Physics 103 – Introduction to Physics I
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Transcript of Physics 103 – Introduction to Physics I
1
Physics 103 – Introduction to Physics I
Motion
Forces
Energy
2
First
Dimensions
Units
Precision
Coordinate Systems
Vectors
Kinematics
Motion Variables
Constant Acceleration
3
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 - .
The physical quantity speed has dimensions of .
2
2
T
L ME
Dimensions
T
L
4
dimension SI cgs Customary
[L] meter(m)
centimeter(cm)
foot(ft)
[T] second(s or sec)
second second
[M] kilogram(kg)
gram(g)
slug or pound-mass
International System (SI)The units of the fundamental dimensions in the SI are
Units
The SI units will be introduced as we go along.
5
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 . In other words, meters yards 44.91100
0.144.91
100
meters
yards
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.]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. Since each conversion factor equals 1, the physical measurement is unchanged, though the numerical value is changed.
meters .yards
meters .
feet
yard
mile
feet miles 6883218
100
4491
3
1
1
52802
Unit Conversions
6
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.
7
The uncertainty in a numerical value may be expressed in terms of a tolerance, as
Alternatively, the uncertainty can be shown in scientific notation simply by the number of digits displayed in the mantissa.
2 digits, the 5 is uncertain.
3 digits, the 0 is uncertain.
005.0273.23
3105.1
31050.1
2100.880053.7965.283.02.37.756
1101414114016112035633 .....
11 10761108118592918861765 . it is or ...
[Notice the ambiguity. Do we speak of the number of significant digits, or ofthe relative “place” of the uncertain digit? That is, should it be 18 or 17.6?]
8
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.
9
The three dimensional Cartesian coordinate system is comprised of three mutually perpendicular, straight axes, commonly denoted x, y, & z or .
[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.
k & ,j ,i
10
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: Aor 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! . . . .
11
x
y
yx
y
x
V
Vtan
VVV
sinVV
cosVV
1
22
The directions defined by the Cartesian coordinate axes are symbolized by unit vectors, . A set of unit vectors that define a coordinate system are called a basis set.
k ,j ,i
A unit vector is a vector of magnitude 1. E.g., , where is the magnitude of the vector . Often, the magnitude of a vector is indicated by the letter without the arrow on top: .
a
aa
ˆ a
a
aa
Components
AA
Two dimensional:
12
An arbitrary vector can be written as a sum of the basis set unit vectors.
kAjAiAA zyx
emweb.unl.edu
Direction cosines
A
Acos
A
Acos
A
Acos
z
y
x
13
Adding vectorsThe sum of two vectors is also a vector.
CBA
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.
Drawn to scale.
kCjCiCC
CBA
CBA
CBA
zyx
zzz
yyy
xxx
14
scalar (or dot) product—result is a scalar; the operation is symbolized by a dot.
The angle is the angle from to . Note: and .
zzyyxx BABABAcosABBA
A
B
ABBA
CABACBA
Vector (or cross) product—result is a another vector; the operation is symbolized by a cross, .
CBA
yxyxzxzxzyzy
zyx
zyx ABBAkABBAjABBAi
BBB
AAA
kji
BAC ˆˆˆ
ˆˆˆ
, direction perpendicular to both and according to the right-hand-rule. [Use the three-finger version.]
sinABCC
A
B
Vector Products
15
Kinematics
Simply describe the motion of an object.
16
The displacement vector, , points from the origin to the present location of the particle. If a particle is at at time and at at some later time , then we say the change in displacement is . Likewise, the elapsed time ortime interval is .
r
1r
1tt
2r
2tt
12 rrr
12 ttt
The average velocity during the time interval is defined to be . It’s the time-rate-of-change in the displacement. In terms of vector components, we’d write
.
tt
rv
t
zv ,
t
yv ,
t
xv zyx
The instantaneous velocity is defined to be
.dt
rd
t
rv
t
lim0
Motion Variables
Similarly, the average acceleration is .
The instantaneous acceleration is
.
