Physics
Chapter 1
Introduction and Vectors
授課老師施坤龍
0
Realms of Physics
Physics provides a nearly unified description of all physical
phenomena.
It's convenient to divide physics into six distinct but related realms.
1
Physics
Fundamental Science

Concerned with the fundamental principles of the Universe

Foundation of other physical sciences

Has simplicity of fundamental concepts
Divided into the following major areas

Classical Mechanics

Relativity

Thermodynamics

Electromagnetism

Optics

Quantum Mechanics
2
The nature of physics
To find the limited number of fundamental laws that govern natural
phenomena
To use these laws to develop theories that can predict the results of
future experiments
Express the laws in the language of mathematics

Mathematics provides the bridge between theory and experiment
Physics is an experimental science in which physicists seek patterns
that relate the phenomena of nature.
The patterns are called physical theories.
A very well established or widely used theory is called a physical law or
principle.
3
Standards of Fundamental Quantities
Standardized systems
Agreed upon by some authority, usually a governmental body
SI – Systéme International (The International System)
Agreed to in 1960 by an international committee
Consists of a system of definitions and standards to describe
fundamental physical quantities
4
Fundamental Quantities and Their Units
Quantity
Length
SI Unit
Meter (m)
Mass
Kilogram (kg)
Time
Second (s)
Temperature
Kelvin (K)
Electric Current
Ampere (A)
Luminous Intensity
Candela (cd)
Amount of Substance
Mole (mol)
5
Quantities Used in Mechanics
In mechanics, three basic quantities are used

Length

Mass

Time
Will also use derived quantities

These are other quantities that can be expressed in terms
of the basic quantities

Example: Area is the product of two lengths

Area is a derived quantity

Length is the fundamental quantity
6
Length
Length is the distance between two points in space.
Units

SI – meter, m
Defined in terms of a meter – the distance traveled by light in a
vacuum during a given time.

The meter is the length of the path traveled by light in a
vacuum during a time interval 1/299 792 458 of a second.
7
Mass
Units

SI – kilogram, kg
Defined in terms of kilogram,
based on a specific Platinum
cylinder
International
kept
at
Bureau
the
of
Weights and Measures
8
Time
Units

seconds, s
Defined in terms of the oscillation of radiation from a cesium
atom.

One second is the duration of 9 192 631 770 periods of the
radiation corresponding to the transition between the two
hyperfine levels of the ground state of the cesium-133 atom.
9
Length Scales
Radius of atomic nucleus ~1 fm = 10-15 m
Radius of atom ~1 Å = 10-10 m
Radius of Earth = 6380 km
Distance Earth-Sun:
1AU = 1.49598 ×1011m
Distance to next star: 4 ly (light year)
1 light year = 9.46 ×1015m
Size of Universe ~ 1.5 ×1010 light years
Cover 41 orders of magnitude
10
Mass Scales
me = 9.11×10-31kg; mp = 1.67×10-27kg;
ME = 5.98×1024kg; MS = 1.99×1030kg
Mass of entire Universe ~ 1051 kg
11
Fundamental and Derived Units
Derived quantities can be expressed as a mathematical
combination of fundamental quantities.
Examples:

Area


Speed


A product of two lengths
A ratio of a length to a time interval
Density

A ratio of mass to volume
12
Prefixes
Prefixes correspond to powers of 10.
Each prefix has a specific name.
Each prefix has a specific abbreviation.
The prefixes can be used with any basic units.
They are multipliers of the basic unit.
Examples:

1.27×109 watts =1.27 gigawatts = 1.27 GW,

2.35×10-9 m = 2.35 nanometers = 2.35 nm.
13
Prefixes Table
Power
Prefix
Abbreviation
Power
Prefix
Abbreviation
10-24
yocto
y
10-1
deci
d
10-21
zepto
z
103
kilo
k
10-18
atto
a
106
mega
M
10-15
femto
f
109
giga
G
10-12
pico
p
1012
tera
T
10-9
nano
n
1015
peta
P
10-6
micro
μ
1018
exa
E
10-3
milli
m
1021
zetta
Z
10-2
centi
c
1024
yotta
Y
14
Dimensional Analysis
Technique to check the correctness of an equation or to assist
in deriving an equation
Dimensions (length, mass, time, combinations) can be treated
as algebraic quantities.

Add, subtract, multiply, divide
Both sides of equation must have the same dimensions.
Any relationship can be correct only if the dimensions on both
sides of the equation are the same.
Cannot give numerical factors: this is its limitation
15
Example: Dimensional Analysis-1
Given the equation: x = 1/2 at2？
Check dimensions on each side：
L = (L/T2) (T2) = L

The equation is dimensionally correct

There are no dimensions for the constant
16
Example: Dimensional Analysis-2
Is this the correct equation for velocity?
v = v0 + ½ at2
Check the dimensions:
(L/T) = (L/T) + (L/T2) (T2) = (L/T) + (L)
The dimensions on the right are not equal, and the
equation is wrong.
17
Example: Dimensional Analysis-3
Suppose we are told that the acceleration a of a particle
moving with uniform speed v in a circle of radius r is
proportional to some power of r, say rn, and some power of v,
say vm. Determine the values of n and m and write the
simplest form of an equation for the acceleration.
Suppose k is a constant, and a = k rn vm
Dimensions on both sides of the equation:
L/T2 = Ln (L/T)m =Ln+m/Tm
From n+m=1 and m = 2 , then n = -1。
So the equation is a = k r-1 v2 = k v2/r
18
等速圓周運動
t=T/4
t=3T/8
t=T/8
V
t=0
t=T/8
V
t=7T/8
r t=0
t=T/2
t=T/4
t=3T/4
t=7T/8
t=5T/8
t=3T/4
V = 2πr/T
t=5T/8
t=3T/8
t=T/2
ac = 2πV/T = 4π2r/T2 = V2/r
19
Conversion of Units
When units are not consistent, you may need to convert to
appropriate ones.
Units can be treated like algebraic quantities that can cancel
each other out.
Always include units for every quantity, you can carry the
units through the entire calculation
Multiply original value by a ratio equal to one

The ratio is called a conversion factor
Example:
25 inches = 25 in * 2.54 cm/in = 63.5 cm
20
EXAMPLE : Is He Speeding?
On an interstate highway in a rural region of Wyoming, a car
is traveling at a speed of 37 m/s. Is the driver exceeding the
speed limit of 75.0 mi/h? What is the speed of the car in km/h?
37 m/s = 37m/s *3600 s/h*1km/1000m = 133 km/h
133 km/h *1mi/1.609km = 82.8mi/h
82.8 mi/h ~ 83 mi/h
exceeding the speed limit
21
Order of Magnitude
An order-of-magnitude estimate of a quantity gives a rough idea of its
magnitude. A quick way to estimate a calculated quantity is to round off
all numbers to one significant figure and then calculate. Your result
should at least be the right order of magnitude; this can be expressed by
rounding it off to the nearest power of 10. In order of magnitude
calculations, the results are reliable to within about a factor of 10
Estimate a number and express it in scientific notation.

The multiplier of the power of 10 needs to be between 1 and 10.
Compare the multiplier to 3.162 ( 10 )

If the remainder is less than 3.162, the order of magnitude is the
power of 10 in the scientific notation.

If the remainder is greater than 3.162, the order of magnitude is one
more than the power of 10 in the scientific notation.
22
Example: How far is a light year
A light-year (ly) is a unit of distance (length) commonly used
in astronomy. A light-year is the distance light travels through
a vacuum in 1 year. The speed of light in a vacuum is c ;
299,792,458 m/s. Find the number of meters in 1 light-year.
1 year = 365days*24hours/days*60mins/hours*60s/mins
= 3.154×107s
1 ly = 3×108m/s* 3.154×107s = 9.46×1015m ~ 1016m
23
Uncertainty in Measurements
There is uncertainty in every measurement – this uncertainty carries over
through the calculations.

May be due to the apparatus, the experimenter, and/or the number of
measurements made

Need a technique to account for this uncertainty
We will use rules for significant figures to approximate the uncertainty in
results of calculations. The uncertainty of a measured quantity is
indicated by its number of significant figures.
For multiplication and division, the answer can have no more significant
figures than the smallest number of significant figures in the factors.
For addition and subtraction, the number of significant figures is
determined by the term having the fewest digits to the right of the
decimal point.
24
Significant Figures
A significant figure is one that is reliably known.
Zeros may or may not be significant.

Those used to position the decimal point are not significant.

To remove ambiguity, use scientific notation.
In a measurement, the significant figures include the first
estimated digit.
25
Significant Figures, examples
0.0057 m has 2 significant figures

The leading zeros are placeholders only.

Write the value in scientific notation to show more clearly: 5.7×10-3
m for 2 significant figures
20.0 m has 3 significant figures

The decimal point gives information about the reliability of the
measurement.
2500 m is ambiguous

Use 2.5×103 m for 2 significant figures

Use 2.50×103 m for 3 significant figures

Use 2.500×103 m for 4 significant figures
26
Significant Figures – Multiplying or Dividing
When multiplying or dividing several quantities, the
number of significant figures in the final answer is the
same as the number of significant figures in the
quantity having the smallest number of significant
figures.
Example: 14.56 m × 3.65 m = 53.1 m2

The 3.65 m limits your result to 3 significant figures.
27
Operations with Significant Figures – Summary
The rule for addition and subtraction are different than the rule for
multiplication and division.
For adding and subtracting, the number of
important consideration.
decimal places is the
For multiplying and dividing, the number of significant figures is the
important consideration.
When adding or subtracting, the number of decimal places in the result
should equal the smallest number of decimal places in any term in the
sum or difference.
Example: 127 cm + 6.23 cm = 133 cm

The 127 cm limits your answer to the units decimal value.
28
Example: The Area of a Dish
A biologist is filling a rectangular dish with growth
culture and wishes to know the area of the dish. The
length of the dish is measured to be 11.65 cm and the
width is measured to be 8.26 cm. Find the area of the
dish.
Area of the dish = 11.65 cm * 8.26 cm = 96.229 cm2
But there are only three significant digits, then
Area of the dish = 96.2 cm2
29
Example : Installing a Carpet
A carpet is to be installed in a rectangular room
whose length is measured to be 13.44 m and whose
width is measured to be 3.25 m. Find the area of the
room.
Area of the room = 13.44 m * 3.25 m = 43.68 m2
But there are only three significant digits, then
Area of the room = 43.7 m2
30
Rounding
Last retained digit is increased by 1 if the last digit dropped is greater
than 5.
Last retained digit remains as it is if the last digit dropped is less than 5.
If the last digit dropped is equal to 5, the retained digit should be
rounded to the nearest even number.
Saving rounding until the final result will help eliminate accumulation
of errors.
It is useful to perform the solution in algebraic form and wait until the
end to enter numerical values.

This saves keystrokes as well as minimizes rounding.
31
Uncertainty of measurement
Every measurement has uncertainty-this uncertainty will continue in the
calculation. It may be due to equipment, experimenter or number of
measurements. A technique is needed to explain this uncertainty.
We use the rule of significant figures to estimate the uncertainty of the
result. The uncertainty of the measurement number is represented by its
significant figure. After multiplying and dividing, the result is the
smallest significant figure between them, which is its effective figure.
After addition and subtraction, the number of significant digits is the
item with the least number of digits to the right of the decimal point as
its significant digits.
32
Reasonableness of Results
When solving problem, you need to check your
answer to see if it seems reasonable
Reviewing the tables of approximate values for
length, mass, and time will help you test for
reasonableness
33
Coordinate Systems
Used to describe the position of a point in space
Coordinate system consists of

A fixed reference point called the origin

Specific axes with scales and labels

Instructions on how to label a point relative to the
origin and the axes
34
Coordinate Systems
Used to describe the position of a point in space
Common coordinate systems of two Dimension are:

Cartesian

Polar
Common coordinate systems of three Dimension are:

Cartesian

Cylindrical

Spherical
35
Cartesian Coordinate System
Also
called
rectangular
coordinate system
x- and y- axes intersect at
the origin O
Points are labeled (x, y)
36
Polar Coordinate System
Origin and reference line are
noted
Point is distance r from the
origin in the direction of angle
, from reference line

The reference line is often
the x-axis.
Points are labeled (r,)
37
Polar to Cartesian Coordinates
Based on forming a right triangle from r and θ
x = r cosθ
r
y
y = r sinθ
If the Cartesian coordinates are known:
=π+
tan-1(y/x)
x
x  r cos 
r = (x2+y2)½ ,
θ = tan-1(y/x)

y  r sin 
if x >0
r  x2  y2
if x<0
y
tan  
x
38
Example
The Cartesian coordinates of a
point in the x-y plane are
(x, y) = (-3.00, -4.00) m, as shown
in the figure. Find the polar
coordinates of this point.
r  x 2  y 2  ( 3.00)2  ( 4.00)2  5.00m
 4.00
o
1 4
  180  tan (
)  180  tan ( )  180o  53o  233o
 3.00
3
o
1
39
Coordinate systems of Three Dimension
Cartesian (x, y z)
Cylindrical (r, ϕ, z)

x = r cos ϕ,

y = r sin ϕ,

z=z

r = (x2+y2)½ ,

ϕ = tan-1(y/x) or π + tan-1(y/x),

z=z
40
Coordinate systems of Three Dimension
Spherical (r, θ, ϕ)

x = r sinθ cosϕ,

y = r sinθ sinϕ,

z = r cosθ

r = (x2+y2+z2)½ ,

θ = cos-1(z/r),

ϕ = tan-1(y/x) or π + tan-1(y/x)
41
Example : Finding the Height of a Tree
You wish to find the height of a tree but cannot measure it
directly. You stand 50.0 m from the tree and determine that a
line of sight from the ground to the top of the tree makes an
angle of 25.0° with the ground. How tall is the tree?
h
tan  
50.0m
o
h  50.0m tan   50.0m tan 25  23.3m
42
Vectors and Scalars
A scalar quantity is completely specified by a single value
with an appropriate unit and has no direction.
A vector quantity is completely described by a number and
appropriate units plus a direction.
Mathematical operations of vectors in this chapter

Addition

Subtraction

Inner Product

Cross product
43
Vector Example
A particle travels from A to C along the path shown by the
blue or green line.

This is the distance traveled and is a scalar.
The displacement is the blue line from A to C

The displacement is independent of the path taken between the two
points.

Displacement is a vector.
44
Vector Notation
Text uses bold with arrow to denote a vector: A
Also used for printing is simple bold print: A
When dealing with just the magnitude of a vector in
print, an italic letter will be used: A or |A|

The magnitude of the vector has physical units.

The magnitude of a vector is always a positive number.
When handwritten, use an arrow: A
45
Equality of Two Vectors
Two vectors are equal if they
have the same magnitude and
the same direction.
A = B if A = B and they point
along parallel lines
All of the vectors shown are
equal.
Allows a vector to be moved to
a position parallel to itself
46
Adding Vectors Graphically
Vector addition is very different from adding scalar quantities, their
directions must be taken into account.
Continue drawing the vectors “tip-to-tail” or “head-to-tail”.
The resultant is drawn from the origin of the first vector to the end of the
last vector.
Measure the length of the resultant and its angle.

Use the scale factor to convert length to actual magnitude.
R=A+B
47
Adding Vectors Graphically
We can use the graphic method below (parallelogram
method) to represent vector addition.
Parallelogram graphics method uses the heads of two vectors
to be joined together to form a parallelogram.
In the figure, the diagonal line from the head to the tail is the
added vector.
R=A+B
48
Adding Vectors Graphically
When you have many vectors, just keep repeating the
process until all are included.
The resultant is still drawn from the tail of the first
vector to the tip of the last vector.
R =A+B +C+D
49
Adding Vectors, Rules
When two vectors are added, the sum is independent
of the order of the addition.

This is the Commutative Law of Addition.
A+ B = B+A
50
Adding Vectors, Rules
When adding three or more vectors, their sum is
independent of the way in which the individual vectors
are grouped.

This is called the Associative Property of Addition.
A+(B +C) =(A+B )+C
51
Negative of a Vector
The negative of a vector is defined as the vector that,
when added to the original vector, gives a resultant of
zero.

Represented as - B
B
-B

B+ (-B) = 0
The negative of the vector will have the same
magnitude, but point in the opposite direction.
52
Subtracting Vectors
Another way to look at subtraction is to find the vector
that, added to the second vector gives you the first
vector.


A + (-B) = C
As shown, the resultant vector points from the tip of
the second to the tip of the first.
53
Multiplying or Dividing a Vector by a Scalar
The result of the multiplication or division of a vector by a
scalar is a vector.
The magnitude of the vector is multiplied or divided by the
scalar.
If the scalar is positive, the direction of the result is the same
as of the original vector.
If the scalar is negative, the direction of the result is opposite
that of the original vector.
54
Unit Vectors
A unit vector is a dimensionless vector with a
magnitude of exactly 1. Unit vectors are used to
specify a direction and have no other physical
significance.
The symbols i, j, and k represent unit vectors
They form a set of mutually perpendicular
vectors in a right-handed coordinate system
The magnitude of each unit vector is 1
|i| = |j| = |k| = 1
55
Components of a Vector, Introduction
A component is a projection of a vector along an axis.
It is useful to use rectangular components.
Ax and Ay are the component vectors of A.

They are vectors and follow all the rules for vectors.
Ax and Ay are scalars, and will be referred to as the components of A.
A = Ax + Ay
These three vectors form a right triangle.
A = Ax i + Ay j
Ax = A cosθ ; Ay = A sinθ
The components are the legs of the right triangle whose hypotenuse is the
length of A.
A
Ax2  Ay2 and
  tan
1
Ay
Ax
or     tan
1
Ay
Ax
56
Adding Vectors Using Unit Vectors
R=A+B
We can use their components to represent
R = (Ax i+ Ay j)+ (Bx i+ By j)
R = (Ax+ Bx) i+ (Ay + By) j
R = Rx i + Ry j
R R R
2
x
2
y
  tan
1
Ry
Rx
Then Rx = Ax + Bx ， Ry = Ay + By
57
Adding Vectors Using Unit Vectors
A = Ax i+ Ay j + Az k
A = (Ax2+ Ay2+Az2)½
A = B ，then
Ax = Bx ， Ay = By ， Az = Bz
R = A + B ，then
Rx = Ax+ Bx
Ry = Ay+ By
Rz = Az+ Bz
58
Three-Dimensional Extension
Using R = A + B
Then
R = (Ax i + Ay j + Az k)+ (Bx i+ By j + Bz k)
R = (Ax+ Bx) i+ (Ay + By ) j+ (Az + Bz ) k
R = Rx i + Ry j+ Rz k
R R R R
2
x
2
y
2
z
Rx
 x  cos
, etc
R
1
So Rx= Ax+Bx, Ry= Ay+By, and Rz = Az+Bz
59
Example : The Sum of Two Vectors
Find the sum of two displacement vectors A=(2.0 i +
2.0 j)m and B=(2.0 i - 4.0 j)m lying in the xy plane?
R = A + B = (2.0+2.0) i + (2.0 -4.0) j
= 4.0 m i -2.0 m j
R = (4.02+(-2.0)2)½ = 4.5 m
θ = tan-1(-2.0/4.0) =-27o
60
Example : The Resultant Displacement-3
A particle undergoes three consecutive displacements: r1 =
(1.5 i + 3.0 j -1.2k) cm; r2 =(2.3 i - 1.4 j -3.6k) cm and r3 = (1.3 i + 1.5 j) cm ， Find the components of the result
displacement and its magnitude.
R = r 1 + r2 + r3
=[(1.5+2.3-1.3) i + (3.0 -1.4+1.5) j+(-1.2-3.6) k] cm
= (2.5 i + 3.1 j - 4.8 k) cm
Rx = 2.5 cm; Ry = 3.1 cm; Rz = -4.8 cm;
R = (2.52+ 3.12 +(-4.8)2)½ = 6.24 cm
61
Example : The Resultant Displacement-1
A girl walks 3 m east and then 4m south. What
is her net displacement?
R = A + B = 3i – 4j
R = (32+ 42)½ = 5.0 m
tanθ = -4/3
θ = -53
o
62
Example : The Resultant Displacement-2
A man walks 5 m at 37o north of east and then 10m at 60o
west of north. What is the magnitude and direction of his net
displacement?
Rx=Ax+Bx=5cos37o–10sin60o = -4.66m
Ry=Ay+By=5sin37o+10cos60o = 8.00m
R = (Rx2+ Ry2)½ =9.26m
tan θ = Ry/Rx =-1.72
θ = 120o
63
Example : The Resultant vectors
Give the vectors A=2i-3j+6k m and B = i +2j-3k m. find (a)
A+B (b) |A+B| (c) 2A-3B?
(a) |A| = (22+32 +62 )½ = 7.00m
|B| = (12+22 +32 )½ = (14)½ = 3.74m
A + B = 7.00m + 3.74m =10.74m
(b) R = A + B = 3i – j + 3k
|R| = (32+12+ 32)½ = (19)½ =4.36m
(c) R = 2A-3B =2(2i -3j+6k) -3( i+2j-3k)
= i - 12 j + 21 k
64
Scalar Product of Two Vectors
The scalar product of two vectors is written as A·B.



It is also called the dot product.
A·B = A B cos θ
θ is the angle between A and B
Applied to work, this means
W = F·∆r = F ∆r cos θ
65
The Scalar (Dot) Product
 
A  B  AB cos 
 
 
A B  B  A
iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  1
iˆ  ˆj  iˆ  kˆ  ˆj  kˆ  0
  
   
A  (B  C)  A  B  A  C

A  Ax iˆ  Ay ˆj  Az kˆ

B  Bx iˆ  B y ˆj  Bz kˆ
 
A  B  ( Ax iˆ  Ay ˆj  Az kˆ)  ( Bx iˆ  B y ˆj  Bz kˆ)
 Ax Bx  Ay B y  Az Bz
66
Example : Angle Between Vectors-1
Find the inner product and the lengths of A = i +2j
and B = 3i-2j +k as well as the angle between these
vectors.
A =(1+22)½ =(5)½ ; B =(32+22+1)½ =(14)½
A ·B = Ax Bx+Ay By+Az Bz=1．3+2．(-2)+0 ．1= -1
cosθ =A．B / A B =-1/[(5)½ *(14) ½ ] =-0.11952
θ =
cos-1(-0.11952)
= 96.9
o
67
Example : Angle Between Vectors-2
Find the inner product and the angle of vectors
A =2 i +3j and B = -i+2j ?
(a) A ·B = 2*(-1) + 3 * 2 = 4
(b) cosθ =A．B / A B =4/[(13)½ *(5) ½ ] =0.496
θ =
cos-1(0.496)
= 60.3
o
68
Example : Work of a particle
When a particle moves on the x-y plane and receives a
constant force F = (5.0 i + 2.0 j) N, and its displacement ∆r is
(2.0 i +3.0 j ) m. What is the work done by this force F?
W = F·∆r = 5*2+ 2 * 3 = 16 N．m
69
The Scalar (Dot) Product
Example : Find the scalar product of A = 8i+2j-3k and B = 3i-6j +4k.
A ·B = Ax Bx + Ay By + Az Bz = 8*3+2*(-6)+(-3)*4=0
Example : Find the angle between A = 2i+j+2k and B = 4i-3j.
cosθ = A ·B / A B = (2*4+1*(-3))/(3*5) =1/3
θ = 70.5o
Example : Drive the law of cosines using the scalar product.
C=A-B
C·C = (A-B)·(A-B) = A2+B2 -2 A·B = C2
C2 = A2+B2-2ABcosθ
70
The law of sines
a
b
c


 2R
sin A sin B sin C
A
b
c
C
a
B
C’ B’
71
Example : Taking a Hike
A hiker begins a trip by first walking 25.0 km southeast from
her car. She stops and sets up her tent for the night. On the
second day, she walks 40.0 km in a direction 60.0° north of
east, at which point she discovers a forest ranger’s tower.
A = 25 (cos45o i - sin45 o j) km
= (17.7 i – 17.7 j) km
B = 40 (cos60o i + sin60 o j)km
= (20.0 i + 34.6 j) km
72
Example : Taking a Hike
The hiker’s displacement R
Rx = Ax + Bx = 17.7 km + 20.0 km = 37.7 km
Ry = Ay + By = -17.7 km + 34.6 km = 16.9 km
then
R = (37.7 i + 16.9 j ) km
|R|=41.3 km
θ =24.1° North of East
73
Example : A Vacation Trip
A car travels 20km due north and then 35km in a direction
60° west of north as shown in Figure. Find the magnitude
and direction of the car’s resultant displacement.
R2 = A2 + B2 – 2AB cos θ
= 202 + 352 – 2*20*35 cos 120o = 2325
R = 48.22 km
R/sin120o =35/sinβ
sinβ =35/48.22*0.866=0.629
β =38.9o west of north
74
Example : A Vacation Trip
A = 20 j
B = -35 sin60o i + 35cos60o j
= -30.3 i + 17.5 j
R = A + B = -30.3 i + 37.5 j
|R| = (30.32+37.52)½ = 48.22 km
ϕ = tan-1(37.5/30.3) =51.1 ° north of west
75
The Vector Product Defined
Given two vectors, A and B
We define a third vector, C = A × B , where the direction of C is
perpendicular to the plane formed by A and B.

C is read as “A cross B”.
The magnitude of vector C is AB sinθ

θ is the angle between A and B.
The quantity AB sinθ is equal to the area of the parallelogram formed by
A and B .
The best way to determine this direction is to use the right-hand rule.
76
Properties of the Vector Product
The vector product is not commutative. The order in which
the vectors are multiplied is important.

To account for order, remember B×A = -A×B
If A is parallel to B; (θ = 0o or 180o), then A×B = 0

Therefore A×A = 0
If A is perpendicular to B, then |A×B| = AB.
The vector product obeys the distributive law.

A× (B+C) = A×B + A×C
77
The Vector Product and Torque
The torque vector lies in a
direction perpendicular to the
plane formed by the position
vector and the force vector.
 
  RF

The torque is the vector (or cross)
product of the position vector and
the force vector.
78
Vector Products of Unit Vectors
iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  0
iˆ  ˆj   ˆj  iˆ  kˆ
ˆj  kˆ  kˆ  ˆj  iˆ
kˆ  iˆ  iˆ  kˆ  ˆj
  
   
A  (B  C)  A  B  A  C

A  Axiˆ  Ay ˆj  Az kˆ

B  Bxiˆ  B y ˆj  Bz kˆ
 
A  B  ( Axiˆ  Ay ˆj  Az kˆ)  ( Bxiˆ  B y ˆj  Bz kˆ)
 Axiˆ  ( Bxiˆ  B y ˆj  Bz kˆ )  Ay ˆj  ( Bxiˆ  B y ˆj  Bz kˆ)  Az kˆ  ( Bxiˆ  B y ˆj  Bz kˆ)
 Ax B y kˆ  Ax Bz ˆj  Ay Bx kˆ  Ay Bziˆ  Az Bx ˆj  Az B y iˆ
 ( Ay Bz  Az B y )iˆ  ( Az Bx  Ax Bz ) ˆj  ( Ax B y  Ay Bx )kˆ
79
Using Determinants
The cross product can be expressed as
 
A  B  ( Ay Bz  Az By )iˆ  ( Az Bx  Ax Bz ) ˆj  ( Ax By  Ay Bx )kˆ
The cross product also can be expressed as
iˆ
ˆj
kˆ
 
A  B  Ax
Ay
Az 
Bx
By
Bz
Ay
Az
By
Bz
iˆ 
Az
Ax
Bz
Bx
ˆj 
Ax
Ay
Bx
By
kˆ
80
Final Properties of the Vector Product
The derivative of the cross product with respect to some
variable such as t is


d   dA   dB
( A  B) 
 B  A
dt
dt
dt
where it is important to preserve the multiplicative order of the
vectors.
81
Vector Product Example
Given A = 2 i + 3 j ; B = -i + 2 j
Find A × B
Result
A × B = (2 i + 3 j) × (-i + 2 j)
= 2 i × (-i + 2 j) + 3 j × (-i + 2 j)
=4k+3k
=7k
82
Torque Vector Example
Given the force and location
F = (2.00 i + 3.00 j) N
R = (4.00 i + 5.00 j) m
Find the torque τ = R × F
R × F = (4.00 i + 5.00 j) × (2.00 i + 3.00 j)
= 4.00 i × (2.00 i + 3.00 j) + 5.00 j × (2.00 i + 3.00 j)
= 12 k - 10 k
= 2 k N ·m
83
Example : Angular Momentum
What is the angular momentum of a particle of mass = 2kg
that is located 15m from the origin in the direction 37o S of
W and has a velocity v=10m/s in the direction 30o E of N.
r = -15 cos37o i - 15 sin37o j = -12 i – 9 j
p = mv = 20 sin30o i + 20 cos30o j = 10 i + 17.3 j
L= r ×p = -117.6 k kg·m2/s
84
Calculus Reminder
d x
e  ex
dx
x
x
e
dx

e
C

d x d ln(a x ) d x ln(a )
d
x ln(a )
a  e
 e
e
 ( x ln a )  a x ln a
dx
dx
dx
dx
x  eln x
d
d ln x
d
d
ln x
x  1  e  e  ln x  x  ln x
dx
dx
dx
dx
d
1
ln x 
dx
x
1
 x dx  ln x  C;  ln xdx  x ln x  x  C
85
Calculus Reminder
Polynomials:
Trig functions:
dx n
 n x n 1
dx
d
sin(ax )  a cos(ax )
dx
n
x
 dx 
1 n 1
x C
n 1
d
cos(ax )  a sin (ax )
dx
d
1
ln( ax ) 
dx
x
Exponential, log:
d ax
e  a eax
dx
Product rule:
d
df ( x)
dg ( x)
[ f ( x) g ( x)] 
g ( x)  f ( x)
dx
dx
dx
Chain rule:
d
df ( y ) dg ( x )
f ( g( x )) 
dx
dy
dx
( y  g ( x ))
86
Physics
Chapter 2
Motion in One Dimension
授課老師施坤龍
0
Types of Motions
Types of Motions: Translation, Rotation, and Vibration.
Translation : A car is traveling on a highway.
Rotation: The Earth spins on its axis.
Vibration: A pendulum moves the back-and-forth.
1
Motion Diagrams
2
Motion Diagrams
A motion diagram can be formed by imagining the stroboscope
photograph of a moving object.
Blue arrows represent velocity.
Green arrows represent acceleration.
3
Position
We will use the particle model. A particle is a point-like object; has
mass but infinitesimal size.
Vectors for one-dimensional motion have only one component, xcomponent for now.
x
Use the symbol x to denote position vector
x
0

All position vectors are measured relative to the origin of the coordinate system.

Vector
x
can be positive or negative (its magnitude is again always positive,
though)
X(t)
Position vector is a function of time


Notation for the time-dependent position vector: x(t)
Notation: Vector x at some specific time t1: x(t1)=x1
X(t1)
t1
t
4
Position and Time
The object’s position is its location with respect to a chosen reference
point. Consider the point to be the origin of a coordinate system.
Only interested in the car’s translational motion, so model as a particle
5
Position and Time
The position-time graph shows the motion of the particle (car). The
smooth curve is a guess as to what happened between the data points.
The table gives the actual data collected during the motion
of the object (car). Positive is defined as being to the right.
Table. Position of the Car at Various Times
Position
t(s)
x(m)
A
0
30
B
10
52
C
20
38
D
30
0
E
40
-37
F
50
-53
6
Displacement
Displacement is defined as the change in position during some
time interval.

Represented as x

x ≡ xf - xi

SI units are meters (m)

x can be positive or negative
Different than distance

Distance is the length of a path followed by a particle.
7
Distance vs. Displacement
Assume a player moves
from one end of the court to
the other and back.
Distance is twice the length
of the court

Distance is always positive
Displacement is zero

 x = xf – xi = 0 since xf = xi
8
Average Velocity
The average velocity is rate at which the displacement occurs.


x f  xi
x
vaverage 

t
t
The x indicates motion along the x-axis.
The dimensions are length / time [L/T]
The SI units are m/s
Is also the slope of the line in the position – time graph
9
Average Velocity
FIGURE Calculation of the slope of the line that connects the points on the
curve at t1＝1.0 s and t2＝1.5 s. The x-component of the average
velocity is given by this slope.
10
Average Speed
Speed is a scalar quantity.

Has the same units as velocity

Defined as total distance / total time:
d
vaverage 
t
The speed has no direction and is always expressed as a
positive number.
Neither average velocity nor average speed gives details about
the trip described.
11
Average Speed and Average Velocity
The average speed is not the magnitude of the average velocity.

For example, a runner ends at her starting point.

Her displacement is zero.

Therefore, her velocity is zero.

However, the distance traveled is not zero, so the speed is
not zero.
12
Instantaneous Velocity
The limit of the average velocity as the time interval
becomes infinitesimally short, or as the time interval
approaches zero.
The instantaneous velocity indicates what is happening
at every point of time.
The general equation for instantaneous velocity is:
Δx
dx
v x  lim

Δt 0 Δt
dt
13
Instantaneous Velocity
The instantaneous velocity is
the slope of the line tangent to
the x vs. t curve.

This would be the green line.
The light blue lines show that
as
t
gets
smaller,
they
approach the green line.
14
Instantaneous Velocity and Speed
At point A, vx is positive

The slope is positive
At point B, vx is zero

The slope is zero
At point C, vx is negative

The slope is negative
The
instantaneous
speed
is
the
magnitude
of
the
instantaneous velocity.
The instantaneous speed has no direction associated with it.
15
A Particle Under Constant Velocity
Constant velocity indicates the instantaneous velocity
at any instant during a time interval is the same as the
average velocity during that time interval.

vx = vx, avg

The mathematical representation of this situation is the
equation.


x f  xi
x
vx 

t
t
or
x f  xi  v x t
Common practice is to let ti = 0 and the equation becomes:
xf = xi + vx t (for constant vx)
16
A Particle Under Constant Velocity
The graph represents the
motion of a particle under
constant velocity.
The slope of the graph is the
value of the constant velocity.
The y-intercept is xi.
17
A Particle Under Constant Speed
A particle under constant velocity moves with a
constant speed along a straight line.
A particle can also move with a constant speed along a
curved path.
This can be represented with a model of a particle
under constant speed.
The primary equation is the same as for average speed,
with the average speed replaced by the constant speed.
d
v 
t
18
Example : Roundtrip
Distance between Des Moines, Iowa, and Iowa City, is listed
as 182.6 km. If we take a round trip Des Moines – Iowa City
– Des Moines, what is the total distance and displacement for
this trip?
(a) d = 182.6 km + 182.6 km = 365.2 km
(b) ΔxDI = 182.6 km; Δ xID= -182.6 km
Δxtotal = 182.6 km + (-182.6 km) = 0 km
19
Example : Average Velocity
You drive a pickup truck along a straight road for 8.4 km at 70 km/h,
then the truck runs out of gasoline and stops. Over the next 30 min, you
walk another 2.0 km farther along the road to a gasoline station. (a)
What is your overall displacement Δx? (b) What is the overall time
interval Δt?
(c) What is your average velocity vavg? Find it both
numerically and graphically.
(a) Δx = Δx1+ Δx2 = 10.4 km.
(b) Δt = Δt1+ Δt2
= 8.4 km/70 km/h + 30/60 h = 0.62 h
(c) vavg = Δx/Δt = 10.4km/0.62h = 16.8 km/h
20
Example : Swimming laps
Suppose a swimmer swims the first 50 m of the 100 m
freestyle in 38.2 s. Once she reaches the far side of the pool,
she turns around and swims back to the start in 42.5 s. What
is her average velocity and speed for the entire lap ?
Δt = 38.2 s + 42.5 s = 80.7 s.
Δx = Δx1+ Δx2 = 50m + (-50m) = 0
d =| Δx1|+| Δx2| = 100m
Vave = ∆x/∆t = 0 m/s
Vspeed = d/∆t = 100 m/80.7 s = 1.24 m/s
21
Example：Average Velocity and Speed
A bird flies east at 10 m/s for 100 m. If then turn
around and flies at 20 m/s for 15 s. Find (a) its
average velocity; (b) its average speed ？
average velocity = (100-20*15)/25 = -8m/s
average speed = (100 +20*15)/(100/10+15)=16 m/s
22
Example : Average velocity
A jogger runs her first 100 m at 5m/s and second 100
m at 4 m/s in the same direction. What is her average
velocity?
∆x = 100 m + 100 m = 200 m
∆t = 100 m/(5 m/s) + 100 m/(4 m/s) =45s
Vav = ∆x/∆t = 200 m/45 s =4.44 m/s
23
Example : Motion of a Jogger
A jogger runs in a straight line, with a magnitude of average
velocity of 5.00 m/s for 4.00 min and then with a magnitude
of average velocity of 4.00 m/s for 3.00 min. (a)What is the
magnitude of the final displacement from her initial
position?(b)What is the magnitude of the her average velocity
during this entire time interval of 7.00 min?
(a) ∆x1 = 5.0 m/s * 60 s/min * 4 min = 1200 m
∆x2 = 4.0 m/s * 60 s/min * 3 min = 720 m
∆x = ∆x1 + ∆x2 =1920 m
(b) Vave = ∆x/∆t = 1920/(7 *60) = 4.57 m/s
24
Example ：Instantaneous Velocity-1
The position of a particle is given by the equation x =3t2 m.
Find the instantaneous velocity at 2 s and 3 s by using (a) a
limiting process, and (b) the derivative of the (parabola)
function.
(a) xi = 3t2 , xf =3(t+ ∆t)2; ∆x = 6t∆t+3∆t2 ; ∆x/∆t = 6t+3∆t
V(2) = lim ∆x/∆t = 6t = 12 (m/s)
V(3) = lim ∆x/∆t = 6t = 18 (m/s)
(b) V(2) =dx/dt = d(3t2)/dt =6t=12 m/s
V(3) =dx/dt = d(3t2)/dt = 6t =18 m/s
25
Example：Instantaneous Velocity -2
A particle moves along the x axis. Its position varies with time
according to the expression x=-4t +2t2, where x is in meters and t is in
seconds. (a) Determine the displacement of the particle in the time
intervals t = 1s to t = 3s? (b) Calculate the average velocity during
these two time intervals ? (c) Find the instantaneous velocity of the
particle at t = 2.5s?
(a) xi = x(t=1)= -2m, xf = x(t=3)=6m
∆x = 8m
(b) Vave = ∆x/∆t = 4 m/s
(c) V = dx/dt = -4 +4t
= -4 +4*(2.5) = 6 m/s
26
Example : Instantaneous Velocity-3
A jet engine moves along an experimental track. We will treat the
engine as if it were a particle. Its position as a function of time is given
by the equation x = 2.1t2 m/s2 + 2.8 m. Determine (a) the displacement
of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s? (b)
the average velocity during this time interval? (c) the magnitude of the
instantaneous velocity at t = 5.00 s.
(a) xi = x(t=3)=21.7m, xf = x(t=5)= 55.3m
∆x = 33.6 m
(b) Vave= ∆x /∆t = 33.6/2
= 16.8 m/s
(c) V = dx/dt = 4.20 t
= 21.0 m/s
27
Example : Average and instantaneous velocity
A cheetah is crouched 20 m to east of an observer. At time t = 0 the cheetah begins to
run due east toward an antelope that is 50 m to the east of the observer. During the
first 2.0 s of the attack, the cheetah’s position varies with time to the equation x = 20
m + 5.0 t2 m/s2. Find (a) the cheetah’s displacement between t1 = 1.0 s to t2 = 2.0 s.
(b) the average velocity during that interval. (c) the instantaneous velocity at t1 = 1.0 s
by taking Δt = 0.1 s, 0.01 s, and 0.001 s. (d) the expression for the cheetah’s
instantaneous velocity vx as a function of time. Find vx at t1 = 1.0 s and t2 = 2.0 s.
(a) x(1) = 25 m; x(2) = 40 m;
Δx = 40 m – 25 m = 15 m
(b) Vave = Δx / Δt = 15 m / 1 s = 15 m/s
(c) x(1) = 25 m; x(1.1) = 26.05 m; VΔt=0.1 = 10.5 m/s
x(1.01) = 25.1005 m; VΔt=0.01 = 10.05 m/s
x(1.001) = 25.010005 m; VΔt=0.001 = 10.005 m/s
Vx(1) = 10.0m/s
(d) Vx = dx / dt = 10 t m/s2 ; Vx(1) = 10 m/s; Vx(2) = 20 m/s
28
Example： Average Velocity and Speed
In the velocity–time graph given in Figure, notice that xA=30
m at tA=0s, xD=-75m at tD=35s, and that xF=-53m at tF=50s.
Find the quantity between a tA and tF ? (a) displacement;
(b)distance; (c) average velocity; (d) average speed.
(a) ∆x = xF - xA = -53-30=-83m
(b) d = 105+22 = 127m
(c) average velocity = -83/50 = -1.66m/s
(d) average speed = 127/50 = 2.54m/s
29
Average Acceleration
Acceleration is the rate of change of the velocity.
a x ,ave
v x , f  v x ,i
v x


t
t f  ti
Dimensions are L/T2
SI units are m/s²
In one dimension, positive and negative can be used
to indicate direction.
30
Instantaneous Acceleration
The instantaneous acceleration is the limit of the
average acceleration as t approaches 0.
v x
dv x
d x
a x  lim


2
dt
dt
t 0 t
2
The term acceleration will mean instantaneous
acceleration.

If average acceleration is wanted, the word average
will be included.
31
Instantaneous Acceleration
The slope of the velocity-time graph is the
acceleration.
The red line represents the instantaneous acceleration.
The brown line is the average acceleration.
32
Acceleration and Velocity
When an object’s velocity and acceleration are in the same
direction, the object is speeding up.
When an object’s velocity and acceleration are in the opposite
direction, the object is slowing down.
Negative acceleration does not necessarily mean the object is
slowing down.

If the acceleration and velocity are both negative, the object is
speeding up.
The word deceleration has the connotation of slowing down.
33
Motion Diagrams
A motion diagram can be formed by imagining the
stroboscope photograph of a moving object.
Blue arrows represent velocity.
Green arrows represent acceleration.
34
Acceleration
FIGURE
(a)The position component x(t) of a
particle moving with constant
acceleration.
(b)Its velocity component v x(t), given
at each point by the slope of the
curve in (a).
(c)Its(constant)component of
acceleration, a x equal to the
(constant) slope of v x(t) .
(c)
35
Acceleration
FIGURE (a) The x vs. t graph for an
elevator cab that moves upward along
an x axis. (b) The v x vs. t graph for the
cab. Note that it is the derivative of the
x vs. t (vx ＝ dx/dt ). (c)The a x vs. t
graph for the cab. It is the derivative of
the v x vs. t (a x ＝d x /dt).
The figures along the bottom suggest
times that a passenger might feel light
and long as the elevator accelerates
downward or heavy and squashed as the
elevator accelerates upward.
36
Example： Car speeding up
A car accelerates along a straight road from rest to 90
km/h in 5.0 s. What is the magnitude of its average
acceleration?
Vi= 0 m/s
Vf = 90km/h = 25m/s
aave = (Vf - Vi )/∆t
= 25/5 =5.0 m/s2
37
Example： Car slowing down
An automobile is moving to the right along a straight
highway, which we choose to be the positive x axis. Then the
driver puts on the brakes. If the initial velocity (when the
driver hits the brakes) is v1 = 15.0 m/s, and it takes 5.0 s to
slow down to v2 = 5.0 m/s, what was the car’s average
acceleration?
Vi= 15.0 m/s
Vf = 5.0 m/s
aave = (Vf - Vi )/∆t
= -10/5 =-2.0 m/s2
38
Example : Acceleration given x(t)
A particle is moving in a straight line so that its position is given by the
relation x = (2.10 m/s2) t2 + (2.80 m). Calculate (a) its average
acceleration during the time interval from t1 = 3.00 s to t2 = 5.00 s, and
(b) its instantaneous acceleration as a function of time.
(a) V = dx/dt = 4.20 t
Vi = V(t=3)=12.6m/s
Vf = V(t=5)= 21.0m/s
aave= ∆V /∆t = 8.40/2
= 4.20 m/s2
(b) a = dV/dt = d(4.20 t)/dt
= 4.20 m/s2
39
Example : Acceleration given v(t)
The velocity of a particle moving along the x axis varies
according to the expression vx = 40 - 5t2, where vx is in
meters per second and t is in seconds. Find (a) the average
acceleration in the time interval t = 0 to t = 2.0 s.(b) the
acceleration at t = 2.0 s.
(a) Vi = V(t=0)=40 m/s
Vf = V(t=2)= 20 m/s
aave= ∆V /∆t = -20/2
= -10 m/s2
(b) a (t=2) = dV/dt = d(40-5t2)/dt=-10t
= -20 m/s2
40
Example：Average Acceleration
At t=0 a car is moving east at 10m/s. Find its average acceleration
between t=0 and each of the following times at which it has the given
velocities; (a) t=2s, 15m/s east; (b) t=5s, 5m/s east; (c) t=10s, 10m/s
west; (d) t=20s, 20m/s west. Find aav between (e) 2 and 10s; (f) 10 and
20s.
(a) aav = (15-10)/2 = 2.5m/s2
(b) aav = (5-10)/5 = -1m/s2
(c) aav = (-10-10)/10 = -2m/s2
(d) aav = (-20-10)/20 = -1.5m/s2
(e) aav = (-10-15)/8 = -3.125m/s2
(f) aav = (-20+10)/10 = -1m/s2
41
Example：Average and Instantaneous Acceleration
Suppose the velocity of a car at any time t is given by the equation vx = 60 m/s + 0.5 t2
m/s3. Find (a) the change of velocity in the time interval t1 = 1.0 s to t2 = 3.0 s. (b) the
average acceleration in this time interval. (c) the instantaneous acceleration at t1 = 1.0
s by taking Δt = 0.1 s, 0.01 s, and 0.001 s. (d) the expression for the instantaneous
acceleration ax as a function of time. Find ax at t1 = 1.0 s and t2 = 3.0 s.
(a) vx(1) = 60.5 m/s; vx(3) = 64.5 m/s
Δvx = 64.5 m/s – 60.5 m/s = 4.0 m/s
(b) aave = Δvx / Δt = 4.0 m/s / 2 s = 2 m/s2
(c) vx(1) = 60.5 m/s; vx(1.1) = 60.605 m; aΔt=0.1 = 1.05 m/s2
vx(1.01) = 60.51005 m/s; aΔt=0.01 = 1.005 m/s2
vx(1.001) = 60.5010005 m; aΔt=0.001 = 1.0005 m/s2
(d) ax = dvx / dt = 1.0 t m/s3 ; ax(1) = 1.0 m/s2; ax(3) = 3.0 m/s2;
42
Graphical Comparison
Given
the
displacement-time
graph (a)
The velocity-time graph is found
by measuring the slope of the
position-time
graph
at
every
instant.
The acceleration-time graph is
found by measuring the slope of
the velocity-time graph at every
instant.
43
Kinematic Equations
Acceleration is the differential
dv x
a x
dt
of velocity with respect to time.
The change in speed is equal to
t
v f  vi   a x dt
the integral of acceleration with
dx
vx
dt
Velocity is the differential of
0
respect to time.
position with respect to time.
t
x f  xi   v x dt
0
Displacement is equal to the
integral of speed with respect
to time.
44
Conceptual Example : Analyzing with graphs
This figure shows the velocity as a function of time for two cars
accelerating from 0 to 100 km/h in a time of 10.0 s. Compare (a) the
average acceleration; (b) instantaneous acceleration; and (c) total
distance traveled for the two cars.
(a) The average acceleration of the two cars in 10.0 seconds is the same
(b) The instant acceleration of car A is larger at the beginning, but the
instant acceleration of car B is larger after 3.0 seconds
(c) The area of ​the A car is larger within 10 seconds.
Therefore, the distance traveled by car A is larger
45
Example : The Use of Areas
A particle is rest at the origin
at t = 0 s, and its acceleration
is 2m/s2 from 0 to 3 seconds;
its acceleration is -2m/s2 from
3 to 6 seconds; try to plot its
speed, time, and displacement
Graph as a function of time.
46
Kinematic Equations
The kinematic equations are useful when you can
model a situation as a particle under constant
acceleration
When the acceleration is zero, (a =0)

vxf = vxi = vx

xf = xi + vx t
The constant acceleration model reduces to the
constant velocity model.
47
Kinematic Equations
For constant ax,
vx , f  vx ,i  ax t
For constant acceleration, this gives you the position of the
particle in terms of time and velocities
1
x f  xi  (vx ,i  vx , f )t
2
Gives final position in terms of velocity and acceleration.
Gives
1
x f  xi  vx ,i t  a x t 2
2
final
velocity
in
terms
of
acceleration
displacement, there are no information about the time.
2
2
x, f
x ,i
x
f
i
v
v
and
 2a ( x  x )
48
The constant acceleration motion
vf= vi +at
vi
vi
t
v f  vi  at
1
x  (vi  v f )t
2
1 2
x  vi t  at
2
1
1
x  (vi  v f )t  (vi  v f )(v f  vi ) / a
2
2
(v 2f  vi2 )

2a
v 2f  vi2  2ax
49
Kinematic Equations
50
Graphical Look at Motion: Time curve
The slope of the curve is the velocity. The curved line indicates
the velocity is changing.

