basics of mathematics and physics for pre

basics of mathematics and physics for pre
BASICS OF MATHEMATICS
AND PHYSICS
FOR PRE-UNIVERSITY COURSE
Minsk BSMU 2017
МИНИСТЕРСТВО ЗДРАВООХРАНЕНИЯ РЕСПУБЛИКИ БЕЛАРУСЬ
БЕЛОРУССКИЙ ГОСУДАРСТВЕННЫЙ МЕДИЦИНСКИЙ УНИВЕРСИТЕТ
КАФЕДРА МЕДИЦИНСКОЙ И БИОЛОГИЧЕСКОЙ ФИЗИКИ
ОСНОВЫ МАТЕМАТИКИ И ФИЗИКИ
ДЛЯ ПОДГОТОВИТЕЛЬНОГО ОТДЕЛЕНИЯ
BASICS OF MATHEMATICS AND PHYSICS
FOR PRE-UNIVERSITY COURSE
Учебно-методическое пособие
Минск БГМУ 2017
2
УДК 51+53(075.8)-054.6
ББК 22.1+22.3я73
О-75
Рекомендовано Научно-методическим советом университета в качестве
учебно-методического пособия 15.06.2016 г., протокол № 10
А в т о р ы: М. В. Гольцев, Л. В. Кухаренко, О. В. Недзьведь, Н. А. Никоненко,
В. Г. Лещенко, Н. И. Инсарова
Р е ц е н з е н т ы: д-р физ.-мат. наук, проф., зав. каф. физики твердого тела Белорусского
государственного университета В. В. Углов; канд. физ.-мат. наук, зав. каф. физики
Белорусского государственного аграрного технического университета В. А. Чернявский
Основы математики и физики для подготовительного отделения = Basics of matheО-75 matics and physics for pre-university course : учеб.-метод. пособие / М. В. Гольцев [и др.]. ‒
Минск : БГМУ, 2017. – 224 с.
ISBN 978-985-567-680-6.
Включает некоторые сведения по элементарной математике и основам дифференциального
исчисления, а также основные разделы школьного курса физики — механику, молекулярную
физику и термодинамику, электричество и магнетизм, оптику, атомную и ядерную физику.
Предназначено для иностранных учащихся подготовительного отделения, обучающихся на
английском языке.
УДК 51+53(075.8)-054.6
ББК 22.1+22.3я73
______________________________________________________
Учебное издание
Гольцев Михаил Всеволодович
Кухаренко Людмила Валентиновна
Недзьведь Ольга Валерьевна и др.
ОСНОВЫ МАТЕМАТИКИ И ФИЗИКИ ДЛЯ ПОДГОТОВИТЕЛЬНОГО
ОТДЕЛЕНИЯ
BASICS OF MATHEMATICS AND PHYSICS FOR PRE-UNIVERSITY COURSE
Учебно-методическое пособие
На английском языке
Ответственный за выпуск М. В. Гольцев
Переводчики: М. В. Гольцев, Л. В. Кухаренко, О. В. Недзьведь, Н. А. Никоненко
Компьютерная верстка Н. М. Федорцовой
Подписано в печать 01.03.17. Формат 60 84/16. Бумага писчая «Снегурочка».
Ризография. Гарнитура «Times». Усл. печ. л. 13,02. Уч.-изд. л. 10,28. Тираж 150 экз. Заказ 117.
Издатель и полиграфическое исполнение: учреждение образования
«Белорусский государственный медицинский университет».
Свидетельство о государственной регистрации издателя, изготовителя,
распространителя печатных изданий № 1/187 от 18.02.2014.
Ул. Ленинградская, 6, 220006, Минск.
© УО «Белорусский государственный
медицинский университет», 2017
ISBN 978-985-567-680-6
3
PREFACE
The textbook is intended for the international students of the preparatory
departments of institutions of higher education studying physics in English
in order to enter the medical universities of the Republic of Belarus.
This textbook contents information on elementary mathematics and
the basics of differential calculus, as well as the main sections of a school
physics course — mechanics, molecular physics and thermodynamics,
electricity and magnetism, optics, atomic and nuclear physics, which are
required for the study of medical and biological physics. The problems and
examples have been selected for the distinct purpose of illustrating the principles
taught in the text and for their practical applications. A list of problems and tests
is placed at the end of every topic. They are in sufficient number to permit
testing at many points and of a teacher‘s choice of problems. The order
of topics, illustrations, and problems have been also selected with the purpose
of leading the student into a clear understanding of the physical phenomena
concerning to biology and medicine. Duly made selection of problems and
examples, conciseness and simplicity of presentation of different topics
contribute to the successful study of the proposed material.
4
THE BASICS OF ELEMENTARY MATHEMATICS
AND DIFFERENTIAL CALCULUS
1. THE BASIC MATHEMATICAL CONCEPTS
AND FORMULAS
1.1. FRACTION. OPERATIONS WITH FRACTIONS.
EXPONENTS AND RADICALS. FACTORING AND EXPANDING
A fraction is an expression of the following form:
a
(a over b), where a —
b
numerator, b — denominator.
A proper fraction is one whose numerator is less than denominator. For
example, 1/2; 1/4; 1/7; 2/3; 2/100 and 36/81 are proper fractions. An improper fraction
is a fraction, whose numerator is equal to or larger than the denominator. Thus,
21
/5, 100/37 and 8/8 are improper fractions.
Operations with fractions.
To reduce a fraction to its lowest terms, divide numerator and denominator
by their highest common factor (or: measure, or: divisor:
ar
br
a
).
b
To reduce a fraction to higher terms, multiply the numerator and
the denominator by the same number:
a
b
ar
.
br
To find the sum (the difference) of two unlike fractions, change them to
like fractions (fractions having their least common denominator) and combine
the numerators:
a
b
c
d
ad cb
.
bd
To find the product of two fractions, multiply the numerators together and
the denominators together:
a c
b d
ac
.
bd
To find the quotient of two fractions, multiply the dividend by the inverted
divisor:
a
b
c
d
ad
.
bc
Equivalent fractions are known as proportions:
a
b
c
d
ad = bc
a
c
b
d
c
a
d
b
d
c
b
.
a
Exponents and Radicals (Roots).
In the expression (an) = с (a to the n-th power is equal to c) the quantity a
is called the base and n is the exponent of the power.
5
A quantity a to the power of m over n is called the n-th root of a
m
an
n m
a
to the m-th power.
The following rules are useful in manipulations with exponents and roots.
a0 = 1
a1 = a
(an)(am) = an+m
(an)(bn) = (ab)n
(an)m = anm
(an)(a–n) = a0 = 1
a
b
an
n
an
bn
m
(a)n
a
n
a
n
an b
a
1
m
n
n
ab
1
an
n
a
n
a
n
b
n
a
b
Factoring and Expanding.
In many cases one needs the inverse operations — Factoring or Expanding.
We can obtain another form of the algebraic expression due to expanding of
powers or a product of items and write down the result as a sum of terms.
Table 1.1
Expanding formulas
1
2
3
4
5
6
7
(a + b)2 = a2 + 2ab + b2
(a – b)2 a2 – 2ab + b2
(a + b) (a – b) = a2 – b2
(a + b)3 a3 + 3a2b + 3ab2 + b3
(a – b)3 a3 – 3a2b + 3ab2 – b3
(a + b)(a2 – ab + b2) a3 + b3
(a – b) (a2 + ab + b2) a3 – b3
Examples:
(a b) 2 (a b) 2
ab
a 2 2ab b 2 a 2 2ab b 2
ab
6
4ab
ab
4.
x
5
y 1 1 y
a
x
5
y 1
y 1
a
6
6
x 5
y 1
a 6
c 3 3 c c 3 c 3 c 3
2m
2n
2m
2n
2m
m n n m m n m n
m
5p
10q
5p
10q
2q p p 2q 2q p 2q p
a 2 16
a 4
8a
4 a
a 2 16
a 4
x2 9 y 2
x 3y
6 xy
3y x
8a
a 4
x2 9 y 2
x 3y
2n
n
2(m n)
m n
2
5 p 10q
2q p
5(p 2q)
2q p
5
a 2 8a 16
a 4
(a 4) 2
a 4
a 4
x 2 6 xy 9 y 2
x 3y
6 xy
x 3y
(x 3 y) 2
x 3y
x 3y
EXERCISES
1. Reduce fractions:
9
;
54
x 4
x 2 16
;
y3 1
.
y 1
2. Perform operations with fractions:
5
7
2
;
7
1
2
2
;
3
3 15
;
5 21
4 12
: .
7 21
3. Simplify expression: 3·(5x + 2) – 10x =
4. Perform factoring operations: a3 – b3 =
b2 – 9 =
(a – b)2 =
a3 – 125 =
5. Perform operations: a4 × a3 =
b8 : b2 =
b9 : b–2 =
1.2. FUNCTIONAL DEPENDENCE. BASIC FUNCTIONS
AND THEIR GRAPHS
Function is the dependence of variable y on the variable x from some set
D, where each variable value x corresponds a single value of the variable y:
y = f(x). The equation above is the mainly used representation of a function; it is
called the function notation.
The variable x is called independent variable or argument. The variable y is
the dependent variable and says that variable y is a function of the variable x. All
7
values of the independent variable x (the set D) are called the domain of
definition. The set of all the values taken by the dependent variable y is called
the range of the function
The graph of the function is the set of all points in the coordinate plane,
the abscissa of which is equal to the argument values, and the ordinate is
the corresponding function values.
Zeros of the function are the values of the argument at which the function
vanishes.
The function is called increasing on some interval I if for any x1, x2 ∈ I
the inequality x1 < x2 corresponds to inequality f(x1) < f(x2). The function is
called decreasing on some interval I if for any x1, x2 ∈ I the inequality x1 < x2
corresponds to inequality f(x1) > f(x2).
The function can be represented in analytical form by formula, in tabular
form by means of tables and in graphical form by graph.
Example:
1) analytical form : y =
2) tabular form:
x
y = f(x)
1
x +2
3
0
2
3
3
6
4
etc.
etc.
3) graphical form:
1.2.1. LINEAR FUNCTION AND ITS GRAPH
A linear function is a function defined by a formula of the form y = kx +b,
where x is the argument, k, b ∈ R. The graph of a linear function is a straight
line (fig. 1.1).
The coefficient k is called the angular coefficient of the straight line.
8
Zero of a linear function: x = −
b
k
b>0
b<0
Fig. 1.1. The graph of a linear function
Direct proportionality is a particular case of a linear function (fig. 1.2).
Direct proportionality is a function that can be set by the formula y = kx,
where x — independent variable, k ≠ 0. Coefficient k is called the coefficient of
direct proportionality.
Fig. 1.2. The graph of a direct proportionality function
EXERCISES
a. Create the graph of the function:
1) y = 3x
2) y = 2x – 1
3) y = –3x + 2
b. Find the x-intercept and the y-intercept for the following linear function
(zeros of a functions). Find the points of functions intersection:
1) y = 4x – 6 and y = –2x
2) y = 2x – 1 and y = –4x + 5
3) y = 3x – 1 and y = –3x + 11
c. Find a linear function that passes through the origin and forming with
the x-axis the following angles:
1) 30º
2) 45º
3) 135º
4) 0º
9
1.2.2. INVERSE PROPORTIONALITY FUNCTION AND ITS GRAPH
An inverse proportionality function is a function defined by a formula of
the form y =
k
, where x is the argument, k ∈ R, k ≠ 0.
x
The domain of this function: x ≠ 0.
The graph of an inverse proportionality function is a hyperbola (fig. 1.3).
Fig. 1.3. The graph of an inverse proportionality function
There is no zeros of the function y = k/x!
If k > 0, the function y = k/x decreases throughout the domain of definition,
when k < 0, the function y = k/x increases in all the field definitions. For a curve
that is the graph of this function, the x-axis and y-axis play the role of
asymptotes.
Asymptote — a straight line which is closer to the points of the curve as
they remove into infinity.
EXERCISES
a. Create the graph of the function:
1) y
3
x
2) y
2
x
1.2.3. QUADRATIC FUNCTION AND ITS GRAPH
A quadratic function is a function defined by a formula of the form:
y = ax2 + bx + с,
where x is the argument, a, b, c ∈ R, a ≠ 0.
The graph of a quadratic function is a parabola (fig. 2.4).
The vertex of the parabola is the point of intersection of the parabola with
its axis of symmetry. The vertex of the parabola y = ax2 + bx + c has coordinates
b
b 2 4ac
(
).
;
2a
4a
10
Fig. 1.4. The graph of a quadratic function
Let‘s consider the function defined by the formula y = ax2 (a ≠ 0) as
a particular case of a quadratic function (fig. 1.5).
Fig. 1.5. The graph of the function defined by the formula y = ax2 (a ≠ 0)
The properties of the function y = ax2:
If x = 0, y = 0, i. e. the graph of the function passes through the origin.
The function graph is symmetrical about the y-axis.
If a > 0, the function decreases on the interval (–∞; 0] and increases on
the interval [0; +∞).
If a < 0, the function increases on the interval (–∞; 0] and decreases on
the interval [0; + ∞).
If a > 0, ymin = 0; if a < 0, ymax = 0.
11
1.2.4. QUADRATIC EQUATIONS. QUADRATIC FORMULA
The quadratic equation can be presented in the following form:
ax2 + bx + c = 0,
where x is a variable, a, b, c are some constants (a ≠ 0).
If a = 1 (i. e., the equation of the form x2 + bx + c = 0), the quadratic
equation is called monic quadratic equation.
b 2 4ac
Quadratic Formula: If ax + bx + c = 0, then x =
.
2a
In the general form of a quadratic equation ax2 + bx + c = 0.
The expression D = b2– 4ac called the discriminant of the quadratic equation
ax2 + bx + c = 0.
If D < 0, the quadratic equation has no roots.
b
2
When D = 0, the quadratic equation has two of the same root: x =
b
If D > 0, the quadratic equation has two roots x1,2 =
b
.
2a
D
.
2a
It is useful to remember the following factoring formula for practical use:
ax2 + bx + c = a (x – x1) ⋅ (x – x2).
Viete Theorem: If x1 and x2 are the roots of the monic quadratic equation
2
x + px + q = 0, then x1 + x2 = –p; x1⋅x2 = q.
Example:
Find the roots of the quadratic equation х2 – 2х – 3 = 0.
2
2
Solution: D = (–2) – 4⋅1⋅(–3) = 4 + 12 = 16 = 4
x1 =
2 4
= –1;
2
Answer: –1; 3.
x2 =
x1,2 =
( 2)
42
2 1
2 4
= 3.
2
EXERCISES
a. Create the graph of the function:
1) y = 3x2 + 2x – 1;
2) y = –x2 + 2;
b. Find the roots of the quadratic equation:
1) x2 + x – 20 = 0;
2) x2 – 8x – 9 = 0;
4) x2 – 6x + –6 = 0;
5) x2 +
1
2
x
= 0;
3
3
3) y = –4x2.
3) 16x2 – 40x + 25 = 0.
6) 2x2 – 5x – 7 = 0.
1.2.5. CUBIC FUNCTION AND ITS GRAPH
Cubic function in mathematics is a numerical function of the following
form
f(x) = ax3 + bx2 + cx + b, x ∈ R,
where a ≠ 0.
12
Generally speaking, a cubic function is a polynomial of the third degree.
The graph of a full cubic function is the following (fig. 1.6).
Fig. 1.6. The graph of a full cubic function
Let‘s consider the function defined by the formula y = ax3 (a ≠ 0). In this
case the graph of the cubic function is a cubic parabola (fig. 1.7).
Fig. 1.7. The graph of the function defined by the formula y = ax3 (a ≠ 0)
Cubic parabola — a plane algebraic curve of the third order. Its canonical
equation in rectangular Cartesian coordinates has the form y = ax3, where a ≠ 0.
The cubic parabola has a center of symmetry at the origin, this point
is the inflection point of the curve. The x-axis is the tangent to the cubic
parabola at that point.
For a > 0 cubic parabola is located in the first and third quarters
of the coordinate, it is an increasing function.
For a < 0 the curve runs in the second and fourth quarters and decreases.
Cubic parabola exists at least one the x-intercept but no more than three
x-intercepts. Axis from 1 to 3 times. Sit means, that the cubic equation
ax3 + bx2 + cx + b = 0 has at least one up to three roots.
13
EXERCISES
a. Create the graph of the function:
1) y = 2x3;
2) y = 3x3.
1.2.6. THE EXPONENTIAL FUNCTION AND ITS GRAPH
An exponential function is a function defined by a formula of the form
y = ax,
where a > 0, a ≠ 1 (fig. 1.8).
Fig. 1.8. The graph of an exponential function
In contexts where the base a is not specified, especially in more theoretical
contexts, the term exponential function is almost always understood to mean
the natural exponential function y = ex (fig. 1.9).
Fig. 1.9. The graph of the natural exponential function y = ex
Let‘s consider the function defined by the formula y = ex (where e is
Euler‘s or irrational number, e = 2.71828) as a particular case of an exponential
function.
The exponential function is used to model a relationship in which
a constant change in the independent variable gives the same proportional
14
change (i. e. percentage increase or decrease) in the dependent variable.
The function is often written as exp(x). The exponential function is widely used
in physics, biology and mathematics.
Example: Radioactivity is one very frequently given example of
exponential decay. The law describes the statistical behavior of a large number
of nuclides, rather than individual atoms.
N(t) = N0 e–λt.
Here N(t) is the quantity at time t, and N0 = N(0) is the initial quantity,
i. e. the quantity at time t = 0, and λ (lambda) is a positive rate called
the exponential decay constant.
1.2.7. LOGARITHM. COMMON AND NATURAL LOGARITHMS.
THE PROPERTIES OF LOGARITHMS. LOGARITHMIC FUNCTION AND ITS GRAPH
If at an exponential function y = ax change the places of x and y we will
have a function x = a y. So y is the power of base a to calculate value x, it means
that y = logax, y is the logarithm of x on base a.
That is, the logarithm of a number x to the base a is that number y which,
as the exponent of a, gives back the number x.
For common logarithms, the base is 10, so if x = 10y, then y = lgx or log x.
The subscript 10 on log10 is usually omitted when dealing with common logs.
Another important base is the exponential base e = 2.71828..., where e is
natural number (or Euler‘s or irrational number). Such logarithms are called
natural logarithms and are written ln.
Thus, if x = ey, then y = lnx.
For any number y, the two types of logarithm are related by the equality:
ln x = 2.3026 log x.
Some simple rules for logarithms:
1. log (xz) = log x + log z,
which is true because if x = 10n and z = 10m, then xz = 10n+m. From
the definition of logarithm, log x = n, log z = m, and log (xz) = n + m; hence,
log(xz) = n + m = log x + log z.
In a similar way, we can show that.
2. log
x
= log x – log z.
z
3. log xn = n log x.
These three rules apply to any kind of logarithm. The function that
assigns to y its logarithm is called logarithm function or logarithmic function
(or just logarithm) y = loga x, if a > 0, a ≠ 1, x > 0.
The graph of a classical logarithmic function y = logax will be
the following (fig. 1.10).
Let‘s consider the graph of a natural logarithmic function y = ln x as
a particular case of a logarithmic function (fig. 1.11).
15
Fig. 1.10. The graph of a classical
logarithmic function
Fig. 1.11. The graph of a natural logarithmic
function
EXERCISES
a. Find the logarithms:
1) log62 + log618;
b. Find x:
1) 3 = log2(15 – x);
2) lg4 + lg25;
3) lg3000 – lg3;
2) 5 = lg(100–x);
3) 6 = ln(e2x).
4) ln e5.
1.2.8. TRIGONOMETRIC FUNCTIONS AND THEIR GRAPHS.
PROPERTIES OF TRIGONOMETRIC FUNCTIONS
Trigonometric functions — math functions of an angle. They are certainly
important when studying geometry, and in the study of periodic processes.
Typically trigonometric functions are defined as the relationship of sides
of a right triangle or the length of certain segments in the unit circle.
An angle can be set in radians (1 rad) and degrees (1º). One radian is equal
to 180/π degrees. Thus, to convert from radians to degrees, multiply by 180/π.
180
π
≈ 57.2958°
1° = 1·
≈ 0,0175 rad.
π
180
Correspondence between the main trigonometric functions presented in
table 1.2.
1 rad = 1·
Table 1.2
Trigonometric functions
Functions
Correspondence
sin
cos
tg or tan
ctg or cot
16
a. Function y = sin x and its graph (fig. 1.12).
Fig. 1.12. The graph of a function y = sin x
b. Function y = cos x and its graph (fig. 1.13).
Fig. 1.13. The graph of a function y = cos x
Trigonometric functions y = sin x and y = cos x are periodical functions
with period T = 2π.
c. Function y = tg x (y = tan x) and its graph (fig. 1.14).
Fig. 1.14. The graph of a function y = tg x (y = tan x)
Trigonometric function y = tg x is periodical function with period T = π.
17
Table 1.3
Trigonometric functions of the main angles
α (rad)
0
α (deg)
0°
sin α
0
cos α
1
tg α
0
ctg α
–
π
6
30°
1
2
3
2
1
3
3
π
4
45°
2
2
2
2
π
2
90°
π
3
60°
3
2
1
2
1
0
2π
3
120°
3
2
1
2
3π
4
135°
2
2
2
2
1
3
–
3
–1
1
1
3
0
1
3
–1
5π
6
150°
1
2
3
2
1
3
3
π
180°
0
–1
0
–
We take an arbitrary right triangle that contains the angle α to define
trigonometric functions of the angle α. We will set the sides of this triangle as:
Hypotenuse is a side opposite right angle (longest side
in the triangle), the c side in this case.
The opposite leg is the side that lies opposite
the angle α. For example, side a is opposite to angle α.
The adjacent leg is the side which is a party angle.
For example, leg b is adjacent to angle α (fig. 1.15).
The sine of angle α is the ratio of the length of
the opposite side a to the length of the hypotenuse с:
sin α
a
.
c
Fig. 1.15. An arbitrary
right triangle
This attitude does not depend on the choice of triangle {ABC} containing
the angle α, since all such triangles are similar.
The cosine of angle α is the ratio of the length of the adjacent side b to
the length of the hypotenuse c: cos α
b
.
c
The sine of one acute angle in the triangle equals the cosine of the second:
sin β
b
c
cos α
b
.
c
The tangent of angle α is the ratio of the length of the opposite side a to
the length of the adjacent side b: tg α
a
.
b
This is easy to see by studying a right triangle and applying
the Pythagorean theorem that in symbolic form the Pythagorean identity can be
written as:
sin2α + cos2α = 1.
18
1.2.9. MAIN TRIGONOMETRIC FORMULAS
sin(α ± β) = sin α cos β ± cos α sin β
cos(α ± β) = cos α cos β ± sin α sin β
tg(α β)
tg α tg β
1 tg α tg β
sin 2α 2sin α cos α
2
cos 2α cos α
tan 2α
2tg α
2tg α
2ctg α
2
1 tg 2α
1 ctg 2α
tg α ctg 2α
2
2
sin α 2cos α
2ctg α
2
1 1 2sin α
2
1 tg 2α ctg 2α 1 ctg α tg α
cos(α β) cos(α β)
sin α sin β
2
cos(α β) cos(α β)
cos α cos β
2
sin(α β) sin(α β)
sin α cos β
2
cos(α β) cos(α β)
tg α tg β
cos(α β) cos(α β)
1 cos2α
sin α
2
2
1 cos2α
cos α
2
2
1 cos2α
tg α
1 cos2α
2
sin α sin β
cos α cos β
cos α cos β
tan α tan β
tg 2α
1 tg 2α
ctg 2α
1 ctg 2α
sin 2α
1 sin 2α
α β
α β
2sin
cos
2
2
α β
α β
2cos
cos
2
2
α β α β
2sin
sin
2
2
sin(α β)
cos α cos β
19
1 tg 2α
ctg 2α 1
1 tg 2α
ctg 2α 1
EXERCISES
a. Simplify the expression:
1) sin 20° cos 40° + cos 20° sin 40°;
3) cos 12° cos 18° – sin 12° sin 18°;
5)
tg 22
tg 23
1 tg 22 tg 23
;
6)
tg 45
2) sin 45° cos 15° + cos 45° sin 15°;
4) cos 98° cos 8° – sin 98° sin 8°;
tg 15
1 tg 15 tg 45
b. Create the graph of the function:
1) y = cos2x;
2) y = sin3x.
c. Calculate:
1) cos120º;
2) cos135º;
.
3) sin75º;
4) tg1105º.
1.3. VECTORS
A geometric vector (or simply a vector) is a geometric object that has
length or magnitude and direction. It can be added to other vectors according to
the rules of vector algebra. A vector is frequently represented in mathematics
and physics by a line segment with a definite direction, or graphically as
an arrow, connecting an initial point A with a terminal point B, and denoted
by AB .
The vector can also be denoted as AB if it represents a directed distance or
displacement from a point A to a point B (fig. 1.16).
Fig. 1.16. The vector AB
Latin word vector means ―carrier‖. A vector is what is needed to ―carry‖
the point A to the point B. The magnitude of the vector is the distance between
the two points and the direction refers to the direction of displacement from A to
B. Mathematical operations on real numbers such as addition, subtraction and
multiplication have close analogues for vectors. Vectors play an important role
in physics. Velocity and acceleration of a moving object as well as forces acting
on it are described by vectors. Although most of other physical quantities do not
represent distances (except, for example, position and displacement), their
magnitude and direction can be still represented by the length and direction
of an arrow. The mathematical representation of a physical vector depends on
the coordinate system used to describe it.
20
1.3.1. VECTOR ADDITION
It is possible to use different rules (methods) for the geometric construction
of the vector addition a + b , but they all give the same result.
a. Triangle rule.
Both vectors are transported parallel to themselves so that the beginning of
one of them coincided with the end of another. Then the vector sum a + b is
defined by a third party resulting triangle, and its beginning coincides with the
beginning of the first vector a and the end of the second vector b (fig. 1.17).
Fig. 1.17. The vector sum a + b
This rule is directly and naturally generalized to add any number of vectors,
moving in the rule of the polygon.
b. Polygon rule.
The beginning of the second vector coincides with the end of the first,
the beginning of the third — with the end of the second, and so on. The sum of
n vectors is a vector with the beginning coinciding with the beginning of
the first vector and the end coinciding with the end of the n-th vector (fig. 1.18).
Fig. 1.18. The sum of n vectors
c. Parallelogram rule.
Both vectors a and b are transported parallel to themselves so that their
beginning matches. Then the vector sum is given by diagonal built on
a parallelogram of them coming from their common base point (fig 1.19).
The rule of the parallelogram is especially useful when there is a need to
represent the vector of amounts immediately applied to the same point, which is
21
applied to both terms — that is, to depict all three vectors having a common
origin.
a – b = (a1 – b1)e1 + (a2 – b2)e2 + (a3 – b3)e3.
Fig. 1.19. Parallelogram rule
1.3.2. VECTOR SUBTRACTION
Subtraction of two vectors can be geometrically defined as follows: to
subtract b from, a place the tails of a and b at the same point, and then draw
an arrow from the head of b to the head of a. This new arrow represents
the vector a – b , as illustrated below on fig. 1.20.
Fig. 1.20. Subtraction of two vectors
Subtraction of two vectors may also be performed by adding the opposite
of the second vector to the first vector, that is a – b a ( b) .
1.3.3. VECTOR MULTIPLICATION (SCALAR MULTIPLICATION)
a. Multiplication by a number.
Multiplication of a vector a by a number n > 0 gives a collinear vector
with length n times greater.
Multiplication of a vector a by a number n < 0 gives the opposite
directional vector with a length of n times more (fig. 1.21).
Fig. 1.21. Multiplication of a vector a by a number n > 0
22
b. The dot product.
For geometric vectors the dot product is defined via their geometric
features and is entered as follows:
ab
a b cos(a, b).
Here to calculate the cosine takes the angle between the vectors, which is
defined as the angle formed by the vectors, if you put them to one point (to
combine them).
1.3.4. VECTOR DECOMPOSITION
An arbitrary vector c can be represented as a sum: c = ma nb , where m
and n are arbitrary numbers, and the triple of vectors c , and are coplanar
(fig. 1.22). It is a decomposition of the vector on a and b components.
If the vectors a and b are not collinear, the submitted decomposition is the only
possible.
Fig. 1.22. Vector decomposition
1.3.5. PROJECTION OF VECTOR ON A COORDINATE AXIS
The projection of the vector AB on the axis l is a number equal to the size
of the segment A1B1 to the axis l, where the points A1 and B1 are the projections
of points A and B on the axis (fig. 1.23).
Fig. 1.23. The projection of the vector AB on the axis l
The projection of the vector on the coordinate axis is a scalar value.
The sign of the projection depends on the direction of the vector relative to
the coordinate axes.
23
Projection may be positive or negative. The projection of the vector AB
on an some axis is called positive if the projection direction coincides with
the direction of the axis from the beginning of the projection up to the end of it.
Let‘s imagine Cx as the projection of the vector c on the axis x (fig. 1.23).
a
b
c
d
e
Fig. 1.24. The projections of the vector c on the axis x
a. if 0° < α < 90°, cx = c·cos α, 0 < cos α < 1, cx > 0.
b. if α = 0°, c ↑↑ OX, cx = c·cos 0°, cx = +c, projection is positive.
c. if α = 90°, c OX, cx = c·cos 90°, cx = 0.
d. if 180° > α > 90°, –1 < cos α < 0, cx < 0.
e. if α = 180°, c ↑↓ OX, cx = c·cos 180°, cx = –c, projection is negative.
EXERCISES
a. Produce an addition of vectors a , b and c :
b. Produce a subtraction of vectors a and b :
c. Produce a decomposition of the vector c on a and b components along
the directions АВ and CD
24
d. Produce a projection of vector a on the coordinate axises x and y for
both cases (1 and 2)
1
2
1.4. ELEMENTARY GEOMETRY FIGURES AND FORMULAS
Triangle. Types of triangles.
Triangles can be classified according to the lengths of their sides:
Equilateral (a), Isosceles (b) and Scalene (с) Triangles (fig. 1.25).
Triangles can also be classified according to their internal angles: Right (a),
Obtuse (b) and Acute (с) Triangles (fig. 1.26).
a
с
b
Fig. 1.25. Triangles classification according to the length of their sides:
a — Equilateral; b — Isosceles; с — Scalene
a
с
b
Fig. 1.26. Triangles classification according to the internal angles:
a — Right; b — Obtuse; с — Acute
The Pythagorean theorem.
In any right triangles, the square of the length of the hypotenuse equals
the sum of the squares of the lengths of the two other sides. If the hypotenuse
has length c, and the legs have lengths a and b (fig. 1.27) then the theorem states
that a2 + b2 = c2.
The area S of the triangle (fig. 1.28): S =
25
1
bh.
2
Fig. 1.27. The Pythagorean theorem
for right triangles
Fig. 1.28. The area S of the triangle
The area S of the parallelogram (fig. 1.29): S = ah = ab sin α.
Fig. 1.29. The area S of the parallelogram
The length L of a circle (fig. 1.30): L = 2πR = πD.
The surface S area of a circle (fig. 1.30): S = πR2.
Fig. 1.30. The surface area S of a circle
A rectangle is any quadrilateral with four right angles. It can also be
defined as a parallelogram containing a right angle (fig. 1.31).
If a rectangle has length a and width b, its area: S = ab, and its perimeter
P = 2(a + b).
26
A rectangle with four sides of equal length is a square (fig. 1.32).
Fig. 1.31. A rectangle
Fig. 1.32. A square
The perimeter P of a square whose four sides have length a is: P = 4a and
d2
the area S is: S = a2 or it can be calculated using the diagonal d: S =
.
2
1.5. LIMIT OF A FUNCTION
The number L is called the limit of a function f(x) at a point p if for any
sequence {xn} D[f] that converges to a point p, the corresponding sequence of
function values {f(xn)} converges to L.
lim f (x ) L .
x
p
It means generally that f(x) can be made as close as desired to L by making
x close enough, but not equal, to p.
1.5.1. LIMITS OF SPECIAL INTEREST
Trigonometric functions
sin x
1 — First wonderful limit
lim
x
x 0
1 cos x
0
lim
x
x 0
Exponential functions
lim (1
x
0
1
x) x
lim 1
x
1
x
x
2.7178282 — Second wonderful limit
e
ex 1
1
lim
x
x 0
c. Logarithmic functions
ln(1 x )
x
0
lim
x
27
1
Examples:
1. We will set the function у = 3x2 + 2x – 5 and the argument x of
the function tends to 2. If the argument of the function takes on the values 2.1;
2.01; 2.001 etc. → 2, the function gets the values 12.43; 11.14; 11.01 etc. → 11.
That is, the sequence of function values has a limit, equals to 11.
We саn write down the following expression: lim (3x 2 2x 5) 11.
x
2
Thus we can simply substitute the value of the argument limit directly to
the to the function expression in order to calculate the value of the function
limit.
x2 4
2. Find the limit of the function y
if x → 2.
x 2
It is not possible to make a direct substitution the value of the argument
limit directly to the to the function expression because we will obtain no sense
0
result such as . We may use factoring formula (a2 – b2) = (a – b)(a + b)
0
x2 4
lim
x 2 x
2
(x 2)(x 2)
2
x 2
lim
x
lim(x
x
2)
2 2
4.
2
EXERCISES
a. calculate the limits of functions:
1. lim(x
x
3
4. lim(5 x
x
4 x 1) ;
2
2
3
7 x) ;
x 4 16
2. lim
;
x
2 x 2
3. lim
x2 9
5. lim
;
x 3 x 3
sinx
ex 1
6. lim
lim
x 0 x
x 0
x
x 1
x2 1
2 x2
;
x 1
1.6. DERIVATIVES AND INTEGRALS
1.6.1. DERIVATIVES. GENERAL RULES
It‘s important to determine how quickly the function y is changing with
variable x in many cases. The derivative of a function represents this
information as an infinitesimal change in the function y = f(x) with respect to its
variable x.
Let‘s consider a function y = f(x) at two points with some values
of argument x0 and x0 + Δx (fig. 1.33). The difference between the points
is an increment of an argument: Δx = x – x0. The increment of function will be:
Δy = y – y0. For continuous functions, if Δx → 0, then Δy → 0. But it is
28
impossible to foretell the value the attitude
y
aspires at unlimited decrease Δx,
x
because it depends on a concrete kind of function y(x).
B
y
y = f(x)
Δv
C
А
y0
dv
D
β
α
Δx = x – x0 = dx
x
Fig. 1.33. Derivative of a function y = f(x) — geometrical sence
Definition: Derivative of the function y(x) in the given point x0 is a limit
of the attitude of an increment of function to an increment of argument at its
unlimited decrease. Derivative of function of one argument is designated: y' or
Δy
dy
Δy
dy
y = lim
, or
lim
.
. Thus:
Δx 0 Δx
dx
dx Δx 0 Δx
Derivative of a function has simple geometrical sense.
From fig. 1.32 it is evident, that the attitude
y
x
BC
AC
tg α,
where α — a slope angle of the secant AB to an axis OX.
If Δx → 0 the point B will move towards point A, then Δx will
unboundedly decrease and approach 0, and the secant AB will approach
the tangent AC. Hence, a limit of the difference quotient is equal to a slope
of a tangent at point A.
Δy
y
lim
lim tgα tgβ.
Δx 0 Δx
Thus, derivative of a function at a point is numerically equal to the tangent
of a corner between the tangent lead to the curve of function in the given point,
and an axis OX, — that is the geometrical sense of a derivative.
Derivative of a function has also a mechanical sense or interpretation.
Let‘s consider a movement of a material point along a coordinate line.
The point displacement during the time interval from t0 till t0 + Δt is equal to:
S(t0 + Δt) – S(t0) = ΔS,
and its average velocity is:
vaver
29
S
.
t
If Δt → 0, an average velocity value approaches the certain value, which is
called an instantaneous velocity v(t0) of a material point at the moment t0.
S dS
vinst lim
S.
dt
t 0 t
An instantaneous velocity vinst = v(t0) is a derivative of a displacement
with respect to time at the moment t0:
vinst = S′(t).
Similarly to this imagination, an instantaneous acceleration a(t0)
is a derivative of a velocity with respect to time at the moment t0:
ainst = v′(t).
There are several certain rules how to calculate a derivative of
elementary functions and its combinations.
Single derivatives of simple functions are calculated with the help of
special table of derivatives (table 1.4). The derivatives of more complicated or
composite functions are calculated easily using special differentiation rules.
Table 1.4
Table of derivatives
Function
Derivative
axa – 1
0
ax ln a
ex
1
x
cos x
–sin x
1
cos2 x
1
sin 2 x
1
1 x2
a
x
C (constant)
ax
ex
ln x
sin x
cos x
tg x
ctg x
arctg x
1.6.2. DIFFERENTIATION RULES
Constant Rule. The derivative of any constant C equals to zero: C′ = 0.
Constant Multiple Rule. The derivative of constant С times a function is
equal to the constant С times the derivative of the function:
(Cu)′ = C·(u)′.
Sum rule: the derivative of a sum is equal to the sum of the derivatives:
(u + v)′ = u′ + v′.
30
Product rule: the derivative of a product of two functions is equal
to the first times the derivative of the second plus the second times the derivative
of the first:
(u·v)′ = u′·v + u·v′.
Quotient rule: the derivative of the quotient of two functions is equal to
the denominator times the derivative of the numerator minus the numerator
times the derivative of the denominator all divided by the square
of the denominator:
u
u v v u
.
v
v2
Chain rule for a composite function:
Composite function consists of the combination of two (or more) functions.
Lets two functions f(x) and g(x) of the composite function f(g(x)) are
obtained by replacing each occurrence of x in f(x) by g(x). Thus, f(g(x)) = f(u),
where u = g(x).
It is necessary to calculate a derivative of each function which
is an argument of another function being a part of a composite function in order
to calculate a composite function derivative. Then they should multiply each
other:
f′(g(x)) = f′(u) · u′(x).
Example 1: y = sin x, and x = (t2 + 3t). Then y = sin(t2 + 3t) is a composite
function of x.
It‘s derivative is y′= (sinx)′·(t2 + 3t)′ = cosx·(2t + 3) = (2t + 3)·cos (t2 + 3t).
Example 2: y = sin3(tg(x2)). It is necessary to make a number
of the conditional steps-mental operations to find the derivative of this
composite function.
1. To determine the number of functions-arguments which are the parts
of a composite function.
1. The power function (cubic function) — the derivative is 3sin2(tg(x2)).
2. sin(tg(x2)) — derivative is equal to cos(tg(x2)).
1
3. tg(x ) — derivative is cos2 x 2 .
2
yx
4. x2 — derivative is equal 2x.
2. To combine the derivatives of the parts of a composite function:
1
3sin 2 (tg(x 2 )) cos(tg(x 2 ))
2 x.
2 2
cos (x )
Example 3. Calculate the function derivative y′, if y = sin(tg( x )),
y is a composite function, y′ = (sin(tg( x )))′ = cos(tg( x ))·(tg( x ))′,·
(tg( x ))′ is also a composite function,
31
y
cos(tg x )
y
cos(tg x )
y
1
cos2 x
1
( x)
1
cos2 x 2 x
cos(tg x )
2 x cos 2 x
EXERCISES
a. Calculate the derivatives of functions:
1. у = 5х3;
2. у = sin26x;
3. у = 3х4·ln 4x;
4.
;
2
9
x
5. у = (sin x + 8x) ;
6. y (e
x ) ln x ;
sin x
7. y
;
8. y = ln(x2 – 4x + 4).
x
1.6.3. MAXIMA AND MINIMA OF FUNCTIONS
The function f(x) has a relative maximum value at point A, if f(A) is
greater than any value in its immediate neighborhood.
The function f(x) has a relative minimum value at point B, if f(b) is less
than any value in its immediate neighborhood. At each of these points
the tangent to the curve is parallel to the x-axis so the derivative of the function
is zero: f′(x) = 0. The term local is used since these points are the maximum and
minimum in this particular region. There may be others outside this region.
If f(x) has a local maximum or minimum at points a and b, and if f'(x)
exists, then f'(x) = 0.
At points immediately to the left of a maximum the slope of the tangent is
positive: f'(x) > 0. While at points immediately to the right the slope is negative:
f'(x) < 0. In other words, at a maximum, f'(x) changes sign from + to –.
At a minimum, f'(x) changes sign from – to + (fig. 1.34).
A point x at which the function has either a maximum or a minimum is
called a critical point.
To find the maximum and minimum values of a function we need:
1. Solve the algebraic equation: f'(x) = 0.
The roots x1, x2, x3 … of this equation are the stationary points.
2. Calculate the second derivative f′′(x) and definite its sign at the points.
If the second derivative is positive at a stationary point is positive
(f′′(x1) > 0), the point x1 is a local minimum; if it is negative (f′′(x2) < 0),
the point x2 is a local maximum; if it is equal to zero: f′′(x3) = 0, it may be no
local extremums. In this case it is necessary to find a sign of the first derivative
on the left side (x < x3) and on the right one (x > x3) from the x3. If the sing on
32
the left side is ―–― and on the right one is ―+‖ there is a local minimum at
the point x3. If the sing on the left side is ―+― and on the right one is ―–‖ there is
a local maximum at the point x3. And if the sing not changes there is no
extrema at this point.
3. Determine a value of function in points of maximum and minimum.
Fig. 1.34. The relative maximum and the relative minimum of the function f(x)
Example. Let f(x) = 2x3 – 9x2 + 12x – 3.
Are there any critical values — solutions to f'(x) = 0 — and do they
determine a maximum or a minimum? And what are the coordinates on
the graph of that maximum or minimum? Where are the turning points?
Solution. f'(x) = 6x2 – 18x + 12 = 6(x2 – 3x + 2) = 6(x – 1)(x – 2) = 0.
Implies: x = 1, x = 2.
Those are the critical values. Does each one determine a maximum or does
it determine a minimum? To answer, we must evaluate the second derivative at
each value.
f'(x) = 6x2 − 18x + 12; f''(x) = 12x – 18 f''(1) = 12 – 18 = –6.
The second derivative is negative. The function therefore has a maximum
at x = 1.
To find the y-coordinate — the extreme value — at that maximum we
evaluate f(1): f(x) = 2x3 – 9x2 + 12x – 3; f(1) = 2 – 9 + 12 – 3 = 2.
The maximum occurs at the point (1, 2).
Next, does x = 2 determine a maximum or a minimum?
f''(x) = 12x – 18; f''(2) = 24 – 18 = 6.
The second derivative is positive. The function therefore has a minimum at
x = 2.
To find the y-coordinate — the extreme value — at that minimum, we
evaluate f(2): f(x) = 2x3 – 9x2 + 12x – 3 f(2) = 16 – 36 + 24 – 3 = 1.
The minimum occurs at the point (2, 1).
33
Here in fact is the graph of f(x):
EXERCISES
a. Determine maxima and minima of functions:
x3
1. y = 2x2 – x4;
2. y = 2 + x – x2;
3. y =
– x;
3
4. y = x·e–x.
1.6.4. DIFFERENTIAL OF A FUNCTION
The differential of a function represents the principal part of the change
in the function y = ƒ(x) with respect to changes in the independent variable.
