Principles of Physics For Scientists and Engineers pdf - Web Education

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Undergraduate Lecture Notes in Physics

Series Editors Neil Ashby William Brantley Michael Fowler Elena Sassi Helmy S. Sherif

For further volumes:

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Undergraduate Lecture Notes in Physics (ULNP) publishes authoritative texts covering topics throughout pure and applied physics. Each title in the series is suitable as a basis for undergraduate instruction, typically containing practice problems, worked examples, chapter summaries, and suggestions for further reading.

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The purpose of ULNP is to provide intriguing, absorbing books that will continue to be the reader’s preferred reference throughout their academic career.

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Hafez A. Radi

•

John O. Rasmussen

Principles of Physics

For Scientists and Engineers

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October University for Modern Sciences and Arts (MSA) 6th of October City

Egypt

John O. Rasmussen

University of California at Berkeley and Lawrence Berkeley Lab Berkeley, CA

USA

ISSN 2192-4791 ISSN 2192-4805 (electronic) ISBN 978-3-642-23025-7 ISBN 978-3-642-23026-4 (eBook) DOI 10.1007/978-3-642-23026-4

Springer Heidelberg New York Dordrecht London

Library of Congress Control Number: 2012947066 Springer-Verlag Berlin Heidelberg 2013

This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. Exempted from this legal reservation are brief excerpts in connection with reviews or scholarly analysis or material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Duplication of this publication or parts thereof is permitted only under the provisions of the Copyright Law of the Publisher’s location, in its current version, and permission for use must always be obtained from Springer. Permissions for use may be obtained through RightsLink at the Copyright Clearance Center. Violations are liable to prosecution under the respective Copyright Law.

The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use.

While the advice and information in this book are believed to be true and accurate at the date of publication, neither the authors nor the editors nor the publisher can accept any legal responsibility for any errors or omissions that may be made. The publisher makes no warranty, express or implied, with respect to the material contained herein.

Printed on acid-free paper

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Solutions to the exercises are accessible to qualified instructors at springer.com on this book’s product page. Instructors may click on the link additional information and register to obtain their restricted access.

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Preface

This book on Principles of Physics is intended to serve fundamental college courses in scientific curricula.

Physics is one of the most important tools to aid undergraduates, graduates, and researchers in their technical fields of study. Without it many phenomena cannot be described, studied, or understood. The topics covered here will help students interpret such phenomena, ultimately allowing them to advance in the applied aspects of their fields.

The goal of this text is to present many key concepts in a clear and concise, yet interesting way, making use of practical examples and attractively colored illustrations whenever appropriate to satisfy the needs of today’s science and engineering students. Some of the examples, proofs, and subsections in this textbook have been identified as optional and are preceded with an asterisk *. For less intensive courses these optional portions may be omitted without significantly impacting the objectives of the chapter. Additional material may also be omitted depending on the course’s requirements.

The first author taught the material of this book in many universities in the Middle East for almost four decades. Depending on the university, he leveraged different international textbooks, resources, and references. These used different approaches, but were mainly written in an expansive manner delivering a plethora of topics while targeting students who wanted to dive deeply into the subject matter. In this textbook, however, the authors introduce a large subset of these topics but in a more simplified manner, with the intent of delivering these topics and their key facts to students all over the world and in particular to students in the Middle East and neighboring regions where English may not be the native lan-guage. The second author went over the entire text with the background of study and/or teaching at Caltech, UC Berkeley, and Yale.

Instructors teaching from this textbook will be able to gain online access from the publisher to the solutions manual, which provides step-by-step solutions to all exercises contained in the book. The solutions manual also contains many tips, colored illustrations, and explanations on how the solutions were derived.

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Acknowledgments from Prof. Hafez A. Radi

I owe special thanks to my wife and two sons Tarek and Rami for their ongoing support and encouragement. I also owe special thanks to my colleague and friend Prof. Rasmussen for his invaluable contributions to this book, and for everything that I learned from him over the years while carrying out scientific research at Lawrence Berkeley Lab. Additionally, I would like to express my gratitude to Prof. Ali Helmy Moussa, Prof. of Physics at Ain Shams University in Egypt, for his assistance, support, and guidance over the years. I also thank all my fellow professors and colleagues who provided me with valuable feedback pertaining to many aspects of this book, especially Dr. Sana’a Ismail, from Dar El Tarbiah School, IGCSE section and Dr. Hesham Othman from the Faculty of Engineering at Cairo University. I would also like to thank Professor Mike Guidry, Professor of Physics and Astronomy at the University of Tennessee Knoxville, for his valuable recommendations. I am also grateful to the CD Odessa LLC for their Concept-Draw software suite which was used to create almost all the figures in this book. I finally extend my thanks and appreciation to Professor Nawal El-Degwi, Professor Khayri Abdel-Hamid, Professor Said Ashour, and the staff members and teaching assistants at the faculty of Engineering at MSA University, Egypt, for all their support and input.

Hafez A. Radi hafez.radi@gmail.com Acknowledgments from Prof. John O. Rasmussen

I would like to thank Prof. Radi for the opportunity to join him as coauthor. I am grateful to the many teachers, students, and colleagues from whom I learned various aspects of the fascinating world of the physical sciences, notably the late Drs. Linus Pauling, Isadore Perlman, Stanley Thompson, Glenn Seaborg, Earl Hyde, Hilding Slätis, Aage Bohr, Gaja Alaga, and Hans-Järg Mang. There are many others, still living, too numerous to list here. I would also like to extend my special thanks to my wife for her support and encouragement.