12
12
tt
vv
t
va
dt
vd
t
va lim
t
0
avrt
, , ,
17
Constant Acceleration
1212
12
12
ttavv
tt
vv
t
va
xxx
xxxx
1212
12
12
ttvxx
tt
xx
t
xv
x
x
1221
22
12
21
22
12
12
12
12
121212
2
22
2
xxavv
xx
vv
vvxx
vva
vv
xx
v
xxtt
xxx
xx
xx
xxx
xxx
21212112
121121
11212
12
12
1212
2
122
and 2
ttattvxx
ttvttav
xttvv
xx
tt
xxvvv
xx
xxxxx
xxx
We have four equations that each relate three of the motion variables.
xxxoxo vvvvxxxxttt 212121 , , , , ,0Commonly,
. vv
v
also then constant, is a If
xxx
x
212
18
Space-time
Mathematically we can treat time and space on the same footing.
The displacement vector in space-time has 4 components.
The scaling factor c is needed to make the units of all 4 componentsthe same, e.g., meters. The geometry of space-time is not Euclidian, but is non-Euclidian. Therefore,
z,y,x,ctr
11 tt ;kkjjii
19
Example: a train traveling on a straight and level track
starting from rest; ends at rest.
2
2
1tatvxx xxoo m sm/s sm/s m x 2 100102
2
11000 2
1
m/s sm/s m/s tavv 2xxox 2010201 m sm/s sm/s m tatvxx 2
xx 7003002
13020100
2
1 22112
02012 xxx a sincem/s vv
m
m/s
m/s m
a
vvxx
xxavv
2x
xx
xxx
75042
200700
2
222
223
23
2322
23
What is the total displacement?Segment 1: We are given the acceleration, elapsed time and initial velocity — vxo = 0 m/s.
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.
Segment 3: For this segment, we know x2, vx2, vx3, and ax, but not .t
time, t (seconds)
acceleration, ax
(m/s2)
0 - 10 2
10 – 40 0
40 - ? -4
20
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 . What are the height of the sandbag and its velocity as functions of time?
down.) going sit' know (We m/s. .s/m v
s/m m m m/s .m/s yyavv
yyavv
2y
22yyoy
oyyoy
420417
41720089252
2
2
220
22
22
ts
m8.9
s
m5tavv
ts
m8.9
2
1t
s
m5m20ta
2
1tvyy
2yyoy
22
2yyoo
What is the y-component of the sandbag’s velocity when it hits the ground?
2y m/s .a 89
Free Fall!
21
Alternative solution for the elapsed time:
s 57.1-or s 59.2s 8.9
42.205
m/s 8.9
m 20m/s 9.44/m 25m/s 5
m/s 8.92
1m/s 5m 20m 0
2
1
2
222
22
2
st
tt
tatvyy yyoo
How long does that take?
s 59.2m/s 8.9
m/s 5m/s 4.202
y
yoyyyoy a
vvttavv
22
Projectile Motion
Constant acceleration, in two dimensions.
js
mia ˆ8.9ˆ0
2
23
tavv o
tgvtavv
vtavv
yoyyoy
xoxxox
)(
ooyo
ooxo
sinvv
cosvv
2
2
1tatvrr oo
22 )(2
1
2
1tgtvytatvyy
tvxx
yooyyoo
xoo
Vector equations:
Component equations:
Notice:The y-component of is ay = -g.
a
24
m .
m/s .
sin53m/s /sm m
a
vvyy
yyavv
2
o22
y
yoyo
oyyoy
152892
4000
2
2222
22
sm/s
sin53m/s m/s
a
vvt
tavvo
y
yoy
yyoy
26.38.9
400
Example:
How long does it take to reach maximum height, ymax?At maximum height, vy = 0 m/s
What is the maximum height?
25
When is the projectile at y = 25m?
s 61.5 and s 910.02
s 1.54)s 52.6()s 52.6(
s 0s 10.5s 52.6
022
02
12
1
22
222
2
2
2
t
tt
a
yyt
a
vt
yytvta
tatvyy
y
o
y
yo
oyoy
yyoo
26
time (s) velocity (m/s)
0.910
5.61
0.23
1.24
y
x
v
v
023
124
.v
.v
y
x
What are the velocity components then, at t = 0.910 s and t = 5.61 s?
tm/s m/s
tm/s m/s tgsinvtavv
m/s m/s vtavv
2
2oooyyoy
oooxxox
8.99.31
)8.9(53sin40)(
1.2453cos40cos
27
Example:
How far does the object travel in the x-direction?
