MP202: Mechanics

Brian Dolan
1. Newtons Laws of Motion
I. Any massive object maintains a state of uniform motion unless acted on by a force.
II. A force, F, applied to any object changes its momentum, p, at a rate
F =
dp
dt
. (1)
III. To every action there is an equal and opposite reaction.
I. Newtons rst law was in fact found by Galileo Galilei and it is often called Galileos
Lay of Inertia. At rst sight it may look as though it is just a special case of the second
law when F = 0, but there are two important aspects to it. Firstly it is not at all obvious
unless friction is eliminated, and this was Galileos insight: once we realise that friction
itself is a force we can include friction among the forces on the left hand side of (1) and in
this sense Newtons rst law is a special case of the second law.
A second, more important point, is that Newtons rst Law is only true in a non-
accelerating reference frame. If we do experiments in a train running on a smooth track
with blinds on the windows then Newtons rst law would not hold inside the train when
it is accelerating, either changing its speed or going round corners at constant speed.
There are funny forces in accelerating reference frames e.g. centrifugal force in a rotating
reference frame. Such forces are only due to the observers motion and are not considered
to be dynamical, they are often called ctitious forces or pseudo-forces. A reference
frame which is not accelerating is called an inertial reference frame because it is a
reference frame in which Galileos law of inertia holds true. From a modern perspective
Newtons rst Law should be viewed as stating that inertial reference frames occupy a very
special place in the theory of mechanics.
II. Newtons second law is perhaps better known in the form when the momentum is
written as p = mv, with m the mass and v the velocity of the object. Then, provided the
mass is constant, we have
F = ma,
where a =
dv
dt
is the acceleration. But this is only correct for constant mass, the form (1)
is more general and is applicable even if the mass changes with time, e.g. a rocket using
fuel at such a rate that its mass decreases signicantly as it accelerates upward.
1
If there is more than one force acting on m then we should add them together using
the rules of vector addition and the force F appearing in Newtons second law is the vector
sum of all the separate forces acting on m.
The true power of Newtons second law only comes to the fore when we know some-
thing about F. The forces on a mass m can depend on the position and velocity of the
mass (e.g. friction depends on velocity), as well as explicitly on time. If the position of
the object as a function of time is denoted by r(t) then, if we know the explicit form of F
as a function of r, v = r,* and t, then Newtons second law, for constant m, is
F(r, r, t) = mr, (2)
which is a dierential equation for r as a function of time.
An example is the equation governing the motion of two masses, m
1
and m
2
, attract-
ing each other gravitationally due to Newtons other famous law his universal law of
gravitation. In this case F is independent of velocity and depends only on the separation
r between the two masses, as an inverse square,
F(r) =
Gm
1
m
2
r
2
r =
Gm
1
m
2
r
3
r,
where r = r/r is a unit vector in the direction r and G = 6.67 10
11
kg
1
m
3
s
2
is
Newtons universal constant of gravitation. If m
1
>> m
2
then we can consider m
1
to be
xed in space and choose our origin to coincide with the position of m
1
, then r is just the
position vector of m
2
and its acceleration is r so, applying Newtons second law to m
2
, we
get
m
2
r =
Gm
1
m
2
r
3
r.
We shall see many other examples of (2) in the following.
Note that if we use a (non-accelerating) reference frame in which r is the position
vector of a point mass then we can transfer to a dierent reference frame, moving with
velocity u relative to the rst, and the position vector in the new reference frame will be
r

= r +ut.
The velocity of the mass in the new reference frame is
r

= r + ut +u
and its acceleration is
r

= r + ut + 2 u.
The accelerations are equal, r

= r, for all t if and only if u = u = 0, i.e. if and only if u

is a constant. In other words the accelerations are equal if and only if the new reference
frame is also non-accelerating. If we transfer to an accelerating reference frame then u
* We shall often denote time derivatives with dots, so v = r =
dr
dt
and a = v = r.
2
must necessarily become a function of time, it is necessarily the case that r

= r, and
Newtons second law will look dierent in the two references frames. Provided we stick to
non-accelerating reference frames then r = r

and the right hand side of Newtons second

law (2) is the same in all non-accelerating reference frames. Using an accelerating (e.g.
a rotating reference frame) is not forbidden but it complicates Newtons second law and
usually makes the analysis more tricky.
III. Newtons third law is often illustrated by considering someone stepping out of a
small boat at a quay side. If the boat is not tied up and you put one leg on the quay side
and push with the other to try and step onto the quay, watch out!
However in other situations Newtons third law is not always so obvious. Consider a
planet orbiting the Sun. The Sun exerts a gravitational force on the planet and Newtons
third law states that the planet must exert an equal and opposite force on the Sun. This
force is in a line between the Sun and the planet but, as the planet is moving this line
is rotating. Since it takes a nite time for any physical eect to pass between the planet
and the Sun, due to the nite speed of light and Einsteins Special Theory of Relativity,

information about the planets position always arrives at the Sun late. If the planet is a
distance R from the Sun and c is the speed of light, no physical eect due to the planet
can possibly inuence the Sun in less than a time t = R/c. Is the force on the Sun due to
the planet in the direction of where the planet is now or where it was at a time t = R/c
earlier? Answers to questions like this lie in the dynamical theory of gravity, developed by
Einstein at the beginning of the 20th century and called the General Theory of Relativity.
The answer is very surprising the very notion of time is dierent for the planet and the
Sun! However the speed of light is so large that, in everyday circumstances over distances
that are not too large we can ignore such eects and treat the gravitational force as being
instantaneous an approximation known as action at a distance.
Most of this course will be dedicated to solving Newtons second law for various kinds
of common forces and we shall ignore the subtleties of General Relativity.
2. One Dimensional Motion
2.1 Forces which depend only on time.
We shall start with a class of problems where F(t) is independent of the position and
velocity of the mass m. This class of problems is perhaps the simplest because we can just
try to integrate Newtons second law directly.
For motion in one dimension we can label the position of the mass m by a single
co-ordinate x and suppose the particle moves along a straight line. The position of m will
in general be a function of time: x(t), e.g. x could be increasing when m moves to the
right and decreasing when m moves to the left along the line. The velocity is v(t) = x(t),
which can be positive or negative depending on the direction of motion of m. Newtons
second law is then
m x = F(t),

The importance of inertial reference frames is central to Special Relativity the

theory is formulated in the class of inertial reference frames.
3
where the force only has one component and is a given function of time. For example if
F(t) = const then m x = F and x = v, so the acceleration
a = v =
F
m
(3)
is a constant and
dv
dt
= a v(t) = at +C
where C is a constant. Setting t = 0 we see that C = v(0), the initial velocity of m.
To save writing we shall denote the initial velocity by v(0) = v
0
(which could be either
positive or negative, depending on the initial direction of ms motion), so
v(t) = at +v
0
changes linearly with time. What we have done here is to integrate equation (3),
v =
dv
dt
= a v(t) =
_
t
0
adt +v
0
= a
_
t
0
dt +v
0
= at +v
0
and v
0
is a constant of integration. We can integrate again to get x(t),
v(t) =
dx
dt
= at +v
0
x(t) =
_
t
0
(at +v
0
) dt +x
0
=
1
2
at
2
+v
0
t +x
0
,
where x
0
is a second constant of integration, x
0
= x(0), which is the initial position of m
at t = 0.
More generally, if F(t) is some given non-trivial function of t (that is less trivial than
the simple case of being just a constant, independent of t) we can try to integrate the
equation
dv
dt
=
F(t)
m
v(t) =
_
t
0
F(t)
m
dt +v
0
=
1
m
_
t
o
F(t)dt +v
0
and, provided the integral
_
t
0
F(t)dt can be evaluated for the given function F(t), we can
evaluate v(t). Once we have v(t) explicitly then we can try the next step and evaluate
x(t) =
_
t
0
v(t)dt + x
0
. For example suppose F(t) = At +B, with A and B constants, is a
linear function. Then
v(t) =
1
2
At
2
+Bt +v
0
is quadratic and
x(t) =
1
6
At
3
+
1
2
Bt
2
+v
0
t +x
0
is cubic.
4
2.2 Forces which depend only on velocity
As a second illustration of the use of Newtons second law, consider the case of a mass
m moving through a uid, eg. air or water, subject to a frictional force which opposes its
motion.
The velocity is v(t) = x(t), which can be positive or negative depending on the
direction of motion of m. For small speeds it usually a good approximation to take the
frictional force to be proportional to the speed, F
friction
= c x = cv, where c is a
constant (called the co-ecient of friction). Then Newtons second law reads
m x = c x m v = cv.
The sign is important here: if m is moving to the right, x is positive and the frictional
force must oppose the motion, i.e. it must be negative; if m is moving to the left, x is
negative and the frictional force must be positive. Hence c must be positive, otherwise m
would accelerate in the same direction as it is moving, which is nonsense. Friction always
makes objects decelerate, F
friction
must always have the opposite sign to that of v. Note
that if v is ever zero then v is also zero and so v does not change with time and it stays
zero, but if v = 0, then it necessarily must change and the speed decreases.
The equation m v = cv is easy to solve for the unknown function v(t). We write it
suggestively as
m
dv
dt
= cv mdv = cvdt m
dv
v
= cdt
as long as v = 0. Suppose that v > 0 and that the speed starts out as v
0
at time t = 0,
then integrate both sides to get
m
_
v
v
0
dv
v
= c
_
t
0
dt m(ln v ln v
0
) = ct,
remembering that m and c are constants and can be taken outside the integrals. Here ln
signies natural logarithms.* Since lnv ln v
0
= ln(v/v
0
) we have
ln
_
v
v
0
_
=
c
m
t
v(t) = v
0
exp
_

c
m
t
_
. (4)
i.e. the velocity decreases exponentially toward zero as time increases.
Having obtained the velocity as a function of time it is now easy to integrate again to
get x(t). Suppose m starts o at t = 0 with speed v
0
from the point x
0
, then
dx
dt
= v
0
exp
_

ct
m
_
dx = v
0
exp
_

ct
m
_
dt
* A subtle point to remember about problems of this type is that, if v < 0, then
_
dv
v
= ln|v| ln |v
0
|.
5
and, integrating again,
_
x
x
0
dx = v
0
_
t
0
exp
_

ct
m
_
dt x x
0
=
v
0
m
c
_
e

ct
m
1
_
.
The x in this last equation is x(t), so
x(t) =
v
0
m
c
_
1 e

ct
m
_
+x
0
. (5)
While the velocity tends exponentially to zero but never reaches zero in any nite
time, m only ever travels a nite distance even as t ,
x(t) x
0

t
v
0
m
c
.
Using (4) to eliminate the exponential in (5) we see that the velocity decreases linearly
with x, until it vanishes,
v(x) = v
0

c
m
(x x
0
). (6)
Notice that as the friction increases (c increases) the distance travelled decreases,
while as the friction decreases (c decreases) the distance travelled increases. The constant
c depends on a number of factors: it is greater in water than in air, for example, and even
greater again in treacle. It also depends on the shape of the mass m, for example racing
cars have spoilers on them to reduce air friction.
In this example the force depends only on v and is independent of t. For problems of
this kind we can nd v(x) directly as follows. Since v =
dx
dt
we can use the Chain Rule for
dierentiation to write
m
dv
dt
= m
dv
dx
dx
dt
= mv
dv
dx
.
So
m v = F(v) mv
dv
dx
= F(v) m
_
vdv
F(v)
=
_
dx.
In this case F(v) = cv is linear, so

m
c
_
dv =
_
dx
m
c
(v v
0
) = x x
0
,
reproducing (6) above.
A slightly more complicated case is when the motion of m is vertical and the ac-
celeration due to gravity must be taken into account. We shall use a co-ordinate y for
vertical motion, reserving x for a horizontal motion when we discuss 2-dimensional prob-
lems later. Measuring y as increasing upwards the force due to gravity is then in the
negative y-direction. In the absence of friction we have
m y = mg (7)
6
as the gravitational force is mg downwards. So
y = g y = gt +v
0
y =
1
2
gt
2
+v
0
t +y
0
,
where v
0
= y(0) is the initial velocity (positive in the upwards direction) and y
0
= y(0)
the initial height, with y = 0 being ground level.
If m is dropped from a height h, then y
0
= h and v
0
= 0, so
v(t) = gt and y(t) = h
1
2
gt
2
and m hits the ground when y(t) = 0, that is at time T with
T =

2h
g
. (8)
If m is thrown upwards from ground level, y
0
= 0, with velocity v
0
> 0, then
y(t) = v
0
t
1
2
gt
2
(9)
and the maximum height is reached when
dy
dt
= 0, that is when
t = t
max=
v
0
/g y
max
=
v
2
0
2g
. (10)
The mass m hits the ground again when y(t) = 0 with t > 0 so, from (9), this happens at
a time T =
2v
0
g
.
Including air friction equation (7) is modied to
m y = mg c y, (11)
with c > 0 so that the frictional force always has the opposite sign to that of y, i.e. it
opposes the motion.
The rst thing to notice about equation (11) is that the acceleration vanishes when
y = mg/c. When the motion is downward with constant velocity, v
term
= mg/c, the
air friction exactly balances the gravitational force and m travels downwards with constant
velocity: |v
term
| is known as the terminal velocity of the falling body.
Now we shall solve equation (11), which is a second order dierential equation for the
unknown function y(t). It is simplest to use v = y and write
m v = mg cv,
which is a rst order dierential equation for the unknown function v(t). We can manip-
ulate this equation as
m
dv
dt
= mg cv mdv = (mg +cv)dt m
_
v
v
0
dv
mg +cv
=
_
t
0
dt
7
and integrate directly to get
ln
_
mg +cv
mg +cv
0
_
=
c
m
t mg +cv = (mg +cv
0
) exp
_

c
m
t
_
,
provided v >
mg
c
between time 0 and time t (this solution works for v
0
either positive or
negative, provided only that v
0
>
mg
c
). So
v(t) =
_
v
0
+
mg
c
_
e

c
m
t

mg
c
= v
0
e

c
m
t

mg
c
_
1 e

c
m
t
_
(12)

mg
c
.
Equation (12) can immediately be integrated to get y(t):
_
t
0
v(t)dt =
_
v
0
+
mg
c
_
_
t
0
e

c
m
t
dt
mg
c
_
t
0
dt =
m
c
_
v
0
+
mg
c
_
_
e

c
m
t
1
_

mg
c
t
and
_
t
0
v(t)dt =
_
t
0
dy
dt
dt = y(t) y(0)
so
y(t) =
m
c
_
v
0
+
mg
c
_
_
1 e

c
m
t
_

mg
c
t +y
0
, (13)
where again y
0
= y(0).
For example if m is dropped from an initial height h, then y
0
= h and v
0
= 0, giving
v(t) =
mg
c
_
1 e

c
m
t
_
and
y(t) =
m
2
g
c
2
_
1 e

c
m
t
_

mg
c
t +h.
Initially, when t << m/c, we can approximate e

c
m
t
1
c
m
t +
1
2
c
2
m
2
t
2
and
v(t) gt, y(t) h
1
2
gt
2
,
so the motion starts out as if there was no air friction, because the velocity is initially zero
and takes time to build up to a point where air friction becomes important.
At late times, when t >> m/c the exponential is negligible so
y(t)
m
2
g
c
2

mg
c
t +h,
and m falls with terminal velocity as if it had been dropped from a higher point h +
m
2
g
c
2
.
8
The mass m will hit the ground at a time T given by
0 =
m
2
g
c
2
_
1 e

c
m
T
_

mg
c
T +h
so
T =
ch
mg
+
m
c
_
1 e

c
m
T
_
. (14)
We cannot get T(h) in closed form, but if
cT
m
>> 1 then we can ignore the exponential,
because it is negligible, and
T =
ch
mg
+
m
c
.
At the other extreme, if
cT
m
<< 1, we can expand the exponential as
e

c
m
t
= 1
ct
m
+
1
2
_
ct
m
_
2
+
and in this case (14) gives
T =
ch
mg
+
m
c
_
1
_
1
cT
m
+
1
2
_
cT
m
_
2
+
__
=
ch
mg
+T
1
2
c
m
T
2
+ ,
0 =
ch
mg

