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

pdF

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

pdF

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