2
2
1tatvxx xoxo 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. The total time, then, is 2x3.26s = 6.52 s. [Verify with .]
ms.s
m.tvx
tvtatvxx
ox
oxxoxo
157526124
002
1 2
2892
100 t.tymax
28
m 1.44s3m/s 8.92
13sm/s 0m 0
2
1
m60s3m/s 20m 0
m/s 4.293s)m/s 8.9(m/s 0
m/s 203m/s 0m/s 20
222
2
2
tatvyy
tvxx
tavv
stavv
yyoo
xoo
yyoy
xxox
What are the velocity & position components at t = 3 seconds?
Example:
29
30
Uniform Circular Motion
va
Curvilinear motion – not in a straight line.Envision an object having, at the moment, a velocity , and subject to an acceleration .
. 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.
31
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.
32
Let the origin be at the center of the circle, as shown.
33
12
12
vvv
rrr
21 rrr 21 vvv
t
varad
0t
vv
rr
Consider two successive displacement and velocity vectors.
By the definition of uniform circular motion,
In the limit as ,
& .
r
v
tr
rv
r
rv
tt
va
r
r
v
v
rad
21
.
The centripetal or radial acceleration is always on a circular arc of radius r. r
v2
Both are isosceles triangles, with the same angle.
34
Second
Dynamics
Newton’s “Laws”
Energy
Momentum
Conservation
35
Dynamics
Relationships among Motion and Force and Energy.
36
Newton’s “Laws” of Motion
“An object in uniform motion remains in uniform motion unlessit 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.”
37
38
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 fourfundamental forces. All other types of forces that we might give a name to are some manifestation of one of the fundamental forces.
2T
L MF
The SI unit of force is the Newton (N).1 N = kg m/s2
Dimensions of force are
39
Fundamental concepts:
i)Space and timeii)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 internalstructure. Any object may be treated as if it were a particledepending on the context.
40
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.
2
1
1
2
m
m
a
a
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.
dt
We write Newton’s 2nd “Law”
in mathematical form:
momentum. the called
cv
vmpquantity The
2
2
1
41
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
Further, if the particle’s mass is unchanging, then
vmp
dt
vdm
dt
42
43
0dt
pd
0p
0p
Should the vector sum of all forces acting on an object be equal to zero, then
and the object is said to be in
Equilibrium
Static equilibrium
Dynamic equilibrium
44
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.
45
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.
Action & Reaction
pair. reaction-action an form N and N
pair. reaction-action an not are N and W
NN
0WN
46
Newton’s Universal “Law” of Gravitation
rr
MMGFg ˆ
212
2112
Any two objects exert gravitational forces on each other, equal in magnitude and opposite in direction.
Take care with the directions. The unit vector points from M1 to M2.The gravitational force on M2 is in the direction, toward M1.
rr
47
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.
21
2221
r
MGaa
gMrr
MMGF
gr
g
Gravitational Field
rr
MGg
21
An object of mass M2 located at experiences a gravitational force.r
rr
MMGgMFg 2
212
In the context of the 2nd “Law”The acceleration due to gravity is
48
Near the Earth’s surface, 22212
1 89s
m.
R
MG
r
MGa
E
Eg
Near the surface of another body, such as the Moon or Mars, theacceleration due to gravity is different, not 9.8 m/s2.
49
gmFg
ggy FNFNF 0
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.
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.
Weight
50
Suppose the object is not in equilibrium.
mgmAN
FmAN
mAFNF
g
gy
Suppose A = -g. Then N = 0. The object is in free fall,but not weightless. The term weightlessis a misnomer.
51
xo
xfxxxx maNFmaFNWF 0030cos
0030sin NmgFmaFNWF oyfyyyy
oox
o FmgFm
aFmgN 30sin30cos1
30sin
Friction
NF Friction Static
NF Friction Dynamic or Kinetic
Sf
Kf
decompose
Friction always opposes the motion of an object, or what’s called the object’s impending motion.