Therefore, there is an acceleration.
The slope gives the acceleration.
The straight line indicates a constant acceleration.
The zero slope indicates a constant acceleration.
51
Equations of Kinematics
V = V0 + at
X = X0 + ½ (V0+V)t
X = X0 + V0t + ½ at2
V 2= V02 + 2a(X-X0)
52
Equations of Kinematics
Vf = V0 + at
Equation 1
In Equation 1, we can obtain the final velocity Vf at any time
t from the initial velocity V0 of an object, the time t, and its
acceleration a.
Xf = X0 + ½ (V0+Vf)t
Equation 2
In Equation 2, we can obtain the position Xfat any time t from
the initial position X0 of the object, the initial velocity V0, the
final velocity Vf, and the time t. This equation is applied
when the acceleration a is not given.
53
Equations of Kinematics
Xf = X0 + V0t + ½ at2
Equation 3
In Equation 3, we can obtain the position Xf at any time from
the initial position X0 of the object, the initial velocity V0, the
acceleration a, and the time t. This equation is applied when
the final velocity Vf is not given.
Vf 2= V02 + 2a(Xf-X0)
Equation 4
In Equation 4, we can obtain the end velocity Vf at any time
from the initial velocity V0 of the object, the acceleration a,
and the displacement (Xf-X0). This equation is applied when
the time t is not given.
54
Example : Accelerating an Electron
An electron in the cathode-ray tube of a television set enters a
region in which it accelerates uniformly in a straight line
from a speed of 3.00 × 104 m/s to a speed of 5.00 × 106 m/s in
a distance of 2.00 cm. For what time interval is the electron
accelerating?
S=0.02 =½ (V0+V)t =½ (3.0 ×104+ 5.0×106)t
t = [0.04/(5.03×106)]
= 7.95×10－9 s
55
Example： Constant Acceleration - 1
Assuming a constant acceleration of a = 4.3 m/s2,
starting from rest, what is the takeoff speed of the
airplane reached after 18 seconds?
V0 = 0 m/s
V = V0 + at = 4.3m/s2 *18 s = 77.4 m/s
56
Example： Constant Acceleration - 2
A car accelerate system with constant acceleration from rest
to 30 m/s in 10 s. It then continues at constant velocity. Find:
(a) its acceleration; (b) how far it travels while speeding up;
(c) the distance it covers while its velocity changes from 10
m/s to 20 m/s.
(a) a = (30-0)/10 = 3m/s2
(b) S = (X-X0) = (V 2-V02)/(2a) = 900/6 =150 m
S = ½ (V0 +V) t = 0.5*30*10 = 150 m
(c) S = (X-X0) = (V 2-V02)/(2a) = (400-100)/6 = 50 m
57
Example： Constant Acceleration - 3
Accelerating from rest, a top fuel racer can reach 148.9 m/s at
the end of a quarter mile (= 402.3 m) run. For this example,
we will assume constant acceleration. (a) What is the value of
this acceleration? (b) How long does it take to complete
quarter mile race from a standing start?
(a)a=(Vf2-V02)/(2S)=148.92/804.6 =27.6 m/s2
(b) t = 2S/(V0+Vf) = 804.6/148.9 = 5.40 s
58
Example： Constant Acceleration - 4
A particle is at 5 m at t=2s and has a velocity v = 10 m/s. Its
acceleration is constant at -4m/s2. Find the initial position at
t=0?
V = V0 + at = 10m/s = V0 - 4m/s2*2s
V0 = 18m/s
X = X0 + ½ (V0+V)t = 5 = X0 + ½ (18+10)*2
X0 = 5m -28m = -23m
59
Example： Constant Acceleration - 5
A car speeding at 144km/h passes a still police car which
immediately takes off in hot pursuit with constant
acceleration. After 10s, the police car to overtake the speeder.
Find the acceleration of the police car?
Sspeeder=40*10 = 400m,
Spolice = 400m = ½ *a *t2
= ½ *a *100
a =8m/s2
60
Example： Constant Acceleration - 6
A car speeding at 45m/s passes a still police car which takes
off in hot pursuit with constant acceleration 3.00m/s2 after 1s
later. When does the speeder get caught?
Sspeeder=45*(t+1) =Spolice = 1.5t2
t2-30t-30= 0
t = 30.97 s = 31 s
61
Example： Constant Acceleration - 7
A speeder moves at a constant 15m/s in a school zone. A
police car starts from rest just the speeder passes it. The
police car accelerates at 2m/s2 until it reaches its maximum
velocity of 20m/s. Where and when does the speeder get
caught?
First 10s, Spolice = ½ *a *t2 = 100m, Sspeeder=15*10= 150m
t =10s +∆t, Spolice = 100 +20*∆t; Sspeeder = 150+15*∆t
Spolice= Sspeeder;
150m -100m = (20m/s-15m/s)*∆t
t = 10s + ∆t = 10s + 50m/(5m/s) = 20s
Stotal = 15m/s * 20s = 300m
62
Example： Constant Acceleration - 8
Two cars approach each other on straight road. Car A moves at 16m/s
and Car B moves at 8m/s. When they are 45m apart, both drivers apply
their brakes. Car A slows down at 2m/s2, while car B slows down at
4m/s2. Where and when do they collide?
VA=16-2t >0; XA=16t-t2
VB = -8+4t<0; Car B stop after 2s;
XB = 45-8t+2t2;
When collide XA=XB; 16t-t2= 45-8t+2t2;
t=3s or 5s > 2s;
XB(t=2s)= 45-8t+2t2 = 37m;
XA=XB=37m= 16t-t2 ; t=2.8s or 13.2s;
So at t=2.8s, two car collide at 37m (distance for Car A moves)
63
Example： Braking distances
Calculate the total stopping distance for an initial velocity of 50 km/h (=
14 m/s ≈ 31 mi/h) and assume the acceleration of the car is -6.0 m/s2 .
The reaction time for the driver is about 0.50 s.
When t=0.5s; S1=14*0.5 = 7m; (V=constant=14)
t > 0.5s; Vi = 14 , Vf = 14-6(t-0.5) =0;
Vf2= Vi2 + 2aS2= 0 = 142 + 2*(-6) S2
S2 = 16.33m;
S =S1 + S2= 23.33m
64
Example： Air bags
Suppose you want to design an air bag system that can protect the
driver at a speed of 100 km/h (60 mph) if the car hits a brick wall.
Estimate how fast the air bag must inflate to effectively protect the
driver. How does the use of a seat belt help the driver?
V = 100 km/h = 27.8 m/s;
d = 1m = Vt = 27.8 t
t = 0.036 s (如果沒有安全帶)
V0 = 27.8 m/s; Vf = 0;
d = 1m = ½ (27.8+0)t
t = 0.07 s (use of a seat belt )
65
Freely Falling Objects
A freely falling object is any object moving freely under the
influence of gravity alone
It does not depend upon the initial motion of the object

Dropped – released from rest

Thrown downward

Thrown upward
The acceleration of an object in free fall is directed
downward, regardless of the initial motion
66
Acceleration of Freely Falling Object
The magnitude of free fall acceleration is g = 9.80 m/s2

g decreases with increasing altitude

g varies with latitude

9.80 m/s2 is the average at the earth’s surface
We will neglect air resistance. Free fall motion is constantly
accelerated motion in one dimension.
Let upward be positive
Use the kinematic equations with ay = g = -9.80 m/s2

The negative sign indicates the direction of the
acceleration is downward
67
Example ： Catch the bill
Emily challenges David to catch a dollar bill as follows. She
holds the bill vertically, as Figure, with the center of the bill
between David’s index finger and thumb. David must catch
the bill after Emily releases it without moving his hand
downward. The reaction time of most people is at best about
0.2 s. Who would you bet on?
When t=0.2s; S= ½ gt2 = 0.2m;
0.2m >the length of bill ;
Then David did not catch the coin
68
Example ：Thrown A ball
A ball is thrown vertically upwards with an initial velocity of
27.0 m/s. (a) Neglecting air resistance, how long is the ball in
the air? (b) What is the greatest height reached by the ball?
(c) In fact, the ball hits a bird on its way up when it has half
of its initial velocity. At what altitude does that happens?
(a) Y = V0t – ½ gt2 =27t-4.9t2 =0 ; t =27/4.9 = 5.51 s
(b) H = (V 2-V02)/(2a)= -V02/(-2g) = 272/19.6 = 37.2 m
(c) H2 = (1/4V02-V02)/(2a)= 0.75* 272/19.6 = 27.9 m
69
Example ：Drop a screw
driver
A construction worker is riding an exterior elevator to the top
of a skyscraper. The elevator’s speed is 4.3 m/s. When the
elevator is 72.4 m above the ground, a screwdriver falls off.
How long does it take the screwdriver to reach the ground?
What is the velocity of the screwdriver as it hits the ground?
a = -g = -9.8 m/s2 ; V = V0 – 9.8 t = 4.3 - 9.8t;
Y = Y0 + V0t – ½ gt2 = 72.4 + 4.3t - 4.9t2
Y = 72.4 + 4.3t - 4.9t2 = 0; t = 4.3 s
V = V0 – 9.8 t = 4.3 - 9.8*4.3 = -37.9 m/s
70
Example： Thrown Stone
A stone thrown from the top of a building is given an initial velocity of 20.0 m/s
straight upward. The stone is launched 50.0 m above the ground, and the stone just
misses the edge of the roof on its way down as Figure. Determine(a) the time at which
the stone reaches its maximum height? (b) the maximum height of the stone? (c) the
velocity of the stone return the launched point? (d) the velocity and position of the
stone at t = 5.0s? (e) the position of the stone at t = 6.0s?
V = V0 – 9.8 t = 20-9.8t; Y = Y0 +V0t – ½ gt2 =20t-4.9t2
(a)At the maximum height, V = 0; t = 20/9.8 =2.04s
(b) H =20*2.04-4.9*2.042 =20.4m
(c) Y=0; t=20/4.9=4.08s; V=20-9.8*4.08= -20m/s
(d)t=5s; V=20-9.8*5=-29m/s; Y=20*5-4.9*25=-22.5m
(e)t=6s; Y=20*6-4.9*36=-56m <-50m;
then Y= -50m; V=0 m/s (The stone has fallen to the ground)
71
Physics
Chapter 3
Motion in Two Dimensions
授課老師施坤龍
0
Galileo Galilei
1564 – 1642
Italian physicist and astronomer
Formulated laws of motion for
objects in free fall
Supported heliocentric universe
1
General Motion Ideas
We will study the vector nature of position, velocity and
acceleration in greater detail, and treat projectile motion and
uniform circular motion as special cases; finally discuss
relative motion in this chapter.
In two- or three-dimensional kinematics, everything is the
same as in one-dimensional motion except that we must now
use full vector notation.

Positive and negative signs are no longer sufficient to
determine the direction.
2
Position and Displacement
The position of an object is described by its position
vector, r .
The displacement of the object is defined as the
change in its position.
  
r  rf  ri
3
Average Velocity
The average velocity is the ratio of the displacement to
the time interval for the displacement.

vave

r

t
The direction of the average velocity is the direction of
the displacement vector.
The average velocity between points is independent of
the path taken.

This is because it is dependent on the displacement, which
is also independent of the path.
4
Instantaneous Velocity
The instantaneous velocity is the
limit of the average velocity as Δt
approaches zero.



r dr
v  lim

dt
t 0 t

As the time interval becomes
smaller, the direction of the
displacement approaches that of
the line tangent to the curve.
5
Instantaneous Velocity
The direction of the instantaneous velocity vector at
any point in a particle’s path is along a line tangent to
the path at that point and in the direction of motion.
The magnitude of the instantaneous velocity vector is
the speed.

The speed is a scalar quantity.
6
Average Acceleration
As a particle moves, the direction of the change in velocity is
found by vector subtraction. ∆V =Vf - Vi
The average acceleration is a vector quantity directed along
∆V.
The average acceleration of a particle as it moves is
defined as the change in the instantaneous velocity vector
divided by the time interval during which that change occurs.
  

v v f  vi
aave 

t t f  ti
7
Instantaneous Acceleration
The instantaneous acceleration is the limiting value of
the ratio ∆v/∆t as ∆t approaches zero.



v dv
a  lim

dt
t 0 t

The instantaneous equals the
derivative of the velocity
vector with respect to time.
8
Two Dimension Motion
Constant Acceleration
V = V0 + at
R = R0 + ½ (V0+V)t
R = R0 + V0t + ½ at2
V2 = V02 + 2a． (R-R0)
The equation of Two Dimension Motion
x-component (horizontal)
y-component (vertical)
ax
ay
vx = vx0 + axt
vy = vy0 + ayt
x = x0 + vx0t + ½ axt2
y = y0 + vy0t + ½ ayt2
vx2 = vx02 + 2ax(x-x0)
vy2 = vy02 + 2ay(y-y0)
9
Two-Dimensional Motion
When the two-dimensional motion has a constant acceleration,
a series of equations can be developed that describe the motion.
These equations will be similar to those of one-dimensional
kinematics.
Motion in two dimensions can be modeled as two independent
motions in each of the two perpendicular directions associated
with the x and y axes.

Any influence in the y direction does not affect the motion
in the x direction.
10
Kinematic Equations
Position vector for a particle moving in the xy plane.
r=xi+yj
The velocity vector can be found from the position
vector.
v = dr/dt = vx i + vy j

Since acceleration is constant, we can also find an
expression for the velocity as a function of time:
Vf = Vi + a t
11
Kinematic Equations
The position vector can also be expressed as a function of
time:

rf = ri + v t + ½ a t2
vf = vi + a t

vxf = vxi + axt

vyf = vyi + ayt
rf = ri + vi t + ½ a t2

xf = xi + vxi t + 1/2 axt2

yf = yi + vyi t + 1/2 ayt2
12
Example : Motion in a Plane
A particle moves in the xy plane, starting from the origin at t = 0 with an
initial velocity having an x component of 20 m/s and a y component of 15m/s. The particle has an acceleration in the x direction, given by
ax=4.0 m/s2. Determine (A) its velocity at any time. (B) its velocity and
speed at t = 5.0 s and the angle of the velocity vector with the x axis.
(C) its position vector at any time t .
(A) Vf = Vi + a t = (20+4t) i – 15 j
Vxf = 20 m/s + 4.0t m/s2, Vyf = -15m/s,
(B) Vf = (40 i – 15 j )m/s； Vxf =40m/s, Vyf= -15m/s,
|Vf| = 42.72 m/s；
θ= tan-1(-15/40)= -20.56o
(C) rf = ri + vi t + ½ a t2；
= (20t+2t2) i - (15t) j
13
Projectile Motion
A projectile is an object moving in two
dimensions under the influence of
Earth's gravity; its path is a parabola.

The acceleration is constant over the
range of motion.

It is reasonable as long as the range
is small compared to the radius of
the Earth.

The effect of air friction is negligible.
14
Projectile Motion
Reference frame chosen

y is vertical with upward positive
Acceleration components

ay = -g and ax = 0
Initial velocity components

υxi = vi cosθ and vyi = vi sinθ
15
Projectile Motion
Kinematic Equations for Ideal Projectile Motion
Quantity
Horizontal
Vertical
Forces
Fx = 0 i
Fy = -mg j
Acceleration Components
ax = 0
ay = -g = -9.8
Velocity Components at t1
v1x = v1 cosθ1
v1y = v1 sinθ1
Position Component Change x2 - x1 = v1x (t2 - t1)
between t1 and t2
y2 - y1 = v1y (t2 - t1)
+ ½ ay (t2 - t1)2
Velocity Component Change v2x - v1x = 0
between t1 and t2
v2y- v1y = ay (t2 - t1)
16
Projectile Motion
We treat the horizontal and vertical
motion independently.
Kinematic Equations for Projectile Motion
Horizontal
ax=0
Vertical
ay=-g=-9.8
vx= vx0
vy= vy0 - gt
x= x0 +vx0 t
y= y0 +vy0 t – ½ g t2
vy2= vy02 - 2g(y-y0)
17
Analyzing Projectile Motion
Consider the motion as the superposition of the motions in the
x- and y-directions.
The actual position at any time is given by:

r f = r i + v i t + ½ g t2
The velocity components for the projectile at any time t are:

vxf = vxi = vi cosθi = constant

vyf = vyi – gt = vi sinθi – gt
The x-direction has constant velocity, ax = 0.
The y-direction is free fall, ay = -g.
18
Projectile Motion – Position
Displacements

xf = vxi t = (vi cosθ)t

yf = vyi t + 1/2ay t2 = (vi sinθ)t - 1/2 gt2
Combining the equations gives:


g
2
y f  (tan i ) x f  [ 2 2 ]x f
2vi cos i
This is in the form of y = ax – bx2 which is the standard
form of a parabola
19
Range and Maximum Height of a Projectile
The range, R, is the horizontal distance of the projectile (y=0).
The maximum height the projectile reaches is H (Vy=0).
υxf = vi cosθi ； vyf = vi sinθi – gt
x f = x i+ vi cosθi t ；yf = y i+ vi sinθi t - ½ gt2
v sin i
H
2g
2
i
2
v sin 2i
R
g
2
i
20
Projectile Motion
The maximum range occurs at θi = 45o .
Complementary angles will produce the same range.
21
Realistic Projectile Motion
As far as ideal projectile motion is concerned, beach balls and
baseballs have the same trajectory
Air resistance: drag force, proportional to v2
Leads to “ballistic curves”
Example: baseballs launched at 35˚ and with 90 or 110 mph
22
Example : A Projectile
A Projectile is fired from the ground with an initial velocity V0 at an
angle θ above the horizontal. Find (a) the time of flight; (b) the
horizontal range R; (c) the Projectile function.
Vx = V0 cosθ;
X = V0 cosθ* t;
Vy = V0 sinθ– gt;
Y = V0 sinθ* t – ½ gt2;
(A)Y=0=V0 sinθ* t – ½ gt2; t = 2V0 sinθ/g;
(B) R =V0 cosθ* t =V0 cosθ*2V0 sinθ/g = V02sin2θ/g
(C) Y = X*tanθ– X2*g/2/(V0cosθ)2
23
Example :Javelin Throwing at the Olympics
An athlete throws a javelin a distance of 80.0 m at the Olympics held at
the equator, where g = 9.78 m/s2. Four years later, the Olympics are
held at the North Pole, where g = 9.83 m/s2. Assuming that the thrower
provides the javelin with exactly the same initial velocity as she did at
the equator, how far does the javelin travel at the North Pole?
R1 =υi2 sin2θ / gequator = 80 m
R2 =υi2 sin2θ / gN_pole
R2 = R1 gequator/gN_pole
= 80*9.78/9.83
=79.6m
24
Example : A projected ball - 1
A ball is projected horizontally at 15m/s from a cliff
of height 20m. Find: (a) its time of flight; (b) its
horizontal range R.
X = 15 t;
Y = 20 – 4.9 t2
(a) 0=20 – 4.9 t2; t = -2.02s or 2.02s; t =2.02s
(b) R = 15 * 2.02 = 30.3m
25
Example : A Projectile ball - 2
A ball is thrown from the top of a building at an angle of 30.0° to the
horizontal and with an initial speed of 20.0 m/s. The height of the
building is 45.0 m. (a) How long is the ball “in flight”?(b) What is the
speed of the ball just before it strikes the ground?
υxf = vi cosθ = 17.32m/s ;
vyf = vi sinθ – gt = 10-9.8t
x f = 17.32t ；
yf = 10t – 4.9 t2
(a) yf = 10t – 4.9 t2 = -45; t =4.22s
(b) t =4.22s;
υxf =17.32m/s ; vyf = -31.4m/s
vf =(17.322+31.42)½ =35.9m/s
θ = tan-1(-31.4/17.3) = -61.1°
26
Example : A Projectile ball - 3
A ball is thrown at 21m/s at 300 above the horizontal from the top of a
roof 16m high. Find (a) the time of flight; (b) the horizontal range R; (c)
the maximum height; (d) the angle at which the ball hits the ground; (e)
the velocity when it is 2m above the roof.
Vx = V0cosθ = 18.2m/s;
X = V0cosθ *t = 18.2* t;
Vy = V0sinθ–gt =10.5-9.8t;
Y = Y0+V0sinθ *t–½ gt2 = 16+10.5t-4.9t2
(A) 0 =16+10.5t-4.9t2; t = -1.03s or 3.17s; t =3.17s
(B) R = 18.2 * 3.17 = 57.7m
(C) Vy = 10.5-9.8t; t =1.07s; Y =16+10.5*1.07-4.9*(1.07)2; Y= 21.6m
(D) Vx=18.2m/s; Vy=10.5-9.8t=-20.6 m/s;
tanθ = -20.6/18.2; θ = -48.50
(E) Vy2= 10.52 – 2* 9.8 * (18 -16) ; Vy = -8.4m/s or 8.4m/s
V = 18.2 i – 8.4 j m/s
or V = 18.2 i + 8.4 j m/s
27
Example : A high fly ball
Suppose a baseball batter hits a high fly ball to the outfield,
directly toward an outfielder and with a launch speed of v0 =
40 m/s and a launch angle of θ0 = 35°. Find (a) its time of
flight (b) its horizontal range R?
Vx = V0cosθ =32.77 m/s; Vy = V0sinθ–gt = 22.94 m/s -9.8*t
X = V0cosθ*t = 32.77*t; Y = V0sinθ*t – ½ gt2 =22.94t -4.9t2
(a)Y=0=V0 sinθ*t – ½ gt2;
t = 2V0 sinθ/g = 80 sin35°/9.8 = 4.68 s
(b) R =V0 cosθ*t =32.77*4.68 = 153.4 m
28
Example : A water-balloon
A boy on a small hill aims his water-balloon slingshot horizontally,
straight at a second boy hanging from a tree branch a distance d away.
At the instant the water balloon is released, the second boy lets go and
falls from the tree, hoping to avoid being hit. Show that he made the
wrong move.
XT =d; YT = – ½ gt2;
XB = V0 t = d;
(The second boy )
YB = – ½ gt2 (water-balloon)
t = d / V0;
YT = –½ gt2 = –½ g(d/V0)2
YB = – ½ gt2 = –½ g(d/V0)2
YT = YB ; The second boy was hitting by the water-balloon.
29
Example : An archerfish
An archerfish shoot a drop of water directly at a beetle. At the same
instant the beetle starts to fall. Show that the beetle will be hit provided
the trajectory of the drop intersects the line of the beetle’s fall. Assume
the beetle at a height H and horizontal distance L from fish.
XB =L; YB = H – ½ gt2;
VDx = V0cosθ;
(beetle)
XD = V0cosθ *t = L; (a drop of water)
VDy = V0sinθ–gt; YD = V0sinθ *t – ½ gt2 (a drop of water)
H = L tanθ; t = L/(V0cosθ);
YD=V0sinθ *t–½ gt2
= L tanθ – ½ gt2 = H – ½ gt2
= YB ; The beetle was hit.
30
Example : The End of the Ski Jump
A ski jumper leaves the ski track moving in the horizontal direction
with a speed of 20.0 m/s as the Figure. The landing incline below her
falls off with a slope of 37.0°. Where does she land on the incline?
Vx = 20 ; X = 20 t;
Vy = - gt; Y = – 4.9t2;
tan(37o)=0.75=4.9t2/20t;
t=3.06s
X=20*3.06=61.2 m;
Y= – 4.9*3.062= – 45.9 m;
d = (61.22+45.92)½ = 76.5 m
31
Example : The Stranded Explorers
If the plane is traveling horizontally at 40.0 m/s at a height of
100 m above the ground, where does the package strike the
ground relative to the point at which it is released?
υxf = 40.0m/s ; vyf = -9.8t
x f = 40.0t ；yf = – 4.9 t2
yf = – 4.9 t2 = -100;
t =4.52s
xf = 40.0*4.52 =181m
32
Example : A rescue helicopter
A rescue helicopter wants to drop a package of supplies to isolated mountain climbers
on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed
of 70 m/s, (a) how far in advance of the recipients must the package be dropped? (b)
Suppose, instead, that the helicopter releases the package a horizontal distance of 400
m in advance of the mountain climbers. What vertical velocity should the package be
given so that it arrives precisely at the climbers’ position? (c) With what speed does
the package land in the latter case?
Vx = 70 ; X = 70 t; Vy =Vy0 - gt; Y =200+Vy0t – 4.9t2;
(A) Vy0=0; Y=0=200 – 4.9t2; t=6.389s;
X=70*6.389=447.2m
(B) X=400= 70 t; t=40/7 =5.71s;
Y=0=200+Vy0t -4.9t2=200+5.71Vy0-160
Vy0 =－7 m/s (↓); Vyf = -7 -9.8*5.71= -63 m/s
(C) Vf2 =V02+2a ·S = 4900+7*7+2*9.8*200;
Vf = 94.18m/s
33
Uniform Circular Motion
Uniform circular motion occurs when an
object moves in a circular path with a
constant speed.
Instantaneous velocity is always tangent to
the circle.
An acceleration exists since the direction of
v2 = v1 +∆v
the motion is changing . This acceleration is
called the centripetal, or radial, acceleration,
and it points toward the center of the circle.
34
Uniform Circular Motion
The period, T, is the time required for one complete revolution.
The speed of the particle would be the circumference of the
circle of motion divided by the period.
V = 2πr/T ；
T = 2πr/V
△r
The centripetal acceleration ac or ar
ac = 2πV/T = 4π2r/T2 = V2/r ；
2

v
ac   rˆ
r
∆r /r = ∆v /v = ∆θ
35
Example : Acceleration of a revolving ball
A 150-g ball at the end of a string is revolving uniformly in a
horizontal circle of radius 0.600 m. The ball makes 2.00
revolutions in a second. What is its centripetal acceleration?
T = 1/2 s = 0.5s ;
V = 2πr/T = 2π*0.6/0.5;
= 7.54 m/s;
a = 4π2r/T2 = 4π2*0.6/(0.5)2
= 94.6 m/s2
36
Example : Ultracentrifuge
The rotor of an ultracentrifuge rotates at 50,000 rpm
(revolutions per minute). A particle at the top of a test tube is
6.00 cm from the rotation axis. Calculate its centripetal
acceleration, in “g’s.”
T = 60/50000 s = 1.2×10-3s = 1.2 ms;
V = 2πr/T = 2π*0.06/1.2ms;
= 314 m/s;
a = 4π2r/T2 = 4π2*0.06/(1.2×10-3)2
=1.64×106 m/s2 = 1.67×105 g
37
Example : Moon’s centripetal acceleration
The Moon’s nearly circular orbit about the Earth has
a radius of about 384,000 km and a period T of 27.3
days. Determine the acceleration of the Moon toward
the Earth.
T = 27.3*86400 s = 2.36×106s ;
a = 4π2r/T2 = 4π2*3.84×108/(2.36×106)2
=2.72×10-3 m/s2
38
Example : A low-altitude satellite
Estimate the period of a low-altitude reconnaissance
satellite. Ignore the effects of air resistance.
r = RE=6.4×106m; ar = g = 9.8m/s2 = V2 /r ;
V2 = gRE = 6.27×107 (m/s)2
V = 7.9×103m/s
T = 2πRE / V = 5.08 ×103 s = 84.8min
39
Example : A particle moving in a circular path
For a particle moving at constant speed in a circular path, the particle’s position is
given by the components: x = A cos(t); y = A sin(t). Suppose their values are A =
2.0 m and ω = 0.50π rad/s. (a) Show that position’s components describes a circular
path. (b) Find the particle’s velocity. (c) Find the particle’s acceleration. (d) Find the
particle’s position in component form at an initial time ti = 0 and at a final time tf = 5.0
s. (e)What is the particle’s displacement over this 5-s interval?
(a) x2 = A2 cos2 (t); y2 = A2 sin2 (t); x2 + y2 = A2
(b) vx= -A sin (t); vy= A cos (t);
(c) ax= -2A cos (t); ay= -2A sin (t);
(d) ti = 0;
xi= 2.0 m ; yi= 0.0 m;
tf = 2.5; xf= 0.0 m ; yf= 2.0 m;
(e)Δr = (xf - xi) i + (yf - yi) j = -2.0 i + 2.0 j
40
Nonuniform Circular Motion
The tangential acceleration causes the change in the speed
of the particle. The radial acceleration comes from a change
in the direction of the velocity vector. The motion would be
under the influence of both tangential and centripetal
accelerations.
41
Nonuniform Circular Motion
If an object is moving in a circular path but at varying
speeds, it must have a tangential component to its
acceleration as well as the radial one.
ar = V2 /r;
at = dV/dt
a = a r + at
a = – V /r r + dV/dt θ
2
a  ar2  at2
42
Different Measurements
The man is walking on the moving beltway. The woman on the beltway
sees the man walking at his normal walking speed.
The stationary woman sees the man walking at a much higher speed. The
combination of the speed of the beltway and the walking.
The difference is due to the relative velocity of their frames of reference.
Observer A measures point P at +5 m from the origin
Observer B measures point P at +10 m from the origin
The difference is due to the different frames of
reference being used.
43
Inertial Reference Frames
A reference frame in which Newton’s first law is valid is called an
inertial reference frame. In an inertial reference frame, a body
subject to no net force will either stay at rest or move at constant
velocity.
Two observers moving relative to each other generally do not agree on
the outcome of an experiment.
However, the observations seen by each are related to one another.
44
Relative Velocity
Reference frame SA is stationary
Reference frame SB is moving to the right relative to SA at
VBA

This also means that SA moves at –VAB relative to SB
Define time t = 0 as that time when the origins coincide
45
Relative Velocity
Each velocity is labeled first with the object, and second with the
reference frame in which it has this velocity.
rPA = rPB + rBA ;
V = dr/dt
VPA = VPB + VBA ;
VAB = -VBA
The positions as seen from the two reference frames are related through
the velocity: rPA= rPB + rBA
The derivative of the position equation will give the velocity equation

VPA= VPB + VBA

VPA is the velocity of the particle P measured by observer A

VPB is the velocity of the particle P measured by observer B
These are called the Galilean transformation equations.
46
Acceleration
Calculating the acceleration gives



dVPA dVPB dVBA


dt
dt
dt
Since vBA is constant, then dvBA/dt=0
Therefore, aPA = aPB
47
Relative Velocity
Here, vWS is the velocity of the
water in the shore frame, vBS is the
velocity of the boat in the shore
frame, and vBW is the velocity of
the boat in the water frame.
The relationship between the three
velocities is:
vBS = vBW + vWS
48
Example : A Boat Crossing a River-1
A motor boat can travel at 10m/s relative to the water. It starts at one
bank of a river that is 100m wide and flows eastward at 5m/s. If the
boat is pointed directly across, find: (a) its velocity relative to the bank;
(b) how far down stream it travels.
(A) VBG = VBR + VRG =(5 i + 10 j ) m/s
VBG = (102+52)½ = 11.2m/s
tanθ = 5/10 = 0.5 ;
θ = 26.50 E of N
(B) t = 100/10 = 10s
D = 5m/s*t = 50m
49
Example : A Boat Crossing a River-2
A motor boat can travel at 10m/s relative to the water. It starts at one
bank of a river that is 100m wide and flows eastward at 5m/s. (a) At
what angle should the boat be headed if it is to travel directly north
across the river, and what is the speed of the boat relative to the Earth?
(b)How long does this take to cross river?
(a) VBG = VBR + VRG
VBG = (102-52)½ = 8.7m/s
sinθ = 5/10 = 0.5
θ = 300 West of North
(b) t = 100/8.7 = 11.5s
50
Example : A Boat Crossing a River-3
The captain of a ferry wants to travel directly across a river
that flows due east with a speed of 1.07 m/s. He starts from
the south bank of the river and wants to reach the north bank
by traveling straight across the river. The boat has a speed of
6.34 m/s with respect to the water. What direction (in
degrees) should the captain steer the boat?
Vbg = Vb + Vr
sinθ = 1.07/6.34 = 0.17
θ = 9.710 West of North
51
Example : An aircraft 1
Airplane moves with a speed of 160 m/s in direction NE. The
wind is blowing with 32.0 m/s in direction W. What is the
velocity vector – speed and direction – of the airplane relative
to the ground?
VPx = Vcos450 =160*(2)–½ = 113m/s
VPy = Vsin450 =160*(2)–½ = 113m/s
VPG = VP +VW = [(113-32) i + 113 j] m/s
= (81 i + 113 j) m/s
VPG = (812+1132)½ = 139 m/s
θ = tan-1(VPGy/ VPGx) = tan-1(113/81)= 54.40 (N of E)
52
Example : An aircraft 2
An airplane flies at an airspeed of 809 km/h at an angle α = 20°
(exactly) south of east. The wind blows at 85.0 km/h toward the
northeast with respect to the ground. What are the plane’s groundspeed
and direction?
VPx = 809 km/h cos20° = 760.2 km/h
VPy = -809 km/h sin20° = -276.7 km/h
VW
θ =45˚
ϕ
α =20˚
VP
VPG
VWx = 85 km/h cos45° = 60.1 km/h
VWy = 85 km/h sin45° = 60.1 km/h
VPG = 820.3 i - 216.6 j (km/h)
VPG = 848.4 km/h;
ϕ = tan-1(VPGy/ VPGx) = tan-1(-0.411) = -14.8°; Direction 14.8° S of E
53
Example : An aircraft 3
The pilot of an aircraft has to get to a point 320km due north in 1 h.
Ground control reports that there is a crosswind of 80km/h toward 370 S
of W. What is the required heading of the plane? What is the speed?
VPA = VPG + VGA = VPG – VAG
Set VPA = P; VPG = R; VAG =W
Px=Rx–Wx=0–(-80 cos370)=64 km/h
Py=Ry–Wy=320–(-80 sin370)=368 km/h
P = 64i + 368j km/h V=| P | =373.5km/h
α = tan-1(Px/ Py) = tan-1(0.174) = 9.90 E of N
54
Example : A Hunting Bow
A deer’s target is 25 m away and moves with 3 m/s from left
to right. A hunting bow shoots an arrow with initial velocity
90 m/s, horizontal direction, exactly aimed at deer’s heart.
Target. Where does the arrow hit?
t = 25m/(90 m/s) = 0.28 s
Y = ½ ·9.8·0.282 m = 0.38 m
X = vt = 3 m/s·0.28 s = 0.84 m
Hit the left rear of the deer's heart by 0.84 m and 0.38 m
below
55
Physics
Chapter 4
The Laws of Motion
授課老師施坤龍
0
Classes of Forces
Contact
forces
involve
physical contact between
two objects
Field forces act through
empty space

No physical contact is
required
1
Fundamental Forces
Gravitational force

Between objects
Fg = G m1m2/r2
Electromagnetic forces

Between electric charges
Fe = ke q1q2/r2
Nuclear force

The force that binds the nucleons to form the nucleus of
an atom
Weak forces

Arise in certain radioactive decay processes
Note: These are all field forces.
2
Force
A force is a push or pull. An object at rest needs a force to get it
moving; a moving object needs a force to change its velocity. Force
is a vector, having both magnitude and direction. The magnitude of a
force can be measured using a spring scale. A Short Catalog of
Forces: Gravitational force, Spring force, Tension force, Normal
force, Friction force, Drag force (空氣阻力), and Thrust (火箭的衝
力).
Forces are what cause any change in the velocity of an object.

Newton’s definition

A force is that which causes an acceleration
3
Tension
We say that a long narrow object that is being pulled taut by
opposing forces is under tension. In order to use Newton’s
laws of motion to analyze the forces and motions of the
objects that are attached to the ends of strings or rods, we
need to understand more about the phenomenon of tension.
4
Normal force
The normal force is not always equal to the
gravitational force of the object.
For example, in this case
∑ Fy = n – Fg – F =0
And n = Fg + F
Fg may also be less than n
5
Force
A spring can be used to calibrate the magnitude of a
force. Forces are vectors, so you must use the rules
for vector addition to find the net force acting on an
object. Doubling the force causes double the reading
on the spring. When both forces are applied, the
reading is three times the initial reading.
The forces are applied perpendicularly to each other.
The resultant (or net) force is the hypotenuse.
Forces are vectors, so you must use the rules for vector
addition to find the net force acting on an object.
6
Newton’s First Law
If an object does not interact with other objects, it is possible to identify
a reference frame in which the object has zero acceleration

This is also called the law of inertia

It defines a special set of reference frames called inertial frames,

We call this an inertial frame of reference
Any reference frame that moves with constant velocity relative to an
inertial frame is itself an inertial frame
A reference frame that moves with constant velocity relative to the
distant stars is the best approximation of an inertial frame

We can consider the Earth to be such an inertial frame although it
has a small centripetal acceleration associated with its motion
7
Newton’s First Law
In the absence of external forces, when viewed from an
inertial reference frame, an object at rest remains at rest and
an object in motion continues in motion with a constant
velocity

Newton’s First Law describes what happens in the
absence of a force

Also tells us that when no force acts on an object, the
acceleration of the object is zero
8
Inertia and Mass
The tendency of an object to resist any attempt to change its velocity is
called inertia. Mass is that property of an object that specifies how
much resistance an object exhibits to changes in its velocity.
Masses can be defined in terms of the accelerations produced by a
given force acting on them:

m1/m2 = a2/a1

The magnitude of the acceleration acting on an object is inversely
proportional to its mass.
Mass is an inherent property of an object, and it is independent of the
object’s surroundings and also independent of the method used to
measure it.
Mass is a scalar quantity.
The SI unit of mass is kg.
9
Mass vs. Weight
Mass and weight are two different quantities.
Weight is not an inherent property of the object. Weight is equal to the
magnitude of the gravitational force exerted on the object.

Weight will vary with location.
Because it is dependent on g, the weight varies with location.

g, and therefore the weight, is less at higher altitudes.

This can be extended to other planets, but the value of g varies from
planet to planet, so the object’s weight will vary from planet to planet.
Example:

wearth = 196 N; wmoon ~ 32.7 N

mearth = 20 kg; mmoon = 20 kg
10
Gravitational Mass vs. Inertial Mass
In Newton’s Laws, the mass is the inertial mass and measures the
resistance to a change in the object’s motion.
In the gravitational force, the mass is determining the gravitational
attraction between the object and the Earth.
Experiments show that gravitational mass and inertial mass have the
same value.
Measuring Gravitational Mass
11
Measuring Mass
We define the Inertial Mass of a system as the
constant of proportionality between acceleration and
the force that causes it.
12
Conceptual Example: Newton’s first law
A school bus comes to a sudden stop, and all of the
backpacks on the floor start to slide forward. What force
causes them to do that?
No force; the backpacks continue moving until stopped by
friction or collision.
13
Conceptual Example : The hockey puck
A hockey puck is sliding at constant velocity across a flat
horizontal ice surface that is assumed to be frictionless.
Which of these sketches is the correct free-body diagram for
this puck? What would your answer be if the puck slowed?
(b)
(c)
14
Newton’s Second Law
The acceleration of an object is directly proportional
to the net force acting on it and inversely proportional
to its mass

Force is the cause of change in motion, as
measured by the acceleration
Algebraically,

a

F
m


  F  ma
15
More About Newton’s Second Law
∑F is the net force

This is the vector sum of all the forces acting on
the object
Newton’s Second Law can be expressed in terms of
components:

∑Fx = max

∑Fy = may

∑Fz = maz
16
Mass and Force
The unit of force in the SI system is the Newton (N).
Note that the pound is a unit of force, not of mass, and can
therefore be transfered to Newtons but not to kilograms.
Unit of Mass and Force
System
Mass
Force
MKS
Kg
Newton(N)= kg·m/s2
cgs
g
dyne(g·cm/s2)
British
Slug
Pound (lb)
Conversion: 1N= 105dyne, 1N= 0.22lb。
17
Measuring Forces
Definition of the standard force unit: One Newton of
force is defined to be the force necessary to impart an
acceleration of 1 m/s2 to the international standard
kilogram.
Measuring force
Force proportion to voltage
18
Conceptual Example : The advantage of a pulley
A mover is trying to lift a piano (slowly) up to a second-story
apartment. He is using a rope looped over two pulleys as
shown. What force must he exert on the rope to slowly lift the
piano’s 2000-N weight?
FT + FT =W = 2000N
FT = 2000N/2 = 1000N
19
Example: Icebergs and Oceans
The density of ice is 0.917 g/cm3, and the density of seawater is 1.024
g/cm3. Only 10.4% of the volume of an iceberg is above the water
surface, whereas 89.6% are below. If the volume of a particular iceberg
above water is 4164.5 m3, what is the magnitude of the net force that
the ocean exerts on this iceberg?
Wice = mice g = ρice Vice g
0.104 Vice = 4164.5 m3
Vice = 4164.5 m3/ 0.104 = 40043.3 m3
ρice= 0.917 g/cm3 = 917 kg/m3
Wice = ρice Vice g =917*40043.3*9.8= 3.6 ×108 N
20
Example : Two railcar
Two railcar, A and B, with masses mA=1.2×104kg and mB=8×103kg can
roll freely on a horizontal track; see figure. A locomotive of mass 105kg
exerts a force F0 on A that produces an acceleration of 2m/s2. Find (a)
F0; (b) the force exerts on A by B.
Railcar A
ΣFx = F0–FAB = mA a = 1.2×104*2 = 2.4×104 N
Railcar B
ΣFx = FBA= mB a = 8×103 *2 = 1.6×104 N =FAB
(a) F0 = FAB + mA a = 2.4×104 + 1.6×104 = 4.0×104 N
(b) FAB = 1.6×104 N
21
Example : Force to stop a car
What average net force is required to bring a 1500-kg car to
rest from a speed of 100 km/h within a distance of 55 m?
Vi = 100km/h = 27.78 m/s
Vf2= Vi2 +2as =27.782 + 2*a*55
a = -7m/s2
F = ma = 1500*7.0 = 10500 N
22
Example：Newton’s Second Law-1
A 1200kg car is stalled on an icy patch of road. Two ropes attached to it
are used to exert forces F1 =800N at 350 N of E and F2=600N at 250 S
of E. What is the acceleration of the car?
ΣF = F1 + F2 = ma
ΣFx = F1cosθ1 + F2cosθ2 = max
ΣFy = F1sinθ1 - F2sinθ2 = may
ax =(800*0.819+600*0.906)/1200=1.00m/s2
ay = (800*0.574 - 600*0.423)/1200=0.17m/s2
a = 1.00i + 0.17j m/s2
a = 1.014 m/s2
θ = tan-1(0.17) = 9.650
23
Example：Newton’s Second Law-2
An electron of mass 9.1×10-31kg has an initial velocity V0 = 106 i m/s. It
enters a region in which it experiences a force F = 8×10-17 j N for a
period 10-8s. What is its velocity as it emerges from the region?
V0 = 106 i m/s
Vy = V0y + (Fy/m)*t = 0 + 8×10-17/9.1×10-31*10-8
= 8.8×105 m/s
V = 106 i + 8.8×105 j m/s
V=[1012+(8.8×105)2]½ =1.33×106m/s
tanθ = Vy / Vx = 0.88
θ = 41.30
24
Example: Still Rings
Gymnast of mass 55 kg hangs from still rings. (a) What is the
tension in each rope? (b) What changes if the rings do not
hang straight down, but at an angle θ relative to the ceiling?
(a) ∑Fy = T+T – mg = 0
T = 0.5mg = 0.5*55*9.8 = 270 N
(b) ∑Fx = T cosθ – T cosθ = 0

∑Fy = T sinθ +T sinθ – mg = 0
T = 0.5mg/sinθ = 0.5mg cscθ
=270 cscθ N



y
x
25
Newton’s Third Law
Newton’s third law: Whenever one object exerts a force on a second
object, the second exerts an equal force in the opposite direction on
the first.
Forces always occur in pairs. A single isolated force cannot exist.
The action force is equal in magnitude to the reaction force and
opposite in direction

One of the forces is the action force, the other is the reaction force

The action and reaction forces must act on different objects and be
of the same type
26
Newton’s Third Law
If two objects interact, the force F12 exerted by object 1 on
object 2 is equal in magnitude and opposite in direction to the
force F21 exerted by object 2 on object 1

F12 = - F21

Note on notation: F12 is the
force exerted by 1 on 2
27
Newton’s Third Law
Rocket propulsion can also be explained using Newton’s
third law: hot gases from combustion spew out of the tail of
the rocket at high speeds. The reaction force is what propels
the rocket.
Note that the rocket does
not need anything to
“push” against.
28
Newton’s Third Law
29
Action-Reaction Examples
The normal force (table on monitor) is the
reaction of the force the monitor exerts on the
table