Definition. The differential dy is defined as a product of function
derivative y′ and an increment (or differential) of an argument dx:
dy = y′·dx.
The argument differential dx is equal to the increment of argument Δх, i. e.
dx = Δх.
Differential dy of function is not equal to its increment Δy but represents its
main part: Δy ≈ dy = y′·dx.
In the fig. 1.31 differential dy corresponds to line segment CD: dy = [CD].
1.6.5. INDEFINITE INTEGRALS. GENERAL RULES
We are able to find the function derivative F′(x) in any case. What about
the reverse operation? Very often it is necessary to find some function F(x)
the derivative of which is equal to the initial function f(x) = F′(x).
Definition. The function F(x) is called an antiderivative function of
the initial function f(x), if the following equation is performed: F′(х) = f(х).
But, for example, if the derivative of expression x3 + 5 is 3x2,
an antiderivative of 3x2 is x3 + 5. At the same time the derivative of x3 + 7 is
also 3x2, another antiderivative of 3x2 is x3 + 7.
Similarly, another antiderivative of 3x2 is x3 + 8 etc. In fact, every
antiderivative of 3x2 has the form x3 + C, where C is an arbitrary constant:
F′(х) = x3 + C.
34
Definition. Indefinite integral of a function f(x) is a set of all its
antiderivatives F(x) of the initial function f(x). The process of calculating an
indefinite integral is called integration.
The symbol f (x )dx is used to indicate the indefinite integral of f(x).
The indefinite integral of any given function is not unique and can differ by up
to a constant. Thus we write, f (x )dx F (x ) C ,
where C is an arbitrary constant known as the constant of integration.
Features of the indefinite integral.
The integral of a sum or difference of functions is the sum or difference of
the individual integrals. (f g )dx
fdx gdx.
We can take the multiplicative constant k outside the integral sign.
kydx k ydx.
The main standard integrals are presented in table 6.1.
Table 6.1
Standard integrals
0 dx C
1 dx x C
α
x dx
xα+1
C (α
α+1
1
cos 2 x
1
dx
tg x C
dx
ctg x C
sin 2 x
1
x
dx ln tg
C
sin x
2
1
x π
dx ln tg
cos x
2 4
1)
1
dx ln x C
x
1
arcsin x C
dx
arccos x C
1 x2
1
arctg x C
dx
arcctg x C
1 x2
ax
x
a dx
C
ln x
sin x dx
cos x C
cos x dx sin x C
C
x
C
a
dx
x
a2 x2
arccos
C
a
1
x
arctg
C
1
a
a
dx
1
x
2
2
a x
arcctg
C
a
a
1
1
a x
dx
ln
C
2
2
2a a x
a +x
1
tg x dx
ln cos x C
ctg x dx ln sin x C
arcsin
Sometimes it is impossible to find an antiderivative which is an elementary
function. There are different methods of integration in this case. The simplest
methods are linear integration and integration by substitution.
Linear integration allows us to break complicated integrals into simpler
ones which correlate table integrals.
35
Example: (5x sinx)dx
5 xdx
5x2
2
sinxdx
cosx C.
Integration by substitution.
The substitution rule is an important tool for finding antiderivatives and
integrals for composite function (like the chain rule for differentiation). It allows
to involve new variables and its differentials with the main aim — to reduce
previous variables for table standart integral with new variables.
Example 1. Find the solution: x x 1dx
By using the substitution t =
x x 1dx
x 1 we obtain
t
x 1
t2
x t
2
dx 2tdt
2t 2 (t 2 1)dt
1
x 1
(t 2 1) t 2tdt
(2t 4 2t 2 )dt
Then use linear integration
4
2t dt
2t 5
5
2
2t dt
2t 3
C
3
Produce reverse substitution and final result is:
x x 1dx
2t 5
5
2t 3
3
C
2
(x 1)5
5
2
(x 1)3
3
Example 2. Find the solution: (1 sinx)3 .
By using the substitution 1 + sinx = t we obtain
1 sin x t
t3 1
3
3
(1 sin x) cos x dx
t dx
C
cos x dx dt
3 1
Produce reverse substitution and final result is:
(1 sin x)4
(1 sin x)3cos x dx
C
4
EXERCISES
C
t4
4
C
a. Calculate the integrals:
sinx e x )dx ;
1. 3 x3dx ;
2. ( x
4. e 2 x 1dx ;
sin 2 x
sinx cos x dx ;
5. e
sin2 x
dx ;
sinx
dx
6.
.
x lnx
3.
1.6.6. DEFINITE INTEGRAL
Integration was introduced as the reverse of differentiation. A more
rigorous treatment would show that integration is a process of adding or
«summation».
36
Consider the graph of the positive function y(x) shown in figure 1.35.
Suppose we are interested in finding the area of the region bounded above by
the graph of y(x), bounded below by the x-axis, bounded to the left by
the vertical line x1 = a, and bounded on the right by the vertical line xn = b.
Fig. 1.35. Calculation of the area under a curve f(x)
One way in which this area can be approximated is to divide it into
a number of rectangles, find the area of each rectangle, and then add up all these
individual rectangular areas. The sum of the areas of all n rectangles is then
n
S ABDС
i 1
yi Δxi .
This quantity gives us an estimate of the area under the curve but it is not
exact. To improve the estimate we must take a large number of very thin
rectangles. So, what we want to find is the value of this sum when n tends to
infinity and Δx tends to zero. We write this value as
n
S ABCD
lim
Δxi
0i 1
yi Δxi
b
ydx.
a
Limit of the sum is called the definite integral of y from x = a to x = b and
b
ydx .
it is written
a
Fundamental theorem of calculus (the Newton–Leibniz formula): Let f(x)
be integrable over the interval [a; b], and suppose there is an antiderivative F(x)
of f(x) over the interval [a; b]. Then, the definite integral with integrand f(x) and
limits a and b is equal to the value of the antiderivative F(b) minus the value of
antiderivative F(a):
b
b
f (x )dx F (x )
F (b) F (a ) .
a
a
37
The notation F (x )
b
a
means the following: at first substitute the upper limit
b into the function F(x) to obtain F(b) and then from F(b) we subtract F(a),
the value obtained by substituting the lower limit a into F(x). This Newton–
Leibniz formula (1.11) allows us to easily solve definite integral, if we can find
the antiderivative function of the integrand.
Unlike the indefinite integral, which is the set of functions, the definite
integral is a numerical value, that represents the area under the curve f(x).
Features of the definite integral:
b
a
f (x )dx
1.
f ( x )dx.
a
b
b
b
2. (f (x )
g(x ))dx
a
a
a
f (x )dx
3.
g(x )dx.
a
0.
a
b
b
4. kf (x )dx
a
b
k f ( x )dx.
a
c
f (x )dx
5.
b
f ( x )dx
a
b
f ( x)dx
a
f (x)dx.
c
Calculation of the area between two curves y1(x) and y2(x).
The definite integral is used for calculation of the area between two curves
y1(x) and y2(x).
At first we must find the crossing points of this curves by solving
the equation: y1(x) = y2(x). If this points are x1 and x2 > x1, we can calculate
the area between the curves (we consider y1(x) > y2(x) at this region):
x2
S
(y1
y2 )dx.
x1
Example 1.
3
xdx. The area under the line is the triangle
Consider the integral
0
(fig. 1.36). The area of any triangle is half its base times the height. It is:
S
1
3 3
2
9
.
2
As expected, the integral yields the same result:
3
x 2 3 32 02 9
9
xdx
0
.
0
2
2
2
2
2
0
38
Fig. 1.36. The area S under the line y = x
Example 2.
Calculate the area S limited the curve у = х2, axis x and lines х1 = –1
и х2 = 2. In fig. 1.37 this area is cross-hatched.
2
x3 2 8
1
9
2
S
x dx
3.
1
3
3
3
3
1
Fig. 1.37. The area S under the curve y = x2
Example 3. Calculate the integral
2
x3 2
8
8
8
2
(4 x )dx (4 x
)
(4· 2
) ( 4· 2
) 2(8
)
2
3
3
3
3
2
32
2
10 .
3
3
EXERCISES
a. Calculate the definite integrals:
9
4
1
4)
4
2
3) x x 9 dx;
2) sin x dx;
x dx;
1)
0
x3dx
03
x
4
4
;
5)
0
0
x dx
x2 9
;
b. Calculate the area between two curves:
1) y = 2x – x2, y = x;
2) y = x2, y = –2x.
39
0
6)
1
x5 (1 x 6 )7dx.
THE BASICS OF PHYSICS
2. KINEMATICS
The study of the object motion, and the related concepts of force and
energy, forms the field of science called mechanics. Mechanics is divided into
three parts: kinematics, dynamics, and statics. Kinematics describes how objects
move, dynamics deals with force and why objects move. Statics is concerned
with the analysis of loads (i. e. forces) acting on and within physical systems
that are in equilibrium.
Kinematics is the branch of mechanics which deals with the study
of the motion without taking into account the factors responsible for producing
motion.
In mechanics the concept (or model) of an idealized particle is used.
It is considered to be a mathematical point (point particle) with no spatial extent
(no size). The particle model is useful in many real situations where we are
interested only in translational motion.
Translational motion means motion in which all particles in the body move
along parallel paths, and with the same velocity and acceleration (fig. 2.1).
a
b
Fig. 2.1. Rectilinear (a) and curvilinear (b) rigid body translation
Therefore the object‘s size is not significant in case of translational motion.
For example, we might consider a billiard ball, or even a spacecraft traveling
toward the Moon, as a point particle, for many purposes.
There are two types of translational motion: rectilinear translation (paths
are straight lines) and curvilinear translation (paths are curved, e. g. circular)
(fig. 2.1).
In this section we consider the simplest two cases of translational motion:
a body motion along a straight line (linear or one-dimensional motion) and
curvilinear motion in a circle path.
40
2.1. MECHANICAL MOTION CHARACTERISTICS
The main task of kinematics is to determine a position of the body at any
instant of time. Any measurement of a body position must be made with respect
to a reference frame. In mechanics a frame of
y
A
reference is represented by a set of coordinate axes
y
(fig. 2.2), therewith the origin (or zero point) of
the frame is often chosen as a reference point.
0
x
x
To locate a body means to find its coordinates (x, y)
Fig. 2.2. Position of a point A relative to the origin of coordinate axes (to
on the plane (x, y)
the reference point). Any point on the plane can be
specified by giving its x and y coordinates (point A in fig. 2.2). In three
dimensions, a z axis perpendicular to the x and y axes is added.
For one-dimensional motion we usually choose
A
the x axis as the line along which the motion takes
0
x
x
place (fig. 2.3). Then the position of a point at any
Fig. 2.3. Position of a point A
moment is given by its x coordinate. If the motion is
for one-dimensional motion
vertical, as with a falling object, we usually use the y
axis.
The position of the moving object is changed with time and hence the x
coordinate is a function of time t (x = f(t)). The dependence x = f(t) is called
the equation of motion.
To describe the motion, such physical characteristics as path, distance,
displacement, velocity and acceleration are introduced.
Path (trajectory) is the curve along which the object moves (line ACDB in
fig. 2.4).
The distance (s) is the actual length of
y
the path followed by the moving object
(length of the line ACDB). It is a scalar
s
quantity.
We need to make a distinction
between the distance a body has travelled
and its displacement. Displacement is
defined as the change in position of
0
x
the object, i. e. it shows how far the object
Fig. 2.4. Distance s (length of the line
is from its starting point (vector
in
ACDB) and the displacement vector
fig. 2.4). The direction of the displacement
vector is from an original position to a final position. Displacement does not
depend on the actual path followed by the object. Only the initial and the final
positions determine the displacement.
Let us consider linear motion of an object during a particular time interval.
Suppose that at some initial time t1, the object is on the x axis at the position x1
41
in the coordinate system shown in fig. 2.5. At some later time, t2, suppose
the object has moved to position x2. The displacement vector shows a change
in the position of the object along the x axis during time interval t = t2 – t1.
From fig. 2.5 it is seen that the magnitude
x2
of the displacement vector r is equal to the change 0 x1
10 20 30 40 50
x
in the x coordinate of the object: r = x = x2 – x1.
The distance travelled s is equal to the magnitude of Fig. 2.5. The displacement
vector
displacement: r = x = x2 – x1 = s.
Note that the distance and the magnitude of the displacement vector are
equal only in case of linear motion in the same direction. In all other cases r < s
(fig. 2.4).
The SI units for the displacement and the distance are meters (m).
Velocity vector describes how fast and in what direction the body moves.
Acceleration vector describes how fast and in what direction the velocity
of the body changes.
Example 2.1. Distance and displacement.
R
A car moves along a circular path from point A to
point B (fig. 2.6). If AB is the diameter of the circle, find
r
A
B
(a) the magnitude of the displacement and (b) the total
Fig. 2.6. Example 2.1
distance travelled.
Solution. The magnitude of the displacement r is equal to the diameter of
the circle: r = AB = 2R.
Distance s is equal to the half of the circle length: s = πR.
B
Example 2.2. Distance and displacement.
The object starts from the ground (point A) and moves
vertically upwards to a maximum height of h (point B) and
h
falls back to the ground (fig. 2.7). Find (a) the magnitude
of the displacement and (b) the total distance travelled.
A
Solution. The displacement is equal to zero and
Fig. 2.7. Example 2.2
the total distance travelled is equal to 2h (s = 2h).
2.2. UNIFORM LINEAR MOTION
Uniform linear motion is a body motion along a straight line with constant
velocity.
Velocity. Velocity
is defined as the rate of a body position change
in a particular direction with respect to time:
r
.
(2.1)
t
Velocity vector has the same direction as the displacement vector .
It is a vector quantity. If the body travels in the same direction and covers equal
42
distances during equal time intervals, then its velocity is said to be a uniform
velocity.
In physics when solving many problems, the magnitudes of various
physical vector quantities should be frequently determined. These magnitudes
are always determined by finding the projection of
the vectors on the chosen coordinate axes.
Let a particle be in a linear motion with
x0
x
a uniform velocity (fig. 2.8). We choose positive
direction of the x axis along the motion direction.
s
A particle is at the position x0 at initial instant of
Fig. 2.8. The velocity vector
time t0 = 0 and at the position x at instant of time t.
From fig. 2.8 it is seen that x component of the velocity and the displacement
vectors are equal to their magnitudes: x = , rx = r. Then the magnitude of
the velocity vector is given by
r
.
t
(2.2)
Taking into account that the magnitude of the displacement vector r is
equal to the distance (s = x – x0) we obtain that the total distance travelled is
equal to magnitude of the velocity vector multiplied by the total elapsed time:
r = s = x – x0 = t.
(2.3)
From the Eq. 2.3 it follows that the position of a particle x at any instant of
time is defined as:
x = x0 + t.
(2.4)
This is the basic equation of a uniform motion.
Figure 2.9 shows a graph of the magnitude of velocity and the distance s
of a particle versus the time in case of a uniform motion. The slope of the s(t)
graph is defined by the velocity magnitude, i. e. tan
= . The larger
the velocity, the larger the angle formed by the s(t) graph with the t axis.
a
b
s
= const
2
2
2
1
1
1
t
Fig. 2.9. A plot of the magnitude of velocity (a) and the distance s (b) of a particle versus
time for case of a uniform linear motion
The SI units of velocity are meter per second (m/s).
43
Example 2.3. Distance.
A car travels along a straight road with constant velocity 80 km/hr. How far
does the car travel after 30 min?
Solution. For solving a problem we use Eq. 2.3. The time interval
t = 30 min = 0.5 hr. The distance travelled at = 80 km/hr is
s=
t = (80 km/hr) (0.5 hr) = 40 km.
Example 2.4. Equation of motion.
A car‘s position as a function of time is given by x = 5 + 30 t. What are (a)
the coordinate, (b) its displacement and distance travelled after 10 s?
Solution.
a) Setting t = 10 s in the equation of motion gives the coordinate x
x = 5 + 30·t = 5 m + 30 m/s·10 s = 305 m.
b) In case of uniform linear motion in the same direction the magnitude of
the displacement vector and the distance are equal and are defined by Eq. 2.3.
Then displacement (distance) during time interval t = 10 s is
r = s = x – x0 = 30 t = 30 m/s 10 s = 300 m.
2.3. NON-UNIFORM LINEAR MOTION
A body has a non-uniform motion if it travels unequal distances during
equal time intervals.
2.3.1. AVERAGE AND INSTANTANEOUS VELOCITY AND SPEED
If the body covers unequal distances during equal time intervals, then
velocity is called a variable velocity. In case of a non-uniform motion
the average velocity
during some time interval is defined as
r
,
(2.5)
av
t
where is the displacement, t is the time interval.
Magnitude of the average velocity vector is given by
r
,
t
av
(2.6)
where r is the magnitude of the displacement vector, t is the time interval.
The average speed of a body s is the total distance travelled divided by
the time interval t:
s
.
t
s
(2.7)
Average speed is a scalar and it is not the magnitude of average velocity
vector.
44
Instantaneous velocity (i. e. the velocity at a specific instant of time or
specific point along the path) is the limit of the average velocity as the time
interval approaches to zero; it is equal to the instantaneous rate of a body
position change with time.
If t
0, then from the Eq. 2.5 it follows that instantaneous velocity is
r
defined as
(2.8)
lim .
t 0 t
The magnitude of the instantaneous velocity is
lim
t
0
r
.
t
(2.9)
The magnitude of the instantaneous speed is always equal to the magnitude
of the instantaneous velocity because the distance travelled and the magnitude of
displacement vector become the same when they are infinitesimally small. Then
we can write
lim
t
where
0
r
t
lim
t
0
s
t
lim
t
0
x
t
dx
,
dt
(2.10)
dx
is the derivative of x with respect to time.
dt
Note that in case of uniform linear motion in the same direction
the magnitude of the velocity vector is speed.
Example 2.5. The average speed and the average
velocity.
A particle moves half of its distance under the angle
2
α1 = 45° with respect to the x-axis at velocity 1 = 5 m/s
and the other half of its distance with the velocity
o
2 = 10 m/s under the angle α2 = 135 (fig. 2.10). (a) What
is the average speed of the particle? (b) What is
1
the average velocity?
x
Solution.
a) For solving a problem we use Eqs. 2.3, 2.6 and
Fig. 2.10. Example 2.5
2.7. The time elapsed to cover the first half of the distance
s
t1
.
2 1
s
The time elapsed to cover the second half of the distance is t2
.
2 2
The time taken to cover total distance is
s
s
s 1 2
Δt = t1 + t2 =
.
2 1 2 2 2
1 2
45
Then the average speed is
s
s
2
s
t s 1 2
1
2
2(5 m/s) 10 (m/s)
15 m/s
1 2
2
6.7 m/s.
1 2
The average speed is 6.7 m/s.
b) The magnitudes of displacement vectors and are r1 = 1t1 = s/2 and
r2 = 2t2 = s/2, respectively. The displacement vectors ,
and
form
rectangular triangle as shown in fig. 2.10, from which it follows
s
2
2
.
r = r1 r1
2
Then the magnitude of average velocity is
r
s
6.7 m/s
s
4.8 m/s.
av
t
2 t
2
2
The average velocity is 4.8 m/s. The values of average speed and average
velocity differ.
NOTE. The difference between the average speed and the magnitude of
the average velocity can occur when the motion is not in the same direction.
Example 2.6. The average speed and the average velocity.
An automobile travels on a straight road for 40 km at 80 km/h. It then
continues in the same direction for another 20 km at 60 km/h. (a) What is the
average velocity of the car during the full trip? (b) What is the average speed?
Solution. A car travels in the same direction, then the magnitude of its
displacement is equal to the distance, i. e. r = s. For solving a problem we use
Eqs. 2.3 and 2.7.
Solving Eq. 2.3 for t and setting 1 = 80 km/hr, s1 = 40 km and
2 = 60 km/hr, s2 = 20 km gives the time elapsed to cover the first (t1) and
the second (t2) part of the distance, respectively
s1 40
t1
0.5 hr,
80
1
s 2 20
t2
0.3 hr.
60
2
The total time taken to cover total distance is
t = t1 + t2 = 0.5 + 0.3 = 0.8 hr.
The total distance travelled is
s = 40 + 20 = 60 km.
Setting s = 60 km and t = 0.8 hr in Eq. 2.7 we obtain
av
r
t
s
t
60 km
0.8 hr
46
75 km/hr.
Both the average velocity and speed have the same value 75 km/hr.
NOTE. Average speed and average velocity have the same magnitude
when the motion is in the same direction.
Example 2.7. Average velocity.
A person walks 50 m at a velocity 1 m/s and then run 60 m for 20 s along
a straight track. What is the average velocity of a person?
Solution. For solving a problem we use Eqs. 2.3 and 2.6 (r = s).
Solving Eq. 2.3 for t and setting 1 = 1.0 m/s, s1 = 50 m gives the time
elapsed to cover the first part of the distance
s
50 m
t1 1
50 s.
1 1 m/s
The total time taken to cover total distance is
t = t1 + t2 = 50 s + 20 s = 70 s.
The total distance travelled is
s = 50 m + 60 m = 110 m.
Setting s = 110 km and t = 70 s in Eq. (2.6) gives:
av
r
t
s
t
110 m
70 s
1.6 m/s.
The average velocity is 1.6 m/s.
2.3.2. AVERAGE AND INSTANTANEOUS ACCELERATION
The velocity generally changes with time either in magnitude or in
direction or in both. In this case the motion of a body is said to be accelerated
(or retarded). The change of velocity with time is measured by a vector quantity
called acceleration .
If velocity changes by unequal amounts during equal time intervals,
the body has a variable acceleration. In case of non-uniformly accelerated
motion the term of the average acceleration is defined. If
is the velocity
at any time t1 and
is the velocity at another time t2, then
is the change in velocity, t = t2 – t1 is the time elapsed. Average acceleration
is defined as the ratio of the change in velocity to the time interval:
aav
.
(2.11)
t
Instantaneous acceleration is defined as the limit of the average
acceleration as the time interval approaches to zero. If t
0, then from
the equation 2.11 it follows that instantaneous acceleration is equal to
a
lim
t
47
0
t
.
(2.12)
A
A
0
0
x
ax > 0
a
ax < 0
b
x
Fig. 2.11. Projections of the acceleration vector along the x axis for case of an accelerated
(a) and retarded (b) linear motion
Magnitude of the instantaneous acceleration vector is given by
d
d dx
a lim
=
=
.
t dt dt dt
t
where
time,
(2.13)
d
is the derivative of the magnitude of the velocity with respect to
dt
d2 x
dt 2
is the second derivative of x with time.
Units of acceleration are meter per second squared (m/s2).
Example 2.8. Average acceleration.
A car accelerates along a straight road from rest to 90 km/h during time
interval equal to 5 s. What is the magnitude of its average acceleration?
Solution. We use Eq. 2.11, where we set t = 5 s. The car starts from rest,
so 1 = 0 m/s. The final velocity is 2 = 90 km/h = 90 103 m/3600 s = 25 m/s.
From Eq. 2.11 the average acceleration is
aav
2
1
t
25 m/s 0 m/s
5s
5 m/s2.
2.4. UNIFORMLY ACCELERATED LINEAR MOTION
If the body travels in some direction and its velocity changes by equal
amounts during equal time intervals, however small these intervals may be, then
its acceleration is said to be a uniform acceleration.
In case of motion with a uniform acceleration, the acceleration is
0
(2.14)
,
t
are the initial and the final velocity of the object, respectively,
a
where
and
t is time interval.
The velocity of the particle that has been accelerated after some elapsed
time t is
.
(2.15)
In this section we consider the case of linear motion. Let us choose
the positive direction of x axis coinciding with the direction of the initial
velocity
(fig. 2.11). To determine the magnitudes of the acceleration and
48
the velocity vectors, let us find the projections of these vectors along x axis.
If the direction of the acceleration vector coincides with the direction
of the initial velocity the projection of the vector along x axis is equal to +a
(ax = +a) (a is the magnitude of the vector ), and ax = –a when the directions of
the vectors and
are opposite (fig. 2.11).
Thus the magnitude of the acceleration vector is defined as:
x
ax
ox
t
,
(2.16)
where ax = a (a is the magnitude of the vector ).
If the velocity increases with time, the acceleration is positive (a > 0), but
if the velocity decreases, the acceleration is negative (a < 0) and it is called
the retardation (motion is decelerated or retarded).
The magnitude of the velocity of the uniformly accelerated particle after
some elapsed time t is given by
(2.17)
x = 0x + ax t.
For case of linear motion x = , 0x = 0, that is,
= 0 a t,
(2.18)
where 0 and are the initial and the final speed of the particle, respectively.
As a check, note that this equation reduces to = 0 at t = 0. As a further
check, take the derivative of the (t) function. Doing so yields
d
dt
a , which is
the definition of a.
Figure 2.12, a shows a plot of the (t) function defined by Eq. 2.18.
a
b
x
α
x0
0
0
Fig. 2.12. A plot of the speed
0
t
t
(a) and the coordinate x (b) of an object moving with constant
acceleration versus time
The function is linear and thus the plot is a straight line. The slope
of the graph is defined by the acceleration magnitude, i. e. tan = a.
Position (coordinate x) of the particle moving with a uniform acceleration
at any instant of time is given by the equation:
at 2
(2.19)
x x0
.
0t
2
49
This is the basic equation of uniformly accelerated motion. The function
defined by Eq. 2.19 is quadratic and thus the plot is curved (fig. 2.12, b).
As a check, note that putting t = 0 yields x = x0. As a further check, taking
the derivative of the function x(t) given by Eq. 2.19 yields to Eq. 2.18.
Distance travelled by a uniformly accelerated particle by a given time.
at 2
(2.20)
s x x0
.
0t
2
By substituting t = ( – 0)/a from the Eq. 2.18 into the equation 2.20, we
obtain speed of a particle after covering a certain distance.
2
= 02 ± 2as,
(2.21)
where 0 is the initial speed, s is the distance. This equation is useful if we do
not know t and are not required to find it. It can be given in other form:
2as = 2 – 02
(2.21)
However other equations can be derived that might be useful in certain
situations. Thus two equations 2.18 and 2.19 can be combined to yield two
additional equations of linear motion with constant acceleration, each of which
involves a different ―missing variable‖.
1. We can eliminate the acceleration a to produce the following equation
x
2. Finally, we can eliminate
0,
1
(
2
x0
0
(2.22)
)t .
obtaining
at 2
(2.23)
.
2
Table 2.1 lists the basic equations of motion with constant acceleration and
the specialized equations which have been derived above. To solve a problem
you can choose an equation for which the only unknown variable is the variable
requested in the problem.
x
x0
t
Table 2.1
Equations of Motion with Constant Acceleration
Equation
= 0 at
Missing quantity
x – x0
Equation Number
2.18
2.19
2.20
t
a
2.21
2.22
2.23
0
50
Example 2.9. Velocity and acceleration.
A particle‘s position on the x axis is given by
x = 5 + 10t + 3t2,
where x in meters and t in seconds. Find (a) the particle‘s instantaneous velocity
at time t = 2s, and (b) the acceleration a.
Solution. To find the velocity function (t) and acceleration we use
Eqs. 2.10 and 2.13.
a) To get the function (t), we differentiate the position function x(t) with
respect to time (Eq. 2.10):
dx d
(5 10t 3t 2 ) 10 6 t.
dt dt
Setting t = 2 s in the above equation gives
= 10 m/s + 6 m/s 2s = 22 m/s.
b) To get the acceleration a, we differentiate the velocity function (t) with
d
a
6 m/s2 .
respect to time (Eq. 2.13):
dt
The particle’s velocity at time t = 2 s is 22 m/s and acceleration is 6 m/s2.
Example 2.10. Acceleration at given x(t).
A particle is moving in a straight line so that its position is given by
the relation x = 5 + 4t2. Calculate (a) its average acceleration during the time
interval from t1 = 2 s to t2 = 5 s, and (b) its instantaneous acceleration as
a function of time.
Solution. To determine acceleration, we first must find the velocity at t1
and t2 by differentiating x(t) (Eq. 2.10). Then we use Eq. 2.11 to find the average
acceleration, and Eq. 2.13 to find the instantaneous acceleration.
a) The velocity at any time t is
dx
dt
Therefore, at t1 = 2 s,
we obtain
aav
1
d
(5 4t 2 ) 8t.
dt
= 8 2 = 16 m/s and at t2 = 5 s,
2
t
1 = 40
m/s 16 m/s
3s
2
= 40 m/s. Thus,
8 m/s2 .
b) The instantaneous acceleration at any time is
a
d
dt
d
(8t ) 8 m/s 2 .
dt
The acceleration in this case is constant.
Example 2.11. Uniformly accelerated linear motion.
A car starts from rest and then it is moving with a constant acceleration
a = 2 m/s2 during a 100 m race. How fast is the car going at the finish line?
51
Solution. We use equation (2.20), where we set
in Eq. 2.20 and solving it for t we obtain
t
2s
2(100 m)
=
a
2 m/s 2
0
= 0. Setting
0
= 0
10 s.
It takes a car 10 s to get a finish line.
2.5. FREELY FALLING OBJECTS
An important example of uniformly
accelerated linear motion is that of an
object falling freely near Earth‘s surface
(fig. 2.13).
Galileo Galilei was the first to
postulate that at a given location on the
Earth and in the absence of air resistance,
all objects fall with the same constant
acceleration. This acceleration is called
the acceleration due to gravity (the freefall acceleration) on the surface of the
Earth, and it is denoted by symbol g. Its
magnitude is approximately g = 9.80 m/s2
(at the surface of Earth). Acceleration due Fig. 2.13. The acceleration due to gravity g
to gravity is a vector as is any acceleration and its direction is downward,
toward the center of the Earth (fig. 2.13). It is independent on the body
characteristics, such as mass, density, or shape; it is the same for all objects.
When dealing with freely falling bodies Eqs. 2.18–2.23 also describe this
motion, where we replace a with the value of g given above. Also, since
the motion is vertical, we refer the motion to the vertical coordinate y axis.
As a rule, the direction of y axis coincides with the direction of the body motion.
We take y0 = 0 unless otherwise specified. Then the equations of the freely
falling body motion (at y0 = 0) are
(2.24)
y = 0y +gyt,
s
g yt 2
0t
2
=
2
0
,
(2.25)
+ 2gys.
(2.26)
2
NOTE. In the above equations the projections of vectors
and are
positive if their directions coincide with the axis OY and they are negative
in the opposite case.
In the special case of a body thrown upward with an initial velocity , its
acceleration is equal to the acceleration due to gravity
(in the absence of air
52
resistance). As the body rises, its velocity decreases until it reaches the highest
point (retarded motion), where its velocity is zero for an instant; then it
descends, with the increasing velocity.
Example 2.12. Falling from a tower.
Suppose that a ball is dropped from a tower 70.0 m
0
high. Calculate (a) how far will it have fallen after time
y
interval t = 2.0 s? (b) how much time it takes for the ball
h
to reach ground. Ignore air resistance. Assume that
the initial velocity is equal to zero. (g = 9.80 m/s2).
y
Solution. Let us take y as positive downward, so the
Fig. 2.14. Example 2.12
acceleration due to gravity g is positive: g = +9.8 m/s2
(fig. 2.14). For solving a problem we use Eq. 2.25, where we set 0 = 0.
a) We set t = 2.0 s in Eq. 2.25:
gt 2 (9.8 m/s 2 )(2s) 2
s
=
19.6 m.
2
2
The ball has fallen a distance of 19.6 m during the time interval t = 2 s.
b) Solving Eq. 2.25 for t and setting s = h = 70 m gives
2h
2 70 m
t
=
14.3 s 3.8 s.
g
9.8 m/s 2
It takes 14.3 s for the ball to reach ground.
Example 2.13. Ball thrown upward.
A person throws a ball upward into the air with an
y
initial velocity of 10.0 m/s (fig. 2.15). Calculate (a) how
maximum high it goes; (b) how much time it takes for
the ball to reach maximum height. Ignore air resistance.
hmax
(g = 9.80 m/s2).
0
Solution. Let us choose y to be positive in
Fig. 2.15. Example 2.13 the upward direction. The acceleration due to gravity is
downward and so it has negative sign (gy = –g = –9.80
2
m/s ) (fig. 2.15). For solving a problem we use Eqs. 2.24 and 2.26:
= 0 – gt, 2 = 02 – 2gs.
We consider the time interval from when the ball leaves the thrower‘s
hand until the ball reaches the highest point. At time t = 0 we have y0 = 0,
0 = 10.0 m/s. As the ball rises, its speed decreases until it reaches the highest
point. At time th (maximum height), the velocity of the ball is equal to zero.
a) To determine the maximum height hmax, we set = 0 in Eq. 2.26 and
2
2
2
(10 m/s) 2
0
solve it for s = hmax:
hmax
= 0 =
5.1 m.
2g
2 g 2(9.8 m/s 2 )
The ball reaches a maximum height of 5.1 m.
53
b) The time th required for the ball to reach its highest point hmax we can
calculate from the Eq. 2.24, where we set = 0. Then
0 = 0 – gth,
0 = 10.0 m/s 1.02 s.
th
g 9.8 m/s 2
The time required for the ball to reach the maximum height is 1.02 s.
2.6. UNIFORM CIRCULAR MOTION
Uniform circular motion occurs when an object (point particle) moves
in a circular path at constant speed.
B
In a uniform circular motion, a particle
covers equal distances within equal interval of
time, but the direction of motion changes at every
point as shown in fig. 2.16. In this case the
A
O
velocity of the body is referred to as linear
R
velocity. A circular motion is an example of an
accelerated motion with a constant speed.
The angular displacement of a particle is
Fig. 2.16. A linear velocity of
measured by the angle covered by the circle a particle rotating about an axis O
radius R and it is subtended at the center of
the circle (fig. 2.16).The fixed axis around which motion takes place is called
axis of rotation (point O in fig. 2.16). In a uniform circular motion, a particle
undergoes the same angular displacement during
A
the same time interval.
l
The angular displacement of a particle can be
R
given in degrees or in radians.
rad
One radian is defined as the angle subtended at
B
O
R
the center of a circle by an arc with the length equal
to the radius of the circle (fig. 2.17). If is the angle
subtended by an arc AB of length l at the center of
Fig. 2.17. A radian definition
a circle of radius R, then
l
.
R
θ (radian)
(2.27)
In one complete rotation or 360°, we have
l
R
2πR
R
1 radian
2π radian.
180
π
54
57 18 .
Angular velocity is the angle described by a particle during a unit time.
Angular velocity is represented by Greek letter (Omega) and is defined as
θ
.
t
ω
(2.28)
The units of angular velocity are radian per second (rad/s).
The relation between the linear and angular
velocities is given by
the following expression:
=R ,
(2.29)
where R is the circle radius.
Change of the linear velocity direction is described by the centripetal
acceleration . Its direction is along the radius R to the circle center (fig. 2.16).
The magnitude of the centripetal acceleration is defined as
2
ac
R
(2.30)
,
where R is the circle radius.
Substituting (2.29) in this equation we obtain the another equation for
the centripetal acceleration:
ac = ω2R.
(2.31)
Time required to complete one rotation is called the time period and
represented by T. The circle length is equal to 2 R, thus time period is given as
T
2π
ω
2πR
2π
.
T
or ω
(2.32)
(2.33)
The number of rotations made by the particle in 1 second is called
frequency of rotation. It is represented by Greek letter (nu):
v
1
.
T
(2.34)
The SI units for frequency of rotation are hertz (Hz).
The angular velocity is related to the frequency of rotation by the following
equation:
=2 .
(2.35)
From the equations 2.30–2.35 the following equations for the centripetal
acceleration can be derived:
2
ac
R
ω2 R
4π2 R
T
2
4π2v 2 R.
(2.36)
Example 2.14.
The platter of the hard drive of a computer rotates at 6000 rev/min
(revolutions per min). (a) What is the angular velocity of the platter?
(b) If the reading head of the drive is located 3.5 cm from the rotation axis, what
is the linear speed of the point on the platter just below it?
55
Solution. a) To find the angular velocity we use Eq. 2.35, where we set
frequency of rotation
= 6000 rev/min = 6000 rev/60 s = 100 Hz. Then
the angular velocity is = 2 = 2 (3.14 rad) (100 Hz) = 628 rad/s.
b) The linear velocity of the point a 3.5 cm out of the rotation axis is given
by the Eq. 2.30, where we set R = 3.5 cm = 3.5 10–2 m:
= R = (3.5 10–2 m) (628 rad/s) = 22 m/s.
PROBLEMS
1. A train 50 meter long passes a bridge 250 m long at the speed of 9 km/h.
How long will it take to completely pass over the bridge? (Answer: 2 min)
2. A ball rolling at the speed of 1 m/s was stopped within 1 meter. What
was the average retardation applied to ball? How long did it take to stop it?
(Answer: 0.5 m/s2, 2 s)
3. A motor car starts from rest and accelerates uniformly for 30 s to a speed
of 72 km/hr. It then moves with uniform velocity and is finally brought to rest in
50 meter with a constant retardation. If the total distance travelled is 950 meter,
find the acceleration, retardation and the total time taken. (Answer: 0.67 m/s2,
4 m/s2, 65 s)
4. A car starts from rest and moves with uniform acceleration. Its velocity
is 25 cm/s after 5 s and 34 cm/s after 8 s. Calculate the distance that it will travel
in tenth second. (Answer: 28.5 cm)
5. A stone is thrown upwards from the surface of earth with initial speed of
5 m/s. What is the maximum height reached by a stone? (Answer: g = 10 m/s2;
1.25 m/s)
6. What is the angular speed of (a) the second (b) the minute and (c)
the hour hand of an analog watch? (Answer: 0.105 rad/s, 1.75 10–3 rad/s,
1.45 10–4 rad/s)
TESTS (Sections 2.1–2.5)
1. Which of following graphs represents motion with uniform speed?
s
s
a
b
0
0
t
c
0
t
d
0
t
56
t
2. Two straight lines drawn at the same displacement
time graph make angles 30° and 45° with X-axis as shown
The ratio of two velocities is:
1
1
a)
b) 1
c)
d) 2
3
2
3
s
0
45°
30°
t
3. A body covers half of its distance with speed ―u‖ and other half with
the speed ― ‖, the average speed of the body is:
a)
u
b)
2
u
c)
2
2u
u
d)
u
2u
4. A body goes from A to B with a velocity 40 km/hr and comes back from
B to A with a velocity of 50 km/hr, the average velocity during the whole
journey is:
a) 45 km/hr
b) 44.4 km/hr
c) zero km/hr
d) 48 km/hr
5. At first half of the distance a body moves with a velocity 40 m/s and at
second half — with a velocity 50 m/s at same direction, so the average velocity
during its whole journey is:
a) 45 m/s
b) zero m/s
c) 44.44 m/s
d) 48 m/s
6. Which of the following statements is not true?
a) velocity, acceleration and displacement are vectors;
b) a vector quantity has only magnitude while scalar has both magnitude
and direction;
c) mass, work, energy, moment of inertia are scalars.
7. Acceleration is the rate of change of
a) speed
b) position
c) velocity
8. A particle is moving with a velocity
changes to 2 m/s. The average acceleration is:
a)
c) (
1
1
t
1
+
m/s2
2)·t
m/s
b)
2
2
1
t
1
d) distance
m/s and after t(s) the velocity
m/s2
d) none of the above
9. An object starting from rest covers distances in direct proportion to
the square of time. Its acceleration is:
a) increasing
b) constant
c) zero
d) none of the above
10. The displacement x of a particle along a straight line at a time t
is x = a0 + a1t + a2t2. The acceleration of the particle is:
a) a0
b) a1
c) 2a2
d) a2
57
11. If the speed of a car increases by 3 times, how does the distance needed
to stop it change:
a) 3 times
b) 6 times
c) 9 times
d) None of the above
12. A body starts from rest and moves with uniform acceleration ―a‖.
The distance covered during the nth second will be:
a) a(n – 1)
b) a(n –
1
)
2
c) a(2n – 1)
d) a(2n + 1)
13. A body starting from rest travels 150 m in 8th second. Its uniform
acceleration is:
a) 15 m/s2
b) 10 m/s2
c) 20 m/s2
d) 30 m/s2
14. An object is projected upwards with a velocity of 100 m/s. It strikes
the ground back in (g = 10 m/s2):
a) 10 s
b) 20 s
c) 5 s
d) 15 s
15. A ball thrown vertically upwards with an initial velocity of 19.6 m/s
returns to thrower in 4 s. The maximum height reached by it is:
a) 9.8 m
b) 44.1 m
c) 19.6 m
d) 26.7 m
16. Two bodies of mass m1 and m2 are dropped from rest from heights h1
and h2. The ratio of their times to reach the ground are:
a) h1 : h2
b) h1 : h2
c) m1 h1 : m2 h2
d) m1h1 : m2h2
TESTS (Section 2.6)
1. A body is said to be in uniform circular motion when its linear velocity:
a) remains constant both in magnitude and direction;
b) remains constant in magnitude but changes in direction;
c) remains constant in direction but changes in magnitude;
d) none of the above.
2. The acceleration of a body performing uniform circular motion with
linear speed in circle of radius R is:
2
a)
R
c)
2
58
d)
2
R
R
3. A car going round a circular path at constant speed
a) has a constant linear velocity;
b) has a constant acceleration;
c) has a constant momentum;
d) none of the above.
R
b)
4. A body moves in a circle of radius R. After completing the circular path
once it returns at the point of start. The displacement of the body is:
a) 2 R
b) R
c) zero
d) R2
5. A body moving in a circle of radius R with constant speed makes n
revolutions per second. Its centripetal acceleration is:
a) 2 nR
b) 4 2n2R
c) n2R
d) 2n2R
6. Two bodies of masses m1 and m2 are moving in concentric circles
of radii r1 and r2 such that the frequencies of revolutions are the same. The ratio
of centripetal accelerations is:
a) R12 : R22
b) R1 : R2
c) R1 : R2
d) R2 : R1
7. In the above problem the ratio of their angular velocities is:
a) R1 : R2
b) R12 : R22
c) 1 : 1
d) R22 : R12
8. The ratio of the angular speeds of minute hand and hour hand of a clock is:
a) 6 : 1
b) 12 : 1
c) 1 : 6
d) none of the above
9. In uniform circular motion which is true?
a) both velocity and acceleration are constant;
b) both acceleration and speed are not constant;
c) both acceleration and speed are constant;
d) both velocity and acceleration are not constant.
10. A particle moves in a plane with a constant speed with its direction
changing. The path of the particle is:
a) straight line;
b) an arc of the circle;
c) a parabola;
d) an ellipse.
11. The length of second‘s hand in a watch is 1 cm. The change in speed of
its tip in 15 seconds is:
π
π
π
a) Zero
b)
cm/s
c)
cm/s
d)
2 cm/s
30
30
30 2
59
3. DYNAMICS
In this Chapter we consider what makes the objects to move. The connection
between the force and motion studies the subject called dynamics.
3.1. NEWTON’S LAWS OF MOTION
Before discussion Newton‘s laws of motion, let us give definitions of some
useful terms.
Inertia. Inertia is the property of a body to maintain its state of rest or
uniform motion in a straight line.
