Physics 103
General Physics I Notes
© J Kiefer 2013
1
Table of Contents
Table of Contents .............................................................................................................. 1
I.
Introduction ............................................................................................................... 3
A.
Dimensions and Units of Measure ........................................................................... 3
1. Physical Dimensions ............................................................................................... 3
2. Systems of Standard Units ...................................................................................... 3
B.
Scientific Notation and Significant Figures ............................................................ 5
1. Significant Figures (Digits)..................................................................................... 5
2. Scientific Notation and Significant Digits .............................................................. 5
C.
1.
2.
II.
Vectors ....................................................................................................................... 6
Coordinate System .................................................................................................. 6
Definition of Scalars & Vectors .............................................................................. 7
Kinematics ........................................................................................................... 11
A.
One Dimensional Motion ........................................................................................ 11
1. Variables of Motion .............................................................................................. 11
2. Motion with Constant Acceleration ...................................................................... 12
B.
Two Dimensional Motion ....................................................................................... 16
1. Projectile motion ................................................................................................... 16
2. Uniform circular motion ....................................................................................... 18
III.
1.
2.
3.
Newton’s “Laws” of Motion................................................................................... 21
First “Law”............................................................................................................ 21
Second “Law” ....................................................................................................... 22
Third “Law” .......................................................................................................... 23
1.
2.
3.
4.
5.
Case Studies ............................................................................................................. 24
Weight & Friction & Tension ............................................................................... 24
Circular Motion ..................................................................................................... 29
Sliding & Pulling .................................................................................................. 31
Restoring Forces ................................................................................................... 34
More than One Object ........................................................................................... 35
A.
B.
C.
Dynamics .............................................................................................................. 21
Relative Motion ....................................................................................................... 38
1. Reference Frames.................................................................................................. 38
2. Accelerated Reference Frames ............................................................................. 40
1
IV.
A.
1.
2.
Physical Work ......................................................................................................... 43
Definition of Physical Work ................................................................................. 43
Two Dimensional Case(s) ..................................................................................... 46
1.
2.
3.
4.
Conservation of Energy .......................................................................................... 47
Conservative & Non-Conservative Forces ........................................................... 47
Potential Energy Functions ................................................................................... 48
Mechanical Energy Conservation ......................................................................... 50
Gravitation ............................................................................................................ 51
1.
2.
3.
Conservation of Momentum .................................................................................. 54
Momentum & Impulse .......................................................................................... 54
Collisions & Explosions ....................................................................................... 56
Center of Mass ...................................................................................................... 60
B.
C.
V.
Rotational Motion & Oscillatory Motion ............................................................. 62
A.
1.
2.
3.
B.
Work and Energy and Impulse and Momentum ............................................. 43
Rotation of a Rigid Body ........................................................................................ 62
Rotational Kinematics ........................................................................................... 62
Rotational Dynamics ............................................................................................. 64
General Motion of a Rigid Body .......................................................................... 70
Oscillatory Motion .................................................................................................. 76
1. Simple Harmonic Motion ..................................................................................... 76
2. Pendulum .............................................................................................................. 79
VI.
A.
Wave Motion ....................................................................................................... 81
1.
3.
4.
Waves ....................................................................................................................... 81
General Wave Properties ...................................................................................... 81
Stretched String ..................................................................................................... 85
Pressure waves--Sound ......................................................................................... 87
VII.
Thermodynamic .................................................................................................. 89
A.
“Heat” Flow ............................................................................................................. 89
B.
“Laws” of Thermodynamics .................................................................................. 89
2
I.
Introduction
We observe the motion of objects in the universe around us, and we experience the
passage of time. Those objects move in three dimensions of space. Mathematically, we
can treat time as a fourth dimension, or axis, giving rise to the term space-time. We
observe further that sometimes an object’s motion changes. We want to understand how
an object’s motion changes, meaning what causes the change in motion.
A.
Dimensions and Units of Measure
1.
Physical Dimensions
The dimension of a physical quantity specifies what sort of quantity it is—space, time,
energy, etc.
a.
Fundamental dimensions
We find that the dimensions of all physical quantities can be expressed as combinations
of a few fundamental dimensions: length [L], mass [M], time [T], and electric charge
[Q].
For example, energy E  
M L
T 
2
2
. The physical quantity speed has dimensions of
L .
T 
b.
Operational definition
Ultimately, the fundamental physical dimensions are defined operationally. That is, we
prescribe how the physical quantity is to be measured. The length of an object is defined
thusly: (i) lay a meter stick next to the object, (ii) count the marks on the meter stick
between the two ends of the object, (iii) that number is the length of the object.
2.
Systems of Standard Units
In the example of an operational definition of length, we used a meter stick. Of course,
this refers to one of the widely accepted units of length, the meter. To facilitate
communication, we all agree to express our measurements in terms of a standard set of
units of measure. A complete set of units would include a standard unit for every
physical dimension.
a.
International system (SI)
The units of the fundamental dimensions in the SI are
dimension
SI
cgs
[L]
meter
centimeter
[T]
second
second
[M]
kilogram
gram
3
Brit
foot
second
slug
These fundamental units are defined operationally. The meter is defined as the distance
1
that light travels in a vacuum in
of a second. In turn, a second is defined to
299792458
be equal to the time required for 9192631770 cycles of microwaves whose frequency is
just right to excite a particular atomic transition in Cesium. The kilogram is still based on
an actual prototype, a chunk of platinum-iridium alloy kept hermetically sealed in France.
The SI units of other physical dimensions will be introduced as we go along. They
almost all are named after a person.
b.
Unit conversions
Now, the amount of a physical quantity remains the same, no matter what system of units
is used to obtain a numerical measure of that quantity. For instance, we might measure
the length of an (American) football field with a meter stick and a yard stick. We’d get
two different numerical values, but obviously there is one field with one length. We’d
say that 100 yards  91.44 meters . In other words,
100 yards
 1.0
91.44 meters
Note: the units are a part of the measurement as important as the number. They
must always be kept together.
Suppose we wish to convert 2 miles into meters. [2 miles = 3520 yards.]
91.44 meters
3520 yards 
 3218.688 meters
100 yards
The units cancel or multiply just like common numerical factors. Since we wanted to
cancel the yards in the numerator, the conversion factor was written with the yards in its
denominator.
Since each conversion factor equals 1, we can do as many conversions as we please—the
physical measurement is unchanged, though the numerical value is changed.
c.
Abbreviations
The names of the standard units also have standard abbreviations. The abbreviation for
meter is m, for second s or sec, for kilogram kg. So each time a new unit is introduced,
so will be its standard abbreviation.
d.
Dimensional analysis
We can check for error in an equation or expression by checking the dimensions.
Quantities on the opposite sides of an equal sign must have the same dimensions.
Quantities of different dimensions can be multiplied but not added together.
4
e.g., a proposed equation of motion, relating distance traveled (x) to the acceleration (a)
and elapsed time (t).
x
Dimensionally, this looks like
1 2
at
2
L  L  T  T 
T 
L  L
2
At least, the equation is dimensionally correct; it may still be wrong on other grounds, of
course.
B. Scientific Notation and Significant Figures
Scientific notation is simply a way of writing very large or very small numbers in a
compact way.
299792485  2.998  10 8
0.0000000010878  1.088  10 9
1.
Significant Figures (Digits)
Instruments cannot perform measurements to arbitrary precision. A meter stick
commonly has markings 1 millimeter (mm) apart, so distances shorter than that cannot be
measured accurately with a meter stick. We report only significant digits—those whose
values we feel sure are accurately measured. There are two basic rules: (i) the last
significant digit is the first uncertain digit and (ii) when combining numbers, the result
has no more significant digits than the least precise of the original numbers.
A third rule is, the exercises and problems in the textbook assume there are 3 significant
digits. Therefore, we never include more than 3 significant digits in our numerical
results, no matter that the calculator displays 8 or 10 or more.
2.
Scientific Notation and Significant Digits
The uncertainty in a numerical value may be expressed in terms of a tolerance, as
23.273  0.005 . Alternatively, the uncertainty can be shown in scientific notation simply
by the number of digits displayed in the mantissa.
1.5  10 3 2 digits, the 5 is uncertain.
1.50  10 3 3 digits, the 0 is uncertain.
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Examples
756.  37.2  0.83  2.5  796.53  800  8.0  10 2 2 digits
3.563  3.2  11.4016  11.  1.1  101 2 digits
5.6    17.59291886  18  1.8  101 2 digits
C. Vectors
Mathematics is a system of logic that we use to discuss physical phenomena. A system
of logic consists of entities and of a list of rules that govern how the entities relate to one
another. A general term for this is group theory. Of course, this is a very general
definition. In physics, entities or concepts are represented mathematically by the
mathematical objects scalars, vectors and sometimes matrices. These mathematical
objects are defined by the rules that govern them.
A particular physical quantity, such as velocity, may be represented by a mathematical
vector. We often say then that velocity is a vector quantity. In the same vein,
temperature is a scalar quantity.
1.
Coordinate System
We measure locations in space relative to a coordinate system. Firstly we select the
origin of coordinates, and then the directions of orthogonal axes. The choice of
coordinate system is entirely at our convenience. However, like a consistent choice of
units, we need to be consistent with our coordinate system. Since the directions shown
by orthogonal axes are mutually perpendicular, components along different axes are
independent of each other.
a.
Cartesian
6
b.
Polar
c.
Three Dimensional
The three dimensional Cartesian coordinate
system is comprised of three mutually
perpendicular, straight axes, commonly
denoted x, y, & z.
The spherical polar coordinate system is
comprised of a radius and two angles, as
shown in the figure. Notice how the polar
coordinates are defined in terms of the
Cartesian system.
Any point in space can be uniquely
specified by listing three numerical
coordinates.
2.
Definition of Scalars & Vectors
a.
Scalar
A scalar is a mathematical entity that has one property, magnitude, only. Temperature,
mass, speed, and energy are scalar quantities. Scalars obey the familiar rules of addition,
multiplication, etc.
7
b.
Vector
A vector is a mathematical entity that possesses two properties, which physically we call
magnitude and direction. Displacement, velocity, acceleration, force, and momentum are
vector quantities.
Vectors are symbolized graphically as arrows, in text by bold-face type or with a
line/arrow on top. A vector may also be represented by an ordered list of numbers, which
are called the components: [t,x,y,z]. Written this way, mathematically the vector would
t 
 x
be called a row matrix. A column matrix is a vertical list of components:   .
 y
 
z
Superscripts and subscripts and implied summation. . . . . . . .
c.
Unit vectors


a
A unit vector is a vector of magnitude 1. E.g., aˆ   , where a is the magnitude of the
a

vector a . Often, the magnitude of a vector is indicated by the letter without the line on

top: a  a .
d.
Components
The directions defined by the Cartensian coordinate axes are symbolized by unit vectors,
iˆ , ˆj , kˆ .
V x  V cos 
V y  V sin 

