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Physics 103 – Introduction to Physics I
Motion
Forces
Energy
1
First
Dimensions
Units
Precision
Coordinate Systems
Vectors
Kinematics
Motion Variables
Constant Acceleration
2
Dimensions
The dimension of a physical quantity specifies what sort of quantity it is—
space, time, energy, etc.
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 either electric charge [Q] or electrical current [A].
For example, energy - E   M L .
T 
2
2
The physical quantity speed has dimensions of L
T
.
3
Units
International System (SI)
The units of the fundamental dimensions in the SI are
dimension
SI
cgs
Customary
[L]
meter
(m)
second
(s or sec)
kilogram
(kg)
centimeter
(cm)
second
foot
(ft)
second
gram
(g)
slug or
pound-mass
[T]
[M]
The SI units will be introduced as we go along.
4
Unit Conversions
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
Suppose we wish to convert 2 miles into meters. [2 miles = 3520 yards.]
The units cancel or multiply just like common numerical factors. Since we want
to cancel the yards in the numerator, the conversion factor is written with the
yards in its denominator.
5280 feet 1 yard 91.44 meters
2 miles 
1 mile

3 feet

100 yards
 3218.688 meters
Since each conversion factor equals 1, the physical measurement is unchanged,
though the numerical value is changed.
Note: the units are a part of the measurement as
important as the number. They must always be kept
together.
5
Precision & Significant 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.
6
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.
756.  37.2  0.83  2.5  796.53  800  8.0  10 2
3.563  3.20  11.4016  11.4  1.14 101
5.6    17.59291886  18  1.8 101 or is it 1.76 101
[Notice the ambiguity. Do we speak of the number of significant digits, or of
the relative “place” of the uncertain digit? That is, should it be 18 or 17.6?]
7
Coordinate Systems
We measure locations in space relative to
a coordinate system. Firstly we select the
origin of coordinates, and then the
directions of orthogonal axes.
Since the directions shown by orthogonal
axes are mutually perpendicular,
components along different axes are
independent of each other.
The commonly used two-dimensional
coordinate systems are the Cartesian and
the plane polar systems.
8
The three dimensional Cartesian coordinate
system is comprised of three mutually
perpendicular, straight axes, commonly denoted
x, y, & z or î , ˆj , & k̂ .
[We’ll talk about those hat-things later.]
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.
9
Vectors
As used in Physics, a scalar is a quantity that has only
one property—a magnitude. Energy, speed,
temperature, and mass are scalar quantities.
A vector is a quantity that has two properties—a
magnitude and a direction. Displacement, velocity,
acceleration, and force are vector quantities.
In text or equations, vectors are denoted with

either a line or an arrow on top, thusly: A or A
In diagrams, a vector is represented by an arrow.
In text books, vectors are often
denoted by bold-faced letters: A .
Weirdly, University Physics uses both
bold-face & an arrow!
A is not the same as A! . . . .
10
The directions defined by the Cartesian coordinate axes are symbolized by unit
vectors, î , ˆj , k̂
. A set of unit vectors that define a coordinate system are
called a basis set.
Two dimensional:
Vx  V cos 
V y  V sin 
Components
V  Vx2  V y2
  tan
1
Vy
Vx

a

aˆ  
a
A unit vector is a vector of magnitude 1. E.g.,
, where
is the
a

magnitude of the vector a . Often, the magnitude of a vector is indicated by
the letter without the arrow on top: a  a
.

A A
11
An arbitrary vector can be written as a sum of the basis
set unit vectors.

A  Axî  Ay ˆj  Az k̂
Direction cosines
Ax
cos  
A
Ay
cos  
A
Az
cos  
A
emweb.unl.edu
12
Adding vectors
The sum of two vectors is also a vector.
  
A B  C
Drawn to scale.
Ax  Bx  C x
Ay  B y  C y
Az  Bz  C z

C  C x î  C y ˆj  C z k̂
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.
13
Vector Products
scalar (or dot) product—result is a scalar; the operation
is symbolized by a dot.
 
A  B  AB cos   Ax Bx  Ay By  Az Bz


The angle  is the angle from A to B .
   