52
xfxxfxx MaFMaFWN 111111 00
0011111 MgNMaFWN yyfyy
xffxxxxfxfxxx maFFFmaRRFWNF 212211222 000
000 12211222 NmgNFmaRRFWNF yyyyfyfyyy
111 NF f
222 NF f
Example: Two objects
53
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).
amWT
mgTmamgTmaWT yyyy 0
mgT 1
22122 aMWTT
gMmMgTTMgTT 2
1
2
102 1212
Cords & Tension
54
amTTWF rightleft
0coscos00 TTTTW rightxleftxx
sin20sinsin0
mgTTTmgTTW rightyleftyy
Example:
No help.
55
20 1
2122p
p
WTTTTTW
MgTMgT 11 0
Pulleys
Apply Newton’s 2nd “Law” to the pulley and to the hanging mass.
56
Case Studies in Applying Newton’s 2nd “Law.”
Circular motion
Inclined plane
Restoring forces—spring & pendulum
Systems of objects
57
58
r
vmmgT
maWT
r
rrr
2
cos
tt
ttt
mamgT
maWT
sin
r
vm)(mgTr
2
1
o180 o0
Circular Motion
ferris wheel
59
R
vmNNmaFNmaFWN srfxfxxx
2
cossincos0sin
sincos0sincos
sfyfyyy
mgNFmgNmaFWN
R
vm
mgs
s
2
cossinsincos
o
2 m m/s
m/s
gR
v
R
vm
mg
20
366.0tan
508.9
4.13
cos
sin
0sin0cos
1
22
2
m/s v ,mR ,s 4.13500
driving ‘round a curve
example
60
2r
MmGFg
r
vm
r
MmG
maF
amF
rg
2
2
r
GMv
circular orbit
Universal “Law” of Gravitation
Newton’s 2nd “law”
Orbital speed
61
amWN
xxxxx maWmaWN 0
0 yyyyy maWNmaWN
coscos
sinsin
mgWW
mgWW
y
x
amFWN
xxxxxxx maFmgmaFWmaFWN sin0
0cos0 mgNmaWNmaFWN yyyyyy
Inclined Plane
62
os kF
o
o
s
s
k
mg
mgk
mgF
amWF
amF
0
0
Restoring Force
Hooke’s “Law”linear restoring force
Typically, we place the origin at theresting length of the spring.
os xxkF
63
amWT
amF
22
coscos ttrrr
vmmgT
r
vmWTmaWT
sinsin0 gamamgmaWT ttttt
Radial and tangential components.
Pendulum
64
111111 amgmTamWT
222 amWNT
xxxxx amWTamWNT 22222 sin0
0cos0 222 WNamWNT yyyy
xaa 21
gmgmmm
aagmWm
a
TWm
aa
agmT
x
1221
11122
1
22
21
11
sin1
sin1
sin1
System of Objects
Newton’s “Laws” apply to each object as well as to the system as a whole.
65
11111111 amgmTamWT
22222222 amgmTamWT
aaa 21 TTT 21
amgmT
amgmT
22
11
gmm
mma
12
12
gmm
mmgmg
mm
mmmgmamT
12
211
12
12111
2
“Atwood’s Machine”
66
uvvdt
Rd
dt
rd
dt
rd
0dt
ud
aa
turrrturrRr
Reference Frames
The position vector of the point Pis written down with respect totwo different reference frames.
The motion variables, as measured by observers in different frames.
67
dt
udaa
uvv
turr
0
amWT
amF
0sinsin WT
0coscos WT
x’:
y’:
amWT
amF
00sin xmaT
0cos WT
x:
y:
dt
duax
dt
udmWW
Apply Newton’s 2nd “Law” in the two reference frames.
Now, bring them together: The perceived weight is different.
68
0 gmFc mAmgFc
mAmggm
An Accelerated Reference Frame
The reading on the scale is F’c. The observer in the elevator interprets that as his/her weight.
In the elevator
Outside the elevator
Adt
Rd
dt
ud
2
2
69
Energy
Work
Kinetic Energy
Potential Energy
70
Physical Work
2
1
sdFW
dzFdyFdxFWz
z
z
y
y
y
x
x
x 2
1
2
1
2
1
22
T
LMLF dimensions The SI unit of work is the Joule (J).