John O. Rasmussen oxras@berkeley.edu

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{

Base SI units

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yotta Y 1024 zeta Z 1021 exa E 1018 peta P 1015 tera T 1012 giga G 109 mega M 106 kilo k 103 hecto h 102 deka da 101 deci d 10-1 centi c 10-2 milli m 10-3 micro l 10-6 nano n 10-9 pico p 10-12 femto f 10-15 atto a 10-18 zepto z 10-21 yocto y 10-24 xviii

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Part I

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Dimensions and Units

₁

The laws of physics are expressed in terms of basic quantities that require a clear definition for the purpose of measurements. Among these measured quantities are length, time, mass, temperature, etc.

In order to describe any physical quantity, we first have to define a unit of measurement (which was among the earliest tools invented by humans), i.e. a mea-sure that is defined to be exactly 1.0. After that, we define a standard for this quantity, i.e. a reference to compare all other examples of the same physical quantity.

1.1 The International System of Units

Seven physical quantities have been selected as base quantities in the 14th General Conference on Weights and Measurements, held in France in 1971. These quantities form the basis of the International System of Units, abbreviated SI (from its French name Système International) and popularly known as the metric system. Table1.1

depicts these quantities, their unit names, and their unit symbols.

Many SI derived units are defined in terms of the first three quantities of Table1.1. For example, the SI unit for force, called the newton (abbreviated N), is defined in terms of the base units of mass, length, and time. Thus, as we will see from the study of Newton’s second law, the unit of force is given by:

1 N= 1 kg.m/s2 (1.1)

When dealing with very large or very small numbers in physics, we use the so-called scientific notation which employs powers of 10, such as:

H. A. Radi and J. O. Rasmussen, Principles of Physics, 3 Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_1,

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4 1 Dimensions and Units

3 210 000 000 m= 3.21 × 109m (1.2)

0.000 000 789 s = 7.89 × 10−7s (1.3)

Table 1.1 The seven independent SI base units

Quantity Unit name Unit symbol

Length Meter m

Time Second s

Mass Kilogram kg

Temperature Kelvin K

Electric current Ampere A

Amount of substance Mole mol

Luminous intensity Candela cd

An additional convenient way to deal with very large or very small numbers in physics is to use the prefixes listed in Table1.2. Each one of these prefixes represents a certain power of 10.

Table 1.2 Prefixes for SI unitsa

Factor Prefix Symbol Factor Prefix Symbol

1024 yotta- Y 10−24 yocto- y 1021 zeta- Z 10−21 zepto- z 1018 exa- E 10−18 atto- a 1015 _peta- _P ₁₀−15 _femto- _f 1012 _tera- _T ₁₀−12 _pico- _p 109 _giga- _G ₁₀−9 _nano- n 106 mega- M 10−6 micro- µ 103 kilo- k 10−3 milli- m 102 hecta- h 10−2 centi- c 101 _deca- _da ₁₀−1 _deci- _d

a_{The most commonly used prefixes are shown in bold face type}

Accordingly, we can express a particular magnitude of force as:

1.23 × 106N= 1.23 mega newtons

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or a particular time interval as:

1.23 × 10−9s= 1.23 nano seconds

= 1.23 ns (1.5)

We often need to change units in which a physical quantity is expressed. We do that by using a method called chain-link conversion, in which we multiply by a conversion factor that equals unity. For example, because 1 minute and 60 seconds are identical time intervals, then we can write:

1 min

60 s = 1 and 60 s

1 min = 1 (1.6)

This does not mean that ₆₀1 = 1 or 60 = 1, because the number and its unit must be treated together.

Example 1.1

Convert the following: (a) 1 kilometer per hour to meter per second, (b) 1 mile per hour to meter per second, and (c) 1 mile per hour to kilometer per hour [to a good approximation 1mi = 1.609 km].

Solution: (a) To convert the speed from the kilometers per hour unit to meters per second unit, we write:

1 km/h= 1 km h × 103m 1 km × 1 h 60× 60 s = 0.2777...m s = 0.278 m/s (b) To convert from miles per hour to meters per second, we write:

1 mi/h= 1 mi h × 1609 m 1 mi × 1 h 60× 60 s = 0.447 m s = 0.447 m/s (c) To convert from miles per hour to kilometers per hour, we write:

1mi/h= 1 mi h × 1.609 km 1 mi = 1.609 km h = 1.609 km/h

1.2 Standards of Length, Time, and Mass

Definitions of the units of length, time, and mass are under constant review and are changed from time to time. We only present in this section the latest definitions of those quantities.

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Length (L)

In 1983, the precision of the meter was redefined as the distance traveled by a light wave in vacuum in a specified time interval. The reason is that the measurement of the speed of light has become extremely precise, so it made sense to adopt the speed of light as a defined quantity and to use it to redefine the meter. In the words of the 17th General Conference on Weights and Measurements:

One Meter

One meter is the distance traveled by light in vacuum during the time interval of 1/299 792 458 of a second.

This time interval number was chosen so that the speed of light in vacuum c will be exactly given by:

c= 299 792 458 m/s (1.7)

For educational purposes we usually consider the value c= 3 × 108m/s. Table1.3lists some approximate interesting lengths.

Table 1.3 Some approximate lengths

Length Meters

Distance to farthest known galaxy 4× 1025 Distance to nearest star 4× 1016 Distance from Earth to Sun 1.5 × 1011 Distance from Earth to Moon 4× 108

Mean radius of Earth 6× 106

Wave length of light 5× 10−7

Radius of hydrogen atom 5× 10−11

Radius of proton 1× 10−15

Time (T)

Recently, the standard of time was redefined to take advantage of the high-precision measurements that could be obtained by using a device known as an atomic clock. Cesium is most common element that is typically used in the construction of atomic clocks because it allows us to attain high accuracy.