  
   
Note: A  B  B  A and A  B  C  A  B  A  C


.
Vector (or cross) product—result is a another vector; the operation is
  
symbolized by a cross,  .
A B  C



C  C  AB sin  , direction perpendicular to both A and B
according to the right-hand-rule. [Use the three-finger version.]
iˆ
  
C  A  B  Ax
Bx
ˆj
Ay
By
kˆ
Az  iˆAy Bz  By Az   ˆj  Ax Bz  Bx Az   kˆAx By  Bx Ay 
Bz
14
Kinematics
Simply describe the motion of an object.
15
Motion Variables

r
The displacement vector, ,
The average velocity during the time


r
interval t is defined to be  v 
.
t
It’s the time-rate-of-change in the
displacement. In terms of vector
components, we’d write
x
y
z
.
 v x 
,  v y 
,  v z 
points from the origin to the
present location of the particle.

r
If a particle is at 1 at time t  t1

r
and at 2 at some later time t  t 2,
then we say the change in
displacement is r  r2  r1 .
Likewise, the elapsed time or
time interval is t  t 2  t1 .
Similarly, the average   
acceleration is  a  v  v2  v1
t
t 2  t1
t
.
t
The instantaneous velocity is defined to be


r dr

v  lim

dt .
t 0 t
.
The instantaneous acceleration is



v dv
a  lim


t
dt
t 0
t
  
t, r , v , a
16
Constant Acceleration
v x v x 2  v x1
 a x 

t
t 2  t1
v x 2  v x1  a x t 2  t1 
t2  t1   x2  x1  2
 vx 
x2  x1
vx 2  v x1
vx 2  v x1
vx22  v x21
ax 

x2  x1
2x2  x1 
2
vx 2  vx1
vx22  vx21  2a x x2  x1 
 v x 
x x2  x1

t t 2  t1
x2  x1   vx  t 2  t1 
 vx 
vx 2  vx1 
x2  x1 
2
and 
vx 2  vx1  t
2
x2  x1
t 2  t1
2  t1   x1 
If ax is constant, then also
 vx 
vx 2  vx1
.
2
vx1  a x t 2  t1   vx1
t2  t1 
2
1
2
x2  x1  vx1 t 2  t1   a x t 2  t1 
2
Commonly,
t1  0, t2  t , x1  xo , x2  x, vx1  vxo , vx 2  vx
We have four equations that each relate three of the motion variables.
17
Space-time
Mathematically we can treat time and space on the same footing.
The displacement vector in space-time has 4 components.

r  ct , x, y , z
The scaling factor c is needed to make the units of all 4 components
the same, e.g., meters. The geometry of space-time is not
Euclidian, but is non-Euclidian. Therefore,
î  î  ˆj  ˆj  k̂  k̂  1; t̂  t̂  1
18
Example: a train traveling on a straight and level track
starting from rest; ends at rest.
time, t
(seconds)
acceleration, ax
(m/s2)
0 - 10
2
10 – 40
0
40 - ?
-4
What is the total displacement?
Segment 1: We are given the acceleration, elapsed time and initial velocity — vxo = 0 m/s.
x  xo  v xo t 
1
a x t 2
2
1
2
x1  0 m  0 m/s 10 s   2 m/s 2  10 s   100 m
2
Segment 2: To find the total displacement at the end of the second segment, we need the
velocity component at the end of the first segment.
v x1  v xo  a x t  0 m/s  2 m/s 2  10 s  20 m/s
vx 2  vx1  20 m/s since ax  0
1
1
2
x2  x1  v x1t  a x t 2  100 m  20 m/s  30 s   0 m/s 2  30 s   700 m
2
2
Segment 3: For this segment, we know x2, vx2, vx3, and ax, but not t .
v x23  v x22  2  a x x3  x2 
2
v x23  v x22
0  20 m/s 
x3  x 2 
 700 m 
 750 m
2  ax
2   4 m/s 2


19
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. We observe that a y  9.8 m/s 2 .
What are the height of the sandbag and its velocity as functions of
time?
1
m
1
m 2
2
y  y o  v yo  t  a y  t  20m  5  t  9.8 2  t
2
s
2
s
m
m
v y  v yo  a y  t  5  9.8 2  t
s
s
Free Fall!
What is the y-component of the
sandbag’s velocity when it hits the
ground?
2
v y2  v yo
 2 a y  y  yo 


2
v y2  v yo
 2a y  y  y0   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. (We know it' s going down.)
20
How long does that take?
v y  v yo  a y t  t 
v y  v yo
ay

 20.4 m/s  5 m/s
 2.59 s
2
 9.8 m/s
Alternative solution for
the elapsed time:
1
y  yo  v yo  t  a y  t 2
2
0 m  20 m  5 m/s  t 
t


1
 9.8 m/s 2  t 2
2


  5 m/s   25 m 2 / s 2  4 4.9 m/s 2  20 m 
9.8 m/s 2

5  20.42
s  2.59 s or - 1.57 s
9.8
21
Projectile Motion
Constant acceleration, in two dimensions.

m ˆ
ˆ
a  0i  9.8 2 j
s
22
Vector equations:
  