Mass. The mass is a measure of the inertia of an object. It is the characteristic
of a body that relates the body‘s acceleration to the net force causing
the acceleration. The more mass a body has, the greater the force needed to
impart it a particular acceleration. It is harder to start it moving from rest, or to
stop it when it is moving. Mass (m) is a scalar quantity. The SI unit for mass is
kilogram (kg).
Force. Force is a push or a pull which changes or
tends to change the state of rest or of uniform motion of
a body in a straight line. We need a force to overcome
inertia of a body. When you push a body, you are
exerting a force on it (fig. 3.1). We often call it contact
force because the force is exerted when one object
comes in contact with another object. On the other hand, Fig. 3.1. A force exerted
we say that an object falls because of the force of on a body by a person
gravity.
Force ( ) is a vector and has both direction and magnitude (fig. 3.1). We
can represent any force on a diagram by an arrow. The direction of the arrow
is the direction of push or pull, and its length is drawn proportional to
the magnitude of force. The SI unit for force is newton (N).
If two or more forces act on a body, we find the net (resultant) force ( ) by
adding them as vectors (see example in fig. 3.2).
m3
m1
m2
Fig. 3.2. The net force
is a vector sum of
60
and
forces (
=
+
)
A single force that has the same magnitude and direction as the calculated
net force would then have the same effect as all the individual forces. This fact
is called the principle of superposition for forces.
Newton’s first law of motion (law of inertia): every body continues in its
state of rest or of uniform motion in a straight line, as long as no net force acts
on it. This is called principle of inertia.
From the Newton‘s first law it is evident that a body by itself is incapable
of changing its state of rest or of uniform motion in a straight line. This
incapability is known as inertia. Hence first law is called law of inertia.
The reference frames in which Newton‘s first law does hold are called
inertial reference frames or inertial frames. Usually approximation is made
that a reference frame fixed on the Earth is an inertial frame. Any reference
frame that moves with constant velocity relative to an inertial frame is also
inertial reference frame.
Reference frames in which the law of inertia is not valid, such as
accelerating reference frames, are called noninertial reference frames or
noninertial frames.
Newton’s second law of motion: acceleration of a body is directly
proportional to the net force acting on it, and is inversely proportional to
the body’s mass. The direction of the acceleration is in the direction of the net
force acting on the body.
F
a
or F ma,
(3.1)
m
where is the acceleration, m is the mass, and is the net force acting on
the body.
When n forces different in magnitude and direction act of the body,
means the vector sum of all the forces acting on the object, which we defined
above as the net (resultant) force. Then
n
i 1
Fi
ma ,
(3.2)
n
where
i 1
Fi
F1 F2
F3
... Fn .
Only external forces, i. e. forces exerted on the body by other bodies, are
to be included.
Newton’s third law of motion:
whenever one object exerts a force on a
m
m1
2
second object, the second exerts an equal
Fig. 3.3. The action
and the reaction force in the opposite direction on the first.
forces exerted by bodies
Forces come in pairs (fig. 3.3).
Commonly one of these forces is called the action force. The other one is
then called reaction force. The term action means the force exerted on the first
61
body by second body (
body by the first body (
), while reaction means force exerted on the second
. Then
F12
F21 or F12
F21 .
(3.3)
Third law of motion can be formulated as ―to every action there is an equal
and opposite reaction‖. But it is very important to remember that the ―action‖
and ―reaction‖ forces are acting on different objects.
3.2. MAIN FORCES
3.2.1. THE GRAVITATIONAL FORCE
Galileo Galilei claimed that all objects dropped near the surface
of the Earth fall with the same acceleration, g, if air resistance is negligible.
The force that causes this acceleration of mass m is the gravitational force Fg,
that can be written as
Fg = mg.
(3.4)
This force is directed down toward the
center of the Earth (fig. 3.4). The gravitational
force (force of gravity) is explained by
Newton’s law of universal gravitation that
states: every particle in the universe attracts Fig. 3.4. The gravitational force
every other particle with a force that is
proportional to the product of their masses and
inversely proportional to the square of
the distance between them. This force acts along
the line joining the two particles (fig. 3.5).
The magnitude of the gravitational force can be
written as
Fg = G·m1m2/r2,
(3.5)
where m1 and m2 are the masses of the two
Fig. 3.5. Newton‘s law of universal
particles, r is the distance between them, and G
gravitation
is a universal constant which has the same
numerical value for all objects (G = 6.67 10–11 N m2/kg2).
3.2.2. GRAVITY NEAR THE EARTH SURFACE
When Eq. 3.5 is applied to the gravitational force between the Earth and
an object at its surface (Fg), m1 becomes the mass of the Earth M, m2 becomes
the mass of the object m, and r becomes the distance of the object from
the Earth‘s center, which is approximately the radius of the Earth R. Thus,
Fg = G·mM/R2.
(3.6)
Combining Eq. (3.4) and (3.6) we obtain
mg = G·mM/R2.
(3.7)
62
Solving Eq. 3.7 for g gives the acceleration of gravity at the Earth‘s
surface:
G M
2
g
9
.
8
m/s
.
(3.8)
R2
Thus, the acceleration of gravity at the surface of the Earth, g, is
determined by the Earth mass M and the radius of the Earth R.
3.2.3. THE FORCE OF ELASTICITY AND HOOKE’S LAW
The elasticity forces are revealed in a response of a body to the action of
the external forces. The external force is called the load.
Changes of shape and (or) dimensions of the sample which is made from
some material under the action of the external forces is called strain
(deformation).
Elasticity is the property of a body to preserve its shape and dimensions
after load removing.
For example, let us consider the case of an axial tension (or compression)
of a body (fig. 3.6). A load tends to stretch (elongate) a body (fig. 3.6, a). Then
absolute deformation (the change in length) is equal to x = l – l0, where l0 is
the original length, l is the elongated length. Relative deformation (strain) is
defined as:
x
(3.9)
ε
.
l0
l0
a
x
b
x
0
Fig. 3.6. The force of elasticity
x
upon tension (a) and compression (b)
The value of is dimensionless, it is usually expressed in percents (%).
When an external force F acts upon a solid body (the body is deformed), a
reaction force (internal force) arises within the body that is equal in magnitude
but opposite in direction to the external force (fig. 3.6). This is a force of
elasticity
and it is sometimes called a ―restoring force‖. According to
Hooke’s law a force of elasticity is directly proportional to deformation:
Fel = –k·x,
(3.10)
where k is the stiffness constant, x is an absolute deformation.
63
The internal force is characterized by the mechanical stress . For an axial
tension (or compression) the value of stress (tensile or compressive stress) is
defined as the average force F per unit cross-sectional area S within the body on
which external force acts:
σ
F
.
S
(3.11)
In the SI system stress is measured in the pascals (symbol Pa), which is
defined as one newton per square meter:
N
σ
Pa .
m2
In terms of strain and stress Hooke's law can be defined as:
σ = E·ε,
(3.12)
where E is called Young’s modulus or the modulus of elasticity.
Thus Hooke’s law states that the stress is directly proportional to the strain
upon elastic deformation.
Substituting Eq. 3.9 and 3.11 for the strain and stress, respectively, in
the Eq. 3.12 we obtain:
F E x
(3.13)
.
S
l0
E S
Then F
x , and from Eq. 3.13 it follows that
l0
E S
(3.14)
k
.
l0
Thus the stiffness constant k is directly proportional to the modulus of
elasticity, and it is dependent on the body shape and dimensions.
3.2.4. NORMAL FORCE
For an object resting on a table, the table exerts
upward force ( ), but it remains stationary (fig. 3.7).
The reason is that the table is compressed slightly
beneath the object, and due to its elasticity, it pushes
up on the object as shown.
Fig. 3.7. The normal force
The force exerted by the table is often called
a contact force, since it occurs when two objects are in contact. The push on an
object from the table is a normal force ( ) (sometimes it is denoted by symbol
). The name comes from the mathematical term normal, meaning
perpendicular: the force on an object from the table is perpendicular to the table
(fig. 3.7). Thus when a body presses against a surface, the surface deforms and
pushes on the body with a normal force that is perpendicular to the surface.
64
3.2.5. TENSION
When a cord (or a rope or other such object) is attached to a body and
pulled taut, the cord pulls on the body with a force directed away from the body
and along the cord (fig. 3.8). The force is often called
a tension force ( ) because the cord is said to be
in a state of tension (or to be under tension), which
means that it is being pulled taut. The tension in
Fig. 3.8. The tension force
the cord is the magnitude T of the force on the body.
A cord is often said to be massless (meaning its mass is negligible
compared to the body‘s mass) and unstretchable.
3.2.6. WEIGHT
Weight of the body ( or ) is the force with which the body acts on a
support or on a cord. The weight force is applied not to the body but to the
support or to the cord (fig. 3.9).
a
b
Fig. 3.9. Weight of the body
In accordance with the Newton‘s third law the weight force
acting on a
support (or on a cord) and the normal force (or tension force ) exerted by a
support on the body (reaction force) are equal in magnitude (W = N and W = T),
and oppositely directed (fig. 3.10):
(3.15)
W
N or W
T.
Fig. 3.10. Weight of the body
and the reaction forces (normal
and tension
forces)
3.2.7. FRICTION
If we either slide or attempt to slide a body over a surface, the motion is
resisted by a bonding between the body and the surface. The resistance is
considered to be a single force called either the frictional force ( ) or simply
friction.
65
We focus our attention on sliding friction, which is usually called kinetic
friction when object slides across a surface. This force is directed along the
surface, opposite to the direction of the motion (fig. 3.11).
Fig. 3.11. The friction force
Sometimes, to simplify a situation, friction is assumed to be negligible
(the surface, or even the body, is said to be frictionless).
The magnitude a frictional kinetic force is given by
Ffr = k N,
(3.16)
where k is the coefficient of kinetic friction, N is the magnitude of the normal
force. The coefficient k is dimensionless and its value depends on the nature of
the two surfaces in contact.
There is also static friction, which refers to a force parallel to the two
surfaces that can arise even when they are not sliding. The force of static friction
depends on various factors and it may change from zero to its maximum value
given by (Ffr)max = s N, where s is the coefficient of static friction generally
being greater than k. However when solving problems it is commonly
suggested that both of these coefficients are equal: s = k.
The static friction force is directed along the surface, opposite to
the direction of the intended motion.
3.2.8. PROBLEM SOLVING
When solving problems involving Newton‘s laws and forces, it is very
important to draw a diagram showing all the forces acting on each object
involved. Such a diagram is called a free-body diagram (fig. 3.12), or force
diagram: choose one object, and draw an arrow to represent each force acting on
it. Include every force acting on that object. Do not show forces that the chosen
object exerts on other objects. Only forces acting on a given object can be
included in the Newton‘s second law equation (Eq. 3.1). If your problem
involves more than one object, a separate free-body diagram is needed for each
object. For the case shown in fig. 3.12, the forces that could be acting are gravity
and contact forces (one object pushing or pulling another, normal force,
friction).
66
Fig. 3.12. A free-body diagram
Newton‘s second law involves vectors, and it is usually necessary to
resolve vectors into components. Choose x and y axis in a way that simplifies
the calculation. Then for each object apply Newton‘s second law to the x and y
components separately (i. e. Fx ma x , F y ma y ).
NOTE. The weight of the body is always applied to the support (or a cord),
and not to the body, this is why it is not included in the Newton’s second law
(force in fig. 3.13). To find the weight of the body, it is necessary to solve
Newton’s second law equation, determine the magnitude of the reaction force
(N or T) which is equal to the weight W, according to the Eq. 3.15.
Fig. 3.13. A box resting on a table
Let us determine the weight of a box ( ) resting on the smooth
(frictionless) horizontal surface of a table (fig. 3.13). The forces acting on
the body are the normal force ( ) and the gravitational force (
) as
shown in fig. 3.13. The box is at rest, so its acceleration
and in
accordance with the Newton‘s second law the net force acting on it must be
zero. Hence
N m g 0 or N
m g.
67
Thus the magnitude of the normal force N = mg. In accordance with
the Newton‘s third law the magnitude of normal force on the box is equal to
the box‘s weight, so W = N = mg. Thus weight of the body is equal to
the magnitude of the gravitational force on the body just in case if it is located
on a resting support (or on a resting cord) with respect to the Earth. If a support
(or a cord) with a body accelerates down or upward, the weight W mg.
Example. 3.1. Weight.
A passenger of mass 70 kg descends in an elevator that accelerates at
2.5 m/s2 downward. He stands on scales that shows in kg. (a) During this
acceleration, what does the scales show? (b) What does the scales show when
the elevator descends at a constant speed of 0.5 m/s?
Solution. We use Newton‘s second law
only in an inertial frame. If the cab accelerates,
then it is not an inertial frame. So we choose
the ground to be our inertial frame and make
any measure of the passenger‘s acceleration
relative to it.
The fig. 3.14 shows all the forces acting
on the passenger. The direction of the
acceleration is downward, so we choose the
positive direction of y axis as down. From
Newton‘s second law it follows
Fig. 3.14. Example 3.1. A free-body
Fg – N = ma.
diagram for a passenger standing on
a) The magnitude of the normal force (N)
the scales
acting on the passenger is given by:
N = Fg – ma = mg – ma = m (g – a) = (70 kg) (9.8 m/s2 – 2.5 m/s2) = 511 N.
So passenger‘s weight is W = N = 511 N. The scales shows in kg
N
511 N
m
52.1 kg.
g 9.8 m/s2
b) When the elevator descends at a constant speed, the acceleration is equal
to zero (a = 0), so by Newton‘s second law we obtain
Fg – N = 0 or N = Fg = mg = (70 kg) 9.8 (m/s2) = 686 N.
Thus passenger‘s weight is W = N = 686 N and the scales shows his mass
70 kg.
Example 3.2. Newton’s second law.
A box of mass m = 5 kg is pulled along a floor by a cord that exerts a force
of magnitude FP = 25 N at the angle 30°. The magnitude of the box‘s
acceleration is 2 m/s2. What is the coefficient of kinetic friction?
Solution. The free-body diagram is shown in fig. 3.15. The forces on the
box are the gravitational force Fg = mg, the normal force exerted by the floor FN,
68
the applied force FP, and the friction force Ffr. We use Newton‘s second law
given by Eq. 3.1.
Let us find projections of the forces
and the acceleration vectors on the
coordinate axes (x and y). We choose the
upward direction as the positive y direction.
In the vertical direction there is no motion
(ay = 0), so Newton‘s second law in
the vertical direction gives
, then
FN + FP·sin α – mg = 0,
where the minus sign means that the
gravitational force Fg acts in the negative y
Fig. 3.15. Example 3.2. Free body
direction (m and g are magnitudes).
diagram for a box pulled by a force
of magnitude Fp at the angle 30°
From this equation we obtain that
the normal force is given by
FN = mg – FP·sin α.
Now we apply Newton‘s second law for the horizontal (x) direction
(positive to the right), and include the friction force:
FP·cos α – Ffr = ma.
Then the frictional force is given by
Ffr = μFN = FP·cos α – ma.
Substituting in this equation expression for FN obtained above gives
the coefficient of kinetic friction ( ).
F fr FP cos α ma 25 N cos 30 (5 kg) (2m/s 2 ) 12.5 3 10
μ
0.3.
FN mg FP sin α (5 kg) (9.8m/s 2 ) 25 N sin 30
49 12.5
The coefficient of kinetic friction
= 0.3.
3.3. CONSERVATION OF MOMENTUM
3.3.1. LINEAR MOMENTUM AND IMPULSE EQUATION
Amount of motion present in body is called linear momentum
(momentum) of the body. It is represented by a vector . Linear momentum of
a body having mass m and moving at any instant of time with a velocity , is
defined as
.
(3.17)
The units for linear momentum in SI system are kg m/s. The adjective
linear is often dropped, but it serves to distinguish from angular momentum.
Since mass m is always a positive scalar quantity, from Eq. 3.17 it follows that
vectors and have the same direction.
69
In mechanics several bodies form system. The forces can be of two types:
external and internal forces. Any force acting on the body system from
the bodies outside the system is called an external force. Forces between bodies
inside the system are called internal forces. Internal forces cannot accelerate
the system. A system is called isolated if the resultant external force acting
on a system of bodies is zero (or no external forces act), the only forces acting
are those between objects of the system.
By substituting the expression for acceleration a
Newton‘s law equation (Eq. 3.1), we obtain
2
1
t
into the second
(3.18)
or
,
(3.18a)
where
and
is the velocity and momentum of the body at some initial
instant of time,
and
is the velocity and momentum of the body after
certain time interval t.
The product of the force multiplied by the time interval during which it acts
upon a body ( ) is called impulse of a force.
A change of momentum of the body is equal to impulse of a force acting on
a body (Eq. 3.18). This relation is called impulse equation.
3.3.2. THE LAW OF CONSERVATION OF MOMENTUM
Now consider, for example, an isolated system of two bodies. Suppose that
the bodies of mass m1 and m2 are moving with the velocities
and
and let
them to interact with each other (fig. 3.16). Let the two bodies continue moving
in the same direction with the velocities
and
after interaction.
a
b
c
Fig. 3.16. Conservation of momentum in collision of two balls:
a — before collision; b — at collision; c — after collision
According to the Newton‘s third law of motion the forces acting on
the bodies are equal in magnitude and oppositely directed, i. e.
Change of momentum of the first body
(3.19)
Change of momentum of the second body
,
(3.20)
where t is the time of acting of a force.
70
From Eq. 3.19–3.20 (at
) we obtain
(3.21)
or
(3.21a)
From Eq. 3.21 it follows that the vector sum of the linear momentum of
two bodies before interaction is equal to a vector sum of the linear momentum of
these bodies after interaction. This is the law of conservation of linear
momentum for two bodies which compose an isolated system.
The law of conservation of momentum states that if the resultant external
force acting on a system of bodies is zero (the system is isolated), the total
linear momentum of the system is conserved (remains constant).
The total linear momentum of the system is a vector sum of the linear
momentum of each body in the system.
Example 3.3. Momentum. Impulse equation.
A golf ball of mass 0.05 kg is hit off the tee at a speed of 50 m/s. The golf
club was in contact with the ball for 2 ms. Find (a) the impulse (momentum)
imparted to the golf ball, and (b) the average force exerted on the ball by
the golf club.
Solution. a) The impulse imparted to the golf ball is given by the Eq. 3.17:
P = m = 0.05 kg 50 m/s = 2.5 kg m/s.
b) We use the impulse equation (Eq. 3.18). The initial momentum of
the ball is equal to zero. Then F t = p.
The average force exerted on the ball by the golf club is
p 2.5 kg m/s
F
1.25 kN.
t
2 10 3 s
Example 3.4. Conservation of momentum.
A railroad car with a mass of 7000 kg traveling at speed of 25 m/s strikes a
second railroad car with a mass of 3000 kg. If the cars lock together as a result
of the collision, what is their common speed immediately after the collision?
Solution. The masses of the railroad cars are m1 = 7000 kg and
m2 = 3000 kg, respectively; initial speed of the first railroad car A = 25 m/s and
of the second car ( B) is equal to zero. Then the initial total momentum is
m1 A + m2 B = m1 A.
Choose the positive direction to the right side as shown in fig. 3.17. After
the collision, the two cars become attached, so they will have the same speed
( ′). Then the total momentum after the collision is
m1 ′ + m2 ′ = (m1 + m2) ′.
According to the law of momentum conservation m1 A = (m1 + m2) ′.
Then
m1 A (7000 kg) (25 m/s)
17.5 m/s.
m1 m2 7000 kg 3000 kg
71
Fig. 3.17. Example 3.4. Collision of railroad cars
NOTE. In case of equal masses (m1 = m2) the mutual speed after collision
is half the initial speed of the first car.
Example 3.5.
A bullet of mass 10 gram is horizontally fired from a gun of mass 5 kg with
a speed of 100 m/s. What is the recoil (backward) speed of the gun?
Solution. We have the mass
of bullet (m =10 g = 0.01 kg) and
the mass of the gun (M = 5 kg);
initial velocities of the bullet ( B)
and gun ( R) are equal to zero.
The final speed of the bullet
B′ = 100 m/s. The positive
direction of bullet is taken from
left to right (fig. 3.18). Let R′ be
Fig. 3.18. Example 3.5. Recoil of a gun
the recoil speed of the gun. Total
momentum of the gun and the bullet before the fire (the gun is at rest) is
m B + M R = 0.
Total momentum of the gun and bullet after it is fired m B′ – M R′.
Negative sign indicates that the direction in which the gun would recoil is
opposite to that of bullet (fig. 3.18). According to the law of conservation of
momentum m B′ – M R′ = 0.
m B 0.01 kg 100 m/s
0.2 m/s.
Then R
M
5 kg
The recoil speed of the gun is 0.2 m/s.
72
PROBLEMS
1. A 75 kg man stands in a lift. What force does the floor exerts on him
when the elevator starts moving upward with an acceleration of 2.0 m/s 2
(assume g = 10.0 m/s2). (Answer: 900 N)
2. Tendons are strong elastic fibers that attach muscles to bones.
To a reasonable approximation, they obey Hooke‘s law. In laboratory tests
on a particular tendon, it was found that, when a 250-g object was hung from it,
the tendon stretched 1.25 cm. What is the stiffness constant of this tendon?
(Answer: 200 N/m)
3. A brick weighing 1 kg is sliding on ice with 2 m/s. It is stopped by
friction in 10 s. Calculate the constant force of friction. (Answer: 0.2 N)
4. Force of 10 N acts on a body for 40 s. Calculate the change
in the momentum of the body. (Answer: 400 kg m/s)
5. You throw a ball with a mass of 0.30 kg against a brick wall. It hits
the wall and rebounds horizontally with the speed of 20 m/s. (a) If the ball is in
contact with the wall for 0.010 s, find the impulse of the net force on the ball
during its collision with the wall. (b) find the average horizontal force that
the wall exerts on the ball during the impact. (Answer: 6 N s, 600 N)
6. While launching a rocket of mass 2 104 kg a force of 5 105 N is applied
for 20 s. What is the velocity attained by the rocket at the end of this interval.
(Answer: 500 m/s)
TESTS
1. Inertia of a body directly depends upon:
a) mass
b) area
c) volume
d) velocity
2. When a body is at rest
a) no force acts on it;
b) the force acting has no contact with it;
c) the forces acting on it balance each other;
d) none of the above.
3. A force of 2 N acting on a certain mass for 6 sec. gives it a velocity
of 6 m/s. The mass is equal to:
a) 0.5 kg
b) 1 kg
c) 2 kg
d) 4 kg
4. A force acting on a body of 10 kg produces in it an acceleration
of 2 m/s2. The force is:
a) 5 N
b) 20 N
c) 10 N
d) none of the above
5. A man of mass m is standing on a lift which is moving downwards with
acceleration a. The weight of the man is:
a) mg
b) m (g + a) c) m (g – a) d) zero
73
6. In the above problem if the downward acceleration of the lift is equal to
the acceleration due to gravity, then the weight of the man is:
a) mg
b) m (g + a) c) m (g – a) d) zero
7. A spring being compressed by 0.1 m develops a restoring force 10 N.
The stiffness constant of the spring is:
a) 100 N/m
b) 10 N/m
c) 1 N/m
d) 1000 N/m
8. The momentum of the system is conserved:
a) always;
b) never;
c) only in the absence of an external force;
d) only when an external force acts.
9. A body of 2 kg is at rest. The impulse required to impart it a velocity of
8 m/s is:
a) 16 N s
b) 40 N s
c) 80 N s
d) none of the above
10. A force of 10 N acts on a body for 5 s. The change in its momentum is:
a) 2 kg m/s
b) 0.5 N s
c) 50 kg m/s d) 500 kg m/s
11. A force of 1 N acts on a body of mass 1 kg. The body acquires an
acceleration of:
a) 1 m/s2
b) 9.8 m/s2
c) 1/9.8 m/s2 d) (9.8)2 m/s2
12. A force of 6 N acts on a body at rest of mass 0.1 kg which acquires a
velocity 30 m/s. The time for which the force acts is:
a) 18 s
b) 5 s
c) 0.5 s
d) 0.3 s
13. A bomb of mass 9 kg explodes into two pieces of mass 3 kg and 6 kg.
The velocity of 3 kg is 16 m/s. The velocity of 6 kg is:
a) 4 m/s
b) 8 m/s
c) 16 m/s
d) 32 m/s
14. A body is moving with uniform momentum of 10 kg m/s. The force
acting on it is:
a) zero
b) 10 N
c) 0.1 N
d) 100 N
15. A body of mass 2 kg moves with an acceleration of 3 m/s 2. The change
in momentum in one second is:
a) 0.67 kg m/s
b) 1.5 kg m/s
c) 6 kg m/s
d) 3 kg m/s
74
4. WORK. POWER. ENERGY
4.1. WORK
Work is done on an object by a force when the object moves through some
distance. The work W done by a constant force F (constant in both magnitude
and direction) is defined as a product of the magnitude of the displacement
(r = d), the force F and cosine of the angle between the directions of the force
and the displacement (fig. 4.1):
W = F d cos .
(4.1)
In SI units work is measured in joule (J): 1 J = 1 N m.
Fig. 4.1. The work done by the force F acting at an angle
to the displacement vector
Work is a scalar quantity; it has only magnitude, which can be positive or
negative.
For the case when the motion and the force are in the same direction
( = 0° and cos = 1) W = F d. When a particular force is perpendicular to
the displacement ( = 90°, cos = 0), no work is done by that force.
In case of linear motion along a straight-line path in the same direction,
the magnitude of displacement is the distance (d = s). Therefore
W = F s cos .
(4.2)
Example 4.1. Work.
A person pulls a 20 kg box 10 m along a horizontal floor by a constant
force F = 100 N acting at 30° angle (fig. 4.1). The floor exerts the opposite force
of friction equal to 60 N. How much work does each of the following forces do
on the box: a) 100-N pull, b) the friction force, c) the normal force from
the floor, and gravity? d) What is the net work done on the box?
Solution. A free-body diagram is similar to the Example 3.2 (fig. 3.15).
75
a) Setting in Eq. 4.1 F1 = 100 N, d = 10 m and = 30° gives the work done
by the force F1
W1 = F1 d cos = (100 N)·(10 m)·cos 30° = 866 J.
b) Setting in Eq. 4.1 F = 60 N, d = 10 m and = 180° gives the work done
by the friction force
W2 = F2 d cos = (60 N)·(10 m)·cos 180° = –600 J.
c) Directions of the normal force (N) and the force due to gravity (Fg) are
perpendicular to the displacement ( = 90°). Then work done by the given forces
is equal to zero (W3 = 0).
d) When several forces act on a body there are two ways to find the net
work. One way is to use Eq. 4.1 to compute the work done by each separate
force.
Then, because work is a scalar quantity, the total work done on the body by
all the forces is the algebraic sum of the quantities of work done by
the individual forces. An alternative way to find the total work is to compute the
net force (i. e. vector sum of the forces) and then use Eq. 4.1. Let us illustrate
both of these approaches.
1) The algebraic sum of the quantities of work done by the person and
the friction force is
Wnet = W1 + W2 = 866 J – 600 J = 266 J.
2) Just the projection of the net force parallel to the displacement vector
does the work on the box. It is given by
Fnet = F1 · cos – F2 = (100 N )·(cos 30°) – 60 N = 86.6 N – 60 N = 26.6 N.
The work done by the net force is
Wnet = Fnet · d · cos = (26.6 N)·(10 m)·cos 0° = 266 J.
4.2. POWER
Power is defined as the rate at which work is done. The average power P is
equal to the work W done divided by the time t it takes to do it:
P
W
.
t
(4.3)
In SI system units, power is measured in joules per second, and this unit is
given a special name, the watt (W): 1W = 1 J/s.
The work done at power of one thousand watts for one hour is equal to one
kilowatt-hour (1 kWh):
1 kWh = (103 W) (3600 s) = 3.6 106 J = 3.6 MJ.
Efficiency. An important characteristic of all engines is their overall
efficiency , defined as the ratio of the useful power output of the engine Pout to
Pout
the power input Pin:
(4.4)
η
.
Pin
76
The efficiency is always less than 1.0 because no engine can create energy,
and in fact, cannot even transform energy from one form to another without
some going to nonuseful forms of energy (friction, thermal energy, etc.). For
example, an automobile engine converts chemical energy released in the burning
of gasoline into mechanical energy that moves the pistons and then the wheels.
But nearly 85 % of the input energy is ―wasted‖ as thermal energy that goes into
the cooling system or out the exhaust pipe, plus friction in the moving parts.
Thus car engines are roughly only about 15 % efficient.
4.3. ENERGY
Energy can be defined as the ability to do work. Energy is measured
in the same units as work: joules in SI units.
An object in motion has the ability to do work and thus can be said to have
energy. The energy of motion is called kinetic energy (from the Greek word
kinetikos, meaning ―motion‖).
To obtain a quantitative definition for kinetic energy, let us consider
a simple rigid object of mass m (treated as a particle) that is moving in a straight
line with an initial speed 1. To accelerate it uniformly to a speed 2, a constant
net force F is exerted on it parallel to its motion over a displacement d.
According to Newton‘s second law:
a
F
.
m
(4.5)
The distance s travelled by the object during the time interval t is defined
as:
2
2
S
2
1
,
2a
where 1 and 2 are the initial and final speed of the object, respectively.
Then the net work done on the object is
2
2
m 22 m 12
2
1
W F s m a
=
.
2a
2
2
(4.6)
(4.7)
4.3.1. KINETIC ENERGY
The quantity Ek
m 2
is defined as the kinetic energy of the object.
2
NOTE. Equation (4.7) derived here for one dimensional motion with
a constant force, is valid in general for translational motion of an object in three
dimensions and even if the force varies.
We can rewrite Eq. 4.7 as:
W = Ek2 – Ek1.
77
(4.8)
Equation 4.8 is a useful result known as the work-energy principle. It states
that the net work done on an object by the net resultant force is equal to
the change in kinetic energy of the object.
If the initial speed of an object 1 = 0 and the final speed 2 = , then
m 2
m 2
W = Ek2 – Ek1 =
(4.9)
0
.
2
2
From Eq. 4.9 it follows that in order to make the body move with some
speed the work should be done upon the body which is equal to its kinetic
energy.
NOTE. The work-energy principle is a very useful reformulation of
Newton’s laws. It tells us that if (positive) net work W is done on an object, the
object’s kinetic energy increases by an amount W. The principle also holds true
for the reverse situation: if the net work W done on an object is negative,
the object’s kinetic energy decreases by an amount W. In other words, a net
force exerted on an object opposite to the object’s direction of motion decreases
its speed and its kinetic energy.
The kinetic energy of a group of objects is the sum of the kinetic energies
of the individual objects.
4.3.2. POTENTIAL ENERGY
Potential energy associated with forces that depend on the position or
configuration of objects relative to the surroundings. Various types of potential
energy can be defined.
The common example of potential energy is gravitational potential energy.
A heavy brick held above the ground has potential energy because of its position
relative to the Earth. The raised brick has the ability to do work, because if it is
released, it will fall to the ground due to the gravitational force, and can do
work.
Let us find the gravitational potential energy of an object near the surface
of the Earth.
Consider the case of falling an object of mass m
h1 h
downwards from height h1 to h2 (fig. 4.2). The
gravitational force on an object of mass m near the
h2
Earth‘s surface is Fg = mg, where g is acceleration due
to gravity. The work done by this gravitational force on Fig. 4.2. The work done by
the gravitational force
an object that falls a vertical distance h = h1 – h2 is
WG = F s = mg (h1 – h2) = –(mgh2 – mgh1) = mgh.
(4.10)
Thus falling an object of mass m from height h requires an amount of work
equal to mgh. And once at height h, the object has the ability to do an amount of
work equal to mgh. We can say that the work done in falling the object has been
stored as gravitational potential energy UG = mgh.
78
We can rewrite Eq. 4.10 as:
WG = –(UG2 – UG1).
(4.11)
Equation 4.11 defines the change in gravitational potential energy when
an object of mass m moves between two points near the surface of the Earth.
Potential energy of an object at the Earth surface is commonly set to zero
(UG2 = 0). Then the gravitational potential energy of an object, UG, at any point
above the surface of the Earth at height h can be defined as
UG = mgh.
(4.12)
Note that the gravitational potential energy is associated with the force of
gravity between the Earth and the mass m.
We consider now elastic potential energy
associated with elastic materials, which includes
a great variety of practical applications. Consider a
a
simple coil spring as shown in fig. 4.3, whose mass
is so small that we can ignore it. When the spring
is compressed and then released, it can do work on
a ball (mass m). Thus the spring-ball system has
potential energy when compressed (or stretched).
b
Like other elastic materials, a spring is described by
Hooke‘s law as long as the displacement x is not too
great. To hold the spring compressed (or stretched)
a distance x from its initial (unstretched) length
c
requires the person‘s hand to exert a force FP = kx
Fig. 4.3. The work done
on the spring (fig. 4.3, a), where к is the spring
by the spring force
stiffness constant. The spring pushes back with
a force Fs = –kx (fig. 4.3, b). A spring force is thus a variable force: it varies
with the displacement of the spring‘s free end. It can be shown that the springball system has potential energy Uel when compressed (or stretched) an amount
kx 2
x from equilibrium:
(4.13)
U el
,
2
where k is the spring stiffness, x is an absolute deformation.
The quantity E equal to the sum of the kinetic energy and the potential
energy of the system at any instant of time is called the total mechanical energy:
E = Ek + U.
(4.14)
If a system is isolated from its environment, there can be no energy
transfers to or from it. For that case, the law of conservation of energy states:
the total mechanical energy of the isolated (closed) system is conserved.
E = Ek + U = const.
(4.15)
This is called the principle of conservation of mechanical energy.
If the kinetic energy Ek increases, then the potential energy U must decrease by
an equivalent amount to compensate. Thus the total energy Ek + U remains
constant.
79
This is one of the most important principles in physics. It is called the law
of conservation of energy and can be stated as follows: The total energy is
neither increased nor decreased in any process. Energy can be transformed
from one form to another, and transferred from one object to another, but
the total amount remains constant.
When the external forces do work on a body, the change in the total
mechanical energy of the system is equal to the work done by external forces
(Wext):
E2 – E1 = Wext.
(4.16)
In case when nonconservative forces (such as a friction force) act within
the closed system we obtain:
E2 – E1 = Wfric.
(4.17)
Example 4.2. Conservation of energy.
A rock with a mass of 2 kg is falling
down due to the Earth‘s gravity from a height
h = 2.5 m above the ground (fig. 4.4). What
are (a) the rock‘s speed when it has fallen to
1.5 m above the ground; (b) the rock‘s speed
just before it hits the ground? (c) the potential
energy of the rock at the moment of release
and the kinetic energy of the rock just before
it hits the ground.
Solution. We apply Eqs. 4.9–4.10 and
the law of conservation of mechanical energy
(Eq. 4.15) with only gravitational force acting
on the rock. We choose the ground as
the reference level. According to the Eq. 4.15 Fig. 4.4. Example 4.2. Note bar graph
the total mechanical energy of the rock at any represents the change in potential U
and kinetic Ek energy
point along the path is constant. Then
m 12
m 22
mgh1
mgh2 .
2
2
where 1 is the rock‘s speed at the position h1 above the ground and 2 is its
speed at some other point h2.
a) At the initial moment (point y1 in fig. 4.4) position is h1 = 2.5 m and
1 = 0. The rock‘s initial kinetic energy in equal to zero and total mechanical
m 22
mgh2 .
energy is equal to potential energy. Hence mgh1
2
We need to find the rock‘s speed at the height h2. Setting 1 = 0, h2 = 1.5 m
and solving for 2 we find
2
2 g (h1 h2 )
2 9.8 m/s 2 (2.5 m 1.5 m)
80
4.3 m/s.
The rock‘s speed 1.5 m above the ground is 4.3 m/s.
b) Just before rock hits the ground (point y2) the height is equal to zero
(h2 = 0) and total mechanical energy is equal to the kinetic energy.
Setting h2 = 0 in the above equation gives
2
2 gh1
2(9.8 m/s 2 )(2.5 m)
7 m/s.
The rock‘s speed just before it hits the ground is 7 m/s.
NOTE. The speed of the rock is independent of the rock’s mass.
c) The potential energy UG of the rock is given by Eq. 4.10. It is equal to
the initial mechanical energy of the rock. Setting h = 2.5 m in Eq. 4.10 we
calculate:
WG = mgh = (2 kg) (9.8 m/s2) (2.5 m) = 49 J.
Just before the rock hits the ground total mechanical energy is equal to
the kinetic energy E = Ek. According to the law of conservation of mechanical
energy it is equal to the initial mechanical energy of the rock, i. e. E = Ek =
WG = 49 J.
NOTE. As the rock falls, the potential energy decreases, but the rock’s
kinetic energy increases to compensate, so that the sum of the two remains
constant. Just before the rock hits the ground all of the initial potential energy
will have been transformed into kinetic energy (fig. 4.4).
PROBLEMS
1. A tow truck pulls a car 5 km along a horizontal roadway using a cable
having a tension of 900 N. How much work does the cable do on the car if it
pulls horizontally? (Answer: 4500 kJ)
2. A body is under the action of a force 5 N moves through 10 m
in a straight line. If work done is 25 J what is the angle at which force acts with
the direction of motion. (Answer: 60°)
3. A factory worker pushes a 50-kg crate a distance of 5 m along a level
floor at constant velocity by pushing horizontally on it. The coefficient of
kinetic friction between the crate and the floor is 0.25. (a) What magnitude of
force must the worker apply? (b) How much work is done on the crate by this
force? (c) How much work is done on the crate by friction? (e) What is the total
work done on the crate? (Answer: 122.5 N, 612.5 J, –612.5J, 0 J)
4. The human heart is a powerful and extremely reliable pump. Each day it
takes in and discharges about 7500 L of blood. Assume that the work done by
the heart is equal to the work required to lift this amount of blood a height equal
to that of the average woman (1.64 m). The density of blood is = 1060 kg/m3.
(a) How much work does the heart do in a day? (b) What is the heart‘s power
output in watts? (Answer: 127.772 kJ, 1.5 W).
81
5. How long will it take a 1250 Watts motor to lift a 400-kg piano to
a sixth-story window 15 m above? (Answer: 47 s)
6. The maximum height a typical human can jump from a crouched start
is about 50 cm. By how much does the gravitational potential energy increase
for a 80-kg person in such a jump? (Answer: 392 J)
7. A 150-g baseball is dropped from a tree 15 m above the ground. (a) With
what speed would it hit the ground if air resistance could be ignored? (b) If it
actually hits the ground with the 10 m/s, what is the average force of air
resistance exerted on it? (Answer: 17.2 m/s, 0.97 N)
8. A spring has a stiffness constant k of 80 N/m. How much must this
spring be compressed to store 40 J of potential energy? (Answer: 1 m)
TESTS
1. Work is:
a) scalar quantity
c) both scalar and vector
b) vector quantity
d) none of the above
2. Power of the body is given:
a) total capacity of doing work
c) product of work and time
b) rate of doing work
d) none of the above
3. Energy is defined as:
a) total capacity of doing work
c) product of work and time
b) rate of doing work
d) none of the above
4. An object is thrown vertically upwards. As it rises its total energy:
a) decreases
b) increases
c) remains constant
d) sometimes decreases, sometimes increases
5. A force 100 N is required to move a body with a velocity of 10 m/s.
The power developed is:
a) 50 watts
b) 1000 watts
c) 10 watts
d) 100 watts
6. A mass of 100 kg rests on a smooth horizontal surface. Energy needed to
accelerate it from rest to a velocity of 10 m/s is:
a) 5000 J
b) 500 J
c) 50000 J
d) 50 J
7. Potential energy cannot be expressed in:
a) J
b) N m
c) N s
d) W s
8. Two bodies of masses m1 and m2 have the same momenta. The ratio of
their kinetic energies is:
a) m1:m2
b) m2:m1
c)
d) m12:m22
9. When velocity of body is doubled, its kinetic energy:
a) doubled
b) remains the same
c) becomes 4 times
d) none of the above
82
10. Work can be:
a) positive only
b) negative only
c) both positive and negative
d) neither positive nor negative
11. A ship of mass 5 107 kg is acted upon a force of 20 104 N by an engine
which moves it through 5 m. If resistance of water is negligible, the speed
of the ship is:
a) 10 m/s
b) 1 m/s
c) 0.2 m/s
d) 5 m/s
12. When mass and velocity of a body is doubled its kinetic energy is
a) doubled
b) four times c) eight times d) sixteen times
13. Work done against friction is:
a) negative
b) positive
c) zero
d) none of the above
14. A body moves a distance of 20 m along straight path when a force of
5/
N acts on it. If work done is 50 J, at what angle with the direction of
motion the force acts?
a) 90°
b) 60°
c) 0°
d) 45°
5. MECHANICAL OSCILLATIONS AND WAVES
5.1. MECHANICAL OSCILLATIONS
Movements or changes of the state of some system, which periodically
(at regular intervals) repeat are called oscillations (or vibrations). If an object
vibrates or oscillates back and forth over the same path, each cycle taking
the same amount of time, the motion is called periodic.
5.1.1. CHARACTERISTICS OF OSCILLATIONS
Characteristics of oscillations are as follows displacement, amplitude,
period and frequency.
The distance x of the object from the equilibrium point at any instant of
time is called the displacement. It is measured in metres m.
The maximum displacement A (the greatest distance from the equilibrium
point) is called the amplitude. The units of amplitude are the same as for
displacement (m).
One cycle is the complete motion from some initial point back to the same
point. The period T is defined as the time required to complete one cycle.
It is measured in seconds (s).
83
The frequency (f or ) is the number of complete cycles per second.
Frequency and period are inversely related:
f
1
.
T
(5.1)
In the SI system frequency is measured in hertz (Hz), where 1 Hz = 1 cycle
per second (s–1).
5.1.2. SIMPLE HARMONIC MOTION
One of the most important types of periodic motion is simple harmonic
motion (SHM). In such motion the displacement x of a particle from its
equilibrium position is described by a sine (or cosine) function of time t:
x(t) = A sin υ = A sin(ωt + υ0)
(5.2)
or
x(t) = A cos υ = A cos(ωt + υ0),
where A is the amplitude, = t + 0 is the phase, 0 is the initial phase,
is the angular frequency.
The phase of oscillations characterizes the state of the vibrational system at
any instant of time.
The above equations are called the equations of simple harmonic
oscillations.
The angular frequency is related to the period T and the frequency of
the motion f by the equation:
=2 f=
2π
.
T
(5.3)
The SI unit of angular frequency is the radian per second (rad/s).
NOTE. SHM is the motion of a body under the influence of elastic or
a similar force that is proportional to the body’s displacement but has
the opposite direction.
Simple harmonic motion can be represented graphically as shown
in fig. 5.1.
A
Time (t)
T
Fig. 5.1. Simple harmonic vibration
84
5.1.3. EXAMPLES OF MECHANICAL OSCILLATIONS
The simple pendulum oscillations.
