Download Study Guide to accompany Physics Including Human Applications

Study Guide
to accompany Physics
Including Human Applications
Fuller/ Fuller/ Fuller
by
Thomas C. Campbell
Illinois Central College
and
Robert G. Fuller
University of Nebraska
Harper & Row, Publishers
New York Hagerstown San Francisco London
All rights reserved. Printed in the United States of America. No part of this book may be
used or reproduced in any manner whatsoever without written permission except in
the case of brief quotations embodied in critical articles and reviews.
For information address Harper & Row, Publishers, Inc., 10 East 53rd Street, New York,
N.Y. 10022.
Standard Book Number: 06-042213-0
TABLE OF CONTENTS:
Introduction and Table of Contents
Preface
Dear Student
Mathematical Skills Review (0)
Human Senses (1)
Unifying Approaches (2)
Kinematics (3)
Forces and Newton's Law (4)
Energy (5)
Momentum and Impulse (6)
Rotational Motion (7)
Fluid Flow (8)
Transport Phenomena (9)
Temperature and Heat (10)
Thermal Transport (11)
Thermodynamics (12)
Elastic Properties of Materials (13)
Molecular Model of Matter (14)
Simple Harmonic Motion (15)
Traveling Waves (16)
Sound and the Human Ear (17)
Optical Elements (18)
Wave Properties of Light (19)
Human Vision (20)
Electrical Properties of Matter (21)
Basic Electrical Measurements (22)
Magnetism (23)
Electromagnetic Induction (24)
Alternating Currents (25)
Bioelectronics and Instrumentation (26)
Quantum and Relativistic Physics (27)
Atomic Physics (28)
Molecular and Solid-State Physics (29)
X-Rays (30)
Nuclear Physics (31)
Applied Nucleonics (32)
PREFACE
We have written this to accompany the general physics textbook, PHYSICS Including
Human Applications, by Fuller, Fuller, and Fuller. This , we think, serves several
purposes. It can be used along with a lecture-recitation course to provide additional
study materials for the students.
On the other hand, for several years we have been using the Fuller 3 manuscript as the
textual material for our Keller Plan physics courses. This can be used as the student
modules for a self-paced physics course. Additional Keller Plan materials to assist
instructors of Keller Plan courses, using this , are available from Harper and Row
Publishers.
In addition, the Fuller3 textual materials were developed for flexible use for a wide
variety of physics courses, including correspondence or independent study courses.
This can be used to assist students in self-study programs for personal improvement.
One result of the flexibility of the Fuller3 materials is the variety of sequences available
for the study of these materials. It is not necessary to go through the textual material in
a linear progression by chapter number. You need only to l follow the prerequisite
statements for each textbook chapter. On the following two pages we have presented
the prerequisite chart for the Fuller3 text.
Please note that we have constructed a Chapter Zero for this from the Mathematical
Background materials in the Appendix of the textbook. We believe it is essential for
students to have sound mathematical skills before beginning a quantitative physics
course. We have made Chapter Zero the prerequisite for the rest of the chapters.
The charts on these two pages illustrate the prerequisite requirements for each chapter
in Part I of Fuller, Fuller, and Fuller. To find the chapter which must be completed
BEFORE you begin to study a later chapter, do the following: a) Locate the chapter title,
b) Trace up the chart following the solid lines noting all the chapter titles encountered
until you arrive at Chapter One, Human Senses. For example, to find the prerequisite
chapters for Temperature and Heat (10), you would encounter chapter titles along two
directions; Energy (5), Forces and Newton's Laws (4), Kinematics (3), Unifying
Approaches (2), and Human Senses along one route plus Transport Phenomena (9)
along the other. Each of these chapters should be completed before you begin Chapter
10, Temperature and Heat.
PREREQUISITE CHART FOR PHYSICS INCLUDING HUMAN APPLICATIONS BY
FULLER/FULLER/FULLER
PART II
Prerequisite Chart (1)
Prerequisite Chart (2)
There are two other aspects of learning physics we would like to mention. While it is
obvious that we learn by doing, it is not so apparent that we learn by interacting with
other persons studying the same material. Nevertheless, our experiences as students
and teachers of physics has convinced us of the importance of peer interactions in
learning new subject material. Some provision which encourages people studying these
materials to work together and talk to each other about these ideas seems to be
essential.
All the ideas and/or principles of physics cannot be presented with equal clarity in any
one book. Students who want to know physics are to be encouraged to study the subject
and not only a textbook. While the Fuller3 text can serve as an introduction to the
concepts of physics, it need not be the only source of physics information consulted.
Students are to be encouraged to use the various references recommended in the
bibliography of the Fuller3 text.
Thomas C. Campbell
Robert G. Fuller
June, 1978
-------------------------*Please note that in several of the tables which outline the Suggested Study Procedure a
dotted line appears. This line separates two levels of competency for that chapter. First,
the basic level appears above the dotted line. Below the line are the more advanced
Chapter Goals and the recommended study outline which accompany them.
Dear Student,
This has been especially written for you to assist in your mastery of physics concepts.
As you study each chapter, we hope you may rely on the material presented here to
resolve some of the questions you may have from the many topics you encounter in the
textbook.
To help you use this , we have attempted to use the same format for each of the thirtytwo chapters (plus APPENDIX). In this way, you will be able to find the same "kinds" of
assistance in each of the chapters. After a chapter or two, you may even anticipate the
sections of the which are most helpful to you and utilize them more extensively than
other sections. Generally, the make-up of each chapter is as follows:
Section One - Chapter Introduction
In the double page opening to each chapter we provide you with a restatement of the
GOALS for each textbook chapter. This is accompanied by a chapter OVERVIEW and a
SUGGESTED STUDY PROCEDURE. This information suggests a realistic study
procedure for you to follow in preparing yourself on the contents of the chapter. The
study procedure does not suggest that you must study everything in the chapter
(although this maybe your style), but outlines an economical and straight-forward
method for mastery of the chapter contents.
The correspondence between the chapter GOALS and the various sections and
problems in the textbook are also summarized in tabular form.
Section Two - Definitions
In most chapters, one of the chapter GOALS deals with Definitions of terms introduced
in the chapter. Although you will find a short definition of each of these terms in the
text GLOSSARY, in this section we provide an extended discussion of them. This
addition should not encourage you to "memorize" our definition but to use the
additional information to formulate additional examples or situations which illustrate
the function of the terms introduced.
Section Three - Answers to Questions Asked in the Text
As you read each text chapter you will encounter numerous questions. Some of these
questions will be at the end of a chapter section, other will be within the narratives
themselves. In this section of the , we provide you with a discussion of these questions.
We hope that you will think seriously about the answers to these questions for yourself
before you look at the answers we have provided.
Section Four - Examples
Many examples are provided in each chapter of the Text. Physics principles are general
in nature and can be applied to a multitude of circumstances and applications. In this
section of the , we present additional examples which illustrate the more quantitative
concepts of each chapter. These problems will help you broaden your knowledge of the
major concepts.
Section Five - Practice Test
The final part of the chapter is a Practice Test. These tests have been included with the
answers to provide you a way to test yourself on your understanding of the information
included in each text chapter. All of these tests have originated from exams we have
given our students while using Fuller3. We hope that you will use the tests in a real
testing situation-time yourself, complete all the test in one sitting, and do not look at the
answers until all the parts are finished. When you are done, check your answers against
those given at the end of the test. If you have difficulties, refer back to a portion of the
text, the , or seek help from your instructor.
Well, now it is up to you. Studying Physics is not an easy chore, but one which requires
good study habits. Don't procrastinate and put off studying on a regular day-to-day
basis. In writing this we hope to give you the basis for developing a regular study
routine (suggested Study Procedure), assistance for potentially difficult areas
(Definitions, Examples), and a self-check assessment of your understanding (Practice
Test). But like any other well-planned lesson guide, it is only as good as the
conscientious effort you make.
Sincerely,
Thomas C. Campbell
Robert G. Fuller
Appendix
MATHEMATICAL SKILLS REVIEW
GOALS 0
Keywords: ; Learning Objectives; Mathematics; Arithmetic; Exponents; Exponentials;
Exponential Functions;
Logarithms; Trigonometry; Graphs; Dimensional Analysis; Cartesian Coordinates;
Significant Digits; Powers Of Ten Notation
When you have mastered the contents of the Appendix you will be able to successfully
demonstrate your mathematical skills in
each of the following eight areas:
1. Powers of Ten Notation
2. Significant Figures
3. Cartesian Graphs
4. Dimensional Analysis
5. Right Triangles
6. Exponents
7. Logarithms
8. Exponential Function
OVERVIEW
This review of basic mathematics is a good place to begin your study of introductory
physics. In the Appendix you will find a
helpful review of the eight basic mathematical skills which you will need during your
study of this text.
SUGGESTED STUDY PROCEDURE
Please note that the math review is broken into eight parts: Powers of Ten Notation,
Significant Figures Cartesian Graphs, Dimensional Analysis, Right Triangles,
Exponents, Logarithms, and The Exponential Function. As you begin your study of this
mathematical review, attempt each of the self-check exercises which begin on the next
page of this . Check your answers against the answers given. If you do not score 100%
on any individual part, refer to the text Appendix for help. The outline below will
provide a quick reference to each of the mathematical skills. Next, turn to the Practice
Test in this . Complete the practice test and check your answers with those given at the
end of the test. In any of the eight mathematical areas where you did not score 100%,
again refer to that particular section in the text Appendix and work through additional
Example Problems and/or Exercises.
------------------------------Chapter Goals Self- Suggested Text Practice
Check Readings (Appendix) Problems
-------------------------------Powers of Ten Notation 1 A.2 (p. 733)
Significant Figures 2 A.3 (p. 734)
Cartesian Graphs 3 A.4 (pp. 735-738)
Dimensional Analysis 4 A.5 (pp. 740, 741)
Right Triangles 5 A.6 (pp. 742-745)
Exponents 6 A.7 (p. 745)
Logarithms 7 A.8 (p. 748)
Exponential Function 8 A.9 (p. 749)
SELF-CHECK EXERCISES
Keywords: Mathematics; Problems; Answers: Powers Of Ten Notation
Powers of Ten Self-Check
Solve each of the following problems, and give the answer in powers of ten notation.
1. 0.252 x 0.000000700/0.0360 = _______.
2. 6.380 x 103 x 5.00 x 104/2.50 x 10 5 = _____.
3. 3.20 x 107 + 6.83 x 106 - 9.90 x 105 = _____.
Powers of Ten Self-Check Answers
If you had difficulty in correctly solving these problems, please study the Section A.2 on
the powers of ten
notation, page 731.
1. 4.90 x 106
2. 1.28 x 1013
3. 3.78 x 107
SIGNIFICANT FIGURES SELF-CHECK
Keywords: Significant Digits
Solve each of the following problems, and give the correct answer in the appropriate
number of significant figures.
1. 0.101 x 1.03/.025 = _____.
2. 4.5 x 103 x 2.5 x 10-2/2.15 x 10-4 = ____.
3. Solve for A; 3x 10-4 A/4.5 x 103 = 1. A = _____.
ANSWERS TO SIGNIFICANT FIGURES SELF-CHECK
If you had difficulty in correctly solving these problems, please study Appendix Section
A.3 on Significant Figures, page 734.
1. 4.13 (3 significant figures)
2. 5.2 x 105 (2 significant figures)
3. 2 x 107 (1 significant figure)
Graphing and Dimensional Analysis Self- Check
Keywords: Graphs; Slope; Units; Conversion Of Units
1. The following table is taken from a drivers manual and shows data for stopping an
automobile on dry
pavement.
------------------------Velocity Thinking Total Stopping
(m/sec) Distance (m) Distance (m)
8.8 6.7 14
13 10 27
18 13 43
23 17 61
27 20 86
--------------------------a. Draw a graph of thinking distance (y-axis) versus velocity (x-axis), and find the slope
of the curve at the
point on the curve where x = 15 m/sec.
b. Draw a graph of the total stopping distance (y axis) versus the velocity (x axis), and
find the slope of the
curve at the point where x = 20 m/sec.
2. We can define length, mass, and time as fundamental dimensions in a system of
measurement. What are
the SI (System International) units for
a. Length
b. Mass
c. Time
The SI units are related to each other by multiples of ten, and the units are represented
by the fundamental
unit with the proper prefix. What are the relationships between the fundamental unit
and the following common
prefixes?
d. The prefix centi-means _______, so one tesla = ______centiteslas. [10-2; 10 2]
e. The prefix milli- means _____, so one liter = _____milliliters. [10-3;10 3]
f. The prefix kilo- means _______, so one watt = ______kilowatts. [103;10 - 3]
Graphing and Dimensional Analysis Self- Check Answers
If you had difficulty in correctly solving these problems, please study Section A.4,
Cartesian Graphs, and
A.5, Dimensional Analysis, on pages 734-741.
1. a. 0.60 sec; b. 4.8 sec
2. a. meter; b. kilogram; c. second; d. 10-2, 102; e. 10- 3, 103; f. 103, 10 -3
RIGHT TRIANGLES SELF-CHECK
Keywords: Trigonometry; Trigonometric Functions
A surveyor wishes to determine the distance between two points A and B, but he
cannot make a direct measurement because
a river intervenes. He steps off a line AC at a 90º angle to AB and 264 meters long. With
his transit, at point C he measures
the angle between line AB and the line formed by C and B. Angle BCA is measured to
be 62º. What is the distance from A to
B?
RIGHT TRIANGLES SELF-CHECK ANSWERS
Distance AB = 497 meters.
If you had difficulty getting this answer, you will find additional information in Section
A.6, Right Triangles, of the appendix.
TRIGONOMETRIC RELATIONSHIPS
While nearly all of the problems in the book can be worked using the definitions of the
sine, cosine, and tangent you have seen
derived for right triangles, there are some relationships between these functions that it
will be useful for you to know. We will
derive them below.
See Fig. (0-1)
From the Pythagorean theorem you learned in high school, you know the relationship
between the three sides of a right triangle;
i.e., the sum of the squares of the two sides is equal to the square of the hypotenuse
x2 + y2 = r2 (1)
Look again at Figure (0-1). Note
y/r = sin q = cos a (4)
x/r = cos q = sin a (5)
Since q + a + 90º = 180º i.e., the sum of the angles of any triangle equals 180º, then
a = 90º - q
The angle a is called the complement of the angle q. From equations (4) and (5) above
notice that the sine of an angle is equal
to the cosine of its complementary angle and the cosine of the angle is equal to the sine
of its complementary angle.
Example: An arrow shot into the air comes vertically down and sticks in the grass on
the side of a hill inclined 68º from
vertical. What portions of the arrow point down the hill and perpendicular to the hill?
See Figure
Now we divide both sides of equation (1) by the square of the hypotenuse, r2,
x2/r2 + y2/r2 = 1
Then recall the definitions of sine and cosine,
sin q = y/r; cosine q = x/r
so x2/r2 cos2q; y2/r2 sin2q
thus x2/r2 + y2/r2 = cos2 q + sin2 q = 1 (2)
The sum of the squares of the cosine and the sine of any angle is equal to one.
In a similar way we can derive a relationship between the tangent, sine and cosine of
any angle.
tan q = y/x. Divide both numerator and denominator by the hypotenuse, r
tan q = (y/r) / (x/r) = sin q/cos q (3)
The tangent of any angle is equal to the ratio of the sine of the angle to the cosine of the
angle.
EXPONENTS SELF-CHECK
Keywords: Exponentials; Powers
1. 86 x 83 = ____.
2. 25 ÷ 2-2 = ____.
3. 103 x 10-3 = ____.
4. 101.5 ÷ 10-5 = ____.
EXPONENTS SELF-CHECK ANSWERS
Keywords: Exponentials; Powers
1. 89
2. 27
3. 1
4. 102
If you had difficulty with any of the above answers, you can find additional assistance
in the Appendix, Section A.7, page 745.
LOGARITHMS SELF-CHECK
Keywords: Powers; Logarithms
Use the properties of Logarithms to solve each of the following problems:
1. 3 x 5 = ____.
2. 3/5 = ____.
3. 84 = ____.
4. Solve the following equation for X: 16 = 4(102X)
LOGARITHMS SELF-CHECK ANSWERS
1. 15
2. .6
3. 4096
If you had difficulty with any of these problems, please study Appendix Section A.8,
page 746.
EXPONENTIAL FUNCTION SELF-CHECK
Use the properties of exponential functions to solve each of the following problems.
1. If the population of a growing country is governed by the relation, N = 100,000.124t,
where t is in years, how long will it take
for the population to double?
2. Predict the total population of the country in 10 years.
EXPONENTIAL FUNCTION SELF-CHECK ANSWERS
1. 5.0 years
2. 350 thousand
If you had difficulty with either of the problems posed above, please turn to Appendix
Section A.9, page 748.
PRACTICE TEST
1. (Powers of Ten Notation and Significant Figures) Solve the following problems using
the powers of ten notation. In each case
write your answer with the appropriate number of significant figures.
a) 731 x 1.009 x 431/0.005 = ____.
b) 4.7 x 10-7 x 2.51 x 105/2.10-3 = ____.
c) Solve for A: 795 x 73.45 x 103 A = 1.0007 x 104 = ____.
2. (Cartesian Graphs and Dimensional Analysis) The following data were taken from a
physics laboratory experiment.
--------------------------Velocity of a
Rolling Coin Time
(m/s) (S)
---------------------------50 0
+15 4
+45 7
+95 11
a) Plot a graph showing velocity and time for this rolling object.
b) Calculate the slope of the curve at (4, 15).
c) Give the proper dimensions to the slope.
3. (Right Triangles) To find the height of a tall building, a physics student steps 75 paces
(each 1 meter) from the base of the
building. Using a ruler at arm's length (1 meter), the student finds that at this distance,
the building appears to be 50
centimeters when compared to the ruler. Determine the approximate height of the
building.
4. (Exponents) Calculate the following using the laws of exponents.
a) 54 ÷ 5-3 = ____.
b) 108 x 10-3 = _____.
c) 123.6 x 12- 4.2 = ____.
5. (Logarithms) Use the properties of logarithms to solve each of the following
problems.
a) 7 x 4 = ____.
b) 20 ÷ 4 = ____.
c) 93 = ____.
d) Solve the following equation for X: 2700 = 300(103X)
6. (Exponential Function) Use the properties of the exponential function to solve the
following problem:
a) A biologist determines that during the month of June, the number of live frogs in a
small pond is governed by the relationship
N = 189e. 075t (t is the number of days). How long will it be until the number of frogs is
three times its initial population?
b) Predict the total population of frogs by the end of the month (30 days).
ANSWERS:
1. a) 6.36 x 107
b) 6 x 101
c) 1.52 x 10-1
2. b) 10
c) m/sec2
3. 38 m
4. a) 57
b) 105
c) 12-.6
5. a) 28
b) 5
c) 728
d) .32
6. a) 14.6 days
b) 1800
Chapter 1
Human Senses
GOALS
Keywords: ; Learning Objectives; Humans; Sensory Systems; Transducers; Energy;
Force; Interactions;
Variables; States Of Systems; Systems; Intensity
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define the following terms, which can be used to describe a relationship between you
and your environment:
system energy
state force
variable field
interaction intensity
transducer
Stimuli
List the major external stimuli that are detectable by humans.
Human Responses
Describe an elementary threshold measurement experiment, and interpret the data
obtained.
Models
Use a mental construct, or model, to explain a common human experience.
PREREQUISITES
You may find it necessary to review your knowledge of the powers of ten notation
before you study this
chapter.
OVERVIEW
Keywords: Sensory Systems; ; Instructions
Without much difficulty you can probably name the five senses of the human body. As
man has learned more about his
surroundings, he has learned more about himself and about his ability to interact with
his environment. In this chapter you will
learn about the physical nature of your human senses.
SUGGESTED STUDY PROCEDURE
When you begin your study of this chapter, be familiar with each of the four Chapter
Goals: Definitions, Stimuli, Human
Responses, and Models. (Please note than an expanded discussion of each term listed
under Definitions is given on the next
page of this .) Next, read Chapter Sections 1.1-1.8. Be sure to attempt some of the more
interesting Simple
Experiments suggested and consider several of the probing questions posed at the end
of most of the sections. (Answers to
these questions are given in the second section of this .)
At the end of the chapter, read the Chapter Summary and complete Summary Exercises
1-4. Now do Algorithmic Problems
1-6 and complete Exercises 1, 2, 3, 4, 7, and 8. Now you should be prepared to attempt
the Practice Test on Human Senses
found in this . This study procedure is outlined below.
-------------Chapter Goals Suggested Summary Algorithmic Exercises &
Text Readings Exercises Problems Problems
-------------Definitions 1.1-1.7 1
Stimuli 1.1-1.3 2 1,3 3,4,7,8
Human
Responses 1.4-1.6 2,3 2,4,5,6 1,2
Models 1.8 4
DEFINITIONS
Keywords: ; Glossary; Transducers; Interactions; Energy; States Of Systems; Variables;
Force; Systems; Fields;
Intensity
SYSTEM
A whole entity.
In marked contrast to the holistic thinking that characterizes much of human activity,
physics uses an analytical technique
that seeks to understand complex situations by dividing them into separate parts which
we call systems. We try to imagine
that a system is a separate part of the universe which we can study without the system
being changed either by our study or
by whatever is happening in the rest of the universe. For example, if we wish to study
the physics of human hearing, we will
define the system of study as the human hearing system and include within it all of the
parts of the body that are necessary
to hear sounds. Of course, such a system is quite unreal since you cannot hear anything
if the other parts of your body
cease to function. Nevertheless, this mental, somewhat artificial, technique of dividing
up the universe into separate systems
has allowed us to make great strides in the study of nature.
VARIABLE
Quantity that is subject to change.
Once you have picked a system for study you can analyze the ways you can change or
vary the characteristics of the
system. For example, you can modify the human hearing system by the use of drugs or
the use of ear plugs. The properties
of a system that we can change or vary we call variables. One big idea of physics is to
find all of the important variables of a
system and see how changes in each of the variables affects the rest of the system. You
will notice once again the
analytical approach, chop a system into parts called variables, and see if we can
understand the whole system by finding the
ways the variables influence one another.
STATE
Particular form or condition of a system.
It is usual for physicists to define the starting point of their study of a system by
attempting to give all of the important
variables of the system some initial value thereby specifying the initial state of the
system. For example, if you wished to
study the influence of various amount of cotton ear plugs on human hearing, you could
define your starting condition, or initial
state, as a well rested, typical human with no wax or obstructions in the ear canals.
TRANSDUCER
A sensor or detector that transforms a change in one physical variable to a change in
another.
In many cases we wish to observe the changes in the properties of a system. Often it is
possible to infer a change in a
system by measuring a related, but different, property. In such a situation we make use
of a transducer, a part of the system
that relates the change of one variable of the system to another. For example, the
eardrum acts to transform the sounds that
fall on your ear into vibrations of the eardrum which are subsequently transformed into
the electrical messages that are sent
to your brain. The eardrum serves as a transducer.
INTERACTION
Action or influence exerted between systems.
If we wish to study the influence of one system on another we say we will study the
interaction between two systems. Then
we will proceed to change the values of variables of one system to see if that causes any
changes in the values of the
variables of the second system, e.g., if we change the noise level in the room where our
human subject is located that will
change the response of a typical person.
ENERGY
Property of a system which causes changes in its own state or state of its surroundings;
measure of ability to do work
(physical).
One way to specify the state of a system is to compute the energy of the system. Energy
is a kind of imaginary property of
a system that physicists have invented to help define in a more holistic way the state of
a system. We never actually
measure energy directly but only calculate it from other variables that we can measure.
In a sense, energy is to a system as
meaning is to a poem. In general, a group of physicists think that they can objectively
agree upon the energy of a system,
while a group of poets are not likely to think that they can objectively agree upon the
meaning of a poem. Does the fact that
ten people can agree upon the same value of an imaginary quantity make it truly
objective?
FORCE
A measure of the strength of an interaction; a push or pull; the effect of a force is to alter
the state of motion of a body. A
vector quantity with a magnitude measured in Newtons (N).
From our earliest experiences as children we know of the interactions between systems
by physical contact. From our
beginnings as humans we have responded to the direct contact of another person.
Throughout our early experiences of
collisions with walls and falling down we know of the effects of direct contact
interactions. We use the term force to designate
the strength of interaction between two systems. We have the idea that a greater force
applied to a system is most likely to
cause a greater change in the system.
FIELD
A region of space characterized by a physical property which has a determinable value
at every point in the region, e.g.,
gravitational field, magnetic field, etc.
Our experiences with direct contact interactions are so convincing that physicists have
invented a concept of a field that can
be used to bring into contact systems that may be widely separated from one another.
The physicist just imagines a field
exists everywhere between the two systems and the field of one system may exert a
force on the second system.
INTENSITY
Energy transported through a unit area in one second.
Another idea that is readily associated in our minds with direct contact interactions is
the transfer of energy from one system
to another, such as the energy of food being transferred to the energy of your body via
the life processes. In a similar way we
can think of a field as having a source which emanates energy to all objects in its field.
We characterize the intensity of a
field by the amount of energy its source emanates per unit of area in one second. If we
are in the field of a light bulb our
eyes interpret the intensity of the bulb in terms of brightness, the more intense the
source, the greater is the brightness of the
light.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Systems; States of Systems; Humans; Anatomy And Physiology;
Biology; Variables
SECTION 1.2 The Human System
1. A human being is a complex set of interacting systems. To properly describe in a
holistic way your present state would
require you to define the different states of your basic body systems such as the
skeletal-muscular system, digestive
system, urinary system, respiratory system, circulatory system, nervous system, and
reproductive system. You are usually
not very aware of the present state of any of these systems even though your brain must
continually monitor the activities
that occur in all parts of your body. To determine the present state of your circulatory
system you could find your pulse rate,
blood pressure, and results of various tests on your blood. Even so, many details of
your circulatory system would be left
undefined.
2. Your present state is determined by the values of many variables of the body systems
listed above. Such variables as
pulse rate, blood pressure, etc., can be specified. However, there is evidence that
personality features such as self-worth or
faith can have an important influence on many of your physiological variables.
Historically such variables as faith and
self-worth are classified as not being physical observables.
SECTION 1.3 The Human Detector
Keywords: ; Answers; Sensory Systems; Interactions; Energy; Electric Fields;
Gravitational Fields; Sound; Light
And Optics; Auditory Systems; Optical Systems; Human Body
3. At this moment you are interacting with your environment through a wide variety of
systems. The force of the chair on your
bottom is a mechanical energy interaction. Your eyes are responding to electromagnetic,
or light, energy. Your ears are
picking up sound energy. Your nose and taste buds sense odors and tastes by chemical
energy. Your body is said to be
immersed in the gravitational and electromagnetic fields of the earth and sun. The
influence of these fields upon you is not
well understood.
4. In general, we can think of the odor, sound, gravitational, and light or
electromagnetic interactions as being
interactions-at-a-distance, while taste of food and the force of the chair may be
envisioned as direct contact interactions.
SECTION 1.4 Direct-Contact Interactions
Keywords: Answers; ; Interactions; Sensory Systems; Tactile Systems; Thermal
Regulatory Systems; Human
Body
5. There are areas of your back, legs, and feet that seem relatively insensitive to touch.
Your finger tips, tongue, and
forehead is quite touch sensitive.
6. By tasting a variety of known salt concentrations it is believed you could train
yourself to be a salt water taster. Just as,
after years of practice, persons are able to perfect their taste sense to become
professional wine tasters.
7. Both the inside of your arm and your lips are particularly thermal sensitive while
your hands and face have fewer nerves to
detect heat and cold than other parts of your body.
SECTION 1.5 Interaction-at-a-Distance
Keywords: ; Answers; Interactions; Action At A Distance; Human Body;
Electromagnetic Waves; Sound;
Gravitational Forces; Sensory Systems
8. You have surely experienced sunshine, music, noise, voices, smells and food today.
Have you climbed some stairs, ridden
a bicycle up a hill, or hiked to the top of a mountain?
9. Sunshine (electromagnetic); music, noise, voices (sound); smells and food (odor);
stairs, hill, mountain (gravitational).
10. Your ears are never shut, so the interaction-at-a-distance most common for non-deaf
humans is probably sound.
SECTION 1.7 Threshold for the Sense of Touch
Keywords: Answers; ; Human Body; Tactile Systems; Thresholds
11. Various whole body variables such as state of restfulness, lack of influence of drugs,
normal blood pressure and blood
sugar concentrations, etc.
12. One source of error is the inability of a subject to tell exactly when a touch is felt.
Doing repeated measurements with
each subject in a varying sequence of locations so that several threshold measurements
are performed at each location
should prove helpful in reducing this uncertainty.
13. Figure 1.3 shows a technique that can be used to determine the distance between
distinct contact points.
SECTION 1.8 Model Building
Keywords: ; Answers; Models; Humans; Sensory Systems; Philosophical Implications
14. In each of these cases a powerful group of persons developed an explanation of the
behaviors of other groups of humans
that defined them as less than human, e.g., savages, enemies of progress, threats to
national security. These mental
constructs, or models, were used to justify a wide variety of inhumane and unjust
treatments.
15. For most scientists, the subject areas of extrasensory perception, of mind-brain
research, and of how the human beings
develop logical thought are still not scientifically ordered. Two areas of science,
molecular biology and astrophysics, have
contributed new concepts to our daily language by popularizing the double helix and
black holes.
PRACTICE TEST
Keywords: ; Problems; Answers; Transducers; Evaluation; Interactions; Thresholds;
Humans; Sensory Systems;
Questions; Heat; Action At A Distance; Tactile Systems; Auditory Systems; Models;
Sound; Contact; Forces; Scientific
Method
1. You are in a swimming pool several meters away from another person also in the
pool. Name below three methods you
might use to gain the person's attention. For each case, identify or describe transducer
you use and mark the interaction as
either a contact (C) or action-at-a-distance (A-D).
Method Transducer Interaction (C or A-D)
1. _____________________________________________________
2. _____________________________________________________
3. _____________________________________________________
2. Imagine that you are taking a shower in a bath with "inadequate plumbing" and find
the shower temperature difficult to
control. As the water temperature changes drastically, please describe the response
your body makes to this variable in terms
of
a. Threshold b. Sensitivity c. Discrimination 3. The different parts of your body respond with different sensitivity to outside
physiological stimuli. Consider the bottom of
the feet and their response to four major stimuli as outlined below. Please rank each
stimuli from 1- body part most sensitive
to stimulus to 4-body part least sensitive to stimulus.
Body Part or Area Tested - Bottom of Feet
Ranking Stimuli Interaction
______ Heat Contact
______ Visible light Action at a distance
______ Ultraviolet light Action at a distance
______ Low frequency sound Action at a distance
4. What is a "model" in physics?
ANSWERS:
1. Shouting, Vocal Chords, A-D Making Waves, Hands & Arms, A-D Splashing Water,
Hands & Arms, C
2. Threshold - Minimum temperature difference detected by the body Sensitivity Body's receptors reaction to various
incoming energy (heat) levels Discrimination - Detection of temperature differences
across the shower spray or ability to tell if
the temperature of the water is changing with time.
3. 1, 4, 2, 3 (other answers accepted under other physical conditions).
4. A model is a specification of a unification of ideas which help to visualize a variety of
experiences. In physics, these
experiences deal primarily with aspects of the physical nature of the world.
Chapter 2
Unifying Approaches
GOALS
When you have mastered the content of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use each term in an operational definition:
equilibrium restoring force conservation laws
inertia oscillatory motion feedback
gradient linear system natural frequency
current superposition resonance
Inertia
Give an example of a physical system that has mechanical, thermal, and electrical
inertia.
Energy Transfer
Explain how you would maximize the transfer of energy at the interface between two
systems.
Superposition
Solve problems making use of the superposition principle-given the proper physical
variables of the
systems.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 1, Human
Senses. If you have
not recently been working with cartesian coordinate graphs and dimensional relations,
you may wish to
review the material on graphs and dimensional analysis in the mathematical
background supplement in the
appendix.
OVERVIEW
Keywords: ; Instructions; Inertia; Superposition Principle; Flow Of Energy
As men have attempted to make sense of physical phenomena, a series of approaches
have emerged. Some of these
approaches are descriptive, some are mathematical, some may be both. In this chapter
you will study some of the basic
approaches used again and again to understand and/or explain the actions of a variety
of physical phenomena.
SUGGESTED STUDY PROCEDURE
In this chapter, you should direct your study toward all four Chapter Goals:
Definitions, Inertia, Energy Transfer, and
Superposition. An expanded treatment of each of the terms listed under Definitions is
given on the next page of this Study
Guide. Next, read all the chapter sections, 2.1-2.10. Please note the questions asked
throughout your reading. Answers to
these questions are given in the following pages of this .
At the end of the chapter, read the Chapter Summary and complete Summary Exercises
1-11. Next, do Algorithmic Problems
1, 2, 3, 4, 7, and 8. Finally, complete Exercises and Problems 2, 3, 4, and 5. Now you
should be prepared to attempt the
Practice Test on Unifying Approaches given in this . This study procedure is outlined
below.
------------------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
------------------------------Definitions 2.1, 2.2, 2.4, 1-6 2
2.9, 2.10
Inertia 2.3 7-9 1, 2 3, 4
Energy Transfer 2.5 10 3, 4 5
Superposition 2.6, 2.7, 2.8 11 7, 8
DEFINITIONS
Keywords: Glossary; ; Equilibrium; Inertia; Kinds Of Force; Oscillations; Oscillatory
Motion; Waves And Sound;
Superposition Principle; Gradient; Conservative Laws; Feedback; Resonance; Frequency
EQUILIBRIUM
All the influences acting on the system are canceled by others, resulting in a balanced
unchanging system.
Nature is comfortable. Everything else being equal, the systems in nature tend to their
lowest possible energy states and to
equilibrium, a state in which all the external disturbances acting on a system are
balanced or canceled out. Perhaps you are
presently seated, relaxed, reading this . You are in a state of equilibrium.
INERTIA
Property of system that is a measure of the system's resistance to change.
If you wish to stand up, the bulk of your body tends to remain seated. The tendency to
stay put is called inertia and is given
a numerical value of mass for systems in which motion is studied. Whereas this
resistance to motion is called mechanical
inertia, the resistance an object offers to the flow of electrical change is called electrical
inertia and the resistance to heat
flow is called thermal inertia.
RESTORING FORCE
Force acting to return a displaced body to equilibrium position.
You may have had the task of trying to move an automobile, or motorcycle, stuck in the
mud or snow. In its stuck condition,
it was in equilibrium: when you pushed it you moved it from its equilibrium position.
When you stopped pushing it may have
rocked back and forth about its starting position. Most systems in equilibrium
experience a force that tends to move them
back to their original position if you only move them a small amount from equilibrium.
This force is called a restoring force.
OSCILLATORY MOTION
Characteristic motion of systems having a linear restoring force.
The action of a restoring force on a system causes the system to rock back and forth
about the original position. This back
and forth motion is called Oscillatory Motion.
LINEAR SYSTEM
A system in which the restoring force is directly proportional to the displacement.
Suppose you attempt to push a stuck vehicle out of the snow or mud. If you find that
you must push twice as hard to move
it twice as far from its original equilibrium position, this system is called a Linear
System. Once this fact is established,
simple addition can be used to estimate the number of pushers (each pushing the same
amount) needed to get your vehicle
out of trouble. If you can push it one-fourth of the way out, then four persons (all
pushing the same direction) should be able
to push it all the way out.
SUPERPOSITION
The resultant effect is equal to the sum of the individual independent effects. A
principle that holds true for linear systems.
When linear and independent systems interact, the total effect of their action can be
found by simply adding the contribution
of each. Please note that an application of this superposition principle in the definition
of linear system was used in the case
of a vehicle stuck in the snow or mud. When one person can push the vehicle onefourth of the way out of its position, four
persons should be able to push it all the way out. Note that if five persons were to push,
three one way and two the other,
the vehicle would move one-fourth of the way from equilibrium.
GRADIENT
Rate of change of a physical quantity relative to position.
What is the present temperature of your surroundings? The thermal conditions of your
surroundings is one of the most
important aspects of maintaining a comfortable environment. We readily notice a
temperature difference when we go from
inside to outside a building in winter or in summer. If a variable, such as temperature,
has different values at two different
locations, then we can talk about gradient, in this case a temperature gradient, existing
between the two locations. We
compute the size of a temperature gradient by calculating the ratio of the difference
between the temperatures at the two
locations to the distance between the locations.
CURRENT
A steady and onward movement of a physical quantity. The ratio of the change in a
quantity to the change in time.
Consider the wall of a home where the inside temperature is much higher than the
outside temperature. Because of the
difference of temperature, we can detect a flow of heat or current from the inside to the
outside which occurs in response to
the temperature gradient. For simple systems the current is a linear function of the
gradient.
CONSERVATION LAWS
There is no change in a physical property with a change in time.
In many life situations we get our bearings from fixed objects, whether they are the
North star, the city hall, or the university
bell tower. Similarly, in physics we look for fixed, or constant, properties of a system to
help us understand it. A property of a
system which does not change is called a conserved property, or quantity. We can then
formulate a conservation law for the
system we are studying. A conservation law, simply put, states that even though time
marches on, a particular property of
the system does not change. It would be difficult to overemphasize the importance of
conservation laws in the growth of
physics. In systems where conservation laws seem to be violated, physicists have
invented new variables to preserve
conservation. Now physicists have a whole list of properties of systems that can be
conserved under certain conditions, such
properties as energy, mass, electrical charge, momentum, atomic number, strangeness,
charm, baryon number, etc. The list
seems to get longer every year, as new conservation laws are invented.
FEEDBACK
A portion of the output from a system is returned as input into the same system.
A common result of positive feedback is the squeal that occurs when an amplifying
system is not correctly used at a musical
event. The result often occurs when the amplifier system speakers are placed close
enough to the microphone to have the
amplified sound picked up by the microphone and circulated through the system again.
NATURAL FREQUENCY
The frequency of a freely vibrating system, i.e., the number of vibrations per second that
are characteristic of the system.
If you strike the middle C key on a piano you excite the natural frequency of vibration
of the stretched wire in the back of the
piano. It oscillates at a frequency of 256 vibrations per second.
RESONANCE
Occurs when the frequency of the external force equals a natural frequency of the
system.
Resonance occurs when the frequency of an external interaction equals the natural
frequency of the system. On a playground
swing, to pump up the amplitude of the swing you must pull on the swing support
chains at the proper time using a frequency
of pulling equal to the frequency with which the swing goes to and fro.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: Answers; ; Superposition; Principle; Feedback; Systems; Equilibrium;
Oscillatory Motion; Oscillations;
Waves And Sound
SECTION 2.3 Inertia
1. Some systems in equilibrium are your (Fuller)3 physics text resting on your study
table and your head resting on your hand
with your elbow on the chair arm in your physics lecture.
2. Both of the above systems may tend to move back to their original position if lifted
slightly. Gravitational interaction and
your muscles will provide the restoring forces.
3. Suppose you can define a lack of knowledge parameter, a misconception parameter,
and a concrete reasoning parameter.
In each type of property, knowledge, concept, and reasoning, your class will have a
characteristic average inertial parameter.
The class may be slow to respond to rapid changes in knowledge, concepts, and
reasoning.
SECTION 2.7 Feedback
4. The input for you is the information you are taking in from your environment. If you
process that information and use it to
modify the information you are taking in by asking a question or writing an explanation
of a comment in your notebook, then
you are using feedback. For your instructor, the input is information that is gathered
from the students by means of questions,
smiles, nods, or snores. The instructor uses feedback from the information to change the
approach being used in the
classroom to respond to student input.
SECTION 2.8 Superposition Principle
5. There are a number of old myths that indicate marriage is not a linear transformation.
"Two can live as cheaply as one," for
example is no doubt false, but it does indicate a common understanding that
superposition does not apply to marriage, i.e.,
the total of the married couple is not the sum of the individual persons.
6. A collection of marbles is a linear system with respect to number, weight, and volume
occupied. A school of fish is linear
with respect to volume occupied and oxygen used. A school of children is not linear
with respect to economic factors.
See Fig.
EXAMPLES
Keywords: ; Worked Examples; Superposition Principle; Speed; Galilean-Newtonian
Relativity; Arithmetic; Scalar
Arithmetic
SUPERPOSITION
1. An Amtrak train is roaring East through Lincoln, Nebraska, at a speed of 100 km/hr.
The conductor on board is collecting
tickets from the back to the front of the train while walking at a speed of 2 km/hour.
Meanwhile, a dining car waiter is serving
hot dishes by rushing back and forth from the kitchen at a speed of 8 km/hour. What
are the speeds of the various moving
objects with respect to Lincoln?
What Data Are Given?
The speed of the train with respect to Lincoln = 100 km/hr. East
The speed of the conductor with respect to the train = 2 km/hr. to the front
The speeds of the waiter with respect to the train = 8 km/hr. to the front = 8 km/hr.
toward the back
What Data Are Implied?
The speeds of all three are assumed to be along the same East- West line, so that linear
superposition holds true.
What Physics Principles Are Involved?
Superposition - in this case, the algebraic sum of the various numbers.
What Equations Are to be Used?
We can write algebraic expressions for this problem by assigning symbols; let st = speed
of the train, sf = speed of object
moving forward on the train and sb = speed of object moving backward on the train.
Algebraic Solutions
The speed of an object with respect to Lincoln sL is given by
sL = st + sf for forward moving objects.
sL = st - sb for backward moving objects.
Numerical Solutions
sL (train) = 100 km/hr. East
sL (conductor) = 100 km/hr. + 2 km/hr. = 102 km/hr. East
sL (waiter) = 100 km/hr. + 8 km/hr. = 108 km/hr. East when he is moving toward the
front
sL (waiter) = 100 km/hr. - 8 km/hr. = 92 km/hr. East when he is moving toward the
back of the train.
Thinking About the Answers
Keywords: Einstein, Albert A.; Humor;
Notice that the speed of the train is so much larger than the speeds of the conductor and
the waiter that their motion relative
to the train does little to change their motion relative to Lincoln. Their motion does
seem important to them and the other
people on the train.
Notice also the analytic style of problem-solving. The final question is answered by
cutting it into small pieces and using
superposition.
See Fig.
PRACTICE TEST
1. Suppose after you have accurately weighed a small sample on an equal arm balance,
you accidentally touch the balance
pan, what happens? Describe the initial state of the balance, and its subsequent
behavior.
2. List the three kinds of inertia of an electric frying pan. Give a specific action you can
perform to verify the existence of
each kind of inertia.
3. On a cold day in January, the inside wall of a home has a temperature of 29ø C.
Twelve centimeters away, the outside
wall has a temperature of - 7ø C. What is the temperature gradient? Which way does
heat flow?
4. Eight college students are deadlocked in tug of war. A ninth student joins the contest.
What physical principle can you use
to predict the outcome? What do you predict will happen? If your prediction does not
come true, how can you explain that?
ANSWERS:
Keywords: Questions; ; Equilibrium; Inertia; Superposition Principle; Harmonic
Oscillations; Oscillations;
Oscillatory Motion; Properties Of Oscillations; Evaluations; Gradients
1. Initial state - equilibrium (no motion); subsequent behavior - restoring force causes
harmonic motion about the balance
position (the equilibrium position).
2. A Mass inertia - Attempt to give the pan a velocity by pulling on it with a spring
balance or to stop once placed in motion.
B. Electrical inertia - Apply a voltage and determine the resistance to the flow of
current.
C. Thermal inertia - Turn on current and place a thermometer in contact with the pan.
Note the time necessary to "warm-up"
or one hot, to "cool- down".
3. GRAD = DT/DX = 29-(-7)/12 = 3ø C/cm. Heat flows from inside to the outside.
4. At the start, the forces exerted by each team place the rope in equilibrium. If another
person is added, the principle of
superposition would predict that either A) his team will win (he exerts a force to help
his team) or B) his team will lose (he
pulled the wrong way, helping the opponents).
Chapter 3 Kinematics
GOALS
When you have mastered the content of this chapter, you will be able to achieve the
following goals:
Definitions
Use the following terms to describe the physical state of a system:
displacement
velocity uniform circular
acceleration motion
uniformly accelerated motion radial acceleration
projectile motion tangential
acceleration
Equations of Motion
Write the equations of motion for objects with constant velocity and for objects with
constant acceleration.
Motion Problems
Solve problems involving freely falling and other uniformly accelerated bodies,
projectile motion, and uniform
circular motion.
Acceleration Effects
List the effects of acceleration on the human body.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 1, Human
Senses, and
Chapter 2, Unifying Approaches. You must also be able to use the properties of right
triangles to solve
problems.
OVERVIEW
Keywords: ; Instructions; Mechanics; Kinematics
As you look over this chapter you will find a large number of algebraic equations.
These expressions are used in describing
the basic motion of objects. In this chapter, the four basic kinds of motion which are
described are 1) Linear Motion (Section
3.6), 2) Uniformly Accelerated Motion (Section 3.7), 3) Projectile Motion (Section 3.8),
and 4) Uniform Circular Motion (Section
3.9).
SUGGESTED STUDY PROCEDURE
As you begin to study this chapter, be familiar with the Chapter Goals: Definitions,
Equations of Motion, Motion Problems,
andAcceleration Effects. (A brief explanation of each of the terms listed under
Definitions is given on the next page of this
.) The initial sections of this chapter require a knowledge of trigonometry. If necessary,
you may want to review
the properties of Right Triangles found in the Appendix, Section A.6. Next, read
Chapter Sections 3.1-3.10. Be sure to work
through the detailed examples given in most of the sections.
At the end of the chapter, read the Chapter Summary and complete Summary Exercises
1-14. Check your answers against
those that are given. If you don't understand a specific answer, try reviewing that
section of the text. Finally, complete
Algorithmic Problems 1, 2, 3, 4, 5, 6, 9, 10, 11, and 12, Exercises 3, 4, 8, 9, 10, and 11, and
Problems 14, 18, and 20. Now
you should be prepared to attempt a Practice Test for Kinematics. This study procedure
is outlined below.
----------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
----------------------Definitions 3.1,3.2,3.3,3.4 1-7 1,2
3.5, 3.6
Equations
of Motion 3.2,3.5,3.6 8-10 3,4,5,9 3, 4
Motion Problems 3.7,3.8,3.9 11-13 6,10,11,12 8,9,10,11
14,18,20
Acceleration
Effects 3.10 14 6
DEFINITIONS
Keywords: ; Glossary; Mechanics; Kinematics; Velocity; Displacement; Acceleration;
Angular Acceleration;
Projectile Motion; Uniform Circular Motion; Circular Motion; Kinds Of Motion
DISPLACEMENT
Change in location of an object specified by both magnitude and direction from a given
origin.
When you raise your coffee cup to your lips you are displacing the cup. If you raise it 30
cm and bring it toward 30 cm then
the total displacement is 30 cm up and 30 cm over, a resultant displacement of about 52
cm at an angle of 45º above the
horizontal from the table to your lips. You will notice that the resultant displacement is
a vector that does not necessarily
point in direction of actual motion but only from the starting position to the final
position.
VELOCITY
Ratio of change in displacement to the change in time.
If you lift your coffee cup (above) straight up in 1/10 second and then pull it horizontal
to your lips, the average velocity is
300 cm/sec up, followed by a horizontal average velocity of 300 cm/sec toward you.
Notice again that the resultant velocity
does not point in the direction of any actual motion, rather the velocity points in the
direction of the change in displacement.
In this case it is 520 cm/sec in a direction 45º above horizontal from the table to your
lips.
ACCELERATION
Ratio of the change in velocity to the change in time.
Any object that you start from rest undergoes a period of acceleration while you are
bringing it up to its final velocity. Notice
that acceleration is a vector that points in the direction of the change in velocity and
may not point in the direction of the
displacement. The acceleration of an object following a curved path always has a
component of the acceleration in the
direction of curvature.
UNIFORMLY ACCELERATED MOTION
Motion of an object whose acceleration is constant, i.e., both the magnitude and
direction of the acceleration remain the same
for all times.
Notice the restrictions that are placed upon motion for it to be of this kind. The change
in velocity must always be in the
same direction and constant. Therefore, the velocity must always be in the same
direction. Hence the change in displacement
is always in the same direction so the displacement occurs along a straight line. There
are no systems you meet in everyday
life that exactly satisfy these conditions. Nevertheless, there are systems, such as freely
falling objects, that are nearly
described by this motion.
PROJECTILE MOTION
An object has constant velocity in one direction and uniform acceleration in a direction
at right angles to the constant velocity.
This motion represents the idealized motion of objects near the surface of the earth
when spin and wind are neglected.
UNIFORM CIRCULAR MOTION
Objects moving in a circle with constant speed.
Think of riding on a merry-go-round. Your speed may be constant but the direction you
are going keeps changing. Perhaps
you start out going East, then North, then West, then South and back toward the East
again. So you go around and around.
Your acceleration is a vector pointing toward the center of the circle.
RADIAL ACCELERATION
Acceleration directed toward center of curvature of motion, i.e., perpendicular to the
curve of the movement of the object in
space.
For uniformly accelerated motion the radial acceleration is zero since the motion is in a
straight line. For uniform circular
motion, all of the acceleration is radial acceleration.
TANGENTIAL ACCELERATION
Acceleration directed along the direction of motion, tangent to the curve of the path of
movement of the object in space.
For uniformly accelerated motion all of the acceleration is tangential acceleration. For
uniform circular motion, none of the
acceleration is tangential acceleration.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: Answers; ; Vectors; Distance; Displacement; Mechanics; Kinematics; Kinds
Of Motion; Graphs;
Acceleration; Time; Uniform Motion; Linear Motion; Area; Slope; Velocity
SECTION 3.1 Introduction
Your present state of motion is probably a state of rest as you move your eyes to read
this page. In contrast, during a race
an Olympic sprinter starts from rest, tries to increase her speed as fast as possible and
then run at top speed to the finish
line. On the other hand, an astronaut moving about in a spacecraft in orbit around the
earth seldom uses his legs to propel
himself around the craft. He seems to hover just about the floor of the space lab and
move around in straight paths between
objects.
SECTION 3.2 Characteristics of Distance and Displacement
In order to add a 3 km vector displacement to a 3 km vector displacement you must
know the angle between the vectors.
What angle between these vectors must you use to obtain a resultant of 3 km? The
numerical answer is given at the end of
Section 3.3 of the textbook.
SECTION 3.5 Characteristics of Motion
In many motion problems the vector nature of displacement and velocity are important
aspects of the problem. To overlook
the "vectorness" of certain properties of physical system is a common mistake among
beginning students of physics. When
reading a problem for the first time, remember that some quantities may be vectors and
must be treated by the special rules
of vector algebra.
SECTION 3.6 Linear Motion
The answers to questions 1-7 are given in the textbook. Notice that for a velocity versus
time graph the area under the curve
represents the displacement and the slope of the curve represents the acceleration.
SECTION 3.7 Uniformly Accelerated Motion
Notice that in the data shown in tables 3.2 and 3.3 the distance down the incline
increases more rapidly than the time; i.e.,
when the time doubles the distance is more than doubled, it is about four times as large.
During the time intervals from 0-1
sec to 0-2 seconds, or from 0-2 seconds to 0-4 seconds, the distance ratios are 12.4/49.2
or 49.2/198.8 for the unweighted
car and 12.0/48.3 or 48.3/198.2 for the weighted car. In both cases, if the time is
doubled, the distance is quadrupled.
Likewise, if the time is tripled, the distance is nine times as much. These data, then,
indicate a quadratic relationship between
distance and time as shown in Equations 3.10. We can conclude that the car is
undergoing uniformly accelerated motion
down the incline. In fact, the weighted and unweighted cars appear to have almost the
same acceleration. An explanation of
the seeming puzzling behavior of more massive objects is discussed in Chapter 4.
If you plot the data given in the Example on page 51 you will obtain a straight line
graph for the velocity versus time with a
positive slop of 2m/s2 and starting at (0, 0). The equation of the line is v = 2t.
SECTION 3.10 Effects of Acceleration
Within the body it seems likely that food is accelerated down the throat and esophagus
to the stomach. It decelerates in the
stomach. The blood is accelerated as it leaves the heart and decelerates on its return trip
to the heart.
EXAMPLES
Keywords: ; Worked Examples; Acceleration; Distance; Time; Projectile Motion;
Acceleration Due To Gravity;
Free Fall; Centripetal Acceleration; Arithmetic; Scalar Arithmetic; Linear Motion;
Angular Acceleration; Circular Motion; Circular
Orbits; Mechanics
MOTION PROBLEMS
1. A physics student accidentally rolled out of bed while dreaming about uniform
circular motion. What was his speed just
before he hit the floor?
What Data Are Given?
An object (a student) of unknown mass falls from rest an unknown distance and hits a
floor.
What Data Are Implied?
If we assume the student falls under idealized conditions, then we can treat his motion
as uniformly accelerated motion down
with an acceleration of 9.8 m/s2. A typical bed is about 50 cm above the floor.
What Physics Principles Are Involved?
If we assumed idealized falling motion, then we can use the principles of uniformly
accelerated motion. The Equations 3.10
are for the special condition of starting from rest with an initial displacement of zero.
Both special conditions can be met by
this example.
What Equations Are to be Used?
In this case we know the acceleration, a = 9.8 m/s2, and the distance of fall, s = 50 cm or
0.50 meters, and we are seeking
the velocity, so we can use the following equation from 3.10:
2 a • s = vf2 (3.10)
where the acceleration a is given by the usual free fall value of 9.8 m/s2, or g.
Algebraic Solution
2gs = vf2
so: vf = (SQR RT) (2gs)
Numerical Solution
vf = (SQR ROOT) [2(9.8 m/s2)(0.50 m)] = (SQR RT) (9.8 m2/s2)
vf = 2.2 m/s
Thinking About the Answer
Notice that the units obtained for the answer, namely meters divided by seconds, are
the proper units for a velocity.
2. A student is skiing down a hill and wishes to take twice as long to reach the bottom,
so she starts up higher on the side of
the hill. How much farther up must she start?
What Data Are Given?
A person of unknown mass is skiing down a slope of unknown incline.
What Data Are Implied?
This problem implies idealized uniformly accelerated motion of unknown acceleration
a.
What Physics Principles Are Involved?
This problem requires the use of the concepts of uniformly accelerated motion for the
special case of starting from rest,
doubling the time of motion, and finding the distance travelled. Equations 3.10 may be
used.
What Equations Are to be Used?
In this case, we infer a starting rest position at a location on the side of the hill. After a
time the skier reaches the bottom.
The problem is to find the distance if the time is to be doubled. The following equation
combines all the known quantities
s = 1/2 at2 (3.10)
Algebraic Solution
Let t1 be the first time, s1 is the first distance, and "a" is the acceleration; then
s1 = 1/2 at12
Let t2 be the second time, twice the first time, so
t2 = 2t1, then the second distance s2 is given by
s2 = 1/2 at22 = 1/2 a(2t1)2
s2 = (1/2 at12)4 = 4s1. So the skier must start up the hill a distance 4s1, four times the
first starting distance.
3. A member of the university girls' basketball team dribbled the length of the court and
at a distance of 15 meters from the
basket, she reached up to a height of 2.5 m and launched a shot with an initial speed of
10 m/s for maximum range. Is it
possible she made a basket?
What Data Are Given?
The distance of the origin of the launch is 15 m from the basket. The height of the origin
of the shot is 2.5 m from the floor
and the initial speed is 12 m/s.
What Data Are Implied?
The height of the basket is 3.0 m (10 ft.) from the floor. The angle of launch for
maximum range is 45º (see page 56).
Idealized projectile motion of a point particle is assumed so we can draw the following
sketch of the problem:
What Physics Principles Are Involved?
The basic equations of projectile motion must be used. The problem is to determine if
the ball can be 3.0 m above the floor
when it is 15 m in a horizontal direction from the launch location that is 2.5 m above the
floor. A combination of the equations
(3.12) and (3.13) are required to solve this problem.
What Equations Are to be Used?
The proper horizontal and vertical displacements must be obtained by the ball to go
through the loop, so we need to use the
horizontal and vertical displacement equations:
horizontal motion: sh = vht (3.12)
vertical motion: sv = vvt + 1/2 gt2 (3.13)
Algebraic Solution
Let us proceed by assuming we know the horizontal displacement sh (which we will set
equal to 15 m) and we can calculate
the initial values for the vertical and horizontal components of the velocity. In the above
equations the two unknown quantities
are the time and the vertical distance. We can proceed by substituting a known
expression, namely sh/vh for the time t and
solving for the vertical displacement;
sv = vv (sh/vh) + 1/2 g (sh/vh)2
where vv = v sin q; vn = v cos q where q is the angle of launch from horizontal, 45º for
this problem,
sv = (v sin q) (sh/(v cos q)) + 1/2g (sh2/v2cos2q)
but sin q /cos q = tan q; so
s v = sh tan q + 1/2 g (sh2/v2cos2q)
Numerical Solution
sh = 15 m
tan q = tan 45º = 1.0, cos q = cos 45º = 1 / ((SQR RT) 2)
g = -9.8 m/s2; v = 12 m/s
sv = (15 m) (1.0) - 1/2 (9.8)m/s2 [(15)2m2/((12)2m2/sec2 (1/2))]
sv = 15 m - (9.8)(225/144)m = -0.3 m
So the shot drops below the basket by 0.8 m, since it should have been at +0.5 meters
instead of -0.3 m.
Thinking About the Answer
This is a fairly complex problem which involves the use of several equations. It is a
good idea on a problem like this one to
check your procedures out carefully.
4. A skylab in orbit 435 km above the surface of the earth completes a trip around the
earth in 93.0 minutes. It is 7.0 minutes
after lift off when the lab finally achieves this orbit. (a) What is the average tangential
acceleration necessary to obtain its
orbital position? (b) What is the radial acceleration when it is in orbit?
What Data Are Given?
The skylab is in an orbit 435 km above the earth, since the earth's radius is 6.38 x 103
km; the radius of the orbit is (6.38 +
0.44) x 103 km or 6.82 x 103 km. The time for one orbit is 93.0 minutes or 5580 seconds.
The time required to accelerate the
skylab into its orbit is 7.0 minutes or 420 seconds.
What Data Are Implied?
The problem implies that the skylab is in a circular orbit. The average tangential
acceleration is the value obtained by dividing
the final tangential velocity by the time required for acceleration.
What Physics Principles Are Involved?
This problem requires you to use the equations for the radial and tangential acceleration
for objects moving in a circular path.