1 2
r  ro  vot  a  t
2
Component equations:
  
v  vo  a  t
v xo  v o cos o
v yo  v o sin o
x  xo  vxot
1 2
1
y  yo  v yot  a y t  yo  v yot  ( g )  t 2
2
2
v x  v xo  a x t  v xo
Notice:
The ycomponent
of a is
ay = -g.
v y  v yo  a y t  v yo  ( g )  t
23
Example:
How long does it take to reach
maximum height, ymax?
At maximum height, vy = 0 m/s
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
What is the maximum height?
v y2  v yo2  2a y  y  yo 
y  yo 
v y2  v yo2
2a y

0 m /s  40 m/s  sin53
 0m
2  9.8 m/s 2
2
2



o 2
 52 .1 m
24
When is the projectile
at y = 25m?
1 2
y  yo  v yo t  a y t
2
1 2
a y t  v yo t   yo  y   0
2
2v yo
2 yo  y 
2
t 
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
 0.910 s and 5.61 s
25
What are the velocity
components then, at
t = 0.910 s and 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)
0.910
5.61
velocity (m/s)
v x  24.1
v y  23.0
vx  24.1
v y  23.0
26
Example:
How far does the object travel in the x-direction?
1 2
x  xo  voxt  a x t
2
We need to know the elapsed time, t.
The total elapsed time is the time it takes to go up plus the time it takes to come down.
Previously, we found that the time to reach maximum height was t = 3.26 s.
1
The total time, then, is 2x3.26s = 6.52 s. [Verify with 0  ymax  0  t  9.8  t 2 .]
2
1 2
x  xo  voxt  a x t  0  voxt  0
2
m
x  voxt  24.1 6.52 s   157 m
s
27
Example:
What are the velocity & position components at t = 3 seconds?
v x  v xo  a x t  20 m/s  0 m/s 2  3s  20 m/s
v y  v yo  a y t  0 m/s  (9.8 m/s 2 )  3s  29.4 m/s
x  xo  v xo t  0 m  20 m/s  3s  60m


1 2
1
2
y  yo  v yo t  a y t  0 m  0 m/s  3s   9.8 m/s 2 3s   44.1 m
2
2
28
29
Uniform Circular Motion
Curvilinear motion – not in a straight line.
Envision an object having, 
at the moment, a velocity
subject to an acceleration a .
.
 , and
v
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, while the perpendicular component affects the
direction of the velocity vector, but does not change its
magnitude. At any instant, the velocity to tangent to the curve.
30
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.
31
Let the origin be at the center of the
circle, as shown.
32
Consider two successive displacement and velocity vectors.

  
r  r2  r1
  
v  v2  v1



v  v1  v2
By the definition of uniform circular motion, r  r1  r2
arad

v

t
In the limit as t  0 ,
 
v  v
&
 
r  r
v r
.
Both
are isosceles triangles, with the same angle.

v
r
v 1 vr vr v 2
arad 



t t r
rt
r
.
The centripetal or radial acceleration is always v 2 on a
circular arc of radius r.
r
33
Second
Dynamics
Newton’s “Laws”
Energy
Momentum
Conservation
34
Dynamics
Relationships among Motion and Force and Energy.
35
Newton’s “Laws” of Motion
“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.]
“The change in motion of an object is directly proportional to the
net external force.”
.
“For every action, there is an equal and opposite reaction.”
36
37
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 particles of matter interact through one or more of these four
fundamental forces.
All other types of forces that we might give a name to are some
manifestation of one of the fundamental forces.
Dimensions of force are
F   M 2L
T 
The SI unit of force is the Newton (N).
1 N = kg m/s2
38
Fundamental concepts:
i) Space and time
ii) Matter and energy
Macroscopic objects—collections of many atoms & molecules.
Molecules—combinations of several atoms; chemical substance.
Atoms—combinations of protons, neutrons & electrons;
chemical element.
Subatomic particles—protons, neutrons, electrons, et al.
A particle is an idealized object that has no shape or internal
structure. Any object may be treated as if it were a particle
depending on the context.
39
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.
Imagine two particles exerting equal and opposite forces on each other.
We observe their accelerations.
a2 m1

a1 m2
 dp
 F  dt
We write Newton’s 2nd “Law”
in mathematical form:

The quantity p 

mv
2
1
v
c2
called the momentum.
40
Classical assumptions:
i) time is independent of space and is absolute.
ii) 3-d space is Euclidian—”flat.”
Unless a particle is moving at a very great speed, its
momentum is approximately


p  mv
Further, if the particle’s mass is unchanging, then

 dp
dv
 F  dt  m dt
41
42
Equilibrium
Should the vector sum of all forces acting on an object be equal to zero, then

dp
0
dt
and the object is said to be in
Static equilibrium

p0
Dynamic equilibrium

p0
43
Isolated body diagram(s)
An isolated body diagram is a sketch of the object only, with arrows indicating each
force acting only on that object.
44
Action & Reaction
Force is an interaction between two material objects. E.g., there is a gravitational
interaction between the Earth and the Moon. They exert forces on each other
of equal magnitudes but opposite directions.