The simple pendulum consists of
a particle of mass m (called the bob of
the pendulum) suspended from one end
of an unstretchable, massless string of
length L that is fixed at the other end
(fig. 5.2). The bob is free to swing back
and forth in the plane of the page, to
the left and right of a vertical line
through the pendulum‘s pivot point. The
forces acting on the bob are the force
from the string T and the gravitational
force Fg, where the string makes an
angle with the vertical line (fig. 5.2). If
the bob swings through only small
Fig. 5.2. A simple pendulum
angles, its motion is approximately
simple harmonic motion. In other words, the restriction is that the angular
amplitude of the motion (the maximum angle of swing) must be small (θ < 5°).
The period T, the frequency f and the angular frequency of the simple
pendulum are defined as
T
2π
L
or f
g
1
2π
g
, ω
L
g
,
L
(5.4)
where L is the length of the pendulum string, g is the acceleration due to gravity.
Spring pendulum oscillations. Spring
pendulum consists of a body of mass m attached
to a horizontal (or vertical) spring as shown in
fig. 5.3. Such block-spring system is called
a linear simple harmonic oscillator (or linear
oscillator), where linear indicates that a restoring
elasticity force Fel is proportional to the
displacement x to the first power (Fel = –kx).
The body moves in simple harmonic motion
under the action of an elastic force Fel once it
has been either pulled or pushed away from
the equilibrium position and released.
The period T, the frequency f and
the angular frequency of the spring pendulum
are defined as
Fig. 5.3. A linear simple harmonic
m
1 k
k
oscillator
(5.5)
T 2π
, f
, ω
,
k
2π m
m
where m is the mass of a body, k is the spring stiffness.
85
Example 5.1. Equation of simple harmonic oscillations.
A body with a mass of 500 g is undergoing harmonic oscillations described
by the following equation x = 10 sin (5 t + π ), where t is in seconds and x in
6
meters. Find (a) the amplitude, (b) the frequency and the period, (c) the position
of a body at the initial instant of time t = 0, (d) the spring stiffness.
Solution. a) We use Eqs. 5.1–5.3, from which it follows that the amplitude
is A = 10 m, the angular frequency is = 5 .
b) The frequency of the oscillations is
the period is equal to T
1
f
f
ω
2π
5π
2π
2.5 Hz. Then
1
0.4 s.
2.5
π
6
c) The position of a body at t = 0 is x(0) = 10 sin ( ) = 5 m.
d) The spring stiffness is k =
2
m = 246.5 0.5 = 123.3 N/m.
Example 5.2. Equation of simple harmonic oscillations.
What is the equation describing the motion of a mass at the end of a spring
which position at the initial instant of time t = 0 is x0 = 5 cm, and whose period
is T = 0.314 s and amplitude is A = 10 cm?
Solution. We use Eq. 5.2: x(t) = A sin(ωt + φ0).
Angular frequency is given by
=
2π
T
2π
0.314
20 rad/s.
At the initial instant of time t = 0 the position of a body is x0 = 5 cm.
Amplitude is A = 10 cm. Then from Eq. (5.2) it follows that
x
5 1
sin 0 = 0
A0 10 2
or
0
=
π
.
6
The equation of simple harmonic oscillations is given by
x = 10 sin (20t +
π
) (cm).
6
5.2. MECHANICAL WAVES
Mechanical wave is the process of the vibrations‘ propagation in the elastic
medium. In other words waves are disturbances which propagate through
a medium. Particles form the elastic medium and their vibrations produce
the wave. Waves can be viewed as a transfer of energy and particles‘ vibrations
through the medium but not the straight-line movement of the particles.
86
The mechanical wave is called longitudinal if the direction of
the displacement of the medium particles is along the direction of the wave
propagation. In a transverse wave the direction of the displacement
of the medium particles is perpendicular to the direction of the wave‘s motion.
Longitudinal mechanical waves can propagate in different media (except of
vacuum), and transverse waves can propagate just in solid.
Waves have peaks and troughs. The height of a peak and depth of
a trough is called the amplitude of the wave (fig. 5.4).
y
A
x
Fig. 5.4. A schematic wavelength representation
The distance between two adjacent peaks (or troughs) is the same. This
distance which is a characteristic of the wave is called the wavelength . The
units are metres m. The wavelength is the distance between any two adjacent
points which are in phase. The wavelength is the distance that the wave
propagates in the medium at speed for the time equal to the period Т:
λ
T
.
(5.6)
f
Acoustic waves are mechanical longitudinal waves, which propagate in
the elastic medium and have frequencies from the lowest ones to 1012–1013 Hz.
Sound (audible) waves have frequencies about 16 Hz to 20 000 Hz. Sound
having frequencies above the range of human hearing is called ultrasound.
Waves having frequencies below 16 Hz are called infrasound.
Speed of acoustic waves is dependent on the properties of the medium
through which they propagate (e. a. temperature, elasticity, density).
In summary a sound is a mechanical wave that moves through a medium as
particles in the medium are displaced relative to each other. The speed of sound
is different in different materials; in general, it is slowest in gases ( = 340 m/s
in air), faster in liquids ( = 1450 m/s in water), and fastest in solids
( = 4900 m/s in iron).
87
Example. 5.3. Wavelength and frequency. (a) What is the range
of the wavelengths of audible sound in air? (b) What is the frequency of
ultrasound corresponding to the wavelength equal to 1 mm? Assume the speed
of sound in air equal to 340 m/s.
Solution. We use the Eq. 5.6 a) The range of audible frequencies is from
about 16 Hz to 20 000 Hz. Then from Eq. 5.6 we obtain
340 m/s
λ1
21.25 m
f1 16 Hz
340 m/s
λ2
17 mm.
f 2 20 000 Hz
The range of the wavelengths of audible sound in air is from 17 mm to 17 m.
b) The frequency of ultrasound in air corresponding to the wavelength
equal to 1 mm is
f
λ
340 m/s
0.34 MHz.
0.001 m
PROBLEMS
1. A particle is executing simple harmonic vibrations with the period
T = 12 s and the initial phase 0 = 0. How much time does it take to travel
a distance equal to half of its amplitude? (Answer: 1 s)
2. A pendulum is first vibrated on the surface of earth. Its period is T.
It is then taken to the surface of moon where acceleration due to gravity is 1/6th
that on earth. What is its period? (Answer: 6T )
3. A spring is loaded with mass m and the period of oscillations is T. It is
then cut into 4 equal parts. What is the period of oscillations of each part?
(Answer: T/2)
4. In diagnostic ultrasound imaging the speed of sound is assumed to be
1540 m/s in soft tissues. If the ultrasound wave has the wavelength of 0.1 mm,
what is its time period and frequency? (Answer: 0.065 s, 15.4 MHz)
TESTS
1. A metal sphere is suspended to a spring. It oscillates with frequency f.
It is then taken to the moon where the acceleration of gravity (g) becomes 1/6th
of its value on the earth. What is the frequency of oscillations?
a) 6f
b) 6 f
c) the same
d) f/6
2. A simple pendulum consists of a bob of a radius r and mass m and its
period is 2 sec. When its bob is replaced by a bob of a mass of 2 m but the same
radius r, the time period of motion is:
a) 4 sec
b) 2 sec
c) 1 sec
d) 8 sec
88
3. A mass m is suspended to a spring of length L and stiffness constant k.
The frequency of vibration is f1. The spring is cut into two equal parts and each
half is loaded with the same mass m. The new frequency f2 is given by:
f1
f
2 f1
a) f 2
b) f 2
c) f 2 1
d) f2 = 2f1
2
2
4. The equation of motion for a body executing simple harmonic vibrations
is given by x = 0.5 sin (10 t + 5). The frequency is given by:
a) 5 Hz
b) 1 Hz
c) 0.5 Hz
d) 5 Hz
5. For a body of mass m attached to the spring the stiffness constant (k) is
given by ( is the angular frequency):
m
a) 2
b) m 2
c) m2
d) m2 2
ω
6. Simple harmonic vibrations is given by x = A sin ( t + 0). When
a particle is at its positive extreme, its phase relative to equilibrium position is:
a)
b) /2
c) zero
d) 2
7. Simple harmonic vibrations is given by x = A sin ( t + 0). If
the body started oscillating from:
3
A c) х = A/2
a) x = 0
b) x
d) x = A/ 2
2
8. Sound waves can not be propagated through:
a) a gas
b) a liquid
c) vacuum
d) a solid
0
= /6,
9. A source of sound vibrates according to the equation x = 0.03 sin 100 t.
The speed of the wave is 1.5 km/s. What is the wavelength of the wave?
a) 60 m
b) 30 m
c) 15 m
d) 45 m
10. The sound generator dipped in sea is sending waves of wavelength
2.5 m and frequency 580 Hz. The speed of sound in sea water is:
a) 1250 m/s b) 1450 m/s
c) 1650 m/s
d) 1050 m/s
11. The speed of sound is the largest in:
a) air
b) water
c) vacuum
d) steel
12. Two persons cannot hear each other on the surface of moon because
the moon has:
a) craters
b) no atmosphere
c) rocks which absorb sound
d) dust suspended all around
13. The range of frequencies audible to human ear is:
a) 16 Hz to 20000 Hz
b) 16 Hz to 2000 Hz
c) 100 Hz to 10000 Hz
d) 16 kHz to 20000 kHz
89
6. STATICS
Statics is concerned with the analysis of loads (i. e. forces) acting on and
within structures that are in equilibrium. Equilibrium means a state of balance.
A body is in static equilibrium when it is at rest relative to a given frame of
reference.
In practice, many structural analyses in biomechanics are performed based
on the assumption of static equilibrium.
In this section we will concern with rigid solid bodies which do not bend,
stretch, or squash when forces act on them.
Let us consider a rigid solid body
(fig. 6.1) which can rotate about an axis that is
perpendicular to the plane of the figure and
passes this plane through point O. Two forces,
and
, act on the body in the plane of
the figure. Such a body is called a lever.
By definition a lever is a rigid solid body
(commonly a bar) free to rotate about a fixed
point called the fulcrum (point O).
The efficiency of the force to cause a rotation Fig. 6.1. Lever arms of the forces
about the fulcrum depends on the torque T
and
(or moment) of the force with respect to point
O. The torque of the force with respect to the point O is defined as
T = F d,
(6.1)
where F is the magnitude of the force, d is the perpendicular distance between
point O and the line of action of the force (i. e. the line along which the force
vector lies) (fig. 6.1).
The distance d is called the lever arm (or moment arm) of force with
respect to point O. The lever arms of the forces
and
in fig. 6.1 are
the distances d1 and d2, respectively. If the line of action of the force passes
through point O, so the lever arm for this force is zero and its torque with
respect to O is zero. Such a force can not cause a rotation of the lever. A force
of a given magnitude has the larger torque when it has the larger lever arm, this
is the principle of levers.
The force
in fig. 6.1 tends to cause counterclockwise rotation about O,
while
tends to cause clockwise rotation. Counterclockwise torques are
commonly considered to be positive and clockwise torques are negative, then
the magnitudes of the torques of the forces
and
about O are
T1 = F1 d1 T2 = –F2 d2.
(6.2)
90
When several torques act on a body, the net torque (or resultant torque Tnet)
is the sum of the individual torques:
n
Tnet
i 1
Ti .
(6.3)
The SI unit of torque is the newton-meter (N m).
6.1. CONDITIONS FOR EQUILIBRIUM
We learned that a particle is in equilibrium (i. e. the particle does not
accelerate) in an inertial frame of reference if the vector sum of all the forces
acting on the particle is zero. For an extended body, considered above,
the equivalent statement is that the center of mass of the body has zero
acceleration if the vector sum of all external forces acting on the body is equal
zero. This is called the first condition for equilibrium.
Fi
or
= 0,
0 (vector form)
= 0,
= 0 (vector projection form).
(6.4)
A second condition for an extended body to be in equilibrium is that
the body must have no tendency to rotate. This condition is based on
the dynamics of rotational motion in exactly the same way that the first
condition is based on Newton‘s first law. A rigid body in equilibrium can not
have any tendency to start rotating about any point, if the algebraic sum
of the torques due to all the external forces acting on the body with respect to
any specified point, is equal to zero. This is the second condition for
equilibrium:
n
i 1
Ti
0.
(6.5)
Levers. Levers are used to lift loads in an advantageous way and to transfer
movement from one point to another. Levers are classified regarding
the relationship between their components (i. e. fulcrum, load and applied
force). There are three classes of levers, as shown in fig. 6.2.
In a Class 1 lever the fulcrum is located between the applied force and
the load. A scissors and lifting a head off the chest are the examples of a Class 1
lever (fig. 6.2). In a Class 2 lever, the fulcrum is at one end of the bar; the force
is applied to the other end; and the load is situated in between. A wheelbarrow
and raising up onto the toes are examples of a Class 2 lever. A Class 3 lever has
the fulcrum at one end and the load at the other. The force is applied between
the two ends. Many of the limb movements, for example, flexion at the elbow
are performed by Class 3 levers.
91
Class 1
Class 2
Class 3
Fig. 6.2. Three classes of levers
According to the Eq. 6.5 condition for equilibrium for all three types of
levers is given by
T1 – T2 = 0 or F1 d1 – F2 d2 = 0.
(6.6)
From Eq. 6.6 it follows that
F1 d 2
d
or F2 F1 1 ,
(6.7)
F2 d1
d2
where d1 and d2 are the lengths of the lever arms as shown in fig. 6.2, F2
is the applied force required to balance a load force F1. If d1 is less than d2,
the applied force F2 is smaller than the load force F1.
The ratio of the load force to the applied force is called the mechanical
F1 d 2
advantage of the lever ( M
). In dependence on the distances from
F2 d1
the fulcrum, the mechanical advantage of a Class 1 lever can be greater or
smaller than one. By placing the load close to the fulcrum, with d1 much smaller
than d2, a very large mechanical advantage can be obtained with a Class 1 lever.
In a Class 2 lever, d1 is always smaller than d2, therefore, the mechanical
advantage of a Class 2 lever is greater than one. The situation is opposite
in a Class 3 lever. Here d1 is larger than d2, therefore, the mechanical advantage
is always less than one.
92
6.2. TYPES OF EQUILIBRIUM
There are three types of static equilibrium: stable, unstable and neutral.
Stable static equilibrium means that with small deviations of the body from
the equilibrium state, forces or moments of forces emerge which tend to return
the body to the state of equilibrium. A ball located at bottom of a spherical
deepening is in a state of stable equilibrium (fig. 6.3).
Fig. 6.3. Different types of equilibrium of the ball on a support:
1 — neutral; 2 — unstable; 3 — stable
Unstable static equilibrium means that, with a small deviation of the body
from the equilibrium state, forces emerge which tend to increase this deviation.
A ball located at the top of a sphere is an example of unstable equilibrium.
Neutral equilibrium means that, with a small deviation, the body remains in
equilibrium. An example is a ball lying on a flat horizontal surface.
Minimum potential energy is correspondent to the state of stable
equilibrium as compared to the adjacent states. Potential energy is larger in
unstable equilibrium state than in the adjacent states. In case of neutral
equilibrium potential energy is the same as in the adjacent states.
6.3. CENTER OF MASS AND CENTER OF GRAVITATION
When an object rotates, or several parts of a system of objects move
relative to one another, there is one point that moves in the same path that
a particle would move if subjected to the same net force. This point is called
the center of mass (abbreviated CM). The general motion of an extended object
(or system of objects) can be considered as the sum of the translational motion
of the CM, plus rotational, vibrational, or other types of motion about the CM.
For symmetrically shaped objects of uniform composition (such as spheres,
cylinders, and rectangular solids) the CM is located at the geometric center of
the object.
93
The center of gravity (CG) of an object is the point
at which the force of gravity can be considered to act
(fig. 6.4). The force of gravity actually acts on all
the different parts or particles of an object, but for
purposes of determining the translational motion of
an object as a whole, we can assume that the entire
weight of the object (which is the sum of the weights of
all its parts) acts at the CG.
Fig. 6.4. Determination
Commonly the center of gravity and the center of of the center of mass of
a flat uniform body
mass are at the same point.
Center of mass or center of gravity of an extended object can be easily
determined experimentally. If an object is suspended from any point, it will
swing (fig. 6.4) due to the force of gravity on it, unless it is placed so its CG lies
on a vertical line directly below the point from which it is suspended.
The algebraic sum of the torques due to the forces of gravity acting on all
particles of the body with respect to the center of gravity is equal to zero.
The subject of statics is important because it allows us to calculate certain
forces acting on (or within) a structure when some of the forces on it are already
known. Let us consider several examples.
Example 6.1. The first class lever.
The arms of a horizontal lever are d1 = 25 cm and d2 = 2 m long at opposite
sides of the fulcrum (fig. 6.2). The shorter arm is loaded with the weight of
500 N at the end. (a) What force should be applied at the lever longer arm to
balance the load? (b) What is the advantage of the lever?
Solution. We use Eq. 6.7. Setting, d1 = 25 cm = 0.25 m, d2 = 2 m, F1 =
500 N, we calculate F2:
d
0.25 m
F2 F1 1 500 N
62.5 N .
d2
2m
F1 d 2
2m
Advantage of the lever is M
8 times.
F2 d1 0.25 m
The force required to balance the load is 62.5 N.
Example 6.2. Balance of a horizontal beam.
A uniform beam of length L and mass m = 2 kg rests on two fulcrums as
shown in fig. 6.5. A uniform block of mass M = 3.5 kg is at rest on the beam at
a distance L/4 from its left end. Find the reactions forces exerted on the beam by
the right and left fulcrums.
Solution. A free-body diagram for the system consisting of the beam and
the block is shown in fig. 6.5. The forces acting on the system are as follows:
the force of gravity of the beam F g,beam mg and of the block F g,block M g
94
and the reaction forces
and
at the left and right ends of the beam,
respectively.
We apply two conditions for equilibrium (Eqs. 6.4–6.5) and solve this
problem in two equivalent ways.
From the Eq. 6.4 (balance of forces) we obtain:
mg M g Fl Fr 0.
Let us choose y axis to be positive in the upward direction (fig. 6.5). Then
for the projections of forces along the y axis we can write:
Fl + Fr – mg – Mg = 0 or Fl = mg + Mg – Fr.
Fig. 6.5. Example 6.2
Then we need to use the Eq. 6.5 (balance of torques). Let us choose
the axis O through the left end of the beam so that the torque of the force is
equal to zero. Then from the balance of torques equation we obtain:
Fr L mg
L
2
Mg
L
4
0.
Solving the above equation for Fr yields:
mg Mg (2m M )g (2 2 kg 3.5 kg) (9.8 m/s 2 )
Fr
18.4 N.
2
4
4
4
95
Let us find the remaining unknown force magnitude Fl from the balance of
forces equation:
Fl = mg + Mg – Fr = (m + M) g – Fr = (5.5 kg) (9.8 m/s2) – (18.4 N) = 35.5 N.
Second solution. This problem can be solved in a different way, applying
the balance of torques equation (Eq. 6.5) about two different axes. Choosing
first an axis through the left end of the beam, as we did above we find
the reaction force at the right end of the beam Fr = 18.4 N.
For an axis passing through the right end of the beam Eq. 6.5 yields
Fl L mg
L
2
Mg
3L
4
0.
Solving for Fl, we find
mg 3Mg (2m 3M )g (2 2 kg 3 3,5 kg) (9.8 m/s 2 )
Fl
35.5 N.
2
4
4
4
It is in agreement with the previous result. Note that the reaction force
magnitudes are not dependent on the beam length, but only on the mass
of the beam and the block.
Example 6.3. Balance of a horizontal beam.
A uniform horizontal rod of length L = 5 m rests on two fulcrums, located
at distances a = 0.8 m and b = 1.2 m from the left and right ends of the rod,
respectively (fig. 6.6). Find the ratio of the reaction forces N2 and N1 exerted on
the rod by the right and left fulcrums.
N1
N2
L
a
O
b
L/2
Fig. 6.6. Example 6.3
Solution. A free-body diagram for the rod is shown in fig. 6.6. The forces
acting on the rod are as follows: the force of gravity of the rod
and
the reaction forces
and . The problem can be solved most easily by using
the torque equation (Eq. 6.5) and by choosing the axis through the point O so
that the torque of the force
is equal to zero.
We calculate torques about the point O. Let d1 =
lever arms of the forces
the rod gives:
L
2
a and d2 =
L
2
b are
, respectively. Condition for equilibrium of
N1 d1 – N2 d2 = 0
96
or
L
L
a
N2
b .
2
2
Solving the above equation for N1/N2 we obtain:
N1
N2
N1
L
2
L
2
a
b
2.5m 0.8m
2.5m 1.2m
1.7m
1.33.
1.3m
Example 6.4. The third class lever. Force exerted by biceps muscle.
How much force must the biceps
muscle exert when a 5.0-kg ball is held
in the hand with the arm horizontal as in
fig. 6.7? The biceps muscle is connected to
the forearm by a tendon attached 5.0 cm
from the elbow joint. Assume that the
mass of forearm and hand together is 2.0
kg and their center of gravity (CG) is
situated as shown in fig 6.7.
Solution. The free-body diagram for
the forearm is shown in fig. 6.7. The forces
Fig. 6.7. Example 6.4
acting on the lever are as follows the force
of gravity of the arm
and the ball , the upward force
exerted by
the muscle, and a force exerted at the joint by the bone in the upper arm (all
assumed to act vertically). We wish to find the magnitude of
. The problem
can be solved most easily by using the torque equation (Eq. 6.5) and by
choosing the axis through the joint so that the torque of the force is equal to
zero.
We calculate torques about the point where Fj acts. Let dM, da db are lever
arms of the forces
and , respectively. Condition for equilibrium of
the lever gives:
FM dM – Fa da – Fb db = 0.
Solving the above equation for FM and setting Fa = ma g and Fb = mb g we
Fa d a Fb db (ma d a mb db )g
obtain:
FM
.
dM
dM
Setting dM = 5 cm = 0.05 m, da = 15 cm = 0.15 m, db = 35 cm = 0.35 m;
ma = 2 kg, mb = 5 kg we calculate the magnitude of the biceps muscle force FM:
(ma d a mb db )g (2 kg)(0.15 m) (5 kg)(0.35 m)
FM
9.8 m/s 2
dM
0.05 m
(41 kg) 9.8 m/s 2
402 N.
97
Note that the force required of the muscle (402 N) is quite large compared
to the weight of the object lifted (Fb = 49 N). Indeed, the muscles and joints of
the body are generally subjected to quite large forces. The position (distance dM)
at which the biceps muscle is connected to the forearm differs in some extent for
different humans. Even small increase in the lever arm of the force exerted by
the muscle the can lead to the noticeable increase in the ability of a human to lift
weights.
PROBLEMS
1. Cylindrical tube of small diameter with mass m = 2 103 kg is lying
on the ground. What the minimum force must be exerted on one of the tube ends
to lift it from the ground? (Answer: 9.8 kN).
2. A uniform beam of mass m = 140 kg is suspended horizontally by two
vertical ropes. Find the tension in each rope, if the distance from the center
of mass of the beam to the ropes are l1 = 3 m, l2 = 1 m, respectively. (Answer:
343 N, 1029 N).
3. A uniform horizontal beam of length L = 4 m rests on two fulcrums,
located at distances a = 0.8 m and b = 1.2 m from the left and right ends of
the beam, respectively. To balance the reaction forces, a load of mass m = 30 kg
is suspended at one of the beam ends. Find the mass of the beam. (Answer:
270 kg)
4. A uniform horizontal beam of mass m = 360 kg and length L = 5 m rests
on two fulcrums, located at distances a = 1.0 m and b = 1.5 m from the left and
right ends of the beam, respectively. A load of some mass is suspended at one of
the beam ends in order to provide the reaction forces exerted on the beam by
both fulcrums to be equal. Calculate the mass of the load. (Answer: 40 kg)
7. FLUID MECHANICS
In this Chapter we consider the materials that are very deformable and can
flow. Such ―fluids‖ include liquids and gases. We concentrate on the main
points of hydrostatics — the branch of fluid mechanics that is related to
the study of fluids at rest in equilibrium state. Hydrodynamics is the study of
fluids in motion.
The three common phases, or states, of matter are solid, liquid, and gas.
We can distinguish these three phases as follows. A solid maintains a fixed
shape and a fixed size. A liquid preserves its volume, but does not maintain
a fixed shape, it takes on the shape of its container. A gas has neither a fixed
shape nor a fixed volume, it will expand to fill its container. Since liquids and
gases do not maintain a fixed shape, they both have ability to flow; they are thus
often referred to fluids.
98
7.1. DENSITY AND PRESSURE
An important property of any material is its density ρ, defined as its mass
per unit volume. If a homogeneous material with a mass m has volume V,
the density is
ρ
m
.
V
(7.1)
Two objects made of the same material have the same density even though
they may have different masses and different volumes. Density is a scalar
quantity; its SI unit is the kilogram per cubic meter (kg/m3).
NOTE. The density of a gas varies considerably with pressure, but
the density of a liquid does not; i. e. liquids are not compressible.
S
Consider the force
acting perpendicular to
the surface with area S (normal force) (fig. 7.1).
Pressure p is defined as magnitude of the normal force
per unit area:
p
F
.
S
(7.2)
Fig. 7.1. A force acting
Pressure is a scalar. The SI unit of pressure
perpendicular to the surface
is pascal (Pa) (after the name of Blaise Pascal)
of area S
2
(1 Pa = N/m ).
The pascal is related to some other common (non-SI) pressure units as
follows:
1 atm = 1.013 105 Pa = 760 torr.
The atmosphere (abbreviated atm) is the approximate average pressure of
the atmosphere at sea level. The torr (named for Evangelista Torricelli, who
invented the mercury barometer) is the pressure exerted by a mercury column of
1 millimeter high (1 torr =1 mm Hg = 133.3 Pa).
7.2. PASCAL’S PRINCIPLE
Solids and fluids transmit forces differently. When a force is applied to one
section of a solid, this force is transmitted to the other parts of the solid
in the same direction. Because of a fluid‘s ability to flow, it transmits a force
uniformly in all directions. Therefore the additional pressure at any point
in a fluid at rest is the same in all directions. This principle is known as Pascal’s
principle (after the name of French scientist Blaise Pascal).
Pascal‘s principle states: the pressure applied to an enclosed
incompressible fluid is transmitted undiminished to every portion of the fluid
and to the walls of its container.
From Pascal‘s principle it follows that the pressure at a point in a fluid in
static equilibrium depends on the depth of that point but not on any horizontal
dimension of the fluid or its container.
99
Let us find quantitatively how the pressure in
S
a liquid of uniform density varies with depth. Consider
a point at a depth h below the surface of the liquid, as
shown in fig. 7.2. The pressure at this depth h is due
to the weight of the column of liquid above it. Thus
the force due to the weight of liquid acting on the area S
is defined as
p
F = mg = ( V)·g = S h g,
(7.3)
where V = S h is the volume of the column of liquid, is
Fig. 7.2. Hydrostatic
the density of the liquid, and g is the acceleration due to pressure at the depth h
gravity. The pressure p due to the weight of liquid is then
p
F
S
ρShg
S
ρ gh.
(7.4)
This pressure is called hydrostatic.
NOTE. The area S doesn’t affect the hydrostatic
pressure. The fluid pressure is directly proportional to
the density of the liquid and to the depth within
the liquid. In general, the pressure at equal depths
within a uniform liquid is the same. The shape of
the container does not matter (fig. 7.3).
In case of a liquid in an open container (e. g.
water in a glass) there is a free surface at the top
exposed to the atmosphere. Then the pressure at
a depth h in the fluid is
p = p0 + ρgh,
where p0 is the atmospheric pressure at the top surface.
Fig. 7.3. The pressure in
a fluid is the same for all
points at the same level
(7.5)
Example 7.1. Pressure. Pascal’s Principle.
Determine the total force and the pressure on the bottom of a swimming
pool with transverse sizes 25.0 m by 12.5 m and depth 2 m (the density of water
= 103 kg/m3). What will be the pressure on the wall of the pool near
the bottom?
Solution. For solving a problem we use Eqs. 7.2 and 7.4. The surface area
of the swimming pool bottom is S = a b, where a and b are the width and
the length of the pool.
Setting = 103 kg/m3, h = 2 m and g = 9.8 m/s2 in Eq. 7.4, we obtain
hydrostatic pressure on the bottom of the swimming pool
p = gh = (103 kg/m3) (9.8 m/s2) (2 m) = 1.96 104 Pa.
From Eq.7.2 it follows that the total force F acting on the pool bottom is
F = p S = p a b. Setting a = 12.5 m, b = 25.0 m, we obtain:
F = (19.6 103 Pa) (12.5 m) (25.0 m) = 6.125 106 N.
100
Pressure is p = 19.6 103 Pa, the force is F = 6.125 106 N.
According to Pascal‘s principle the pressure is the same for all points at
equal depth, thus the pressure on the wall of the pool near the bottom is
p = 19.6 103 Pa.
7.3. ARCHIMEDES’ PRINCIPLE AND BUOYANCY
7.3.1. ARCHIMEDES’ PRINCIPLE
Archimedes‘ principle states: when a body is completely or partially
submerged in a fluid, a buoyant force ( ) from the surrounding fluid acts on
the body (fig. 7.4). The force is directed upward and has a magnitude equal to
the weight of the fluid that has been displaced by the body1:
Fb = mf · g = ρf · Vf · g,
(7.6)
where mf is the mass of the fluid displaced by the body; f is the fluid density,
Vf is the submerged volume of a body, g is an acceleration due to gravity.
The buoyant force occurs because the pressure in a fluid increases with
depth.
Objects submerged in a fluid appear to weigh less than they do when
outside the fluid. For example, a large rock that you would have difficulty lifting
off the ground can often be easily lifted from the bottom of a stream. When
the rock breaks through the surface of the water, it suddenly seems to be much
heavier. Many objects, such as wood, float on the surface of water. These are
examples of buoyancy (fig. 7.4). In each example, the force of gravity is acting
downward. But in addition, an upward buoyant force is exerted by the liquid.
a
b
c
Fig. 7.4. Examples of buoyancy:
a — a body is partially submerged; b — a body is completely submerged; c — a body sinks
(V0 is the volume a body, Vf is the submerged volume of a body)
By ―fluid displaced‖, we mean a volume of fluid equal to the submerged volume of the object
(or that part of the object that is submerged).
101
1
The weight of a body on which a buoyant force acts is given by
the following expression:
W = mg – Fb = (m – f Vf) g.
(7.7)
The net downward force acting on a body of a mass m submerged in a fluid
is the difference between the gravitational force and the buoyant force:
F = Fg – Fb = mg – f Vf ·g = ρ·Vg – f Vf ·g,
(7.8)
where and f are the body and fluid density, correspondently; V is the body
volume and Vf is the fraction of a body volume which is submerged.
From Eq. 7.8 it follows that whether a body sinks or floats in liquid
depends on its density. If its density is greater than that of liquid ( > f), then
Fg > Fb and a body sinks. When the magnitude Fb of the (upward) buoyant force
acting on the body is equal to the magnitude Fg of the (downward) gravitational
force, a body floats in a fluid.
7.3.2. CONDITION FOR FLOTATION
Using Eq. 7.8 we obtain a condition at which a body floats at the surface of
a liquid:
mg = f Vf g or ρ·Vg = f Vf g.
(7.9)
From the Eq. 7.9 a condition for flotation of bodies can be expressed as
follows:
ρ Vf
(7.10)
.
ρf
V
Example 7.2. Buoyancy.
A wood block of density 800 kg/m3 floats in a liquid of density 1200
kg/m3. The block has height h = 6 cm. By what depth is the block submerged?
Solution. The block floats when the buoyant
force is equal to the gravitational force (fig. 7.5).
To find depth h we use Eq. 7.10. The volume of
the displaced liquid is V = h S, where S is the area
of the block surface submerged in the liquid.
From
Eq.
(7.10)
it
follows
ρ
ρf
Vf
hf S
hf
V
h S
h
.
Then the depth h by which the block
submerged is defined by the following equation:
hf
Setting
f
ρ
h.
ρf
= 1200 kg/m3,
Fig. 7.5. Example 7.2.
Forces acting on a wood
block submerged in a liquid
= 800 kg/m3, h = 6 cm = 0.06 m, we obtain hf:
102
hf
ρ
h
ρf
0.8 103 kg/m3
3
3
6 10
2
m
4 10
2
m
0.04 m.
1.2 10 kg/m
The depth by which the block is submerged is 4 cm.
PROBLEMS
1. Water stands 12.0 m deep in a storage tank whose top is open to
the atmosphere. What is the pressure at the bottom of the tank? (Answer:
2.2 105 Pa)
2. What is the difference in hydrostatic blood pressure (mm Hg) between
the top of the head and the bottom of the feet of 1.70 m tall person standing
vertically? (the density of blood blood = 1.06 103 kg/m3). (Answer: 133 mm Hg)
3. If the force on the tympanic membrane (eardrum) increases by about
1.5 N above the force from atmospheric pressure, the membrane can be
damaged. When a man goes scuba diving in the sea, below what depth could
damage to his eardrum start to occur? The eardrum is typically 8.2 mm
in diameter (the density of seawater = 1.03 103 kg/m3). (Answer: 2.8 m)
4. A manometer tube is partially filled with water. Both arms of the tube
are open to the air. Oil (which does not mix with water) is poured into the left
arm of the tube. Find the ratio between the heights of oil and water columns
in the left and right arms correspondently over the level of oil–water interface
( oil = 0.86 103 kg/m3) (Answer: 1.16).
5. A geologist finds that a Moon rock with an actual mass 9 kg has a mass
of 6 kg when completely submerged in water. What is the density of the rock?
(Answer: 3 103 kg/m3)
TESTS
1. SI units of pressure is
a) torr
b) atm
c) Pa
d) bar
2. Hydrostatic pressure depends on liquid
a) density
b) viscosity c) volume
d) none of the above
3. A body floats in liquid when
a) Fg > Fb
b) Fg < Fb
c) Fg = Fb
d) none of the above
4. What fraction of a piece of iron will be submerged when it floats
in mercury? ( iron = 7.8 103 kg/m3, mercury = 13.6 103 kg/m3)
a) 0.28
b) 0.42
c) 0.57
d) 0.65
103
8. FUNDAMENTALS OF KINETIC THEORY OF GASES
8.1. ASSUMPTIONS OF KINETIC MOLECULAR THEORY
OF GASES
The kinetic theory of gases is the study of the structure and properties of
substance which describes a gas as a large number of small particles (atoms or
molecules) in permanent random motion. There are following assumptions of
this model:
A gas consists of large number of small particles (atoms or molecules),
which are in permanent random motion. The random motion means that any
molecule can move in any direction with different speed at any moment.
The molecules undergo elastic collisions with each other. The interaction
forces between gas molecules are negligible except during a collision, because
the average separation between particles is great compared with their
dimensions.
The molecules undergo elastic collisions with the walls of their container
and produce a pressure on the walls.
This properties of gas molecules explain many phenomena, as well known
Brownian motion, diffusion and etc.
A diffusion is the movement of atoms and
molecules from a region of high concentration to
a region of low concentration. For example, some
particles are dissolved in a glass of water. At first,
the particles are all near one side of the glass. If the
particles all randomly move around — diffuse — in
Fig. 8.1. Diffusion
the water, then the particles will eventually become
distributed randomly and uniformly (fig. 8.1).
8.2. AMOUNT OF SUBSTANCE, MOLAR MASS
Amount of substance (chemical amount) n is a quantity that measures
the amount of ensemble of atoms, molecules or other particles. The SI unit for
amount of substance is the mole. The mole is defined as the amount of substance
that contains the Avogadro‘s number (NA = 6.02·1023) of its elementary particles
(atoms or molecules). Such amount of atoms is contained in 12 g of the isotope
carbon-12.
Amount of substance in moles can be determined as:
N
(8.1)
n
,
NA
where N is number of molecules in substance.
104
Molar mass M is a mass of 1 mole of a given substance, its physical
characteristic equal to the mass m of substance per its amount of substance n:
m
.
n
M
(8.2)
The SI unit for molar mass is kg/mol. However, for both practical and
historical reasons, molar masses are almost always measured in g/mol.
Molecular mass can be determined from the following relation:
M
(8.3)
m0
.
NA
8.3. IDEAL GAS. GAS PRESSURE
An ideal gas is a theoretical gas composed of many randomly moving
point particles that do not interact except when they collide elastically.
At normal conditions such as standard temperature and pressure, most real
gases such as hydrogen, helium, neon, nitrogen, oxygen behave qualitatively
like an ideal gas.
As gas particles are constantly moving, they are also constantly colliding
with the walls of their container and provide the forces pushing this walls
(fig. 8.2). The gas pressure is the sum of these forces F divided by the area S of
the container wall:
F
.
S
p
(8.4)
Fig. 8.2. Gas molecules colliding with container walls
The molecule-kinetic theory of ideal gas shows that the gas pressure on
the container walls is determined by formula:
1
cm
3 0
p
where c
2
,
(8.5)
N
is concentration of gas molecules (a number of molecules per unit
V
volume), m0 is mass of molecule,
the molecule.
105
2
is the mean square speed of
As an average kinetic energy of one molecule is E
m0
2
2
, the gas
pressure depends on the kinetic energy of gas molecules:
p
2
2 m0
c
3
2
2
= cE .
3
(8.6)
The equation (8.6) is a very important in the kinetic theory because it
relates pressure, a macroscopic property, to the average kinetic energy per
molecule which is a microscopic property. The SI unit of pressure is a pascal
(Pa) which is a newton per square meter (1 Pa = 1 N/m2). Other common units
of pressure are the atmosphere (atm) and millimeter of mercury (mmHg):
1 atm = 760 mmHg = 101325 Pa ≈ 105 Pa.
8.4. TEMPERATURE AS A MEASURE OF KINETIC ENERGY
OF MOLECULES
From the equation (8.6) we can find the average kinetic energy of gas
molecules as
E
3 pV
.
2N
(8.7)
The special experiments have shown that at any constant temperature
pV
the value
is the same for any gas and it is proportional to the gas
N
temperature. English physicist William Thomsom, 1st Baron Kelvin suggested
the absolute temperature scale T in which
pV
(8.8)
kT .
N
and k is called Boltzmann constant (k = 1.38 · 10–23 J/K).
Zero of Kelvin temperature scale defines the absolute zero, a hypothetical
temperature at which all molecular movement stops and gas pressure drops to
zero. All actual temperatures are above absolute zero.
The average kinetic energy of molecules of the monoatomic gas depends
on the absolute temperature according to the formula:
E
m0
2
2
3
= kT .
2
From (8.9) the mean square speed of gas molecules is:
3kT
3 RT
2
=
.
mss
m0
M
where R = NA·k = 8.31
(8.9)
(8.10)
J
is the constant called the universal gas
K mol
constant, M = m0NA is molar mass.
106
Therefore, ideal gas pressure in absolute temperature scale is:
p = ckT.
(8.11)
Usually we use a Celsius temperature scale (fig. 8.3). According to this
scale the temperature difference between the reference temperatures of
the freezing and boiling points of water is divided into 100 degrees. The freezing
point is taken as 0 Celsius degrees and the boiling point as 100 Celsius degrees.
Fig. 8.3. Relation between Celsius and Kelvin temperature scales
The intervals for the Celsius scale and for Kelvin scale are the same
1 K = 1 °C, but the absolute zero 0 K = –273.15 °C.
Thus, the conversion between these temperatures is:
TK ≈ TC + 273.
(8.12)
8.5. IDEAL GAS LAW
An ideal gas state can be characterized by three main values: pressure p,
volume V, and absolute temperature T. The Ideal Gas Law (the equation of state
for an ideal gas) is general relation between these variables at fixed quantity of
gas. As an ideal gas pressure (8.11) is p = ckT, where c
N nN A
=
, we can
V
V
write:
p
n
N kT
V A
n
RT ,
V
(8.13)
The most frequently introduced form of this equation is:
pV = nRT or pV =
m
RT ,
M
(8.14)
where m is mass, M is molar mass and n is amount of gas moles.
According to the equation of state, one of the thermodynamic variables p,
V or T may always be expressed as the function of the other two values.
107
Example 8.1.
A helium balloon, assumed to be a perfect sphere, has a radius of 18 cm.
At room temperature (20 °C), its internal pressure is 1.05 atm. Find the number
of moles of helium in the balloon and the mass of helium needed to inflate
the balloon to these values.
Solution:
V
4 3
πr
3
4
π(0.18 m)3 0.0244 m3.
3
P = 1.05 atm = 1.064 · 105 N/m2.
T = 20 °C = (20 + 273) K = 293 K.
R = 8.31 J/(mol·K).
Thus
n
(1.064 105 N/m 2 )(0.0244 m3 )
1.066 mol.
(8.31 J/mol K)(293 K)
PV
RT
The atomic mass of helium M = 4 g/mol, so gas mass is
m
n M
1.66 mol
4g
mol
4.26 g
4.26 10
3
kg.
8.6. ISOMETRIC PROCESSES
Isometric processes are thermodynamic processes during which the amount
of substance and one of the state variables — pressure, volume or
temperature — remain unchanged.
1. Isothermal process is carried out with a fixed amount of gas at constant
temperature. In this case, the product of an ideal gas pressure and volume is
always constant (Boyle’s law):
PV = const = nRT
or
p1V1 = p2V2,
(8.15)
where n = const, T = const.
A plot of p versus V at constant temperature for an ideal gas is a hyperbolic
curve called an isotherm (fig. 8.4). Each point on the curve represents the state
of the system at a given moment — that is, at pressure p and volume V. At a
lower temperature another isothermal process is represented by a lower curve
A′B′ (the product pV is less when T is less).
Fig. 8.4. Isotherms for an ideal gas at two different temperatures
108
2. At isobaric process the pressure of a fixed amount of gas is kept
constant. During the isobaric process the gas volume is proportional to gas
temperature:
nR
V
T const T ,
(8.16)
p
where n = const, p = const.
It is known as Charles’ law:
V V2
V
(8.17)
const or 1
.
T
T1 T2
This process is represented by a straight line on the volume-temperature
diagram called isobar and coming out of the origin of coordinates (fig. 8.5).
3. At isochoric process the gas volume does not change. The pressure of
gas of fixed mass and fixed volume is directly proportional to the absolute
temperature of gas. This law is expressed mathematically as:
p
nR
T
V
const T ,
(8.18)
where n = const, V = const, or as Gay-Lussac’s law:
p1 p2
(8.19)
.
T1 T2
On pressure-temperature diagram of the isochoric process is represented by
a straight line called isochor (fig. 8.6).
Fig. 8.5. VT-diagram of isobaric process
Fig. 8.6. pT-diagram of isochoric process
Example 8.2.
The gas pressure in cylinder is 4.40 kPa at 25°C. At what temperature in °C
will it reach a pressure of 6.50 kPa?
Solution. Since a cylinder volume is constant, V1 = V2, then
P1 P2
,
T1 T2
T P 298 K 6.50 Pa
T2 1 2
440 K 167 C.
P1
4.40 kPa
109
TESTS
1. According to the kinetic theory, the velocity of molecules increases
with the:
a) rise in temperature;
b) fall in temperature;
c) neither increases nor decreases.
2. According to the kinetic theory, the collision between the molecules of
gas are:
a) perfectly inelastic;
b) partially elastic;
c) perfectly elastic;
d) none of the above.
3. According to the gas kinetic theory the molecules:
a) repel each other;
b) collide with each other elastically;
c) move with uniform velocity;
d) are massless particles.
