simple pendulum

Suppose S is an inertial frame with coordinatesx, y and z. Consider a second frame S′ withcoordinates x′ = x, y′ = y and z′ = z + h(t).because of gravity a particle in S feels anacceleration z = −g. However a particle in S′

feels z′ = −g − h. E.g., if S′ is freely falling,h = vt − 1

2gt2, then particles in S′ experience novertical acceleration. In this frame there is noeffective gravity.

We now want to consider a simple pendulum oflength whose pivot is oscillating in a verticaldirection z = A cos(ΩT ) where, for the moment,T is the time coordinate. Thus the effectivegravity is g + AΩ2 cos(ΩT ). Allowing for friction,the equation of motion is

d2θ

dT 2+ K

dθ

dT+

1

(g + AΩ2 cos(ΩT )

)sin θ = 0.

For further study it is sensible to use dimensionlessvariables

t =T√/g

, ω =Ω√g/

, a =A

, k =

K√g/

.


The equation of motion then becomes

θ + kθ + (1 + aω2 cos ωt) sin θ = 0. (133)

The extra a-term can change completely thedynamics. Note that because of the t-dependencethis is not a system with one degree of freedom,even though it is one-dimensional. Nor is itconservative. The k-term takes energy out ofthe pendulum. The a-term can add or subtractenergy, sometimes both in the same motion.

The conventional picture suggests that after fixingthe parameters k, a and ω, and specifying theinitial data θ = θ0 and θ = ω0, then the solutionis determined.

Consider first the case where k = 0 and θ is smallso that we can replace sin θ by θ. Choosing a newdimensionless time coordinate τ = ωt we have

d2θ

dτ2+ (α + β cos τ)θ = 0,

where α = ω−2 and β = a. This is Mathieu’sequation which has been extensively studied. In


particular it is known that for certain values of αand β the exact solution θ(t) ≡ 0 is unstable.

The shaded region corresponds to values of α andβ for which the exact solution θ(t) ≡ 0 is unstable.Note in particular there is a region of instabilitynear α = 1

4, β 1, which corresponds to smalla and ω being twice the natural frequency of thependulum. Then θ = 0 is unstable; an initiallysmall disturbance will lead to a swinging motionof steadily increasing amplitude.

Let us now return to the general case. Figure 1shows the behaviour of a pendulum starting fromθ = π, θ = 0.01, with a = 0.2 and ω = 10 for0 � t � 200π. The upright position θ = π is, in


this case stable!

3.13 3.135 3.14 3.145 3.15 3.155−0.02

−0.015

−0.01

−0.005

0

0.005

0.01

0.015

0.02

Figure 1: a = 0.2, omega = 10, theta0 = pi, omega

0 = 0.01

theta

thet

adot

Figure 2 corresponds to the same data, except thatθ = −0.01 initially, and we are using a differentvalue for ω.


−2 −1 0 1 2 3 4−3

−2

−1

0

1

2

3

thet

adot

theta


0 = −0.01

Now the pendulum drops to θ = 0 and oscillateswildly before dissipation causes the solution tosettle at θ = 0. Both of these pictures arecompletely different to the phase portrait of thesimple pendulum discussed earlier. The orbitappears to self-intersect but this is just an illusion.In fact the orbit is the curve (t, θ(t), θ(t)) in R3


which never self-intersects. What is being shownis the projection onto a t = const. plane.

−1.5 −1 −0.5 0 0.5 1 1.5−1.5

−1

−0.5

0

0.5

1

1.5

Figure 3: a = 0.1, omega = 2, theta0 = 0.01, omega

0 = 0

theta

thet

adot

Figures 3 and 4 illustrate realistic limit cycles. Infigure 3 the bob is released from rest at θ = 0.01,but in this case θ = 0 is unstable equilibrium. Thependulum oscillates periodically about its lowestposition but never comes to rest. Energy is being


pumped in at the pivot at a rate which balancesthe frictional dissipation.

2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

thet

adot

theta


0 = 1.3

In figure 4 the bob is released from θ = π withθ = 1.3 and moves to an elaborate but stablelimit cycle.


The pendulum has one more secret to reveal.When we solve a system like

θ + kθ + sin θ = 0, θ(0) = θ0, θ(0) = ω0,

we know that the solution θ(t, θ0, ω0) is acontinuous function not only of t but also ofthe initial data θ0 and ω0. This continuity isessential for solving differential equations on acomputer because floating point numbers (R\Z)can only be stored approximately. If we store say8 significant figures we expect to get a reasonableapproximation to the exact solution.

However certain non-linear systems are chaotic,i.e.,

1. the solution curves are very sensitive to the initialconditions,

2. they never settle into an equilibrium or repeatingpattern.