The magnitude of the average tangential acceleration is given by the following
equation:
at = (vt final - vt initial) / time
and the direction is toward the larger value of the tangential velocity. The radial
acceleration has a magnitude v2/r as derived
on page 58 and points towards the center of the path.
What Equations Are to be Used?
The two equations given above will be needed plus a knowledge of the basic properties
of circles and velocity.
radial acceleration = (velocity)2/(radius) (3.21)
tangential acceleration = (vf - vi) / time
tangential velocity = circumference / time for one cycle = 2p(radius) / time
Algebraic Solution
(a) at = (vf - vi) / time of acceleration; but vi = 0 since the skylab starts from rest. Let
t = time of acceleration.
vt = (2p(r)) / time for one cycle; let T = time for one cycle
for this problem, tangential acceleration = at = (2pr) / (T/t) = (2pr) / tT
(b) radial acceleration = av = vf2/r = (2pr/T)2/r = (4p2r2) / rT2 = (4p2r) / T2
Numerical Solution
t = 420 seconds; T = 5580 seconds
r = 6.82 x 103 km = 6.82 x 106 m
(a) tangential acceleration = at = (2p (6.82 x 106 m)) / (4.2 x 102 x 5.58 x 103s)
at = 1.8 x 101 m/s2 = 18 m/s 2 towards increasing velocity.
(b) radial acceleration ar = ar = [(4p2)(6.82 x 106 m)] / (5.58 x 103 s)2
ar = 8.6 m/s2, towards the center of the earth.
Thinking About the Answer
Notice how this problem combines many different concepts, your knowledge of the
properties of circles from geometry, as
well as the derived quantities from Section 3.9 in the book. This is typical of problems in
physics. Any knowledge you gained
from your previous experiences with mathematics or nature can be useful in solving
physics problems. Notice that both
answers came out with the correct units for acceleration, meters per second squared,
m/s2.
4. Below we have given a step-by-step problem which we think will help you compare
and contrast two different types of
motion near the surface of the earth. Work the problem out in the spaces provided and
then compare your answers with ours.
(Show your work in this space)
At the same instant in time, two slingshots are shot from the same location on the
ground. The first slingshot shoots its
projective vertically up into the air at a velocity of 98 m/s. The second slingshot shoots
its projectile at an angle of 30º above
the horizontal with a velocity of 196 m/s.
(a) How long does it take the first projectile to reach the top of its
trajectory?
(b) How long does it take the second projectile to reach
the top if its trajectory?
(c) Give the position of the first
projectile when it reaches the top of its path. (Hint: Assume the
start position to be 0, 0 of a coordinate system and give the location
in terms of its x, y position.)
(d) Give the position of the second
projectile at the top of its path. (Hint: Same as in the previous
problem.)
(e) What is the velocity of the first projectile at the top
of its path?
(f) What is the velocity of the second projectile at the
top of its path?
(g) How long does it take the first projectile to
return to the ground?
(h) How long does it take the second projectile
to return to the ground?
(i) Give the location of the point of impact
when the first projectile hits the ground. (See hint in part c.)
(j) Give the location of the second projectile when it strikes the ground.
(See hint in part c.)
(k) Tabulate the results of your calculations
below and sketch the actual path of each projectile. Discuss the
similarity and differences between the motion of the two projectiles.
See Table.
See Fig.
(l) What is the acceleration of the first projectile at its maximum point?
(m) What is the acceleration of the second projectile at its
maximum point?
ANSWERS:
(a) 10 seconds; (b) 10 seconds; (c) 0,490 m; (d) (1670 m, 490 m); (e) 0 m/s; (f) 167 m/s
horizontal; (g) 20 seconds; (h) 20
seconds; (i) 0, 0; (j) 3340 m, 0 m; (k)
See fig.
See Fig.
(l) 9.8 m/s2, downward; (m) 9.8 m/s2, downward.
PRACTICE TEST
Keywords: Problems; Questions; Answers; ; Mechanics; Kinematics; Evaluations; Time
; Graphs; Projectile
Motion; Velocity ; Acceleration Due To Gravity ; Acceleration
1. A car traveling 54 meters/sec skids uniformly to rest in 6.0 seconds after the brakes
are applied.
a. What is the car's acceleration in meters/sec2?
b. How many meters did the car skid in coming to rest?
2. An American space team in their command module is attempting to dock with the
orbiting Skylab Spacecraft. They
approach the craft along a straight course. The graph below gives the separation
distance as a function of time.
See Fig.
a. Find the initial relative velocity of the two spacecrafts at zero seconds.
b. What occurs at 200 seconds?
c. What is the average velocity of relative motion during the first 400 sec. of the interval
shown?
3. A golfer finds himself confronted with the golf shot diagrammed below. He has
calculated that he is 90.0 meters from the
green and knows that he must clear a 12 meter tree which stands half-way (and on a
line) between the ball and the green.
Hitting the ball with a nine iron gives the ball a velocity of 32 meters/sec at an angle q =
45º.
a. What will be the ball's time of flight assuming that it does not strike the tree?
b. Assuming that the ball is driven directly toward the green, will the ball clear the tree?
Prove your answer.
See Fig
4. Early test with the Rocket Sled illustrated that man was capable of withstanding large
units of acceleration and
deceleration. The test pilot was strapped to a padded chair and the sled was accelerated
rapidly from rest and then
decelerated rapidly from a high velocity to rest. If the accelerations were of the order of
5 g's, compare the physiological
effects of the acceleration with those of deceleration.
ANSWERS:
1. -9 m/sec2, 162 meters
2. - 25 m/sec, collision, -10 m/sec
3. 4.6 sec, yes! by 13.14 meters
4. For horizontal accelerations, both acceleration and decelerations have approximately
the same effect: progressive tightness
in chest, loss of peripheral vision, blurred vision, difficulty in speaking. Deceleration
has all these with reduced chest
pressure.
Chapter 4 Forces and Newton's Law
GOALS
When you have mastered the concepts of this chapter, you will be able to achieve the
following goals.
Definitions
Define each of the following terms, and use each term in an operational definition:
force coefficient of friction
weight centripetal force
frictional force weightlessness
Newton's Laws
State Newton's laws of motion and gravitation.
Resolution of Forces
Given the force acting on a system, draw a force diagram and/or resolve forces into
their components
and/or solve for an unknown force.
Newton's Second Law Problems
Given any two of the three variables in Newton's second law, solve for the third.
Centripetal Force Problems
Given any three of the four variables in the centripetal force equation, calculate the
value of the fourth
variable.
PREREQUISITES
Before you begin this chapter you should have achieved the goals of Chapter 3,
Kinematics, including
uniformly accelerated motion and uniform circular motion.
OVERVIEW
Keywords: ; Instructions; Mechanics; Dynamics; Newton's Laws; Force
The statements which we call Newton's Laws of Motion are considered to be a central
feature of the study of the motion of
objects. These laws generalize much of our common experiences with forces and their
influence on the state of motion of
objects. As you read and consider the Laws of Motion presented in this chapter, make a
point to relate the meaning of the
laws to real experiences you have had with bicycles, automobiles, electric trains, etc.
SUGGESTED STUDY PROCEDURE
As you begin your study of this module, begin by carefully reading each of the Chapter
Goals: Definitions, Newton's Laws,
Resolution of Forces, Newton's Second Law Problems, and Centripetal Force Problems.
Remember that on the next page of
this you will find a paragraph concerning each of the terms listed under Definitions.
Next, read text sections
4.1-4.7. Remember that answers to questions asked in the text are given in the second
section of this . As you
read, be sure to consider the examples given. If you have difficulties with the
mathematical treatment of forces and their
components, refer to the initial chapter of this (Chapter 0).
At the end of the chapter, read the Chapter Summary and complete Summary Exercises
1-17. Now do Algorithmic Problems
1, 2, 5, 6, 7, 8, 9, 10, and 11. Check your answers to each problem carefully. Now
complete Problems and Exercises 1, 2, 5,
8, 10, 12, 14, 26, 27, and 28.
Now turn to the Examples portion of this and consider the additional problems and
questions presented. After
you have completed this task, you should be prepared to attempt the Practice Test on
Forces and Newton's Laws. This study
procedure is outlined below.
------------------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
------------------------------Definitions 4.1, 4.2, 4.3, 1-6
4.4,4.5,4.7
Newton's Laws 4.2, 4.3 7-11 2, 9, 10 8
Resolution of
Forces 4.2, 4.4 12, 13 2, 10, 11 1, 2, 5, 10,
12, 26
Newton's Second
Law Problems 4.2, 4.3, 4.4, 14-17 1,5,6,7 8,28
4.6
Centripetal
Force Problems 4.6 8 14,27
DEFINITIONS
Keywords: ; Glossary; Mechanics; Dynamics; Force; Weight; Friction; Centripetal Force;
Kinds Of Force
FORCE
A measure of the strength of an interaction; a push or pull; the effect of a force is to alter
the state of motion of a body. A
vector quantity with a magnitude measured in newtons (N).
The strength of the earth's gravitational field is measured by the weight of an object.
Weight is a force acting vertically
downward.
WEIGHT
The product of the mass and the acceleration due to gravity.
Another way to think of g is to treat it as the strength of the force of gravity near the
surface of the earth. Then the weight is
a downward, vertical force.
FRICTIONAL FORCE
Force that opposes the motion of an object.
In all of the cases of motion you commonly encounter there is a force acting to oppose
the motion. When you are coasting
down a slope on your bicycle, what are the forces that tend to slow you down? All of
them can be classified as frictional
forces.
COEFFICIENT OF FRICTION
Ratio of the maximum frictional force to the net force pressing the surface together.
Try rubbing two sheets of paper together, then try rubbing two sheets of sandpaper
together. Which requires the greater force
to slide them along? This illustrates the larger coefficient of friction of sandpaper. If you
wet the sandpaper, what happens to
the force required to slide the paper along?
CENTRIPETAL FORCE
The unbalanced force acting on the body towards the center of rotation in circular
motion.
The radial acceleration of an object going around in a circle is proportional to centripetal
force.
WEIGHTLESSNESS
The gravitational attraction of the earth for an object in motion around the earth is
precisely equal to the centrifugal force of
the body in its orbit. When you are in such an orbiting spacecraft floating objects and
persons are the rule rather than the
exception. Ask your physics instructor to show you some of the interesting films made
from NASA photographs.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: Answers; ; Mechanics; Dynamics; Newton's Laws; Newton's First Law;
Newton's Second Law;
Newton's Third Law; Force; Inertia; Acceleration; Mass
SECTION 4.4 Force of Friction
Suppose you think of oil as composed of many smooth planes loosely packed together.
Then the oil could slide easily along
the planes and would greatly reduce friction in those directions. This is not a very good
model for oil. Can you think of a
better one? At least this smooth planes model for oil offers an explanation for the
reduction of friction in certain directions.
Many frictional forces are essential. They enable us to place ourselves in a fixed position
and remain there even when
subjected to small external forces. If your chair were resting on a frictionless floor, the
slightest breeze would blow you away
from the table where you are studying physics. That would be too bad, wouldn't it?
Indeed, it would be difficult to live as you presently do in a frictionless world. If we got
started moving, how would we stop?
GENERAL COMMENTS
Keywords: Newton's First Law; Newton's Second Law; Newton's Third Law; Inertia;
Acceleration; Mass; Exposition
(valuable insights included in the , free of charge)
NEWTON'S LAWS OF MOTION
as given in Latin in his famous book PRINCIPIA.
Lex. I - Corpus omneperseverare in statu suo quiesendi vel movendi uniformiter in
directum, nisi quatenus illuda viriblus
impressis cogitur.
Lex. II - Mutationem motus, proportionalem esse vi motrici impressae, et fieri secujdum
lina lineam rectam qua vis illa
imprimitur.
Lex. III - Actioni contrarium semper et aequalem esse reactionem; sive corporum
duorum actiones in se mutuo semper esse
aequales et in partes contrarias dirigi.
COMMENTS ON NEWTON'S FIRST LAW
The inertia of a body may be defined as that property of a body that tends to resist a
change in its state of rest or motion.
Mass is defined as a quantitative measure of inertia.
COMMENTS ON NEWTON'S SECOND LAW
Another way of stating the classical version of Newton's second law is: when a body is
acted upon by a constant force, its
resulting acceleration is proportional to the force and inversely proportional to its mass;
a = F/m
COMMENTS ON NEWTON'S THIRD LAW
It is important to note that the action force and the reaction force in Newton's third law
of motion act on different bodies.
EXAMPLES
Keywords: ; Worked Examples; Mechanics; Dynamics; Vectors; Arithmetic;
Trigonometry; Newton's Laws;
Newton's First Law; Newton's Second Law; Newton's Third Law; Centripetal Force;
Free Body Diagrams
RESOLUTION OF FORCES
1. A beanbag chair is at rest on a driveway inclined at 22ø. The chair has a weight of
68.0 newtons. (a) Draw the diagram of
the chair and label all of the forces acting on the chair. (b) Calculate the values for all
forces, etc., for this situation.
What Data Are Given?
The weight of the chair (68.0 N) and the angle of incline of the plane (22ø) are given.
The velocity of the chair is zero
throughout.
What Data Are Implied?
The weight is a force acting vertically down in a region where the strength of the earth's
gravitational force produces an
acceleration of 9.80 m/s2. The chair is not moving because of the presence of friction, so
a frictional force is implied in this
problem. Likewise, the plane is supporting the chair so a normal force is acting on the
chair because the plane is implicit in
this problem.
What Physics Principles Are Involved?
Both the second and third laws of Newton are necessary to understand this problem as
well as knowledge of the vector
nature of force. Before we go on to the algebraic solution of this problem we are ready
to use a force diagram to answer part
(a) and get us ready to solve part (b).
(a) Force Diagram
See Fig.
A free body diagram can be drawn by omitting all real objects and only showing the
forces,
See Fig.
What Equations Are to be Used?
We can make use of Newton's Second Law. We know the velocity remains constant (at
zero) so the acceleration is zero;
thus, the net force on the chair must be zero. That means the vector sum of the forces on
the chair must be zero;
W+N+f=0
Algebraic Solution
The vector equation can be written as two separate scalar equations, a vertical
component equation and a horizontal
component equation.
Vertical forces = 0 = W = N cos q - f sin q where positive is downward. (1)
Horizontal forces = 0 = N sin q - f cos q where positive is to the right. (2)
These equations can be combined to solve for the magnitudes of the two unknown
forces f and N.
Rearrange Equation (2) and divide by cos q:
f = N sin q / cos q) = N tan q (3)
Rearrange equation (1), divide by cos q and substitute the friction for f,
N = (W/cos q) - (f sin q/cos q) = (W/cos q) - N tan2 q (4)
Now combine the terms containing the normal force N, so
N (1 + tan2 q) = W/cos q or
N = W / [cos q (1 + tan2 q)] (5)
Substitute this expression into Equation (3) to obtain
f = W tan q / [cos q (1 + tan2q] (6)
Numerical Solution
W = 68.0 N; q = 22ø, so cos q = 0.927; tan q = 0.404.
N = 68.0 N / [0.927(1 + (.404)2)] = 63.1 N perpendicular to the driveway.
f = N tan q = (63.1 N) (.404) = 25.5 N, parallel and up the driveway.
Thinking About the Answer
Notice that both answers come out in the correct units, newtons. The normal force is
much larger than the friction force. This
seems reasonable for a low angle driveway. Notice as the angle decreases in Equations 5
and 6 the normal force approaches
a value of W and the friction force goes to zero. Is that what you expect?
NEWTON'S SECOND LAW PROBLEMS
2. Assume the beanbag chair given in the previous problem is lying at rest on an icy
(friction ÷ zero) driveway. What
happens? Give quantitative results.
What Data Are Given
Same as in the previous problem except now the friction force f is equal to zero.
What Data Are Implied?
Same as in the previous problem, except the frictional force is zero. Motion along the
driveway can be deduced.
What Physics Principles Are Involved?
As before, Newton's 2nd and 3rd laws. Now Newton's 2nd law shows that the beanbag
chair will undergo some acceleration
down the driveway since the net force cannot be zero. This is clearly seen by removing
the frictional force from the previous
force diagram. Then you are left with See Fig.
It is not possible for the weight and the normal force to add to zero. Hence, the chair
must experience some net force and
some acceleration.
What Equations Are to be Used?
The vector equation for the 2nd law in this case is given by
ma = W + N (7)
However, we know the chair can only move along the driveway since there is no force
acting to lift the chair. Hence we can
transform Equation (7) into a scalar equation by considering only motion and forces
that act parallel to the driveway. The
normal force has no components parallel to the driveway. The weight has only one
component, W sin q, acting down the
driveway, so we can write
ma = W sin q {motion down the driveway} (8)
Algebraic Solution
Remember, only the weight W, g, and q are given, so we need to replace the mass on
the left side of Equation (8) by W/g,
thus
(W/g)a = W sin q or
a = g sin q (9)
Numerical Solution
g = 9.80 m/s2, sin 22ø = 0.375
a = (9.80)(0.375) = 3.68 m/s2, down the driveway.
Thinking About the Answer
Note the units are correct, and that if the slope of the driveway is decreased to zero
degrees the acceleration goes to zero;
i.e., the chair remains at rest, does that seem reasonable to you?
CENTRIPETAL FORCE PROBLEMS
3. An 80.0 kg earth satellite circles the earth once every 94.2 minutes at a mean altitude
of 522 km. Use 6.38 x 106 m as
the radius of the earth. (a) What is the gravitational force acting on the satellite while in
orbit? (b) What fraction of its force at
the surface of the earth is it? (c) If you were riding inside the satellite what forces would
you experience?
What Data Are Given?
The mass of the satellite is 80.0 kg. Data which can be used to calculate its velocity and
the radius of its circular path are
given.
What Data Are Implied?
The problem intends for you to assume the path is a circle with the center of the earth as
the center of the path, hence the
radius of the path is 6.38 x 106 m + 522 km = 6.38 x 106 + 0.52 x 106 = 6.90 x 106 m.
What Physics Principles Are Involved?
You need to recognize that the gravitational force must have the same magnitude as the
centripetal force. Furthermore, for an
observer inside the satellite the gravitational force will appear to be canceled by the
centrifugal force and the condition of
weightlessness will be experienced.
What Equations Are to be Used?
The equation for centripetal force is needed.
F = mv2/r (4.6)
In order to calculate the velocity v we need to divide the circumference of the orbit by
the time required for one pass around
it.
v = 2p(radius)/time for one orbit (10)
Algebraic Solutions
(a) In orbit Fo = mv2/r but v = 2pr/t
Fo = m4p2r2) / rt2 = 4p2mr / t2 (11)
(b) On the surface Fe = mg
Ratio of orbit force to surface force is given by F0/Fe = 4p2r / gt2 (12)
Numerical Solution
r = 6.90 x 106 m
t = 94.2 minutes = 5.65 x 103 seconds
m = 80.0 kg; g = 9.80 m/s2
(a) Putting these values in Equation (11)
Fo = (80.0 kg)4p2(6.90 x106) / (5.65 x 103)2 = Fo = 6.82 x 102 kg/s2
Fo = 6.82 x 102 N
(b) Then Fo/mg = 0.87 which is 87%.
In summary, the force on the satellite in orbit is 682 N. Since its weight on earth is 784
N, the orbital force is 87% of its
weight on the earth. As an observer inside the satellite you would observe the effects of
the gravitational force being balanced
by a centrifugal force and you would experience weightlessness.
Thinking About the Answers
Notice again that the units come out correctly, 1 newton is a kilogram meter per second
squared. According to Equation (12)
the ratio of the orbit force to the surface of the earth force is independent of the mass of
the object since the symbol (m)
does not appear in Equation (12). Does that seem reasonable to you?
PRACTICE TEST
Keywords: ; Problems; Answers; Questions; Evaluations; Gravity; Law Of Universal
Gravitation; Mechanics;
Dynamics; Weight; Free Body Diagrams; Anatomy And Physiology; Applications;
Newton's Laws; Newton's First Law;
Newton's Second Law; Newton's Third Law; Centripetal Force; Circular Motion
1. Newton's Universal Law of Gravitation applies to all massive objects, including
planets, automobiles, and people. However,
many people misrepresent the gravitational law. Use your knowledge of Newton's Law
and write a short correction for the
following statements.
a. Because of its larger mass, the sun attracts the earth with a greater force than the earth
attracts the sun.
b. The earth's rotation produces our gravitation. If the rotation were to stop, we would
drift into space.
c. A body taken to an altitude of about 100 miles has no weight because the
gravitational force does not extend out into
space beyond the atmosphere.
2. The Fig. shown below shows a modified Russell traction apparatus for femoral
fixation.
See Fig.
a. Construct a force diagram isolating point B showing all the forces which are at that
point.
b. Find the total horizontal force applied to the leg at B by the apparatus when a 4
kilogram mass is suspended at A.
3. A 500 kilogram subcompact car is parked on top of a long straight incline whose
inclination is 30ø from the horizontal.
a. Draw a free body diagram isolating the car and showing all the forces holding the car
in equilibrium.
b. Assume that the brakes are released and the car rolls (without friction) down the
plane. Using Newton's second law, predict
the car's acceleration.
4. Astronaut Owen Garriott (mass = 65 kilograms) is placed aboard a spinning device
used to determine his ability to
withstand high "g" forces.
See Fig.
a. If the apparatus is revolving at 1 revolution per second, what force does Garriott
experience due to this rotation motion?
b. How many times his weight is this force? (how many "g" 's does he experience?)
ANSWERS:
1. a) Newton's Universal Law states that these forces must be exactly equal in
magnitude, opposite in direction.
b) Rotation of the earth has little effect on our weight. Actually, if rotation were to cease,
our apparent weight (as measured
by a spring balance on the surface of the earth) would increase slightly.
c) The gravitational force extends indefinitely into space. In a space ship which is
circling the earth, we say an astronaut is
weightless because in the condition of "falling" around the earth, he exerts no force on a
spring balance in his ship. Actually,
his actual weight (attraction to the earth) is 95% of the force at the earth's surface.
2. 69.6 N
3. 5 m/s2
4. 25,600 N, 39 g's
Chapter 5 Energy
GOALS
Keywords: Learning Objectives; ; Mechanics; Energy; Simple Machines; Levers; Work;
Kinetic Energy; Potential
Energy; Power; Efficiency; Conservation Of Energy
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
work lever systems
energy theoretical mechanical advantage
potential energy actual mechanical advantage
kinetic energy efficiency power
Conservative and Nonconservative Systems
Establish the difference between a conservative and a nonconservative system.
Conservation of Energy
Explain the principle of conservation of energy.
Energy Problems
Apply the principle of conservation of energy to solve mechanics problems.
Lever Systems
Determine the theoretical mechanical advantage of human body lever systems.
Efficiency
Calculate the efficiency of a machine or human in action.
Power
Determine the power required for a given process or activity.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 3,
Kinematics, and Chapter 4,
Forces and Newton's Laws.
OVERVIEW
Keywords: ; Instructions; Mechanics; Energy; Simple Machines
Newton's three Laws of Motion provide a useful model for viewing the motion of
objects we encounter in our daily experience.
However, the model becomes difficult to apply in situations where the forces involved
vary significant with time and
displacement. To provide a parallel assessment procedure, the concept of work and
energy was introduced. This alternate
model for viewing our environment allows for a simple solution to many problems.
SUGGESTED STUDY PROCEDURE
To begin your study of this chapter, become familiar with the following Chapter Goals:
Definitions, Conservation of Energy,
Energy Problems, Lever Systems, Efficiency and Power. A discussion of each of the
terms listed under Definitions can be
found in the first section of this chapter. Next, read text sections 5.1-5.13 and study the
examples provided at
the end of each section. Please remember that the answer to questions asked in the text
are given in this
chapter, section two.
At the end of the chapter, read the Chapter Summary and complete Summary Exercises
1-7 and 11-15. Check your answers
against those provided. Now, complete Algorithmic Problems 1-6 and check each
answer. Then do Exercises and Problems 1,
2, 3, 5, 7, 12, 15, 18, 19, and 22. For additional work on the goals of this chapter, see the
third section of this
chapter. Finally, attempt the Practice Test. If you have difficulty with any of the
concepts, refer to the appropriate chapter
section for additional work. This study procedure is outlined below.
-----------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
-----------------Definition 5.1, 5.2, 5.3 1-7 1,2
Conservation of 5.4, 5.5, 5.6, 11-14 15
Energy 5.7, 5.8, 5.11
Energy Problems 5.4, 5.5, 5.6, 1, 3, 6 3, 5, 12,
5.8, 5.9, 5.13 18, 19
Lever Systems 5.12 5, 6, 13 4 7
Efficiency 5.10 7, 14 22
Power 5.9 15 2, 5
DEFINITIONS
Keywords: Glossary; ; Potential Energy; Kinetic Energy; Mechanics; Energy; Power;
Simple Machines; Levers;
Efficiency
WORK
Applied force produces a displacement; magnitude of work done is given by product of
force component parallel to the
displacement and the magnitude of the displacement.
Note that work as used in physics requires a force component acting in the direction of
motion as well as motion. So, for
example, the force that a string exerts on a whirling object connected to it is always
perpendicular to the direction of motion
so the work done by the force is zero.
ENERGY
Property of a system which causes changes in its own state or state of its surroundings;
measure of ability to do work
(physical).
Energy is an intrinsic property of a system. We always compute energy from other
measured properties of a system. Energy
is a scalar quantity. There are many forms of energy; mechanical, electrical, chemical,
thermal, etc. There are two kinds of
mechanical energy; potential and kinetic.
POTENTIAL ENERGY
Ability to do work as a result of position or configuration.
A most common form of potential energy is the energy which results from changing the
height of objects above the earth.
Have you ever wondered why it is more tiring to climb up stairs than to come down
them? The changes in potential energy
that occur may be an answer.
KINETIC ENERGY
Energy of motion of a body.
The idea that the kinetic energy of a moving object is proportional to the square of the
speed is well-illustrated by data about
automobiles. The distance required to stop a moving automobile increases with the
square of the speed, as does the fatality
rate in accidents.
POWER
Time rate of doing work or using energy.
The increased power of large automobile engines is most manifest in the ability of an
automobile to have its speed rapidly
increased.
LEVER
A simple machine. Essentially a rigid body with a fixed pivot.
A lever is only one kind of simple machine. Others, such as the wheel and axle, the
inclined plane, and the screw, are only
briefly mentioned in the textbook.
THEORETICAL MECHANICAL ADVANTAGE
Ratio of distance through which the applied force acts to the distance through which the
load moves.
ACTUAL MECHANICAL ADVANTAGE
Ratio of the load to the applied force.
EFFICIENCY
Ratio of work accomplished to work supplied.
The efficiency of a simple machine is also equal to the ratio of the actual mechanical
advantage to the theoretical mechanical
advantage.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Work; Energy; Mechanics; Gravitational Potential Energy;
Conservative Forces;
Nonconservative Forces; Potential Energy; Dynamics
SECTION 5.2 Work
Any force you apply to an object which has a component which is not in the direction of
motion is not doing work. When you
lift an object you do work, but as you carry the object to a new location, the vertical
force you exert to hold up the box does
not do any work as long as the motion is in a horizontal direction.
SECTION 5.3 Energy
The chemical energy stored in a battery is converted to electrical energy to operate a
portable radio and then the electrical
energy is converted to the sound energy which you hear coming from the radio.
SECTION5.5 Potential Energy
Whenever you lift or lower an object near the surface of the earth you change its
gravitational potential energy. The potential
energy of water that can flow down from high elevations is converted into electrical
energy through the use of hydroelectric
stations.
SECTION 5.7 Conservative and Nonconservative Forces
1. Each time the boy slides up a hill he will rise a smaller amount. Each trip he loses
some of his energy to frictional,
non-conservative forces working on his sled. Finally he will come to rest in the valley
between the hills.
2. The total energy of a non-conservative system will always decrease as the nonconservative forces dissipate the energy of
the system.
3. All real systems are non-conservative. It is only in our imaginations that we have a
system which never suffers a loss of
energy to dissipative forces.
4. The final equilibrium state of a system will be its state of lowest energy.
EXAMPLES
Keywords: Worked Examples; ; Mechanics; Energy; Work; Potential Energy; Kinetic
Energy; Work-Energy
Theorem; Trigonometry; Friction; Conservation Of Energy; Simple Machines; Levers;
Efficiency; Free Body Diagrams;
Dynamics; Anatomy And Physiology; Arithmetic; Pulleys; Screws; Inclined Planes;
Respiratory System; Applications; Time;
Power
ENERGY PROBLEMS
1. A professional skier of 60 kg mass is on a ski jump ramp like the one shown See Fig.
5-1. (not recommended for
beginners!). At point C the ramp is pointed up at an angle of 45ø. Point A is 40 meters
above the lowest point of the ramp
and point C is 10 meters above the lowest point. The distance along the ramp from
point A to point C is 81 meters. Assume
the potential energy of the skier at point B is zero.
(a) What is the potential energy of the skier at point A?
(b) Assuming no friction, what is his speed when he gets to point B if he starts from rest
at A?
(c) Suppose for artificial snow the force of friction is a constant equal to 5 newtons. How
much energy does the skier lose in
going from A to C?
(d) What is the maximum height to which the skier can rise above B if he starts from
rest at A on artificial snow, and jumps
off the ramp at point C?
What Data Are Given
The mass of the skier = 60 kg; distance travelled on the ramp = 81 m; starting elevation
= 40 m; friction force = 0 for part (b)
and 5 N for parts (c) and (d). The angle of incline of the ramp is 45ø.
What Data Are Implied?
It is assumed that the skier does no work on himself through the use of his muscles. In
addition, the only dissipative force
considered is that artificial snow frictional force. The gravitational acceleration constant
is assumed to be 9.80 m/s2.
What Physics Principles Are Involved?
This problem involves the use of the work-energy theorem, the conservation of energy,
and the expressions for gravitational
potential energy, and kinetic energy.
What Equations Are to be Used?
Kinetic energy = KE = 1/2 mv2 (5.2)
Potential energy = PE = mgh (5.4)
Work done = F ¥ d (5.1)
Work-energy theorem; net work on a system = change in energy
W = DKE + DPE (5.9)
Projectile motion concepts from Chapter 4.
Algebraic Solutions
Let m = mass of skier; h = starting height of point A.
(a) Potential energy at A = mghA
(b) Speed at B; conservation of energy gives
PEB + KEB = PEA + KEA (5.11)
since friction is neglected for this part of the problem. Both the kinetic energy at A and
the potential energy at B are zero.
Can you explain why?
KEB = PEA
1/2 mvB2 = mgh
The Speed at B = vB = SQR RT[2gh] (1)
(c) Energy loss = work done by friction
= friction force times the distance
(d) Maximum height is found by applying the conservation of energy concepts to his
motion through the air. At point C he has
a kinetic energy,
KEc = mghA - mghC - work done by friction
where hA and hC are the heights of points A and C respectively. His speed at C is given
by SQR RT[2KEC]/m, so his
horizontal velocity,
vhorz = SQR RT[2KEC)/m] cos q (2)
where q is the angle of the ramp. At the peak of the jump path his vertical component of
velocity will equal zero so his total
energy will be equal to his potential energy plus his kinetic energy because of his
constant velocity in the horizontal direction.
This total energy must be equal to the energy he head when he left the ramp at point C.
Energy at peak of jump = kinetic energy at point C = KEC
mghpeak + 1/2 mv2 horizontal = KEC = mghA - mghC - friction work (3)
Since everything in this equation is known expect the peak height, hpeak it can be
found by combining Equations (2) and (3)
hpeak = (KEC / mg) - (v2horz / 2g) = (KEC / mg) - (2KECcos2 q / 2mg)
hpeak = (KEC / mg)(1 - cos2 q) (4)
Numerical Solutions
(a) Potential energy at A = (60 kg)(9.8 m/s2)(40 m) = 2.4 x 104 m2kg/s2
PEA = 2.4 x 104 joules
(b) vB = SQR RT[2(9.8)(40) m2/s2] = 1.2 x 102 m/s
(c) Energy loss = (5 N)(81 m) = 405 joules
(d) Kinetic energy at C = (60)(9.8)(40 - 10) J - 405 joules
KEC = 1.72 x 104 J
hpeak = [1.72 x 104 / (60)(9.8)] (1 - cos2 45ø)
hpeak = 1.46 x 101 m ÷ 15 meters above point C
Thinking About the Answers
You should check to see that all of the units came out correctly. You can see in this
problem that the friction force made
little difference in the energy at point C and the height of the skier peak. In equation (4)
we can see that if the angle q is
zero, the height of the peak is zero and if the angle q is 90ø the peak height will be a
maximum. In this particular case, since
the angle is 45ø the launch velocity is equally divided between its horizontal and
vertical components and the kinetic energy
at the peak is one-half the total kinetic energy at point C. Did you notice that the zero
point for potential energy was shifted to
point C for answering part (d)? Can you rework the problem without making that shift?
LEVER SYSTEMS
Keywords: Simple Machines; Screws; Pulleys; Worked Examples
2. Lever systems are only one of several kinds of simple machines. Shown below, in
addition to the lever, are the crank and
axle, inclined plane, pulley system, and screw jack. Using the symbols shown on each
machine, calculate the actual
mechanical advantage, the theoretical mechanical advantage, and the efficiency of each
machine.
See Fig.
What Data Are Given?
In each case a load of W newtons is moved by the application of a force of F newtons.
The distances for both the load W
and the applied force F are given.
What Data Are Implied?
Any explicit calculation of frictional forces is not done. Instead all of the frictional forces
are included in the value of W that
can be moved by the force F.
What Physics Principles Are Involved?
The definitions of theoretical mechanical advantage, actual mechanical advantages, and
efficiency are needed.
What Equations Are to be Used?
Theoretical mechanical advantage (TMA) = distance applied force moves/distance
move load (5.15) Actual mechanical
advantage (AMA) = load W/force F (5.16)
Efficiency (e) = work done by the machine/energy supplied to the machine (5.14)
Algebraic Solutions
(i) AMA = W/F; TMA = 2pl/2pr = l/r; e =Wr/Fl
(ii) AMA = W/F; TMA = d/h; e = W ¥ h/F ¥ d
(iii) AMA = W/F; TMA = l/d; e = Wd/Fl; same for all three types of lever
(iv) AMA = W/F; TMA = l/d; e = Wd/Fl
Note for the case shown for every 1 m the force moves, the W goes up only 1/4 m; so
TMA = 4. This is determined by the
geometry of the pulley system. The system shown has 4 ropes supporting the bottom
pulleys so the TMA is 4. Can you use
the geometry of a pulley system to derive this algorithm for computing the TMA of a
pulley system?
(v) AMA = W/F; TMA = 2pl/p; e = Wp/2pFl
EFFICIENCY
Keywords: Efficiency; Work; Respiratory System; Worked Examples
3. A mile runner felt that he needed to eat 1.3 ounces (37 gm) of a candy bar every time
he ran a mile in 4 1/2 minutes.
Estimate his energy efficiency. Use Table 5.1 and Table 5.2 for the constants you need.
What Data Are Given?
The time of running and the amount of candy eaten, 4.5 minutes and 37 gm
respectively.
What Data Are Implied?
To calculate efficiency of energy use you need to calculate the energy input and the
energy used in a useful way. You can
use Table 5.1 and your knowledge that sugar is a carbohydrate to convert 37 gm of
sugar to its approximate energy value
since its energy conversion value is 4 kcal/gam. Likewise, you can use Table 5.2 to
estimate the rate at which the runner
uses energy, 11.4 kcal/minute.
What Physics Principles Are Involved?
The definition of energy efficiency must be used in conjunction with some concepts
implied by the data in Table 5.1 and
Table 5.2 . You can take the product of the amount of food eaten in grams times the
energy conversion factor ( Table 5.1) to
compute the total energy input. You can take the product of the running times the
energy use rate ( Table 5.2 ) to calculate
the use of energy.
What Equations Are to be Used?
Efficiency = Useful work done/energy input
Energy Input = (mass of food) (energy conversion factor)
Energy Consumption = useful work = (time of activity) (energy use rate)
Algebraic Solution
E = U.W.D. / E.I. = (time)(kcal/min) / (mass)(kcal/gm)
Numerical Solution
E = (4.5 min)(11.4 kcal/min) / (37 gm)(4 kcal/gm) = 51.3 kcal/148 kcal = 35%
Thinking About the Answer
This runner seems to be 35% efficient. That means that 65% of the energy input is not
used in running. Where does it go?
POWER
Keywords: Power; Work; Time; Respiratory System; Human Anatomy And Physiology;
Biology; Worked Examples
4. What are the power ratings in watts of a person engaged in various activities? You
may use the values in Table 5.2 for
this problem.
What Data Are Given?
One person engaged in various human activities.
What Data Are Implied?
The energy use rate for human sitting, resting, etc., are given in Table 5.2 The watt is an
energy use rate of 1 joule per
second.
What Physics Principles Are Involved?
You need to use the definition of power as given in Equation 5.12.
Power = Work Done/Time to Do it (5.12)
What Equations Are to be Used?
Power = Work/Time = Energy Used/Time (5.12)
The data in Table 5.2 are already given in units of energy (kcal) per unit time (minutes)
and only need to be converted to
joules and seconds respectively.
Algebraic Solution
Power (Watts) = Energy Used (joules) / Time (seconds)
joules = (4187 J/kcal)(no. of kcal)
seconds = 60 sec./min. (time in minutes)
Power (watts) = (4187/60) (kcal/min.) = 69.8 (kcal/min.)
Numerical Solutions - Data from Table 5.2
Power (watts) sitting = 69.8 (2.0 kcal/min.) = 140 W
Power (watts) resting = 69.8 (1.2 kcal/min.) = 84 W
Power (watts) sleeping = 69.8 (1.1 kcal/min.) = 77 W
Power (watts) walking = 69.8 (3.8 kcal/min.) = 270 W
Power (watts) bicycling = 69.8 (6.9 kcal/min.) = 470 W
Power (watts) swimming = 69.8 (8.0 kcal/min.) = 560 W
Power (watts) skiing = 69.8 (9.0 kcal/min.) = 630 W
Power (watts) running = 69.8 (11.4 kcal/min.) = 796 W
Power (watts) climbing up stairs = 69.8 (12.0 kcal/min.) = 838 W
Keywords: Power; Mechanics; ; Energy; Humor
See Fig.
PRACTICE TEST
Keywords: ; Problems; Answers; Mechanics; Potential Energy; Kinetic Energy; Friction;
Simple Machines;
Pulleys; Efficiency; Levers; Anatomy And Physiology; Power; Muscular Systems
1. A professional skier of 100 kg mass is on a ski ramp See Fig. (not recommended for
beginners). At point C the ramp is
pointed straight up. Point A is 40 meters above the ground. Point C is 10 meters above
the ground. The distance from point
A to point C along the ramp is 150 meters. The potential energy at B is zero.
________a. What is the potential energy of the skier at point A?
________b. If the friction force along the path is a constant 50 N, what is the work done
by friction in going from point A to
point C?
________c. What is the kinetic energy of the skier at point C?
________d. What is the maximum height h, above the end of the ramp (point C) to
which the skier can rise in his trajectory?
2. A force of 6 N is required to raise a weight of 30 N by means of a pulley system. If the
weight is raised 1 meter in one
second while the applied force moves through a distance of 8 m _______a. What is the actual mechanical advantage of the pulley system?
_______b. What is the theoretical mechanical advantage of the pulley system?
_______c. What is the efficiency of this system?
_______d. What is the power dissipated in the system?
3. The human arm See Fig. is being tested for mechanical advantage. In this test, the
biceps arm action is utilized with a
weight of 10 Newtons. In the configuration shown, the force of the muscle is found to
be 85 Newtons.
_______a. Find the TMA of the arm.
_______b. Find the AMA of the arm.
_______c. Calculate the arm's efficiency.
ANSWERS:
1. 40,000 J, 7500 J, 22,500 J, 22.5 meters above C.
2. 5, 8, 62.5%, 18 watts
3. .125, .12, .94
Chapter 6 Momentum and Impulse
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
impulse elastic collision
impulsive force inelastic collision
momentum rocket propulsion
Impulse Problems
Use the relationship between impulse and change in momentum to solve problems.
Conservation of Momentum
Explain the principle of conservation of momentum.
Collision Problems
State the difference in conditions between an elastic impact and an inelastic impact, and
use both kinds of
conditions to solve problems.
Momentum and Energy Problems
Apply the principles of conservation of momentum and conservation of energy to solve
problems.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 3,
Kinematics, Chapter 4,
Forces and Newton's Laws, and Chapter 5, Energy.
OVERVIEW
Keywords: ; Instructions; Mechanics; Momentum; Impulse; Collisions
After studying and using the conservation of energy principle in solving mechanical
problems, you can appreciate the
advantages of finding conservation principles. Like the conservation of energy, the
quantity called momentum is also
conserved in active systems. The application of the conservation of momentum
principle is helpful in considering a host of
mechanical problems; e.g., impact and collisions between large and/or small physical
objects.
SUGGESTED STUDY PROCEDURE
When you begin to study this chapter, carefully read the following Chapter Goals:
Definitions, Impulse Problems, Conservation
of Momentum, Collision Problems, and Momentum and Energy Problems. A more
complete discussion of each of the
Definitions terms is found in the next section of the . Next, read text sections 6.1, 6.2,
6.3, 6.4, and 6.6. As you
read, take careful note of the fact that momentum is a Vector quantity. This is illustrated
by the vector nature of equation
(6.2) and in example 4. Also, notice that in considering simple collisions between two
objects, that although the momentum is
always conserved, energymay not be conserved. This fact is illustrated in section 6.4
and by the example at the end of the
section. Also remember that the answers to questions asked in the text reading are
answered in section 3 of this Study
Guide.
Next, read the Chapter Summary and complete Summary Exercises 1-14. Then complete
Algorithmic Problems 1-6. Then
complete Problems and Exercises 1, 2, 4, 6, 9, and 10. For additional practice, more
example problems follow in this Study
Guide chapter. Finally, attempt the Practice Test included in the . If you have
difficulties, please refer to the
particular section of the text for additional practice. This study procedure is outlined
below.
----------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
----------------Definitions 6.1, 6.2, 6.3, 1-5
6.4, 6.6
Impulse
Problems 6.2 6, 7 1, 2, 4 1, 2
Conservation 6.3 8, 9 3, 5
of Momentum
Collision 6.4 10, 11 3,5 4,6
Problems
Momentum and 6.4 12-14 6 9,10
Energy Problems
DEFINITIONS
Keywords: Glossary; ; Mechanics; Momentum; Collisions; Inelastic; Collisions; Elastic
Collisions; Rockets;
Propulsion; Impulse
IMPULSE
Product of the net force and the time interval over which it acts is called the impulse of
the force.
Note that the impulse of a force is a vector quantity that can be calculated from two
measured variables, the force and the
time.
IMPULSIVE FORCE
is usually thought of as a force which lasts a short time but reaches a large value (N).
MOMENTUM
Product of the mass of a body and its velocity, a vector property whose direction is the
same as the direction of the velocity.
Note that momentum, like energy and impulse, is a derived quantity that can be
calculated from the measured quantities of
mass and velocity.
ELASTIC COLLISION
Both momentum and kinetic energy of bodies involved in a collision are the same
before and after impact.
In most collisions we see in everyday life there is some loss of energy, but if the energy
loss is small, then we see a good
approximation of an elastic collision. Perhaps the best example is a collision between
two steel ball bearings.
INELASTIC COLLISION
Bodies stick together during impact; momentum is conserved, but kinetic energy is not
conserved.
Most collisions we see such as between participants in sporting events, or between
automobiles, fall into this category. Part
of the energy is used to change the shape of the participant or the automobile.
ROCKET PROPULSION
Practical example of conservation of momentum.
A small amount of fuel is ejected at very high velocity, so the large rocket moves
forward in a manner that the total
momentum of the system of rocket and fuel remains constant.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: Answers; ; Mechanics; Dynamics; Momentum; Impulse
SECTION 6.1 Introduction
The word momentum is found in many different contexts in daily living. Please listen
for the use of this word outside of
physics and compare its meanings and connotations with those you learn in physics.
SECTION 6.2 Momentum and Impulse
Most examples of the forces we exert on objects in our surroundings are impulsive
forces: we push open a door, we slide a
box, we pedal a bicycle with a series of impulsive forces.
EXAMPLES
Keywords: ; Mechanics; Dynamics; Momentum; Conservation Of Momentum; Impulse;
Force; Vectors;
Trigonometry; Collisions; Elastic Collisions; Friction; Kinetic Energy; Worked Examples;
Arithmetic
IMPULSE PROBLEMS
1. A 2000 kg automobile was moving with a constant velocity to the East at 12 m/s. It
was struck by a tornado and 7.0
seconds later was headed northeast at 40 m/s.(a) What was the impulse that acted on
the automobile? (b) What was the net
force exerted by the tornado?
What Data Are Given?
Mass of automobile = 2000 kg
Initial velocity = 12 m/s east
Final velocity = 40 m/s northeast
Time elapsed = 7.0 seconds
What Data Are Implied?
In order to calculate the net force you can assume the force exerted by the tornado was a
constant lasting for 7 seconds.
What Physics Principles Are Involved?
You must use two concepts, the definition of impulse and the relationship between
impulse and change in momentum.
What Equations Are to be Used?
Impulse = (Force) (time) (1)
Impulse = change in momentum = Dp (6.2)
Algebraic Solutions
See Fig.
Let x be the east direction, then y stands for north
Initial momentum = pi = mvi in the x direction
Final momentum = Pf = mvi in the northeast direction
Change in momentum = mvf - mvi
This is a vector equation. Now let us convert this to equations for the x and y
components of the momenta using q as the
angle between the final and initial momenta.
change in x component of momentum = DPx = mvf cos q - mvi
change in y component of momentum = DPy = mvf sin q - 0
We can now use Equation 6.2 to find the impulse.
x - component of the impulse = DPx = mvf cos q - mvi (2)
y - component of the impulse = DPy = mvf sin q (3)
The angle f the impulse makes with the x - direction (east) is given by
tan = Py/Px = mvf sinq / (mvf cos q - mvi) (4)
The force exerted by the tornado = impulse/time
Numerical Solutions
x - component of impulse = (1.0 x 103 kg)(40 m/s)sin 45ø = 5.7 x 104 kgm/s
y - component of impulse = (2.0 x 103 kg)(40 m/s)cos 45ø - (2.0 x 103)(12 m/s) = 3.3 x
104 kgm/s
f = arctan 3.3 x 104/5.7 x 104 = 30ø north of east
Magnitude of the total impulse = SQR RT[(5.7 x 104)2 + (3.3 x 104)2] = 6.6 x 104 kgm/s
Net force = impulse/time = 6.6 x 104/7.0 = 9.4 x 103 N at 30ø north of east
Thinking About the Answers
You may notice that the algebraic form of the answers is simplified because the xdirection was chosen to be in the direction
of the initial momentum, so the y component of momentum was initially zero. This is a
good strategy to follow in solving
problems that involve vectors. If you are free to choose an arbitrary coordinate system
choose one that makes the algebraic
manipulations in the problem easier. Note also that the impulse is not in the direction of
the final momentum. Rather the
impulse is in the direction of the CHANGE IN MOMENTUM. Beware of the vector
properties of both impulse and momentum!
Many physics students have missed exam equations because they have added or
subtracted momenta as if they were
scalars.
COLLISION PROBLEMS
2. A 104 kg tight end has caught a short pass from the quarterback and heads for the
end zone at a speed of 8.9 m/s at an
angle of 75ø with the yard lines. A small defensive back (77 kg) running parallel to the
yard line at a speed of 9.6 m/s dives
through the air, collides with the tight end, and holds on to him for dear life. What is
the final momentum of the two entangled
football players?
What Data Are Given?
Let us take the tight end to be subscript 1; ml = 104 kg initial vl = 8.9 m/s 275ø
defensive back, subscript 2; m2 = 77 kg; v2 = 9.6 m/s 20ø
What Data Are Implied?
The problem does not give the exact direction of the motion of the two players with
respect to each other. Let us assume the
tight end is trying to avoid the defensive player so they are both running toward the
same sideline. Then a top view of the
problem may be drawn as follows:
See Fig.
What Physics Principles Are Involved?
We can solve this problem using the conservation of momentum.
What Equations Are to be Used?
We can use the equation that applies to the special case of when the two colliding
objects stick together.
m1v1 + m2v2 = (m1 + m2)vf (6.15)
Algebraic Solutions
In component form, picking the v2 direction as the x - direction; and q as the angle
between v1 and v2,
m1v1 cos q + m2v2 = (m1 + m2) vfx = Pfx (5)
m1v1 sin q = (m1 + m2)vfy (6)
The magnitude and direction of the final momentum are given by
Pf = SQR RT[P2fx + P2fy] = SQR RT[(m1v1 cos q + m2v2)2 + (m1v1 sin q)2]
Numerical Solution
Pfx = (104 kg)(8.9 m/s) cos 75ø + (77 kg)(9.6 m/s) = 9.8 x 102 kgm/s
Pfy = (104 kg)(8.9 m/s) sin 75ø = 8.9 x 102 kgm/s
Pf = 1.3 x 103 kgm/sec; f = 42ø
Thinking About the Answer
The final magnitude is greater than the initial momentum of either player. The final
direction is more towards the sidelines than
the original direction of the tight end but more towards the end zone than the original
direction of the defensive back. All of
these features of the answer fit our observations of such collisions on a football field.
MOMENTUM AND ENERGY PROBLEMS
3. In the preceding problem, what fraction of the initial energy was lost in the collision?
What happened to that energy?
What Data Are Given?
The initial masses and velocities and the final mass.
What Data Are Implied?
The fact that the two players stick together after the collision makes it a completely
inelastic collision.
What Physics Principles Are Involved?
We know the total momentum is conserved so the final kinetic energy can be calculated
and then the fractional energy lost
can be computed.
fractional energy loss = initial energy - final energy/initial energy
What Equations Are to be Used?
We can combine the equations from the previous problem with an equation to calculate
the loss in kinetic energy.
1/2 m1v12 + 1/2 m2v22 - 1/2 (m1 + m2)v2f = loss in KE (6.16)
Algebraic Solution
Fraction loss of kinetic energy =
[1/2 m1v12 + 1/2 m2v2 - 1/2 (m1 + m2)vf2]/(1/2 m1v12 + 1/2 m2v22)
m1v1 cos q + m2v2 = (m1 + m2)vfx (5)
m1v1 sin q = (m1 + m2)vfy (6)
Final kinetic energy = Pf2 / 2(m1 + m2)
= [(m1v1 cos q + m2v2)2 + (m1v1 sinq)2] / 2(m1 + m2) (7)
This equation (7) is derived from the relationships that the final momentum Pf is given
by (m1 + m2)vf and the final kinetic
energy is given by 1/2(m1 + m2)vf2, so
1/2 (m1 + m2)vf2 = KEf
= Pf2 / 2(m1 + m2) = (m1 + m2)2 vf2 / 2(m1 + m2)
Numerical Solution
Initial kinetic energy of the tight end = 1/2(104 kg)(8.9 m/s)2 = 4.1 x 103 J.
Initial kinetic energy of the defensive back = 1/2(77 kg)(9.6 m/s)2= 3.5 x 103 J
Total initial kinetic energy = 7.6 x 103 J.
Final kinetic energy = Pf2 / 2(m1 + m2)
= (1.3 x 103 kg)2 / 2(104 + 77)kg = 4.7 x 103 J
Fractional kinetic energy loss = (7.6 x 103 - 4.7 x 103) / 7.6 x 103 = 0.38
Thinking About the Answer
So 38% of the initial kinetic energy is lost in this collision. This energy is used to change
the shape of the players' bodies,
uniforms and padding. It may even produce changes in the bodies of the players to
cause bruises and muscular aches and
pains.
4. What are the impulses that act on the two players in the collision discussed in
problem 2?
What Data Are Given?
Same as 2 above.
What Data Are Implied?
Changes in momentum result from the impulse of the collision.
What Physics Principles Are Involved?
The change in momentum is equal to the impulse acting on the system. Newton's Third
Law of Action - reaction pairs - can
also be used to determine the impulse on the tight end after we have calculated the
impulse on the defensive player, or vice
versa. The principle of superposition can also be used.
What Equations Are to be Used?
Impulse = change in momentum (6.2)
Final momentum of two players = sum of individual player's final momentum
(superposition)
Algebraic Solution
Let us pick the defensive player and calculate his change in momentum because of the
collision. His final momentum is a
fraction m2/m1 + m2 of the total final momentum,
P2f = m2Pf / (m1 + m2) (8)
His initial momentum was P2i = m2v2 and was chosen to be the x-direction.
Impulse = final momentum - initial momentum
= P2f - P2i = [m2 / (m1 + m2)]Pf - m2v2
Numerical Solutions
From the numerical answer to problem (2) above:
P2f = [77kg (1.3 x 103) / (104 kg + 77 kg)] kgm/s 242ø
P2f = 5.5 x 102 kgm/s 242ø
= { 4.1 x 102 kgm/s in the x-direction
= { 3.7 x 102 kgm/s in the y-direction
P2i = (77 kg)(9.6 m/s) = 7.4 x 102 kgm/s in the x -direction
change in momentum
= { 4.1 x 102 - 7.4 x 102 kgm/s in the x -direction
{ 3.7 x 102 - 0 in the y-direction
= { -3.3 x 102 kgm/s in the x-direction
{ 3.7 x 102 kgm/s in the y-direction
So DP = SQR RT[(-3.3 x 102)2 + 3.7 x 102)2] = 5.0 x 102 kgm/s
The angle of DP is arctan(3.7 x 102/-3.3 x 102) = 132ø
The impulse on the defensive back is 5.0 x 102 kgm/s 2132ø
The impulse on the tight end is 5.0 x 102 kgm/s 2 -48ø
Relative Directions
See Fig.
Thinking About the Answer
Keywords: Conservation Of Momentum; Humor
Notice that this impulse is five times that needed for a fatality from a head injury as
given in example 2 in section 6.2.
See Fig.
PRACTICE TEST
Keywords: Problems; Answers; Evaluations; Mechanics; Dynamics; Momentum;
Collisions; Elastic Collisions; Inelastic
Collisions; Force; Time Inelastic Collisions; Energy; Kinetic Energy; Conservation Of
Momentum; Velocity
1. A small foreign car (mass = 500 kilograms) is traveling at 5.0 m/sec East when it
makes a head-on collision with a brick
wall. Following the initial impact, the car bounces straight back away from the wall
with a velocity of 2.0 m/sec.
See Fig.
________(a) Find the initial momentum of the foreign car.
________(b) Calculate the final momentum of the car.
________(c) If the impact with the wall lasted .02 seconds, what was the average force
applied to the car during the impact?
________(d) Calculate the percentage of initial kinetic energy lost during the collision.
2. Consider the problem above for the case when the collision is perfectly elastic.
________(a) What would be the final velocity of the car?
________(b) Calculate the average force exerted on the car by the wall if the impact time
remained .02 seconds.
________(c) What percentage of the initial kinetic energy is lost during the elastic
collision?
ANSWERS:
1. 2500 kgm/s, 1000 kgm/sec, 1.8 x 105N, west, 84% lost
2. 5 m/s, 2.5 x 105 N, 0%
Chapter Rotational Motion (7):
STUDY
GUIDE
Citation: H. Q. Fuller, R. M. Fuller and R. G. Fuller, to Accompany Physics Including
Human Applications.
(Harper and Row, New York, 1978). Permission granted by the authors.
7 Rotational Motion
Keywords: ; Mechanics; Learning Objectives; Rotational Kinematics; Rotational Motion;
Rotational Dynamics;
Torque; Center Of Mass; Moment Of Inertia; Rotational Kinetic Energy; Angular
Momentum; Angular Acceleration; Angular
Displacement; Angular Velocity; Angular Acceleration; Statics
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
angular displacement center of mass
angular velocity moment of inertia
angular acceleration rotational kinetic
uniformly accelerated angular energy
motion angular momentum
torque
Equilibrium
State the conditions for static equilibrium.
Rotational Motion
Write the equations for rotational motion with constant angular acceleration.
Rotational Kinematics
Solve problems for systems with a fixed axis of rotation using the principles of
rotational kinematics.
Rotational Dynamics
Solve problems using the principles of rotational dynamics, for systems with fixed axes
of rotation,
including conservation of energy and conservation of angular momentum.
Equilibrium Problems
Solve problems involving conditions of static equilibrium.
PREREQUISITES
Before beginning this chapter you should be familiar with Chapter 4, Forces and
Newton's Laws, and
Chapter 5, Energy. The quantitative aspects of rotational motion are very similar to
those of kinematics
(Chapter 3).
OVERVIEW
Keywords: ; Instructions; Mechanics; Rotational Kinematics; Rotational Dynamics;
Statics
This chapter may seem long because it contains both Rotational Kinematics and
Rotational Dynamics. The essential features
of each of these concepts are summarized and compared to the linear case in the table
on page 153 (Rotational Kinematics)
and table 7.1, page 167 (Rotational Dynamics). The chapter will be easier to handle if
you keep these subdivisions in mind.
SUGGESTED STUDY PROCEDURE
When you begin to study this chapter, read the following Chapter Goals: Definitions,
Equilibrium, Rotational Dynamics, and
Equilibrium. An expanded discussion of each of the terms listed asDefinitions can be
found in the next section of this Study
Guide chapter. Next, read Chapter Sections 7.1-7.9. As you read, be sure to note the
parallels between rotational motion and
linear motion. The table on page 153 summarizes the expressions for rotational
kinematics, table 7.1 on page 167
summarizes rotation dynamics. As you read, remember that the answers to all
questions posed in the text sections are
answered in the second section of this chapter.
At the end of the chapter, read the Chapter Summary and completeSummary Exercises
3, 9, 10, 11, 12, and 15. Next, do
Algorithmic Problems 1, 3, 4, 5, and 6 and complete Problems and Exercises 5, 6, 7, 8, 9,
15, 16, 17, and 23. For additional
practice on rotational motion problems, see section three of this for more Examples.
Finally, attempt the Practice
Test. If you have difficulties with any part of the test, seek extra work from the
appropriate text section. This study procedure
is outlined below.