V  V x iˆ  V y ˆj
V  V x2  V y2
  tan 1
e.
Adding vectors
The sum of two vectors is also a vector.
a
  
A B  C
x
 
 
, a y , a z  b x ,b y ,b z  c x , c y , c z
8

Vy
Vx
Graphical method:
Vectors are represented by arrows, drawn to scale. Place the tail of the 2nd vector on the
head of the 1st, preserving the relative orientations. The resultant vector extends from the
tail of the 1st to the head of the 2nd vector.
Component method
Decompose both vectors along the same set of coordinate axes. The components are
scalars, and so can be added in the usual way of scalars.
Ax  Bx  C x
Ay  B y  C y

C  C x iˆ  C y ˆj
f.
Multiplying vectors
scalar (or dot) product—result is a scalar; the operation is symbolized by a dot.
 
A  B  AB cos   Ax Bx  Ay B y


The angle  is the angle from A to B .


   
  
   
Note: A  B  B  A and A  B  C  A  B  A  C .
9
b x 
 
As a matrix multiplication, A  B  a x , a y a z b y   a x bx  a y b y  a z bz . The magnitude
 
bz 
2  
squared of a vector, A  A  A  a x2  a y2  a z2 .


Vector (or cross) product—result is a another vector; the operation is symbolized by a
cross,  .
  
A B  C
Since the result is a vector, it has magnitude and direction. The rule for evaluating a
cross product is as follows:



C  C  AB sin  , direction perpendicular to both A and B according to the righthand-rule.
Alternatively, the cross product can be evaluated by use of a determinant, thusly:
iˆ
  
C  A  B  Ax
Bx
ˆj
Ay
By
kˆ
Az  iˆAy B z  B y Az   ˆj  Ax B z  B x Az   kˆAx B y  B x Ay 
Bz
Note: This must be done in Cartesian coordinates.
 
 
Evidently, A  B   B  A .
g.
Multiplication of a vector by a scalar
A vector may be multiplied by a scalar. This affects the magnitude of the vector, but
does not affect its direction. The exception to this rule is multiplication by –1. That
leaves the magnitude unchanged, but reverses the direction.

m  A  max , ma y , maz


10
II. Kinematics
Kinematics refers simply to the description of motion. We will be studying the
mathematical representation of the motion of a point mass. A point mass is an idealized
object having no extent, nor volume, nor internal structure.
A. One Dimensional Motion
The motion is constrained to lie along a single straight line: back & forth, up & down,
left & right, etc. The coordinate system consists of a single line, with components on
either side of an origin.
1.
Variables of Motion
We describe the motion of a particle with four motion variables: time, displacement,
velocity, and acceleration. We will be looking for a mathematical means of describing
the relationships among these variables—an equation of motion.
a.
Displacement

The displacement vector, r , points from the origin to the present location of the particle.


If a particle is at r1 at time t  t1 and at r2 at some later time t  t 2 , then we say the
  
change in displacement is r  r2  r1 . Likewise, the elapsed time is t  t 2  t1 .
b.
Velocity


r
The average velocity during the time interval t is defined to be  v 
. It’s the
t
time-rate-of-change in the displacement. In terms of vector components, we’d write
x
y
z
 v x 
,  v y 
,  v z 
.
t
t
t



r dr
The instantaneous velocity is defined to be v  lim
.

dt
t 0 t
c.
Acceleration
  

v v2  v1
Similarly, the average acceleration is  a 
.

t t 2  t1



v da
The instantaneous acceleration is a  lim
.

dt
t 0 t
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d.
Motion graphs
2.
Motion with Constant Acceleration
a.
An Equation of Motion
To be specific, let’s say that the motion is along the x-axis. Then we will deal with xcomponents only. To develop equations of motion, we begin with the definition of
acceleration.
 a x 
v x v x 2  v x1

t
t 2  t1
We might solve for v x 2 as a function of elapsed time and of v x1 .
v x 2  v x1   a x  t 2  t1 
If we say that t1  0 , and v x1  vo , and  a x  a x , then v x (t )  vo  a x  t .
Or we might begin with the definition of velocity. . .
 v x 
x
t
But because the acceleration is assumed to be constant, the average velocity can be
written also as
 v x 
v x 2  v x1
.
2
12
Set ‘em equal
x v x 2  v x1

t
2
v  v x1
x 2  x1  x 2
 t
2
Next, we might substitute for the velocity component. . .
x2  x1 
v x1  a x t  v x1
1
 t  x1  v x1t  a x t 2 .
2
2
Finally, we could eliminate t from the equations by solving the definition of the
acceleration for t .
t 
x 
v x 2  v x1
 ax 
v x 2  v x1 v x 2  v x1 v x22  v x21

2
ax
2a x
We have four possible equations, each involving three of the motion variables. They are
valid in cases of constant acceleration only.
Example: a train traveling on a straight and level track
time, t (seconds) acceleration, ax (m/s2)
0 - 10
2
10 – 40
0
40 - ?
-4
We might compute the total displacement from time t = 0 until the train comes to rest,
given that the train starts also from rest. Firstly, we set up the coordinate system, in this
case just the x-axis.
We have three different accelerations, so we find the displacement for each segment.
Segment 1: We are given the acceleration, elapsed time and initial velocity—vxo = 0 m/s.
1
So we choose the equation of motion x  xo  v xo t  a x t 2 .
2
x1  xo  v xo t 
1
1
2
a x t 2  0 m  0 m/s  10 s   2 m/s 2  10 s   100 m
2
2
13
Segment 2: To find the displacement during the second segment, we need the velocity
component at the end of the first.
v x1  v xo  a x t  0 m/s  2 m/s 2  10 s  20 m/s
Then,
x 2  x1  v x1 t 
1
1
2
a x t 2  100 m  20 m/s  30 s   0 m/s 2  30 s   700 m
2
2
And, of course v x 2  v x1  20 m/s .
Segment 3: For this segment, we know x2, vx2, vx3, and ax, but not t .
v x23  v x22  2  a x x3  x 2 
2
v x23  v x22
0  20 m/s 
x3  x 2 
 700 m 
 750 m
2  ax
2   4 m/s 2

b.

Vertical motion in a uniform gravity field
A freely falling object is one on which only the gravitational force is acting. Such an
object experiences an acceleration toward the center of the Earth. The magnitude of the
acceleration due to gravity varies slightly from place to place, and with altitude.
However, we will use the average magnitude g = 9.81 m/s2 and pretend the acceleration
is constant and uniform.
Example: A hot air balloon is rising at a constant speed of 5 m/s. At time zero, the
balloon is at a height of 20 m above the ground and the passenger in the balloon drops a
sandbag, which falls freely straight downward.
What are the height of the sandbag and its velocity as functions of time? How long does
it take for the sandbag to reach the ground?
Firstly, set up the coordinate system and draw a picture.
14
Secondly, select the appropriate equations.
1
y  y o  v yo  t  a y  t 2
2
v y  v yo  a y  t
t (s)
y (m) vy (m/s)
0
20
5
0.25 20.9
2.55
0.5
21.3
0.1
1.0
20.1
-4.8
2.0
10.4
-14.6



We could answer the second question numerically, by continuing the table until y  0 m .
That’s perfectly legitimate. However, in this case we have an exact expression for the
magnitude of the velocity at y = 0 m.
2
v y2  v yo
 2a y  y  y o 


2
v y2  v yo
 2a y  y  y 0   5 m/s   2   9.8 m/s 2  0 m  20 m   417 m 2 / s 2
2
v y  417 m 2 / s 2  20.4 m/s
Notice that we had to decide, based on our physical intuition, what algebraic sign to
attach to the velocity component. Finally, we use this final velocity to solve for the
elapsed time.
v y  v yo  a y t  t 
v y  v yo
ay

0 m/s  20 m/s
 2.59 s
 9.8 m/s 2
Now, there is an alternative, one-step solution.
1
y  y o  v yo  t  a y  t 2
2
1
0 m  20 m  5 m/s  t   9.8 m/s 2  t 2
2

t



  5 m/s   25 m 2 / s 2  4 4.9 m/s 2  20 m 
9.8 m/s

5  20.42
s  2.59 s or - 1.57 s
9.8
We cannot have a negative elapsed time, physically, so we ignore the negative root.
15
B. Two Dimensional Motion
The general approach is to decompose the vector equations of motion into component
equations. We obtain thereby a system of simultaneous equations to solve for the motion
along the coordinate axes.
1.
Projectile motion
The motion of a freely falling body in two or three dimensions is often called projectile
motion.
a.
Set up


At any instant, the projectile has displacement r , velocity v , and constant acceleration

a . Using the coordinate system shown in the figure, we’ll have a x  0 m/s and
a y   g . We are assuming that the projectile is initially at the origin of coordinates, and
that there is no other force, such as air resistance, acting.
The equation for the velocity vector is
 

v  vo  a  t  v xo ,v yo   a x , a y  t .
Decompose
v x  v xo  a x t  v xo
v y  v yo  a y t  v yo  ( g )  t
The components of the initial velocity are
v xo  vo cos  o
v yo  vo sin  o
The equation for the displacement vector is
  