1
2
c
m
T
2
+ .
Ignoring terms of order c
2
or higher, which are indicated by the dots, this reduces to
0 =
ch
mg

1
2
c
m
T
2
T
2
=
2h
g
which is the answer we got before (8), when we analysed the case with no friction (c = 0).
If m is thrown upwards from ground level at t = 0, with initial speed v
0
> 0, it reaches
a maximum height when, using (12),
dy
dt
= v(t) = 0
_
v
0
+
mg
c
_
e

c
m
t

mg
c
= 0 t =
m
c
ln
_
mg
cv
0
+mg
_
. (15)
Putting this value of t into (13), with y
0
= 0, we get the maximum height
y
Max
=
m
c
_
v
0
+
mg
c
_
_
1
_
mg
cv
0
+mg
__
+
m
2
g
c
2
ln
_
mg
cv
0
+mg
_
=
mv
0
c

m
2
g
c
2
ln
_
cv
0
+mg
mg
_
. (16)
Note that, when
ct
m
is small, we can expand the exponential in (15) as
e

c
m
t
= 1
c
m
t +
1
2
_
c
m
t
_
2
+
9
to give
v
0
_
1
c
m
t +
_
+
mg
c
_
1
c
m
t +
1
2
_
c
m
t
_
2
+
_

mg
c
= 0
v
0
gt +
ct
m
_
1
2
gt v
0
_
+ = 0.
In the limit of zero viscosity, c 0, we are left with
v
0
= gt t =
v
0
g
which is the answer we got before with c = 0, when we ignored friction, (10). We can also
nd y
Max
for small c, using
ln
_
cv
0
+mg
mg
_
= ln
_
1 +
cv
0
mg
_
=
cv
0
mg

1
2
_
cv
0
mg
_
2
+
1
3
_
cv
0
mg
_
3
+
in (16) to give
y
Max
=
mv
0
c

m
2
g
c
2
_
cv
0
mg

1
2
_
cv
0
mg
_
2
+
1
3
_
cv
0
mg
_
3
+
_
=
1
2
v
2
0
g

1
3
_
cv
3
0
mg
2
_
+ .
Again c = 0 gives the previous result (10), y
Max
=
v
2
0
2g
, and now we see that including a
small amount of air friction reduces the maximum height by an amount proportional to
v
3
0
.
2.3 Forces which depend only on position.
The next example of the application of Newtons laws is when the force is independent
of velocity and time and depends only on position, so F(r). Newtons second law is then
m v = F(x).
As an example consider Hookes law, where F = kx is proportional to x, with k > 0 a
constant. If m is pushed away from x = 0 then F will always push it back towards x = 0
again. With v = x we have
m x = kx x =
k
m
x
which has the general solution
x(t) = A
0
cos(
0
t
0
) (17)
where A
0
0 and 0
0
< 2 are constants, called the amplitude and the phase of the
motion. The position oscillates as a function of time, between x = A
0
and x = A
0
,
with angular frequency
0
=
_
k
m
(the frequency, in Hertz, is

0
2
). Oscillatory motion of
10
this kind is called Simple Harmonic Motion and is very common in Nature, provided the
amplitude is not too large.
2
0
t
x(t)
0
A
0
A
0

When the force depends only on position we can associate an energy with it, called the
potential energy, and the total energy, kinetic plus potential, is conserved. The potential
energy is derived as follows,
F(x) = m
dv
dt
vF(x) = mv
dv
dt
=
m
2
d(v
2
)
dt

dx
dt
F(x) =
m
2
d(v
2
)
dt
. (18)
Dene
U(x) :=
_
x
F(x)dx F(x) =
dU
dx
so, using the Chain Rule,
dU
dt
=
dx
dt
dU
dx
=
dx
dt
F(x).
Using this in (18) we get

dU
dt
=
m
2
d(v
2
)
dt
or
d
dt
_
1
2
mv
2
+U(x)
_
= 0,
hence the total energy,
E =
1
2
mv
2
+U(x)
is constant during the motion. U(x) is called the potential energy for the motion. For
example Hookes law has
F(x) = kx =
d
dx
_
k
2
x
2
_
11
and the potential U(x) =
1
2
kx
2
is quadratic. With x = A
0
cos(
0
t
0
),
v = x = A
0

0
sin(
0

0
) and
E =
1
2
_
mA
2
0

2
0
sin
2
(
0
t
0
) +kA
2
0
cos
2
(
0
t
0
)
_
=
1
2
mA
2
0

2
0
,
since
2
0
= k/m. Hence the energy is a constant proportional to the amplitude squared.
x
E
A A
0 0
EU(x)= mv
U(x)
2
1 2
The mass oscillates between x = A
0
and x = A
0
. At any point |x| A
0
the
dierence between E and U(x) manifests itself as a non-zero velocity v =
_
2
_
EU(x)
_
m
,
which vanishes at the end points x = A
0
. The speed is a maximum |v| =
_
2E
m
at x = 0.
Knowing the shape of the potential energy U(x) as a function of x gives a qualitative
understanding of the possible motion of a particle.
For example in the next gure if a particle has energy E then the region x < x
1
is
forbidden, the particle will oscillate between x = x
1
and x = x
2
if placed between x
1
and
x
2
; the region x
2
< x < x
3
is forbidden; and the particle will accelerate to indenitely
large x if placed at any point x > x
3
.
The particle has no forces on it at any point x
0
where the derivative of U(x) vanishes,
F(x
0
) = U

(x
0
) = 0. The particle can be placed at rest at any such point and it will
remain at rest in a state of equilibrium. If U

(x
0
) = 0 and x
0
is a minimum of U, so
U

(x
0
) > 0, then F(x) = U

(x) < 0 at any point x > x

0
close to x
0
on the right and
F(x) = U

(x) > 0 at any point x < x

0
close to x
0
on the left. Hence any displacement
away from x
0
will result in a force that pushes the particle back to x
0
. In a situation like
this x
0
is a point of stable equilibrium. If x
0
is a maximum of U, so U

(x
0
) < 0, then
12
F(x) = U

(x) > 0 at any point x > x

0
close to x
0
on the right and F(x) = U

(x) < 0
at any point x < x
0
close to x
0
on the left. Hence any displacement away from x
0
will
result in a force that pushes the particle further away from x
0
. In a situation like this x
0
is a point of unstable equilibrium. If U

(x
0
) = U

(x
0
) = 0 then x
0
is said to be a point of
neutral equilibrium.
x
E
x
x
2
1
x
3
V(x)
2.4 Forced Harmonic Oscillator
Frequently in physics we encounter situations in which a simple harmonic oscillator is
driven by an external force. For example a child on a swing is a simple harmonic oscillator
which will oscillate with some natural angular frequency
0
if left alone but if someone
pushes the swing then they are applying a time-dependent external force F(t). In such
situations Newtons second law reads
m x = kx +F(t) m x +kx = F(t)
where the total force on the right hand side of the rst equation is the sum of the force due
to Hookes law and the external force. For concreteness suppose the external force is of the
form F(t) = F
0
cos(t) where F
0
and are constants ( is called the forcing frequency)
so the equation governing the motion is
m x +kx = F
0
cos(t). (19)
Our task is to solve this equation for the unknown function x(t). Let us try looking for a
solution of the form
x(t) =

Acos(t)
with

A a constant. Then x =

Asin(t) and x =
2

Acos(t), so equation (19) is

m
2

Acos(t) +k

Acos(t) = F
0
cos(t) (F
0
+m

A
2
k

A) cos(t) = 0.
13
We want this to be true for all t, which xes

A to be

A =
F
0
k m
2
=
F
0
m(
2
0

2
)
(20)
where
0
=
_
k
m
is the frequency of the unforced oscillator, that is the frequency found in
section 2.3 when F(t) = 0. So we have found a solution of (19), namely
x(t) =
F
0
m(
2
0

2
)
cos(t). (21)
Note that

A is positive for <
0
, negative for >
0
and diverges for =
0
. *
The magnitude of

A, A() = |

A|, is called the amplitude of the oscillation. The mass

m oscillates about x = 0 with maximum displacement equal to the amplitude A() and
angular frequency equal to that of the forcing frequency .
We can rewrite (21) in terms of the amplitude by introducing a phase into the argu-
ment of the cosine to take care of the sign,

0
=
_
0, for <
0
, for >
0
.
Then (21) can be written as
x(t) =
F
0
m|
2
0

2
|
cos
_
t
0
_
= A() cos
_
t
0
_
. (22)
* Strictly speaking this solution is invalid when the forcing frequency equals the
natural frequency
0
but we can see that the amplitude of the oscillations gets arbitrarily
large as approaches
0
. This is the phenomenon of resonance and it is very common
in oscillatory systems. We shall tame the innity at =
0
in the next section where we
shall consider a more realistic model by including friction.
14
m
2
0
0
F
0
A( )

0
()
The amplitude and phase are plotted above, as functions of the forcing frequency .
For <
0
the motion of m is in phase with the driving force but for >
0
it is 180

out of phase with the driving force.

We have found a solution to (19) but this is not the only one! We already know from
section 2.3 that
x(t) = A
0
cos(
0
t
0
) (23)
15
satises
m x +kx = 0. (24)
Hence adding (21) to (23) gives another solution of (19), since
x(t) =
F
0
m(
2
0

2
)
cos(t) +A
0
cos(
0
t
0
) (25)
m x +kx = F
0
cos(t) + 0 = F
0
cos(t).
For a given external driving force, with F
0
and xed, the solution (25) has two indepen-
dent constants, A
0
and
0
. By varying the constants A
0
and
0
in (25) we can change the
initial position and velocity. For example suppose the mass starts o initially at x(0) = 0
and with initial speed zero, x(0) = 0. Then
x(0) =
F
0
m(
2
0

2
)
+A
0
cos(
0
) (26)
and
x(t) =
F
0
m(
2
0

2
)
sin(t)
0
A
0
sin(
0
t
0
)
x(0) =
0
A
0
sin
0
= 0. (27)
Equation (26) tells us that A
0
= 0 and (27) then forces sin(
0
) = 0. Choosing
0
in the
range 0
0
< 2 then leaves only two possibilities,
0
= 0 and
0
= , giving cos(
0
) = 1
and cos(
0
) = 1 respectively. Suppose <
0
, then we can clearly satisfy (26) by setting
A
0
=
F
0
m(
2
0

2
)
and cos(
0
) = 1. For >
0
, choose A
0
=
F
0
m(
2

2
0
)
and cos(
0
) = +1.
In either case (25) becomes
x(t) =
F
0
m(
2
0

2
)
_
cos(t) cos(
0
t)
_
=
2F
0
m(
2
0

2
)
_
sin
_
(
0
)t
2
_
sin
_
(
0
+ )t
2
__
.
Note that for (
0
)t << 1 small, so sin
_
(
0
)t
2
_

(
0
)t
2
,
x(t)
F
0
t
m(
0
+ )
sin
_
(
0
+ )t
2
_
.
In particular if =
0
+ with /
0
<< 1 we are close to resonance and
x(t)
F
0
t
2m
0
sin(
0
t)
so the system oscillates with frequency
0
and the amplitude grows linearly with time.
Eventually the amplitude becomes so large that Hookes law breaks down and equation
(19) is no longer valid.
16
2.6 Damped Harmonic Oscillator.
In the previous section we saw that the amplitude of the motion of a forced harmonic
oscillator diverges when the forcing frequency equals the natural frequency
0
of the
oscillator. This is not physical as the amplitude can never be innite. A more realistic
model should include friction. If there is a frictional force proportional to velocity, c x
with c > 0, then Newtons second law (19) is modied to
m x = kx c x +F
0
cos(t) m x +c x +kx = F
0
cos(t). (28)
For simplicity we shall rst study the case with no forcing, F
0
= 0,
m x +c x +kx = 0. (29)
This is an example of a second order linear dierential equation for the unknown function
x(t). It is second order because it involves the second derivative of x(t) and it is linear
because it only involves the function x(t) and its derivatives linearly. This last feature
is very important and makes the problem tractable. Non-linear dierential equations,
involving x
2
(t) or sin(x(t)) are much harder and we shall not consider them in this course.
Before going on to nd explicit solutions of (29) we rst prove a useful little result
that will help in the analysis. Suppose x
1
(t) are x
2
(t) are two solutions of (29) so we have
that
m x
1
+c x
1
+kx
1
= 0
m x
2
+c x
2
+kx
2
= 0.
Now let x(t) = a
1
x
1
(t) +a
2
x
2
(t) with a
1
and a
2
arbitrary constants. Then
m x +c x +kx = m(a
1
x
1
+a
2
x
2
) +c(a
1
x
1
+a
2
x
2
) +k(a
1
x
1
+a
2
x
2
)
= a
1
(m x
1
+c x
1
+kx
1
) +a
2
(m x
2
+c x
2
+kx
2
) = 0.
This means that, given two solutions of (29) any linear combination of them, with constant
co-ecients, is another solution. In fact it can be shown, but will not be proven here, that
it is sucient to nd two independent solutions (excluding the trivial case x = 0, which
is clearly a solution) and any solution of (29) is a linear combination of them (x
1
and x
2
are independent if they are not constant multiples of each other, i.e. x
2
= a
1
x
1
is not
independent of x
1
and so). This leads to the following result:
Theorem
The most general solution x(t) of (29) is given by rst nding any two independent
solutions, x
1
(t) and x
2
(t), and then taking the linear combination
x(t) = a
1
x
1
(t) +a
2
x
2
(t)
where a
1
and a
2
are arbitrary constants.
17
Notice the solution (25) that we found for (19) also had two independent constants,
called A
0
and
0
there.
This is extremely useful. There is in fact an innite number of solutions of (29), since
there is an innite number of pairs (a
1
, a
2
), but we only need to nd two independent
ones and we have them all! The two arbitrary constants a
1
and a
2
in the general solution
of the dierential equation (29) are associated with the fact that we lose information in
dierentiating a function we throw away any constant part so the solution of a dier-
ential equation is not unique. In a second derivative we lose two pieces of information, the
constant part and the linear part. The arbitrary constants a
1
and a
2
represent two pieces
of information about the function x(t) that are lost in (29). The solution of a dierential
equation is not unique. We can however restrict to a unique solution by specifying more
information, such as initial conditions for the motion. For example if we know the initial
position and velocity of m, at t = 0, then this xes the two constants a
1
and a
2
uniquely.
Suppose we have found two solutions x
1
(t) and x
2
(t) and we are told that m starts out
from x
0
= x(0) at t = 0 with velocity v
0
= x(0). This tells us that
x(0) = a
1
x
1
(0) +a
2
x
2
(0) = x
0
, and x(0) = a
1
x
1
(0) +a
2
x
2
(0) = v
0