----------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
----------------Definitions 7.1,7.2,7.3 3,10,11,12 1,3,4
Equilibrium 7.4 9 6,7,8
Rotational 7.5,7.6,7.7, 5,6 9,23
Dynamics 7.8,7.9
Equilibrium 7.4 15 5,6,7,8,
Problems 15,16,17
DEFINITIONS
Keywords: ; Glossary; Mechanics; Angular Displacement; Angular Velocity; Angular
Acceleration; Torque; Moment
Of Inertia; Rotational Kinetic Energy; Rotational Dynamics; Rotational Kinematics;
Rotational Motion; Angular Momentum
ANGULAR DISPLACEMENT
Angle between two positions of a rigid body rotating about a fixed axis. Usually given
in radians.
You are probably most likely to give angular displacements in degrees, such as "Make a
45ø turn," rather than "Turn p/4
radians." So you need to practice the conversion from radians to degrees and vice versa
so that both ways of measuring
angles are familiar to you.
ANGULAR VELOCITY
Time rate of change of angular displacement. Usually given in radians per second.
The most common way of talking about angular velocity in daily life is probably in
turns of revolutions per second or per
minute. Do you know what the 33 1/3 for a phonograph record means? It stands for 33
1/3 r.p.m. (revolutions per minute). Do
you know how to convert that to an angular speed measured in radians per second?
ANGULAR ACCELERATION
Time rate of change of angular velocity. Usually given in radians per second per
second.
When the rate of rotation of an object is changing then it has an angular acceleration. If
the rate of rotation is increasing, then
the angular acceleration is positive.
UNIFORMLY ACCELERATED ANGULAR MOTION
Rotational motion with a constant angular acceleration.
Common examples of this kind of motion seem scarce except for systems that roll under
the action of gravity, such as a
yo-yo rolling down its string.
TORQUE (moment of force)
A vector product of a force and its perpendicular distance to the point of rotation. A
right-hand rule can be used to determine
its direction.
You can note that the direction of the torque vector is exactly the direction
wherenothing is happening since the acting force
and the distance to the rotation point form a plane perpendicular to the torque.
CENTER OF MASS
The balance point of an object; the force of gravity (weight) produces zero torque about
the center of mass.
You have noticed since your early childhood that even large objects can be supported at
one point. There are several kinds of
children's toys that make use of an unusual shape to stably balance on a point. What can
you deduce about the location of
the center of mass of such toys?
MOMENT OF INERTIA
Measure of the ability of a body to resist a change in rotation. It depends upon the
distribution of its mass relative to an axis
of rotation.
If most of the mass of an object is far from the axis of rotation, then the moment of
inertia is large. Hence, a large baseball
bat is easier to swing if you choke up on the handle. You are moving more of the mass
of the bat closer to the center of
rotation, your hands, so that you reduce its moment of inertia for the swing.
ROTATIONAL KINETIC ENERGY
Energy of rotation equal to one-half the product of moment of inertia and angular
velocity squared.
Objects that rotate about an axis fixed in space have this kind of kinetic energy.
ANGULAR MOMENTUM
The product of moment of inertia and angular velocity gives the magnitude of the
angular momentum. A right-hand rule can be
used to determine its direction.
It is a straight forward matter to show that a system may have a constant linear
momentum while its angular momentum
changes, but the reverse case is not possible. Consider the case of motion of a system of
two equal forces acting in
opposite directions but separated by some distance. What properties of such a system
are constant? What properties
change?
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Rotational Kinematics; Mechanics; Rotational Dynamics;
Rotational Kinetic Energy; Torque;
Angular Momentum
SECTION 7.1 Introduction
The most common way of talking about rotation is in terms of the number of complete
rotations or revolutions an object or
system makes. This is often done in terms of revolutions per minutes or rotations per
second.
You can start an object rotating by applying some forces to the object that are some
distance away from the point of rotation,
or axis, and do not point exactly toward the axis.
We can specify the angular motion of a system in terms of its angular displacement,
angular velocity, angular acceleration,
moment of inertia, rotational kinetic energy, and angular momentum.
Photographs, page 154
Notice the line of action of the applied torque with respect to the point of rotation; i.e.,
the point of contact between the spool
and the table.
EXAMPLES
Keywords: ; Dynamic Equilibrium; Worked Examples; Angular Acceleration;
Rotational Kinematics; Uniform
Circular Motion; Angular Velocity; Angular Displacement; Rotational Kinetic Energy;
Moment Of Inertia; Conservation Of
Angular Momentum; Rotational Dynamics; Arithmetic; Statics; Muscular System;
Human Anatomy And Physiology; Muscular
Systems; Free Body Diagrams; Torque
ROTATIONAL KINEMATICS
1. A physics student is earning some extra money for his college expenses by serving as
a teacher's aide at a nursery
school. One day on the playground he notices a child (24 kg) playing on the merry-goround which has an off-center set of
hand and foot bars so that it can be started from rest by a person sitting on the merrygo-round. He makes the following
observations: the child starts the merry-go-round from rest and pumps it up to a
constant rotation rate in 12 seconds, after
which time the child passes the observer each 1.4 seconds. The diameter of the merrygo-round is 5 meters. Quantitatively
describe the child's rotation.
What Data Are Given?
The mass of the child is 24 kg. The period of angular acceleration is 12 sec and the time
required for one complete rotation
of the child is 1.4 sec. If the child is assumed to be sitting on the rim of the merry-goround, then the child is 2.5 meters from
the axis of rotation.
What Data Are Implied?
In order to calculate the child's angular acceleration, let us assume that the motion is
uniformly accelerated angular motion
(u.a.a.m.) until the merry- go-round reaches a constant speed of rotation. It then is a
special case of u.a.a.m.; i.e., the
angular acceleration is zero.
What Physics Principles Are Involved?
You can begin the problem by using the concepts of u.a.a.m. as given on page 150, or
summarized in the right-hand column
of the table on page 153. You can also calculate the work done by the child and the
kinetic energy of the child using
concepts from Section 7.7.
What Equations Are to be Used?
Angular velocity; w = 2p / T where T is the period of rotation.
For u.a.a.m.; a = w / t where t is the time of acceleration (7.3)
the angle turned during acceleration = 1/2 a t2 (7.8)
Kinetic Energy = (Iw2) / 2 = (mr2w2) / 2
for a point mass m located at a distance r from the axis of rotation (7.26)
Work = change in kinetic energy (7.25)
Algebraic Solutions
The equations above are all written with terms we may wish to calculate on the lefthand side of each equation.
Numerical Solutions
Final angular velocity = (2p rad.) / (1.4 sec.) = 45 rad/sec
Angular acceleration = (4.5 rad/sec.) / (12 sec.) = 0.37 rad/s2
Angle of rotation during acceleration = 1/2 (0.37)(12)2
q = 27 radians
Angular displacement of the child at any time at 12 seconds where t is the time from
rest,
q = 27 rad. + (4.5 rad/s)(t - 12) where t > 12 s.
Distance travelled in meters = gq = (2.5 m)q
Kinetic energy of the child = 1/2 (24 kg)(2.5 m)2(4.5 rad/s)2
= 1.5 x 103 J
Work done to rotate the child = 1.5 x 103 J, this neglects the rotational kinetic energy of
the merry-go-round.
ROTATIONAL DYNAMICS
2. Consider a playground merry-go-round at rest and a child (24 kg) runs at a speed of
6.2 m/s tangential to the rim of the
merry-go-round on 3.0 meter radius and jumps on it. If the moment of inertia of the
merry-go-round is 1.44 x 102 kgm2, what
is the final angular speed of the system? What is the final kinetic energy of the system?
Is any energy lost? If so, what
happens to it?
What Data Are Given?
The mass, speed, and distance of the child from the axis of rotation are given. The
moment of inertia of the merry-go-round is
given.
What Data Are Implied?
The visual arrangement of the problem is implied by the setting of this problem. Can
you picture it in your mind?
See Fig. 7.1
This problem also implies that friction be neglected.
What Physics Principles Are Involved?
The concept of the conservation of angular momentum can be used to analyze the
motion of the combined system of child
and merry-go-round. From the instant the child jumps from the ground to land on the
merry-go-round, the total angular
momentum is constant since there are then no external torques acting on the combined
child - merry-go-round system if the
frictional forces on the system are neglected. It seems as if conservation of energy can
also be assumed from the instant the
child jumps up to the merry-go-round. There are no external forces acting on the
system.
What Equations Are to be Used?
The angular momentum is a constant
Iiwi = Ifwf (7.30)
Total energy is a constant, perhaps
1/2 mvi2 = 1/2 Ifwf2
To use these equations you need to know that the magnitude linear velocity v can be
related to the magnitude angular
velocity w about a fixed axis is the perpendicular distance from the velocity to the axis
is v by the following equation.
v = rw (7.12)
Algebraic Solutions
The initial angular momentum of the child - merry-go-round system is equal to the
angular momentum of the child just as she
jumps on the merry-go-round because the merry-go-round is initially at rest and so has
an angular momentum of zero.
Initial angular momentum = Li = mcvcv c (1) where the subscript c stands for the child
of mass, speed, and distance from axis
of rotation of m, v, and r respectively.
Final angular momentum = Lf
= ImWm + IcWc (2)
where the subscript m applies to the merry-go-round. Since the child and the merry-goround are rotating together in the final
state the angular speed of the child and the merry-go-round are the same.
Wm = WC (3)
The child can be treated as a rotating point mass, so
Ic = mcrc2.
Then, from conservation of angular momentum,
Lf = Li
Wm(Im + mcrc2) = mcvcrc (4)
so Wm = (mcvcrc) / (Im + mcrc2)
Is energy conserved?
The initial energy = KEi = 1/2 mcvc2 (5)
The final energy = KEf = 1/2 ImWm2 + 1/2 IcWc2 = 1/2 Wm2 (Im + Ic) (6)
We can substitute the above expression for Wm, then
KEf = 1/2 [(mcvcrc) / (Im + mcrc2)]2 ¥ [Im + mcrc2] = 1/2 [(mc2vc2rc2) / (Im + mcrc2)]
Let us set the initial and final kinetic energies equal
1/2 mcvc2 = 1/2 [(mc2vc2rc2) / (Im + mcrc2)] (5)
Divide by 1/2 mcvc2
1 = (mcrc2) / (Im + mcrc2) which will be true if Im = 0.
The energy will be conserved only if the moment of inertia of the merry- go-round is
neglected. What happens to the energy
when Im is not treated as zero? What explanation can you give of the failure of the law
of conservation of energy to be true
for this system?
Numerical Solutions
Initial angular momentum = (24 kg)(6.2 m/s)(3.0 m) = 4.5 x 102 kgm2/s
Final angular speed = Wm = (4.5 x 102 kgm2/s) / (1.44 x 102 kgm2 + 24 kg (3.0 m)2)
= (4.5 x 102) / (3.6 x 102) ¥ rad/sec
= (5.0 rad/s) / 4.0 = 1.3 rad/s.
Initial kinetic energy = 1/2 (24 kg)(6.2 m/s)2 = 4.6 x 102 J
Final kinetic energy = 1/2 (1.44 x 102)(5/4)2 + 1/2 (24)(3.0)2(5/4)2 = 2.8 x 102 J
Fraction of kinetic energy lost = (4.6 x 102 J - 2.8 x 102J) / (4.6 x 102 J) = 0.39
EQUILIBRIUM PROBLEMS
3. What is the force applied by the quadriceps tendon to the lower portion of the human
leg to hold it up at a 45ø angle from
horizontal? Assume the values of the variables as shown in the following mechanical
analog to the human leg.
What Data Are Given?
See Figure 7.2
What Data Are Implied?
Use the strength of the earth's gravitational field to be g = 9.8 m/s2. Then a generalized
force diagram can be drawn as
follows
See Fig. 7.3
Thinking About the Answer
Notice how just the simple statement that the child jumps on the merry-go-round, with
its implication of a completely inelastic
collision means that 39% of the initial kinetic energy is lost. What happens to it?
You are invited to postulate a different final situation. Assume the child makes a
completely elastic collision with the
merry-go-round. What then is the final state of the child and of the merry-go-round?
This means that the condition that the
final angular velocity of the child and the merry-go-round must no longer be equal, so
Equation (3) is no longer true. Can you
solve the completely elastic collision case? Hint: Equations (1), (2), (5), and (6) are still
valid, but they must be solved
simultaneously for Wm and Wc.
[Answers: rotation speed of the merry-go-round = 2.5 rad/s. speed of the child = 1.1
m/s still along the tangent line]
If you wish you can extend the elastic collision calculations and compute the impulsive
torque on the merry-go-round and the
impulse on the child. Would he be hurt by an elastic collision with the merry-go-round?
What Physics Principles Are Involved?
The system must satisfy the conditions for equilibrium, p. 157.
Sum of forces = zero
Sum of torques = zero
What Equations Are to be Used?
There is an unknown force L exerted by the upper leg bone of the hinge at the knee. If
we sum the torques about point A,
then we can eliminate the force L and solve for F.
Sum of torques = 0 = -Fd sin a + W2 (1/2)cos q + WF (1 cos q + F/2) (6)
where a negative sign indicates a counter-clockwise rotation about the point A.
Algebraic Solution
Solve for F in Equation 6
F = [W1 1/2 cos q + WF (1 cos q + f/2)] / (d sin a) (7)
Numerical Solution
F = [((4 kg)(9.8 m/s2)(0.22 m) cos 45ø + (2 kg)(9.8 m/s2) (0.44 cos 45ø + 0.09)) / ((0.10 m)
sin (arctan 0.5))]
F = (6.1 + 7.9) / (4.5 x 10-2) = 31 N
Thinking About the Answer
You can see from Equation (7) that the smaller you make the angle q; i.e., the more
nearly you hold your lower leg out
horizontally, the greater is the force applied by the quadriceps tendon. Try lifting your
leg and feeling of this tendon as you do.
What can you infer from what you feel? Does our mechanical analog seem to be a good
approximation for the way your leg
really works?
PRACTICE TEST
Keywords: Problems; ; Answers; Simple Mechanics; Statics; Force; Angular Velocity;
Torque; Moment Of Inertia;
Rotational Equilibrium; Mechanics; Tension; Angular Acceleration; Evaluations;
Rotational Dynamics
1. The six meter uniform bar has a weight of 20 Newtons. It is placed so that one end
rests on the side of a table, and a 15
Newton weight is placed 1 meter from its center.
See Fig.
a. What force if applied to point D will keep the bar in equilibrium?
b. With the bar in equilibrium, what force acts on the bar at point A?
2. The bicycle wheel shown below has a mass of 5 kilograms and a moment of inertia of
.5 kg m2 about its center. The
wheel is supported above the ground so that it can rotate freely and be driven by the
chain C. The driving sprocket has a
radius of 10 cm.
a. If the chain is producing an angular acceleration a = 5 rad/sec2, what is the tension
in the driving chain?
b. If the wheel starts at rest, what will be its angular velocity after 3 sec?
See Fig.
ANSWERS:
1. 20 N, 15 N
2. 2.5 N, 15 rad/sec
Chapter Fluid Flow (8):
Citation: H. Q. Fuller, R. M. Fuller and R. G. Fuller, to Accompany Physics Including
Human Applications.
(Harper and Row, New York, 1978). Permission granted by the authors.
8 Fluid Flow
Keywords: ; Learning Objectives; Fluids; Fluid Statics; Density; Pressure; Buoyant
Force; Buoyancy; Specific
Gravity; Viscosity; Fluid Dynamics; Streamline Flow; Archimedes' Principle; Bernoulli's
Principle; Pascal's Principle;
Poiseuille's Law; Fluid Motion
GOALS
When When you have mastered the contents of this chapter, you you will be able to
achieve the following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
fluid buoyant force
density streamline flow
specific gravity viscosity
pressure (absolute and gauge)
Fluid Laws
State Pascal's law of hydrostatic pressure, Archimedes' principle of buoyancy,
Bernoulli's equation for the conservation of
energy in a fluid, and the law of conservation of fluid flow.
Fluid Problems
Solve problems making use of the principles of fluids and conservation laws.
Viscous Flow
Use Poiseuille's law of viscous flow to solve numerical problems.
PREREQUISITES
Before you begin this chapter, you should be able to solve problems that use energy
concepts (see Chapter 5).
OVERVIEW
Keywords: ; Instructions; Fluids; Fluid Statics; Fluid Dynamics; Fluid Motion
The movement of liquid or gaseous substances is an important consideration for
medical doctors and air conditioning
mechanics, as well as children using a soda straw. There seems to be endless examples
of how we are dependent on the
flow of fluids for making the work we do easier and more complete.
SUGGESTED STUDY PROCEDURE
To begin your study of this chapter, read the following Chapter Goals: Definitions,
Fluid Laws, Fluid Problems, and Viscous
Flow. Expanded discussion of each term under Definitions is found in the next section
of this chapter.
Next, read text sections 8.1-8.13. Remember that answers to the questions asked in these
sections are given in this Study
Guide chapter. Now read the Chapter Summary and complete Summary Exercises 1-12.
Now do Algorithmic Problems 1, 2, 3,
and 4, and complete Exercises and Problems 2, 3, 4, 5, 7, 8, 9, 15, 17, and 21. For
additional experience, select from the
other Exercises and Problems and/or consider the Examples given in the third section
of this chapter. Now you
should be prepared to attempt the Practice Test on Fluid Flow at the end of this
chapter. Remember to wait until
you have worked the problem before looking at the answer. Seek assistance in areas
where you do not score 100%. This
study procedure is outlined below.
-----------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
-----------------------Definitions 8.1,8.2,8.3, 1,2,3,4,
8.4 5,6,7
Fluid Problems 8.5,8.6,8.7, 8,9,10,11 2 2,3,4,5,
8.8 7,8,9,15
Viscous Flow 8.10,8.11,8.12 12 1,3,4 17,21
8.13
DEFINITIONS
Keywords: Glossary; ; Fluids; Density; Specific Gravity; Pressure; Buoyancy; Buoyant
Force; Streamline Flow;
Viscosity; Fluid Statics; Fluid Dynamics; Fluid Motion
FLUID
A substance which takes the shape of its container and flows from one location to
another.
The class of all materials that are fluids includes both liquids, such as water, and gases,
such as air. The class of
incompressible fluids is called liquids.
DENSITY
The ratio of the amount of matter contained in an object to the amount of space
occupied by the object is called its density.
In the SI system the density of water at a temperature of 4ºC is defined as exactly 1000
kilograms of mass for each cubic
meter of water, or 1 gm per cubic centimeter of water.
SPECIFIC GRAVITY
The ratio of the density of a substance to the density of water.
The specific gravity of an object that sinks in water is a number larger than 1.0. Note
that specific gravity is a number with no
units.
PRESSURE
The normal force acting on a unit area is the pressure.
Since all of the objects near the surface of the earth are usually acted upon by the
pressure of the earth's atmosphere of air,
we often measure only the excess pressure of a system. We then call it a gauge pressure.
We must add the atmospheric
pressure to the gauge pressure to obtain the value of the absolute pressure.
BUOYANT FORCE
The resultant of all the pressure forces acting on an object submerged in a fluid.
In general, the buoyant force pushes up on an object making its weight in a fluid less
than it would be in a vacuum. The
greater the density of the fluid in which you weigh the object the larger is the buoyant
force.
STREAMLINE FLOW
Every particle in a flow which passes a given point will flow through all the points in
the line of flow.
Such flow has no swirls or turbulence. In general it is only approximated in real fluids
by low speed flow in smooth-walled
containers.
VISCOSITY
The tendency of a real fluid to resist flow is known as viscosity. It is the internal friction
of the fluid. The liquid coefficient of
viscosity is a relative measure of liquid friction and is equivalent to the ratio of the force
per unit area to the change in
velocity per unit length perpendicular to the direction of flow.
The most common example of the importance of viscosity is probably in the motor oils
used to lubricate automobile and
motorcycle engines. The greater the viscosity of a fluid the greater is its resistance to
flow. Cold syrup has a greater
resistance to pouring than warm syrup. Viscosity of many liquids is strongly
temperature dependent.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Fluids; Density; Viscosity; Dimensional Analysis; Units; Viscous Flow;
Friction; Pressure; Pressure
In Fluids; Measurement Of Pressure; Answers; Fluids; Viscosity; Archimedes' Principle;
Force; Fluid Statics; Fluid Dynamics;
Poiseuille's Law; Medicine And Law
SECTION 8.1 Introduction
The circulation of air in a building to provide proper temperature of the habitation of
human beings is an example of using fluid
flow to transfer energy from one location to another. The flow inertia of a fluid is
measured by its density. The loss of energy
in the flow of a liquid is called viscosity.
SECTION 8.3 Density
The dimensions of density are M/L3. To covert kg/m 3 to gm/cm3 divide by 1000 or
103.
SECTION 8.4 Force on Fluids
If you try to push your hand along the surface of a fluid, parallel to its surface, you force
the fluid to travel along with your
hand with very little resisting force. As you slid your hand along a solid surface you
feel a frictional force opposing the motion
of your hand.
SECTION 8.5 Pressure
Keywords: Pressure in Fluids; Pressure Measurement; Answers
Because of the persistent use of the English system of units, a common pressure unit is
pounds per square inch. The
pressure for inflation of bicycle tires is given in lb. per square inch. Tire pressures are
intended to be gauge pressures.
Pressure is also given in atmospheres, 1 atmosphere of pressure being equal to the
pressure caused by the usually
atmospheric pressure on the surface of the earth. The weather bureau gives the pressure
in inches of mercury; i.e., the height
of a column of mercury that is supported by the pressure of the atmosphere.
SECTION 8.6 Pressure of a Liquid in a Column
We can use Equation 8.4 in conjunction with our knowledge of the density of mercury
13.6 x 103kg/m3 and the gravitational
constant 9.8 m/s2 to calculate the standard atmospheric pressure in newtons per square
meter, P = (13.6 x 10 3 kg/m3) (9.80
m/s2) (0.760 m) = 1.01 x 105 N/m2.
SECTION 8.9 Archimedes Principle
Keywords: Archimedes' Principle; Density; Answers; History; Worked Examples
Two ways to measure the density of a liquid are (1) direct measurement of the mass of a
known volume of the liquid using a
container whose volume has been calibrated, or (2) float an object of known mass and
volume in the liquid and measure the
portion of the object that remains above the surface of the liquid. In the second method
Archimedes Principle is used to
calculate the weight of displaced liquid.
To perform some sample calculations from the traditional Archimedes story let us
assume the crown was a hollow cylinder of
56 1/2 cm circumference, 0.50 cm in thickness and averaged 8.0 cm in height, then the
volume of the crown is approximately
V ˜ 2prh(Dt) = 2p (9 cm)(8 cm)(0.5 cm)
v ˜ 226cm3
See Fig.
If the crown were pure gold it would have a mass of 4.4 kg. If the crown were pure lead
it would have a mass of 2.6 kg. One
possibility is that Archimedes stepped into a completely full bath tub causing the water
to overflow. He then realized he could
use the same technique by catching the overflowing water to find the volume of the
crown. He could then perform a
straight-forward comparison of the density of the crown to the density of gold.
However, mythology has it that Archimedes
was extremely brilliant, so he no doubt worked out the Greek equivalent to Equation
(8.7) on the way running to the palace;
i.e.
SG = A / (A-W) (1)
SECTION 8.12 Poiseuille's Law of Viscous Flow
Keywords: Poiseuille's Law; Medicine and Health; Answers
Example 2. The flow of the I.V. liquid will be determined by its viscosity, the diameter
of the injection needle and the pressure
head, h. If the diameter is reduced by one-half, then by Equation 8.19 the flow is
reduced to 1/16th its original value so the
pressure head would have to be increased to 16 h!
EXAMPLES
Keywords: ; Fluids; Worked Examples; Fluid Statics; Fluid Dynamics; Fluid Motion;
Arithmetic; Conservation Of
Energy; Viscous Flow; Poiseuille's Law; Medicine And Health; Cardiovascular Systems;
Friction; Anatomy And Physiology;
Streamline Flow; Pressure; Bernoulli's Principle; Velocity
FLUID PROBLEMS
1. The water flowing from a garden hose is passed through a nozzle and then sprayed
into the air. Assume the hose is in a
horizontal position and the pressure and velocity of the water in the hose are 3.0 x 105
N/m2 and 2.6 m/s respectively. If the
nozzle reduces the effective area by 84%; i.e., area of nozzle = (0.16) area of hose, what
happens to the water pressure and
water velocity in the nozzle?
What Data Are Given?
Phose = 3.0 x 105 N/m2;
vhose = 2.6 m/s;
Anozzle = 0.16 Ahose
What Data Are Implied?
The height of the hose and the nozzle are the same. The water may be treated as an
ideal, streamline flow fluid.
What Physics Principles Are Involved?
We can make use of the conservation of liquid volume (Equation 8.8) and the
conservation of energy (Equation 8.14).
What Equations Are to be Used?
vhAh = vnAn (8.8)
Ph + pghh + 1/2 pvh2 = P n + pghn + 1/2 pvn2 (8.14)
where the subscripts h and n stand for hose and nozzle respectively.
Algebraic Solution
We are given that the elevation of the hose and the nozzle are the same, so we can
proceed as follows.
The velocity in the nozzle Vn = (Ahvh) / An (2)
The velocity is increased by a factor equal to the ratio of the areas.
We can substitute Equation (2) into equation (8.14) and solve for the pressure in the
nozzle; noting that pghh = pghn
Pn = Ph + 1/2 p(vh2 - vn2)
Pn = Ph + 1/2 pvh2 ((1 - Ah2) /An2) (3)
NOTE! The area of the hose Ah is larger than the area of the nozzle An so the second
term on the right-hand side of equation
(3) is negative. The pressure in the nozzle must be less than the pressure in the hose!
Numerical Solution
vn = (Ah/0.16 Ah)(2.6 m/s) = 16 m/s
Pnozzle = (3.0 x 105) + 1/2 (103 kg/m3) (2.6 m/s)2 (1 - (Ah/0.16 Ah)2)
= 3.0 x 105 + 1/2 (6.8 x 103)
Pn = 3.0 x 10 5 - 1.3 x 105 = 1.7 x 105 N/m2
Thinking About the Answers
The velocity in the nozzle is increased by more than six times while the nozzle pressure
is 57% of the pressure in the hose.
Notice that the high velocity region is also the region of low pressure.
VISCOUS FLOW
2. When taken from a refrigerator (4ºC) cold human blood has a viscosity of 8
centipoise. At room temperature (20º) its
viscosity is 4 centipoise. Assume the viscosity of blood changes linearly with
temperature and that the blood flow was proper
for cold blood for an IV arrangement as shown in Figure 8.15 with a bottle elevation of
ho. Explain how the elevation of the
bottle should be changed as the blood warms up to maintain a constant flow.
What Data Are Given?
The viscosity of blood at 4ºC is 8 centipoise. The initial elevation head for the flow of the
blood is ho.
What Data Are Implied?
It is assumed that Poiseuille's Law is applicable. The viscosity of the blood can be
related to the temperature by a linear
function. Since the viscosity at two temperatures is known and the general form of the
equation must be viscosity =
(constant)1 temperature + (constant)2 the quantitative relationship between viscosity
and temperature can be derived,
h(centipoise) = -1/4 t(ºc) + 9 (4)
for the temperature region 4 Û t Û 20.
What Physics Principles Are Involved?
The flow of human blood is viscous flow, so Poiseuille's Law may be applied, Equation
8.19.
What Equations Are to be Used?
rate of flow = p/8h (P1 - P2/L) R4 (8.19)
and Equation (4) h = -t/4 + 9 (4)
pressure difference = P1 - P2 = pgh (8.4)
Algebraic Solution
In order for the rate of flow to be a constant as the viscosity of the blood changes the
pressure, or elevation head, must be
changed. Since all the other variables except h and h are constant, then n and h must be
changed so that they maintain a
constant ratio;
rate of flow = (p/8) (R4/L) (pgh/h)
At the beginning for cold blood (h = 8 centipoise) the elevation is ho; so
h/h = ho / (8 centipose) = h / (9 - t/4)
Solving for h; h = (ho/8) (9 - t/4) = 9ho/8 - ho/32 t (5)
Numerical Solutions
We can draw a graph of the elevation of the bottle as a function of the temperature of
the blood.
See Fig.
Thinking About the Answer
As the blood becomes "thinner," its viscosity is diminished and the bottle is lowered to
lower the elevation head in order to
maintain a constant flow.
3. The flow of a viscous liquid through a tube is constricted by a narrow rubber hose
connection. As the connection ages, the
rubber expands and the diameter of the constriction increases by 8%. How much does
the flow rate increase?
What Data Are Given?
For the normal constriction there is some standard rate of flow. The diameter of the
constriction increases from d to 1.08 d.
What Data Are Implied?
It is assumed that the conditions are correct for Poiseuille's Law to apply.
What Physics Principles Are Involved?
The streamline flow of a viscous liquid as described by Poiseville's Law, Equation 8.19,
are assumed to apply.
What Equation is to be Used?
rate of flow = (p/8h) (P1 - P2/L)R4 (8.19)
Algebraic Solution
All of the variables of the system are constant except for the diameter of the tubing; so
(rate of flow)1 ∞ (radius)14 = ((diameter)14) / 16
so rate of flow ∞ (diameter)4
2nd flow rate / 1st flow rate = (second diameter)4 / (first diameter)4 (6)
Numerical Solution
2nd flow rate / 1st flow rate = (1.08 d)4 / d4 = (1.08)4 = 1.36
Thinking About the Answer
A relatively small increase in the size of the tube (8%) makes a large (36%) increase in
the flow rate.
PRACTICE TEST
Keywords: ; Problems; Answers; Fluids; Fluid Statics; Fluid Dynamics; Fluid Motion;
Pascal's Principle; Static
Equilibrium In Fluids; Force; Pressure; Bernoulli's Principle; Weight; Flight;
Cardiovascular Systems; Anatomy And
Physiology; Evaluations; Poiseuille's Law; Hydraulics
1. A schematic diagram representing a hydraulic jack pictured below is being used to
lift the front of a car off the ground. A
force of 10,000 Newtons is required.
See Fig.
a. What force at B is required on the small piston to produce the needed force of 10,000
N?
b. Under ideal conditions, what force F is required at the end of the 34 cm lever arm to
produce this force?
2. A jet airplane is flying at an elevation of 10 kilometers where the air has a density of
4.25 x 10-1 kg/m3 and a pressure of
2.7 x 104 N/m2. The velocity of the air along the bottom of the wing is 20 m/sec. Given
the dimension of the wing as shown
below:
See Fig.
a. What is the velocity of the air along the top of the wing? (Assume that the air flow
along the top and the bottom of the
wing requires the same time.)
b. What is the magnitude of the unbalanced pressure acting up on the wing?
c. If the airplane has a total wing area of 150 m2, what weight can the airplane support?
3. A person with hardening of the arteries can survive an effective diameter decrease of
his arteries of 20%. Under these
conditions, how much must the blood pressure increase to keep the rate of blood flow
constant?
Hint: (Assume that the viscosity of blood remains constant and that the blood flow is
unchanged and use Poiseuille's Law in a
modified form)
See Fig.
ANSWERS:
1. 2,000 N, 118 N
2. 25 m/s, 49 N/m2, 7,200 N
3. 2.4 times the initial pressure
Chapter
Transport Phenomena (9):
GOALS
Keywords: ; Learning Objectives; Continuity Equations; Transport Theory; Thermal
Physics
After you have mastered the contents of this chapter you will be able to achieve the
following goals:
Transport Equation
Write the quantitative equation for the transport process of a system whose variables
are given.
Continuity
State the continuity equation for a system, and explain the flow properties of a system
in terms of that equation.
Transport Problems
Use algebraic and graphical methods to solve transport problems for one dimensional
systems.
PREREQUISITES
Before beginning this chapter, you should have achieved the goals of Chapter 1, Human
Senses, and Chapter 2, Unifying
Approaches.
OVERVIEW
Keywords: ; Instructions; Thermal Physics; Transport Theory
Scientists have long recognized the symmetry and simplicity of nature. As you consider
the basic aspects of flow (called
Transport Phenomena) please notice the basic uniformity of the expressions utilized for
each case. Thus, explaining heat
flow, electrical charge flow, water flow, and diffusion are related by a basic model for
flow.
SUGGESTED STUDY PROCEDURE
Before you begin to study this model, be familiar with two of the Chapter Goals:
Transport Equation and Transport Process.
These two chapter goals are summarized by the equations under AlgorithmicProblems
in the Chapter Summary. Please note
the form of the Transport Equations: flow (9.1 and 9.3) and current density (9.4, 9.15,
and 9.16). For a brief explanation of
each of these equations and an example of how each might be used, see the following
section of this .
Next, read text sections 9.1-9.7. Carefully consider the Examples and Questions given at
the end of the sections. Remember
to look in this for answers to the questions asked in the text. Now read the Chapter
Summary and complete
Summary Exercises 1, 2, 3, 6, 7, and 8. Check your answers carefully. Now do
Algorithmic Problems 1-4 and complete
Exercises and Problems 4, 5, 9, and 11. For additional practice with the concepts
presented in this chapter, turn to the
Examples section of this chapter. Now you should be prepared to attempt the Practice
Test on Transport
Phenomena provided at the end of this chapter. If you have difficulties with any of the
questions, refer to this
study procedure for additional assistance. This study procedure is outlined below.
-------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
--------------Transport 9.1, 9.2, 9.3, 9.4, 1, 2, 3,
Equations 9.5, 9.6, 9.7
Transport 9.2, 9.5 6, 7, 8 1, 2, 3, 4 4, 5, 9, 11
Problems
----------Continuity 9.6 4, 5 3
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Gradients; Temperature; Transport Theory; Fluid Motion; Rate Of Flow;
Heat; Diffusion; Units;
Thermal Conductivity; Thermal Transport; Current Density
SECTION 9.1 Introduction
The water flows down a slope. There is a difference in elevation from one place to
another. The greater the difference in
elevation the faster is the flow of water.
The ladle is a different temperature at different locations. This temperature difference
may be thought of as the cause of a
flow of heat. The cup is less susceptible to the flow of heat than is the silver ladle.
We can talk about an electric potential being of different value at different locations in
an electric circuit, so we call it a flow
of electricity, or electric current.
SECTION 9.2 Temperature Differences and Gradients
Example - The gradient of the height in the easterly direction is greatest; i.e., most
positive, on the west side of the hill. It is
least; i.e., most negative, on the east side of the hill. The gradient of the height in the
easterly direction is the negative of the
gradient of the height on the westerly direction. The gradient of height in the southerly
direction is greatest on the north side,
least on the south side, and zero on the top of the hill. The water flowing down the hill
will go fastest in the steepest place;
i.e., where the contour lines are the closest together, which occurs on the southeast
portion of the hill.
As time goes on the temperatures at various places on the ladle handle will increase
until constant values are obtained and
the heat flow becomes stable.
1. A negative value of the temperature gradient means that the gradient points from a
high temperature region to a low
temperature region.
2. The temperature gradient is a larger negative number at 4 cm than at 3 cm. It is a
smaller negative number at 5 cm than
at 3 cm.
3. The gradient is just the slope of the line tangent to the curve at the point of interest.
The relative size of the gradient can
be found by looking at the steepness of these tangent lines.
4. The gradx T has its largest, or least negative, value at the distance of 7 cm. Gradx T is
smallest, or most negative at 4
cm. It is not zero any place between zero and seven.
SECTION 9.4 Heat Flow
5. Heat can escape from the portion of the handle directly into the air.
6. Yes, your hand on the rod will be hotter than a hand in air the same distance from the
fire.
SECTION 9.5 Water Flow
7. The water flow between 150 and 200 seconds appears to be at a constant rate of about
5 liters/50 sec. or 0.1 liter/second.
8. The fact that the amount of water in the bucket stays the same after 300 seconds can
be explained by a condition where
the inflow and outflow of water become equal. It can be obtained by (1) turning off the
water so both inflow and outflow are
zero or (2) filling the bucket then the inflow equals the outflow as the bucket overflows.
SECTION 9.7 How to Increase the Flow
9. The units of current density for heat flow are joules per square meter per second.
10. The temperature gradient in SI units is measured in degrees Celsius per meter.
11. The coefficient of thermal conductivity must have the units of heat current divided
by temperature gradient, joules/ meter
øC sec.
12. The various values for the temperature gradients as estimated from the graph See
Fig. 9.2 are -3øC/cm, -9øC/cm, and
-7øC/cm at the locations of 1 cm, 3 cm, and 5 cm respectively. The current densities can
be found by multiplying the
temperature gradients by the coefficient of thermal conductivity 420 J/møCs. The
current densities are then 1 x 105 J/m2s, 4 x
10 5 J/m2s, and 3 x 105 J/m2s respectively.
SECTION 9.8 Diffusion
13. From Equation 9.16 the diffusion coefficient D must have the same units as the ratio
of the mass current density J
(moles/m2 sec) to the concentration gradient Dc/Dx (moles/m4); so D has the units of
m2/sec.
14. Equation 9.19 has consistent units since the gradient of the concentration is
multiplied by the ratio of D2/D, it is the same
as being multiplied by D as in Equation 9.16.
WORD STATEMENT OF EQUATIONS
The rate of heat flow (I) is equal to the change in the quantity of heat (DH) divided by
the time required to produce such a
change (Dt), I = DH/Dt (9.1)
The rate of matter flow (I) is equal to the change in the quantity of matter (DQ) divided
by the time required to produce such a
change, I =DQ/Dt (9.3)
The current density (J) is the amount of something that flows through a unit area in a
unit of time. Hence, if the area is of
size A, the time is of duration t, and the amount of something is Q; J = Q/At. (9.4)
The change of the mass density of a substance per unit time is equal to the change in
current density per unit length, Dp/Dt
= D J/l (9.13)
For incompressible fluids, the density does not change, soDp/Dt = 0, and the current
density cannot change from one location
in the liquid to another one.
The thermal energy current through an object Jx is equal to the negative of the
coefficient of thermal conductivity of the object
times the temperature gradient across the object.
Jx = -KH DT/Dx (9.15)
The matter current of diffusive flow (J) is equal to the negative product of the diffusion
constant D and the gradient of the
mass concentration.
J = -D Dc/Dx (9.16)
EXAMPLES
Keywords: Gradients; Worked Examples; ; Oxygen; Continuity; Equations; Anatomy
And Physiology;
Cardiovascular Systems; Transport Theory; Rate Of Flow; Cells; Current Density;
Arithmetic
TRANSPORT PROBLEMS
1. While studying the flow of oxygen in living systems a pathologist quick froze a
human blood cell and then used a
microprobe to analyze the concentration of oxygen at various locations in the cell at one
instant in time. These data are
shown below.
See Fig.
(a) Where are regions of maximum oxygen concentration gradients? Minimum oxygen
concentration gradients?
(b) As time goes on, if the cell were not frozen, explain in words what you expect to
happen to the oxygen concentration in
the blood cell.
(c) By using a microprobe around the outside of the warmed cell the scientist measured
a maximum oxygen diffusion current
to be 2.3mmoles/mm2 sec. Where do you think that oxygen current is flowing? What is
its direction of flow? Estimate a value
of the diffusion coefficient for oxygen for this blood cell.
What Data Are Given?
The oxygen concentrations are given at various locations in a plane.
What Data Are Implied?
It is assumed that gradient and current calculations are appropriate for the scale of sizes
of the various quantities given in
this problem. This includes the assumption that superposition is valid for this system.
What Physics Principles Are Involved?
This problem involves the use of the definitions of gradient and current and their
assumed linear relationship as expressed in
Equation 9.16.
What Equations Are to be Used?
concentration gradient = change in concentration/change in location =Dc/Dx (1)
matter current density through an area = change in concentration/ area ¥ time (2)
matter current density = J = -D Dc/Dx (3)
Algebraic Solutions
This is a problem without a closed algebraic solution, you need to determine numerical
results from the given data.
Numerical Solutions
(a) Regions of maximum gradient will occur with the maximum increase in oxygen
concentration while the x and y value is
increasing. The maximum gradient in the x direction occurs for y = 0.0, between x = 0.4
and 0.6, then
gradx c = (0.9 - 0.1)/(0.6 - 0.4) = 0.8/0.2 = 4 micromoles/ cm3 ¥ mm = 40 mm/cm4
= 4 x 10-5 moles/cm4 = 4 x 103 moles/(meter)4
The maximum gradient in the y - direction occurs when x = 1.0 and between y = 0.2 and
0.4.
grady c = (1.7 - 0.8)/(0.4 - 0.2) = 0.9/0.2 ÷ 5 micromoles/cm3 ¥ mm = 5 x 103 moles/m4
The minimum gradient will occur with the maximum decrease in oxygen concentration
while the x or y value is increasing.
The minimum gradient in the x - direction occurs for y = 0.4, and x is between 1.0 and
1.2
gradx c = (0.4 - 1.7)/(1.2 - 1.0) = -1.3/0.2 = -7 micromoles/cm3 ¥ mm = 7 x 103
moles/m4
The minimum gradient in the y - direction occurs for x = 0.8 and between y = 0.4 and
0.6,
grady c = (1.1 - 2.3)/(0.6 - 0.4) = -1.2/0.2 = -6 micromoles/cm3 ¥ mm = -6 x 103
moles/m4
(b) If the walls of the cell are impervious to the flow of oxygen, then the total amount of
oxygen in the cell must remain a
constant. There would be flow of oxygen inside the cell from regions of high
concentration; e.g., (0.8, 0.4), to the regions of
low concentration around the cell boundary. The final average uniform concentration
would be about 0.8mm/cm3. If the walls of
the cell will allow oxygen to flow in and out, then the oxygen concentration in the cell
will finally become equal to the
concentration of oxygen in the environment. In both cases, the internal gradients of
oxygen concentration will become zero.
(c) The largest gradient is absolute magnitude is gradx c for y = 0.4 and 1.0 ó x ó 1.2, so
it seems most likely that the
maximum flow of oxygen occurs in response to that gradient, in the positive x direction.
Jx = -D Dc/Dx = -D (-7 x 103 moles/m4) (9.16)
Jx = (2.3 x 10-6 moles/mm2 ¥ sec) = (2.3 moles/m2 ¥ sec) = 7 x 103 moles D/m4
so D ÷ (3 x 10-3 m2)/sec.
See Fig.
In studying problems of flow
You need to find where to go
Students radiant
Follow the gradient
But end up not above but below!
PRACTICE TEST
Keywords: Thermal Transport; Transport Theory; Gradients; Problems; Answers; ;
Evaluations; Gradients; Rate Of
Flow; Thermal Conductivity
A metal rod is placed in a flame and the temperature at various locations along the rod
is measured at various times.
See Fig.
The following data were obtained.
See Fig.
1. On graph paper plot these data as different curves so you can calculate the
temperature gradient from your graph. Label
your axes.
2. What is the temperature gradient at the 40 cm location at a time of 6 minutes?
3. If the Coefficient of thermal conductivity of the rod is 4W/øC cm, calculate the rate of
flow of heat along the rod.
(Cross-sectional area = 1.5 cm2)
Chapter
Temperature and Heat (10):
GOALS
Keywords: ; Learning Objectives; Temperature; Thermometers; Heat; Thermal
Expansion; Units; Heat; Heat
Equivalent; Mechanical Equivalent; Specific Heats; Heat Capacity; Latent Heats;
Calorimetry; Heat Of Combustion
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it an operational definition:
temperature mechanical equivalent of heat
thermometer heat capacity
heat specific heat
linear expansion latent heat of fusion
volumetric expansion latent heat of vaporization
calorie heat of combustion
Calorimetry
Solve problems in calorimetry.
Gas Laws
Solve problems using the gas laws involving the pressure, volume, and temperature of
a confined gas.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 5, Energy,
and Chapter 9,
Transport Phenomena.
OVERVIEW
Keywords: ; Heat; Thermal Physics; Instructions
Heat is a form of energy. As this energy moves from one object to another, we may
detect physical changes in both the
object losing heat energy and the object gaining heat energy. This chapter deals with
many of these basic changes and the
rules used to describe them.
SUGGESTED STUDY PROCEDURE
This chapter places emphasis on three primary Chapter Goals: Definitions, Calorimetry,
and Gas Laws. Please read these
chapter goals carefully. For an expanded discussion of the terms listed under
Definitions, please turn to the next page of this
Chapter. Next, read text sections 10.1-10.9. Be sure to check the answer to each
question in the text readings.
You will find these answers in the second section of this chapter.
Now read the Chapter Summary and complete Summary Exercises 1-12. Then do
Algorithmic Problems 1-9 and check your
answers carefully against the answers given in the text. Now do Problems and Exercises
1, 2, 4, 9, 10, 14, and 18. For
additional examples of problems from this text chapter, please see the Examples section
of this chapter. Now
you should be prepared to attempt the Practice Test over Temperature and Heat. Check
your answers with those given. If you
have difficulties with any of the problems, refer to the appropriate text section and the
suggested study procedure. This study
procedure is outlined below.
---------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
---------------Definitions 10.1, 10.2, 10.3, 1-10 1, 2, 3, 4, 1, 2, 4
10.4, 10.5, 10.7, 5, 6, 7, 8
10.8
Calorimetry 10.6 11 9, 10
Gas Laws 10.9 12 9 14, 18
DEFINITIONS
Keywords: ; Glossary; Thermal Physics; Temperature; Thermometers; Heat; Thermal
Expansion; Units; Heat
Equivalent; Mechanical Equivalent; Heat Capacity; Specific Heats; Latent Heats; Heat
Of Combustion
TEMPERATURE
The relative hotness or coldness of an object.
The human body contains many temperature sensors to inform you of the temperature
of your body and its environment.
THERMOMETER
A device or transducer which can be used to measure the temperature of a system.
The most common kinds of thermometer are mercury in glass and red-colored alcohol
in glass.
HEAT
Thermal work, a process by which the internal energy of a system is changed.
LINEAR EXPANSION
Change in length accompanied by a change in temperature.
Most materials increase in length when their temperatures are increased.
VOLUMETRIC EXPANSION
The change in volume accompanied by a change in temperature.
Most substances expand when they are heated. Water at 4øC expands if it is heated and
if it is cooled!
CALORIE
The mean amount of heat required to increase the temperature of one gram of water
one degree Celsius.
MECHANICAL EQUIVALENT OF HEAT
The amount of mechanical energy which is equivalent to one unit of heat or thermal
work.
In the SI units 4.186 joules is equivalent to 1 calorie. Your (Fuller)3 textbook weighs
about 13 Newtons. If you lift it about 32
cm you will have done about 4.19 joules of work. That amount of work would raise the
temperature of 1 cubic centimeter of
water one degree Celsius.
HEAT CAPACITY
Product of the mass of a body in grams and specific heat; i.e., amount of heat required
to raise the temperature of the body
one degree Celsius.
SPECIFIC HEAT
Heat energy required to raise the temperature of a mass of material one Celsius degree.
Compared to many other substances, See Table 10.3, water has a large specific heat.
Water is a good substance to use for
the storage of thermal energy.
LATENT HEAT OF FUSION
Heat required to change unit mass of material from the solid state to the liquid at the
melting point.
Consider how high you would have to lift your copy of (Fuller)3 to do the amount of
work equivalent to the latent heat of
fusion of one cubic centimeter of ice at 0øC. More than 25 meters!
LATENT HEAT OF VAPORIZATION
Heat required to convert unit mass of material from liquid state to vapor state at the
boiling point temperature.
Guess how high you would have to lift your copy of (Fuller)3 to do an amount of work
equivalent to the amount of energy
required to convert one cubic centimeter of hot water at 100øC to steam at 100øC. About
173 meters! That's much higher
than the state capitol building in Lincoln, Nebraska (which is about 130 meters tall)!
HEAT OF COMBUSTION
Energy produced per unit measure of quantity of material for complete oxidation.
Which would you rather oxidize, spinach or beef steak? You will get about 20 times
more energy from the steak as from an
equal mass of spinach.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Temperature; Thermal Expansion; Heat Of Combustion;
Absolute Zero; Temperature
Scales; Thermal Physics
SECTION 10.2 Temperature
Absolute zero in Celsius is -273.16øC. In Fahrenheit, absolute zero is -459.69øF.
SECTION 10.4 Volumetric Expansion
Let us consider a cube of side L0, then its volume is V0 at a temperature T0. Now let us
heat the cube to a new temperature
DT degrees hotter than T0. Now we calculate the new volume V in terms of the
volumetric coefficient of expansion b,
V = V0 + DV = V0 +bV0DT = V0(1 +bDT) (1)
But we can use the linear measurement to calculate this same volume.
V = L3 = (L0 +aL0DT)3 = L03 (1 + aDT)3 = V0 (1 + aDT)3
= V0 (1 + 3aDT + 3a2DT2 +a3DT3)
If a is much smaller than 1; then a2 << a; so neglect the a2 and a3 terms:
V ÷ V0 (1 + 3aDT) (2)
By comparing equations (1) and (2) we see
b ÷ 3a
SECTION 10.8 Heat of Combustion
At this date the cost of butter is about $.80 per pound and gasoline costs about $.65 per
gallon.
Energy from butter = 716 kcal/100gm x 454 gm/1 lb. x 1 lb./$.80
= 4060 kcal per dollar
Energy from gasoline = 1150 kcal/100 gm x 4546 cm3/gallon x 1 gallon/$0.65 x 0.68
gm/1 cm3
= 54700 kcal per dollar
Butter costs 13 times as much as gasoline per unit of energy.
EXAMPLES
Keywords: Worked Examples; ; Heat; Calorimetry; Anatomy And Physiology;
Cardiovascular System; Volume;
Pressure In Gases; Latent Heats; Heat Of Combustion; Charles' Law; Boyle's Law; Heat
Equivalent; Mechanical Equivalent;
Thermal Physics
CALORIMETRY
1. A nurse withdraws 50 cc's of blood from the arm of a patient. He wants to keep the
sample for a while, so he puts it in a
refrigerator and cools the sample to 2øC. How much heat in calories must be removed
from the blood to do this? Assume for
this problem that 50 cc's of blood has a mass of 52 gm; assume that the specific heat of
blood is the same as the specific
heat of water; assume that the body temperature of the patient is 98.6øF or 37øC. Be
sure to give the correct units.
What Data Are Given?
Mass of blood = 52 gm; initial temperature of blood = 37øC; final temperature of blood
= 2øC, specific heat of blood = 103
calories per kg.
What Data Are Implied?
None, it is all given.
What Physics Principles Are Involved?
The basic Calorimetry equation (10.9) is all that is needed.
What Equations Are to be Used?
DQ = mcDT (10.9)
Algebraic Solution
Heat Removed = DQ = mcDT
Numerical Solution
DQ = (52 gm)(1 cal/gm)(2øC - 37ø) = -1.8 x 103 calories
Thinking About the Answer
The change in heat, DQ, is negative because heat is lost from the system of blood.
2. Heat is required to change ice (solid) to water (liquid) without changing its
temperature. In fact, it requires 80 calories of
heat to change 1 gram of ice at 0øC to 1 gm of water at 0øC. Furthermore, heat is
required to change water (liquid) to steam
(gas) without changing its temperature. In fact, under standard conditions it requires
540 calories of heat to change 1 gm. of
water at 100øC to 1 gm. of steam at 100øC.
(a) How much total heat is required to warm and convert a 1 gm piece of ice at 0øC to
steam at 100øC under standard
conditions?
(b) What percentage of the total heat is used to change the 100øC water to 100øC steam?
What Data Are Given?
1 gram of ice exists at 0øC.
What Data Are Implied?
During the heating process, no heat will be lost from the ice - water - steam system to its
surroundings.
What Physics Principles Are Involved?
The basic concepts of Calorimetry (Equation 10.9) and changes in state (Section 10.7) are
needed.
What Equations Are to be Used?
Heating the water = DQ = mcDT (10.9)
Melting the ice = DQf = Lfm (3)
Vaporizing the water = DQv = Lvm (4)
Algebraic Solution
(a) total heat = DQf + DQ +DQv = mLf + mLDT + mLv (5)
(b) percentage for vaporization = mLv/total heat (6)
Numerical Solutions
(a) Qtotal = (1)(80) + (1)(1)(100 - 0) + (1)(540) = 720 calories
(b) Percentage = 540/720 = 75%
Thinking About the Answer
Notice how much energy is required to convert the liquid water to the gaseous water,
75% of all the energy!
3. About how many hamburgers would a 91 kg (200 lb) man have to eat to enable him
to climb to the top of the Nebraska
capitol building, about 131 meters high? An average hamburger has a fuel value of
about 250 kilocalories. The human body
has a 10% efficiency; i.e., 90% of the fuel value of food is used to maintain the body,
only 10% is available for work.
What Data Are Given?
Mass = 91 kg; height to be lifted = 131 meters, efficiency = 10%, heat of combustion =
250 kcal/hamburger.
What Data Are Implied?
The energy intake must be 10 times the amount used because the useful work is only
1/10 of the total energy input. Assume
g = 9.8 m/s2. It is assumed that the total fuel value of the hamburgers is used by the
human body.
What Physics Principles Are Involved?
The mechanical equivalent of heat is what is needed for this problem. The potential
energy needed or the work done, to climb
to the top of the Nebraska capitol building.
What Equations Are to be Used?
Work done in climbing = DP.E. = mgh
Number of hamburgers = work done in kilocalories / (efficiency)250kcal/hamburger
Algebraic Solution
N = W = mgh [W(1 kcal/4186 J)/250 kcal/lb.]/(efficiency)
Numerical Solution
W = (91 kg)(9.8 m/s2)(131 m) = 1.2 x 105 J
W = 28 kcal
Energy needed = W/efficiency = 28/.1 = 280 kcal.
N = 280 kcal / 250 kcal/lb. = 1.1 hamburgers.
During his climb up to the top he would use up about the energy of one hamburger for
climbing and for maintaining his body
conditions.
Thinking About the Answer
If you assume his other food is used to maintain his body conditions and he ate an extra
hamburger, in order to keep from
gaining weight he would have had to climb to the top of the capitol building 9 times.
This shows why dieting rather than
exercise is the easiest way to maintain, or reduce, body weight.
GAS LAWS
4. A weather balloon contains 280 m3 of helium gas at sea level where the atmospheric
pressure is 1.01 x 105 N/m2 and the
temperature is 27øC. Calculate the volume of the balloon when it reaches an altitude of
16 km where the pressure is 1.00 x
104 N/m2 and the temperature is -55øC.
What Data Are Given?
V1 = 280 m3; P1 = 1.01 x 10 5 N/m2; T1 = 27øC or 300øK; P2 = 1.00 x 104 N/m2, and T2
= -55øC or 218øK.
What Data Are Implied?
That the balloon skin applies no extra pressure to the gas on the inside of the balloon, so
the gas laws can be applied
directly using the given data. It is assumed helium is an ideal gas.
What Physics Principles Are Involved?
The combined laws of Boyle's and Charles' for ideal gases can be used.
What Equations Are to be Used?
P1V1/T1 = P2V2/T2 (10.15)
Algebraic Solution
V2 = (T2/T1)(P2/P1) V1
Numerical Solution
V2 = (218øK/300øK) x (1.01 x 105/1.00 x 104) (280 m3)
V2 = 2060 m3
Thinking About the Answer
The effect of the greatly reduced pressure to allow the balloon to expand is much more
important than the reduced
temperature that causes the gas in the balloon to have a reduced volume. The reason for
this, of course, is the fact that
absolute zero is many degrees below the range of temperatures occurring in this
problem.
Suppose absolute zero were only -60øC instead of -273øC. How would the results of
this problem be changed?
See Fig.
SPECIAL PRACTICE PROBLEM FOR FUN
4. A well-known secret fact about Robinson Crusoe is that he was an amateur
entomologist. While shipwrecked on his island
every day at noon (sun directly overhead) he counted the number of ants coming out of
the largest anthill in about one minute
(70 beats of his heart). He noticed that on hot days the ants were more active than on
cold days, so he constructed his own
Robinson Crusoe temperature scale. He assumed there was a linear relationship
between the number of ants he counted and
the temperature. He made his own simple temperature scale, zero ants was 0øRC and
100 ants was 100øRC. After he was
rescued he desired to calibrate his ant hill from known meteorological data and to fill in
the missing numbers in his daily log.
Fill in the missing numbers in the table below:
Meteorological Data for the temperature of his island at noon.
See Fig.
ANSWERS:
The relationship between the two temperature scales is given by T(RC) = 2.5 T(øF) - 150.
So 86øF = 65øRC; 92øF = 80øRC;
72øF = 30øRC. Do you think this is a creepy problem?
PRACTICE TEST
Keywords: Problems; Answers; Evaluations; ; Temperature; Thermal Physics;
Temperature Scales; Fahrenheit
Scale; Celsius Scale; Pressure In Gases; Heat; Thermal Expansion; Charles' Law; Boyle's
Law; Calorimetry
1. Find the Celsius temperatures corresponding to the following common Fahrenheit
temperatures.
a. Room temperature (68øF)
b. Human body temperature (98.6øF)
c. Cold winter temperature (-4øF)
d. Ice melts (32øF)
e. Water boils (212øF)
2. An automobile tire at 32øF has a gauge pressure of 24 lb/in 2 (then its absolute
pressure is 2.6 x 105 N/m2). Immediately
after running at high speed on the interstate highway, the tire pressure is measured to
be 32 lb/in2 (its absolute pressure is
3.2 x 105 N/m2). What is the temperature of the tire, assuming the volume of the tire
remained constant?
3. On a hot summer day (40øC) a physics student fills the 72 liter gasoline tank of her
car with cool gasoline (10øC) from a
self-service pump. Why will the gasoline overflow the tank? How much gasoline will
flow out of the tank? (Coefficient of
volumetric expansion of gasoline = 0.90 x 10-3/øC)
4. A 5 kilogram lead sphere falls from a 100 meter high building and lands on a hard
concrete sidewalk. If all the heat energy
created at impact is assumed to be retained by the sphere, answer the following
questions.
________a. How much heat is generated at the impact?
________b. What is the final temperature of the lead if its initial temperature was 22øC?
(Specific heat of lead = .13 J/gmøC).
ANSWERS:
1a. 20øC b. 37øC c. -20øC d. 0øC e. 100øC
2. 60øC
3. Increase in the temperature of the gasoline causes volumetric expansion; 1.9 liters
4a. 5,000 J b. 29.7øC
Chapter
Thermal Transport (11):
GOALS
Keywords: ; Thermal Physics; Learning Objectives; Thermal Transport; Conduction;
Convection; Radiation; Heat
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define the following terms, and use them in an operational definition:
conduction
radiation
convection
Energy Transfer Problems
Solve numerical problems that involve a transfer of energy, temperature gradients, and
conduction,
convection, or radiation.
Living System Thermal Properties
Explain basic thermal effects in living systems.