−10 −5 0 5 10 15 20 25−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

thet

adot

theta

Figure 5: a = 0.5, omega = 1.1, theta0 = pi +0.00001 x n, omega

0 = 0

Figure 5 shows a pendulum with a = 0.5, ω = 1.1,ω0 = 0 and θ0 = π+n×10−5 where n = 1, 2, 3, 4corresponding to the black, blue, green and redcurves respectively. Clearly the orbits stay closeonly for a short time. If one continues beyondt = 200π the orbits wiggle and loop with noobvious pattern. Thus if my computer had only 4


digit accuracy the evolution of such a pendulumwould be essentially random and unrepeatable. Ofcourse this is only a toy model, but prediction is aserious problem for weather forecasters, for theirequations are genuinely chaotic

How can one tell whether a system is chaotic?One necessary condition is that the equations benon-linear. A second such condition is that thephase space must have more than two dimensions.A PAS cannot exhibit chaos. Suppose that in aPAS, a phase path starts at (x0, y0) and cannotleave a bounded area in phase space. Then thePoincare–Bendixson theorem asserts thatthe phase orbit must eventually either

1. terminate at an equilibrium point,

2. return to (x0, y0) giving a closed path,

3. approach a limit cycle.

None of these possibilities is chaotic.

These conditions are however not sufficient, andthere is a very simple example, extending chapter


9 to demonstrate this. Consider a rigid body,centroid O, subject to no external forces. Thenby theorem 16 the rate of change of the angularmomentum H0 with respect to an inertial frameS is zero, (

dH0

dt

)S

= 0.

By theorem 12, the rotating axes theorem, asseen in a frame S′ fixed in the body (with suffixS′ suppressed)

H0 + ω × H0 = 0.

Choose S′ so that the inertia tensor is diagonal,IO = diag(A, B,C), and set ω = (ω1, ω2, ω3)T .Then H0 = (Aω1, Bω2, Cω3) and

Aω1 =(B − C)ω2ω3,

Bω2 =(C − A)ω3ω1,

Cω3 =(A − B)ω1ω2.

These are clearly 3-dimensional and nonlinear, butthey cannot exhibit chaos.


Exercise 16. Show that the kinetic energy Tand the norm (squared) of the angular momentum

T =12(Aω1

2 + Bω22 + Cω3

2),

|H|2 =A2ω12 + B2ω2

2 + C2ω32,

are both constants of the motion.

You can study chaos further in the Part II andIII courses on dynamical systems, and the lectureschedules give useful references. The not-so-simple pendulum is based on some exercises in

• D. Acheson, From calculus to chaos: anintroduction to dynamics, Oxford, 1997,

which also cites some more popular references.


Hamiltonian dynamics

Most of this course has been pretty similar toA-level school courses. In reality dynamics ismore sophisticated than this. Suppose we havea system which can be described by coordinatesand angles. We unite them in the concept ofgeneralized coordinates traditionally denotedq1, q2, . . . , qn Suppose we can form the kineticenergy and potential energy

T =T (t, q1, q2, . . . , qn, q1, q2, . . . , qn)

U =U(t, q1, q2, . . . , qn, q1, q2, . . . , qn).

We then form the lagrangian function

L(t,q, q) = T − U.

There is a powerful theory of dynamics based onthe lagrangian function, but I will not describethat here, because there is a brilliant stimulatingaccount in the Feynman lectures in physics,volume 2, chapter 19, which will also be coveredin the Part IB methods course. Instead I want tolook at the hamiltonian extension.


For each q1 the corresponding generalizedmomentum is

pi =∂L

∂qi.

Finally we form the hamiltonian function

H(t,q,p) =n∑

i=1

piqi − L,

where the generalized velocities are eliminated interms of the generalized momenta. Then theequations of motion of the system, Hamilton’sequations take the beautiful form

dqi

dt=

∂H

∂pi,

dpi

dt= −∂H

∂qi.

In almost all situations H = T + U , the totalenergy, so that for any system if you can writedown the total energy, eliminate the velocitiesin terms of the momenta, then you have theequations of motion of the system.


Example 81. [Hamiltonian derivation of CFP equatioWe study a system for which the kinetic andpotential energies are given by

T =12m(r2 + r2θ2), U = −GMm/r.

We set q1 = qr = r, q2 = qθ = θ, so that thelagrangian is

L =12m(q2

r + qr2q2

θ) + GMm/qr.

Then pr = ∂L/∂qr = mqr = mr, theconventional momentum. However pθ =∂L/∂qθ = mqr

2qθ = mr2θ, the angularmomentum. (This is why p is a generalizedmomentum.)

Next we form H = T + U as a function of the q’sand p’s,

H =pr

2

2m+

pθ2

2mqr2− GMm

qr.


The first set of Hamilton’s equations dqi/dt =∂H/∂pi are uninteresting (in this very simpleexample). They assert r = pr/m and θ =pθ/(mr2). However the second set has realcontent.

Firstly dpr/dt = −∂H/∂qr asserts mr = mrθ2 −GMm/r2. This is true, and reminds us that thisformalism knows all about curvilinear coordinatesystems.

Secondly dpθ/dt = 0 since H is independent ofqθ = θ.

More generally we note that if H does notdepend on qi then pi is a constant. This is asimple instance of Noether’s theorem: everysymmetry implies a conserved quantity.

Here is another example. Systems with one degreeof freedom do not involve the time coordinateexplicitly, and so they are invariant under timetranslations. This is a symmetry and theconserved quantity is energy. Further examplesoccur in the Dynamics and General Relativitycourses at Part II and beyond.


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