_
x
0
v
0
_
=
_
x
1
(0) x
2
(0)
x
1
(0) x
2
(0)
__
a
1
a
2
_
,
so, since we know x
0
and v
0
, we can solve for a
1
and a
2
provided the matrix
_
x
1
(0) x
2
(0)
x
1
(0) x
2
(0)
_
is invertible:
a
1
=
x
0
x
2
(0) v
0
x
2
(0)
x
1
(0) x
2
(0) x
2
(0) x
1
(0)
a
2
=
v
0
x
1
(0) x
0
x
1
(0)
x
1
(0) x
2
(0) x
2
(0) x
1
(0)
,
which xes the constants a
1
and a
2
uniquely in terms of the initial conditions.*
An example of this procedure is the solution presented in section 2.3, with c = 0 and
no friction,
x +
2
0
x = 0,
with
2
0
=
k
m
. It is easy to check that x
1
(t) = cos(
0
t) and x
2
(t) = sin(
0
t) are two
independent solutions of this equation. Hence the general solution is
x(t) = a
1
cos(
0
t) +a
2
sin(
0
t).
Indeed the solution (17)
A
0
cos(
0
t
0
) = A
0
_
cos(
0
t) cos
0
+sin(
0
t) sin
0
_
= A
0
cos
0
cos(
0
t)+A
0
sin
0
sin(
0
t)
is of precisely this form, with a
1
= A
0
cos
0
and a
2
= A
0
sin
0
. Imposing, for example, the
initial condition that the mass m starts o a distance d from the origin with zero velocity,
x(0) = d and x(0) = 0, we see that
d = x(0) = a
1
and 0 = x(0) =
0
a
2
* This does not work if x
1
(0) x
2
(0) x
2
(0) x
1
(0) = 0.
18
so a
1
= d and a
2
= 0, giving the unique solution
x(t) = d cos(
0
t)
with these initial conditions.
Returning now to the case with friction (29) we need to nd two independent solutions.
To do this we shall use a mathematical trick which is often useful in problems involving
oscillations, we shall use complex numbers. First we try to nd solutions of the form
x(t) = e
t
,
with a constant. Using this form in (29) gives
(m
2
+c +k)e
t
= 0,
since x = e
t
and x =
2
e
t
. For this to be true for all t it must be the case that
m
2
+c +k = 0.
Notice what this has achieved, we have succeeded in turning a dierential equation for an
unknown function x(t) into an algebraic equation for one unknown constant which is
easy to solve. Divide through by m and let = c/m > 0 so the equation becomes

2
+ +
2
0
= 0
which has two solutions

=

_

2
4
2
0
2
.
We have thus found two solutions of the dierential equation, namely
x
1
(t) = e

+
t
and x
2
(t) = e

t
,
and the general solution is a linear combination of these with constant co-ecients, which
we shall call C
1
and C
2
here,
x(t) = C
1
e

+
t
+C
2
e

t
.
Notice however that these solutions are very dierent depending on whether
2
4
2
0
is
positive, negative or zero.
i) If the friction is small so c is not large, in particular if < 2
0
, then let =
_

2
0

2
4
giving

2
i
so
x(t) = e

t
2
(C
1
e
it
+C
2
e
it
),
19
with
e
it
= cos(t) +i sin(t).
At rst sight x(t) now looks complex, but this need not be the case. We can get a real
solution simply by choosing C
1
and C
2
to be complex numbers with C
2
the complex
conjugate of C
1
, so C
2
= (C
1
)

. This gives a solution with two real constants in it: denote

the real part of C
1
by
a
1
2
and the imaginary part by
a
2
2
(so C
1
=
a
1
ia
2
2
, C
2
=
a
1
+ia
2
2
)
then
x(t) = e

t
2
__
a
1
ia
2
2
_
_
cos(t) +i sin(t)
_
+
_
a
1
+ia
2
2
_
_
cos(t) i sin(t)
_
_
= e

t
2
(a
1
cos(t) +a
2
sin(t))
is a linear combination of two real solutions, namely e

t
2
cos(t) and e

t
2
sin(t).
Alternatively we can express this in terms of two constants* A

> 0 and 0

< 2
dened as A
2

= a
2
1
+a
2
2
and tan

= a
2
/a
1
, so a
1
= A

cos

and a
2
= A

sin

, giving
x(t) = A

t
2
cos(t

). (30)
This is oscillatory motion with an amplitude A

t
2
that decreases to zero exponen-
tially with time. The mass starts out a distance d = A

cos

from the origin and oscillates

with angular frequency =
_

2
0


2
4
, losing energy with time due to friction, the decay
constant in the exponential

2
being proportional to the co-ecient of friction,

2
=
c
2m
.
Behaviour like this is common in many oscillating systems, e.g. the oscillations of a
clock pendulum will decrease with time and eventually cease if the clock is not wound up.
The motion looks something like the following:
* The subscript on A

and

is just to distinguish them from the constants A

0
and

0
that were introduced in the solution of the undamped oscillator, = 0.
20
t
x(t)
ii) If the friction is large enough and > 2
0
then the roots

are both real and

negative and the general solution is
x(t) = a
1
e

+
t
+a
2
e

t
with a
1
and a
2
real. The rst term decays faster than the second and after a while, when
t >> 1/
+
,
x(t) a
2
e

t
to a very good approximation. If the mass starts o at t = 0 a distance d = a
1
+ a
2
> 0
from the origin it just relaxes exponentially to x = 0. It never overshoots and x never
becomes negative because the friction is strong enough to dampen any oscillations.
iii) = 2
0
is a critical value that separates oscillating motion, case i) above, with
completely damped motion, case ii) above. When this happens
+
=

= /2 and we
only have one solution,
x(t) = a
1
e

t
2
.
In this case, and only in this case, a second solution of (29) is x = te

t
2
and the general
solution is then
x(t) = a
1
e

t
2
+a
2
te

t
2
.
If the mass starts at t = 0 from x = d with zero velocity, then a
1
= d and

2
a
1
+a
2
= 0,
so
x(t) = d
_
1 +
t
2
_
e

t
2
.
The mass relaxes back to x = 0 as shown below. This case is known as critically damped
and can be important in many engineering situations: for example in a sensitive voltmeter
21
we want the needle to be fast and responsive, so the friction should be low, but we also
want oscillations to be damped so that the needle settles down to a stationary position as
soon as possible after the voltage is varied, the optimal damping is critical.
x(t)
t
2.7 Forced Damped Harmonic Oscillator
Now consider a damped harmonic oscillator that is driven by an external force of
frequency , equation (28). Dividing through by m this is
x + x +
2
0
x =
F
0
m
cos(t). (31)
In section 2.4, for = 0, we found a solution by trying x =

Acos(t), but this doesnt
quite work here. Instead we try the slightly more general form
x = Acos(t ),
with a constant (the = 0 case was written this way in equation (22)). Substituting
this guess into (31) gives

2
Acos(t ) Asin(t ) +
2
0
Acos(t ) =
F
0
m
cos(t)
A(
2
0

2
) cos(t ) A sin(t ) =
F
0
m
cos(t)
A
_
(
2
0

2
) cos + sin
_
cos(t) +A
_
(
2
0

2
) sin cos
_
sin(t)
=
F
0
m
cos(t).
22
We can satisfy this for all t by setting
A =
F
0
m{(
2
0

2
) cos + sin }
(32)
and
(
2
0

2
) sin cos = 0 tan =

2
0

2
. (33)
Simple trigonometry allows us to calculate sin and cos from the gure below
2
2
2
( ) +
0
2 2 2

0

2
to get
sin =

_
(
2

2
0
)
2
+
2

2
, cos =

2
0

2
_
(
2

2
0
)
2
+
2

2
and these can be used in (32) to give
A =
F
0
m
_
(
2

2
0
)
2
+
2

2
.
When the friction vanishes, = 0 and = 0, this reduces to (20), provided we choose
_
(
2

2
0
)
2
=
2

2
0
. When the friction is non-zero the amplitude is now nite when
=
0
and we see that friction removes the innity that we found at resonance in the
friction free case.
F
0

m
2
0
A( )
0

23
The amplitude peaks at a frequency

Max
=
_

2
0

2
2
=
_

2
4
,
where
dA()
d
= 0. For small damping, this very close to the natural undamped frequency

0
.
So we have found a solution, namely
x(t) =
F
0
m
_
(
2

2
0
)
2
+
2

2
cos(t ) (34)
with determined by (33). However this is not the only solution. In solving equation
(19) for the forced harmonic oscillator without damping we found that the most general
solution, (25), had two arbitrary constants, an amplitude A
0
and a phase
0
, which we then
xed by choosing specic initial conditions (A
0
cos(
0
t
0
) remember was the general
solution of the undamped equation with no forcing, (24)). A similar situation applies
here. The general solution of the damped oscillator equation with no forcing is (30),
A

t
2
cos(t

). Adding this to the solution of (31) that we have just found, namely
(34), gives
x(t) = A

t
2
cos(t

) +
F
0
m
_
(
2

2
0
)
2
+
2

2
cos(t ). (35)
This is a solution of (31), since
x + x +
2
0
= 0 +
F
0
m
cos(t) =
F
0
m
cos(t).
Equation (35) is in fact the most general solution of the forced damped harmonic oscillator
equation (31). It has two arbitrary constants, an amplitude A

and a phase

which, as
for the undamped oscillator, can be xed by choosing specic initial conditions. Notice
however that, because of the damping factor e

t
2
, the rst term dies away with time and
becomes negligible for t >>
2

, at late times the solution becomes insensitive to the initial

conditions and always settles down to (34). Perturbations of (34) induced by the rst term
on the right hand side of (35) are called transients.
Let us focus on the late time behaviour (34),
x(t) =
F
0
m
_
(
2

2
0
)
2
+
2

2
cos(t ).
This has an amplitude
A() =
F
0
m
1
_
(
2

2
0
)
2
+
2

2
=
F
0
m
0
1
_
_

_
2
+Q
2
24
0
2
m
F
0
Q=10
Q=5
Q=3

A( )
Q=1
0

/2

Q=10
Q=1
Q=5
Q=3
( )

0
25
and a phase
= tan
1
_

2
0

2
_
= tan
1
_

0
Q
_

2
0

2
_
_
= tan
1
_
_
(/
0
)
Q
_
1 (

2
0
)
_
_
_
,
where we have dened Q :=

0

which is called the quality factor of the oscillator. A small

Q means that the oscillator is heavily damped, a large Q means that there is very little
damping, with Q the undamped case. The amplitude and phase, as functions of the
forcing frequency , are plotted above, for some dierent values of Q.
The amplitude of the late-time behaviour oscillations (34) is independent of time, but
energy is being dissipated due to friction. This is because the external force is continually
supplying energy to the oscillator to compensate for the frictional dissipation. We can
determine the rate at which energy is supplied from the power absorbed by the oscillator,
using the formula Power = Force V elocity. The external force is
F
0
cos(t)
while the velocity is
x =
F
0
m
_
_

2
0
_
2
+
2

2
sin(t )
so the power is
P(t) =
F
2
0
m
_
_

2
0
_
2
+
2

2
cos(t) sin(t )
=
F
2
0
m
_
_

2
0
_
2
+
2

2
_
cos(t) sin(t) cos() cos
2
(t) sin()
_
.
This is a complicated oscillating function of time and it is more useful to calculate the
average power, P, over a single cycle of period T =
2

, which is
P =
1
T
_
T
0
P(t)dt =
F
2
0
2m
_
_

2
0
_
2
+
2

2
sin() =
F
2
0
2m
0
Q
1
_
_

_
2
+Q
2
_.
The average power absorbed by the oscillator per cycle is shown below, as a function of
the forcing frequency , for dierent values of Q. The larger Q is the higher and narrower
the peak, indicating that an oscillator with large Q will absorb a lot of energy only at
driving frequencies close to the natural frequency of the undamped oscillator,
0
.
26
P( )

Q=10
Q=5
Q=3
Q=1
We have been studying (31) which is a special case of a more general class of problems,
which we write as
x + x +
2
0
x =
F(t)
m
(36)
with F(t) some specied function of time. Dierential equations of this type are called lin-
ear, second-order, inhomogeneous equations: linear because the function x, and its deriva-
tives, only appear linearly in the equation, there is nothing like x
2
or sin( x) in the equation,
which would be non-linear; second-order because the highest derivative of x appearing in
the equation is a second-order derivative; inhomogeneous because the presence of the func-
tion F(t) on the right hand side means that, if we have a solution x(t) then simply shifting
t to get x(t +c), with c a constant, is in general not another solution this is in contrast
to the homogeneous equation
x + x +
2
0
x = 0 (37)
for which, having found a solution x(t), x(t +c) is another solution.
We shall describe some useful mathematical results for solving equations like (36). We
already know from section 2.6 that the problem of nding the most general solution of a
homogeneous equation is greatly simplied by rst nding any two independent solutions
and then taking linear combinations of them. This reduces the amount of work considerably
because we only need to nd two independent solutions and we are done. It is not quite
so simple for an inhomogeneous equation, but it is not much harder either. We rst note
27
that, if x
1
(t) is a solution of (37) and x
2
(t) a solution of (36) then x(t) = x
1
(t) + x
2
(t) is
another solution of (36), since*
x + x +
2
0
x = ( x
1
+ x
2
) +( x
1
+ x
2
) +
2
0
(x
1
+x
2
)
= ( x
1
+ x
1
+
2
0
x
1
) + ( x
2
+ x
2
+
2
0
x
2
) = 0 +
F(t)
m
=
F(t)
m
.
We shall state, without proof, the following theorem:
Theorem
The most general solution x(t) of (36) is given by rst nding any solution of (36)
and adding to this the most general solution of (37).
So, assuming we can nd any two independent solutions, x
1
(t) and x
2
(t), of (37) and
any solution, x
3
, of (36) then the most general solution of (36) is
x(t) = a
1
x
1
(t) +a
2
x
2
(t) +x
3
(t), (38)
with a
1
and a
2
arbitrary constants. In this construction x
3
(t) is called a particular solution
of (36).

The problem now reduces to nding two independent solutions of (37) and one par-
ticular solution of (36). We have already done the rst part in section 2.6 to nd two
solutions of (37) try a solution of the form x = e
t
which reduces the dierential equation
to an algebraic one,

2
+ +
2
0
= 0

2

_

2
4

2
0
.
If
2
> 4
2
both roots are real and the general solution of the homogeneous equation is
x = a
1
e

+
t
+a
2
e

t
;
if
2
< 4
2
the roots are complex, one being the complex conjugate of the other,

2
i with =
_

2
0


2
4
, and the general solution is
x = e

2
t
_
a
1
cos(t) +a
2
sin(t)
_
;
* Note that this would not be true if x
1
(t) and x
2
(t) were both solutions of (36), since
we would then have x + x +
2
0
x = 2
F(t)
m
, which is not equation (36).