N  and N form an action - reaction pair.


W and N are not an action - reaction pair.


N   N
 
N W  0
45
Newton’s Universal “Law” of Gravitation

M 1M 2
Fg 12  G
rˆ
2
r12
Any two objects exert gravitational
forces on each other, equal in
magnitude and opposite in direction.
r̂
Take care with the directions. The unit vector points from M1 to M2.
The gravitational force on M2 is in the
direction, toward M1.
 r̂
46
Gravitational Field
Let’s say M1 is at the origin of coordinates. The presence of M1
gives rise to a gravitational field that extends outward into space.

M1
g  G 2 r̂
r

An object of mass M2 located at r
experiences a gravitational force.


M 1M 2
Fg  M 2 g  G
r̂
2
r
In the context of the 2nd “Law”
The acceleration due to gravity is


M 1M 2
Fg  G
r̂  M 2 g
2
r
M1
ar  a g  G 2
r
47
Near the Earth’s surface, a g  G
M1
ME
m

G

9
.
8
r122
RE2
s2
Near the surface of another body, such as the Moon or Mars, the
acceleration due to gravity is different, not 9.8 m/s2.
48
Weight
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.


Fg  mg
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.
F
y
N  Fg  0  N  Fg
49
Suppose the object is not in equilibrium.
F
y
N  Fg  mA
N  mA  Fg
N  mA  mg
Suppose A = -g. Then N = 0.
The object is in free fall,
but not weightless. The term weightless
is a misnomer.
50
Friction
Friction always opposes
the motion of an object, or
what’s called the object’s
impending motion.
decompose
Kinetic or Dynamic Friction Ff   K N
Static Friction Ff   S N
Fx  Wx  N x  F fx  max  F cos 30o  0  0  N  max
Fy  Wy  N y  F fy  ma y  F sin 30o  mg  N  0  0
N  mg  F sin 30 o  a x 


1
F cos 30 o   mg  F sin 30 o
m

51
Example: Two objects
N1x  W1x  F f 1x  Ma1x  0  0  F f 1  Ma1x
N1 y  W1 y  F f 1 y  Ma1 y  N1  Mg  0  0
F f 1  1 N 1
Ff 2  2 N 2
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
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
52
Cords & Tension
T1  mg
 

T  W  ma
T y  W y  ma y  T  mg  ma y  0  T  mg
2T2  T1  W2  Ma2
2T2  T1  Mg  0  T2 
1
T1  Mg   1 m  M g
2
2
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 (as long as we ignore friction).
53
Example:
  


F

W

T

T

left
right  ma
Wx  Tleftx  Trightx  0  0  T cos   T cos   0
No help.
mg
Wy  Tlefty  Trighty  0  mg  T sin   T sin   0  T 
2 sin 
54
Pulleys
Apply Newton’s
2nd “Law” to the
pulley and to the
hanging mass.
 W p  T2  T2  T1  0  T2 
T1  W p
2
T1  Mg  0  T1  Mg
55
Case Studies in Applying Newton’s 2nd “Law.”
Circular motion
Inclined plane
Restoring forces—spring & pendulum
Systems of objects
56
57
Circular Motion
ferris wheel
Tr  Wr  mar
Tt  Wt  mat
v2
Tr  mg cos   m
r
Tt  mg sin   mat
  180o
  0o
v2
Tr  mg( 1 )  m
r
58
driving ‘round a curve
v2
N x  Wx  F fx  max   N sin   0  F f cos   mar   N sin    s N cos   m
R
example
mg
s  0, R  50m, v  13.4 m/s
cos    s sin 
mg
v2
sin   0  m
cos


0
R
2
2
2
v


sin

v
13
.
4
m/s
m


cos  gR 9.8 m/s 2  50 m
R
N y  Wy  F fy  may  N cos   mg  F f sin   0  N 
mg
sin    s cos  
cos    s sin 
  tan 1 0.366
  20 o
59
circular orbit
Universal “Law” of
Gravitation
Fg  G
Mm
r2


F
  ma
Newton’s 2nd “law”
Fg  mar
Mm
v2
G 2 m
r
r
Orbital speed
v
GM
r
60
Inclined Plane
 

N  W  ma
N x  Wx  max  0  Wx  max
N y  Wy  may  N  Wy  ma y  0
Wx  W sin   mg sin 
W y  W cos   mg cos 
  