Normal means perpendicular, in this case
The action (Earth on monitor) force is equal in
magnitude and opposite in direction to the
reaction force (the monitor exerts on the Earth)
The normal force and the force of gravity are
the forces that act on the monitor
30
Conceptual Example : Third law clarification.
Michelangelo’s assistant has been assigned the task of moving a block
of marble using a sled. He says to his boss, “When I exert a forward
force on the sled, the sled exerts an equal and opposite force backward.
So how can I ever start it moving? No matter how hard I pull, the
backward reaction force always equals my forward force, so the net
force must be zero. I’ll never be able to move this load.” Is he correct?
Because the acting force and the reaction force exist in
different objects, the net force of the resultant force on the
sled is not zero, so the sled will move.
31
Conceptual Example : What exerts the force to move a car
What exerts the force to move a car ？
The engine makes the wheels go around. But if the tires are
on slick ice or deep mud, they just spin. Friction is needed.
On firm ground, the tires push backward against the ground
because of friction. By Newton’s third law, the ground pushes
on the tires in the opposite direction, accelerating the car
forward.
32
Example : Horse and cart system
The horse knowledgeable in the ways of the world, protests
that the harder he pulls forward, the harder the cart will pull
backward. Hence, there is no point in wasting his effort.
Explain why the cart does move forward.
FHC & FCH: Force between horse and cart
FHR & FRH:Force between horse and ground
FCR: Friction between the car and the ground
FHR – FCR = (mH + mC) a
FHR is greater than FCR , then the cart accelerates forward
33
Objects in Equilibrium
Particle in equilibrium Σ F = 0

If a particle maintains a constant velocity (including a
value of zero), the forces on the particle balance and
Newton’s Second Law becomes.
If the acceleration of an object that can be modeled as a
particle is zero, the object is said to be in equilibrium
Mathematically, the net force acting on the object is zero
Σ F = 0 since a =0
Σ Fx = 0 , Σ Fy = 0. and Σ Fz = 0
34
Objects in Equilibrium
A lamp is suspended from a chain of negligible mass.
The forces acting on the lamp are:

the downward force of gravity

the upward tension in the chain
Applying equilibrium gives
ΣFy= 0  T – Fg =0  T = Fg
35
Example: Accelerometer
A small mass m hangs from a thin string and can swing like a pendulum.
You attach it above the window of your car as shown. What angle does
the string make (a) when the car accelerates at a constant a = 1.20 m/s2,
and (b) when the car moves at constant velocity, v = 90 km/h?
ΣF = FT + m g = ma
ΣFx = FT sinθ = m a
ΣFy = FT cosθ– m g = 0
(a) θ = tan-1(a/g) = tan-1(0.122) = 7o
(b) θ = tan-1(a/g) = 0o
36
Objects in Equilibrium 1
As the figure, three strings are tied together. They are on top of a
circular table and meet exactly in the center of the table. Each of the
strings hangs over the edge of the table, and a weight is supported from
it as shown. The masses m1 = 3.9 kg and m2 = 5.2 kg are known. The
angle α = 74.2 between strings 1 and 2 is also known. Find mass m3 ?
ΣF = F1 + F2 + F3 =0
ΣFx = m1g +m2g cosα + F3x =0
ΣFy = m2g sinα + F3y =0
F3x = - m1g - m2g cosα = -3.9*9.8 - 5.2*9.8 cos74.2
= -52.095 N
F3y = - m2g sinα = - 5.2*9.8 sin74.2 = -49.034 N
|F3|= 71.54 N = m3g; m3 = 7.3 kg
37
Example：Objects in Equilibrium 2
A picture frame of weight 20N is suspended by two ropes as
shown in Figure. Find the tensions in the ropes given that θ1
= 300 and θ2 = 450.
ΣF = T1 + T2 + W =0
ΣFx =T1cos300 -T2cos450 =0
0.866 T1 – 0.707T2= 0;
ΣFy =T1sin300 +T2sin450-20 = 0
0.5T1+0.707T2 – 20 = 0
T1 = 14.7N
T2 = 17.9N
38
Example : A Traffic Light at Rest
A traffic light weighing 122 N hangs from a cable tied to two other
cables fastened to a support as Figure. The upper cables make angles of
37.0 and 53.0 with the horizontal. These upper cables are not as
strong as the vertical cable and will break if the tension in them exceeds
100 N. Does the traffic light remain hanging in this situation, or will
one of the cables break?
ΣF = T1 + T2 + W =0
ΣFx =T2cos530-T1cos370 = 0.6T2 – 0.8T1= 0;
T1sin370 +T2sin530-122 = 0
0.6T1+0.8T2 – 122 = 0
T1 = 73.2N
T2 = 97.6N； T1 & T2 < 100 N
39
Example : Pulling a box.
We applied a horizontal force T to pull a box with
mass m. Find (a) the acceleration of the box(b) the
normal force of the box？
ΣF = m a
ΣFx = T = m a ；
ΣFy = n – m g = 0
(a) a = T / m
(b) n = m g
40
Example : Pulling the mystery box
A friend pulls the 10.0-kg box by the attached cord along the
smooth surface. The magnitude of the force exerted is = 40.0
N, and it is exerted at a 30.0°. Calculate (a) the acceleration
of the box, and (b) the magnitude of the upward force FN
exerted by the table on the box.
ΣF = m a
ΣFx = 40 cos30o= 10 a ；
ΣFy = FN + 40 sin30o – 98 = 0
(a) a = 34.64/10 = 3.46 m/s2
(b) FN = 98 – 20 = 78 N
41
Example : An Accelerating Hockey Puck
A hockey puck having a mass of 0.30 kg slides on the frictionless,
horizontal surface of an ice rink. Two hockey sticks strike the puck
simultaneously, exerting the forces on the puck shown in Figure. The
force F1 has a magnitude of 5.0 N, and force F2 has a magnitude of 8.0
N. Determine both the magnitude and the direction of the puck’s
acceleration.
ΣF = ma
ΣFx =8.0 cos60o+ 5.0 cos20o = 8.7N
ΣFy = 8.0 sin60o– 5.0 sin20o = 5.2 N
ax = 8.7/0.3 = 29 m/s2
ay = 5.2/0.3 = 17.3 m/s2
a = (292+17.32)½ = 33.8 m/s2
θ = tan-1(17.3/29) = 31o (positive x-axis)
42
Example : Two boxes connected by a cord
Two boxes, A and B, are connected by a lightweight cord and are
resting on a smooth table. The boxes have masses of 12.0 kg and 10.0
kg. A horizontal force of 40.0 N is applied to the 10.0-kg box. Find (a)
the acceleration of each box, and (b) the tension in the cord connecting
the boxes.
ΣF = (mA+ mB) a
ΣFAx = 40 – T = 10 a ；
ΣFAy = nA – 98 = 0
ΣFBx = T = 12 a ；
ΣFBy = nB – 117.6 = 0
(a) a = F /(mA+mB) = 1.82 m/s2
(b) T = mB a = 21.82 N
43
Example : One Block Pushes Another
Two blocks of masses m1 and m2, with m1 > m2, are placed in contact
with each other on a frictionless, horizontal surface, as Figure. A
constant horizontal force F is applied to m1 as shown. Find (a) the the
acceleration of the system. (b) the contact force between the two blocks.
(c) |P21| = |P12|？
ΣF = (m1+ m2) a
ΣF1x = F – P21 = m1 a ； ΣF1y = n1 – m1g = 0
ΣF2x = P12 = m2 a ；
F = (m1+ m2) a
ΣF2y = n2 – m2g = 0
(x-axis)
(a) a = F /(m1+m2)
(b) P12 = m2 a = Fm2/(m1+m2)
(c) P21 = F–m1 a = F–m1F/(m1+m2) = Fm2/(m1+m2) = P12
44
Example : How Much Do You Weigh in an Elevator?
You have most likely been in an elevator that accelerates
upward as it moves toward a higher floor. In this case, you
feel heavier. Are yon heavier, why?
ΣFy= T – Fg = ma ； Fg = mg
W=T=Fg+ma= m(g+a)
a >0； W = mg(1+a/g) > mg
45
Example : Weighing a Fish in an Elevator
A person weighs a fish of mass m on a spring scale attached
to the ceiling of an elevator as Figure. Find (a) that if the
elevator accelerates either upward or downward? (b) the scale
readings for a 40.0-N fish if the elevator moves with an
acceleration ay = ±2.00 m/s2?
ΣFy= T – Fg = ma ； Fg = mg
(a) W=T=Fg+ma= m(g+a) = mg(1+a/g)
(b) W= m(g+a) = mg(1+a/g)
ay=2 m/s2; W= 40(1+2/9.8) =48.2N
ay=-2 m/s2; W= 40(1-2/9.8) =31.8N
46
Example：A skier on the Incline
A skier of mass 60kg slides down an icy slope which is inclined at 200
to the horizontal. Find her acceleration and the force exerted on her by
the slope. Use the following coordinate systems: (a) the x axis is
horizontal; (b) the x axis points down along the incline.
ΣF = N + W = ma ; W = mg
(a) ΣFx = Nsinθ + 0 = ma cosθ
ΣFy = Ncosθ - mg = -ma sinθ
(b) ΣFx = 0 + mg sinθ = ma
ΣFy = N – mg cosθ = 0
N = mg cosθ = 550N
a = g sinθ = 9.8 sin(200) = 3.3 m/s2
47
Example : Box slides down an incline
A box of mass m is placed on a smooth incline that makes an angle θ
with the horizontal. (a) Determine the normal force on the box. (b)
Determine the box’s acceleration. (c) Evaluate for a mass m = 10 kg and
an incline of θ = 30°?
ΣF = N + W = ma
ΣFx = 0 + mg sinθ = ma
ΣFy = N – mg cosθ = 0
(a) a = g sinθ
(b) N = mg cosθ
(c) a = g sinθ = 9.8 sin300 = 4.9 m/s2
N = mg cosθ = 84.9 N
48
Example : The Runaway Car
A car of mass m is on an icy driveway inclined at an angle 
as Figure. Find (a) the acceleration of the car ? (b) Suppose
the distance of the car moving is d. How long does it take,
and what is the car’s speed?
ΣF = N + W = ma
ΣFx = 0 + mg sinθ = ma
ΣF y = N – mg cosθ = 0
(a) a = g sinθ
(b) d = ½ at2 ；
t = (2d/gsinθ)½
vf2 = vi2 + 2ad
vf = (2gd sinθ) ½
49
Example：A sled on the Incline
A sled of mass 8kg is on a frictionless slope inclined at 350 to the
horizontal. It is pulled by a rope whose tension is 40N and which makes
an angle of 200 with the slope, as figure. Find the acceleration of the
sled and the normal force due to the incline？
ΣF = T + N + W = ma
ΣFx =Tcosα – Wsinθ = ma
ΣFy = Tsinα + N – Wcosθ = 0
W = 78.4N,
T = 40N
θ = 350, α = 200
a = -0.92 m/s2;
N = 50.5N
50
Example : The Atwood Machine
When two objects of unequal mass are hung vertically over a
frictionless pulley. Determine the magnitude of the
acceleration of the two objects and the tension in the
lightweight cord.
ΣF1y = T – m1 g = m1 a
(1)
ΣF2y = m2 g – T = m2 a
(2)
(a) a = (m2-m1) g /(m1+m2)
(b) T = m1 ( g + a )
= 2m1m2 g /(m1+m2)
51
Example : Elevator and counterweight
A system of two objects suspended over a pulley by a flexible cable is
sometimes referred to as an Atwood’s machine. Here, let the mass of
the counterweight be 1000 kg. Assume the mass of the empty elevator
is 850 kg, and its mass when carrying four passengers is 1150 kg. For
the latter case calculate (a) the acceleration of the elevator and (b) the
tension in the cable.
ΣF1y = m1g – T = 11270 – T =1150 a (1)
ΣF2y = T – m2g = T – 9800 = 1000 a (2)
(a) a = (m1-m2) g /(m1+m2) =1470/2150
=0.684m/s2
(b) T = m2 ( g + a ) = 10484 N
52
Example : Acceleration of Two Object system
A ball of mass m1 and a block of mass m2 are attached by a lightweight
cord that passes over a frictionless pulley of negligible mass as Figure.
The block lies on a frictionless incline of angle θ. Find the magnitude of
the acceleration of the two objects and the tension in the cord.
F1x =T –m1g = m1 a
(1)
F2x = m2g sinθ –T = m2 a
(2)
F2y = N2 – m2g cosθ = 0
(3)
(1) & (2) m2g sinθ– m1g = (m1+m2)a
(a) a = [m2sinθ– m1]g/(m1+m2)
(b) T = m1g +m1 a = m1m2(sinθ +1)g/(m1+m2)
53
Example：Three blocks on the Wedge
Three blocks with masses m1=3kg, m2=2kg, and m3=1kg are connected
by two ropes, one of which hangs over a light, frictionless pulley as
shown in figure. Find the acceleration of the blocks and the tension on
the ropes. (θ=300) ?
Block 1 ΣFx = W1–T1 = m1a
Block 2 ΣFx = T1–T2-W2sinθ = m2a
Block 3 ΣFx = T2–W3sinθ = m3a
W1–(W2+W3)sinθ = (m1+m2+m3)a
a = [m1–(m2+m3)sinθ]g/(m1+m2+m3)
= [3–(2+1)sin300]*9.8/(3+2+1)
= 2.45 m/s2
T1 = W1–m1a = 3*9.8 – 3*2.45 = 22.05N
T2 = W3sinθ +m3a = 9.8sin300 + 2.45 = 7.35N
54
Example：Two blocks on the Wedge
Two blocks with masses m1=7kg and m2=3kg are connected with a rope
and moves on a wedge as figure. Given that θ1=370, θ2=530, the
acceleration of the blocks and the tension in rope.
F1x = m1g sinθ1 – T = m1 a
(1)
F1y = N1 – m1g cosθ1 = 0
(2)
F2x = T – m2g sinθ2 = m2 a
(3)
F2y = N2 – m2g cosθ2 = 0
(4)
(1) & (3) m1g sinθ1– m2g sinθ2 = (m1+m2)a
(2) a = [m1sinθ1– m2sinθ2]g/(m1+m2) = [7*0.6-3*0.8]*9.8/(7+3)
= 1.77m/s2
T = m2g sinθ2 +m2 a = 23.52+5.31 = 28.83N
55
Example：Wedge and Block system
A block of mass m is placed on a wedge of mass M that is on a
horizontal table. All surfaces are frictionless. Find the acceleration of
the wedge.
Wedge ΣFx = N1sinθ = MA
Block
(1)
ΣFx = -N1sinθ = m (A - a` cosθ)
(2)
ΣFy = N1cosθ – mg = -m a` sinθ
(3)
(1)&(2) (m + M)A = ma` cosθ
(4)
(2)&(3) mg sinθ = ma` - mA cosθ
(5)
(3)&(4) A = mg sinθ cosθ / (M + m sin2θ)
56
Physics
Chapter 5
Applications of Newton’s Laws
授課老師施坤龍
0
Forces of Friction
When an object is in motion on a surface or through a viscous medium,
there will be a resistance to the motion. This is due to the interactions
between the object and its environment. This resistance is called the force
of friction. Friction is always present when two solid surfaces slide along
each other. The microscopic details are not yet fully understood.
The coefficient of friction depends on the surfaces in contact.
The force of static friction is generally greater than the force of kinetic
friction.
1
Forces of Friction
The direction of the frictional force is opposite the direction of motion
and parallel to the surfaces in contact.
The coefficients of friction are nearly independent of the area of contact.
Static friction acts to keep the object from moving.
As long as the object is not moving, ƒs = F
If F increases, so does ƒs .
If F decreases, so does ƒs .
ƒs  µ s n

The equality holds when the surfaces are on the verge of slipping.
The force of kinetic friction acts when the object is in motion.
Although µ k can vary with speed, we shall neglect any such variations.
ƒk = µ k n
2
Forces of Friction
Friction is proportional to the normal force.

ƒs ≤ µ s n and ƒk= µ k n

µ is the coefficient of friction

These equations relate the magnitudes of the forces; they are not
vector equations.

For static friction, the equals sign is valid only at impeding motion,
the surfaces are on the verge of slipping.

Use the inequality for static friction if the surfaces are not on the
verge of slipping.
Friction is a force, so it simply is included in the ΣF in
Newton’s Laws. The rules of friction allow you to determine
the direction and magnitude of the force of friction.
3
Some Coefficients of Friction
μs
μk
Rubber on concrete
1.0
0.8
Steel on steel
0.74
0.57
Aluminum on steel
0.61
0.47
Glass on glass
0.94
0.4
Copper on steel
0.53
0.36
Wood on wood
0.25-0.5
0.2
Waxed wood on wet snow
0.14
0.1
Waxed wood on dry snow
-
0.04
Metal on metal (lubricated)
0.15
0.06
Teflon on Teflon
0.04
0.04
Ice on ice
0.1
0.03
Synovial joints in humans
0.01
0.003
4
Conceptual Example : To push or to pull a sled
Your little sister wants a ride on her sled. If you are on flat
ground, will you exert less force if you push her or pull her?
Assume the same angle θ in each case.
pull the sled
5
Example : Friction: static and kinetic
Our 10.0-kg mystery box rests on a horizontal floor. The coefficient of
static friction is 0.40 and the coefficient of kinetic friction is 0.30.
Determine the force of friction acting on the box if a horizontal external
applied force is exerted on it of magnitude:
(a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N.
(a) 0 N;
(b) 10N;
(c) 20N;
(d) 38N; (e) 29.4N
6
Example： Friction of a block
A 5kg block is on a horizontal surface for which μs=0.2 and
μk=0.1. It is pulled by a 10N force directed at 530 above the
horizontal, find the of friction on the block given that: (a)
rest; (b) it is moving.
N = mg – F sin530 = 5*9.8-10*0.8 = 41.0N
Fs(max) = Nμs = 8.2N
(a) Since Fx=10 cos530=6.0N<Fs
then Ffr = 6.0N
(b) Fk = Nμk = 4.1N = Ffr
7
Conceptual Example : A box against a wall
You can hold a box against a rough wall and prevent it from
slipping down by pressing hard horizontally. How does the
application of a horizontal force keep an object from moving
vertically?
F – FN = 0
Ffr – mg = 0
FN µ s ≧ Ffr = mg
F = FN ≧ mg/µ s
8
Friction Example
The block is sliding down the plane, so friction acts up the
plane
This setup can be used to experimentally determine the
coefficient of friction
µ s = tanθc

For µ s, use the angle where the block just slips

For µ k, use the angle where the block slides down at a constant
speed
9
Example： Determine the coefficient of friction
Suppose a block is placed on a rough surface inclined relative to the
horizontal Figure. The incline angle is increased until the block starts to
move. Show that you can obtain s by measuring the critical angle c at
which this slipping just occurs. If we change the incline angle to θk, such
that the block moves down the incline with constant speed . Find µ k？
(a) Fx = mg sinθc –fs = 0 ; Fy = n –mg cosθc = 0
fs = n µ s = mg cosθc µ s;
mg sinθc – mg cosθc µ s = 0
µ s = tanθc
(b) Fx = mg sinθk – fk = 0 ; Fy = n –mg cosθk = 0
fk = n µ k = mg cosθk µ k;
mg sinθk – mg cosθk µ k = 0
µ k = tanθk
10
Example : The skier
This skier is descending a 30° slope, at constant
speed. What can you say about the coefficient of
kinetic friction?
Fx = mg sinθk – fk = 0 ;
Fy = n –mg cosθk = 0
fk = n µ k = mg cosθk µ k;
mg sinθk – mg cosθk µ k = 0
µ k = tanθk = 0.577
11
Example : The Sliding Hockey Puck
A hockey puck on a frozen pond is given an initial speed of
20.0 m/s. If the puck always remains on the ice and slides 115
m before coming to rest, determine the coefficient of kinetic
friction between the puck and ice.
Fx =–fk = m a ;
Fy = n – mg = 0
fk = n µ k = mg µ k;
a = – g µk
vf2 = vi2 + 2ad = vi2 – 2 g µ k d = 0
µ k = vi2/(2gd ) = 400/2254 =0.177
12
Example：The coefficient of friction between two blocks
A block of mass m1=2kg is placed on a block of mass m2=4kg. The
lower block is on frictionless horizontal surface and is subject to a force
F0=30N as figure. Find the minimum value of the coefficient of friction
such that m1 does not slide on m2.
Block 1
f = m1a1
Block 2
F0 – f = m2a2
a1 = a 2 = a
F0 = (m1 + m2) a;
a = 30/(4+2) = 5m/s2
fs(max) = N1µ s = m1gµ s > m1a
µ s > a /g = 5/9.8 = 0.51
13
Example : Pulling against friction
A 10.0-kg box is pulled along a horizontal surface by a force
of 40.0 N applied at a 30.0° angle above horizontal. The
coefficient of kinetic friction is 0.30. Calculate the
acceleration.
Fx =40 cos30 °–fk = 10 a ;
Fy = FN +40 sin30 °– 10g = 0
fk = FN µ k = 78*0.3=23.4;
a = (34.64-23.4)/10 = 1.12 m/s2
14
Example : The Skidding Truck
The driver of an empty speeding truck slams on the brakes and skids to a
stop through a distance d. (a) If the moving mass is doubled, what would
be its skidding distance if it starts from the same initial speed? (b) If the
initial speed of the truck is halved, what would be the skidding distance?
(a) Fx =–fk = (2m) a ; Fy = n –(2mg) = 0
fk = n µ k = (2mg)µ k;
a = – g µk
vf2 = vi2 + 2ad1 = vi2 – 2 g µ k d1 = 0
d1 = vi2/(2gµ k ) = d
(b) Fx =–fk = m a ; Fy = n –mg = 0
fk = n µ k = mg µ k; a = – g µ k
vf2 = (vi/2)2 + 2ad2 = vi2/4 – 2 g µ k d2 = 0
d2 = vi2/(8gµ k ) = d / 4
15
Example : Two boxes and a pulley
Two boxes with mass mA=5kg and mB=2kg are connected by a cord
running over a pulley. The coefficient of kinetic friction between box A
and the table is 0.20. We wish to find the acceleration, a, of the system,
which will have the same magnitude for both boxes assuming the cord
doesn’t stretch. As box B moves down, box A moves to the right.
FB =mBg–T = mB a =19.6–T = 2 a
FAx =T – fk = mA a = 5 a
FAy = NA – mAg = NA – 49 =0
fk = NA µ k = 9.8N;
(4)
19.6 – 9.8= 7 a
(a) a = 1.4 m/s2
(b)T = 19.6 – 2*1.4 = 16.8 N
16
Example：The system of ball and block
A block of mass m2 on a rough, horizontal surface is connected to a ball
of mass m1 by a lightweight cord over a lightweight, frictionless pulley
as Figure. A force of magnitude Fat an angle  with the horizontal is
applied to the block as shown, and the block slides to tile right. The
coefficient of kinetic friction between the block and surface is k
Determine the magnitude of the acceleration of tile two objects.
F1 =T –m1g = m1 a
(1)
F2x = Fcosθ –T –fk = m2 a
(2)
F2y = N + Fsinθ– m2g = 0
(3)
fk = N µ k =(m2g–Fsinθ) µ k ;
(4)
Fcosθ –m1g –(m2g–Fsinθ) µ k = (m1+m2)a
(a) a =[F(cosθ+µ ksinθ)–(m1+m2µ k) g] /(m1+m2)
(b)T=m1g+m1a=[F(cosθ +µ ksinθ)+m2(1–µ k)g]m1/(m1+m2)
17
Example : A ramp, a pulley, and two boxes
Box A, of mass 10.0 kg, rests on a inclined at 37°. It is connected by a cord, which
passes over a pulley, to a second box B, which hangs freely. (a) If the coefficient of
static friction is 0.40, determine what range for mass B will keep the system at rest. (b)
If the coefficient of kinetic friction is 0.30, and mB = 10.0 kg, determine the
acceleration of the system.
NA=mAg cos37°=78.4;
fs =NA µ s=31.36N;
fk=NA µ k = 23.52N;
(a) FB =mBg –T =0
FAx =T –mAg sin37°– fs =T –58.8– 31.36=0
mB ≦(58.8+31.36)/9.8 kg = 9.2 kg
FB =T–mBg =0
FAx =mAg sin37°– T –fs =58.8– T –31.36=0
mB ≧ (58.8–31.36)/9.8 kg = 2.8 kg
(b) mBg –T =98–T =10 a; T –58.8– 23.52=10 a
a = 0.784 m/s2
18
Example Two blocks and pulley system
Two blocks with masses m1=7kg and m2=4kg are connected with a rope
and move on two surfaces of a right angle wedge as shown in figure.
Given that θ1=370, θ2=530, μs=0.2, and μk=0.1, find the acceleration of
the blocks and the tension of rope.
m1: N1 =7*9.8*0.8=54.88N; f1s=54.88*0.2=10.98N; f1k=5.49N;
m2: N2 =4*9.8*0.6=23.52N; f2s=23.52*0.2=4.70N; f2k=2.35N;
m1:m1g sinθ1 =7*9. 8*0.6=41.16N;
m2:m2g sinθ2 =4*9. 8*0.8=31.36N;
f1k+f2k = 7.84N<41.16-31.36=9.8N<f1s+f2s=15.7N
If the system was originally stationary, the system remains rest; if the
system has begun to move, then a =( 9.8-7.84)/(7+4)= 0.18m/s2
T= m2g sinθ2 +m2g cosθ2µ k +m2 a = 39.2*0.86+ 0.72 =34.43N
19
Uniform Circular Motion
The force causing the centripetal acceleration
is sometimes called the centripetal force. It is
a force acting in the role of a force that causes
a circular motion.
This force is always directed radially inward, toward the
center. A force, FC , is directed toward the center of the circle.
This force is associated with an acceleration, ac .
The particle must be subject to a centripetal force of
magnitude:
v2
 Fc  mac  m r
20
Uniform Circular Motion
The centripetal force: ΣF = FC = m aC = － mV2/r r
The centripetal acceleration: aC = 2πV/T = 4π2r/T2 = V2/r
Centrifugal force (center-fleeing): This is a misconception.
For example: When twirling a stone in a circular motion by a string, the
outward force on the hand and inward pull on the stone are equal and
opposite forces but they act on different bodies.
The centripetal force is directed toward the center of the circle. It causes
a change in the direction of the velocity vector.
If the force vanishes, the object would move in
a straight-line path tangent to the circle.
21
Example： A conical pendulum - 1
A small object of mass m is suspended from a string of length
L. The object revolves in a horizontal circle of radius r with
constant speed v, as in Figure. Find (a) the centripetal
acceleration ? (b) the speed v? (c) the period ?
ΣFx=T sinθ = m a = m V2/r
ΣFy=T cosθ – mg = 0
(a) a= g tanθ
(b) V = (g r tanθ)½ =(gL sinθ tanθ)½ =(gL sin2θ /cosθ)½
(c) T = 2π r/V = 2π[r/(g tanθ)]½ = 2π[Lcosθ /g]½
22
Example：A conical pendulum - 2
Estimate the force a person must exert on a string attached to
a 0.150-kg ball to make the ball revolve in a horizontal circle
of radius 0.600 m. The ball makes 2.00 revolutions per
second. Ignore the string’s mass.
ac= 4π2r/T2 =94.75
Fc =m ac = 14.21N
mg = 1.47N
T = (14.212+1.472)½
=14.29N
23
Example： A conical pendulum - 3
A puck of mass 0.500 kg is attached to the end of a cord 1.50
m long. The puck moves in a horizontal circle as shown in
Figure. If the cord can withstand a maximum tension of 50.0
N, what is the maximum speed at which the puck can move
before the cord breaks?
ΣFx=T sinθ= m a = m V2/r
ΣFy=T cosθ – 4.9 = 0
50 sinθ = 49.76 ; sinθ = 0.995
r = L sinθ
V = (TLsin2θ /m)½ = (50*1.5*0.9952/0.5)½ = 12.2 m/s
24
Horizontal (Flat) Curve
The force of static friction supplies the centripetal force. The
maximum speed at which the car can negotiate the curve is V
= (μsgR)½ 。
ΣFx= fs = m ac = m V2/R
ΣFy= N– mg = 0
m V2/R = fs ≦ Nμs = mgμs
V≦ (μsgR)½
25
Example：The maximum safe speed - 1
A 1 500-kg car moving on a flat, horizontal road negotiates a
curve as Figure. If the radius of the curve is 35.0 m and the
coefficient of static friction between the tires and dry
pavement is 0.523, find the maximum speed the car can have
and still make the turn successfully.
ΣFx= fs = m ac = m V2/R
ΣFy=N– mg = 0
m V2/R = fs ≦ N μs = mg μs
V≦ (μsgR)½ = (0.523*9.8*35)½ = 13.4 m/s
26
Example：The maximum safe speed - 2
A 1000-kg car rounds a curve on a flat road of radius 50 m at
a speed of 15 m/s (54 km/h). Will the car follow the curve, or
will it skid? Assume: (a) the pavement is dry and the
coefficient of static friction is μs = 0.60; (b) the pavement is
icy and μs = 0.25.
ΣFx= fs = m V2/R
ΣFy= N – mg = 0
m V2/R = fs ≦ Nμs = mg μs
(a) V≦(μs gR)½ =(0.6*9.8*50)½ =17.1m/s 安全
(b) V≦(μs gR)½ =(0.25*9.8*50)½ =11.1m/s 不安全
27
Example： The minimum coefficient of friction - 1
A small coin is placed at the rim of a turntable of radius 15cm
which rotates at 30rev/min. Find the minimum coefficient of
friction for the coin to stay on.
ΣFx= fs = m V2/R； ΣFy=N– mg = 0
fs = Nμs = mgμs = ma
frequency = 30 rev/min = 0.5rev/s
T = 1/frequency = 1/0.5 = 2s
a = 4π2R/T2 = 1.4775m/s2
μs = a/g = 0.15
28
Example： The minimum coefficient of friction - 2
In a carnival ride called the rotor, people stand on a ledge
inside a large cylinder that rotate about a vertical axis. When
it reaches a high enough rotational speed. The ledge drops
away. Find the minimum coefficient of friction for the people
not to slide down. Take R=2m and T=2s.
ΣFx= N = m V2/R；
ΣFy= fs – mg = 0
fs = N μs = μs mV2/R = mg
V = 2 πR/T = 2π
μs = gR/V2 = 0.5
29
Highway Curves: banked
Banking the curve can help keep cars from skidding.
In fact, for every banked curve, there is one speed at
which the entire centripetal force is supplied by the
horizontal component of the normal force, and no
friction is required. This occurs when:
ΣFx= N sinθ = m V2/R
ΣFy= N cosθ – mg = 0
tanθ = V2/(gR)
V = (gR tanθ)½
30
Example：The angle for an highway - 1
The road is icy highway curves are banked. Suppose that a
car of mass m moves at a constant speed v of 20 m/s around a
curve, now banked, whose radius R is 190 m. What bank
angle makes reliance on friction unnecessary?
ΣFx= N sinθ = m V2/R
ΣFy= N cosθ – mg = 0
tanθ =V2/(gR) = 400/(9.8*190)=0.2148
θ = tan-1[V2/(gR)] = tan-1(0.2148) =12o
31
Example：The angle for an highway - 2
A civil engineer wishes to redesign the curved roadway ,
especially if the road is icy or wet. Suppose the designated
speed for the ramp is to be 13.4 m/s (30.0 mi/h) and the
radius of the curve is 35.0 m. At what angle should the curve
be banked?
ΣFx= N sin θ = m V2/R
ΣFy= N cos θ – mg = 0
tanθ = V2/(gR) = 180/(9.8*35)=0.52
θ = tan-1[V2/(gR)] = tan-1(0.52) =27.6o
32
Example：The angle for an highway - 3
A civil engineer wishes to redesign the curved roadway ,
especially if the road is icy or wet. What is this angle for an
expressway off-ramp curve of radius 50 m at a design speed
of 50 km/h?
ΣFx= N sin θ = m V2/R
ΣFy= N cos θ – mg = 0
tanθ = v2/(gR) = 193/(9.8*50)=0.394
θ = tan-1[V2/(gR)] = tan-1(0.394) =22o
33
A car turns round in a sloping friction road
A car of mass m rounds a circle R that is banked at θ to the horizontal.
The coefficient of static friction for this road is μs. Find the maximum
safe speed V at which the car travel.
ΣFx = N sinθ + f cosθ = m V2/R
ΣFy = N cosθ - f sinθ – mg = 0
f ≦ Nμs ; N ≧ mg/(cosθ - μssinθ)
V2 ≦ gR(sinθ + μscosθ)/(cosθ - μssinθ)
V ≦ [gR(sinθ + μscosθ)/(cosθ - μssinθ)]½
= [gR(μ +tanθ)/(1-μ tanθ
s
s
)]½
34
Example：The maximum safe speed - 4
A car of mass 1000kg rounds a circle 100m that is banked at
370 to the horizontal. The road is slippery, so the coefficient
of static friction is only 0.1. Find the maximum safe speed at
which the car travel.
ΣFx = N sinθ + f cosθ = m V2/R
ΣFy = N cosθ - f sinθ – mg = 0
f ≦ Nμs ; N ≧ mg/(cosθ - μssinθ)
V2 ≦ gR(sinθ + μscosθ)/(cosθ - μssinθ)
= 9.8*100*(0.6+0.08)/(0.8-0.06)
=925.5
V ≦ 30.4m/s = 109.5 km/h
35
Example：The maximum safe speed - 5
If the coefficient of friction between the racetrack surface and the tires
of the racecar is s = 0.620 and the radius of the turn is R = 110.0 m,
what is the maximum speed with which the driver can take this curve
banked at 21.1˚?
ΣFx = N sinθ + f cosθ = m V2/R
ΣFy = N cosθ - f sinθ – mg = 0
f ≦ Nμs ; N ≧ mg/(cosθ - μssinθ)
V2 ≦ gR(sinθ + μscosθ)/(cosθ - μssinθ)
= 9.8*110*(0.36+0.62*0.933)/(0.933-0.62*0.36)
= 1425
V ≦ 37.8m/s = 136 km/h
36
Ferris Wheel
The normal and gravitational forces act
in opposite direction at the top and
bottom of the path.
At the bottom of the wheel, the upward
force experienced by the object is
greater than its weight
v2
nbot  mg (1 
)
gR
At the top of the path, the force exerted
on the object is less than its weight
2
v
ntop  mg (1 
)
gR
37
Example : Riding the Ferris Wheel - 1
A child of mass m rides on a Ferris wheel as shown in Figure. The child
moves in a vertical circle of radius 10.0 m at a constant speed of 3.00
m/s. Determine (a) the force exerted by the seat on the child at the
bottom of the ride. (b) the force exerted by the seat on the child at the
top of the ride.
(a) ΣF = N – mg= mV2/R
Nbot = mg + mV2/R = mg[1+V2/(gR)]
= mg[1+9/98] = 1.09mg
(b) ΣF = mg N = mV2/R
Ntop = mg – mV2/R = mg[1–V2/(gR)]
= mg[1-9/98] = 0.91mg
38
Example : Riding the Ferris Wheel - 2
A 50kg woman is on a ferris wheel of radius 9m that rotates
in a vertical circle of 6 rev/min. What is the magnitude of the
apparent weight when she is halfway up?
ΣFx = N2 = ma = mV2/R
ΣFy = N1- mg = 0
T=10s; V = 2πr/T = 5.65 m/s
N1 =50*9.8 = 490N
N2 =mV2/r = 178N
N = (N12+ N22)½ = 521N
39
Non-Uniform Circular Motion
The acceleration and force have
tangential components.
Fr
produces the centripetal
acceleration
Ft produces
acceleration
the
tangential
The total force is



 F   Fr   F t
40
Vertical Circle with Non-Uniform Speed
Σ F = Fr + Ft
Σ Fx = Ft = mg sinθ = m dVθ/dt
Σ Fy = Fr =Tθ – mg cosθ = mVθ2/R
at = dVθ /dt = g sinθ
Tθ = mVθ2/R + mg cosθ
41
Vertical Circle with Non-Uniform Speed
The gravitational force exerts a tangential
force on the object.
The tension at any point can be found.
v2
T  mg (
 cos  )
gR
The tension at the bottom is a maximum.
Tbot  mg (
2
v bot
gR
 1)
The tension at the top is a minimum.
Ttop  mg (
If Ttop = 0, then
2
v top
gR
 1)
vtop  gR
42
Example : Keep Your Eye on the Ball
A small sphere of mass m is attached to the end of a cord of
length R and set into motion in a vertical circle as illustrated
in Figure. Determine the tangential acceleration of the sphere
and the tension in the cord at any instant when the speed of
the sphere is v and the cord makes an angle  with the
vertical.
Σ Fx = Ft = mg sinθ = m dVθ/dt
Σ Fy = Fr =Tθ–mg cosθ = mVθ2/R
(a) at = dVθ/dt = g sinθ
(b) Tθ = mVθ2/R + mg cosθ
Tbot = mVbot2/R + mg; Ttop = mVtop2/R – mg
43
Example : Revolving stone
A stone attached to the end of a rope moves in a vertical
circle solely under the influence of gravity and the tension in
the rope. Find the tension in the rope at the following points:
(a) at the lowest point; (b) at the highest point; (c) when the
rope is at angle θ to vertical.
Σ Fx = Ft = mg sinθ = m dVθ/dt
Σ Fy = Fr =Tθ–mg cosθ = mVθ2/R
(a) Tbot = mVbot2/R + mg
(b) Ttop = mVtop2/R – mg
(c) Tθ = mVθ2/R + mg cosθ
44
Motion with Resistive Forces
Motion can be through a medium.
The medium exerts a resistive force, R, on an object moving
through the medium. The magnitude of R depends on the
medium. The direction of
R is opposite the direction of
motion of the object relative to the medium.
We will discuss only two cases

R is proportional to v


Good approximation for slow motions or small objects
R is proportional to v2

Good approximation for large objects
45
R Proportional to v
The resistive force can be expressed as
R=-bv
b depends on the property of the medium, and on the
shape and dimensions of the object. The negative
sign indicates R is in the opposite direction to v
Analyzing the motion results in
dv
ma  mg  bv  m
dt
dv
b
a
g v
dt
m
46
R Proportional to v
Σ F = mg –bv = m a = m dv/dt
a = dv/dt = g –(b/m)v
Set v = vsp(special sol.)+ vg(general sol.)
The general solution Vsp is a solution at t∞,
dvsp/dt = 0 ; vsp = mg/b = VT (Terminate speed)
v = vsp+ vg ; substitute to a = dv/dt = g –(b/m)v
dvg/dt = –(b/m)vg ; dvg/vg = –(b/m) dt ; ln(vg) = C–(b/m)t
vg = ec e－bt/m = A e－bt/m ; v = vsp+ vg = mg/b + A e－bt/m
v(t)=mg/b(1－e－bt/m)=VT(1－e－t/τ)；τ = m/b is time constant.
47
R Proportional to v2
For objects moving at high speeds through air, the resistive
force is approximately proportional to the square of the speed
R = 1/2 DρAv2

D is a dimensionless empirical quantity that is called the
drag coefficient

ρ is the density of air

A is the cross-sectional area of the object

v is the speed of the object
48
R Proportional to v2, example
Analysis of an object falling through air accounting for air
resistance
1
F  mg  DAv 2  ma
2
dv
DA 2
a
g
v
dt
2m
R
v
mg
The terminal speed will occur when the acceleration goes to
zero. Solving the equation gives
VT 
2mg
DA
R
vT
mg
49
Drag Force and Terminal Speed
Object
Terminal Speed (m/s) 95% Distance (m)
Shot (form Shot put)
145
2500
Sky diver(typical)
60
430
Baseball
42
210
Tennis ball
31
115
Basketball
20
47
Ping-Pong ball
9
10
Raindrop
(radius=1.5mm)
7
6
Parachutist(typical)
5
3
50
Some Terminal Speeds
Mass(kg)
Cross Section
Area (m2)
VT (m/s)
75
0.70
60
Baseball (radius 3.7 cm)
0.145
4.2*10-3
43
Golf ball (radius 2.1 cm)
0.046
1.4*10-3
44
Hailstone (radius 0.50cm)
4.8*10-4
1.3*10-3
14
Raindrop (radius 0.20cm)
3.4*10-5
7.9*10-3
9.0
Object
Skydiver
51
Example : Falling Coffee Filters
Table presents typical terminal
Terminal speed and resistance R
Filter’s No. VT (m/s)
speed data from a real experiment
using these coffee filters as they
fall through the air.
At the terminal speed, the upward
resistive
force
balances
downward gravitational force.
R = mg
the
Resistance
R(N)
1
1.01
0.0161
2
1.40
0.0322
3
1.63
0.0483
4
2.00
0.0644
5
2.25
0.0805
6
2.40
0.0966
7
2.57
0.1127
8
2.80
0.1288
9
3.05
0.1449
10
3.22
0.1610
52
Coffee Filters, Graphical Analysis
Graph of resistive force and terminal speed does produce a
straight line.
The resistive force is proportional to the object’s speed.
53
Coffee Filters, Graphical Analysis
Graph of resistive force and terminal speed does not produce a
straight line.
The resistive force is proportional to the square of the object’s
speed.
54
Resistive Force on a Baseball – Example
The object is moving horizontally through the air.
The resistive force causes the ball to slow down.
Gravity causes its trajectory to curve downward.
The ball can be modeled as a particle under a net force.

Consider one instant of time, so not concerned about the
acceleration
Analyze to find D and R
2mg
1
D 2
and R  DAV 2
VT A
2
55
EXAMPLE : A Sphere Falling in Oil
A small sphere of mass 2.00 g is released from rest in a large
vessel filled with oil. The sphere approaches a terminal speed
of 5.00 cm/s. Determine (a) the time constant . (b) the time
interval required for the sphere to reach 90.0% of its terminal
speed.
VT = mg/b = 0.002*9.8/b = 0.05m/s; b =0.392 Ns/m
(a) τ = m/b = 0.002/0.392 = 5.1*10-3 s
(b) V = VT(1－e－t/τ) =0.9VT ; e－t/τ = 0.1
t = -τ ln(0.1) = 2.3*5.1*10-3 =11.7*10-3 s =11.7ms
56
Example : Resistive Force Exerted on a Baseball
A pitcher hurls a 0.145-kg baseball past a batter at 40.2 m/s
(90 mi/h). Find the resistive force acting on the ball at this
speed.
Terminal Speed’s Table 2, the baseball
VT=43m/s, A=0.0042 m2
D=2mg/(ρAVT2)= 2*0.145*9.8/(1.2*0.0042*432 )
= 0.305
R = ½ DρAv2 = 0.5*0.305*1.2*0.0042*40.22
= 1.2 N
57
Example : Sky Diving
Area of parachute = 42.7 m2, drag coefficient 0.63, density of
air = 1.15 kg/m3, mass of the man + equipment = 76.4 kg. (a)
What is the terminal velocity? (b) What is the magnitude of
the drag force at terminal velocity?
(a) VT =[2mg/(DρA)]½
=[2*76.4*9.8/(0.63*1.15*42.7)]½
= 6.95 m/s
(b) ∑F = Fd - Fg = Fd - mg = 0
Fd = mg = 76.4*9.8 = 750 N
58
Physics
Chapter 6
Work and Energy
授課老師施坤龍
0
Energy Review
Kinetic Energy

Associated with movement of members of a system
Potential Energy

Determined by the configuration of the system

Gravitational and Elastic Potential Energies have been
studied
Internal Energy

Related to the temperature of the system
1
Work Done by a Constant Force
The work done by a constant force is defined as the distance
moved multiplied by the component of the force in the
direction of displacement.
 
W  F  r  Fr cos 

The work done by a force on a moving object is zero when
the force applied is perpendicular to the displacement.

A force does no work on the object if the force does not
move through a displacement.
2
Work Done by a Constant Force
The sign of the work depends on the direction of the force
relative to the displacement.


Work is positive when projection of F onto ∆r is in the same
direction as the displacement.
Work is negative when the projection is in the opposite direction.
Work is a scalar quantity.
The unit of work is a joule (J) in the SI system

1 joule = 1 newton . 1 meter = kg ∙ m²/ s²
3
Work Example
As long as this person does not lift or lower the
bag of groceries, he is doing no work on it.
The force he exerts has no component in the
direction of motion.
The normal force n and the gravitational force
mg do no work on the object.

cosθ = cos 90° = 0
The force F is the only force that does work on
the object.
4
Work Is An Energy Transfer
This is important for a system approach to solving a problem.
If the work is done on a system and it is positive, energy is
transferred to the system.
If the work done on the system is negative, energy is
transferred from the system.
If a system interacts with its environment, this interaction can
be described as a transfer of energy across the system
boundary.