PREREQUISITES
Before beginning this chapter, you should have achieved the goals of Chapter 9,
Transport Phenomena, and
Chapter 10, Temperature and Heat.
OVERVIEW
Keywords: Learning Objectives; ; Heat; Thermal Physics; Thermal Transport
From Chapter Nine you should recall that three Transport Phenomena were
highlighted. In this chapter, the flow or transport of
heat is considered in greater detail. Please note that the chapter deals with three major
forms of Thermal Transport:
Conduction (Section 11.2), Convection (Section 11.3), and Radiation (Section 11.4).
SUGGESTED STUDY PROCEDURE
To begin your study of this chapter, please read all the Chapter Goals: Definitions,
Energy Transfer Problems, and Living
System Thermal Properties. For an expanded treatment of each of the terms listed under
Definitions, turn to the next section
of this chapter.
Next, read text sections 11.1-11.5. Even though the algebra becomes involved in
example 2 in section 11.2, try to go through
the problem as you may find the result surprising. As you read the text sections,
consider the questions asked. The answers
to these questions are answered in this chapter.
Now read the Chapter Summary and complete Summary Exercises 1-9 and do
Algorithmic Problems 1-4. Next, complete
Exercises and Problems 1, 4, 6, 7, and 10. Additional examples of the major concepts
presented in this chapter are found in
the Examples section of this chapter.
Now you should be prepared to attempt the Practice Test in the at the end of this
chapter. Check your answers.
If you have difficulties, please refer to the specific section of this chapter again. This
study procedure is outlined below.
-----------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
----------------Definitions 11.1, 11.2, 11.3 1-5 4, 6, 10
11.4
Transfer 11.2, 11.3, 11.4 6 1-4 1, 7
Problems
Living System 11.5 7-9
Thermal Properties
DEFINITIONS
Keywords: Glossary; ; Conduction; Radiation; Convection; Thermal Physics; Thermal
Transport; Heat
CONDUCTION
Transfer of thermal energy by the interaction of neighboring portions of a substance at
rest.
Conduction is a most effective transfer mechanism in solids, particularly in metals.
CONVECTION
Transfer of thermal energy by the motion of matter as in liquids and gases.
Convection arises because of the different densities of warm and cool fluids which
cause buoyant forces in the earth's
gravitational field. In the weightless environment of space, convection ceases to be an
energy transfer mechanism.
RADIATION
Transfer of energy by electromagnetic fields.
Microwave cooking is a most recent example of using radiation to heat objects.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: Microwaves; Answers; Applications;
SECTION 11.6 Microwave Cooking
The explanation of microwave cooking taken from the cookbook of the Litton Systems
may seem more vivid but it certainly
carries some wrong implications. Food does not "attract" microwaves anymore than you
would say that sunglasses attract
light. Sunglasses absorb light that would otherwise strike your eyes. Foods, especially
fats and sugars, absorb the
microwaves that pass through other materials. The absorption process leads to the
transformation of microwave,
electromagnetic energy into thermal energy. The Consumer Reports description is more
in accord with our present day
scientific model to explain these phenomena.
EXAMPLES
Keywords: Worked Examples; ; Thermal Physics; Thermal Transport; Conduction;
Thermal Conductivity;
Arithmetic; Convection; Radiation; Stefan-Boltzmann Law; Cooling; Heat
ENERGY TRANSFER PROBLEMS
1. The wooden handle of a frying pan has an area of 10 cm2 and is 15 cm long. If the pan
is at 120øC and the free end of
the handle is at 30øC, how much heat will flow through the handle in 10 minutes?
(Kwood = 0.15 Wm-1 deg-1)
What Data Are Given?
The area of the handle, the high and low temperatures, the distance between those two
temperatures, the time, and the
thermal conductivity of the material.
What Data Are Implied?
It is assumed that no heat escapes from the handle out the sides, but that heat only
flows along the handle. The problem can
be treated as a one-dimensional problem.
What Physics Principles Are Involved?
The concepts of thermal conduction are needed.
What Equations Are to be Used?
We can make use of the one dimensional thermal conductivity equation 11.2
H = -KA (DT/DX) (11.2)
Algebraic Solution
H = -KA (DT/DX) where H is in joules/second
Heat flow in 10 minutes = 600 H
Numerical Solution
Heat Flow in 10 minutes = 600 sec. x (-0.15/m ¥ deg)(J/sec) x (10 x 10-4m2) x [(30 120)øC/0.15 m]
= 600 (-.15)(10-3) (-90/.15) joules
= 54 J
Thinking About the Answer
Notice that the units for the answer, joules, are correct.
2. In a room of temperature 20øC, two identical open pans, one containing ice water, the
other boiling water, are placed in
front of an air-circulating fan. If the pans contain equal amounts of water, compute the
ratio of the initial rates of their
temperature changes. Is the ratio a positive or negative number?
What Data Are Given?
The temperature of the room in a situation of forced air convection.
What Data Are Implied?
The two pans are assumed to be in identical situations with respect to the forced air
circulation. The only difference between
the two pans is their initial starting temperatures. One pan has a temperature 0øC. The
other pan has a temperature of
100øC.
What Physics Principles Are Involved?
This problem can be solved using Newton's law of cooling; i.e., the rate of change of
temperature of an object is proportional
to the difference between the temperature of the object and its surroundings.
What Equations Are to be Used?
The heat capacity equation, Q = mcDT, (Equation 10.9). The forced convection equation,
rate of thermal energy flow = H = -K1 (Ts - Ta) (11.6)
Algebraic Solution
Let Tc be initial temperature of the cold water
Let Th be initial temperature of the hot water
for cold water Hc = -K1 (Tc - Ta)
for hot water Hh = -K1 (Th - Ta). The rate of heat flow of an object is related to the
temperature change of an object by
Equation 10.9 D0 = mc DT, but H =DQ/Dtime.
so: H = DQ/Dt = mcDT/Dt
Since the masses and specific heat of the two systems of water are equal, then
DTc/Dt = -K1 (Tc - Ta) andDTh/Dt = -K1 (Th - Ta)
The ratio of the initial rates of temperature change are given by
(DTc/Dt) / (DTh/Dt) = (Tc - Th) / (Th - Ta)
Numerical Solution
(DTc/Dt) / (DTh/Dt) = (0ø - 20ø)C / (100ø - 20ø)C = -20/80 = -1/4
The ratio is a negative number because one container is warming up so its DT/Dt is
positive and the other container is
cooling down so its DT/ Dt is negative. Hence the ratio is negative.
Thinking About the Answer
Notice that the answer for this question is a pure number with no units because it is a
ratio of similar quantities. It is always
a negative number. For the case given the hot water is cooling down four times as fast
as the cold water is heating up.
3. In a cryogenics (low temperature) experiment the temperature of the sample is
decreased from liquid helium temperatures
(4øK) to 1øK. How much is the rate of heat loss by radiation decreased?
What Data Are Given?
The absolute temperature of the sample is given.
What Data Are Implied?
Any heat radiated to the samples from the environment is neglected.
What Physics Principles Are Involved?
The Stefan-Boltzmann equation for radiation.
What Equations Are to be Used?
Rate of energy radiated = P = sAT4 (11.8)
Algebraic Solution
Ti = initial temperature; Tf = final temperature
Pf/Pi = Tf4/Ti4 = (Tf/Ti)4 (2)
Numerical Solution
Pf/Pi = (1øK/4øK)4 = 1/256 = 3.9 x 10-3
Thinking About the Answer
You can notice that even though the temperature is only changed 3ø, there is a large
change in the amount of energy lost by
radiation.
4. Suppose you double the absolute temperature of an object. How much will the three
types of heat transfer change? (Hint:
Assume the ambient room temperature is one-half the original temperature of the
object.)
What Data Are Given?
The initial temperature of the object is Tc. The initial ambient temperature is T0/2. The
final temperature of the object is 2T0.
What Data Are Implied?
The situation is such that all of the first order heat transfer equations are valid,
Equations (11.2), (11.6), and (11.8).
What Physics Principles Are Involved?
The basic concepts of conduction, convection, and radiation are needed.
What Equations Are to be Used?
Conduction H = -KA DT/DX (11.2)
Convection H = -K1(Ts - Ta) (11.6)
Radiation P = sAT4 (11.8)
Algebraic Solution
Let Ti = initial temperature, Tf = final conduction temperature, and Ta = ambient temp.
Conduction:
Hi a (Ti - Ta) / Dx and Hf a (Tf - Ta) / Dx
so ratio Hf/Hi = (Tf - Ta) / (Ti - Ta) (3)
Convection:
Hi a (Ti - Tu) and Hf a (Tf - Ta)
ratio Hf/Hi = (Tf - Ta) / (Ti - Ta) (4)
Radiation:
Pi a Ti4 - Ta4; i.e., heat lost by the object goes at T4, but heat received from the
environment goes at Ta4, so the net rate of
radiated energy is the difference in the fourth power of the temperatures.
Pf a Tf4 - Ta4
ratio Pf/Pi = (Tf4 - Ta4) / (Ti4 - Ta4) (5)
Numerical Solutions
Conduction
Hf/Hi = (2T0 - T0/2) / (T0 - T0/2) = (3/2)/(1/2) = 3
Convection
Hf/Hi = (2T0 - T0/2) / (T0 - T0/2) = (3/2)/(1/2) = 3
Radiation
Pf/Pi = [(2T0)4 - (T0/2)4] / [T04 - (T0/2)4] = (255/16)/(15/16) = 15.9
Thinking About the Answer
Once again you see the strong temperature dependence of radiation as a form of energy
transfer.
See Fig.
PRACTICE TEST
Keywords: ; Evaluations; Thermal Physics; Questions; Answers; Heat; Convection;
Conduction; Radiation;
Anatomy And Physiology; Thermal Transport
1. A hot cup of black coffee sits in a cool room in a black coffee cup.
a. Name and describe the thermal processes which are actively involved in the transfer
of heat from the cup to the room.
b. Explain how the coffee, the cup, or the room environment could be altered to reduce
the thermal transport by each of the
processes named in Part A.
2. A modern home is insulated with 15.2 cm (6 inches) of glass wool (K = .042). How
many cm thick must a brick wall be
constructed (K = .147) to achieve the same insulating factor?
3. The human body has an intricate set of regulating systems which act together to keep
the core of the body at a constant
temperature of 37ø C.
a. What is the typical temperature of the body at the skin's surface? Does the
temperature vary over the skin's surface? (If it
does, give an example showing the magnitude of the variation.)
b. The major core regulating process is by "forced convection". Explain briefly how this
process works.
ANSWERS:
1. Convection (to room air), conduction (to room air and to table), radiation (to each
object in the room which is lower in
temperature and to the walls, ceiling, and floor). To change; convection (reduce air
currents), conduction (insulate cup),
radiation (change color of cup and/or coffee or room walls as increase temperature of
room walls.
2. 53.2 cm
3. 33øC, Yes, about 3øC less at exposed extremities: vasodilation regulates the flow of
blood and therefore the direction of
heat flow (see section 11.5).
Chapter
Thermodynamics (12):
STUDY
GUIDE
Citation: H. Q Fuller, R. M. Fuller and R. G. Fuller, to Accompany Physics Including
Human Applications. (Harper
and Row, New York, 1978). Permission granted by the authors.
12 Thermodynamics
GOALS
Keywords: ; Learning Objectives; Thermal Physics; Heat Pumps; Thermodynamics;
Heat Engines; Carnot Cycle;
Efficiency; Thermodynamic Processes; Adiabatic Processes; Isobaric Processes;
Isothermal Processes; Isochoric Processes;
PV Diagrams; The Laws Of Thermodynamics
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
PV diagram efficiency of a heat engine
isochoric process Carnot cycle
isobaric process refrigerator
isothermal process coefficient of performance
adiabatic process of a refrigerator
heat engine
Laws of Thermodynamics
State three laws of thermodynamics, and explain the operation of a physical system in
terms of these laws.
Thermodynamics Problems
Solve problems consistent with the laws of thermodynamics.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 5, Energy,
and Chapter 10,
Temperature and Heat.
OVERVIEW
Keywords: ; Instructions; Thermal Physics; Thermodynamics
The problem of converting heat energy into useful work energy is an important
consideration for an industrial society. The
Three Laws of Thermodynamics stated in this chapter express our understanding of
how thermal processes operate. As you
look over this chapter, please note the importance of the idea of system and the
idealized processes by which a system can
be changed.
SUGGESTED STUDY PROCEDURE
As you begin your study of this chapter, carefully read the three Chapter Goals:
Definitions, Laws of Thermodynamics, and
Thermodynamic Problems. If you need additional assistance with any of the terms
listed under Definitions, refer to the next
section of this chapter.
Next, read chapter sections 12.1-12.11. Be sure to read the examples given at the end of
most of these text sections.
Answers to most questions posed during your reading are discussed in the second
section of this chapter.
Now read the Chapter Summary and complete Summary Exercises 1-7. Then do
Algorithmic Problems 1-5. Now complete
Exercises and Problems 1, 3, 5, 6, 11, 12, 17, 19, and 21. For additional work, see the
Examples section of this .
Now you should be prepared to attempt the Practice Test at the end of this chapter. If
you have difficulty with
any of the concepts, check the appropriate section of the text for further help.
---------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
---------------------Definitions 12.1,12.2,12.7, 1,2,3
12.8 4,5
Laws of 12.4,12.9 6
Thermodynamics
Thermodynamics 12.3,12.5,12.6, 7 1,2,3,4, 1,3,6,11,
Problems 12.10,12.11 5 12,17,19,
21
DEFINITIONS
Keywords: ; Glossary; Thermal Physics; Heat Pumps; Thermodynamics; Heat Engines;
Carnot Cycle; Efficiency;
Thermodynamic Processes; Adiabatic Processes; Isobaric Processes; Isothermal
Processes; Isochoric Processes; PV
Diagrams; The Laws Of Thermodynamics
P-V DIAGRAM
Graphical representation of a process using pressure-volume axes.
See Fig.
The work done during a process is shown as the area under the curve.
ISOCHORIC PROCESS
Takes place at constant volume.
After hot foods are sealed in glass containers during canning, the cooling of a constant
volume of material helps to seal the
can shut.
ISOBARIC PROCESS
Takes place at constant pressure.
The expansion of a heated gas in a flexible container open to the atmosphere illustrates
this process.
ISOTHERMAL PROCESS
Takes place at constant temperature.
The slow compression of a gas can occur at constant temperature.
ADIABATIC PROCESS
A process that takes place with no change in thermal energy in the system.
When we make a rapid change in the pressure of the air in a bicycle tire, we can think of
it as an adiabatic process.
HEAT ENGINE
Absorbs a quantity of energy from higher temperature reservoir, does work, and rejects
a quantity of energy to lower
temperature reservoir, and returns to its original state.
An automobile engine is a heat engine. It makes use of the thermodynamic properties of
combustion and gases to convert
the chemical energy of gasoline into the kinetic energy of the automobile.
EFFICIENCY OF HEAT ENGINE
Ratio of useful work output divided by energy input.
If we can improve the efficiency of our automobile engines we can save many gallons of
gasoline.
CARNOT CYCLE
Cycle of an ideal engine which includes an isothermal expansion at higher temperature,
an adiabatic expansion, an isothermal
compression at lower temperature, and an adiabatic compression.
REFRIGERATOR
A heat engine that operates by the input of work when energy is extracted from a lower
temperature reservoir and transferred
to a higher temperature reservoir.
COEFFICIENT OF PERFORMANCE OF A REFRIGERATOR
Ratio of heat absorbed to the amount of work supplied to the refrigerator.
EXAMPLES
Keywords: PV Diagram; Isochoric Process; Isothermal Process; Isobaric Process;
Worked Examples; The First Law Of
Thermodynamics; Ideal Gas Law; Arithmetic; Thermal Physics; Thermodynamics;
Thermodynamic Processes; Work; Internal
Energy
THERMODYNAMIC PROBLEMS
1. A container sealed by a moveable piston holds 0.908 moles of an ideal gas. The gas is
originally at a pressure of 1.01 x
105 N/m 2, a temperature of 300øK and a volume of 22.4 liters. The gas was heated at
constant volume until the temperature
reached 750øK. Then the gas was allowed to expand isothermal until it reached its
original pressure. Then the gas was
compressed isobarically back to its original state. Draw a P-V diagram for the three
processes. Calculate the pressure,
volume, and temperature of the gas at the end of each process. Calculate the work done,
the change in internal energy and
the amount of heat added to or subtracted from the system during each of the
processes. What is the efficiency of this cycle
of three processes?
What Data Are Given?
The initial conditions for the 0.908 moles of gas are given: P1 = 1.01 x 105 N/m2; V1 =
22.4 x 103 cm3 = 2.24 x 10-2 m3; T1
= 300øK. The highest temperature T 2 = 750øK = T3. The three processes between the
initial state, the second state, the third
state, and the return to the initial state are all given.
What Data Are Implied?
The fact that the confined gas is an ideal gas allows us to use equation (12.6), (12.8), and
(12.12) to calculate properties of a
system where Cv = 3/2 R; Cp = 5/2 R.
What Physics Principles Are Involved?
The first law of thermodynamics can be used in conjunction with the ideal gas laws to
solve this problem.
What Equations Are to be Used?
Process 1 = isochoric compression
P2 = (P1T2)/T1 (10.15)
DU = DQ = ncvDT (12.6)
Process 2 - isothermal expansion P3 = (P2V2)/V3 (10.15)
DQ = DW = PDV = P2V2ln (V3/V2) (12.12)
Process 3 - isobaric compression V1 = (P3V3T1)/(P1T3) (10.15)
DQ = ncpDT (12.8)
DU = ncvDT (12.6)
Algebraic Solution
See Fig.
(a) The P-V diagram - let T2 be given as some constant times T1, T2 = kT1; then P2 =
kP1; V2 = V1
Then T3 = T2; but P3 = P1; so V3 = kV1.
DQ12 = ncv (T2 - T1);
DQ23 = P2V2 ln (V3/V2)
DQ31 = ncp (T1 - T3)
DU12 = DQ12
DU23 = 0
DU31 = ncv (T1 - T3)
efficiency = (DW23 +DW31) / DW23
Numerical Solution
P2 = (1.01 x 105 N/m2) (750øK/300øK) = 2.53 x 105 N/m2
V2 = V1 = 2.24 x 10-2 m3
V3 = 5.60 x 10-2 m3
DQ12 = (0.908 mole)(3/2)(8.31 J/mole øK) x (750øK - 300øK) = 5.09 x 103 J
DW23 = DQ23 = (2.53 x 105 N/m2)(2.24 x 10-2m3) x ln (5.60 x 10-2)/(2.24 x 10-3) = 5.19
x 103 J
DQ31 = (0.908 moles)(5/2)(8.31 J/moleøK) x (300 - 750)øK = -8.49 x 103 J
DU31 = (0.908 moles)(3/2)(8.31 J/mole øK)(300 - 750)øK = -5.09 x 103 J
DW31 = P1(V1 - V3) = (1.05 x 105 N/m2)(2.24 x 10-2 m3 - 5.60 x 10-2m3)
DW31 = -3.40 x 103 J
Efficiency = (5.19 x 103 - 3.40 x 103) J/ (5.19 x 103 J) = 34.0%
The French cyclist, Sadi Carnot, In a race against a Peugeot Whose driver Michael
Smashed his cycle Which lives on in
spite of the blow...
See Fig.
PRACTICE TEST
Keywords: ; Problems; Answers; Evaluations; Thermal Physics; Thermodynamics;
Carnot Cycle; Heat Engines;
Isochoric Processes; Isobaric Processes; Isothermal Processes; Thermodynamic
Processes; The First Law Of
Thermodynamics; The Second Law Of Thermodynamics; The Laws Of
Thermodynamics; Questions; Internal Energy; Work
1. A Carnot engine operates between two temperatures of 600ø and 300øC.
a. Calculate this engine's efficiency.
b. If the engine absorbs 1000 Joules of heat energy at the higher temperature, how
much heat is exhausted at the lower
temperature?
c. How much mechanical work is produced by the engine per cycle?
2. A heat engine is operated through the cycle process AçBçCçDçA as illustrated below.
See Fig..
a. Identify each process as isochoric, isobaric, or isothermal
BçC
CçD
DçA
b. Calculate the work done by the system during each cycle and give the net work done
per cycle.
WAçB = Joules
WBçC = Joules
WCçD = Joules
WDçA = Joules
Net work done = Joules (per cycle)
c. If 60 x 105 Joules of heat energy are absorbed during the thermal process A-B, what
internal energy change occurs during
this process?
3. Is it possible to cool a kitchen during the summer by leaving the door of the
refrigerator open? Defend your answer using
the Laws of Thermodynamics.
ANSWERS:
1. a. 34 b. 660 c. 340
2. a. isochoric, isobaric, isochoric
b. 1.05 x 106, zero, -3 x 105, zero, +7.5 x 105
c. 49.5 x 105 Joules
3. No, the heat produced in running the electric motor for cooling exceeds the heat
removed from the air inside the refrigerator
compartments. This arrangement would be possible only if the system was vented to
allow for pumping the heat to the
outside.
The first law predicts that this is true; the kitchen system has energy added in the form
of electrical energy. Thus, when
converted to heat, the overall effect is to raise temperature. The second law also predicts
the temperature increase; for any
system, the entropy either increases or stays constant. Thus, another system (i.e.,
outdoors) must be involved to cool (or
lower the entropy) one system.
Chapter
Elastic Properties of Materials (13): to Accompany Physics Including
Human
Applications:
13 Elastic Properties of Materials
GOALS
Keywords: ; Learning Objectives; Mechanics; Elasticity; Young's Modulus; Bulk
Modulus; Stress; Strain; Shear
Modulus; Pressure In Solids
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
elastic body Young's modulus
stress bulk modulus
strain modulus of rigidity
elastic limit
Hooke's Law
State Hooke's law.
Stress and Strain
Calculate the strain and stress for various types of deformation.
Elasticity Problems
Solve problems involving the elastic coefficients.
PREREQUISITES
Before you begin this chapter, you should have achieved the goals of Chapter 4, Forces
and Newton's Law,
Chapter 5, Energy, and Chapter 8, Fluid Flow.
OVERVIEW
Keywords: Elasticity; Mechanics; ; Instructions
This is a short but important chapter concerning the reaction of materials to deforming
forces. The recognized method for
reporting these results for various materials is a number called Modulus, which is
simply the ratio of stress to strain. Thus
Young's Modulus, the Modulus of Rigidity, and the Bulk Modulus rate a material's
reaction to forces producing elongation,
change in volume, or sheer distortion respectively.
SUGGESTED STUDY PROCEDURE
Before you begin to study this chapter, be familiar with theChapter Goals: Definitions,
Hooke's Law, Stress and Strain, and
Elasticity Problems. An expanded discussion of each of the terms listed under
theDefinitions goal can be found in the
Definitions section of this chapter.
Next, read text sections 13.1-13.5, and consider the Example problems discussed. Please
note the three major measures of
elastic properties of matter outlined in table 13.1 on page 297. These are Young's
Modulus, the Modulus of Rigidity and the
Bulk Modulus. At the end of the chapter, read the Chapter Summary and complete
Summary Exercises 1-11. Check your
answers against those given on page 301. Now do Algorithmic Problems 1-5 and do
Exercises and Problems 1, 2, 5 and 6.
For additional work, turn to the Examples section of this chapter and complete each
problem. Now you should be
prepared to attempt the Practice Test found at the end of this chapter. If you have
difficulty with any of the
answers, please refer to the appropriate text section for additional assistance.
-----------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
----------------Definitions 13.1,13.2 1-9
Hooke's Law 13.3 3
Stress
and Strain 13.4,13.5 1,2,4
Elasticity
Problems 13.4,13.5 10,11 5 1,2,5,6
DEFINITIONS
Keywords: Glossary; ; Mechanics; Elasticity; Stress; Strains; Young's Modulus; Bulk
Modulus; Pressure In
Solids; Shear Modulus
ELASTIC BODY
Any material or body which is deformed by an applied force and returns to its original
shape after the distorting force is
removed.
We often think of elastic materials as the ones most easily distorted from the original
shape, such as rubber bands. In
physics, a high elastic material requires a large force to produce a distortion. This notion
is contrary to the common use of
the word.
STRESS
Ratio of the applied force to the area.
This is the force applied to change the shape of an object.
STRAIN
Ratio of change in a given physical dimension to the original dimension; i.e., change in
length to original length, or change in
volume to original volume.
This is the measure of the changes in shape of an object acted upon by a stress.
ELASTIC LIMIT
Limit of distortion for which deformed body returns to original shape after deforming
force is removed.
Many students have their lives so stretched by college that they cannot return to their
original life. Have they exceeded their
elastic limit?
YOUNG'S MODULUS
Elastic constant of proportionality for a linear deformation.
For most solids this is a large number ~1010 N/m2. To stretch most solids by even 1%
we would need to apply a pressure of
about 1000 times the pressure of the atmosphere.
BULK MODULUS
Elastic constant of proportionality for a deformation of volume.
This can be measured by squeezing an object in a hydrostatic press and measuring its
change in volume.
MODULUS OF RIGIDITY
Elastic constant of proportionality for a shear deformation.
Some materials, such as graphite, shear much more easily in one direction than in
others.
EXAMPLES
Keywords: ; Worked Examples; Elasticity; Mechanics; Hooke's Law; Bulk Modulus;
Elastic Potential Energy;
Shear Modulus; Mechanics; Elasticity
ELASTICITY PROBLEMS
1. A measuring device is able to apply a force of 1.0 x 10 5 N. A student wishes to use
the device to study the elastic
properties of a 1 cm cube of steel. What is the student likely to find?
What Data Are Given?
The applied force = 1.0 x 105 N. The material being studied is steel whose elastic
coefficients are given in Table 13.1.
What Data Are Implied?
It is assumed that the applied force is not too great so as to exceed the elastic limit of the
steel sample.
What Physics Principles Are Involved?
We can make use of the basic definitions of the linear, bulk, and shear deformations.
Then we can predict what will happen in
each case.
What Equations Are to be Used?
Linear deformation DL/L = (l/Y) (F/A) (13.1)
Bulk deformation DV/V = (-l/B)P (13.7)
Shear deformation f = (l/n)(F/A) (13.8)
Algebraic Solutions
These are all given above, since we wish to predict the kinds of deformations that will
occur in each case.
Numerical Solutions
1. Suppose the student applies the 1.0 x 105 N force across the ends of the cube to
stretch it. Then its length will be
changed
DL/L = (l/Y)(F/A) = (m2/ (10.0 x 1010 N)) x ((1.0 x 105 N)/(1 x 10-4 m2))
DL/L = 5.0 x 10-3. The new length = 1.005 cm.
2) Suppose the student uses the device to apply a 1.0 x 10 5 N force to each of the six
faces of the cube; then the volume
of the cube will be decreased.
DV/V = (-l/B)P = ((-1)m2)/(16.0 x 1010N) = (1.0 x 105 N)/10-4m2
DV/V = -6.3 x 10-3; new volume = 9.937 x 10-7 m3
3) Suppose the student uses the device to apply a 1.0 x 10 5 N force tangent to the top of
the cube while holding the bottom
fixed, then the cube will be deformed by an angle f where
f = ((1 m2)/(8.0 x 1010 N)) x ((1.0 x 105 N)/(1 x 10-4 m2)) = 1.25 x 10-2 radians
f = 0.72ø; so the sides of the cubes are inclined 0.72ø from vertical.
Thinking About the Answers
For which case is the elastic energy the greatest? In each case Hooke's Law is obeyed so
the energy is of the form
E = 1/2 kDx2 but Dx = F/k
so E = 1/2 F2/k; The energy is inversely proportional to the elastic constant for a
constant force, so the most elastic energy
is stored in the shear case (3) above. From where does the extra energy come?
See Fig.
PRACTICE TEST
Keywords: Questions; Problems; Evaluations; Answers; Young's Modulus; Stress;
Strain; Bulk Modulus; Elasticity;
Mechanics; Pressure In Solids
1. Which of the three elastic moduli (Young's, Bulk, or Rigidity) are most important in
each of the case below?
a. A front car tire is deformed as the car rounds a turn on a flat dry road.
b. In the use of a bicycle pump, a cyclist forces down on the pump handle to force the
air into the flat tire.
c. A workman uses a screwdriver to tighten a screw. (Consider your answer for the
steel shaft of the screwdriver.)
d. A rubber band is stretched to fit around a pile of loose papers.
2. Design an experiment for finding the elasticity constant for a rubber band. What
equipment is needed? What measurements
must be made? What results do you anticipate?
3. In an experimental test, the following data was collected for stretching a rubber "tiedown" strap:
cross-sectional area, A = 5.3 cm2
length, L = 75 cm
elongating force, F = 10 Newtons
length increase, Dx = 4 cm
a. Calculate the stress
b. Calculate the strain
c. Find a value for Young's Modulus for rubber.
ANSWERS
1. Rigidity, Bulk, Rigidity, Young's
2. Your experiment should include known weights to be used to stretch the band. As
the band stretches as each weight is
added, the length of stretch should be measured. The elasticity constant in the ratio of
weight to length of stretch. The results
should give a constant K until the band is stretched to near its limits of elasticity.
3. Stress = 1.9 x 104 N/m2, strain = .05, Young's Modulus = 3.8 x 105 N/m2
Chapter
Molecular Model of Matter (14):
14 Molecular Model of Matter
Keywords: ; Learning Objectives; Fluids; Fluid Statics; Surface Tension; Capillarity;
Cohesion; Adhesion;
Osmosis; Fluid Motion; RMS Velocity; Gases; Macroscopic Description Of Ideal Gases;
Properties Of Fluids; Molecular
Model Of Matter
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
rms velocity surface tension
adhesion capillarity
cohesion osmosis
Molecular Model
Explain the properties of fluids using the molecular model.
Problems
Solve problems that relate the gas laws to the properties of the gas and that involve
surface tension,
capilarity, and osmosis.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 4, Forces
and Newton's Laws,
Chapter 5, Energy, Chapter 6, Momentum and Impulse, Chapter 10, Temperature and
Heat, and Chapter 13,
Elastic Properties of Materials
OVERVIEW
Keywords: ; Instructions; Fluids; Molecular Model Of Matter
Molecules are extremely small. Only the very largest molecules have been
photographed and only with indirect methods using
an electron microscope. The model scientists use to describe molecules and their
movements and interactions is only an
ideal representation. However, this model helps us gain an understanding for
phenomena we can observe for a large number
of molecules acting together.
SUGGESTED STUDY PROCEDURE
Begin this chapter by reading the following Chapter Goals: Definitions, Molecular
Model, and Problems. For additional
discussion of the terms listed under the goal of Definitions, turn to theDefinitions
section of this chapter.
Now, continue by reading text sections 14.1-14.11. In section 14.3 you may have
difficulties following all the assumptions and
mathematical steps. Even though this may be the case, pay close attention to the results
as expressed by equations (14.14),
(14.15), and (14.18). The example on page 312 will help clarify the use of the results
derived in this section.
At the end of the chapter, read the Chapter Summary and complete Summary Exercises
1-13. Check your answers carefully
against those provided. Next, do Algorithmic Problems 1-8 again checking your
answers with those provided. Now do
Exercises and Problems 1, 4, 5, 6, 9, 10, and 12. For additional work on the contents of
this chapter, see the Examples
section of this chapter. Now you should be prepared to attempt the Practice Test on
Molecular Model of Matter
provided at the end of this chapter. If you have difficulties with any of the concepts
involved, please refer to the
appropriate section of the text or this for more assistance. This study procedure is
outlined below.
------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
------------------Definitions 14.1,14.6,14.7 1-6
14.11
Molecular Model 14.2,14.3,14.4, 7
14.5
Problems 14.8,14.9,14.10 8-13 1-8 1,4,5,6,
9,10,12
DEFINITIONS
Keywords: ; Glossary; Fluids; Fluid Statics; Adhesion; Cohesion; Surface Tension;
Capillarity; Osmosis; Fluid
Motion; RMS Velocity
ROOT MEAN SQUARE VELOCITY
The square root of the sum of the squares of the velocities divided by the number of the
particles.
If we have two particles, one with a velocity of -20 m/s and the other with +20 m/s,
their mean velocity is zero; their rms is
+20 m/s. If we have two particles, one with velocity +50 m/s, the other with a velocity
of +100 m/s; their mean velocity is
+75m/s; their rms velocity is 56 m/s.
ADHESION
Attraction of unlike molecules.
The new epoxy cements have brought a whole era of adhesives for daily use.
COHESION
Attraction of like molecules.
Cohesive forces in solids are stronger than cohesive forces in liquids.
SURFACE TENSION
Force necessary to break a unit length of surface.
Have you ever seen a water spider "skate" along on the surface of a pond?
CAPILLARITY
The rise of water in a capillary above the water level in the surrounding area.
This arises when a liquid adheres to the walls of a narrow tube.
OSMOSIS
Motion of fluid through a semipermeable membrane until the potential energy on the
two sides of the membrane is the same.
Osmosis is extremely common in biological systems.
EXAMPLES
Keywords: ; Worked Examples; Momentum; Kinetic Theory; Pressure In Gases;
Microscopic Description Of Ideal
Gases; Ideal Gas Law; Avogadro's Number; Boltzmann Constant; Molecular Model Of
Gases; Algebra
KINETIC THEORY OF GASES
1. Assume that all the molecules in a gas are travelling in the x direction (1/2 of the
molecules are headed in the plus x
direction and 1/2 of the molecules are headed in the minus x direction). Assume that all
the molecules have the same
magnitude of velocity, v, and that n molecules strike a wall of area A in a time t.
(a) From what volume do the molecules come that reach the area A of the wall in time t?
(Sketch this volume on the figure
below.)
See fig.
(b) Assume that all n molecules strike the wall head-on, are perfectly elastic, and
rebound with a velocity v in the opposite (or
- x) direction. What is the impulse imparted to the wall? (mass of each molecule = m)
(c) What is the pressure on the wall?
(d) In part (a) above, what is the number of molecules per unit volume that produce the
bombardment of the wall?
(e) In a closed container which has a volume V and contains a total of N molecules the
number of molecules per unit volume
is N/V. Make the correct substitution into your answer of part (c) and find an
expression for the pressure of a gas in terms of
N and V as well as m and v.
What Data Are Given?
Ther are n molecules of mass m travelling with velocity v in the positive x direction that
strike an area A of the wall
perpendicular to the x-axis in a time of t.
What Data Are Implied?
The molecules are perfectly elastic and obey the laws of classical physics during the
motion and collisions.
What Physics Principles Are Involved?
The fundamental laws of motion and collisions will be needed.
What Equations Are to be Used?
distance = velocity ¥ time
impulse = change in momentum = force ¥ time
pressure = force / area
Algebraic Solution
(a) In a time t, the molecules moving with a velocity v will travel a distance of vt. So the
n molecules that hit the wall in time
t had to be within a distance vt of the wall.
(b) For each molecule p = -mv = (mv) = -2mv
Impulse of wall on molecule = -2mv
Impulse of a molecule on wall = +2mv
Total impulse on the wall = n(2mv) = 2nmv
(c) Pressure = force / area = (Impulse/time) / area = (2nmv) / (At)
(d) Number per unit volume = number / volume = (rn) / (Avt) = rn
then pressure = 2rnmv2 (1)
(e) In a closed container only 1/6 of the molecules will have a positive x component of
velocity since there are six equivalent
orthogonal directions, +x, -x, +y, -y, +z, -z. So the r in Equation (1) is only 1/6 N/V;
thus
pressure = 2 (1/6 x N/V)mv2 = 1/3 ((Nmv2) / (v)) (2)
2. By the middle of the nineteenth century, the ideal gas law was written as
PV = nRT (3)
where the three variables are the pressure P (dynes/cm2), the volume V(cm3), and the
temperature T (øK). The quantity n is
the number of moles of gas in the sample (dimensionless) and R is a universal constant
simply called the gas constant. The
experimental value of the gas constant R is
R = 8.31 x 107 ergs/øK where an erg = 1 dyne ¥ cm
Later work by Avogadro led to a re-expression of the same law as
PV = NkT
where the quantity N is the total number of molecules in the sample of gas and k is
Boltzmann's constant. The experimental
value for k is
k = 1.38 x 10-16 ergs/øK
(a) Derive an expression for N in terms of k, n, and R.
(b) Avogadro's number, No, is the number of molecules in a mole (N/n). From the
above data, determine the numerical value
of No.
Thinking About This Problem
The problem does not involve any profound knowledge of physics; rather straightforward manipulation of Equations (3) and (4)
is all that is necessary.
pV = nRT = NkT
(a) so N = nR/k
(b) Avogadro's number No = N/n = R/k = (8.31 x 107 ergs/øK)/(1.38 x 10-16 ergs/øK)
= 6.02 x 1023
PRACTICE TEST
Keywords: Questions; Problems; Answers; Evaluations; Fluids; Fluid Statics; Surface
Tension; Adhesion; Cohesion; Molecular
Model Of Liquids
1. In an interesting physics demonstration, a young physics teacher placed a thin film of
water between two flat panes of
glass. The result is a bond which holds the two pieces of glass firmly together. Is this an
example of adhesion or cohesion?
Explain your answer as if you were teaching the class.
2. The rms speed of oxygen (O2) molecules at 27øC is 482 m/sec. Assume that this is an
ideal gas for the following
questions. (The Boltzmann constant is 1.38 x 10-23 J/molecule- øk)
a. Find the mass of the oxygen molecule in kilograms.
b. Calculate the energy of an individual oxygen molecule.
c. If the energy is to be reduced by a factor of 3, what reduction in temperature is
required?
3. The surface tension of an unknown liquid is measured at room temperature with the
apparatus shown below. First the
system is balanced using a small counter balance m. After the ring of radius r (10 cm) is
placed in the unknown liquid, an
additional mass of Dm must be added to pull the circular wire from the liquid. If m =
50.2 grams and Dm = 18.8 grams,
calculate the surface tension of the liquid at room temperature. (Express your answer in
dynes/cm.)
See Fig.
ANSWERS:
1. Adhesion. Water molecules adhere (are attracted) to glass molecules. The forces
between individual water molecules are
cohesive forces.
2. 5.34 x 10-26 kg, 6.2 x 10-21 Joule, Reduce temperature by 200øK.
3. 7.1 dynes/cm
Chapter
Simple Harmonic Motion (15):
Citation: H. Q Fuller, R. M. Fuller and R. G. Fuller, to Accompany Physics Including
Human Applications.
(Harper and Row, New York, 1978). Permission granted by the authors
15 Simple Harmonic Motion
Keywords: Learning Objectives; ; Waves And Sound; Oscillations; Simple Harmonic
Motion; Oscillatory Motion;
Period; Amplitude; Phase; Damped Oscillations; Frequency; Properties Of Oscillations
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
period frequency
simple harmonic motion restoring force
amplitude damping
phase angle
UCM and SHM
Correlate uniform circular motion and simple harmonic motion.
SHM Problems
Solve problems involving simple harmonic motion.
Energy Transformations
Analyze the transfer of energy in simple harmonic motion.
Superposition
Explain the application of the principle of superposition to simple harmonic motion.
Natural Frequencies
Calculate the natural frequencies of solids from their elastic moduli and density values.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 5, Energy,
Chapter 7, Rotational Motion, and
Chapter 13, Elastic Properties of Materials.
OVERVIEW
Keywords: Instructions; ; Waves And Sound; Oscillations; Properties Of Oscillations;
Oscillatory Motion
Examples of repeated motion are all around us. Examples include a child in a swing
and the up and down motion of a fish
bobber. Most repeated motions are either uniform, simple harmonic motion, or reduce
to nearly this type of motion for small
amplitudes. In this chapter you will learn how repeated motion of this type is treated
mathematically.
SUGGESTED STUDY PROCEDURE
Begin your study by carefully reading the following Chapter Goals: Definitions, UCM
and SHM, SHM Problems, and Energy
Transformations. For an expanded discussion of the terms listed under the goal of
Definitions, see the Definitions section of
this chapter.
Now, read text sections 15.1-15.4 looking carefully at the examples given. The
mathematics in Section 15.4 become
somewhat difficult, but the use of Figure 15.5 will help you visualize the meaning of the
equations. Remember to look in the
second section of this for the answers to all questions posed throughout the reading
sections.
At the end of the chapter, read the Chapter Summary and complete Summary Exercises
1-14. Now do Algorithmic Problems
1-6 and check your answers carefully. Next, complete Exercises and Problems 2, 5, 7, 12,
and 21. For additional work on the
concepts presented in this chapter, turn to the Examples section of this . Now you
should be prepared to attempt
the Practice Test provided at the end of this chapter. If you have difficulties with any
of the problems, please
refer to that particular part of the text or to the sections of this chapter. This study
procedure is outlined below.
------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
------------------Definitions 15.1 1-4
UCM and SHM 15.2,15.3 5-14 4 2
SHM Problems 15.2,15.3,15.4 1-3,5,6 5,7,12,21
------------------------------------------------Superposition 15.6 15,16 9
Natural 15.5,15.6 17,18 8
Frequencies
DEFINITIONS
Keywords: ; Glossary; Waves And Sound; Oscillatory; Properties Of Oscillations;
Oscillatory Motion; Period;
Simple Harmonic Motion; Amplitude; Phase; Frequency; Damped Oscillations
PERIOD
The time required for a system to go through one complete cycle of motion and usually
measured in seconds.
The period of motion of a second hand on a clock is 60 seconds as it takes 60 seconds for
a second hand to go completely
around the face of a watch. The period of the hour hand is 12 hours, since it makes two
complete cycles in 24 hours.
SIMPLE HARMONIC MOTION
If the restoring force is proportional to the displacement and oppositely directed, the
system will execute SHM when displaced
from equilibrium.
The motion of a simple pendulum through small angles is a very familiar example of a
system that executes simple harmonic
motion.
AMPLITUDE
Maximum displacement from equlibrium position.
The amplitude of motion of a pendulum is the maximum angle from the vertical to
which the pendulum swings.
PHASE ANGLE
Angular displacement at the starting time.
It is common for us to start a SHM system moving by displacing it from its rest position.
The initial location determines the
value of the phase angle.
FREQUENCY
Number of complete oscillations per second.
If you have a SHM system that requires 8 seconds to make a complete cycle, then that
system goes 1/8th of a cycle in one
second. We say its frequency is 1/8 cycles per second or 1/8 hertz.
RESTORING FORCE
Force acting to return a displaced body to equilibrium position.
DAMPING
A loss of energy during vibration because of friction. It shows its pressure by a decrease
in the amplitude of vibration. In all
real, isolated oscillating systems the friction forces finally bring the system to rest in its
position of lowest energy.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Waves And Sound; Oscillations; Oscillatory Motion; Simple
Harmonic Motion
SECTION 15.1 Introduction
The universality of simple harmonic motion, the simplest kind of periodic motion, is
striking. It has been suggested that the
pervasive existence of simple harmonic motion is one of the miracles provided by the
nature of our universe. A mathematical
derivation can be used to show how the universality of SHM arises (see Taylor's
Theorem in the enrichment section 15.7).
The simplicity of SHM and the completeness with which we can treat SHM, makes it a
favorite model for physicists. It is
widely used to explain atomic phenomena such as the vibration of atoms in solids and
the oscillations of atomic nuclei.
EXAMPLES
Keywords: Worked Examples; ; Waves And Sound; Oscillations; Oscillatory Motion;
Simple Harmonic Motion;
Hooke's Law; Newton's Second Law; Kinetic Energy; Potential Energy; Period;
Frequency; Young's Modulus; Energy;
Resonance; Resonant Frequencies; Properties Of Oscillations; Temperature; Arithmetic
SHM PROBLEMS
1. Of all the possible forces in the universe, there is only one force that results in simple
harmonic motion. (a) What is the
equation for that force? (b) What is this force called? (c) Sketch the magnitude of this
special force IFI as a function of
displacement. (d) Sketch the force F as a function of the displacement. (e) Explain in
words the relationship between this
force and the magnitude and direction of the acceleration of the body upon which it
acts.
What Data Are Given?
Consider an ideal system which undergoes simple harmonic motion in one dimension.
What Data Are Implied? 222 Assume that friction can be neglected and that the
amplitude and period have constant, finite
values.
What Physics Principles Are Involved?
The problem only involves the dynamics of SHM as discussed in Section 15.3.
What Equations Are to be Used?
Newton's Second Law of Motion and Hooke's Law are all that is required for this
problem.
F = ma (4.1)
F (Hooke) = - kx (13.3)
Algebraic Solution
(a) The equation for the force is Hooke's Law (Equation 13.3) Force = -kx where k is a
positive constant and x is the
displacement.
(b) This force is called a Hooke's Law force on a linear restoring force.
(c) See Fig. 15.1
IFI = I-kxI, so
F = kx for x ò 0
F = -kx for x ó 0
(d) F = -kx
See Fig.
(e) Combining Equations (4.1) and (13.3) we obtain the equation of motion for SHM
ma = -kx (1)
so a = -k/mx where k and m are both positive constants. So the acceleration of the body
is opposite in direction to the
displacement. It is directly proportional to the displacement with a proportionality
constant equal to k/m.
2. Draw a graph of the total energy, kinetic energy, and potential energy of a onedimensional system that undergoes simple
harmonic motion.
What Data Are Given?
Consider an ideal system in one dimension, say x, which uses SHM.
What Data Are Implied?
Neglect friction and consider a finite system.
What Physics Principles Are Involved?
The energy concepts discussed in Section 15.4
What Equations Are to be Used?
Potential Energy = 1/2 kx2 (15.12)
Kinetic Energy = 1/2 mv2 = 1/2 m(k/m (A2 - x2)) (15.16)
Total energy = KE + PE = 1/2 kA2 = 1/2 mvmax2 (15.14 & 15.17)
Algebraic Solution
See Fig.
3. A 8 kg monkey grabbed onto an elastic grapevine hanging from a jungle tree. The
vine stretched 16 m. (a) What force was
applied to the vine by the monkey? (b) What is the effective force constant k of the vine,
assuming the applied force did not
stretch it beyond its elastic limit? (c) Write Newton's Second Law of Motion equation for
the motion of the monkey up and
down. (d) What is the frequency in Hertz of the up and down motion? (e) Suppose the
monkey wishes to reach down 2
meters from its equilibrium position to catch a banana. What is the total energy of the
monkey as it reaches the banana with
respect to an assumed total energy of zero at the equilibrium position? (f) What is the
velocity of the monkey as it passes
through the equilibrium position? (g) Calculate the period of oscillation and sketch the
motion of the monkey as a function of
time. Show the units on both the vertical and time axes.
What Data Are Given?
A vertical unstretched vine will stretch down 16 m when an 8 kg mass is hung onto it.
That mass is then displaced 2 m
and allowed to oscillate up and down.
What Data Are Implied?
The equlibrium position, 16 m below the starting position of the monkey, is to be taken
as the zero energy elevation. Ideal
SHM is assumed near the surface of the earth where g = 9.8 m/s2.
What Physics Principles Are Involved?
This problem requires the coordination of the kinematics, dynamics, and energy
relationships of SHM, Sections 15.2, 15.3,
and 15.4.
What Equations Are to be Used?
F = w = mg (4.5)
F = -kx (13.3)
F = 1/2p ¥ (SQR RT)(k/m) (15.8)
Espring = 1/2 kA2 = 1/2 mv2 + 1/2 kx2 (15.14)
PEgravity = mgh (5.5)
Vmax = ñwA (15.17)
T = 2p ¥ (SQR RT)(m/k) = 1/f (15.9)
x = A cos (wt + f) (15.2)
Algebraic Solutions
See Fig.
(a) Force on the vine = mg
(b) Force constant = mg / (elongation of the vine)
(c) ma = -ky where y is the change in elevation from the elongated position of the vine
(d) f = 1/2p ¥ (SQR RT)k/m Hz
(e) Total Energy = 1/2 kA2 - mgA at the bottom
(f) V = ñwA
(g) T = 1/f sec.
Numerical Solutions
(a) Force on the vine = (8 kg)(9.8 m/s2) = 7.8 x 101 N
(b) Force constant = (7.8 x 101 N)(1.6 x 10 1 m) = 4.9 N/m
(c) acceleration = ((-4.9 N/m)/8 kg)y = (0.61 N/kg.m)y
(d) f = 0.12 Hz
(e) E = 1/2 (4.9 N/m)(2 m)2 - (8)(9.8)(2) = -150 Joules
(f) Vmax = ñwA = I2pfA = ñ 2(p)(.12) (2 m) = 0.75 m/s
(g) T = 1/f = 8.0 seconds
See Fig.
Thinking About the Answer
Notice how the need to also consider the gravitational potential energy changes this
problem. In other kinds of SHM problems
negative energies would not be possible. Consider shifting the zero in gravitational
potential energy to the bottom of the
motion - what is the energy at the equilibrium position then?
NATURAL FREQUENCIES
4. A bell manufacturer constructed a copper bell with a resonance frequency of 440 Hz.
He found that if he made it of lead (Y
lead = 1.7 x 1010N/m2) it would have the same resonance frequency only near liquid
nitrogen temperature (~73øK). Estimate
the temperature coefficient for Young's modulus for lead and the room temperature
resonance frequency for the lead bell.
What Data Are Given?
Ylead = 1.7 x 1010 N/m2 at normal temperatures. Ycopper = 12 x 1010 N/m2 at normal
temperatures (from Table 13.1), copper
normal temperature ÷ 20øC ÷ 293øK; LN2T ÷ 73øK
What Data Are Implied?
The densities of the materials are needed to determine natural frequencies. From Table
8.1 density of lead = 11.3 x 103 kg/m
3; density of copper = 8.9 x 103kg/m3.
What Physics Principles Are Involved?
The dependence of the natural frequency of a solid is given by Equation 15.18. In this
problem let us consider only the
compressional waves in the materials so that the natural frequency depends upon the
square root of the ratio of Young's
modulus to the density.
The temperature coefficient can be found by calculating the Young's modulus value at
73øK; then find the ratio
temperature coefficient = change in Y's modulus / change in temperature
What Equations Are to be Used?
f0 = (1/2p) ¥ (SQR RT)(Young's Modulus / ((density)(area)) (15.19)
temperature coefficient = change in Young's Modulus / change in temperature (2)
Algebraic Solutions
Since all the dimensions of the two bells are the same, the ratio of the natural
frequencies will only depend upon the ratio of
the square roots of Young's modulus to density.
fPb/fcu = (SQR RT)[(YPb/rPb)/(Ycu/rcu)] = (SQR RT)[(YPb ¥ rcu)/ (rPb ¥ Ycu)] (3)
temperature coefficient = (YR. T. - YLN2T)/R. T. - LN2T (4)
Numerical Solutions
fPb = (SQR RT)[((1.7 x 1010) ¥ (8.9 x 103))/ ((11.3 x 103) x (12 x 1010))]
To have the same frequency the ratio of Young's modulus to density must be the same,
so
Young's Modulus Pb at 73øK / (11.3 x 133 kg/m3) = (12 x 1010 N/m2)/(8.9 x 103
kg/m3)
Young's modulus Pb at 73øK = 15.2 x 1010 N/m2
temperature coefficient = (1.7 x 1010 - 15.2 x 1010) / (293øK - 73øK)
= (-13.5 x 1010 N/m2) / 220øK
temperature coefficient = (-6.3 x 108 N/m2) / øK
Thinking About the Answer
Notice that the temperature coefficient is negative. As the temperature increases, the
Young's modulus decreases. This is
typical of many materials. They become less elastic as they are heated. Use the
molecular theory of matter to explain this
phenomenon.
PRACTICE TEST
Keywords: Problems; Answers; Evaluations; Graphs; Velocity; Anatomy And
Physiology; Uniform Circular Motion; Velocity;
Period; Displacement; Waves And Sound; ; Oscillations; Oscillatory Motion; Properties
Of Oscillations Energy
1. A patient's heart of effective mass 0.05 kg has a maximum displacement given by Z =
0.02 sin3pt where Z is in meters
and t is in seconds.
a. Sketch a graph of the motion of this heart on the axis provide below.
See Fig.
b. Find the position of the heart at .75 seconds.
c. Give the frequency of oscillation of the heart.
d. What is the total energy of the heart?
e. What is the maximum displacement of the heart?
f. Calculate the maximum velocity of the heart at Z = 0.
g. The projection of uniform circular motion is simple harmonic motion. On the axis
below, specify the requirements of a
mass undergoing uniform circular motion so that the projection of the motion on the Z axis will be exactly the same as the
heart above.
See Fig.
2. In another monitoring session, the patient above is again observed. The position of
the heart is again a periodic function
and is illustrated below.
See Fig.
a. Write the equation of the relation found in this graph.
b. What is the maximum energy of the heart in this situation?
ANSWERS:
1. a. See Fig.
b. .014 meter
c. 3/2 hertz
d. .009 J
e. .02 M
f. .19 m/s g.
g. See Fig.
2. Z = .03 cos 5pt, .19 J
Chapter
Traveling Waves (16):
16 Traveling Waves
Keywords: Learning Objectives; Waves And Sound; Waves; Behavior Of Waves;
Properties Of Waves; Frequency;
Wavelength; Amplitude; Longitudinal Waves; Transverse Waves; Phase; Intensity;
Reflection; Refraction; Superposition
Principle; Interference; Diffraction; Standing Waves; Types Of Waves; Fourier
Expansion; Dispersion
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms as it is used in physics, and use the term in an
operational definition:
frequency refraction
wavelength superposition principle
amplitude interference
longitudinal wave diffraction
transverse wave standing wave
phase Fourier's theorem
intensity dispersion
reflection
Wave Forms
Sketch a longitudinal wave and a transverse wave.
Wave Problems
Solve wave problems involving the relationships that exist between the different
characteristics of waves.
Superposition and Fourier's Theorem
Use the superposition principle and Fourier's theorem to explain the wave form of a
complex wave.
Standing Waves
Use the superposition principle to explain the formation of standing waves in different
situations.
Inverse Square Law
Use the inverse square law to calculate the intensity of a wave emanating from a point
source.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 5, Energy,
Chapter 13, Elastic Properties of
Materials, and Chapter 15, Simple Harmonic Motion.
OVERVIEW
Keywords: Waves And Sound; Waves; Instructions;
The motion of waves is a fascinating topic. Whenever you see a wave of water move
along the surface of a quiet lake,
you are observing the movement of energy. Most of the cases of energy transported by
waves cannot be seen but are
detected through our other senses. Examples include both sound and light. This chapter
will expand your knowledge of
simple harmonic motion (Chapter 15) to include a description of wave motion. As you
look over the chapter you should
note that in general waves can be classified as Transverse or Longitudinal depending
upon their mode of vibrating.
SUGGESTED STUDY PROCEDURE
When you begin your study of this chapter, please concentrate your attention on the
following Chapter Goals: Definitions,
Wave Problems, and Inverse Square Law. For an expanded discussion of each of the
terms listed under Definitions, see
theDefinitions section of this chapter. Now, read text sections 16.1-16.15. Be sure to
note that the
description of wave motion is included in Sections 16.4-16.6. The remaining parts of the
Chapter describe many of the
properties of waves. Answers to the questions you encounter in your reading can be
found in the second section of this
chapter.
Next, read the Chapter Summary and complete Summary Exercises 1-6 and 8. Check
your answers carefully. Now, do
Algorithmic Problems 1, 3, 5, and 9, and Exercises and Problems 1, 2, 4, 5, 6, and 13. For
additional work with the
important concepts introduced in this chapter, turn to the Examples section of this
chapter. Now you should
be prepared to attempt the Practice Test included at the end of this chapter. Be sure to
check your answers
after you have considered the problems posed. Refer to the text section or to this
chapter for additional help
if needed.
-----------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
-------------------Definitions 16.1-16.3, 1-3,4,6,
16.7-16.13 7
Wave Problems 16.3-16.6,16.14 5 1,3,5 1,2,4,5,
6
Inverse Square 16.15 8 9 13
Law
--------------------------------------------------Wave Forms 16.4,16.8, 4 5,6
16.11,16.12
Superposition 16.8,16.12 6 9,10
and Fourier's
Theorem
Standing Waves 16.11 7 16
DEFINITIONS
Keywords: ; Glossary; Waves And Sound; Waves; Behavior Of Waves; Types Of
Waves; Properties Of
Waves; Frequency; Wavelength; Amplitude; Longitudinal Waves; Transverse Waves;
Phase; Intensity; Reflection;
Refraction; Superposition Principle; Interference; Diffraction; Standing Waves;
Dispersion; Fourier Expansion
FREQUENCY
The number of complete oscillations made by a wave in one second is called its
frequency. Frequency is measured in
complete oscillations, or cycles, per seconds. The unit is named hertz, one hertz means
one cycle per second.
WAVELENGTH
The distance a wave moves during one time period.
As waves travel out from the place a pebble hits still water the distance between
successive peaks is one wavelength.
AMPLITUDE
The maximum displacement of the oscillating system from its equilibrium position.
Large ocean waves are those of the greatest amplitude.
LONGITUDINAL WAVES
The individual particle vibrates in a direction parallel to the direction of propagation of
the wave.
The shaking back and forth in the direction of the stretching of a Slinky toy creates
longitudinal waves along the Slinky.
TRANSVERSE WAVE
The plane of vibration is perpendicular to the direction of propagation of the wave.
The shaking up and down of the end of a Slinky in a plane perpendicular to the length
of the toy creates transverse
waves that travel along the toy.
PHASE
The starting position of a wave with respect to the equilibrium position.
Two locations of a traveling wave that differ in phase by 180ø must be one-half of a
wavelength apart.
INTENSITY
Energy transported through a unit area in one second.
If you make the amplitude of a traveling wave twice as large you make its intensity four
times as large.
REFLECTION
A phenomenon which occurs at the interface between two media and the wave from the
interface is in the same media as
the incident wave.
An echo is a reflected sound wave that comes back to your ear some time after it first
fell on your ear.
REFRACTION
A phenomenon which occurs at the interface between two media and the direction of
the wave in the second medium is
different than in the incident medium.
When a water wave passes through a region of abrupt change in depth from shallow
water to deep water, its direction of
propagation is changed.
SUPERPOSITION PRINCIPLE
The resultant effect is equal to the sum of the individual independent effects. A
principle that holds true for linear
systems.
Most complex systems in physics are analyzed into simpler linear systems. Then the
resultant is the sum of the
individual parts. A free body diagram is an example of the superposition concept
applied to a problem of mechanical
equilibrium.