4. Brownian motion is:
a) discontinuous;
b) not random;
c) regular;
d) due to molecular collision.
5. According to kinetic theory of a gas the kinetic energy of a gas is:
a) proportional to the square root of its temperature;
b) independent of its absolute temperature;
c) proportional to its absolute temperature;
d) proportional to cube of absolute temperature.
6. Absolute zero is the temperature at which:
a) all molecular motion ceases;
b) some molecules are at rest;
c) none of the above.
7. The absolute zero is expressed as:
a) 273 ºC;
b) –273 ºC; c) –273 K;
d) 373 ºC.
8. A gas is at one atmosphere. To what pressure it should be subjected
at constant temperature so as to have 1/4th of its initial volume.
a) 1/4 atm;
b) 2 atm;
c) 3 atm;
d) 4 atm.
110
9. If the temperature of air is increased from 20 ºC to 200 ºC, the increase
in kinetic energy will be:
a) 1.6 times; b) 2 times;
c) 3.4 times; d) 10 times.
10. The temperature at which the kinetic energy of the gas will be half of
the kinetic energy at room temperature of 27 ºC is:
a) 13.5 ºC;
b) –27 ºC;
c) 123 ºC;
d) –123 ºC.
11. An ideal gas at 27ºC is heated at constant pressure so as to double its
volume. The temperature of the gas will be:
a) 54 ºC;
b) 327 ºC;
c) 450 ºC;
d) 600 ºC.
12. In which process PV-diagram is a straight line parallel to the volume
axis?
a) isothermal;
b) isochoric;
c) isobaric;
d) none of these.
13. Isothermal relation for 1g of a gas is:
a) PV = RT;
b) PV = constant;
c) P/T = constant;
d) V/T = constant.
14. An ideal gas is expanded isothermally, its temperature will:
a) increase;
b) decrease;
c) remain the same;
d) become zero.
15. Gas occupies 100 ml volume at 104 Pa pressure. If during isothermal
process the pressure is changed to 103 Pa the volume of the gas will be:
a) 10 ml;
b) 50 ml;
c) 200 ml;
d) 1000 ml.
PROBLEMS
16. What is the pressure inside 38 L container holding 105 kg of argon gas
at 20 °C? (Answer: 168 MPa)
17. If 61.5 L of oxygen at 18°C and an absolute pressure of 245 kPa are
compressed to 48.8 L and at the same time the temperature is raised to 56 °C,
what will the new pressure be? (Answer: 349 kPa)
111
9. THERMAL PHENOMENA.
BASICS OF THERMODYNAMICS
Thermodynamic is the theory of thermal phenomena which is not
considered atomic and molecular structure of bodies.
A thermodynamic system is the content of a macroscopic volume in space,
along with its walls and surroundings; it undergoes thermodynamic processes
according to the principles of thermodynamics. This system can be described by
thermodynamic variables such as temperature, internal energy and pressure.
9.1. INTERNAL ENERGY. WORK OF GAS. FIRST LAW
OF THERMODYNAMICS
The total sum of the energies of all molecules in an object is called its
internal energy. It includes total kinetic energy of the molecule‘s motion and
total potential energy of their interactions. Energy can be transferred between
a molecule system and its surroundings in two ways:
a) work W done by the system;
b) heat Q transfer, which occurs between the bodies with different
temperature.
The work produced by gas is equal to
W = Fgas · Δl·cos α = p·S·Δl·cos α = p·ΔV.
(9.1)
If the gas expands, ΔV ≡ V2 – V1 > 0 and gas work is positive W > 0.
F gas ) is negative in this case. Vice versa
The work of external force ( F ext
if the gas is compressed by the external force the ΔV ≡ V2 – V1 < 0 and gas work
is negative W < 0, but the work of external force is positive.
The heat Q is transferred from body of higher temperature to the body of
lower temperature. The transferred heat Q > 0 if thermal energy enters the body
and Q < 0 when this energy leaves it. The SI unit of heat quantity is Joule.
The First law of thermodynamics: the changing of the internal energy ΔU
of molecular system is due to heat Q transfer and to the work W done by
the system:
ΔU = Q – W.
(9.2)
In this formula Q > 0 when thermal energy enters the system and Q < 0
when this energy leaves the system.
The First law of thermodynamics (9.2) may be written in the form:
Q = ΔU + W.
(9.3)
The internal energy of n moles of an ideal monatomic (one atom per
molecule) gas is equal to
(9.4)
Thus, the internal energy of an ideal gas depends only on temperature and
the number of moles of gas.
112
The changing of internal energy ΔU of an ideal monatomic gas is equal to:
.
(9.5)
9.2. FIRST LAW OF THERMODYNAMICS AT DIFFERENT
PROCESSES
At isothermal process the T is constant and internal energy of gas doesn‘t
change: ΔT = 0 and ΔU = 0. First Law is:
Q = W.
That means that all heat energy transfers to a work.
At isobaric process gas pressure is fixed but the gas volume and
temperature may changes and First Law is:
Q = ΔU + W,
one part of transferred heat goes to doing a work by system and the another part
goes to increasing the internal energy of gas.
At this process W = p·ΔV = nR·ΔT.
For ideal monoatomic gas ΔU = 1.5·nR·ΔT, so at this case
Q = 2.5nR·ΔT.
At isochoric process gas volume is fixed and gas doesn‘t make a work:
W = p·ΔV = 0. So the First Law in this case is Q = ΔU, and for monoatomic gas
Q = ΔU = 1.5nR·ΔT.
All heat transfers to internal energy. It is the best process for body heating.
There is a process, when the heating is absent: Q = 0. This process is called
―adiabatic process‖. In this case First Thermodynamics Law has a form:
ΔU + W = 0 or ΔU = –W.
It means that the gas internal energy decreases (and gas temperature
decreases too) if gas adiabatically extends (W > 0) and vice versa, gas
temperature and internal energy increases if gas is adiabatically compressed
(W < 0).
Example 9.1.
2500 J of heat is added to a system, and 1800 J of work is done on the
system. What is the changing of internal energy of the system?
Solution. The heat added to the system is Q = 2500 J. The work W done by
the system is –1800 J. Why the minus sign? Because 1800 J done on the system
(as given) equals –1800 J done by the system. Hence:
ΔU = Q – W = 2500 J – (–1800 J) = 2500 J + 1800 J = 4300 J.
9.3. HEAT TRANSFER, TYPES OF HEAT TRANSFER
The transfer of heat normally occurs from a higher temperature object to
a lower temperature object. Heat transfer changes the internal energy of both
objects involved. There are three different ways to transfer energy from one part
113
of system to another: conduction, convection, and radiation. In practical
situations, any two or all three may be operating at the same time.
Conduction is heat transfer by means of molecular agitation within
a material without any motion of the material as a whole. If one end of a metal
rod is at a higher temperature, then energy will be transferred along the rod
toward the colder end. It happens due to the higher speed particles will collide
with the slower ones with a transfer of energy to them.
Convection is heat transfer by mass motion of a fluid such as air or water
when the heated fluid moves away from the source of heat, carrying energy with
it. Convection above a hot surface occurs because hot air expands, becomes less
dense, and rises. Hot water is likewise less dense than cold water and rises,
causing convection currents which transport energy.
Thermal radiation is energy transfer by the emission of electromagnetic
waves which carry out energy away from the emitting object. All life on Earth
depends on radiation from the Sun, which consists of visible light plus many
other wavelengths that the eye is not sensitive to, including much of the infrared
radiation.
9.4. AMOUNT OF HEAT. SPECIFIC HEAT
The amount of heat Q required to change the temperature of a given
material is proportional to the mass m of the material and to the temperature
change ΔT. It can be expressed by the equation:
Q = cmΔT,
(9.6)
where с is a quantity characteristic of the material which is called a specific
heat.
If ΔT = T2 – T1 > 0, the Q > 0 too (process of body heating), if
ΔT = T2 – T1 < 0 than Q < 0 (process of body cooling).
The specific heat c is the amount of heat per unit mass required to raise the
temperature by one Kelvin. Specific heat is essentially a measure of how thermally
insensitive a substance is to the addition of energy. The greater a material‘s
specific heat, the more energy must be added to a given mass of the material to
cause a given temperature change. The SI unit of specific heat is J/kg·K.
9.5. PHASE CHANGES
A phase change is the transformation of a thermodynamic system from one
phase or state of matter to another one by heat transfer. The term is most
commonly used to describe transitions between solid, liquid and gaseous states
of matter. The graph below (fig. 9.1) shows the matter temperature T changes
with the heat Q transfer.
114
Fig. 9.1. Change temperature of a substance during the heating
Melting is a physical process that results in the phase change from solid to
liquid. Inverse process, the transition from liquid to solid, is called
crystallization. The melting point is the temperature at which state of
a substance changes from solid to liquid. At the melting point the solid phase
and liquid one exist in a heat equilibrium. At the melting point, although
the substance is still being heated, there is a time when the temperature does not
change and the graph is horizontal. During this time, all the extra heat which is
being added goes to overcome the force of attraction between the particles
of the solid as it turns into a liquid.
The heat involved in a change of solid-liquid phase depends on the total
mass of the substance m as:
Q = λm,
(9.7)
where λ is the heat of fusion.
Heat of fusion is the heat required to change 1.0 kg of a substance from
the solid to the liquid state at melting temperature. The heat of fusion of water is
340 kJ/kg.
At any phase of the matter state the amount of heat Q required to change
the temperature of the material is expressed by the equation (9.6):
Qi = ci m·ΔT,
where сi is a specific heat of the corresponding state of a matter.
Vaporization is a phase transition from the liquid phase to vapour.
If conditions allow the formation of vapour bubbles within a liquid,
the vaporization process is called boiling. Boiling is a phase transition from
the liquid phase to gas phase that occurs at or above the boiling temperature.
Direct conversion from solid to vapour is called sublimation.
At the boiling point bubbles of vapour are formed within the whole liquid
volume. The boiling point of a liquid depends on the applied pressure; at sea
115
level water boils at 100 °C, at higher altitudes the temperature of the boiling
point is lower.
The heat involved in a liquid-vapor phase change at the boiling point is
determined by the formula:
Q = rm.
(9.8)
where r is the heat of vaporization. It is the heat required to change 1.0 kg
of a substance from the liquid state to the vapor phase at the boiling point.
For water it is 2260 kJ/kg.
9.6. THE HEAT BALANCE EQUATION
A closed system is an isolated system if no energy in any form passes
across its boundaries. Because the total energy of the system cannot change,
the heat lost by one part of the system is equal to the heat gained by the other
part and sum of all these heat transfers is equal to zero:
Q1 + Q2 + … + Qn = 0.
(9.9)
It is the heat balance equation for isolated system.
For not isolated system the heat balance equation is
Q1 + Q2 + … + Qn = Qext,
(9.10)
where Qext is an external heat transferred to system.
Example 9.2.
How much heat input is needed to raise the temperature of an 10-kg vat
made of iron filled with 20 kg of water from 10 °C to 90 °C?
Solution. The specific heat of iron is c1 = 450 J/kg·K°.
j
(90 10) K 360 kJ.
Thus: Q1 c1m1 T 10 kg 450
kg K
The specific heat of water is cw = 4200 J/kg·K°.
j
Qw cwmw T 20 kg 4200
(90 10) K 6720 kJ.
kg K
Q Q1 Qw 360 kJ 6720 kJ 7080 kJ.
Example 9.3.
A 0,5-kg chunk of ice at –10°C is placed in 3 kg of water at 20 °C. At what
temperature and in what phase will the final mixture be? Ignore any heat flow to
the surroundings, including the container.
Solution. First, check to see if the final state will be all ice, a mixture of ice
and water at 0 °C, or all water.
To bring the 3 kg of water at 20 °C down to 0 °C would require an energy
release of:
Qw = mwcw(20 °C – 0 °C) = 3 kg · 4200 J/kg · C° · 20 °C = 252 kJ.
116
On the other hand, to raise the ice temperature from –10 °C to 0 °C would
require:
Qice = micecice(0 °C – (–10 °C)) = 0.5 kg · 2100 J/kg · C° · 10 °C = 10.5 kJ.
To change the ice to water at 0°C would require:
For a total:
Qice + QF = Qw;
10.5 kJ + 167 kJ = 177 kJ.
This is not enough energy to cool the 3 kg of water from 20 °C to 0 °C, so
that mixture will stay water, somewhere between 0 °C and 20 °C.
To determine the final temperature T:
micecice(0 °C – (–10 °C)) + λmice + micecw(T – 0 °C) = mwcw(20 °C – T)
or
Qice + QF + T(micecice + mwcw) = mwcw · 20 °C.
Solving for T we obtain: T = 5 °C.
9.7. THERMAL EXPANSION. COEFFICIENT OF LINEAR
AND VOLUME EXPANSION
Most substances expand when heated and contract when cooled. However,
the amount of expansion or contraction varies, depending on the material.
The change in length Δl is directly proportional to the change in temperature ΔT
and the original length of the object l0:
Δl = αl0ΔT,
(9.11)
where α is the proportionally constant, which is called the coefficient of linear
expansion for the particular material and has units of K–1.
We can write this proportionality in another form:
l = l0(1 + αΔT).
(9.12),
The change in volume of a material that undergoes a temperature change is
given by a similar relation:
ΔV = βV0ΔT,
(9.13),
where ΔT is the change in temperature, V0 is the original volume and β is
the coefficient of volume expansion. The units of β are K–1.
Example 9.4.
The steel bed of a suspension bridge is 200 m long at 20 °C. If the extremes
of temperature to which it might be exposed are –30 °C to +40 °C, how much
will it contract and expand?
Solution. For steel α = 12 · 10–6 (K)–1.
When the temperature increases to 40 °C:
Δl1 = αl0ΔT = 12 · 10–6 (K)–1 · 200 m · (40 °C – 20 °C) = 4.6 · 10–2 m.
When the temperature decreases to –30 °C:
Δl2 = αl0ΔT = 12 · 10–6 (K)–1 · 200 m · (–30 °C – 20 °C) = –12 · 10–2 m.
The total range the expansion is: Δl = 4.6 · 10–2 m + 12 · 10–2 m ≈ 17 cm.
117
TESTS
1. The first law of thermodynamics is concerned with the conservation of:
a) number of molecules;
b) energy;
c) number of moles;
d) temperature.
2. The addition of heat to a system appears as:
a) only increase in internal energy;
b) partly increase in internal energy and partly work done by the system;
c) only work done by the system;
d) all of these;
e) none of these.
3. In the equation for first law of thermodynamic ΔU = Q + W, the ΔU
represents:
a) change in internal energy;
b) change in external energy;
c) both of them;
d) none of these.
4. The internal energy of a ideal gas does not change during:
a) isothermal process;
b) isobaric process;
c) isochoric process;
d) none of these.
PROBLEMS
1. A steel railroad track has a length of 30 m when the temperature is 0 °C.
What is its length when the temperature is 40 °C? The coefficient of linear
expansion for steel is equal to 12·10–6 K–1. (Answer: 30,013 m)
2. An aluminum sphere is 8.75 cm in diameter. What will be its change in
volume if it is heated from 30 °C to 180 °C? The coefficient of volume
expansion for aluminum is equal to 75·10–6 K–1. (Answer: 3,9 cm3)
3. How much external work can be done by a gas when it expands from
0.003 m3 to 0.04 m3 in volume under a constant pressure of 400 kPa? (Answer:
14800 J)
4. An engineer wishes to determine the specific heat of a new metal alloy.
A 0,15-kg sample of the alloy is heated to 540 °C. It is then quickly placed in
0.4 kg of water at 10 °C, which is contained in a 0.2-kg aluminum calorimeter
cup. The final temperature of the system is 30.5 °C. Calculate the specific heat
of the alloy. (Answer: 497 J/kg·°C )
5. To what temperature will 8700 J of heat raise 3 kg of water that is
initially at 10 °C? (Answer: 10,7°C)
6. When a 290-g piece of iron at 180 °C is placed in a 95-g aluminum
calorimeter cup containing 250 g of glycerin at 10 °C, the final temperature
is observed to be 38 °C. Estimate the specific heat of glycerin. (Answer:
2,3·103 J/kg·°C)
7. How much heat must be absorbed by ice of mass m = 720 g at –10 °C to
take it to the liquid state at 15 °C? (Answer: 300 kJ)
118
10. ELECTRICITY
10.1. ELECTRIC CHARGE
There are two types of observed electric charge, which are designated as
positive and negative. The symbol for charge is ―q‖. The SI unit of charge is
coulomb (C): 1 C = 1 A·1 sec.
The smallest charge found in nature is the charge of proton, it is given
the symbol ―e ― and is often referred to as the elementary charge: e = 1,6·10–19 C.
The charge on the electron is ―–e‖. Electric charge is thus said to be quantized
(existing only in discrete amounts: 1e, 2e, 3e, etc.). Any net charge q (negative
or positive) can be determined as: q = n · e, and n = 1, 2, 3…
10.2. LAW OF CONSERVATION OF ELECTRIC CHARGE
The law of conservation of electric charge makes sense for the isolated
system that does not interact with or receive charge from other systems.
In an isolated system, the total electric charge of the system is equal to
the algebraic sum of all electric charges q1, q2, … qi located in the system:
q1, q2, … qi = const.
The law states that the total electric charge in an isolated system always
remains constant, regardless of other possible changes within the system.
10.3. COULOMB’S LAW
When the electric charges have likely signs there is a repulsive force
between them and, opposite, when the charges are unlikely, there is attractive
force between them. The force between two charged small spheres was studied
by Coulomb. Coulomb‘s Law states that the electrostatic force F in vacuum
between two point charges q1 and q2 is directly proportional to the product
of the magnitude of the charges and inversely proportional to the square
of the distance r between them:
qq
F k 1 22 ,
(10.1)
r
2
9 N m
where k 9 10
is the electrostatic constant.
C2
The direction of forces is always along the line joining the two point
charges, and it is attractive if the charges are opposite and repulsive
if the charges are like.
This formula (10.1) only applies to point charges (spatial size negligible
compared to other distances) or spherically charges when they are at rest.
119
The electrostatic constant k in equation (10.1) is often written in terms
of another constant, ε0, called the permittivity of free space (i. e. vacuum).
It is related to k by: k = 1/4πε0. So in air or vacuum Coulomb‘s law can then be
1 q1q2
written as
F
,
4πε0 r 2
where
ε0
1
4πk
8.85 10
12
C2
N m2
.
(10.2)
If the charges are situated in a medium of permittivity relative ε, then
the magnitude of the interaction force Fm between them will be less:
1 q1q2
(10.3)
Fm
.
4πε 0 εr 2
Dividing equation (10.2) by (10.3) one can obtain:
F
F
.
(10.4)
ε 1 or Fm
ε
Fm
The value εa = ε·ε0 is the absolute permittivity of the medium.
Coulomb‘s law describes the electrostatics force between two charges at rest.
Example 10.1. Electric force on electron by proton.
Determine the magnitude of the electric force on the electron of a hydrogen
atom exerted by the single proton (q2 = +e) that is its nucleus. Assume
the electron ―orbits‖ the proton at its average distance of r = 0.53·10–10 m.
qq
Solution. We use Coulomb‘s law F k 1 22 with r = 0.53·10–10 m and
r
q1 = q2 = 1.6 · 10–19 C (ignoring the signs of the charges):
(9.0 109 N m 2 /C2 )(1.6 10 19 C)(1.6 10 19 C)
F
8.2 10 8 N.
10
2
(0.53 10 m)
The direction of the force on the electron is toward the proton, since
the charges have opposite signs and the force is attractive.
Example 10.2. Which charge exerts the greater force?
Two positive point charges, q1 = 50 μC and q2 = 1 μC, are separated by
a distance l. Which is larger in magnitude, the force that q1 exerts on q2, or
the force that q2 exerts on q1?
Fig. 10.1. Example 10.2
Solution. From Coulomb‘s law, the force on q1 exerted by q2 is
qq
F12 k 12 2 .
l
120
qq
The force on q2 exerted by q1 is F21 k 22 1 , which is the same magnitude.
l
The equation is symmetric with respect to the two charges, so F21 = F12.
NOTE. Newton’s third law also tells us that these two forces must have
equal magnitude.
10.4. THE ELECTRIC FIELD. THE ELECTRIC FIELD STRENGTH
Electric field is said to exist in the region of space around a charged object:
the source of electric field is a charge. The presence of an electric field may be
detected with another charge. When a positive test charge q0 is placed near
a charge q, which is the source of electric field, an electrostatic force F will act
on the test charge (fig. 10.2) and F ~ q0.
Fig. 10.2. A test charge q0 is placed near a charge q, which is the source of electric field,
an electrostatic force F acts on the test charge
But the ratio
E
F
q0
(10.5)
does not depend on the test charge q0 and is called the electric field strength or
electric field intensity (or electric field).
The electric field strength E is a vector whose direction is the direction
of the force acting on a positive test charge q0 placed at the point, and whose
magnitude is the force per unit charge.
Thus E has SI units of Newtons per Coulomb (N/C). If q0 is positive, F and
E will point in the same direction. If q0 is negative, F and E point in opposite
direction (fig. 10.3).
Fig. 10.3. If q0 is positive, F and E will point in the same direction. If q0 is negative, F and E
point in opposite direction
121
The electric field strength E at a distance r from single point charge q can
be written as:
qq
F
q
Е
k 0 2 k
.
(10.6)
q0
q0 εr
ε r2
If the electric field strength E is due to more than one charge (q1, q2, q3, …
qn), the individual electric field strengths (call them E1, E2, … En) due to each
charge are added as vectors to get the total electric field strength E at any point
of field (fig. 10.4). It is the principle of superposition:
E E1 E2 E3 ... En .
(10.7)
Fig. 10.4. At any point P, the total electric field strength due to the charges q1 and q2 equals
the vector sum of electric field strengths of the charges: E = E1 + E2. The direction of
the individual electric field strengths is the direction of the force on a positive test charge
Example 10.3. E at a point between two charges.
Two point charges are separated by a distance of 10.0 cm. One has a charge
of –25 μC and the other +50 μC. (a) Determine the direction and magnitude
of the electric field at a point P between the two charges that is 2.0 cm from
the negative charge (fig. 10.5, a). (b) If an electron (mass = 9.11·10–31 kg) is
placed at rest at P and then released, what will be its initial acceleration
(direction and magnitude)?
Fig. 10.5. Example 10.3. In (b) we don‘t know the relative lengths of E1 and E2 until we do
the calculation
122
Solution. The electric field at P will be the vector sum of the fields created
separately by q1 and q2. The field due to the negative charge q1 points toward q1,
and the field due to the positive charge q2 points away from q2. Thus both fields
point to the left as shown in Figure b and we can add the magnitudes of the two
fields together algebraically, ignoring the signs of the charges. In fig. 10.5, b we
use Newton‘s second law (F = ma) to determine the acceleration, where F = qE.
a) Each field is due to a point charge as given by E = kq/r2. The total field is
E
q
k 12
r1
q
k 22
r2
q
k ( 12
r1
q2
25 10 6 C
9
2 2
) (9.0 10 N m /C )(
r2 2
(2.0 10 2 m)2
50 10 6 C
(8.0 10 2 m) 2
)
6.3 108 N/C.
b) The electric field points to the left, so the electron will feel a force to
the right since it is negatively charged. Therefore the acceleration a = F/m
(Newton‘s second law) will be to the right. The force on a charge q in an electric
field E is F = qE. Hence the magnitude of the acceleration is
F qE (1.60 10 19 C)(6.3 108 N/C)
a
1.1 1020 m/s 2 .
31
m m
9.11 10 kg
NOTE. By carefully considering the directions of each field (E1 and E2)
before doing any calculations, we made sure our calculation could be done
simply and correctly.
Example 10.4. E above two point charges.
Calculate the total electric field (a) at point A and (b) at point B in fig. 10.6
due to both charges, q1 and q2.
Fig. 10.6. Calculation of the electric field at points A and B for Example 10.4
123
Solution. The calculation is much like that of Example 9.5, except now we
are dealing with electric fields instead of force. The electric field at point A
is the vector sum of the fields EA1 due to q1, and EA2 due to q2. We find
the magnitude of the field produced by each point charge, then we add their
components to find the total field at point A. We do the same for point B.
a) The magnitude of the electric field produced at point A by each
of the charges q1 and q2 is given by E = kq/r2, so
q (9.0 109 N m 2 /C2 )(50 10 6 C)
E A1 k 2
1.25 106 N/C,
2
r
(0.60 m)
E A2
k
q
(9.0 109 N m 2 /C2 )(50 10 6 C)
2
2
5.0 106 N/C.
r
(0.30 m)
from A away from q2, as shown; so the total electric field at A, EA has
components
E Ax E A1cos300 1.1 106 N/C,
E Ay
E A2sin300
4.4 106 N/C.
Thus the magnitude of EA is
EA
(1.1)2
(4.4)2 106 N/C 4.5 106 N/C,
and its direction is ϕ given by tan ϕ = EAy /EAx = 4.4/1.1 = 4.0, so ϕ = 76°.
b) Because B is equidistant from the two equal charges (40 cm by
the Pythagorean theorem), the magnitudes of EB1 and EB2 are the same; that is,
q (9.0 109 N m 2 /C2 )(50 10 6 C)
EB1 EB 2 k 2
2.8 106 N/C.
2
r
(0.40 m)
Also, because of the symmetry, the y components are equal and opposite,
and so cancel out. Hence the total field EB is horizontal and equals
EB1cosθ + EB2cosθ = 2EB1cosθ.
From the diagram, cosθ = 26 cm/40 cm = 0.65. Then
EB
2 EB1cosθ
2(2.8 106 N/C)(0.65) 3.6 106 N/C,
and the direction of EB is along the +x direction.
NOTE. We could have done part (b) in the same way we did part (a). But
symmetry allowed us to solve the problem with less effort.
Since the electric field strength is a vector, it is sometimes referred to as
vector field. Lines of electric field strength are a convenient way of visualizing
the electric field. Lines of electric field strength indicate the direction
of the force due to the given field on a positive test charge. For a positive point
charge, the electric field strength lines are directed radially outward from
the charge (fig. 10.7). For a negative point charge they point radially inward
toward the charge because that is the direction the force would be on a positive
124
test charge in each case. Since the electric field strength is the electric force per
unit charge, the electric field strength lines are sometimes called lines of force.
Fig. 10.7. The electric field strength lines near a single positive point charge and negative one.
Electric field lines extend away from positive charge (where they originate) and toward
negative charge (where they terminate)
The number of lines starting on a positive charge, or ending on a negative
charge, is proportional to the magnitude of the charge. Notice that near
the charge, where the electric field strength is greatest, the lines are closer
together. This is a general property of electric field strength lines: the closer
the lines are together, the stronger the electric field strength in that region.
The uniform electric field is electric field where the vector E is constant
everywhere in magnitude and direction. Thus electric fields are drown with
parallel, equally spaced electric field strength lines.
Example 10.5. Electron accelerated by electric field.
An electron (mass m = 9.1·10–31 kg) is accelerated in the uniform field E
(E = 2.0·104 N/C) between two parallel charged plates. The separation of
the plates is 1.5 cm. The electron is accelerated from rest near the negative plate
and passes through a tiny hole in the positive plate. (a) With what speed does it
leave the hole? (b) Show that the gravitational force can be ignored. Assume
the hole is so small that it does not affect the uniform field between the plates.
Solution. We can obtain the electron‘s velocity using the kinematic
equations, after first finding its acceleration from Newton‘s second law, F = ma.
The magnitude of the force on the electron is F = qE and is directed to the right.
a) The magnitude of the electron‘s acceleration is
mg
(9.1 10
31
kg)(9.8 m/s 2 ) 8.9 10
30
N.
Between the plates E is uniform so the electron undergoes uniformly
accelerated motion with acceleration
(1.6 10 19 C)(2.0 104 N/C)
a
3.5 1015 m/s 2 .
31
9.1 10 kg
125
It travels a distance x = 1.5·10–2 m before reaching the hole, and since its
initial speed was zero, we can use the kinematic equation, υ 2 = υ02 + 2ax, with
υ0 = 0:
2ax
2(3.5 1015 m/s 2 )(1.5 10 2 m) 1.0 107 m/s.
There is no electric field outside the plates, so after passing through
the hole, the electron moves with this speed, which is now constant.
b) The magnitude of the electric force on the electron is
qE
(1.6 10
19
C)(2.0 104 N/C)
3.2 10
15
N.
The gravitational force is mg (9.1 10 31 kg)(9.8 m/s 2 ) 8.9 10 30 N
which is 1014 times smaller! Note that the electric field due to the electron does
not enter the problem (since a particle cannot exert a force on itself).
PROBLEMS
1. Calculate the magnitude of the force between two 2.50 C point charges
3.0 m apart in air. (Answer: 6.25·10+9 N)
2. The force between two charges in free space is 5 N. What will the force
between them be if they are in a medium of relative permittivity 2? (Answer:
2.5 N)
3. Calculate the magnitude of the electric force between an iron nucleus
(q = +26e) and its innermost electron if the distance between them is 1.5·10–12 m.
(Answer: 2.66·10–3 N)
4. Particles of charge +88, –55, and +70 μC are placed in a line fig. 10.8.
The center one is 0.75 m from each of the others. What are the net force on each
charge due to the other two? (Answer: 52.8 N, 15.84 N, 36.9 N)
Fig. 10.8
5. Find the magnitude and direction of the electric field strength at
points A and B in fig. 10.9 due to the two positive charges (q = 7 μC). (Answer:
4.5·10+6 N/C)
Fig. 10.9
6. How many electrons make up a charge of 100 μC? (Answer: 6.25·10+14)
7. Four equal point charges of +3 μC are placed at the four corners of
a square that is 40 cm on a side. Find the force on an one of the charges?
(Answer: 0.97 N)
126
8. What must the charge (sign and magnitude) of a particle of mass 5g be
for it to remain stationary when placed in a downward-directed electric field of
magnitude 800 N/C? (Answer: –0.0000613 C)
9. Two charges of +1μC and –1μC are placed at the corners of the base of
an equilateral triangle. The length of a side of the triangle is 0.7 m. Find
the electric field strength at the apex of the triangle. (Answer: 18.4 kN/C)
TESTS
1. The conservation of electric charge implies that:
a) charge can‘t be created;
b) charge can‘t be destroyed;
c) the number of charged particles in the universe is constant;
d) simultaneous creation of equal and opposite charges is permissible.
2. Two charges are placed at a certain distance apart. If a dielectric slab is
placed between them, what happens to the force between the charges?
a) decreases;
b) increases;
c) remains unchanged;
d) may increase or decrease depending on the nature of the dielectric.
3. Coulomb‘s law is given by F = kq1q2rn, where n is:
a) 1/2; b) –2;
c) 2;
d) –1/2.
10.5. ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE
Any charged body has an electric potential energy Wpot in electric field,
which is directly proportional to the magnitude of the charge q0.: Wpot ~ q0. But
the ratio Wpot/q0 doesn‘t depend on the charge magnitude placed in the electric
field. The ratio Wpot/q0 is the energy characteristic of the electric field and is
Wpot
υ
.
called electric potential υ:
(10.8)
q0
Electric potential υ is the potential energy per unit charge at a point in an
electric field. Electric potential υ is a scalar characteristic of an electric field.
Unit of electric potential is Volt (V) (1 Volt = 1 Joule per Coulomb (J/C)).
The electric potential υ at a distance r from a single point charge is
q
q
υ k
.
(10.9)
ε r 4πεε0r
The potential from a collection of n charges is the algebraic sum
of the potential due to each charge separately (this is much easier to calculate
than the net electric field strength, which would be a vector sum). Potential due
to a group of point charges q1, q2, q3,… qn can be found as:
υ = υ1 + υ2 + υ3 + … + υn.
(10.10)
127
Example 10.6. Potential due to the two charges.
Calculate the electric potential (a)
at point A in fig. 10.10 due to the two
charges shown, and (b) at point B.
Solution. The total potential at
point A (or at point B) is the sum of
the potentials at that point due to each of
the two charges q1 and q2. The potential
due to each single charge is given by
q
φ k .
r
Fig. 10.10. Example 10.6
We do not have to worry about directions because electric potential is
a scalar quantity. But we do have to keep track of the signs of charges.
a) We add the potentials at point A due to each charge q1 and q2:
φA
φ A2 φ A1 k
q2
r2 A
k
q1
.
r1A
where r1A = 60 cm and r2A = 30 cm. Then
(9.0 109 N m 2 /C2 )(5.0 10 5 C) (9.0 109 N m 2 /C)( 5.0 10 5 C)
φA
0.30 m
0.60 m
1.50 106V 0.75 106V 7.5 105V .
b) At point B, r1B = r2B = 0.40 m, so
(9.0 109 N m 2 /C 2 )(5.0 10
φ B φ B 2 φ B1
0.40 m
(9.0 109 N m 2 /C)( 5.0 10
0.40 m
5
C)
5
C)
0V .
NOTE. The two terms in the sum in (b) cancel for any point equidistant
from q1 and q2 (r1B = r2B). Thus the potential will be zero everywhere on
the plane equidistant between the two opposite charges. This plane where φ is
constant is called an equipotential surface.
If all the points of a surface are at the same electric potential, then
the surface is called an equipotential surface. In case of an isolated point charge,
all points equidistant from the charge are at same potential. Thus, equipotential
surfaces in this case will be a series of concentric spheres with the point charge
as their centre (fig. 10.11). The potential will however be different for different
spheres.
Work A done by electric field in moving a unit positive charge is equal to
potential energy difference:
А = Wpot1 – Wpot2 = q0 (φ1 – φ2) = q0U.
(10.11)
128
The potential difference U between point 1 and point 2 is:
A
U φ1 φ 2
.
q0
(10.12)
Fig. 10.11. The equipotential lines and electric field lines
The potential difference U between two points in an electric field is defined
as the amount of work A done in moving a unit positive charge q0 from one
point to the other. The unit of potential difference U is Volt (1 V = 1 J/1 C). The
electric potential υ of an electric field at infinity from charge is equal to zero.
Suppose a positive test charge q0 is moved from point 1 to point 2
in a uniform electric field between the two charged plates (fig. 10.12).
Fig. 10.12. The charge q0 is moved from point 1 to point 2 in a uniform electric field between
the two charged plates
The work A done by an electric force F or ―field‖ in moving a positive test
charge +q0 along the electric field line at a distance d = x2 – x1 is:
A F d q0 E d.
(10.13)
129
Thus, in uniform electric field the relation between potential difference U
and electric field strength E is the following:
U = E·d,
(10.14)
where d is the distance, parallel to the field lines, between two points.
The equation (10.14) shows that the unit for electric field strength E is volt
per meter. Thus, the following relation among units is valid: 1 N/C = 1 V/m.
The electric force is found to be a conservative force. Therefore, the work
A done by the electrostatic force does not depend upon the path chosen to move
charge from point 1 to point 2, but it is determined by the potential difference U
between point 1 and point 2.
Example 10.7. Work required to bring two positive charges close
together.
What minimum work must be done by an external force to bring a charge
q0 = 3.00 μC from a great distance away (take r = ∞) to a point 0.500 m from
a charge q = 20.0 μC?
Solution. To find the work we can not simply multiply the force times
distance because the force is not constant. Instead we can set the change in
potential energy equal to the work required of an external force, and equation
А = –(Wpot1 – Wpot2) = –q0(φb – φa). We get the potentials υb and υa using
q
φ k .
r
The work done by the electric field is equal to the change in potential
kq kq
energy:
A
q0 (φ b φa )
q0 (
),
rb ra
where rb = 0.500 m and ra = ∞. The right-hand term within the parentheses is
zero (1/∞ = 0) so
(8.99 109 N m 2 /C2 )(2.00 10 5 C)
A
(3.00 10 6 C)
1.08 J .
0.500 m
NOTE. The electric field does negative work in this case. In order to bring
the charge to this point, the external force would have to do work A= +1.08 J,
assuming no acceleration of the charge.
Example 10.8. Electron in TV tube.
Suppose an electron in a cathode ray tube is accelerated from rest through
a potential difference υb – υa = υba = +5000 V (fig. 10.13). (a) What is
the change in electric potential energy of the electron? (b) What is the speed of
the electron (m = 9.1·10–31 kg) as a result of this acceleration? (c) Repeat for
a proton (m = 1.67·10–27 kg) that accelerates through a potential difference
φba = –5000 V.
130
Fig. 10.13. Example 10.8. Electron accelerated in TV tube
Solution. The electron, accelerated toward the positive plate, will decrease
in potential energy by an amount ΔW = qφba. The loss in potential energy will
equal its gain in kinetic energy (energy conservation).
a) The charge on an electron is q = –e = –1.6·10–19 C. Therefore its change
in potential energy is
ΔW = qφba = (–1.6·10–19 C).(+5000 V) = –8.0·10–16 J.
The minus sign indicates that the potential energy decreases. The potential
difference υba has a positive sign since the final potential υb is higher than
the initial potential υa. Negative electrons are attracted toward a positive
electrode and repelled away from a negative electrode.
b) The potential energy lost by the electron becomes kinetic energy K.
From conservation of energy, ΔK + ΔW = 0, so ΔK = –ΔW:
1
m
2
2
0
q(φb φ a )
qφba ,
where the initial kinetic energy is zero since we are given that the electron
started from rest. We solve for υ:
2qφba
2( 1.6 10 19 C)(5000 V)
4.2 107 m/s.
31
m
9.1 10 kg
c) The proton has the same magnitude of charge as electron, through of
opposite sign. Hence for the same magnitude of υba we expect the same change
in W, but a lesser speed since the proton‘s mass is greater. Thus:
ΔW = qφba = (+1.6·10–19 C)(–5000 V) = –8.0·10–16 J
and
2qφba
m
2(1.6 10
19
C)( 5000 V)
1.67 10
27
9.8 105 m/s.
kg
NOTE. The electric potential energy does not depend on the mass, only on
the charge and voltage. The speed does depend on m.
131
Example 10.9. Uniform electric field obtained from voltage.
Two parallel plates are charged to produce a potential difference of 50 V.
If the separation between the plates is 0.050 m, calculate the magnitude of
the electric field in the space between the plates.
Fig. 10.14. Example 10.9
Solution. We apply equation U = E·d to obtain the magnitude of E, assumed
uniform. The electric field magnitude is E = U/d = (50 V/0.050 m) = 1000 V/m.
PROBLEMS
1. A 5.0g object carries a net charge of 3.8 μC. It acquires a speed υ when
accelerated from rest through a potential difference U. A 2.0 g object acquires
twice the speed under the same circumstances. What is its charge? (Answer:
6.08 μC)
2. A particle with a charge of +8 nC is in a uniform electric field E directed
to the left. It is released from rest and moves to the left. After it has moved 3 cm,
its kinetic energy is found to be 5·10–6 J. (a) What work was done by the electric
force? (b) What is the potential of the starting point with respect to the endpoint?
(c) What is the magnitude of E? (Answer: a) 5·10–6 J; b) 625 V; c) 20833 N/C)
3. The potential energy of an electron (q = –1.6·10–19 C) increases by
3.3·10–15 J when it moves 3.5cm parallel to a uniform electric field. What
is the magnitude of the electric field through which the electron passes?
(Answer: 5.9·105 N/C)
TESTS
1. The work needed to move a –7.0 μC charge from ground to a point
whose potential is +6.00 V higher is:
a) –4.2·10–5 J;
b) +4.2·10–5 J;
c) 10–4 J;
d) –1.2·10–5 J.
132
2. State which of the following is correct:
a) Joule = Coulomb · Volt;
b) Joule = Coulomb / Volt;
c) Joule = Volt / Ampere;
d) Joule = Volt · Ampere.
3. How much kinetic energy will an electron gain (in joules) if it falls
through a potential difference of 21 000 V in a TV picture tube.
a) 3.10+15 J;
b) 3.4.10–15 J;
c) 2.10–13 J;
d) 3.4 J.
4. An electric field of 640 V/m is desired between two parallel plates
11.0 mm apart. A voltage which should be applied is:
a) 700 V;
b) 0.7 V;
c) 7.04 V;
d) 10 V.
5. The potential difference which is needed to give a helium nucleus
(q = 3.2·10–19 C) 48 keV of kinetic energy is:
a) 2.4·104 V;
b) 4.8·103 V;
c) 24 kV;
d) 103 V.
6. What is the potential energy of two equal positive charges of 1 μC each
held 1 m apart in air?
a) 9·10–3 J;
b) 9·10–3 C/V;
c) 1 J;
d) zero.
7. How strong is the electric field between two parallel plates 5.0 mm apart
if the potential difference between them is 110 V?
a) 220·102 V/m;
b) 110 V/m;
c) 22·10–3 V/m;
d) 22·103 V/m.
8. An electron of mass m and charge e travels from rest through a potential
difference of U volts. The final velocity of the electron is:
2mU
2eU
2eU
2mU
;
;
a)
b)
c)
d)
;
.
m
e
e
m
10.6. CAPACITORS. CAPACITANCE.
ELECTRIC ENERGY STORAGE
A capacitor is a device used to store electric charge, and usually consists
of two conducting plates placed near each other but not touching. The parallel
plate capacitor shown in fig. 10.15 has two identical conducting plates with
a surface area S, separated by a distance d (with dielectric material between
the plates, for example air). The capacitor is symbolized by a set of parallel lines.
If a voltage is applied to a capacitor by connecting it to a battery as in
fig. 10.15, the capacitor becomes charged quickly: one plate acquires a negative
charge, the other an equal amount of positive charge. Each battery terminal and
the plate of the capacitor connected to it are at the same potential: hence the full
battery voltage appears across the capacitor. It is found that the amount of
charge q acquired by each plate is proportional to the magnitude of the potential
difference U between them:
Q = CU.
(10.15)
133
Fig. 10.15. Parallel-plate capacitor connected to a battery (a). The same electric circuit is
shown using symbols (b)
The constant of proportionality, C, in this relation is called the capacitance
of the capacitor. Capacitance C is equal to the amount of charge q stored per
volt. The unit of capacitance is coulombs per volt and this unit is called a farad
(1 F = 1 C/1 V). A 1-farad capacitor would be able to store 1 coulomb (a very
large amount of charge) with the application of only 1 volt. One farad is, thus,
a very large capacitance. Most capacitors have capacitance in the range of 1 pF
(picofarad = 10–12 F) to 1μF (microfarad = 10–6 F). The capacitance С does not
depend on q or U. It depends only on the size, shape, and relative position
of the two capacitor plates, and also on the material that separates them.
The capacitance of a parallel plate capacitor is given by formula:
C
ε ε0
S
.
d
(10.16)
where S is the area of one capacitor plate in square meters, and d is the distance
between the plates in meters; the constant ε0 is the permittivity of free space and
ε is a dielectric constant of a medium between the plates.
Example 10.10. Capacitor calculations.
(a) Calculate the capacitance of a parallel-plate capacitor whose plates are
20 cm × 3.0 cm and are separated by a 1.0 mm air gap. (b) What is the charge
on each plate if a 12 V battery is connected across the two plates? (c) What is
the electric field between the plates? (d) Estimate the area of the plates needed to
achieve a capacitance of 1 F, given the same air gap d.