This will result in a change in the amount of energy stored in the
system.
5
Conceptual Example: Does the Earth do work on the Moon
The Moon revolves around the Earth in a nearly circular orbit, with
approximately constant tangential speed, kept there by the gravitational
force exerted by the Earth. Does gravity do (a) positive work, (b)
negative work, or (c) no work at all on the Moon?
(c) no work, the force is always perpendicular to the
displacement
6
Example : Mr. Clean
A man cleaning the floor pulls a vacuum cleaner with a force
of magnitude F= 50.0 N at an angle of 30.0° with the
horizontal as shown in Figure. The vacuum cleaner is
displaced 3.00 m to the right. Calculate the work done by the
50.0-N force on the vacuum cleaner.
W =F d cosθ = 50.0 * 3.00* cos30.0°= 130.0 J
7
Example : Using the dot product
The force shown has magnitude FP = 20 N and makes an
angle of 30° to the ground. Calculate the work done by this
force, using the dot product, when the wagon is dragged 100
m along the ground.
W = 20*100*cos300 = 1732J
8
Example : Work of a particle
When a particle moves on the x-y plane and receives a
constant force F = (5.0 i + 2.0 j) N, and its displacement ∆r is
(2.0 i +3.0 j ) m. What is the work done by this force F?
W = F·∆r = 5*2+ 2 * 3 = 16 N．m
9
Example : Work on a backpack
(a) Determine the work a hiker must do on a
15.0-kg backpack to carry it up a hill of
height h = 10.0 m, as shown. Determine also
(b) the work done by gravity on the
backpack, and (c) the net work done on the
backpack. For simplicity, assume the motion
is smooth and at constant velocity.
(a) WH = 15.0*9.8*10 = 1470J
(b) Wg = 15.0*9.8*(-10) = -1470J
(c)Wnet = WH + Wg = 0 J
10
Example : Does the Ramp Lessen the Work Required?
A man wishes to load a refrigerator onto a truck using a ramp
at angle  as shown in Figure. He claims that less work
would be required to load the truck if the length L of the
ramp were increased. Is his claim valid?
WF = mg L sinθ
Wg = -mgL sinθ
Wnet = WF + Wg = 0
11
Example : Work done on a crate
A person pulls a 50-kg crate 40 m along a horizontal floor by
a constant force FP = 100 N, which acts at a 37° angle as
shown. The floor is smooth and exerts no friction force.
Determine (a) the work done by each force acting on the crate,
and (b) the net work done on the crate.
(a) FN =mg - FP cos370 = 50*9.8 - 60 = 430 N
WF = 100*40*cos370 = 3200J
WN = 430*40*cos900 = 0J
Wg = 50*9.8*40*cos900 = 0J
(b) Wnet = WF + WN + Wg = 3200 J
12
Example：Work of A skier
A skier of mass m=40kg is given a displacement of 20m along a slope
inclined at θ=150 to the horizontal. The tension in the towrope is
T=250N and acts at an angle α=300 to the incline. Given μk=0.1, find
the work done by each force and the net work on the skier.
N = mg cosθ – T sinα = 379N -125N = 254N
WT = T s cos300 = 4330J
Wf = -f s = -Nμks = -508J
WN = N s cos900 = 0
Wg = -mgs sin150 = -2030J
Wnet = WT + Wf + WN + Wg = 1790J
13
Example： A simple pendulum
A horizontal force F very slowly lifts the bob of a simple
pendulum from a vertical position to a point at which the
string makes an angle θ0 to the vertical. The magnitude of the
force is varied so that the bob is essentially in equilibrium at
all times. What is the work done by the force on the bob?
ΣFx = F – T sinθ = 0
ΣFy = T cosθ – mg = 0
F = mg tanθ
dW=F‧ds = Fxdx = mg tanθ dx = mgdy
W=
y0
∫0
mg dy = mgy0 = mgL(1-cosθ0)
14
Work Done by a Varying Force
To use W = F Δx, the force must be constant, so the
equation cannot be used to calculate the work done
by a varying force. Assume that during a very small
displacement,
Δx,
F
is
constant.
For
that
displacement, W ~ FΔx . For all of the intervals,
xf
W   Fx x
xi
Let the size of the small displacements approach
xf
zero. Since W  lim  Fx x  
x 0
Therefore,
W 

xf
xi
xi
xf
xi
Fx dx
Fx dx
The work done is equal to the area under the curve
between xi and xf.
15
Work Done By Multiple Forces
If more than one force acts on a system and the system can be
modeled as a particle, the total work done on the system is
the work done by the net force.
W
 Wnet 
 (  F )dx   
xf
i
xi
xf
xi
Fi dx
In the general case of a net force whose magnitude and
direction may vary.
W
 Wext 
 ( F
xf
xi
ext
)dx
The subscript “ext” indicates the work is done by an external
agent on the system.
16
Example : Calculating Total Work Done from a Graph
A force acting on a particle varies with x as shown in Figure.
Calculate the work done by the force on the particle as it
moves from x = 0 to x = 6.0 m.
(a)W0-4 =5.0 N*4.0 m= 20 J
(b)W4-6 =½ 5.0 N*2.0 m=5 J
(c)W0-6 = 20 J +5 J = 25 J
17
Example: Force as a function of x
A robot arm that controls the position of a video camera in an
automated surveillance system is manipulated by a motor that
exerts a force on the arm. The force is given by F(x) = F0
[1+x2/(6x02)] ; where F0 = 2.0 N, x0 = 0.0070 m, and x is the
position of the end of the arm. If the arm moves from x1 =
0.010 m to x2 = 0.050 m, how much work did the motor do?
0.05
W = ∫ 0.01 F(x) dx
= 2.0*[(0.05-0.01) +(0.053-0.013) /(8.82*10-4)]
= 0.361 J
18
Hooke’s Law
The force exerted by a spring is given by: Fs = - k x
Where k is the spring constant.
The vector form of Hooke’s Law is
Fs = Fx i = - kx i

When x is positive, F is negative.

When x is 0, F is 0

When x is negative, F is positive.
19
Work Done by a Varying Force
Calculate the work as the block moves from xi to xf :
Fs = - k x
Ws =
xf
∫ xi
Fs (x) dx=
xf
∫x i
-kx dx
= -½ k ( xf2-xi2 )
= ½ k ( xi2-xf2 )
Fp = -Fs = k x
Work done by the external force Fp :
Wp =
xf
∫ xi Fp(x)
dx =
xf
∫xi
kx dx
= ½ k ( xf2-xi2 )
20
Example : Measuring k for a Spring
A common technique used to measure the force constant of a spring is
demonstrated by the setup in Figure. The spring is hung vertically, and
an object of mass m is attached to its lower end. Under the action of the
“1oad” mg. the spring stretches a distance d from its equilibrium
position. (a) If a spring is stretched 2.0 cm by a suspended object
having a mass of 0.55 kg, what is the force constant of the spring? (b)
How much work is done by the spring on the object as it stretches
through this distance?
(a) k = mg/d = 0.55*9.8/0.02 = 270 N/m
(b)Ws =½ k ( xi2-xf2 ) =135(0-0.022 )
= -0.054 J
21
Example : Work done on a spring
(a) A person pulls on a spring, stretching it 3.0 cm, which
requires a maximum force of 75 N. How much work does the
person do? (b) If, instead, the person compresses the spring
3.0 cm, how much work does the person do?
k = 75/0.03 = 2500 N/m
(a) Wp = ½ k (xf2-xi2) = 1250[(0.03)2 – 02] = 1.125J
(b) Wp = ½ k (xf2-xi2) = 1250[(-0.03)2 – 02] = 1.125J
22
Example：Work of a spring
A block of mass 100g is attached to the end of a spring whose
spring constant is k =40N/m. The block slides on a horizontal
surface for which μk=0.1. The spring is extended by 5cm and
then released. Find (a) the work done by the spring up to the
point at which it is compressed by 3cm; (b) the net work done
on the block up to this point?
Wsp = ½ k (xi2-xf2) = 20[(0.05)2 – (-0.03)2] = 0.032J
Wf = -fks = -0.1*9.8*0.1*(0.08) = -0.0078J
Wnet = Wsp + Wf = 0.024J
23
Kinetic Energy
Kinetic Energy is the energy of a particle due to its
motion.

K = ½ mv2

K is the kinetic energy

m is the mass of the particle

v is the speed of the particle
A change in kinetic energy is one possible result of
doing work to transfer energy into a system.
24
Kinetic Energy
Calculating the work:
W 
xf
xi
 Fdx  
xf
xi
madx  
xf
xi
dv
m dx
dt
vf
W   mvdv
vi
1 2 1 2
W  mv f  mvi
2
2
25
Kinetic Energy and the Work-Energy Principle
This means that the work done is equal to the change
in the kinetic energy:
Wnet
1 2 1 2
 K  mv f  mvi
2
2
If the net work is positive, the kinetic energy
increases.
If the net work is negative, the kinetic energy
decreases.
26
Example : Work on a car, to increase its kinetic energy
How much net work is required to accelerate a 1000-
kg car from 20 m/s to 30 m/s?
W = ½ mvf2 - ½ mvi2 = 500*302-500*202
= 2.5×105 J
v1 =20m/s
v2 =30m/s
27
Example : A Block Pulled on a Frictionless Surface
A 6.0-kg block initially at rest is pulled to the right
along a frictionless, horizontal surface by a constant
horizontal force of 12 N. Find the block’s speed after
it has moved 3.0 m.
W = Fs= 12*3.0 = 36.0
= ½ mvf2 - ½ mvi2 = 3 vf2
vf = (12)½ = 3.464 m/s
28
Example : Kinetic energy and work done on a baseball
A 145-g baseball is thrown so that it acquires a speed
of 25 m/s. (a) What is its kinetic energy? (b) What
was the net work done on the ball to make it reach
this speed, if it started from rest?
(a) K= ½ mvf2 = ½ *0.145*252 = 45.3 J
(b) W = ½ mvf2 - ½ mvi2 = 45.3 J
29
Example : Work to stop a car
A car traveling 60 km/h can brake to a stop within a distance
d of 20 m. If the car is going twice as fast, 120 km/h, what is
its stopping distance? Assume the maximum braking force is
approximately independent of speed.
v1 = 60km/h = 16.67 m/s; v2 = 120km/h = 33.33 m/s;
W1 = 0- ½ mv12 = F*S1 = F*20
W2 = 0- ½ mv22 = F*S2 ； S2 = 4 S1 = 80m
30
Energy Transfer
Work is a method of energy transfer
Work has the effect of transferring energy between
the system and the environment

If positive work is done on the system, energy is
transferred to the system

Negative work indicates that energy is transferred
from the system to the environment
31
Conservation of Energy
Energy is conserved

This means that energy cannot be created or destroyed

If the total amount of energy in a system changes, it can
only be due to the fact that energy has crossed the
boundary of the system by some method of energy
transfer
The Work-Kinetic Energy theorem is a special case of
Conservation of Energy
32
Problems Involving Kinetic Energy
When kinetic friction is involved in a problem, you
must use a modification of the work-kinetic energy
theorem
∑Wother forces – ƒk d= ∆K

The term ƒk d is the work associated with the frictional
force

Also, ∆Eint = ƒk d when friction is the only force acting
in the system
33
Example： A block on a rough surface
A block of mass 4kg is dragged 2m along
a horizontal
surface by a force F=30N acting at 530 to the horizontal. The
initial speed is 3m/s and μk=1/8. Find (a) the change in
kinetic energy of this block; (b) its final speed?
N = mg – F sinθ = 4*9.8–30*0.8=15.2N
(a) ∆K = Wnet = WF + Wfr
∆K = 30*2*cos530-15.2*(1/8)*2 = 32J
(b) ∆K = ½ mVf2 - ½ mVi2 = 2Vf2–2*9=32
Vf = 5m/s
34
Example : : A Block-Spring System
A block of mass 1.6 kg is attached to a horizontal spring that has a force
constant of 1.0  103 N/m as Figure. The spring is compressed 2.0 cm
and is then released from rest. Calculate (a) the speed of the block as it
through the equilibrium position x = 0 if the surface is frictionless. (b)
the speed of the block as it passes through the equilibrium position if a
constant friction force of 4.0 N retards the block’s motion from the
moment it is released.
Ws=½ kxi2-½ kxf2 = 500(0.02)2 = 0.2J
=½ mvf2-½ mvi2 = 0.8vf2
(a) vf = 0.5 m/s
(b) Wnet = Ws + Wf = 0.2+(-4)*0.02 = 0.12J =0.8 vf2
vf = 0.387 m/s
35
Example：block and Spring on an Incline
A block of mass m=0.2kg is held against, but not attached to
a spring (k=50N/m) which is compressed by 20cm, as figure.
When released, the block slides 50cm up to the rough incline
before come to rest. Find: (a) the force of friction; (b) the
speed of the block just as it leaves the spring.
Wsp + Wg+ Wfr = 0
(a) ½ kA2 -mgd sinθ -ffrd=0
25*0.04-0.2*9.8*0.5*0.6 -0.5*ffr=0
ffr = 0.824N
(b) ½ mV2 =½ kA2 - mgA sinθ -ffrA
0.1V2 = 1 -0.2*9.8*0.2*0.6 -0.824*0.2
V = 2.44 m/s
36
Example : A block Pulled on a Rough Surface
(a) A 6.0-kg block initially at rest is pulled to the right along a
horizontal surface by a constant horizontal force of 12 N. Find the
speed of the block after it has moved 3.0 m if the surfaces in contact
have a coefficient of kinetic friction of 0.15. (b) Suppose the force F is
applied at an angle  as shown in Figure b. At what angle should the
force be applied to achieve the largest possible speed after the block has
moved 3.0m in to the right?
(a) Wnet = F d－ mg μk d = 9.54J = 3 vf2
vf = 1.78 m/s
(b) n=mg－F sinθ ；fk= nμk =(mg－F sinθ )μk
Wnet = Fd cosθ － (mg－F sinθ )μk d =Kf
dKf /dθ =0= -Fd sinθ + F cosθμk d
tanθ=μk ; θ =tan-1(μk); θ =tan-1(0.15) =8.5o
37
Power
The time rate of energy transfer is called power
The average power is given by
W
Pave 
t
The instantaneous power is the limiting value of the average
power as Δt approaches zero
W dW
P  lim

dt
t 0 t
This can also be written as
 


dW
dr
P
F
 F v
dt
dt
38
Power
Power can be related to any type of energy transfer
In general, power can be expressed as
dE
P
dt
dE/dt is the rate at which energy is crossing the boundary of
the system for a given transfer mechanism
The SI unit of power is watt:1watt = 1 joule/s = 1 kg . m2/s3
A unit of power in the US Customary system is horsepower

1 hp =550 ft .lb/s = 746 W
Units of power can also be used to express units of work or
energy. 1 kWh = (1000 W)(3600 s) = 3.6×106 J
39
Example: Stair-climbing power
A 60-kg jogger runs up a long flight of stairs in 4.0 s.
The vertical height of the stairs is 4.5 m. (a) How
much energy did this require? (b) Estimate the
jogger’s power output in watts and horsepower.
(a) E =mgh =60*9.8*4.5 = 2646 J
(b) P =∆E/∆t= 2646/4 = 661.5 watt =0.887 hp
40
Example : Power Delivered by an Elevator Motor
A 1 600-kg elevator is carrying passengers having a combined mass of
200 kg. A constant friction force of 4 000 N retards its motion upward
as in Figure. (a) What is the minimum power delivered by the motor to
lift the elevator at a constant speed of 3.00 m/s? (b) What power must
the motor deliver at the instant the speed of the elevator is v if it is
designed to provide an upward acceleration of 1.00 m/s2?
(a) T = mg +F = 1800*9.8+4000 = 21640N
P = Fv = 21640*3 = 64920 watt
(b) T = mg +F+ma = 23440N
P = Fv = 23440v
P = Fv = 23440*3 = 70320 watt
41
Example: The elevators of Taipei 101 tower
The Taipei 101 tower has the fastest elevators in the world today. The elevator’s ascent
is so fast that a special pressure-control system was installed. After the express
elevator is loaded, it quickly reaches a constant speed of 16.6 m/s. Assume the
elevator and its passengers have a mass of 4800 kg and there is a counterweight that
reduces the effective weight of the elevator by 40%. Also assume there is a constant
5000-N kinetic friction force acting between the elevator and the shaft. Find the power
supplied by the elevator’s motor.
∑Fy = FT – Fg - Fk = 0;
FT = Fg + Fk = 0.6mg + Fk
= 0.6*4800*9.8 +5000 = 3.3 × 104 N
P = Fv = 3.3 × 104 N * 16.6m/s
= 5.5 × 105 W
42
Example: Power to Keep a Car Moving
A car travels at 60.0 mile/h on a level road. The car has a
drag coefficient of 0.33 and a frontal area of 2.2 m2. How
much power does the car need to maintain its speed? Take the
density of air to be 1.29 kg/m3.
v = 60 mile/h = 60*1.609 km/h = 26.8 m/s
Fd = ½ DρAv2 = 0.5*0.33*1.29*2.2*26.82 = 336.8N
P = Fd·v = 336.8*26.8 = 9026 W = 9.03 kW
43
Example : Power needs of a car
Calculate the power required of a 1400-kg car under the following
circumstances: (a) the car climbs a 10° hill (a fairly steep hill) at a
steady 80 km/h; and (b) the car accelerates along a level road from 90 to
110 km/h in 6.0 s to pass another car. Assume that the average retarding
force on the car is FR = 700 N throughout.
(a) v= 80km/h =22.22 m/s;
F = mg sinθ + FR = 3083N
P = Fv = 6.85*104 watt = 91.8 hp
(b) v1= 90km/h =25m/s ; v2=110km/h =30.6m/s
F=1400*(30.6-25)/6+700 = 1996N
P = Fvave = 5.55*104 watt = 74.4 hp
44
Example : Average and Instantaneous Power
A horizontal cable accelerates a suspicious package across a frictionless
horizontal floor. The amount of work that has been done by the cable’s
force on the package is given by W(t)=(0.20 J/s2)t2. (a) What is the
average power due to the cable’s force in the time interval t1=0 s to t2 =
10 s ? (b) What is the instantaneous power P due to the cable’s force at
t3=3.0 s, and is P then increasing or decreasing?
W0=0J； W10= 20 J
(a) Pave = (W10- W0)/∆t = 20/10 = 2.0 watt
(b) P = dW/dt = 0.4 t =1.2 watt ；
dP/dt = 0.4 >0; P increasing
45
Physics
Chapter 7
Potential Energy
授課老師施坤龍
0
Potential Energy
Potential energy is energy determined by the configuration of a system in
which the components of the system interact by forces.

The forces are internal to the system.

Can be associated with only specific types of forces acting between
members of a system
There are many forms of potential energy, including:

Gravitational

Elastic

Electromagnetic

Chemical

Nuclear
One form of energy in a system can be converted into another form.
1
Potential Energy
An object can have potential energy by virtue of its
surroundings.
Familiar examples of potential energy:

A wound-up spring

A stretched elastic band

An object at some height above the ground
2
Gravitational Potential Energy
Assume the book in fig. is allowed to lift.
There is no change in kinetic energy since
the book starts and ends at rest.
Gravitational potential energy is the energy
associated with an object at a given location
above the surface of the Earth.


Wnet  Fapp  r
 ( mgˆj )  [( x f  xi )iˆ  ( y f  yi ) ˆj ]
 mgy f  mgyi
3
Gravitational Potential Energy
The quantity mgy is identified as the gravitational potential
energy, Ug.

Ug = mgy
Units are joules (J)
Is a scalar
Work may change the gravitational potential energy of the
system.

Wext = ∆ug
Potential energy is always associated with a system of two or
more interacting objects.
4
Elastic Potential Energy
The force the spring exerts (on a block) is Fs = - kx
The work done by an external applied force on a spring-block
system is

W = ½ kxf2 – ½ kxi2

The work is equal to the difference between the initial
and final values of an expression related to the
configuration of the system.
Elastic Potential Energy is associated with a spring,
Us = ½ k x2
5
Elastic Potential Energy
This expression is the elastic potential energy:
Us = ½ kx2
The elastic potential energy can be thought of as
the energy stored in the deformed spring.
The stored potential energy can be converted
into kinetic energy.
Observe the effects of different amounts of compression of the spring.
The elastic potential energy stored in a spring is zero whenever the
spring is not deformed (U = 0 when x = 0). The elastic potential energy
is a maximum when the spring has reached its maximum extension or
compression.
The elastic potential energy is always positive.

x2 will always be positive.
6
Example : The Proud Athlete and the Sore Toe
A trophy being shown off by a careless athlete slips from the
athlete’s hands and drops on his toe. Choosing floor level as
the y = 0 point of your coordinate system, estimate the
change in gravitational potential energy of the trophy-Earth
system as the trophy falls. Repeat the calculation, using the
top of the athlete’s head as the origin of coordinates.
(a) U1 = mg y1 =2.0*9.8*0.5 = 9.8 J
(b) U2 = mg y2 =2.0*9.8*0.03 = 0.588 J
(c) ∆U = U2 – U1 = -9.21 J
7
Example : Potential energy changes for a roller coaster
A 1000kg roller-coaster car moves from point 1 to point 2 and then to
point 3. (a)What is the gravitational potential energy at points 2 and 3
relative to point 1? That is, take y=0 at point 1. (b)What is the change in
potential energy when the car goes from point 2 to point 3? (c)Repeat
parts (a) and (b), but take the reference point (y=0) to be at point 3.
(a) U2=mg(y2 - y1)=1000*9.8*10 = 9.8*104 J
U3=mg(y3 - y1)=1000*9.8*(-15)=-1.47*105 J
(b) ∆U = U3 – U2 = - 2.45*105 J
(c) U1=mg(y1 – y3)=1000*9.8*15= 1.47*105 J
U2=mg(y2 – y3)=1000*9.8*(25)=2.45*105 J
∆U = U3 – U2 = – U2 = - 2.45*105 J
8
Example : Potential energy of A Hammer
A construction worker is repairing the roof of a house. His hammer of mass 0.55 kg
slips from his hand and falls through a basketball hoop to the driveway below. The
basketball hoop is 3.0 m above the ground and 5.7 m below the roof where the
hammer left the worker’s hand. Place the origin of an upward-pointing y axis on the
driveway. (a) What is the change in the potential energy of the system? (b) Now use
the basketball hoop as the origin of the y axis and find the change in potential energy.
(a) UD = 0; UR = mgyR = 0.55*9.8*(3.0+5.7) = 46.9 J;
ΔU = UD - UR = -46.9 J
(b) UD = mgyD = 0.55*9.8*(-3.0) = -16.2 J;;
UR = mgyR = 0.55*9.8*5.7 = 30.7 J;
ΔU = UD - UR = -46.9 J
9
Conservative Forces
The work done by a conservative force on a particle moving
between any two points is independent of the path taken by
the particle.
The work done by a conservative force on a particle moving
through any closed path is zero. A closed path is one in which
the beginning and ending points are the same.
Examples of conservative forces: Gravity, and Spring force.
The work done by a conservative force around any closed
path is zero.
WA12 +WB21 =0
10
Nonconservative Forces
A non-conservative force does not satisfy the conditions of
conservative forces.
Non-conservative forces acting in a
system cause a change in the mechanical energy of the
system.
The work done against friction is greater
along the red path than along the blue
path.
Because the work done depends on the
path, friction is a non-conservative force.
11
Conservation of Mechanical Energy
The mechanical energy E is defined as the sum of kinetic
energy K and potential energy U : E = K + U
If only conservative forces are doing work, the total
mechanical energy of a system neither increases nor
decreases in any process. It stays constant—it is conserved.
And its conservation: ∆K + ∆U =0
∆E = E2 - E1 = 0
E2 = K2 + U2 = K1 + U1 = E1
12
Conservation of Mechanical Energy
The mechanical energy of a system is the algebraic sum of
the kinetic and potential energies in the system

Emech = K + Ug + Usp

Ef = Ei
The statement of Conservation of Mechanical Energy for an
isolated system is Kf + Ugf + Uspf = Ki+ Ugi + Uspi

An isolated system is one for which there are no energy
transfers across the boundary
13
Conservation of Mechanical Energy
Conservation of Mechanical Energy
E = K + Ug = ½ mv2 + mgy
14
Conservation of Mechanical Energy
Conservation of Mechanical Energy
E = K + Usp = ½ mv2 + ½ kx2
In general, the Conservation of Mechanical Energy
Kf + Ugf + Uspf = Ki+ Ugi + Uspi
15
Conservation of Mechanical Energy, example
Look at the work done by the book as it falls from some
height to a lower height
Won book = ∆Kbook
Also, W = mgya – mgyb
So, ∆K = - ∆Ug
16
Example – Ball in Free Fall
Determine the speed of the ball at a height y above the ground.

Use energy instead of motion

Apply the principle of Conservation of Mechanical
Energy


Kf + Ugf = Ki + Ugi

Ki = 0, the ball is dropped

Solve for vf
v f  2 g (h  y )
17
Conceptual Example : Speeds on two water slides
Two water slides at a pool are shaped differently, but start at
the same height h. Two riders, Paul and Kathleen, start from
rest at the same time on different slides. (a) Which rider, Paul
or Kathleen, is traveling faster at the bottom? (b) Which rider
makes it to the bottom first? Ignore friction and assume both
slides have the same path length.
(a) WPaul=∆KPaul=W Kathleen =∆K Kathleen
Their final speed are the same.
(b) V Kathleen >VPaul , at the begining
；Kathleen gets to the bottom first 。
18
Example : Ball in Free-Fall
A ball of mass m is dropped from rest at a height h above the
ground as in Figure. Determine (a) the speed of the ball when
it is at a height y above the ground. (b) the speed of the ball at
y if at the instant of release it already has an initial speed vi at
the initial altitude h.
Ef = Kf + Ugf = Ki + Ugi= Ei
(a) Ef = mgy+½ mvf2 = Ei = mgh
vf = [2g(h-y)]½
(b) Ef = mgy+½ mvf2 = Ei= mgh +½ mvi2
vf = [vi2+2g(h-y)]½
19
Example : Falling rock
If the original height of the rock is y1 = h = 3.0 m, calculate
the rock’s speed when it has fallen to 1.0 m above the ground.
Ef = Kf + Ugf = Ki + Ugi= Ei
Ef = mgy+½ mvf2 = Ei = mgh
vf = [2g(h-y)]½
= [2*9.8(3-1)]½ = [39.2]½
= 6.26 m/s
20
Example : Pole vault
Estimate the kinetic energy and the speed required for a 70-
kg pole vaulter to just pass over a bar 5.0 m high. Assume the
vaulter’s center of mass is initially 0.90 m off the ground and
reaches its maximum height at the level of the bar itself .
Ei= Ki + Ugi= Kf + Ugf =Ef
mgy1+½ mv12 = mgh
v1= [2g(h-y)]½ = [2*9.8(5-0.9)]½
= 8.96 m/s
21
Example： A projected ball
A ball is projected from the top of a cliff of height H with an
initial speed V0 at some angle θ above the horizontal. Discuss
its motion in term of (a) the equation of kinematics, and (b)
the conservation of mechanical energy.
Ei= Ki + Ugi= Kf + Ugf =Ef
Ei = ½ mV02 + mgH
Ef = ½ mV2
Ei = Ef
V2 = V02 + 2gH
22
Example：Atwood machine
Two blocks with masses m1=3kg and m2=5kg are connected
by a string that slides over two frictionless pegs. Initially m2
is held 5m off the floor while m1 is on the floor. The system is
then release. At what speed does m2 hit the floor?
Ki + Ui = Kf + Uf
½ (m1+m2)V2 + m1gh = 0 + m2gh
V = [2(m2-m1)gh/(m1+m2)]½
= [2*2*9.8*5/8]½ = 4.95m/s
23
Example：A simple pendulum - 1
This simple pendulum consists of a small bob of mass m suspended by
a cord of length L.The bob is released at t=0, where the cord makes an
angle θ=θ0 to the vertical. (a) Describe the motion of the bob in terms of
kinetic energy and potential energy. Then determine the speed of the
bob (b) as a function of position θ as it swings back and forth, and (c) at
the lowest point of the swing. (d) Find the tension in the cord.
Ei = mgL(1-cosθ0)=½ mV2 + mgL(1-cosθ) = Ef
Σ Fy = T – mg cosθ = mV2/L
(a) mgL(1-cosθ0)=½ mV2 + mgL(1-cosθ)
V =[2gL(cosθ -cosθ0)]½
(b) θ =0; V =[2gL(1-cosθ0)]½
(c) T = mg(3 -2cosθ0)
(d) T = mg(3cosθ -2cosθ0)
24
Example： A simple pendulum - 2
The bob of a simple pendulum of length L=2m has a mass m=2kg and a
speed V=1.2m/s when the string is at 35° to the vertical. Find the
tension in the string at: (a) the lowest point in its swing; (b) the highest
point.
Σ Fy = T – mg cosθ = mV2/L
E = ½ mV2 + mgL(1-cos θ)= 8.5J
(a) E = ½ mVmax2 + 0 = 8.5
Vmax = 2.9m/s
T = mg + mVmax2/L = 19.6 + 8.5 = 28.1N
(b) The highest point V=0; E = 0+ mgL(1-cosθmax)= 8.5J
cosθmax= 0.783;
T = mg cosθmax =15.3N
25
Example: A Grand Entrance
You are designing apparatus to support an actor of mass 65 kg who is to
“fly” down to the stage during the performance of a play. You attach the
actor’s harness to a 130-kg sandbag by means of a lightweight steel
cable running smoothly over two pulleys as in Figure. Let us call the
initial angle that the actor’s cable makes with the vertical . What is the
maximum value  can have before the sandbag lift off the floor?
Ei = magR(1-cosθ) = Ef = ½ mavf2
Vf2 = 2gR(1-cosθ)
T = mag +mavf2/R = mbg =2mag
2(1-cosθ) =1; θ = 60o
26
Example : The tension in the cord
A 0.150-kg ball on the end of a 1.10-m-long cord is swung in a vertical
circle. (a) Determine the minimum speed the ball must have at the top
of its arc so that the ball continues moving in a circle. (b) Calculate the
tension in the cord at the bottom of the arc.
Ttop = mVtop2/R – mg ≥0
(a) Vtop = (gR)½ = (9.8*1.1)½ = 3.28 m/s
(b) E=Ktop+Utop = Kbot+Ubot
= ½ mgR + 2mgR = ½ mVbot2
Vbot = (5gR)½
Tbot = mVbot2/R + mg = 6mg = 8.82 N
27
Example：A Spring and a block
A block of mass m1=0.8kg is attached to a spring whose spring constant
k=20N/m. It slides on a frictionless surface. The spring is extended
12cm and then released. Find: (a) the maximum speed of the block? (b)
its velocity when the spring is compressed by 8cm? (c) at what point the
kinetic and potential energy is equal? (d) at what point is the speed half
its maximum value?
E = K + U = ½ mV2 + ½ kx2 ; E = ½ kA2 = 0.144J;
(a) Ei = ½ kA2; Ef = ½ mVmax2; Vmax = (k/m)½ A = 0.6m/s;
(b) E = ½ mV2 + ½ kx2 = 0.4V2 +10*(-0.08)2 = 0.144 J
V2 = (0.144 -0.064)/0.4 = 0.2; V = 0.45m/s
(c) K = U = E/2 = 0.072 = ½ kx2 ; x = ± 0.085m
(d) V = ½ Vmax = ± 0.3m/s; U = E-K = 0.144 – 0.036 = ½ kx2
x = (0.0108)½ = 0.104m
28
Example: Toy dart gun
A dart of mass 0.100 kg is pressed against the spring of a toy
dart gun. The spring (with spring stiffness constant k = 250
N/m and ignorable mass) is compressed 6.0 cm and released.
If the dart detaches from the spring when the spring reaches
its natural length (x = 0), what speed does the dart acquire?
E = K+Usp = ½ mV2 + ½ kx2
Ei = ½ kxi2 = Ef = ½ mVf2
Ei = 125(0.06)2 =0.45= 0.05Vf2 = Ef
Vf = 3.0 m/s
29
Example : Two kinds of potential energy
A ball of mass m = 2.60 kg, starting from rest, falls a vertical distance h
= 55.0 cm before striking a vertical coiled spring, which it compresses
an amount Y = 15.0 cm. Determine the spring stiffness constant of the
spring. Assume the spring has negligible mass, and ignore air
resistance. Measure all distances from the point where the ball first
touches the uncompressed spring (y = 0 at this point).
E =K+Ug+Usp= ½ mV2+mgy+½ ky2
Ei = mgh= Ef = mgy3+½ ky32
Ei = 2.6*9.8*0.55=2.6*9.8*(-0.15) +½ k(-0.15)2
k = 2*2.6*9.8*(0.55+0.15)/0.152= 1585 N/m
30
Example: The Spring-Loaded Popgun
The launching mechanism of a popgun consists of a trigger-released
spring . The spring is compressed to a position yA = –0.120 m, and the
trigger is fired. The projectile of mass m= 35.0 g rises to a position yc=
20.0 m above the position at which it leaves the spring as position yB
=0. Find (a) the spring constant? (b) the speed of the projectile as it is in
the equilibrium position yB=0?
E = K+Ug+Usp = ½ mV2 + mgy + ½ ky2
(a) EC = mgyC = EA = mgyA+ ½ kyA2
k = 2mg(yC - yA)/yA2
=2*0.035*9.8[20-(-0.12)]/0.122 =958 N/m
(b) EC=mgyC= ½ mVB2+mgyB+½ kyB2=EB
VB = (2gyC)½ = (392)½ = 19.8 m/s
31
Example：Two blocks and spring system
Two blocks with masses m1=2kg and m2=3kg hang on either side of a
pulley. Block m1 is on an incline (θ=30°) and is attached to a spring
whose constant is 40N/m. The system is released from rest with spring
at its natural length. Find: (a) the maximum extension of the spring; (b)
the speed of m2 when the extension is 0.5m.
∆K + ∆Ug + ∆Usp = 0
(a) 0+(-m2gD+m1gDsinθ)+½ kD2=0
D = 2g/k*(m2-m1sinθ) = 0.98m
(b) ½ (m1+m2)V2+(-m2gd+m1gdsinθ)+½ kd2=0
V = [(14.7-4.9-5)/2.5]½ = 1.39m/s
32
Mechanical Energy and Nonconservative Forces
If friction or air resistance is present, mechanical energy of
the system is not conserved
Use energy with non-conservative forces instead

The difference between initial and final energies equals the energy
transformed to or from internal energy by the non-conservative
forces
In general, if friction is acting in a system:

∆Emech = ∆K + ∆U = -ƒkd

∆U is the change in all forms of potential energy

If friction is zero, this equation becomes the same as Conservation
of Mechanical Energy
33
Example 1 : Nonconservative Forces
∆Emech = ∆K + ∆U
∆Emech =(Kf – Ki) + (Uf – Ui)
∆Emech = (Kf + Uf) – (Ki + Ui)
Ki = 0 ;
Ui = mgh
Kf = 1/2 mvf2 ;
Uf = 0
∆Emech = 1/2 mvf2 – mgh = -ƒkd
34
Example 2 : Nonconservative Forces
Without friction, the energy continues to be transformed
between kinetic and elastic potential energies and the total
energy remains the same
If friction is present, the energy decreases

∆Emech = -ƒkd
35
Example： A crate on a conveyor belt
A crate of mass m is dropped onto a conveyor belt that moves at a
constant speed V. The coefficient of kinetic friction is μk. (a) What is
the work done by friction? (b) How far does the crate move before
reaching its final speed? (c) When the crate reaches its final speed, how
far has the belt moved?
(a) Wf = ∆K = ½ mV2
(b) f = Nμk = mgμk
Wf = fd = mgμk d =½ mV2
d = V2/(2gμk)
(c) V = at ; d = ½ at2 = ½ Vt ;
Since the belt’s speed is fixed, for time interval t,
S = Vt = 2d = V2/(gμk);
36
Example：A crate slides down a ramp
A 3.00-kg crate slides down a ramp at a loading dock. The
ramp is 1.00 m in length and is inclined at an angle of 30.0°
as shown in Figure. The crate starts from rest at the top and
experiences a constant friction force of magnitude 5.00 N.
Use energy methods to determine the speed of the crate when
it reaches the bottom of the ramp.
∆Emech= ∆K + ∆Ug= ½ mVf2 – mgh = -ƒkd
Vf = [2(gh - ƒkd/m)]½
= [2(9.8*0.5-5*1.0/3)]½
= 2.54m/s
37
Example：A crate slides up a ramp
We want to slide a 12.0-kg crate up a 2.5 m long ramp inclined at 30.0°.
A worker gives the carte a initial speed 5.0 m/s at the bottom and let it
go. But the crate slides only 1.6 m up the ramp, stops, and slides back
down. (a) Find the magnitude of the friction force acting on the crate,
assuming it is a constant. (b) How fast is the carte moving, when it
reaches the bottom of the ramp?
(a) ∆Emech= ∆K+ ∆Ug = mgh–½ mVi2=-ƒkd
= 12*9.8*0.8–½ *12*52 = -56 J = -ƒk*1.6;
ƒk = 35N
(b) ∆Emech= mgh–ƒkd = ½ mVf2
= 12*9.8*0.8 –35*1.6 = 38.08 J = 6 Vf2
Vf = 2.52 m/s
38
Example： A child on an irregularly curved slide
A child of mass m takes a ride on an irregularly curved slide of height h
= 2.00 m as in Figure. The child starts from rest at the top. (a)
Determine the speed of the child at the bottom, assuming that no
friction is present.(b) If a friction force acts on the 20.0-kg child and he
arrives at the bottom of the slide with a speed vf = 3.00 m/s, by how
much does the mechanical energy of the system decrease as a result of
this force?
∆Emech= ∆K+ ∆Ug = ½ mVf2–mgh=-ƒkd
(a) ƒk =0；Vf =[2gh]½ = 6.26m/s
(b) ∆Emech= ½ mVf2–mgh = 90J -392J
= -302 J
39
Example： The coefficient of kinetic friction
A block of mass m sliding along a rough horizontal surface is traveling
at a speed VD when it strikes a massless spring head-on and compresses
the spring a maximum distance X. If the spring has stiffness constant k,
determine the coefficient of kinetic friction between block and surface.
∆Emech= ∆K + ∆Us= ½ kX2 – ½ mVD2 = -ƒkd ;
d=X; ƒk= mgμk
½ kX2+mgμkX–½ mVD2 =0
μk = (VD2/2gX) – (kX/2mg)
40
Example：A compressed Spring - 1
A horizontal spring has spring constant k = 360 N/m. (a) How much
work is required to compress it from its uncompressed length (x = 0) to
x = 11.0 cm? (b) If a 1.85-kg block is placed against the spring and the
spring is released, what will be the speed of the block when it separates
from the spring at x = 0? Ignore friction. (c) Repeat part (b) but assume
that the block is moving on a table and that some kind of constant drag
force FD = 7.0 N is acting to slow it down, such as friction.
(a) ∆W = ½ kx2 =180*(0.11)2= 2.18J
(b) ∆K+ ∆Us=0 ；½ mV2 – ½ kxmax2=0
V = [k/m]½ xmax = 1.53 m /s
(c) ∆Emech= ∆K+ ∆Us=½ mV2–½ kxmax2=-ƒkd
0.925V2=2.178 –7*0.11; V = 1.23 m/s
41
Example：A compressed Spring - 2
A block of mass 0.800 kg is given an initial velocity vA = 1.20 m/s to
the right and collides with a light spring of force constant k = 50.0 N/m
as Figure. (a) If the surface is frictionless, calculate the maximum
compression of the spring after the collision. (b) If a constant force of
kinetic friction acts between the block and the surface with k = 0.500
and if the speed of the block just as it collides with the spring is vA =
1.20 m/s. what is the maximum compression in the spring?
∆Emech= ∆K+ ∆Us = – ½ mVA2 +½ kxmax2=-ƒkd
(a) ƒk =0；xmax =[m/k]½ VA= 0.152m
(b) ƒk =3.92N；d = xmax ；
½ kxmax2+ƒkd –½ mVA2 =0
25xmax2 +3.92xmax – 0.576=0
xmax = 0.0924 m or –0.249 m (不合)
42
Example: Block Pushed off a Table
A block of mass 1.35 kg is joined to a spring with k = 560 N/m and
attached a wall. The spring is initially compressed by 0.11 m. The block
slides d = 0.65 m across table ( = 0.16), then falls h = 0.76 m down to
the ground. With what speed will the block land on the floor?
Ei = ½ kx2 + mgh = 0.5*560*0.112+1.35*9.8*0.76 = 13.443 J
ΔW = –mgμd = - 1.35*9.8*0.16*0.65 = -1.376 J
Ef =½ mv2 =Ei+ΔW =½ kx2+mgh–mgμd = 12.067 J
V =(12.067*2/1.35)½ = 4.23 m/s
43
Example：block and Spring on an Incline - 1
A block of mass m is attached to a spring and moves on a rough incline
as figure. Initially, the block is at rest with the spring un-extended. A
force F acting at an angle α to the incline pulls the block. Write the
modified form of the work-energy theorem.
E = K + Ug + Usp = ½ mV2 + mgh + ½ kx2
Ei = Ki + Ugi + Uspi = 0
∆Emech= ∆K+ ∆Us = Wnc;
Wnc = Fx cosα – ffr x
Ef = Ei + Wnc = Wnc = Fx cosα – ffr x
= ½ mV2 + mgx sinθ + ½ kx2
44
Example：block and Spring on an Incline - 2
A block of mass 2kg is attached to a spring constant is k =8N/m. The
block slides on an incline for which μk=1/8 and θ =37o. If the block
starts at rest with the spring un-extended. What is its speed when it has
slid a distance d=0.5m down the incline?
∆Ug = mg sinθ (-d)= -5.88J;
∆Usp = ½ kd2 = 1.0 J
Wnc = - μk(mgcosθ)d = -0.98J
∆K+ ∆U =V2-5.88+1 = Wnc= -0.98;
∆K= V2 = 3.9
V =1.97 m/s = 2 m/s
45
Example：block and Spring on an Incline
A block of mass m=0.2kg is held against, but not attached to a spring (k
= 50N/m) which is compressed by 20cm, as figure. When released, the
block slides 50cm up to the rough incline before come to rest. Find: (a)
the force of friction; (b) the speed of the block just as it leaves the
spring.
Ef – Ei = Wnc
(a) mgd sinθ – ½ kA2 = -ffrd
0.2*9.8*0.5*0.6-25*0.04 = -0.5*ffr
ffr = 0.824N
(b) ½ mV2 + mgA sinθ – ½ kA2 = -ffrA
0.1V2 = 1 -0.2*9.8*0.2*0.6 -0.824*0.2
V = 2.44 m/s
46
Conservative Forces and Potential Energy
When the conservative force of a system is F, we can obtain
the potential energy difference of the two points from the
work done on the displacement of two points in space, and
we can define the potential energy of the system. Since F is a
conservative force, the energy of the system is conserved,
ΔK + ΔU = 0；then ΔU = – ΔK.
xf
xi
W=∫ F·ds = Δ K = –ΔU= –(Uf–Ui) = Ui–Uf

If the force F is in the same direction as the displacement
ds, then ΔU is negative and the potential energy of the
system decreases.
47
Conservative Forces and Potential Energy
If the potential energy U(r) of a system is known, by the
differential of the potential energy to any displacement ds, we
can obtain the projection of the conservative force in this
direction; when the object moves from the position r to r+ds,
the potential energy changes ΔU
∆U = U(r+ds) - U(r) = －∆W =－F·ds
We consider ds=dxi at first, then ∆U=U(r+dxi)-U(r) = - F·ds
= -Fxdx; Fx= -∆U/dx = -dU/dx; similarly consider ds = dy j,
and ds = dz k, then Fy = -dU/dy; Fz = -dU/dz; F=-▽U
48
Conservative Forces and Potential Energy
Look at the case of the gravitational force:
dU g
d
Fy  
  (mgy )  mg
dy
dy

This is the expression for the vertical component of the
gravitational force
Look at the case of a deformed spring:
dU s
d 1 2
Fs  
  ( kx )  kx
dx
dx 2

This is Hooke’s Law and confirms the equation for U
U is an important function because a conservative force can
be derived from it.
49
Potential Energy for Gravitational Forces
Generalizing gravitational potential energy uses Newton’s
Law of Universal Gravitation:

GM E m
Fg  
rˆ
2
r
GM E m
The potential energy then is U g   r
The result for the earth-object system can be extended to any
two objects:
Gm1m2
Ug  
r
For three or more particles, this becomes
U total
m1m2 m1m3 m2 m3
 U 12  U 13  U 23  G (


)
r12
r13
r23
50
Electric Potential Energy
Coulomb’s Law gives the electrostatic force between two
particles
 kq1q2
Fe  2 rˆ
r
This gives an electric potential energy function of
kq1q2
Ue 
r
51
Example : The Change in Potential Energy
A particle of mass m is displaced through a small vertical
distance y near the Earth’s surface. Show that the general
expression for the change in gravitational potential energy
reduces to the familiar relationship Ug = mgy.
Ug(r)= －GmEm/r ; rj = ri + y ≒ RE
Ug = Ug(rj ) － Ug(ri) = －GmEm/rj + GmEm/ri
= GmEm(rj-ri)/(rirj) = m GmEy/(rirj)
= m (GmE/RE2) y
= mg y
52
Energy Diagrams and Equilibrium
Motion in a system can be observed in terms of a
graph of its position and energy.
In a spring-mass system example, the block
oscillates between the turning points, x = ±xmax.
The block will always accelerate back toward x = 0.
The x = 0 position is one of stable equilibrium.