N  W  F  ma
N x  Wx  Fx  max  0  Wx  F  max  mg sin   F  max
N y  Wy  Fy  may  N  Wy  0  may  N  mg cos   0
61
Restoring Force
Hooke’s “Law”
linear restoring force
Fs  k     o 
Fs  k  x  xo 


F  ma

 

Fs  W  ma
 Fs  mg  0
 k    o   mg  0
mg

 o
k
Typically, we place the origin at the
resting length of the spring.
62
Pendulum


F  ma

 

T  W  ma
Radial and tangential components.
vt2
vt2
Tr  Wr  mar  T  W cos   m  T  mg cos   m
r

Tt  Wt  mat  0  mg sin   mat  at  g sin 
63
System of Objects
 

T  W1  m1a1  T  m1 g  m1a1
  

T  N  W2  m2 a2
Tx  N x  Wx  m2 a2 x  T  0  W2 sin   m2 a2 x
Ty  N y  Wy  m2 a2 y  0  N  W2 cos   0
a1  a 2 x
T  m1 g  a1 
a1  a2 x 
a1 
1
W2 sin   T  
m2
Newton’s “Laws” apply to
each object as well as to the
system as a whole.
1
W2 sin   m1 g  a1   a1  1 m2 g sin   m1 g 
m2
m1  m2
64
“Atwood’s Machine”
 

T2  W2  m2 a2  T2  m2 g  m2 a2
 

T1  W1  m1a1  T1  m1 g  m1a1
a1  a2  a
T1  T2  T
T  m1 g  m1 a
 T  m2 g  m2 a
a
T  m1a  m1 g  m1
m2  m1
g
m2  m1
m2  m1
2m1m2
 g  m1 g 
g
m2  m1
m2  m1
65
Reference Frames
The position vector of the point P
is written down with respect to
two different reference frames.
        
r  R  r  r  ut  r  r  r  ut
The motion variables, as measured
by observers in different frames.



dr dr dR
  


 v  v  u
dt dt dt

du
0
dt
 
a  a
66
  
r   r  ut
  
v  v  u

  du
a  a 
dt
Apply Newton’s 2nd “Law” in the two reference
frames.


F

m
a

 

T  W  ma


  ma 
F

 

T  W   ma   0
x’:
y’:
T sin   W  sin   0
x:
T sin   0  max  0
T cos  W  cos  0
y:
T cos W  0
du
dt



du
W W  m
dt
ax 
Now, bring them together:
The perceived weight is different.
67
An Accelerated Reference Frame


2
du d R 
 2 A
dt
dt
In the
elevator
Outside the
elevator
Fc  mg   0
Fc  mg  mA
mg   mg  mA
The reading on the scale is F’c. The observer in
the elevator interprets that as his/her weight.
68
Energy
Work
Kinetic Energy
Potential Energy
69
Physical Work
 
W   F  ds
2
1
x2
y2
z2
x1
y1
z1
W   Fx dx   Fy dy   Fz dz
F L  M  L
T 
2
dimensions
2
The SI unit of work is the Joule (J).
m2
1 J  1 kg 2
s
70
constant force
 
W  F  s  Fs cos 
 
W  Ff  s  Ff s cos180o   Ff s
71
 
dW  F  ds  Fx dx
force varies with position
Wnet   Fx x
x2
W   F x dx
x1
x2
k 2 
k 2 k 2
W   Fdx    kxdx    x   x1  x 2
2
 2  x1 2
x1
x1
x2
x2
72
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
The SI unit for power is the Watt: 1W  1
J
s
 
W
 P 
 F v
t
. A kilowatt is 1000 Watts.
The electric bills often mention kilowatt-hours.
That’s one kilowatt times one hour = 3.6x106 Joules.
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 106W
or
P  F  d / t  10 5 N  100m / 5 sec  2  10 6 W
73
Kinetic Energy
An increment of work is done during an incremental
displacement. We assume that the applied force is
constant during the incremental displacement.
W  Fx x  m


vx
v v  v
1
1
1
x  m x 2 x 1x t  m v22x  v12x  mv22x  mv12x
t
t
2
2
2
2
Work-Energy Theorem
Kinetic energy
W
1 2 1 2
mv2 x  mv1x  K
2
2
1 2
K  mv
2
74
Examples:


Wg  Fg  r  Fgx x  Fgy y  0  mgy
Wg  mg y  K 
v
1 2 1 2
mv  mvo
2
2
2
1 2

mg

y

mvo 

m
2

Keep in mind: y  y  yo
Energy is a scalar; it has no directional
information.
75
Every force acting
does some work.
The block slides up the
Incline from xo to x.
W g   Mg  x  xo sin 
W f   F f  x  xo     K N  x  xo 
W N  N  x  xo cos 90 o  0
WT  T  x  xo cos 0 o  T  x  xo 
W  Wg  W f  WN  WT
W  K 
1 2 1 2
mv  mvo
2
2
76
Potential Energy
Conservative forces are those for
which the work done during a
displacement is independent of the
path followed. Call the work done
by a conservative force, WC.
 