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Since 1967, the International System of Measurements has been basing its unit of time, the second, on the properties of the isotope cesium-133(133₅₅Cs). One of the transitions between two energy levels of the ground state of 133₅₅Cs has an oscil-lation frequency of 9 192 631 770 Hz, which is used to define the second in SI units. Using this characteristic frequency, Fig.1.1shows the cesium clock at the National Institute of Standards and Technology. The uncertainty is about 5× 10−16 (as of 2005). Or about 1 part in 2× 1015. This means that it would neither gain nor lose a second in 64 million years.

One Second

One second is the time taken for the cesium atom 133₅₅Cs to perform 9 192 631 770 oscillations to emit radiation of a specific wavelength

Fig. 1.1 The cesium atomic clock at the National Institute of Standards and Technology (NIST) in Boulder, Colorado (photo with permission)

Table1.4lists some approximate interesting time intervals.

Table 1.4 Some approximate time intervals

Time intervals Seconds

Lifetime of proton (predicted) 1× 1039

Age of the universe 5× 1017

Age of the Earth 1.3 × 1017

Period of one year 3.2 × 107

Time between human heartbeats 8× 10−1 Period of audible sound waves 1× 10−3 Period of visible light waves 2× 10−15 Time for light to cross a proton 3.3 × 10−24

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Mass (M)

The Standard Kilogram

A cylindrical mass of 3.9 cm in diameter and of 3.9 cm in height and made of an unusu-ally stable platinum-iridium alloy is kept at the International Bureau of Weights and Measures near Paris and assigned in the SI units a mass of 1 kilogram by international agreement, see Fig.1.2.

One Kilogram

The SI unit of mass, 1 kilogram, is defined as the mass of a platinum-iridium alloy cylinder kept at the International Bureau of Weights and Measures in France.

Fig. 1.2 The standard 1 kilogram of mass is a platinum-iridium cylinder 3.9 cm in height and diameter and kept under a double bell jar at the International Bureau of Weights and Measures in France

Accurate copies of this standard 1 kilogram have been sent to standardizing lab-oratories in other countries. Table1.5lists some approximate mass values of various interesting objects.

A Second Standard Mass

Atomic masses can be compared with each other more precisely than the kilogram. By international agreement, the carbon-12 atom,12₆C, has been assigned a mass of 12 atomic mass units (u), where:

1 u= (1.660 540 2 ± 0.000 001 0) × 10−27 kg (1.8) Experimentally, with reasonable precision, all masses of other atoms can be measured relative to the mass of carbon-12.

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Table 1.5 Mass of various objects (approximate values)

Object Kilogram

Known universe (predicted) 1× 1053 Our galaxy the milky way (predicted) 2× 1041

Sun 2× 1030 Earth 6× 1024 Moon 7× 1022 Small mountain 1× 1012 Elephant 5× 103 Human 7× 101 Mosquito 1× 10−5 Bacterium 1× 10−15 Uranium atom 4× 10−25 Proton 2× 10−27 Electron 9× 10−31

1.3 Dimensional Analysis

Throughout your experience, you have been exposed to a variety of units of length; the SI meter, kilometer, and millimeter; the English units of inches, feet, yards, and miles, etc. All of these derived units are said to have dimensions of length, symbolized by L. Likewise, all time units, such as seconds, minutes, hours, days, years, and centuries are said to have dimensions of time, symbolized by T. The kilogram and all other mass units have dimensions of mass, symbolized by M. In general, we may take the dimension (length, time, and mass) as the concept of the physical quantity.

From the three fundamental physical quantities of length L, time T, and mass M, we can derive a variety of useful quantities. Derived quantities have different dimen-sions from the fundamental quantities. For example, the area obtained by multiplying one length by another has the dimension L2. Volume has the dimension L3. Mass density is defined as mass per unit volume and has the dimension M/L3. The SI unit of speed is meters per second (m/s) with the dimension L/T.

The concept of dimensionality is important in understanding physics and in solv-ing physics problems. For example, the addition or subtraction of quantities with different dimensions makes no sense, i.e. 2 kg plus 8 s is meaningless. Actually, phys-ical equations must be dimensionally consistent. For example, the equation giving the position of a freely falling body (seeChap. 3) is giving by:

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x= v◦t+1

2g t 2

(1.9)

where x is the position (length),v_◦the initial speed (length/time), g is the acceleration due to gravity (length/time2), and t is time. If we analyze the equation dimensionally, we have: L= L T× T + L T2× T 2_{= L + L (Dimensional analysis)}

Note that every term of this equation has the dimension of length L. Also note that numerical factors, such as 1₂ in Eq.1.9, are ignored in dimensional analysis because they have no dimension. Dimensional analysis is useful since it can be used to catch careless errors in any physical equation. On the other hand, Eq.1.9may be correct with respect to dimensional analysis, but could still be wrong with respect to dimensionless numerical factors.

If we had incorrectly written Eq.1.9as follows:

x= v_◦t2+1

2g t (1.10)

Then, by analyzing this equation dimensionally, we have:

L= L T× T 2₊ L T2 × T _{(Dimensional analysis)} = L T × T × T + L T× T× T

and finally we get: L= LT+L

T (Dimensional analysis)

Dimensionally, Eq1.10is meaningless, and thus cannot be correct.

Example 1.2

Use dimensional analysis to show that the expressionv = v_◦+ at is dimensionally correct, where v and v_◦ represent velocities, a is acceleration, and t is a time interval.

Solution: Since L/T is the dimension of v and v_◦, and the dimension of a is L/T2, then when we analyze the equation v = v_◦+ at dimensionally, we have:

L T = L T + L T2× T (Dimensional analysis) =L T + L T× T× T

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and finally we get: L T = L T+ L T (Dimensional analysis)

Thus, the expressionv = v_◦+ at is dimensionally correct.

