Download 1 Lecture 1 - School of Physics and Astronomy

Dynamics
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Lecture 1: Introduction and dynamics in one dimension
Introduction:
The goal of this course is twofold: first to acquire good understanding of Classical Mechanics and second to
master the subject of ordinary di↵erential equations. You will see that these two subjects are strongly interlinked,
and indeed, historically the physical and mathematical aspects developed together.
This course can be regarded as the first course of mathematical physics or theoretical physics. At the same time,
the tools you will acquire here are universal; they will serve you whatever you do.
By the end of the course you should be able to analyse a dynamical problem yourself. This process involves three
steps:
• Using the basic principles (Newton’s laws), one translates the physics problem into mathematical equations.
These would usually be di↵erential equations describing the motion of the system plus appropriate initial
conditions.
• In the second step one solves the equations to determine the motion at all times. There are various ways
to arrive at this solution, some numerical, some analytic, some involve systematic approximations and others
exact. In this course we will concentrate on simple equations that can be solved exactly.
• In the final step one analyses the solution to get a physical picture.
The recommended texts for classical mechanics are:
1. “Classical Mechanical,” R. Douglas Gregory, Cambridge University Press.
2. “Analytical Mechanics,” Fowles/Cassiday, Sounders College Publishing.
On the mathematical methods side (di↵erential equations, for example) you may use
1. “Mathematical Methods for Physics and Engineering,” K. F. Riley, M. P. Hobson, S. J. Bence, Cambridge
University Press.
2. “Foundation Mathematics,” K. F. Riley and M. P. Hobson, Cambridge University Press.
3. “Mathematical Methods in the Physical Sciences,” Mary L. Boas, Published by John Wiley and Sons, Inc.
We will not closely follow any book, but it may be good to use books in parallel. There are also some resources on
the web. For example, for di↵erential equations I recommend the following three sites:
http://www.aw-bc.com/ide/idefiles/navigation/main.html
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2006/syllabus/
http://eqworld.ipmnet.ru/en/solutions/ode.htm
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The course structure:
This course, called ‘Dynamics’ may either be taken as part of the 20 point course ‘Dynamics and Vector Calculus’
or as a stand alone 10 point course.
In Dynamics we have two hours of lectures every week (weeks 1-11), Tuesday and Friday 11:10-12:00. These
lectures take place in Lecture Theatre A of JCMB. Each student should also attend one two-hour workshop every
week on weeks 2-11. You need to choose between the Tuesday 14:00-16:00 and Wednesday 09:00-11:00 workshops
(both run in JCMB 3207). In case you have questions beyond what can be dealt with in or after the lectures, please
email me at [email protected], and we’ll arrange a meeting in my office, JCMB 4417.
This is a course on problem solving. It therefore involves a lot of problem solving. You must
devote to it sufficient time and energy within the workshops and at home.
A problem sheet will be issued on every Friday, starting on week 1. Every week we will have three separate sets
of problems:
• The first set - preparation problems - you are expected to work on before the workshops, either alone or in a
small group. These will usually be straightforward questions, but if you encounter any problems do not hesitate
to ask for help during the workshop.
• The second set - workshop problems - you are expected to work on during the workshops. Discuss them with
your friends and get help from the tutors and from me when needed. If you do not finish it in the workshop continue at home. Workshops will be held on weeks 2-11.
• The third set - hand-in problems - you must do individually. You are expected to work on it in your free time.
This work should be handed in for assessment as explained below.
Continuous assessment will be based on the weekly Homework Worksheets. This work ought to be handed in the
School of Physics and Astronomy Teaching Office, JCMB Room 4315, by lunchtime (12:30) on the Thursday, two
weeks after you have received the worksheets. Work which is handed in late will not be marked.
The continuous assessment mark will be computed based on the best 8 out of 9 worksheets and will amount to 20
percent of the final mark.
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Dynamics in one dimension:
In physics we are trying to model the world in the simplest way that still captures the essentials. Modelling
involves simplifications and approximations. The important thing is that we can develop tools to quantify how good
these approximations are.
In this course we start right away by considering point particles. This means that we will be always neglecting the
dimensions of bodies as compared to the distances they traverse (Rigid body dynamics is a very interesting subject,
but it is out of the scope of this course).
Our second simplification is considering motion in one dimension only. We will relax this assumption later on in
the course, and discuss motion in two and three dimensions.
Our third simplification is that the particle considered has a constant mass m. This will also be relaxed later on
when we consider varying mass problems such as rockets motion.
Considering a point particle in one dimension, we denote the position at time t by the function x(t). The velocity
v is obtained by the first time-derivative, the rate of change of the position:
v(t) = ẋ =
dx(t)
dt
Note that throughout the course time derivatives will be denoted by a dot. Further, the acceleration is given by the
rate of change of the velocity,
dv(t)
d2 x(t)
a(t) =
= ẍ =
.
dt
dt2
Classical Mechanics is also called Newtonian Mechanics. It is based on the following three Newtons laws:
• The first Newton law states that a particle continues moving at constant velocity v (which can of course be
zero) unless acted upon by an external force.
• The second law states that a body, when acted upon by a net external force F , would accelerate according to:
F = mẍ
(18)
where the proportionality constant is called the inertial mass, or simply the mass of the body.
• The third law states that the mutual forces between two bodies (action and reaction) are opposite in direction
and equal in magnitude. We will develop the consequences of this third low later on in the course.
The second law is the key to determining the motion. As we shall see, if one knows the initial position and
velocity, the second Newton law uniquely determines the position of the particle x(t) at any give time t. This
complete determinism is an essential element of classical physics – it is violated however in the quantum world.