1
r  ro  vo t  a  t 2 .
2
16
Decompose
x  xo  v xo t
1
1
y  y o  v yo t  a y t 2  y o  v yo t  ( g )  t 2
2
2
b.
Example
i) The maximum height occurs when v y  0 m/s . So we choose the equation involving y,
vy, and ay, and solve for y.
2
v y2  v yo
 2a y  y  y o 
y  yo 
2
v y2  v yo
2a y

0 m 2 /s 2  40 m/s  sin53 o
 0m
2  9.8 m/s 2



2
 52.1 m
ii) The elapsed time to reach that maximum height is
v y  v yo  a y t
t
v y  v yo
ay
0 m/s  40 m/s  sin53 o
 3.26 s
 9.8 m/s

iii) We might ask, at what time(s) is the projectile at a specified height, say y = 25 m?
y  y o  v yo t 
1
a yt 2
2
1
a y t 2  v yo t   y o  y   0
2
2v yo
2 y o  y 
t2 
t
0
ay
ay
t 2  6.52 s  t  5.10 s 2  0 s 2
t
 (6.52 s )  (6.52 s ) 2  4  5.1 s 2
2
17
 0.910 s and 5.61 s
The significance of the two roots is that the projectile achieves the height of 25 m on the
way up and on the way down.
iv) The velocity components at t = 0.910 s and at t = 5.61 s.
v x  v xo  a x t  vo cos  o  40 m/s  cos 53o  24.1 m/s
v y  v yo  a y t  vo sin  o  ( g )  t  40 m/s  sin 53o  (9.8 m/s 2 )  t
 31.9 m/s  9.8 m/s 2  t
time (s) velocity (m/s)
0.910
v x  24.1
v y  23.0
5.61
v x  24.1
v y  23.0
c.
Example
v x  v xo  a x t  20 m/s
v y  v yo  a y t  0 m/s  (-g)  t  9.8 m/s  2s  19.6 m/s
2.
Uniform circular motion
a.
Curvilinear motion


Envision an object having, at the moment, a velocity v , subject to an acceleration, a .
We might decompose the acceleration into components parallel to and perpendicular to
the velocity vector. The parallel acceleration component affects the speed of the object,
18
while the perpendicular component affects the direction of the velocity vector, but does
not change its magnitude.
b.
Circular motion
Uniform circular motion refers to motion on a circular path at constant speed. While the
magnitude of the velocity is constant, the velocity vector is not constant. The same is
true of the acceleration vector—its magnitude is constant but its direction is not.
However, the acceleration is always directed toward the center of the circular path. The
component of acceleration parallel to the velocity vector is zero. The acceleration
component directed toward the center of the circle is called the centripetal acceleration.
What acceleration is necessary to cause uniform circular motion? Let the origin be at the
center of the circle, as shown.
Consider two successive displacement and velocity vectors.
  
r  r2  r1
  
v  v 2  v1




By the definition of uniform circular motion, r  r1  r2 and v  v1  v2 and

 
v
 
a rad 
. In the limit as t  0 , v  v and r  r .
t
19
Representing these differences graphically, we can see by similar triangles that
v r

v
r
v vr v 2
a rad 


t rt
r
The magnitude of the centripetal acceleration always
v2
.
r
The centripetal acceleration is the perpendicular acceleration component for an object
executing curvilinear motion since we can suppose the object is following a circular arc
at any time, but perhaps with inconstant radius of curvature.
20
III. Dynamics
A. Newton’s “Laws” of Motion
In effect, Newton’s “Laws” define force and give two of the attributes of matter.
1.
First “Law”
“An object in uniform motion remains in uniform motion unless it is acted upon by an
external force.”
In this context, uniform motion means moving with constant velocity.
a.
Force
In other words, if the velocity of an object is changing (whether magnitude, direction, or
both), then a force is acting on the object. A force is an external influence that changes
the motion of an object, or of a system of objects.
We find that there are four fundamental forces in nature, gravity, electromagnetic force,
and the strong and weak nuclear forces. All other types of forces that we might give a
name to are some manifestation of one of the fundamental forces.
Force is a vector quantity having magnitude and direction. The dimensions of force are:
F   M L
T 
2
The SI unit of force is the Newton (N); the common British unit is the pound (lb).
Note that for historical reasons, the gravitational force may be called the weight.
b.
Inertia & mass
Two of the attributes of matter are i) resists changes in its motion—matter has inertia, and
ii) a force acts between any two pieces of matter—material objects or particles exert
forces on each other.
The quantitative measure of inertia is called the inertial mass of a particle. The SI unit of
mass is the kilogram (kg). The inertial mass of a particle is measured by observing the
acceleration that results from a known applied force. It is observed that if the same force
a
m
is applied to two objects, then 2  1 . So, we see that inertial mass is a relative
a1 m2
21
thing—we have to pick some object to define to be 1 kg, and compare all other objects
with that standard.
2.
Second “Law”
The Second “Law” makes quantitative the relation between applied force and the
resulting change in motion.
a.
Statement
“The change in motion of an object is directly proportional to the net external force.”
 d mv 
Symbolically,  F 
.
dt

 d mv 

dv
m
 ma .
If the mass of the object is unchanging, then  F 
dt
dt
This is the famous eff equals em ay.
Note that the acceleration results from the vector sum of all forces acting on the object.
b.
Momentum

The quantity mv is called the momentum. Like velocity, momentum is a vector quantity.
c.
Equilibrium

dp
 0 and
Should the vector sum of all forces acting on an object be equal to zero, then
dt


the object is said to be in static ( p  0 ) or dynamic ( p  0 ) equilibrium.
d.
Free body diagram(s)
To help us avoid confusion when identifying all the forces acting on an object or particle,
we draw what are called free body diagrams. A free body diagram is a sketch of the
object only, with arrows indicating each force acting only on that object. For instance,
here’s a block resting on an inclined surface. We sketch the incline and the block,
showing coordinate axes and the angle of incline and whatever might be given, such as
initial velocities, etc. Then, to one side, we draw a second sketch, which includes the
block only, as if it were isolated.
22
3.
Third “Law”
“For every action, there is an equal and opposite reaction.”
In the old days, the word action was used in the sense that we now use the word force.
a.
Action & reaction
Force is an interaction between two material objects. There is a gravitational interaction
between the Earth and the Moon. They exert forces on each other of equal magnitudes
but opposite directions.
b.
Example(s)

i) Consider a block resting on a table. The forces acting on the block are its weight, W ,

and the contact force, N , exerted by the table on the block. In turn, the block exerts a

contact force, N  , on the table. A contact force is the force one object exerts on another
by virtue of being in contact with that second object, i.e., touching. It is the force with
which two objects are pressing together. In fact, on the microscopic scale, two objects
are not really “touching” each other.


N and N  form an action-reaction pair, because they are acting on different objects.




We would never add N and N  together. On the other hand, N and W are both acting
on the block. We add them to obtain the net force on the block.
 


The reason this is confusing is that N + W may = 0 and N = - N  .
23
ii) Consider a block being dragged along a frictionless surface.
Decompose the forces acting on the block.
x: Fx  N x  Wx  ma x  Fx  0  0  ma x  F cos 30 o  ma x
y: Fy  N y  Wy  ma y  Fy  N  mg  m  0  F sin 30o  N  mg  0
Notice that in this case, the reaction of the block on the surface is N    N , but
N  mg  F sin 30 o .
B.
Case Studies
1.
Weight & Friction & Tension
a.
Weight & scales
Weight is the term we use to refer to the force of gravity near the Earth’s surface, or near
a planetary body’s surface, or near a moon’s surface, etc. In such a situation, the
magnitude of the weight is W  mg , and it’s directed downward.
We do not measure weight of an object directly. Instead, we place the object on a scale.
The number we read off of the scale is actually the contact force exerted upward by the
scale on the object. If the object is in equilibrium, then we infer that the weight has the
same magnitude. N  W  0  W  N
24
b.
Friction
When two objects slide against one another, there is a resistive force, called friction. On
the microscopic scale, it arises from the electromagnetic forces that act between
molecules and atoms.
Macroscopically, we observe that the frictional force F f is different if there is relative
motion between the two objects, or the two surfaces. If there is no relative motion, then
we have static friction, F f   s N , where N is the contact, or normal, force and  s is the
coefficient of static friction. If there is relative motion, then we have dynamic or kinetic
friction, F f   K N , where  K is the coefficient of kinetic friction.
In either case, the direction of the frictional force is opposite to the motion (kinetic) or
opposite to the pending motion (static). Its direction is always tangent to the surfaces at
the point or area of contact.
The value of the coefficient of friction depends on the nature of the two surfaces at which
the friction is acting. See the table in the textbook. Notice that the static friction does not
have a single value, but can be any value from zero (if N = 0) up to some maximum, at
which point the objects start to slide, and the friction becomes kinetic friction. Notice,
too, that  s   K in every case.
i)
Decompose the forces acting on the block.
x: Fx  Wx  N x  F fx  max  F cos 30o  0  0  N  max
y: Fy  Wy  N y  F fy  ma y  F sin 30o  mg  N  0  0
Let’s say F is given, then there are two unknowns, N and ax. Fortunately, we have two
equations. We can solve the y-equation for N and substitute that into the x-equation.
N  mg  F sin 30 o  a x 


1
F cos 30 o   mg  F sin 30 o
m
25

ii) Now suppose a second block is resting atop the block which is resting on the floor.
It is important to see that the applied force F is acting only on the lower block (m), and
not on the upper block (M). Let’s call the blocks 1 (upper) and 2 (lower). A free body
diagram of block 1 might look like this:
On the other hand, a free body diagram of block 2 will look like this. (The block 1 is still

present to show the origin of the contact force between the blocks). N 1 is the contact
force exerted on block 1 by block 2. So, the contact force exerted by block 1 on block 2


is  N 1 . The floor exerts a contact force on block 2, N 2 , but not on block 1, ‘cause block
1 is not touching the floor!
There are frictional forces between the blocks, and between block 2 and the floor, but
they are not the same.
We apply Newton’s 2nd “Law” to each of the blocks. (Notice the action-reaction pairs.)
Block 1
x: N1x  W1x  F f 1x  Ma1x  0  0  F f 1  Ma1x
y: N1 y  W1 y  F f 1 y  Ma1 y  N1  Mg  0  0
26
Block 2
x: Fx  N 2 x  W2 x  F f 2 x  R f 1x  R1x  ma2 x  Fx  0  0  F f 2  F f 1  0  ma2 x
y: Fy  N 2 y  W2 y  F f 2 y  R f 1 y  R1 y  ma2 y  Fy  N 2  mg  0  0  N1  0
The frictional forces would be found by setting F f 1  1 N 1 and F f 2   2 N 2 . The
coefficients of friction are 1 between the blocks and  2 between the floor and block 2.
We see, for instance, that a1x  a2 x and that N 2  mg , but mg  Fy  N1 ! And so on.
c.
Tension
Tension is a force that tends to pull apart. In our common introductory examples, we
consider a cord or rope, but more generally any material object, such as a rod or beam or
wooden board can be subject to tension.
The ideal cord is massless, non-stretchable and perfectly flexible. This means that it can
sustain tension, but cannot resist compression along its length. It means also that the
tension in the cord is the same throughout its entire length.
i) a mass hanging at rest from a cord
 