W   F  ds
2
1
The potential energy function is defined
thusly:
WC  U
x2
y2
z2
x1
y1
z1
WC   Fx dx   Fy dy   Fz dz
For x-components
To derive the potential energy
function for a specified force, we
evaluate the work.
x2
x2
x1
x1
U 2    Fx dx  U 1    Fx dx
x2
WC   Fx dx  U  U 2  U 1 
x1
Because we are interested in potential
energy changes, we can set the zero of
potential energy for convenience.
77
Conversely, we can derive the force components from the potential
energy function:

U x , y , z 
x

Fy   U  x , y , z 
y

Fz   U  x , y , z 
z

F  Fx î  Fy ˆj  Fz k̂
Fx  
For a spring:
For uniform gravity:
Gradient operator




F  U x , y , z   î U  ˆj U  k̂ U
x
y
z
Fx  
d 1 2
 kx   kx
dx  2

Fy  
d
mgy  mg
dy
78
WC  U
Spring:
1 2 1 2
1 2
WC   Fx dx   F cos180 dx   kxdx   kx  k 0    kx
2
2
2

0
0
0
x
x
x
0
1 2
U s  kx
2
79
Gravity:
y2
y2
y1
y1
WC   Fy dy    mgdy  mgy2  mgy1   U 2  U1 
U g  mgy
80
Mechanical Energy
E  K U
Wtotal  Wother  Wc  K
Wtotal  Wother  U  K
Wother  K  U  E
If only conservative external forces are acting, then the total
mechanical energy of a system is conserved.
E  Wother  0
81
Gravity and spring restoring forces are conservative. Friction is
non-conservative.
Let us say that a number of forces do work on an object.
Wtotal  Wother  WC  K  U  I
K  U  I  0
E  I  0
The I is the change in internal energy of the object.
Typically it is manifested as an increase in temperature.
E.g., friction causes a decrease in mechanical energy & an
increase in internal energy.
82
Near the Earth’s surface,
K 2  U 2  K1  U 1
1 2
1 2
mv2  mgy2  mv1  mgy1
2
2
Spring
K 2  U 2  K1  U 1
1 2 1 2 1 2 1 2
mv2  kx2  mv1  kx1
2
2
2
2
83
Recall the “Law” of Gravitation
Gm1m2
Fg 
r2
G  6.67  10
r
N  m2
kg 2
12
m1
m2
F12 
Fg 
11
Gm1m2
r122
GM earthm
GM earth

mg

g

2
2
Rearth
Rearth
84
Gravitational potential energy
  r2 Gm1m2
 Gm1m2 Gm1m2 
  U (r )
Wg   Fg  dr  
dr  

2
r
r1 
 r2
r1
r1
r2
U (r )  
Gm1m2
r
Notice that we are setting U = 0 at 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 (-).
85
Escape from Earth
Define “escape”
1
Gm1m2 1
Gm1m2
m2 v22 
 m2 v12 
2
r2
2
r1
00 
vo2 
r1  Rearth , v1  vo , v 2  0, & r2  
1 2 GM earthm
mvo 
2
Rearth
2GM earth
Rearth
Nm 2
24
2  6.67 10

5
.
97

10
kg
2
kg
4 m
4 miles


1
.
12

10

2
.
49

10
6.38 106 m
s
hour
11
vo 
2GM earth
Rearth
Notice the direction does not matter!
(Assuming the direction is not straight
down.) Nor does the mass m.
Compare the escape velocity with the
orbital speed of the space shuttle in a
circular orbit of altitude 325 km. It’s
about 7.6x103 m/s (17000 mph).
86
Impulse & Momentum
Go back to the definition of acceleration.
Impulse
v x
 a x 
t
vx
mt   ax  mt 
t
 Fx  t  mvx 2  mvx1
J x  Fx  t  p x
In general