Any movement away from this position results
in a force directed back toward x = 0.
Configurations of stable equilibrium correspond to
those for which U(x) is a minimum.
x = xmax and x = -xmax are called the turning points.
53
Energy Diagrams and Unstable Equilibrium
Fx = 0 at x = 0, so the particle is in equilibrium.
For any other value of x, the particle moves away from the
equilibrium position.
This is an example of unstable equilibrium.
Configurations of unstable equilibrium correspond to those for
which U(x) is a maximum.
54
Conservative Force and Potential Energy Function
Conservative Force for a Hypothetical Potential Energy
Function
(r > r2): Fr > 0, repulsive force.
(r = r2): Fr = 0, unstable equilibrium
(r2> r > r0): Fr < 0, attractive force.
(r = r0): Fr = 0, stable equilibrium.
(r0> r ): Fr > 0, repulsive force.
55
Energy Diagrams
potential well of depth –Uo
(E >0): unbound state
(E <0): bound state
The binding energy of a particle in a bound state is the
minimum energy requirement to make it an unbound state.
56
Potential Energy in Molecules
There is potential energy associated with the force between
two neutral atoms in a molecule which can be modeled by
the Lennard-Jones function.
U ( x )  4 [(

x
)12  (

x
)6 ]
Find the minimum of the function (take the derivative and set
it equal to 0) to find the separation for stable equilibrium.
The graph of the Lennard-Jones function shows the most
likely separation between the atoms in the molecule (at
minimum energy).
57
Example : Force and Energy on an Atomic Scale
The potential energy associated with the force between two
neutral atoms in a molecule can be modeled by the LennardJones potential energy function : U(x)=4ε[(/x)12-(/x)6]， ε
= 1.51  10–22J ， = 0.263 nm. Find the most likely distance
between the two atoms.
dU/dx=4ε[-1212/x13+66/x7] =0
xeq6 = 26； xeq = (2)1/6
xeq = (2)1/6(0.263nm)
= 2.9510–10m
58
Physics
Chapter 8
Momentum and Collisions
授課老師施坤龍
0
The Effective Position—Center of Mass
There is a special point in a system or object,
called the center of mass, that moves as if all of
the mass of the system is concentrated at that
point. The general motion of an object can be
considered as the sum of the translational motion
of the CM, plus rotational, vibrational, or other
forms of motion about the CM. The system will
move as if an external force were applied to a
single particle of mass M located at the center of
mass.
1
The Center of Mass
The center of gravity is the point at
which the gravitational force can be
considered to act. It is the same as the
center of mass. The center of gravity
can be found experimentally by
suspending an object from different
points. The CM need not be within the
actual object—a doughnut’s CM is in
the center of the hole.
2
The Center of Mass
The coordinates of the center of mass are RCM =XCM i +YCM j +ZCM k

RCM 
X CM 


m
r
 ii
i
M
 mi xi
i
M
; YCM 
m y
i
i
M
i
; Z CM 
m z
i i
i
M
M is the total mass of the system.
For an extended object, we imagine making it up of tiny particles, each of
tiny mass, and adding up the product of each particle’s mass with its
position and dividing by the total mass. In the limit that the particles
become infinitely small, this gives:

r dm
xdm
ydm
zdm





RCM 
; X CM 
; YCM 
; Z CM 
M
M
M
M
3
Notes on Various Densities
Volumetric Mass Density → mass per unit volume:
ρ=m/V
Surface Mass Density → mass per unit thickness of a sheet
of uniform thickness, t : σ = ρ t
Linear Mass Density → mass per unit length of a rod of
uniform cross-sectional area: λ = m / L = ρ A
4
Example : CM of three guys on a raft
Three people of roughly equal masses m on a lightweight
(air-filled) banana boat sit along the x axis at positions xA =
1.0 m, xB =5.0 m, and xC=6.0 m, measured from the left-hand
end. Find the position of the CM. Ignore the boat’s mass.
XCM= (mA xA+mB xB+mC xC)/(mA+mB+mC)
=m (xA+xB+ xC)/3m
= (1.0+5.0+6.0)/3
= 4.0 m
5
Example: Three particles in 2-D -1
Three particles, each of mass 2.50 kg, are located at the
corners of a right triangle whose sides are 2.00 m and 1.50 m
long, as shown. Locate the center of mass.
RCM= (mA rA+mB rB+mC rC)/(mA+mB+mC)
= (m(0, 0)+m(2, 0)+m(2, 1.5))/3m
= (4m, 1.5m)/3m
= (4/3, 0.5) m
6
Example: Three particles in 2-D-2
A system consists of three particles located at the corners of a
right triangle as in Figure. Find the center of mass of the
system.
RCM= (m1 r1+m2 r2+m3 r3)/(m1+m2+m3)
= (2m(d, 0)+m(d+b, 0)+4m(d+b, h) )/7m
= (7md+5mb, 4mh)/7m
= (d + 5b/7, 4h/7)
7
Example: Three particles in 2-D-3
Three particles of masses mA = 1.2kg, mB = 2.5kg and mC =
3.4kg form an equilateral triangle of edge length a=140cm.
Where is the center of mass of this three-particle system?
Particle
Mass
Location of CM
A
1.2 kg
RA=(0, 0)cm
B
2.5 kg
RB=(140, 0)cm
C
3.4 kg
RC=(70, 121)cm
RCM= (mA rA+mB rB+mC rC)/(mA+mB+mC)
= (1.2(0, 0)+2.5(140, 0)+3.4(70, 121))/7.1
= (350+238, 411.4)/7.1
= (83, 58) cm
8
Example: Four particles in 2-D
Find the CM of the four point masses as figure.
RCM= (m1 r1+m2 r2+m3 r3+m4 r4)/(m1+m2+m3+m4)
= (2(3, -1)+4(3, 3)+5(-4, 4)+1(-3, -2))/12
= (-5, 24)/12
= (-5/12, 2) m
9
Example : CM of L-shaped flat object-2
A thin rod of length 3L is bent at right angles at a distance L
from one end (see Fig.). Locate the CM with respect to the
corner. Take L=1.2 m.
m1 = m;
m2 = 2m
RCM= (m1 r1+m2 r2)/(m1+m2)
= (2m*(0, L) + m* (L/2, 0))/3m
= (L/6, 2/3L)
10
Example : CM of L-shaped flat object-1
Determine the CM of the uniform thin L-shaped construction
brace shown.
RCM= (m1 r1+m2 r2)/(m1+m2)
= (2.06*0.2*(1.03, 0.1) +0.2*1.48* (1.96, -0.74))/0.708
= ((0.42436, 0.0412) + (0.58016, -0.21904))/0.708
= (1.418, -0.251) m
11
Example : CM of U-shaped flat object
The U-shaped object pictured in Fig. has outside dimensions
of 100 mm on each side, and each of its three sides is 20 mm
wide. It was cut from a uniform sheet of plastic 6.0 mm thick.
Locate the center of mass of this object.
Object
Mass
Location of CM
Left bar
MA
RA=(10, 50, 3)mm
Right bar
MB
RB=(90, 50, 3)mm
Bottom bar
MC
RC=(50, 10, 3)mm
RCM= (mA rA+mB rB+mCrC)/(mA+mB+mC)
= (2000(10, 50, 3)+2000(90, 50, 3)+1200(50, 10, 3))/5200
= (50, 41, 3) mm
12
Example : Crescent-Shaped Object
Figure shows a uniform metal plate P of radius 2R from
which a disk of radius R has been stamped out (removed) in
an assembly line. Using the xy coordinate system shown,
locate the center of mass of the plate.
Plate
Mass
Location of CM
P
mP=?
XP=?
S
mS = -R
XS = -R
C
mC = mP+mS
XC = 0
mP=0.75mC ;
mS=0.25 mC;
XCM = (mP*XP+ mS*(-R))/ mC = 0
= 0.75*XP-0.25*R
XP = R/3 ,
YP =0
13
Example : CM of a thin rod - 1
(a) Show that the center of mass of a rod of mass M and length L lies
midway between its ends, assuming the rod has a uniform mass per unit
length.(b) Suppose a rod is nonuniform such that its mass per unit
length varies linearly with x according  = ax, where a is a constant.
Find the x coordinate of the center of mass as a fraction of L.
(a) λ= M/L; dm = λdx ;
L
L
Xcm = ∫0 x dm/M = ∫0 λxdx/ λL = (½ λL2)/λL = ½ L
L
(b) λ= ax; dm = λdx ; M = ∫ dm = ∫0 axdx = ½ aL2
L
L
Xcm = ∫ 0 xdm/M = ∫ 0 x(ax)dx/(½ aL2)
= (1/3aL3)/(½ aL2)
= 2/3 L
14
Example : CM of a thin rod - 2
Determine the CM of the rod assuming its linear mass density
λ (its mass per unit length) varies linearly from λ = λ0 at the
left end to double that value, λ = 2λ0, at the right end.
dm = λ(x) dx ; λ(x) = λ0 (1+x/L)
L
L
M = ∫ 0 dm = ∫0 λ0(1+x/L) dx = λ0L+ ½ λ0L = 3/2 λ0L
L
L
Xcm = ∫ 0 x dm/M = ∫0 xλ0(1+x/L)dx/(3/2λ0L)
= (½ λ0L2+1/3 λ0L2)/(3/2λ0L)
= 5/9 L
15
Example：The CM of a Right Triangle
The sign is of the triangular shape shown in Figure. The
bottom of the sign is to be parallel to the ground. Find the
center of mass ?
A = ½ a b ； σ = M/A = 2M/(ab) ； y = (b/a)x；
a
a
Xcm = ∫ xdm/M = ∫ 0 xσy dx/M = ∫0 σ(b/a)x2dx/M
= 1/3 σa2b/(½ σab) = 2/3 a
b
Ycm = ∫ y dm/M = ∫ 0 yσ(a-x)dy/M
b
= ∫0 σ(ay-ay2/b)dy/(½ σab)
= (1/2-1/3)σab2/( ½ σab) = 1/3 b
16
Example：The CM of Semicircular rod
Find the CM of a semicircular rod of radius R and linear
density λ kg/m as shown in Fig.
S = π R； λ = M/S = M/(πR) ； Xcm = 0
p
Ycm = ∫ ydm/M = ∫0 R sinθ λ Rdθ/M
p
=
λR2
∫ 0 sinθ dθ/(λπR)
=
λR2
p
(-cosθ)|0 /(λπR)
=2R/π
17
Example：The CM of solid cone
Find the CM of a uniform solid cone of height h and semi-
angle α, as in Fig.
dV = πx2dy；x=yR/h； ρ = M/V； Xcm = 0
h
V= ∫0
πx2dy
h
2
2
=R /h ∫
0
πy2dy = πR2h/3
h
Ycm = ∫ ydm/M = ∫0 yρπx2dy/(ρπR2h/3)
=
h 3
2
2
ρπR /h ∫ y dy/(ρπR2h/3)
0
= 3h/4
18
Linear Momentum
The linear momentum of a particle or an object that can be
modeled as a particle of mass m moving with a velocity v is
defined to be the product of the mass and velocity: p = m v
Linear momentum is a vector quantity

Its direction is the same as the direction of v
The dimensions of momentum are ML/T
The SI units of momentum are kg ·m / s
Momentum can be expressed in component form:

px = m vx
py = m vy
pz = m vz
19
Newton’s Second Law and Momentum
Newton called the product m v the quantity of motion of the
particle. Newton’s Second Law can be used to relate the
momentum of a particle to the resultant force acting on it.
Σ F = m a = m dv/dt = d(mv)/dt = dp/dt
Conservation of momentum can be expressed mathematically

 




in various ways Ptotal  P1  P2  constant; or P1i  P2i  P1 f  P2 f
In component form, the total momenta in each direction are
independently conserved


 P1   P2
system
system
Conservation of momentum can be applied to systems with
any number of particles
20
Conceptual Example: Falling on or off a sled
(a) An empty sled is sliding on frictionless ice when Susan
drops vertically from a tree above onto the sled. When she
lands, does the sled speed up, slow down, or keep the same
speed? (b) Later: Susan falls sideways off the sled. When she
drops off, does the sled speed up, slow down, or keep the
same speed?
(a) The sled will slow down.
(b) The sled keeps going at the same speed
21
Example： A fireworks
A fireworks box with mass m=6.0 kg slides with speed v=4.0
m/s across a frictionless floor in the positive direction along
an x axis. It suddenly explodes into two pieces. One piece,
with mass mA=2.0 kg, moves in the positive direction along
the x axis with speed vA=8.0 m/s. What is the velocity of the
second piece, with mass mB?
Psys = PA 2 + PB 2 = mA vA + mB vB
mB =6kg-2kg=4kg
Psys = 6kg*4m/s= 2kg*8m/s + 4kg* vB
vB =2.0m/s i
22
Example : Can We Ignore the Kinetic Energy of the Earth?
A ball of mass m=200g is dropped from rest at a height
h=1.3m above the ground, and rebound. we claimed that we
can ignore the kinetic energy of the Earth, why?
V =(2gh)½ = 5.0 m/s (Ball landing speed)
∆p=0.2*10 =2 kg m/s (momentum change)
VE = 3×10-25 m/s (The speed of the Earth)
The kinetic energy of the Earth K = 3×10-25 J. We can ignore.
23
Example : The Arche
A 60-kg archer stands at rest on frictionless ice and fires a
0.50-kg arrow horizontally at 50 m/s. With what velocity
does the archer move across the ice after flung the arrow?
ptot = 0.5*50 + 60 V = 0
V = -25/60 = -0.42 m/s
24
Example : Rifle recoil -1
Calculate the recoil velocity of a 5.0-kg rifle that shoots a
0.020-kg bullet at a speed of 620 m/s.
ptot = 0.02*620 + 5 VR = 0
VR = -12.4/5 = -2.48 m/s
25
Example : Rifle recoil-2
A 3.24-kg Winchester Super X rifle, initially at rest, fires a
11.7-g bullet with a muzzle speed of 800 m/s. (a) What is the
recoil velocity of the rifle? (b) What is the ratio of the kinetic
energies of the bullet and the rifle?
(a) mbvb+MRVR= 0.0117*800+3.24 VR=0
VR=-2.89 m/s
(b) Kb= ½ mb vb2 = 0.5*0.0117*(800)2 = 3744J
KR= ½ mR vR2 = 0.5*3.24*(2.89)2 = 13.5J
KR/ Kb = 13.5/3744 =0.36%
26
Example：A moving cart
we find that the cart of mass mA=2.84kg is moving from left
to right with an initial x-component of velocity of vi=1.78
m/s. After two bricks of total mass mB= 4.26kg are placed
gently on the cart, the combined masses continue to move
more slowly from left to right, find the system velocity vf？
The conservation of momentum, then
Σ p = 2.84*1.78 = (2.84+4.26)*Vf
Vf = 5.0552/7.1= 0.712m/s
27
Example : Nucleus decays
An unstable nucleus decays into three particles. Particles 1 and 2 each have mass 2m,
and particle 3 has mass m. We wish to use two detectors to observe particles 1 and 2.
In the center-of-mass frame, the nucleus initially has no kinetic energy. When the
nucleus decays, the three particles have the velocities shown in Figure and total
kinetic energy K = 17/8mv2. Particles 1 and 2 have the same speed vα , and particle 3’s
speed is (2)½ v. Find the angles θ1 and θ2.
pi = pf = p1 + p2 + p3 =0
pfy = p1y+p2y+p3y= 2mvαsinθ1- 2mvαsinθ2=0 ; θ1= θ2 = θ ;
pfx = p1x+p2x+p3x= 2mvαcosθ+2mvαcosθ–m(2)½ v = 0 ;
cos θ = (2)½ v/(4vα)
K = ½ m [(2)½ v]2+½ (2m)(vα)2+½ (2m)(vα)2 = mv2 +2mvα2 =17/8 mv2 ; vα =3/4 v
cos θ = (2)½ v/(4vα) = (2)½ /3; θ1= θ2 = θ = 61.9o
28
Impulse and Momentum Change
During a collision, objects are deformed due to
the large forces involved.
Since F = dp/dt,
Integrating
pf
∫ pi
we can write dp = F dt,
dp = pf –pi = ∆p =
tf
∫ti
Fdt = I
I = Fave∆t = ∆p
This equation expresses the impulse-momentum
theorem: The impulse of the force acting on a
particle equals the change in the momentum of
the particle
 tf 

I   Fext dt  ptot
ti
29
Example : How Good Are the Bumpers?
An automobile of mass 1500 kg collides with a wall .The
initial and final velocities of the automobile are -15.0m/s and
2.60m/s. If the collision lasts for 0.150 s. Find the impulse
due to the collision and the average force exerted on the
automobile?
I = ∆p = 1500*[2.6-(-15)] i
=2.64×104 i kg．m/s
Fave = ∆p /∆t = 1.76×105 N i
30
Example : Force of a tennis serve
For a top player, a tennis ball may leave the racket on the
serve with a speed of 55 m/s (about 120 mi/h). If the ball has
a mass of 0.060 kg and is in contact with the racket for about
4 ms (4×10-3 s), estimate the average force on the ball. Would
this force be large enough to lift a 60-kg person?
∆p = 0.06*55-0=3.3 kg．m/s
Fave = ∆p/∆t = 825 N
Fg=60*9.8=588 N < Fave , the force can lift a 60=kg person.
31
Example : Karate blow
Estimate the impulse and the average force delivered by a
karate blow that breaks a board a few cm thick. Assume the
hand moves at roughly 10 m/s when it hits the board.
I = ∆p=1*[10-0] =10.0 kg．m/s
Assume the compression of a board is1cm,
then ∆t=2×10-3s
Fave = ∆p/∆t = 5.0*103 N
32
Example : Kick the soccer
A soccer has a mass 0.4 kg. Initially it is moving to the left at 20.0 m/s,
but then it is kicked. After the kick it is moving at 45.0 upward and to
the right eith speed 30.0 m/s. Find the impulse of the net force and the
average net force, assuming the collision time is 0.010 s.
Ix = mΔVx= m (V2x–V1x)
=0.4*[(30*cos45o) -(-20)] =16.5 kg m/s
Iy = m ΔVy= m (V2y–V1y)
=0.4* 30*sin45o = 8.5 kg m/s
Fx =Ix /∆t= 16.5/0.01 = 1650 N
Fy =Iy /∆t= 8.5/0.01 = 850 N
(a) I = (16.5 i + 8.5 j ) kg m/s ; |I| =18.56 kg m/s; θ =tan-1(Iy/Ix) = 27.3o
(b) F = (1650 i + 850 j ) N; |F| =1856 N; θ =tan-1(Fy/Fx) = 27.3o
33
Example : Ball and Bat - 1
A pitcher throws a fastball, which crosses the plate with a speed of 40.2
m/s and an angle of 5.0 relative to the horizontal. The batter hits it for
a home run, launching it with 49.2 m/s at an angle of 35.0 to the
horizontal. Mass of the baseball 0.145 kg. What is the magnitude of the
impulse the baseball receives from the bat?
ΔVx=V2x–V1x=[(49.2m/s)(cos35o) -(-40.2 m/s )(cos5o) ]
=80.35 m/s.
ΔVy=V2y–V1y =[(49.2m/s)(sin35o) ) -(-40.2 m/s )(sin5o) ]
= 31.72 m/s
I = m(ΔVx, ΔVy) = 0.145(80.35, 31.72) kg m/s
I = (11.65 i+4.6 j ) kg m/s; |I| = 12.53 kg m/s; θ =tan-1(Iy/Ix) = 21.55o
34
Example : Ball and Bat - 2
A pitched 140 g baseball, in horizontal flight with a speed v1 of 39.0 m/s, is struck by a
bat. After leaving the bat, the ball travels in the opposite direction with speed v2, also
39.0 m/s. (a) What impulse I acts on the ball while it is in contact with the bat during
the collision? (b) The impact time ∆t for the baseball bat collision is 1.20 ms. What
average net force acts on the baseball? (c) Now suppose the collision is not head-on,
and the ball leaves the bat with a speed v2 of 45.0 m/s at an upward angle of 30o.What
now is the impulse on the ball?
(a) Ix =p2x–p1x=(0.140kg)(39.0m/s)-(0.140kg)(-39.0m/s) =10.9 kg m/s
(b) <Fxnet >= ∆px/∆t =10.9/0.0012 =9080 N
(c)Ix=p2x–p1x=(0.14kg)[(45m/s)(cos30o) -(-39.0 m/s)]
=10.92 kg m/s.
Iy=p2y–p1y=(0.14kg)[(45m/s)(sin30o) -(0 m/s)]
=3.15 kg m/s.
I = (10.9 i+3.15j ) kg m/s; |I| = 11.4 kg m/s; θ =tan-1(Iy/Ix) = 16o
35
Example： The collision of two carts
A moving cart coming from the left has a mass of 1.8kg and an initial velocity of 0.3 m/s. It then collides
with a stationary cart to its right with a mass of 0.8kg. After the collision the 1.8kg cart slows down and the
0.8kg cart moves away from it at a brisk velocity as Fig. (a) Use the measured impulse curves shown in
Fig. to find the approximate collision times when the collision involves contact between a metal hook and a
deformable rubber stopper (case a) (b) Also use the measured impulse curves to compare the maximum
forces experienced by the initially stationary cart for the two types of collisions (metal–rubber and metal–
metal). (c) Use the measured impulse curves to estimate the magnitude of the momentum transferred to the
right cart during each type of collision.
(a) (case a) 22×10-3 s. (case b) 15×10-3 s.
(b) Ix=p2x–p1x= ∫ Fx dt = Fx ∆t
(case a) Ix=½ Fx peak ∆t=0.5*40* 22×10-3 =0.4Ns
Ix =p2x–p1x= p2x– 0 = 0.4kg m/s;
p2x= 0.4kg m/s
(case b) Ix=½ Fx peak∆t=0.5*49* 15×10-3 =0.4Ns
Ix =p2x–p1x= p2x– 0 = 0.4kg m/s;
p2x= 0.4kg m/s
36
Collisions – Characteristics

dPsys

Fnet 
0
dt

Psys  cons tan t (for an isolated system)


Psys1  Psys 2 (for an isolated system)
37
Collisions – Characteristics
We use the term collision to represent an event during which two
particles come close to each other and interact by means of forces.
The time interval during which the velocity changes from its initial to
final values is assumed to be short. The interaction force is assumed to
be much greater than any external forces present.
In an elastic collision, momentum and kinetic energy are conserved.
Perfectly elastic collisions occur on a microscopic level. In macroscopic
collisions, only approximately elastic collisions actually occur.
In an inelastic collision, kinetic energy is not conserved although
momentum is still conserved. If the objects stick together after the
collision, it is a perfectly inelastic collision.
Momentum is conserved in all collisions.
38
Newton’s Laws and Momentum Conservation
Conservation of momentum can also be derived from
Newton’s laws. A collision takes a short enough time that we
can ignore external forces. Since the internal forces are equal
and opposite, the total momentum is constant.
ΣFnet = 0; ΣPi(before) = ΣPf (after).
39
Newton’s Laws and Momentum Conservation
During a collision, measurements show that the total
momentum does not change:
PA1 + PB1 = PA2 + PB2 = constant
mAVA +mBVB = mAUA +mBUB
The initial and final velocities are related according to
the principle of the conservation of linear momentum.
40
Elastic Collisions
For the conservation laws of Energy and Momentum：
½ mAUA2+ ½ mBUB2=½ mAVA2+½ mBVB2
mAUA + mBUB = mAVA + mBVB
UA-UB = - (VA-VB)
VCM ≡ (mAVA+mBVB)/(mA+mB)
UA=[(mA-mB)VA+2mBVB]/(mA+mB) = 2VCM - VA
UB=[2mAVA+ (mB-mA)VB]/(mA+mB) = 2VCM -VB
41
Elastic Collisions
Example of some special cases

mA = mB – the particles exchange velocities

When a very heavy particle collides head-on with a very
light one initially at rest, the heavy particle continues in
motion unaltered and the light particle rebounds with a
speed of about twice the initial speed of the heavy particle

When a very light particle collides head-on with a very
heavy particle initially at rest, the light particle has its
velocity reversed and the heavy particle remains
approximately at rest
42
Example ：Two block’s elastic collision
A block of mass m1=2kg and initial velocity u1=4i m/s makes
a one-dimensional elastic collision with a block of mass
m2=3kg moving at u2=2i m/s. Find their final velocities?
m1v1+m2 v2 = m1u1+m2u2 = 2*4+3*2 =14 kgm/s
Vcm =(m1v1+m2 v2)/(m1+m2) =14/5 = 2.8 m/s
V1 = 2*2.8 – 4 = 1.6 m/s;
V2 = 2*2.8 – 2 = 3.6 m/s;
43
Example : A nuclear collision
A proton (p) of mass 1.01 u (unified atomic mass units)
traveling with a speed of 3.60×104 m/s has an elastic head-on
collision with a helium (He) nucleus (mHe = 4.00 u) initially
at rest. What are the velocities of the proton and helium
nucleus after the collision? Assume the collision takes place
in nearly empty space.
Vcm = mpvp/(mp+mHe) = 7.26×103m/s
up = 2 mpvp/(mp+mHe) – vp= -2.15×103m/s
uHe=2 mpvp/(mp+mHe) =1.45×104m/s
44
Example : Equal masses
Billiard ball A of mass m moving with speed vA collides head-on with
ball B of equal mass. What are the speeds of the two balls after the
collision, assuming it is elastic? Assume (a) both balls are moving
initially (vA and vB), (b) ball B is initially at rest (vB = 0).
(a) Vcm = (mAvA+mB vB) /(mA+mb) = (vA+vB) /2
uA = 2 (vA+vB) /2– vA= vB
uB = 2 (vA+vB) /2– vB = vA
(b) Vcm = mAvA /(mA+mB) = vA /2
uA = 2 vA/2– vA= 0
uB = 2 vA /2– 0 = vA
45
Example : Unequal masses, target at rest
A very common practical situation is for a moving object (mA) to strike a second
object (mB, the “target”) at rest (vB = 0). Assume the objects have unequal masses, and
that the collision is elastic and occurs along a line (head-on). (a) Derive equations for
vB’ and vA’ in terms of the initial velocity vA of mass mA and the masses mA and mB. (b)
Determine the final velocities if the moving object is much more massive than the
target (mA >> mB). (c) Determine the final velocities if the moving object is much less
massive than the target (mA << mB).
(a) Vcm = mAvA/(mA+mB)
uA = 2 mAvA/(mA+mB) – vA= (mA–mB)vA /(mA+mB)
uB = 2 mAvA/(mA+mB) – 0 = 2mAvA/(mA+mB)
(b) uA = (mA–mB)vA /(mA+mB) ≈ vA
uB = 2mAvA/(mA+mB) ≈ 2vA
(c) uA = (mA–mB)vA /(mA+mB) ≈ - vA
uB = 2mAvA/(mA+mB) ≈ 0
46
Example : A Two-Body Collision with a Spring
A block of mass m1=1.6kg initially moving to the right with a speed of 4.0m/s on a
frictionless, horizontal track collides with a light spring attached to a second block of
mass m2=2.1kg initially moving to the left with a speed of 2.5m/s. The spring constant
is 600 N/m. Find (a) the velocities of the two blocks after the collision. (b) the velocity
of block 2 during the collision, at the instant block 1 is moving to the right with a
velocity of 3.0m/s as Figure. (c) the distance the spring is compressed at that instant.
(a) m1u1+m2u2 = m1v1+m2 v2 = 1.6*4+2.1*(-2.5) = 1.15 kg m/s
Vcm =(m1v1+m2 v2)/(m1+m2) =1.15/3.7 = 0.311 m/s
u1 = 2*0.311 – 4 = -3.38 m/s; u2= 2*0.311 – (-2.5) = 3.12 m/s;
(b) 1.6*3+2.1*u2 = 1.15 kg m/s ； u2= (1.15-4.8)/2.1 = -1.74 m/s
(c) U = ½ kx2= 300x2 = Ei – Ef
Ei=0.8*42+1.05*(-2.5)2 =19.363
Ef=0.8*32+1.05*(-1.74)2=10.379
x = (8.984/300)½ = 0.173 m
47
Two-Dimensional Collision
The momentum is conserved in all directions
After the collision, the momentum in the xdirection is m1u1 cosθ + m2u2 cosϕ = m1v1
After the collision, the momentum in the ydirection is m1u1sinθ - m2u2sinϕ = 0
The negative sign is due to the component of
the velocity being downward.
If the collision is elastic, apply the kinetic
energy equation.
½ m1u12 + ½ m2u22 = ½ m1v12
Need one more parameter to solve u1, u2, θ, ϕ
48
Example : Proton-Proton Collision - 1
A proton collides elastically with another proton that is initially at rest.
The incoming proton has an initial speed of 3.5105 m/s and makes a
glancing collision with the second proton. After the collision, one
proton moves off at an angle of 37° to the original direction of motion
and the second deflects at an angle of  to the same axis. Find the final
speeds of the two protons and the angle .
mv1 = mu1 + mu2 ;
v1 = u1 + u2
(1)
½ mv12 = ½ mu12 + ½ mu22 ; v12 = u12 + u22 (2)
(1) ·(1); v12 = u12 + u22 +2u1 ·u2
(2) & (3); u1 ·u2 = 0 ;
0.8u1+0.6u2= 3.5×105;
(3)
ϕ =53o
0.6u1-0.8u2= 0;
u1 = 3.5×105*0.8 =2.8×105m/s
u2 = 3.5×105*0.6 =2.1×105m/s
49
Example : Proton-Proton Collision - 2
A proton moving at speed u1= 50 km/s makes an elastic
collision with another proton initially at rest. Given that
θ=30°, find v1, v2 and ?
mv1 = mu1 + mu2 ;
v1 = u1 + u2
(1)
½ mv12 = ½ mu12 + ½ mu22 ; v12 = u12 + u22 (2)
(1) ·(1); v12 = u12 + u22 +2u1 ·u2
(3)
(2) & (3); u1 ·u2 = 0 ; ϕ =60o
0.866u1+0.5u2= 5.0×104;
0.5u1-0.866u2= 0;
u1 = 5.0×104*0.866 =4.33×104m/s
u2 = 5.0×104*0.5 =2.5×104m/s
50
Example : Proton-Proton Collision - 3
A proton traveling with speed 8.2×105 m/s collides elastically with a
stationary proton. One of the protons is observed to be scattered at a
60° angle. At what angle will the second proton be observed, and what
will be the velocities of the two protons after the collision?
mv1 = mu1 + mu2 ;
v1 = u1 + u2
(1)
½ mv12 = ½ mu12 + ½ mu22 ; v12 = u12 + u22 (2)
(1) ·(1); v12 = u12 + u22 +2u1 ·u2
(3)
(2) & (3); u1 ·u2 = 0 ; ϕ =30o
0.5u1+0.866u2= 8.2×105;
0.866u1-0.5u2= 0;
u1 = 8.2×105*0.5 =4.1×105m/s
u2 = 8.2×105*0.866 =7.1×105m/s
51
Example : Billiard ball collision in 2-D
Billiard ball 1 moving with speed v1 = 3.0 m/s in the +x direction strikes
an equal-mass ball 2 initially at rest. The two balls are observed to
move off at 45° to the x axis, ball 1 above the x axis and ball 2 below.
That is, θ= 45° and = -45 °. What are the speeds of the two balls after
the collision?
mv1 = mu1 + mu2 ;
v1 = u1 + u2
(1)
½ mv12 = ½ mu12 + ½ mu22 ; v12 = u12 + u22 (2)
(1) ·(1); v12 = u12 + u22 +2u1 ·u2
(3)
(2) & (3); u1 ·u2 = 0 ; ϕ =45o
0.707u1+0.707u2= 3.0;
0.707u1-0.707u2= 0;
u1 = 3.0*0.707 =2.12 m/s
u2 = 3.0*0.707 =2.12m/s
52
Inelastic Collisions
With inelastic collisions, some of the initial kinetic energy is
lost to thermal or potential energy.
Kinetic energy may also be gained during explosions, as
there is the addition of chemical or nuclear energy.
A completely inelastic collision is one in which the objects
stick together afterward, so there is only one final velocity.
m1V1i + m2V2i = (m1 + m2) Vf
53
Example : Railroad cars collide
A 10,000-kg railroad car, A, traveling at a speed of 24.0 m/s
strikes an identical car, B, at rest. If the cars lock together as a
result of the collision, what is their common speed
immediately after the collision?
MAvA=(MA+ MB) Vf= 2.4*105 = 20000* Vf
Vf=12.0 m/s
54
Example : Two automobiles collide-1
An 2000-kg car stopped at a traffic light is struck from the
rear by a 1000-kg car. The two cars become entangled,
moving along the same path as that of the originally moving
car. If the smaller car were moving at 18.0 m/s before the
collision, what is the velocity of the entangled cars after the
collision?
m2v2 = (m1+m2)Vf
Vf = m2v2 /(m1+m2)
= 1000*18/(2000+1000) = 6.0 m/s
55
Example : Two automobiles collide-2
A 2000-kg Cadillac limousine moving west at 10 m/s collides with a
1000-kg Honda Prelude moving east at 26 m/s. The collision is
completely inelastic and takes place on an icy (frictionless) patch of
road. (a) Find their common velocity after the collision. (b) what is the
fractional loss in kinetic energy?
(a) P =m1V1+m2V2= 2000*(-10)+1000*(26)=6000
=(m1+m2)V =3000*V
V =2m/s (to east)
(b)Ef = ½ (m1+m2)V2 = 6000J
Ei = ½ (m1V12+m2V22) = 438000J
(Ef-Ei)/Ei= -0.99 =-99%
56
Example : The Ballistic Pendulum - 1
The ballistic pendulum is an apparatus used to measure the speed of a
fast-moving projectile such as a bullet. A projectile of mass m1 is fired
into a large block of wood of mass m2 suspended from some light wires.
The projectile embeds in the block, and the entire system swings
through a height H. How can we determine the speed of the projectile
from a measurement of H?
m1v1 = (m1+m2)V； v1 = (1+m2/m1) V
½ (m1+m2)V2 = (m1+m2)gH
V=(2gH)½
v1 = (1+m2/m1)(2gH)½
57
Example : The Ballistic Pendulum - 2
A ballistic pendulum, for measuring the speed of a bullet. Suppose that
a bullet of mass m=10g and speed u is fired into a block of mass M=2kg
suspended. The bullet embeds in the block and raises it by a height
H=5cm. (a) Determine u from H? (b) What is the thermal energy
generated?
m1v1 = (m1+m2)V； v1 = (1+m2/m1) V
½ (m1+m2)V2 = (m1+m2)gH ；V=(2gH)½ = 0.99m/s
(a) v1 = (1+m2/m1)(2gH)½ = (1+200)*0.99 = 200m/s
(b) Q = Ki-Kf =½ m1v12 -½ (m1+m2)V2
= 0.005*40000 –1.005*0.992
= 199 J
58
Example : Collision at an Intersection - 1
A 1 500-kg car traveling east with a speed of 25.0 m/s collides at an
intersection with a 2 500-kg van traveling north at a speed of 20.0 m/s
as shown in Figure. Find the direction and magnitude of the velocity of
the wreckage after the collision, assuming that the vehicles stick
together after the collision.
P = m1V1 + m2V2= (m1+m2) Vc =(37500 i + 50000 j) kg m/s
Vc =(9.375 i + 12.5 j) m/s
(a) θ = tan-1(12.5/9.375) = 53.1o
(b) Vc = (9.3752+12.52)½
= 15.625 m/s
59
Example : Collision at an Intersection - 2
A 950kg Chevrolet moving east collides with a 1350kg Ford moving
north as figure. Although the brakes were applied, the cars collided
together. The skid marks after the collision were straight and 6m long,
in a direction 37o N of E. The coefficient of kinetic friction is 0.6. Was
either car exceeding 15m/s speed limits?
ffr=-(m1+m2)gμk= (m1+m2)a; a = -gμk
0 = V02+2ad=V02-2gμk d
V0=(2gμkd)½ =(2*9.8*0.6*6)½ =8.5m/s
Px=(m1+m2) V0cos37o=1.564*104=950*v1
v1=16.46m/s
Py= (m1+m2) V0sin37o=1.173*104 =1350*v2
v2=8.69m/s
60
Example : Two pucks collide
A puck of mass m1=3kg has a initial velocity of 10m/s at 20o of S of E.
A second puck of mass m2=5kg has a velocity of 5m/s at 40o of W of N.
They collide and stick together. Find their common velocity after the
collision.
P = m1V1 + m2V2= (m1+m2) Vc
Px = 3*10*cos20o-5*5*sin40o=12.12
Py =-3*10*sin20o+5*5*cos40o=8.89
Vx = 12.12/(3+5) = 1.52 m/s
Vy = 8.89/(3+5) = 1.11 m/s
Vc = (1.52, 1.11) m/s
|Vc| =1.88 m/s at 36o N of E
61
Example: Skaters Embrac
Two skaters collide and embrace, “sticking” together after impact, as
suggested by Fig., where the origin is placed at the point of collision.
Alfred, whose mass mA is 83 kg, is originally moving east with speed vA
= 6.2 km/h. Barbara, whose mass mB is 55 kg, is originally moving
north with speed vB =7.8 km/h. What is the velocity vsys 2 of the couple
after they collide?
Psys = mA vA + mB vB=(mA + mB) vc =(83*6.2 i +55*7.8 j )
= (514.6 i + 429 j) kg km/h
vc=(mA vA+mB vB)/(mA+mB)
= (3.73 i + 3.1 j) km/h
|vc|=(3.732+3.12)½ = 4.86 km/h; θ = tan-1(3.1/3.73) = 39.8o
62
Motion of a System of Particles
Assume the total mass, M, of the system remains constant. We can also
describe the momentum of the system and Newton’s Second Law for the
system. The velocity of the center of mass of a system of particles is
VCM = dRCM/dt = Σmivi /M
The momentum can be expressed as: Ptot = Σmivi = MVCM
The acceleration of the center of mass can be found by differentiating the
velocity with respect to time: aCM = dVCM/dt = Σmiai /M
The acceleration can be related to a force
ΣFext = ΣFi = Σmiai = MaCM = dPtot/dt
The total linear momentum of a system of particles is constant if no
external forces act on the system.


ptot  Mvcm  cons tan t ( when  Fext  0 )
63
Example : Exploding Project
A projecti1e fired into the air suddenly explodes into several fragments.
(a) What can be said about the motion of the center of mass of the
system made up of all the fragments after the explosion? (b) If the
projectile did not explode, it would land at a distance R from its launch
point. Suppose the projectile explodes and splits into two pieces of
equal mass. One piece lands at a distance 2R from the launch point.
Where does the other piece land?
(a) The center of mass of the fragments follows the same parabolic
path the projectile 。
(b) This piece right back at the launch point!
64
Example : The Exploding Rocket
A rocket is fired vertically upward. At the instant it reaches an altitude
of 1 000 m and a speed of vi = 300 m/s, it explodes into three fragments
having equal mass. One fragment moves upward with a speed of v1 =
450 m/s following the explosion. The second fragment has a speed of v2
= 240 m/s and is moving east tight after the explosion. What is the
velocity of the third fragment immediately after the explosion?
Mvi = M/3 v1 + M/3 v2 + M/3 v3
v3 = 3 vi - v1 - v2
= 3*300 j – 450 j - 240 i
= ( - 240 i + 450 j) m/s
65
Example :Walking on a boat
A man m1=60kg is at the rear of a stationary boat of mass
m2=40kg and length 3m, which can move freely on the water.
The front of the boat is 2m from the deck. What happens
when the man walks to the front?
Xcm= (m1x1+ m2x2)/(m1+ m2) = (60*5+40*3.5)/100
= 4.4m
Xcm= [m1d+ m2(d+1.5)]/(m1+ m2)=4.4m
= (60*d+40*d+60)/100
d = 3.8m
66
Example : Two ball collision
Two ball with masses m1=3kg and m2=5kg have initial velocities
v1=v2=5m/s in the direction as figure. They collide at the origin. Find(a)
the velocity of CM? (b) the position of the CM 2s after collision?
(a)Vcmx= (m1v1x+ m2v2x)/(m1+ m2)
= (3*-5*cos37o+ 5*0)/8
= -1.5m/s
Vcmy= (m1v1y+ m2v2y)/(m1+ m2)
= (3*-5*sin37o+ 5*5)/8
= 2.0m/s
(b) rcmx= Vcm*t = (-3i + 4j) m
67
Example : A man walks at the platform
A 75kg man sits at the rear end of a platform of mass 25kg and length
4m, which moves initially at 4m/s over a frictionless surface. At t=0, he
walks at 2m/s relative to the platform and then sits down at the front
end. During the walking period, find displacements of: (a) the platform,
(b) the man, and (c) the center of mass.
100*4 = 75(2+V)+25V； V = 2.5m/s
∆t = 4m/(2m/s) = 2s
(a) ∆xp = V∆t = 5m
(b) ∆ xm = (2+V)∆t = 9m
(c) ∆xCM = VCM∆t = 8m
68
Rocket Propulsion
The operation of a rocket depends upon the law of conservation of
linear momentum as applied to a system of particles, where the system
is the rocket plus its ejected fuel. The initial mass of the rocket plus all
its fuel is M +∆m at time ti and velocity v.
The initial momentum of the system is (M + ∆m)v.
At some time t + ∆t, the rocket’s mass has been reduced to M and an
amount of fuel, ∆m has been ejected
The rocket’s speed has increased by ∆v
The final momentum of the system is M(v + ∆v) + ∆m(v + vrel)
(M + ∆m)v =M(v + ∆v) + ∆m(v + vrel) ;
M∆v + ∆mvrel= 0
69
A System with Mass Exchange—A Rocket and Its Ejected Fuel
Psys 1 = Psys 2 = 0 = dm( vrel)+(M-dm)dv
(M>>dm)
dm( vrel)+(M)dv = 0
-( vrel) dm/dt=M dv/dt ;
dM=-dm,
( vrel) dM/dt=M a = - R vrel (the thrust of the rocket engine)
dm( vrel)+(M)dv = 0 dv = vrel dM/M
dv = -vrel dM/M
v2
∫v1
dv =
M2
-vrel ∫M
1
dM/M
v2 = v1 + vrel ln(M1/M2)
70
Rocket Propulsion
Because the gases are given some momentum when they are
ejected out of the engine, the rocket receives a compensating
momentum in the opposite direction
Therefore, the rocket is accelerated as a result of the “push”
from the exhaust gases
In free space, the center of mass of the system (rocket plus
expelled gases) moves uniformly, independent of the
propulsion process
71
Rocket Propulsion
The basic equation for rocket propulsion is
Mi
v f  vi  veln(
)
Mf
The increase in rocket speed is proportional to the speed of
the escape gases (ve)

So, the exhaust speed should be very high
The increase in rocket speed is also proportional to the
natural log of the ratio Mi/Mf

So, the ratio should be as high as possible, meaning the
mass of the rocket should be as small as possible and it
should carry as much fuel as possible
72
Thrust
The thrust on the rocket is the force exerted on it by
the ejected exhaust gases
dv
dM
Thrust  Ma  M
 ve
dt
dt
The thrust increases as the exhaust speed increases
The thrust increases as the rate of change of mass
increases

The rate of change of the mass is called the burn
rate
73
Example : Washing a car
Water leaves a hose at a rate of 1.5 kg/s with a speed of 20
m/s and is aimed at the side of a car, which stops it. What is
the force exerted by the water on the car?
Thrust = dM/dt vrel = 1.5kg/s*20m/s
=30 kg．m/s2
Thrust = 30 N
74
Example : Fire Hose
What is the magnitude of the force that acts on a firefighter
holding a fire hose that ejects 360 liters of water per minute
with a muzzle velocity of 39.0 m/s?
dM/dt = ρ dV/dt = 1.0 kg/liter * 360 liters/min
= 360 kg/min = 6.0 kg/s
F = dM/dt vrel = 6.0 kg/s * 39.0 m/s
=234 kg．m/s2 =234 N
75
Example : Fighting a Fire
Two firefighters must apply a total force of 600 N to steady a
hose that is discharging water at the rate of 3600 L/min.
Estimate the speed of the water as it exits the nolzze.
dV/dt = 3600 L/min = 3.60 106cm3/60s = 6.0104 cm3/s
dM/dt = 1g/cm3 * 6.0104 cm3/s = 6.0104 g/s = 60 kg/s
Thrust = 600 N = dM/dt vrel = 60*vrel
vrel = 10 m/s
76
Example:Rocket
Let’s assume that we have a rocket that has a mass of 5000
kg without fuel. When the rocket is completely fueled, it has
a mass of 25,000 kg. The exhaust of the rocket has a speed of
5 km/s. If the rocket starts from rest and burns all of its fuel,
what is its final speed?
vf = vi + vrel ln(M1/M2) = 0+5000 ln(25000/5000)
=8000 m/s
77
Example : A Rocket in Space
A rocket in free space has a speed of 3.0103m/s relative to
the Earth. Its engines are turned on, and fuel is ejected in a
direction opposite the rocket’s motion at a speed of
5.0103m/s relative to the rocket. (a) What is the speed of
the rocket relative to the Earth once its mass is reduced to one
half its mass before ignition? (b) What is the thrust on the
rocket if it burns fuel at the rate of 50 kg/s?
(a) v2 = v1 + vrel ln(M1/M2) = 3000+5000 ln(Mi/(½ Mi))
= 6500 m/s
(b) Thrust = dM/dt vrel = 50 *5000 = 2.50105N
78
Example : Rocket Thrust
A rocket whose initial mass M1 is 850 kg consumes fuel at the rate R
=2.3 kg/s. The speed vrel of the exhaust gases relative to the rocket
engine is 2800 m/s. (a) What thrust does the rocket engine provide? (b)
What is the initial acceleration of the rocket launched from a
spacecraft? (c) Suppose that the mass M2 of the rocket when its fuel is
exhausted is 180 kg. What is its speed relative to the spacecraft at that
time? Assume that the spacecraft is so massive that the launch does not
alter its speed.
(a) Thrust = dM/dt vrel = 2.3 *2800 = 6440N
(b) a = Thrust / M1 =6440 N / 850 kg = 7.6 m/s2
(c) v2 = v1 + vrel ln(M1/M2) = 2800 ln(850/180)
= (2800 m/s) ln(4.72) = 4300 m/s
79
Physics
Chapter 9
Newton’s Theory of Gravity
授課老師施坤龍
0
Newton’s Law of Gravitation
A large amount of data had been collected by 1687.

Isaac Newton provided “Newton’s Law of Gravitation”.
If the force of gravity is being exerted on objects on Earth,
what is the origin of that force?
Newton’s realization was that the
force must come from the Earth.
He further realized that this force must be what keeps the
Moon in its orbit.
1
Newton’s Law of Gravitation
Every particle in the Universe attracts every other
particle with a force that is directly proportional
to the product of their masses and inversely
proportional to the square of the distance between
them

Gm1m2
F12  
r̂12
2
r12
G is the universal gravitational constant and
equals 6.673×10-11N·m2/kg2
2
Newton’s Law of Gravitation
In 1789 Henry Cavendish measured G.
The two small spheres are fixed at the ends of a light
horizontal rod.
Two large masses were placed near the small ones.
The angle of rotation was measured by the deflection of a
light beam reflected from a mirror attached to the vertical
suspension.
3
Newton’s Law of Gravitation
The magnitude of the force acting on an object
of mass m in freefall near the Earth’s surface
is mg. This can be set equal to the force of
universal gravitation acting on the object.
M Em
ME
mg  G 2  g  G 2
RE
RE
If an object is some distance h above the
Earth’s surface, r becomes RE + h.
ME
g G
( RE  h) 2
This shows that g decreases with increasing
altitude.
As r → , the weight of the object approaches
zero.
高度 h (公
里, km)
重力加速度
g(m/s2)
1000
7.33
2000
5.68
3000
4.53
4000
3.70
5000
3.08
6000
2.60
7000
2.23
8000
1.93
9000
1.69
10000
1.49
50000
0.13
∞
0
4
Gravitational Field
Use the mental representation of a field

A source mass creates a gravitational field throughout the space
around it

A test mass located in the field experiences a gravitational force
The gravitational
 field is defined as:
 Fg
GM E
g
  2 rˆ12
m
r

At the surface of the earth, r = RE and g = 9.80 m/s2
5
Example ：Can you attract another person gravitationally?
A 50-kg person and a 70-kg person are sitting on a bench
close to each other. Estimate the magnitude of the
gravitational force each exerts on the other.
The distance between those two person is 0.5m
The magnitude of the gravitational force F
F = Gm1m2/r2 =6.67 × 10-11*50*70/(0.5)2
= 9.34 × 10-7N ~ 10-6N
6
Example ： Force on the Moon
Find the net force on the Moon (mM=7.35×1022 kg) due to the
gravitational attraction of both the Earth (mE=5.98×1024 kg)
and the Sun (mS=1.99×1030 kg), assuming they are at right
angles to each other.
RME= 3.84×108 m; RMS= RES= 1.496×1011 m;
FME= GmMmE/RME2 =1.99×1022 i N
FMS= GmMmS/RMS2 =4.34×1020 j N
FM = (1992+4.342)½ ×1020 N =2.0×1022 N
θ = tan-1(199/4.34) =88.75o
7
Example ： The net force of Three particles
Three point particles with mass m1=4kg, m2=2kg, and
m3=3kg are at the corners of an equilateral triangle of side
L=2m, as figure. Find the net force on m2.
F21 = Gm2m1/L2 = 1.33×10-10N
F23 = Gm2m3/L2 = 1.01×10-10N
F2x = -F21 cos60o + F23 cos60o = -1.6×10-11N
F2x = -F21 sin60o - F23 sin60o = -2.03×10-10N
F2 = -(1.6 i + 20.3 j ) ×10-11N
8
Example : An Earth Satellite
A satellite of mass m moves in a circular orbit about the Earth
with a constant speed v and at a height of h = 1 000 km above
the Earth’s surface as in figure. Find the orbital speed of the
satellite?
Fg = GmmE/r2 = mv2/r
v = (GmE/r)½
= [6.67×10-11*5.98 ×1024/(7.38×106)]½
= 7360 m/s
9
Example ：Gravity for a uniform rod
Find the field strength at a point along the axis of a thin rod
of length L and mass m, at a distance d from one end.
λ = m/L; dm = λ dx ; dg = Gdm/x2
g=∫
Gdm/x2
L+d
= ∫ d Gm/(Lx2)dx
= (Gm/L) [1/d–1/(L+d)]
= Gm /[d·(L+d)]
d >> L, g －> Gm /d2
10
Example ：Gravity for a semicircular ring
Find the field strength at the center of a thin semicircular ring
of radius R and mass m, as shown in figure. The linear mass
density is λ kg/m.
λ = m/(πR); dm = λ Rdθ ; dgy = Gdm/R2*sinθ
p
gy = ∫ dgy = ∫ 0 G λ Rsinθdθ/R2
p
= Gλ/R∫ 0 sinθdθ
= 2Gλ/R
= 2Gm/[πR2]
11
Example ：Gravity for a uniform sphere
Prove a uniform spherical mass distribution attracts an external point
particle as if all mass were concentrated at its center.
σ = M/(4πR2);
dM= σ(2πRsinϕ)(Rdϕ)= σ 2πR2sinϕdϕ
s2=R2+r2–2Rrcosϕ; 2sds =2Rrsinϕdϕ
dM/s = (2πσR/r)ds ; cosθ =(s2+r2-R2)/(2sr)
dFx = -GmdMcosθ/s2 = - πσGmR(s2+r2-R2)/(sr)2ds
Fx =
∫-GmdMcosθ/s2
=
-πσGmR/r2∫
r+R
r-R
[1+(r2-R2)/s2]ds
= -πσGmR/r2 [2R+(r2-R2)(-1/(r+R)+1/(r-R) ]
= -4πσGmR2/r2 = -GmM/r2
12
Example ：Gravity inside a sphere
How does the field strength vary inside a uniform solid
sphere of density ρ kg/m3 and radius R?
在半徑r外的球殼質量對此點的重力互相抵消
ρ = 3M/(4πR3);
M(r) = ρ 4π r3/3
g(r) = GM(r)/r2 = 4πGρr/3
= GMr/R3
13
Johannes Kepler
1571 – 1630
German astronomer
Best known for developing laws
of planetary motion

Based on the observations of
Tycho Brahe
14
Kepler’s Laws
Kepler’s First Law

Each planet in the Solar System moves in an elliptical
orbit with the Sun at one focus
Kepler’s Second Law

The radius vector drawn from the Sun to a planet sweeps
out equal areas in equal time intervals
Kepler’s Third Law

The square of the orbital period of any planet is
proportional to the cube of the semimajor axis of the
elliptical orbit
15
Kepler’s Laws of Planetary Motions
Kepler’s laws describe planetary motion.
1. The orbit of each planet is an ellipse, with the Sun
at one focus.
16
Notes About Ellipses
F1 and F2 are each a focus of the ellipse.