Solution. The capacitance is found by using equation C = ε0S/d. The charge
on each plate is obtained from the definition of capacitance q = CU. The electric
field is uniform, so we can use equation for the magnitude E = U/d. In (d) we
use equation C = ε0 S/d again.
134
a) The area S = (20·10–2 m)·(3.0·10–2 m) = 6.0·10–3 m2. The capacitance C
3 2
S
12 2
2 6.0 10 m )
C ε0
(8.85 10 C /N m
53 pF.
is then
d
1.0 10 3 m
b) The charge on each plate is q = CU = (53·10–12 F)·(12 V) = 6.4·10–10 C.
c) For a uniform electric field, the magnitude of E is
U
12 V
E
1.2 104 V/m.
3
d 1.0 10 m
d) We solve for S in equation C = ε0 S/d and substitute C = 1.0 F and
d = 1.0 mm to find that we need plates with an area
Cd
(1F )(1.0 10 3 m)
8 2
S
10
m .
ε0 8.85 10 12 C2 /N m2
NOTE. This is the area of a square of 104 m = 10 km on a side!
Example 10.11.Capacitance of an axon.
Do an order- of magnitude estimate for the capacitance of an axon 10 cm
long of radius 10 μm. The thickness of the membrane is about 10–8 m, and
the dielectric constant is about 3.
Solution. We model the membrane of an axon as a cylindrically shaped
parallel-plate capacitor, with opposite charges on each side. The separation
of the ―plates‖ is the thickness of the membrane, d ~ 10–8 m. We first calculate
the area of the cylinder
S = 2πrl = 6.28 · 10–5 m · (0.1 m) ≈ 6 · 10–6 m2.
and then can use equation C = ε0 S/d to find the capacitance. The area S is
the area of a cylinder of radius r and length l:
The capacitance C is then
C
εε 0
S
d
3 8.85 10
12
C2 /N m 2
6.0 10 6 m 2
8
10
8
F.
10 m
10.7. THE EQUIVALENT CAPACITANCE
Capacitors are found in many electric
circuits. Capacitors can be connected together in
various ways. Two common ways are in series, or
in parallel. A circuit containing three capacitors
connected in parallel is shown in fig. 10.16.
When capacitors are connected in parallel,
the equivalent capacitance is the sum of
the individual capacitances:
Ceq = C1 + C2 + C3.
(10.17)
Fig. 10.16. Capacitors in parallel
135
The net effect of connecting capacitors in parallel is thus to increase the
capacitance. This makes sense because we are essentially increasing the area of
the plates where charge can accumulate.
When capacitors are connected in series (fig. 10.17), the reciprocal of the
equivalent capacitance equals the sum of the reciprocals of the individual
capacitances:
1
1
1
1
.
(10.18)
Ceq C1 C2 C3
Fig. 10.17. Capacitors in series
10.8. ENERGY STORED IN A CAPACITOR
A charged capacitor stores electric energy. The energy stored in a capacitor
is equal to the work done to charge the capacitor. Energy stored in a capacitor is
electrical potential energy Wpot, and it is thus related to the charge q and voltage
U on the capacitor. From the definition of capacitance C, the energy stored
in a capacitor can be written in different forms:
1
1
q2
(10.19)
Wpot
qU
CU 2
.
2
2
2C
If we divide the stored energy by the volume of the capacitor, we find
the electric energy per unit volume of capacitor of C; this result is valid for any
electric field:
wpot
electric energy density
1
εε0 E 2 .
2
(10.20)
Example 10.12. Equivalent capacitance.
Determine the capacitance of a single capacitor that will have the same
effect as the combination shown in fig. 10.18. Take C1 = C2 = C3 = C.
Solution. First we find the equivalent capacitance of C2 and C3 in parallel,
and then consider that capacitance in series with C1. Capacitors C2 and C3 are
connected in parallel, so they are equivalent to a single capacitor having
capacitance: C23 = C2 + C3 = 2C.
136
Fig. 10.18. Example 10.12
This C23 is in series with C1, fig. 10.18, so the equivalent capacitance of
1
1
1
1
1
3
.
the entire circuit, Ceq, is given by
Ceq C1 C23 C 2C 2C
Thus Ceq = 3/2 C.
NOTE. Hence the equivalent capacitance of the entire combination is
Ceq = 3/2 C.
Example 10.13. Energy stored in a capacitor.
A camera flash unit stores energy in a 150 μF capacitor at 200 V. (a) How
much electric energy can be stored? (b) What is the power output if nearly all
this energy is released in 1.0 ms?
Solution. We use equation for energy in the form Wpot
1
CU 2 because we
2
are given C and U.
The energy stored is Wpot
1
CU 2
2
1
(150 10
2
6
F )(200V ) 2
3.0 J.
If this energy is released in 1/1000 of a second, the power output is
P = W/t = (3.0 J) / (1.0·10–3 s) = 3000 W.
PROBLEMS
1. The membrane that surrounds a certain type of living cell has a surface
area of and 5.0·10–9 m2 and a thickness of 1.0·10–8 m. Assume that
the membrane behaves like a parallel plate capacitor and has a dielectric
constant of 5.0. (a) The potential on the outer surface of the membrane is
+60.0 mV greater than that on the inside surface. How much charge resides on
the outer surface? (b) If the charge in part (a) is due to K+ ions (charge +e), how
many such ions are present on the outer surface? (Answer: a) 1.3·10–12 C;
b) 8.1·106)
2. A heart defibrillator delivers 4.00·102 J of energy by discharging
a capacitor initially at 1.00·104 V. What is its capacitance? (Answer: 8.00 μF)
3. Find the total capacitance for three capacitors connected in series, given
their individual capacitances are 1.000, 5.000, and 8.000 μF. (Answer: 0.755 μF)
137
TESTS
1. SI unit of capacitance is:
a) Coulomb/Volt = Farad (F);
b) F·V;
c) F/A;
d) F·N.
2. The dielectric with the permittivity ε filled the space between the plates
of capacitor. Capacitance C of the capacitor is determined by:
a) ε 0
S
;
d
S
d
b) ε 0ε ;
c) ε
d
;
S
S
d
d) ε .
3. The capacitance of a parallel plane capacitor depends on:
a) the type of metal used;
b) the thickness of the plates;
c) the potential applied across the plates;
d) the separation between the plates.
4. The two plates of a capacitor hold +2500 μC and –2500 μC of charge,
respectively, when the potential difference is 950 V. What is the capacitance?
a) 2.6·10–6 F;
b) 2.6·106 F;
c) 3.8·10–5 F;
d) 2.6 μF.
5. Two parallel plates are separated by 2 cm. If the potential difference
between them is 20 V, then the electric field between them is:
a) 100 N/C;
b) 1000 N/C;
c) 2000 N/C;
d) zero.
6. A 12000 pF capacitor holds 28.0·10–8 C of charge. What is the voltage
across the capacitor?
a) 23.3 V;
b) 2.33 V;
c) 23 kV;
d) none of the above.
7. A parallel plate capacitor is charged from a battery. After charging,
the battery is disconnected. Which of the following increases, when the plates
of the capacitor are moved apart?
a) charge;
b) capacitance;
c) potential; d) none of these.
8. What is the equivalent
the combination shown in fig. 10.19?
a) C; b) 2C; c) 4C; d) C/4.
capacitance
of
Fig. 10.19. Test 8
9. In fig. 10.20, what is the
capacitance between points P and Q?
a) 6/11 μF;
b) 2/11 μF;
c) 24/17 μF;
d) none of these.
effective
Fig. 10.20. Test 9
138
10.9. AN ELECTRIC CURRENT
In the previous Chapters we have been studying static electricity: electric
charges at rest. In this Chapter we begin our study of charges in motion.
It is known the charges can move in electric field.
A directed flow of charged particles is called an electric current. There are
two conditions for an electric current existence:
1. The presence of free charges.
2. The presence of an electric field.
The materials are divided into two categories: conductors and insulators
(dielectrics). Bodies which allow the charges to pass through are called
conductors, e. g. metals, human body, Earth etc. Bodies which do not allow
the charges to pass through are called insulators, e. g. glass, mica, ebonite,
plastic etc. Nearly all natural materials fall into one or the other of these two
very distinct categories. However, a few materials (notably silicon and
germanium) fall into an intermediate category known as semiconductors.
From the atomic point of view, the electrons in an insulating material are
bound very tightly to the nuclei. In a good conductor, on the other hand, some
of the electrons are bound very loosely and can move about freely within
the material (although they cannot leave the object easily) and are often referred
to as free electrons or conduction electrons. In a semiconductor, there are very
few free electrons, and in an insulator, almost none.
10.10. DIRECT CURRENT. OHM’S LAW. RESISTANCE
To produce an electric current it is necessary to apply a potential difference
at conducting material. One way of producing a potential difference along a wire
is to connect its ends to the opposite terminals of a battery (fig. 10.21).
Fig. 10.21. A simple electric circuit (a). Schematic drawing of the same circuit (b)
On any diagram of a circuit a battery is symbolized by the symbol
139
.
The electric current is defined as the net amount of charge flowing through
cross section of a conductor per unit time. Thus, the average current I is
defined as:
q
,
t
I
(10.21)
where Δq is the amount of charge that passes through the conductor during
the time interval Δt.
Electric current is measured in amperes (abbreviated A). When the flow of
charge past any cross section is 1 coulomb per second, the current is 1 ampere.
Thus, 1 A = 1 C/1 s. Ampere is a large unit for current. In practice smaller units
are often used, such as the milliampere (l mA = 10 –3 A) and microampere
(1 μA = 10–6 A).
Example 10.14. Current is flow of charge.
A steady current of 2.5 A exists in a wire for 4.0 min. (a) How much total
charge passed by a given point in the circuit during those 4.0 min? (b) How
many electrons would this be?
Solution. Current is flow of charge per unit time, I = Δq/Δt, so the amount
of charge passing a point is the product of the current and the time interval.
To get the number of electrons (b), we divide the total charge by the charge on
one electron.
a) Since the current was 2.5 A, or 2.5 C/s, then in 4.0 min (= 240 s)
the total charge that flowed past a given point in the wire was
Δq = IΔt = (2.5 C/s)(240 s) = 600 C.
b) The charge on one electron is 1.60.10–19 C, so 600 C would consist of
600 C
3.8 1021 electrons.
19
1.60 10 C/electron
Current is a scalar quantity. The arrows in fig. 10.21, b do not indicate
vectors: they merely show direction of flow along a conductor, not a direction in
space. The direction of the current is taken to be the direction of the flow of
positive charge, even if the actual charge carriers are negative and move in the
opposite direction.
George Simon Ohm established the relationship between potential
difference and current, which is known as Ohm‘s Law. The law states that
the steady current I flowing through a conductor is directly proportional to
the potential difference U between the two ends of the conductor:
I
U
,
R
(10.22)
where the constant R is known as the resistance of a conductor. The SI unit for
resistance is called the ohm and is abbreviated Ω: 1 Ω = 1 V/1 A.
140
In a circuit diagram a resistor and a resistance is represented by
the symbol. Resistance R depends on the conductor geometrical size (fig. 10.22),
the material the conductor is made of, and the temperature of the conductor.
Fig. 10.22. Current I is driven by a potential difference U applied between the ends
of a conductor of length l and cross section S
It is found experimentally that the resistance of a conductor R is directly
proportional to the length of the conductor l and is inversely proportional to its
area of cross section S:
R
ρl
,
S
(10.23)
where ρ, the constant of proportionality, is called specific resistance or electrical
resistivity of the material of the conductor and depends on the material used.
If l = l m, S = l m2, then ρ = R. The electrical resistivity ρ of a material is
defined as the resistance R offered to current flow by a conductor of unit length l
having unit area of cross section S. The unit of ρ is Ω·m. It is a constant for
a particular material. The electrical resistivity ρ of a material depends somewhat
on temperature.
The reciprocal of electrical resistivity ρ, is called electrical conductivity σ:
1
σ
.
(10.24)
ρ
The unit of conductivity σ is Ω–1 m–1.
For most conductors, a temperature increase causes an increase in
resistance. If the temperature change is not too great, the resistance R of metals
usually increases nearly linearly with temperature. An empirical relationship for
the temperature dependence of the resistance R of metals is given by formula:
R R0[1 α(T T0 )],
(10.25)
where R0 is the resistance at some reference temperature T0 (such as 0 °C or
20 °C); R is the resistance at a temperature T; α is the temperature coefficient of
resistance.
Resistance R is a property of an object, whereas resistivity ρ is a property
of a material.
Example 10.15. Current is flow of charge.
A steady current of 2.5 A exists in a wire for 4.0 min. (a) How much total
charge passed by a given point in the circuit during those 4.0 min? (b) How
many electrons would this be?
141
Solution. Current is flow of charge per unit time, I = Δq/Δt, so the amount
of charge passing a point is the product of the current and the time interval.
To get the number of electrons (b), we divide the total charge by the charge on
one electron.
a) Since the current was 2.5 A, or 2.5 C/s, then in 4.0 min (= 240 s)
the total charge that flowed past a given point in the wire was
Δq = IΔt = (2.5 C/s)(240 s) = 600 C.
b) The charge on one electron is 1.60.10–19 C, so 600 C would consist of
600 C
3.8 1021electrons.
19
1.60 10 C/electron
Example 10.16. Speaker wires.
Suppose you want to connect your stereo to
remote speakers (fig. 10.23). (a) If each wire must
be 20 m long, what diameter copper wire should
you use to keep the resistance less than 0.10 Ω per
wire? (b) If the current to each speaker is 4.0 A,
what is the potential difference, or voltage drop,
across each wire? The resistivity of copper is
1.68·10–8 Ω·m.
Solution. We use equation R
ρl
to get
S
the area S, from which we can calculate the wire‘s
radius using S = πr2. The diameter is 2r. In (b) we
can use Ohm‘s law, U = IR.
ρl (1.68 10 8 m) (20 m)
S
3.4 10 6 m 2 .
Fig. 10.23. Example 10.16
R
0.10
The cross-sectional area S of a circular wire is S = πr2. The radius must
S
then be at least r
1.04 10 3 m 1.04 mm.
π
The diameter is twice the radius and so must be at least 2r = 2.1 mm.
b) From V = IR we find that the voltage drop across each wire is
U = IR = 4.0 A · 0,10 Ω = 0.40 V.
NOTE. The voltage drop across the wires reduces the voltage that reaches
the speakers from the stereo amplifier, thus reducing the sound level a bit.
Example 10.17. Stretching changes resistance.
Suppose a wire of resistance R could be stretched uniformly until it was
twice its original length. What would happen to its resistance?
142
Solution. If the length l doubles, then the cross-sectional area S is halved,
because the volume (V = Sl) of the wire remains the same. From R
that the resistance would increase by a factor of four 2/
1
2
ρl
we see
S
4.
Example 10.18.
A uniform copper wire, having mass of 2.23·10–3 kg carries a current of
1 A, has potential difference of 1.7 V across its ends. Find its length and area
of cross section. The wire is now uniformly stretched to double its length, find
the new resistance. Density of copper is 8.92·103 kg·m–3 and its resistivity is
1.7·10–8 Ω·m.
Solution. Suppose l and S be the length and area of cross section of
the given copper wire. Volume V of the wire is V = Sl
mass
2.23 10 3 kg
V Sl
density 8.92 103 kg m 3
or
10 6 3
m .
4
Sl
Also, resistance of the copper wire is given by
or
1.7V
1.7 .
1A
l R
1.7
10 6 3
Combining Sl
m and we get
4
S ρ 1.7 10 8
R
Also, S = 5·10–8 m2. l 2
ρl
S
l
108 m 1,
S
U
I
10 6 3
m 108 m1
4
Since volume remains constant, V
Then R1
ρl1
S1
ρ
2l
S /2
R1
4ρ
l
S
4 1.7 10
Sl
25 m2 or l
5 10
Ω m 5m
8
m
2
m
5 m.
S1l1, where l1 = 2l and S1
4 R, where R
8
108 m 1.
S
.
2
ρl
S
6.8
.
Example 10.19. Resistance thermometer.
The variation in electrical resistance with temperature can be used to make
precise temperature measurements. Platinum is commonly used since it is
relatively free from corrosive effects and has a high melting point. Suppose at
20.0 °C the resistance of a platinum resistance thermometer is 164.2 Ω. When
143
placed in a particular solution, the resistance is 187.4 Ω. The temperature
coefficient of resistivity of platinum α is equal to 3.927·10–3 (Co)–1. What is
the temperature of this solution?
Solution. Since the resistance R is directly proportional to the resistivity ρ,
we can combine equation R
ρ l
with equation R R0[1 α(T T0 )] to find R
S
as a function of temperature T, and then solve that equation for T.
We use the equation R R0[1 α(T T0 )].
Here R0 = ρ0l/S is the resistance of the wire at T0 = 20.0 °C. We solve this
equation for T and find
R R0
187.4
164.2
T T0
20.0 C
56.0 C.
αR0
3.927 10 3 (Co ) 1(164.2 )
NOTE. Resistance thermometers have the advantage that they can be used
at very high or low temperatures where gas or liquid thermometers would be
useless.
10.11. RESISTORS IN SERIES AND IN PARALLEL
There are active and passive elements of electrical circuit. Active elements
can generate energy (voltage and current sources, batteries), passive ones cannot
generate energy (resistors, capacitors and inductors). A resistor is a circuit
element that dissipates electrical energy (usually as heat). Devices that are
modeled by resistors: light bulbs, heating elements (stoves, heaters, etc.), long
wires.
When two or more resistors are connected end to end as shown
in fig. 10.24, they are said to be connected in series.
Fig. 10.24. Resistors in series
Any charge that passes through R1 in fig. 10.24 will also pass through R2
and then R3. Hence the same current I passes through each resistor I = I1 = I2 = I3.
We let U represent the potential difference (voltage) across all three
resistors in fig. 10.24. We assume all other resistance in the circuit can be
144
ignored, so U equals the terminal voltage supplied by the battery. We let U1, U2
and U3 be the potential differences across each of the resistors, R1, R2 and R3,
respectively, as shown in fig. 10.24. From Ohm's Law U = IR one can write
U1 = IR1, U2 = IR2 and U3 = IR3. Because the resistors are connected end to end
the total voltage U is equal to the sum of the voltages across each resistor:
U1 + U2 + U3 = υ0 – υ1 + υ1 – υ2 + υ2 – υ3 = υ0 – υ3 = U.
Thus,
U = U1 + U2 + U3 = IR1 + IR2 + IR3.
(10.26)
Let‘s determine the equivalent (or effective) single resistance Req of
the series combination. The equivalent single resistance Req is related to U by
U = IReq.
Hence, IReq = IR1 + IR2 + IR3,
or
Req = R1 + R2 + R3.
(10.27)
Thus, the equivalent resistance of a number of resistors in series
connection is equal to the sum of the resistance of individual resistors.
Consider resistors of resistances R1, R2 and R3 are connected in parallel,
so that the current from the source splits into separate branches as shown
in fig. 10.25.
Fig. 10.25. Resistors in parallel
In parallel circuit the total current I that leaves the battery splits into three
separate paths. Let I1, I2 and I3 be the currents through each of the resistors,
R1, R2 and R3, respectively. Because electric charge is conserved, the current I
flowing into a junction A (where the different wires or conductors meet) must
equal the current flowing out of the junction. Thus,
I = I1 + I2 + I3.
(10.28)
When resistors are in parallel, the potential difference U across each
resistor is the same (U = U1 = U2 = U3). Applying Ohm‘s Law to each resistor
one can write:
U
U
U
(10.29)
I1
,
I2
,
I3
.
R1
R2
R3
145
Let‘s determined what single resistor Req will draw the same current I as
these three resistances in parallel:
I
Thus,
or
U
.
Req
(10.30)
U
Req
U
R1
U
R2
U
,
R3
(10.31)
1
Req
1
R1
1
R2
1
.
R3
(10.32)
Thus, when a number of resistors are connected in parallel, the sum
of the reciprocal of the resistance of the individual resistors is equal
to the reciprocal of the effective resistance of the combination.
Example 10.20. Circuit with series and parallel resistors.
How much current is drawn from the battery shown in fig. 10.26, a?
a
b
Fig. 10.26. Circuit for Example 10.20 (a).
Fig. 10.26. (b) Equivalent circuit, showing
the equivalent resistance of 290 Ω for the
two parallel resistors in fig. 10.6, a
Solution. The current I that flows out of the battery all passes through
the 400 Ω resistor, but then it splits into I1 and I2 passing through the 500 Ω and
700 Ω resistors. The latter two resistors are in parallel with each other. We look
for something that we already know how to treat. So let‘s start by finding
the equivalent resistance, Rp, of the parallel resistors, 500 Ω and 700 Ω. Then we
can consider this Rp to be in series with the 400 Ω resistor.
The equivalent resistance, Rp, of the 500 Ω and 700 Ω resistors in parallel
is given by
1
1
1
0.0020 1 0.0014 1 0.0034 1.
R p 500
700
This is l/Rp, so we take the reciprocal to find Rp. It is a common mistake to
forget to take this reciprocal. Notice that the units of reciprocal ohms, Ω–1, are
a reminder.
146
1
Rp
290 .
0.0034 1
This 290 Ω is the equivalent resistance of the two parallel resistors, and is
in series with the 400 Ω resistor as shown in the equivalent circuit of fig. 10.26, b.
Thus
To find the total equivalent resistance Req, we add the 400 Ω and 290 Ω
resistances together, since they are in series, and find
Req = 400 Ω + 290 Ω = 690 Ω.
The total current flowing from the battery is then
U 12.0 V
I
0.0174 A 17 mA.
Req 690
10.12. ELECTRIC ENERGY AND ELECTRIC POWER
If I is the current flowing through a conductor of resistance R in time Δt,
then the quantity of charge flowing is Δq = IΔt. If the charge Δq flows between
two points having a potential difference U, then the work ΔW done in moving
the charge is the product of potential difference U and the charge Δq:
ΔW = U·Δq = U·IΔt.
(10.33)
The unit of the work W is Joule (J).
Then, electric power P is defined as the rate of doing electric work and is
equal to the product of current I and voltage U:
P
W
t
(10.34)
IU .
For resistors the electric power P can be written as:
P
IU
I (IR )
2
I R
U
U
R
U2
.
R
(10.35)
When resistors are in parallel, the potential difference U across each
U2
resistor is the same and electric power P can be found as: P
.
R
When resistors are connected in series, the same current I passes through
each resistor and electric power P can be found as: P = I2R.
The SI unit of electric power P is the Watt (W): 1 W = 1 J/s.
A resistor dissipates power when a current passes through it. The energy is
released in the form of heat. For a steady current I, the amount of heat Q
produced in time Δt is equal to the work ΔA done in moving the charges:
Q = PΔt = UIΔt.
(10.36)
2
For a resistance R,
Q = I RΔt
(10.37)
U2
(10.38)
Q
t.
R
147
The above relations were experimentally verified by Joule and are known
as Joule‘s Law of heating. By equation (10.37) Joule‘s law implies that when
current flows through a conductor, the heat produced Q is directly proportional
to the square of the current I, directly proportional to resistance R of a conductor
and the time Δt of passage of current. Also by equation (10.38), the heat
produced Q is directly proportional to the square of the voltage U and inversely
proportional to resistance R for a given U. This is also known as Joule heat that
is dissipated in R.
Example 10.21. Headlights.
Calculate the resistance of a 40-W
automobileheadlight designed for 12 V.
Solution. We solve equation P = U2/R for R:
U 2 (12 V) 2
R
3.6 .
Fig. 10.27. Example 10.21
P
(40 W)
NOTE. This is the resistance when the bulb is burning brightly at 40 W.
When the bulb is cold, the resistance is much lower, as we saw in equation
(10.25). Since the current is high when the resistance is low, lightbulbs burn out
most often when first turned on.
Example 10.22.
A 100 W bulb and a 400 W bulb are joined in parallel to the mains. Which
bulb will draw more current?
Solution. Let U be the voltage of the mains and I1, I2 are the currents
through the two bulbs. In case of parallel combination of the bulbs, U is
the same for these two bulbs: P1 = I1·U
and
P2 = I2·U,
I1 P1 100 W 1
or
I 2 P2 400 W 4
I1 1
or I1 I 2 .
I2 4
NOTE. 400 W bulb will draw more current.
Example 10.23.
An electric kettle draws a current of 10 A when connected to the 230 V
mains supply. Calculate: (a) the power of the kettle; (b) the energy produced in
5 minutes; (c) the rise in temperature if all the energy produced in 5 minutes is
used to heat 2 kg of water. (Specific heat capacity of water c = 4200 J kg–1 K–1.)
Solution. We use equation P = IU to calculate power P of the kettle:
a) P = IU = 10 A × 230 V = 2300 W = 2.3 kW.
b) Heat energy produced in 5 minutes: Q = Pt = 2300 W × 5 × 60 s =
690 000 J.
148
c) Heat Q produced is absorbed by water: Q = energy gained by water.
We solve equation Q = mcΔT for ΔT:
Q
690 000 J
T
82.1 K .
mc 2 kg 4200 J kg 1K 1
NOTE. Rise in temperature of the water is 82.1 K.
10.13. ELECTROMOTIVE FORCE. OHM’S LAW
FOR A COMPLETE CIRCUIT
To maintain a steady current, there must be a device (such as a battery or
an electric generator) in the circuit wherein the potential rises along the direction
of the current. For the potential to rise along the direction of the current there
must be a source which converts some form of nonelectric energy (chemical,
mechanical, or light, for example) to electrical energy. Such a device is called
a source of electromotive force or of emf. An emf device is a (charge pump)
device that maintains a constant potential difference between a pair of terminals.
The symbol ε is usually used for emf. The emf ε of a device is the work Af per
unit charge q that the nonelectrostatic force does in moving charge from low
potential terminal to high potential terminal:
Af
ε
.
(10.39)
q
The unit of the emf = ε is Volt (V).
A battery itself has some resistance, which is called its internal resistance:
it is usually designated r. A real battery is modeled as if it were a perfect emf ε
in series with a resistor r as shown in fig. 10.28. Since this resistance r is inside
the battery one can never separate it from the battery.
Fig. 10.28. A battery in a complete circuit
149
The work Af that the nonelectrostatic force does in moving charge from low
potential terminal to high potential terminal is given by:
Af = ε·q = ε(I·t).
(10.40)
The work Af can be also written as (if all work Af is converted into
the heating of the conductor А = Q):
Af = I2(R + r)·t.
(10.41)
Comparing equations (10.20) and (10.21) gives:
ε = I(R + r).
(10.42)
For a complete circuit, Ohm‘s Law assumes the following form:
ε
I
R r
.
(10.43)
where Rt = R + r is the total resistance of the entire circuit, which is equal to
the sum of the external resistance R of the circuit and the internal resistance r of
the source of emf.
When connected to a load resistance R, so that a current I flows through
the circuit there is an internal drop in voltage equal to (I·r). Thus the terminal
voltage of the battery is therefore:
U = ε – Ir.
(10.44)
The potential drop across the external resistance R can be written as:
U = IR.
(10.45)
The internal resistance r of the source of emf is low. When the external
resistance R of the circuit drops (R → 0), current becomes very high and can
cause damage. The largest amount of current in circuit is called the short-circuit
current Ish:
Ish
ε
.
r
(10.46)
Example 10.24. Battery with internal resistance.
A 65.0 Ω resistor is connected to the terminals of a battery whose emf is
12.0 V and whose internal resistance is 0.5 Ω. Calculate: (a) the current in
the circuit, (b) the terminal voltage of the battery, Uab, (c) the power dissipated
in the resistor R and in the battery‘s internal resistance r.
Fig. 10.29. Example 24
Solution. We first consider the battery as a whole, which is shown in
fig. 10.29 as an emf ε and internal resistance r between points a and b. Then we
apply U = IR to the circuit itself.
150
a) From the equation relating emf ε to terminal voltage, we have
Uab = ε – Ir. We apply Ohm‘s law to this battery and the resistance R of
the circuit: Uab = IR. Hence IR = ε – Ir or ε = I(R + r), and so
PR
I 2R
(0.183 A)2 (65.0
)
b) The terminal voltage is
U ab ε Ir 12.0 V (0.183 A)(0.5
I 2R
c) The power dissipated in R is PR
and in r is Pr
I 2r
(0.183 A) 2 (0.5
)
2.18 W,
) 11.9 V.
(0.183 A)2 (65.0
)
2.18 W,
0.02 W .
10.14. ALTERNATING CURRENT
When a battery is connected to a circuit,
the current moves steadily in one direction.
This is called a direct current, or dc. Electric
generators at electric power plants, however,
produce alternating current, or ac.
Sometimes capital letters are used, DC and
AC. An alternating current reverses direction
many times per second and is commonly
sinusoidal, as shown in fig. 10.30. The charges
in a wire first move in one direction and then
in the other. The current supplied to homes
and businesses by electric companies is ac
throughout virtually the entire world. It is
easier and cheaper to produce as well as
transmit AC than DC.
Fig. 10.30. (a) Direct current,
An alternating current is one that
(b) Alternating current
changes continuously in magnitude and
periodically in direction. The alternating currents varying according to harmonic
law have the most important practical significance. It is represented by a sine
curve (fig. 10.30, b). The mathematical form of an alternating voltage as
a function of time is:
U U0sin2πft U0sinωt .
(10.47)
The voltage oscillates between +U0 and –U0, and U0 is referred to as
the peak voltage. U is the instantaneous value of voltage at an instant of time t.
The frequency f is the number of complete oscillations made per second
(measured in Hertz, the unit ―hertz‖ means cycles per second 1/sec); and
ω = 2 f — is angular frequency (radians/sec). In many countries, f = 50 Hz is
used.
151
From equation U = IR, if a voltage U exists across a resistance R, then
the current I through the resistance is
I
U
R
U0
sinωt
R
I 0sinωt .
(10.48)
The quantity I0 = U0/R is the peak current. The current is considered
positive when the electrons flow in one direction and negative when they flow in
the opposite direction. It is clear from fig. 10.30, b that an alternating current is
as often positive as it is negative. Thus, the average current is zero. This does
not mean, however, that no power is needed or that no heat is produced
in a resistor. Electrons do move back and forth, and do produce heat. Indeed,
the power transformed in a resistance R at any instant is
(10.49)
P I 2 R I 02 Rsin 2ωt .
Because the current is squared, we see that the power is always positive, as
graphed in fig. 10.31. The quantity sin2ωt varies between 0 and 1; and it is not
too difficult to show that its average value is 1/2 as indicated in fig. 10.31.
Fig. 10.31. Power delivered to a resistor in an AC circuit
Thus, the average power developed, is
P
1 2
I R.
2 0
(10.50)
Since power can also be written P = U2/R = (U02 R) sin2 ωt, we also have
that the average power is
1 U 02
(10.51)
P
.
2 R
The average or mean value of the square of the current or voltage is thus
what is important for calculating average power: I 2
1 2
I 0 and U 2
2
1 2
U .
2 0
The square root of each of these is the rms (root-mean-square) value of
the current or voltage:
I0
I rms
I2
0.707 I 0 ,
(10.52)
2
152
U2
U rms
U0
(10.53)
0.707U0 .
2
The rms values of U and I are sometimes called the effective values. They
are useful because they can be substituted directly into the power formulas,
equations (10.50, 10.51), to get the average power:
P
P
1 2
2
I 0 R I rms
R;
2
2
1 U 02 U rms
.
2 R
R
(10.54)
(10.55)
Thus, a direct current whose values of I and U equal the rms values of I
and U for an alternating current will produce the same power. Hence it is usually
the rms value of current and voltage that is specified or measured. For example,
in a country, standard line voltage is 120 V ac. The 120 V is Urms; but the peak
voltage U0 is
U0
2U rms 170 V .
In much of the world (Europe, Australia, Asia) the rms voltage is 240 V, so
the peak voltage is 340 V.
Example 10.25. Hair dryer.
(a) Calculate the resistance and the peak current in a 1000-W hair dryer
connected to a 120-V line; (b) What happens if it is connected to a 240-V line in
Britain?
2 I rms . Then
Solution. We are given P and Urms, so Irms = P/Urms, and I 0
we find R from U = IR.
a) We solve P I rnsU rms for the rms current:
I rms
Then I 0
P
U rms
1000 W
8.33 A.
120 V
2 I rms 11.8 A.
U rms 120 V
14.4 .
I rms 8.33 A
b) When connected to a 240 V line, more current would flow and
the resistance would change with the increased temperature. But let us make
an estimate of the power transformed based on the same 14.4 Ω resistance.
2
U rms
(240 V) 2
4000 W.
The average power would be P
R
(14.4 Ω)
The resistance is R
NOTE. This is four times the dryer’s power rating and would undoubtedly
melt the heating element or the wire coils of the motor.
153
PROBLEMS
1. A potential difference of 400 volt is applied across the ends of
a conductor of resistance 80. Calculate the number of electrons flowing through
it in one second. Charge of the electron is 1.6·10–19 C. (Answer: 3.125·1019)
2. A wire of resistance 1 Ω is drawn out so that its length is increased by
twice its original length. Find its new resistance. (Answer: 4 Ω)
3. At room temperature (27.0 °C) the resistance of a heating element is
100 Ω. What is the temperature of the element if the resistance is found to be
117 Ω, given that the temperature coefficient of the material of the resistor is
1.70·10–4 °C–1. (Answer: 1027 °C)
4. Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is
the total resistance of the combination? If the combination is connected to
a battery of emf 20 V and negligible internal resistance, determine the current
through each resistor, and the total current drawn from the battery. (Answer:
20/19 Ω, 10 A, 5 A, 4 A, 19 A)
5. A storage battery of emf 8.0 V and internal resistance of 0.5 Ω is being
charged by a 120 V DC supply using a series resistor of 15.5 Ω. What is
the terminal voltage of the battery during charging? What is the purpose of
having a series resistor in the charging circuit? (Answer: 11.5 V)
6. A bulb of 484 Ω is producing light when connected to 220 V main
supply. What is the electric power of the bulb? (Answer: 100 W)
7. An electric heater having resistance equal to 5 Ω is connected to electric
source. If it produces 180 J of heat in one second, find the potential difference
across the electric heater. (Answer: 30 V)
8. An ac voltage, whose peak value is 180 V, is across a 210 Ω resistor.
What is the value of the rms and peak currents in the resistor? (Answer: 127 V,
606 mA)
TESTS
1. Electric current flows when:
a) there is some potential difference;
b) there is some resistances;
c) atoms arrange themselves;
d) all of the above.
2. Ohm‘s law deals with the ratio between:
a) current and potential difference;
b) capacity and charge;
c) potential and capacity;
d) induced potential and flux.
154
3. When the resistance of the conductor is increased then the current will:
a) increase;
b) decrease;
c) remains the same;
d) none of these.
4. If the current I flowing through conductor under an applied potential U,
then according to Ohm‘s law:
a) U ~ I2;
b) U ~ 1/I;
c) I ~ U;
d) none of these.
5. A wire has resistance R. Another wire identical to the first but having
twice the diameter, has a resistance of:
a) R/4;
b) 4/R;
c) R/2;
d) 2R.
6. Resistance of a conductor depends on its:
a) length;
b) volume;
c) area of cross-section;
d) temperature.
7. The resistance of a coil with a current of 12 A at 120 V is:
a) 0.1 Ω;
b) 10 Ω;
c) 1440 Ω;
d) 132 Ω.
8. When the temperature of a metallic conductor is increased its resistance:
a) always decreases;
b) always increases;
c) remains the same;
d) None of these.
9. The reciprocal of resistance is:
a) specific resistance;
b) effective resistance;
c) conductance;
d) current.
10. Net resistance joined in series is:
a) sum of reciprocal of individual resistance;
b) sum of all the individual resistance;
c) product of resistances;
d) reciprocal of product of resistances.
11. The length of the conductor is halved. Then its conductivity be;
a) halved;
b) doubled; c) unchanged;
d) quadrupled.
Fig. 10.32. Test 12
12. What is the equivalent resistance
between points P and Q in the circuit shown in
fig. 10.32?
a) 1 Ω;
b) 2 Ω;
c) 3Ω;
d) None of these.
13. The temperature coefficient of the resistance of a wire is 0.00125 °C–1.
At 300 °K its resistance is 1 Ω. The resistance will become 2 Ω at a temperature:
a) 1154 K;
b) 1100 K;
c) 1127 K;
d) 1400 K.
14. There are three resistance 2, 3, 5 Ω connected in parallel to a battery of
10 V of negligible resistance. The potential drop across 3 Ω resistance is:
a) 2 V;
b) 5 V;
c) 3V;
d) 10 V.
155
15. The electron (e = 1.6·10–19 C) in hydrogen atom circles round
the proton with a speed of 2.18·106 m/s in an orbit radius 5.3·10–11 m. What is
current constituted by it?
a) 1.05 mA;
b) 10.5 mA;
c) 0.105 mA;
d) none of the above.
16. Assuming that e = 1.6·10–19 C, the number of electron passing per sec.
through a wire carrying 1 A of current is:
a) 0.625·1019;
b) 1.5·1019;
c) 1.6·1019;
d) 6.25·1016.
17. An electric heater operating at 220 V boils 5 liter of water in 5 minutes.
If it is used on 110 V, it will boil the same amount of water in:
a) 10 minutes;
b) 20 minutes;
c) 15 minutes;
d) 25 minutes.
18. A constant voltage is applied between the two ends of a uniform
metallic wire. Some heat is developed in it. The heat developed is doubled if:
a) the length of the wire is doubled;
b) the radius of the wire is doubled;
c) both the length and radius of the wire are doubled;
d) both the length and radius of the wire are halved.
19. A coil has a resistance 20 Ω at 0 °C and 21 Ω at 200 °C. What is
the temperature coefficient of metal used for the coil:
a) 2·10–4 K–1;
b) 2·10–4 K–1;
c) 8·10–4 K–1;
d) 1·10–4 K–1.
20. An electric lamp is marked 100 W. It is working on 200 V. The current
through the lamp is given as:
a) 5A;
b) 2 A;
c) 0.5 A;
d) 1.0 A.
21. Two electric bulbs whose resistances are in the ratio 1 : 2 are connected
in series to a constant voltage source. The powers dissipated in them have
the ratio:
a) 1 : 2;
b) 1 : 1;
c) 2 : 1;
d) 4 : 1.
22. Two electric bulbs whose resistances are in the ratio 1 : 2 are connected
in parallel to a constant voltage source. The powers dissipated in them have
the ratio:
a) 1 : 2;
b) 1 : 1;
c) 2 : 1;
d) 4 : 1.
23. Efm is measured in
a) Joule;
b) Joule · Coulombs;
c) Joule / Coulomb;
d) Joule / Coulomb/metre.
24. The emf of a battery is 3V and internal resistance 0.2 Ω. The difference
of potential at the terminals of battery when connected across the external
resistance of 1 Ω is:
a) 1.67 V;
b) 2.5 V;
c) 2.67 V;
d) 3.67 V.
156
25. What is the peak current in a 2.8-kΩ resistor connected to a 120 V rms
ac source.
a) 60.6 mA; b) 60 A;
c) 6 A;
d) 606 mA.
26. The peak value of an alternating current passing through a 1500 W
device is 4.0 A. What is the rms voltage across it?
a) 530 mV;
b) 530 V;
c) 50 V;
d) 350 V.
27. A heater coil connected to a 240 V ac line has a resistance of 40 Ω.
What is the average power used?
a) 1440 W;
b)140 W;
c) 1000 W;
d) 1400 W.
11. MAGNETIC FIELD
As well known all charges create electric fields, and these fields can be
detected by other charges resulting in electric force. However, when charges are
moved they create a new completely different field. This is the magnetic field.
So any moving charge or an electric current produces a magnetic field
in the surrounding space. In turn the magnetic field exerts a force only on
moving charge or electric current and on the magnetized bodies.
The magnetic field is characterized by its magnitude B and direction.
The magnetic field B is a vector and is represented by lines called lines
of magnetic field. The vector B direction is tangent to this line at any point
of the field. The strength of the magnetic field is proportional to the closeness of
the lines.
11.1. THE MAGNETIC FIELD PRODUCED
BY ELECTRIC CURRENT
At first let‘s look a magnetic fields created by two main form of electric
current: a magnetic fields of a straight current and a of current-carrying coil
(solenoid) that illustrated in fig. 11.1.
A straight wire with electric current I going through it produces a magnetic
field going in circles around it (fig. 11.1, a). The magnitude B of this magnetic
field is equal to
μμ0 I
ВА
,
(11.1)
2πb
where I is a value of electric current, b is a distance from the wire, μ is the
magnetic permeability of medium and μ0 = 4π·10–7 H/m is magnetic constant
(the magnetic permeability of vacuum).
157
B
B
a
b
Fig. 11.1. Magnetic fields B of a straight current-carrying wire (a) and a current-carrying coil
(solenoid) (b)
A solenoid is many loops of wire with a current going through. When
the length of the solenoid is much larger than its radius, the magnetic field inside
a solenoid is strong and uniform and is weak outside. The field lines inside the
solenoid are nearly parallel, uniformly spaced, and close together (fig. 11.1, b).
The magnitude B of the magnetic field in a solenoid is equal to
В
μμ0 I
n,
2r
(11.2)
where I is a value of electric current, r is a radius of a loop, n is an amount of
the loops.
Magnetic field direction can be determined using the right hand rule
(fig. 11.2). The right-hand rule gives the direction of the magnetic field lines that
encircle a current-carrying wire. If the wire is grasped in the right hand with
the thumb in the direction of the current, the fingers will curl in the direction of B.
Fig. 11.2. Right-hand rule for determining the direction of B
158
11.2. FORCE ON AN ELECTRIC CURRENT IN A MAGNETIC FIELD
(AMPERE’S FORCE)
It was found that a magnetic field exerts a force on a current-carrying wire.
The force F (it is named after Ampere) on
a wire carrying a current I with length l in
a uniform magnetic field В is given by formula:
F = I · l · B sin θ.
(11.3)
where θ is the angle between the current direction
and the magnetic field B.
Equation (11.3) serves as a practical
definition of B: В
F
, if θ = π/2.
I· l
The SI unit for magnetic field В is the tesla
(T), 1 T = 1 N/A·m. Another unit sometimes used
Fig. 11.3. Current-carrying wire to specify magnetic field is the gauss (G):
of length l in a magnetic field
1 G = 10–4 T. It is necessary to note that
the magnetic field of the Earth at its surface is about 0.5 G = 50 μT.
The direction of the force F is given by another right-hand rule, as
illustrated in fig. 11.4, c. Orient your right hand so that outstretched fingers can
point in the direction of the conventional current I, and when you bend your
fingers they point in the direction of the magnetic field lines B. Then your
thumb points in the direction of the force F on the wire.
As one can see from equation (11.3), the force F depends on the angle θ
between the current direction and the magnetic field. When the current is
perpendicular to the field lines (θ = 90°) the force is strongest Fmax = IBl. When
the wire is parallel to the magnetic field lines (θ = 0°) there is no force at all
(F = 0). At other angles, the force F is proportional to sinθ.
a
b
c
Fig. 11.4. (a) Force on a current-carrying wire placed in a magnetic field B; (b) same, but
current I reversed; (c) right-hand rule for setup in (b)
Example 11.1. Magnetic force on a current-carrying wire.