J  p   Fdt


Conservati on of momentum, if F  0, then p  0.
87
A ball bounces straight off a wall.
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
88
One dimensional elastic collision


p1  p2
mAv A1  mB vB1  mAv A2  mB vB 2
K1  K 2
1
1
1
1
mAv A21  mB vB21  mAv A2 2  mB vB2 2
2
2
2
2
Inelastic collision—kinetic energy is not conserved.
K2  K1  I
89
Two
dimensional
elastic
collision


p1  p2
m Av Ax1  mB vBx1  m Av Ax 2  mB vBx 2
m Av Ay1  mB vBy1  m Av Ay 2  mB vBy 2
K1  K 2
1
1
1
1
m Av A21  mB vB21  m Av A2 2  mB vB2 2
2
2
2
2
90
“Perfectly” inelastic collision


p1  p2
m Av Ax1  mB vBx1  m A  mB   vx 2
m Av Ay1  mB vBy1  m A  mB   v y 2


p1  p2
mA  mB  v1  mAv A2  mB vB 2
91
92
Center of Mass

m
r

 ii
rcm 
 mi

m
v

d 
 ii
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 v .
cm
The individual parts of the system may exert forces on each other, but those
do not affect the motion of the center of mass.
93
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

F

F

m
a

M
a
  ext  int  i i
cm
Because of Newton’s
3rd

“Law”,  Fint . 0
Consequently, the center of mass of a system of particles moves like a particle
of mass M.
.


 Fext  Macm
94
Rockets & Rain Drops
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.


F
Then the impulse delivered to the mass, m, is ext t  p
Further suppose that during that interval




p  mv2  mv2  m  mv1

 
 
p  mv2  v1   mv2  v1 
  
v  v2  v1
t , the mass changes by an amount m .
We may as well just let m + Δm
be m at this point.
  
V  v2  v2

.



Fext t  m  mv  mV




 

p  mv1  v  v1   m V  v




p  mv  mv  mV






p  mv  mv  mV  m  m v  mV




Fext t  p  mv  mV



v m 
dv  dm
Fext  m

V m
V
t t
dt
dt
95
Recap:

v
V
is the velocity of the object (rocket or rain drop),
is the velocity of the
m relative to the object, and
dm
is the absolute value of the time rate of change in the mass of the object.
dt
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
In real life, there is always gravity, friction, air
resistance, etc.
96
Third
Rotation
Vibration
Wave Propagation
97
Rotation
Rigid body
Arc length, radians
There are 2 radians in 360o.
s
s  r or   in radians
r
name
definition
angular displacement, 
angular velocity component, 
A rigid body is one in which all the
rij are constant.
angular acceleration, 


d
dt
d d 2
 2
dt
dt
98
Equations of rotational motion
1
2
   o  o t   t 2
  o   t
 2   o2  2    o 
  o
  
s  r
2
vt  r
at  r
vt2
ar 
 r 2
r
Example:
t  3 s
o  0
  234 (radians)
radians
  108
s
  constant
o  unknown

   o   2     60 radians
radians




20
t
t
t
3
s2
s2
  
 234  0 radians
radians

 78
t
3
s
s
99
Moment of Inertia
I   mi ri2
I
 r dm    r dxdydz
2
Volume
2
Volume
The dimensions of moment of inertia are [M][L]2.
100
101
Rotational Dynamics
Angular kinetic energy & angular momentum
Kr 
1
1
1
mi vi2   mi ri 2 2  I 2

2
2
2
î
  
Lrp x
mvx
Rigid body:
ˆj
y
mvy
k̂
z
mvz
torque

 



 
dL d r  p   dp dr   dp

 r    p  r  0  r F
dt
dt
dt dt
dt


 
dL
  r F 
dt


 

 
L   Li   ri  pi   i mi i    mi i2  I 



 dL
d

I
I
dt
dt
102
103
Unwinding
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?
F
y
 ma y
T  mg  ma
F
y
N  Mg  T  0
  I

a
R
2 m g  a 
1
m
a
g  4.36 2
M 
M
s

1 

 2m 





1
T  m g 
g   21.8 N

M  

1


 

 2m  

a 1
a 1
m
  I  I  MR 2  10kg0.5m(4.36 2 )  10.9 Nm
R 2
R 2
s
a
0
T  R  I
a
R
TR 2
TR 2
2T
I  TR  0  a  TR  



R
I
I
M
0.5MR 2
T  m(a  g )  0  T  m( g  a)
a
R
104
Roll down an incline


F  ma
Mg sin   f  Mg sin   N  Ma x
N  Mg cos   0  N  Mg cos 
a x  g sin    cos  


  I
(We' ll compute torques about the center of mass.)