INTERFERENCE
The superposition of two or more waves with constant phase differences produces
interference.
Two waves that interfere constructively will have a maximum amplitude greater than
either one alone. Two waves that
interfere destructively will have a reduced maximum amplitude.
DIFFRACTION
Occurs if a wave front encounters an object and there results a superposition of wave
fronts.
The bending of sound waves around corners is a result of the diffraction of sound
waves.
STANDING WAVES
These result from the superposition of two traveling waves of equal frequency moving
in opposite directions in the
medium and 180ø out of phase. For a given physical system standing waves result at its
resonance frequency.
The sounds that originate from the vibrating strings of a guitar represent the standing
waves of the strings.
FOURIER'S THEOREM
Any wave form may be produced by the superposition of sine waves of specific
wavelengths and amplitudes.
DISPERSION
The speed of a traveling wave may be different for different wavelengths of the wave.
This difference results in the spread
of the various wavelengths of a complex traveling wave as it passes through various
media.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: Answers; Waves And Sound; Waves; Behavior Of Waves; Reflection;
Refraction;
SECTION 16.7 Reflection and Refraction
Most of our common experiences with reflection and refraction occur with light. Your
use of a mirror to check on your
appearance is a use of reflection. Have you ever noticed how shallow a very clear lake
appears to be? That deception
results from refraction.
If the energy reaching a surface is to be conserved, then the incident energy must equal
the energy leaving the surface.
Since both reflected and transmitted waves are leaving the surface, then
Incident Energy = Reflected Energy + Transmitted Energy.
Divide both sides by the incident energy and multiply by 100 to convert to percentage,
then
100% = % Reflected Energy + % Transmitted Energy.
SECTION 16.13 Dispersion
If air were a dispersive medium for sound, then the different frequencies, or pitches, of
sounds would travel at different
speeds. Suppose, for example, that low frequency waves, bass notes, travelled faster
than high frequency waves,
soprano notes. Then students seated at the back of a large lecture hall would hear
different sounding words than the
students seated near the front!
EXAMPLES
Keywords: Worked Examples; Waves And Sound; Waves; Velocity; Amplitude;
Frequency; Properties Of Waves;
Arithmetic; Sound; Intensity Of Sound; Properties Of Sound; Power; Units; Watts;
Functions
WAVE PROBLEMS
1. The equation for a travelling wave y(m) when x is in meters and t is in seconds is
given by
y = 10 cos(4x + 200t).
(a) What is the amplitude of this wave?
(b) What is its wavelength?
(c) What is its frequency?
(d) What is its velocity?
(e) Sketch the wave at t = 0.
(f) Sketch the wave at t = p/800 sec.
(g) In which direction is the wave moving?
What Data Are Given?
The standard travelling wave equation with numerical values is given.
What Data Are Implied?
The wave is transverse with amplitude in the y - direction and motion along the x - axis.
What Physics Principles Are Involved?
The basic definitions of the various aspects of travelling waves as presented in Sections
16.4 and 16.5 are needed.
What Equations Are to be Used?
y = A sin 2pft - (2px / l) (16.4)
vwave = lf (16.2)
Algebraic Solution
The problem involves primarily the recognition of the various terms in Equation 16.4
and identifying their values from the
numbers given in the problem. Once the wavelength l and the frequency f have been
identified from Equation 16.4, then
they can be multiplied together to obtain the wave velocity as shown in Equation 16.2.
Numerical Solution
y(m) = 10 cos (4x + 200t) and y = A sin (2pft - (2px / l)) (16.4)
(a) So the amplitude of the wave A = 10 meters.
(b) The wavelength of the wave, 2p / l = 4; sol = p/2 meters
(c) The frequency of the wave, 2pf = 200; so f = 100/p Hz.
(d) The wave velocity v = lf = (p/2)(100/p) = 50 m/s.
See Fig.
(g) The peaks of the wave have moved to the left in p/800 seconds, so the wave is
moving in the negative x - direction.
The trans ocean swimmer named Dave, Got caught by a traveling
wave. The waves w, Nearby Bodega, Did cause his watery
grave.
See Fig.
INVERSE SQUARE LAW
2. A sound wave of pressure amplitude 3 x 10-2 N/m 2 is eminating from an 80 watt hifi speaker. (a) What is the
amplitude of the sound wave at a distance of 3 meters from the speaker? (b) What is the
intensity of the sound 3 meters
from the speaker? (c) What is the amplitude of the sound wave at a distance of 10 m
from the speaker? (d) What is the
intensity of the sound at a distance of 10 m from the speaker?
What Data Are Given?
The pressure amplitude is given as 3 x 10-2 N/m 2. The power output of the hi-fi
speaker is given as 80 watts.
What Data Are Implied?
It is implied that the sound wave eminates in a spherically uniform way and that all of
the power output of the speaker is
carried away by the sound wave.
What Physics Principles Are Involved?
The definition of amplitude and the inverse square law are needed for this problem.
What Equations Are to be Used?
Intensity = E / (4pr2t) (16.11)
Algebraic Solutions
Amplitude is independent of distance from the source for ideal, undamped systems.
Intensity = (E/t) / (4pr2) = Power / (4pr2)
Numerical Solutions
(a), (c) Amplitude = 3 x 10-2 N/m2 at all distances.
(b) Intensity3 = 80 W / (4p(3 m)2) = 80 W / 36pm2 = 0.707 W/m2
(d) Intensity10 = 80 W / (4p(10 m)2) = 80 W / 400p = 0.064 W/m2
Thinking About the Answer
The answer to (d) should be [(3)2 / (10)2] times the answer to (b); dividing (d) by (b) we
obtain 0.090 which is equal to 32
/ 102.
PRACTICE TEST
Keywords: ; Problems; Questions; Answers; Evaluations; Amplitude; Frequency;
Velocity; Reflection; Phase;
Sound; Intensity Of Sound; Functions; Waves; Waves And Sound; Properties Of Waves;
Behavior Of Waves
1. The motion of a wave moving along a string is described by the equation y(m) = 10
cos (4.0 x - 200t) where x is in
meters and t is in seconds. For this wave find the following:
a. Amplitude =
b. Wavelength =
c. Frequency =
d. Velocity =
e. In which direction is the wave moving?
2. If the positive pulse shown below is sent down a taut string, describe the pulse as
itreflects from a rigid post.
See Fig.
3. A large bell is located in a tall church tower is rung for several minutes. Local physics
students are measuring sound
levels in the open field nearby. Their measurements show that at 100 meters, the
intensity of the ringing bell sound is 10
-6 watts/m2. What would be the intensity of sound received by a person standing only
3 meters from the bell tower?
See Fig.
ANSWERS:
1. (a) 10 meter, (b) 1.6 meter (c) 32/sec, (d) f = 51 m/sec, (e) left to right
2. 180ø phase change
3. 1.1 x 10-3 watts / m2
Chapter
Sound and the Human Ear (17): to Accompany Physics Including Human
Applications:
17 Sound and the Human Ear
Keywords: ; Learning Objectives; Waves And Sound; Sound; Properties Of Sound;
Intensity Of Sound; Amplitude;
Frequency; Wavelength; Acoustical Impedance; Infrasonic; Ultrasonic; Standing Waves;
Beats; Doppler Effect; Waves;
Properties Of Waves; Behavior Of Waves; Distortion
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms and use it in an operational definition:
sound wave infrasonic
intensity of sound wave ultrasonic
amplitude distortion
frequency standing wave
wavelength beats
acoustic impedance Doppler effect
decibel
Sound Wave Form
Graph the amplitude of a sound wave in pressure versus space or time coordinates.
Sound Level
Calculate the sound level in decibels (dB).
Sound Intensity
Solve sound intensity and sound amplitude problems.
Sound Application
Explain the use of sound in selected medical applications.
Hearing System
Describe the properties of the human sound detection system.
Doppler Effect
Solve Doppler-effect and standing-wave problems.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 5, Energy,
Chapter 15, Simple
Harmonic Motion, and Chapter 16, Traveling Waves, respectively.
OVERVIEW
Keywords: Waves And Sound; Waves; ; Instructions
Our bodies have a delicate sound receiving apparatus, the human ear. Each day we
utilize our sense of hearing to go about
our regular routine. In this chapter you will discover more about the nature of both the
sound which carries information to the
ear and about the ear as a sound transducer.
SUGGESTED STUDY PROCEDURE
As you begin to study this chapter, be familiar with the following Chapter Goals:
Definitions, Sound Level, Sound Intensity,
Sound Applications, and Hearing System. For an expanded discussion of the terms
listed under Definitions, turn to the
Definitions section of this chapter.
Next, read text sections 17.1-17.9 and 17.11-17.13. Answers to the questions asked in the
text readings can be found in the
second section of this chapter. Be sure to note one of the important ideas introduced in
section 17.4 is the
development of a logarithmic scale for describing sound intensity. If you have
difficulties using the logarithmic form
suggested, you should refer to chapter 0 of this and to the Appendix, Section A.8.
At the end of the chapter, read the Chapter Summary and complete Summary Exercises
1-8, 12, and 13. Check your answers
carefully against those provided. Now do Algorithmic Problem 1 and Exercises and
Problems 3-7 and 14. Be careful to check
your answers carefully. For assistance, refer back to the appropriate text section. For
additional exercises with the concepts
presented in this chapter, refer to the Examples section of this chapter.
Now you should be prepared to attempt the Practice Test on Sound and the Human Ear
provided at the end of this Study
Guide chapter. If you are unsuccessful with any of the questions, you should refer back
to the text or to this
chapter for assistance. This study procedure is outlined below.
------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
------------------Definitions 17.1,17.2,17.5- 1-8
17.9,17.11-17.13
Sound Level 17.4 1 5,6,7,14
Sound Intensity 17.3 3,4
The Hearing 17.12 13
System
Sound 17.3,17.5,17.13 12
Applications
--------------------------------------------------Sound Wave Form 17.2 9-11 1,2
Doppler Effect 17.10 14,15 2,3,4 8,9,15
DEFINITIONS
Keywords: Glossary; ; Waves And Sound; Sound; Properties Of Sound; Intensity Of
Sound; Amplitude;
Frequency; Wavelength; Acoustical Impedance; Infrasonic; Ultrasonic; Standing Waves;
Beats; Doppler Effect; Waves;
Properties Of Waves; Behavior Of Waves; Distortion
SOUND WAVE
A longitudinal pressure wave which travels through matter.
This definition answers the question "Is there sound if no human is around to hear it?"
What is the answer, according to this
definition?
INTENSITY OF SOUND WAVES
The energy per unit area per second transmitted by sound waves.
The usual intensity units are watts / (meter)2, and typical sounds, such as normal
conversation have an intensity of about 1 x
10-6 W/m 2.
AMPLITUDE
The maximum pressure obtained during the periodic variations of pressure in a sound
wave is the amplitude.
Typical pressure amplitudes of sound waves are of the order of 10-2 N/m2. This can be
converted to an equivalent
displacement amplitude using Equations 17.3 and 17.2,
A = Po / Wrvs
For typical values, then A is of the order of 10-9 m.
FREQUENCY
The number of complete pressure oscillations that occur in a sound wave in one second.
Common audible sounds have frequencies in the tens and hundreds of oscillations per
second. The dominate frequency of
middle C on a piano is about 250 Hz; so the third C above middle C has a frequency of
about 2000 Hz and the third C below
middle C has a frequency of about 30 Hz.
WAVELENGTH
When we picture sound as a spatial distribution of periodic pressure oscillations then
we can measure the distance between
adjacent regions of equal pressure to find the wavelength of the sound wave.
Typical (audible) sounds have wavelengths in the range from about 10 meters (35 Hz)
to about 1/10 meter (3500 Hz).
ACOUSTIC IMPEDANCE
The inertia of a medium in its response to pressure oscillations is called its acoustic
impedance. It is measured in kg/m2 ¥ s.
The acoustic impedance of air is about 102 kg/m2 ¥ s, while the acoustic impedance of
water is about 106 kg/m2 ¥ s.
DECIBEL
The sound level is measured on a logarithmic scale of intensity ratios using a unit called
decibel. The equation used to
calculate the sound level in decibels is Equation 17.6.
Average audible sound levels range from 40 dB in the average room to about 110 dB for
a rock music concert. Human
discomfort is caused by sound levels above 120 dB.
INFRASONIC
Sounds which have frequencies below the usual human auditory limit of 20 Hz are
called infrasonic sounds.
The effects of infrasonic sounds on living systems is not well understood.
ULTRASONIC
Sounds which have frequencies above the usual human auditory limit of 20,000 Hz are
called ultrasonic sounds.
Much human tissue is transparent to ultrasonic waves. These waves can be used to
observe internal features of the human
body.
DISTORTION
A sound wave is distorted by any media which do not obey Hooke's Law.
If the output of a sound amplifier system is not proportional to the input to the system
for all frequencies, then the output
from the system will be distorted. Many electronically amplified music groups use
distortions caused by amplifying high
intensity sounds to create their unique sounds.
STANDING WAVE
The superposition of traveling waves of the same wavelength and opposite velocities
results in waves that do NOT appear to
move.
Typical guitar string vibrations are standing waves.
BEATS
If two pure sinusoidal sound waves of nearly equal frequencies occur in the same
region of space, then they may be heard as
a number of periodic fluctuations in the sound intensities called beats.
If you excite two tuning forks, one of frequency 440 Hz and the other of 444 Hz, you can
hear periodic intensity fluctuations
of the sounds with a frequency of 4 Hz.
DOPPLER EFFECT
Moving sources of sound waves and moving receivers of sound waves will cause
apparent shifts in the frequencies of the
observed sounds.
You may have noticed the apparent change in pitch of a train whistle or automobile
siren as it went passed you. Can you
recall how the pitch changed?
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Waves And Sound; Waves; Behavior Of Waves; Diffraction; Behavior Of
Sound; Sound; Infrasonic;
Ultrasonic; Frequency
SECTION 17.1 Introduction
Sound originates from vibrating objects. Sound travels around objects of typical sizes by
diffraction. The vibrating ruler on
your desk can be used to produce different frequencies of sounds by changing the
length of the ruler that extends beyond the
edge of the desk.
From the definition of sound given above you will notice that the presence of a human
observer is not required for sound to
exist.
SECTION 17.5 Infrasonics and Ultrasonics
Ultrasound readily penetrates much human tissue with little loss of energy. Hence
ultrasonic waves can pass through tissue
without damage. When focused into a small volume of absorptive tissue the ultrasonic
energy can be deposited to produce
changes in the tissue.
Ultrasound is a high frequency pressure wave. This wave causes high frequency (~
40,000 Hz) oscillations in the molecules
in tissue. One model would be to assume that the high frequency pressure waves cause
changes in the chemical structure of
tissue, thereby changing the biological properties of the tissue.
EXAMPLES
Keywords: ; Worked Examples; Intensity Of Sound; Doppler Effect; Watts; Units;
Frequency; Power; Functions;
Arithmetic; Waves And Sound; Sound; Behavior Of Sound; Properties Of Sound
SOUND LEVEL AND SOUND INTENSITY
1. A railroad train has a whistle that emits an isotropic sound of 300 watts. The train is
approaching you along a straight track
that passes within 5 meters of where you are standing. The train whistle begins to blow
when the train is 100 meters from you
and continues to blow until the train is passed its nearest position to you. What is the
lowest sound level of the whistle that
you hear? What is the highest level of the whistle that you hear?
What Data Are Given?
Whistle power = 300 watts; farthest distance = 100 m; nearest distance = 5 m
What Data Are Implied?
The inverse square law can be used since the sound is equally intense in all directions.
What Physics Principles Are Involved?
The definition of sound level and the inverse square law.
What Equations Are to be Used?
dB = 10 log (I/Io) (17.6)
I = (E/t)/(4pr2) (16.11)
Algebraic Solutions
Lowest level occurs when the whistle is a maximum distance from the observer rmax.
dBlowest = 10 log (Ilowest / Io) where
Ilowest = Power / (4pr2max)
Highest level occurs when the whistle is the minimum distance from the observer, rmin.
dBmax = 10 log (Imax / Io) where
Imax = Power / (4pr2min)
Io = 10-12 W/m2
Numerical Solution
Ilowest = 300W / (4p(100)2) =
2.4 x 10-3 W/m2; dBlowest = 94 db
Imax = 9.5 x 10-1 W/m2;
dBmax = 120 dB
Thinking About the Answer
Notice that the ratio of the distances is 20 to 1, so the ratio of intensities is 1 to 400, yet
the dB value only changes from 94
dB to 120 dB.
DOPPLER EFFECT
2. The operator of the world's largest ferris wheel in Vienna, Austria, with a radius of 30
meters needs to determine how fast
his ferris wheel turns at its maximum speed. Since the operator has perfect pitch and his
assistant has taken physics, they
decide to utilize the Doppler shift for making this measurement.
The assistant rides the wheel and blows a concert A on a pitch pipe. As the wheel
reaches its maximum velocity, the
operator stands near the bottom and listens for the frequency while the operator is
approaching. The data collected is as
follows; observed frequency of approach is 444 vibrations/sec, velocity of sound is 345
meters/sec. Utilize the information to
find the tangential velocity of ferris wheel.
What Data Are Given?
The radius of the wheel = 30 meters, the observed frequency of approach = 444
vibrations/sec., at rest frequency = 440
vibrations/sec., and the velocity of sound = 345 m/sec.
What Data Are Implied?
This problem implies that the frequency difference (moving - at rest) can be detected
with enough accuracy that the Doppler
relationship can be utilized to solve for the velocity of the source.
What Physics Principles Are Involved?
The Doppler effect relationship for source moving, observer at rest.
What Equations Are to be Used?
f1 = (f(vs + vo)) / (vs - v) (17.11)
This is a general equation for relative approach. In our problem, the observer's velocity,
vo = 0. Therefore, equation 17.11 is
reduced in this case to the following:
f1 = (f(vs)) / (vs - v)
Algebraic Solution
Since we are attempting to solve for v, the velocity of the source, we must rearrange the
relationship above solving for v.
Please note that vs is the speed of sound. Then,
f1(vs - v) = f(vs)
f1vs - f1v = f(vs)
f1v = f1(vs) - f(vs)
v = (f1(vs) - f(vs)) / f1
= vs (1 - (f/f1))
Numerical Solution
v = 345 (1 - 440 / 444)
v = 2.8 m/s
See Fig.
Thinking About the Answer
Note that the answer seems reasonable; if the moving frequency was closer to 440
vibrations/sec, the calculated velocity
would be closer to zero. If the observed frequency was much higher than 440
vibrations/sec, the calculated velocity would be
increased as expected. Can you predict what frequency the operator would hear when
the assistant was moving away from
him?
See Fig.
PRACTICE TEST
Keywords: ; Problems; Questions; Evaluations; Answers; Intensity Of Sound; Power;
Watts; Units; Ear; Sensory
Systems; Anatomy And Physiology; Energy; Transducers; Feedback; Acoustical
Impedance; Waves And Sound; Sound;
Ultrasonic; Frequency; Properties Of Sound
1. The sound intensity measured in a large factory is .10 watts/m2
a. What is the level of this sound in decibels?
b. If through the use of sound absorbing material the sound intensity is reduced to 1/2
its original value, what would be the
new decibel level?
2. The block diagram below represents the various parts of the human ear system.
See Fig.
Identify the parts of the ear labeled A, B, and C above, and describe the form of the
energy input and energy output of each
part (i.e., mechanical, electrical, chemical, etc.)
Part of the Ear Energy Input Energy Output
A.
B.
C.
3. There are several medical applications for Ultrasonic sound.
a. Give the approximate frequency range for ultrasonic waves.
b. Outline the one particular use of ultrasonic waves in neurosonic surgery.
ANSWERS:
1. a. 110 dB b. 107 dB
2. A. Ear drum (outer ear) Sound (Mechanical) Mechanical
B. Bones (middle ear) Mechanical Mechanical
C. Eustachian tube(inner ear) Mechanical Electrical
3. a. Any greater than 20,000 Hz
b. Ultrasonics are focused on a particular part of the brain thus destroying tissue at focal
point without harming other
surrounding tissue
Chapter
Optical Elements (18):
STUDY
GUIDE
Citation: H. Q Fuller, R. M. Fuller and R. G. Fuller, to Accompany Physics Including
Human Applications. (Harper
and Row, New York, 1978). Permission granted by the authors.
18 Optical Elements
Keywords: ; Learning Objectives; Light And Optics; Images; Nature And Propagation
Of Light; Behavior Of Light;
Reflection; Ray Diagrams; Internal Reflection; Refraction; Index Of Refraction; Ray
Model Of Light; Mirrors; Focus; Lenses;
Lens Aberration; Spherical Aberration; Magnification; Optical Instruments; Cameras;
Microscopes; Fiber Optics
GOALS
When you have mastered the content of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms and use it in an operational definition:
light ray optical axis
object distance converging optical elements
image distance diverging optical elements
index of refraction real image
reflection coefficient virtual image
internal reflection magnification
focal point aberrations - chromatic and
spherical
Ray Diagrams
Draw ray diagrams for some common optical systems.
Lens and Mirror Equations
Apply the basic equations for lenses and mirrors to optical systems with one or two
components.
Optical Devices
Explain, using physical principles, the operation of a reading glass, camera, microscope,
and fiber optics.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 16,
Travelling Waves.
OVERVIEW
Keywords: ; Instructions; Light; Light And Optics; Nature And Propagation Of Light;
Optical Components
Light travels rapidly through free space in a straight line. When light rays are deviated
from this straight line motion, either
reflection or refraction is usually involved. In this chapter you will be introduced to the
reflection properties of plane and
curved mirrors plus the refraction properties of single thin lenses and combinations of
lenses. The image producing properties
of each of these phenomena will be outlined with application examples.
SUGGESTED STUDY PROCEDURE
As you begin to study this chapter, be familiar with these Chapter Goals: Definitions,
Ray Diagrams, Lens and Mirror
Equations, and Optical Devices. Remember to look at the next section of this for an
expanded discussion of
each of the terms listed under Definitions. Next, read chapter sections 18.1-18.9.
Remember that answers to questions asked
in the text are answered in thesecond section of this Chapter. As you read, pay special
attention to the many
figures which illustrate the properties of reflection, refraction, and image formation.
Also, be sure to practice the rules given on
page 408 and 412 for drawing ray diagrams for curved mirrors and thin lenses,
respectively. Although Section 18.9 is short,
the information presented about cameras is important. See the examples section of this
for additional
information.
At the end of the chapter, read the Chapter Summary and completeSummary Exercises
1-16. Now do Algorithmic Problems
1-4 and completeExercises and Problems 1, 8, 9, 10 and 12. For additional practice with
the concepts presented in this
chapter, work through the additional problems provided in the Examples section of this
chapter. Now you should
be prepared to attempt the Practice Test given at the end of this chapter. If you are
unsuccessful with any part,
refer to that portion of the text. This study procedure is outlined below.
---------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
---------------------Definitions 18.1,18.4, 1, 2, 3,4, 1,2 1
18.5 5,6,7
Ray Diagrams 18.2,18.3, 8,13,14 8,9
18.6,18.7
Lens & Mirror 18.6,18.7 9,10,11 3 8,9,10,12
Equations
Optical Devices 18.8,18.9 12,15,16 4 13,20
-----------------------------------------------------18.10,18.11 14 5,6 14,15
18.12
DEFINITIONS
Keywords: Glossary; ; Light And Optics; Images; Nature And Propagation Of Light;
Behavior Of Light; Reflection;
Ray Diagrams; Internal Reflection; Refraction; Index Of Refraction; Ray Model Of Light;
Mirrors; Focus; Lenses; Lens
Aberration; Spherical Aberration; Magnification; Optical Instruments; Cameras;
Microscopes; Fiber Optics
LIGHT RAYS
Radii of spherical waves. Imaginary lines drawn in the direction of the light wave
propagation.
These rays are of special help when drawing diagrams for locating images formed by
concave mirrors and convex lenses.
OBJECT DISTANCE
The axial distance from the center of a lens or mirror to the object position.
IMAGE DISTANCE
The axial distance from the center of a lens or mirror to the image position.
INDEX OF REFRACTION
Ratio of the velocity of light in a vacuum to the velocity of light in the refractive
material.
The index number gives a rating of a transparent material's ability to bend light as it
transverses from outside to inside the
material. Diamond has a high ability to bend light (n = 2.4) whereas water has less
ability to bend light (n = 1.33).
REFLECTION COEFFICIENT OF LIGHT
For normal incidence, the fraction of light intensity reflected.
The reflection coefficient depends upon the relative comparison of the index of
refractions of the materials on either side of
the reflection boundary.
INTERNAL REFLECTION (total)
When angle of incidence is equal to or greater than critical angle qc, light is totally
reflected back into incident medium.
This reflection occurs only when light traveling in a medium of high index of refraction
is incident on a boundary where a
medium of lower index of refraction exists.
FOCAL POINT
The location on the axis of a lens or mirror where parallel light converges to a point.
In the special cases of the convex mirror and the concave lens, the focal points are
located at a point on the axis where
parallel rays Seem to the observer to have originated.
OPTICAL AXIS
The line of symmetry through the center of a lens or mirror.
The focal point and the center of curvature point are both located on the optical axis.
CONVERGING OPTICAL ELEMENTS
Optical elements which converge parallel rays of light to a focus and can produce a real
image. Both the concave mirror and
the double convex lens are of this type.
DIVERGING OPTICAL ELEMENTS
Optical elements which diverge parallel rays of light and form virtual images. Both the
convex mirror and the double concave
lens are examples of this type.
REAL IMAGE
An image formed by a mirror or a lens which can be displayed on a screen. Real light
rays actually converge to form this
image.
VIRTUAL IMAGE
An image formed by either a mirror or a lens which can be seen but cannot be displayed
on a screen. These images seem to
exist but real light rays do not converge to produce them.
MAGNIFICATION
The ratio of image size to object size. The ratio of the angle subtended by the image with
lens to an angle subtended by the
object at the near point.
The magnification value may be used to predict how large (or small) an image will
appear in relation to the original object.
CHROMATIC ABERRATION
The effect produced by the dependence of focal length on the frequency (color) of light.
This effect is mainly of concern with lenses. As light of a different color (wavelength) is
bent by the lens, different colors are
bent by different amounts. Thus the focal length will be different for different colors.
SPHERICAL ABERRATION
The effect resulting in different focal lengths for off-axis rays when incident on spherical
lenses or mirror surfaces.
Even for a lens which is perfectly spherical, a true focal point will be difficult to
produce. Since rays bent by the outer edges
of the lens or mirror are bent more than rays along the optical axis, these rays will be
influenced more.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: Light And Optics; Answers; Nature And Propagation Of Light; Magnifying
Glasses; Microscopes; Lenses; Mirrors;
Types Of Mirrors; Refraction; Index Of Refraction; Answers; ; Optical Components;
Cameras; Microscopes;
Reflection; Wavelength; Geometric; Optics
SECTION 18.2 Straight-Line Wave Propagation and Ray Diagrams
Let us draw your attention to the relative wavelengths of sound waves and light waves.
Typical audible sounds have
wavelength of the order of 1 meter. The wavelength of typical visible light is 5 x 10-7 m.
So sound waves are about 10 million
times as long as light waves. Consider the properties of waves discussed in Chapter 16
and propose various phenomena that
might occur because of the great difference in wavelength between sound and light.
SECTION 18.6 Reflection from Spherical Mirrors
You will notice that is only for the case shown in Figure 18.11 (a) do the actual rays pass
through the image and the image is
located on the object side of the mirror. Hence, only the image in (a) is a real image and
can be shown on a screen. For both
(b) and (c) of Figure 18.11, the images are behind the mirror. The rays must be extended
behind the mirror to intersect at the
location of the image. A virtual image cannot be shown on a screen, but must be viewed
through an optical system, such as
the human eye, capable of constructing an image from the rays that appear to come
from the virtual image.
SECTION 18.7 Lenses
The power of a lens to refract and focus rays depends upon the ratio of the index of
refraction of the lens to the index of
refraction of the surrounding medium. If the index of refraction of the media is
increased then the focusing power of the lens
is decreased. The focal length of the lens is lengthened.
SECTION 18.8 Magnifier or Reading Glass
Examples:
1. You can use Equation 18.14 for calculating the magnification of a typical reading glass
for a person with a near point 25
cm from the person's eye.
M = 1 + 25 (cm) / f (cm) = 1 + 25/10 = 3.5
How is the magnification changed if your near point is 15 cm?
2. You can use Equation 18.14 to find the focal length of a lens if you know the
magnification.
M1 = 1 + 25/f; so f = 25/(M-1) = 25/7 = 3.6
An experimental way to determine the focal length of a lens is to measure the distance
from the lens to a real image of a
distance object. Use Equation 18.13 to justify this experimental procedure. How distant
must the object be to determine the
focal length within ñ 10% using this method?
SECTION 18.9 Cameras
The total energy that enters a camera lens and falls on the film is proportional to the
product of the lens area, the light
intensity and the time;
Energy = Area x Intensity x Time
f - number = focal length / diameter of lens aperture
Area = p/4 (diameter)2 =p/4 x (f.1./(f - no.))2
The area is proportional to the inverse square of the f - number.
Table of Values
f - no. 1.4 2.0 2.8 4 5.6 8 11 16 22
103 x inverse 510 250 130 63 32 16 8.3 3.9 2.1
square of f-no.
To keep the total energy constant the product of the area and the exposure time must be
held constant.
Time = constant / (inverse sq. of the f-no.) = constant ¥ (f-no.)2
If an exposure time of 1/100 of a second is correct for an f - number of 1.4, then the
approximate exposure table is as below
f-number 1.4 2.0 2.8 4 5.6 8 11 16 22
exposure .01 .02 .04 .08 .16 .32 .61 1.26 2.42
Hence, the exposure time doubles each f - number, or the energy per unit area is
decreased by 50% each time.
SECTION 18.11 The Compound Microscope
The image you see in a microscope is enlarged, inverted, and virtual. You cannot show
a microscope image on a screen
without adding an additional lens to make a real image on a screen. In order to
photograph a microscope image you need to
use a camera with a focusing lens on it.
EXAMPLES
Keywords: Worked Examples; ; Lenses; Mirrors; Types Of Mirrors; Types Of Lenses;
Thin Lenses; Ray
Diagrams; Magnification; Light And Optics; Focus; Optical Components; Algebra;
Geometric Optics
LENS AND MIRROR EQUATIONS
1. An object is placed a distance equal to twice the focal length in front of (a) a diverging
spherical mirror and (b) a diverging
spherical lens. Locate the image in each case. Characterize the image and compute the
magnification.
What Data Are Given?
Two different optical elements: a diverging (convex) mirror and a diverging (concave)
lens. According to the sign convention
given in the text, both of them will have negative focal lengths, say - f.
What Data Are Implied?
The conditions are assumed to be appropriate for Equations (18.8) and (18.13) to be
used.
What Physics Principles Are Involved?
The basic physics of geometric optics was used to derive the equations for this problem.
What Equations Are to be Used?
1/p + 1/q = 1/f (18.8 or 18.13)
m = -q/p (18.6 or 18.9)
Algebraic Solution
p = 2f; then if f is a positive number
1/2f + 1/q = 1/-f; 1/q = -1/f - 1/2f = (-2 -1)/2f
q = -2f/3; so q is virtual and less than f.
m = -(-2f/3)/2f = 1/3; positive m indicates an upright, virtual image
Ray Diagram Solutions
See Fig.
Thinking About the Answer
Even though these two different optical elements give the same results, notice the
difference in the location of the image. In
one case the image is behind the optical element and in the other case the image is in
front of the optical element. If you
have a chance in a laboratory experiment, compare these two optical elements by using
them to view different objects.
2. Two thin lenses each of focal length +0.20 m are 0.20 m apart. An object is located 0.40
m left of the left lens. Find the
magnification and location of the final image using a ray diagram and appropriate
equations.
What Data Are Given?
The location of the object = 0.40 m. The focal lengths of the two lens = ñ 0.20 m. The
distance between the two lens = 0.20
m.
What Data Are Implied?
The lens equations can be used so the object is appropriate to the sizes of the lenses.
What Physics Principles Are Involved?
The basic principles of geometric optics are used as given in Equations 18.9 and 18.13.
What Equations Are to be Used?
1/p + 1/q = 1/f (18.13)
m = -q/p (18.9)
Algebraic Solution
let f1 = focal length of left lens, f2 = focal length of right lens, d = distance between the
lenses, and p1 = object distance
from left lens
1/q1 = 1/f1 - 1/p1; q1 = (f1p1)/(p1 - f1) (1) where q1 is the location of the first image.
p2 = object distance with respect to the right lens = d - q.
then q2 = (f2p2)/(p2 - f2) = f2(d - d1)/(d - q1 - f2) (2)
Total magnification = -q1/p1 - q2/p2 = (f1/(p1 - f1)) (f2/(d - q1 - f2)) (3)
Numerical Solution
q1 = (0.20)(0.40)/(0.40 - 0.20) = 0.40 m
thus p2 = 0.20 m - 0.40 m = -0.20 m
final image = q2 = ((0.20)(-0.20))/(-0.20 - 0.20) = -0.04/-.40 = 0.1 m to the right of the
right lens
magnification = (0.20/(0.40 - 0.20)) (0.20/(-0.20 - 0.20))
magnification = 1(0.20/-0.40) = -1/2
Ray Diagram
inverted; real: reduced image
See Fig.
See Fig.
It is during the study of light, With
mirrors and lenses and sight, That a person can tell, The old wisdom of
Snell, His sine law makes the rays bend just right.
Thinking About the Answer
For multiple lens or mirror problem it is important to pay careful attention to the sign
convention used to determine which
terms are treated as positive and negative. The use of a ray diagram to check your
answer is a helpful technique.
See Fig.
Oh,
the mirror in Tom's new boutique, Can transform even Zorba the Greek!
From the fat to the thin, From the neck to the shin, It can make you and
me look so sleek!
PRACTICE TEST
Keywords: Problems; Answers; Evaluations; ; Ray Diagrams; Optical Components;
Optical Instruments; Cameras;
Magnification; Images; Mirrors; Light And Optics; Geometric Optics
1. An object (represented by the arrow) is placed on the axis in front of a concave mirror
as shown below. The focal point is
20 cm. from the mirror and the object is placed at 10 cm.
See Fig.
a. Calculate the image distance.
b. Draw the principal ray diagram to locate the image formed by the lens.
c. Describe the image: Is it real or virtual, upright or inverted? Calculate the
magnification of this mirror system.
2. A classroom slide projector contains a converging lens of focal length 10.0 cm. It
projects an image on a screen that is
2.50 m from the lens.
a. What is the distance between the slide and the lens?
b. What is the magnification of the image?
c. Calculate the width of the image of a slide 35 mm wide formed on the screen.
3. Suppose you have a camera with a 10 cm. focal length.
a. What is the diameter of the lens aperture for an F number of 2.8 (f/2.8 setting)?
b. With this lens the proper exposure time for a given scene is 1/100 sec. at the setting
f/2.8. What would be the exposure
time for the same scene at a f/5.6 setting?
ANSWERS:
See Fig.
1. -20 cm; virtual, upright, larger, m = +2
2. 10.4 cm, 24, 84 cm
3. 3.6 cm, .04 s
Chapter
Wave Properties of Light (19):
19 Wave Properties of Light
Keywords: ; Learning Objectives; Light And Optics; Nature And Propagation Of Light;
Interference; Diffraction;
Dispersion; Polarization; Birefringent Materials; Refraction; Double Refraction; Optical
Activity; Fermat's Principle; Coherence;
Lasers; Resolution Of Optical Instruments; Gratings; Holography; Thin Film; Wave-Like
Nature
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
interference optical path length
diffraction optical activity
dispersion coherent
polarization noncoherent
birefringence
Application of Wave Properties
Explain the physical basis for:
thin film colors polarimetry
diffraction grating spectrometry holography
resolving power of optical systems
Problems Involving Wave Properties
Solve problems involving interference, diffraction, and polarization.
Optical Activity
Design an experimental system capable of measuring the optical activity of a solution.
Lasers
Compare the laser with other light sources in terms of their optical characteristics.
PREREQUISITES
Before you begin this chapter you should have achieved the goals of Chapter 16,
Traveling Waves, and
Chapter 18, Optical Elements.
OVERVIEW
Keywords: ; Instructions; Light And Optics; Nature And Propagation Of Light;
Behavior Of Light; Wave-Like
Nature
Under most day-to-day conditions, we see light behave as though it were a wave.
Primarily, these wave phenomena include
interference, diffraction and polarization. In this chapter, you will be introduced to the
quantitative aspects of these phenomena
in addition to many applications of these phenomena.
SUGGESTED STUDY PROCEDURE
When you begin to study this chapter, read the following Chapter Goals carefully:
Definitions, Applications of Wave Properties,
Problems Involving Wave Properties, and Lasers. An expanded discussion of each of
the terms listed under Definitions can
be found in the next section of this . Next, read text sections 19.1-19.5 and 19.7-19.11
and 19.14. Answers to
questions posed in each section are discussed in this chapter, section three.
At the end of the chapter, read the Chapter Summary and complete Summary Exercises
1-8. Next, do Algorithmic Problems
1-6 and Exercises and Problems 1-5, 7, and 17. For more work with the concepts of this
chapter, see the Examples section
of this .
Now you should be prepared to attempt the Practice Test provided at the end of this
chapter. If you have
difficulties with any individual problem, refer to that particular part of the text or this
chapter for assistance. This
study procedure is outlined below.
-------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
-------------------Definitions 19.1
Applications of 19.2,19.3,19.4, 1,2 2 1,2
Wave Properties 19.5;19.7,19.8,
19.9,19.10,19.11,
and 19.14
Problems 19.3,19.5,19.7, 3,4,5 3,5,6 3,4,5,7,
Involving Wave 19.10 6,7,8 17
Properties
Lasers 19.13 10
---------------------------------------------Optical Activity 19.12 9 7
DEFINITIONS
Keywords: ; Glossary; Interference; Diffraction; Dispersion; Polarization; Birefringent
Materials; Optical Activity;
Fermat's Principle; Coherence; Light And Optics; Behavior Of Light; Wave-Like Nature;
Refraction; Double-Refraction
INTERFERENCE
The superposition of two or more waves with constant phase differences produces
interference.
In a case when two waves meet in phase, the interference will be constructive. When the
waves meet out-of-phase, the
interference will be destructive.
DIFFRACTION
Occurs if a wave front encounters an object and there results a superposition of wave
fronts.
Diffraction is the bending of waves around objects. The most common example of
diffraction is the spreading out (bending) of
a wave which must pass through a narrow opening.
DISPERSION
A medium will produce a dispersion of waves if the wave velocity is a function of the
frequency.
As light passes through glass, different wavelengths (frequencies) are slowed by
different amounts. Thus the wavelengths do
not move the same distance in the exact same amount of time.
POLARIZATION
The process by which a light wave loses its random orientation of transverse
propagation and only a single direction of
transverse propagation is allowed.
BIREFRINGENCE
Property of crystals that have different velocity of light for different directions.
This property allows double refraction of light.
OPTICAL PATH LENGTH
Equals thickness of sample multiplied by index of refraction for wavelength in the
sample.
More generally, this is the length from source to screen travelled by a light wave.
OPTICAL ACTIVITY
Phenomenon produced by certain materials that rotate the plane of polarization as light
passes through.
COHERENT
Light emitted from a laser has each wave in phase with the other waves emitted.
NONCOHERENT
Light emitted from an ordinary incandescent bulb is randomly produced, thus the
waves are not in phase.
ANSWER TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Interference; Thin Films; Polarization; Light And Optics;
Behavior Of Light; Nature And
Propagation; Wave-Like Nature; Index Of Refraction
SECTION 19.3 Interference
One condition necessary for interference between waves is that the frequency be the
same for all the waves. Most common
light sources emit a wide range of frequencies of light. So it may not be so obvious to
the casual observer of natural events
that interference occurs. The color patterns of an oil film on the surface of water and of
some insect wings can be attributed
to interference.
Most light sources are polychromatic and noncoherent, so if the light waves are all
present in the same region of space, a
detector will measure the superposition of all the wave amplitudes.
As shown in Figure 19.1, if two equal waves destructively interfere, then the resultant
amplitude is zero. Such a case for two
traveling waves would not hold true for all directions in space. The energy conservation
concept applies to a total system, not
just to some individual waves within the system. If the waves are added up over the
whole system, then energy is conserved.
SECTION 19.4 Effective Optical Path Lengths
Since the index of refraction in a vacuum is by definition exactly one, then the error for
not correcting for the index of
refraction of air is equal to (1.0003 - 1.0000)/1.00 = 33 x 10-4 or 0.03%. That seems like a
very small difference, right? But
consider the number of complete wavelengths of yellow-green light that extend 1 cm in
a vacuum compared to 1 cm of air.
The wavelength of yellow-green light is, say 550.06 nm; then 18,180 wavelengths extend
for 1.00000 centimeters in a
vacuum, but that is 18,185 wavelengths in 1 cm of air, a difference of 5 wavelengths.
SECTION 19.5 Thin-Film Interference Patterns
The wings of insects can be studied using interference techniques. If you have available
a monochromatic source of light
whose wavelength you can vary, then you could study the various interference patterns
for different wavelengths of light.
What results might you expect?
SECTION 19.10 Polarization
We may construct almost any model for the emission of light from natural objects that
we like, but as long as isotropic
randomness of microscopic systems is preserved, as seems to be required by the second
law of thermodynamics,
unpolarized light would be most common from typical emission systems.
One interesting way to study polarization effects in your daily environment is to tip
your head 90ø while wearing Polaroid
sunglasses. With your head in its usual vertical position the plane of polarization of
Polaroid sunglasses is vertical to screen
out the intense horizontally-polarized reflected light. By tipping your head sideways
you permit the horizontally polarized light
to reach your eyes. Some strain patterns in automobile windows are observable with
Polaroid sunglasses.
As shown in Figure 19.13, the fact that Rayleigh scattering is more pronounced for the
short wavelength blue light than for red
light can be used to explain the redness of sunsets and sunrises. The direct light falling
on your eyes from the sun has more
of its blue light scattered away so it appears more red than when directly overhead.
This is a result of the increased distance
the light travels through the earth's atmosphere at sunrise or sunset, see the figure
below.
See Fig. 19.13.
The sunset you see almost daily, Is attributed to Mister Rayleigh.
That reddish hue, Comes from blue, That colors the sky, oh so gaily.
EXAMPLES
Keywords: ; Light And Optics; Wave-Like Nature; Interference; Phase Shifts; Index Of
Refraction; Wavelength;
Diffraction; Gratings; Polarization; Nature And Propagation Of Light; Algebra
PROBLEMS INVOLVING WAVE PROPERTIES
1. What wavelengths of light will be most strongly reflected by an oil (noil - 1.40) film
2.80 x 10-7 mm thick (a) floating on
water (nwater = 1.33) or (b) on a glass plate with an index of refraction (nglass = 1.50).
What Data Are Given?
The thickness of the oil film is 2.80 x 10-7 mm. The indices of refraction of the three
media, oil, water, and glass are given as
1.40, 1.33, and 1.50 respectively.
What Data Are Implied?
Assume the angle of incident of the light is 0ø; i.e., the incoming light is perpendicular
to the oil interfaces. The most strongly
reflected waves are the ones which undergo constructive interference.
What Physics Principles Are Involved?
This problem uses the ideas of interference and phase shift upon reflection as discussed
in Section 19.5 for constructive
interference.
What Equations Are to be Used?
Constructive Interference with no phase shift upon reflection at 2nd surface
2nt = (m/2)l m = 1, 3, 5 (19.5)
Constructive interference with a 180ø phase shift upon reflection at both surfaces.
2nt = l/2 (m + 1) m = 1, 3, 5 (1)
Algebraic Solutions
Note that in going from a region of lower index of refraction to higher index of
refraction the reflection wave undergoes a
phase change of 180ø or l/2.
Case (a) air - oil - water - In this case there is a phase change at the air - oil surface only,
thus
2nt = (m/2) l m = 1, 3, 5 (19.5)
Case (b) air - oil - glass - In this class there are phase changes at the air - oil and oil glass surfaces, thus
2nt = (l/2)(m + 1) m = 1, 3, 5, ... (1)
Numerical Solutions
(a) 2(1.40)(2.80 x 10-7) = (l1) / 2 = (3l2) / 2 = (5l3) / 2 = . . .
l1 = 1.57 x 10-6 m;
l2 = 5.23 x 10-7 m;
l3 = 3.14 x 10-7 m;
l1 = 1570 nm;
l2 = 523 nm;
l3 = 314 nm; . . .
(b) 2(1.40)(2.80 x 10-7) = l1 = 2l2 = 3l3 = ...
l1 = 7.84 x 10-7m; l2 = 3.92 x 10-7 m; l3 = 2.61 x 10-7 m ...
l1 = 784 nm; l2 = 392 nm; l3 = 261 nm ...
2. A diffraction grating has 1250 lines per centimeter (a) Over what range of angles will
it spread the first order visible
spectrum (350 nm to 700 nm)? (b) If you use this grating as a monochromator to
produce light of 588 nm, what other
wavelengths occur at the same angle?
What Data Are Given?
The distance between the slits = d = 1 cm/1250 = (10-2 m)/1.25 x 103 = 8.00 x 10-6 m
wavelength's range; 350 nm ó l ó 700 nm
What Data Are Implied?
The problem asks you to calculate the angles of constructive interference from a
diffraction grating, see Figure 19.6,
assuming normal (90ø) incident light.
What Physics Principles Are Involved?
The problem uses interference from a diffraction grating as discussed in Section 19.8.
What Equation is to be Used?
d sin q = ml; m = 1, 2, ... (19.3)
Solutions
(a) The smallest angle will occur for the smallest wavelength; for first order m = 1
sin qmin = l/d = 350 nm / 8.00 x 10-6 m = (3.5 x 10-7) / (8.0 x 10-6 m) = 4.38 x 10-2
The largest angle will occur for the largest wavelength
sin qmax = 700 nm / 8.00 x 10-6 m = (7.0 x 10-7) / (8.0 x 10-6 m) = 8.75 x 10-2
q = 5.0ø
So the first order visible spectrum is spread between the angles of 2.5ø and 5.0ø with
respect to the incident angle. Compare
these angles with the example on p. 439. What accounts for the differences?
(b) From Equation 19.3 you can see that the angle q will be the same for all constant
products for a given grating. Hence the
first order 588 nm light will coincide with second order 294 nm; third order 196 nm;
fourth order 147 nm, sixth order 98 nm,
seventh order 84 nm; etc. None of these wavelengths are visible, other than 588 nm.
However, if you are using a grating
monochromator with transducers that have a wider spectral response than the human
eye, care must be taken to avoid
multiple orders.
3. A lens is used to convert the light emitted from a light bulb into a plane wave of light
traveling in the positive x - direction.
Three polarizers are arranged along the x - axis with their axes of polarization along the
y- axis, 30ø from the y - axis, and
90ø from the y - axis, as shown below. Assume that the intensity before passing through
the first polarizer is Io. Find the
intensity, amplitude, and direction of polarization at (a) point A, (b) point B, and (c)
point C.
See Fig.
What Data Are Given?
The initial intensity Io and the angles of polarization of three polarizers.
What Data Are Implied?
It is to be assumed that the light at the beginning is unpolarized of intensity Io,
associated with an electromagnetic wave of
amplitude Eo.
What Physics Principles Are Involved?
The concepts of polarization as related to a transverse wave of amplitude Eo are needed
for this problem. See Section 19.10
and Figure 19.11.
What Equations Are to be Used?
E = Eocos q (2)
IaE2 = Eo2 cos2 q (3)
I = Io cos2 q (19.11)
Algebraic Solutions
(a) The amplitude of the wave is unchanged by an ideal polarizer; so EA = Eo.
Since the light at the beginning is completely unpolarized we can resolve it into two
components that are right angles to one
another. Then half of the intensity at Point A is one-half Io, 1/2 Io. The direction of
polarization is parallel to the y - axis.
(b) The light now polarized along the y - axis of amplitude Eo and intensity Io/2, comes
to the second polarizer oriented 30ø
from the y - axis. The amplitude after passing through the second polarizer is Eo cos q
or Eo cos 30ø for this case. The
intensity is reduced by a factor of cos2 q or cos2 30ø for this case.
At Point B:
EB = EA cos 30ø = 0.87 Eo
IB = IAcos2 30ø = 0.75 IA = 0.38 Io
The angle of polarization is 30ø from the x - axis.
(c) The light is now polarized at an angle of 30ø to the y - axis, with amplitude 0.87 Eo
and intensity 0.75 Io. It now comes to
a polarizer at an angle of 60ø to its plane of polarization. So the amplitude at point C is
Eb cos2 qBC or EB cos 60ø or 0.87 Eo
cos 60ø.
The intensity at Point C = IB cos2qBC = IB cos2 60ø
At Point C; E = 0.87 Eo cos 60ø = 0.44 Ec
IC = IB cos260ø = 0.38 Io (0.25) = 0.10 Io
the polarization is parallel to the z - axis.
Thinking About the Answer
Notice that the first and third polarizers are at right angles to one another. If they were
the only polarizers in the system NO
light would pass through the system. The insertion of a third polarizer between the two
crossed polarizers allows some light to
pass through the system. In this case, it is at an angle of 30ø with respect to the first
polarizer and 10% of the initial light
gets through the system. Is that the maximum amount you can get through such a
system? What is the best angle for the
middle polarizer in order to obtain the maximum intensity through the system?
When dealing with Rayleigh's criteria,
The students displayed mass
hysteria. For tests they had cram'da,
One point two times lambda, And
such terms as hypermetropia.
PRACTICE TEST
Keywords: ; Problems; Answers; Evaluations; Questions; Phase Shifts; Reflection;
Diffraction; Coherence;
Lasers; Gratings; Nature And Propagation Of Light; Wave-Like Nature
1. A thin air wedge is formed from two microscope slides and a thin piece of masking
tape. The wedge is illuminated with
green light.
See Fig.
Will the wedge at point A appear dark (black) or bright (green)? Explain your answer
by treating the "wedge" as a thin film.
2. If one looks at a light bulb filament through a piece of sheer curtains the pattern seen
is that of a diffraction grating. The
first order pattern is centered at 0.01 radians with respect to the central maximum. If a
wavelength of 500 nm is used,
calculate the thread spacing in the sheer curtains.
3. How does the light from a laser source differ from the light emitted by an ordinary
source, like an incandescent bulb?
ANSWERS:
1. At A, wedge appears dark. Light entering wedge undergoes 180ø phase change and
is reflected. At bottom surface,
reflected light undergoes no phase change. Therefore, since the optical path approaches
zero, the two reflections will be 180ø
out of phase and produce destructive interference.
2. 5 x 10-3 cm
3. Light from ordinary sources is incoherent, or not produced in a constant phase
relationship. Emitted light is randomly
produced in an incandescent bulb. Normal light sources (white) contain many
wavelengths which taken together produce a
nearly white color. Laser light is both monochromatic (single wavelength or color) and
coherent (produced with a single phase
relationship).
See Fig.
Chapter
Human Vision (20):
STUDY
GUIDE
20 Human Vision
Keywords: ; Learning Objectives; Biology; Anatomy And Physiology; Sensory Systems;
Optical Systems; Eye;
Lenses; Study Of Contemporary Life
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Human Vision
Characterize the physical parameters that are significant in human vision.
Visual Defects
Explain the causes and corrections for the visual defects of myopia, hypermetropia,
presbyopia and astigmatism.
Characteristics of Vision
Define the following terms:
visual acuity accommodation
scotopic vision dichromat
photopic vision
Corrective Lenses
Solve problems involving visual defects and their corrections using lenses.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 18,
Optical Elements, and Chapter 19, Wave
Properties of Light.
OVERVIEW
Keywords: ; Instructions; Biology; Anatomy And Physiology; Sensory Systems; Optical
Systems; Eye; Study Of
Contemporary Life; Lenses
In the simplest sense, the human eye is a single lens, refractive system with a retina as a
screen. But in practice, the
operation of the eye is amazing. In this chapter, you will learn about many of the
intricate and delicate operations performed
by eye parts. In addition, optically defective eyes will be classified and corrected using
additional converging and/or diverging
eye glasses.
SUGGESTED STUDY PROCEDURE
As you begin to study this chapter, become familiar with the following Chapter Goals:
Human Vision, Visual Defects,
Characteristics of Vision and Corrective Lenses. Please note that the terms listed under
the goal ofCharacteristics of Vision
are discussed under the first section of this , Definitions. Next, read text sections 20.120.7. Answers to the
questions you encounter in your reading can be found in the second section of this
chapter.
Next, turn to the end of the chapter and read the Chapter Summary and complete
Summary Exercises 1-10. Check your
answers carefully with those provided. Now do Algorithmic Problems 1 and 2, and
completeExercises and Problems 1, 2, 5,
and 12. For additional work with the concepts presented in this chapter, see the
Examples section of this
chapter.
Now you should be prepared to test your understanding of the important parts of this
chapter. Attempt the Practice Test
provided at the end of this chapter. If you have difficulties with any of the individual
parts of the test, see that
part of the text or refer to this chapter. This study procedure is outlined below.
----------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
----------------------Human Vision 20.1,20.2,20.3 1 1 1
Visual Defects 20.3,20.4 2,3
Characteristics 20.1, 20.2,20.5, 4-9
of Vision 20.6, 20.7
Corrective 20.4 10 2 2,5,12
Lenses
DEFINITIONS
Keywords: ; Glossary; Resolving Power; Eye; Optical Systems; Sensory Systems;
Biology; Colors; Focus;
Anatomy And Physiology; Study Of Contemporary Life
VISUAL ACUITY OF THE EYE
The minimum separation between two equidistant point objects that can be resolved as
separate objects.
The resolving power of the eye allows a driver to detect that an oncoming light source is
actually two headlights of an
approaching automobile.
SCOPTIC VISION
Vision of the dark-adapted eye.
The rod cells of the retina are more concentrated on the periphery around the fovea and
are more sensitive to low light
intensity.
PHOTOPIC VISION
Vision of bright-light adapted eye.
The cone cells of the retina are more concentrated in the center fovea and are most
sensitive to high light intensity.
ACCOMMODATION
The ability of the eye to focus on objects between the near point and far point of the eye.
The maximum change in power of the eye as it focuses on distant objects and then on
near objects. The accommodation is
measured in diopters and is produced by the change in power of the eye lens.
DICHROMAT
Color vision resulting from only two of the three pigments required for normal color
vision.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Resolving Power; Eye; Rayleigh Criterion; Optical Systems;
Biology; Sensory Systems;
Ear; Anatomy And Physiology; Auditory Systems; Study Of Contemporary Life;
Systems In Organisms; Wavelength;
Frequency; Perception
SECTION 20.5 The Receptive Eye
The relative sensitivity of the human eye is shown as normalized to the most sensitive
wavelength in figure 20.4. You will
notice that the eye is sensitive to light that is within about 40% of the most sensitive
wavelength. Let us compare that to the
frequency response of the human ear.
Ear Eye
Lowest frequency 2x101Hz 4.3x1014Hz
Most sensitive 2x103Hz 5.5x1014Hz
Highest frequency 2x104Hz 7.5x1014Hz
Compared to Figure 20.4 the relative sensitivity of the human eye is a broad peak with a
maximum at 2000 Hz. The ear
stretches over three orders of magnitude in its frequency response!
Notice also that monochromates are able to function in our society with little loss in
information input. A person who could
hear only one frequency would appear to have great difficulty with human
communication.
SECTION 20.6 The Perceptive Eye
If you use the Raleigh criterion for acuity of the human eye, assuming an iris size of
3mm and light of 550mm wavelength,
the angular separation between just resolvable sources of light is about 2x10-4 radians.
This is a larger angle, than the angle
between the cones. For some conditions human vision seems to be Rayleigh limited
which changes only a small amount for
visible light.
EXAMPLES
Keywords: ; Optical Instruments; Lens Aberration; Correction Methods; Thin Lenses;
Glasses; Lenses; Worked
Examples; Eye; Light And Optics; Algebra
CORRECTIVE LENSES
1. An elderly person can see distinctly with unaided vision only objects between 64cm
and 540cm from her eyes. What kind
of eyeglasses does she require? For what distances does she not see objects clearly when
she is wearing her spectacles?
What data are given?
Persons near point = 64 cm.
Persons far point = 540 cm.
What data are implied?
The use of simple lens equations are assumed to be adequate. There is an implication
that the person needs bifocal eye
glasses since both her near point and far points are not the normal values of 25 cm. and
infinity.
What physics principles are involved?
This problem is just an additional application of the physics principles discussed in
Section 20.4.
What equations are to be used?
1/p + 1/q = 1/f (18.13)
Solutions
In order to compensate for the person's inability to see near objects she needs to have a
lens that will put the virtual, upright
image of an object at 25cm at her near point of 64cm.
1/25 + 1/-64 = 1/f; so f = +41 cm.
To compensate for her farsightedness she needs a lens of +2.4 diopters.
To compensate for her nearsightedness, she needs a lens that will make objects at
infinity appear to be at 540cm. This
clearly demands a diverging lens because an object a long distance from a converging
lens has an image at its focal point.
So the image will be a virtual image
1/ì - 1/q = -(1/f)
1/ì - 1/5.4m = 1/f; f = -5.4m
She needs a lens of -0.19 diopters. In general such eyeglasses are composed of a lower
portion of positive power and an
upper portion of negative power.
For objects farther away from her than 38 cm, the positive power portion of her lens will
give her a virtual image outside of
her far point of 540 cm. So for all objects between 38 cm and 64 cm she will not see them
clearly when she is wearing her
spectacles.
PRACTICE TEST
Keywords: ; Evaluations; Questions; Problems; Answers; Eye; Optical Systems; Sensory
Systems; Anatomy And
Physiology; Optical Instruments; Glasses; Lens Aberrations; Biology
1. The block diagram below represents the various parts of the human eye.
See Fig.
Identify the parts of the eye involved in each of the systems A, B, and C above and
describe the functions of each in
processing light information.
Parts of the Eye Function
A.
B.
C.
2. A person, looking at an eye chart that contains an arrangement of lines similar to
those at right, finds that some of the
lines appear blurred to him or of unequal intensity.
See Fig.
A. What visual defect does this person probably have?
B. What is the usual cause of this defect?
C. How can this defect be corrected?
3. Typical optical data for an elderly man is shown below.
See Fig.