A wire carrying a 30 A current has a length l = 12 cm between the pole
faces of a magnet at an angle θ = 60° (fig. 11.5). The magnetic field is
159
approximately uniform at 0.90 T. We ignore the field beyond the pole pieces.
What is the magnitude of the force on the wire?
Fig. 11.5. Example 11.1. Current-carrying wire in a magnetic field. Force on the wire is
directed into the page
Solution. The force F on the 12-cm length of wire within the uniform field
B is F = IlBsinθ = (30 A) · (0.12 m) · (0.90 T) · (0.866) = 2.8 N.
Example 11.2. Measuring a magnetic
field.
A rectangular loop of wire hangs
vertically as shown in fig. 11.6. A magnetic
field B is directed horizontally, perpendicular
to the wire, and points out of the page at all
points as represented by the symbol
. The
magnetic field B is very nearly uniform along
the horizontal portion of wire ab (length
l = 10.0 cm) which is near the center of the
gap of a large magnet producing the field.
The top portion of the wire loop is free of the Fig. 11.6. Measuring a magnetic field B
field. The loop hangs from a balance which measures a downward magnetic
force (in addition to the gravitational force) of F = 3.48·10–2 N when the wire
carries a current I = 0.245 A. What is the magnetic field at the center of magnet?
Solution. Three straight sections of the wire loop are in the magnetic field:
a horizontal section and two vertical sections. We apply F = IlBsinθ to each
section and use the right-hand rule. The magnetic force on the left vertical
section of wire points to the left; the force on the vertical section on the right
points to the right. These two forces are equal and in opposite directions and so
add up to zero. Hence, the net magnetic force on the loop is that on
the horizontal section ab, whose length is l = 0.100 m. The angle θ between B
and the wire is θ = 90°, so sinθ = 1. Thus
F
3.48 10 2 N
B
1.42 T.
I l (0.245 A) (0.100 m)
160
NOTE. This technique can be a precise means of determining magnetic
field strength.
Since a current in a wire creates its own magnetic field, two current
carrying wires placed close together exert magnetic forces on each other.
Parallel wires with current flowing in the same direction, attract each other and
parallel wires with current flowing in the opposite direction, repel each other
(fig. 11.7).
a
b
Fig. 11.7. Antiparallel currents (in opposite direction) exert a repulsive force on each other
(a). Parallel currents in the same direction exert an attractive force on each other (b)
The force between two parallel wires carrying a current is used to define
the SI unit of current is Ampere: one Ampere is definite as that current flowing
in each of two long parallel conductors 1m apart, which result in a force of
exactly 2·10–7 N/m of 1 m length of each conductor.
11.3. FORCE ON AN ELECTRIC CHARGE MOVING IN A MAGNETIC
FIELD (LORENTZ’S FORCE)
If a particles of charge q moves through a magnetic field B with a velocity
υ then a force F (named after Lorentz) acts on it:
F q Bsinθ,
(11.4)
θ is angle between υ and В. The force F is greatest when the particle moves
perpendicular to В (θ = 90°): Fmax q B.
The force F is zero, if the particle moves parallel to the field lines B
(θ = 0°). The direction of the force F is perpendicular to the magnetic field B
and to the velocity υ of the particle. It is given again by a right-hand rule (for
q > 0): you orient your right hand so that your outstretched fingers point along
the direction of the particle‘s velocity υ and when you bend your fingers they
must point along the direction of B. Then your thumb will point in the direction
of the force F. This is true only for positively charged particles, and will be ―up‖
for the positive particle shown in fig. 11.8. For negatively charged particles,
the force is in exactly the opposite direction ―down‖ in fig. 11.8.
161
Fig. 11.8. Force on charged particles due to a magnetic field is perpendicular to the magnetic
field direction. If υ is horizontal, then F is vertical
Example 11.3. Magnetic force on ions during a nerve pulse.
Estimate the magnetic force due to the Earth‘s magnetic field on ions
crossing a cell membrane during an action potential. Assume the speed of
the ions is 10–2 m/s.
Solution. Using F = q υ B, set the magnetic field of the Earth to be roughly
B ≈ 10–4 T, and the charge q ≈ e ≈ 10–19 C.
F ≈ (10–19 C) · (10–2 m/s) · (10–4 T) = 10–25 N.
NOTE. This is an extremely small force. Yet it is thought migrating
animals do somehow detect the Earth’s magnetic field, and this is an area of
active research.
Example 11.4. Electron’s path in a uniform magnetic field.
An electron travels at 2.0·107 m/s in a plane perpendicular to a uniform
magnetic field of 10 mT. Describe its path quantitatively.
Solution. An electron at point P (fig. 11.9) is moving to the right, and
the force on it at this point is downward as shown (use the right-hand rule and
reverse the direction for negative charge). The electron is thus deflected toward
the page bottom. A moment later, say, when it reaches point Q, the force is still
perpendicular to the velocity and is in the direction shown.
Because the force is always perpendicular to υ, the magnitude of υ does not
change — the electron moves at constant speed. If the force on a particle is
always perpendicular to its velocity υ, the particle moves in a circle (fig. 11.9)
and has a centripetal acceleration a = υ2/r.
We find the radius of curvature using Newton‘s second law. The force is
given by F = q υ B as sin θ = 1.
F = ma.
162
Fig. 11.9. Example 11.4. Force F exerted by a uniform magnetic field on a moving charged
particle (in this case, an electron) produces a circular path
m 2
We insert F and a into Newton‘s second law: q B
.
r
We solve it for r and find the radius of circle and period T of rotation
m
2πr 2πm
r
,
T
.
qB
qB
The radius of circle is directly proportional to speed of a charge, period T
of a charge rotation doesn‘t depend on a speed magnitude and both values are
inverse proportional to magnitude of magnetic field. To get r we put in
the numbers:
(9.1 10 31 kg) (2.9 107 m/s)
r
1.1 10 2 m 1.1 cm.
19
(1.6 10 C) (0.010 T)
NOTE. Thus a charged particle moves in a circular path with constant
centripetal acceleration in a uniform magnetic field.
PROBLEMS
1. a) What is the force per meter of length on a straight wire carrying
a 7.40 A current when perpendicular to a 0.90 T uniform magnetic field?
b) What if the angle between the wire and field is 45.0°? (Answer: a) 6.7 N/m;
b) 4.7 N/m)
2. How much current is flowing in a wire 4.20 m long if the maximum
force on it is 0.900 N when placed in a uniform 0.0800 T field? (Answer: 2.68 A)
3. In a magnetic field with B = 0.5 T, for what path radius, will an electron
circulate at 0.1 the speed of light (c = 3·108 m/s)? What will be its kinetic
energy? Given mass of electron 9.1·10–31 kg, its charge –1.6·10–19 C. (Answer:
3.41·10–4 m; 4.095·10–16 J)
163
TESTS
1. SI unit of the magnetic field B is:
a) Oersted;
b) Tesla;
c) Gauss;
d) Maxwell.
2. Current is passed through a straight wire. The lines of magnetic field
produced by it are:
a) circular and endless;
b) oval in shape but endless;
c) straight but endless;
d) None of these.
3. In an uniform magnetic field, lines of magnetic field are:
a) inclined;
b) parallel;
c) circular:
d) perpendicular.
4. When the direction of current is opposite in two parallel wires placed
near each other, they will:
a) attract each other;
b) repel each other;
c) neither attract nor repel;
d) sometimes attract sometimes repel.
5. Force acting on a moving charge in a magnetic field varies with
the velocity υ as:
a) 1/υ; b) υ;
c) υ2;
d) υ½.
6. Force acting on a moving charge in a magnetic field is maximum when
velocity of charge is inclined to field at:
a) 0°;
b) 60°; c) 30°; d) 90°.
7. A wire of length 2 m carries a current of 10 A. What is the force acting
on it when it is placed at an angle 45° to the uniform magnetic field of 0.15 T:
3
N.
a) 1.5 N;
b) 3N; c) 3 2 N ;
d)
2
8. An electron is moving parallel to the magnetic field B with velocity υ.
The force acting on it is:
a) Beυ;
b) Be/υ; c) zero;
d) eυ/B.
9. The force acting on a charge q moving with velocity υ in the magnetic
field B is given by:
a) q/ υB;
b) υB/q;
c) qυB;
d) υ/Bq.
10. An electron moving with velocity 106 ms–1 enters a magnetic field and
describes a circle of radius 0.1 m, the magnetic field B is:
a) 1.8·10–4 T;
b) 5.5·10–5 T;
c) 1.4·10–5 T;
d) 14·10–5 T.
11. An electron is moving with velocity 3·107 ms–1 perpendicular to
the magnetic field of 2.0 T. The magnitude of the force acting on it is:
a) 96·10–12 N;
b) 9.6·10–12 N;
c) 0.96·10–12 N;
d) none of the above.
164
12. A particle having charge q enters with a uniform velocity υ
in the magnetic field B, the radius of the path in which it moves is (m is
the mass of particle):
a) mυ/Bq;
b) Bυ/mq;
c) mB/qυ;
d) none of the above.
11.4. ELECTROMAGNETIC INDUCTION AND FARADAY’S LAW
We know that there are two ways in which electricity and magnetism are
related:
1) an electric current produces a magnetic field;
2) a magnetic field exerts a force on an electric current or moving electric
charge.
Here we deal with the reverse phenomena i. e. the production of an electric
current from a magnetic field. This phenomenon of producing an electric current
from a magnet or a magnetic field is known as electromagnetic induction.
Faraday has found that a steady magnetic field produces no current in
a conductor (fig. 11.10, c), but a changing magnetic field can produce an electric
current (Fig. 11.10, a, b). Such a current is called an induced current. When
the magnetic field changes, a current flows as if there were a source of emf in
the circuit. Therefore an induced emf is produced by a changing magnetic field.
Fig. 11.10. (a) A current is induced when a magnet is moved toward a coil. (b) The induced
current is opposite when the magnet is moved away from the coil. Note that the galvanometer
zего is at the center of the scale and the needle deflects left or right, depending on
the direction of the current. In (c) no current is induced if the magnet does not move relative
to the coil. It is the relative motion that counts: the magnet can be held steady and the coil
moved, which also induces an emf
165
Fig. 11.10 shows that if a magnet is moved quickly into a coil of wire,
a current is induced in the wire. If the magnet is quickly removed, a current is
induced in the opposite direction (B through the coil decreases). Furthermore, if
the magnet is held steady and the coil of wire is moved toward or away from
the magnet, again an emf is induced and a current flows. Motion or change is
required to induce an emf. It doesn‘t matter whether the magnet or the coil
moves. It is their relative motion that counts.
11.5. FARADAY’S LAW OF INDUCTION
Faraday investigated quantitatively what factors influence the magnitude of
the emf induced. He found first of all that the more rapidly the magnetic field
changes, the greater the induced emf. But the emf is not simply proportional to
the rate of change of the magnetic field, B. Rather the emf is proportional to
the rate of change of the magnetic flux, Ф, passing through the circuit or loop of
area S. Magnetic flux Ф for a uniform magnetic field is defined as
Ф B S BScosα,
(11.5)
here B┴ is the component of the magnetic field В perpendicular to the face of
the loop, and α is the angle between В and normal to the surface of the loop
(fig. 11.11).
Fig. 11.11. Determining the magnetic flux through a flat loop of wire of area S
The unit of magnetic flux Ф is called a weber: 1 Wb = 1 T·m2. Hence, one
Weber may be defined as the amount of magnetic flux produced by uniform
magnetic field of 1 Tesla normal to an area of 1 meter2.
With this definition of flux Ф one can write down the results of Faraday‘s
investigations: the emf εi induced in a circuit is equal to the rate of change of
magnetic flux through the circuit:
Ф
.
t
εi
166
(11.6)
This fundamental result is known as Faraday's law of induction, and is
one of the basic laws of electromagnetism.
If the circuit contains N loops that are closely wrapped so the same flux
passes through each, the emfs induced in each loop add together, so
εi
N
Ф
.
t
(11.7)
Example 11.5. A loop of wire in a magnetic field.
A square loop of wire of side l = 5.0 cm is in a uniform magnetic field
B = 0.16 T. What is the magnetic flux in the loop (a) when B is perpendicular to
the face of the loop and (b) when B is at an angle of 30° to the area S of
the loop? (c) What is the magnitude of the average current in the loop if it has a
resistance of 0.012 Ω and it is rotated from position (b) to position (a) in 0.14 s?
Solution. We use the definition ФB = BS to calculate the magnetic flux.
Then we use Faraday‘s law of induction to find the induced emf in the coil, and
from that the induced current (I = ε/R).
The area of the coil is S = l2 = (5.0·10–2 m)2 = 2.5·10–3 m2, and the direction
of S is perpendicular to the face of the loop.
a) B is perpendicular to the coil‘s face, and thus parallel to S, so
ФB = BS cos0o = (0.16 T)·(2.5·10–3 m2)·l = 4.0·10–4 Wb.
b) The angle between B and S is 30°, so
ФB = BS cosθo = (0.16 T)·(2.5·10–3 m2)·cos 30° = 3.5·10–4 Wb.
c) The magnitude of the induced emf is
I
ε
R
The current is then I
3.6 10 4 V
0.030 A 30 mA .
0.012
ε 3.6 10 4 V
0.030 A 30 mA .
R
0.012
The minus signs in equation (11.6) is placed there to remind us in which
direction the induced emf εi acts. Experiments show that an induced emf εi gives
rise to a current whose magnetic field opposes the original change in flux.
This is known as Lenz’s law.
Let‘s apply Lenz‘s law to the relative motion between a magnet and a coil
in fig. 11.10. The changing flux through the coil induces an emf εi, which
produces a current in the coil. And this induced current produces its own
magnetic field. In fig. 11.10, a the distance between the coil and the magnet
decreases. The magnetic field (and number of field lines), and therefore the flux
through the coil increases. The magnetic field of the magnet points upward.
To oppose this upward increase, the magnetic field inside the coil produced by
the induced current points downward. Thus, Lenz's law tells us that the current
moves as shown (use the right-hand rule). In fig. 11.10, b the flux decreases
(because the magnet is moved away), so the induced current produces an upward
167
magnetic field through the coil that is ―trying‖ to maintain the status quo. Thus
the current in fig. 11.10, b is in the opposite direction from fig. 11.10, a.
According to equation (11.5) induced emf εi can be produced by any one of
the following methods:
by changing the magnetic field В;
by changing the area S of the circuit;
by changing the angle α, i. e. the relative orientation of the field В and
area S.
Consider induced emf εi which is produced by changing the area S of the
circuit (fig. 11.12). Assume that a uniform magnetic field В is perpendicular to
the area bounded by the U-shaped conductor and the movable rod resting on it.
Fig. 11.12. A conducting rod is moved to the right on a U-shaped conductor in a uniform
magnetic field В that points out of the page
If the rod moves at a speed υ, it travels a distance Δx = υΔt in a time Δt.
Therefore, the area of the loop increases by an amount ΔS = l Δx = l υΔt
in a time Δt. By Faraday's law there is an induced emf εi whose magnitude
is given by:
Ф B S Bl t
(11.8)
εi
Bl .
t
t
t
This equation (11.8) is valid as long as В, l and υ are mutually
perpendicular. An emf εi induced by a conductor moving with a velocity υ at
an angle α with a magnetic field B is given by:
εi
Bl sinα .
(11.9)
An emf εi induced on a conductor moving in a magnetic field is sometimes
called motional emf.
Right hand rule is used to determine the direction of the induced efm εi of
a conductor moving in a magnetic field (fig. 11.13). If the first finger points in
the direction of the magnetic field, the thumb points in the direction of motion of
the conductor, then the central finger points in the direction of the induced efm.
168
Fig. 11.13. Right hand rule for determining the direction of the induced efm εi of a conductor
moving with velocity υ in a magnetic field B
Example 11.6. Pulling a coil from a magnetic field.
A 100-loop square coil of wire, with side l = 5.00 cm and total resistance
100 Ω, is positioned perpendicular to a uniform 0.600 T magnetic field, as
shown in fig. 11.14.
It is quickly pulled from the field at constant speed (moving perpendicular
to B) to a region where B drops abruptly to zero. At t = 0, the right edge of
the coil is at the edge of the field. It takes 0.100 s for the whole coil to reach
the field-free region. Find: a) the rate of change in flux through the coil;
b) the emf and current induced; c) how much energy is dissipated in the coil?
d) What was the average force required (Fext)?
Fig. 11.14. Example 11.6. The square coil in a magnetic field B = 0.600 T is pulled abruptly
to the right to a region where B = 0
Solution. We start by finding how the magnetic flux, ФB = BS, changes
during the time interval Δt = 0.100 s. Faraday‘s law then gives the induced emf
and Ohm‘s law gives the current.
169
a) The area of the coil is S = l2 = (5.00·10–2 m)2 = 2.50·10–3 m2. The flux
through one loop is initially ФB = BS = (0.600 T)·(2.50·10–3 m2) = 1.50·10–3 Wb.
After 0.100 s, the flux is zero. The rate of change in flux is constant (because
the coil is square), equal to
ФB
t
0 (1.50 10 3 Wb)
0.100 s
1.50 10 2 Wb/s.
b) The emf induced in the 100-loop coil during this 0.100 s interval is
ε
N
ФB
t
( 100) ( 1.50 10 2 Wb/s) 1.50 V.
The current is found by applying Ohm‘s law to the 100-Ω coil:
I
ε 1.50 V
1.50 10 2 A 15.0 mA.
R 100
By Lenz‘s law, the current must be clockwise to produce more B into
the page and thus oppose the decreasing flux into the page.
c) The total energy dissipated in the coil is the product of the power
(P = I2R) and the time t:
E = Pt = I2Rt = (1.50·10–2 A)2·(100 Ω)·(0.100 s) = 2.25·10–3 J.
d) We can use the result of part (c) and apply the work-energy principle:
the energy dissipated E is equal to the work A needed to pull the coil out of
the field. Because A = Fd where d = 5.00 cm, then
A 2.25 10 3 J
F
0.0450 N.
d 5.00 10 2 m
Alternate Solution (d) We can also calculate the force directly using
F = I·l·B, which here for constant B is F = IlB. The force the magnetic field
exerts on the top and bottom sections of the square coil of fig. 11.14 are in
opposite directions and cancel each other. The magnetic force Fm exerted on
the left vertical section of the square coil acts to the left as shown because
the current is up (clockwise). The right side of the loop is in the region where
B = 0. Hence the external force, to the right, needed to just overcome
the magnetic force to the left (on N = 100 loops) is
Fext = NIlB = (100)·(0.0150 A)·(0.0500 m)·(0.600 T) = 0.0450 N,
which is the same answer, confirming our use of energy conservation above.
Example 11.7. Does a moving
airplane develop a large emf?
An airplane travels 1000 km/h in
a region where the Earth‘s magnetic
field is about 5·10–5 T and is nearly
vertical (fig. 11.15). What is the
potential difference induced between
the wing tips that are 70 m apart?
Fig. 11.15. Example 11.7. A moving airplane
170
Solution. We consider the wings to be a 70-m-long conductor moving
through the Earth‘s magnetic field. We use equation (11.9) to get the emf.
Since υ = 1000 km/h = 280 m/s, and
B , we have εi Bl
εi = Blυ = (5·10–5 T)·70 m · 280 m/s ≈ 1 V.
Example 11.8. Electromagnetic blood-flow measurement.
The rate of blood flow in our body‘s vessels can be measured using
the apparatus shown in fig. 11.16, since blood contains charged ions. Suppose
that the blood vessel is 2.0 mm in diameter, the magnetic field is 0.080 T, and
the measured emf is 0.10 mV. What is the flow velocity υ of the blood?
Fig. 11.16. Example 11.8. Measurement of blood velocity from the induced emf
Solution. The magnetic field B points horizontally from left to right
(N pole toward S pole). The induced emf acts over the width l = 2.0 mm
of the blood vessel, perpendicular to B and υ (fig. 11.16). We can then use
equation εi = Blυ to get υ. We solve for υ in equation εi = Blυ:
εi
(1.0 10 4 V)
0.63 m/s.
Bl (0.080 T) (2.0 10 3 m)
11.6. SELF-INDUCTANCE
Let us consider circuit contains a coil of N turns (fig. 11.17). When
a changing current passes through the coil (or solenoid), a changing magnetic
flux is produced inside the coil, and this in turn induces an emf in that same coil
(fig. 11.17). This induced emf opposes the change in flux (Lenz‘s law). For
example, if the current through the coil is increasing, the increasing magnetic
flux induces an emf that opposes the original current and tends to retard its
increase. If the current is decreasing in the coil, the decreasing flux induces an
emf in the same direction as the current, thus tending to maintain the original
current.
171
Fig. 11.17. The current I in the circuit causes a magnetic field B in the coil and hence
a magnetic flux Ф through the coil. When the current changes, the flux Ф changes also and
a self-induced emf appears
Self-inductance occurs when a changing current in a circuit results in
an induced emf that opposes the change in the circuit itself. Self-inductance
occurs because some of the magnetic flux produced in a circuit passes through
that same circuit.
For a coil carrying current I, there is a magnetic field produced around it.
The value of B at each point is proportional to the current. Therefore
the magnetic flux Ф passing through the every loop of the coil is also
proportional to the current I in the coil:
Ф ~ I or Ф = LI,
(11.10)
where L is constant of proportionality and is called the self-inductance or
inductance.
If current I is equal to 1 (unit) then from equation (11.10) L = Ф. Hence,
the self-inductance L of a circuit is equal to the flux linked with it when a unit
current flows through the circuit.
According to Faraday‘s law the emf εi induced in a coil of self-inductance
L is:
Ф
I
(11.11)
εi
N
L
.
t
t
Let ΔI/Δt be equal to a unit then the emf εi = –L. Hence, self-inductance L
of a circuit is numerically equal to the induced emf εi set up in it, when the rate
of change of current through the circuit is unity. The self-inductance L is
measured in henries H:
1 Volt
1 Henry
.
1 ampere/sec
The magnitude of L depends on the geometry and on the presence of
a ferromagnetic material.
Circuits always contain some inductance, but often it is quite small unless
the circuit contains a coil of many turns. A coil that has significant selfinductance L is called an inductor.
Inductance is shown on circuit diagrams by the symbol:
.
172
Example 11.9. Solenoid inductance.
(a) Determine a formula for the self-inductance L of a tightly wrapped and
long solenoid containing N turns of wire in its length l and whose crosssectional area is S. (b) Calculate the value of L if N = 100, l = 5.0 cm,
S = 0.30 cm2 and the solenoid is air filled.
Solution. To determine the inductance L, it is usually simplest to start with
NФ
equation L
, so we need to first determine the flux Ф.
I
The magnetic field inside a solenoid (ignoring end effects) is constant:
B = μonI, where the number of loops per unit length n = N/l.
NФ
I
The flux is Ф = BS = μoNIS/l, so L
b) Since μo = 4π·10–7 T·m/A, then
L
(4π 10
7
μ0 N 2S
.
l
T m/A) (100)2 (3.0 10
5.0 10
2
5
m2 )
7.5 μH.
m
NOTE. Magnetic field lines “stray” out of the solenoid (see fig. 11.1, b),
especially near the ends, so our formula is only an approximation.
Example 11.10.
A solenoid that has a length equal to 25.0 cm, a radius equal to 0.800 cm,
and 400 turns is in a region where a magnetic field of 600 G exists and makes an
angle of 60° with the axis of the solenoid. (a) Find the magnetic flux through the
solenoid. (b) Find the magnitude of the average emf induced in the solenoid if
the magnetic field is reduced to zero in 1.40 s.
Solution. We can use its definition to find the magnetic flux through
the solenoid and Faraday‘s law to find the emf induced in the solenoid when
the external field is reduced to zero in 1.4 s.
a) Express the magnetic flux Ф through the solenoid in terms of N, B, S,
and θ: Ф = NBS cos θ = NBπR2 cos θ.
Substitute numerical values and evaluate Ф:
Ф = 400·(60.0 mT)·3.14·(0.00800 m) 2·cos 600 = 2.41 mWb.
b) Apply Faraday‘s law to obtain the emf induced in the solenoid:
Ф
0 2.41 mWb
εi
1.72 mV.
t
1.40 s
11.7. MUTUAL INDUCTION
If two coils of wire are placed near each other, as in fig. 11.18, a changing
current in one will induce an emf in the other. According to Faraday‘s law,
the emf ε2 induced in coil 2 is proportional to the rate of change of magnetic flux
passing through it. This flux is due to the current I1 in coil 1 (which is called
173
the primary coil), and it is often convenient to express the emf in coil 2 (which is
called the secondary coil) in terms of the current in coil 1.
Fig. 11.18. A changing current in one coil will induce a current in the second coil
We let Ф21 be the magnetic flux in each loop of coil 2 created by
the current in coil 1. If coil 2 contains N2 closely wrapped loops, then N2 ·Ф21 is
the total flux passing through coil 2. If the two coils are fixed in space, N2·Ф21 is
proportional to the current I1 in coil 1; the proportionality constant is called
the mutual inductance, M21, defined by
N 2Ф21
(11.12)
M 21
.
I1
The emf ε2 induced in coil 2 due to a changing current in coil 1 is, by
Faraday‘s law,
Ф21
ε2
N2
.
(11.13)
t
Let us combine this with equation (11.12) rewritten as Ф21 = M21 I1 / N2
and obtain
I
ε2
M 21 1 .
t
This relates the change in current in coil 1 to the emf it induces in coil 2.
The mutual inductance of coil 2 with respect to coil 1, M21, is a ―constant‖
in that it does not depend on I1; M21 depends on ―geometric‖ factors such as
the size, shape, number of turns, and relative positions of the two coils, and also
on whether iron (or some other ferromagnetic material) is present. For example,
if the two coils in fig. 11.18 are farther apart, fewer lines of flux can pass
through coil 2, so M21 will be less.
Suppose we consider the reverse situation: when a changing current in
coil 2 induces an emf in coil 1. In this case,
I
ε1
M12 2 ,
t
where M12 is the mutual inductance of coil 1 with respect to coil 2. It is possible
to show that M12 = M21.
174
Hence, for a given arrangement we do not need the subscripts and we can
let M = M12 = M21, so that
I
(11.14a)
ε1
M 2
t
I
and
(11.14b)
ε2
M 1.
t
The SI unit for mutual inductance is the henry (H), where
1 H = l V·s/A = l Ω·s.
Example 11.11.
Two coils have mutual inductance of 1.5 H. If the current in the primary
circuit is raised to a value of 50 A in one second after closing the circuit, what is
the induced efm in the secondary?
Solution. Here, mutual inductance of the coils M = 1.5 H. Rate of change
I 50 A
50 A/s.
of current in the primary is:
t 1s
Then the induced efm in the secondary is:
I
ε M
1.5 H 50 A/s 75 V.
t
Example 11.12. Solenoid and coil.
A long thin solenoid of length l and cross-sectional area S contains N1
closely packed turns of wire. Wrapped around it is an insulated coil of N2 turns,
fig. 11.19. Assume all the flux from coil 1 (the solenoid) passes through coil 2,
and calculate the mutual inductance I1.
Fig. 11.19. Example 11.12. A long thin solenoid of length l and cross-sectional area S
contains N1 closely packed turns of wire. Wrapped around it is an insulated coil of N2 turns
Solution. We first determine the flux produced by the solenoid, all of
which passes uniformly through coil N2, using equation for the magnetic field
inside the solenoid:
B μ0
N1
I1,
l
where n = N1/l is the number of loops in the solenoid per unit length, and is
the current in the solenoid.
175
The solenoid is closely packed, so we assume that all the flux in the solenoid
stays inside the secondary coil. Then the flux Ф21 through coil 2 is
N
Ф12 BS μ 0 1 I1S.
l
N 2Ф21 μ 0 N1N 2 S
Then the mutual inductance is M
.
I1
l
NOTE. We calculated M21; if we had tried to calculate M12, it would have
been difficult. Given M12 = M21 = M, we did the simpler calculation to obtain M.
Note again that M depends only on geometric factors, and not on the currents.
11.8. ENERGY STORED IN A MAGNETIC FIELD
Since an inductor of inductance L in a circuit serves to oppose any change
in the current I through it, work must be done by an external source such as
a battery in order to establish a current in the inductor. One can conclude
that energy can be stored in an inductor. The role played by an inductor
in the magnetic case is analogous to that of a capacitor in the electric case.
The power P, or rate at which an external emf εext works to overcome the selfinduced emf εi and pass current I in the inductor is
Wext
t
PL
I εext .
If only the external emf and the inductor are present, then εext = –εi which
implies
PL
Wext
t
I εi
IL
I
.
t
If the current is increasing with ΔI/Δt > 0, then P > 0 which means that
the external source is doing positive work to transfer energy to the inductor.
Thus, the internal energy UB of the inductor is increased. On the other hand,
if the current is decreasing with ΔI/Δt < 0, we then have P < 0. In this case,
the external source takes energy away from the inductor, causing its internal
energy to go down. The total work done by the external source to increase
the current form zero to I is then:
1 2
Wext
LI .
2
This is equal to the magnetic energy stored in the inductor:
1 2
WB
LI .
2
The above expression is analogous to the electric energy stored in
a capacitor:
WE
1 q2
.
2C
176
We comment that from the energy perspective there is an important
distinction between an inductor and a resistor. Whenever a current I goes
through a resistor, energy flows into the resistor and dissipates in the form of
heat regardless of whether I is steady or time-dependent (recall that power
dissipated in a resistor is PR = IUR = I2R). On the other hand, energy flows into
an ideal inductor only when the current is varying with ΔI/Δt > 0. The energy is
not dissipated but stored there; it is released later when the current decreases
with ΔI/Δt < 0. If the current that passes through the inductor is steady, then
there is no change in energy since PL = LI(ΔI/Δt) = 0. It makes sense to say
there is no energy in inductor with no current.
The energy stored in the magnetic field generated by the current flowing
through the inductor can be also given by:
LI 2 IФ Ф 2
(11.15)
W
,
2
2
2L
as Ф = LI.
If the current that passes through the inductor is steady, then change in
energy since PL = LI(ΔI/Δt) = 0. It makes sense to say there is no energy in
inductor with no current.
The energy per unit volume or magnetic energy density is:
1 B2
(11.16)
wB energy density
,
2 μμ0
where μ0 is permeability of free space (μ0 = 4π·10–7 T·m/A), μ is the magnetic
permeability of the material.
1
This equation is analogous to that for an electric field: w E
εε 0 E 2 .
2
Example 11.13. Energy stored in solenoid.
Calculate the energy associated with the magnetic field of a 200-turn
solenoid in which a current of 1.75 A produces a magnetic flux of 3.70·10–4 Wb
in each turn.
Solution. The energy W stored in the solenoid when it is carrying a current
1 2
I is W
LI . To determine the inductance L, it is usually simplest to start
2
NФ
with equation L
I
L = (200)·(3.70·10–4)/1.75 = 4.23·10–2 H.
Thus, the energy W stored in the solenoid is:
LI 2 (4.23 10 2 H) (1.75 A) 2
W
0.065 J.
2
2
177
Example 11.14. Energy density.
Compare the energy density stored in Earth‘s electric field near its surface
to that stored in Earth‘s magnetic field near its surface.
Solution. We can compare the energy density stored in Earth‘s electric
field to that of Earth‘s magnetic field by finding their ratio. We‘ll take Earth‘s
magnetic field to be 0.3 G and its electric field to be 100 V/m.
The energy density in an electric field E is given by: wE
1
ε0 E 2 .
2
The energy density in a magnetic field B is given by: wB
1 B2
.
2 μ0
Express the ratio of wB to wE to obtain:
wB
wE
0.3 G
B2
μ 0ε 0 E
2
1T
2
104 G
(4π 10 7 N/A 2 ) (8.854 10 12 C2 /N m 2 ) (100 V/m) 2
8.09 103.
11.9. LC CIRCUIT AND ELECTROMAGNETIC OSCILLATIONS
In any electric circuit, there can be three basic components: resistance R,
capacitance C, and inductance L, in addition to a source of emf. Let‘s consider
an LC circuit, one that contains only a capacitance C and an inductance, L,
fig. 11.20. This is an idealized circuit in which we assume there is no resistance.
Let us suppose the capacitor in fig. 11.20, A is initially charged so that one plate
has charge q0 and the other plate has charge –q0, and the potential difference
across it is U = q/C. Suppose that at t = 0, the switch is closed (fig. 11.20, B).
The capacitor immediately begins to discharge.
A
B
Fig. 11.20. (A) an LC sircuit: a capacitor C be charged q0 and connected to an inductor L.
(B) The switch is closed in the LC circuit. At the instant shown, the current is increasing so
the polarity of induced emf in the inductor is as shown
178
The moment the circuit is completed, the charge on the capacitor starts
decreasing, giving rise to current in the circuit. Let q and I be the charge and
current in the circuit at time t. Since ΔI/Δt is positive, the induced emf in L will
have polarity as shown, i. e. Ub < Ua.
We know that a capacitor C and an inductor L can store electrical and
magnetic energy, respectively. When a capacitor (initially charged) is connected
to an inductor, the charge on the capacitor and the current in the circuit exhibit
the phenomenon of electromagnetic oscillations similar to oscillations in
mechanical system.
Let us now try to visualize how this electrical oscillation takes place
in the circuit. Fig. 11.21, a shows a capacitor with initial charge q0 connected to
an ideal inductor. The electrical energy stored in the charged capacitor is
WE
a
1 q02
2 C
1
CU 02 .
2
c
b
d
e
Fig. 11.21. Electromagnetic oscillations in LC circuit
Since, there is no current in the circuit, energy in the inductor is zero
(WB = 0). Thus, at times t = 0, t = T/2, t = T and so on (where T = 1/ν is
the period; and angular frequency ( ω 2πν
1
) all the energy is stored in
LC
the electric field of the capacitor:
W
WE
1 q0 2
.
2 C
At t = 0, the switch is closed and the capacitor starts to discharge
(fig. 11.21, b). As the current increases, it sets up a magnetic field in
the inductor and thereby, some energy gets stored in the inductor in the form of
magnetic energy:
1 2
WB
LI .
2
As the current reaches its maximum value I0 (at t = T/4, 3T/4 and so on,
as in fig. 11.21, c), all the energy is stored in the magnetic field of the inductor
(WE = 0):
1
W WB
LI 0 2 .
2
179
You can easily check that the maximum electrical energy equals
the maximum magnetic energy. The capacitor now has no charge and hence no
energy. The current now starts charging the capacitor, as in fig. 11.21, d. This
process continues till the capacitor is fully charged (at t = T/2) (fig. 11.21, e).
But it is charged with a polarity opposite to its initial state in fig. 11.21, a.
The whole process just described will now repeat itself till the system reverts to
its original state. Thus, the energy oscillates between being stored in the electric
field of the capacitor and in the magnetic field of the inductor. The total energy
W is constant, and energy is conserved (fig. 11.22).
Fig. 11.22. Energy WE and WB stored in the capacitor and the inductor as a function of time.
Note how the energy oscillates between electric and magnetic. The dashed line at the top is
the (constant) total energy: W = WE + WB
Natural frequency ν of oscillation of the LC circuit is given by:
1
.
2π LC
(11.17)
At any time the sum of electric and magnetic energy stored in the capacitor
and in the inductor has the constant value that is equal to the maximum electrical
energy in the capacitor and to the maximum magnetic energy in the inductor:
W
q2
2C
1 2
LI
2
1
CU 02
2
1 2
LI 0 .
2
Example 11.15. LC circuit.
A 1200 pF capacitor is fully charged by a 500 V dc power supply. It is
disconnected from the power supply and is connected, at t = 0, to a 75 mH
inductor. Determine: a) the initial charge on the capacitor; b) the maximum
current; c) the frequency ν and period T of oscillation; d) the total energy
oscillating in the system.
Solution. We use the analysis above, and the definition of capacitance
q = CU.
a) The 500 V power supply, before being disconnected, charged
the capacitor to a charge of q0 = CU = (1.2·10–9 F)·500 V = 6.0·10–7 C.
180
b) The maximum current, Imax, is
q0
6.0 10 7 C
I max ωq0
63 mA.
LC
0.075 H (1.12 10 9 F)
ω
1
c) The frequency v can be obtained from:
17 kHz,
2π 2π LC
and the period T is T
1
6.0 10 5 s.
d) Finally the total energy is W
q02
2C
(6.0 10 7 C)2
9
1.5 10 4 J .
2(1.2 10 F)
PROBLEMS
1. The magnetic field perpendicular to a single 15.6 cm diameter circular
loop of copper wire decreases uniformly from 0.550 T to zero. If the wire is
2.05 mm in diameter, how much charge moves past a point in the coil during
this operation? (Answer: 4.21 C)
2. What is the mutual inductance of a pair of coils, if a current change
of 6 A in one coil causes the flux in the second coil of 2000 turns to change by
12·10–4 Wb. (Answer: 0.4 H)
3. A coil that has a self-inductance of 2.00 H and a resistance of 12.0 Ω
is connected to an ideal 24.0 V battery. (a) What is the steady-state current?
(b) How much energy is stored in the inductor when the steady-state current is
established? (Answer: 2.0 A; 4.0 J)
4. An LC circuit has an inductance of 3.0 mH and a capacitance of 10 μF.
Calculate (a) the angular frequency and (b) the period of oscillation. (Answer:
5800 rad/s; 1.1 ms)
5. A capacitor C = 25 μF is charged to voltage of 200 V and then
discharges on inductance L = 10 mH. Calculate the whole energy oscillated
in the circuit, the maximum value of current and oscillation frequency. (Answer:
0.5 J; 10 A; 318 Hz)
TESTS
1. The unit of magnetic flux in SI system is:
a) Oersted;
b) Henry;
c) Tesla;
d) Weber.
2. The varying magnetic field through a conductor produces electromotive
force. This is in accordance with:
a) Faraday‘s law;
b) Lenz‘s law;
c) Laplace‘s law;
d) Ampere‘s law.
181
3. A magnet is taken towards a coil and is moved (i) quickly (ii) slowly,
then the induced emf is:
a) larger in first case;
b) larger in second case;
c) equal in both cases;
d) None of the above.
4. ‖The induced emf is always in such a direction so as to oppose
the change that causes it‖ is called:
a) Lenz‘s law;
b) Faraday‘s law;
c) Kirchoff‘s law;
d) Joule‘s law.
5. The expression for the induced emf contains a negative sign εi
Ф
.
t
What is the significance of the negative sign?
a) the induced emf opposes the changes in the magnetic flux;
b) the induced emf is produced only when the magnetic flux decreases;
c) the induced emf is opposite to the direction of the flux;
d) none of above.
6. The induced emf in a coil rotating in a magnetic field is maximum when
the angle between the plane of the coil and direction of the field is:
a) π/4;
b) zero;
c) π/2;
d) some angle other than mentioned above.
7. If line of magnetic field is parallel to a surface, then the magnetic flux
through the surface is:
a) small but not zero;
b) infinite;
c) zero;
d) large but not infinite.
8. The unit of self-inductance in SI system is:
a) Henry;
b) Tesla;
c) Weber;
d) Oersted.
9. When a conductor of length l is moved perpendicular to a uniform
magnetic field B with uniform velocity υ, the emf induced is:
a) B l υ;
b) B l/υ;
c) B υ/l;
d) none of these.
10. An airplane is moving north horizontally with a speed of 720 km/h at
a place where vertical component of earth‘s field is 0.5·10–4 T. What
is the induced emf set up between the tips of the wings 10 m apart?
a) 1.0 V;
b) 0.1 V;
c) 10 V;
d) 0.01 V.
182
12. GEOMETRICAL OPTICS
12.1. THE RAY MODEL OF LIGHT
In a transparent homogeneous medium light travels in straight lines. This
ray model is useful in describing many aspects of light such as reflection,
refraction, and the formation of images by mirrors and lenses.
The speed of light in vacuum and in air is c = 3·108 m/s. In other
transparent materials the light speed is always less than in vacuum. The ratio
of the speed of light in vacuum to the speed in a given material is called
the absolute index of refraction n of this material:
n
c
.
υ
(12.1)
The refraction index n is different for various materials and it is never less
than one. The refractive index of a medium is a measure of light speed in
the medium:
υ
c
.
n
(12.2)
The relation between light speeds in two mediums is called the relative
refractive index of the second medium with respect to the first one n21:
n2 υ1
(12.3)
n21
.
n1 υ2
Reflection Law. When light strikes the surface of an object, some
of the light is reflected. The rest can be absorbed by the object (and transformed
to thermal energy) or transmitted through (in case of transparent medium like
glass or water).
When a light ray strikes a flat surface divided two media (fig. 12.1), we
define the angle of incidence α as the angle between an incident ray and
the normal (perpendicular) to the surface, and the angle of reflection, γ as
the angle between the reflected ray and the normal. The law of reflection is:
the incident ray, the normal to the surface and the reflected ray lie in the same
plane; the angle of reflection equals to the incidence angle:
α = γ.
(12.4)
Fig. 12.1. Law of reflection
183
12.2. IMAGE FORMATION BY A FLAT MIRROR
When light is incident upon a rough surface, even microscopically rough
such as this page, it is reflected in many directions, as shown in fig. 12.2. This is
called the diffuse reflection (fig. 12.2). Reflection from a mirror is known as
specular reflection (fig. 12.3) (―Speculum‖ is Latin for mirror.)
Fig. 12.2. Diffuse reflection
Fig. 12.3. Specular reflection
A flat mirror is one with a smooth flat reflecting surface. Fig. 12.4 shows
how an image is formed by a plane mirror according to the ray model. Each ray
that reflects from the mirror and enter the eye appear to come from a single
point (called the image point) behind the mirror, as shown by the dashed lines.
That is, our eyes and brain interpret any ray that enters an eye as having traveled
straight-line path. The point from which each ray seems to come is one point on
the image. For each point on the object, there is a corresponding image point.
Fig. 12.4. Formation of a virtual image by a plane mirror
The image appears as far behind the mirror as the object is in front.
The perpendicular distance from mirror to image (the image distance) is equals
the perpendicular distance from object to mirror (the object distance). From
the geometry, we can also see that the height of the image is the same as that of
the object.
This image would not appear on paper or film placed at the location of
the image, therefore it is called a virtual image. This is to distinguish it from
a real image in which the light does really pass through the image and which
therefore could appear on film or on a while sheet of paper or screen placed at
the position of the image. Our eyes can see both real and virtual images, as long
as the diverging rays enter our pupils.
184
12.3. REFRACTION. SNELL’S LAW
Refraction is the bending of a light ray when it
enters a medium where the refractive index is
different (fig. 12.5). The amount of bending depends
on the indices of refraction of the two media and is
described quantitatively by Snell’s Law:
– the incident ray, the normal to the surface and
the refrected ray lie in the same plane;
– the ratio of the sine of the angle of incidence α
to the sine of the angle of refraction β is equal to
Fig. 12.5. Snell‘s Law
the reciprocal of the ratio of the refractive indices:
sinα n2
(12.5)
,
sinβ n1
where n1 is the absolute index of refraction of the first medium; n2 is the
absolute index of refraction of the second medium.