 
 N   g   f  I
0  0  f R   MgR cos  
5
2
  
g
cos 
R
2
MR 2
5
Now, if the ball is to roll without
slipping, what must be true?
The friction must be just right
such that
a
 x
R
1
1
2
2
Kinetic energy: K  2 Mv  2 I
2
No slipping. . . .
 v  1 1
1
12
K  Mv 2 
MR 2      Mv 2
2
25
 R   2 5
105
Gyroscope
Pulley

Lo  Lo iˆ

 
L
    riˆ   Mgkˆ  0iˆ  Nkˆ 
t

L
  rMgˆj
t


106
107
108
Static Equilibrium
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  T sin 60 o  0

T  4m  sin 67 o  20kg  g  4m  sin 53o  10kg  g 
Fy  T y  Fgy  Fby  0
Fy  T cos 60  20kg  g  10kg  g  0
o

T  g  b  0
Fx  Tx  Fgx  Fbx  0
4m
 sin 53o  0
2
3.68m  T  626.13Nm  156.53Nm  0
T  213N
Fx  T sin 60 o  213 N  0.866  184 N
Fy  30kg  g  T cos 60 o  294 N  213N  0.5  188 N
109
Oscillation
yt   A sin 2ft   
A is the amplitude, the maximum displacement either side
of equilibrium.
f is the frequency of oscillation, in cycles/second (Hz).
 is a phase factor, which depends on the initial y at t = 0.
T
1
f
  2 f radians/s
y  A sin  t   
dy
v
  A cos t   
dt
dv
a
  2 A sin  t   
dt
For an object bouncing on a spring:
F  ma

 ky  m   2 y

k
m

110
Mechanical energy of an
oscillator
Again, think of an object oscillating
horizontally on a spring. The mechanical
energy is
1 2 1 2
mv  ky
2
2
1
1
1
1
2
E  m  0  kA2  mA  k  0
2
2
2
2
1
1
1
2
E  mv2  ky2  mA
2
2
2
E
Notice E is proportional to the square of the amplitude.
111
pendulum
  mg sin   I  m 2
 mg sin   m 2
g

   sin 
Assume the oscillations are small
sin   
Physical Pendulum
  I
 mg sin   I
d 2
mg
  2 

I
dt

mg
I
g
  

d 2
g



2

dt

g

112
113
Wave Propagation
A wave is a disturbance in an elastic medium which travels, or
propagates through the medium. The wave is intangible. The medium
itself does not travel, but only oscillates back and forth. So there is not
a net transport of matter from place to place.
However, a wave transports energy from place to place, through the
medium.
Waves come in many forms, all with certain common properties.
There are waves in a plucked string, seismic waves, sound waves,
electromagnetic waves. These are different sorts of disturbances
propagating in different sorts of media.
In this course, we will consider the common properties.
114
115
Wave Motion
x t 
y x, t   Af   
 T 
y is the displacement from
equilibrium at position x and
at time t. f is an unspecified
function.
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.
116
Harmonic wave – a wave of a particular shape that repeats
itself. It’s periodic.
  x t 

 x  



y x, t   A cos 2     A cos 2f   t 
   T 

c
 



k
2

  2f
yx, t   A coskx   t 
Each point in the medium (x) is displaced from equilibrium (y).
As time passes, the pattern is shifted by a phase factor   ct ;
the wave pattern moves through the medium.
117
superposition
reflection
“Standing waves”
y x, t   y1 x, t   y 2 x, t 
y x, t   A coskx   t   A coskx   t 
y x, t   2 A sin kxsin  t 
cosa  b  cos a cos b  sin a sin b
118
Stretched string
c
F
m
L
Only vibrations that “fit” in the length of the string will persist. This is an example of
resonance. Every physical system has “natural” modes at which it will vibrate. The
natural modes depend on the physical properties of the system: mass, elasticity, size.
We saw this same phenomenon with the spring and the pendulum.
119
“Beats”
In this case, two waves are traveling in the same direction,
but with slightly different frequencies.
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

 

120
Spectrum
 2

y   Ai cos x  2f i t 
i
 i

121
.
Energy
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 restoring force constant.
It’s related to the frequency by
1
1
K
E  Ky 2  dmv 2

2
2
dm
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
Over one cycle, the cosine-squared and sine-squared average to 1
2
. The total mass of the medium spanning one cycle (or one wavelength) is 
, where  is the mass per unit length of the medium.


E   2 f 2   2 f 2  2 2 f 2  A2 f 2
2
2
In terms of the wave speed, c, E  2  c f A
.The energy flux is the power transported through the medium by the wave:
P  fE  2 2  c f 2 A2
The intensity is the power pr unit area through which the power is transported:
I
P
 2 2  c f 2 A 2
a
, were

is the mass per unit volume.
122
Pressure waves--Sound
Compression, or longitudinal waves.
Medium oscillates parallel to direction of propagation.
Pressure amplitude, yp.
Speed of sound waves depends on density, pressure, temperature &
elasticity of the medium.
Doppler effect. . . .
deciBels. . .
123
124
The “Laws” of Thermodynamics
125