They are located a distance c from the center.

The sum of r1 and r2 remains constant.
The longest distance through the center is the major axis.
a is the semi-major axis.

The shortest distance through the center is the minor axis.
b is the semi-minor axis.

r1
r2
17
Notes About Ellipses
The eccentricity of the ellipse is defined as e = c /a.

For a circle, e = 0

The range of values of the eccentricity for ellipses is 0<e<1.

The higher the value of e, the longer and thinner the ellipse.
The Sun is at one focus.

It is not at the center of the ellipse.

Nothing is located at the other focus.
r1
r2
Aphelion is the point farthest away from the Sun.

The distance for aphelion is a + c.

For an orbit around the Earth, this point is called the apogee.
18
Notes About Ellipses
Perihelion is the point nearest the Sun.

r1
r2
The distance for perihelion is a – c.

For an orbit around the Earth, this point is called the perigee.
A circular orbit is a special case of the general elliptical orbits, and it is a
direct result of the inverse square nature of the gravitational force.
Elliptical (and circular) orbits are allowed for bound objects.

A bound object repeatedly orbits the center.

An unbound object would pass by and not return.

These objects could have paths that are parabolas (e = 1) and
hyperbolas (e > 1).
19
Orbital Eccentricity Examples
Mercury’s orbit has the highest eccentricity of any planet,
eMercury = 0.21.
Comet Halley’s orbit has a much higher eccentricity.

e = 0.97.
20
Kepler’s Laws of Planetary Motions
2. An imaginary line drawn from each planet to the
Sun sweeps out equal areas in equal times.
21
Kepler’s Second Law
It is a consequence of conservation of angular momentum for
an isolated system. Consider the planet as the system.
Model the Sun as massive enough compared to the planet’s
mass that it is stationary. The gravitational force exerted by the
Sun on the planet is a central force. The force produces no
torque, so angular momentum is a constant.
  
 
L  r  P  M p r  v  constant
22
Kepler’s Second Law
Geometrically, in a time dt, the radius vector r sweeps out the area dA,
which is half the area of the parallelogram .
1   1  
dA  r  dr  r  v dt
2
2
Its displacement is given by
 
dr  v dt
Mathematically, we can say
dA
L

 constant
dt 2 M P
The radius vector from the Sun to any planet sweeps out equal areas in
equal times.
The law applies to any central force, whether inverse-square or not.
23
Kepler’s Laws of Planetary Motions
3. The square of a planet’s orbital period is
proportional to the cube of its mean distance from
the Sun.
v
GM
4p r
ac 
 2 
2
r
r
T
2pr
v
T
2

4p  3
2
r
T  
 GM 
2
2
24
Kepler’s Laws of Planetary Motions
Planet
Mercury
Radius Mass
(106m) (1024kg)
2.44
0.330
Escape r-orbit Eccent T-orbit
v (km/s) (109 m)
(years)
4.3
57.9
0.205
0.241
T2/r3
(10-19s2/m3)
2.97
Venus
6.05
4.87
10.4
108.2
0.007
0.615
2.99
Earth
6.38
5.97
11.2
149.6
0.017
1
2.97
Mars
3.40
0.642
5.0
227.9
0.094
1.88
2.98
Jupiter
71.5
1,890
59.5
778.6
0.049
11.9
2.97
Saturn
60.3
568
35.5
1433
0.057
29.4
2.99
Uranus
25.6
86.8
21.3
2872
0.046
83.8
2.95
Neptune
24.8
102
23.5
4495
0.009
164
2.99
Pluto*
1.20
0.012
1.1
5870
0.249
248
2.96
Sun
696
1,990,000
618
-
-
-
25
Kepler’s Third Law
Can be predicted from the inverse square law
Start by assuming a circular orbit.
The gravitational force supplies a centripetal force.
Ks is a constant
GM Sun M Planet M Planet 4p 2 r
FP 

2
r
T2
2

4p  3
2
3


T 
r  KS r

 GM Sun 
26
Kepler’s Third Law
This can be extended to an elliptical orbit.
Replace r with a.

Remember a is the semi-major axis.

2

 3
4
p
2

a  K Sun a 3
T 
 GM Sun 
Ks is independent of the mass of the planet, and so is valid for
any planet.
If an object is orbiting another object, the value of K will
depend on the object being orbited.
For example, for the Moon around the Earth, KSun is replaced
with Kearth.
27
Example, Mass of the Sun
Using the distance between the Earth and the Sun, and the
period of the Earth’s orbit, Kepler’s Third Law can be used to
find the mass of the Sun.
4p r

2
GT
2 3
M Sun
Similarly, the mass of any object being orbited can be found if
you know information about objects orbiting it.
Irregularities in planetary motion led to the discovery of
Neptune, and irregularities in stellar motion have led to the
discovery of many planets outside our solar system.
28
Conceptual Example : Catching a satellite
You are an astronaut in the space shuttle pursuing a
satellite in need of repair. You find yourself in a
circular orbit of the same radius as the satellite, but
30 km behind it. How will you catch up with it?
You have to drop into a lower orbit to speed up; when
you get ahead of the satellite you need to slow down
and get back into the higher orbit.
29
Example, Geosynchronous Satellite
A geosynchronous satellite appears
to remain over the same point on
the Earth.
The gravitational force supplies a
centripetal force.
Consider the satellite as a particle
under a net force and a particle in
uniform circular motion.
You can find h or v.
30
Satellite Orbits
View from above
North Pole, June 23
2004
Low orbit satellites:
Hubble, space station,
Iridium system, spy
sat.s
GPS, research sat.s
Geostationary
communication sat.s
31
Example : The Sun’s mass determined
Determine the mass of the Sun given the Earth’s
distance from the Sun as rES = 1.5×1011 m.
TE = 365*86400 = 3.1536×107 s
MS = 4π2/G*(RES3/TE2)
= 4π2/(6.67×10-11)*(1.5×1011)3/(3.16×107)2
= 2.00×1030 kg
32
Example :Where is Mars?
Mars’ period (its “year”) was first noted by Kepler to be
about 687 days (Earth-days), which is (687 d/365 d) = 1.88 yr
(Earth years). Determine the mean distance of Mars from the
Sun using the Earth as a reference.
(RMS/RES)3 =(TM/TE)2
RMS = RES (TM/TE)2/3 = 1.496×1011× (1.88) 2/3
= 2.28×1011 m
33
Example : Comet Halley
Comet Halley moves in an elliptical orbit around the sun. Its
distance from the sun at the perihelion and aphelion are
8.75×107 km and 5.26×109 km, respectively. Find the
orbitical semi-major axis, eccentricity, and period.
a = (8.75×1010 + 5.26×1012)/2 = 2.67×1012m
rp = a –c= a –ae = a(1-e)
e = 1 - rp/a = 1 - 8.75×1010/ 2.67×1012 = 0.967
T = 2p [a3/(GmS)]½
= 2p [(2.67×1012)3/(6.67×10-11*1.99 ×1030)]½
= 2.38 ×109 s = 75.5 years
34
Example : Geosynchronous satellite
A geosynchronous satellite is one that stays above the same point on the
Earth, which is possible only if it is above a point on the equator. Such
satellites are used for TV and radio transmission, for weather
forecasting, and as communication relays. Determine (a) the height
above the Earth’s surface such a satellite must orbit, and (b) such a
satellite’s speed. (c) Compare to the speed of a satellite orbiting 200 km
above Earth’s surface.
(a) RSa = [GMET2/(4π2)]1/3 =4.22×107m;
h=RSa-RE=3.6×107m
(b) VSa= 2πRSa /T = 2π×4.22×107/86400 = 3070 m/s
(c) v = (GME/r)½ =(6.67×10-11×5.98×1024/(6.58×106))½
= 7786 m/s
35
Example : Lagrange point
Joseph-Louis Lagrange discovered five special points in the vicinity of
the Earth’s orbit about the Sun where a small satellite (mass m) can
orbit the Sun with the same period T as Earth’s (= 1 year). One of these
“Lagrange Points,” called L1, lies between the Earth (mass mE) and Sun
(mass mS), on the line connecting them. That is, the Earth and the
satellite are always separated by a distance d. If the Earth’s orbital
radius is RES, then the satellite’s orbital radius is (RES-d). Determine d.
T2 =RES3 4π2/(GMS)
FSa = GMSm/(RES-d)2- GMEm/d2
= m4π2(RES-d)/T2 = GMSm(RES-d)/RES3
MS/(RES-d)2 - MS(RES-d)/RES3 = ME/d2
d ~ 10-2 RES= 1.5×109 m
36
Gravitational Potential Energy
Near the Earth’s surface, the gravitational potential energy
function was U = mgy for a particle-Earth system.

This was valid only when the particle is near the Earth’s surface
The gravitational force is conservative.
The change in gravitational potential energy of a system is
defined as the negative of the internal work done by the
gravitational force on that member during the displacement.
 
rf
U  U f  U i    F  ds    F (r )dr
ri
U f  Ui  
rf
ri
 GM E m
1 1
dr  GM E m(  )
2
r
ri rf
37
Gravitational Potential Energy
As a particle moves from A to B, ∆U.
GM E m GM E m
U f Ui  (

)
ri
rf
Choose the zero for the gravitational potential energy at ri = 


This means Ui = 0 where ri = 
U(r) ≡ -GMEm/r

This is valid only for r ≥ RE

It is not valid for r < RE.

U is negative because of the

choice of Ui.
38
Gravitational Potential Energy
Graph of the gravitational potential energy U
versus r for an object above the Earth’s surface.
The potential energy goes to zero as r approaches
infinity.
The absolute value of the potential energy can be thought of as the
binding energy.
If an external agent applies a force larger than the binding energy, the
excess energy will be in the form of kinetic energy of the particles when
they are at infinite separation.
39
Systems with Three or More Particles
The total gravitational potential energy of the system is the
sum over all pairs of particles.
Each pair of particles contributes a term of U.
Assuming three particles:
m3
U total  U12  U13  U 23
 m1m2 m1m3 m2m3 

 G 


r13
r23 
 r12
r13
m1
r23
r12
m2
The absolute value of Utotal represents the work needed to
separate the particles by an infinite distance.
40
Energy and Satellite Motion
Assume an object of mass m moving with a speed v in the
vicinity of a massive object of mass M.

M >> m
Also assume M is at rest in an inertial reference frame.
The total energy is the sum of the system’s kinetic and
potential energies.
Total energy E = K +U
1 2 GMm
E  mv 
2
r
In a bound system, E is necessarily less than 0.
41
Energy in a Circular Orbit
An object of mass m is moving in a circular
orbit about M.
The gravitational force supplies a centripetal
force.
GMm GMm
GMm
E


2r
r
2r
The total mechanical energy is negative in the case of a
circular orbit. The kinetic energy is positive and is equal to
half the absolute value of the potential energy.
The absolute value of E is equal to the binding energy of the
system.
42
Energy in an Elliptical Orbit
For an elliptical orbit, the radius is replaced
by the semi-major axis.
GMm
E
2a
The total mechanical energy is negative.
The total energy is constant if the system is
isolated.
Both the total energy and the total angular
momentum of a gravitationally bound, twoobject systems are constants of the motion.
43
Escape Speed from Earth
An object of mass m is projected upward from the Earth’s
surface with an initial speed, vi.
Use energy considerations to find the minimum value of the
initial speed needed to allow the object to move infinitely
far away from the Earth.
This minimum speed is called the escape speed.
vesc
2GM E

RE
Note, vesc is independent of the mass of the object. The result is
independent of the direction of the velocity and ignores air resistance.
44
Escape Speed
The Earth’s result can be
extended to any planet.
vesc
2GM

R
The table at right gives
some escape speeds from
various objects.
Planet
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto*
Moon
Sun
Escape Speed (km/s)
4.3
10.4
11.2
5.0
59.5
35.5
21.3
23.5
1.1
2.3
618
45
Escape Speed
Complete escape from an object is not really possible.

The gravitational field is infinite and so some gravitational
force will always be felt no matter how far away you can
get.
This explains why some planets have atmospheres and others
do not.

Lighter molecules have higher average speeds and are more
likely to reach escape speeds.
46
Example ：Gravity for a uniform sphere
Prove a uniform spherical mass distribution attracts an external point
particle as if all mass were concentrated at its center.
σ = M/(4πR2);
dM= σ(2πRsinϕ)(Rdϕ)=2πσR2sinϕdϕ
s2 = R2 + r2 – 2Rrcosϕ; sds = Rr sinϕdϕ
dM/s = (2πσR/r)ds
U = ∫ -GmdM/s = -Gm2πσR/r
r+R
∫ r–R
ds
= -Gm2πσR/r [(r+R) –(r–R) ]
= -GmM/r
Fr = -dU/dr = -GmM/r2
47
Orbits
As a spacecraft fires its engines, the exhausted fuel
can be seen as doing work on the spacecraft-Earth
orbit Therefore, the system will have a higher
energy. The spacecraft cannot be in a higher
circular orbit, so it must have an elliptical orbit.
Larger amounts of energy will move the spacecraft
into orbits with larger semimajor axes
If the energy becomes positive, the spacecraft will
escape from the earth. It will go into a hyperbolic
path that will not bring it back to the earth.
48
Orbits
The spacecraft in orbit around
the earth can be considered to be
in a circular orbit around the sun
Small perturbations will occur

These
correspond
to
its
motion around the earth

These are small compared
with the radius of the orbit
49
Example : A Spacecraft in an Circular Orbit
Imagine that you are in a spacecraft in circular orbit around
the Earth, at a height of 300 km from the surface. You fire
your rocket engines, and as a result the magnitude of the
mechanical energy of the spacecraft-Earth system decreases
by 10.0%. What is the greatest height of your spacecraft
above the surface of the Earth in your new orbit?
Ei = - ½ mGmE/(RE+h) = - ½ mGmE/(6.68×106 )
Ef = -½ mGmE/(6.68×106)*0.9= -½ mGmE/(7.42×106)
h max= 2*7.42×106 -6.68×106-6.38×106 = 1.78×106m
= 1.78×103 km
50
Example : A Satellite in an Circular Orbit
A 1000 kg satellite moves in a circular orbit 300 km above the Earth surface. (a)What
is its speed, period, and radial acceleration? (b)How much work must be done to the
satellite to put it orbit? (c) How much additional work would have to be done to make
satellite escape the Earth?
F = mGmE/r2 =mV2/r = m4p2r/T2 = m ac ; r = rE+ h= 6.68 ×106 m
(a) V = (GmE/r)½ =(6.67 ×10-11*5.98 ×1024/ 6.68 ×106 )½ = 7720 m/s
T = [4p2r3/(GmE)]½ = 2p r/V = 2p *6.68 ×106/7720= 5440 s = 90.6 min
ac = V2/r = 77202/(6.68 ×106 ) = 8.92 m/s2
(b)Ef = - GmmE/(2r) = -6.67 ×10-11*1000*5.98 ×1024/ 1.336 ×107 = -2.98 ×10-10 J
Ei = - GmmE/rE = -6.67 ×10-11*1000*5.98 ×1024/6.38×106 = -6.24 ×10-10 J
W =Ef - Ei = -2.98 ×10-10 - (-6.24 ×10-10 ) = 3.26×10-10 J
(c) Ei = -2.98 ×10-10 J ; Ef = 0 J
W =Ef - Ei = 0 - (-2.98 ×10-10 ) = 2.98×10-10 J
51
Example : A Satellite in an Elliptical Orbit
A satellite moves in an elliptical orbit about the Earth as the Figure. The
minimum and maximum distances from the surface of the Earth are 400
km and 3 000 km, respectively. Find the speeds of the satellite at
apogee and perigee.
Lp = rpmvp = ramva= La; rpvp = rava
rp=6380 +400=6780; ra=6380 +3000=9380;
vp /va= ra/rp= 1.38; a = ½ (rp+ra) = 8080
Ep=-mGmE/rp+½ mvp2 =-mGmE/ra+½ mva2=Ea=-mGmE/(2a)
vp=[GmE(2/rp-1/a)]½ = 8.27×103 m/s;
va =[GmE(2/ra-1/a)]½ =5.98×103 m/s
52
Black Holes
A black hole is the remains of a star that has collapsed under
its own gravitational force.

The core of the star must have a mass greater than 3
solar masses.
The escape speed for a black hole is very large due to the
concentration of a large mass into a sphere of very small
radius.

If the escape speed exceeds the speed of light, c, radiation
cannot escape and it appears black.
53
Black Holes
The critical radius at which the escape speed equals c is called the
Schwarzschild radius, RS = 2GM/c2 .
The imaginary surface of a sphere with this radius is called the event
horizon.

This is the limit of how close you can approach the black hole and
still escape.
54
Black Holes
There is evidence that supermassive black holes exist at the
centers of galaxies. These have masses much higher than the
mass of the Sun. For example, there is strong evidence of a
supermassive black hole at the center of the Milky Way that
has a mass of 2 – 3 million solar masses.
Although light from a black hole cannot escape, light from
events taking place near the black hole should be visible
If a binary star system has a black hole and a normal star, the
material from the normal star can be pulled into the black hole.
55
Black Holes
When a black hole feeds, it emits a lot of light: from the accreting disk
of material that swirls around its mid-plane, from the hot corona of gas
surrounding the black hole, and sometimes from fast-moving jets
emitted from its poles. These different components all have different
temperatures, which means their light peaks at different wavelengths.
This means that we can use the shape of a black hole’s spectrum to
learn about the physical geometry of an accreting black hole system.
56
Gravitational Waves
Gravitational waves are disturbances in the curvature
of spacetime, generated by accelerated masses, that propagate
as waves outward from their source at the speed of light.
They were proposed by Henri Poincaré in 1905 and
subsequently predicted in 1916 by Albert Einstein on the
basis of his general theory of relativity.
The first indirect evidence for the existence of gravitational
waves came from the observed orbital decay of the Hulse–
Taylor binary pulsar, which matched the decay predicted by
general relativity as energy is lost to gravitational radiation.
57
Gravitational Waves
In 1993, Russell A. Hulse and Joseph Hooton Taylor
Jr. received the Nobel Prize in Physics for this discovery.
The first direct observation of gravitational waves was not
made until 2015, when a signal generated by the merger of
two black holes was received by the LIGO gravitational wave
detectors in Livingston and in Hanford. The 2017 Nobel
Prize in Physics was subsequently awarded to Rainer
Weiss, Kip Thorne and Barry Barish for their role in the
direct detection of gravitational waves.
58
Physics
Chapter 10
Rotational Motion
授課老師施坤龍
0
Rigid Object
A rigid object is one that is non-deformable

The relative locations of all particles making up
the object remain constant

All real objects are deformable to some extent, but
the rigid object model is very useful in many
situations where the deformation is negligible
1
Angular Position
Axis of rotation is the center of the disc
Choose a fixed reference line.
P(r, θ)
Point P is at a fixed distance r from the origin.

A small element of the disc can be modeled as a particle at P.
Polar coordinates are convenient to use to represent the position of P.
P is located at (r, θ) where r is the distance from the origin to P and θ is
the measured counterclockwise from the reference line.
As the particle moves, the only coordinate that changes is θ
As the particle moves through θ, it moves though an arc length s.
The arc length and r are related:


θ = s/r
360o

1rad 
 57.3o ;  (rad) 
 (deg)
o
2π rad
180
2
Angular Displacement and Angular velocity
Angular displacement θ : θ = θ2 - θ1
The average angular velocity ωavg is defined as the total angular
displacement divided by time:
ωavg = θ/t = (θ2- θ1)/(t2-t1)
The instantaneous angular velocity:
 d 2
  lim


 2f
t 0
t
dt
T
Units of angular speed are radians/sec

rad/s or s-1 since radians have no dimensions
Angular speed will be positive if θ is increasing (counterclockwise)
Angular speed will be negative if θ is decreasing (clockwise)
3
Direction of Angular velocity
The angular velocity vector points along the axis of rotation,
with the direction given by the right-hand rule. If the
direction of the rotation axis does not change, the angular
acceleration vector points along it as well.
4
Angular Acceleration
The average angular acceleration is the rate at which the
angular velocity changes with time:
 
 2  1
t


t
The instantaneous angular acceleration:
 d
  lim

dt
t 0 t
Units of angular acceleration are rad/s² or s-2 since radians
have no dimensions.
5
Rotational Kinematic Equations
The kinematic expression for the rigid object under constant
angular acceleration are of the same mathematical form as
those for a particle under constant acceleration.
Substitutions from translational to rotational are
all with constant angular accelation 
 x → θ
 f   i  t
 v → ω
1 2
 f   i   i t  t
 a → α
2
 2f   i2  2 ( f   i )
1
 f   i  ( i   f )t
2
6
Comparison Between Rotational and Linear Equations
Rotational and Linear Equations
constant angular acceleration
constant acceleration
α = constant
a = constant
ωf = ωi + αt
vf = vi + at
θf = θi + ωi t + ½ αt2
xf = xi + vi t + ½ at2
θf =θi + ½ (ωi + ωf)t
xf = xi + ½ ( vi + vf )t
ωf2 = ωi2 + 2α(θf - θi)
vf2 = vi2 + 2a(xf - xi)
7
Relationship between linear and angular quantities
The distance between any point P on a rigid body with the
center of rotation O is R. The point P is at the position
of (R, 0) at t=0. The displacement s of the point P with an
angular displacement θ, then s = Rθ.
Linear velocity V of point P with an angular velocity ω,
V = ds/dt = Rdθ/dt = Rω.
Tangent acceleration at of point P with an angular acceleration α,
at = dV/dt = Rα
Normal acceleration ar of point P with an angular velocity ω,
ar = Rω2
The total linear acceleration a of point P, a = (at2 + ar2)½ = R(α2 + ω4)½
8
Conceptual Example : Is the lion faster than the horse?
On a rotating carousel or merry-go-round, one child sits on a
horse near the outer edge and another child sits on a lion
halfway out from the center. (a) Which child has the greater
linear velocity? (b) Which child has the greater angular
velocity?
(a) The child sits on a horse
(b) The angular velocities are the same.
9
Example : Birds of prey—in radians
A particular bird’s eye can just distinguish objects that
subtend an angle no smaller than about 3×10-4 rad. (a) How
many degrees is this? (b) How small an object can the bird
just distinguish when flying at a height of 100 m?
(a) θ = 3×10-4rad *57.3o/rad = 0.017o
(b)ℓ= Rθ = 100*3×10-4
=0.03m = 3cm
10
Example : Rotating Wheel
A wheel rotates with a constant angular acceleration of 3.50
rad/s2. (a) If the angular speed of the wheel is 2rad/s at t=0,
through what angle does the wheel rotate between t=0 and t =
2s? (b) What is the angular speed of the wheel at t = 2.0 s?
(a) ∆θ = θf - θi = ωi t + ½ αt2
= 2*2+ ½ *3.5*4 = 11 rad
(b) ωf = ωi + αt
= 2 +3.5*2 = 9.0 rad/s
11
Example : Hard drive
The platter of the hard drive of a computer rotates at 7200
rpm . (a) What is the angular velocity of the platter? (b) If the
reading head of the drive is located 3.00 cm from the rotation
axis, what is the linear speed of the point on the platter? (c) If
a single bit requires 0.50 μm of length along the direction of
motion, how many bits per second can the writing head write
when it is 3.00 cm from the axis?
(a) ω = 7200rpm =120rps = 120*2π rad/s =754 rad/s
(b) V = rω = 0.03*754 = 22.62 m/s
(c) N= vt /ℓ= 22.62*1/(5×10-7) = 4.5×107 = 45M bits
12
Example : Rotating Disk
The rotational position (t) of a reference line on the disk is given by θ(t)
=-(1.00 rad)-(0.600 rad/s)t +(0.250 rad/s2)t2. (a) Graph the rotational
position of the disk versus time from t =-3.0 s to t=6.0 s (b) At what
time t min does (t) reach the minimum value shown in Fig.? What is
that minimum value? (c) Graph the rotational velocity of the disk
versus time from t =-3.0 s to t =6.0 s.
(a) θ(t) = -1.0 - 0.6 t + 0.25 t2
(b) ω = dθ/dt = -0.6 +0.5t = 0; t =1.2s θ(t)有最小值
θ(1.2) = -1-0.72 +0.36 = -1.36 rad = -77.9o
(c) ω(t) = -0.6 +0.5t
13
Example : Given ω as function of time
A disk of radius R = 3.0 m rotates at an angular velocity ω =
(1.6 + 1.2t) rad/s, where t is in seconds. At the instant t = 2.0
s, determine (a) the angular acceleration, and (b) the speed v
and the components of the acceleration a of a point on the
edge of the disk.
(a) α = dω/dt =1.2 rad/s2
(b) V = Rωf = 3.0*4.0 = 12.0 m/s
at = Rα = 3.0*1.2 = 3.6 m/s2
ar = Rω2 = 3.0*42 = 48.0 m/s2
14
Example : A carousel
A carousel is initially at rest. At t = 0 it is given a constant angular
acceleration α = 0.060 rad/s2, which increases its angular velocity for
8.0 s. At t = 8.0 s, determine the magnitude of the following quantities:
(a) the angular velocity of the carousel; (b) the linear velocity of a child
located 2.5 m from the center; (c) the tangential (linear) acceleration of
that child; (d) the centripetal acceleration of the child; and (e) the total
linear acceleration of the child.
(a) ω = 0.060 rad/s2 * 8.0 = 0.48 rad/s
(b) v = Rω = 2.5*0.48 = 1.2 m/s
(c) at = Rα = 2.5*0.06 = 0.15 m/s2
(d) ar = Rω2 = 2.5*0.482 = 0.576 m/s2
(e) a =(at 2+ar2)½ =0.595 m/s2 ； θ =tan-1(0.15/0.576)=15o
15
Example： A flywheel (1)
A flywheel of radius 20 cm starts from rest, and has a
constant angular acceleration of 60 rad/s2. Find: (a) the
magnitude of net linear acceleration of a point on the rim
after 0.15s; (b) the number of revolutions completed in 0.25s.
(a) at=αr=12 m/s2,
ar=ω2 r=16.2 m/s2
Net linear acceleration
a = (at2 + ar2)½ = 20.2 m/s2
(b) θ = ½αt2 = 1.88 rad = 0.3 rev
16
Example : A flywheel (2)
The angular position θ of a 0.36 m diameter flywheel is given by θ(t) = (2.0 rad/s2) t3.
(a) Find θ in radians and in degrees, at t =2.0 s and t=5.0 s. (b) Find the distance that a
particle on the wheel rim moves over the time interval from t =2.0 s to t=5.0 s (c) Find
the angular velocity, in rad/s and rev/min, over that time interval. (d) Find the
instantaneous angular velocity at t =2.0 s and t=5.0 s. (e) Find the average angular
acceleration between t =2.0 s and t=5.0 s. (f) Find the instantaneous angular
acceleration at t =2.0 s and t=5.0 s.
(a) θ(2.0s) =16.0 rad = 16.0 rad * 360o/(2 rad) = 916.73o
θ(5.0s) =250.0 rad = 250.0 rad * 360o/(2 rad) = 14323.94o
(b) s = r Δθ= 0.18*(250.0-16.0) = 42.12 m
(c) ωave = Δθ/Δt = 234/3.0 = 78 rad/s
= 78 rad/s *1.0 rev/(2 rad)*60s/1.0 min = 744.85 rev/min
(d) ω(t) = dθ/dt = 6 t2; ω(2.0s) = 24.0 rad/s ; ω(5.0s) = 150.0 rad/s
(e) αave = Δ/Δt = (150.0-24.0)/3.0 = 42.0 rad/s2
(f) α(t) = d/dt = 12 t; α(2.0s) = 24.0 rad/s2; α(5.0s) = 60.0 rad/s2
17
Rotational Kinetic Energy
An object rotating about some axis with an angular speed, , has
rotational kinetic energy even though it may not have any translational
kinetic energy. Each particle has a kinetic energy: Ki = 1/2 mivi2
Since the tangential velocity depends on the distance, ri, from the axis
of rotation, we can substitute vi = ω ri
Now let’s discuss the kinetic energy of several point particles
n
1 n
1 n
2
K   K i   mi vi   mi ri 2 2
2 i 1
2 i 1
i 1
If we assume that these particles keep their distances fixed with respect
to each other (solid object, all moving with the same angular velocity)
we can write
n
1 n
1
1
K   mi ri 2 2  ( mi ri 2 ) 2  I 2
2 i 1
2 i 1
2
Where I is the moment of inertia given by
n
1
1
I   mi ri 2 ; compare K linear  mv 2  K circular  I 2
2
2
i 1
18
Rotational Kinetic Energy, Moments of Inertia
The dimensions of moment of inertia are ML2 and its SI units are kg.m2
We can calculate the moment of inertia of an object more easily by
assuming it is divided into many small volume elements, each of mass
n
∆mi I  lim  ri 2 m
i 1
For a continuous rigid object, imagine the object to be divided into
many small elements, each having a mass of Δmi.
We can rewrite the expression for I in terms of ∆m.
n
I  lim  ri 2 m   r 2 dm
m 0 i 1
With the small volume segment assumption,
I   r 2 dV
If r is constant, the integral can be evaluated with known geometry,
otherwise its variation with position must be known.
19
Rotational Kinetic Energy, Moments of Inertia
A cylinder, a disk, and a ring with same mass. The moments
of inertia about the central axis depend on how the mass is
distributed relative to the axis:
IC >IB>IA
20
Parallel-Axis Theorem
Parallel-Axis Theorem: I = ICM + MD 2

I is the moment of inertia of any non-centroid axis of
rotation

ICM is the moment of inertia of the center of mass

D is the distance between the above two parallel axes
I = Σmiri2 = Σmi|D+ ri’|2
=ΣmiD2+2D‧Σmiri’ + Σmi|ri’|2
=MD 2 +2D‧0+ ICM
(O’為質心)
=MD 2 + ICM
21
Perpendicular-Axis Theorem
Perpendicular-Axis Theorem: Iz = Ix + Iy

The object is a flat plate

The z-axis is the coordinate axis perpendicular to the plate

The x-axis and the y-axis are a pair of vertical coordinate
axes on the plate
For the mass of any mi on the plate:
dIx = miyi2; dIy = mixi2 ;
dIz = miri2 = mi(xi2+yi2) = dIx + dIy
Iz = Ix + Iy
22
Example： Moments of inertia for molecules diatomic
Many molecules diatomic, dumbbell-like structure. Let us find the
moments of inertia about four axes. We treat as points particles with
m1=3kg and m2=5kg. Take d1=1m and d2=2m.
Axis A : IA = m1d12 + m2d22 = 3*1 +5*4 = 23 kg m2
Axis B : IB = m102 + m2(d1 + d2)2 = 3*0 +5*9 = 45 kg m2
Axis C : IC = m1(d1 + d2)2 + m202= 3*9 +5*0 = 27 kg m2
Axis D : ID = m102 + m202 = 0 kg m2
23
Example： Moments of inertia for four point masses
Four point masses lie at the corner of a rectangle with sides
of length 3m and 4m, as figure. Find the moments of inertia
about each diagonals. Take M=1kg.
The distance from particle to diagonal d
d=3*4/5=2.4m
IA = 4Md2 + 2Md2 = 6Md2 = 34.56 kg m2
IB = Md2 + 3Md2 = 4Md2 = 23.04 kg m2
24
Example : The Oxygen Molecule
Consider the diatomic oxygen molecule O2, which is rotating in the xy
plane about the z axis passing through its center, perpendicular to its
length. The mass of each oxygen atom is 2.661026kg, and at room
temperature, the average separation between the two oxygen atoms is
d=1.211010m. (a) Calculate the moment of inertia of the molecule
about the z axis. (b) A typical angular speed of a molecule is 4.60 1012
rad/s. If the oxygen molecule is rotating with this angular speed about
the z axis, what is its rotational kinetic energy?
I = m(d/2)2+ m(d/2)2 = ½ * 2.66  1026 * (1.211010) 2
= 1.95  1046 kg m2
K = ½ I ω2 = ½ *1.95  1046*(4.60 1012) 2
= 2.06  1021J
25
Example : Four Rotating Object
Four small spheres are fastened to the corners of a frame of negligible
mass lying in the plane. (a) If the rotation of the system occurs about
the y axis, as in Figure, with an angular speed . find the moment of
inertia Iy about the y axis and the rotational kinetic energy about this
axis. (b) Suppose the system rotates in the xy plane about the z axis
through O (Fig.). Calculate the moment of inertia about the z axis and
the rotational energy about this axis.
(a) Iy = Ma2 + Ma2 = 2Ma2
Ky = ½ Iyω2 = Ma2ω2
(b) Iz = 2Ma2 + 2mb2 = 2(Ma2+mb2 )
Kz = ½ Iz ω2 = (Ma2+mb2 )ω2
26
Moment of Inertia of Continuous Bodies
27
Example：Moment of inertia of the rod
Find the moments of inertia of a thin rod of mass M and
length L about the axis perpendicular to the rod.(a) about the
axis pass CM? (b) about the axis at one end?
(a) λ =M/ L; dm = λdx ;
Icm = ∫
x2dm=
L
(b) I’ = ∫0
L/2
-L/2
∫
L/2
λx2dx =1/3λx3|-L/2
= 1/12 λL3 = 1/12 ML2
L
2
x dm= ∫
0
λx2dx =1/3 λx3|L0 = 1/3 λL3 = 1/3 ML2
= 1/12 ML2 + M(L/2)2 = Icm + M(L/2)2
28
Example：A uniform rectangular plate
Find the moment of inertia of the uniform rectangular plate
with mass M and sides a and b about the axis pass CM?
σ = M/(ab); dm = σ dx dy ;
Icm = ∫
b/2 a/2
2
r dm= ∫ ∫
-b/2 -a/2
b/2
-b/2
=σ∫
σ(x2+y2)dxdy
(1/12 a3+a y2)dy
= 1/12 σ (a3b+ab3) = 1/12 σab (a2+b2)
= 1/12 M(a2+b2)
29
Example：A ring
A thin ring of radius R and mass M with uniform mass distribution, try
to find its moment of inertia passing through the center of mass, (a) The
axis of rotation is perpendicular to the ring? (b) The axis of rotation is
parallel to the ring
(a) Ia = ∫ r2dm= ∫ R2dm = MR2
(b) λ =M/(2πR); dm= λRdθ ; dI=R2sin2θdm= λ R3sin2θ dθ


Ia = ∫dI = 2∫0 λR3sin2θ dθ = λR3[θ - ½ sin(2θ)]|0
= λπR3 = ½ MR2
30
Example： A uniform solid cylinder
A uniform solid cylinder has a radius R, mass M, and
length L. Calculate its moment of inertia about its
central axis
σ = M/(πR2); dm = σ2πrdr ;
I=∫
=2
r2dm=
R
∫0
r2σ2πrdr
R
4
πσ(r /4)| =
0
R
0
= 2πσ∫ r3dr
2πσ(R4/4) = ½ πσR4
= ½ MR2
31
Example： A uniform hollow cylinder
A uniform hollow cylinder has a inner and outer radius R1
and R2, mass M, and length L. Calculate its moment of inertia
about its central axis ?
σ = M/(πR22- πR12); dm = σ2πrdr ;
I=∫
=
r2dm=
R2
∫R
1
r2σ2πrdr
R2
4
2πσ(r /4)| =
R1
R2
= 2πσ∫R r3dr
1
2πσ (R24-R14)/4
= ½ πσ(R24-R14) = ½ πσ(R22-R12)(R22+R12)
= ½ M(R12+R22)
32
Example：Moment of inertia of a cylinder rod
A cylinder rod has radius R, length L, and mass M. The mass
is uniformly distributed, as shown in the figure. Try to find its
moment of inertia passing through the axis of mass center
perpendicular to the cylinder surface?
λ = M/L; dm = λdx ; dI = dm(¼ R2+ x2)
I = ∫ dI = ∫(¼
R2+
x2)dm
L/2
-L/2
= ∫ (¼ R2+ x2)λdx
L/2
= ¼ λR2L+1/3 λx3|-L/2
=¼ λR2L+1/12 λL3
= ¼ MR2 + 1/12 ML2
33
Example：Sphere shell
A thin spherical shell has the radius R and mass M with
uniform mass distribution. Try to find the moment of inertia
of its axis of rotation passing through the center of mass?
σ= M/(4πR2); dm = σ 2πRsinθ Rdθ ;
dI = R2sin2θ dm = σ 2πR4sin3θ dθ
I = ∫ dI =
=
σ2πR4∫
σ2πR4 (-cosθ

0
sin3θ dθ
+1/3

3
cos θ)|
0
= 8/3σπR4
= 2/3 MR2
34
Example：Uniform Sphere-1
What is the moment of inertia of a solid sphere of radius R
and mass M with uniform distribution for its axis of rotation
passing through the center of mass ?
ρ = M/(4/3πR3); dm = ρ4πr2dr;
dI = 2/3 r2dm = 8/3 ρπr4dr
R
I = ∫ dI =8/3 ρπ ∫0 r4dr
= 8/15 ρπR5
= 2/5 MR2
35
Example ：Uniform Sphere-2
What is the moment of inertia of a solid sphere of radius R
and mass M with uniform distribution for its axis of rotation
passing through the center of mass ?
ρ = M/(4/3πR3); dm = ρπ(R2- z2)dz;
dI = ½ (R2- z2)dm = ½ ρπ(R2- z2)2dz
R
I = ∫ dI =½ ρπ∫-R
(R2- z2)2dz
R
=½ ρπ∫-R (R4- 2R2z2 +z4)dz
= ½ ρπ(2R5- 4/3 R5 +2/5R5) = 8/15ρπR5
= 2/5 MR2
36
Example : Rotating Rod
A uniform rod of length L and mass M is free to rotate on a
frictionless pin through one end. The rod is released from rest
in the horizontal position. (a) What is the angular speed of the
rod at its lowest position? (b) Determine the tangential speed
of the center of mass and the tangential speed of the lowest
point on the rod in the vertical position.
(a) I = 1/3ML2;
Ei = Ui = Mg(L/2) = Ef = Kf = ½ Iω2 = 1/6 ML2ω2
ω =(3g/L)½
(b) V1= r1 ω = (L/2) (3g/L)½ = ½ (3gL)½
V2= r2 ω = L(3g/L)½ = (3gL)½
37
Example : Parallel axis
Determine the moment of inertia of a solid cylinder of radius
R0 and mass M about an axis tangent to its edge and parallel
to its symmetry axis.
(a) σ = M/(πR2); dm = σ2πrdr ;
R
R
R
I = ∫ r2dm= ∫0 r2σ2πrdr = 2πσ ∫0 r3dr= 2πσ (r4/4)|0
= 2πσ(R4/4) = ½ πσR4 = ½ MR2
(b)平行軸定理
Irim = Icm + MR2 = 3/2 MR2
38
Torque
To make an object start rotating, a force is needed; the
position and direction of the force matter as well.
The perpendicular distance from the axis of rotation to the
line along which the force acts is called the lever arm R⊥.
R1 is the lever arm for F1; R2 is the lever arm for F2; R3 is the
lever arm for F3; the lever arm for F4 is zero.
39
Torque
The torque is the rotational analog of force: force causes
linear acceleration; torque causes angular acceleration.
The “turning ability” of a force about an axis or pivot is
called its torque. Only the perpendicular component of the
force F⊥ contribute to the turning effect. The lever arm R⊥ is
the perpendicular distance from the origin (pivot or axis) to
the line of action of the force F.
40
 
  R F
Torque

  RF  R F
  RF sin 
Although torque has the same dimension as energy, these two concepts
are unrelated.
Energy is a scalar, whereas torque is a vector.
The SI units of torque are N.m.

Although torque is a force multiplied by a distance, it is very
different from work and energy.