211
s
mkg J
2
71
cosFssFW
sFsFsFW fo
ff 180cos
constant force
72
xFW xnet
2
1
x
x
dxxFW
22
21
2
222
2
1
2
1
2
1
xk
xk
xk
dxkxFdxWx
x
x
x
x
x
force varies with positiondxFsdFdW x
73
tvF
t
WP
s
JW 11
WmNFvvFP 65 102sec/2010
WmNtdFP 65 102sec5/10010/
Power is the rate at which work is done. If W is the work done during an elapsed time,
, then the average power during that interval is
The SI unit for power is the Watt:
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
Power
. A kilowatt is 1000 Watts.
or
74
Kinetic Energy
21
22
21
22
12
2
1
2
1
2
1
2 xxxxxxxx
x mvmvvvmtvv
t
vmx
t
vmxFW
An increment of work is done during an incremental displacement. We assume that the applied force is constant during the incremental displacement.
KmvmvW xx 21
22 2
1
2
1Work-Energy Theorem
2
2
1mvK Kinetic energy
75
22
2
1
2
1 og mvmvKymgW
2
2
12omvymg
mv
Examples:
ymgyFxFrFW gygxgg 0
Energy is a scalar; it has no directional information.
oyyy Keep in mind:
76
o
ooT
ooN
oKoff
og
xxTxxTW
xxNW
xxNxxFW
xxMgW
0cos
090cos
sin
TNfg WWWWW
22
2
1
2
1omvmvKW
Every force acting does some work.
The block slides up the Incline from xo to x.
77
Potential Energy
2
1
12
x
x
xC UUUdxFW
2
1
2
1
12
x
x
x
x
x
x dxFUdxFU
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.
The potential energy function is defined thusly:
UWC For x-components
To derive the potential energy function for a specified force, we evaluate the work.
Because we are interested in potential energy changes, we can set the zero of potential energy for convenience.
2
1
sdFW
dzFdyFdxFWz
z
z
y
y
y
x
x
xC 2
1
2
1
2
1
78
kFjFiFF
z,y,xUz
F
z,y,xUy
F
z,y,xUx
F
zyx
z
y
x
Gradient operator
Uz
kUy
jUx
iz,y,xUF
kxkxdx
dFx
2
2
1
mgmgydy
dFy
Conversely, we can derive the force components from the potential energy function:
For a spring:
For uniform gravity:
79
222
00
0
0 2
10
2
1
2
1180cos kxkkxxdxkxdFxdFW
xxx
xC
2
2
1kxU s
UWC
Spring:
80
1212
2
1
2
1
UUmgymgymgdydyFWy
y
y
y
yC
mgyU g
Gravity:
81
Mechanical Energy
UKE
If only conservative external forces are acting, then the total mechanical energy of a system is conserved.
EUKW
KUWW
KWWW
other
othertotal
cothertotal
0 otherWE
82
Gravity and spring restoring forces are conservative. Friction is non-conservative.
Let us say that a number of forces do work on an object.
0
0
IE
IUK
IUKWWW Cothertotal
The 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.
I
83
1212
22
1122
2
1
2
1mgymvmgymv
UKUK
21
21
22
22
1122
2
1
2
1
2
1
2
1kxmvkxmv
UKUK
Near the Earth’s surface,
Spring
84
Recall the “Law” of Gravitation
221
r
mGmFg 2
2111067.6
kg
mNG
m 1 m 2
r 12
22earth
earth
earth
earthg R
GMgmg
R
mGMF
212
2112 r
mGmF
85
r
mGmrU
rUr
mGm
r
mGmdr
r
mGmrdFW
r
r
r
r
gg
21
1
21
2
212
21
)(
)(2
1
2
1
Gravitational potential energy
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 (-).
86
2211 & ,0 v, , rvvRr oearth
hour
miles.
s
m.
m.
kg.kg
Nm.
R
GMv
R
GMv
R
mGMmv
r
mGmvm
r
mGmvm
earth
eartho
earth
eartho
earth
eartho
446
242
211
2
2
1
21212
2
21222
104921012110386
109751067622
2
2
100
2
1
2
1
Escape from EarthDefine “escape”
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).
87
Impulse & Momentum
t
va x
x
xxx
xxx
xx
ptFJ
mvmvtFt
vtmatm
12
.0 then ,0 if momentum, ofon Conservati pF
Impulse
In general dtFpJ
Go back to the definition of acceleration.
88
12 xxx mvmvp
Let’s say that m = 0.5 kg, vx1 = 40 m/s, and vx2 = -20 m/s.