Example 1.3

A particle moves with a constant speedv in a circular orbit of radius r, see the figure below. Given that the magnitude of the acceleration a is proportional to some power of r, say rm, and some power of v,say vn, then determine the powers of r andv.

v a

r

Solution: Assume that the variables of the problem can be expressed mathemat-ically by the following relation:

a= k rmvn,

where k is a dimensionless proportionality constant. With the known dimensions of r,v, and a we analyze the dimensions of the above relation as follows:

L T2 = L m_× L T n =Lm+n Tn (Dimensional analysis)

This dimensional equation would be balanced, i.e. the dimensions of the right hand side equal the dimensions of the left hand side only when the following two conditions are satisfied:

m+ n = 1,

and n= 2.

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12 1 Dimensions and Units Therefore, we can rewrite the acceleration as follows:

a= k r−1v2= k v

2

r

When we later introduce uniform circular motion inChap. 4, we shall see that

k= 1 if SI units are used. However, if for example we choose a to be in m/s2and

v to be in km/h, then k would not be equal to one.

1.4 Exercises

Section 1.1 The International System of Units

(1) Use the prefixes introduced in Table1.2to express the following: (a) 103lambs, (b) 106 bytes, (c) 109 cars, (d) 1012 stars, (e) 10−1Kelvin, (f) 10−2 meter, (g) 10−3ampere, (h) 10−6newton, (i) 10−9kilogram, (j) 10−15second.

Section 1.2 Standards of Length, Time, and Mass

Length

(2) The original definition of the meter was based on distance from the North pole to the Earth’s equator (measures along the surface) and was taken to be 107m. (a) What is the circumference of the Earth in meters? (b) What is the radius of the Earth in meters, (c) Give your answer to part (a) and part (b) in miles. (d) What is the circumference of the Earth in meters assuming it to be a sphere of radius 6.4 × 106_{m? Compare your answer to part (a)}

(3) The time of flight of a laser pulse sent from the Earth to the Moon was measured in order to calculate the Earth-Moon distance, and it was found to be 3.8 × 105km. (a) Express this distance in miles, meters, centimeters, and millimeters. (4) A unit of area, often used in measuring land areas, is the hectare, defined as 104m2. An open-pit coal mine excavates 75 hectares of land, down to a depth of 26 m, each year. What volume of Earth, in cubic kilometers, is removed during this time?

(5) The units used by astronomers are appropriate for the quantities they usually measure. As an example, for planetary distances they use the astronomical

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unit (AU), which is equal to the mean Earth-Sun distance(1.5 × 1011m). For stellar distances they use the light-year(1 ly = 9.461 × 1012km), which is the distance that light travels in 1 yr(1 yr = 365.25 days = 3.156 × 107s) with a speed of 299 792 458 m/s. They use also the parsec (pc), which is equal to 3.26 light-years. Intergalactic distances might be described with a more appropriate unit called the megaparsec. Convert the following to meters and express each with an appropriate metric prefix: (a) The astronomical unit, (b) The light-year, (c) The parsec, and (d) The megaparsecs.

(6) When you observe a total solar eclipse, your view of the Sun is obstructed by the Moon. Assume the distance from you to the Sun(ds) is about 400 times the distance from you to the Moon(dm). (a) Find the ratio of the Sun’s radius to the Moon’s radius. (b) What is the ratio of their volumes? (c) Hold up a small coin so that it would just eclipse the full Moon, and measure the distance between the coin and your eye. From this experimental result and the given distance between the Moon and the Earth(3.8 × 105km), estimate the diameter of the Moon.

(7) Assume a spherical atom with a spherical nucleus where the ratio of the radii is about 105. The Earth’s radius is 6.4 × 106m. Suppose the ratio of the radius of the Moon’s orbit to the Earth’s radius(3.8 × 105km) were also 105. (a) How far would the Moon be from the Earth’s surface? (b) How does this distance compare with the actual Earth-Moon distance given in exercise 6?

Time

(8) Using the day as a unit, express the following: (a) The predicted life time of proton, (b) The age of universe, (c) The age of the Earth, (d) The age of a 50-year-old tree.

(9) Compare the duration of the following: (a) A microyear and a 1-minute TV commercial, and (b) A microcentury and a 60-min TV program.

(10) Convert the following approximate maximum speeds from km/h to mi/h: (a) snail(5 × 10−2km/h), (b) spider (2 km/h), (c) human (37 km/h),(d) car

(220 km/h), and (e) airplane (1,000 km/h).

(11) A 12-hour-dial clock happens to gain 0.5 min each day. After setting the clock to the correct time at 12:00 noon, how many days must one wait until it again indicates the correct time?

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14 1 Dimensions and Units (12) Is a cesium clock sufficiently precise to determine your age (assuming it is

exactly 19 years, not a leap year) within 10−6s ? How about within 10−3s ? (13) The slowing of the Earth’s rotation is measured by observing the occurrences

of solar eclipses during a specific period. Assume that the length of a day is increasing uniformly by 0.001 s per century. (a) Over a span of 10 centuries, compare the length of the last and first days, and find the average difference. (b) Find the cumulative difference on the measure of a day over this period.

Mass

(14) A person on a diet loses 2 kg per week. Find the average rate of mass loss in milligrams every: day, hour, minute, and second.

(15) Density is defined as mass per unit volume. If a crude estimation of the average density of the Earth was 5.5 × 103kg/m3and the Earth is considered to be a sphere of radius 6.37 × 106m, then calculate the mass of the Earth.