The units for torque are reported in N.m and not changed to Joules.
41
Net Torque
The force F1 will tend to cause a counterclockwise rotation about O.
The force F2 will tend to cause a clockwise rotation about O.
τnet = τ1 + τ2 = F1R1 – F2R2
Forces can cause a change in linear motion

Described by Newton’s Second Law
Forces can cause a change in rotational motion

The effectiveness of this change depends on the force and the
moment arm

The change in rotational motion depends on the torque
42
Torque Vector Example
Given the force and location
F = (2.00 i + 3.00 j) N
R = (4.00 i + 5.00 j) m
Find the torque τ = R × F
R × F = (4.00 i + 5.00 j) × (2.00 i + 3.00 j)
= 4.00 i × (2.00 i + 3.00 j) + 5.00 j × (2.00 i + 3.00 j)
= 12 k - 10 k
= 2 k N ·m
43
Example : The Net Torque on a Cylinder
A one-piece cylinder is shaped as Figure, with a core section protruding
from the larger drum. The cylinder is free to rotate around the central
axis shown in the drawing. A rope wrapped around the drum, of radius
R1, exerts a force F1 to the right on the cylinder. A rope wrapped around
the core, of radius R2, exerts a force F2 downward on the cylinder. (a)
What is the net torque acting on the cylinder about the rotation axis? (b)
Suppose F1 = 5.0 N, R1 = 1.0 m, F2 = 6.0 N, and R2 = 0.50 m. What is
the net torque about the rotation axis and which way does the cylinder
rotate if it starts from rest?
(a) τnet = τ1 + τ2 = -F1R1 + F2R2
(b) τnet =-F1R1+F2R2=-5*1+6*0.5=-2.0Nm
Rotate clockwise around the z-axis (moment is negative)
44
Example : Torque on a compound wheel
Two thin disk-shaped wheels, of radii R1 = 30 cm and R2 = 50
cm, are attached to each other on an axle that passes through
the center of each, as shown. Calculate the net torque on this
compound wheel due to the two forces shown, each of
magnitude 50 N.
τnet = τ1 + τ2
= F1R1 - F2R2cos30o
= 50* 0.3 – 50*0.5*0.866
= － 6.65 N m
45
Example：Net torque of three forces
Three forces F1, F2 and F3 act on a rod at distances r1, r2, and
r3 from the pivoted end, as figure. Find the torque due to each
force about the pivot.
τ1 = －R1F1cosα
τ2 = R2F2sinβ
τ3 = R3F3cosγ
τnet = τ1 + τ2 + τ3
= －R1F1cosα + R2F2sinβ + R3F3cosγ
46
Example：Support force
A uniform rod of weight W1=35N is supported at its ends as
shown in Figure. A block of weight W2=10N is placed onequarter of the distance from one end. What are the forces
exerted by the supports?
Σ Fy = N1 + N2 – 35 – 10 = 0
τnet = N1 *L –10* ¾ L – 35* ½ L= 0
N1 = 25 N
N2 = 20 N
47
Example : The Leading Ladder
A uniform ladder of length L and mass m rests against a
smooth, vertical wall. If the coefficient of static friction
between ladder and ground s = 0.40, find the minimum
angle ϕmin such that the ladder does not slip.
Σ Fx = fs－N2= 0; fs = N1s
Σ Fy = N1－mg = 0
Στ =N2Lsinϕmin－½ mgL cosϕmin=0
= mgsLsinϕmin －½ mgL cosϕmin
tanϕmin = 1/(2s);
ϕmin= tan-1[1/(2s)] = tan-1[1.25] = 51o
48
Example : Standing on a Horizontal Beam
A uniform horizontal beam of length 8.00 m and weight 200 N is
attached to a wall by a pin connection. Its far end is supported by a
cable that makes an angle of 53.0 with the horizontal. If a 600-N man
stands 2.00 m from the wall, find the tension in the cable and the force
exerted by the wall on the beam at the pivot.
Σ Fx = Nx－T cos53o= 0;
Σ Fy = Ny+ T sin53o -200-600 = 0
Στ =T*8sin53o－200*4-600*2 = 0
T = 312.5 N;
Nx = 187.5 N; Ny = 800 － 250 = 550 N
N = 581 N; ϕ = tan-1(550/187.5) = 71.1o
49
Example : Hinged beam and cable
A uniform beam, 2.20 m long with mass m = 25.0 kg, is mounted by a
small hinge on a wall. The beam is held in a horizontal position by a
cable that makes an angle θ = 30.0°. The beam supports a sign of mass
M = 28.0 kg suspended from its end. Determine the components of the
force F that the (smooth) hinge exerts on the beam, and the tension T in
the supporting cable.
Σ Fx = Nx－T cos30o= 0;
Σ Fy = Ny+ T sin30o -245-274.4 = 0
Στ =T*2.2sin30o－245*1.1-274.4*2.2 = 0
T = 793.8 N;
Nx = 687.4 N; Ny =245+274.4－ 396.9 = 122.5 N
N = 698.2 N; ϕ = tan-1(122.5/687.4) = 10.1o
50
Example : Force exerted by biceps muscle-1
As figure, the muscle is attached at a point about 4cm from
the socket which acts as a pivot point. If a weight of 50N is
held in the hand, what is the tension in the muscle? Assume
the forearm is a horizontal uniform rod of weight 15N and
length L=30cm. The force exerted by the muscle acts at 10o
to the vertical.
Στ = Tcosθ ·d – N1L/2 – N2L=0
Tcos10o·0.04–15·0.15 –50·0.3=0
T = 438N
51
Example : Force exerted by biceps muscle-2
How much force must the biceps muscle exert when a 5.0-kg
ball is held in the hand (a) with the arm horizontal, and (b)
when the arm is at a 45° angle? The biceps muscle is
connected to the forearm by a tendon attached 5.0 cm from
the elbow joint. Assume that the mass of forearm and hand
together is 2.0 kg and their CG is as shown.
(a) Στ =T·0.05–19.6·0.15 –49·0.35=0
T = 401.8N
(b) Στ =T·0.035–19.6·0.105 –49·0.245=0
T = 401.8N
52
Example：The critical speed
A Car goes around an un-banked curve of a radius R at speed
V. Find the critical speed at which it tends to overturn.
f1 + f2 = mV2/R
N1 + N2 – mg = 0
－ (f1 + f2)H － N1 D/2+N2D/2= 0
N1 = m[g/2 － V2H/(RD)]
When N1=0; inner wheel lost contact with the road.
Vmax2 = ½ gRD/H;
R =50m, D=1.5m, H=0.5m,
then Vmax = 27.5m/s = 100km/h
53
Torque and Angular Acceleration
Consider a particle of mass m rotating in a circle of
radius r under the influence of tangential force .
The tangential force provides a tangential acceleration:
Ft = mat

The radial force causes the particle to move in a circular path.
The magnitude of the torque produced by a force around the center of the
circle is

τ = Ft r = (mat) r
The tangential acceleration is related to the angular acceleration

Στ = Σ(mat)r = Σ (mrα)r = Σ (mr2)α
Since mr2 is the moment of inertia of the particle,

Στ = Iα
54
Torque and Angular Acceleration
Consider the object consists of an infinite number of mass elements dm
of infinitesimal size.
Each mass element rotates in a circle about the origin, O.
Each mass element has a tangential acceleration.
From Newton’s Second Law

dFt = (dm)at
The torque associated with the force and using the angular acceleration
gives

dτ ext = rdFt = atr dm = αr2dm
Finding the net torque

2
2



r
dm


r
 ext 
 dm  I

This becomes Σ τ = Iα
55
Example：The angular acceleration of the rod
A uniform rod of length L and mss M is pivoted freely at one
end. (a) what is the angular acceleration of the rod when it is
at angle θ to the vertical? (b) What is the tangential linear
acceleration of the free end when the rod is horizontal? The
moment of inertia of a rod about one end is 1/3ML2.
(a) τ = ½ mgLsinθ = Iα = 1/3mL2α
α = 3gsinθ/(2L)
(b) at = Lα = 3gsinθ/2
56
Example: An Atwood Machine with a Massive Pulley
The pulley has mass M and radius R. There are two objects
with mass m1 and m2, on the end of string both sides, suppose
the string does not slide on the pulley. Calculate the
magnitude of the acceleration of the two objects.
F1 = m1g– T1 = m1 a
F2 = T2 – m2g = m2 a
τ net = RT1 –RT2 = Iα
I = ½ MR2;
a = Rα
T1–T2= Iα/R = ½ Ma
a = (m1 – m2)g/(m1+m2+½ M)
57
Example : The falling block-1
A pulley has mass M=4 kg and radius R=0.5 m. It rotates freely on a
horizontal axis, as in figure. A block of mass m=2 kg hangs by a string
that is tightly wrapped around the pulley. (a) What is the angular
velocity of the pulley 3 s after the block is released? (b) Find the speed
of the block after it has fallen 1.6 m. Assume the system starts at rest.
F = mg– T = m a ; τ net = RT = Iα
I = ½ MR2;
a = Rα ; T= Iα/R = ½ Ma ;
a = mg/(m+½ M) =4.9 m/s2;
α = a/R = 9.8 rad/s2
(a) ω = αt = 29.4 rad/s
(b) V2 = V02+ 2aS = 2*4.9*1.6;
V = 4m/s
58
Example : The falling block-2
A pulley with mass M=2.5 kg and radius R=20 cm, mounted
on a fixed horizontal axle. A block with mass m=1.2 kg hangs
from a massless cord that is wrapped around the rim of the
disk. Find the acceleration of the falling block, the rotational
acceleration of the disk, and the tension in the cord.
F = mg– T = m a ; τ net = RT = Iα
I = ½ MR2;
a = Rα ; T= Iα/R = ½ Ma ;
(a) a = mg/(m+½ M) =4.8 m/s2
(b) α = a/R = 24 rad/s2
(c) T = ½ Ma = 6N
59
Example： A flywheel
A flywheel of mass M=2kg and radius R=40cm rotates freely
at 600rpm. Its moment of inertia is ½ MR2. A brake applies a
force F=10N radial inward at the edge as figure. If the
coefficient of friction is μk=0.5, how many revolutions does
wheel make before coming to rest?
f = μkF =5N ; ω0 = 20π rad/s
τ = - f R = Iα = ½ MR2α
α = -2 μkF /MR = -12.5 rad/s2
ω2 = ω02 + 2α∆θ = 0 ;
∆θ = 16π2 rad = 8π revolutions = 25.13 圈
60
Example：The Rolling sphere
A sphere of mass M and radius R that rolls without slipping
down an incline. Its moment of inertia is 2/5MR2. (a) Find the
linear acceleration of the CM. (b) Which is the minimum
coefficient of friction required for the sphere to roll without
slipping.
Fx = mg sinθ– f = ma = mRα ;
τnet = Rf = Iα = 2/5 mR2α
f =N μs = mg cosθ μs = 2/5 mRα;
(a) α = 5gsinθ/(7R) ; a = Rα = 5/7gsinθ
(b) μs = f /N = (2/7mgsinθ)/(mgcosθ) = 2/7 tanθ
61
Angular Momentum
The instantaneous angular momentum L of a particle relative to the
origin O is defined as the cross product of the particle’s instantaneous
position vector r and its instantaneous linear momentum P.
  
L  r P
The SI units of angular momentum are (kg.m2)/ s.
The rigid object is a non-deformable system.
Each particle of the object rotates in the xy plane
about the z axis with an angular speed of ω
The angular momentum of an individual particle is Li = mi ri2 ω
L and ω are directed along the z axis.





2
L   Li   ri  mi vi   mi ri   I
i
i
i
62
Angular Momentum
Consider a particle of mass m located at the vector position r and moving
with linear momentum p . Find the net torque:
Στ = r ×ΣF = r × dp/dt.
Since dr/dt × p = 0
Σ τ = r ×ΣF = d(r × p)/dt= dL/dt
Torque and angular momentum are rotational analogs of force and
momentum.
The total angular momentum of a system of particles is defined as the
vector sum of the angular momenta of the individual particles.

Ltot = L1 + L2 +…+Ln = Σ Lj
Differentiating with respect to time

dLtot/dt = Σ(dLj/dt) = Στj
63
Rotational Dynamics
Rotational and Translational motion
Rotational
Kinetic Energy KR = ½ Iω2
Translational
K = ½ mv2
Equilibrium
Στ=0
ΣF=0
Newton’s Law
Στ=Iα
Σ F = ma
Newton’s Law
Σ τ = dL/dt
Σ F = dP/dt
Momentum
L = Iω
P = mv
Conservation
Li = Lf
Pi = Pf
Power
P = τ·ω
P = F·v
64
Conceptual Example : Bicycle wheel
Suppose you are holding a bicycle wheel by a handle connected to its
axle. The wheel is spinning rapidly so its angular momentum points
horizontally as shown. Now you suddenly try to tilt the axle upward (so
the CM moves vertically). You expect the wheel to go up (and it would
if it weren’t rotating), but it unexpectedly swerves to the right! Explain.
你用一向上的力，因此產生一向右的力矩
65
Example： Angular momentum of a disk
A disk of mass M and radius R is rotating at angular velocity
ω about an axis perpendicular to its plane at a distance R/2
from the center, as figure. What is its angular momentum?
The moment of inertia of a disk about the central axis is ½
MR2
I = ½ MR2 +M(R/2)2 = ¾ MR2
L=I ω = ¾ MR2 ω
66
Example： Angular momentum of a particle
What is the angular momentum of a particle of mass
= 2kg that is located 15m from the origin in the
direction 37o S of W and has a velocity v=10m/s in
the direction 30o E of N.
r = -15 cos37o i -15 sin37o j = -12 i – 9 j
p =mv= 20 sin30o i+20 cos30o j = 10 i + 17.3 j
L = r ×p = -117.6 k kg·m2/s
67
Example： The motion of projectile
Show τ = dL/dt can be applied to the motion of
projectile.
r=xi +yj
F = -mg j
τ = r × F = -mgx k
L = r × P = r × mv
dL/dt = d(r × mv)/dt = mr × dv/dt
= -mgx k = τ
68
Example : The Atwood Machine
Consider again the Atwood machine with the massive
pulley in Example. Determine the acceleration of the
two objects using an angular momentum approach.
ω = V/R;
I = ½ MR2
L=Rm1V+Rm2V+I ω =R(m1+m2+½ M)V
τ net = R(m1-m2)g = dL/dt
= R(m1+m2+½ M)a
a = (m1 – m2)g/(m1+m2+½ M)
69
Example： The linear acceleration
Two blocks with masses m1 and m2 are connected by a rope
that passes over a pulley of radius R and mass M; as figure.
Find out the linear acceleration of the blocks. There is no
friction.
V = Rω; a = Rα; I = ½ MR2
L=R (m1+m2)V+ ½ MR2ω
=(m1+m2)R2ω+(M/2)R2ω
τ=m1gR=dL/dt=(m1+m2+M/2)R2α
a = Rα = m1g/(m1+m2+M/2)
70
Conservation of Angular Momentum
In the absence of an external torque,
angular momentum is conserved:
τ = dL/dt =0 ，L = Iω = constant
Ii ωi = If ωf

Net torque = 0, if the system is isolated.
The total angular momentum of a rotating
object remains constant if the net external
torque acting on it is zero.
71
Conservation of Angular Momentum
The total angular momentum of a system is constant in both
magnitude and direction if the net external torque acting on the
system is zero.

Net torque = 0 means that the system is isolated.

This is the basis of the angular momentum version of the
isolated system model


 
Ltot  constant or Li  L f
For a system of particles,


Ltot   Ln  constant
72
Conservation Law Summary
For an isolated system (1) Conservation of Energy:

Ei = Ef
(2) Conservation of Linear Momentum:



pi  p f
(3) Conservation of Angular Momentum:

 
Li  L f
73
Conceptual Example : Spinning bicycle wheel
Your physics teacher is holding a spinning bicycle wheel while he sits
on a stationary frictionless turn-chair. What will happen if the teacher
suddenly flips the bicycle wheel over so that it is spinning in the
opposite direction?
Angular momentum is conserved, so the teacher will start
spinning in the direction the wheel was spinning originally.
74
Example : A Revolving Puck - 1
A small mass m attached to the end of a string revolves in a
circle on a frictionless tabletop. The other end of the string
passes through a hole in the table. Initially, the mass revolves
with a speed v1 = 2.4 m/s in a circle of radius R1 = 0.80 m.
The string is then pulled slowly through the hole so that the
radius is reduced to R2 = 0.48 m. What is the speed, v2, of the
mass now?
τ = 0; L1=L2= R1mv1= R2mv2
0.8*2.4 = 0.48*v2
v2 = 4.0 m/s
75
Example : A Revolving Puck - 2
A puck of mass m on a horizontal, frictionless table is
connected to a string that passes through a small hole in the
table. The puck is set into circular motion of radius R, at
which time its speed is vi. (a) If the string is pulled from the
bottom so that the radius of the circular path is decreased to r,
what is the final speed vf of the puck? (b) Is the kinetic
energy of the puck conserved in this process?
(a) τ = 0; Li=Lf= Rmvi= rmvf
vf = vi R/r
(b) Kf/Ki = (½ mvf2) /(½ mvi2)
= (R/r) 2
76
Example：Rotating platform - 1
A man stands on a platform that rotates at 0.5rev/s. With arms
outstretched he holds two 4-kg blocks at a distance of 1m from the axis
of rotation －which passes through him. He then reduces the distance of
the blocks from axis to 0.5m. Assume that the moment of inertia of the
system “man + platform” is constant at 4kg·m2. (a) What is the new
angular velocity? (b) What is the change in kinetic energy?
Ii = 4+2mri2=12kg·m2 ; If = 4+2mrf2 = 6kg·m2
ωi = π rad/s
(a) Ii ωi = If ωf = 12π = 6ωf ; ωf = 2π rad/s
(b) Ki = ½ Ii ω i2 =6π2 J; Kf = ½ If ω f2 =12π2 J
∆K = Kf - Ki = 6π2 J = 60J
77
Example：Rotating platform - 2
A man stands on a stationary platform with a spinning bicycle wheel in
his hands, as figure. The moment of the inertia of the man plus platform
is IM=4 kg·m2, and for the bicycle wheel it is IB=1 kg·m2. The angular
velocity of the wheel is 10 rad/s counterclockwise as viewed from
above. Explain what occurs when the man turn the wheel upside down.
Li = +IB ωB =10 kg·m2/s;
Lf = -IB ωB +IM ωM =IB ωB;
IM ωM = 2 IB ωB = 20 kg·m2/s
ωM = 5 rad/s
78
Example : Bullet strikes cylinder edge
A bullet of mass m moving with velocity u strikes and becomes
embedded at the edge of a cylinder of mass M and radius R0. The
cylinder, initially at rest, begins to rotate about its symmetry axis, which
remains fixed in position. Assuming no frictional torque, what is the
angular velocity of the cylinder after this collision? Is kinetic energy
conserved?
(a) Li = muR = (½ MR2+mR2)ωf = Lf
ωf = mu/(½ MR+mR)
(b) Ki = ½ mu2
Kf = ½ If ωf2 = ½ (½ MR2+mR2)[mu/(½ MR+mR)]2
=½ m2u2/(½ M+m)
∆K = Kf - Ki = － ¼ mMu2/(½ M+m)
79
Example ：A disk-shaped platform
A man of mass m=80 kg runs at a speed u=4 m/s along the tangent to a
disk-shaped platform of mass M=160 kg and radius 2 m. The platform
is initially at rest but can rotate freely about and axis through its center.
(a) Find the angular velocity of the platform after the man jumps on. (b)
He then walks to the center, find the new angular momentum. Treat the
man as a point particle.
(a) Li=muR=80·4·2= 640=(½ MR2+mR2)ω1=640ω1 = L1
ω1 = mu/(½ MR+mR) =1 rad/s
(b) L2=(½ MR2+mR2)ω2= 640kg·m2/s
L2 = ( ½ MR2)ω2 = 320ω2= L1
ω2 = 2 rad/s
80
Example : Running on a circular platform
Suppose a 60-kg person stands at the edge of a 6.0-m-diameter circular
platform, which is mounted on frictionless bearings and has a moment
of inertia of 1800 kg·m2. The platform is at rest initially, but when the
person begins running at a speed of 4.2 m/s (with respect to the Earth)
around its edge, the platform begins to rotate in the opposite direction.
Calculate the angular velocity of the platform.
Li = Lf = 0
0 = muR + Iωf
= 60*4.2*3 + 1800 ωf
ωf = -0.42 rad/s
81
Example : Clutch - 1
The two plates have masses M1 = 6.0 kg and M2 = 9.0 kg, with equal radii R0 = 0.60
m. They are initially separated. Plate MA is accelerated from rest to an angular velocity
ωi = 7.2 rad/s in time Δt = 2.0 s. Calculate (a) the angular momentum of M1, and (b)
the torque required to have accelerated M1from rest to ωi. (c) Next, plate M2, initially
at rest but free to rotate without friction, is placed in firm contact with freely rotating
plate M1, and the two plates both rotate at a constant angular velocity ωf, which is
considerably less than ωi. Why does this happen, and what is ωf?
(a) I1 = ½ M1 R2 =1.08 ; I2 = ½ M2 R2 =1.62
Li = I1 ωi = 7.78 kg·m2/s
(b) α = Δω1/Δt =3.6; τ = Iα = 3.89 N·m
Lf = (I1 + I2)ωf = (1.08+ 1.62)ωf = Li = 7.78
ωf = 2.88 rad/s
82
Example : Clutch - 2
A disk of moment of inertia 4 kg·m2 is spinning freely at 3
rad/s. A second disk of moment of inertia 2 kg·m2 slides
down a spindle and they rotate together. (a) What is the
angular velocity of the combination? (b) What is the change
in kinetic energy of the system?
(a) Lf = (I1 + I2) ωf = I1 ωi = Li
ωf = 4.0*3.0/6.0 = 2.0 rad/s
(b) Ki = ½ I1 ωi 2 = 18.0 J;
Kf = ½ (I1+ I2)ωf2 = 12.0 J
∆K = Kf - Ki = － 6.0 J
83
Example： Kepler’s second law
According to Kepler’s second law of planetary motion, the
line joining the sun to a planet sweeps out equal area in equal
time intervals. Show that this is a consequence of the
conservation of angular momentum. (Kepler’s second law)
dA= ½ r (vdt) sinθ = ½ |r × vdt |
dA/dt = ½ rv sinθ
=½ |r × P|/m = |L|/2m
No net torque is applied on the planet, then the angular
momentum must be conserved, and dA/dt is also conserved.
84
Example : Rotating Period of a Neutron Star - 1
A neutron star is believed to form from the inner core of a larger star
that collapsed, under its own gravitation, to a star of very small radius
and very high density. Before collapse, suppose the core of such a star
is the size of our Sun (r ≈ 7×105 km) with mass 2.0 times as great as the
Sun, and is rotating at a frequency of 1.0 revolution every 100 days. If it
were to undergo gravitational collapse to a neutron star of radius 10 km,
what would its rotation frequency be? Assume the star is a uniform
sphere at all times, and loses no mass.
Ii = 2/5 Mri2 =7.8×1047kg·m2 ;
If = 2/5 Mrf2 =1.6×1038kg·m2
ωi = 7.3×10-7 rad/s
Ii ωi = If ωf = 5.7×1041 = 1.6×1038ωf ;
ωf = 3.56×103 rad/s = 567 rev/s
85
Example : Rotating Period of a Neutron Star - 2
A star undergoes a supernova explosion. The material left
behind forms a sphere of radius 8.0  106 m just after the
explosion with a rotation period of 15 h. This remaining
material collapses into a neutron star of radius 8.0 km. What
is the rotation period T of the neutron star?
Ii = 2/5 Mri2 =2.56×1013M ; If = 2/5 Mrf2 =2.56×107M ;
ωi = 1.16×10-4 rad/s
Ii ωi = If ωf = 2/5 Mri2ωi = 2/5 Mrf2ωf
ωf = ωi(ri/rf)2 = 1.16×102 rad/s = 18.5 rev/s
Tf = Ti(rf/ri)2 = 5400/106 = 0.054 s
86
Rolling Object
The red curve shows the path moved by a point on the rim of the object.

This path is called a cycloid.
The orange line shows the path of the center of mass of the object.
In pure rolling motion, an object rolls without slipping.
In such a case, there is a simple relationship between its rotational and
translational motions.
87
Pure Rolling Motion, Object’s Center of Mass
The translational speed of the center of mass is
vCM
ds
d

R
 R
dt
dt
The linear acceleration of the center of mass is
aCM
Rolling
dvCM
d

R
 R
dt
dt
motion
can
be
modeled
as
a
combination of pure translational motion and
pure rotational motion.
The contact point between the surface and the
cylinder has a translational speed of zero (c).
88
Why Does a Rolling Sphere Slow Down?
A rolling sphere will slow down and stop rather than roll
forever. What force would cause this?
If we say “friction”, there are problems:
The frictional force has to act at the point of
contact; this means the angular speed of the
sphere would increase.
Gravity and the normal force both act through
the center of mass, and cannot create a torque.
No real sphere is perfectly rigid. The bottom
will deform, and the normal force will create
a torque that slows the sphere.
89
Total Kinetic Energy of a Rolling Object
The total kinetic energy of a rolling object is the sum of the
translational energy of its center of mass and the rotational
kinetic energy about its center of mass.
K = ½ MVCM2 + ½ ICMω2

½ MVCM2 : the translational kinetic energy

½ ICMω2 : the rotational kinetic energy

Vcm = Rω
K = ½ (MR2+ICM)ω2 = ½ (M+ICM/R2)VCM2
90
Total Kinetic Energy of a Rolling Object
When using conservation of energy, both rotational and
translational kinetic energy must be taken into account.
All these objects have the same potential energy at the top,
but the time it takes them to get down the incline depends on
how much rotational inertia they have.
91
Total Kinetic Energy of a Rolling Object
Accelerated rolling motion is possible only if friction is present between
the sphere and the incline.

The friction produces the net torque required for rotation.

No loss of mechanical energy occurs because the contact point is at
rest relative to the surface at any instant.

In reality, rolling friction causes
mechanical energy to transform
to internal energy.

Rolling friction is due to
deformations of the surface
and the rolling object.
92
Total Kinetic Energy of a Rolling Object
Apply Conservation of Mechanical Energy:

Let U = 0 at the bottom of the plane

Kf + U f = Ki + Ui

Kf = ½ (ICM/R2)vCM2+½ MvCM2 = ½ (M+ICM/R2)VCM2

Ui = Mgh

Uf = Ki = 0
VCM =[2gH/(1+[ICM/(MR2)])]½
93
Example : Sphere Rolling Down an Incline
(a) If the object in Figure is a solid sphere, calculate the speed
of its CM at the bottom. (b) Determine the magnitude of the
translational acceleration of the CM.
(a) Ef = ½ (M+ICM/R2)VCM2 = Ei = MgH
= ½ M(1+2/5) VCM2
VCM = (10gH/7)½
(b) a = ½ (Vcm,f2 - Vcm,i2)/(xf - xi)
= ½ (Vcm,f2 - Vcm,i2)/(Hcscθ) = ½ (10gH/7)*sinθ/H
= 5/7 g sinθ
94
Example : The Rolling object
Show that the velocity of a point on the rim of a wheel that
rolls without slipping is perpendicular to the line joining to
the point of contact.
rB = R sinθ i + R (1+ cosθ) j
VB = VC + Vt
= V i +(V cosθ i －V sinθ j )
= V(1+cosθ) i －V sinθ) j
rB ·VB = 0 , ∴ rB ⊥ VB
95
Work in Rotational Motion
Find the work done by F on the object
as it rotates through an infinitesimal
distance ds = r dθ
dW=F·ds=[(Fsinϕ)r]dθ= τdθ
The radial component of the force does
no work because it is perpendicular to
the displacement.
96
Work in Rotational Motion
To derive the work-energy theorem for rotational motion, we
first express torque in a convenient form. Using the chain
rule we have: τ = Iα = I dω/dt = Idω/dθ dθ/dt = Iωdω/dθ
We use this result in dW=τdθ and integrate to find
f
W=∫i τdθ
f
=∫i
I ωdω = ½ I(ωf2－ ωi2) = Kf－Ki
The work done by a torque on a rigid body rotating about a
fixed axis leads to a change in its rotational kinetic energy.
The net work done is the sum of the translational and
rotational kinetic energies.
W = ∆K + ∆KR
97
Power in Rotational Motion
The rate at which work is being done in a time
interval dt is
P = dW/dt = τ dθ/dt = τ ω = τ ·ω
This is analogous to P = F·v in a linear system.
98
Example : A Block Unwinding form a Wheel
The wheel in Figure is a solid disk of mass M = 2.00 kg and
radius R = 30.0 cm. The suspended block has a mass m =
0.500 kg. If the suspended block starts from rest and
descends to a position 1.00 m lower, what is its speed when it
is at this position?
W = mgh= 0.5*9.8*1=4.9 J
I = ½ MR2 ; V = Rω ;
W = ∆K +∆KR = ½ mVf2 + ½ Iωf2
= (½ m + ¼ M)Vf2 = ¾ Vf2
Vf = (4.9*4/3)½ = 2.56 m/s
99
Example：A block, a spring and a pulley
A block of mass m=4kg is attached to a spring (k=32N/m)
by a rope that hangs over a pulley of mass M=8kg. If the
system starts from rest with the spring un-extended, find the
speed of the block after it falls 1m. I = ½ MR2 for the pulley.
I = ½ MR2 ;
V = Rω;
½ mV2 + ½ Iω2 + ½ kx2 – mgx =0
½ (m+ ½ M)V2 + ½ kx2 – mgx =0
4V2 + 16 – 39.2 = 0
V = 2.41m/s
100
Example：Power of A motor
A motor rotates a pulley of radius 25cm at 20rpm. A rope
around the pulley lifts a 50kg block, as figure. What is the
power output of the motor?
ω = 20*2π/60 = 2π/3
τ = 50*9.8*0.25 =125 Nm
P = τ ω = 125*2π/3 = 26 W
101
Example : Rotating Sculpture
A rigid sculpture consists of a thin hoop (of mass m and radius R=0.15
m) and a thin radial rod (of mass m and length L=2.0 R), arranged as
shown in Fig. The sculpture can pivot around a horizontal axis in the
plane of the hoop, passing through its center. (a) In terms of m and R,
what is the sculpture’s rotational inertia I about the rotation axis? (b)
Starting from rest, the sculpture rotates around the rotation axis from
the initial upright orientation of Fig. What is its rotational speed about
the axis when it is inverted?
(a) L=2R; I1 = 1/12 ML2 + M(L/2+R)2 = 13/3 MR2
I2 = ½ MR2
I = I1 + I2 = (29/6)MR2
(b) W=Mg(L+2R)=4MgR
=½ Iω2 = (29/12) MR2ω2
ω = [48g/(29R)]½ = 10.4 rad/s
102
A spinning top
A spinning top will process around its point of contact with a
surface, due to the torque created by gravity when its axis of
rotation is not vertical. The only external forces acting on the
top are the normal force and the gravitational force. The
right-hand rule indicates that the torque is in the xy plane.
The direction of the angular momentum is along the axis of
symmetry.
τ = r × F = r × mg
ωp=dθ/dt = mgrcmsinϕ/(Iω sinϕ)
= mgrcm/(Iω)
103
Gyroscope
The torque results in a change in
angular momentum in a direction
perpendicular to the axle.

The axle sweeps out an angle dθ in
a time interval dt.
The direction, not the magnitude,
of the angular momentum is
changing .
The
gyroscope
experiences
precessional motion.
104
Gyroscope
The angular momentum : L = I ω ；
dL= Ldθ = Iωdθ
= τ dt = Mgrcmdt
ωp = dθ/dt = Mgrcm/(Iω)
ωp the angular velocity of
precession of a toy gyroscope, need ωp << ω 。
105
Example : A precessing gyroscope
A top view of a spinning gyroscope disk is shown as the figure. The
pivot is at O. (a) From the figure, is the precession clockwise or
counterclockwise? (b) How fast is the gyroscope precession, if the disk
is spinning around the axle at 30.0 rev/s? (c) If the gyroscope takes 4.0
s for a revolution of precession, what is the angular speed of the disk?
(a) L is to the left. The torque τ is toward the top of
the page, so the dL/dt is also toward the top of page.
As the shown, the precession is clockwise.
(b) I = ½ ma2 ; a = 0.03 m; rcm = 0.02 m
ωp = mgrcm/(Iω) = mgrcm/(½ ma2ω) = 2grcm/ (a2ω)
=2* 9.8*0.02/(0.032*60) =2.31 rad/s
(c) ωp = 0.5 ;
 = 2grcm/ (a2ωp) =2* 9.8*0.02/(0.032*½ ) = = 277.42 rad/s
106
Physics
Chapter 11
Oscillations
授課老師施坤龍
0
Oscillations
Periodic motion is the repeating motion of an object in which it continues
to return to a given position after a fixed time interval.
The repetitive movements are called oscillations.
A special case of periodic motion called simple harmonic motion will be
the focus.

Simple harmonic motion also forms the basis for understanding
mechanical waves.
Oscillations and waves also explain many other phenomena quantity.

Oscillations of bridges and skyscrapers

Radio and television

Understanding atomic theory
1
Simple Harmonic Motion
A block of mass m is attached to a spring, the block is free to move on a frictionless
horizontal surface. When the spring is neither stretched nor compressed, the block is at
the equilibrium position, x = 0.
Such a system will oscillate back and forth if disturbed from its equilibrium position.
Hooke’s Law states Fs = - kx

Fs is the restoring force.

It is always directed toward the equilibrium position.

Therefore, it is always opposite the displacement from equilibrium.

k is the force (spring) constant.

x is the displacement.
2
Simple Harmonic Motion
When the block is displaced from the equilibrium point and released, it is
a particle under a net force and therefore has an acceleration.
The force described by Hooke’s Law is the net force in Newton’s Second
Law.
F = -kx = ma ;
a = -(k/m)x
The acceleration is proportional to the displacement of the block.
The direction of the acceleration is opposite the direction of the
displacement from equilibrium.
An object moves with simple harmonic motion whenever its acceleration
is proportional to its position and is oppositely directed to the
displacement from equilibrium.
3
Simple Harmonic Motion
The acceleration is not constant.

Therefore, the kinematic equations cannot be applied.

If the block is released from some position x = A, then the
initial acceleration is –kA/m.

When the block passes through the equilibrium position, a
= 0.

The block continues to x = -A where its acceleration is
+kA/m.
4
Simple Harmonic Motion
F = -kx = ma ;
a = -(k/m)x
Choose x as the axis along which the oscillation occurs.
Acceleration a = d2x/dt2 = -(k/m)x = -ω2x
We let ω2 = k/m, then d2x/dt2 = -ω2x
x(t) = A1cos(ωt) + A2sin(ωt) = A cos(ωt+ϕ)
A1 and A2 are constants which are determined by the initial
conditions. The constants A and ϕ will be also determined by
initial conditions; A is the amplitude, and ϕ gives the phase
of the motion at t = 0.
5
Simple Harmonic Motion
Displacement is measured from the
equilibrium point.
Amplitude is the maximum displacement.
A cycle is a full to-and-fro motion.
Period is the time required to complete
one cycle.
Frequency is the number of cycles
completed per second.
6
Simple Harmonic Motion
The period, T, of the motion is the time interval required for the
particle to go through one full cycle of its motion.

The values of x and v for the particle at time t equal the values of x
and v at t + T.

k
m
2
m
T
 2

k
The inverse of the period is called the frequency. The frequency
represents the number of oscillations that the particle undergoes per
unit time interval.
1 
1
f  

T 2 2
k
m
Units are cycles per second = hertz (Hz).
7
Simple Harmonic Motion
Simple harmonic motion is one-dimensional and so directions can be
denoted by + or - sign. The velocity and acceleration for simple
harmonic motion can be found by differentiating the displacement:
x(t )  A cos(t   )
dx
v
 A sin(t   )
dt
d 2x
a  2   2 A cos(t   )
dt
The velocity is 90o out of phase with the
displacement and the acceleration is 180o out
of phase with the displacement.
8
Simple Harmonic Motion
Initial conditions at t = 0 are x(0)= A and v(0) = 0.
This means ϕ = 0.
The acceleration reaches extremes of ± ω2A at x =
±A.
The velocity reaches extremes of ± ωA at x = 0.
Initial conditions at t = 0 are x(0)=0 and v(0) = vi .
This means ϕ = - π/2 .
The graph is shifted one-quarter cycle to the right
compared to the graph of x(0) = A.
9
Simple Harmonic Motion
Because the sine and cosine functions oscillate between ±1, we can
easily find the maximum values of velocity and acceleration for an
object in SHM.
vmax
k
 A 
A
m
amax
k
 A A
m
2
The block continues to oscillate between –A and +A.

These are turning points of the motion.
The force is conservative.
In the absence of friction, the motion will continue forever.

Real systems are generally subject to friction, so they do not
actually oscillate forever.
10
Simple Harmonic Motion
If the spring is hung vertically, the only change is in the
equilibrium position, which is at the point where the spring
force equals the gravitational force.
F = - ky - mg = md2y/dt2
Set y’=y+mg/k, then
-ky’= md2y’/dt2, it is a SHM equation.
y’ (t) = Acos(ωt+ϕ) ; ω =(k/m)½
y(t) = Acos(ωt+ϕ) - mg/k
11
Example : A vibrating floor
A large motor in a factory causes the floor to vibrate at a
frequency of 10 Hz. The amplitude of the floor’s motion near
the motor is about 3.0 mm. Estimate the maximum
acceleration of the floor near the motor.
ω =2πf = 62.8 rad/s
amax= ω2A = 62.832*0.003 = 11.8 m/s2
12
Example : Car springs - 1
When a family of four with a total mass of 200 kg step into their 1200kg car, the car’s springs compress 3.0 cm. (a) What is the spring
constant of the car’s springs, assuming they act as a single spring? (b)
Determine the period and frequency of this car with total mass 1400 kg
after hitting a bump.
(a) F = 200*9.8 = k*0.03
k = 6.5×104N/m
(b) T = 2π/ω =2π(m/k)½
= 6.28*(1400/ 6.5×104)½ = 0.92s
f = 1/T = 1.09 Hz
13
Example : Watch Out for the Potholes
A car with a mass of 1300 kg is constructed so that its frame
is supported by four springs. Each spring has a force constant
of 20000 N/m. Two people riding in the car have a combined
mass of 160 kg. Find the frequency of vibration of the car
after it is driven over a pothole in the road.
Ftot = Σkix= kx; k=Σki = 80000 N/m
Mtot= 1300+160 =1460kg
f= (k/m)½ /(2π) = (80000 /1460)½ /(2π) = 1.18 Hz
14
Example : Loudspeaker
The cone of a loudspeaker oscillates in SHM at a frequency of 262 Hz
(“middle C”). The amplitude at the center of the cone is A = 1.5 × 10-4
m, and at t = 0, x = A. (a) What equation describes the motion of the
center of the cone? (b) What are the velocity and acceleration as a
function of time? (c) What is the position of the cone at t = 1.00 ms?
ω =2πf =1646 rad/s
(a) x(t) = 1.5 × 10-4 cos(1646t) m
(b) v(t) = dx/dt = -0.25 sin(1646t) m/s
a(t) = dv/dt = -410 cos(1646t) m/s2
(c) x(0.001s) = 1.5 × 10-4 cos(1.646) m=-1.2 × 10-5 m
15
Example：The oscillation function
The position of a particle moving along the x-axis is given by
x=0.08 sin(12t+0.3) m, where t is in seconds. (a) What are
the amplitude and period of the motion? (b) Determine the
position, velocity, and acceleration at t=0.6 s.
(a) A=0.08 m
T = 2π/12=0.524 s.
(b) x(t=0.6) = A sin(ωt+ϕ) = 0.075 m
v(t=0.6) = Aωcos(ωt+ϕ) = 0.333 m/s
a(t=0.6) = - ω2A sin(ωt+ϕ) = -10.8 m/s
16
Example : A Block-Spring System - 1
A 2-kg block is attached to a spring for which k=200 N/m. It is held at
an extension of 5 cm and then release at t=0. Find: (a) the displacement
as a function of time; (b) the velocity when x=+A/2; (c) the acceleration
when x=+A/2.
(a) k=200 N/m, m=2 kg, ω=10 rad/s, T= π/5 s, A= 0.05m
x(t) = A cos(ωt) = 0.05 cos(10t)
(b) v(t) = - ωA sin(ωt) = -0.5 sin(10t)
x(t) = A cos(ωt) = A/2; cos(ωt) = 0.5; sin(ωt) = ±0.86
v(t) = ±0.43 m/s
(c) a(t) = - ω2A cos(ωt) = -100*0.05*0.5 = -2.5m/s2
17
Example : A Block-Spring System - 2
A 200-g block connected to a light spring for which the force constant
is 5.00 N/m is free to oscillate on a frictionless, horizontal surface. The
block is displaced 5.00 cm from equilibrium and released from rest.
Find (a) the period of its motion? (b) the maximum speed of the block?
(c) the maximum acceleration of the block? (d) the position, velocity,
and acceleration as functions of time?
(a) k=5 N/m, m=0.2 kg, ω=5 rad/s, T= 2π/5 s =1.257s
(b) vmax= ωA =5*0.05 = 0.25 m/s
(c) amax= ω2A =25*0.05 =1.25 m/s2
(d) x(t) = A cos(ωt+ϕ) = 0.05 cos(5t) m
v(t) = - ωA sin(ωt+ϕ) = -0.25 sin(5t) m/s
a(t) = - ω2Acos(ωt+ϕ) = -1.25 cos(5t) m/s2
18
Example : A Block-Spring System - 3
A spring stretches 0.15m when a 0.30kg mass is gently attached to it. The spring is
then set up horizontally with the 0.30kg mass resting on a frictionless table. The mass
is pushed so that the spring is compressed 0.10m from the equilibrium point, and
released from rest. Determine: (a) the spring stiffness constant k and angular
frequency ω; (b) the amplitude of the horizontal oscillation A; (c) the magnitude of the
maximum velocity vmax; (d) the magnitude of the maximum acceleration amax of the
mass; (e) the period T and frequency f; (f) the displacement x as a function of time;
and (g) the velocity at t = 0.15s.
(a) k=mg/x = 0.3*9.8/0.15=19.6N/m; ω =(k/m)½ =8.08 rad/s
(b) A = 0.1m
(c) vmax= ωA = 8.08 *0.1 =0.808 m/s
(d) amax= ω2A = 8.082 *0.1 = 6.53 m/s2
(e) T= 2π/ω = 0.777s; f = 1/T =1.29Hz
(f) x(t) = -0.10 cos(8.08t) m
(g) v(t) = 8.08*0.10 sin(8.08*0.15) = 0.756 m/s
19
Example : Initial Conditions - 1
Suppose the initial position xi and initial velocity vi of a
harmonic oscillator of known angular frequency ω are given;
that is, x(0) = xi and v(0) = vi. Find general expressions for
the amplitude and the phase constant in terms of these initial
parameter.
x(t) = A0cos ωt +A1sin ωt = Acos(ωt+ϕ)
x(0) = A0= xi = Acosϕ;
v(t) = - ωA0sin ωt+ ω A1cos ωt = -ωAsin(ωt+ϕ);
v(0) = ω A1= vi = -ωA sinϕ; A1= vi/ω
A=(A02+ A12)½ = [xi2+ (vi/ω)2]½
ϕ = -tan-1(A1/A0) = -tan-1(vi/ωxi)
20
Example : Initial Conditions - 2
A spring with spring constant 56.0 N/m has a mass of m
=1.00 kg attached to its end. The mass is pulled +5.5 cm from
its equilibrium position and pushed so that it receives an
initial velocity of –0.32 m/s. (a) What is the equation of
motion for this oscillation? (b) What is the amplitude of this
oscillation?
(a) x(t) = A0cos ωt +A1sin ωt ; ω = (k/m)½ = 7.48 rads/s
A0 = 0.055; ωA1= -0.32; A1= -0.32/7.48 =-0.0428=-0.043;
x(t) =0.055 cos7.48t -0.043 sin7.48t
(b) A = (A02+A12)½ = 0.07 m
21
Example : Initial Conditions - 3
A spring stretches 0.15m when a 0.30kg mass is gently attached to it.
(where ω = 8.08 s-1 ) The spring is compressed 0.10m from equilibrium
(x0 = -0.10m) but is given a shove to create a velocity in the +x
direction of v0 = 0.40 m/s. Determine (a) the phase angle ϕ, (b) the
amplitude A, and (c) the displacement x as a function of time, x(t).
k=19.6N/m; ω =(k/m)½ = 8.08 rads/s
x(t) = A0cos ωt +A1sin ωt = Acos(ωt+ϕ);
A0 = -0.1; ωA1= 0.4; A1= 0.4/8.08 =0.0495;
x(t) = -0.1 cos ωt +0.0495 sin ωt = 0.112cos(ωt +3.6)
(a) ϕ = -tan-1(v0/ωx0) =3.6 rads
(b) A = 0.112 m
(c) x(t) = 0.112 cos(ωt+3.6) m
22
Example : Initial Conditions - 4
A spring with spring constant 5.0 N/m has a mass of m=0.2 kg attached
to its end. The mass is pulled +5.0 cm from its equilibrium position and
pushed so that it receives an initial velocity of –0.1 m/s. Determine (a)
the period T? (b) the maximum velocity? (c) the maximum acceleration?
(d) the displacement, velocity, and acceleration as a function of time?
(a) ω=5 rad/s, T= 2π/5 s =1.257s
(b) x(t) = A0cosωt+A1sinωt = Acos(ωt+ϕ); A0=0.05; ωA1=-0.1;
A1=-0.02; A=(A02+ A12)½ =0.0539m; ϕ =-tan-1(A1/A0)=0.121π
vmax= ωA =5*0.0539 = 0.269 m/s
(c) amax= ω2A =25*0.0539 =1.35 m/s2
(d) x(t) = A cos(ωt+ϕ) = 0.0539 cos(5t+0.121π) m
v(t) = - ωA sin(ωt+ϕ)= -0.269 sin(5t+0.121π) m/s
a(t) = - ω2A cos(ωt+ϕ)= -1.35 cos(5t+0.121π) m/s2
23
Example : Initial Conditions - 5
A 0.2kg block is attached to a spring for which k=5N/m. At t=π/10s, the
spring has a compression of 6 cm and block velocity v= -40cm/s. (a)
What is the displacement as a function of time; (b) What is the first time
at which the velocity is positive and 60% of its maximum value?
(a) k=5 N/m, m=0.2 kg, ω =5 rad/s, T= 2π/5 s
x(t) = A cos(ωt+ϕ); v(t) = -ωA sin(ωt+ϕ)
x(π/10) = A cos(π/2+ϕ) = -A sin(ϕ) = -0.06
v(π/10) = -5A sin(π/2+ϕ) = -5A cos(ϕ) = -0.4
tan(ϕ) = ¾ ; ϕ = 37o; A = 0.1
x(t) = 0.1 cos(ωt+ 37o)
(b) v(t) = -ωA sin(ωt+ϕ)= -0.5 sin(ωt + 37o)
sin(ωt+37o) = -0.6; ωt = π;
t = π/5 = 0.628s
24
Energy of the SHM Oscillator
Mechanical energy is associated with a system in which a particle
undergoes simple harmonic motion.
Assume the surface is frictionless, the system is isolated.

This tells us the total energy is constant.
The kinetic energy can be found by

K = ½ mv2 = ½ mω2A2sin2(ωt + ϕ)

Assume a massless spring, so the mass is the mass of the block.
The elastic potential energy can be found by

U = ½ kx2 = ½ kA2cos2(ωt + ϕ)
The total energy is E = K + U = ½ kA2 = ½ mω2A2
25
Energy in SHM, summary
26
Energy in the Simple Harmonic Oscillator
The total energy is, therefore, ½ k A2
And we can write:
E = K + U = ½ mv2 + ½ kx2 = ½ kA2
This can be solved for the velocity as a function of
position:
v = ±ω[A2 -x2]½ = ± ωA[1-(x/A)2]½ = ±vmax[1-(x/A)2]½
Where ω2 = k/m, and vmax = ωA.
27
Conceptual Example : Doubling the amplitude
Suppose this spring is stretched twice as far (to x = 2A).What
happens to (a) the energy of the system, (b) the maximum
velocity of the oscillating mass, (c) the maximum
acceleration of the mass?
(a) The total energy is proportional to the square of the
amplitude, so it goes up by a factor of 4.
(b) The velocity is doubled.
(c) The acceleration is doubled.
28
Example : A Block-Spring Energy
A 2-kg block is attached to a spring for which k=200 N/m. It
is held at an extension of 5 cm and then release at t=0. x(t) =
0.05 cos(10t) (a) Find K, U, and E at t = π/15s. (b) What is
the speed at x = A/2.
(a) x(t) = A cos(ω t); x(π/15) = 0.05 cos(2π/3) =-0.025m
v(t) = - ωA sin(ωt); v(π/15) = -0.5sin(2π/3) =-0.43m/s
K = ½ mv2; K = 0.1875J; U = ½ kx2; U = 0.0625J
E = ½ kA2 = 0.25J
(b) x(t) = A cos(ωt) = A/2; cos(ωt) =1/2; sin(ωt) = ±0.86
v(t) = -ωA sin(ωt) =±0.43 m/s
29
Example：The oscillation function
For the simple harmonic oscillation of Example (where
k=19.6N/m, A = 0.1m, x = -(0.1 m) cos8.08t, and v = (0.808
m/s) sin8.08t), determine (a) the total energy, (b) the kinetic
and potential energies as a function of time, (c) the velocity
when the mass is 0.050 m from equilibrium, (d) the kinetic
and potential energies at half amplitude (x = ± A/2).
(a) E = ½ kA2 = 9.80×10-2 J.
(b) U = ½ kx2 = (9.80×10-2 J) cos2 8.08t,
K = (9.80×10-2 J) sin2 8.08t,
(c) x=0.05m ; cosωt=-1/2; sinωt= ± 0.866;v = ± 0.70 m/s
(d) U = 2.45×10-2 J, K = E – U = 7.35×10-2 J
30
Example : Oscillations on a Horizontal Surface - 1
A 0.5kg cart connected to a light spring for which the force constant is
20N/m oscillates on a friction-less, horizontal air track. Calculate (a) the
maximum speed of the cart if the amplitude of the motion is 3.0cm ? (b)
the velocity of the cart when the position is 2.0cm? (c) the kinetic and
potential energies of the system when the position of the cart is 2.0cm?
(a) ω =(20/0.5)½ = 6.32 rads/s; x(t) = 0.03 cos(6.32t);
v(t) = - ω A sin(ωt) = -0.19 sin(6.32t); vmax = 0.19 m/s
(b) x(t) = 0.03cos(ωt) = 0.02; cos(ωt) =2/3; sin(ωt) =±0.745
v(t) = - ωA sin(ωt) = ±0.141 m/s
(c) K= ½ mv2 = ½ *0.5*0.1412 = 0.005J;
U= ½ kx2 = ½ *20*0.022 = 0.004J
31
Example : Oscillations on a Horizontal Surface - 2
A 0.50kg cart connected to a light spring for which the force
constant is 20N/m oscillates on a friction-less, horizontal air
track. xi = 3.0cm and vi = –0.10 m/s. What are the total energy
of the system, the new amplitude and maximum speed of the
cart?
(a) E = ½ mv02 + ½ kx02=½ *0.5*0.01+½ *20*9×10-4
= 0.0115 J
(b) A = [2E/k]½ =[0.023/20]½ = 0.0339 m
(c) vmax = [2E/m]½ = [0.023/0.5]½ = 0.214 m/s
32
SHM and Circular Motion
If we look at the projection onto the x axis of an object moving in a
circle of radius A at a constant speed vm = ωA, we find that the x
component of its velocity varies as:
vx = vm (1-x2/A2)½ = ωA (1-x2/A2)½
This is identical to SHM.
The circle is called a reference circle.

For comparing simple harmonic motion and uniform circular
motion.
Take P at t = 0 as the reference position.
Line OP makes an angle ϕ with the x axis at t = 0.
33
SHM and Circular Motion
The particle moves along the circle with constant
angular velocity ω
OP makes an angle θ with the x axis.