Note that a constant multiple, a

3
x
3
(t) with a
3
a constant, of x
3
(t) is not a solution
of (36) unless a
3
= 1, since
a
3
x
3
+a
3
x
3
+a
3

2
0
x
3
= a
3
F(t)
m
.
28
if
2
= 4
2
the general solution of the homogeneous equation is
x =
_
a
1
+a
2
t
_
e

2
t
.
Finally we must nd particular solutions, x
3
(t), of the inhomogeneous equation (36). The
form of these depends on the functional form of F(t), we have already studied the case
where F(t) = F
0
cos(t) is an oscillating function and here we shall simply list a few other
possibilities.
1) F(t) a polynomial in t of order k with F(0) = 0: try x
3
(t) a polynomial of degree k
and x the co-ecients by requiring that x(t) satisfy (36). An example with k = 2 is
x + 5 x + 6x = 6t
2

1
3
.
We obtain two independent solutions of the corresponding homogeneous equation
x + 5 x + 6x = 0
as we did before, try x(t) = e
t
for which
(
2
+ 5 + 6)e
t
= 0 ( + 3)( + 2) = 0 = 2 or 3,
which gives two independent solutions, x
1
(t) = e
3t
and x
2
(t) = e
2t
.
For a particular solution try x = At
2
+ Bt + C, with A, B and C constants. Then
x = 2At +B and x = 2A, so
x+5 x+6x = 2A+5(2At+B)+6(At
2
+Bt+C) = 6At
2
+(10A+6B)t+(2A+5B+6C).
demanding that this equals 6t
2

1
3
requires, equating co-ecients,
6At
2
+ (10A+ 6B)t + (2A+ 5B + 6C) = 6t
2

1
3
6A = 1, 10A+ 6B = 0, 2A+ 5B + 6C =
1
3
,
so A = 1, B =
5
3
and C = 1, so a particular solution is
x = t
2

5
3
t + 1.
Of course this is only one solution, it is not the most general one, but we can use it
as x
3
(t) in (38) to get the most general solution, namely
x(t) = a
1
e
3t
+a
2
e
2t
+t
2

5
3
t + 1.
29
Determining the two constants a
1
and a
2
requires more information, for example,
suppose we are told that x(0) = 1 and x(0) = 0. Then
x(0) = a
1
+a
2
+ 1 = 1 a
1
= a
2
and
x(t) = 3a
1
e
3t
2a
2
e
2t
+ 2t
5
3
x(0) = 3a
1
2a
2

5
3
= 0
a
1
=
5
3
, since a
2
= a
1
,
and the unique solution satisfying these initial conditions is
x(t) =
5
3
_
e
2t
e
3t
_
+t
2

5
3
t + 1.
If F(0) = 0 then this will not quite work. If F(t) is a polynomial of degree k with
F(0) = 0, then t = 0 is a root of the equation F(t) = 0 and F(t) must be of the form
F(t) = tG(t) where G(t) is a polynomial of degree k1. In this case the form x(t) = ty(t),
where y(t) is a polynomial of degree k whose co-ecients are determined by substitution
into (36), will give a solution, provided G(0) = 0.
2) If F(t) is of the form sin(t) or cos(t) then try a solution of the form
x(t) = Acos(t) +Bsin(t),
with A and B constants which can be determined by substituting this form in (36).
We have already seen an example of this in section 2.7 and we will not go through it
again. Note however that this will not work in the special case = 0 when =
0
,
in this case we need a dierent substitution and
x(t) = t
_
Acos(
0
t) +Bsin(
0
t)
_
will work.
30
3. Two Dimensional Motion
We now consider motion in 2-dimensions. We shall use Cartesian co-ordinates (x, y)
and denote positions in 2-dimensional space by vectors. Vector quantities will be indicated
by bold-face letters and the position denoted by x. Using an orthonormal basis (i, j), with
i a unit vector in the x-direction and j a unit vector in the y-direction, so an arbitrary
point in the 2-dimensional plane is denoted by x = xi + yj. In general the position of a
moving particle will be a function of time, and x(t) traces out a curve in the 2-dimensional
plane whose points can be labelled by the values of t at the time when the particle is
at that point. The co-ordinates are then functions of time (x(t), y(t)), but i and j are
constant vectors, so the velocity is
v =
dx
dt
= xi + yj,
which is a vector tangent to the curve x(t) at the time t.
i
j
O
y
x
v
= x + y x i j
The acceleration of the particle is
x = xi + yj,
so Newtons second law for a particle of mass m is
F = F
x
i +F
y
j = m x = m( xi + yj),
where F
x
is the x-component and F
y
the y-component of the force. Equating components
gives two equations
F
x
= m x F
y
= m y.
31
If we know the explicit form of F
x
and F
y
in a given physical problem we can try to solve
for x(t) and y(t) and hence predict the motion of m.
3.1 Projectiles
As a rst example we consider projectiles in the absence of friction. Let x label
a horizontal direction and y the vertical direction. The gravitational force is vertically
downwards so we take F
y
= mg and F
x
= 0, giving
F = mg j = m x
or, in components,
m y = mg y = g, m x = 0 x = 0.
Taken individually we have already solved both of these equations, if the initial position
is given x(0) = x
0
, y(0) = y
0
and the initial velocity is x(0) = v
x,0
, y(0) = v
y,0
with x
0
,
y
0
, v
x,0
and v
y,0
given constants, then the subsequent motion is
x(t) = v
x,0
t +x
0
, y(t) =
1
2
gt
2
+v
y,0
t +y
0
.
These are the equations for a parabola, written in parametric form with parameter t. As
a concrete example, suppose m is thrown from the origin, x
0
= y
0
= 0 with speed v
0
at
angle to the horizontal at t = 0, then v
x,0
= v
0
cos and v
y,0
= v
0
sin , then
x(t) =
_
v
0
cos()t
_
i +
_

1
2
gt
2
+v
0
sin()t
_
j
and it hits the ground again at the time T when
y(T) = 0 T =
2v
0
sin
g
=
2v
y,0
g
during which time it has travelled a distance
x(T) = v
0
cos()T =
2v
2
0
cos sin
g
=
v
2
0
sin(2)
g
=
2v
x,0
v
y,0
g
.
If a hurler can hit a sliotar with a maximum speed v
0
, then he should hit it at an angle
=

2
= 45

above the horizontal, so that sin(2) = 1, if he wants to maximise the

distance that it will travel before it hits the ground again (this conclusion will be modied
when air friction is taken into account). The maximum height is achieved, for a xed v
0
and , when
dy
dt
= 0 gt +v
0
sin = 0 t =
v
0
sin
g
32
at which point
y
Max
=
1
2
g
_
v
0
sin
g
_
2
+v
0
sin()
_
v
0
sin
g
_
=
1
2
(v
0
sin )
2
g
=
v
2
y,0
2g
.
Again, for a xed v
0
, the best one can do is
y
Max
=
v
2
0
2g
,
when =

2
.
0
x
y

x
max
y
max
v
Now we shall consider the same problem when friction is included. As before suppose
the frictional force is such as to oppose the motion and is proportional to the velocity.
This can be represented in vector notation as
F
Friction
= cv = c x,
with c > 0 a positive constant. Adding this to the gravitational force Newtons second law
now reads
m x = mgj c x
or, in components
m x = c x, m y = mg c y.
33
Again, taken individually we have already solved equations like these.. Using the same
initial conditions as above, when air friction was ignored, set
x(0) = y(0) = 0 and x(0) = v
0
cos , y(0) = v
0
sin
we can use the results of section 2.2, namely (5), with v
0
replaced by v
0
cos and x
0
= 0,
to express the solution for x(t) as
x(t) =
mv
0
cos
c
_
1 e

ct
m
_
.
For y(t) we use the solution (13), with v
0
replaced by v
0
sin and y
0
= 0, to give
y(t) =
m
c
_
v
0
sin +
mg
c
__
1 e

ct
m
_

mg
c
t.
A plot of the trajectory, for xed v
0
and dierent is shown below. Note that the
curves are not symmetric about their maximum and the trajectory that extends farthest to
the right is not the line with =

4
, but rather has <

4
. The angle, , that maximises
the horizontal distance traveled depends on c, but it is always less than

4
for c > 0.
x
y
=5/16
=/8
=3/16
=/4
34
3.2 Central Forces
An important problem concerning motion in two dimensions is when there is a force
acting on a mass m which is always directed toward a xed point in the two dimensional
plane. For example consider a particle carrying an electric charge Q
1
moving in the electric
eld of another charge, Q
2
, which is xed in space inside a dielectric material, such as air
or water. Choosing the origin to coincide with the xed position of Q
2
the Coulomb force
on Q
1
is always directed toward or away from the origin, O, and is inversely proportional
to the square of the distance from Q
1
to O. For example, if Q
1
and Q
2
are the same sign
the force is repulsive and equals
F =
Q
1
Q
2
4
r
r
2
=
Q
1
Q
2
4
r
r
3
,
where r is the position of Q
1
relative to O and r =
r
r
is a unit vector in the same direction
as r. The constant goes under the fancy name of the electric permittivity of the medium
it determines the strength of the Coulomb force in the dielectric: a small value of
means that the Coulomb force is weak.
Another example is Newtons universal law of gravitation which states that the attrac-
tive gravitational force between two masses, m and M say, is again inversely proportional
to their separation and in the same direction as a line joining the masses,
F = GmM
r
r
2
= GmM
r
r
3
,
where G = 6.67 10
11
kg
1
m
3
s
2
is Newtons universal constant of gravitation and r is
the vector between the positions of m and M. If one of the masses, say M, is much larger
than the other then the larger mass can be considered to be at rest and we can take M to
be xed at the origin with r the position vector of m.
Both of these situations are called central force problems, because in each case the
force is directed to one central point, which can be chosen to be the origin. For central force
problems it is usually convenient to use two dimensional polar co-ordinates, (r, ), dened
as x = r cos , y = r sin . Thus r
2
= x
2
+y
2
and the radial co-ordinate, r =
_
x
2
+y
2
is
the distance of the point r from the origin while = tan
1
_
y
x
_
, with 0 < 2, is the
angle between the point r and the x-axis, is called the azimuthal angle of the point r.
35
O x
y

r
e
r
e

It is useful to dene two mutually orthogonal unit vectors, e

r
and e

with e
r
.e
r
=
e

.e

= 1 and e
r
.e

= 0. The rst, e
r
, is a unit vector in the same direction as r, so
e
r
= r =
r
r
and the second, e

, is a unit vector in the direction that r traces out when is

increased, keeping r constant, i.e. e

is a unit tangent vector at the point r, in the anti-

clockwise direction, to a circle of radius r centred on the origin. In terms of the Cartesian
basis, i and j, used before we see from the gure below that
e
r
= cos i + sin j, e

= sin i + cos j (39)

i = cos e
r
sin e

, j = sin e
r
+ cos e

.
e
r

j
i

For a moving point mass r(t) is a function of time, so r(t) and (t) are also functions
of time. An extra subtlety with polar-coordinates however is that e
r
and e

also depend
on time. Since i and j are constant vectors we have
e
r
=
d
dt
(cos i + sin j) =

(sini + cos j) =

e

=
d
dt
(sin i + cos j) =

(cos i sin j) =

e
r
.
Hence the velocity of m is
v =
dr
dt
=
d
dt
(re
r
) = r e
r
+r e
r
= r e
r
+r

e

. (40)
v
r
= r is the radial component of the velocity and v

= r

the angular component.
36
The acceleration is
a =
d
2
r
dt
2
= v = r e
r
+ r e
r
+
d
dt
(r

) e

+r

e

= r e
r
+ r

e

+ ( r

+r

) e

r
2

2
e
r
= ( r r

2
) e
r
+ (r

+ 2 r

) e

. (41)
Hence the acceleration has radial component a
r
= r r

2
and angular component a

=
r

+ 2 r

.
For a central force F is by denition in the radial direction, F = F e
r
with F the
magnitude of the force,

and Newtons second law immediately leads to the important

conclusion that the angular acceleration is zero,
F = ma F e
r
= m
_
( r r

2
) e
r
+ (r

+ 2 r

) e

_
.
Equating the co-ecients of the two basis vectors we conclude that
F = m( r r

2
), 0 = m(r

+ 2 r

).
In particular
d
dt
_
mr
2

_
= rm(r

+ 2 r

) = 0
so mr
2

is a constant, called the angular momentum of m. The momentum of m at any
time is a vector p = mv and the angular momentum is also a vector, dened to be
L = m(r v).
For motion conned to a two-dimensional plane m(r v) is a vector pointing perpendic-
ularly out of the plane. Dening k to be a unit vector in the z-direction, so k = i j, we
have
m(r v) = m(r e
r
) ( r e
r
+r

e

) = mr
2

(e
r
e

) = mr
2

k.
The angular momentum always points in the same direction and has magnitude L = mr
2

.
Sometimes it will be convenient to use l = L/m = r
2

, the angular momentum per unit
mass.
So, for a particle moving under a central force, the angular momentum is constant.
this is because the only thing that changes angular momentum is a torque and for a
force directed toward the origin the torque about the origin is zero. In general only the
combination l = r
2

is constant, not necessarily r or

independently. Thus if r decreases

must increase and vice versa.

Having established that angular momentum is conserved we now only have to think
about the radial part of the acceleration,
r r

2
= r r
_
l
2
r
4
_
= r
l
2
r
3
.

The plus sign is for a repulsive force, the minus sign for an attractive force.
37
Newtons second law then gives
F = ma F = m
_
r
l
2
r
3
_
.
This is true for any central force, but we shall now specialise to the important case where
F(r) depends only on the distance of m from the origin, as in Coulombs law or Newtons
universal law of gravitation, where F(r) 1/r
2
.
For Newtons universal law of gravitation, for example, with M xed at the origin
(which is a good approximation only if M >> m, eg. a planet going round the Sun or a
satellite orbiting the Earth),
F(r) =
GmM
r
2
= m
_
r
l
2
r
3
_
r =
GM
r
2
+
l
2
r
3
. (42)
It is remarkable that m drops out of the equation completely! In fact m plays two
completely dierent roles here: in F = ma it is the inertial mass of the moving particle
while in F =
GmM
r
2
it is the gravitational mass of the particle (analogous to electric charge
in Coulombs law). There is no a priori reason why the inertial mass and the gravitational
mass of a body should be the same, but all the experimental observations imply that they
are.

The experimental observation that inertial and gravitational are the same is one of
the cornerstone assumptions of Einsteins general theory of relativity, that describes the
dynamics of gravitational elds that change with time, but we shall not go into that here
we shall use the Newtonian description of gravity in which all gravitational elds are
static.
Now let us focus on solving (42). We proceed by analogy with a problem we have
already studied, particle motion in one dimension only. Consider a particle with unit mass
moving in one dimension and label the position of the particle by r(t), with r constrained
so that r > 0. Then the equation
r =
GM
r
2
+
l
2
r
3
is exactly the same as that of a unit mass particle moving under a force
F
Eff
=
GM
r
2
+
l
2
r
3
.
This is called the eective force for the original problem, because it is not the real force in
two dimensions, that only involved
GM
r
2
. Nevertheless we can use the same technique as
before to solve the one dimensional problem. Introduce an eective potential U
Eff
(r) by
F
Eff
=
dU
Eff
dr
so, for motion under gravity,
U
Eff
=
GM
r
+
l
2
2r
2
.