T  W  ma
Decompose (we have only the vertical direction, anyhow.)
y: T y  W y  ma y  T  mg  ma y  0  T  mg
27
ii) a mass hanging from two cords
Again, we only have one dimension. But, we have two objects—the hanging mass, m,
and the pulley, M. As before, T1  mg . For the pulley,
 


2T2  T1  W2  Ma 2
y: 2T2  T1  Mg  0  T2 
1
T1  Mg   1 m  M g
2
2
We have made use of the fact that the tension is the same T2 on either side of the pulley.
iii) a rope stretched across a chasm
The forces on the hanging mass are
 

W  Tleft  Tright  ma
Decompose
x: Wx  Tleftx  Trightx  0  0  T cos   T cos   0
y: W y  Tlefty  Trighty  0  mg  T sin   T sin   0  T 
28
mg
2 sin 
2.
Circular Motion
Earlier, we described uniform circular motion, defining the centripetal acceleration, etc.
Now, we examine how applied forces produce circular motion. We saw that for circular
v2
motion, the radial acceleration had to be equal to
. That is the a in the F = ma! Any
r
number of forces may cause the centripetal acceleration.
a.
Ferris wheel
Imagine one of the cars on a vertical Ferris wheel. Consider the forces acting on the car
at some arbitrary position on the circle. The velocity vector is always tangential to the
circle, but may not be of constant magnitude.
A free body diagram for the car--
Rather than x- and y-components, we have radial and tangential components.
Tr  Wr  mar
Radial component:
Tr  mg cos   m
Tangential component:
v2
r
Tt  Wt  mat
Tt  mg sin   mat
29
At the highest (   180 o ) and lowest (   0 o ) points, we have special cases. In those
cases, the tension is entirely radial because there is no tangential component of the
weight.
Tr  mg (1)  m
v2
r
We might be given v, r, and m and be asked to find the Tr.
b.
Banked curve
Consider an automobile driving around a banked curve. In cross section, it looks like an
object on an inclined plane. Viewed from above, it’s an object traveling on a circular arc,
perhaps with constant speed.
In connection with potential sliding up or down the incline, the friction is static friction if
the car is not sliding. Decompose:
x:
N x  Wx  F fx  max   N sin   0  F f cos   mar   N sin    s N cos   m
y: N y  W y  F fy  ma y  N cos   mg  F f sin   0  N 
v2
R
mg
cos    s sin 
Make the substitution for N in the x-component equation:
2
mg
sin    s cos   m v
cos    s sin 
R
Now, for instance, say  s  0, R  50m, v  13.4 m/s ; what must the angle  be if the car
is not to slide? We are looking for the limiting case. Notice that the mass of the car
appears on both sides of the equation, so it will divide out.
30
mg
v2
sin   0  m
cos   0
R
2
13.4 m/s 2
sin 
v


cos  gR 9.8 m/s 2  50 m
  tan 1 0.366
  20 o
c.
Merry-go-round, or Orbit
Consider a satellite moving around the Earth in a circular orbit at constant speed.
According to Newton’s “Law” of Gravitation, the Earth exerts a force on the satellite
Mm
equal in magnitude to Fg  G 2 , where G is the universal gravitational constant, M is
r
the mass of the Earth, m is the mass of the satellite, and r is the distance between the
satellite and the center of the Earth. In Newton’s Second “Law”


 F  ma
Fg  mar
G
Mm
v2

m
r
r2
Notice that the mass of the satellite divides out! We might solve for v in terms of the
orbit radius, r.
v
GM
r
The larger the orbit, the slower the satellite moves.
3.
Sliding & Pulling
a.
Inclined plane
Consider an object, such as a box or trunk, resting or sliding on an inclined plane surface.
That is, the surface is straight and flat, but inclined at an angle to the horizontal.
31
We apply Newton’s 2nd “Law” to the mass, m.
 

N  W  ma
It is convenient to orient the coordinate axes parallel and perpendicular to the inclined
surface, because we expect the mass to slide along that incline.
Decompose
x: N x  Wx  max  0  Wx  max
y: N y  W y  ma y  N  W y  ma y
What are the components of the weight this time?
Wx  W sin   mg sin 
W y  W cos   mg cos 
The components of the weight vector form a triangle similar to the triangle formed by the
inclined surface.
What if there is friction? What if there is an additional applied force?
32
We apply Newton’s 2nd “Law” to the mass, m.
  

N  W  F  ma
It is convenient to orient the coordinate axes parallel and perpendicular to the inclined
surface, because we expect the mass to slide along that incline.
Decompose
x: N x  Wx  Fx  max  0  Wx  F  max  mg sin   F  max
y: N y  W y  ma y  N  W y  0  ma y  N  mg cos   0

Had there been friction, the F f would have been parallel to the inclined surface, and we

would use the y-component equation to find N , whence F f  N .
b.
Pulleys & cords
Imagine an object hanging from an ideal cord that is wound around a set of two pulleys.
We assume the cord is perfectly flexible and does not stretch, and has no mass of its own.
This means that the tension within the cord is the same along its entire length.
Newton’s 2nd “Law” applies to each object. We’ll just need vertical components, so for
T
the pulley T2  T2  T1  0  T2  1 . For the hanging mass, T1  Mg  0  T1  Mg .
2
In other words, we find that we need to pull on the cord with a force (T2) of only half the
hanging weight (W) to support it. That’s why we have pulleys. They provide mechanical
advantage, similarly to the use of a lever.
33
4.
Restoring Forces
a.
Spring
If a spring is stretched by an external force, the spring exerts a force in the opposite
direction tending to restore it to its original length. For instance, when a mass is hung on
a spring, the spring is stretched by the weight of the mass. The spring exerts a force
upward on the mass. On the other hand, when an external force compresses a spring
(shortens its length), the spring pushes back.
A linear restoring force is such that Fs  k  x  xo  . This is the magnitude; the direction
is always such as to restore the spring to its resting length, the length of the spring when
no external force is acting on it. The proportionality constant, k, is called the force
constant.
Applying Newton’s 2nd “Law” to the figure above, with the hanging mass in equilibrium,


F  ma




Fs  W  ma
 Fs  mg  0
 k  x  xo   mg  0
x
mg
 xo
k
Notice that we have taken the +x direction to be in the direction of increasing springlength. So, x is positive downward in this case. Often we find it convenient to place the
origin at the end of the unstretched spring. Therefore, xo = 0.
34
b.
Pendulum
If a mass hung on a light cord is pulled to one side, and released, it will tend to swing
back toward its starting point. One way to analyze the pendulum is to regard it as a case
of circular motion. We have radial and tangential components:


F  ma

 

T  W  ma
radial: Tr  Wr  mar  T  W cos   m
vt2
v2
 T  mg cos   m t
r

tangential: Tt  Wt  mat  0  mg sin   mat  at  g sin   0
5.
More than One Object
a.
Inclined plane and pulley
35
We will ignore the mass of the pulley, and also assume that the pulley has no friction.
The inclined surface also is frictionless. There are two bodies in the system, and we draw
free-body diagrams of each of them.
Notice that a + direction of motion is indicated on the sketch. We are allowed to use
different coordinate axes for the two bodies, for our own convenience, as long as we
remain consistent with the (+/-) directions.
For m1, there is only vertical motion.
 

T  W1  m1 a1  T  m1 g  m1 a1
For m2, there are x- and y-components.
  

T  N  W2  m 2 a 2
x: Tx  N x  Wx  m2 a2 x  T  0  W2 sin   m2 a2 x
y: T y  N y  W y  m2 a 2 y  0  N  W2 cos   0
What we appear to have is 3 unknowns and 2 equations: T, a1, and a2x. However,
because the two masses are connected by a non-stretchable cord, a1  a 2 x ! So, we can
solve for T and for a1.
T  m1 g  a1 
a1  a 2 x 
1
W2 sin   T   a1  1 W2 sin   m1 g  a1   a1  1 W2 sin   m1 g 
m2
m2
m1  m2
36
b.
Atwood’s machine
Consider two masses hung over a massless, frictionless pulley by an ideal cord.
Once again we have two objects, with a free-body diagram for each object.
Newton’s 2nd “Law” applies to each object.



m1: T1  W1  m1 a1  T1  m1 g  m1 a1



m2: T2  W2  m2 a 2  T2  m2 g  m2 a 2
Notice the + direction is defined to be clockwise around the pulley. Too, since the pulley
has no mass nor friction, and the cord is ideal, a1  a2  a and T1  T2  T (magnitude).
37
T  m1 g  m1 a
 T  m2 g  m2 a
2 unknowns & 2 equations. . . .
a
C.
m2  m1
m  m1
2m1m2
 g and T  m1a  m1 g  m1 2
 g  m1 g 
g
m2  m1
m2  m1
m2  m1
Relative Motion
Query: To what extent does it matter, physically, where we stand while we observe the
motion of an object?
1.
Reference Frames
We always measure the position and velocity and acceleration vectors relative to a
convenient coordinate system—an origin and coordinate axes. The coordinate system is
called a reference frame.
a.
Position vectors
Consider a point in space. Its position is the vector that runs from the origin to the point.