(16) A carbon-12 atom(12₆C) is found to have a mass of 1.992 64 × 10−26kg. How many atoms of12₆C are there in: (a) 1 kg? (b) 12 kg? (This number is Avogadro’s number in the SI units.)

(17) A water molecule(H2O) contains two atoms of hydrogen (1₁H), each of which has a mass of 1 u, and one atom of oxygen (16₈O), that has a mass 16 u, approx-imately. (a) What is the mass of one molecule of water in units of kilograms? (b) Find how many molecules of water are there in the world’s oceans, which have an estimated mass of 1.5 × 1021kg?

(18) Density is defined as mass per unit volume. The density of iron is 7.87 kg/m3, and the mass of an iron atom is 9.27×10−26kg. If atoms are cubical and tightly packed, (a) What is the volume of an iron atom, and (b) What is the distance between the centers of two adjacent atoms.

Section 1.3 Dimensional Analysis

(19) A simple pendulum has periodic time T given by the relation:

T = 2πL/g

where L is the length of the pendulum and g is the acceleration due to gravity in units of length divided by the square of time. Show that this equation is dimensionally correct.

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(20) Suppose the displacement s of an object moving in a straight line under uniform acceleration a is giving as a function of time by the relation s= kamtn, where k is a dimensionless constant. Use dimensional analysis to find the values of

the powers m and n.

(21) Using dimensional analysis, determine if the following equations are dimen-sionally correct or incorrect: (a)v2 = v2_◦+ 2a s, (b) s = s_◦+ v_◦t+ 1₂a t2,

(c) s= s_◦cos kt, where k is a constant that has the dimension of the inverse of time.

(22) Newton’s second law states that the acceleration of an object is directly propor-tional to the force applied and inversely proporpropor-tional to the mass of the object. Find the dimensions of force and show that it has units of kg·m/s2in terms of SI units.

(23) Newton’s law of universal gravitation is given by F= Gm1m2/r2, where F is the force of attraction of one mass, m1, upon another mass, m2, at a distance r. Find the SI units of the constant G.

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Vectors

₂

When a particle moves in a straight line, we can take its motion to be positive in one specific direction and negative in the other. However, when this particle moves in three dimensions, plus or minus signs are no longer enough to specify the direction of motion. Instead, we must use a vector.

2.1 Vectors and Scalars

A vector has magnitude and direction, examples being displacement (change of position), velocity, acceleration, etc. Actually, not all physical quantities involve direction, examples being temperature, mass, pressure, time, etc. These physical quantities are not vectors because they do not point in any direction, and we call them scalars.

A vector, such as a displacement vector, can be represented graphically by an arrow denoting the magnitude and direction of the vector. All arrows of the same direction and magnitude denote the same vector, as in Fig.2.1a for the case of a displacement vector.

The displacement vector in Fig.2.1b tells us nothing about the actual path taken from point A to B. Thus, displacement vectors represent only the overall effect of the motion, not the motion itself.

Another way to specify a vector is to determine its magnitude and the angle it makes with a reference direction, as in Example 2.1.

H. A. Radi and J. O. Rasmussen, Principles of Physics, 17 Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_2,

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Fig. 2.1 (a) Three vectors of the same direction and magnitude represent the same displacement. (b) All three paths connecting the two points A and B correspond to the same displacement vector

A1 B1 A B A2 B2 A B (a) (b) Example 2.1

A person walks 3 km due east and then 2 km due north. What is his displacement vector?

Solution: We first make an overhead view of the person’s movement as shown in Fig.2.2. The magnitude of the displacement d is given by the Pythagorean theorem as follows:

d =(3 km)2_{+ (2 km)}2_{= 3.61 km}

The angle that this displacement vector makes relative to east is given by:

tanθ = 2 km

3 km = 0.666... Then:θ = tan−1(0.666...) = 33.69◦

Thus, the person’s displacement vector is 56.31◦east of north.

Fig. 2.2 Start End 3 km 2 km d θ E N W S

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2.2 Properties of Vectors 19

2.2 Properties of Vectors

In text books, it is common to use boldface symbols to identify vectors, such as

A, B, etc., but in handwriting it is usual to place an arrow over the symbol, such as,

→

A,B→, etc. Throughout this text we shall use the handwriting style only and use the

italic symbols A, B, etc. to indicate the magnitude of vectors.

Equality of Vectors

The two vectors A→ andB→are said to be equal if they have the same magnitude, i.e. A= B, and point in the same direction; see for example the three equal vectors

AB, A1B1, and A2B2in Fig.2.1a.

Addition of Vectors

Of course, all vectors involved in any addition process must have the same units. The rules for vector sums can be illustrated by using a graphical method. To add vector B→to vector A→, we first draw vector A→on graph paper with its magnitude represented by a convenient scale, and then draw vectorB→to the same scale with its tail coinciding with the arrow head ofA→, see Fig.2.3a. This is known as the triangle method of addition. Thus, the resultant vector R→is the red vector drawn from the tail of A→to the head ofB→and is shown in the vector addition equation:

→

R =→A +B→, (2.1)

which says that the vector R→is the vector sum of vectors A→andB→. The symbol +

in Eq.2.1and the words “sum” and “add” have different meanings for vectors than they do in elementary algebra of scalar numbers.

(a) (b) A A B A B

=

+ R A B

=

+ R A B B = + R A B

Fig. 2.3 (a) In the triangle method of addition, the resultant vector →R is the red vector that runs from the tail of →A to the head of →B. (b) In the parallelogram method of addition, the resultant vec-tor →R is the red diagonal vector that starts from the tails of both →A and →B. This method shows that→A +→B =→B +→A

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An alternative graphical method for adding two vectors is the parallelogram rule of addition. In this method, we superpose the tails of the two vectors A→and→B; then

the resultant →R will be the diagonal of the parallelogram that starts from the tail of both →A and→B (which form the sides of that parallelogram), as shown in Fig.2.3b.