Strictly speaking all the experiments tell is that the inertial mass and the gravitational
mass are proportional to each other, we then chose a convention for G so that they are
equal.
38
The rst term is just the usual potential energy for a 1/r
2
force, but the second term
requires some explanation it is due to the rotation of m about the origin in two dimensions
and is related to centrifugal force. To understand the signicance lets rst turn o gravity
and ignore the rst term, we are now considering the motion of a particle in two dimensions
moving under no force whatsoever, so the motion should be in a straight line. Nevertheless,
unless the particle happens to be heading straight for the origin, it will have some non-zero
angular momentum l = r
2

= 0 and
U
Eff
=
l
2
2r
2
F
Eff
=
l
2
r
3
and, in the eective one dimensional motion, the particle experience a repulsive force from
the origin, proportional to 1/r
3
, called the centrifugal barrier. This force diverges as r 0
and the particle can never get to the origin but that is what we expect because it is not
moving towards the two-dimensional origin if l = 0. Picture the motion as below,
O
r
min
r
Eff
E
r
min
U (r)
in two dimensions the particle moves in a straight line, but when only r is considered the
particle comes in from large r, reaches a point of closest approach to the origin and then
recedes again. In the eective one dimensional problem the particle is repelled from the
origin by an inverse cube force. In the absence of the gravitational force the total energy
is just the knietic energy
E =
1
2
mv
2
=
1
2
m( r
2
+r
2

2
) =
1
2
m
_
r
2
+
l
2
r
2
_
,
and the point of closest approach to the origin, r
min
, is easily seen from the eective poten-
tial as being the point at which r = 0 where E =
m
2
l
2
r
2
min
=
1
2m
L
2
r
2
min
, hence the minimum
distance from the origin depends on both the energy E and the angular momentum L,
r
min
=
L

2mE
=
L
mv
.
When the angular momentum is zero the particle is heading straight for the origin and
will pass through O, so r
min
= 0.
39
Now return to the full problem with r =
GM
r
2
+
l
2
r
3
. We treat this as an eective one
dimensional problem for a particle of unit mass, with
U
Eff
=
GM
r
+
l
2
2r
2
.
Note that the potential energy is negative when r >
l
2
2GM
and has a minimum when
dU
Eff
dr
= 0, that is at
r =
l
2
GM
,
where U
Eff
=
1
2
_
GM
l
_
2
.
r
Eff
E
E
E
min

+
U (r)
The total energy per unit mass is conserved,
E =
r
2
2

GM
r
+
l
2
2r
2
(43)
and the behaviour of m is very dierent for E > 0 and E < 0. For E = E
+
> 0, m is
repelled from the origin and moves out to indenitely large r. For E = E

< 0, m is
trapped in a potential well and moves between a minimum and a maximum value of r. In
the limiting case of E = E
min
=
1
2
_
GM
l
_
2
, r is constant at r =
l
2
GM
.
40
It should not be forgotten that, if l = 0, the full two dimensional motion involves
rotation with

= l/r
2
= 0. So the fact that r is constant for E = E
min
=
1
2
_
GM
l
_
2
does not mean that M is not moving in two dimensions, it only means that

= l/r
2
=
(GM)
2
l
3
is constant, so m moves in a circle of radius
l
2
GM
with angular velocity
(GM)
2
l
3
. For
E
min
< E < 0 the particle orbits around the origin, moving in and out between two
extreme values of r this is called a bound orbit. For E 0 the particle will eventually
move o to arbitrarily large r in an unbound orbit.
We can go further and nd the geometrical shape of the orbit by solving equation (43)
for the unknown function r(t),
r
2
= 2E +
2GM
r

l
2
r
2
.
One way to proceed is to use l = r
2

to change the independent variable from t to and
nd r() instead of r(t). From the chain rule r =
dr
d
d
dt
=
l
r
2
dr
d
giving
l
2
r
4
_
dr
d
_
2
= 2E +
2GM
r

l
2
r
2
. (44)
A simplication arises from using u() = 1/r(), with
du
d
=
1
r
2
dr
d

_
du
d
_
2
=
1
r
4
_
dr
d
_
2
,
and (44) can be re-written
_
du
d
_
2
=
2E
l
2
+
2GMu
l
2
u
2
=
_
u
GM
l
2
_
2
+
2E
l
2
+
_
GM
l
2
_
2
. (45)
Dierentiating with respect to gives
_
du
d
__
d
2
u
d
2
_
=
_
du
d
__
u
GM
l
2
_

d
2
u
d
2
=
_
u
GM
l
2
_
.
This is nothing more than our old friend the harmonic oscillator equation! Shifting the
origin of u to u = u
GM
l
2
we have simply
d
2
u
d
2
= u
with the most general solution
u = A
0
cos(
0
) u = A
0
cos(
0
) +
GM
l
2
.
41
The constant A
0
can be related to the energy using (45),
_
du
d
_
2
= A
2
0
sin
2
(
0
) = A
2
0
cos
2
(
0
) +
2E
l
2
+
_
GM
l
2
_
2
A
2
0
=
2E
l
2
+
_
GM
l
2
_
2
.
Choosing A
0
> 0 we have
r() =
1
_
_
GM
l
2
_
2
+
2E
l
2
cos(
0
) +
GM
l
2
.
If E < 0, the numerator never vanishes and r() is nite for 0 < 2 giving a bound
orbit, as expected. If E =
1
2
G
2
M
2
l
2
, then r =
l
2
GM
is a constant and the orbit is a circle.
for
1
2
G
2
M
2
l
2
< E < 0, r oscillates between a minimum value
r
min
=
1
GM
l
2
+
_
_
GM
l
2
_
2
+
2E
l
2
when cos(
0
) = 1 and a maximum
r
max
=
1
GM
l
2

_
_
GM
l
2
_
2
+
2E
l
2
when cos(
0
) = 1, returning to the same value for + 2. The fact that
r() = r( + 2) means that m returns to exactly the same point in space after every
revolution around M and the orbit is said to be closed. Not all types of central force give
closed orbits, a 1/r
3
force does not, for example. Indeed the only two kinds of central
force that lead to closed orbits are inverse square, 1/r
2
, and quadratic r
2
. Without losing
any generality of the solution we can choose = 0 to be the angle at which r = r
min
, ie.

0
= 0, then we have the solution of the orbit equation as
r() =
l
2
GM
1
_
1 + cos()
_, (46)
where we have dened
:=
_
1 +
2El
2
G
2
M
2
.
For
1
2
G
2
M
2
l
2
< E < 0 we have 0 < < 1 and equation (46) is actually the equation for
an ellipse, with one focus centred on the origin and with eccentricity . We have thus
established that Keplers First Law of planetary motion, that the planets move in ellipses
with the Sun at one focus, follows from Newtons inverse square law of gravitation.
If = 1, then r is innite for = and (46) is the equation of a parabola, such as the
orbit of many irregular comets. If > 1 (E > 0) then r is innite when = cos
1
(1/),
which occurs at two dierent values of in the range < < and this corresponds to
42
a trajectory that comes in from innity and swings round M to recede back out to innity
in less than one complete revolution. Typical orbits are shown below:
0.5 0 -0.5 -1 -1.5
1
-2
0.5
-2.5
-0.5
0
-1
4
-1
2
0
0
-2
-4
-3 -2 -4 -10
-4
-8
2
0 -2 -4 -6
4
0
-2
=1
=1.2
=0.6
cos (5/6)
1
GM/(E)
l 2 /(E)
The planets all have orbits with small eccentricities, the largest is that of Mercury with
= 0.21. The next largest is Mars with = 0.093 which is dicult to distinguish from a
circle just by looking at it it requires careful measurement to see that it is not a perfect
circle.
3.3 Conservative Forces
We saw in section 2.3 that, for frictionless motion in one dimension for a force F(x)
that depends only on position, we can dene a potential energy U(x) such that the total
energy, kinetic plus potential, is conserved and the force is minus the derivative of the
potential, F(x) =
dU
dx
. The situation in two (or three) dimensions is a little more subtle,
if the components of a force, F
x
(x, y) and F
y
(x, y), depend only on position we cannot
conclude that the force is derivable from a potential. To see this suppose that F =
F
x
(x, y) i +F
y
(x, y) j is derivable from a potential, so
F
x
=
U
x
, F
y
=
U
y
.
Then

y
F
x
=
F
x
y
=

2
U
yx
,
x
F
y
=
F
y
x
=

2
U
xy
43
and, assuming that U is at least twice dierentiable,

x
F
y
=
y
F
x
, or
x
F
y

y
F
x
= 0 F = 0,
so the curl of F must vanish. This is a necessary condition on F for a potential to exist.
An example of a force depending only on position for which no potential exists is a central
force that depends on the direction,
F = F
r
(r, )e
r
,
with

F
r
= 0. Because the force is central there is no torque, F

= 0, but if the force were

derivable from a potential we would have F
r
(r, ) =
r
U(r, ) and

F
r
=

r
U = 0
requires

U = 0, which then implies that F

= 0, contrary to our assumption that the

force is central. Hence no potential exists for such a force (angular momentum is still
conserved though, since there is no torque).
If the force can be derived from a potential then energy is conserved. Suppose a
potential exists then dene the total energy to be
E =
m
2
v.v +U(x, y).
Its time derivative is
dE
dt
= m
dv
dt
.v +
dx
dt
U
x
+
dy
dt
U
y
= F.v +v.U = v.(F+U) = 0,
where U =
x
U i +
y
U j is the gradient of U(x, y).
The existence of a potential has an interesting consequence which is perhaps not
immediately obvious. Consider the work done, W
12
, by an external force F on a particle
m moving between two position r
1
and r
2
along a specied curve, C
12
. For an innitesimal
displacement dr at a point r of the curve the work done is F.dr and the total work done
on m in moving from r
1
to r
2
along the curve is
W
12
=
_
C
12
F.d r =
_
r
2
r
1
F.d r = m
_
r
2
r
1
d v
dt
.vdt =
m
2
_
r
2
r
1
d
dt
_
v
2
_
dt =
m
2
(v
2
2
v
2
1
),
where we have used d r =
d r
dt
dt = vdt and v
2
, v
1
are the speeds at the end points r
2
and
r
1
respectively. The last expression here shows that the work done is just the dierence
between the kinetic energies at r
2
and that at r
1
and, in general, this will depend on the
path taken between r
1
and r
2
. Now consider two dierent paths C
12
and C

12
between r
1
and r
2
and suppose m rst travels from r
1
to r
2
along C
12
and then back from r
2
to r
1
along C

12
The dierence
_
C
12
F.d r
_
C

12
F.d r =
_
C
F.d r
44
is an integral around the closed loop consisting of C
12
followed by C

12
, which we shall
call C, and we remind ourselves that C is a closed curve by using the symbol
_
C
( )d r
for the integral. Now let us assume that F is such that F = 0, then we can use Stokes
theorem to conclude
_
C
F.d r =
_
S
(F).d S = 0
where S is the area enclosed by C and dS is a vector normal to the surface whose magnitude
is an innitesimal area element at the point interior to S where the integrand is being
evaluated. This means that
F = 0
_
C
12
F.d r =
_
C

12
F.d r,
i.e. the dierence in kinetic energy between r
1
and r
2
is independent of the path taken
between them.
r
2
r
C
12
1
r
2
r
C
12
2
r
1
r
12
C
12
C
C
C
1
S
Another way of seeing this is to use conservation of energy directly,
E =
m
2
v
2
1
+U(r
1
) =
m
2
v
2
2
+U(r
2
)
m
2
v
2
2

m
2
v
2
1
= U(r
2
) U(r
1
),
so the dierence of the kinetic energies depends only on the points r
2
and r
1
, not on any
path between them. Indeed
_
C
12
F.d r =
_
C
12
U(r).d r = [U(r)]
r
2
r
1
= U(r
2
) U(r
1
).
So an alternative way of characterising a conservative force is to say that the work
done in transporting a particle under the inuence of a conservative force around any
closed loop is always zero.
45
3.4 Rotating Reference Frames
Sometimes it is useful to analyse natural phenomena from the point of view of a
reference frame rotating with constant angular velocity, for example the Earth is rotating
once on its axis every 24 hours so a laboratory xed to the surface of the Earth is rotating.
A rotating reference frame is not an inertial reference frame and this aects the way that
the dynamics is described. We can still use Newtons laws of motion in a rotating reference
frame, we just have to be careful with the analysis. Restricting to two dimensional motion
for the moment we can describe vectors in a reference frame rotated about a xed point
O by an amount by using the basis
i

= cos i + sin j, j

= sin i + cos j,
where i and j are xed unit vectors, associated with Cartesian co-ordinates (x, y) in an
inertial reference frame.
O
i
j
i
j

x
y
x

y
r
We can use either (i, j) or (i

, j

) as a basis for two dimensional vectors, eg. for the position

vector
r = x i +y j
r

= x

+y

.
In fact
x

= r cos( ) = x cos +y sin , y

= r sin( ) = x sin +y cos ,

and r = r

, the position of a point is described by the same vector in both reference frames,
though the co-ordinates are dierent because the bases are dierent.
If is a constant then (x

, y

) are Cartesian co-ordinates in an inertial reference frame

and (i

, j

) is a basis in this inertial reference frame. If is not a constant but a function

of time, (t), then (x

, y

) are not co-ordinates for an inertial reference frame, nevertheless

they are still perfectly good co-ordinates, they are just not inertial. We can still use them,
but Newtons second law will look a little unfamiliar using these co-ordinates. For example
suppose (x

, y

) are co-ordinates in a reference frame that rotates with constant angular

46
velocity = , so is a linear function of t. We can always choose the x

-direction so
that (0) = 0 and then (t) = t. Now we have
i

= cos(t) i + sin(t) j, j

= sin(t) i + cos(t) j,
x

= x cos(t) +y sin(t), y

= x sin(t) +y cos(t). (47)

We shall refer to (x, y) as static co-ordinates and (x

, y

) as rotating co-ordinates.
O
i
j
t
i
j
The rotating basis vectors i

and j

are not constant, we nd

di
dt

= sin(t) i+ cos(t) j = j

,
dj
dt

= cos(t) i sin(t) j = i

. (48)
Now consider a particle of mass m following a trajectory r(t), but watched by an
observer in the rotating reference frame using co-ordinates (x

, y

),
r

(t) = x

(t) i

+y

(t) j

= x(t) i +y(t) j = r(t). (49).

The rotating observer, using co-ordinates (x

, y

), would measure the velocity of m to be

v

= x

+ y

(50)
and the acceleration to be
a

= x

+ y

.
In a rotating reference frame we expect centrifugal forces. For example if m is initially
at rest in the rotating frame, at x

(0) = x
0
and y

(0) = 0 say with x

(0) = y

(0) = 0, then
it must be moving in the static reference frame. Its initial position in the static reference
frame is x(0) = x
0
, y(0) = 0 and it is moving with speed x
0
in the positive y-direction.
If there are no forces on m in the static frame then m continues to move with constant
47
velocity in that frame, v = x
0
j, and its subsequent position in the static frame is x = x
0
,
y = x
0
t. Its motion in the rotating frame is then given by (47) to be
x

= x
0
cos(t) +tx
0
sin(t), y

= x
0
sin(t) +tx
0
cos(t),
x

=
2
tx
0
cos(t), y

=
2
tx
0
sin(t),
x

=
3
tx
0
sin(t) +
2
x
0
cos(t), y

=
3
tx
0
cos(t) +
2
x
0
sin(t).
Thus even though m experiences no forces in the static reference frame, there is an ac-
celeration in the rotating reference frame. For example at t = 0, when v

= 0, we see
that
x

=
2
x
0
, y

= 0 a

(0) =
2
x
0
i

=
2
r

(0).
There is an apparent force in the rotating reference frame, the centrifugal force, which at
t = 0 is F

Cent
= m
2
r

, though the story gets a little more complicated when v

= 0.
Returning to the case of a general motion we have, using (48) and (49), since r = r

r = r

= x

+x

+ y

= ( x

)i

+ ( y

+x

)j

= v

(y

),
r = r

= ( x

)i

+ ( x

)j

+ ( y

+ x

)j

( y

+x

)i

(51)
= ( x

2 y

2
x

)i

+ ( y

+ 2 x

2
y

)j

= a

2
r

2( y

),
To interpret this suppose that there are no forces on m in the static frame, then a = r = 0.
Now r = r

, so r

= 0 and
ma

= m
2
r

+ 2m( y

) (52).
The rst term on the right hand side is the centrifugal force due to the rotation of the
reference frame, but what is the meaning of the second term? To interpret the last term we
rst write it in a more concise form. Note that the denition of angular velocity requires
specifying not only the magnitude of the velocity, , but also the axis of rotation, which in
the two dimensional examples considered here is the axis perpendicular to the (x, y)-plane,
k = i j, or equivalently the axis perpendicular to the (x

, y

)-plane since k = i

.
Angular velocity is actually a vector which here is = k. Now, using (50),
v

= ( x

+ y

) k = ( x

+ y

)
and so (52) can be written as
ma

=
2
r

+ 2m(v

).
The last term here, 2m(v

) is called the Coriolis force. Like centrifugal force it is

not a real physical force and exists only in rotating reference frames as a consequence of
48
the rotation it is an example of a pseudo-force. Nevertheless it has real physical conse-
quences, it is responsible for the spiral shapes of tropical storms, which spiral anti-clockwise
in the northern hemisphere and clockwise in the southern hemisphere. To understand this
imagine a circular area of low pressure surrounded by a region of higher pressure. A wind
will blow and air will start to ow radially in towards the centre of the low pressure, with
velocity v

as measured relative to the surface of the Earth. But the Earth is rotating in an
anti-clockwise direction about an axis from the south pole to the north pole, so points
out of the Earths surface in the northern hemisphere and into the surface in the southern
hemisphere. In the picture below, the wind starts to move radially with velocity v