The two position vectors, r and r  , point to the same point, yet they have different
magnitudes and directions. The origin of the “primed” reference frame is a point in

space, too, and has a position vector, R , relative to the
“unprimed”
reference frame. The


rule for vector addition tells us the relation between r and r  . Further, it may be that the

“primed” frame is moving with relative velocity, u , with respect to the “unprimed”
frame.
  
  
  
r  R  r   r  ut  r   r   r  ut
38
Notice that the relative velocity is measured in the “unprimed” frame of reference.
Further, we’ll label the reference frames as the S-frame and S’-frame rather than spell out
the words “primed” and “unprimed”.
b.
Relative velocities
The time rate of change in position is the velocity.



  
dr  dr dR


 v  v  u
dt
dt dt

du
 0 , whence
At this point we’ll assume that
dt


dv  dv
 

 a  a
dt
dt
These three equations together constitute the Galilean Transformation of coordinates.
Here’s an example. Let the S-frame be fixed to the Earth’s surface while an S’-frame is
pained on the wall inside the cabin of a jet liner flying overhead. We’d say that the S’frame is moving with velocity u with respect to the S-frame. Now, as shown in the
sketch, we choose the coordinate axes, both sets, to make things as simple as possible.
For instance, the orientations of the two frames are the same. We often imagine that the
two origin coincide at time t = 0. We get a simplified sketch of the situation. Keep in
mind that whatever object is located at the point P is not attached to either coordinate
system. It moves through space, and we are observing that motion from one reference
frame or another.
39
2.
Accelerated Reference Frames
a.
Inertial reference frame
An inertial reference frame is one in which Newton’s “Laws” of motion are valid—
observed accelerations result from observable physical forces, such as a spring, or
gravity, or tension in a cord, etc.
b.
Accelerated reference frame

du
 0.
Non-inertial reference frames are accelerated reference frames. That is,
dt
  
r   r  ut
  
v  v  u

  du
a  a 
dt
i) Consider a railroad boxcar. Inside, from the ceiling, hangs ball of mass, m, on a cord.
Imagine two observers, one standing outside the boxcar and the other standing inside the
boxcar. Both apply Newton’s 2nd “law” to the hanging mass.
As seen by the observer inside the car:
40


  ma 
F

 

T  W   ma   0
Decompose
x’: T sin   W  sin   0
y’: T cos  W  cos  0
Now here’s the thing. We normally think that the direction that an object hangs on a cord
defines the vertical. The gravitational force is presumed opposite in direction to the
tension.
As seen by the observer outside the boxcar:


F  ma

 

T  W  ma
Decompose:
x: T sin   0  max  0
y: T cos W  0
Now we bring the two views together by noting that a x 



du

W W m
dt
41
du
. Evidently,
dt
The two observers interpret the weight differently—different in magnitude and in
direction.
c.
On the rotating Earth
Consider a plumb bob hanging at rest, and let the
origin of our coordinate system be at the bob, so
 
r   r  0 . The “real” physical forces on the bob
are gravity and the tension in the cord. We have


also   0 and r   0 .
 
As seen in the rotating frame: T   W   0 . Since
we use plumb bobs to determine the vertical
direction, we’d say that T   mg   0 . However,

since the Earth is rotating, W 
does not point toward the center of
the Earth. As seen in an inertial
frame, Newton’s 2nd “Law”
 

appears as T  W  mao , where
ao 
v2

  2 is the centripetal
 

acceleration, directed toward the Earth’s rotational axis. Now, T  T   W  , so we can






write  W   W  mao  0 , or W   W  mao . The deviation from the true vertical
(pointing toward the center of the Earth), is a function of the latitude,  , since
sin e sin 

ao
g
sin e sin 

g
 2
sin e
sin 

g
R cos  2
R 2
sin( 2 )
2g 
Our scales actually measure W   mg  . However, since the Earth rotates slowly, g   g .
The same would not be true on a planet that rotates much faster than the Earth. See for
instance the science fiction novel A Mission of Gravity by Hal Clement.
sin e 
42
IV.
Work and Energy and Impulse and Momentum
A. Physical Work
In Physics and engineering, the word work has a specific definition. It’s basically equal
to a force times the displacement that occurs while the force is acting.
1.
Definition of Physical Work
a.
Path integral
Imagine a particle moving along a curved path through space. As it moves, an external
force acts on the particle. The vector ds is an infinitesimal displacement along the
path—it’s always tangent to the path.
 
W   F  ds
2
1
x2
y2
z2
x1
y1
z1
W   Fx dx   Fy dy   Fz dz
Work has dimensions of F L   M 
L
T 
2
2
. The SI unit of work is the Joule (J).
m2
. The British unit of energy is the foot-pound (ft-lb). The British Thermal
s2
Unit (BTU) is also used. Since the scalar product is involved, the result is a scalar.
1 J  1 kg
b.
One dimensional cases
i) constant force
 
W  F  s  Fs cos
For instance, friction acting on a sliding block:
43
 
W  F f  s  F f s cos180 o   F f s


Notice that if F  0 or s  0 or   90 o or 270 o , then W = 0. No work is done.
ii) varying force
We suppose that Fx = constant during an infinitesimal displacement, x . A tiny bit of
work is done by the force W  Fx x . The total work done during a displacement from
x1 to x2 is the sum
Wnet   Fx x .
If we graphed F vs x, we’d see that this is the integral of F(x) from x = x1 to x = x2.
In the limit as x  0 , the sum becomes an integral
x2
W   F x dx
x1
As an example of a force that varies with x, consider a spring.
44
The spring exerts a force on the block (mass = m) F x   kx , where x is the
displacement of the block from the equilibrium position (at the unstretched length of the
spring).
x2
k 
k
k
W   Fdx    kxdx    x 2   x12  x22
2
 2  x1 2
x1
x1
x2
x2
This is the work done by the spring on the block.
c.
Kinetic energy
At the same time, Fx  max . During the infinitesimal displacement, the elapsed time is
t . We assumed that the force is constant during that t , so a x 
W  Fx x  m

v x
.
t

v x
v v  v
1
1
1
x  m x 2 x 1x t  m v22x  v12x  mv22x  mv12x
t
t
2
2
2
2
What we see is that there is a quantity that changes while the work is being done. We
call that quantity the kinetic energy. It depends on the mass of the object and its speed.
1 2 1 2
mv2 x  mv1x  K is called the Work-Energy Theorem. When a
2
2
force does work on an object, it changes the object’s kinetic energy. [Check the
dimensions of K.]
The equation W 
d.
Power
Power is the rate at which work is done. If W is the work done during an elapsed time,
t , then the average power during that interval is
 
W
 P 
 F v
t
J
. A kilowatt is 1000 Watts. The electric
s
bills often mention kilowatt-hours. That’s one kilowatt times one hour = 3.6x106 Joules.
The SI unit for power is the Watt: 1W  1
Imagine a locomotive engine dragging a train along a straight track at a constant speed of
20 m/sec. Let’s say the engine exerts a force of 105 N and pulls the train 100 m. The
locomotive engine expends

P  F  v  Fv  105 N  20m / sec  2 106 W
P  F  d / t  10 5 N  100m / 5 sec  2  10 6 W
45
2.
Two Dimensional Case(s)
a.
Projectile
A free-falling projectile is acted upon only by gravity. As the projectile falls, gravity
does work on it, changing its kinetic energy.


dWg  Fg  dr
y
Wg    mgdy  mg y  K 
yo
1 2 1 2
mv  mvo
2
2
We might solve for v.
v
2
1 2
  mgy  mvo 
m
2

Notice that this gives us only the speed; we have to decide on the direction based on our
physical intuition.
Notice that all that matters are the magnitudes of the initial and final velocities, not their
directions! Energy has no direction; it’s a scalar.
b.
Inclined plane
An object on an inclined plane is acted upon by a number of different forces, gravity,
friction, the normal force, perhaps a tension in a cord or some other applied force. Each
and every force does work on the object as it moves on the inclined surface, from xo to x.
46
Wg   Mg  x  xo sin 
W f   F f  x  xo     K N  x  xo 
WN  N  x  xo cos 90 o  0
WT  T  x  xo cos 0 o  T  x  xo 
The total work done by the forces is just the sum of these
W  W g  W f  W N  WT
This total work in turn equals the change in the object’s kinetic energy that took place
while it slid along the surface.
W  K 
1 2 1 2
mv  mvo
2
2
B.
Conservation of Energy
1.
Conservative & Non-Conservative Forces
a.
Conservative forces
The work done by a conservative force is independent of the path taken from the initial
position to the final position. On the other hand, if the work done does depend on the path
taken, then the force is a non-conservative force, or a dissipative force.
47
b.
Potential energy
For a conservative force, the work done during a displacement depends only on the
coordinates of the beginning and end points.
We define a function, U, such that (for a one-dimensional case)
x2
WC   Fx dx  U  U 2  U 1 
x1
The function U is a function of x, and is called the potential energy function.
To obtain U for a given conservative force, solve the expression for the work done by the
force, W, for U2, and let U1 be zero. Because we are interested only in the change in
potential energy, we can set the zero anywhere we like.
x2
x2
x1
x1
U 2    Fx dx  U 1    Fx dx
c.
Dissipative forces
A dissipative force is a force like friction. The work done by a frictional force depends
on the distance an object slides from point 1 to point 2. If a longer path is followed
between points 1 and 2, then more work is done. We cannot define a potential energy
function for a dissipative force.
2.
Potential Energy Functions
a.
Spring
F  kx , where x is the change from the resting length of the spring-- x     o .
The work done by a spring is
1

1
1
Ws   Fx dx   F cos180 dx   kxdx   kx2  k 0 2    kx2
2
2
2

0
0
0
x
x
x
0
According to the definition of a potential energy function, Ws  U , so evidently,
48
Us 
b.
1 2
kx
2
Uniform gravity
y2
y2
y1
y1
W   Fy dy    mgdy  mgy2  mgy1   U 2  U 1 
We identify the potential energy function for a uniform gravitational field to be
U g  mgy
c.
Working backwards
We defined the potential energy in terms of work done by a force. The potential energy,
then, is related to the integral of the force over a displacement. The converse process is
to derive the force expression by taking the derivative of a given potential energy
function.
1 2
kx , where
2
x is the displacement from the equilibrium position of an object. Take the derivative. . .
Thus, the potential energy function for an elastic, or spring, force is U ( x) 
Fx  
d 1 2
 kx   kx
dx  2