Then the change in momentum is
s
mkg
s
m
s
m20-kg mvmvp xxx
30405.012
This is the impulse on the ball! The ball exerts an equal and opposite impulse on the wall. If the impact lasts st 01.0 , then the average force on the wall is
N s
mkg
sp
tF xx 300030
01.0
11
A ball bounces straight off a wall.
89
22
22
21
21
21
2211
21
2
1
2
1
2
1
2
1BBAABBAA
BBAABBAA
vmvmvmvm
KK
vmvmvmvm
pp
One dimensional elastic collision
Inelastic collision—kinetic energy is not conserved. IKK 12
90
22
22
21
21
21
2211
2211
21
2
1
2
1
2
1
2
1BBAABBAA
ByBAyAByBAyA
BxBAxABxBAxA
vmvmvmvm
KK
vmvmvmvm
vmvmvmvm
pp
Two dimensional elastic collision
91
211
211
21
yBAByBAyA
xBABxBAxA
vmmvmvm
vmmvmvm
pp
221
21
BBAABA vmvmvmm
pp
“Perfectly” inelastic collision
92
93
i
iicm m
rmr
Pvmmv
m
vmr
dt
dv
iiicm
i
iicmcm
cmvMP
P
cmv
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.
If no net external force acts on any part of the system, then
is constant, and so is .
The individual parts of the system may exert forces on each other, but those do not affect the motion of the center of mass.
Center of Mass
94
cmiiext aMamFFF
int
cmext aMF
Because of Newton’s 3rd “Law”, . 0 intF
.
The sum of all forces acting on all parts of the system is
On the other hand, if one or more external forces acts on the system, then
cmv
is not constant.
Consequently, the center of mass of a system of particles moves like a particle of mass M.
95
ptFext
122 vmmvmvmp
1212 vvmvvmp
12 vvv
22 vvV
Vmvmvmp
vVmvvvmp
11
VmvmmVmvmvmp
VmvmmtFext
VmvmptFext
dt
dmV
dt
vdmV
t
m
t
vmFext
Suppose the total mass of a moving object is not constant.
Rockets & Rain Drops
extF
t extF
Say the net external force acting on an object (such as a rocket or a rain drop) is .
Assume that during a short time interval, , the is approximately constant.
Then the impulse delivered to the mass, m, is .
t mFurther suppose that during that interval , the mass changes by an amount .
We may as well just let m + Δm
be m at this point.
96
dt
dmV
dt
vdm
0
dt
dmV
dt
vdm
dt
dm
m
V
dt
vd
.
v
V
m
dt
dm
m
v
V
Recap:
is the velocity of the object (rocket or rain drop),
is the velocity of the relative to the object, and
is the absolute value of the time rate of change in the mass of the object. Actually, we have to be careful of the directions of things. As derived here, if
is leaving the object, then the object is losing mass and
is in the opposite direction as .
Consider a rocket in the absence of gravity or any other external force.
In real life, there is always gravity, friction, air resistance, etc.
97
Third
Rotation
Vibration
Wave Propagation
98
Rotation
dt
d
2
2
dt
d
dt
d
angular displacement,
angular velocity component,
angular acceleration,
name definition
Arc length, radians
There are radians in 360o.
Rigid body
2
radiansin or r
srs
A rigid body is one in which all therij are constant.
99
2
2
2
1
22
2
o
oo
o
oo
t
t t
22
rr
va
ra
rv
rs
tr
t
t
unknown
constant
108
(radians) 234
0
3
o
s
radians
st
o
s
radians
s
radians
t78
3
0234
2220
3
602
s
radians
s
radians
ttto
Equations of rotational motion
Example:
100
2ii rmI
VolumeVolume
dxdydzrdmrI 22
The dimensions of moment of inertia are [M][L]2.