Vector addition has two important properties. First, the order of addition does not matter, and this is known as the commutative law of addition, i.e.

→

A +B→=B→+A→ (Commutative law) (2.2)

Second, if there are more than two vectors, their sum is independent of the way in which the individual vectors are grouped together. This is known as the associative law of addition, i.e.

→

A + (→B +C→) = (A→+→B) +C→ (Associative law) (2.3)

The Negative of a Vector

The negative of a vector B→is a vector with the same magnitude which points in the opposite direction, namely−→B, see Fig.2.4a. Therefore, when we add a vector and its negative we will get zero, i.e.

→

B + (−B→) = 0 (2.4)

Adding−B→to A→has the same effect of subtracting B→from →A, see Fig.2.4b, i.e. → S =A→+ (−→B) =A→−B→ (2.5) (a) (b) B -B

=

-B -S A B A

Fig. 2.4 (a) This part of the figure shows vector→B and its corresponding negative vector−→B, both of which have the same magnitude but are opposite in direction. (b) To subtract vector→B from vector→A, we add the vector−→B to vector→A to get→S =→A −→B

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2.2 Properties of Vectors 21

Example 2.2

A car travels 6 km due east and then 4 km in a direction 60◦north of east. Find the magnitude and direction of the car’s displacement vector by using: (a) the graphical method, and (b) the analytical method.

Solution: (a) Let A→ be a vector directed due east with magnitude A= 6 km and B→be a vector directed 60◦north of east with magnitude B= 4 km. Using graph paper with a reasonable scale and a protractor, we draw the two vectors →

A and→B; then we measure the length of the resultant vector R→. The

measure-ments shown in Fig.2.5indicates that R= 8.7 km. Also, the angle φ that the resultant vectorR→makes with respect to the east direction can be measured and will giveφ = 23.4◦. Fig. 2.5 ° Start End 6 km 4k m θ E N W S 60° A B R φ sin 60 B

(b) The analytical solution for the magnitude of →R can be obtained from geom-etry by using the law of cosines R = √A2_{+ B}2_{− 2AB cos θ as applied to an} obtuse triangle with angleθ = 180◦− 60◦= 120◦, see exercise (10b). Thus:

R=A2_{+ B}2_{− 2 AB cos θ}

=(6 km)2_{+ (4 km)}2_{− 2(6 km)(4 km) cos 120}◦ =(36 + 16 + 24) (km)2_{= 8.72 km}

The angle that this displacement vectorR→makes relative to the east direction, see Fig.2.5, is given by:

sinφ = B sin 60 ◦ R = 4 km sin 60◦ 8.72 km = 0.397 Then:φ = sin−1(0.397) = 23.41◦.

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2.3 Vector Components and Unit Vectors

Vector Components

Adding vectors graphically is not recommended in situations where high precision is needed or in three-dimensional problems. A better way is to make use of the projections of a vector along the axes of a rectangular coordinate system.

Consider a vector A→ lying in the xy-plane and making an angle θ with the positive x-axis, see Fig.2.6. This vector →A can be expressed as the sum of two vectors→Ax and

→

Aycalled the rectangular vector components of →

A along the x-axis and y-axis, respectively. Thus:

→

A =→Ax+A→y (2.6)

Fig. 2.6 A vector→A in the xy-plane can be presented by its rectangular vector components→Axand→Ay, where →_A ₌→_A x+→Ay θ A x y Ay Ax o

From the definitions of sine and cosine, the rectangular components ofA→, namely Axand Ay, will be given by:

Ax = A cos θ and Ay= A sin θ , (2.7) where the sign of the components Axand Aydepends on the angleθ.

The magnitudes Axand Ayform two sides of a right triangle that has a hypotenuse of magnitude A. Thus, from Ax and Aywe get:

A= A2 x + Ay2 and θ = tan−1 Ay Ax (2.8)

The inverse tan obtained from your calculator is from −90◦< θ < 90◦. This may lead to incorrect answer when 90◦< θ ≤ 360◦. A method used to achieve the correct answer is to calculate the angleφ such as:

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2.3 Vector Components and Unit Vectors 23

φ = tan−1_|A y|/|Ax|

(2.9)

Then, depending on the signs of Axand Ay, we identify the quadrant where the vector →

A lies, as shown in Fig.2.7.

x y |Ax| A o x y o x y o x y o |Ay| |Ax| A |Ay| θ |Ax| A |Ay| |Ax| A |Ay| θ θ Ax positive Ax negative Ay negative Ax positive Ay positive Ax negative Ay negative Ay positive φ φ φ θ Quadrant I Quadrant II

Quadrant III Quadrant IV

θ ≡ φ

Fig. 2.7 The signs of Axand Aydepend on the quadrant where the vector→A is located

Once we determine the quadrant, we calculateθ using Table2.1.