A. What is the power of accommodation of this man's eye without glasses?
B. What is the power of accommodation of his optical system, with glasses?
C. What type of eyeglasses does the man wear?
D. Calculate the power of these eyeglasses.
ANSWERS:
1. A) Cornea, lens, muscles (refracts and focus of light on retina, muscles contract
changing curvature and power of eye lens)
B) Iris (adjusts the opening of the eye thus controlling the intensity of light admitted)
C) Rods & Cones - Retina (transducers which convert light energy to electrical nerve
impulses which are imparted to the
brain via the optic nerve)
2. Astigmatism, nonuniform curvature of the cornea, using a cylindrical lens which is
produced to compensate for the defect
in curvature of the cornea.
3. .93 diopters, 4 diopters, bifocals,-.4 diopters and +2.7 diopters
Chapter
Electrical Properties of Matter (21): to Accompany Physics Including
Human
Applications:
21 Electrical Properties of Matter
Keywords: ; Learning Objectives; Electrostatics; Electricity And Magnetism; Electric
Fields; Potential Difference;
Capacitance; Dielectrics; Capacitors; Equipotential Lines; Electric Potential Energy;
Electric Potential; Electric Charge;
Charges Moving In Electric Fields; Gradients; Applications
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
dielectric constant equipotential surfaces
electrical field dipole
potential gradient capacitance of a capacitor
potential difference
Coulomb's Law
Apply the basic model of an electrostatic field, and use Coulomb's law to calculate the
force on one, two,
or three given point charges.
Potential Gradient
Apply the gradient to electrical field phenomena.
Moving Charged Particles
Explain the motion of charged particles in an electric field.
Capitance
Solve capacitance and capacitor problems, including the use of a capacitor as a means of
storing electrical
energy.
Applications of Electrostatics
List a number of applications of electrostatic principles to daily living and to medical
equipment.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 2,
Unifying Approaches,
Chapter 4, Forces and Newton's Laws, Chapter 5, Energy, and Chapter 9, Transport
Phenomena.
OVERVIEW
Keywords: ; Electricity And Magnetism; Electrostatics; Instructions
The fundamental laws of electricity are abstract. As you look at this chapter you will
find reference to "point" charges and
"fields." These and other strange ideas have been introduced by scientists in an effort to
explain the action in and around
groups of charged particles interacting among themselves and with other groups of
charges. Although some of the concepts
seem difficult, most will be important throughout your study of electricity (e.g., force,
work and voltage).
SUGGESTED STUDY PROCEDURE
Begin your study of this chapter by reading the following Chapter Goals: Definitions,
Coulomb's Law, Moving Charged
Particles and Capacitance. In the first section of this you will find a discussion of each
of the terms listed under
the goal of Definitions.
Next, read text sections 21.1-21.4, 21.8 and 21.9. Be careful to note the vector nature of
the electrical force as illustrated in
the Examples on page 476, whereas electrical potential is a scalar quantity see Figure
21.5. Questions you encounter in your
reading of the above-mentioned sections will be answered in the second section of this
chapter.
Now turn to the end of the chapter and read the Chapter Summary and
completeSummary Exercises 1, 4, 5, 7-11, and 14-16.
Next, do Algorithmic Problems 1-6 and Exercises and Problems 2, 4, 5, 8, 10, and 15. For
additional work with the concepts
introduced in this chapter, see the Examples section of this .
Now you should be prepared to attempt the Practice Test provided in the last section of
this chapter. When you
finish the entire test, check your answers carefully against those given. If you have
difficulty with any part of the test, refer
back to a specific portion of the text or to a part of this chapter for additional
assistance. This study procedure
is outlined below.
---------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
---------------------Definitions 21.1,21.4 1,4,5 1
Coulomb's Law 21.2 9,10,&11
Moving Charged 21.3 14&15 2 2,4,5,10
Particles
Capacitance 21.8,21.9 7,8,&16 3,4,5,6 8,15
Applications of 21.5,21.7 17 17,18
Electrostatics
DEFINITIONS
Keywords: Glossary; Farads; Electrostatics; Electricity And Magnetism; Electric Fields;
Potential Difference; Capacitance;
Dielectrics; Capacitors; Equipotential Lines; Electric Potential Energy; Electric Potential
DIELECTRIC CONSTANT
The ratio of the permitivity of a material to permitivity of a vacuum.
Materials with high dielectric constants (like water), tend to be highly polar molecules.
Thus when an external electric field is
established, the material becomes polarized and acts to reduce the overall field within
the substance.
ELECTRIC FIELD
A property of space defined as the electric force per unit charge at each point.
If the electric field is given for each point in space surrounding a charge distribution,
the force on any single charge brought
into that region can be found by multiplying the magnitude of the charge by the electric
field value. The direction will be in the
direction of the field if the charge is positive.
POTENTIAL GRADIENT
The rate of change of potential in space; a vector quantity oriented in the direction of
maximum change of potential.
The potential gradient is identical in magnitude to the electric field in a region of space.
POTENTIAL DIFFERENCE
If a charge is forced to move through a region of space against an electrical force, the
work done in Joules per coulomb of
charge is defined as the potential difference in volts. If a +1 coulomb charge is moved
through a battery from the negative
terminal to the positive terminal and 12 joules of work are done, then the battery has a
potential difference of 12 volts.
EQUIPOTENTIAL SURFACE
The surface defined by constant potential coordinates.
Moving a charge along a line or surface of euipotential requires no work because no net
voltage change occurs.
DIPOLE
Equal but opposite charge distributions separated by some small distance.
CAPACITANCE
Defined as the ratio of charge to voltage for a system of conductors 1 farad =
coulomb/volt.
Any arrangement of surfaces capable of storing electrical charges.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: Answers; Electricity And Magnetism; Electrostatics; Charges In Electric
Fields; Electric Forces; Electric Fields;
Newton's Second Law; Worked Examples; Electrical Potential; Electric Potential Energy;
Equipotential Lines; Electrophoresis;
SECTION 21.3 Electrical Forces Acting on Moving Charges
Examples-1. On particles of charge -e in an upward vertical electric field E the force is
acting downward. The downward force
has a constant magnitude and direction, eE downward. Since there is no horizontal
force on the charges, if they enter the
electric field with a constant horizontal velocity, they will continue to maintain their
constant horizontal motion. In addition they
will feel a constant downward force. To what other type of motion is this similar? If you
combine Newton's Second Law with
the equations given in Section 3.8 what do you notice? The motion of this particle in this
electric field is exactly like idealized
projectile motion of an object near the surface of the earth except the gravitational force
is increased by the addition of the
downward electric force of magnitude eE. The path of the projectile is a downward
curved parabola.
This is a good place to show the relative strength of the electric field when compared to
the gravitational field near the
surface of the earth. Let us consider two of the so-called fundamental particles, the
electron and the proton. The electron is
attributed a charge of - 1.6 x 10-19C and a mass of 9.1 x 10-31 kg. The proton is ascribed
a charge of +1.6 x 10-19C and a
mass of 1.7 x 10-27 kg. A very weak electric field is of the order of 1 V/m. Right now
you are probably seated a few meters
from a 110 volt electric line so you are sitting in an electric field on the order of 10 to 50
volts/meter, which you are not able
to detect. So a low field of 1 V/m is well below a level of human detection, nevertheless
the electric force on either an
electron or a proton in such a field is many times larger than the gravitational force on
these particles, for the electron Fe/Fg = qE/mg = ((1.6 x 10-19C)(1 V/m))/((9.1 x 10-31kg) x (9.8 m/s2)) = (1.6 x 10-19N)
/ (8.9 x 10-30N)
Fe/Fg = 1.8 x 1010 (1)
The electric force is 18 billion times the gravitational force acting on an electron in a
weak electric field near the surface of
the earth! for the proton Fe/Fg = qE/mg = ((1.6 x 10-19C)(1 V/m))/((1.7 x 10-27kg) x (9.8 m/s2) = (1.6 x 10-19N)
/ (1.7 x 10-26N)
Fe/Fg = 9.4 x 106 (2)
The electric force on a proton in a weak electric field near the surface of the earth is
almost 10 million times the gravitational
force on the proton!
So you can see why it is possible to neglect the gravitational aspects of problems that
involve the electric forces on
fundamental particles. The gravitational forces on fundamental particles are so small for
experiments carried out on the
surface of the earth that they can be neglected.
2. The speed of migration of a constituent of a fluid in an electric field will be
determined by such properties as the electric
charge of the constituent, the mass of the constituent, and the resistance to flow of the
constituent through the fluid.
SECTION 21.4 Electric Potential
Question - Let us begin by assuming that there exists an equipotential surface Vo and at
some point A on that surface there
is an electric field (vector)E which is not perpendicular to the surface at point A. Then
we can resolve (vector)E into two
components, one component Eê which is perpendicular to the surface and ET which is
tangential to the surface. The
tangential component of E is able to do work on an electric charge and move it from
point A to another point along the
surface, say point B. Since the tangential component of (vector)E did work on the charge
to move it from point A to point B,
then point B must have a different electric potential than point A. But A and B are both
points on the same equipotential
surface. We have reached a contradiction. Something we have assumed to be true must
not be true. The item that turns out
to be false is the existence of a non-zero tangential component (vector)E. Therefore ET
must be zero, so (vector)E is
perpendicular to the equipotential surface.
EXAMPLES
Keywords: Worked Examples; ; Electricity And Magnetism; Electrostatics; Electric
Forces; Coulombs Law;
Vectors; Algebra; Electric Fields; Electric Potential; Potential Difference; Gradients;
Capacitors; Circuits Containing
Capacitors; Conservation Of Energy; Conservation Of Charge; Capacitance
COULOMB'S LAW
1. Find the magnitude and direction of the total force acting on a +4.0C point charge at
the origin if a +16.0C point charge is
located at 4.0 m along the x-axis and a +9.0C charge is located at 3.0 m along the y-axis.
What data are given?
The numerical values of this problem are shown in the sketch below at points 1, 2, and
3:
See Fig. 21-1.
What data are implied?
The three charges are point charges, i.e. have no spatial dimensions, so we can use
Coulomb's law to calculate the force
between any two of the charges. Then the total force on any particle can be found by
superposition.
What physics principles are involved?
We need to use Coulomb's law to find the vector forces between the charges. Then we
will use the rules of vector algebra to
add the individual vectors to find the total force.
What equations are to be used?
Coulomb's Law
(vector)F12 = (kq1q2)/ (r2) r12 (21.5)
Algebraic Solutions
Let us begin by finding the total force acting on a charge Q2 located at point 2.
(vector)F12 = (kQ1Q2)/ (r12)2 r12 (negative y-direction) (3)
(vector)F32 = (kQ3Q2)/ (r32)2 r34 (negative x-direction) (4)
(vector)F2 = (vector)F12 + (vector)F32.
This is in case the two vectors (vector)F12 and (vector)F32 are at an angle q = 90ø with
respect to one another. So we must
use the vector algebra to add them.
The magnitude of F2 = (SQR RT)(F122 + F322)
if q2 = angle of F2 downward from the negative x- axis; then
tan q2 = ((kQ1Q2)/(r12)2)/ (kQ3Q2)/(r32)2)
=Q1(r32)2/Q3(r12)2 (5)
See Fig. 21-2
Now let us find the total force on a charge Q1 located at point 1 inFigure 21-1 above.
(vector)F21 = (KQ2Q1)/ (r21)2 r21 (positive y-direction) (6)
(vector)F31 = (KQ3Q1)/ (r31)2 r32 (an angle b from the positive y-axis) (7)
See Fig. 21-3.
The two forces acting on charge Q1 are not at right angles, but we can use the
knowledge of force components to add them
together. The forces (vector)F21 and (vector)F31 in Figure 21-3 at equivalent to the
following three forces:
See Fig. 21-4.
Thus the magnitude of the total force = (SQR RT)((F31sin b)2 + (F21 + F31cos b)2) (8) if
q1 is the angle from the positive
y-axis then
tan q1 = (F31sinb)/(F21 + F31cosb)
= [((kQ3Q1)/(r31)2) x sinb]/ (kQ2Q1/(r21)2 + kQ3Q1/(r31)2 cosb)
tan q1 = (Q3 sinb/(r31)2) / ((Q2/(r21)2) + (Q3 cosb/r31)2)
This expression can be rewritten to eliminate the angle b if you wish. The size of b is
determined by the distances r21 and r31
in the Figure 21-1, so
sinb = r31/((SQR RT)(r312 + r212)
and
cosb = r21/((SQR RT)(r312 + r212)
These expressions can be put into Equation (9) to find tan q1 in terms of Q2, Q3, r21,
and r31.
Notice how messy the algebra becomes for forces which are not at right angles or
colinear to one another.
In a similar way we can find the total force of the charge Q3. We resolve the forces into
components, add the components,
and use vector addition to find the magnitude of the total force. We use trigonometry to
determine the direction of the total
force. At this point we think it is more instructive to proceed with the actual numerical
solution than to go farther with the
more general algebraic solutions.
Numerical Solutions
Before we begin to add the forces let us calculate all the forces and their directions using
Coulomb's Law:
Force on charge Q1 from Q2 = (vector)F21 = ((9.0 x 109Nm2/C)(9.0C)(4.0C))/(3.0m)2
r21 (vector)F21 = 3.6 x 1010 N in the
positive y-direction.
Force on charge Q2 from Q1 = (vector)F12 = 3.6 x 1010N in the negative y-direction
Force on charge Q3 from Q2 = (vector)F23 = ((9.0 x 109Nm2/C)(16.0C)(4.0C))/(4.0m)2
(vector)F23 = 3.6 x 1010N in the positive y-direction
Force on charge Q2 from Q3 = (vector)F32 = 3.6 x 1010N in the negative y-direction
The distance between charges Q1 and Q3 = (SQR RT)((3.0)2 + (4.0)2) = 5.0m
Force on Q1 from Q3 = (vector)F31 = ((9.0 x 109Nm2/C)(16.0C)(9.0C))/(5.0m)2 N
(vector)F31 = 5.2 x 1010N at an angleb from the positive y-axis where tan b = 4.0/3.0; b
= 53ø
Force on Q3 from Q1 = (vector)F13 = 5.2 x 1010 at an angle b from the negative y axis, or
an angle a from the positive
x-axis; a = 37ø.
See Fig. 21-5.
Now we can use vector addition and trigonometry to find the numerical values for the
magnitudes of the three total forces
(vectors)F1,F2 and F3 and the three angles q1, q2, and q3.
Magnitude of (vector) F1 = (SQR RT) ((3.6 x 1010N + 5.2 x 1010N cos53ø)2 + (5.2 x
1010N sin 53ø)2))
= (SQR RT) ((3.6 x 1010 + 3.1 x 1010)2 + (4.2 x 1010)2 N)
= (SQR RT)((6.7 x 1010)2 + (4.2 x 1010)2) = (SQR RT)(6.72 + 4.22) x 1010N
F1 = 7.9 x 1010N
angle q1 = tan-1 (4.2 x 1010/6.7 x 1010) = 32ø
Magnitude of (vector)F2 = (SQR RT)((3.6 x 1010)2 + (3.6 x 1010)2)
= 5.1 x 1010N
angle q2 = tan-1 (3.6 x 1010/3.6 x 1010) = 45ø
Magnitude of (vector)F3 = (SQR RT) ((3.6 x 1010 + 5.2 x 1010cos37ø)2 + (5.2 x
1010sin37ø)2)
= (SQR RT)((3.6 x 1010 + 4.2 x 1010)2 + (3.1 x 1010)2)
= (SQR RT)((6.8 x 1010)2 + (3.1 x 1010)2) = (SQR RT)(6.82 + 3.12) x 1010N
F3 = 7.5 x 1010N
q2 = tan-1 (3.1 x 1010/ 6.8 x 1010) = 25ø
Thinking about the answer
Consider the three charges as rigidly attached to one another by massless rods so that
they form a rigid body. What will the
total force on the body be from the sum of the forces (vectors)F1, F2, and F3? Can you
think of a way to answer this
question without using algebraic or numerical computations? (Hint: See Section 4.2 in
the textbook)
POTENTIAL GRADIENT
2. Use the potential gradient concept to estimate the sizes of the environmental electric
fields in which you live, e.g. under an
electric blanket, in a typical room of a house, under a high voltage (380,000 volts)
transmission line.
What data are given?
Not much, three typical situations, with the voltage given in only one; Vtransmission
lines = 380,000V.
What data are implied?
In the United States of America, except for certain high power electrical devices, all the
electric potentials in buildings and
homes are 110 volts. You need to know typical distances to calculate gradients. You are
only a few centimeters from an
electric blanket, a few meters from the electrical wires in a house, and a few tens of
meters from a transmission line.
What physics principles are involved?
The basic concept of the electric fields as the gradient of the electric potential is all that
is needed for this problem, see
Section 21.4 in the textbook.
What equation is to be used?
E = -DV/Ds (21.23)
Solutions
Since we are only interested in the size of the electric field we will omit the negative
sign.
Electric Field under an Electric Blanket ÷ 110V / 7cm ÷ 20 V/cm ÷ 2000 V/m
Electric Field in a House ÷ 110V / 3m ÷ 40 V/m
Electric Field under a Transmission Line ÷ 380,000V / 10m ÷ 38000 V/m
Thinking about the answers
You see that you are always living in some manmade electric fields. What are the effects
of these fields on living organisms?
Little is really known. This is presently an area of active research. Why don't you plan a
career to study this topic?
CAPITANCE
3. A 3.0mF capacitor and a 6.0mF capacitor are connected in parallel and that
combination is connected in series to a 4.0mF
capacitor. This group of three capacitors is connected across a 24 V battery. a) Draw the
circuit, labeling all the elements by
their specified values. b) Calculate the equivalent capacitance of the 3.0mF and 6.0mF
capacitors. c) Calculate the equivalent
capacitance of the combination of three capacitors. d) Calculate the voltage across each
capacitor and e) Calculate the
energy stored in each capacitor.
What data are given?
This question can be answered by answering part (a) of this question. So let us draw the
circuit diagram and label the parts.
See Fig. 21-6.
What data are implied?
It is assumed that these are ideal capacitors that obey the conditions necessary for the
deviations in Section 21.9 of the
textbook to be valid.
What physics principles are involved?
The concepts of conservation of energy and conservation of electric charge can be used
to derive equations for combining
various combinations of capacitors.
What equations are to be used?
For a general system of three capacitors.
Combined in Series:
Vtotal = V1 + V2 + V3 (21.31)
1/Ctotal = 1/C1 + 1/C2 + 1/C3 (21.33)
Qtotal = Q1 = Q2 = Q3 (21.30)
Combined in Parallel:
Vtotal = V1 = V2 = V3 (21.34)
Ctotal = C1 + C2 + C3 (21.36)
Qtotal = Q1 + Q2 + Q3 (21.35)
Energy Stored in a Capacitor:
E = 1/2 CV2 (21.39)
Algebraic Solution
Let us begin by combining capacitors C2 and C3 and replace them by one capacitor C23,
then
C23 = C2 + C3 (11)
The circuit then looks as follows:
See Fig. 21-7.
We can combine C1 and C23 to obtain C
1/C = 1/C1 + 1/C23;
C = (C1C23)/(C1 + C23) (12)
V = V1 + V23 (13)
The total charge Q is given by
Q = VC = (V (C1C23))/(C1 + C23) (14)
The separate charges Q1 and Q23 will be equal to Q since C1 and C23 are in series.
V1 = Q1/C1 = Q/C1 = VC/C1 = VC23/(C1 + C23) (15)
V23 = Q23/C23 = Q/C23 = VC/C23 = VC1/(C1 + C23) (16)
The energy stored in the two capacitors can also be calculated
E1 = 1/2 C1V12 = (C1V2(C23)2)/ (2(C1 + C23)2) (17)
E23 = 1/2 C23V232 = (C23V2(C1)2)/ (2(C1 + C23)2) (18)
We have completed the calculations we can make for the simplified circuit shown in
Figure 21-7. We can now complete the
calculations for the more complex circuit given in the problem. Since none of the values
we have calculated for C1 will
change if we consider the two parallel capacitors C2 and C3 which are equivalent to C23
we need not do any further
calculations for C1. We do need to find the charge and energy stored for both C2 and
C3. Since C2 and C3are in parallel we
already know the voltage across them.
V2 = V3 = V23 = (VC1)/(C1 + C23) (19)
where C23 = C2+C3 from Equation (11).
The charge on each capacitor is given by:
Q2 = C2V2 = (C2VC1)/(C1 + C23) (20)
Q3 = C3V3 = (C3VC1)/(C1 + C23) (21)
The energy stored in each capacitor can be calculated also
E2 = 1/2 C2(V2)2 = 1/2 C2 x (V2(C1)2)/ ((C1 + C23)2) (22)
E3 = 1/2 C3(V3)2 = 1/2 C3 x (V2(C1)2)/ (C1 + C23))2 (23)
We are finished. We have an algebraic solution for every part of this question: part (b) is
answered by Equation (11); part (c)
is answered by Equation (12): part (d) is answered by Equations (15) and (19); part (e) is
answered by Equations (17), (22)
and (23).
Numerical Solutions
part (b) Equivalent capacitance of 6.0mF and 3.0mF in parallel
C23 = (3.0)+(6.0)mF = 9.0mF
part (c) Equivalent capacitance of all three capacitors
1/C = 1/C1 + 1/C23 = 1/(4.0mF) + 1/(9.0 mF); C = 2.8mF
part (d)
Q = CV = (2.8mF)(24V) = 6.7 x 10-5C
V1 = Q/C1 = (6.7 x 10-5)/4 F = 17V
V23 = 7.4V = V2 = V3
part (e)
E1 = 1/2C1(V1)2 = 1/2(4.0)(17)2 = 5.8 x 10-4 joules
E2 = 1/2C2(V2)2 = 1/2(6.0)(7.4)2 = 1.6 x 10-4 joules
E3 = 1/2C3(V3)2 = 1/2(3.0)(7.4)2 = 8.2 x 10-5 joules
Thinking about the answers
You will notice how keeping two significant figures in each of the answers leads to
small peculiarities in some of the answers.
We know that V1 and V23 should add up to 24V. If you add the numbers calculated
separately you must add 17V and 7.4
Volts.
How can you check the answer to part (e)? The total energy in the three capacitors
should be equal to the sum of the three
energies.
Etotal = 1/2CV2 = 1/2 (2.8mF)(24V)2 = 8.0 x 10-4 joules
add up the values obtained in part (e)
5.8 x 10-4 + 1.6 x 10-4 + 0.82 x 10-4 = 8.2 x 10-4 joules
The small difference of 2 parts in 80 arises from the use of two significant figures in the
calculations.
PRACTICE TEST
Keywords: Problems; Answers; Questions; Evaluations; Electricity And Magnetism; ;
Electric Forces; Electric
Fields; Coulomb's Law; Electrical Potential ; Electric Potential Energy; Potential
Difference; Capacitors; Capacitances; Circuits
Containing Capacitors; Field Lines
1. A proton (qp = +1.6 x 10-19 coulomb) and an electron (qe = -1.6 x 10-19 coulombs) are
stationary and are separated by a
distance of 5 x 10-10 meters (typical hydrogen atom separation).
See Fig.
A. Calculate the magnitude and the direction of the force exerted on both the proton
and the electron due to the presence of
the other. Show the direction of both of these forces on thediagram above.
B. Sketch the electrostatic field which exists between the charges by drawing several
lines of force between the charges in
the diagram above.
2. An electron (qe = -1.6 x 10 coul and me = 9.11 x 10-31 kg) is accelerated through a
series of three sets of parallel plates.
Each set of plates (A, B, and C) has a similar voltage differential (100 volts) but a
progressively increasing plate separation
(10.0 cm, 20.0 cm, and 30.0 cm). (See the diagram shown below).
A. Predict the electron's velocity when it leaves plate set A, at point x.
B. What change in electron velocity occurs between points x and x1?
C. Calculate the electron's energy at y and again at z.
D. If a fourth set of plates were added, can you predict the final velocity of the electron?
3. Three capacitors are connected in a series arrangement as indicated by thediagram
shown below.
After the circuit was connected, a physics student finds that the lmf capacitor stores a
charge of 27.5mcoul.
A. What is the charge stored on each of the other capacitors?
B. Find the voltage of the battery.
C. Calculate the voltage drop across the 2.00mF capacitor.
D. How much energy is stored by the entire series configuration of capacitors?
ANSWERS:
See Fig.
1. 9.2 x 10-10 N(AH reaction)
2. 5.9 x 106m/s, none, 8.3 x 106m/s, 10.2 x 106m/s, yes, 11.8 x 106m/s
3. A) 27.5 mcoul, B) 50 volts, C) 13.8 volts, D) 688 x 10-6 Joules
Chapter
Basic Electrical Measurements (22): to Accompany Physics Including
Human
Applications:
22 Basic Electrical Measurements
Keywords: ; Learning Objectives; Electricity And Magnetism; Electric Current; Amps;
Ohms; Resistance;
Resistivity; EMF; Peltier Effect; DC Circuits; DC Instruments; Seebeck Effect;
Thermocouples; Piezoelectric Effect;
Conductivity; Circuits Containing Resistors; Ohm's Law; Potential Difference; Electric
Power; Applications; Anatomy And
Physiology; DC Current
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
ampere electromotive force (emf)
electrical conductivity Seebeck effect
ohm Peltier effect
resistivity piezoelectric effect
Resistors
Determine an equivalent resistance for a series or parallel combination of resistors.
Ohm's Law
Solve problems using the relationship among resistance, potential difference, and
electric current-that is, apply Ohm's law to
simple circuits.
Power Loss
Solve problems for instantaneous power in resistive elements obeying Ohm's law.
DC Circuits
Analyze direct-current circuits consisting of resistances and sources of emfs, and find
the currents, terminal potential
difference of sources of emfs, potential drops, and power developed in circuit elements.
DC Instruments
Explain the basic principle of operation of direct-current instruments: potentiometer,
Wheatstone bridge, and thermocouple.
Bioelectricity
Describe some application of electricity in human physiology.
PREREQUISITES
Before you begin this chapter, you need to have mastered the basic concepts of Chapter
5, Energy, Chapter 11, Thermal
Transport, and Chapter 21, Electrical Properties of Matter.
OVERVIEW
Keywords: ; Electricity And Magnetism; Electric Current
Most of the every day applications of electricity involve the flow of electric charges
through circuits. In this chapter, many of
the basic properties of electrical circuits will be discussed. From a simple discussion of
Ohm's and Joule's Law in section
22.4, more sophisticated circuits are introduced. The remaining parts of the chapter deal
with either electrical measuring
devices or electrical physiology.
SUGGESTED STUDY PROCEDURE
To begin your study of this chapter, read the following Chapter Goals: Definitions,
Resistors, Ohm's Law, Power Loss, DC
Circuits, and Bioelectricity. Each term listed under the Goal of Definitions is discussed
in the first section of this
chapter.
Next, you should carefully read through the following chapter sections and work
through the examples provided at the end of
each: 22.1-22.4, 22.8-22.11, and 22.13. Questions you encounter during your reading are
answered in the second section of
this chapter.
At the end of the chapter, read the Chapter Summary and do Summary Exercises 1-11,
14 and 16. Check your answers
carefully and compare them with those given at the end of the section. Next, do
Algorithmic Problems 2-7 and Exercises and
Problems 3, 7, 13, 17, and 18. Now, for more practice with the concepts of DC Circuits,
complete each of the problems
presented in the Examples section of this chapter.
After completing the above procedure, you should be prepared to attempt the Practice
Test on Basic Electrical Measurements
found at the end of this chapter. After you have completed each part, check your
answers against those given at
the end of the test. If you were unsuccessful in any of the areas, refer back to the text or
to this chapter for
additional assistance. This study procedure is outlined below.
------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
------------------Definitions 22.1,22.2,22.8 1-8
22.9,22.10
Resistors 22.3 9 2,3,4
OHM's Law 22.4 10 5 3,7
Power Loss 22.4 11 7 17,18
DC Circuits
Bioelectricity 22.11,22.13 14,16 13
--------------------------------------------------DC Circuits 22.14 12 16
DC Instruments 22.5,22.6,22.7 13 8,9,10,12
DEFINITIONS
Keywords: ; Electricity And Magnetism; Electric Current; Amps; Conductivity;
Resistance; Ohms; Resistivity;
EMF; Seebeck Effect; Thermocouples; Peltier; Effect; DC Circuits; DC Instruments;
Piezoelectric Effect; Glossary; DC
Current
AMPERE
A measurement of the flow of charges through an electrical conductor. The passage of
one coulomb of electric charge in one
second is one ampere.
ELECTRICAL CONDUCTIVITY
The proportionality constant between current density and potential gradient.
OHM (unit of resistance)
The OHM is the resistance that has a potential difference of one volt across it when it
carries a current of one ampere.
Each individual component of an electrical circuit will display a unique resistance to the
flow of electrical charges.
RESISTIVITY
The reciprocal of conductivity with units of ohms-meter.
This constant reflects a general resistance property for a material like copper or lead. It
is not dependent upon the length
and/or cross-sectional arrangement of a circuit component.
ELECTROMOTIVE FORCE (emf)
The potential difference across a source of electrical energy with no current flow.
An emf of a battery is a voltage measurement when the current flow through the
battery is zero. Thus the terminal voltage
has a maximum value.
SeeBECK EFFECT
The physical process that produces the emf generated by junctions of dissimilar metals
when they are heated or cooled, (e.g.,
thermocouples).
PELTIER EFFECT
The process characterized by the generation or absorption of heat at a junction of
dissimilar metals when a current passes
through the junction. (The reverse of the Seebeck effect.)
An electric current flowing through the junction of dissimilar metals produces either a
liberation or absorption of heat
(depending upon the direction of the current) in direct proportion to the quantity of
charge passing through the junction.
PIEZO ELECTRICITY
The ability of certain crystals (e.g., quartz) to generate an emf when subjected to
mechanical strain.
A common crystal microphone operates as a mechanical energy-electrical energy
transducer using this effect.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Electricity And Magnetism; Electric Current; Energy; Electric
Power; Resistivity;
Resistance; DC Current; Amps; Electrons; Joule's Law; Electrical Equivalent; Ohm's
Law; Galvanometers; Wheatstone
Bridges; Piezoelectric Effect; Threshold; DC Circuits; DC Instruments; Circuits
Containing Resistors; Conductors
SECTION 22.1 Introduction
The proliferation of electrical appliances has become a hallmark of the United States.
Many electrical energy supply systems
have a procedure for charging less per unit of energy the more energy you use. The
economics of the system has
encouraged the development of a vast array of electrical devices from electric
toothbrushes to electric garage door openers.
A typical person in a single family house may use about 10 units of electrical energy per
day. The units used to measure
electrical energy by the electrical energy supply systems are kilowatt hours rather than
joules. There are 3.6 x 106 joules in
one kilowatt hour.
SECTION 22.2 Electrical Charges in Motion
The resistivity r (rho) is typically measured in the units of ohm ¥ centimeters.
Questions - 1 and 2. Since the electrical charge of an electron is 1.6 x 10-19C, if 6.25 x
1018 electrons passed a location in
one second, there would be an electrical current of one ampere passing that location. If
the current density were 1A/m 2, then
6.25 x 1018 electrons would be passing through an area of 1 square meter or 104 square
centimeters. The current density of
1A/m2 is equal to 6.25 x 1014 electrons per square centimeter.
3. If the potential difference is linear with distance along the circuit then the total
potential difference from end to end divided
by the total length will be equal to the ratio of the change in voltage to change in
displacement for any portion of the circuit.
If we rewrite Equation 22.3 in terms of the electric field it becomes J = sE.
4. A good conductor will have a small value of resistivity.
SECTION 22.3 Sources of Electrical Energy
If you measure the internal resistance of a battery then you find low resistance means a
good battery and high resistance of
the same kind of battery means a bad, or dead, battery.
5. If various portions of electrostatic systems are connected by conducting materials,
then electric charge can always flow in
response to electric potentials and no large electrostatic potentials can develop.
SECTION 22.4 Electric Circuits: Ohm's Law and Joule's Law
6. Equation 22.6 is a statement of Ohm's Law, V=IR. Suppose you have two resistances,
R1 and R2, in parallel with one
another and you put a voltage V across them. There will be a current of I1=V/R1
through the first resistance R 1 and there will
be a current of I2=V/R2 through the second resistance. The total current in the circuit
will then be equal to I1+I2. What
equivalent resistance R do you need to hook across the voltage V in place of R1 and R2
that will draw the same amount of
current, namely I1+I2? To answer this question let us set the potential difference across
the equivalent resistance equal to the
potential difference across each separate parallel resistor, V = R (I1+I2) = I1R1 = I2R2.
We can use the right sides of these
equations to derive equations for the currents I1 = V/R1 and I2 = V/R2, so V = R
(V/R1+V/R2);
1 = R(1/R1 + 1/R2) = R((R2 + R1) / R1R2)
(R1R2) / (R1 + R2) = R (1)
The equivalent resistance of two parallel resistors is the product of the resistances
divided by the sum.
7. We can rewrite the Joule's Law expression of the power dissipated in a circuit by
using Ohm's Law to replace the current in
Equation 22.23 by the ratio V/R. Then
Power = (V/R)2R = V2/R (2)
8. One experiment that can be used to relate the electrical energy to mechanical energy
is to place a heating coil of wire in
an insulated container that holds a known volume of water. Turn on the current and
measure the potential difference, the
current, the time, and the change in temperature of the water. You can then check to see
if the VIt product is equal to the
volume times change in temperature times 4.19 joules/milliliter for the water.
SECTION 22.5 Galvanometers
9. The voltage across a 1000W galvanometer measuring a current of 10-9A is 10-6 volts.
10. A perfect ammeter would have a resistance zero.
11. A perfect voltmeter would have an infinite resistance.
12. In the circuit where the ammeter is in series with the resistance, the voltmeter
reading is the voltage drop across both the
unknown resistance and the ammeter. But the reading of the ammeter is the correct
reading for the current through the
unknown resistance. The voltmeter reading must be reduced by an amount equal to the
current times the ammeter resistance,
then divided by the ammeter current reading to find the correct value of the unknown
resistance.
R = (V - IARA) / IA (3)
In the circuit where the voltmeter is in parallel with the unknown resistance the
ammeter reading is the current through both
the voltmeter and the unknown resistance. So the current reading of the ammeter must
be reduced by the amount of current
in the voltmeter, then divided into the voltmeter reading to determine the correct value
of the unknown resistance,
R = V / (IA - V/RV) (4)
SECTION 22.7 The Wheatstone Bridge
13. When the Wheatstone Bridge is balanced there must be no current through the null
detector so points (a) and (b) of the
circuit must be at the same potential. Thus the potential difference across R1 and R2
must be the same, i.e. V1 = V2 or iaR1
= ib R2. Similar reasoning can be used to deduce that iaR3 = ib Rx.
14. If we assume the current detector can act as an infinite resistor to measure and
voltage difference between (a) and (b),
then we can write down the voltage difference from the top of the emf to point (b) by
two routes; through R1 and the current
detector and through R2. The potential difference must be the same in each case
i1R1-Vab = i2R2
If we do not allow for any current flow through the detector then
E = i1(R1 + R3) = i2(R2 + Rx)
so Vab = i1R1 - i2R2
Vab = ((e / R1 + R3) R1) - ((e / R2 + Rx) R2)
SECTION 22.10 The Piezoelectric Effect
15. A high input resistance device that draws very little current must be used with
piezoelectric crystals.
16. The use of a periodic voltage to cause mechanical vibrations in a piezoelectric crystal
is an example of a resonance
phenomena. The frequency of the applied voltage must be the same as the natural
frequency of the piezoelectric crystal.
SECTION 22.13 Electric Thresholds and Effects
17. Typical amounts of static electricity are not able to produce any appreciate current
flow through humans so they are not
usually dangerous.
18. The energy dissipated in an electric shock is proportional to the square of the current
times the resistance times the time.
The resistance of an object is inversely proportional to its area. The combination of these
concepts, Equations 22.7 and
22.23, leads to the relationship
damage ° (current2/area) x time
EXAMPLES
Keywords: Worked Examples; ; Electricity And Magnetism; Ohm's Law; DC Circuits;
Electric Current; Circuits
Containing Resistors; Voltage; Resistance; Electric Power; Kirchhoff's Rules; Joules's
Law; Electrical Equivalent; Conservation
Of Energy; DC Current
RESISTORS
1. (a) What is the equivalent resistance of n equal resistors in parallel? (b) Combine this
result with the Least Common
Multiple concept from mathematics to devise a technique for finding the equivalent
resistance for unequal resistors in parallel.
(c) Use the technique to find the resistor equivalent to 20W, 30W, and 50W in parallel.
What data are given?
The number of equal resistors = n, an integer greater than zero.
What data are implied?
The circuit obeys Ohm's Law.
What physics principles are involved?
The conservation of electric charge and energy are used to derive the basic equations
that are needed for this problem.
What equations are to be used?
We will use Ohm's Law to derive some other equations.
V = IR
Solutions
See Fig. 22-1.
Assume there is a voltage drop V across each of these resistors of value R.
Then a current I flows through each branch of the circuit given by I=V/R. But the
current in the rest of the circuit must be
equal to the sum of all the currents in parallel branches of the circuit, so
Itotal = I + I +...I = nI since
there are n branches, each carrying a current I. Thus we have a voltage of V and a
current of nI, what must the resistance
be? The equivalent resistance Req must be
Req = (V/nI) but (V/I) = R, so
Req = R/n (5)
The equivalent resistance of n equal resistors in parallel is the resistance of one resistor
divided by n, the number of resistors
in parallel.
(b) Now take a circuit of unequal resistors R1, R2 and R3. We know from Equation (5)
the resistor R1, for example, can be
considered the equivalent of some member, say n1, of larger equal resistances, say R, in
parallel. Likewise R2 can be
equivalent to n2 of the larger resistances R in parallel and R3 equal to n3 of the R's in
parallel. Then we can replace each of
the unequal resistors by some numbers, n1, n2 and n3 respectively, of equal resistors
and find the equivalent resistance of R1
, R 2 and R3 in parallel using Equation (5).
Req = R / (n1 + n2 + n3) (6)
See Fig. 22-2.
How can we find the value of R and of n1, n2 and n3? Let us assume the R1, R2 and R3
have no common factors, then the
Least Common Multiple of the three is just their product.
L.C.M. = R1R2R3 (7)
If we choose the LCM of R1, R2 and R3 for the value of R then we know that n1, n2 and
n3 will all be integer numbers.
Choose R in ohms = R1R2R3
Then n1 = (R1R2R3) / R1 = R2R3; n2 = R1R3 and n3 = R2R1
Thus Req = R / (n1 + n2 + n3) = (R1R2R3) / (R2R3 + R1R3 + R2R1) (8)
(c) To see how this works let us consider the specific case of a 20W resistor in parallel
with a 30W resistor in parallel with a
50W resistor.
See Fig. 22-3.
so n20 = 300/20 = 15; n30 = 300/30 = 10; n50 = 300/50 = 6
So these three resistors in parallel are equivalent to 15+10+6, or 31, branches of 300W
resistors in parallel, thus
Req = 300/31 = 9.7 W
Thinking about the answers
Look back over this procedure. Can you show that this result is equivalent to Equation
(22.16)? Which technique do you
prefer, the use of the sum of reciprocals, as in Eq. 22.16, or the use of the LCM as shown
here? Take your pick, they give
the same result.
OHM'S LAW
2. A 3W resistor and a 6W resistor are connected in parallel. This combination is
connected in series with a 4W resistor; then
the group of three resistors are connected in parallel with a 12W resistor. These four
resistors are connected to the terminals
of a 12V battery with an internal resistance of 2W. (a) Draw the circuit, labeling all the
elements by their specified values. (b)
Calculate the resistance of the combination of 4 resistors. (c) Calculate the current
through each portion of the circuit. (d)
Calculate the voltage drop across each portion of the circuit and show that energy is
conserved in the circuit.
What data are given?
The answer to this question can also answer part (a) of this question, so let us draw a
circuit diagram and label at the parts.
See Fig. 22-4.
What data are implied?
It is assumed that all aspects of this circuit satisfy Ohm's Law.
What physics principles are involved?
The conservation of energy and electric charge are combined with Ohm's Law to derive
all of the circuit equations to be used.
What equations are to be used?
Vterminal = e-Ir (22.8)
V = IR (22.6)
Series Stotal = SIRi (22.9)
Series Reff = R1 + R2...Rn (22.10)
Parallel I = i1 + i2 + i3 (22.11)
-1 -1 -1 -1
Parallel Reff = R1 + R2 +...Rn (22.16)
Numerical Solutions
Rather than work out a complete algebraic solution for this problem before putting in
numbers, let us work out algebraic
solutions to each part of the problem, then obtain a numerical result and carry the
numerical result from one part of the
problem to another.
(b) To solve this portion of the problem we must begin with the 3W and 6W parallel
combination so that we can combine that
with the 4W resistor to find the equivalent resistance of those in parallel with the 12W
resistor.
See Fig.
So the equivalent resistance of the top part of Figure 22-4 is 2W + 4W, or 6W.
Finally,
See Fig.
The equivalent resistance of the four resistors is 4W.
(c) To calculate the current through each portion of the circuit we will find the current
through the battery first and then trace
its subdivision through a portion of the circuit. The 12V battery is connected to 4W of
external resistance plus having an
internal resistance of 2W, so
Ibattery = 12V / (4W + 2W) = 2A
When the 2A current leaves the battery it meets an equivalent parallel circuit that
contains 6W in parallel with 12W. The
resistance (6W) is half as much in one branch so the current in that branch must be
twice as much as in the other (12W)
branch; thus
I12 = [2/3]A; Itop branch = [4/3]A
All of the [4/3]A will pass through the 4W resistor in the top branch. Note that the
[4/3]A also comes to a parallel portion of
the circuit where the ratio of resistances in the two branches is 2 to 1 (6W to 3W). The
current through the 3W resistor will be
twice the current in the 6W resistor. Thus
I3 = [8/9]A; I6 = [4/9]A
Of course, the current through the 2W resistance of the battery is the total of 2A. Notice
how current is conserved at each
branch point in the circuit.
See Fig. 22-5
(d) We can find the voltage drops across each portion of the circuit by using Ohm's
Law, V=IR.
V3 = (8/9 A)(3) = 8/3 Volts
V6 = (4/9 A)(6) = 8/3 Volts
V4 = (4/3 A)(4) = 16/3 Volts
V12 = (2/3 A)(12) = 8 Volts
V2 = (2A)(2) = 4 Volts
Conservation of energy
Potential across the 3W and 6W in parallel = 8/3 V.
Potential across the 12W and 3W resistor combination = 8V.
Potential across all resistances = 12V = Emf of the battery.
POWER LOSS
3. Find the power loss in each portion of the circuit in Problem 2 above.
What data are given?
See Figure 22-4 and the answers to part (c) and (d) of question 2.
What data are implied?
The circuit elements are ohmic.
What physics principles are involved?
The use of Joule's Law as discussed in Section 22.4.
What equation is to be used?
Joule's Law: P = VI = I2R (22.23)
Solutions
P3 = V3I3 = (8/3 V)(8/9 A) = 64/27 W = 2(10/27) W = 2.4 W
P6 = V6I6 = (8/3 V)(4/9 A) = 32/27 W = 1(5/27) W = 1.2 W
P4 = V4I4 = (16/3 V)(4/3 A) = 64/9 W = 7 1/9 W = 7.1 W
P12 = V12I12 = (8 V)(2/3 A) = 16/3 W = 5.3 W
P2 = V2I2 = (4 V)(2 A) = 8 W
Power Output of the Battery = eI = (12V) (2A) = 24W
Is that equal to the total power loss of the complete system?
Power Loss = 2.4 + 1.2 + 7.1 + 5.3 + 8.0 = 24W
Thinking about the answers
The power supplied by the battery has to equal the power loss in the circuit for
electrical energy to be conserved in this
system. From where does the electrical energy come for this circuit?
Please note that if you were given a circuit and asked to find the power loss in each
circuit element you would have to go
through the procedures used in problem 2 first to find the current in each branch of the
circuit, then use P=I2R to find the
power dissipated in each circuit element. A procedure which may be lengthy, but if
done systematically will yield the correct
results.
PRACTICE TEST
Keywords: ; Problems; Questions; Answers; Evaluations; DC Current; DC Circuits;
Electric Current; Resistance;
Resistivity; Kirchhoff's Rules; Voltage; Energy; Circuits Containing Resistors; Ohm's
Law; Medicine And Health;
Cardiovascular Systems; Anatomy And Physiology; Applications; Electric Power
1. Briefly describe the difference between resistance (OHMS) and resistivity (OHMmeters). Give an example of how each
term is used in a practical situation.
2. The circuit outlined by the diagram below was used to show the electrical flow in a
DC circuit. The 1 OHM resistor is the
internal resistance of the 75 volt battery.
See Fig.
Using this information, find the following:
A. The equivalent resistance of the circuit.
B. The current measured by the ammeter.
C. The voltage indicated by the voltmeter.
D. The "Terminal" voltage of the battery in the circuit.
3. Three heat lamps (120 volts, 200 watts) are to be connected to a 120 volt electrical
outlet. Circuit 1 shows the bulbs
connected in parallel.
See Fig. Circuit 1.
A. What is the voltage drop across each bulb in circuit 1?
B. What total current is required from the outlet where circuit 1 is connected?
C. Find the cost for operating the circuit for a time of 3 hours if the charge is $1.05 for
each KWH.
4. An important application of electricity in human physiology is the ECG. A typical
output signal is shown below.
See Fig.
A. What is this signal called and what does it control?
B. Briefly describe the electrical origins of the signal and what the function of each part
of the wave serve.
ANSWERS:
1. OHM - Rating given to a part of an electrical circuit offering a resistance to the flow of
charges. For a 1 volt potential
which can produce a current of 1 Amp, the resistance is 1 OHM.
OHM - Meter - A basic property of a material substance rather than an individual
device. The property is independent of any
size and/or shape. The rating can predict the ohmic resistance of a material after being
formed to a certain size and shape.
2. 25W, 3 Amp, 36 volts, 72 volts
3. 120 volt - AC, 5 Amps, 9 cents ($.09)
4. Electrocardiagram (ECG) - Heart, the electric origin of the signal is found within the
center of the brain. The function is to
synchronize the heart's pumping action: P (atrial changing dipole signal), QRS
(Ventricular changing dipole), T (Ventricular
dipole restoration).
Chapter
Magnetism (23):
STUDY
GUIDE
Citation: H. Q Fuller, R. M. Fuller and R. G. Fuller, to Accompany Physics Including
Human Applications.
(Harper and Row, New York, 1978). Permission granted by the authors
Keywords: ; Learning Objectives; Magnetic Fields; Electricity And Magnetism;
Magnetic Properties Of Matter;
Charges Moving In Magnetic Fields; Magnetic Fields Of Currents; Devices Using
Magnetic Fields And Current; Hall Effect
GOALS
When you have mastered the content of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
magnetic field current sensitivity
magnetic forces ferromagnetism
ampere
Biot-Savart Law
Apply the basic relationship between current and its associated magnetic field.
Magnetic Forces on Moving Particles
Explain the motion of a charged particle in a uniform magnetic field.
Magnetic Interactions
Discuss the interaction of magnetic fields.
Electric and Magnetic Fields
Explain the difference between the behavior of charged particles in electric and
magnetic fields.
Magnetic Field Applications
Explain such applications as: electromagnetic pump, focusing of charged particles by a
magnetic field, DC
electric meters, motors, and the Hall effect.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 4, Forces
and Newton's Laws.
Chapter 21, Electrical Properties of Matter, and Chapter 22, Basic Electrical
Measurements.
OVERVIEW
Keywords: ; Electricity And Magnetism; Magnetism; Instructions
As you skim over the contents of this chapter, you will surely notice that the description
of magnetism is similar to that of
Electrostatics; e.g., both produce interacting forces of attraction and repulsion and
therefore can be modeled using the field
concept. Be sure at the same time to notice that these concepts are basically different
and that the interaction between
magnetism and electric charges occurs only when there is relative motion between the
magnetic field and the electric charge.
SUGGESTED STUDY PROCEDURE
To begin your study of this chapter, read each of the Chapter Goals: Definitions, BiotSavart Law, Magnetic Forces on Moving
Particles, Magnetic Interactions, Electric and Magnetic Fields, and Magnetic Field
Applications. An extended discussion of
each of the terms listed under Definitions is given in the next section of this chapter.
Next, read text sections 23.1-23.8 and 23.10-23.13. Be sure to be cautious of the vector
product (cross product) nature of the
magnetic field - moving charged particle interaction (e.g., equation 23.2, and Figure
23.9). Answers to the questions you
encounter as you read are answered in the second section of this chapter.
Now turn to the end of the chapter and read the Summary and complete Summary
Exercises 1-15. Next, do Algorithmic
Problems 1-5 and complete Exercises and Problems 2, 3, 4, 6, 8, 9, and 11. For additional
exercises with the concepts of
this chapter, turn to the Example section of this .
Now you should be prepared to attempt the Practice Test on Magnetism provided at
the end of this chapter.
When you have completed the test, check your answers with those given. If you were
unsuccessful with any of the concepts,
refer to either a particular section of the text or this chapter. This study procedure is
outlined below.
------------------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
------------------------------Definitions 23.1, 23.2, 23.3 1, 2, 3,
4, 5
Biot-Savart Law 23.4 6,7 3,4,5 4,6,8
Magnetic Forces 23.5 8,9 2 2,3,9
on Moving Particles
Magnetic 23.7,23.8 10 1 11
Interactions
Electric and 23.3 11,12,13
Magnetic Fields
Magnetic Field 23.6,23.10, 14,15
Applications 23.11,23.12,
23.13
DEFINITIONS
Keywords: Magnetic Fields; Magnetic Forces; Amps; Kinds Of Magnetism; Glossary; ;
Magnetism; Electricity
And Magnetism; Magnetic Fields And Currents; Magnetic Fields Of Currents
MAGNETIC FIELD
The magnetic (induction) field, B, is a quantity introduced to explain interaction
between moving charges. (Units are weber/m2
).
In a region of space which is occupied by a magnetic field, a vector can be drawn which
represents the force a unit north
pole would receive if placed at that point. The force exerted on a moving charged
particle is more complicated and is in a
direction perpendicular to both the magnetic field and the velocity of the particle (see
magnetic force).
MAGNETIC FORCE
The force on a moving charge (+q) with velocity (v) in a magnetic induction field (B) has
the value of (q x v xB).
The right-hand (vector-product) rule is used to determine the direction of this force. See
Fig. 23.3.
AMPERE
The amount of current carried by two parallel conductors placed 1 meter apart in a
vacuum so that the force between them is
2 x 10-7 Newtons per meter of length.
CURRENT SENSITIVITY
The proportionality constant between the current in a coil and its angle of deflection in
a magnetic field.
For highly sensitive galvanometers, k is the order of 0.00001mA/mm or 10-11
amp/mm. For less sensitive meters, k is a
larger number. Please note that the term sensitivity is a technical term meaning the
reciprocal of sensitiveness.
FERROMAGNETISM
The result of cooperative interactions of magnetic dipoles in such as iron, nickel, and
cobalt. The result is very strong internal
magnetic fields.
Introducing an iron (ferromagnetic) core in an air core solenoid can increase the
magnetic field produced by the same current
by a factor 500 to 1000 times. This increase is attributed to the alignment of individual
Fe atoms each with small magnetic
dipole moments.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: Answers; Electricity And Magnetism; Magnetism; Devices Using Magnetic
Fields And Currents;
Magnetic Properties Of Matter; Magnetic Fields And Currents
SECTION 23.1 Introduction
Today someone rung the doorbell of my house, the plunger which causes the ring was
activated by an electromagnet. I have
some messages attached to the front of our home refrigerator with small magnets. I
used an electric blender to mix up a
solution. The blender motor operates through the interaction of electric and magnetic
forces. As I drove our automobile the
alternator generated electrical energy, by means of a magnetic field, to recharge the
battery. The door of our refrigerator is
held shut by a magnetic strip. These are five ways that magnetism has played a part in
my life today. How about yours?
In each of these cases the magnetic field of the earth is much smaller than the field that
influenced my life. The earth's
magnetic field is about 10-4 times the typical values of magnetic fields in motors, etc.
The exact role of the earth's magnetic field in various living systems is still not well
understood. It seems to play a role in the
migration patterns of animals, but its effects on humans is not known. With modern
low field sensitive detectors, (discussed
in section 23.13), we are beginning a more systematic study of the presence of weak
fields and their possible effects on
living systems.
EXAMPLES
Keywords: ; Worked Examples; Biot-Savart Law; Magnetic Fields; Charges Moving In
Magnetic Fields; Algebra;
Magnetic Fields Of Currents; Magnetism; Electricity And Magnetism; Magnetic Forces;
Centripetal Force; Circular Motion
BIOT-SAVART LAW
1. What is the magnetic induction B at the origin from a line carrying a current I along
the negative x-axis to a position -a,
around a circle in a clockwise manner to the negative y-axis at a point -a, and then out
to a long distance.
What data are given?
The current I follows a path as shown below:
See Fig. 23-1.
What data are implied?
The external aspects of this circuit which supply the emf are far, far away from the
origin of this coordinate system.
What physics principles are involved?
The Biot-Savart relationship between electric current and magnetic induction can be
used.
What equations are to be used?
Biot-Savart Law
B = mo/4p (iDlsinq/r2) (23.3)
Solution
The angleq in Equation 23.3 is the angle between the current direction and the direction
from the element of current iDl to the
point where the magnetic induction is to be computed. Since the origin is in line with
the straight line portions of the current
circuit in Fig. 23-1, the angle q is zero or 180ø, so sinq = 0 for the current along the
negative x-axis and the negative y-axis.
Thus the only portion that makes a contribution to the magnetic field at the origin is the
circular portion. It has a length of
(3/4) (2pa), or 3pa/2. In addition the origin is at right angles and a distance a from all
parts of the circular portion. We can put
these values in Equation 23.3 to find the magnitude of B:
B = mo/4p[(I)(3pa/2)sin 90ø / a2]
B = mo / 4p3p I / 2a) = 3moI / 8a (1)
What is the direction of B for this problem?
Thinking about the answer
Notice that the answer is just 3/4 of the value of the field at the center of a loop as given
in Equation 23.4. Does that seem
reasonable? Why?
How can you determine the direction of the field in this case? Use a right rule or the
vector product to show that the B points
in the negative Z-direction.
MAGNETIC FORCES ON MOVING PARTICLES
2. A beam of electrons with speed 2.0 x106 m/s is to be deflected 90ø by a magnet as
shown in Fig. 23-2. (e = -1.6 x10-19C,
m = 9.1 x10- 31kg). a) What must be the direction of B? b) What is the force on the
electrons when in the magnetic field? c)
What is the magnitude of B? d) What is the speed of the electrons after they leave the
region of the magnetic field?
What data are given?
Electron speed = 2.0 x106 m/s
Radius of path = 0.20 m
Charge of electrons = -1.6 x10-19C
Mass of electrons = 9.1 x10-31kg.
What data are implied?
The problem assumes that the path of the electrons is in a plane perpendicular to the
magnetic field, not as shown in Figure
23.7(b) in the textbook.
What physics principles are involved?
The problems require that you know the force exerted on a moving charge by a
magnetic field and the centripetal force on a
particle moving in a circular path.
What equations are to be used?
Magnetic force Fm = Bqvsinq (23.2)
Centripetal force F = mv2/r (4.6)
Algebraic solutions
The magnetic force will have to be equal to the centripetal force,
Fm = Bqvsinq = mv2/r
so: B = mv / rqsinq (2)
Furthermore, the B field will have to point in a direction so that the force on negatively
charged particles will point toward the
center of their circular path, in Figure 23-2 that will be toward the lower left hand
corner of the magnetic pole face.
Numerical solutions
Since the plane of the electrons path is perpendicular to the magnetic field the angleq is
90ø, so sinq = 1 in Equation (2).
(a) Using the right hand rule, if you turn your fingers into the page from the direction of
v your thumb points away from the
center of curvature of the path which is the direction the force of the field into the page
would be on positive charges.
However, electrons have negative charge, so a magnetic field into the page would force
them into the curved path as shown if
it has a magnitude given by Equation (2).
(b) The force on the electrons = mv2/r
=[(9.1 x 10-31)(2.0 x 106m/s)2] / (0.20 m) = 1.8 x 10-17 N
(c) B = 1.8 x 10-17N/qv = 1.8 x 10-17N / [(1.6 x 10-19C)(2.0 x 106m/s)] = 5.7 x 105W/m2
(d) The speed of the electrons remains unchanged, only the direction is changed.
Thinking about the answers
Notice that the magnetic field exerts a force on a moving electric charge which is always
perpendicular to the direction of
motion. Hence, no net work is done on the moving particle and the energy of the
particle is not changed by its interaction with
the magnetic field.
PRACTICE TEST
Keywords:: Evaluations; ; Problems; Answers; Questions; Electricity And Magnetism;
Magnetism; Magnetic
Forces; Charges Moving In Magnetic Fields; Biot-Savart Law; Field Lines; Devices Using
Magnetic Fields And Currents;
Pumps; Magnetic Fields
1. A section of the large electrical element inside an oven carries a large current. At the
peak of the AC cycle, the current is
7 amperes. The element has a radius of .6 mm. Assume that this 7 ampere current is
flowing into the cross-section of the
conductor shown.
a. Find the magnetic field produced by the section of wire at a distance of 25 centimeters
from its center. (Given: B =m0I/2pr.)
b. Sketch the magnetic field lines which surround the conductor.
2. An electron with a charge of -1.6 x 10-19 coulomb enters a magnetic field from the
right and follows the path shown.
Please indicate the direction of the magnetic field present in this region. Write a short
statement explaining how you arrived at
your result.
3. The drawing shows a square chamber which is to be the basis for an electromagnetic
pump. The liquid to be moved can
enter from the tube on the left, and leave via the tube on the right. What electrical
components are needed in and/or around
the square chamber area to operate the pump? Please draw a diagram showing the
placement of the equipment and write a
short paragraph explaining how and/or why it works as a pump.
ANSWERS:
1. 5.6 x 10-6 W/m2 See Fig.
2. See Fig.
The B field must be uniform and point into the pages in the region shown. Now using
the right-hand rule with the fingers
pointing to the "left" in the direction of the positive current flow (opposite to the
direction of the negative charge velocity), the
fingers must be rotated down into the page toward the magnetic field vector to get a
force as shown.