Moment of Inertia
101
102
2222
2
1
2
1
2
1 IrmvmK iiiir
zyx mvmvmv
zyx
kji
prL
Frdt
pdrp
dt
rd
dt
pdr
dt
prd
dt
Ld
0
dt
LdFr
Rotational Dynamics
Angular kinetic energy & angular momentum
torque
ImmprLL iiiiiiii
2
Idt
dI
dt
LdRigid body:
103
104
Unwinding
mamgT
maF yy
0
0
TMgN
Fy
R
a
)(0)(
2
5.00
2
22
agmTgamTM
T
MR
TR
I
TR
I
RTRaTR
R
aI
Nms
mmkg
R
aMR
R
aII
Ng
mM
gmT
s
mg
mM
aM
agma
9.10)36.4(5.0102
1
2
1
8.21
21
1
36.4
21
12
22
2
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?
R
aIRT
I
105
Roll down an incline
cos2
55
2cos 00
mass.) ofcenter about the torquescompute ll(We'
cossin
cos0cos
sinsin
2
R
g
MRMgRRf
I
I
ga
MgNMgN
MaNMgfMg
amF
fgN
x
x
Now, if the ball is to roll without slipping, what must be true?
The friction must be just right such that
R
ax
Kinetic energy:22
2
1
2
1 IMvK
2
2
22
5
1
2
1
5
2
2
1
2
1Mv
R
vMRMvK
No slipping. . . .
106
Gyroscope
Pulley
jrMgt
L
t
LkNikMgir
iLL oo
ˆ
ˆˆ0ˆˆ
ˆ
107
108
109
Static Equilibrium
0102060cos
0
060sin
0
gkggkgTF
FFTF
TF
FFTF
oy
bygyyy
ox
bxgxxx
NT
NmNmTm
mgkgmgkgmT ooo
bgT
213
053.15613.62668.3
053sin2
41053sin42067sin4
0
NNNTgkgF
NNTFo
y
ox
1885.021329460cos30
184866.021360sin
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.
110
Oscillation
ftAty 2sin
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.
fT
1 f 2
t Adt
dva
t A dt
dyv
t Ay
sin
cos
sin
2
m
k
ymky
maF
2
radians/s
For an object bouncing on a spring:
111
222
22
22
2
1
2
1
2
1
02
1
2
1
2
10
2
12
1
2
1
AmkymvE
kAmkAmE
kymvE
Mechanical energy of an oscillator
Notice E is proportional to the square of the amplitude.
Again, think of an object oscillating horizontally on a spring. The mechanical energy is
112
sin
sin
sin2
2
g
mmg
mImg
sin
g
dt
d
g
2
2
g
I
mg
dt
d
Img
I
2
2
sin
I
mg
pendulum
Assume the oscillations are small
Physical Pendulum
113
114
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.
115
116
Wave Motion
i) wave speed is a property of the medium.ii) shape of the wave pulse is unchanged as it travelsiii) two or more wave pulses that exist at the same place & time in a medium add—superimpose.
T
txAftxy
,
y is the displacement from equilibrium at position x and at time t. f is an unspecified function.
117
t
c
xfA
T
txAtxy
2cos2cos,
f
k
2
2
t kxAtxy cos,
Harmonic wave – a wave of a particular shape that repeats itself. It’s periodic.
Each point in the medium (x) is displaced from equilibrium (y). As time passes, the pattern is shifted by a phase factor ;the wave pattern moves through the medium.
ct
118
t kxAtxy
t kxAt kxAtxy
txytxytxy
sinsin2,
coscos,
,,, 21
bababa sinsincoscoscos
“Standing waves”
superpositionreflection
119
LmF
c
Stretched string
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.
120
tff
tff
Ay
tf tf Ay
yyy
22cos
22cos2
2cos2cos
2121
21
21
In this case, two waves are traveling in the same direction, but with slightly different frequencies.
“Beats”
121
Spectrum
i
ii
i tfxcosAy 22
122
Energy
dm
K
tkxtkxAdmE
tkxdmAtkxAdmE
dmvKyE
sin cos2
1
sin2
1 cos
2
12
1
2
1
2222
222222
22
2
1
22222222 2 fAfffE
222 A f c E
2222 Af c fEP
2222 Af c a
PI
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
.
Over one cycle, the cosine-squared and sine-squared average to
. The total mass of the medium spanning one cycle (or one wavelength) is
, where is the mass per unit length of the medium.
In terms of the wave speed, c,
.The energy flux is the power transported through the medium by the wave:
The intensity is the power pr unit area through which the power is transported:
, were is the mass per unit volume.
123
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. . .
124
125
The “Laws” of Thermodynamics