Or, take the potential energy for uniform gravity (up is positive).
49
Fy  
d
mgy  mg
dy
We can extend this to two or three dimensions, with some hypothetical potential energy
function, U(x,y,z).
d
U  x, y , z 
dx
d
F y   U  x, y , z 
dy
d
Fz   U  x, y, z 
dz

F  F iˆ  F ˆj  F kˆ
Fx  
x
y
z
These three expressions are summed up in one by using what’s called the gradient
operator.

d
d
d
F  U x, y, z   iˆ U  ˆj U  kˆ U
dx
dy
dz
Qualitatively, the force is in the “downhill direction” of the potential energy function.
3.
Mechanical Energy Conservation
a.
Mechanical energy
The mechanical energy is defined to be E  K  U , the sum of kinetic and potential
energies. Of course, we add them only at the same position or the same time.
Now, if only conservative forces are acting, then WC  U  K . Evidently,
K  U  E  0
We have what is called a Conservation Principle or “Law.” If only conservative external
forces are acting, then the total mechanical energy of a system is conserved.
Dissipative forces, such as friction, do not conserve the total mechanical energy. In the
particular case of friction, some of the mechanical energy of the system is dissipated, or
“lost” in raising the temperature of the system.
50
b.
Examples
i) gravity near the Earth’s surface
K 2  U 2  K1  U 1
1 2
1
mv2  mgy2  mv12  mgy1
2
2
Notice that energy is a scalar quantity, so it will give us information about the magnitude
of velocity, but not the direction.
ii) spring
K 2  U 2  K1  U 1
1 2 1 2 1 2 1 2
mv2  kx2  mv1  kx1
2
2
2
2
4.
Gravitation
a.
Universal “Law” of Gravitation
Fg 
Gm1m2
r2
The direction is along the radius, r, and always attractive.
G is the universal gravitational constant, G  6.67  10 11
N  m2
.
kg 2
The gravitational masses of the two objects are m1 and m2. It turns out that the
gravitational mass is the same as the inertial mass, though that need not have been so.
r
12
m1
m2


Two objects exert a gravitational force on each other, equal and opposite. F12   F21
51
Near the surface of the Earth, Fg 
b.
GM earthm
GM earth
.
 mg  g 
2
2
Rearth
Rearth
Potential energy
The work done by the force of gravity while one mass is displaced from r1 to r2 is
r2
r2
r1
r
Wg   Fg  dr  
U (r )  
 Gm1 m2 Gm1 m2
Gm1 m2
dr  

2
r
r1
r
2


  U (r )


Gm1 m2
r
Let’s take a closer look at our (+/-) signs. The force is in the –r direction. The
displacement is in the +r direction, and r is always (+). The anti-derivative gives a (-)
sign. The W is negative change in U. Hence, U is (-).
For a system of objects, the total potential energy is the sum over pairs:
c.
Orbit
Consider two objects in orbit around each other, such as the Earth and the Moon. The
total mechanical energy of the system is
E
Gm1m2
1
1
m1v12  m2 v22 
2
2
r12
52
d.
Escape velocity
Here’s a question: How fast must an object, starting on or near the Earth’s surface, be
moving in order to escape from the earth’s gravity? We use the conservation of
mechanical energy to answer the question.
i) What constitutes escape? We’ll say that if the object just comes to rest as r   , it
has escaped.
ii) We’ll assume that once started, the object coasts—no rocket engines, etc.
iii) We have, therefore, r1  R , v1  vo , v2  0, & r2   .
Gm1 m2 1
Gm1 m2
1
m2 v 22 
 m2 v12 
2
r2
2
r1
00 
vo2 
GM easrthm2
1
m2 vo2 
2
Rearth
2GM earth
Rearth
Nm 2
2  6.67  10
 5.97  10 24 kg
2
m
miles
kg
 1.12  10 4  2.49  10 4
6
s
hour
6.38  10 m
11
vo 
2GM

R
m
will escape from
s
the Earth. Notice, too, that the direction of the initial velocity does not matter (as long as
it’s not straight down or something like that).
Any object, starting at the Earth’s surface with a speed of 1.12 10 4
A black hole is an object that is so dense that the escape velocity exceeds even the speed
of light. Therefore, even light cannot escape, and the object would be invisible to an
external observer.
53
C.
Conservation of Momentum
1.
Momentum & Impulse
a.
Impulse
Recall the definition of acceleration:  a x 
v x
. Multiply both sides by mt . . .
t
v x
t
 Fx  t  mvx 2  mvx1
mt   a x  mt 
On the left hand side is what is called the impulse—the force multiplied by the elapsed
time. On the right hand side is the change in momentum during the same elapsed time.


Extended to two or three dimensions, the impulse is J  F  t , and momentum is


p  mv . Analogous to the Work-Energy Theorem, we have the Impulse-Momentum


Theorem: J  p .
Impulse and momentum are vector quantities. If the force varies with time, we have to
integrate over time.
We define an impulsive force to be a force that acts for a very short time, and is much
larger in magnitude than other forces acting. Usually, we do not know the exact timedependence of the impulsive force. Therefore, rather than integrate the force over time,

we measure the  p that results from the impulsive force.
Example: A person of 70 kg mass approaches the ground (wearing a parachute) at
m
m
v1  9.9 . If he/she lands stiff-legged, v  v 2  0 in the space of t  0.002 s .
s
s
What is the average impulsive force on the person? (That is, average F over 0.002 s.)
We have one-dimensional motion, with downward being (+).
54
F t  p
 mv2  mv1
F  t  mv2  mv1 
F
 3.36  10 5 N
If the parachutist allows his/her legs to bend on landing, the t  0.05 s and
F 
mv1
 1.39  10 4 N
t
This is the reason we want our car fenders to crumple when the car runs into something.
The crumpling has the effect of prolonging the collision thereby reducing the average
force exerted on the car. The deformation of the fenders also dissipates some of the
kinetic energy.
Example: What is the force exerted on a wall when a ball is thrown against it?

We don’t know F t  during the impact, nor even the duration of the impact for that

matter. We can evaluate  p .
p x  mvx 2  mvx1
Let’s say that m = 0.5 kg, vx1 = 40 m/s, and vx2 = -20 m/s. Then the change in momentum
is

m
m
kg  m
p x  mvx 2  mvx1  0.5 kg - 20  40   30
s
s
s

This is the impulse on the ball! The ball exerts an equal and opposite impulse on the
wall. If the impact lasts t  0.01 s , then the average force on the wall is
Fx  
1
1
kg  m
p x 
30
 3000 N
t
0.01 s
s
55
b.
Conservation of momentum
Now, if the net external force is zero, then


J  p

0  p
The momentum of an isolated system is constant; it’s conserved.
If we divide the impulse by the time, then we get Newton’s 2nd “Law”
 p
 dp d (mv )
F


F
 t   dt  dt
2.
Collisions & Explosions
a.
Collisions
In a simple, idealized collision, we imagine two objects which collide, exerting only an
impulsive force on one another during the collision. Visualize two billiard balls. We
presume that no other net force is acting on the two balls, so the system of two balls is
isolated—the total momentum is conserved.


p2  p1  0
Momentum is a vector; the total momentum is the vector sum of the momenta of the two
balls.
We distinguish between elastic collisions and inelastic collisions. An elastic collision
conserves the total kinetic energy as well as the momentum. In an inelastic collision, the
total kinetic energy decreases to some degree. In a perfectly inelastic collision, the two
objects stick together, and become one.


To sum up: for elastic collisions, p  0 and K  0 . For inelastic collisions, p  0 ,
but K  I  0 . The I stands for the change(s) in kinetic energy of the objects. The
objects grow warmer, or possibly emit sound waves, etc.
56
i) elastic (1 dimensional)


p1  p 2
m A v A1  m B v B1  m A v A2  m B v B 2
K1  K 2
1
1
1
1
m A v A21  m B v B21  m A v A2 2  m B v B2 2
2
2
2
2
Because we can choose a reference frame to suit our convenience, we can usually set
vB1  0 .
ii) elastic (2-dimensional)
57


p1  p 2
m A v Ax1  m B v Bx1  m A v Ax 2  m B v Bx 2
m A v Ay1  m B v By1  m A v Ay 2  m B v By 2
K1  K 2
1
1
1
1
m A v A21  m B v B21  m A v A2 2  m B v B2 2
2
2
2
2
iii) plain inelastic


p1  p 2
m A v Ax1  m B v Bx1  m A v Ax 2  m B v Bx 2
m A v Ay1  m B v By1  m A v Ay 2  m B v By 2
K1  I  K 2
1
1
1
1
m A v A21  m B v B21  I  m A v A2 2  m B v B2 2
2
2
2
2
ii) perfectly inelastic
In a perfectly inelastic collision, in effect v A2  vB 2 .


p1  p 2
m A v Ax1  m B v Bx1  m A  m B   v x 2
m A v Ay1  m B v By1  m A  m B   v y 2
58
b.
“Explosions”
We might regard an explosion as a reverse perfectly inelastic collision. A single object
separates into two or more objects. The conservation of momentum still holds.
p1  p 2
m A  mB   v1  m A v A2  mB v B 2
c.
Rockets
Suppose the total mass of a moving object is not constant. Say the net external force

acting on an object (such as a rocket or a rain drop) is Fext . Assume that during a short

time interval, t , the Fext is approximately constant. Then the impulse delivered to the
mass, m, is


Fext t  p .
Further suppose that during that interval t the mass changes by an amount m . The
change in momentum that results is




p  mv2  mv2  m  mv1 .

 
 
p  mv2  v1   mv2  v1 
We want to rewrite this in terms of the change in velocity of the mass, m, and the relative
  
  
velocity of the m and m . Namely, v  v2  v1 and V  v 2  v 2 .
 