0
, but
as it moves inwards its tangential velocity is too high at smaller radii for the motion to
remain radial it moves o to one side and misses the centre. This is due to the Coriolis
acceleration, which is initially 2(v

0
). The paths followed are the curved trajectories.

v
vx
v
0
0

4. Three Dimensional Motion

To describe motion in three dimensions we need a basis of three vectors. We can use a
basis appropriate to Cartesian co-ordinates (x, y, z), with unit vectors i, j and k in the x,
y and z directions respectively, with i j = k, but for some problems it is more convenient
to use an orthonormal basis adapted to spherical polar co-ordinates (r, , ). These are
dened as shown in the diagram below. In terms of Cartesians
x = r cos sin , y = r sin sin, z = r cos
and
e
r
= sin cos i + sin sin j + cos k,
e

= cos cos i + cos sin j sin k, (53)

e

= sin i + cos j.
49
Some calculation involving vector products shows that
e
r
e

= e

,
e

= e
r
,
e

e
r
= e

.
The position of a particle in three dimensions, relative to the origin O, is then
r = x i +y j +z k = r e
r
.
To calculate the velocity and the acceleration in a basis adapted to polar co-ordinates
we must bear in mind that e
r
, e

and e

are not constant vectors, they depend on time

because they depend on the position r which is a function of time.
e
r
e

z
r
r cos( )

rsin( )

O
k
e
e

y
i
j
x
Orthogonal vectors associated with polar co-ordinates. The direction e

associated with a point r is always

perpendicular to k and is shown twice, at the point r itself and also in the xy plane.
Using (53) gives
e
r
=

e

+

sin e

,
e

e
r
+

cos e

,
e

(sin e
r
+ cos e

),
and
r = r e
r
+r e
r
= re
r
+r

+r

sin e

, (54)
r = ( r r

2
r

2
sin
2
)e
r
+ (2 r

+r

2
sin cos )e

(55)
+ (2 r

sin + 2r

cos +r

sin )e

.
50
As an example consider a central force problem for a particle of mass m moving under
the inuence of a force F = F(r, , ) e
r
that is always directed toward the origin in three
dimensions,
F(r, , ) e
r
= mr.
In particular F has no component in the e

and e

directions and so there is no acceleration

in these directions either and (55) gives
2 r

+r

2
sin cos = 0 2 r

sin + 2r

cos +r

sin = 0. (56)
The second of these equations implies that
d
dt
_
r
2

sin
2

_
= 0 l := r
2

sin
2

is a constant, so we can eliminate

from the rst using

2
=
l
2
r
4
sin
4

to show that
d
dt
_
r
2

_
= r
2

2
sin cos =
l
2
cos
r
2
sin
3

.
The interpretation of these equations is the following: r sin is the distance of the mass
m from the z-axis, so l is the angular momentum per unit mass about the z-axis this
is a constant for a central force, since there is no torque. Referring to the diagram on
the previous page, r
2

is the angular momentum per unit mass about the e

-axis, but in
general this is not a constant if e

is moving (that is if l = 0).

A solution of equations (56) is always given by setting = /2, then it is consistent
to have stay at this value, with

= 0, since cos = 0 and now l = r
2

is the total
angular momentum per unit mass. The motion then occurs entirely in the (x, y)-plane and
equations (54) and (55) reduce to
r = r e
r
+r

, e

= ( r r

2
) e
r
+ (r

+ 2 r

) e

which reproduces (40) and (41). Hence the central force problem for a single particle can
always be reduced to the two-dimensional problem that we have already studied.
Angular momentum is actually a vector of course and another way to derive this result
is to use vector notation. The linear momentum of m is dened to be the vector
p = mv = m
dr
dt
and the angular momentum about O is
L = r p = m(r v) = m
_
r
dr
dt
_
.
51
Note that L is always at right angles to the plane dened by r and v at any moment of
time. The rate of change of angular momentum is also a vector
dL
dt
= m
_
dr
dt

dr
dt
_
+m
_
r
d
2
r
dt
2
_
= 0 +r
_
m
d
2
r
dt
2
_
= r F.
The combination
:= r F
is called the torque, it represents a force acting on m that tends to twist it about the
origin. For a central force F is parallel to r, the torque vanishes, and hence the angular
momentum L is a constant vector. Since L is perpendicular to the plane in which r and
v lie and L is constant, r and v must always lie in the same two dimensional plane for a
central force. The initial position and velocity of m specify a two dimensional plane and we
are free to choose our axes so that this plane is the (x, y)-plane, with L in the z-direction.
The subsequent motion of m is always conned to the (x, y)-plane and we can treat the
whole problem as two dimensional. In particular the analysis of the Kepler problem and
planetary orbits that was presented in section 3.2 was completely general and applies to
the full three dimensional situation.
5. Systems of Particles
Consider a system of N particles, labelled by an index i = 1, , N, each with a
constant mass m
i
and moving in three dimensions with positions r
i
. The velocity of each
particle is v
i
= r
i
and the momentum p
i
= m
i
v
i
. Newtons second law gives
F
i
= p
i
= m
i
v
i
= m
i
r
i
where the force on the i-th particle is F
i
. In general F
i
can be split into two parts there
could be a force external to the system of particles acting on each particle, which we shall
denote by F
(e)
i
(such as the force of gravity, for example), but there could also be forces
between the particles. If particle j exerts a force F
ji
on particle i, then the total force on
particle i will be
F
i
= F
(e)
i
+

j=i
F
ji
= m
i
r
i
,
where the sum extends over all the particles in the system except for particle i, because it
does not exert a force on itself.
Note that, if F
ji
is the force that particle j exerts on particle i, then Newtons third
law implies that the force that particle i exerts on particle j, F
ij
, should be equal and
opposite, ie.. F
ji
= F
ij
.
The total force on the system of particles is then
F
(e)
=
N

i=1
F
i
=
N

i=1
_
F
(e)
i
+

j=i
F
ji
_
=
N

i=1
F
(e)
i
+
N

i=1
_

j=i
F
ji
_
=
N

i=1
F
(e)
i
,
52
where in the last equality we have used the fact that the double sum always contains both
F
ij
+ F
ji
for each pair of particles and these just cancel. So Newtons third law tells us
that the forces between all the particles cancel out and the total force is just the sum of
the external forces only.
5.1 Centre of mass motion
To analyse the motion of the particles in more detail it is useful to consider the position
of each particle relative to the centre of mass of the system. The centre of mass of the
system is the point
R :=

N
i=1
m
i
r
i

N
i=1
m
i
=

N
i=1
m
i
r
i
M
,
where M =

N
i=1
m
i
is the total mass of the system of particles. Let r

i
be the position of
particle i relative to the centre of mass,
r
i
= R+r

i
.
Then

R =

N
i=1
m
i
r
i
M
=

N
i=1
F
(e)
i
M
=
F
(e)
M
,
so
F
(e)
=
N

i=1
F
(e)
i
= M

R (57)
and the centre of mass of the entire system moves in the same way as a point particle with
mass M =

N
i=1
m
i
moving under the inuence of a force F
(e)
=

N
i=1
F
(e)
i
. In terms of
the total momentum of the system
P =
N

i=1
m
i
r
i
=
d
dt
_
N

i=1
m
i
r
i
_
=
d
dt
_
MR
_
= M

R
so (57) can be written
F
(e)
=

P,
since M is constant. Note the total momentum P =

N
i=1
m
i
r
i
=

N
i=1
p
i
is just the sum
of the individual momenta of each particle, p
i
= m
i
r
i
.
In particular if the sum of the external forces vanishes, F
(e)
=

N
i=1
F
(e)
i
= 0 then
the centre of mass momentum P is constant, as is the velocity of the centre of mass
V =

R =
1
M
P.
5.2 Angular momentum
Now consider the total angular momentum of the system of particles about the origin,
O. The angular momentum of particle i about O is r
i
p
i
and the total angular momentum,
L, of the system about O is the sum of the individual angular momenta of each particle,
L =
N

i=1
r
i
p
i
.
53
This can be decomposed into the angular momentum of the system about is own centre
of mass and the angular momentum of the centre of mass about O. Using r
i
= R+r

i
we
have

N
i=1
m
i
r

i
= 0 since, by denition MR =

N
i=1
m
i
r
i
,
MR =
N

i=1
m
i
r
i
=
N

i=1
m
i
R+
N

i=1
m
i
r

i
= MR
N

i=1
m
i
r

i
= 0.
Now
r
i
= R+r

i
v
i
= r
i
=

R+ r

i
= V+v

i
,
where v

i
= r

i
is the velocity of particle i relative to the centre of mass of the system.
Hence
L =
N

i=1
m
i
_
r
i
v
i
_
=
N

i=1
m
i
_
R+r

i
_

_
V+v

i
_
=
N

i=1
m
i
(RV) +
N

i=1
m
i
_
Rv

i
_
+
N

i=1
m
i
_
r

i
V
_
+
N

i=1
m
i
_
r

i
v

i
_
=
N

i=1
m
i
(RV) +
N

i=1
m
i
_
r

i
v

i
_
= M(RV) +
N

i=1
m
i
_
r

i
v

i
_
,
where in the last equation we have used the fact that

N
i=1
m
i
rp
i
= 0, which implies

N
i=1
m
i
v

i
= 0. Now the momentum of particle i relative to the centre of mass is p

i
= m
i
v

i
and the total momentum of the system is P =

N
i=1
p
i
so
L = (RP) +
N

i=1
_
r

i
p

i
_
.
Thus the total angular momentum consists of two separate pieces
L = (RP) +
N

i=1
_
r

i
p

i
_
:= (RP) +L

, (58)
where the rst term is the angular momentum of the centre of mass of the system about
the origin O and the second term, L

:=

N
i=1
_
r

i
p

i
_
, is the angular momentum of the
system of particles about its own centre of mass.
The rate of change of the total angular momentum is

L =
d
dt
_
N

i=1
r
i
p
i
_
=
N

i=1
d
dt
_
r
i
p
i
_
=
N

i=1
_
( r
i
p
i
) + (r
i
p
i
)
_
=
N

i=1
(r
i
F
i
),
54
since p
i
is parallel to r
i
so r
i
p
i
= 0. Breaking F
i
up into its two parts,
F
i
= F
(e)
i
+

j=i
F
ji

N

i=1
(r
i
F
i
) =
N

i=1
_
r
i
F
(e)
i
_
+
N

i=1
_
r
i

j=i
F
ji
_
_
the sum

N
i=1
_
r
i

_
j=i
F
ji
_
_
consists of sums of pairs
_
r
i
F
ji
_
+
_
r
j
F
ij
_
=
_
r
i
r
j
_
F
ji
,
since Newtons third law implies that F
ij
= F
ji
. It is often (but not always)* the case
that the force between any two particles is in the same direction as a straight line between
the particles, so F
ji
r
i
r
j
, in which case
_
r
i
r
j
_
F
ji
= 0. For example this would be
true for electrostatic or gravitational forces between the particles. When this is the case
(and only then) we have

L =
N

i=1
_
r
i
F
(e)
i
_
=
N

i=1

(e)
i
,
where
(e)
i
= r
i
F
(e)
i
is the external torque about the origin on particle i. The total
external torque about O,
(e)
, on the whole system of particles is the sum of the torques
on the individual particles,

(e)
=
N

i=1

(e)
i
,
so

L =
(e)
(59)
and the rate of change of the total angular momentum about the origin is equal to the
total torque about O on the system of particles. In particular if the total torque vanishes
then the total angular momentum is a constant.
Equations (57), (58) and (59) are very useful because they allow the motion of the
system to be analysed in two parts, the motion of the centre of mass and the motion of the
particles relative to the centre of mass. Using (57) and (59) we can deal with the centre
of mass motion as though we had a single particle of mass M at the point R experiencing
a force F
(e)
and a torque
(e)
about O.
5.3 Energy
The total kinetic energy of the system, which we shall denote by T, is the sum of the
individual kinetic energies of each particle,
T =
1
2
N

i=1
m
i
v
2
i
=
1
2
N

i=1
m
i
_
V +v

i
_
.
_
V+v

i
_
* An exception is magnetic forces between moving charged particles.
55
=
1
2
N

i=1
m
i
V.V+
N

i=1
m
i
V.v

i
+
1
2
N

i=1
m
i
v

i
.v

i
=
1
2
M(V.V) +
1
2
N

i=1
m
i
v

i
.v

i
, (60)
where we have again used

N
i=1
m
i
v

i
= 0. The rst term in the last equation here is the
kinetic energy of the centre of mass and the second term is the total kinetic energy of the
individual particles relative to the centre of mass, which can be thought of as the internal
kinetic energy of the system viewed as a whole.
If the external forces are conservative then there is a potential energy function U
(e)
i
(r
i
)
for each F
(e)
i
in Cartesian components, with r
i
= x
i
i +y
i
j +z
i
k,
F
(e)
i,x
=
U
(e)
i
(r
i
)
x
i
, F
(e)
i,y
=
U
(e)
i
(r
i
)
y
i
, F
(e)
i,z
=
U
(e)
i
(r
i
)
z
i
, i = 1, . . . , N.,
where x
i
, y
i
and z
i
are the Cartesian co-ordinates of particle i.
If the internal forces are conservative then there is a potential energy U
ij
(r
i
, r
j
) asso-
ciated with every pair (i, j) of particles,
F
ji,x
=
U
ij
x
i
, F
ji,y
=
U
ij
y
i
, F
ji,z
=
U
ij
z
i
.
It is the same potential U
ij
(r
i
, r
j
) that gives rise to both F
ji
and F
ij
= F
ji
, so it must
be the case that
U
ij
x
i
=
U
ij
x
j
,
U
ij
y
i
=
U
ij
y
j
,
U
ij
z
i
=
U
ij
z
j
,
which implies that U
ij
(r
i
, r
j
) is really only a function of r
i
r
j
and not of r
i
and r
j
separately, and we express this by writing U
ij
(r
i
r
j
). Note that, since r
i
r
j
= r

i
r

j
,
U
ij
(r
i
r
j
) = U
ij
(r

i
r

j
) and the internal potential energies are independent of the
position of the centre of mass R.
The total potential energy of the system is the sum of the external potential energies
U
(e)
i
(r
i
) and the internal potential energies U
ij
(r
i
r
j
),
U =
N

i=1
U
(e)
i
+

(i,j)
U
ij
(61)
where the second sum is over all possible pairs (i, j).
5.4 Rigid Body Motion
If the system of particles constitutes a rigid body, such as a rock or crystal, then the
distances between all the particle pairs |r

i
r

j
| are xed, as is the distance of particle i
56
from the centre of mass, r

i
= |r

i
|. Our analysis will be based on equation (57), (58) and
(59)
F
(e)
= M

R,
(e)
=

L, L = M(RV) +L

(62)
with L

N
i=1
r

i
p

i
the angular momentum about the centre of mass, R.
For simplicity we shall consider a two dimensional rigid body, a solid at object like
a at piece of metal. Choose the object to lie in the (x, y)-plane. The conguration of the
object is completely specied by giving the position of the centre of mass and one angle to
determine its orientation relative to the (x, y)-axes. For the latter we choose a xed line
in the body passing through the centre of mass, eg. a line parallel to the x-axis at a given
time, then this line will rotate if the body is rotating and the rotation angle , which will
be a function of time in general, determines the orientation of the body at any later time.
Denoting the centre of mass by R = X i +Y j the conguration of the body at any time
t is completely specied if we know X(t), Y (t) and (t).
x
y
O
R
O
O

Since the body is rigid every particle rotates about the centre of mass, which we shall
denote by O