Table 2.1 Calculatingθ from φ according to the signs of Ax and Ay

Sign of Ax Sign of Ay Quadrant Angleθ

+ + I θ = φ

− + II θ = 180◦− φ

− − III θ = 180◦+ φ

+ − IV θ = 360◦− φ

Unit Vectors

A unit vector is a dimensionless vector that has a magnitude of exactly one and points in a particular direction, and has no other physical significance. The unit vectors in

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the positive direction of the x, y, and z axes of a right-handed coordinate system are often labeled→i,→j, and→k, respectively; see Fig.2.8. The magnitude of each unit vector equals unity; that is:

|→i | = |→j | = |→k| = 1 (2.10) x y o x y o OR Z Z i k j i j k

Fig. 2.8 Unit vectors→i,→j, and→k define the direction of the commonly-used right-handed coordinate system

Consider a vector →A lying in the xy-plane as shown in Fig.2.9. The product of the component Axand the unit vector→i is the vector→Ax= Ax

→

i, which is paral-lel to the x-axis and has a magnitude Ax. Similarly,A→y= Ay→j is a vector parallel to the y-axis and has a magnitude Ay. Thus, in terms of unit vectors we writeA→

as follows:

→

A = Ax→i + Ay→j (2.11)

Fig. 2.9 A vector→A in the xy-plane can be represented by its rectangular components Ax and Ayand the unit vectors →_{i and}→_j_{, and can be written} as→A = Ax→i + Ay→j A x y Axi o Ay j

This method can be generalized to three-dimensional vectors as: → A = Ax → i + Ay → j + Az → k (2.12)

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2.3 Vector Components and Unit Vectors 25 We can define a unit vector→n along any vector, say,A→, as follows:

→ n =

→

A

A. (2.13)

Adding Vectors by Components

Suppose we wish to add the two vectors A→= Ax → i + Ay → j andB→= Bx → i + By → j using the components method, such as:

→

R =→A+B→

= (Ax→i + Ay→j ) + (Bx→i + By→j )

= (Ax+ Bx)→i + (Ay+ By)→j (2.14) If the vector sum R→is denoted by→R = Rx

→ i + Ry

→

j, then the components of the resultant vector will be given by:

Rx= Ax+ Bx

Ry= Ay+ By

(2.15)

The magnitude of R→can then be obtained from its components or the components of A→and B→using the following relationships:

R= R2 x+ R2y = (Ax+ Bx)2+ (Ay+ By)2 (2.16)

and the angle that R→makes with the x-axis can also be obtained by using the fol-lowing relationships: θ = tan−1Ry Rx = tan−1Ay+ By Ax+ Bx (2.17)

The components method can be verified using the geometrical method, as shown in Fig.2.10.

IfA→= Ax→i +Ay→j +Az→k and→B = Bx→i +By→j +Bz→k, then we can generalize the previous case to three dimensions as follows:

→ R =A→+→B = (Ax+ Bx) → i + (Ay+ By) → j + (Az+ Bz) → k = Rx → i + Ry → j + Rz → k (2.18)

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Fig. 2.10 Geometric representation of the sum of the two vectors→A and→B, showing the relationship between the components of the resultant→R and the

components of→A and→B x y o i j B R A_x Rx B_x B_y A_y R_y A θ Example 2.3

Find the sum of the following two vectors: →

A = 8→i + 3→j →

B = −5→i − 7→j

For convenience, the units of the two vectors have been omitted, but for instance, you may take them to be kilometers.

Solution: The two vectors lie in the xy-plane, since there is no component in the z-axis. By comparing the two expressions ofA→and→B with the gen-eral relations A→= Ax → i + Ay → j and →B = Bx → i + By → j we see that, Ax= 8,

Ay= 3, Bx= − 5, and By= − 7. Therefore, the resultant vector R is obtained by using Eq.2.14as follows:

→ R =→A +B→= (Ax+ Bx) → i + (Ay+ By) → j = (8 − 5)→i + (3 − 7)→j = 3→i − 4→j That is: Rx= 3 and Ry= −4. The magnitude of

→ R is given according to Eq.2.16as: R= R2 x + Ry2= (3)2_{+ (−4)}2₌√₉_{+ 16 =}√₂₅_{= 5}

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2.3 Vector Components and Unit Vectors 27 while the value of the angle θ that→R makes with the positive x-axis is given according to Eq.2.17as:

θ = tan−1 Ry Rx = tan−1 −4 3 = 360◦− tan−1 4 3 = 360◦− 53◦= 307◦

where we used Table2.1to calculateθ in case of negative Ry(Q IV).

2.4 Multiplying Vectors

Multiplying a Vector by a Scalar

If we multiply vector A→by a scalar a we get a new vector→B, i.e.

→

B = aA→ (2.19)

The vector B→ has the same direction as A→ if a is positive but has the opposite direction if a is negative. The magnitude of B→is the product of the magnitude of A→

and the absolute value of a.

The Scalar Product (or the Dot Product)

The scalar product of the two vectors A→and→B is denoted by A→• B→ and is defined as:

→

A •B→= AB cos θ (2.20)

where A and B are the magnitudes of the two vectorsA→andB→, and θ is the angle

between them, see Fig.2.11. The two anglesθ and 360◦− θ could be used, since their cosines are the same. As we see from Eq.2.20, the result ofA→•B→is a scalar quantity, and is known as the dot product from its notation. Also, we get:

→ A •B→= AB cos θ = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ +AB if θ = 0◦ 0 ifθ = 90◦ −AB if θ = 180◦ (2.21)

According to Fig.2.11, the dot product can be regarded as the product of the magnitude of one of the vectors with the scalar component of the second along the direction of the first. That is:

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A B A B A B θ θ θ cos A

θ

cos B

θ

Fig. 2.11 The left part shows two vectors →A and →B, with an angle θ between them. The middle and the right parts show the component of each vector along the other

→

A • B→= (A cos θ)B = A(B cos θ) (2.22) This indicates that scalar products obey the commutative and associative laws, so that:

→

A • →B =B→•A→ (Commutative law) (2.23)

→

A • (B→+C→) =A→• →B +A→•C→ (Associative law) (2.24) By applying the definition of dot product to the unit vectors→i,→j, and→k, we get the following:

→

i • →i =→j • →j =→k • →k = 1 →

i • →j =→i • →k =→j • →k = 0 , (2.25) where the angle between any two identical unit vectors is 0◦and the angle between any two different unit vectors is 90◦.