The first thing we need to do in a given physics problem is to determine the net force on any given body. F can
depend on the position x and on the velocity ẋ. Consider a simple example of a constant force:
F =f
=)
constant acceleration: ẍ(t) = f /m
(19)
at any given time t. An example where constant force is relevant is Galilean Gravity, the force by which the Earth
attracts massive bodies downwards, where f /m = g = 9.8 m sec 1 . We will learn about the more general case of
Newtonian Gravity in later lectures, and eventually be able to compute trajectories of satellites and planets. For
now, let’s consider the constant force. Integrating (19) from t = 0, where we assume an initial velocity ẋ(0), up to
some later time t, we find:
Z
Z
t
ẋ(t)
ẋ(0) =
t
d⌧ ẍ(⌧ ) =
0
d⌧ f /m = f t/m .
0
So at any time t
ẋ(t) = ẋ(0) + f t/m ,
Further integration from t = 0 to t yields:
Z t
Z t h
i
1
x(t) x(0) =
d⌧ ẋ(⌧ ) =
d⌧ ẋ(0) + f ⌧ /m = ẋ(0)t + f t2 /m
2
0
0
so finally we have
1
x(t) = x(0) + ẋ(0)t + f t2 /m .
2
Of course, in the trivial case of f = 0 we obtain motion at constant velocity x(t) = x(0) + ẋ(0)t in agreement with
the first Newton law.
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We see that we can describe the position at any time t starting from the equation of motion (EOM) ẍ(t) = f /m
and two initial conditions. The EOM is a second order Di↵erential Equation, as it involved the second derivative
of the function we wish to determine x(t). We have seen that at each integration stage we must fix one constant of
integration: starting from a second order equation it is obvious we should need two initial conditions. We will see
that this is a completely general feature of di↵erential equations: the order of the equation determines the number
of initial conditions. EOM in Classical Mechanics are second order, so if we have n particles we need 2n initial
conditions.
Let us now consider more complicated forces. On what can the force depend? Most often it depends on the
position x, sometimes also on the velocity ẋ or directly on the time t.
Consider for example a spring. According to Hooke’s law the spring elastic force depends linearly on the
displacement x x0 , namely,
F (x) = k (x x0 ) ,
(20)
Note that this is a restoring force: it acts against the displacement (k > 0), always attempting to decrease the
absolute value of the displacement. The corresponding EOM is
⇣
⌘
mẍ(t) = k x(t) x0 .
(21)
A general solution is given by
x(t) = x0 + A cos(!t
);
!=
p
k/m
Again, we need to fix two parameters (A and ) and we do that using the initial conditions. This oscillatory
motion is called Harmonic motion, where ! is the characteristic frequency of the system. We will encounter the
same equation later on in the course when we analyse the Pendulum. Furthermore, we will consider more general
scenarios where additional forces show up,
F (x, ẋ, t) =
|
k (x x0 )
k 0 ẋ + fe (t) ,
{z
} | {z }
| {z }
elastic
friction
(22)
external
which obviously lead to a more complicated di↵erential equation:
⇣
⌘
mẍ(t) = k x(t) x0
k 0 ẋ(t) + fe (t).
We will learn how to solve such equations.
Energy conservation:
Energy can be stored in di↵erent ways, and total energy conservation is a general physics principle. However,
Mechanical energy is not always conserved: for example, friction converts kinetic energy to heat, decreasing the total
amount of mechanical energy.
In contrast, many forces conserve the mechanical energy; they are referred to as conservative forces. Such
forces merely convert between potential V and kinetic T energy, such that the sum remains constant:
T +V =E
(23)
In one dimension conservative forces can be defined as ones that may be expressed as (minus) the derivative of some
potential function V (x) with respect to the position x:
F (x) =
dV (x)
dx
(24)
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Obviously, velocity-dependent forces, such as friction, fall outside this category, while constant forces or elastic forces
are conservative. For example to obtain the elastic restoring force of (22) we define:
V (x) =
k
(x
2
x0 ) 2 .
(25)
Note that this potential has a minimum at x = x0 , where the spring is not stressed nor stretched.
In general, for one-dimensional conservative forces we may express the EOM as:
mẍ +
dV (x)
=0
dx
(26)
Multiplying this equation by ẋ we get:
mẋẍ + ẋ
which yields
2
dV (x)
=0
dx
3
7
d 6
6 1 mẋ2 +V (x)7 = 0
4
5
dt |2 {z }
(27)
T
Where we identified the first term as the kinetic energy T . Upon integration this equation obviously implies that
the total mechanical energy, T + V = E, is invariant in time. This is so for any conservative force. The energy
conservation condition, namely,
1
mẋ2 + V (x) = E
(28)
2
can be an alternative starting point for determining the solution, equivalent to the EOM (26). The advantage of
(28) is that it leads to a first order equation, as it does not involve the acceleration ẍ. This reduction of order
was achieved by identifying a constant of motion, the energy E. Often one may identify additional constants of
motion. For example, in spherically symmetric problems, the angular momentum is also a constant of motion. We
will see this explicitly later in the course.
Dimensions: All physical quantities have dimensions. In mechanics there are three fundamental dimensions length,
time and mass. Dimensional analysis is very often useful, and it can be done without specifying units. For example,
one can verify that the kinetic energy T and the potential energy V have the same dimensions:
[T ] = [V ] = mass ⇥ (length)2 ⇥ (time)
2
as must be the case (note that terms appearing in a sum always have the same dimensions).