, with the same angular velocity. In an innitesimal time interval, t, we have

=

t, |r

i
| = r

i
= r

t v

i
= |v

i
| = r

,
and v

i
.r

i
= 0. The angular velocity of particle i about O

is dened to be the vector

=
v

i
r

i
k =

k,
in the z-direction if

> 0 and in the z-direction if

< 0. The three vectors r

i
, v

i
and
are mutually orthogonal for a rigid body and, using vector product notation,
v

i
= r

i
. (63)
r
r
r

i
i
i
(t)
(t+ t)
O

57
The angular momentum about the centre of mass is
L

=
N

i=1
r

i
p

i
=
N

i=1
m
i
_
r

i
v

i
_
=
N

i=1
m
i
_

_
r

i
_
2

_
r

i
.
_
r

i
_
=
N

i=1
m
i
_
r

i
_
2
.
The constant
I :=
N

i=1
m
i
_
r

i
_
2
is a property of the body and is called the moment of inertia of M.* The equations (62)
governing the motion now read
F
(e)
= M

R,
(e)
=

L, L = M(RV) +I. (64)
It should be emphasised that the moment of inertia depends on the point about which
it is calculated. If we calculate the moment of inertia about the origin O,
I
O
=
N

i=1
m
i
(r
i
.r
i
) =
N

i=1
m
i
(R+r

i
).(R+r

i
)
=
N

i=1
m
i
_
R
2
+ 2R.r

i
+ (r

i
)
2
_
= MR
2
+ 2R.
_
N

i=1
m
i
r

i
_
+I
= MR
2
+I,
hence
I
O
= MR
2
+I (65)
a result known as the parallel axes theorem.
The total energy of a rigid body follows from (60) and (63), with
v

i
.v

i
= ( r

i
).( r

i
) =
2
(r

i
)
2

i=1
m
i
v

i
.v

i
=
2
N

i=1
m
i
(r

i
)
2
=
2
I.
* For a very large number of particles we can replace the discrete sum with a two
dimensional integral,

N
i=1
m
i

_
(r

)dS

, where (r

) is the mass per unit area at the

point r

and dS

is an innitesimal area element. For example, using Cartesian co-ordinates

I =
_
Area of body
(x

, y

)dx

dy

.
58
The total energy is therefore
E =
M
2
V.V+
I
2
., (66)
where the second term is the energy due to the rotation of the rigid body about its centre
of mass, its rotational energy. The total energy is therefore sum of the kinetic energy of
the centre of mass plus the rotational energy around the centre of mass.
5.5 The Compound Pendulum
As an example of the use of these equations we shall consider the motion of a compound
pendulum, that is a pendulum that is not just a point mass on the end of a taut string or
light rod but one that has a more general mass distribution. For simplicity we stay in two
dimensions and consider a pendulum that consists of a at shape pinned at a xed point
O about which it is free to rotate, without friction, in a vertical plane the same plane
as the object itself lies. The total force on the object is the force due to gravity plus the
force due to the pin that keeps O xed. The former acts on every point of the body while
the latter only acts directly on the point O but its eect will be transmitted to other parts
of the body because it is rigid. Denoting the angle between the vertical and the centre of
mass by , then the value of at any time uniquely determines the conguration of the
whole system at that time and our task is to determine (t) as a function of time.

R
r
O
m g
Mgj
j
i
r
i
i
F
O
Calculating

R directly using (57) is tricky because we would need to know the force
acting on the hinge, F
O
, to determine F
(e)
and F
O
changes as the pendulum swings.
Since we dont yet know F
O
, it is easier to use (58) and (59). For almost all the parti-
cles the only external force on particle i is the force of gravity acting vertically downwards,
F
(e)
i
= m
i
gj (the only exception to this is the point O, where the force F
O
acts). Hence
the total torque about O is

(e)
i
= r
i
F
(e)
i
= m
i
g(r
i
j)
59
(F
O
does not contribute to the torque about O because it acts precisely at the point O).
So the total external torque is

(e)
= g
N

i=1
m
i
_
r
i
j
_
= g
N

i=1
m
i
_
_
R+r

i
_
j
_
= g
_
N

i=1
m
i
_
(Rj) g
N

i=1
_
m
i
r

i
_
j
= gM(Rj),
since

N
i=1
_
m
i
r

i
_
= 0 from the denition of the centre of mass. This means that the total
external torque acts as though it were acting on a point mass M at the centre of mass R
of the body and (59) gives

(e)
= gM(Rj) =

L.
Now
L = M(RV) +I
and
V = R RV = R( R) = R
2
(R.)R = R
2
,
as R. = 0, so
L = (MR
2
+I) = I
O
,
where I
O
is the moment of inertia about the pivot O and we have used the parallel axis
theorem (66).
We now have

(e)
= Mg(Rj) =
dL
dt
= I
O
.
So
Mg(j R) = I
O
= I
O

k.
Now since
R = X i +Y j = R(sin i cos j)
this is
I
O

k = MgR(j i) sin = (MgRsin ) k

=
_
MgR
I
O
_
sin .
For small oscillations, << 1 and sin so

=
_
MgR
I
O
_

60
and

=
2
0
with
2
0
=
MgR
I
O
. This is again the harmonic oscillator equation with general
solution
(t) = A
0
cos(
0
t
0
).
The natural frequency of the pendulum is
0
=
_
MgR
I
O
. Note that for a simple pendulum,
with all the mass M concentrated at the point R, I = 0, I
O
= MR
2
and
0
reduces to
_
g
R
which is a result with which you should be familiar the frequency of a simple pendulum
is independent of the mass and inversely proportional to the square root of the length. We
see here that a compound pendulum yields a frequency that depends on how the mass is
distributed.
6. Lagrangian Formulation of Mechanics
6.1 Constrained Systems
A rigid body is an example of a constrained system. In general a system of N particles
in three dimensions requires 3N co-ordinates to specify its conguration uniquely, for
example (x
i
, y
i
, z
i
), i = 1, . . . , N are 3N co-ordinates. But for a rigid body these are
not all independent degrees of freedom as there are constraints between the co-ordinates,
|r
i
r
j
|
2
= const. In general if there are k independent constraints among the 3N co-
ordinates there are only 3N k independent degrees of freedom in the problem.
Here are some examples of constrained systems:
1) Two point masses, m
1
and m
2
, xed at the end of a light, rigid rod of length l, free
to move in three dimensions . r
1
and r
2
involve 6 Cartesian co-ordinates, but there
is one constraint, |r
1
r
2
| = l, so there are only 5 degrees of freedom in the problem.
For example we could use the co-ordinates of the centre of mass X =
m
1
x
1
+m
2
x
2
m
1
+m
2
,
Y =
m
1
y
1
+m
2
y
2
m
1
+m
2
and Z =
m
1
z
1
+m
2
z
2
m
1
+m
2
, together with two angles to specify the direction
in which the rod lies as our 5 degrees of freedom.
m m
2 1
l
If the rod and masses moved in two dimensions there would only be 3 degrees of
freedom, not 5. Two Cartesian co-ordinates for the centre of mass and one angle to
specify the orientation of the rod.
2) A bead of mass m slides on a frictionless wire in a two dimensional plane. The
Cartesian co-ordinates of the bead are (x, y), but they are not independent. If the
wire is bent into a shape specied by some function y = f(x), then there is only one
degree of freedom in the problem. We could use x as the degree of freedom or we could
use 2-dimensional polar co-ordinates with x = r cos , y = r sin and choose as the
independent degree of freedom, with r given by r() =
_
x
2
+y
2
=
_
x
2
+f
2
(x).
61
r

In both these examples the 3N k degrees of freedom that we use to describe the
motion need not be simple Cartesian co-ordinates, they are more general quantities that
are called generalised co-ordinates. We shall denote generalised co-ordinates by q

, =
1, . . . , 3N k in three dimensions. ( = 1, . . . , 2N k in two dimensions)*
Let us consider example 2 above in a little more detail. The wire is assumed to be
frictionless and this means that there are no frictional forces opposing the motion of the
bead, nevertheless there are still forces acting on the bead, the forces of constraint that
force it to follow the wire. In the absence of friction the constraint forces on the bead are
always normal to the wire, at the point where the bead is at any given time. Suppose
the equation describing the shape of the wire is y = f(x), then under an innitesimal
displacement (x, y) of the bead, y =
df(x)
dx
x = f

x.
t
n
x
y

F
n
v
The direction tangent to the wire at this point is x = x i + y j and the unit vector
tangent to the wire is
t =
x i +y j
_
(x)
2
+ (y)
2
=
i +
y
x
j
_
1 +
_
y
x
_
2

x0
i +f

j
_
1 + (f

)
2
.
* Unconstrained motion in two dimensions can be thought of as motion in three dimen-
sions with k = N constraints, 2N = 3N N.
62
The unit normal to the wire, n, is at right-angles to this
n =
_
f

i + j
_
1 + (f

)
2
_
,
where we choose the sign so that n is the same direction as F
n
. When the constraining
force is normal to the wire F
n
.t = 0, or
F
n
.x = 0. (67)
Of course F
n
.x is just the work done on the bead by the constraining force when the bead
moves through an innitesimal displacement x so this is telling us that the constraining
forces do no work.
The velocity of the bead is
v = x i + y j = x(i +f

j) =
_
x
_
1 + (f

)
2
_
t,
where we have used y = xf

. The constraint force is in the n-direction

F
n
= F
n
n =
F
n
_
1 + (f

)
2
(f

i + j), (68)
where F
n
is the magnitude of the constraining force.
Suppose the wire is in a vertical plane, with x along the horizontal axis and y up the
vertical axis. Then the gravitational force on m is mg j and Newtons second law reads
m x = mg j +F
n
or, in components,
x =
F
n
m
f

_
1 + (f

)
2
, y = g
F
n
m
1
_
1 + (f

)
2
. (69)
Obviously we need to know F
n
in order to solve these equations, but they are not inde-
pendent because of the constraint. Since y = f(x) the chain rule gives
y =
df
dx
x = f

x, and y = f

x
2
+f

x,
and x =
F
n
m
f

1+(f

)
2
then gives
y = f

x
2
+f

x = f

x
2

F
n
m
(f

)
2
_
1 + (f

)
2
= g
F
n
m
1
_
1 + (f

)
2
f

x
2
= g
F
n
m
_
1 + (f

)
2

F
n
m
=
f

x
2
+g
_
1 + (f

)
2
.
63
We can eliminate F
n
from the x equation in (69) to give
x
_
1 + (f

)
2
_
= f

_
f

x
2
+g
_
, (70)
but this is still going to be a very hard equation to solve for anything except the simplest
functions f(x).
Notice that the constraint force depends on x, but it is still a conservative force, it
does no work on the bead (67). Lets check that energy is conserved. The kinetic energy,
T, is
T =
m
2
_
x
2
+ y
2
_
=
m
2
x
2
_
1 + (f

)
2
_
. (71)
Choose the zero of potential energy to be at y = 0, then the potential energy due to the
gravitational force is U(x) = mgy = mgf(x) and the total energy is
E = T +U =
m
2
x
2
_
1 + (f

)
2
_
+mgf(x)

dE
dt
= m x x
_
1 + (f

)
2
_
+m x
2
_
f

x
_
+mgf

x
= m x
_
x
_
1 + (f

)
2
_
+f

_
x
2
_
f

_
+g
__
= 0,
which vanishes due to the equation of motion (70). Hence the energy is indeed constant.
There is a quick way of getting at (70) without ever introducing the constraint force.
For an unconstrained free particle moving under the inuence of a force, F = U, the
kinetic energy, T( x, y) =
m
2
_
x
2
+ y
2
_
is a function of two independent quantities, x and
y. The x-component of the momentum, p
x
= m x can be obtained from T by partial
dierentiation with respect to x, keeping y xed,
p
x
=
T
x
= m x,
and the x-component of Newtons second law can be written as
m x = p
x
=
d
dt
_
T
x
_
=
x
U.
For the bead on the wire y is not independent of x, rather y = f

(x) x, and
T(x, x) =
m
2
x
2
_
1 + (f

)
2
_
depends on x and x. Dene the generalised momentum as being the partial derivative of
T(x, x) with respect to x, keeping x xed,

x
=
T
x
= m x
_
1 + (f

)
2
_
.
Then the rate of change of the generalised momentum is

x
= m x
_
1 + (f

)
2
_
+ 2m x
2
f

.
64
Now notice that the equation of motion (70) gives

x
= m x
_
1 + (f

)
2
_
+ 2m x
2
f

= m x
2
f

mgf

.
The right hand side is not quite the negative of the derivative of the potential energy with
respect to x,
x
U = mgf

, but the extra term

m x
2
f

=
T
x
is the partial derivative of T(x, x) with respect to x with x held xed. Hence Newtons
second law for the bead on the wire can be written as

x
=

x
_
T(x, x) U(x)
_
.
The function
L(x, x) := T(x, x) U(x)
is called the Lagrangian for the system (after the French mathematician Joseph-Louis
Lagrange, (1736-1813), who was the rst person to formulate mechanics in this way).
Since U(x) is independent of x the generalised momentum can equally well be obtained
from L as

x
=
T
x
=
L
x
.
In terms of the Lagrangian, Newtons second law for the bead reads
d
dt
_
L
x
_
=
L
x
. (72)
Equation (72) is completely equivalent to Newtons second law for the constrained motion
of the bead and has the advantage that it can be obtained purely from a knowledge of
T(x, x) and U(x) we never need to see the force of constraint, F
n
, explicitly. Three
points to note here are:
1) The Lagrangian L = T U is not the same as the energy E = T +U (unless of course
U = 0). The Lagrangian is not a constant of the motion in general.
2) The derivative with respect to t here is the total derivative, it acts on both x(t) and
x(t).
3) A subtle point has been slipped in here. The Euler-Lagrange equation assumes that x
and x are independent of each other, whereas for the actual motion of course x =
dx
dt
and x is not independent, it is derived from x(t) once x(t) is known. The philosophy
of the Euler-Lagrange equation is to treat both x and x as independent until we have
completely solved for the motion, they are varied separately in L(x, x). Note that
the initial position and velocity x
0
= x(0) and v
0
= x(0) are independent, they must
both be given to solve for the subsequent motion. There is still only one generalised
co-ordinate in this problem, and that is x.
65
6.2 Lagrangian Formulation
More generally for a system with n = 3N k independent generalised co-ordinates
q

, with = 1, . . . , n, we can express the equations of motion in Lagrangian form if we

know the potential energy U(q) and the kinetic energy T(q, q). Then the Lagrangian is
L(q, q) = T(q, q) U(q) (73)
and Newtons second law can be written
d
dt
_
L
q

_
=
L
q

, (74)
which are known as the Euler-Lagrange equations of motion. There are n generalised
momenta,

=
L
q

, one for each generalised co-ordinate, and there are n independent

equations of motion in (74).
The Lagrangian formulation of Newtons equations is an extremely useful way of
describing systems with constraints, though it is also useful for systems without constraints
but whose dynamics is best described using generalised co-ordinates rather than Cartesian
co-ordinates, polar co-ordinates for example. We only need to determine the number of
independent degrees of freedom of the system and identify useful generalised co-ordinates
and then calculate the Lagrangian as a function of our generalised co-ordinates. The
dynamical equations follow from (74). All of the dynamics follows from a single function
of the generalised co-ordinates and their time derivatives.
Here are some examples:
1) A very simple example is a single particle of mass m moving in three dimensions under
the inuence of a conservative force arising from a potential U. Using Cartesian co-
ordinates (x, y, z)
T =
m
2
_
x
2
+ y
2
+ z
2
_
and the force has components
F
x
=
U
x
, F
y
=
U
y
, F
z
=
U
z
.
The Lagrangian is
L = T U =
m
2
_
x
2
+ y
2
+ + y
2
_
U(x, y, z).
The momenta are
p
x
=
L
x
= m x, p
y
=
L
y
= m y, p
z
=
L
z
= m z
66
and Newtons equations follow from (74),
p
x
=
U
x
, p
y
=
U
y
, p
z
=
U
z
.
For Cartesian co-ordinates with no constraints Lagranges equations are just another
way of writing Newtons second law,
2) Sometimes Cartesians are not the best description of a dynamical system. Take a
central force in three dimensions, where the force depends only on the distance from
the origin. Then polar co-ordinates are more suited to the symmetry of the problem.
In three dimensional polar co-ordinates the velocity is (54) so the kinetic energy of a
particle of mass m is
T =
m
2
_
r
2
+r
2