 
 
p  mv1  v  v1   m V  v1  v  v1






p  mv  mv  mV  m  m v  mV





The impulse, then, is Fext t  m  mv  mV . We may as well just let m + m be
m at this point.



Fext t  mv  mV
Divide by t ;



v m 
dv  dm
Fext  m

V m
V
.
t t
dt
dt
59


Recap: v is the velocity of the object (rocket or rain drop), V is the velocity of the m
dm
relative to the object, and
is the absolute value of the time rate of change in the mass
dt
of the object. Actually, we have to be careful of the directions of things. As derived

here, if m is leaving the object, then the object is losing mass and v is in the opposite

direction as V . Consider a rocket in the absence of gravity or any other external force.

dv  dm
0m
V
dt
dt

 dm
dv
m
 V
.
dt
dt


dv
V dm

dt
m dt
3.
Center of Mass
a.
Position

rcm 

m r
m
i i
i
60
b.
Velocity

mi v i

d 

vcm  rcm 
dt
 mi 


vcm   mi   mi vi  P
The total momentum of a system of particles is equal to the total mass of the system


times the velocity of the center of mass. P  Mvcm If no net external force acts on any


part of the system, then P is constant, and so is vcm . The individual parts of the system
may exert forces on each other, but those do not affect the motion of the center of mass.
c.
Acceleration
On the other hand, if one or more external forces acts on the system, then vcm is not
constant.
The sum of all forces acting on all parts of the system is





 F   Fext   Fint   mi ai  Macm

However, because of Newton’s 3rd “Law”,  Fint  0 .


 Fext  Macm
Consider a projectile that fragments in mid flight.
61
V. Rotational Motion & Oscillatory Motion
Until now, we basically ignored the fact that physical objects occupy volume, and have
shapes. We treated everything as if it were a particle—an object with mass but no
volume or internal structure.
A. Rotation of a Rigid Body
A rigid body is an object whose shape and dimensions (length, width, height, radius, etc.)
do not change. The distances between any two mass elements that comprise the body do
not change.
When a rigid body rotates about an axis, all parts of the body rotate at the same rate.
However, because different parts of the body are farther from or closer to the axis, all
parts do not have the same linear, or tangential speed.
1.
Rotational Kinematics
a.
Rotational variables
In the rotational equations of motion, we always use radians as the angular measure,
rather than degrees. Recall that the radian as an angular measure is the ratio between the
arc length and the radius of the arc.
62
By convention, Greek letters are used to represent angular variables.
name
definition
angular displacement, 
angular velocity component,

angular acceleration, 
b.

d
dt

d d 2
 2
dt
dt
Equations of rotational motion
The equations for cases of constant angular acceleration are the same as those for
constant acceleration. Just imagine all of those divided through by a radius.
1
2
   o  ot   t 2
  o   t
 2   o2  2    o 
  o
  
2
Every point in and on the body executes circular motion.
s  r
v t  r
a t  r
ar 
vt2
 r 2
r
Example: a rigid body, such as a disk, rotates about an axis perpendicular to the plane of
the disk.
63
t  3 s
o  0
  234 (radians)
radians
s
  constant
  108
The average angular velocity is   
 234  0 radians
radians

 78
.
t
3
s
s
The angular acceleration is

   o   2     60 radians
radians



 20
2
t
t
t
3
s
s2
[Since α is constant,   
2.
Rotational Dynamics
a.
Moment of inertia
o  
2
.]
A moment of ____________ is equal to __________ times a squared distance from a
point, or axis. The moment of inertia is mass times radius squared from a point.
For individual, or discrete particles, I   mi ri 2 . For an object having a continuous mass
density,  , we have to integrate over the volume of the object.
I
r
2
Volume
 r
dm 
Volume
The dimensions of moment of inertia are [M][L]2.
64
2
dxdxdz
Examples. . . .
65
Parallel Axis Theorem: The moment of inertia is just a sum of scalars. Therefore, we
can divide an inconveniently shaped body into sections that are easier to handle. By a
similar token, we can construct the moment of inertia of a body about an arbitrary axis
from the moment about an axis passing through the center of mass of the body.
66
b.
Kinetic energy
Each mass element that comprises the rigid body is moving in a circle with speed
vi  ri  . We could add up a total kinetic energy for all the mass elements:
Kr 
1
1
1
mi vi2   mi ri 2 2  I 2

2
2
2
Notice that the angular velocity is common to all the mass elements.
c.
Angular momentum
Consider a mass, m, with momentum, p . Relative to a point O, the mass has an angular
momentum equal to
  
Lrp
iˆ
x
mv x
67
ˆj
y
mv y
kˆ
z
mv z
The magnitude is L  rp sin  , while the direction of the angular momentum vector is
obtained with the right-hand-rule.
d.
Torque
The time-rate-of-change of the angular momentum is

 



 
dL d r  p   dp dr   dp

r

pr
0 rF
dt
dt
dt dt
dt
Define the torque, or moment of force, about the point O

   dL
  rF 
dt
The magnitude of the torque about the point O is rF sin  and the direction is obtained
from the right-hand-rule.
e.
Conservation of total angular momentum
For a system of particles or of rigid bodies, the total angular momentum of the system
about the point O is the vector sum of all the individual angular momenta, about the same
point O.


Ltotal   Li
68
If the system is isolated, so that the net external force is zero, then Ltotal is conserved.
That means both in direction and in magnitude.
A rotational collision:
A particle of mass m = 0.05 kg and speed v = 25 m/s collides with and sticks to the edge
of a uniform solid disk of mass M = 1 kg and radius R = 0.3 m. If the disk is initially at
rest on a frictionless axis through its center, what is its angular velocity after the
collision?


The total angular momentum of the system is conserved. L2  L1 .
mR 2  I  mvR  0  mR 2 

f.
1
MR 2  mvR
2
m
v
radians
 7.58
1
R
s
m M
2
Work done by a torque
F is the force applied a the point P.
ds is a short displacement.
r is the position vector from the selected axis, O, to the point P.
 is the angle between r and F .
 
dW  F  ds  F sin  rd    d
The work done on a rigid body by an external torque is equal to the change in rotational
kinetic energy that results.
W  K r 
The power expended is P 
1 2 1 2
I 2  I1
2
2
dW
  .
dt
69
3.
General Motion of a Rigid Body
a.
Rolling along
When a rigid body is both translating and rotating, we can divide its total kinetic energy
into two parts.
K
1
1
2
Mv cm
 I 2
2
2
Here, the angular speed is about the center of mass.
b.
Case studies
i)
Consider an ideal cord wound around a solid cylinder of radius R = 0.5 m and mass
M = 10 kg. The cylinder is set on a horizontal axis and mass of m = 4 kg is hung on the
free end of the cord. What’s the torque experienced by the cylinder and what’s the
acceleration downward of the mass, m?
First thing we do is assume that the cord does not slip on the cylinder.
We have a system of two bodies, to which we apply Newton’s 2nd “Law.”
The mass, m:
F
y
 ma y
T  mg  ma
70
The cylinder, M:
F
y
0
N  Mg  T  0
  I
T  R  I
a
R
The tension would tend to cause a clockwise rotation, so it produces a negative torque.
a
Because the cord does not slip on the cylinder,   . We have therefore two equations
R
and two unknowns.
a
R
TR 2
TR 2
2T
I  TR  0  a  TR  


2
R
I
I
M
0.5MR
T  m(a  g )  0  T  m( g  a)
2mg  a 
1
m
a
g  4.36 2
M
s

M 
1 

 2m 






1
T  m g 
g   21.8 N

M  

1 


2
m

 

a
ii)
Rolling down an incline
71
72
73
c.
Gyroscope & precession

Lo  Lo iˆ

 

L
    riˆ   Mgkˆ  0iˆ  Nkˆ 
t

L
  rMgˆj
t

Notice that L is not down, but sideways. The spinning disk does not fall, but rather
precesses in a horizontal arc.


4.
Static Equilibrium
a.
Conditions for equilibrium of a rigid body
For a rigid body to be in equilibrium, both the translational and rotational accelerations
must be zero. Therefore, the sum of the external forces and the sum of the external
torques must both be zero.


F
 i  Macm  0


 i  I  0
The torques must be computed about the same axis, but the choice of axis is arbitrary—it
need not be necessarily through the center of mass.
74
b.
Case studie(s)
i) A uniform beam of length r = 4 m and mass 10 kg supports a 20 kg mass as shown.
The beam in turn is supported by a taut wire. What’s the tension in the wire?
The beam is in static equilibrium. Therefore, the sum of forces on the beam is zero, and
the sum of torques exerted by those forces is zero.
Fb is the weight of the beam itself.
Fx  Tx  Fgx  Fbx  0
Fx  T sin 60 o  0
Fy  T y  Fgy  Fby  0
Fy  T cos 60 o  20kg  g  10kg  g  0
To compute torques, we need to specify an
axis of rotation. We choose one that will
eliminate some of the unknowns, for
instance the end of the beam where it joins
the wall. The force F exerts zero torque about that point. Then the sum of torques
reduces to:
T  g b  0
T  4m  sin 67 o  20kg  g  4m  sin 53o  10kg  g 
4m
 sin 53o  0
2
3.68m  T  626.13 Nm  156.53Nm  0
T  213N
We return to the force component equations in order to find the force the wall exerts on
the beam.
Fx  T sin 60 o  213N  0.866  184 N
Fy  30kg  g  T cos 60 o  294 N  213N  0.5  188 N
ii) ladder leaning against a wall
The mass of the ladder is M.
The angle is θ.
The length of the ladder is L.