3. The fluid must be a good conductor. Then the two important components are a)
voltage source (v) connected to the sides
of the pump which are parallel to the plane of the page. Then, a magnetic field is
needed which provides a field B pointing
down.
See Fig.
As positive ions drift toward the negatively charged front pump face, a force (F = q x v)
is produced toward the liquid exit.
Chapter
Electromagnetic Induction (24):
Keywords: ; Learning Objectives; Electricity And Magnetism; Electromagnetic
Induction; Faraday's Law;
Inductance; Self Inductance; Lenz's Law; Henry; Applications Of Electromagnetic
Induction; Electric Generators
GOALS
When you have mastered the content of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
Faraday's induction law Lenz's law
self-inductance henry
Applications of Induction
Explain the physical basis for the operation of each of the following:
AC generator
electromagnetic damping
search coil
Faraday's Law Problems
Solve problems involving Faraday's induction law.
Lenz's Law
Predict the correct directions for induced current flow in electromagnetic induction
phenomena.
Inductors
Solve problems involving combinations of inductors.
PREREQUISITES
Before beginning this chapter, you should have achieved the goals of Chapter 22, Basic
Electrical Measurements, and
Chapter 23, Magnetism.
OVERVIEW
Keywords: Instruction; Electricity And Magnetism; ; Electromagnetic Induction
Much of our modern technology is based on the fact that an electrical current is
produced by the relative motion between a
magnetic field and a current conductor. In generalizing this phenomenon, Faraday
found a simple rule for predicting the
voltage induced. But predicting the magnitude of the voltage produced by applying
equation (24.2) is only part of the solution;
Lenz's Law predicts the direction of the induced current.
SUGGESTED STUDY PROCEDURE
Start your investigation of this chapter by reading the following Chapter Goals:
Definitions, Applications of Induction, Faraday's
Law Problems, andLenz's Law. An expanded discussion of each of the terms listed
under the Definitions goal can be found in
the next section of this chapter.
Next, read chapter sections 21.1 - 21.5. As you read, be sure to pay particular attention
to the relative directions of motion,
magnetic field, and induced current in the examples given. In each case, the relative
directions are predicted by Lenz's Law,
Section 24.2. For more practice with Lenz's Law, see the Examples section of this
chapter.
Now turn to the end of the chapter and read the Chapter Summary and complete
Summary Exercises 1-10. Next, do
Algorithmic Problems 1-5. Check your answers carefully against those given. Now do
Exercises and Problems 1-6, 8, and 12.
For additional work with the concepts introduced in this chapter, see the Examples
section of this chapter.
After you have completed this study procedure you should be prepared to attempt the
Practice Test provided at the end of
this chapter. If you have difficulty with any of the problems, refer to your text or to
this chapter for
further assistance. This study procedure is outlined below.
------------------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
------------------------------Definitions 24.1 1, 2, 3, 4
Applications of 24.3, 24.4, 24.8 7 5, 6
Induction
Faraday's Law 24.2 8, 9, 10 1, 2 1, 2, 3
Problems
Lenz's Law 24.2 3 4, 12
Inductors 24.5, 24.6 5, 6 4, 5
-----------------------------------------------------24.7, 24.8 11 8
DEFINITIONS
Keywords: Glossary; Electricity And Magnetism; Electromagnetic Induction; Faraday's
Law; Inductance; Self Inductance;
Lenz's Law; Henry
FARADAY'S LAW OF INDUCTION
(law of electromagnetic induction) Predicts that a change in magnetic flux in time
produces an induced emf.
Relative motion between a closed loop conductor (which contains charges) and a
magnetic field will produce an induced emf if
the motion causes a change in either the magnetic field intensity around the conductor
or the area enclosing the field or both.
SELF-INDUCTANCE
Self inductance, L, determines induced emf in a coil of wire due to change of current
through the wire.
The increasing or decreasing field around a coil, even though the field is due to the
current in its own windings, will produce a
current in the windings. The direction of the induced current will always act to inhibit
an increasing current or sustain a
decreasing current.
LENZ'S LAW
Predicts that the induced emf of Faraday's law gives a current that opposes the flux
change producing the emf.
This law is derived from the conservation of energy principle. If Lenz's Law were not
true, it would be possible to build a
perpetual motion machine.
HENRY
The unit for inductance (self and mutual) is the henry.
Large inductors have inductance values of about .5 henries.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Electricity And Magnetism; Electromagnetic Induction;
Faraday's Law; Applications Of
Electromagnetic Induction
SECTION 24.1 Introduction
This chapter discusses the phenomena that are explained by a model of electricmagnetic interactions by which a rapidly
fluctuating magnetic field causes changes in the electrical properties of circuits in those
magnetic fields. The static or noise
on the radio, the signal message on the radio, and the generation of AC voltage all are
understood by means of this model.
SECTION 24.3 Applications of Faraday's Law
The more turns in a search coil the larger will be the angular deflection for a given value
of magnetic induction, Equation 24.4.
So the sensitivity of the detection process is at least partly determined by the number of
turns in the search coil. However,
the number of turns in the coil cancels out of the equation, number 24.5, which is used
to calculate the size of an unknown
field in relation to a known field.
SECTION 24.8 Applications of Electromagnetic Induction
Look up from this and look around to see how many applications of electro-magnetic
induction (EI) you can
deduce?
For example, I am working at a desk in my study. The fluorescent lamp uses EI to start
it. An electric calculator is plugged
in to an AC-DC converter nearby. The telephone on the desk is an application of EI. The
electric typewriter motor is an
application of EI. In fact, a large portion of all of our electrical devices make some use of
at least one result of
electromagnetic induction.
EXAMPLES
Keywords: ; Worked Examples; Lenz's Law; Electricity And Magnetism;
Electromagnetic Induction; Faraday's
Law; Magnetic Flux; Electric Power; Inductors; Inductance; Conservation Of Energy;
Conservation Of Charge
FARADAY'S LAW PROBLEMS
1. A square coil of N turns, edge a, and resistance R is moved through a B field at a
constant speed v. Assume the coil
moves in the xy plane and the motion of the coil is in the positive y- direction. The B
field is constant and parallel to the Z
axis for all values of y less than zero. The B field is zero for values of y greater than zero.
Assume that the coil is initially
completely in the magnetic field. (a) Express the magnetic flux f linking the coil in terms
of B, a, v and/or y, where y is the
coordinate of the forward edge of the moving coil. Sketch the flux as a function of y to
show any variation you predict. (b)
Express the induced emf e in terms of these same variables and sketch it as a function of
y also. (c) Find the power and
total energy dissipated in the coil as it passes out of the B field. (d) What force must be
applied to the coil to keep it moving
at a constant speed?
What data are given?
Let us draw a diagram of this problem so we have a clear picture in our minds as to
what the physical setting really is.
See Fig. 24-1.
What data are implied?
Idealized motion is assumed. Friction is neglected. The magnetic field only exists in the
negative y portion of the xy plane
with the positive y half of the xy plane being entirely void of any magnetic fields.
What physics principles are involved?
Part (a) requires the definition of magnetic flux f.
Part (b) requires the use of Faraday's Law of induction.
Part (c) requires the use of the expression of the electric power dissipated in a
circuit.
Part (d) requires the use of the definition of power and
the special case of a system moving with constant velocity.
What equations are to be used?
Magnetic flux f = BAcosq (24.1)
Faraday's Law e = -Df/Dt (24.2)
Power Dissipated P = e2/R = I2R = Energy/Time (22.23)
Power = Fv (5.13)
Solutions
Part (a). When the coil is completely in the magnetic field B, i.e. the leading edge is still
in the negative y portion of the plane
then the flux f is just equal to B times the area of the coil
f = Ba2 (1)
As the leading edge of the coil emerges from the field the area of the coil is reduced. The
area is reduced by the amount y of
the coil that sticks out of the field
See Fig. 24-2.
See Fig. 24-3.
Part (b). Now we can write the expression for the magnitude of the induced emf in a
coil of N turns
e = ND(f) / Dt = ND(Ba2 - Bay) / Dt (4)
Notice that the only thing in the numerator that changes in time is the value of y; so
Equation (4) is equivalent to
e = NBaDy / Dt = NBav (5)
since the change in y with respect to time is just the definition of speed, for O < y < a.
For all other values of y there is no
change in the flux as time changes since the flux is constant through the coil except for
O < y < a.
See Fig. 24-4.
Part (c). The power dissipated in the coil
P = e2/R = N2B2a2v2 / R (6)
Since the power is a constant, the total energy will be equal to the power multiplied by
the time required for the coil to leave
the field.
time to leave field = a/v = distance/speed (7)
Energy = (Power)time = N2B2a2v2/R (a/v)
Energy = N2B2a3v/R (8)
Part (d). Since the coil is moving with constant speed, the power is equal to the force
times the speed.
Power = Fv (5.13)
So Force = Power/v = N2B2a2v2 / Rv = N2B2a2v / R (9)
Thinking about the answers
This problem illustrates the essential aspect of Faraday's Law - if the magnetic flux is
not changing there is no induced
electromotive force. So it is only as the coil goes from one region of constant flux to
another that there is an induced emf.
Try using your knowledge of the units of the various quantities in Equations 5, 6, 8 and
9 to verify that they are correct
dimensionally.
LENZ'S LAW
2. A square wire loop is rotated upward from the positive x-axis about one edge that
lies along the y-axis in a constant B field
that points in the upward Z-direction. What are the directions of current flow when the
coil is in each of the four portions of
space defined by the xy plane and the yz plane?
What data are given?
See Fig. 24-5.
What data are implied?
This problem seeks only a directional answer, so it need only be assumed that the coil is
always moving in the same
direction but not necessarily with the same angular speed.
What physics principles are involved?
Lenz's law is needed in conjunction with the definition of magnetic flux.
What equations are to be used?
No equations are used. The notion needed is that the current induced in the coil is in a
direction so that its magnetic field
opposes the changes in the magnetic flux linking the coil.
Solution
Let us designate the position of the coil by the location of the edge of the coil that is
rotating in a circle about the y-axis,
shown by the line p p' in See Fig. 24-5.. As the coil rotates up from the positive x-axis,
what is happening to the flux linking
the coil? Since f is the product of B times the component of the area perpendicular to B
and B is constant, we need to focus
our attention on the component of the area perpendicular to the field. This component
of the area is maximum when the coil
lies in the xy plane and zero when the coil is rotated into the yz plane. As the coil is
being rotated up from the x-axis the
area perpendicular to B is being reduced so the flux is decreasing. According to Lenz's
Law then the magnetic field from the
induced current will try to increase the flux so the field caused by the induced current
will be upward in the positive Z-direction
so the induced current around the coil will be such that the current is directly from p to
p' by use of a right hand rule, See
Fig. 23.4 in the textbook. We can write out the answer to this problem as follows:
See Figure.
Thinking about the answers
Notice that the direction of induced current is determined by the direction of the
changes in the flux. In this case the flux is
always in the positive Z-direction, but the direction of current flow changes.
INDUCTORS
3. If you have three inductors of value 3H, 5H and 7H respectively, (a) how many
different ways can you connect them
together in a circuit always using all three of them? (b) what are the effective inductance
values for each connection?
What data are given?
You have a 3.0H inductor, a 5.0H inductor, and a 7.0H inductor connected in various
ways.
What data are implied?
You may assume it is a single wire circuit, i.e. each combination hooks into the circuit at
only two points, since multiple wire
circuits are beyond the scope of our treatment.
What physics principles are involved?
We will use the conservation of energy and charge as discussed in Section 24.7 to derive
the equations for combining
inductors.
What equations are to be used?
Series connections: Leff = L1 + L2 + L3 (24.11)
Parallel connections: 1/Leff = 1/L1 + 1/L2 + 1/L3 (24.12)
Algebraic solutions
Assume you have three inductors of values L1, L2, and L3. How many different ways
can you connect them.
Of course you can connect them all in series, or you can connect them all in parallel.
That makes two different combinations.
You can connect any one of them in series with the other two in parallel. That adds
three more combinations.
Finally you can connect any two of them in series in parallel with the third one. That
adds three more combinations for a total
of eight different combinations as shown below:
See Fig. 24-6.
The effective inductances of each combination are
(A) Series Leff = L1 + L2 + L3 (10)
(B) Parallel Leff = L1L2L3 / (L1L2 + L2L3 + L3L1) (11)
(C) Two in parallel: Leff = L1 + L2L3 / L2 + L3 (12)
(D) Two in parallel: Leff = L2 + L3L1 / (L1 + L3) (13)
(E) Two in parallel: Leff = L3 + L1L2 / (L1 + L2) (14)
(F) Two in series: Leff = L3(L1 + L2) / (L1 + L2 + L3) (15)
(G) Two in series: Leff = L2(L1 + L3) / (L1 + L2 + L3) (16)
(H) Two in series: Leff = L1(L2 + L3) / (L1 + L2 + L3) (17)
Numerical solutions
Let L1 = 3.0H; L2 = 5.0H; L3 = 7.0H. Then the numerical values for the eight
combinations above are(A) Leff = 3.0H + 5.0H + 7.0H = 15H
(B) Leff = (3.0)(5.0)(7.0) / [(3.0)(5.0) + (5.0)(7.0) + (3.0)(7.0)] = 1.5H (C) Leff = 3.0 +
(5.0)(7.0) / (5.0 + 7.0) = 5.9H
(D) Leff = 5.0 + (3.0)(7.0) / (3.0 + 7.0) = 7.1H
(E) Leff = 7.0H + (3.0)(5.0) / (3.0 + 5.0) = 8.9H
(F) Leff = (7.0)(3.0 + 5.0) / (3.0 + 5.0 + 7.0) = 3.7H
(G) Leff = (5.0)(3.0 + 7.0) / 15.0 = 3.3H
(H) Leff = (3.0)(5.0 + 7.0) / 15.0 = 2.4H
Thinking about the answers
Notice that the dimensions of each answer come out to be henries to the first power.
The values range from a high value for
all of them in series (A) to a low value of all of them in parallel (B).
PRACTICE TEST
Keywords: ; Evaluations; Problems; Answers; Questions; Electromagnetic Induction;
Electricity And Magnetism;
Electric Generators; Faraday's Law; Applications Of Electromagnetic Induction;
Induced Current; Devices Using Faraday's
Law; Induced EMF; Inductors; Inductance; Lenz's Law; Energy
1. The drawing below represents the top view of a simple generator.
Sketch the initial direction of current flow through the coil.
See Fig.
b. What type of current will be generated by this mechanism? Sketch a graph showing
how it will vary with time.
c. Name four design changes which will result in producing increased current output
for the above apparatus.
2. A large rectangular shaped core of 50 turns of copper wire is pulled quickly from an
intense magnetic field of 2500 Gauss
which is confined to an area of 25 cm by 25 cm. A voltmeter is connected into the turns
of wire to measure the induced
voltage.
See Fig.
a. If the coil of wire is pulled from the position shown with a velocity of 100 cm/ sec,
find the voltage induced as indicated by
the voltmeter.
b. Show the direction of the induced current in the coil of wire.
3. The large coil below has an inductance of .35 henry. This inductor is connected
through a switch to a 70 volt batter. An
ammeter is also placed in the circuit as shown below.
See Fig.
a. Explain the initial reaction of the inductor at the instant the switch is closed.
b. When the current reaches its maximum value, what energy will be stored in the
inductor if the ohmic resistance of the
inductor is 35 OHMS?
c. Explain the initial reaction of the inductor at the instant when the switch is opened.
ANSWERS:
1. Clockwise; AC; speed of rotation, strength of B, number of turns in armature, area of
coil.
2. 3.1 volt, clockwise
3. Inertial resistance to the increased current, .7 Joule, inertial resistance to the decrease
current flow tends to maintain flow
or produce a spark at the switch which releases the energy stored in the magnetic field.
Chapter
Alternating Currents (25):
Keywords: ; Learning Objectives; Electricity And Magnetism; AC Current; Impedance;
Reactance; Resonance;
Quality Factor; Electromagnetic Oscillations; Measurements In AC AC Circuits; AC
Circuits; Phasors; Transformers; RMS
Voltage; RMS Current
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms and use it in an operational definition:
effective values of current and voltage power factor
reactance resonance
impedance Q-factor
AC Circuits
Solve alternating-current problems involving resistance, inductance, and capacitance in
a series circuit.
Phasor Diagrams
Draw phasor diagrams for alternating current circuits.
Transformer
Explain the operation of the transformer.
AC Measurements
Describe the use of alternating currents in physiological measurements.
PREREQUISITES
Before you begin this chapter you should have achieved the goals of Chapter 22, Basic
Electrical Measurements, and Chapter
24, Electromagnetic Induction.
OVERVIEW
Keywords: ; Instructions; Electricity And Magnetism; AC Current
The electrical energy utilized in the homes of America is delivered by the use of AC
circuits. In this chapter you will learn
about the nature of the AC circuit and its energy equivalent relationship with the DC
circuit. An in-depth study of the AC
circuit: additional components such as inductors and capacitors give the series AC
circuit many interesting phase
relationships and the property of resonance as outlined in Section 25.5. Then in Section
25.8, the Transformer and Its
Application, you will study one of the major advantages in using AC circuits.
SUGGESTED STUDY PROCEDURE
The key Chapter Goals in this chapter are Definitions, AC Circuits, Transformer, and
AC Measurement. When you begin your
study of this chapter, be familiar with each of these goals. Remember that each of the
terms listed under Definitions is
discussed in some detail in the first section of this chapter.
Next, read Chapter Sections 25.1-25.9. Answers to the questions you encounter in your
reading are answered in the second
section of this chapter. In section 25.9, pay particular attention to Table 25.1.
Now turn to the end of the chapter and read the Chapter Summary and complete
Summary Exercises 1-6 and 8-10. Next, do
Algorithmic Problems 1-4, 6, and 7, and Exercises and Problems 1-3, 5, 9, 11, 12, and 15.
Check your answers to each of
these questions against the answers given. If you have difficulty, refer back to the
correct text section. For more practice with
the concepts introduced in this chapter, see the Examples section of this chapter.
Now you should be prepared to attempt the Practice Test on Alternating Currents
provided at the end of this
chapter. Complete the entire test before you check your answers. If you have difficulty
with any part of the test, refer to either
the appropriate section of the or the text. This Study Procedure is outlined below.
-----------------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
-----------------------------Definitions 25.1 1, 2, 3, 4
AC Circuits 25.2, 25.3, 5, 6 1, 2, 3, 1, 2, 3,
25.4, 25.5 4, 6 5, 9
Transformers 25.8 8 7 11, 12
AC Measurements 25.9 9, 10 15
----------------------------------------------------Phasor Diagrams 25.4 7 5 7, 8
DEFINITIONS
Keywords: Glossary; ; Electricity And Magnetism; AC Current; Impedance; Reactance;
Resonance; Quality
Factor; Electromagnetic Oscillations; Measurements In AC AC Circuits; AC Circuits;
Phasors; RMS Voltage; RMS Current
EFFECTIVE VALUES OF CURRENT & VOLTAGE
The values of AC current and AC voltage which will produce the identical heating in a
resistor as a DC circuit. These values
are referred to as RMS values meaning Root Mean Square.
REACTANCE
The proportionality factor (measured in ohms) between the current and voltage for
capacitors and inductors in AC circuits.
Two typical kinds of reactance are capacitive reactance (XC) and inductive reactance
(XL),
IMPEDANCE
Just as ohms law for DC circuits defines resistance as the ratio of voltage (v) to the
current (I), for AC circuits, the
impedance of an AC circuit is the ratio of the effective voltage to the effective current.
POWER FACTOR
Equals the cosine of the phase angle, f, between voltage and current in an AC circuit.
When f = 0ø, the maximum power is dissipated in the AC circuit.
RESONANCE
Occurs when the frequency of the external force equals a natural frequency of the
system.
The condition present in an RLC circuit when XL = XC .
Q-FACTOR
A measure of the sharpness of resonance.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; AC Circuits; Transformers; RMS Current; RMS Voltage;
Measurements In AC Circuits;
Quality Factor; Resonance; Energy; Inductors; Electricity And Magnetism
SECTION 25.1 Introduction
In alternating current circuits the electrical voltage is continuously changing since the
voltage is given by a periodic function
of time such as sin wt where w is the frequency. In the United States of America 60Hz
or 120p radians/second is the usual
frequency of AC circuits. Since the sine function changes its sign from positive to
negative and back, this kind of electrical
circuit is called an alternating current circuit.
Usually electrical devices with no moving parts will work on either AC or DC.
However the transformer is a major exception.
This device with no moving parts is used to connect two portions of a circuit by means
of linking, fluctuating electromagnetic
fields. They are highly efficient devices for changing one voltage into another.
SECTION 25.2 Nomenclature Used for Alternating Currents
Since we specify, in common usage, the effective, or rms, value of an AC current or
voltage the heating effect of AC and DC
circuits are the same for equal currents -- as long as we used rms values for the AC
circuit parameters. Remember it means
that the actual current and voltages in the AC circuit vary by plus or minus 140% of the
stated AC values, e.g. for the usual
110V line in a house the voltage maximum varies from - 155V to + 155V.
SECTION 25.5 Resonance
Questions - 1. The maximum current in a resonance circuit is determined by the
effective resistance of the circuit. At
resonance the current will be given by the voltage divided by the resistance.
2. At any instant in time the total energy of an AC circuit must add up to the input
energy. But consider capacitors and
inductors as devices that are able to store energy for some periods of time. Then at a
time when energy has been stored in a
capacitor it would be possible to take energy from the capacitor and from the emf itself
to increase, for a short time, the
energy stored in the inductor.
SECTION 25.7 Q-Factor
Questions - 3. The energy stored in an inductor is given by 1/2L I 2 and the energy
dissipated in a circuit is I2R. Clearly then
the ratio of energy stored to energy dissipated is proportional to the ratio of L to R. But
L/R is also proportional to the
Q-factor. In fact, if you divide the ratio of the energy stored to energy dissipated by the
period of a cycle then you have the
following:
[(1/2 L I2) / (I2R)] / T = (L/2R) / (1/f) = fL / 2R = wL / 4pR = Q / 4p
so the ratio of energy stored to energy dissipated per cycle is proportional to Q, in fact,
it is equal to Q divided by 4p.
EXAMPLES
Keywords: Worked Examples; ; AC Current; AC Circuits; Impedance; Reactance;
Frequency; LRC Circuits;
Quality Factor; Electricity And Magnetism; Phasors; Conservation Of Energy;
Conservation Of Charge; Resistance; Ohm's
Law; Measurement In AC Circuits; Graphs
AC CIRCUITS
1. Consider a series AC circuit which contains a resistor of 100 ohms, a capacitor of 16.7
microfarads, and an inductor of 240
millihenries connected to a 100 volt (rms) variable frequency power supply.
(a) Calculate the resistance, capacitive reactance, inductive reactance, impedance,
current, and phase angle for frequencies
of 15.9 Hz, 47.7 Hz, 79.6 Hz, 111 Hz, and 143 Hz.
(b) Draw the following graphs for this circuit
(i) Capacitive reactance versus frequency
(ii) Inductive reactance versus frequency
(iii) Resistance versus frequency
(iv) (XL - XC) versus frequency
(v) Z versus frequency
(vi) Phase angle versus frequency
(c) For what frequency is the phase angle equal to zero?
(d) Draw the phasor diagram for each of the above frequencies.
(e) Calculate the Q factor for this circuit and draw a graph of the current versus the
frequency. Show the half-width on the
graph and illustrate the Q factor.
What data are given?
The circuit as shown below is given:
See Fig. 25-1.
What data are implied?
All of the circuit components are assumed to be ideal. The power supply has zero
effective impedance. The resistor has zero
reactance and the reactances have zero resistance.
What physics principles are involved?
Ohm's Law and the conservation of energy and charge as applied to AC circuits as
discussed in Section 25.4 of the text.
What equations are to be used?
Capacitive Reactance XC = 1 / 2pfC ohms (25.11)
Inductive Reactance XL = 2pfL ohms (25.9)
Resistance R = R ohms; independent of f
Total Reactance = XL - XC = 2pfL - 1 / 2pfC
Impedance = SQR RT[R2 + (XL - XC)2] (25.19)
Phase Angle; tan q = (XL - XC) / R (25.20)
Zero Phase Angle occurs when XL = XC; wo = 1 / SQR RT[LC] (25.23)
Q-Factor Q = woL / R (25.24)
Current = Applied Voltage / Z (25.12)
Solutions
To solve this problem we can now just enter the correct numbers from the problem into
these various equations. Rather than
show all of the arithmetic, which is easily accomplished with any hand calculator, let us
just collect our numerical results in a
Table.
See Figs. 25-2.
(c) Phase angle = zero at wo = 500 rad/s; f = 79.6 Hz.
(d) w = 99.9 rad/s w = 300 rad/s
See Fig.
(e) Q Factor = woL / R = (500)(0.240) / 100 = 1.2
w2 - w1 = wo/Q = 500/1.2 = 417 rad/s
See fig.
Thinking about the answers
Notice that in general the curves are not symmetrical about the resonance frequency.
This occurs because the impedance
does not depend in a symmetric way on the frequency when the reactance changes
from being primarily capacitive to
primarily inductive. The curves we have drawn are characteristic of all AC series
circuits.
PRACTICE TEST
Keywords: ; Evaluations; Answers; Problems; Questions; AC Current Transformers;
Voltage; RMS Current;
Resistors; Anatomy And Physiology; Cardiovascular System; Resistance; LRC Circuits;
AC Circuits; Electricity And
Magnetism; Measurements In AC Circuits
1. The apparatus shown below is used to show the properties of the transformer. The
primary and secondary windings have a
common core of heavy iron. The input is from a 120 volt - AC household line.
See Fig.
________a. If the voltmeter shows 6 volts, how many turns are in the secondary
winding?
b. If the resistor (R) in the external secondary circuit is 12 ohms, what RMS current will
be measured by the ammeter? What
maximum IMAX = "peak" current? (Show your calculations.)
IRMS =_______ IMAX =
2. Most electrical signals generated by the body are AC in nature. One common
example is the voltage detected in a heart
E.C.G. measurement.
________a. What is the approximate voltage of this body generated signal?
________b. Explain briefly what electrical hazards exists for a patient while this
measurement is being made.
3. The patient below is inadvertently connected in a series with a 50.0 volts of a 60.0
cycle AC line while being monitored by
an ECG unit.
See Fig.
Assume that the body in this situation acts like a series circuit of R = 2000 OHMS and
the ECG unit provides an inductance
of .100 henry and a capacitance of 0 2.00mfd.
________a. At 60.0 cycles what is the equivalent impedance of this unfortunate circuit?
________b. What current will flow through the patient?
________c. Is this current dangerous? Explain your answer.
ANSWER:
1. 50 turns, Irms = .5 amp, IMAX = .71 amps
2. millivolts, most electrical hazards are associated with electrical leaks in equipment. If
the patient touches a ground (wet
floor, faucet, metal bed post, etc.) a complete circuit could result and allow current to
flow through the body. Leakage currents
higher than 10-3 amps are considered dangerous.
3. 2380 Ohms, .02 amps. It is the current through the heart that is dangerous. At this
level, 20 milliamps could be a problem,
especially if the voltage is continuous, allowing the body's effective resistance to reduce
as the moving current alters the
body's resistance.
Chapter
Bioelectronics and Instrumentation (26): to Accompany Physics
Including Human
Applications:
Citation: H. Q Fuller, R. M. Fuller and R. G. Fuller, to Accompany Physics Including
Human Applications.(Harper
and Row, New York, 1978). Permission granted by the authors.
26 Bioelectronics and Instrumentation
Keywords: ; Learning Objectives; Feedback; Electronics; Diodes; Oscilloscopes;
Amplifiers; Electricity And
Magnetism; Stability; Noise; Signal Processing; Transistors
GOALS
When you have mastered the content of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms and use each term in an operational definition:
feedback interference noise
linearity signal-to-noise ratio
amplification stability
frequency response
Electronic Devices
Explain the basis of operation and potential use for each of the following:
diode rectifier differential amplifier
transistor amplifier oscilloscope
operational amplifier
Oscilloscopes
Evaluate the specifications provided for commercially produced oscilloscopes
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 21,
Electrical Properties of Matter, Chapter 22,
Basic Electrical Measurements, Chapter 23, Magnetism, and Chapter 25, Alternating
Currents.
OVERVIEW
Keywords: ; Electricity And Magnetism; Electronics; Instructions
The advent of transistors has greatly altered the effectiveness and the use of monitoring
instrumentation. In this chapter you
will learn about the basic electronic circuitry and about the important functions
performed by transistors.
SUGGESTED STUDY PROCEDURE
The three goals of this chapter are: Definitions, Electric Devices, andOscilloscopes. First,
read each of these Chapter Goals
carefully. An extended discussion of each of the terms listed under the goal of
Definitions can be found in the first section of
this chapter.
Next, read Chapter Sections 26.1-26.6. As you read, be sure to make good use of the
figures provided. Figures 26.2, 26.3,
and 26.6 are especially important to your understanding of the contents of this chapter.
Now turn to the end of the chapter. Read the Chapter Summary and complete
Summary Exercises 1-10. Next, do Algorithmic
Problems 1 and 2 and complete Exercises and Problems 1-3. Check each of your
answers carefully against those given in
these sections. For more work with the concepts presented in the chapter, now turn to
the Examples section of this Study
Guide chapter.
Now you should be prepared to attempt the Practice Test provided at the end of this
chapter. Complete the
entire test before you check your answers. If you have difficulty with any of the
concepts, check back to your text or to a
section of this for further assistance. This study procedure is outlined below.
-------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
-------------------Definitions 26.1, 26.2 1
Electronic 26.2,26.3, 2,3,4,5, 1,2 1,2,3
Devices 26.4,26.5 6,7,8,9
Oscilloscopes 26.6 10
DEFINITIONS
Keywords: ; Glossary; Feedback; Electronics; Noise; Signal Processing; Amplifiers;
Stability; Electricity And
Magnetism
FEEDBACK
The portion of the output from a system which is returned as input into the same
system.
LINEARITY
Refers to the condition when the output (0) has a linear relationship with the input (I).
If the input signal is being amplified by a factor of 5, a linear condition will assure that
each output signal is five times the
input signal.
AMPLIFICATION
The ratio of output signal to input signal of a device.
The input signal to be amplified might be current, voltage, or power.
FREQUENCY RESPONSE
Refers to the range of input frequencies that yield linear outputs for an instrument or
device.
INTERFERENCE NOISE
A random signal from the environment which interferes with the reception and
monitoring of a signal. A typical problem in
laboratory practice is the interference noise from typical 60 Hz equipment like lights,
xerox machines, motors, etc.
SIGNAL-TO-NOISE-RATIO
Refers to the ratio of the signal amplitude to the undesirable noise signal in a measuring
system.
In practice, reproducible measurements require that this ratio be approximately 10.
STABILITY
Refers to the ability of a system to return to a previously established stable base line
when the input is zero.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: Answers; ; Oscilloscopes; Anatomy And Physiology; Cardiovascular
Systems; Electronics; Electricity
And Magnetism; Applications
SECTION 26.6 Oscilloscope
Questions - 1. To observe an unusual pulse shape such as a heart beat we would
probably want to set the oscilloscope to
trigger on the incoming (external) signal. We probably want to see two or three
complete cycles of the heart beat, let's say
three. Since there are 72 beats in one minute; 3 beats will take 1/24 of a minute or 2 1/2
seconds. Typical oscilloscopes
have a 10 centimeter horizontal sweep distance, so the 2 1/2 seconds should occupy 10
centimeters, or 0.25 seconds/cm.
Compare this answer with the heart beat traces in Figure 22.15.
EXAMPLES
Keywords: Worked Examples; Electronics; Electricity And Magnetism; ; Oscilloscopes
OSCILLOSCOPES
1. Suppose you are using a photomultiplier tube with an output impedance of 50 ohms
to study a very fast light pulse. You
estimate the light pulse will have a half width of 8 nanoseconds and cause the
photomultiplier to put out a current of 25
milliamps. What oscilloscope specifications should you state if you wish to observe the
light pulse on the screen of an
oscilloscope?
What data are given?
The impedance = 50 ohms. The time duration is 8 nanoseconds. The current output is 25
mA.
What data are implied?
The photomultiplier is connected directly into the input terminals of the oscilloscope. It
can be assumed that we will want to
trigger the oscilloscope on the incoming signal.
What physics principles are involved?
The physical characteristics of the oscilloscopes discussed in Section 26.6 of the text.
Solution
First we note that an oscilloscope is a kind of voltmeter so we need to determine the size
of the voltage pulse we are hoping
to observe. The photomultiplier is expected to put out a current of 25 mA across a 50
ohm resistor so the voltage pulse will
have a peak of about 1.25 volts. So a voltage sensitivity of about 200 mV/cm would give
us a voltage pulse about 6 cm high
on the oscilloscope screen.
A time base that would spread out the pulse would be best, so a sweep speed of 1
ns/cm would give us a pulse 8 cm wide
at half maximum. That is probably too wide for photographing the pulse so 2 ns/cm
would be better.
We need a oscilloscope with a fast rise time, able to track this fast signal. It would be
good to have a rise time much faster
than the rise time we expect from the voltage pulse. A rise time of 1 ns would be good.
A time period of 1 ns corresponds to
a frequency of 109 Hz or 1 GHz. In addition we will probably want to trigger on the
external signal which will have a voltage
pulse of 1.25 V so we need a trigger sensitivity less than that. To avoid oscillations setup by the high frequency components
of this voltage pulse we should terminate the input cable with a 50 ohm resistor at the
oscilloscope.
Thinking about the answer
There are a great variety of oscilloscopes and except for the general use introductory
ones they all are designed for special
uses. The kind of oscilloscope you select for use needs to be matched to your needs.
PRACTICE TEST
Keywords: ; Evaluations; Answers; Problems; Questions; Electronics; Electricity And
Magnetism; Diodes;
Oscilloscopes; Frequency
1. Medical monitoring devices which rely upon electric instrumentation are sometimes
limited in the range of measurements
which can be made. One of the limitations is called frequency response.
a. Briefly describe the term frequency response and explain why it is important in
making electronic measurements.
b. What is the frequency response range of a typical oscilloscope? Explain what this
range implies about the use of
oscilloscopes for making measurements.
2. The diode circuit below is connected to a 60 cycle, AC source producing 4 volts, rms.
See Fig.
a. Trace the signal shown on the screen of oscilloscope A.
b. Trace the signal shown on the screen of oscilloscope B.
c. Name and briefly describe at least two uses of the diode in electronic
instrumentation.
ANSWERS:
1. a) Many signals to be measured and monitored are oscillating. The matching of this
frequency to the frequency response of
the measuring device assures an accurate and reliable measurement. The frequency
response in the range of frequencies to
which a system (electronic or mechanical) can respond without distortion.
b) Typical bandwidth: (0-15 megahertz, 0-16 x 106 hertz)
The oscilloscope is a very versatile instrument for making measurements. It can
respond to nearly any frequency and
represent that signal visually as well as accurately.
2. See Fig.
c) Rectifier temperature transducer diode is used extensively.
Chapter
Quantum and Relativistic Physics (27): to Accompany Physics Including
Human
Applications:
Citation: H. Q Fuller, and R. M. Fuller and R. G. Fuller, to Accompany Physics
Including Human Applications.
(Harper and Row, New York, 1978). Permission granted by the authors
27 Quantum and Relativistic Physics
Keywords: ; Learning Objectives; Modern Physics; Quantum Theory; Planck Theory;
Planck's Constant; Planck's
Quantum Hypothesis; Planck's Radiation Law; Wien's Displacement Law; Photons;
Photon Theory Of Light; Photoelectric
Effect; Work Functions; Photoelectric Equation; Theory Of Relativity; Mass Increase;
Theory Of Special Relativity; Length Of
Contraction; Time Dilation; Relativistic Mechanics; Mass And Energy; Quantum
Mechanics; De Broglie's Hypothesis;
Heisenberg Uncertainty Principle; Compton Effect; Wave-Particle Duality; Tunneling
Effect
GOALS
When you have mastered the contents of this chapter you should be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
Planck's constant length contraction
Planck's radiation law time dialation
Wien's law mass-energy equivalence
photon Compton scattering
photoelectric effect complementarity principle
work function deBroglie wave
relativistic mass uncertainty principle
Quantum and Relativistic Problems
Solve problems involving Wien's law, photoelectric effect, Compton scattering,
deBroglie waves, the
uncertainty principle, and relativistic physics formulations.
Tunnel Effect
Define and explain the physical significance of the tunnel effect.
PREREQUISITE
You should have mastered the goals of Chapter 4, Forces and Newton's Laws, Chapter
5, Energy, and
Chapter 21, Electrical Properties of Matter, before starting this chapter.
OVERVIEW
Keywords: ; Instructions; Quantum Theory; Modern Physics; Theory Of Special
Relativity; Theory Of Relativity
Until the end of the nineteenth century, science viewed the basic quantities of a particle
(energy and momentum) as being
continuous. With new phenomena being investigated (radiation, photoelectric effect,
etc.) it became necessary to alter this
basic view to allow each of these quantities to change only in discrete (quantized)
amounts. In this chapter, you will read
about several of these phenomena and to study the hypotheses which result from the
change.
SUGGESTED STUDY PROCEDURE
This chapter places emphasis on three Chapter Goals: Definitions, Quantum and
Relativistic Problems, and Tunnel Effect.
When you begin to study this chapter, read these goals carefully. Remember, a
discussion of the terms listed under the goal
of Definitions can be found in the first section of this chapter.
Now read text sections 27.1-27.9. As you read, please note the importance of the initial
work by Plank on the electromagnetic
radiator, and Einstein on the Photoelectric Effect. A summary of Einstein's Theory of
Relativity is found in Section 27.4. Also,
be sure that you deal seriously with section 27.9 and work carefully through the
examples provided.
At the end of the chapter, read the Chapter Summary and complete Summary Exercises
1-18. Check your answers carefully
against those given on page 623. If you have difficulty, refer to the section referencing
given. Now do Algorithmic Problems
1-7, and Exercises and Problems 1-4, 6-10 and 20. For more work with the important
concepts of this chapter, turn to the
Examples section of this chapter. Be sure to see the tunnel effect example included in
this section.
Now you should be prepared to attempt the Practice Test provided at the end of this
chapter. Be sure to work all
the test parts before you check your answers. For more assistance, refer back to the text
or to this procedure.
This study procedure is outlined below.
------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
------------------Definitions 27.1,27.2 1-11 1 1-4
Quantum & 27.2,27.3,27.4, 12-17 2-7 6-10,20
Relativistic 27.5,27.6,27.7,
Problems 27.8
Tunnel Effect 27.9 18 (see
example in
)
DEFINITIONS
Keywords: Glossary; ; Modern Physics; Quantum Theory; Planck Theory; Planck's
Constant; Planck's Quantum
Hypothesis; Planck's Radiation Law; Wien's Displacement Law; Photons; Photon
Theory Of Light; Photoelectric Effect; Work
Functions; Photoelectric Equation; Theory Of Relativity; Mass Increase; Theory Of
Special Relativity; Length Of Contraction;
Time Dilation; Relativistic Mechanics; Mass And Energy; Quantum Mechanics; De
Broglie's Hypothesis; Heisenberg
Uncertainty Principle; Compton Effect; Wave-Particle Duality; Tunneling Effect
PLANCK'S CONSTANT (h)
The proportionally constant for energy quanta equal to 6.64 x 10-34 Joule/sec.
According to Plank's hypothesis, radiation can only have energy given by En = nhf,
where h is Plank's constant.
PLANCK'S RADIATION LAW
Predicts the thermal radiation intensity as a function of wavelength for a black body at
temperature T (øK).
WEIN'S DISPLACEMENT LAW
Predicts the relationship between the absolute temperature and wavelength for
maximum thermal radiation for a perfect
radiator.
PHOTON
The quantum of electromagnetic energy given by the product of Planck's constant and
the frequency of the radiation.
Einstein first suggested the possibility for photons when discussing the photoelectric
effect in 1905.
PHOTOELECTRIC EFFECT
The phenomena that results in electron emission from materials when radiated with
electromagnetic radiation.
As light is incident in various metal surfaces, electrons are ejected from the metal
surface. Study of the cause and effect
relationships found in this experiment led to the photon theory for light.
WORK FUNCTION
The energy required to free an electron from the surface of a metal.
RELATIVISTIC MASS
The special theory of relativity predicts that the mass of an object increases as its speed,
v, approaches the speed of light,
c.
LENGTH CONTRACTION
Another consequence of the special theory of relativity. It predicts that the apparent
length of an object (measured along the
direction of motion) becomes shorter as the objects speed approaches the speed of light
(c).
TIME DILATION
The special theory of relativity predicts that moving clocks run slow.
MASS-ENERGY RELATION
Einstein's special theory of relativity predicts the famous mass-energy equivalence
given by E = mc2.
Another consequence of the special theory of relativity requires that we consider mass
and energy as two forms of the same
physical quantity. These forms are related through the famous energy-mass equation
which involves the speed of light.
COMPTON EFFECT
The phenomena that results in scattered electrons when photons of incident radiation
interact with weakly bound electrons in
metals.
The loss of energy of the scattered photons (or increase in length) occurs in descrete
quantum amounts depending upon the
angle of scattering.
COMPLEMENTARITY PRINCIPLE
This principle defined by N. Bohr uses both the wave and particle nature of matter and
radiation.
Is light a particle or a wave? The model we choose depends on the nature of the
experiment. This dual description is not a
conflict, but these models complement one another.
DE BROGLIE WAVES
The matter waves that were postulated by de Broglie to have a wave length equal to
Planck's constant divided by the
momentum of the particle.
These wave characteristics have been detected for small particles such as electrons.
Electron diffraction and electron
interference experiments have shown that the dual particle-wave model can be
extended to include even larger particles.
UNCERTAINTY PRINCIPLE
Heisenberg's formulation of the limits of simultaneous knowledge of position and
momentum and energy and time in modern
quantum theory.
The two forms of this principle (Equations 27.16 and 27.17) express the upper limits of
preciseness which we may know the
position, velocity and/or energy of an observed particle.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Wave-Particle Duality; Modern Physics; Quantum Theory;
Quantum Mechanics; De
Broglie's Hypothesis
SECTION 27.6 Complementary Principle
In the political-economic areas the Marxist-Leninist political- economic model with its
emphasis on the collective structure of a
society is in sharp contrast to the classical theory of capitalism with its strong political
and economic emphasis on individual
rights and enterprise. Today it seems likely that most countries have "mixed" economies
that combine features of both
models. The strict adherents of each model claim the other model as the deadly enemy
of themselves and the future of
humankind. Can we think of these two models as two aspects of a complementarity
principle?
SECTION 27.7 The deBroglie Wave
The thermal neutrons are useful for studying crystals structures since their deBroglie
wavelength and range are such as to
produce good diffraction patterns from crystal lattices.
X-rays of the proper wavelength for diffraction from crystals are of greater penetration
in solids and do not readily diffract from
the lower mass atoms found in organic crystals.
Electrons of proper wavelength for diffraction from crystals are of such low energy and
penetration that they produce
diffraction patterns from surface layers or thin films of matter.
EXAMPLES
Keywords: Worked Examples; ; Modern Physics; Quantum Theory; Quantum
Mechanics; Planck's Quantum
Hypothesis; Photon Theory Of Light; Photelectric Effect; Photons; Electrons; Compton
Effect; Conservative Laws;
Conservation Of Energy; Conservation Of Momentum; Wavelength; De Broglie's
Hypothesis; Algebra; Heisenberg Uncertainty
Principle; Time Dilation; Wien's Displacement Law; Theory Of Special Relativity;
Distance; Time; Velocity
QUANTUM AND RELATIVISTIC PROBLEMS
1. Photoelectric Effect. When a copper surface is irradiated with 254 nm light from a
mercury arc it is found that the
photocurrent can be stopped by a potential of 0.24 volts. (a) What is the energy of the
incident photons? (b) What is the
energy of the photo electrons? (c) What is the work function in eV of copper? (d) What
is the threshold wavelength for
copper?
What data are given?
The incident light has l = 254 nm. The photoelectrons are stopped by a potential of 0.24
volts.
What data are implied?
A typical experimental set-up as shown in Figure 27.2 in the textbook is assumed.
What physics principles are involved?
Einstein's theory of the photoelectric effect, as discussed in Section 27.3 of the text, will
be needed.
What equations are to be used?
E = hf = hc / l = KE + W (27.6)
KE = qV (21.16)
Solutions
(a) Incident Photon Energy = hf = hc / l
= ((6.63 x 10-34 J ¥ s)(3 X 1017 nm/sec)) / 254 nm
= 7.83 x 10-19 J = 4.89 eV
(b) The energy of the photoelectrons is given by the stopping potential, so KE of
photoelectrons = 0.24eV = 3.8 x 10-20J.
(c) Work function in eV = 4.89eV - 0.24eV = 4.65eV. It is the difference between the
incident photon energy and the kinetic
energy of the photoelectrons.
(d) Threshold wavelength = hc / E = hc / 4.65eV
= [(6.63 x 10-34 J ¥ s) x (1eV/(1.6 x 10-19 J)) x 3.00 x (1017 nm/s)] / 4.65eV
= 267 nm.
Thinking about the answer
The two wavelengths of light that are mentioned in this problem are in the ultraviolet
part of the spectra. Neither of them is
visible to humans. Copper is not a photoelectron emitter in visible light.
2. Compton Scattering. A 0.0104 nm photon is scattered through 60ø by a collision with
an electron. (a) What happens? (b)
What is the final wavelength of the photon? (c) What is the final velocity of the electron.
What data are given?
The initial energy of the photon is given as 0.0104 nm and the scattered angle is given as
60ø.
What data are implied?
It is assumed that the electron is essentially at rest when struck by the photon.
What physics principles are to be used?
The conservation of momentum and energy for particles of light was done by Compton,
see Section 27.5 in the book.
What equations are to be used?
Wavelength shift l' - l = (h / (moC)) ¥ (1 - cos f) (27.14)
Momentum conservation hf/c - hf'/c cos f = Pe cos q (1)
hf'/c sin f = Pe sin q (2)
Algebraic solutions
Part (b) is found by using equation 27.14. To find an expression for the velocity of the
electron we can combine Equations (1)
and (2). To find the angle of electron departure, divide Equation (2) by Equation (1)
where l = c / f,
tan q = (h/l' sin f) / (h/l - ((h cos f/l'))
tan q = (l sin f) / (l' - l cos f) (3)
Similarly to find the magnitude of the electron velocity, square Equations (1) and (2)
and add them together, remember sin2q +
cos 2q = 1
Pe2sin2q + Pe2cos2q = (h2/l'2) sin2f + (h/l - h/l' cos f)2
If the electron does not travel at relativistic speeds, then
(me)2(ve)2 ÷ Pe2 = h2/l'2 + h2/l2 - (2h2)/ll' cos f
(ve)2 ÷ h2/(me)2 ¥ (1/l2 + 1/l'2 - ((2cosf) / ll')
ve ÷ h/me ¥ (SQR RT) (1/l2 + 1/l'2 - ((2cosf) / ll') (4)
Numerical solutions
(a) When the photon is treated as a particle then this phenomena can be discussed as if
it were a typical classical collision
problem. The incident object (a photon) collides with a massive, stationary object (an
electron) and both go off after the
collision in different directions. Both energy and momentum are conserved for the
collision process so if the initial energy of
the incident particle and its scattered angle are known all the other results of the
collision can be calculated.
(b) The calculation of the final wavelength of the scattering photon is straight forward
using equation 27.14:
l' = 0.0104nm + (2.43 x 10-12 (1 - cos 60ø))
= 10.4 x 10-12 + 1.2 x 10-12
lø = 1.16 x 10-11 m.
(c) The calculation of the angle and magnitude of the electron velocity after the collision
is done using Equations (3) and (4).
tan q = (1.04 x 10-11 sin 60ø) / (1.16 x 10-11 - 1.04 x 10-11 cos 60ø)
tanq = 1.41
q = 54.6ø
ve = (6.63 x 10-34J ¥ S)/(9.1 x 10-31kg) ¥ (SQR RT)[1/(1.04 x 10-11)2 + 1/(1.16 x 10-11)2 ((2 cos 60ø)/(1.04 x 10-11)(1.16 x 10
-11))]
= 7.29 x 10-4 ¥ (SQR RT)(9.25 x 1021 + 7.43 x 1021 + 8.29 x 1021)
ve = 7.29 x 10-4 ¥ (SQR RT)(2.50 x 1022)
ve = 1.15 x 108 m/s
Thinking about the answers
You can see that the scattered photon has a longer wavelength, hence a lower energy
than the incident photon. During the
scattering process the photon loses energy and the electron gains energy. In the above
problem the incident photon lost about
10% of its energy in the collision. Can you verify this value for the photon energy loss.
Since the value obtained for ve is
near the speed of light, the classical approximation used to derive Equation (4) is not
valid. The momentum equation 27.9
should have been used.
3. deBROGLIE WAVES. Derive an expression for the wavelength associated with the
molecules of a gas whose absolute
temperature is TøK. Then calculate the wavelength associated with oxygen molecules at
room temperature (300øK).
What data are given?
The absolute temperature of the gas is TøK. For the specific case of the oxygen
molecules T = 300øK.
What data are implied?
The gas can be assumed to obey the kinetic theory of ideal gases. The mass of the gas
molecules must be known, call it m.
We can calculate the mass of the oxygen molecule from the periodic chart (p. 754) and
Avogadro's number.
1 gram molecular wt. of 02 = 32.0 gm and contains 6.02 x 1023 molecules.
mo2 = 32.0 gm / 6.02 x 1023 = 3.20 x 10-2 kg / 6.02 x 1023 = 5.32 x 10-26 kg.
What physics principles are involved?
The properties of an ideal gas, Section 14.3, must be combined with the deBroglie
hypothesis, Section 27.7.
What equations are to be used?
The kinetic energy of a gas molecule 1/2mv2 = 3/2kT (14.15)
deBroglie wavelength l = h/p = h/mv if v << c (27.15)
Algebraic solution
The rms velocity of a gas molecule is found from Equation 14.15
1/2 mv2 = 3/2 kT (5)
v = (SQR RT)((3kT)/m)
So the deBroglie wavelength for a gas molecule is given by
l = h/[m ¥ (SQR RT)((3kT)/m)] = h/(SQR RT)(3mkT)
Numerical solution
For room temperature oxygen T = 300ø
m = 5.32 x 10-26kg
lo2 = 6.63 x 10-34J¥s/(SQR RT)[3(5.32 x 10-26kg)(1.38 x 10-23 J/moleøK)(300øK)]
lo2 = 2.58 x 10-11 meters
Thinking about the answer
This wavelength is much smaller than any usual laboratory diffraction grating spacing
and will not be easily detected by any
usual means.
4. UNCERTAINTY PRINCIPLE. Consider a proton trapped in a nucleus of 6.0 x 10-15m
in diameter. Suppose the potential
barrier holding the proton in the nucleus is 24 MeV. (a) What is the uncertainty in the
momentum of the proton? (b) What is
the uncertainty in the energy of the proton. (c) What is the minimum energy the proton
needs to escape from the nucleus? (d)
What is the lifetime of the proton in this nucleus?
What data are given?
The uncertainty in the position of the proton is 6.0 x 10-15m. The height of the potential
barrier is 24 MeV.
What data are implied?
The classical, nonrelativistic relationships between momentum and energy are good
approximations in this problem. The mass
of the proton is 1.7 x 10-27 kg.
What physics principles are involved?
This problem uses the Heisenberg Uncertainty Principle.
What equations are to be used?
DP Dx ò h (27.16)
DE Dt ò h (27.17)
Solutions
(a) Uncertainty in momentum ÷ h/Dx = (6.63 x 10-34 J ¥ s)/(6.0 x 10-15m)
DP ÷ 1.1 x 10-19N ¥ s
(b) Uncertainty in energy = (DP)2/2m = (1.1 x 10-19)2 / (2(1.7 x 10-27)) = 3.6 x 10-12 J
= 22.4 MeV.
(c) Minimum Energy to Escape = 24 MeV - 22.4 MeV = 1.6 MeV
(d) Lifetime ÷ h / DE ÷ (6.63x 10-34J ¥ s) / (3.6 x 10-12J) ÷ 1.8 x 10-22 seconds.
5. A Space Ship traveling at 0.6 c passes a "stationary" observer. Both the observer and
the spaceship are provided with
identical timers, which they start the instant they pass. Each timer is wired to a light,
which flashes after ten seconds has
elapsed.
(a) Which light does the stationary observer believe flashes first?
(b) Which light does the spaceship crew believe flashes first?
(c) How much time does the stationary observer believe has elapsed on the observers
own timer when the one on the
spaceship flashes?
(d) How much times does the spaceship crew believe has elapsed on the stationary
timer when their own spaceship light
flashes?
(e) How far apart are the two timers when the one on the spaceship flashes according to
the stationary observer?
(f) How far apart are the two timers when the one on the spaceship flashes according to
the spaceship crew?
(g) How long does it take the light from the first spaceship light flash to reach the
stationary observer, according to the
observers timer?
(h) How long does it take the light from the first spaceship light flash to reach the
stationary observer, according to the
spaceship crew?
(i) What time does the spaceship crew believe is shown on the stationary timer for the
time of flight?
(j) From your above results fill out the table below:
See Fig.
(k) What do the two reporters agree upon? disagree upon?
What data are given?
The spaceship travels at 0.6c (2.4 x 108m/s) past the observer. Lights are flashed every
10 seconds according to a clock
connected to the light flasher.
What data are implied?
The conditions of special relativity are satisfied.
What physics principles are involved?
The concept of time dilation is used in conjunction with the classical concepts of
distance, time, and velocity.
What equations are to be used?
Time dilation Dt = Dto / (SQR RT)(1 - (v2/c2)) (27.8)
Distance = vt (3.2)
Solution
(a) The stationary observer knows that moving clocks run slow. So the observer can
calculate how long 10 seconds will seem
to be in the spaceship, according to the observers clock.
(SQR RT)(1 - (v2/c2)) = (SQR RT)(1 - (0.6c/c)2) = (SQR RT)(1 - .36) = (SQR RT)(.64) = 0.8
So a spaceship 10.0 seconds would appear to be 10/0.8 or 12.5 seconds on the observers
clock. The observer believes the
spaceship light flashes after the observers light has already flashed.
(b) The spaceship crew believes that moving clocks run slow, but according to them the
observer is the one who is moving.
Consequently when 10 seconds has passed on their clock the observers clock will show
a number smaller than 10 seconds,
in fact they compute a time of (0.8) (10 sec) = 8 seconds will show on the observers clock
when their light flashes. Which,
according to them, occurs a full two seconds on the observers clock before the observers
light flashes.
So both parties believe the other light flashes after their own light has already flashed.
(c) We answered above, 12.5 seconds.
(d) We answered above, 8.0 seconds.
(e) According to the observer the spaceship has been traveling at 0.6c for 12.5 seconds
when the spaceship light flashes, so
d = (0.6c) (12.5), d = 7.5c.
(f) According to the spaceship crew they have been traveling for 10 seconds when their
light flashes so they are d = (0.6c)
(10 sec) or 6c from the observer when their light flashes.
(g) Since, according to the observer, the light was 7.5c away and light travels at a speed
c it will take 7.5 seconds to reach
the observer.
(h) The spaceship crew has a tough problem to solve. They are 6c away from an object
which is going away from them at a
speed of 0.6c. So they send a message to that moving object at a speed of c, how long
will it take to catch up? Let t' be the
time to catch up; then 0.6c t' is the distance the moving object travels in t' seconds, that
distance plus 6c will be equal to the
distance the message travels which is ct'
0.6ct' + 6c = ct'
6 = 0.4t'
t' = 15 seconds.
So according to the spaceship crew it will take 15 seconds for their light flash to reach
the observer.
(i) However, the spaceship crew knows that the observers clock runs slow, so the 15
seconds on their clock will appear as
(15sec)(0.8) or 12 seconds on the observers clock.
(j) We can now fill in the table:
See Fig.
(k) The reporters agree on the time of arrival of the flash at the observers clock. They
disagree about intermediate computed
quantities.
6. Wien's Law - (a) If the absolute temperature of an object is doubled what happens to
its emitted radiations? (b) What is a
low temperature limit for the emission of visible light?
What data are given?
(a) The absolute temperature of the emitted body is doubled.
What data are implied?
(b) Let us take the low energy limit for visible light to be 700 nm.
What physics principles are involved?
This problem makes use of Planck's Law for black body radiation as expressed in
Wien's Law.
What equation is to be used?
Wien's Law: lmax T = 2.88 x 10-3møK (27.5)
Solutions
(a) If the absolute temperature is doubled, then the peak wavelength of the light is
halved. If the original wavelength is red,
the final wavelength would be in the violet, the total spectrum would become whiter.
This explains the fact that white hot
objects are hotter than red hot objects.
(b) lmax ÷ 700 nm = 7 x 10-7 m
T = (2.88 x 10-3øK) / (7x10-7) = 4.1 x 103øK = ÷ 3800øC
This is a relatively low temperature for a hot object, about 1600ø cooler than the sun's
surface.
PRACTICE TEST
Keywords: ; Evaluations; Answers; Questions; Problems; Modern Physics; Theory Of
Special Relativity; Length
Contraction; Mass Increase; Time Dilation; Relativistic Mechanics; Postulates Of Special
Relativity; Photon Theory Of Light;
Photons ; Photoelectric Effect; Quantum Theory; Tunneling Effect; Kinetic Energy
1. Einstein's theory of Relativity was a milestone in the development of scientific
thought in the 20th century.
a. State the two postulates upon which Einstein based his special theory of relativity.
Postulate 1:
Postulate 2:
b. Briefly describe the consequences of observers in one frame of reference measuring
time, distance, and mass in a frame
which is moving at speeds approaching the velocity of light.
2. Light of wavelength 200 nm falls on an aluminum surface where 4.20 eV's are
required to remove electrons.
a. What is the kinetic energy of the fastest emitted photoelectrons?
b. What is the stopping potential for these photoelectrons?
c. What is the cutoff wavelength for aluminum?
3. An alpha particle in a nucleus can be thought of as being trapped in a "potential well"
that has a barrier, height of 8.0 MeV
(8.0 x 106 eV) and a barrier width and well width of 1.0 x 10-14 m as shown.
a. Calculate the uncertainty in the energy of the alpha particle. (m = 6.7 x 10-27 kg)
See Fig.
b. What minimum energy must the alpha particle have to tunnel through this potential
barrier?