2
+r
2
(sin )
2

2
_
.
The potential for a central force is a function of r only, U(r), so F = F
r
e
r
with
F
r
=
dU
dr
. The Lagrangian is
L =
m
2
_
r
2
+r
2

2
+r
2
(sin )
2

2
_
U(r)
giving generalised momenta

r
=
L
r
= m r,

=
L

= mr
2

,

=
L

= mr
2
(sin)
2

while
L
r
= mr
_

2
+ (sin )
2

2
)
dU
dr
,
L

= mr
2
(sin cos )

2
,
L

= 0.
The Euler-Lagrange equations are thus

r
= m r = mr
_

2
+ (sin)
2

2
)
dU
dr
,

= m
d
dt
_
r
2

_
= mr
2
(sin cos )

2
,

= m
d
dt
_
r
2
(sin )
2

_
= 0.
The middle equation is satised by setting =

2
= const, since then

= 0 and
cos = 0. The motion of the particle lies in a two dimensional plane, which we can
choose to be the plane =

2
(ie. the (x, y)-plane). The last equation then implies
that ml := mr
2

= const, this is conservation of angular momentum (we do not use
L for total angular momentum in this section, in order to avoid confusion with the
Lagrangian). Finally, using l in the rst equation above gives
r =
ml
2
r
3

dU
dr
,
67
the rst term on the right hand side representing centrifugal acceleration. All this
obtained in a few lines from the Lagrangian!
3) As a last example we consider a system that has constraints. Let two masses, m
1
and
m
2
, be connected together by a light inextensible string of length l that hangs over
a light circular pulley of radius R in a vertical plane, which is free to rotate about a
horizontal axis through its centre. We assume the string does not slip on the pulley.
R
m g

y
2
1
y
m g
1
2
The vertical distances y
1
of mass m
1
and y
2
of mass m
2
downward from the axis of
the pulley are not independent, but are constrained by y
1
+ R + y
2
= l, so there is
only one degree of freedom, we shall choose to use y = y
1
, and
y
2
= l y
1
R y
2
= y
1
.
Ignoring the masses of the string and the pulley, the kinetic energy is just the sum of
the kinetic energies of m
1
and m
2
,
T =
m
1
2
y
2
2
+
m
1
2
y
2
2
=
m
1
+m
2
2
y
2
.
The potential energy is the sum of the potential energies of m
1
and m
2
,
U = g(m
1
y
1
+m
2
y
2
) = g
_
m
1
y +m
2
(l Ry)
_
= g(m
1
m
2
)y +const.
Ignoring the constant in the potential the Lagrangian is
L = T U =
1
2
(m
1
+m
2
) y
2
g(m
1
m
2
)y.
68
The generalised momentum is

y
=
L
y
= (m
1
+m
2
) y
and the Euler-Lagrange equations are
d
dt
_
L
y
_
=
L
y

y
= (m
1
+m
2
) y = g(m
1
m
2
),
so the acceleration
y = g
_
m
1
m
2
m
1
+m
2
_
is constant, y increases if m
1
> m
2
and decreases if m
2
> m
1
. Suppose m
1
> m
2
and
the system is released from rest at t = 0 with y(0) = 0 then the subsequent motion is
y(t) =
g
2
_
m
1
m
2
m
1
+m
2
_
t
2
,
at least until y reaches l R.
It is even easy to include the mass and moment of inertia of the pulley. If the pulley
has mass M and is of constant density =
M
R
2
, then its moment of inertia about the
central axis is
I = 2
_
R
0
(r
2
)rdr =
2
4
R
4
=
1
2
MR
2
.
Measure the angle of rotation from the vertical by , with = 0 when y = 0, then
is not an independent degree of freedom, y = R. The kinetic energy of the rotating
pulley is then given by (66), with V = 0 since the centre of the pulley is xed in space,
T
Pulley
=
I
2

2
=
I
2
y
2
R
2
=
M
4
y
2
and we add this to the Lagrangian,
L =
1
2
_
m
1
+m
2
+
M
2
_
y
2
+g(m
1
m
2
)y.
The equation of motion is now
y = g
_
m
1
m
2
m
1
+m
2
+
M
2
_
,
which is easily solved.
69
6.3 Time Dependent Constraints*
The Lagrangian formulation of mechanics is an extremely ecient way of describing
dynamics in the presence of constraints and conservative forces, even when the constraints
depend on time.
First consider a simple example with no constraints. Take N identical particles each
of mass m moving in three dimensions, with Cartesian co-ordinates x
I
, i = 1, , 3N,
under the inuence of a conservative force,
F
I
=
U
x
I
,
with potential U(x
I
). We can use the Cartesian co-ordinates themselves as generalised
co-ordinates, then the kinetic energy is
T =
m
2
3N

I=1
x
I
x
I
(75)
so the Lagrangian is
L( x
I
, x
I
) =
m
2
3N

I=1
x
I
x
I
U(x
I
).
The generalised momenta are the ordinary momenta in this case,
p
I
=
L
x
I
=
T
x
I
= m x
I
and the kinetic energy depends only on x
I
, not on x
I
, so
L
x
I
=
U
x
I
.
The Euler-Lagrange equations then read
d
dt
_
L
x
I
_
=
_
L
x
I
_

dp
I
dt
=
_
U
x
I
_
= F
I
,
which simply reproduces Newtons second law.
The Lagrangian formulation even allows for time dependent constraints. Consider a
system of N particles, labelled by an index i = 1, . . . , N, moving in three dimensions.
There are 3N degrees of freedom and we shall use 3N Cartesian co-ordinates which will
be labelled by x
I
, I = 1, . . . , 3N. Let there be k constraints and choose n = 3N k
* The material in this section was not covered in the lectures and is included for further
interest.
70
generalised co-ordinates q

, with = 1, . . . , n. The Cartesian co-ordinates labelling the

particle positions are then not all independent degrees of freedom, they can be expressed
as functions of q

and possibly t itself, if the constraints depend on time, x

I
(q

, t). Then
x
I
=
n

=1
x
I
q
a
q

+
x
I
t
(76)
is in general a function of q

, q

and t, x
I
( q

, q

, t). However its dependence on q

is
particularly simple, it is linear in q

,
x
I
q
a
=
x
I
q
a
. (77)
Now decompose the force the force on each particle into an external force F
(e)
, eg. gravity,
and a constraint force F
(c)
. Newtons second law is
F = F
(e)
+F
(c)
= p
where p
I
= m x
I
, assuming for simplicity that each particle has the same mass (this is not
necessary). Then under a virtual displacement x for which the constraints do no work,
F
(c)
.x = 0, we have
0 = (F p).x = (F
(e)
+F
(c)
p).x = (F
(e)
p).x.
So, for virtual displacements,
(F
(e)
p).x = 0.
Note that it is not true that F
(e)
p = 0, unless of course there are no constraints and
F
(c)
= 0.
If the external forces are conservative there is a potential energy function, U(x
I
), for
which
F
(e)
I
=
U
x
I
and we can write
N

I=1
_
U
x
I
+
dp
I
dt
_
x
I
= 0.
Virtual displacements cannot be in arbitrary directions, they must correspond to variations
in the generalised co-ordinates q

+q

, so the variation in x
I
is constrained to be
of the form
x
I
=
n

=1
x
I
q

.
However the q

are arbitrary, so
N

I=1
_
U
x
I
+
dp
I
dt
__
x
I
q

_
= 0. (78)
71
Now L = T( x
I
, x
I
) U(x
I
) with the kinetic energy given by (75), and using (77),
N

I=1
p
I
x
I
q

=
N

I=1
L
x
I
x
I
q

=
N

I=1
L
x
I
x
I
q

=
L
q

,
so
N

I=1
dp
I
dt
x
I
q

=
d
dt
_
N

I=1
p
I
x
I
q

I=1
p
I
d
dt
_
x
I
q

_
(79)
=
d
dt
_
L
q

I=1
p
I
d
dt
_
x
I
q

_
.
While
N

I=1
U
x
I
x
I
q

=
U
q

=
L
q

+
T
q

=
L
q

+
N

I=1
T
x
I
x
I
q

=
L
q

+
N

I=1
p
I
x
I
q

. (80)
Putting (79) and (80) into (78) we get
d
dt
_
L
q

L
q

+
N

I=1
p
I
_
x
I
q

d
dt
_
x
I
q

__
= 0. (81)
Now the magic is, using (76),
x
I
q

=
n

=1
_

2
x
I
q

_
q

+

2
x
I
q

t
and
d
dt
_
x
I
q

_
=
n

=1
_

2
x
I
q

_
q

+

2
x
I
tq

,
so every term in the last summation in (81) vanishes leaving the Euler-Lagrange equations
for the constrained system
d
dt
_
L
q

L
q

= 0.
6.4 Variational Formulation and the Action
72
In static equilibrium a system always seeks out a conguration that minimises the
energy, in particular statics requires that the kinetic energy vanishes and the potential
energy be minimised with respect to vatriations of particle positions,
U
q

= 0.
Amazingly the Euler-Lagrange equations can be obtained from a similar principle, but it
is not the energy that is minimised. Historically the origins of this concept are related to a
minimisation principle in optics known the principle of least time a ray of light travels
in such a way as to minimise the time it takes to travel between any two given points.
Consider a beam of light traveling across a planar interface from a point A in one
medium (eg. air) in which the speed of light is v
1
, to a point B in a dierent medium (eg.
water) in which the speed of light is v
2
. What trajectory will minimise the time taken for
the light to travel from A to B? The light will travel in a straight line in medium 1 and a
straight line in medium 2, but we can vary the point O to try and minimise the time.
2

1
O
x
y
y
1
1
2
v
2
v
1
A
B
2
x
d

Since A and B are xed y

1
and y
2
are xed and x
1
+x
2
= d, but we can vary x
1
and
x
2
by moving the point O though only one of them is independent as x
2
= d x
1
. The
73
time taken for the light to travel from A to O, t
1
, is the length of AO, which is
_
x
2
1
+y
2
1
,
divided by the speed of light in the medium 1,
t
1
=
_
x
2
1
+y
2
1
v
1
.
Similarly the time taken for light to travel from O to B is
t
2
=
_
x
2
2
+y
2
2
v
2
=
_
(d x
1
)
2
+y
2
2
v
2
.
Hence the total time to travel from A to B is
T = t
1
+t
2
=
_
x
2
1
+y
2
1
v
1
+
_
(d x
1
)
2
+y
2
2
v
2
.
Now y
1
and y
2
are xed and we can minimise T(x
1
) by varying x
1
and demanding that
dT
dx
1
= 0, so
dT
dx
1
=
x
1
v
1
_
x
2
1
+y
2
1

(d x
1
)
v
2
_
(d x
1
)
2
+y
2
2
=
x
1
v
2
_
(d x
1
)
2
+y
2
2
(d x
1
)v
1
_
x
2
1
+y
2
1
v
1
v
2
_
x
2
1
+y
2
1
_
(d x
1
)
2
+y
2
2
.
This vanishes, for nite x
1
, only when
x
1
v
2
_
(d x
1
)
2
+y
2
2
= (d x
1
)v
1
_
x
2
1
+y
2
1
(82)
x
1
v
2
_
x
2
2
+y
2
2
= x
2
v
1
_
x
2
1
+y
2
1

x
1
_
x
2
1
+y
2
1
v
2
=
x
2
_
x
2
2
+y
2
2
v
2
sin
1
v
2
= sin
2
v
1
,
or
sin
1
sin
2
=
v
1
v
2
,
which is Snells law for refraction! Snells law follows from the assumption that the light
travels in a manner that minimises the time taken to go from A to B (we leave it as an
exercise to show that this is a minimum and not a maximum, ie. check that
d
2
T
dx
2
1
> 0 when
x
1
is given by (82). This way of viewing refraction, as minimising the travel time of a light
beam, is known in optics as Fermats principle or the principle of least time. The law of
reection can be derived the same way.
We are about to show that Newtons 2nd law, in the Euler-Lagrange formulation
(74), can also be derived by extremising a certain quantity known as the action. Suppose
a particle travels from a point with generalised co-ordinates q

1
at time t
1
to a point q

2
at a time t
2
along a trajectory q

(t), with q

1
= q

(t
1
) and q

1
= q

(t
1
). The action, S,
74
is dened to be the integral of the Lagrangian with respect to time along the trajectory
q

(t),
S[q] =
_
t
2
t
1
L(q, q)dt.
The notation S[q] is a conventional way of expressing the notion that the action depends on
the path, q

(t), taken between the initial point q

1
and the nal point q

2
. The action is not
a function of time, time has been integrated over in the denition of S, but it does depend
on the path taken between the xed points q
1
and q
2
. It is a function of a function, called
a functional, and is written S[q] to emphasise the fact that it depends on the function q(t).
Now suppose we vary the path, keeping the end points xed.
q
1
2
q(t)
q (t )
q (t )
1

2
q+ q

Then in general S[q] will change. If the path is varied from q(t) to q(t) + q(t), and also
q(t) q(t) + q(t), where q(t) is an innitesimally small function with q(t
1
) = 0 and
q(t
2
) = 0 so the end points do not move, then the change in the action will be
S[q] = S[q +q] S[q] =
_
t
2
t
1
L(q +q, q + q)dt
_
t
2
t
1
L(q, q)dt
=
_
t
2
t
1
n

=1
_
L
q

+
L
q

_
dt +
=
n

=1
_
t
2
t
1
_
L
q

+
L
q

_
dt + , (83)
where the dots indicate quantities of order q
2
, which we shall ignore for innitesimal q.
The second term under the summation on the right hand side can be integrated by
parts using,
L
q

=
d
dt
_
L
q

d
dt
_
L
q

_
q

,
so
_
t
2
t
1
L
q

dt =
_
t
2
t
1
d
dt
_
L
q

_
dt
_
t
2
t
1
d
dt
_
L
q

_
q

dt
=
_
L
q

_
t
2
t
1

_
t
2
t
1
d
dt
_
L
q

_
q

dt
.
75
But q(t
1
) = q(t
2
) = 0, since the end points are assumed to be held xed, so the integrated
term in square brackets above vanishes and
_
t
2
t
1
L
q

dt =
_
t
2
t
1
d
dt
_
L
q

_
q

dt.
Using this in (83) we nd
S[q] =
n

=1
_
t
2
t
1
_
L
q

d
dt
_
L
q

__
q

dt + .
The variation of S[q] then vanishes, to rst order in q, for any q(t) with q
1
= 0 xed at
t
1
and q
2
= 0 xed at t
2
, if and only if
L
q

d
dt
_
L
q

_
= 0,
that is S[q] = 0 if and only if the Euler-Lagrange equations are satised. The actual
trajectory of the particle is such as to extremise the action. In fact the action is a maximum
when q(t) is the correct physical trajectory, rather than a minimum, but this just means
that S is minimised. This is clear from a static situation where the kinetic energy is zero,
so L = U, and the second term above, involving the total time derivative, also vanishes
then
L
q

=
U
q

and a minimum of U is a maximum of L, and hence a maximum of S.

76