At some time, the angle between OP and the x axis
will be θ = ωt + ϕ
The points P and Q always have the same x coordinate.
x(t) = A cos θ = A cos(ωt + ϕ)
This shows that point Q moves with simple harmonic
motion along the x axis.
34
SHM and Circular Motion
The angular speed of P is the same as the angular
frequency of simple harmonic motion along the x axis.
Point Q has the same velocity as the x component of
point P. The x-component of the velocity is
vx = dx/dt = - ωA sin(ωt + ϕ)
The acceleration of point P on the reference circle is
directed radially inward. P ’s acceleration is a = ω2A
The x component is
ax = d2x/dt2 = - ω2Acos(ωt + ϕ) = - ω2x
This is also the acceleration of point Q along the x axis.
35
Simple Pendulum
A simple pendulum also exhibits periodic motion. It consists of a
particle-like bob of mass m suspended by a light string of length L.
The motion occurs in the vertical plane and is driven by gravitational
force. The motion is very close to that of the SHM oscillator.

If the angle is <10o
In the tangential direction,
Σ Fx = -mg sinθ =md2x/dt2 = md2s/dt2
Σ Fy =T -mg cosθ = mV2/L
s = Lθ ; d2θ/dt2 = -(g/L) sinθ
For small values of θ <10o, sinθ ～ θ ; d2θ/dt2 = -(g/L)θ = -ω2θ
The motion is the same as for SHM.
36
Simple Pendulum
d2θ/dt2 = -(g/L)θ = -ω2θ
θ = θmax cos(ω t + ϕ)
Angular velocity ω = (g/L)½
The period of this simple pendulum is
T = 2π(L/g)½
The period of a simple pendulum is only
related to gravitational acceleration and this
pendulum length, but has nothing to do with
the pendulum's mass and amplitude; but the
amplitude cannot be too large (θmax<10o ).
37
Simple Pendulum
38
Example : Measuring g
A geologist uses a simple pendulum that has a length of 37.10
cm and a frequency of 0.8190 Hz at a particular location on
the Earth. What is the acceleration of gravity at this location?
T = 1/f =1/0.819 = 1.221s = 2π(L/g)½
g = 4π2L/T2
= 4*9.87*0.371/1.2212
= 9.825 m/s2
39
Example：A simple pendulum - 1
Christian Huygens (1629–1695), the greatest clock- maker in
history, suggested that an international unit of length could be
defined as the length of a simple pendulum having a period
of exactly 1 s. How much shorter would our length unit be if
his suggestion had been followed?
T = 2π(L/g)½ = 1.00 s
L = T2g/4π2
= 1*9.8/(4*9.87)
= 0.248 m
40
Example：A simple pendulum - 2
What if Huygens had been born on another planet? What
would the value for g have to be on that planet such that the
meter based on Huygens’s pendulum would have the same
value as our meter?
g = 4π2L/T2
= 4*9.87*1.00/1.002
= 39.5 m/s2
41
Example：A simple pendulum - 3
The angular displacement of a simple pendulum is given by
θ(t) =0.05π sin(2πt+π/6) rad. The mass of bob is 0.4kg.
Calculate: (a) the length of the simple pendulum; and (b) the
velocity of the bob at t=0.25s.
(a) θ0=0.05π;ϕ= π/6; ω = 2π; ω2 = 4π2 = g/L;
L =g/(4π2) = 0.248m
(b) s = Lθ;
v(t) = ds/dt = L dθ/dt = 0.25*2π*0.05πcos(2πt+π/6)
v(0.25) = 0.025π2 cos(π/2+π/6) = -0.125 m/s
42
Example：A simple pendulum - 4
A simple pendulum in a clock is displaced to θi = 5.00°
(8.73×10-2 rad) and released from rest. When the pendulum
passes through its equilibrium position, the bob’s speed is
0.300 m/s. What are the length, angular frequency, and period
of the pendulum?
=(g/L)½ ; θ(t) =π/36 cos(t) = 8.73×10-2 cos(t);
v(t) = -Lπ/36 sin(t)= -(gL)½ * 8.73×10-2 sin(t)
vmax = (gL)½ θmax= (gL)½ * 8.73×10-2 = 0.3 m/s
L = (vmax/θmax)2/g = (0.3/8.73×10-2)2/9.8 =1.20 m
 = (g/L)½ = (9.8/1.2)½ = 2.86 rad/s
T = 2π/ = 2π/2.86= 2.20 s
43
Example: Restricted Pendulum
A pendulum of length 45.3 cm is hanging from the ceiling
and is restricted in its motion by a peg that is sticking out of
the wall 26.6 cm directly below its pivot point. What is its
period of oscillation?
Solve separately for each side ;
L1 = 45.3 cm ; L2 = 45.3 cm - 26.6 cm = 18.7 cm
T = (T1 + T2)/2 = π[(L1/g)½ + (L2/g)½ ]
= π[(0.453/9.8)½ + (0.187/9.8)½ ]
= 1.11 s
44
Physical Pendulum
If a hanging object oscillates about a fixed axis that does not pass
through the center of mass and the object cannot be approximated as a
point mass, the system is called a physical pendulum.

It cannot be treated as a simple pendulum.
The gravitational force provides a torque about an axis through O.
The magnitude of the torque is mgd sinθ
τ = -mgd sinθ = I α = Id2θ/dt2
Assuming θ is small,
d2θ/dt2 = -(mgd/I) sinθ ～-(mgd/I) θ = - ω2θ
Angular velocity ω = (mgd/I)½ ; I = ICM + md2
The period of this physical pendulum is T = 2π(I/mgd)½
45
Example : A Swinging Rod
A uniform rod of mass m and length L is freely pivoted at one
end. (a) What is the period of its oscillation? (b) What is the
length of a simple pendulum with the same period?
d = L/2
(a) I = 1/3mL2;
T = 2π(I/mgd)½ = 2π[(2L/3g)]½
(b) T = 2π[(2L/3g)]½
Leq = I/md = 2/3 L
46
Example : A Swinging Sign
A circular sign of mass M and radius R is huge on a nail from
a small loop located at one edge. After it is on the nail, the
sign oscillates in a vertical plane. Find the period of
oscillation if the amplitude of the motion is small.
I =ICM+Md2= ½ MR2+MR2 =3/2MR2
T = 2π(I/mgd)½ = 2π[(3R/2g)]½
47
The Torsional Pendulum
A torsional pendulum is one that twists rather than
swings. The restoring torque is τ = - κθ
τ = - κθ = Iα = Id2θ/dt2
d2θ/dt2 = - (κ/I) θ = - ω2 θ ;
ω2 = κ /I
(κ is a constant that depends on the wire.)
Period T = 2π(I/κ)½
48
Importance of Simple Harmonic Oscillators
Simple harmonic oscillators are good models of a wide variety
of physical phenomena.
Molecular example

If the atoms in the molecule do not move too far, the forces
between them can be modeled as if there were springs
between the atoms.

The potential energy acts similar to that of the SHM
oscillator.
49
Damped Oscillations
In many real systems, non-conservative forces are present.

This is no longer an ideal system (the type we have dealt with so far).

Friction and air resistance are common non-conservative forces.
In this case, the mechanical energy of the system diminishes in time, the
motion is said to be damped.
One example of damped motion occurs when an object is attached to a
spring and submerged in a viscous liquid.
The retarding force can be expressed as

R = - bv

b is a constant

b is called the damping coefficient
50
Damped Oscillation
retarding force R= -bv
ΣF=-kx-bv=- kx-bdx/dt=ma=md2x/dt2
m d2x/dt2 + b dx/dt + k x= 0; when b2 < 4mk
x(t) = Ae-γtcos(ω′t+ϕ) = Ae-bt/(2m)cos(ω′t+ϕ)
ω′=[k/m- b2/(4m2)]½
=[ω02 - b2/(4m2)]½
γ = b/(2m) ; ω02 = k/m
It is underdamped
51
Damped Harmonic Motion
If the restoring force is such that b/2m < ω0, the system is said to be
underdamped. When b reaches a critical value bc such that bc/2m= ω0,
the system will not oscillate. The system is said to be critically damped.
If the restoring force is such that b/2m > ω0, the system is said to be
overdamped.
m d2x/dt2 + b dx/dt + k x= 0;
If b2 = 4mk; (red line) critically damped
x(t)=Ae-γt+Bte-γt;
γ = b/(2m)
If b2 > 4mk ; (blue line) it is overdamped
x(t)=Ae-αt+Be-βt
α =b/(2m)+[b2/(4m2) - k/m]½ ;
β =b/(2m)-[b2/(4m2) - k/m]½ ;
52
Example : Simple pendulum with damping
A simple pendulum has a length of 1.0 m. It is set swinging
with small-amplitude oscillations. After 5.0 minutes, the
amplitude is only 50% of what it was initially. (a) What is the
value of γ for the motion? (b) By what factor does the
frequency, f’, differ from f, the undamped frequency?
(a) e-(γ300) = ½ ; γ =ln2/300 = 2.3×10-3 s-1
(b) ω0=(g/L)½ =3.13 s-1
ω′ = [3.132- (2.3×10-3)2]½ =3.13(1- ½ *5.4×10-7 )
(ω0- ω′)/ω0 = 2.7×10-7
53
Example :The damping spring
A 0.5kg block is attached to a spring (k=12.5 N/m). The
damped frequency is 0.2% lower than the natural frequency.
(a) What is the damping constant? (b) How does the
amplitude vary in time? (c) What is the critical damping
constant?
(a) ω0 = (k/m)½ = 5 rad/s; ω = 0.998 ω0 = 4.99 rad/s
γ2 = (b/2m)2 = (ω02 – ω2) ; γ = 0.316 s-1
b = 2mγ = 0.316 kg/s
(b) A(t) = A0 e-0.316t
(c) The critical damping constant bc = 2m ω0 = 5kg/s
54
Example: Critical Damping
A spring with a spring constant of k=1.0 N/m, has a mass m=1.0 kg
attached to it, and moves in a medium with damping constant b=2.0
kg/s. The mass is released from rest at a position of +5 cm from
equilibrium. Where will it be after a time of 1.75 s?
b2 = 4kg2/s2 = 4*1.0N/m*1.0kg = 4km ; If b2 = 4mk; critically damped;
x(t)=Ae-γt+Bte-γt; γ = b/(2m) = 1.0 s-1;
v(t) = dx/dt = - γAe-γt+Be-γt - γBte-γt = (B- γA)e-γt - γBte-γt
x(0) = 0.05m ; v(0) = 0; A = 0.05 m ; B = γA = 0.05 m/s
x(t) = 0.05 m*(1+γt)e-γt
x(1,75) = 0.05 m*(1+1.75)e-1.75 = 0.0239 m = 2.39 cm
55
Forced Oscillations
Forced vibrations occur when there is a periodic driving
force. This force may or may not have the same period as the
natural frequency of the system.
If the frequency is the same as the natural frequency, the
amplitude can become quite large. This is called resonance.
56
Forced Oscillations
ΣF=-kx-bv+F0cos(ωt) = ma = md2x/dt2
x(t)=Acos(ωt)+Bsin(ωt) =A0cos(ωt+ ϕ0);
v(t)= -ωAsin(ωt)+ ωBcos(ωt) ;
a(t)= -ω2[Acos(ωt) +Bsin(ωt)];
A0 = F0/[m2(ω02- ω2)2+b2ω2]½
ω0 = (k/m)½
ϕ0 = tan-1[ m(ω02- ω2)/(ωb)]
Q = mω0/b
When ω=ω0, the oscillatory system is at "resonance“.
57
Resonance
When the frequency of the driving force is near the natural
frequency (ω ≒ ω0) an increase in amplitude occurs.
This dramatic increase in the amplitude is called resonance.
The natural frequency ω0 is also called the resonance
frequency of the system.
At resonance, the applied force is in phase with the velocity
and the power transferred to the oscillator is a maximum.

The applied force and v are both proportional to sin(t + ϕ).

The power delivered is F·v

This is a maximum when the force and velocity are in phase.

The power transferred to the oscillator is a maximum.
58
Example : Forced Oscillations
Suppose we use a driving force of the form F0cos(ωt) to compensate for
the mechanical energy lost by their damped harmonic oscillator. We set
the driving frequency equal to the natural frequency of their oscillator.
Recall that the disk’s mass is 0.100 kg and that the time constant of
their damped oscillator is 1.5 s. If the period T = 2.0 s and the amplitude
A0 = 0.40 m of the motion is the same as that of their simple harmonic
oscillator, find F0.
A0 = F0/{m[(ω02- ω2)2+ b2ω2/m2]½ } = F0/ [m2(ω02- ω2)2+b2ω2]½ ;
when ω = ω0 = 2π/T = π; A0 =F0/(bω)
τ = 2m/b = 1.5 s ; b = 2m/τ = 0.133 N·s/m
F0 = bωA0 = 0.133* π *0.4 = 0.168 N
59
Physics
Chapter 12
Fluid Mechanics
授課老師施坤龍
0
States of Matter
Solid

Has a definite volume and shape
Liquid

Has a definite volume but not a definite shape
Gas – unconfined

Has neither a definite volume nor shape
All of the previous definitions are somewhat artificial
More generally, the time it takes a particular substance to
change its shape in response to an external force determines
whether the substance is treated as a solid, liquid or gas
1
Volume and Density
A system is also characterized by its density. Suppose you
have several blocks of copper, each of different size. Each
block has a different mass m and a different volume V.
The most important such parameter is the ratio of mass to
volume, which we call the mass density ρ :
ρ = m/V
The SI units of mass density are kg/m3. Nonetheless , units of
g/cm3 are widely used.
2
Densities of some common substances
Material
Density (kg/m3) Material
Density (kg/m3)
Helium gas
0.179
Aluminum
2.70×103
Air (1 atm, 200 C)
1.29
Iron, steel
7.80×103
Oxygen gas
1.43
Brass
8.60×103
Ethanol
0.81×103
Copper
8.90×103
Benzene
0.90×103
Silver
10.8×103
Ice
0.92×103
Lead
11.3×103
Water
1.00×103
Mercury
13.6×103
Seawater
1.03×103
Gold
19.3×103
Blood
1.06×103
Platinum
21.4×103
Glycerine
1.26×103
Neutron star
1018
3
Example: The weight of a roomful of air
Find the mass and weight of the air at 200 C in a room with a
4.0 m × 5.0 m floor and a ceiling 3.0 m high, and the mass
and weight of an equal volume of water.
V = 4.0 m × 5.0 m × 3.0 m = 60 m3
Mair = ρairV = 1.29 kg/m × 60 m3 = 77.4 kg
Wair = 77.4 kg * 9.8 m/s2 = 758.5 N
Mw = ρwV = (103 kg/m3) × 60 m3 = 6.0 × 104 kg
Ww = 6.0 × 104 kg * 9.8 m/s2 = 5.88 × 105 N
4
Fluids
A fluid is a collection of molecules that are randomly
arranged and held together by weak cohesive forces
and by forces exerted by the walls of a container
Both liquids and gases are fluids
5
Forces in Fluids
A simplification model will be used

The fluids will be non-viscous

The fluids do no sustain shearing forces

The fluid cannot be modeled as a rigid object
The only type of force that can exist in a fluid is one that is
perpendicular to a surface
The forces arise from the collisions of the fluid molecules with the
surface

Impulse-momentum theorem and Newton’s Third Law show the
force exerted
6
Pressure
The pressure, P, of the fluid at the
level to which the device has been
submerged is the ratio of the force
to the area
F
P
A
7
Pressure
Pressure is a scalar quantity

Because it is proportional to the magnitude of the force
Pressure compared to force

A large force can exert a small pressure if the area is very
large
Units of pressure are Pascals (1 Pa = 1
N/m2)
N
1Pa  1 2
m
Pressure is a scalar and force is a vector
The direction of the force producing a pressure is
perpendicular to the area of interest
8
Atmospheric Pressure
The atmosphere exerts a pressure on the surface of
the Earth and all objects at the surface
Atmospheric pressure is generally taken to be
1 atm = 1.013×105 Pa = Po
Atmospheric pressure is often quoted in non-SI units
1 atm = 760 torr = 760 mm Hg = 29.92 in. Hg
= 14.7 lb/in2
9
Measuring Pressure
The pressure in a fluid is measured with a pressure
gauge.
The fluid pushes against some sort of spring, and the
spring's displacement is registered by a pointer on a dial.
The spring is calibrated by a known force.
The force due to the fluid presses on the top of the piston
and compresses the spring. The force of the fluid exerts
on the piston is then measured.
Many pressure gauges, such as tire gauges and the
gauges on air tanks, measure not the actual or absolute
pressure p but what is called gauge pressure. The gauge
pressure Pg is the pressure in excess of 1 atm. That is,
Pg= p - 1 atm or p = Pg + 1 atm.
10
Pressure Measurements
Invented by Torricelli. A long closed tube is filled with mercury and
inverted in a dish of mercury

The closed end is nearly a vacuum
Measures atmospheric pressure as
Po = Hggh
1 atm = 0.760 m (of Hg)
A device for measuring the pressure of a gas contained in a vessel
One end of the U-shaped tube is open to the atmosphere
The other end is connected to the pressure to be measured
Pressure at B is Po+ gh
11
Conceptual Example : Suction
A student suggests suction-cup shoes for Space Shuttle
astronauts working on the exterior of a spacecraft. Having
just studied this Chapter, you gently remind him of the
fallacy of this plan. What is it?
Suction cups work because of air pressure, and there isn’t any
air where the shuttle orbits.
12
Example : An underwater pressure gauge
An underwater pressure gauge reads 60 kPa. What is its
depth?
Pg = P - 1 atm = (1 atm +ρgd) 一1 atm =ρgd
60000 Pa = 1000*9.8*d
d = 60000/(1000*9.8) = 6.12 m
13
Example : The Water Bed
The mattress of a water bed is 2.00 m long by 2.00 m wide and 30.0 cm
deep. Find (a) the weight of the water in the mattress? (b) the pressure
exerted by the water bed on the floor when the bed rests in its normal
position? Assume the entire tower surface of the bed makes contact
with the floor.
M  V  1000kg / m *1.2m  1.20  10 kg
3
3
3
Mg  1.20  10 kg * 9.8m / s  1.18  10 N
3
2
4
1.18  10 N
3
P
 2.94  10 Pa
2
4.00m
4
14
Pressure and Depth
Fluids have pressure that vary with depth
If a fluid is at rest in a container, all portions of
the fluid must be in static equilibrium
All points at the same depth must be at the same
pressure

Otherwise, the fluid would not be in equilibrium
Examine the darker region, assumed to be a fluid

It has a cross-sectional area A

Extends to a depth h below the surface
Three external forces act on the region
15
Pressure and Depth
The liquid has a density of ρ

Assume the density is the same throughout the fluid

This means it is an incompressible liquid
The three forces are:

Downward force on the top, PoA

Upward on the bottom, PA

Gravity acting downward, mg

The mass can be found from the density:

m = ρV = ρAh
16
Pressure and Depth
Since the fluid is in equilibrium,

∑Fy = 0 gives PA – PoA – mg = 0
Solving for the pressure gives
P = Po + gh
The pressure P at a depth h below a point in the liquid at
which the pressure is Po is greater by an amount gh
If the liquid is open to the atmosphere, and Po is the pressure
at the surface of the liquid, then Po is atmospheric pressure
The pressure is the same at all points having the same depth,
independent of the shape of the container
17
Conceptual Example : Finger holds water in a straw
You insert a straw of length ℓ into a tall glass of water. You place your
finger over the top of the straw, capturing some air above the water but
preventing any additional air from getting in or out, and then you lift the
straw from the water. You find that the straw retains most of the water.
Does the air in the space between your finger and the top of the water
have a pressure P that is greater than, equal to, or less than the
atmospheric pressure P0 outside the straw?
There must be a net upward force on the water
in the straw to keep it from falling out;
therefore the pressure in the space above the
water must be less than atmospheric pressure.
18
Example : Pressure at a faucet
The surface of the water in a storage tank is 30 m above a
water faucet in the kitchen of a house. Calculate the
difference in water pressure between the faucet and the
surface of the water in the tank.
Δ P = ρgh =1000*9.8*30=2.94×105N/m2
19
Example : A Pain in Your Ear
Estimate the force exerted on your eardrum due to the water
when you are swimming at the bottom of a pool that is 5.0 m
deep.
Pbot  P0  gh  1000 * 9.8 * 5.0  4.9  10 Pa
4
F  ( Pbot  P) A  4.9  104 *1.0  104  4.9 N  5N
20
Example - Pressure Below the Surface
A submarine’s maximum diving depth is 500 m. What is the
pressure on the hull at that depth?
P  P0  gh;
  1030kg / m
USS Seawolf
3
P  1.0110 Pa  1030kg / m * 9.8m / s * 500m
5
3
2
P  5.15 10  51atm
6
21
Example : Force on aquarium window
Calculate the force due to water pressure exerted on a 1.0 m ×
3.0 m aquarium viewing window whose top edge is 1.0 m
below the water surface.
2
1
F = ∫Pℓdy = ∫ ρgyℓdy
= ½ ρgℓy2|21
= 500*9.8*3*3
= 44100 N
22
Example : The Force on a Dam
Water is filled to a height H behind a clam of width w.
Determine the resultant force on the dam.
P  gh  g ( H  y )
dF  PdA  g ( H  y ) wdy
F   PdA  
H
0
1
2
g ( H  y ) wdy  gwH
2
23
Example : The Force on a Dam 2
What if you were asked to find this force without using
calculus? How could you determine its value?
1
Pavg  ( Ptop  P bottom) / 2  (0  gH ) / 2  gH
2
1
1
2
F  Pavg A  gH * Hw  gwH
2
2
24
Barometric Altitude Relation for Gases
In the derivation of the depth pressure relationship, we have
made use of the incompressibility of liquids
However, if our fluid is a gas, we cannot make this
assumption
We start again with a thin layer of fluid in a fluid column
The pressure difference between bottom and top surface is
still given by the weight of the thin layer of fluid divided by
the area
F
mg
Vg
 ( Ah) g
p    


  gh
A
A
A
A
25
Barometric Altitude Relation for Gases
The negative sign comes from the fact that the pressure
decreases with increasing altitude, because the weight of the
fluid column above is reduced
So far nothing is different from the derivation of the
incompressible case
However, for compressible fluids we find that the density is
proportional to the pressure

p

 0 p0
Strictly this relationship is only true for ideal gases
26
Barometric Altitude Relation for Gases
Combining our two equations gives us
p   gh;

p
 ;
0 p0
g0
p
  g  
p
h
p0
Taking the limit of h0, we find the equation
g 0
dp

p
dh
p0
We separate the variables for this differential equation, then
we get solution of this differential equation is
p(h)  p0e
 hg 0/ p0
This equation is known as the barometric pressure formula. It
relates altitude to pressure in gases. It applies as long as the
temperature does not change with altitude and gravitation is
constant
27
Example: Air Pressure on Mount Everest
What is the air pressure on the top of Mount Everest?
We use the barometric pressure formula
p(h)  p0e

hg 0
p0
 p0e

h
p0 /( g 0)
The density of and pressure of air at sea level
0  1.229 kg/m 3 ;
p0  1.01105 Pa
We can find the constant term in the barometric pressure
formula
5
p0
1.0110 Pa

 8377m
3
2
0 g 1.229 kg/m  9.8m / s
28
Example: Air Pressure on Mount Everest
For the Earth’s atmosphere we can write an expression for the
pressure at a given height as
p(h)  p0e

h
8377
The height of Mount Everest is 8850 m (29,035 ft) so the
pressure at the summit is
p(8850)  p0e

8850
8377
 0.347 p0  35 kPa
Which is 35% of the pressure at sea level.
This low air pressure is dangerous if you are not acclimated.

Everest climbers take months to acclimate and many still
have trouble
29
How Close?

p

 0 p0
Since
the barometric pressure formula
also means
 ( h)   0 e
p(h)  p0e


hg 0
p0
hg 0
p0
 p0e

h
8377
30
Example : Elevation effect on atmospheric pressure
(a) Determine the variation in pressure in the Earth’s atmosphere as a
function of height y above sea level, assuming g is constant and that the
density of the air is proportional to the pressure. (This last assumption is
not terribly accurate, in part because temperature and other weather
effects are important.) (b) At what elevation is the air pressure equal to
half the pressure at sea level?
(a) ρ/ρ0 = P/P0;
dP = ρgdy;
dP/P = -(ρ0/P0) gdy
P = P0e-(ρ0gy/P0)
(b) P = ½ P0 ; e-(ρ0gy/P0) = ½ ;
y =P0 ln2/(ρ0g) = 1.013×105 *0.69/(1.29*9.8)=5553 m.
31
Pascal’s Law
The pressure Po in a fluid depends on depth
A change in pressure at the surface must be transmitted to
every other point in the fluid.
This is the basis of Pascal’s Law
Named for French scientist Blaise Pascal
A change in the pressure applied to a fluid is transmitted to
every point of the fluid and to the walls of the container
P1  P2 ;
F1 F2

A1 A2
32
Pascal’s Law
This is a hydraulic press. A large output force can be applied
by means of a small input force
The volume of liquid pushed down on the left must equal the
volume pushed up on the right
Since the volumes are equal,
A1 x1 = A2 x2
Combining the equations,

F1 x1 = F2 x2 which means W1 = W2

This is a consequence of Conservation of Energy
Pascal’s Law, Other Applications: Hydraulic brakes, Car
lifts, Hydraulic jacks, Forklifts
33
Example : The Car Lift
In a car lift used in a service station, compressed air exerts a
force on a small piston that has a circular cross section and a
radius of 5.00 cm. This pressure is transmitted by a liquid to a
piston that has a radius of 15.0 cm. What force must the
compressed air exert to lift a car weighing 13300 N? What air
pressure produces this force?
A1
 0.05
F1 
F2 
13300 N  1478 N
2
A2
 0.15
2
F1
1478 N
5
P

 1.88  10 Pa
2
2
A1  0.05 m
34
Example : Lifting a Car
A car lift at a garage is filled with oil of density 900 kg/m3. We want to
lift a car with mass 1500 kg to a height of 2.50 m. The lift works by
pumping oil through a small hole of diameter 5 cm and lifting the car
with a piston with a diameter of 25 cm. How much pressure (gauge) do
we need in the hose connecting the pump to the lifting cylinder?
Fout = mg = 1500 *9.8 = 14700 N
Ain = π(0.025)2= 0.00196 m2 ; Aout = π(0.125)2 = 0.0491 m2
Pout = mg/Aout = 14700 N/ 0.0491 m2 = 2.99 ×105 Pa
Pin = Pout + ρgh = 2.99 ×105 Pa +900*9.8*2.5
= 3.21×105 Pa
35
Archimedes
ca 289 – 212 BC
Greek mathematician, physicist
and engineer
Computed the ratio of a circle’s
circumference to its diameter
Calculated the areas and volumes
of various geometric shapes
Famous for buoyant force studies
36
Buoyant Force
The buoyant force is the upward force exerted by a fluid on
any immersed object
The object is in equilibrium
There must be an upward force to balance the downward
force
The upward force must equal (in magnitude) the downward
gravitational force
The upward force is called the buoyant force
The buoyant force is the resultant force due to all forces
applied by the fluid surrounding the object
37
Archimedes’ Principle
Any object completely or partially submerged in a fluid experiences an
upward buoyant force whose magnitude is equal to the weight of the
fluid displaced by the object. This is called Archimedes’ Principle
The pressure at the top of the cube causes a downward force of Pt A
The pressure at the bottom of the cube causes an upward force of Pb A
B = (Pb – Pt ) A = mg
38
Archimedes's Principle: Totally Submerged Object
An object is totally submerged in a fluid of density f
The upward buoyant force is B = fgVf = fgVo
The downward gravitational force is w = mg = ogVo
The net force is: B-w = (f -o) gVoj
If the density of the object is less than the density of the fluid, the
unsupported object accelerates upward
If the density of the object is more than the density of the fluid, the
unsupported object sinks
The motion of an object in a fluid is determined by the densities of the
fluid and the object
39
Archimedes’ Principle: Floating Object
The object is in static equilibrium
The upward buoyant force is balanced by the downward
force of gravity
Volume of the fluid displaced corresponds to the volume of
the object beneath the fluid level
0 V

 f V0
The fraction of the volume of a floating object that is below
the fluid surface is equal to the ratio of the density of the
object to that of the fluid
40
Archimedes’ Principle, Crown Example
Archimedes was (supposedly) asked, “Is the crown gold?”
Weight in air = 7.84 N
Weight in water (submerged) = 6.84 N
Buoyant force will equal the apparent weight loss

Difference in scale readings will be the buoyant force
∑F = B + T2 - Fg = 0
B = Fg – T2

Weight in air – “weight” submerged
Archimedes’ Principle says B = gV
Then to find the material of the crown:
crown = mcrown in air / V
41
Conceptual Example : Two pails of water
Consider two identical pails of water filled to the brim. One
pail contains only water, the other has a piece of wood
floating in it. Which pail has the greater weight?
Both weigh the same; if both pails were full to the brim
before the wood was put in, some water will have spilled out.
42
Example : Recovering a submerged statue
A 70-kg ancient statue lies at the bottom of the sea. Its
volume is 3.0 × 104 cm3. How much force is needed to lift it?
The buoyant force is equal to the weight of the water, which
is 300 N; therefore the force needed to lift the statue is 390 N
(whereas its weight is 690 N).
43
Example : Is the crown gold? 1
When a crown of mass 14.7 kg is submerged in water, an accurate scale
reads only 13.4 kg. Is the crown made of gold?
F  B T
2
 Fg  0
B  Fg  T2  144.06  131.32  12.74 N
mc mc g
mc g
mc g w
c 



Vc Vc g B /  w
B
144.06 N * 1000kgm3
c 
 11.3  103 kgm3
12.74 N
The density of gold is 19.3  103 kg/m3.
Either the crown was hollow, or it was not made of pure gold.
44
Example : Is the crown gold? 2
Archimedes supposedly was asked to determine whether a crown made
for the king consisted of pure gold. According to legend, he solved this
problem by weighing the crown first in air and then in water as shown
in Figure. Suppose the scale read 7.84 N when the crown was in air and
6.84 N when it was in water. What should Archimedes have told the
king?
 F  B  T2  Fg  0
B  Fg  T2  7.84  6.84  1.00 N
c 
mc mc g
mg
m g

 c  c w
Vc Vc g B /  w
B
7.84 N * 1000kgm3
c 
 7.84  103 kgm3
1.00 N
The density of gold is 19.3  103 kg/m3.
Either the crown was hollow, or it was not made of pure gold.
45
Example : Is the crown gold? 3
Suppose the crown has the same weight but is indeed pure
gold and not hollow. What would the scale reading be when
the crown is immersed in water?
B   wVw g   wVc g   w
mc g
c
7.84 N
B  1000kgm
 0.406 N
3
3
1.93  10 kgm
T2  Fg  B  7.84 N  0.406 N  7.43N
3
46
Example : A Titanic Surprise
An iceberg floating in seawater as shown in Figure is extremely
dangerous because most of the ice is below the surface. This hidden ice
can damage a ship that is still a considerable distance from the visible
ice. What fraction of the iceberg lies below the water level?
f 
Vdisp
Vice

ice
sea water
917kgm3

 0.89 or 89%
3
1030kgm
Therefore, the visible fraction of ice above the
water’s surface is about 11%. It is the unseen
89% below the water that represents the danger
to a passing ship.
47
Example : Hydrometer calibration
A hydrometer is a simple instrument used to measure
the specific gravity of a liquid by indicating how
deeply the instrument sinks in the liquid. This
hydrometer consists of a glass tube, weighted at the
bottom, which is 25.0 cm long and 2.00 cm2 in cross-
sectional area, and has a mass of 45.0 g. How far
from the end should the 1.000 mark be placed?
The hydrometer has a density of 0.900 g/cm3; it will
float with 25.0 cm*0.9= 22.5 cm submerged. The
mark should be placed 22.5 cm from the bottom.
48
Example : Helium balloon 1
What volume V of helium is needed if a
balloon is to lift a load of 180 kg (including
the weight of the empty balloon)?
The buoyant force, which is the weight of the
displaced air, must be equal to at least the
weight of the helium plus the weight of the
balloon. This gives 160 m3. More volume
would be needed at higher altitude, where the
density of air is less.
49
Example: Helium balloon 2
A party balloon is filled with helium. The balloon has a
volume of 0.150 m3. The density of air is 1.30 kg/m3. The
density of helium is 0.179 kg/m3. The mass of the balloon
empty is 50.0 g. What is the net upward force on the balloon?
The buoyant force is
FB  airVballoomg  1.30kg / m3  0.15m3  9.8m / s 2  1.911N
The weight of the balloon is
Wballoom  mballoomg  0.050 kg  9.8m/s 2  0.49 N
The weight of the helium is
Whelium  mheliumg  0.179 kg/m 3  0.150m3  9.8m/s 2  0.263N
Therefore the net upward force is
Fnet  FB  Wballoom  Whelium  1.91N  0.49 N  0.263N  1.16 N
50
Examples of Flow
Laminar Flow
Laminar to
Turbulent Flow
Turbulent Flow
51
Types of Fluid Flow
Laminar flow

Steady flow

Each particle of the fluid follows a smooth path

The paths of the different particles never cross each other

The path taken by the particles is called a streamline
Turbulent flow
An irregular flow characterized by small whirlpool like
regions
Turbulent flow occurs when the particles go above some
critical speed
52
Ideal Fluid Flow
The motion of real-life fluids is complicated
Here we will make some simplifying assumptions that will allow us to
reach some relevant conclusions concerning the motion of fluids
There are four simplifying assumptions made to the complex flow of
fluids to make the analysis easier

Incompressible flow – the density remains constant

Non-viscous flow – internal friction is neglected

Laminar flow (steady) – the velocity of each point remains constant

Irrotational flow – the fluid has no angular momentum about any point;
no turbulence
The first two assumptions are properties of the ideal fluid
The last two assumptions are descriptions of the way the fluid flows
53
Equation of Continuity
The path the particle takes in steady flow is a streamline
The velocity of the particle is tangent to the streamline
No two streamlines can cross
Consider a fluid moving through a pipe of nonuniform size (diameter)
The particles move along streamlines in steady flow
The mass that crosses A1 in some time interval is the same as the mass
that crosses A2 in that same time interval
54
Equation of Continuity
Analyze the motion using the nonisolated system in a steady-
state model
Since the fluid is incompressible, the volume is a constant
ρ1A1v1 =ρ2 A2v2

This is called the equation of continuity for fluids

The product of the area and the fluid speed at all points
along a pipe is constant for an incompressible fluid
The speed is high where the tube is constricted (small A)
The speed is low where the tube is wide (large A)
55
Example : Watering a Garden
A gardener uses a water hose 2.50 cm in diameter to fill a 30.0-L bucket. The gardener
notes that it takes 1.00 min to fill the bucket. A nozzle with an opening of crosssectional area 0.500 cm2 is then attached to the hose. The nozzle is held so that water
is projected horizontally from a point 1.00 m above the ground. Over what horizontal
distance can the water be projected?
3
3
dV
30
.
0

10
cm
 30 L min 1 
 500cm 3 / s
dt
60.0s
dV / dt
500cm 3 / s
v1 

 102cm / s  1.02m / s
2
2
A1
 (2.50 / 2) cm
A1
A1
 (2.50 / 2) 2 cm 2
v2 
v1 
v1 
1.02m / s  10.0m / s
2
A2
A2
0.500cm
2h
2 * 1.00
t

 0.452 s
g
9.80
S  v2t  10.0 * 0.452  4.52m
56
Example : Blood flow
In humans, blood flows from the heart into the aorta, from
which it passes into the major arteries. These branch into
the small arteries (arterioles), which in turn branch into
myriads of tiny capillaries. The blood returns to the heart
via the veins. The radius of the aorta is about 1.2 cm, and
the blood passing through it has a speed of about 40 cm/s. A
typical capillary has a radius of about 4×10-4 cm, and blood
flows through it at a speed of about 5×10-4 m/s. Estimate
the number of capillaries that are in the body.
Use the equation of continuity, assuming the density of
blood is constant; the total area of the capillaries is the area
of one capillary multiplied by the number of capillaries.
This gives about 7×109 capillaries.
57
Example : Heating duct to a room
What area must a heating duct have if air moving 3.0 m/s
along it can replenish the air every 15 minutes in a room of
volume 300 m3? Assume the air’s density remains constant.
Think of the room as a wider part of the duct, and apply the
equation of continuity. This gives an area of 0.11 m2.
58
Daniel Bernoulli
1700 – 1782
Swiss
mathematician
and
physicist
Made important discoveries
involving fluid dynamics
Also worked with gases
59
Bernoulli’s Equation
As a fluid moves through a region where its speed and/or
elevation above the Earth’s surface changes, the pressure in
the fluid varies with these changes
The relationship between fluid speed, pressure and elevation
was first derived by Daniel Bernoulli
Consider the two shaded segments
The volumes of both segments are equal
The net work done on the segment is W=(P1 – P2) V
Part of the work goes into changing the kinetic energy and
some to changing the gravitational potential energy
60
Bernoulli’s Equation
The change in kinetic energy:

∆K = ½ m v22 - ½ m v12

There is no change in the kinetic energy of the unshaded
portion since we are assuming streamline flow

The masses are the same since the volumes are the same
The change in gravitational potential energy:

∆U = mgy2 – mgy1
The work also equals the change in energy
Combining:
W = (P1 – P2)V= ½ m v22 - ½ m v12 + mgy2 – mgy1
61
Bernoulli’s Equation
Rearranging and expressing in terms of density:
P1 + ½ ρ v12 + ρ g y1 = P2 + ½ ρ v22 + ρ g y2
This is Bernoulli’s Equation and is often expressed as
P + ½ ρ v2 + ρ g y = constant
When the fluid is at rest, this becomes P1 – P2 = ρgh which is
consistent with the pressure variation with depth we found
earlier
The general behavior of pressure with speed is true even for
gases

As the speed increases, the pressure decreases
62
Curve Ball
A baseball thrown with spin curves
An analysis with Bernoulli’s principle leads us in the wrong
direction
63
Curve Ball - Magnus Effect
The Magnus effect explains the curving of the baseball
Air molecules collide with fast moving boundary layer air
molecules and transfer more energy when moving against the
motion of the boundary layer than when moving with the
motion of the boundary layer
64
Why Does an Airplane Fly?
Mass of fully loaded and fueled 747: 350 tons
Weight: 3.4 MN
Maximum combined engine thrust: 0.9 MN (!!!)
65
Applications of Fluid Dynamics
Streamline flow around a moving
airplane wing
Lift is the upward force on the wing
from the air
Drag is the resistance
The lift depends on the speed of the
airplane, the area of the wing, its
curvature, the angle between the wing
and the horizontal
66
Lift – General
In general, an object moving through a fluid experiences lift
as a result of any effect that causes the fluid to change its
direction as it flows past the object
Some factors that influence lift are

The shape of the object

Its orientation with respect to the fluid flow

Any spinning of the object

The texture of its surface
67
Atomizer
A stream of air passes over one end
of an open tube
The other end is immersed in a
liquid
The
moving
air
reduces
the
pressure above the tube
The fluid rises into the air stream
The liquid is dispersed into a fine
spray of droplets
68
Example: Bernoulli’s Equation
We squeeze the bulb of a perfume atomizer, causing air to
flow horizontally across the opening of a tube that extends
down into the perfume. If the air is moving at 50.0 m/s, what
is the pressure difference at the top of the tube?
Before we squeze the blub, v0  0
1 2
1 2
p0  p0  v0  p  v
2
2
1 2
p  p0  v
2
1 2 1.30kg / m3  (50.0m / s ) 2
| p  p0 | v 
 1630 Pa
2
2
69
Example : A hot-water heating system
Water circulates throughout a house in a hot-water heating system. If
the water is pumped at a speed of 0.5 m/s through a 4.0-cm-diameter
pipe in the basement under a pressure of 3.0 atm, what will be the flow
speed and pressure in a 2.6-cm-diameter pipe on the second floor 5.0 m
above? Assume the pipes do not divide into branches.
A1
 0.022
v2 
v1 
0.5m / s  1.18m / s
2
A2
 0.013
1
1
2
P1  v1  gh1  P2  v22  gh2
2
2
1
5
P2  3 * 1.013  10  1000 * (0.52  1.182 )  1000 * 9.8 * 5
2
P2  2.54  105 N / m 2  2.5atm
Get the flow speed as 1.2 m/s and the pressure as 2.5 atm.
70
Example : The Venturi Tube
The horizontal constricted pipe illustrated in Figure, known as a Venturi
tube, can be used to measure the flow speed of an incompressible fluid.
Determine the flow speed at point 2 of Figure if the pressure difference
P1 –P2 is known.
1
1
2
P1  v1  P2  v22
2
2
A2
v1 
v2
A1
1 A22 2
1
P1   2 v2  P2  v22
2 A1
2
v2  A1
2( P1  P2 )
 ( A12  A22 )
71
Example : Torricelli’s Law
An enclosed tank containing a liquid of density  has a hole in its side
at a distance y1 from the tank’s bottom. The hole is open to the
atmosphere, and its diameter is much smaller than the diameter of the
tank. The air above the liquid is maintained at a pressure P. Determine
the speed of the liquid as it leaves the hole when the liquid’s level is a
distance h above the hole.
1 2
P0  v1  gy1  P  gy2
2
2( P  P0 )
v1 
 2 gh

As P  P0 ; v1  2 gh
72
Example : Torricelli’s Law 2
What if the position of the hole in Figure could be adjusted
vertically? If the tank is open to the atmosphere and sitting on
a table, what position of the hole would cause the water to
land on the table at the farthest distance from the tank?
2 y1
t
g
v1  2 gh  2 g ( y2  y1 )
x f  v1t  2 y1 ( y2  y1 )  2 ( y1 y2  y12 )
dx f
dy1

y2  2 y1
( y1 y2  y )
1
y1  y2
2
2
1
0
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Viscosity
If you have ever been drifting on a gentle river in a boat, you may have
noticed that you were floating faster with the stream in the middle of
the river than very close to the bank
If water were an ideal fluid in laminar flow in this river, then it should
make no difference how far away from shore you are
It turns out that water is not quite an ideal fluid
Instead it has some “sticky-ness”, called viscosity
For water, the viscosity is quite low, but for oil it is significantly higher,
and it is even higher yet for substances like honey, which flows very
slowly
Viscosity causes the fluid streamlines at the surface to partially stick to
the boundary, and for neighboring streamlines to partially stick to each
other
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Viscosity
The resulting velocity profile for the streamlines is sketched
in the right panel below
The velocity profile is parabolic, with the velocity
approaching zero at the walls and reaching its maximum
value in the center
Note that this flow is still laminar, with the streamlines all
flowing parallel to each other
75
Viscosity
The standard procedure of measuring the viscosity of a fluid
is to use two parallel plates of area A and fill the gap of width
h between them with the fluid
Then one drags one of the plates across the other and
measures the force F that is required
The resulting velocity profile between the plates is linear, as
shown below
F = ηAv/ℓ
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Viscosity
The viscosity  is then defined as the ratio of the force per
unit area divided by the velocity difference between top and
bottom plate over the distance between the plates
F/A
Fh


v / h Av
The unit of viscosity is then the unit of pressure (force per
unit area) multiplied by time, Pa·s
Often this unit is also called a poiseuille (PI)
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Viscosity
It is very important to realize that the viscosity of any fluid
depends strongly on the temperature
You can see an example for this temperature dependence in
the kitchen
If you store olive oil in the fridge and then pour it from the
bottle, you can see how slowly it flows
But heat the same olive oil up in a pan, and then it flows
almost as easily as water
In the design of motor oils, the temperature dependence of
the viscosity is of great concern, and the goal is to have a
temperature profile that is as flat as possible
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Viscosity
In the table below, we list some typical values of viscosities for
different fluids
All values are quoted at room temperature, except for the viscosity of
blood, which we quote at the human body temperature (37 C = 98.6 F)
Incidentally, the viscosity of blood increases by about 20% during a
human’s life, and the average value for men is slightly higher than for
women (4.7·10-3 Pa·s vs. 4.3·10-3 Pa·s)
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Viscosity
The viscosity of a fluid enters if we want to determine how much fluid
can flow through a pipe of given radius r and length
Hagen (1839) and Poiseuille (1840) found independently of each other
that Q, the volume of fluid that can flow per time unit, is
r 4 p
Q
8
Here p is the pressure difference between the two ends of the pipe
As expected, the flow is inversely proportional to the viscosity and the
length of the pipe
Most significantly, though, it is proportional to the fourth power of the
radius of the pipe
In particular, we can consider a blood vessel as such a pipe. This helps
us understand the severity of the problem associated with the clogging
of arteries
80
Research Frontiers
In laminar flow the streamlines of a fluid
follow smooth paths
In contrast, for a fluid in turbulent flow
vortices form, detach, and propagate
We have already mentioned that the laminar
ideal fluid flow or laminar viscous fluid
flow transitions into turbulent flow when
flow velocities exceed a certain value
This effect is impressively visible in rising
cigarette smoke, which undergoes a
transition from laminar flow to turbulent
flow
But what is the criterion that decides if flow
is laminar or turbulent?
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Research Frontiers
The answer lies in the so-called Reynolds number
It is defined as the ratio of typical inertial forces to viscous
forces and thus is a purely dimensionless number
The inertial force has to be proportional to the density  and
the typical velocity of the fluid v , because F = dp/dt
according to Newton’s Second Law
The viscous force is proportional to the viscosity  and
inversely proportional to the characteristic length scale L
Here is the formula for the Reynolds number
v L
Re 

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Research Frontiers
For flow through a pipe with circular cross section, this
length scale is the diameter of the pipe, L = 2r
Rule of thumb:
Re less than 2,000 => laminar flow,
Re higher than 4,000 => turbulent flow
For Reynolds numbers inside the interval from 2,000 to 4,000
the flow character depends on many fine details of the exact
configuration, and engineers try hard to avoid this region in
their designs because of its essential unpredictability
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Research Frontiers
The true power of the Reynolds number lies in the fact that
flows in systems, which have the same geometry and the
same Reynolds number, behave similarly to each other
This scaling allows one to reduce the typical length scales or
velocity scales and build scale models of boats or airplanes
and test their performance in relatively modest scale water
tanks or wind tunnels
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Research Frontiers
Beyond scale models, modern research on fluid flow and turbulence
relies heavily on computer models
Hydrodynamics modeling is employed in an incredible variety of
physical systems
Of course there are the applications that come to mind right away, such
as the performance and aerodynamics of cars, airplanes, rockets, and
boats
But hydrodynamic modeling is also utilized in the collision of atomic
nuclei at the highest energies available with modern accelerators, or in
the modeling of supernova explosions
Just in the year 2005, the experimental groups working at the
Relativistic Heavy Ion Collider in Brookhaven, NY, proclaimed that
they have discovered that gold nuclei show the characteristics of a
perfect non-viscous fluid, when smashed into each other at the highest
available energies
85
Research Frontiers
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