ANSWERS:
1. Postulate 1: The laws of physics are the same in all inertial reference frames.
Postulate 2: The speed of light (c) in a vacuum is measured as a constant by all
observers regardless of their relative
motion.
consequences: length contraction, time dialation, effective mass increase
2. K.E.MAX = 2.00 ev (3.20 x 10-19J), 2.00 volt, 295 nm
3. 3.3 x 10-13J (2.0 Mev, 6.0 Mev)
Chapter
Atomic Physics (28):
STUDY
GUIDE
Citation: H. Q Fuller, R. M. Fuller and R. G. Fuller, to Accompany Physics Including
Human Applications.(Harper
and Row, New York, 1978). Permission granted by the authors.
28 Atomic Physics
Keywords: Learning Objectives; ; Modern Physics; Atomic Physics; Quantum Theory;
Atomic Number; Bohr
Model; Bohr Radius; Models Of The Atom; Quantum Numbers; Atomic Energy Levels;
Atomic Spectra; Lasers; Pauli
Exclusion Principle; Lambert's Law; Spectroscopy; Atomic Structure And Properties
GOALS
After you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms and use it in an operational definition:
atomic number energy level
Bohr radius spectra
quantum number laser
Bohr Model
State the assumptions and predictions of the Bohr model of the hydrogen atom. Deduce
the concept of ionization potential.
Pauli Exclusion Principle
Use the Pauli exclusion principle and the four quantum numbers; n, l, ml, andms in the
periodic chart of the elements
Bohr Model Problems
Solve problems using the Bohr equations for hydrogen.
Energy Level Diagrams
Use energy level diagrams to explain the emission and absorption of radiation by
atomic systems.
Spectrometry and Lambert's Law
Sketch a simple spectrometer, and discuss how it can be used in measuring the
concentration of an element in a solution.
Laser
Explain the principle of laser action.
PREREQUISITES
Before you begin this chapter you should have achieved the goals in Chapter 4, Forces
and Newton's Laws, Chapter 5,
Energy, Chapter 21, Electrical Properties of Matters, and Chapter 27, Quantum and
Relativistic Physics.
OVERVIEW
Keywords: ; Instructions; Modern Physics; Quantum Theory; Atomic Physics
The introduction of Bohr's model for the atom was a significant advancement for
science. Although this model does not
explain much of the fine detail of more complicated atoms, this model was successful in
representing the hydrogen atom and
explaining how electrons behave in and around the nucleus. In this chapter you will
find the derivation of the Bohr model and
will study the atomic effects which are examples of its success.
SUGGESTED STUDY PROCEDURE
Before you begin to study this chapter, be familiar with the following Chapter Goals:
Definitions, Bohr Model, Bohr Model
Problems, Energy Level Diagrams, and Spectronomy and Lambert's Law. More
complete discussion of the terms listed under
Definitions can be found in the first section of this Chapter.
Next, read text sections 28.1-28.8 and 28.10. As you read the derivation of the Bohr
model in section 28.3, be sure to note
the major hypotheses (Postulate 1 and 2) and follow carefully through the derivation on
page 632. The Example on page 633
will help you see how the model can be used in a problem situation.
At the end of the chapter, read the Chapter Summary and completeSummary Exercises
1-3 and 5-12. Now do Algorithmic
Problems 1-4 and complete Exercises and Problems 3, 5, 8, 9, and 10. For more practice
with the concepts introduced in this
chapter, see the Examples section of this .
Now you should be prepared to attempt the Practice Test provided at the end of this
chapter. If you have
difficulties with a part of the test, refer back to the appropriate text section or to this
study procedure. This study procedure is
outlined below.
-------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
-------------------Definitions 28.1,28.2,28.4,
28.10
Bohr Model 28.3 1,2,3
Bohr Model 28.3 5,6,7 1,2 3
Problems
Energy Level 28.5 5,6,8,9 5
Diagrams
Spectrometry 28.6,28.7,28.8 10,11,12 3,4 8-10
& Lambert's Law
---------------------------------------------------Pauli Exclusion 28.4 4 7,11
Principle
Laser 28.10
DEFINITIONS
Keywords: ; Glossary; Modern Physics; Quantum Theory; Atomic Physics; Models Of
The Atom; Bohr Model;
Atomic Number; Bohr Radius; Quantum Numbers; Atomic Energy Levels; Atomic
Spectra; Lasers; Atomic Structure And
Properties
ATOMIC NUMBER
The number of protons (or electrons) in an atom. The symbol Z represents the atomic
number.
The total electric charge of the nucleus of an atom is equal to +Z.
BOHR RADIUS
The radius of the first Bohr orbit for the hydrogen atom 0.53 x 10-10m.
This represents the distance from the nucleus (proton) to the electron when the electron
has a minimum energy (ground
state).
QUANTUM NUMBER
The discrete nature of quantum theory is characterized by four quantum numbers (n, l,
ml, ms) that determine the energy
structure and quantum state of atoms.
Notice that quantum numbers can have only certain discrete values, n, l, and ml are
always integers; ms is either +1/2 or
-1/2.
ENERGY LEVEL
The allowed energy values predicted for an atom by quantum theory.
In the original work of Bohr the condition of an atom in which, at least for an instant, it
had a definite value for its energy was
called a stationary state. Since each stationary state has a characteristic energy we also
call them energy levels.
SPECTRA
Each atom has a unique set of emission lines, these spectra are the "fingerprints" of
atoms.
The discovery by Sir Isaac Newton that white light consisted of a continuous mixture of
all the colors lead to a study of the
light emitted by many light sources. The study of spectra was highly developed before
the formulation of the modern quantum
theory of matter.
LASER
An acronym for light amplifier by stimulated emission of radiation. A highly
monochromatic coherent light source.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Lasers; Atomic Excitation; Models Of The Atom; Bohr Model;
Bohr Electron Energy;
Atomic Physics; Quantum Theory; Modern Physics
SECTION 28.1 Introduction
Perhaps a laser is the most common, visible device that illustrates quantum properties
of matter. However, since the quantum
model is the best model we have there is a sense in which every phenomenon is an
example of the quantum model.
The excitation of atoms in the gases in a fluorescent light results in the emission of
electromagnetic radiation.
SECTION 28.3 Bohr's Model for the Hydrogen Atom
The negative values for the energy of the hydrogen atom indicate that the electron
states are bound, or finite, states for the
electrons.
The energy difference between the n=3 and n=1 states of hydrogen is given by
E3 - E1 = -13.6 (1/32 - 1/12) eV = -13.6 (-8/9) eV = 12.1 eV
The transition from n=5 to n=2 permits the emission of energy given by
E2 - E5 = -13.6 (1/22 - 1/52) = -2.86 eV
EXAMPLES
Keywords: ; Worked Examples; Modern Physics; Quantum Theory; Models Of The
Atom; Bohr model; Quantum
Numbers; Bohr Electron Energy; Paschen Series; Hydrogen; Lambert's Law; Beer's Law;
Hydrogen Spectra; Spectroscopy;
Atomic Physics
BOHR MODEL PROBLEMS
1. Calculate the wavelengths of the three lines of the Paschen (n=3) series of spectra for
hydrogen.
What data are given?
Principle quantum number = 3.
What data are implied?
The usual Bohr model postulates are adequate for this system.
What physics principles are involved?
The basic concepts of the Bohr model for the hydrogen atom.
What equations are to be used?
Em - En = -13.6 (1/m2 - 1/n2) eV (28.5)
E = hc / l (27.1)
note: l E = 1240 eV/mn (30.2)
Solutions
The Paschen series is formed by electron transitions from higher n values to n=3. The
first three Paschen lines will thus be
for transitions from the n=4, 5, and 6 levels down to n=3.
E4 - E3 = -13.6 (1/42 - 1/32) eV = 0.661eV
E5 - E3 = -13.6 (1/52 - 1/32) eV = 0.967eV
E6 - E3 = -13.6 (1/62 - 1/32) eV = 1.13eV
Thinking about the answers
None of the Paschen lines are visible but they lie in the infrared portion of the
electromagnetic spectra.
2. What is the approximate quantum number n for a hydrogen orbit large enough to be
observed under a microscope?
What data are given?
A hydrogen atom is being inspected under a microscope.
What data are implied?
Typical quality microscope can resolve objects about 0.0005 millimeters or 5 x10-7
meters in size.
What physics principles are involved?
The Bohr model for the hydrogen atom.
What equations are to be used?
rn = (0.53 x 10-10) n2 (28.3)
Solution
Let us take the diameter of the orbit to be 5 x10-7m then r = 2.5 x10-7m
2.5 x 10-7m = (0.53 x 10-10) n2
4.7 x 103 = n2
69 ÷ n
Thinking about the answer
The energy of an electron in the n=69 state is about 3 x 10-3eV. The energy of a visible
photon that you would need to reflect
off of the electron in order to see it has an energy ~ 3eV. What other reasons can you
give to explain why no one has seen
an electron?
SPECTROMETRY AND LAMBERT'S LAW
3. A technician is mixing silver into molten glass to make partially reflecting mirrors.
She knows that an 8% silver glass only
transmits 88% of the incident light for a given thickness of glass. For twice the thickness
of glass she wishes to have a
glass transmit only 50% of the light. What percentage silver glass does she need for this
glass?
What data are given?
For a thickness of glass x the transmission is 88% when the silver concentration is 8%.
What data are implied?
It is assumed the conditions for Beer's Law to be valid are met.
What physics principles are involved?
The basic concept of absorption of energy as it passes through matter, as expressed in
Lambert's Law.
What equations are to be used?
Lambert's Law I = Ioe-mx (28.9)
Beer's Law m = aC (1)
Algebraic solution
Let xo be the initial thickness, To the initial percent transmission, x1 the second
thickness, and T1 the final transmission, Co
the initial concentration and C1 the final concentration.
To/100 = I/Io = e -a Cx = e -a Coxo (2)
T1 / 100 = e -a C1x1 (3)
Take the logarithms to the base e of both sides of Equations (2) and (3).
ln (To/100) = -aCoxo (4)
ln (T1/100) = -aC1x1 (5)
Then (ln(To/100)) / (ln (T1/100)) = (Coxo) / (C1x1) (6)
So by knowing To, T1, Co, xo, and x1 you can calculate C1.
Numerical solution
To = 88%;
T1 = 50%;
xo = xo;
x1 = 2xo;
Co = 8%
(ln (0.88)) / (ln (0.50)) = (8% xo) / (2C1xo)
0.184 = 8% / 2C1
C1 = 8 / (2(.184)) = 21.7% silver glass
Thinking about the answer
Notice the logarithmic dependence on concentration makes it difficult to estimate
correct answers by linear extrapolation.
PRACTICE TEST
Keywords: ; Evaluations; Answers; Questions; Problems; Bohr Model; Hydrogen;
Hydrogen Spectra; Modern
Physics Quantum Theory; Atomic Physics; Models Of The Atom; Spectral Absorption;
Medical Diagnosis
1. The remarkable Bohr model was designed to explain the structure and stability of the
atom, and to explain the
characteristic line spectra produced when the atom was excited.
a. Briefly state the two postulates of the Bohr model for the atom.
b. Does the Bohr model successfully predict the observed spectra of most atoms?
Explain your answer.
2. The first four energy levels for the Hydrogen Atom are: -13.6 ev (ground state), -3.40
ev, -1.51 ev, and -.85 ev.
a. Predict the wavelength of the radiation produced when an excited atom in the n = 4
state returns to the ground state.
b. What wavelength of incident radiation will be absorbed by the hydrogen atom and
produce an atom from the ground state
to jump into the n = 3 state?
c. Predict the energy of the n = 7 state.
3. A lab technician uses a colorimeter to conduct blood sugar studies. She uses the
following calibration data:
Solution Readout Current (mA)
Distilled Water 300.
20.0 ppm sugar in water 30
a. When a patient's morning sample is placed in the colorimeter a readout current of
20.0 mA is obtained. What is the blood
sugar concentration in this patient's morning sample?
b. When the same patient's evening sample is placed in the colorimeter, the scale
reading is 150 mA. What is his evening
concentration of blood sugar?
ANSWERS:
1. a) *Electrons are contained in stationary states whose energy level is predicted by
quantum conditions.
*Radiation is emitted or absorbed when an electron makes a transition from one
stationary state to another
b) For the relatively uncomplicated hydrogen atom, the theory is remarkably accurate.
With more complex atoms containing
many orbiting electrons, the theory does not predict the nature of the fine structure
found in spectral analysis. Therefore the
use of the theory is limited.
2. 98 nm, 100 nm, -.28 ev's
3. 23.4 ppm, 6.0 ppm
Chapter
Molecular and Solid-State Physics (29): to Accompany Physics Including
Human
Applications: Citation: H. Q Fuller, R. M. Fuller and R. G. Fuller, to
Accompany Physics Including Human Applications.(Harper and Row, New York,
1978). Permission granted
by the authors.
29 Molecular and Solid-State Physics
Keywords: ; Learning Objectives; Modern Physics; Solid State Physics; Molecules;
Bonding In Molecules; Ionic
Bonds; Covalent Bonds; Molecular Spectra; Fluorescence; Phosphorescence;
Bioluminescence; Semiconductors;
Superconductivity; Molecular Spectra; Nuclear Magnetic Resonance; Solids; Band
Theory Of Solids
GOALS
When you have mastered the content of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use each term in an operational definition:
ionic bonding extrinsic semiconductor
covalent bonding semiconductor
fluorescence superconductivity
phosphorescence nuclear magnetic resonance
bioluminescence
Band Theory of Solids
State the band theory of solids, and use it to explain the optical and electrical properties
of different solids.
Molecular Absorption Spectra
Explain the basis of vibrational and rotational absorption spectra of molecules.
Solid-State Problems
Solve problems involving vibrational and rotational energy states of molecules and the
band theory of
solids.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 21,
Electrical Properties of
Matter, Chapter 27, Quantum and Relativistic Physics, and Chapter 28, Atomic Physics.
OVERVIEW
Keywords: ; Instructions; Modern Physics; Solid State Physics; Molecules
Investigation has shown that there are basically four types of bonding which hold
atoms together in solids; ionic, covalent,
Van der Walls, and metallic bonding. These bonding models are primarily electrical in
nature and predetermine many of the
macroscopic properties observed in solids. These properties include hardness, chemical
activity, electrical activity, thermal
activity, and boiling point. In this chapter you will be introduced to several general
properties of solids and to applications
which utilize these unique characteristics.
SUGGESTED STUDY PROCEDURE
Start your study of this chapter by reading these Chapter Goals: Definitions, Band
Theory of Solids, and Molecular Absorption
Spectra. An extended discussion of many of the terms listed under the Definitions goal
are provided in the first section of this
chapter.
Next, read text sections 29.1-29.18. As you read sections 29.4 and 29.3, please note the
simplistic nature of the "dumbbell"
model and the influence of the quantum nature found in the expressions for Eviband
Erot. Questions which you encounter in
reading are answered in the second section of this chapter.
Now turn to the end of the chapter and read the Chapter Summary and complete
Summary Exercises 1-9. Next, do
Algorithmic Problems 1 and 2 and Exercises and Problems 1 and 2. For more work with
the concepts introduced in this
chapter, work through the problems given in the Examples section of this chapter.
Now you should be prepared to attempt the Practice Test located at the end of this
chapter. If you have
difficulty with any part of the test, refer back to a specific text section or to this for
further assistance. This
study procedure is outlined below.
-------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
-------------------Definitions 29.1,29.2,29.3, 1-6 1,2
29.6,29.7,29.14
29.15, 29.16,
29.17
Band Theory of 29.8,29.9,29.10, 7
Solids 29.11,29.12,
29.13
Molecular 29.4,29.5,29.18 8,9 1,2
Absorption Spectra
--------------------------------------------Solid State 29.4,29.5 10,11 3,4 4,5,8
Problems
DEFINITIONS
Keywords: Glossary; ; Modern Physics; Solid State Physics; Molecules; Bonding In
Molecules; Ionic Bonds;
Covalent Bonds; Molecular Spectra; Fluorescence; Phosphorescence; Bioluminescence;
Semiconductors; Superconductivity;
Molecular Spectra; Nuclear Magnetic Resonance; Solids; Band Theory Of Solids
IONIC BONDING
Attractive force between positive and negative ions makes a strong bond for ionic
materials like Na+ Cl-.
To gain some appreciation for the size of the electric forces in ionic materials remember
that the largest electric fields in your
usual environment are on the order 104V/m near high voltage transmission lines. The
electric fields in ionic materials are a
million times larger than those, i.e. 1010V/m.
COVALENT BONDING
The bond produced by shared electrons between atoms making up covalent systems,
e.g., H2.
Compounds studied in organic chemistry are primarily covalent materials in which the
carbon-hydrogen covalent bonding is
most prevalent.
FLUORESCENCE
The light emitted by a substance while it is irradiated. Fluorescent light has longer
wavelength than incident light and ceases
when incident light is removed.
The common fluorescent light fixture makes use of this phenomena for producing
visible light.
PHOSPHORESCENCE
Characterized by light emitted after incident radiation is removed from the source
material.
The various glow-in-the-dark objects are examples of phosphorescent materials.
Bioluminescence
Certain biological systems produce light as a consequence of their biochemical
processes.
The cool light emitted by insects such as fireflies is a common example of
bioluminescence.
EXTRINSIC SEMICONDUCTOR
A semiconductor that has been doped with a donor material (n type) or an acceptor
material (p type).
A donor is thought of as an element with one more electrons in its outer most electron
shell than the host semiconductor. An
acceptor is an element which one less electron in its outer most electron shell than the
host semiconductor. Look at a
periodic table (p. 754-755 in the text) and list the elements that will act as acceptors in
germanium or silicon, as donors.
SEMICONDUCTOR
A material whose room temperature electrical conductivity (s ÷ 1 ohm-1 m-1) is much
greater than the conductivity of
insulators (s ÷ 10-17 ohm-1 m-1) and much less than the conductivity of the good
metallic conductors (s ÷ 10 4 ohm-1 m-1).
SUPERCONDUCTIVITY
The phenomena characterized by materials that show resistivity going to zero at a
critical temperature between absolute zero
(OøK) and 30øK.
It is possible to construct a model of superconductivity which predicts the existence of
room temperature superconductors,
but no one has found one yet.
NUCLEAR MAGNETIC RESONANCE
The phenomena characterized by energy level resonance associated with the proton
magnetic dipole in materials.
The extreme sensitivity of detection systems making use of resonance have made such
systems common in the basic
research laboratories around the world.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Modern Physics; Solid State Physics; Solids; Bonding In Solids;
Ionic bonds; Conductors;
Semiconductors; Band Theory Of Solids
SECTION 29.1 Introduction
In Chapter 28 we developed a nuclear model to explain the properties of atoms. If we
put many atoms close together to form
a solid object what aspects of our atomic model will we need to change? How will
photons of light interact with atoms closely
packed together in solids? Presumably the transition by electrons from one energy level
to another will result in the emission
or absorption of photons.
SECTION 29.8 Solids
Ionic bonding would seem to lead to the formation of light, brittle solids. The weaker
bonding mechanisms should allow for
softer, more pliable solids.
Stainless steel is an alloy especially developed for use where a strong metal which
would not rust is required.
SECTION 29.10 Conductors
Question - 1. In a conductor there are always electrons free to move in the conduction
band. These electrons can move in
response to any energy gradient, either electrical or thermal. Hence, materials which are
excellent electrical conductors would
also be good thermal conductors. The best specific example is silver.
SECTION 29.12 Intrinsic Semiconductors
Question - 2. Intrinsic semiconductors have a band gap of the order of 1eV. Visible light
has an energy of the order of 2eV.
The photons of visible light can be absorbed by electrons in the valence band of
semiconductors and lifted into the
conduction band. Hence we predict that intrinsic semiconductors will be opaque to
visible light and will conduct electricity
readily while exposed to bright visible light.
EXAMPLES
Keywords: ; Worked Examples; Solid State Physics; Modern Physics; Molecules;
Molecular Spectra; Molecular
Vibration; Molecular Rotation; Spectral Absorption; Quantum Numbers; Moment Of
Inertia
MOLECULAR ABSORPTION SPECTRA
1. The vibrational and rotational lines of the molecule H Cl are found at about l = 3.3 x
10-4 cm. (a) What is the force
constant for the vibrational motion of H Cl? (b) What is the moment of inertia of H Cl?
(c) What are the splittings we could
expect between the various quantum levels for vibrational and rotational spectra in eV?
(d) What are the splittings (in eV) we
could expect between the H Cl35 lines and the H Cl37 lines?
What data are given?
The H Cl molecule has ground state vibrational and rotational energy levels at l = 3.3
x10-4 cm = 3.3 x10-6 m = 3,300 nm.
What data are implied?
Natural occuring chlorine occurs with two different masses, one Cl35 has a gram
molecular mass of 35 gm and the other Cl37
has a gram molecular mass of 37 gm. The natural abundances gives chlorine an average
gram molecular mass of 35.5 gm.
What physics principles are involved?
The properties of rotating and vibrating quantum mechanical systems as discussed in
Sections 29.4 and 29.5 of the textbook.
What equations are to be used?
Evib (n + 1/2)(h/2p)(k/m)1/2, where m = (M1M2) / (M1 + M2) (29.2)
Erot = J(J + 1)(h/2p)2 (1/2I) (29.3)
Solutions
(a) Use an average mass for chlorine; M2 = 35.5, then
mHCL ÷ ((1)(35.5) / (1 + 35.5)) x 1.66 x 10-27 kg = 1.61 x 10 -27kg
Eo ÷ hc / lo = (1/2)(h/2p) x (SQR RT)(k / 1.61 x 10-27)
(3 x 108 m/s) / (3.3 x 10-6m) = 1/(4p) x (SQR RT)(k / 1.61 x 10-27)
11.4 x 1014/s = (SQR RT)(k / 1.61 x 10-27)
1.30 x 1030 = (SQR RT)(k / 1.61 x 10-27)
2.1 x 103 N/m = k, the force constant for H Cl.
(b) Erot ÷ hc / lo ÷ 1(h/2p)2 x (1/2I)
C / lo = h / (8p2I)
(3.0 x 108m/s) / (3.0 x 10-6m) = (6.63 x 10-34J ¥ S) / (8p2I)
I = 9.2 x10-50 kg m2
(c) Ground State Energy = hc / lo = ((6.63 x 10-34)(3.0 x 108)) / 3.3 x 10-6
Eo = 6.03 x10-20 J = 0.377eV
For vibrational levels the levels are evenly split, so
Eo = Evib ÷ 1/2hf; DEo = hf = 2Eo
energy splittings for vibration states ÷ 2Eo ÷ 0.75eV For rotational states the splittings
change
Say J = 0 to J = 1 transition is Eo
Then J = 1 to J = 2 transition is 2Eo
and J = 2 to J = 3 transition is 6Eo and so on.
(d) The effective mass of the molecule changes slightly
m35/m37 = [((1)(35))/(1+35)) / ((1)(37)/(1+37))]
m35/m37 = 0.9722 / 0.9737 = 99.85%
Thus the Evib would change by less than 15 parts in 10,000.
PRACTICE TEST
Keywords: ; Evaluations; Answers; Questions; Problems; Modern Physics; Solid State
Physics; Molecules;
Molecular Spectra; Fluorescence; Phosphorescence; Solids; Band Theory Of Solids;
Semiconductors; Molecular Vibrations;
Molecular Rotations
1. Distinguish between the following terms:
a) Florescence
b) Phosphorescence
c) Illuminescence
2. Use the band theory of solids to explain how a pure, intrinsic semiconductor (like
germanium) can become a p-type
semiconductor. Your explanation should utilize a diagram showing the details of the
band theory.
3. The energy levels predicted for a molecule are as follows:
Evib = (n + 1/2) (h/2p)(K/m)1/2 (vibrational)
Erot = J (J + 1) h2/2I (rotational)
a) Describe the mechanical model used to formulate each of these mathematical
models.
b) Identify the important parts of each of the mathematical models.
ANSWERS:
1. a) Excited molecule gives up some of its energy through vibrational contact with
other molecules. In a lower energy state,
the molecules then emit a photon (longer l) to return to a ground state.
b) Molecules involved are excited and return to the ground state through an
intermediate state with a long lifetime. These
molecules then emit photons at various times to return to the ground state. This slow
nature gives a characteristic "afterglow."
c) The emission of light due to the excitation of electrons. Lasers and light-emitting
diodes used in calculators are examples
of illuminescence.
2. See Fig.
A pure semiconductor when doped with a substance that has one less electron than the
semiconductor gains acceptor states
or holes in the forbidden gap region near the valence band. Electrons from the valence
band can then move into the acceptor
states (by tunneling or by increased thermal energy) where they are trapped.
3. See Fig.
The molecule is like a dumbbell with a spring connecting each atom.
b) h - planck's constant
k - effective spring constant
m - effective mass of the system where n = h/2p
n - quantum no. for vibration
J - quantum no. for rotation
I - moment of inertia
Chapter
X-Rays:
30 X-Rays
GOALS
Keywords: ; Learning Objectives; Modern Physics; Radioactivity; X-Rays; Absorption
Coefficients; Bremsstrahlung
Radiation; X-Ray Spectra; Types Of Radioactivity; Radiation Interactions; Radiation
Absorption
When you have mastered the material in this chapter, you will be able to:
Definitions
Define each of the following terms, and use it in an operational definition:
hard and soft x rays
Bremsstrahlung
characteristic x rays
absorption coefficient
X-ray Problems
Solve problems involving the generation, absorption, and detection of x rays.
X-ray Interactions
List and discuss the interactions of x rays with matter- particularly those with humans.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 21,
Electrical Properties of
Matter, Chapter 27, Quantum and Relativistic Physics, and Chapter 28, Atomic Physics.
OVERVIEW
Keywords: ; Instructions; Modern Physics; Radioactivity; Types Of Radioactivity; XRays
When high speed electrons (greater than 10,000 eV) strike a solid target (e.g., copper)
high energy, short wavelength photons
are produced. These photons are called "x rays." Because of their high energy, x rays
have remarkable penetrating power. In
this chapter you will read about the production, absorption, and detection of x rays.
Also, because x rays can produce
damage to biological tissue, the biological effects of x rays will be discussed.
SUGGESTED STUDY PROCEDURE
Before you begin your study of this chapter, be familiar with the following Chapter
Goals: Definitions, X-Ray Problems, and
X-Ray Interactions. An extended discussion of each of the terms listed under the
Definitions goal are included in the first
section of this chapter.
Next, read chapter sections 30.1-30.9. As you read please keep in mind that X rays are
not electrons, but high-energy
photons. Thus, from the photon theory, x rays have both particle and wave
characteristics. Also, be cautious of equation
30.3. If you do not recognize the exponential nature of the equation, look at Appendix,
Section A9.
Now turn to the end of the chapter and read the Chapter Summary and complete
Summary Exercises 1-9. Now, do
Algorithmic Problems 1-3 and Exercises and Problems 1-3, 5, 6, 11-14, and 16. Be sure
that you compare your answers to
each of these problems to those given. If you need assistance with any part, turn back to
a portion of the text or to this
chapter. This study procedure is outlined below.
--------------------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
-------------------------------Definitions 30.1,30.2,30.6 1,2,3, 14, 16
30.7,30.8 4,5,6
X-Ray 30.2,30.8 7,8 1,2,3, 1,2,3,
Problems 5, 6
X-Ray 30.3,30.4,30.5 9 11,12,13
Interactions
---------------------DEFINITIONS
Keywords: Glossary; Modern Physics; Radioactivity; X-Rays; Absorption Coefficients;
Bremsstrahlung Radiation; X-Ray
Spectra; Types Of Radioactivity; Radiation Interactions; Radiation Absorption;
HARD X RAYS
The higher energy and greater penetrating x rays.
SOFT X RAYS
The lower energy and lesser penetrating x rays.
BREMSSTRAHLUNG
"Break radiation" is the name given to the continuous x-ray spectrum from an x-ray
source.
In most medical uses of x rays a metal plate such as an aluminum disc is used to absorb
the low energy Bremsstrahlung
radiation as it leaves the x- ray tube.
CHARACTERISTIC X-RAYS
The x-ray production due to inner electron energy state transitions that result in x-ray
photons.
The inner electron energy shells are often labeled by the letters K, L, M, N, ... instead of
the value of the principle quantum
number n = 1, 2, 3, 4, ... respectively. Thus the various characteristic x-rays are
sometimes labeled by the final state of the
electron after x-ray production, such as Ka where the K means the final state of the
electron is the K or n=1 state and alpha
means the electron originated at the first level above K, i.e. the L or n=2 state. Hence,
when an electron goes down from an
n = 2 state to an n = 1 state of an atom a Ka characteristic x ray is emitted.
ABSORPTION COEFFICIENT
The physical parameter that characterizes the absorption of electromagnetic radiation (x
rays) per unit length of absorber.
Typical values of absorption coefficients for x rays from ~10 -3 cm-1 for air to 1 cm-1 for
water to ~103 cm- 1 for gold,
platinum and lead.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Modern Physics; Radioactivity; Types Of Radioactivity; X-Rays
SECTION 30.1 Introduction
The fact that x rays penetrate body tissue while being absorbed by bone and that x rays
will expose photographic emulsions
enable us to use x rays to photograph the interior portions of human beings.
Since x rays are photons of a few thousand electron volts and since the electrons in color
television sets are accelerated
through electric potentials of several thousands of volts, x rays will be produced
whenever the high energy electrons strike
suitable targets.
The detection of x rays first occurred because they expose photographic emulsions.
Now more quantitative measurements of
x-ray intensities are made using ionization chambers and scintillation counters.
EXAMPLES
Keywords: Worked Examples; ; Modern Physics; X-Rays; Radioactive Sources;
Conservation Of Energy;
Wavelength
X- RAY PROBLEMS
1. An x- ray tube is operated at a peak voltage of 250 kV and a current of 1.0mA. Then
the exposure rate is 2.5 R/hr at a
distance of 1.5 m from the target. (a) What is the minimum wavelength of x-ray
radiation emitted from this tube? (b) If the
current is increased to 20. mA what is the exposure rate at 1.5 m from the target. (c) At
what distance for the increased
current can the exposure be reduced back to 2.5 R/hr.
What data are given?
Accelerating voltage = 250 kV = 2.5 x 105 V
Current = 1.0 mA = 1.0 x 10-3 A
Exposure = 2.5 R/hr at a distance of 1.5 m.
What data are implied?
We will assume an inverse square law relationship between exposure and distance.
What physics principles are involved?
The conservation of energy as expressed in Equation 30.1 can be used. The definitions of
radiation units (Section 30.6) and
the inverse square law will also be used.
What equations are to be used?
lV = 1240 eV.nm (30.2)
I ° (1/d2) (16.12)
Solutions
(a) lmin = 1240eV-nm / 2.5 x 105 V
lmin = 4.96 x 10-3 nm ÷ 5.0 x 10-12 m.
(b) If the current is increased to 20 mA then the power of the tube is increased by a
factor of 20, so the intensity of x-ray
radiation from the tube is increased by a factor of 20, so the exposure rate is increased
by a factor of 20 to a value of 50
R/hr.
(c) In order to reduce the exposure rate back to 2.5 R/hr the distance from the target
must be increased
(Dose) (distance)2 = constant
(50 R/hr) (1.5m)2 = 2.5 R/hr (X2)
((SQR RT)(20)) (1.5 m) = x
6.7 meters = x
Thinking about the answers
Remember dose rates increase linearly with the tube current. High current, high voltage
x-ray sources ought to be avoided. If
you can keep a large distance away from sources of radiation you can reduce your
exposure.
PRACTICE TEST
Keywords: ; Evaluations; Answers; Questions; Problems; X-Rays; X-Ray Spectra;
Atomic Spectra; Modern
Physics; Bremsstrahlung Radiation; Radioactivity; Applications; Of Radioactivity;
Radiation Therapy; Energy; Radiation
Damage; Radiation Effects; Medicine And Health; Radiation Interactions; Radiation
Absorption
1. A target of copper will produce X-rays when bombarded with high speed electrons.
A typical graph showing the X-ray
spectrum produced is shown below.
See Fig.
a) What approximate electron energy (in eV's) is required to produce X-rays from
a target like copper?
b) What produces the Bremsstrahlung? What is the origin of the two "peaks"
called the characteristic X-ray spectral lines?
2. A technician operates an X-ray machine at 200 kV peak for therapeutic purposes.
a) Find the energy and wavelength of the most energetic X-rays emitted.
b) This machine operates at an exposure rate of 1.0 R/min at a distance of 1.0 meter
from the machine. The recommended dosage for the technician (for a 40
hour week) is 1.0 x 10-4 R/min. How thick should the
lead shield be for protecting the technician? (Given: The density of
lead (r) = 11.4 gram/cm3. The ratio (m/p)
for lead at 200 KeV = 0.80 cm2/gm.)
3. The interactions of radiation with living matter depends upon many factors and are
therefore complex. In general, absorbed
radiation produces changes in cells.
a) Rate the following as most sensitive (MS), sensitive (S) or less sensitive (LS)
to the effects of radiation.
____ bone marrow cells ____ lymphoid cells
____ bone cells ____ blood vessel cells
____ muscle cells
b) What properties of cells and their response to radiation allows the radiation
treatment of cancer to be effective?
ANSWERS:
1. 20- 50 KeV (The accelerating voltage must be approximately 20,000 to 50,000 volts.)
The Bremsstrahlung is white noise or
background radiation produced by the deceleration of the electrons upon impact with
the target. The "peaks" are produced by
the excited target atoms. This energy allows target electrons to return to a ground state
after excitement and produce similar
radiation frequencies.
2. 6.2 x 10-12 m, 200 KeV, 1.0 cm
3. MS, LS, LS, MS, LS. Radiation treatment has shown that cells which reproduce
rapidly are the most sensitive to
obstruction. Thus rapid growing cancerous cells can be selected for destruction with
proper levels of radiation which can allow
healthy cells to survive.
Chapter
Nuclear Physics (31):
STUDY
GUIDE
Citation: H. Q Fuller, R. M. Fuller and R. G. Fuller, to Accompany Physics Including
Human Applications.(Harper
and Row, New York, 1978). Permission granted by the authors.
31 Nuclear Physics
Keywords: ; Learning Objectives; Modern Physics; Nuclear Physics And Chemistry;
Atomic Mass Number;
Isotopes; Isobars; Isofones; Nuclear Binding Energy; Radioactivity; Nuclear
Characteristics; Nuclear Emission; Alpha
Particles; Beta Particles; Gamma Rays; Radioactive Decay; Decay Law; Half-Life;
Nuclear Fission; Nuclear Fusion;
Applications Of Radioactivity; Radioactive Dating
GOALS
When you have mastered the contents of this chapter, you should be able to achieve the
following goals:
Definitions
Define each of the following terms, and use each term in an operational definition:
atomic mass number alpha, beta, and gamma
isotope radiation
isobar half-life
isotone nuclear fission
nuclear binding energy nuclear fusion
radioactivity
Carbon-14 Dating
Describe and explain the physical basis of carbon-14 dating.
Radioactivity Problems
Solve problems involving radioactivity.
Binding-Energy Problems
Solve nuclear binding-energy problems.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 5, Energy,
and Chapter 28,
Atomic Physics.
OVERVIEW
Keywords: ; Instructions; Modern Physics; Nuclear Physics And Chemistry
Almost all of the mass of the atom is located in the nucleus. Investigations into the
nature of nuclear matter has led to
several models which help us understand many observable nuclear properties. These
observations include nuclear binding
energies, natural radioactivity, fission, and fusion. In this chapter you will be
introduced to the basis of nuclear physics and
several phenomenalogical applications which result.
SUGGESTED STUDY PROCEDURE
When you begin your study of this chapter, you should be familiar with the following
Chapter Goals: Definitions, Carbon-14
Dating, Radioactivity Problems, and Binding Energy Problems. A further discussion of
each term listed under the Goal of
Definitions section of this chapter.
Next, read text sections 31.1-31.6 and 31.9-31.11. As you read, take special note of Table
31.1 which outlines the relative
strength of the four interactional forces. Be sure to look carefully at the Example on
page 692 and 693 and on page 698.
Answers to questions you encounter during your reading will be discussed in the
second section of this chapter.
Now turn to the end of the chapter and read the Chapter Summary and do Summary
Exercises 1-6. Then do Algorithmic
Problems 1-4 and Exercises and Problems 1-3, 13, 20, 22, and 23. After you complete
each of these exercises, check your
answer against those that are given. For more work with the concepts introduced in this
chapter, turn to the Examples section
of this chapter and work through the problems discussed.
Now you should be prepared to attempt the Practice Test given at the end of this
chapter. Work completely
through the test before checking your answers. For assistance with any part, see specific
text sections or refer again to this
chapter. This study procedure is outlined below.
-------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
-------------------Definitions 31.1,31.2,31.3, 1 1,2 1
31.5,31.9-31.10
Carbon-14 31.6 2 23
Dating
Radioactivity 31.6 3,4 3,4 20,22
Problems
Binding-Energy 31.4 5,6 2,3,13
Problems
DEFINITIONS
Keywords: Glossary; ; Modern Physics; Nuclear Physics And Chemistry; Atomic Mass
Number; Isotopes;
Isobars; Isofones; Nuclear Binding Energy; Radioactivity; Nuclear Characteristics;
Nuclear Emission; Alpha Particles; Beta
Particles; Gamma Rays; Radioactive Decay; Decay Law; Half-Life; Nuclear Fission;
Nuclear Fusion
ATOMIC MASS NUMBER (A)
The number of nucleons (protons and neutrons) in an atomic nucleus.
The atomic mass number of elements ranges from 1 for hydrogen to 260 for hahnium.
ISOTOPE
Refers to atoms that have the same atomic number but different numbers of neutrons.
Since the chemical properties of materials are determined by their electronic structure,
hence their atomic number, isotopes
have the same chemical properties.
ISOBAR
Nuclei that have the same atomic mass number, A, are called isobars.
Since the atomic number, rather than the atomic mass number, determines the chemical
properties of a material isobars will
have quite different chemical properties, e.g. neon, sodium, and magnesium.
ISOTONE
Nuclei with equal numbers of neutrons are called isotones.
The chemical properties of isotones is quite different, consider boron, carbon and
nitrogen.
NUCLEAR BINDING ENERGY
The energy that binds neutrons and protons in a nucleus is equal to the mass-energy
difference between separate nucleons
and the particular nucleus.
The Einstein equation that relates mass to energy is used to relate the mass difference
between the separate masses to the
combined mass to the binding energy, see Eq. 31.4.
RADIOACTIVITY
The name given to the process occurring when unstable decay with characteristic
radiation.
This was discovered in 1896, barely one generation ago. Modern human life has
probably changed more since the discovery
of radioactivity than all previous changes combined. Has radioactivity been the cause of
this tremendous change?
ALPHA RADIATION
The radiation that consists of helium nuclei emitted by heavy radioactive nuclei.
The alpha particle was used in the scattering experiments conducted by Rutherford, et.
al. which lead to the nuclear model for
matter.
BETA RADIATION
The radiation that results in electron or positron emission from a radioactive nucleus.
The beta emission process which did not seem to conserve energy lead Fermi to
propose a new particle, not experimentally
verified for nearly 20 years! Such confidence in conservation laws!
GAMMA RADIATION
Radiation of high energy electromagnetic waves.
Gamma radiation has a high penetration ability. It is difficult to shield persons from
gamma emission processes.
RADIOACTIVE HALF-LIFE
The time required for one-half the radioactive nuclei in a sample to decay.
The half- lives of materials vary widely from a small fraction of a second for very
unstable nuclei to millions of years for the
more stable radioactive nuclei.
NUCLEAR FISSION
The process occurring when a heavy unstable nucleus splits to form two lighter nuclei
with the release of considerable energy
and a few neutrons. The process is exploited in nuclear reactors.
This was the first man made nuclear process. It seems to not occur anywhere without
human intention.
NUCLEAR FUSION
The process that occurs when two light nuclei get close enough together that the nuclear
force causes them to fuse to form a
heavier nucleus with a release of energy. The basic source of energy in stars.
This is the basic natural nuclear process and has so far not been duplicated in a
scientific laboratory for any extended period
of time.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: ; Answers; Nuclear Physics And Chemistry; Nucleus; Nuclear Binding
Energy; Nuclear Stability;
Nuclear Fusion; Mass And Energy
SECTION 31.1 Introduction
The basic physics of this chapter is contained in the explanation of two facts. If a heavy
nucleus is divided into two smaller
nuclei by a fission process, the final mass of the nuclei is less than the initial mass of the
heavy nucleus. If two light nuclei
are fused together to form a larger nucleus the mass of the final large nucleus is less
than the total mass of the two initial
light nuclei. That is, in both processes, splitting heavy nuclei and joining light nuclei,
mass is lost. That mass loss appears as
its equivalence in energy according to Einstein's E=mc2 equation.
SECTION 31.2 The Nucleus
The six carbon isotopes contain six protons and four, five, six, seven, eight, and nine
neutrons respectively.
SECTION 31.3 Properties of the Nucleus
The great density of nuclear matter seems much greater than we could predict from the
forces of interaction. The forces of
interaction for atoms and bulk materials are primarily electromagnetic. The nuclear
force of interaction is about 100 times the
electromagnetic force yet the nuclear matter is a thousand billion times as dense as an
atom.
SECTION 31.4 Nuclear Binding Energy
The binding energy per nucleon is seven times greater for helium than it is for
deuterium. It seems safe to predict that helium
is a much more stable nucleus than deuterium.
SECTION 31.6 Stable and Unstable Nuclei
The neutrino prediction of Fermi ranks as a great triumph for the conservation of
energy principle. Today in studying the
increasing maze of "elementary" particles basic conservation principles are used to
guide experimental studies.
SECTION 31.10 Fusion
The binding energy per nucleon graph (Figure 31.1) gives us a way of predicting which
nucleons would be appropriate for
fission or for fusion. The most stable nuclei are near an atomic mass number of 55.
Clearly only nuclei with mass numbers
much larger than 55 are suitable candidates for fission. In a similar way only nuclei with
atomic mass numbers much less
than 55 are suitable candidates for fusion.
6
Look at the location of 3Li. What do you predict about that nucleus?
EXAMPLES
Keywords: ; Worked Examples; Modern Physics; Nuclear Physics And Chemistry;
Mass And Energy;
Conservation Of Energy; Nuclear Interactions; Nuclear Emission; Atomic Mass
Number; Nuclear Binding Energy; Conservative
Laws; Radioactivity; Radioactive Decay; Decay Law; Half-Life; Exponential Decay;
Conservation Of Mass
NUCLEAR REACTION PROBLEMS
1. The energy released in the radioactive decay
238 4 234
92U ç 2He + 90Th
is 4.27MeV. Calculate the mass of 234Th.
What data are given?
238U mass = 238.0498 amu
4He mass = 4.00260 amu
Energy Released = 4.27MeV
What data are implied?
This is an isolated system with no external energy input.
What physics principles are involved?
Conservation of energy using the Einstein relation.
What equation is to be used?
E = mc2 (31.4)
Solution
Initial Energy = 238.0498 amu
Final Energy = 4.0026 amu + 4.27 MeV + mass of Th
Since 1 amu = 931 MeV; 4.27 MeV = .0046 amu.
Final Energy = 4.0072 amu + mass of the Th
mass of Th = 238.0498 - 4.0072
234Th = 234.0428 amu.
2. In the slow neutron reaction
1 10 7
0n + 5B ç 3Li + ? + Q
(a) Determine the unknown component of the reaction and its energy released.
What data are given?
The basic nuclear equation is given.
What data are implied?
There is one component missing from the equation.
What physics principles are involved?
Conservation of atomic number and atomic mass number can be used to determine the
unknown component. Then
conservation of energy using the Einstein equation can be used to find the energy
released.
What equations are to be used?
E = mc2 (31.4)
Solution
There are 11 amu's shown on the left hand side of the equation and 5 units of atomic
number. There are 7 and 3 respectively
on the right hand side of the equation. That leaves 4 amu and 2 units of atomic number
not counted.
4
Thus the particle must be 2X, where X must be helium.
4
So the missing component is an alpha particle 2He.
1
mass of 0n = 1.00867 amu
10
mass of 5B = 10.01295
--------------total = 11.02162 amu
7
mass of 3Li = -7.01601
-------------4.00561
4
mass of 2He = -4.00260 amu
Q = .00301 amu = 2.80 MeV
RADIOACTIVITY PROBLEMS
3. The half life of Rn222 is 3.82 days. (a) What is the decay constant of this isotope in sec1? (b) If a sample of Rn222
contains 1016 atoms at time t=0, how many Rn222 atoms will be left after 25 days?
What data are given?
t1/2 = 3.82 days = 3.30 x 105 seconds
No = 1016 atoms
What physics principles are involved?
The exponential decay of radioactive nuclear is basic to this problem.
What equations are to be used?
N = Noe-lt (31.12)
t1/2 = 0.693 / l (31.13)
Solutions
(a) The decay constant can be found from the half-life using Equation 31.13.
l = 0.693 / t1/2 = 0.693 / (3.30 x 105 sec)
lRn222 = 2.10 x 10-6 sec-1
(b) The number of atoms at time t is given by Equation 31.12
N = (1016)e - (25 days = 24 hr/d x 60 min/hr = 60 sec/min)l
N = 1016e - (2.16 x 106)(2.10 x 10-6)
n = 1016 e-4.54 = 1.07 x1014 atoms
BINDING-ENERGY PROBLEMS
7
4. What is the binding energy of 3Li whose atomic mass is 7.01822 amu.?
What data are given?
Atomic number = 3 (number of protons)
Atomic mass number = 7 (number of nucleons)
Atomic mass = 7.01822 amu (includes 3 electrons)
What data are implied?
7
The 3Li nucleus consists of 3 protons and 4 neutrons.
What physics principles are involved?
The Einstein equation is combined with conservation of mass equations to compute
nuclear binding energies.
What equations are to be used?
BE = (Zmp + Nmn)c2 - Mc2 (31.7)
Solution
71
The 3Li atom is equivalent to 3 1H atoms
and 4 neutrons. Their separate masses are
1
3 1H = 3(1.007825) = 3.023475 amu
4 n = 4(1.008665) = 4.034660 amu
-----------------7.058135 amu
7
3Li -7.01822
----------------------------BE = .03992 amu BE = 37.19 MeV
7
There are 7 nucleons in 3Li so the binding energy per nucleon is 5.312 MeV/nucleon.
PRACTICE TEST
Keywords: ; Evaluations; Answers; Problems; Questions; Modern Physics;
Radioactivity; Radioactive Decay;
Half-Life; Decay Law; Nuclear Physics And Chemistry; Nuclear Binding Energy;
Nuclear Interactions
1. a) The half-life of C-14 is 5730 years. The activity of C-14 from a living sample is 16.0
counts per minute. If the activity of
carbon from an ancient wood sample is found to be 1.88 counts/min, what is the age of
the sample?
b) Comment on the "accuracy" of this dating method.
2. Given the following nuclear reaction:
174
1H + 3Li ç ? 2He + ?
a) Complete the reaction by filling in the blanks with proper numbers and/or symbols.
b) Calculate the amount of energy released in this reaction.
1
1H = 1.00813 amu,
7
3Li = 7.01613 amu,
4
2He = 4.00386 amu,
and 1 amu = 931 MeV.
ANSWERS:
1. 17,800 yr. In actual practice, the 1.88 counts/min may be less than the normal
"background count" due to natural occurring
disintegration. Since normal variations in background occur, it becomes difficult to
distinguish between background and
radiation attributed to the sample itself. Thus the possibility of error is great.
4
2. 2 2He + Q, Q = 15.4 MeV
Chapter
Applied Nucleonics (32):
32 Applied Nucleonics
Keywords: Learning Objectives; ; Modern Physics; Nuclear Physics And Chemistry;
Nuclear Fission; Moderators;
Chain Reactions; Fission; Reactors; Nuclear Fusion; Plasmas; Magnetic Confinement;
Half-Life; Detection Of Radiation;
Nuclear Medicine; Medical Diagnosis
GOALS
When you have mastered the content of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use each term in an operational definition:
fission thermonuclear reaction
chain reaction plasma
critical mass "magnetic bottle"
moderator tracer
breeder reactor biological half-life
neutron activation effective half-life
fusion
Fission Reactors
Explain the operation of a nuclear fission reactor.
Radiation Detectors
Compare different nuclear radiation detectors in their uses.
Radiation Problems
Solve problems involving radiation application using biological half-life and dosage
data.
Tracer Applications
Outline the use of radioactive tracers in medical applications.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 21,
Electrical Properties of
Matter, Chapter 27, Quantum and Relativistic Physics, Chapter 30. X Rays, and Chapter
31, Nuclear
Physics.
OVERVIEW
Keywords: ; Instructions; Modern Physics; Nuclear Physics And Chemistry
Our modern society places an ever-increasing demand on energy acquisition. During
the 1980's, nuclear energy from the
processes of both fission and fusion will surely provide additional energy for your use.
The uses of nuclear energy will no
doubt continue to be important; i.e., medical and tracer applications. In this chapter you
will read about the conversion of
nuclear energy into electrical energy and some of the problems which complicate the
process. In addition, the problem of
containing new and used nuclear materials will be presented.
SUGGESTED STUDY PROCEDURE
When you begin to study this final chapter, you should be familiar with the following
chapter Goals: Definitions, Fission
Reactors, Radiation Problems and Tracer Application. A further discussion of the terms
listed under Definitions can be found
in the Definitions section of this chapter.
Next, read chapter sections 32.1-32.5 and 32.11. In addition to the important energy
issue, be sure to note the contribution
nuclear energy has made in other areas like medicine. Tables 32.1 and 32.2 will
highlight some specific uses of radioactive
materials.
Now, turn to the end of the chapter and read the Chapter Summary and complete
Summary Exercises 1-7 and 10. Next, do
Algorithmic Problem 1 and Exercises and Problems 1, 2, 7, and 8. For more work with
the concepts of this chapter, now turn
to the Examples section of this chapter.
Now you should be prepared to attempt the final Practice Test provided in this . If you
have difficulties with any
part of the test, you should refer to the appropriate text or section for assistance. This
study procedure is outlined
below.
-------------------Chapter Goals Suggested Summary Algorithmic Exercises
Text Readings Exercises Problems & Problems
-------------------Definitions 32.1,32.3,32.4 1,2,3,4,
5,6
Fission 32.2 7 1
Reactors
Radiation 32.11 1 2,7,8
Problems
Tracer 32.5 10
Applications
---------------------------------------------------Radiation 32.6,32.7,32.8 8 5
Detectors 32.9,32.10
DEFINITIONS
Keywords: Glossary; ; Modern Physics; Nuclear Physics And Chemistry; Nuclear
Fission; Moderators; Chain
Reactions; Fission; Reactors; Nuclear Fusion; Plasmas; Magnetic Confinement; Half-Life;
Detection Of Radiation; Nuclear
Medicine; Medical Diagnosis
FISSION
The dividing up of heavy nuclei to obtain energy.
This process, which creates heavy radioactive waste products, is the basic process in all
nuclear power generation.
CHAIN REACTION
The fission reaction sustained by neutrons from preceding nuclear fission reactions. The
basis of nuclear fission reactions and
bombs.
In some fission processes excess neutrons are emitted which can be used to trigger the
fission of nearby nuclei, if there are
enough nearby nuclei.
CRITICAL MASS
The necessary mass of fissionable material to sustain a chain reaction.
Enough nearby nuclei to continue a fission process is called a critical mass.
MODERATOR
The material used in fission reactors to slow down fast neutrons so that a controlled
chain reaction can be maintained.
BREEDER REACTOR
A nuclear fission reactor that produces additional fissionable fuel material in addition
to usable nuclear energy.
At this time breeder reactors are still not widely used in the generation of electrical
energy.
NEUTRON ACTIVATION
The process resulting in induced radioactivity of samples subjected to neutron
bombardment.
Neutron activation analysis has lead to the solution of some spectacular crimes, such as
in the sale of fraudulent paintings.
FUSION
Combining light nuclei to obtain energy.
This appears to be the fundamental natural source of all the energy from the sun and
stars.
THERMONUCLEAR REACTION
The name given to all fusion reactions and the energy source of stars.
The fusion process is one of intense heat, more than 100,000,000 degrees celsius.
PLASMA
The state of matter consisting of completely ionized atoms and electrons as
characterized by fusion reactions.
Plasma physics is now a widely recognized research specialty. Plasma physicists have
made use of very high power pulsed
lasers beams as energy sources.
MAGNETIC BOTTLE
The magnetic field confinement system used to contain plasma in fusion reactors.
Have you ever thought of having a container for particles with no "walls"? This is a
spectacular example of
interaction-at-a-distance.
TRACERS
Radioactive tracers are samples that contain radioactive isotopes of biochemically active
materials. Their activity can be
traced in a biological system. (See Table 32.1)
BIOLOGICAL HALF-LIFE
The time required for one-half the biochemically active sample to be eliminated by
natural processes.
Since about 20% of the air in your lungs is expelled in each breath the biological half-life
of the stale air in you lungs is about
the time required for you to take three breaths, (4/5)3 ÷ 1/2.
EFFECTIVE HALF-LIFE
The resultant of radioactive and biological half-life.
The magnitude of the effective half-life of a tracer is always less than the smaller of its
biological or its radioactive half-life.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
Keywords: Answers; ; Nuclear Physics And Chemistry; Modern Physics; Nuclear
Fission; Half-Life; Plutonium;
Ecology; Nuclear Energy; Fossil Fuels
SECTION 32.1 Introduction
As the fossil fuel supply continues to be used at an ever increasing rate the use of
nuclear fuels becomes more and more
common. What are the fundamental processes of nuclear fuels? What are the benefits?
the risks? These are questions that
all of us consider. We will have to decide on the energy sources of the future.
SECTION 32.2 Nuclear Fission
Since plutonium-239 only has a half-life of 24,000 years and the older rocks on the
surface of the earth date back at least a
million years, that's almost 42 half-lives of plutonium-239, or one part in 3 1/2 trillion
parts of any original plutonium-239 would
still be around. It seems rather unlikely we would find any natural occurring
plutonium-239.
EXAMPLES
Keywords: Worked Examples; ; Nuclear Physics And Chemistry; Half-Life; Biology;
Alpha Particles; Radiation
Absorption; Radioactivity; Radioactive Decay; Radium
RADIATION PROBLEMS
32
1. The isotope 15P passes through the liver with a half-life of 18 days.
32
The effective half-life of 15P in the liver is 8.0 days.
32
What is the radioactive half-life of 15P?
What data are given?
Tb = 18 days; T = 8.0 days
What physics principles are involved?
The biological and radioactive half-lifes of an isotope may be combined to yield an
effective half-life.
What equation is to be used?
1/T = 1/Tb + 1/Tr (32.1)
Solution
We need to solve Equation 32.1 for Tr
1/8.0 days = 1/18 days + 1/Tr
Tr = 14.4 days
Thinking about the answer
Notice that the effective half life is always less than even the biological or the
radioactive half-life.
2. Assume that a man has incorporated within the bones of his body 10mCi of the alpha
emitter, 276Ra which has a
radioactive half-life of 1622 years. The energy of the primary alpha particle is 4.79 MeV.
Assume a uniform distribution of the
radium in 7.0 kg of bone. What is the total dose per day from the alpha particles of 226
Ra?
What data are given?
Activity = 10mCi = 10 x 3.70 x 104 dis/sec.
= 3.7 x105 dis/sec
Energy = 4.79 MeV / alpha = 4.79 MeV/dis.
Mass = 7.0 kg = 7000 gm.
What data are implied?
The dose unit is the rad which is equivalent to the absorption of 10-5 J/gm or 6.25 x
1013 eV/gm = 6.25 x 107 MeV/gm.
What physics principles are involved?
All of the energy of the alpha particles will be absorbed by the bone. We only need to
compute the total alpha particle kinetic
energy and divide by the bone mass.
What equations are to be used?
Dose = Energy Absorbed / Unit Mass in rad
Solution
Dose = (4.79 MeV / alpha) x (3.70 x 105 alphas/sec.) = 1.77 x 106 MeV/sec
Daily Dose = 1.77 x 106 MeV/sec x 60 sec/min x 60 min/hr x 24 hr/day = 1.53x1011
MeV/day
To convert the daily dose of 1.53 x 1011 MeV/day radiation unit of rads, we need to
divide it by the mass of the absorber
Daily Dose / Mass = (1.53 x 1011 MeV) / 7000 = 2.19 x 107 MeV/gm
But 1 rad = 6.24 x 107 MeV/gm
Daily Dose in rads = (2.19 x 107 MeV/gm) x (1rad / (6.25 x 107 MeV/gm))
Daily Dose = 0.350 rads
Thinking about the answer
Such a person will receive an annual dose of about 125 rads. Recall that the relative
biological effectiveness (RBE) of 5 MeV
alpha particles is 20; then this person will get a radiation dose of 2500 rem per year.
Look at Table 30.1 and you will see that
this is much larger than typical background radiation levels.
PRACTICE TEST
Keywords: ; Evaluations; Questions; Answers; Problems; Nuclear Physics And
Chemistry; Modern Physics;
Nuclear Fission; Nuclear Energy; Fission Reactors; Radioactivity; Radioactive Decay;
Half-Life; Nuclear Medicine; Medical
Diagnosis
1. A nuclear fission reactor produces electrical energy from release nuclear energy.
a) Make a block diagram showing the three main parts of the fission reactor generating
station needed to convert the nuclear
energy into electrical energy.
b) Inside the reactor, uranium fuel divides (fissions) and forms a chain reaction with
other atoms. How does this reaction
remain "self-sustaining? How is the reaction controlled?
2. The effective half-life of K42 used for radioactive muscle tracing is .5 days. The
radioactive half-life is also .5 days. Find
the biological half-life of K42.
3. A doctor wishes to study a defective thyroid by using a radioactive tracer. Please
name the isotope which would be used
and outline the procedure explaining the important steps in making this analysis.
ANSWERS:
1. See Fig.
Each fission produces two slow neutrons. If the fuel density is large enough to make it
highly probable that these two
neutrons will hit and "split" two additional atoms, each of these will in turn produce 2
slow neutrons, etc.
2. Infinite
3. 131I After an injection of 131I, the thyroid gland is monitored with a giger counter.
The rate of increased activity is related
to the activity of the gland.