F  0

  0
The forces on the ladder are the normal contact forces and
friction exerted by the wall and the floor and gravity, which
75
acts at the center of mass of the ladder. Each of those forces exert a torque about some
selected axis. Typically, the selected axis is either the top or bottom of the ladder, or
both. It may be the ladder’s center of mass, as well.
If, for instance, the wall is not “smooth”, then there are potentially 4 unknown force
components, and therefore 4 equations are needed.
F
F


x
0
y
0
O
0
O
0
Since this is a two-dimensional situation, we indicate the directions of the torque by +/signs.
B. Oscillatory Motion
Here, we speak of things that go back and forth in a regular way.
1.
Simple Harmonic Motion
Harmonic means there is a sine or cosine in it.
a.
Description
Let y be the displacement of a particle from its equilibrium position. Then by definition,
if the particle is executing simple harmonic motion its position as a function of time is
yt   A sin 2 f t   
A is the amplitude, the maximum displacement either side of equilibrium.
f is the frequency of oscillation, in cycles/second.
 is a phase factor, which depends on the initial y at t = 0.
Graphically,
We find sometimes other parameters useful, such as
76
T is the period of the oscillation. T 
1
.
f
 is the angular frequency, in radians/second.   2 f
[This was called the angular velocity, and had a line on top in the context of rotational
motion. Here, omega is a scalar, not a vector.]
b.
Equations of motion
y  A sin  t   
dy
  A cos t   
dt
dv
a
  2 A sin  t   
dt
v
Notice that a   2 y ; the acceleration is proportional to the displacement from
equilibrium, but in the opposite direction. This is exactly what we saw with springs,
Fy  ky . In Newton’s 2nd “Law”
77
F  ma

 ky  m   2 y


k
m
Evidently, linear restoring forces give rise to simple harmonic motion.
c.
Energy
The total mechanical energy of a simple harmonic oscillator of mass, m, is
1 2 1 2
mv  ky
2
2
1
1
1
1
2
E   0  kA2  mA  k  0
2
2
2
2
1
1
1
2
E  mv 2  ky 2  mA
2
2
2
E
Given v, we can find the corresponding y; given y, we can find the corresponding v.
Notice that the total energy is proportional to the square of the amplitude of the
oscillation.
d.
Applying initial conditions to the equations of motion
y  A sin  t   
v  A cos t   
Suppose that at t = 0, y = yo and v = vo. Then
y o  A sin  
vo  A cos 
tan   
vo
 yo
 2 y o2   2 A 2 sin 2  
vo2   2 A 2 cos 2  
A
y o2 
vo2
2
Given the position and velocity of the mass at one time (usually t = 0) we are able to
specify its motion exactly at any other time.
78
2.
Pendulum
a.
Non-simple pendulum
Consider a mass hanging on the end of a light string, cord or stiff rod. Two forces act on
the mass, the tension in the string, cord, or rod and gravity. The string, cord or rod
constrains the mass to move on a circular arc of radius equal to the length of the string,
cord or rod.
The force of gravity exerts a torque on the mass, about the pivot point of the pendulum
   mg sin   I  m 2
 mg sin   m 2
g
   sin 

Notice that the mass has divided out. Notice that the torque is restoring, tending to return
the mass to   0 .
b.
Simple pendulum
Now we make a simplification by assuming that the pendulum swings in a small arc. In
that case, sin    , in radians.
g

  
d 2
g
 
2

dt
g
. The frequency, or period,

of a simple pendulum depends only on its length (and on g), not on how much mass is
hung on the end.
This exactly simple harmonic motion, if we identify  
79
c.
“Physical” pendulum
Of course, all pendula are physical, except for politico-socio-economic pendula. The
“Physical” pendulum is a pendulum in which the mass is not concentrated at the end of a
string, cord, or rod.
In the equation of motion, we use the moment of inertia. Assuming small angular
displacement (a simple “physical” pendulum, as it were),
  I
 mg sin   I

d 2
mg


2
I
dt
In this case, the length,  , is the distance from the pivot point to the center of mass of the
mg
pendulum. We can identify the angular frequency as before:  
.
I
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VI.
Wave Motion
A. Waves
A mechanical wave is a disturbance that propagates through a material medium.
Electromagnetic waves are moving electromagnetic fields and do not require a material
medium to propagate—they can propagate through a vacuum. All forms of waves have
certain “wave-like” properties in common. [Propagate is a word meaning to travel, but it
means more than that. It implies that the wave is traveling by reproducing itself.] Here,
we will be discussing mechanical waves.
1.
General Wave Properties
a.
Wave pulse
i) wave speed is a property of the medium.
ii) shape of the wave pulse is unchanged as it travels
iii) two or more wave pulses that exist at the same place & time in a medium add—
superimpose.
Mathematically, we write the displacement of the medium at position x and at time t as
x t 
y x, t   Af   
 T 
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b.
Periodic waves
A periodic wave consists of a series of identical wave pulses, so the wave pattern is
repeating. The easiest periodic wave mathematically is the sine wave (or cosine).
The amplitude is A, the wavelength is  ; the frequency is f; the wave speed is c   f .
As the wave passes a specified point in space, the medium there executes simple
harmonic motion.
  x t 

 x 
y x, t   A cos 2      A cos 2f   t  
   T 

c
 



The (+) sign yields a wave traveling from right to left; the (-) sign a wave traveling form
2
left to right. We define the wave number it be k 
.

yx, t   A coskx   t 
c.
Transverse and longitudinal waves
The direction of the disturbance of the medium may be either perpendicular to the
direction of wave-travel, or it may be parallel to the direction of wave-travel. The former
is called a transverse wave, while the latter is called a longitudinal (or compression)
wave. For the time being, we won’t concern ourselves with the distinction, but waves in
a stretched string are transverse, while sound waves are longitudinal.
d.
Refection
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e.
Superposition
If two or more wave disturbances exit at the same place in a medium and at the same
time, they simply add to give a net displacement of the medium. It is entirely possible
then, to have waves traveling in a medium that superimpose to give a zero net
displacement of the medium.
i) standing waves
Consider two sine waves traveling in a medium in opposite direction and having the same
amplitude, A.. If their wavelengths are the same, they will superimpose to form what is
called a standing wave pattern.
Every point in the medium oscillates with a constant amplitude. At some points, spaced
one half wavelength apart, the medium does not oscillate—the amplitude is zero. These
points are called nodes. Midway between the nodes are the antinodes—points whose
amplitude of oscillation is a maximum—twice the amplitude of the individual waves.
The standing, or stationary wave pattern does not move through the medium.
yx, t   y1 x, t   y 2 x, t 
cosa  b  cos a cos b  sin a sin b
yx, t   A coskx   t   A coskx   t 
yx, t   2 A sin kxsin  t 
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ii) beats
Consider two waves traveling through a medium, in the same direction but slightly
different frequencies. They will superimpose to form a disturbance whose amplitude
rises and falls periodically. These patterns are called beats. The frequency of the beats is
equal to the difference in the two original frequencies. The effective frequency within
f  f2
the envelope is f e  1
, the average of the two separate frequencies.
2
y  y1  y 2
y  Acos2 f1t   cos2 f 2 t 

f  f2  
f  f2 
y  2 A cos 2 1
t  cos 2 1
t 
2
2

 

f.
Energy transport
While the medium in which the wave propagates does not flow from one place to
another, the wave disturbance nonetheless carries energy from one place to another.
Each mass element, dm, of the medium executes simple harmonic motion. K is the
K
restoring force constant. It’s related to the frequency by  
.
dm
1 2 1
Ky  dmv 2
2
2
1
1
E  dm 2 A 2 cos 2 kx   t   dmA2 2 sin 2 kx   t 
2
2
1
E  dm 2 A 2 cos 2 kx   t   sin 2 kx   t 
2
E


1
. The total mass of the
2
medium spanning one cycle (or one wavelength) is  , where  is the mass per unit
length of the medium.
Over one cycle, the cosine-squared and sine-squared average to
E   2 f 2   2 f 2  2 2 f 2  A 2 f 2
In terms of the wave speed, c, E  2 2  c f A2 . The energy flux is the power
transported through the medium by the wave: P  fE  2 2  c f 2 A2 . The intensity is
P
the power per unit area through which the power is transported: I   2 2  c f 2 A 2 ,
a
where  is the mass per unit volume.
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3.
Stretched String
a.
Tension
One end of a stretched string is fixed, as if tied to a wall. That end cannot be displaced.
If the free end is disturbed by an outside force, then a pulse is generated which then
propagates along the string, until it encounters the wall. The fixed end cannot move, so a
reflected pulse is created which travels back along the string, but in reversed orientation.
Regardless of the exact shape of the pulse, it travels with the same speed. We observe
F
that the wave speed in a stretched string is c 
. By extension to three-dimensional
m
L
media, c 
intermolecular forces
.
mass density
We can derive the wave speed in the string by applying Newton’s 2nd “Law” to a short
segment of the string.
y x, t   A coskx   t 
y
v y  x, t  
 A sin kx   t 
t
2 y
a y x, t   2   2 A coskx   t    2 y
t
But also,
2 y
 k 2 A coskx   t   k 2 y
x 2
whence
2 y k 2 2 y

x 2  2 t 2
This is the wave equation for a wave travelling along the x-axis with speed v 
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
k
 f .
b.
Normal modes
Because of the boundary conditions imposed on the string, only certain waves are able to
persist in the string. In particular, since the ends of the string are fastened, they cannot
move—the ends must be nodes. Only waves for which the string length equals an
integral number of half-wave lengths will persist in the string. Put another way, the
string will vibrate at only certain discrete frequencies, corresponding to those allowed
wavelengths. The allowed frequencies are called normal modes of the string. The
normal modes are also known as the string’s natural frequencies of vibration. Every
physical system has natural vibration modes, based on its dimensions and elasticity. For
a string, those are its length and its tension and its mass per unit length.
The normal modes frequencies for the string are f n 
v
n

nv
, where v is the wave
2L
speed in the string.
c.
Spectra
An arbitrary wave form may be expressed as a superposition of sine waves having
differing amplitudes. This exactly analogous to expressing a function as a Fourier Series.
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The Fourier components are just the normal modes, fn. The chart of the amplitudes is
what we call the spectrum.
y t    An sin 2f n t 
n
4.
Pressure waves--Sound
a.
Longitudinal waves
b.
Pressure amplitude
c.
Properties of the medium, speed of sound
d.
Doppler Effect
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VII. Thermodynamic
A.
“Heat” Flow
1.
Temperature
a.
b.
2.
Energy Transport
a.
Mechanisms
b.
R-values
B.
“Laws” of Thermodynamics
1.
Energy Changes
2.
Entropy
3.
Heat Engines
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