When two vectors are written in terms of the unit vectors→i,→j, and→k, then to get their dot product, each component of the first vector is to be dotted into each component of the second vector. After that, we use Eq.2.25to get the following:

→

A • →B = (Ax→i + Ay→j + Az→k) • (Bx→i + By→j + Bz→k)

= AxBx+ AyBy+ AzBz (2.26)

Thus, from Eqs.2.20 and 2.26, we can generally write the dot product as follows:

→

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2.4 Multiplying Vectors 29

Example 2.4

Find the angle between the vectorA→= 8→i +3→j and the vector B= −5→i −7→j . Solution: Since A= A2 x+ Ay2and B= B2

x+ By2, then using the dot product given by Eq.2.20we get:

→

A • B→= AB cos θ =82_{+ 3}2_×₍₋₅₎2_{+ (−7)}2_cos_θ = 8.544 × 8.60 cos θ

= 73.5 cos θ

Keeping in mind that there is no component for A→andB→along the z-axis, we can find the dot product from Eq.2.26as follows:

→

A •B→= AxBx+ AyBy+ AzBz

= 8 × (−5) + 3 × (−7) + 0 × 0 = −61

Equating the results of the last two steps to each other, we find:

73.5 cos θ = −61 Thus: θ = cos−1 −61 73.5 = 146.1◦.

The Vector Product (or the Cross Product)

The vector product of the two vectors A→and →B is denoted by A→×B→and defined as a third vector C→whose magnitude is:

C= AB sin θ , (2.28)

whereθ is the smaller angle between A→andB→(hence, 0≤ sin θ ≤ 1). The direction ofC→is perpendicular to the plane that contains both →A and B→, and can be determined

by using the right-hand rule, see Fig.2.12. To apply this rule, we allow the tail of →

A to coincide with the tail of →B, then the four fingers of the right hand are pointed

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right thumb is the direction of C→, i.e. the direction of →A ×→B. Also, the direction of

→

C is determined by the direction of advance of a right-handed screw as shown in Fig.2.12. Right-hand rule Direction of Advance A B C =A B θ

Fig. 2.12 The vector product→A×→B is a third vector→C that has a magnitude of AB sinθ and a direction perpendicular to the plane containing the vectors→A and→B. Its sense is determined by the right-hand rule or the direction of advance of a right-handed screw

The vector product definition leads to the following properties:

1. The order of vector product multiplication is important; that is: →

A×B→= −(B→×A→) (2.29)

which is unlike the scalar product and can be easily verified with the right-hand rule.

2. If A→is parallel to →B (that is,θ = 0◦) or →A is antiparallel to →B (that is,θ = 180◦), then:

→

A×→B = 0 (if →A is parallel or antiparallel to B→) (2.30) 3. If A→is perpendicular to B→, then:

|A→×→B| = AB (if A→⊥B→) (2.31) 4. The vector product obeys the distributive law, that is:

→

A × (B→+C→) =A→×→B +A→×C→ (Distributive law) (2.32) 5. The derivative of A→×B→with respect to any variable such as t is:

d dt( → A×B→) =A→×d → B dt + dA→ dt × → B (2.33)

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2.4 Multiplying Vectors 31 6. From the definition of the vector product and the unit vectors→i, →j, and→k, we

get the following relationships: →

i ×→i =→j ×→j =→k×→k = 0 (2.34) →

i × →j =→k, →j × →k =→i, →k × →i =→j (2.35) The last relations can be obtained by setting the unit vectors→i ,→j , and →k on a circle, see Fig.2.13, and rotating in a clockwise direction to find the cross product of one unit vector with another. Rotating in a counterclockwise direction will involve a negative sign of the cross product of one unit vector with another, that is:

→

i ×→k = −→j, →k ×→j = −→i, →j ×→i = −→k (2.36)

Fig. 2.13 The clockwise and counterclockwise cyclic order for finding the cross product of the unit vectors→i,→j, and→k

k + + +

-j i

7. When two vectors A→andB→are written in terms of the unit vectors→i , →j , and →

k, then the cross product will give the result: → A×→B = (Ax → i + Ay → j + Az → k) × (Bx → i + By → j + Bz → k) = (AyBz− AzBy) → i + (AzBx− AxBz) → j + (AxBy− AyBx) → k (2.37)

This result can be expressed in determinant form as follows:

→ A×B→= → i →j →k AxAy Az BxBy Bz = → i AyAz ByBz − → j Ax Az Bx Bz + → k Ax Ay Bx By (2.38)

Principles of Physics For Scientists and Engineers pdf - Web Education

Undergraduate Lecture Notes in Physics

Hafez A. Radi

John O. Rasmussen

Principles of Physics

For Scientists and Engineers

Preface

Contents

{

Part I

Dimensions and Units

1

1.1

The International System of Units

1.2

Standards of Length, Time, and Mass

Length (L)

Time (T)

Mass (M)

A Second Standard Mass

1.3

Dimensional Analysis

1.4

Exercises

Section 1.1 The International System of Units

Section 1.2 Standards of Length, Time, and Mass

Time

Mass

Section 1.3 Dimensional Analysis

Vectors

2

2.1

Vectors and Scalars

2.2

Properties of Vectors

Equality of Vectors

Addition of Vectors

=

=

The Negative of a Vector

=

2.3

Vector Components and Unit Vectors

Vector Components

Unit Vectors

Adding Vectors by Components

2.4

Multiplying Vectors

Multiplying a Vector by a Scalar

The Scalar Product (or the Dot Product)

θ

θ

The Vector Product (or the Cross Product)

₁

₂