12.4. PHENOMENON OF TOTAL INTERNAL REFLECTION
When the light is travelling from medium with bigger refractive index
to the medium with smaller refractive index (for example, from water to air),
the angle of refraction is greater than angle of incidence. If the angle of
incidence increases, the angle of refraction approaches to 90°. The angle of
incidence at which the angle of refraction is equal to 90° is called the critical
angle αcr. For angles of incidence greater than αcr, there is no refracted ray, all
of the incident light is reflected (fig. 12.6). This effect is called total internal
reflection. The formula for critical angle αcr is derived from Snell‘s Law:
n2
n
(12.6)
sinαcr
sin 90 = 2 .
n1
n1
Fig. 12.6. To a total internal reflection
The phenomenon of total internal reflection is used for the fiber optics.
185
12.5. THIN LENSES. RAY TRACING
A thin lens is a transparent object with two refracting surfaces.
The thickness of lens is negligible compared to radii of curvature of these
surfaces. A straight line passed through the curvature centers of these surfaces is
the principal axis of the lens.
The lens can be convex or concave. If the lens is convex, a parallel beam of
light passing through the lens is focused to a point on the axis called focal point.
In this case the lens is called a converging lens. The focal point placed at
the principal axis of the lens is called a principal focal point. The distance of
the principal focal point from the center of the lens is called the focal length.
If the lens is concave, a parallel beam of light passing through the lens is
diverged. This lens is called a diverging lens. The focal point F of a diverging
lens is defined as the point from which refracted rays seem to be emanating.
The distance from F to the lens is called the focal length, just as for
a converging lens (fig. 12.7).
Fig. 12.7. Convex and concave lenses and its principal focuses
The image produced by a converging lens can be constructed using just two
of three rays (fig. 12.8):
1. A ray which is parallel to the optical axis refracts through principal focal
point behind the lens.
2. A ray which passes through the principal focal point in front of the lens
refracts parallel to the optical axis.
3. A ray which passed through the optical center of the lens does not refract
at all.
Fig. 12.8. Principal rays for converging lenses
186
The point where the refracted rays are crossing is the image of the object
point. Actually, any two of these rays will suffice to locate the image point, but
drawing the third ray can serve as a check.
Converging lens can form both real and virtual image depending on
the location of object with respect to foсal distance of lens. Diverging lens can
form virtual images only.
For diverging lens the image produced by two of three rays (fig. 12.9):
1) a ray passing through the optical center of the lens;
2) a ray parallel to the principal axis, which refracts through the lens and
appears to have come from the principal focus;
3) a ray heading towards the principal focus (on the opposite side of
the lens) is refracted by the lens goes parallel to the principal axis.
Fig. 12.9. Principal rays for diverging lenses
The real image is located on the opposite side of the lens and it may be
projected on a screen or film. The virtual image is located on the same side of
the lens as the object and can‘t be projected on a screen or film. But the eye does
not distinguish between real and virtual images — both are visible.
12.6. THE THIN LENS EQUATION. MAGNIFICATION
Optical power is the degree to which a lens converges or diverges light.
It is equal to the reciprocal of the focal length of the device:
1
(12.7)
D
.
F
The shorter the focal length, the stronger the refraction in the lens and
the larger the value of the optical power. For converging lenses the optical
power is positive, while for diverging lenses it is negative. The most common
unit of the optical power measurements is diopter (D): 1 D = 1 m–1.
187
The thin lens equation relates the object distance, image distance and focal
1 1
1
length:
(12.8)
.
d
f
F
where d is the distance (measured along the axis) from the object to the lens; f is
the distance (measured along the axis) from the image to the lens; F is the focal
length of the lens (fig. 12.10).
Fig. 12.10. Scheme for the thin lens equation
When using this equation, signs are very important. Distance d from
the object to the lens is always positive. Distance f from the image to the lens is
positive for real images and negative for virtual ones. Focal length F is positive
for converging lenses and negative for diverging ones.
The magnification of the lens is given by:
H
f
(12.9)
M
,
h d
where H is a size of an image; h is a size of an object.
A magnifying glass (also called a hand lens) is a convex lens that is used to
produce a virtual magnified image of an object (fig. 12.11).
Fig. 12.11. Magnifying glass
188
Magnification of the magnifying glass can be found as:
d0
,
F
f
d
M
(12.10)
where d0 is the distance of the best vision (d0 = 25 cm), F is focal length.
Example 12.1.
What is the position and the size of the image of a 7.6-cm-high leaf placed
at 1m from a +50-mm-focal-length camera lens?
Solution.
d = 1 m = 100 cm;
F = 50 mm = 5 cm;
f=?
H=?
1 1 1
1
1
, f = 5.26 cm.
f F d 5 cm 100 cm
The magnification is: M
H
h
f
d
5.26 cm
100 cm
0.0526.
So: H = M·h = 0.0526·7.6 cm = 0.4 cm.
The image is 4 mm high.
Example 12.2.
An object is placed 10 cm from a 15-cm-focal-length converging lens.
Determine the image position and size.
Solution.
d = 10 cm;
F = 15 cm;
f=?
H=?
1 1 1
1
1
1
, f = –30 cm.
f F d 15 cm 10 cm
30 cm
Because f is negative, the image must be virtual and located on the same
side of the lens as the object. The magnification:
M
H
h
f
d
30 cm
10 cm
3.
The image is three times as large as the object and it is upright.
TESTS
1. A light ray has an angle of incidence of 34°. The reflected ray will make
what angle with the reflecting surface?
a) 34°;
b) 56°;
c) 66°;
d) 74°.
189
2. The critical angle for diamond (n = 2.42) submerged in water (n = 1.33) is:
a) 17°;
b) 24°;
c) 33°;
d) 49°.
3. Calculate the index of refraction for a substance in which light travels at
1.97 · 108 m/s.
a) 1.97;
b) 0.66;
c) 1.42;
d) 1.52.
4. The critical angle of zircon is 31°. Which of the following incident
angles would result in total internal reflection?
a) 17°;
b) 34°;
c) 42°;
d) A and C; e) B and C.
5. A object is placed between F and 2F for a diverging lens. The virtual
image will be located:
a) between F and 2F;
b) between the lens and F;
c) farther than 2F;
d) there is insufficient information to answer the question.
6. The focal length of a converging lens is 15 cm. An object is placed
40 cm away from the lens. The image will be:
a) smaller and real;
b) larger and real;
c) the same size and real;
d) smaller and virtual;
e) larger and virtual.
PROBLEMS
1. A sharp image is located 373 mm behind a 215-mm-focal-length
converging lens. Find the object distance by calculation. (Answer: 508 mm)
2. It is desired to magnify reading material by a factor of 2.5X when
a book is placed 9 cm behind a lens. Describe the type of image this would be.
What type of lens is needed? What is the power of the lens in diopters?
(Answer: 6.7 D upright, magnified; converging lens)
3. An object is located 1.5 m from an 8-D lens. By how much does
the image move if the object is moved 0.9 m closer to the lens? (Answer:
0.02 m)
4. What is the focal length of a magnifying glass of 3.8X magnification
for a normal eye? (Answer: 6.6 cm)
190
13. THE WAVE NATURE OF LIGHT
13.1. ELECTROMAGNETIC WAVES SPECTRUM
The electromagnetic spectrum is distribution of electromagnetic waves
according to their wavelength or frequency. The electromagnetic spectrum
includes radio waves, infrared, visible and ultraviolet light, X-rays and
γ-radiation. The wavelength and frequency are different for different type of
electromagnetic waves, but a speed of these waves in air and vacuum is constant
for all of them and equal to c
3 108
m
.
s
Radio waves have wavelengths λ > 1 mm; they are used to transmit radio
and television signals. Infrared (IR) is invisible waves with wavelengths
1 mm > λ > 760 nm. Most of the radiation emitted by heated objects is infrared.
Visible light (760 nm > λ > 400 nm) is a small part of the electromagnetic
spectrum that human eye can respond. Ultraviolet (UV) is an electromagnetic
radiation with a wavelength from 400 nm to 80 nm, shorter than that of visible
light but longer than X-rays. UV radiation is present in sunlight. A small dose of
ultraviolet radiation is beneficial to humans, but larger doses cause cataracts and
skin cancer. X-rays have great penetrating power and are used extensively
in medical applications as a diagonistic tool. Their wavelength range is
80 nm > λ > 10–5 nm. γ-radiation have wavelengths of less than λ < 10–5 nm.
They are more penetrating than X-rays. Gamma rays are generated by
radioactive atoms.
13.2. LIGHT INTERFERENCE
Light is a transverse, electromagnetic wave that can be seen by humans.
An electromagnetic wave consists of oscillating electric and magnetic fields.
An electromagnetic wave traveling along an x axis has an electric field E and
a magnetic field B with magnitudes that depend on x and t:
E
E0 sin ω(t
x
),
υ
B
B0 sin ω(t
x
),
υ
(13.1)
B0 = E0/υ or E0 = B0·υ,
where E0 and B0 are the amplitude values of the electric field strength and
the magnetic induction respectively; υ
ω(t
x
) is the phase, ω = 2πυ is
υ
the angular frequency; t is a time; υ is the velocity, x is the coordinate.
The wave nature of light was first illustrated through experiments on
diffraction and interference.
191
The interference is superimposition of waves resulting in a steadily in time
non uniform distribution of wave energy, with a maxima and minima of light
intensity.
To produce the interference of light it is necessary to have two waves or
two wave sources with the same frequency and constant phase difference. They
are called coherent waves or coherent sources correspondently.
Two independent sources of light cannot be coherent because the waves
emitted by them would not have the same phase or constant difference in phase.
However if the light from the single light source is splited in two beams, these
beams may be the coherent waves of light.
For two coherent waves or sources of light the distribution of energy
in the surrounding medium is not uniform at all points. There are certain
regions, where the intensity of light is maximum. Also, there are other regions,
where the intensity of light is minimum. Thus the energy due to sources of light
is redistributed. This phenomenon of redistribution of energy in a medium due
to superimposition of coherent waves of light is called the interference of light.
At the any point where superimposition of coherent waves occurs
the amplitude of oscillator of the electrical field is equal to
E12
E22
2 E1 E2cos( υ),
(13.2)
where E1 and E2 are the electrical amplitude of the first and second waves,
Δυ = (υ1 – υ2) — their phase difference.
If these waves are not coherent their phase difference changes in time very
quickly and (cosΔφ) changes from (–1) to (+1) about 109 times per second! Its
average value is equal to zero, so the human eye can see the average amplitude
E
only: E
E12 E22 and average intensity I = I1 + I2.
But if these waves are coherent, their phase difference is constant in time
so the amplitude of resulting oscillation is constant in time too and defined by
(13.2).
At the points where Δυ = 2πk (k = 0, ±1, ±2 …) cos Δυ = +1, so
the amplitude of an electrical field oscillator will be maximum:
Emax
E12
E22
2 E1 E2
E1
(13.3a)
E2 .
Hence the intensity of light at such points becomes maximum too:
I
I1
I2
2 I1 I 2 .
This is called constructive interference.
At some other points, where Δυ = π(2k + 1) (k = 0, ±1, ±2 …) will be
cos Δυ = –1, so the amplitude of an electrical field oscillators will be minimum:
E12
Emin
E22
2 E1 E2
E1
E2 .
Hence the intensity of light at such points is minimum:
I
I1
I2
192
2 I1 I 2 .
(13.3b)
This is called destructive interference.
If E1 = E2 the destructive interference intensity is equal to zero (see 13.3, b).
There are no light energy losses at the interference of light. The loss of
energy at the points of destructive interference appears as the increase of energy
at the points of constructive interference.
A simple experiment of the interference of light was demonstrated by
Thomas Young in 1801. It provides solid evidence that light is a wave.
The light from the single light source goes through two slits S1 and S2 that
serves as a coherent sources. At the screen one has seen the interference pattern
called fringes, consisting of alternating light and dark bars, where
correspondently maximum and minimum of wave interference has occurred.
The path difference of the two rays coming through two slits to the point M
(fig. 13.1) on the screen is:
Δd = d2 – d1,
(13.4)
and phase difference is:
υ
2π
d.
λ
(13.5)
Fig. 13.1. Light interference
If the path difference equals a whole number of wavelengths, then
constructive interference takes place and a bright fringe will appear on
the screen:
Δd = kλ (Δυ = 2πk), k = 0, ±1, ±2 …
(13.6)
The value of k is called the order of the interference fringe. The first order
(k = 1), for example, is the first fringe on each side of the central fringe (which
is at Δυ = 0, k = 0).
If the ray‘s paths differ by a half number of wavelengths, destructive
interference occurs and dark fringes will appear:
λ
d (2 k 1) , if υ (2k 1)π, k 0, 1, 2 ...
(13.7)
2
193
In daily life, we observe that a thin layer of an oil spread over water surface
and a thin soap film appear colored in sunlight. This production of colors is due
to interference of light.
13.3. DIFFRACTION OF LIGHT
The first convincing wave theory for light was established in 1678 by
Dutch physicist Christian Huygens. Huygens‘ wave theory is based on
a geometrical construction and allows to tell where a wave front will be at any
time in the future if we know its present position. Huygens‘ principle is:
All points on a wave front serve as point sources of a coherent spherical
secondary wavelets. Due to its interference the next new position of the wave
front will be a surface tangent to these secondary wavelets.
Huygens‘ principle is particularly useful for analyzing what happens when
waves impinge on an obstacle and the wave fronts are partially interrupted.
Huygens‘ principle predicts that wave bends behind an obstacle.
This phenomenon of bending around obstacles by light waves and its
spreading into regions of geometrical shadow of an object is called diffraction.
Each element of a wave front starting from a source of light becomes the source
of secondary coherent wavelets, which spread out into space. The phenomena of
diffraction occurs due to interference of the secondary wavelets that forms a new
wave front (fig. 13.2).
Fig. 13.2. Single slit diffraction
A large number of equally spaced parallel slits is called a diffraction
grating. Grating can be made by precision machining of very fine parallel lines
on a glass plate. The untouched spaces between the lines serve as the slits.
The sum of the slit width a and the opaque section b is called the period grating d:
d = a + b.
(13.8)
194
We assume parallel rays of light are incident on the grating as shown in
fig. 13.3. The slits are narrow enough to produce the coherent diffracted beams,
and interference can occur.
Fig. 13.3. Diffraction grating
Principal maxima diffraction grating occurs under condition:
d sin υ = kλ, k = 0, ±1, ±2 …
(13.9)
Suppose the light striking a diffraction grating is not monochromatic. Then
for all diffraction orders except k = 0, each wavelength will produce a maximum
at a different angle υ. If white light strikes a grating, the central (k = 0)
maximum will be a sharp white peak. But for all other orders, there will be
a distinct color spectrum (fig. 13.4).
k=2
k=1
k=0
k = –1
k = –2
Fig. 13.4. Simplified diagram of diffraction of white light on a diffraction grating
195
13.4. DISPERSION OF LIGHT
Color is related to the wavelengths or frequencies of the light. Visible light
has wavelengths in air in the range of about 400 nm (violet) to 750 nm (red), this
is known as the visible spectrum.
White light passing through a prism is separated into different colors
(fig. 13.5). The colors in the order of decreasing deviation are violet, indigo,
blue, green, yellow, orange and red. The deviation is maximum for violet color
and minimum for red color.
Fig. 13.5. The visible spectrum
The phenomenon of dependence of light speed (or the index of refraction)
in a material on the wave frequency (or wavelength) is called dispersion of
light.
This phenomenon goes to the splitting of white light into its spectrum, on
passing through a prism (fig. 13.6). This happens because the index of refraction
of a material depends on the wavelength. White light is a mixture of all visible
wavelengths, and when it incident on a prism, the rays of different wavelength
are bent to varying degrees. Because usually the index of refraction is greater for
the shorter wavelengths, violet light is bent the most and red the least, as
indicated. Rainbows are a spectacular example of light dispersion by drops of
water.
Fig. 13.6. Dispersion of light
As light moves from one medium into another where it travels with
a different speed, the frequency remains the same. The wavelength changes as
the speed changes. When the light goes from air into a material with index of
refraction n, the wavelength becomes:
196
λ
,
(13.10)
n
where λ is the wavelength in vacuum or air and λn is the wavelength in
the material with index of refraction n.
λn
TESTS
1. Which of the following types of electromagnetic waves have the longest
wavelength:
a) radio waves;
b) visible light;
c) infrared rays;
d) X-rays.
2. Monochromatic light enters from one medium to the other. Which one of
the following properties does not change:
a) amplitude;
b) velocity;
c) wavelength;
d) frequency.
3. What happens to pattern in Young‘s experiment when the monochromatic
source is replaced by the white light source?
a) all bright fringes become white;
b) all bright fringes get colored from violet to red;
c) only the central fringe is white, all other fringes are colored;
d) no fringes are observed.
4. In Young‘s experiment the sources should be:
a) incoherent;
b) coherent;
c) of any two colors;
d) of any frequency.
5. Huygens‘ principle of secondary waves:
a) allows to find focal length of a thick lens;
b) is a geometrical method to find a wave front;
c) is used to determine the velocity of light.
6. Two sources are said to be coherent if the waves produced by them have
the same:
a) wavelength;
b) amplitude;
c) amplitude and wavelength;
d) wavelength and constant phase difference.
7. The bending of light around the corners of an obstacle is called:
a) dispersion;
b) refraction;
c) diffraction;
d) interference.
8. The dispersion of light in a medium implies that:
a) lights of different wavelengths have the different speeds;
b) lights of different frequencies have the different speeds;
197
c) refractive indices are different for different wavelengths;
d) all the above.
9. When rays from sun passes through a glass prism, the emergent beam
shows all colors of the rainbow. This is due to the phenomenon of:
a) scattering;
b) dispersion;
c) diffraction;
d) polarization.
10. When white light passes through a glass prism, we get spectrum on the
other side of the prism. In the emergent beam the ray which is deviated most is:
a) red ray;
b) violet ray;
c) yellow ray;
d) green ray.
PROBLEMS
1. The two slits in Young‘s experiment for producing interference fringes
are 0.51 mm apart. Interference fringes of width 0.2 cm are observed on a screen
placed 200 cm away from the slits. Find the wavelength of light. (Answer:
510 nm)
2. If a diffraction grating has 600 slits per mm and the second diffraction
maximum is observed at 30° of the central maximum, find the wave length of
light. (Answer: 417 nm)
3. A diffraction grating has 300 slits per mm, wave length of light is
λ = 500 nm. How many diffraction maxima will be observed at the screen?
(Answer: 13)
13.5. QUANTUM PROPERTIES OF LIGHT.
PHOTOELECTRIC EFFECT
In 1990 German physicist Max Planck proved that light has both particle
and electromagnetic wave properties at the same time. A propagation of light
can be described by a wave with frequency ν and wavelength λ = υ/ν. And
according to the quantum theory light has particle properties in their interaction
with matter. Each particle of light is called the photon (a ―particle‖ concept) and
has energy E, that relates to wave frequency (a ―wave‖ concept) by
c
(13.10)
E h
h .
λ
The proportionality constant h is called Plank‘s constant. It has
the numerical value h = 6.63·10−34 J·s = 4.14·10−15 eV·s. The electron volt (eV)
is a small unit of energy:
1 eV = 1.6·10−19 J.
The quantum properties of light have been proved by a photoelectric effect.
198
The photoelectric effect is phenomenon
of emission of electrons from substances
when light of suitable wavelength fall at
them (fig. 13.7). Photoelectric effect occurs
in many materials, but it is most easily
observed with metals.
This effect can be observed using
the device shown in fig. 13.8. A metal plate
P and a smaller electrode C are placed inside
Fig. 13.7. Photoelectric effect scheme
an vacuum glass tube, called a photocell.
Electrodes are connected to an ammeter and a battery. When the photocell is in
the dark, the ammeter reads zero. But when light of suitable wavelength
illuminates the plate, the ammeter indicates a current flowing in the circuit. This
is the photoelectric current which is produced when electrons ejected from plate
C reach the plate P.
Fig. 13.8. Device for photoelectric effect
The plate P can be given a positive potential. If this potential is increased,
more and more photoelectrons are able to reach P. When all photoelectrons
emitted from C are able to reach the plate P the photoelectric current is
maximum and called the saturation current (fig. 13.9).
Fig. 13.9. Photoelectric current
199
If a negative potential is applied to the plate P, the photoelectrons get
retarded. As negative potential of P is increased, more and more electrons are
stopped. When the potential of P is so much negative that even the fastest and
most energetic electrons fail to reach P. The photoelectric current will reduce to
zero. The minimum negative potential that reduce the photoelectric current to
zero is called stopping potential. The photoelectric current will be cut off, when
the photoelectron with maximum kinetic energy is stopped:
2
mυmax
(13.11)
eV0
,
2
where e is charge of electron, V0 is the stopping potential, m is mass of electron,
υmax is maximum velocity of electron.
This relation is known as Einstein’s photoelectric equation:
m υ2
(13.12)
hv
+Wout ,
2
Wout is some minimum energy required just to get an electron out through
the surface. Wout is called the work function (work of electron exit). Note that
the photoelectric effect does not occur if the frequency is below a certain cutoff
frequency υmin or, equivalently, if the wavelength is greater than the
corresponding cutoff wavelength:
hc
(13.13)
λ max
,
Wout
where c is speed of light.
Example 13.1.
Calculate the energy of a photon of blue light, λ = 450 nm in air (or
vacuum).
Solution. E
c
h
λ
h
6.63 10
34
J s 3 108 m/s
4.5 10
7
4.4 10
19
J.
m
Example 13.2.
What is the kinetic energy and the speed of an electron ejected from
a sodium surface whose work function is Wout = 2.28 eV when illuminated by
light of wavelength (a) 410 nm, (b) 550 nm?
Solution. For λ = 410 nm, the energy of the photons:
E
h
h
c
λ
4.85 10
19
J
3.03 eV.
This energy is greater than Wout, then electrons will be ejected with varying
amounts of kinetic energy, with a maximum of
Ek = hv – Wout = 3.03 eV – 2.28 eV = 0.75 eV = 1.2·10–19 J.
200
υmax
For λ = 550 nm, E
2 Ek
m
h
2 1.2 10
9.11 10
h
c
λ
19
31
3.61 10
J
5.1 105 m/s.
kg
19
J 2.26 eV.
Since this photon energy is less than the work function, no electrons are
ejected.
TESTS
1. Photoelectric effect reveals the:
a) wave nature of radiation;
b) particle nature of radiation;
c) both wave as well particle nature;
d) none of these.
2. The current in a photocell:
a) decreases with increase of intensity of incident light;
b) increases with increase of intensity of incident light;
c) decreases with increase of frequency of incident light;
d) increases with increase of frequency of incident light.
3. The threshold wavelength for sodium is 500 nm. Its work function is:
a) Wout = 4·10–19 J;
b) Wout = 2·10–19 J;
c) Wout = 6·10–19 J;
d) Wout = 8·10–19 J.
4. The photoelectric cutoff wavelength of certain metal is 200 nm.
The maximum kinetic energy of photoelectrons released by a wavelength of
300 nm is:
a) 2 eV;
b) 3 eV;
c) 4 eV;
d) no electron can be ejected.
5. Photoelectric cell is device which converts:
a) electrical energy into light energy;
b) light energy into electrical energy;
c) electrical energy into sound energy;
d) light energy into elastic energy.
PROBLEMS
1. The work function of metal is 3.45 eV. Calculate what should be
the maximum wavelength of light that can eject photoelectrons from the metal.
(Answer: 360 nm)
201
14. ATOMIC PHYSICS
14.1. RUTHERFORD’S MODEL OF ATOM
Rutherford obtained an important insight
into the structure of atom by means of
performing experiments on the scattering of
α-particles on thin gold foil. On the basis of this
experiment Rutherford suggested that the atom
must consist of a tiny but massive positively
charged nucleus, containing over 99.9 % of
the mass of the atom, surrounded by electrons
some distance away. The electrons would be
moving in orbits about the nucleus — much as
the planets move around the Sun — because if
they were at rest, they would fall into the Fig. 14.1. Rutherford‘s ―planetary‖
model of the atom
nucleus due to electrical attraction (fig. 14.1).
Rutherford‘s experiments suggested that the nucleus must have a radius of
about 10–15 m. The radius of atoms was estimated to be about 10–10 m (if
the nucleus were the size of a baseball, the atom would have the diameter of
a big city several kilometers across). So an atom would be mostly empty space.
Based wholly on classical physics, the Rutherford model was unable to
explain stability of atom. According to the Rutherford model, electrons orbit
the nucleus, and since their paths are curved the electrons are accelerating.
Hence they should give off light like any other accelerating electric charge.
Since light carries off energy and energy is conserved, the electron‘s own energy
must decrease to compensate. Hence electrons would be expected to spiral into
the nucleus.
14.2. BOHR’S THEORY OF THE HYDROGEN ATOM
Rutherford‘s model was replaced in a few
years by the Bohr‘s atomic model, which
incorporated some early quantum theory.
According to Bohr‘s atom model the electrons
could only orbit the nucleus in particular
circular orbits with fixed angular momentum
and energy (fig. 14.2). They were not allowed
to spiral into the nucleus, because they could
not lose energy in a continuous manner; they
could only make quantum leaps between fixed
energy levels.
202
Fig. 14.2. The Bohr‘s atom model
Bohr’s theory of the hydrogenic (one-electron) atom is based on
the following postulates:
1. An atom can exist in certain allowed or stationary states, with each state
having a definite value for its total energy E1, E2, E3 … En (fig. 14.3). When
the atom is in one of these states it is stable and does not radiate energy.
Fig. 14.3. The Bohr atom (a). Energy level diagram (b)
2. An atom emits or absorbs energy only when an electron moves from one
the stable state with energy En to another stable state with energy Ek.
In a transition from its initial state to its final state, a photon is either emitted (if
En > Ek) or absorbed (if En< Ek ) and the energy hν of the photon is equal to
the difference in the energy of the two states:
(14.1)
h
En Ek .
Example of the light emission is illustrated in fig. 14.4. The electron jumps
from an stationary orbit of higher energy E2 to an stationary orbit of lower
energy E1 and a photon of energy hν = E2 – E1 is emitted.
Fig. 14.4. Emission of photon
The electron can absorb energy from some source and jump from a lower
energy level to a higher energy level and then emits energy jumping from
a higher energy level to a lower energy level as shown in the following fig. 14.5.
203
Fig. 14.5. The various ways of an energy absorption and energy emission
Thus from the Bohr model of the atom follows that electrons exist only in
the certain energy levels within an atom. The electron energy in these levels has
well defined values and electrons jumping between them must absorb or emit
the energy equal to the difference between them. The energy emitted as
the electron moves to a lower energy level can be in the form of a photon
(a particle of light). The wavelength λ of the emitted light can be related to its
energy:
hc
(14.2)
h
E.
λ
14.3. ENERGY STATES OF A HYDROGEN ATOM
The above postulates can be used to calculate allowed energies of the atom
for different allowed orbits of the electron. The theory developed should be
applicable to hydrogen atoms and ions having just one electron. Thus, within
the Bohr atom framework, it is valid for He+, Li++, Be3+ etc.
Energy states of the hydrogen atom for different allowed orbits of
the electron can be described by equation:
E0
13 ,6 eV
(14.3)
En
.
2
2
n
n
where n is the principal quantum number, that labels the orbit radii and also
the energy levels, E0 = –13.6 eV. The lowest energy level or energy state has
energy E1, and is called the ground state. The higher states, E2, E3, and so on,
are called excited states. The fixed energy levels are also called stationary
states.
Notice that although the energy for the larger orbits has a smaller numerical
value, all the energies are less than zero. Thus, E2 = –3.4 eV is a higher energy
than E1 = –13.6 eV. Hence the orbit closest to the nucleus (r1) has the lowest
energy E1 = –13.6 eV. If an electron is free and has kinetic energy, then the total
energy E > 0. Since E > 0 for a free electron, then an electron bound to an atom
needs to have E < 0. The minimum energy required to remove an electron from
204
an atom initially in the ground state is called the binding energy or ionization
energy. The ionization energy for hydrogen has been measured to be 13.6 eV,
and this corresponds precisely to removing an electron from the lowest state,
E1 = –13.6 eV, up to E = 0 where it can be free.
It is useful to show the various possible energy values as horizontal lines on
an energy-level diagram. This is shown for hydrogen in fig. 14.6. The electron
in a hydrogen atom can be in any one of these levels according to Bohr‘s theory.
Fig. 14.6. Hydrogen energy diagram illustrating Lyman, Balmer and Paschen series formation
The radiation of atoms that do not interact with one another consists of
separate spectral lines. The emission spectrum of atoms is accordingly called
a line spectrum. The atomic spectra show the energy structure of atoms
therefore the studying of these spectra served as a key to cognition of
the structure of atoms. It was noted first of all that the lines in the spectra of
205
atoms are arranged not chaotically, but are combined into groups or, as they are
called, series of lines:
Ek
En
h
E0 1
h n2
1
k2
,
(14.4)
where n = 1, 2, 3, 4,… ; k = n + 1, n + 2, n + 3 … Equation (14.4) is
the generalized Balmer formula.
Once in an excited state, an atom‘s electron can jump down to a lower
state, and give off a photon in the process. This is, according to the Bohr model,
the origin of the emission spectra. The vertical arrows in fig. 14.6 represent
the transitions or jumps that correspond to the various observed spectral lines.
The group of spectral lines that corresponds to transitions from any higher
energy levels to certain low level forms spectral series. There are some spectral
series in hydrogen emission spectrum:
1. The Lyman series of lines corresponds to transitions or ―jumps‖ that
bring the electron down to the ground state E1 (n = 1) from any exited energy
levels k ≥ 2 (where n and k are the principal quantum numbers of the states).
The lines of the Lyman series are located in the ultraviolet range of
the spectrum. The frequencies of the Lyman series are obtained from formula
(14.3) if n = 1 and k = 2, 3, 4, 5, …:
E0
1
(14.5)
(1
),
h
k2
where k = 2, 3, 4, 5 …
2. The Balmer series is characterized by the electron transitions from any
exited energy levels k ≥ 3 to the second energy level E2 (n = 2), where n and k
are the principal quantum numbers of the states. The spectral lines associated
with this series are located in the visible part of theelectromagnetic spectrum.
The frequencies of the Balmer series can be represented in the form:
Ek E2 E0 1 1
(14.6)
(
),
h
h 4 k2
where k = 3, 4, 5, 6 …
3. The Paschen series is the emission lines corresponding to an electron
transitions from k ≥ 4 to the third energy level E3 (n = 3). The lines of
the Paschen series are located in the near infrared range of the spectrum.
The frequencies of the Paschen series are given by formula:
E0 1 1
(14.7)
(
),
h 9 k2
where k = 4, 5, 6, 7 …
206
15. PHYSICS OF ATOMIC NUCLEUS
An atomic nucleus consists of elementary particles called nucleons.
A proton is a positively charged particles of mass 1.673·10 –27 kg and charge
1.6·10–19 C. A neutron is electrically neutral and its mass is 1.675·10–27 kg.
A species of nucleus is represented as nuclear symbol AZ X , where X is
the chemical symbol of the element, Z is the atomic number (the number of
protons inside the nucleus) and A is the mass number (the number of nucleons).
The mass number of the nucleus can be written as:
A = Z + N.
(15.1),
where N is the number of neutrons.
Nuclei with the same number of protons but different neutron numbers are
isotopes of one another.
The nuclear size depends on species of nucleus; it grows through
the periodic table. The nuclear radius R and the atomic mass number A are
related by formula:
(15.2)
R R0 3 A ,
–15
where R0 = 1.2·10 m — is the radius of hydrogen nuclear (proton).
The nuclear charge is due to the protons contained in it. Each proton has
a positive charge of ep = 1.6·10–19 C. Thus the nuclear charge is equal to:
q = Ze,
(15.3)
where Z is the atomic number of the nucleus (the amount of protons in nucleus).
The unit of energy commonly used in atomic and nuclear physics is
the electron volt (eV):
1 eV = 1.6·10–19 C · 1 V = 1.6·10–19 J.
Neutrons held together in the nucleus by the strong or nuclear forces.
There are extremely short range forces confined only to the nucleus. At short
distances, the nuclear forces are much stronger than electrostatic force of
repulsion of protons inside the nucleus. Hence the nucleus is stable.
15.1. BINDING ENERGY
The amount of work required to be done to separate the nucleons an infinite
distance apart is called the binding energy of the nucleus. The total mass of
a stable nucleus is always less than the sum of the masses of its separate protons
and neutrons: Mnucleus < (Zm
Nm ) . The difference between the mass of
p
n
a nucleus and the sum of the masses of protons and neutrons constituting it is
called the mass defect:
ΔM = (Zm
(15.4)
Nmn ) – Mnucleus.
p
207
In formation of any nucleus a certain mass disappears. According to
Einstein‘s theory, this mass defect must be appearing in the form of energy
responsible for binding the nucleons together:
2
2
E = ΔM · c2 = (Zm
(15.5)
Nmn ) ·c – Mnucleus · c .
p
The binding energy of a nucleus divided by the number of nucleons (or
mass number A) is binding energy per nucleon:
E
(15.6)
ε
.
A
Fig. 15.1 shows the binding energy per nucleon as a function of A
for stable nuclei. The curve rises as A increases and reaches a plateau at about
8.7 MeV per nucleon above A ≈ 40. Beyond A ≈ 80, the curve decreases slowly,
indicating that larger nuclei are held together a little less tightly than those in
the middle of the Periodic Table.
Fig. 15.1. Binding energy per nucleon as a function of mass number A
15.2. NUCLEAR REACTIONS
A nuclear reaction is the transformation of a stable atomic nucleus into the
unstable one by bombarding it with a suitable particle.
A nuclear reaction can be represented symbolically as follows:
A
4
A 3
1
Q.
(15.7)
ZX
Z+1Y
1p
2 He
Here the parent nucleus X is struck by a helium nucleus to give a daughter
nucleus Y and proton. In this nuclear reaction Q represents the total kinetic
energy change in the reaction. It is called reaction energy or Q-value of
the reaction. It may be positive or negative. For Q > 0, the reaction is said to be
exothermic; energy is released in the reaction, so the total kinetic energy is
208
greater after the reaction than before. If Q is negative (Q < 0), the reaction is
said to be endothermic: the final total kinetic energy is less than the initial
kinetic energy, and an energy input is required to make the reaction happen.
In any nuclear reaction total energy, electric charge and nucleon number
are conserved.
15.3. RADIOACTIVITY. ALPHA, BETA, AND GAMMA RADIATION
Nuclear stability depends on the atomic number Z and on the number of
neutrons N. The light atomic nuclei contain practically as many neutrons as
protons (N/Z = 1). They are the most stable. In case N/Z >1.6 the atomic nuclei
are unstable and undergo a radioactive decay.
Many unstable isotopes occur in nature, and such radioactivity is called
natural radioactivity. Other unstable isotopes can be produced in the laboratory
by nuclear reactions; these are said to have artificial radioactivity. Radioactive
isotopes are sometimes referred to as radioisotopes or radionuclides.
The most common types of radiation are called alpha, beta, and gamma
radiation, named after the first three letters of the Greek alphabet. Gamma rays
are very high-energy photons whose energy is even higher than that of X-rays.
Beta particles are electrons, identical to those that orbit the nucleus, but they are
created within the nucleus itself. Alpha particles are simply the nuclei of helium
atoms, that is, every α-particle consists of two protons and two neutrons bound
together.
These types of radiation were classified according to their penetrating
power. Alpha radiation can barely penetrate a piece of paper. Beta radiation can
pass through as much as 3 mm of aluminum. Gamma radiation is extremely
penetrating: it can pass through several centimeters of lead and still be detected
on the other side.
Each type of radiation has a different charge and hence is bent differently
in a magnetic field (fig. 15.3); α-rays are positively charged, β-rays are
negatively charged, and γ-rays are neutral.
When a nucleus emits an α-particle, the remaining nucleus will be different
from the original: it has lost two protons and two neutrons. Alpha decay
proceeds according to the following scheme:
A
A 4
4
(15.8)
ZX
Z 2Y
2 α γ.
A nucleus that decays spontaneously by emitting an electron or a positron
(a positively charged particle with the mass of an electron) is said to undergo
beta decay. In beta-minus (β–) decay, an electron is emitted by a nucleus:
A
ZX
A
Z+1Y
209
0
1β
v.
(15.9)
Fig. 15.3. Deflection of alpha, beta and gamma radiation in magnetic field
In beta-plus (β+) decay, a positron is emitted by a nucleus:
A
ZX
A
Z 1Y
0
1β
v.
(15.10)
The symbol ν represents a neutrino, a neutral particle which has a very
small mass, that is emitted from the nucleus along with the electron or positron
during the decay process; v is antineutrino.
Beta-decay is accompanied by the interconversion between neutrons and
protons inside a nucleus.
15.4. RADIOACTIVE DECAY LAW
A macroscopic sample of any radioactive isotope consists of a vast number
of radioactive nuclei. These nuclei do not all decay at one time. Rather, they
decay one by one over a period of time. This is a random process: we can not
predict exactly when a given nucleus will decay. But we can determine, on
a probabilistic basis, approximately how many nuclei in a sample will decay
over a given time period, by assuming that each nucleus has the same
probability of decaying in each second that it exists.
Following to the radioactive decay law the number of undecayed nuclei N
decreases exponentially with time t:
N = N0e–λt,
(15.11)
where λ is a constant characteristic of the given radioactive substance and known
as the decay constant, N0 is the initial number of undecayed nuclei at the time
t = 0.
210
There are two common time characteristics of how long any given type of
radionuclides lasts. One measure is the half-life T1/2 of a radionuclide, which is
the time at which a half of the initial number of nuclei N0 decays:
ln2
(15.12)
T1 / 2
.
λ
The other measure is the mean life τ, which is the time at which N0 has
been reduced to e–1 of their initial values:
1
(15.13)
τ
.
λ
15.5. NUCLEAR FISSION
Nuclear fission is a process of breaking up the heavier nuclei into lighter
ones with sufficient mass defect, which appears in the form of a tremendous
amount of energy. For example, if a massive nucleus like uranium-235 absorbs
a low energy (also called thermal) neutron, it breaks apart with the release of
energy. This phenomenon was named nuclear fission because of its resemblance
to biological fission (cell division).
In a typical 235U fission event, a 235U nucleus absorbs a thermal neutron,
producing a compound nucleus 236U in a highly excited state. This nucleus
undergoes fission, rapidly emits three neutrons and splits into two fission
fragments 141Ba and 92Kr in a typical case:
235
1
236
141
92
(15.14)
3 01n 200 MeV .
92 U
92 U
56 Ba
36 Kr
0n
A tremendous amount of energy is released in a fission reaction because
the mass of 236U is considerably greater than the total mass of the fission
fragments plus released neutrons.
235
The energy released in a fission of 1g of 92 U is equal to 82.000 MJ and
corresponds to the burning of 3300 kg of coal or 2000 kg of gasoline.
Neutrons released in each fission could be used to create a chain reaction.
That is, the three secondary neutrons produced in the reaction bring about
235
the fission of three more 92 U atoms and produce 9 neutrons, which in turn, can
235
bring about the fission of nine 92 U atoms and so on. This process multiplies as
shown schematically in fig. 15.3.
211
Fig. 15.3. Chain reaction
15.6. ELEMENTARY PARTICLES
Elementary particles are the particles which have no further structure.
They are the ultimate building blocks of matter. The various elementary
particles have been divided into the following four groups:
1. Photons.
2. Leptons.
3. Mesons.
4. Baryons.
Photons are field particles linked with electromagnetic forces. Every
photon is a quantum of radiation with no charge and no rest mass. The energy of
a photon is E = hv. Every photon moves with the velocity of light.
Leptons are particles whose masses are smaller than masses of mesons.
Important members of this group are electron, positron and neutrino. Electron is
a particle of mass 9.1·10–31 kg. It carries a charge –1.6·10–19 C, which is taken as
a unit negative charge. It is a stable particle. Positron is anti-particle of electron.
Mesons have rest mass 250–1000 times bigger than electron. They are
regarded as particles of strong interaction. Most of them owe their existence to
cosmic rays.
Baryons have the rest mass equal to or greater than that of proton.
The members of this class are proton and neutron.
212
TESTS
1. Rutherford scattering of α-particles by atom shows that:
a) the atom as a whole is positively charged;
b) the atom consists of uniformly distributed positive and negative charged
particles;
c) there is no charged particles inside the atom;
d) the atom has a very small positively charged core at the center.
2. In Bohr‘s model of atom stationary orbits are postulated:
a) in accordance with classical theory of electromagnetism;
b) to meet the condition for dynamic equilibrium of electrons;
c) to meet the condition that the electrons moving in these orbits do not
radiate energy;
d) none of the above.
3. According to Bohr‘s atomic model:
a) electrons radiate energy only when it jumps to another orbit;
b) an atom has heavy, positively charged nucleus;
c) electrons can move only in particular orbits;
d) all of the above statements are true.
4. Two elements having same number of protons but different number of
neutrons are called:
a) isobars;
b) isotopes; c) isomers;
d) isotones.
5. When the electron jumps from orbit n1 to n2 orbit, the energy radiated is
given by:
a) hv = E2 – E1;
b) hv = E1 – E2;
c) hv = E2 + E1;
d) hv = E2/E1.
6. The spontaneous emission of high energy particles from the nucleus of
the atom is called:
a) radioactivity;
b) photoelectricity;
c) thermoelectricity;
d) nuclear fusion.
7. The radius of the nucleus is directly proportional to (A = mass number):
a) A2;
b) A ;
c) 3 A ;
d) A3.
8. If radiations from radioactive substance pass through an electromagnetic
field, then:
a) all are deflected;
b) only γ-rays are deflected;
c) only α- and β-rays are deflected;
d) only α-rays are deflected.
213
9. The γ-radiation consists of:
a) photons;
b) electrons;
c) protons;
d) neutrons.
10. The atomic number ‗Z‘ of the nucleus is:
a) number of protons in it;
b) number of neutrons in it;
c) number of electrons round it;
d) number of deuterons.
11. Alpha-particles consist of:
a) electrons;
b) helium nuclei;
c) protons;
d) none of these.
12. Beta-particles consist of:
a) high speed moving electrons;
b) protons;
c) neutrons;
d) photons.
13. A nucleus composed of:
a) electrons and protons;
b) neutrons and protons;
c) electrons and neutrons;
d) electrons, protons and neutrons.
14. The nucleus
a) 92 protons;
c) 238 nucleons;
238
92 U
has all the following except:
b) 146 neutrons;
d) 146 electrons.
7
2
8
15. In the reaction: 3 Li 1 H 3 Li X, X is:
a) proton;
b) neutron;
c) photon;
d) α-particle.
214
REFERENCES
1. Англо-русский словарь математических терминов. М. : Мир, 1994.
2. Англо-русский физический словарь / сост. Д. М. Толстых [и др.]. М. : Советская
энциклопедия, 1972.
3. Дорожкина, В. П. Английский язык для математиков : учеб. / В. П. Дорожкина.
2-е изд., доп. и перераб. М. : МГУ, 1986.
4. Коваленко, Е. Г. Англо-русский математический словарь / Е. Г. Коваленко. М. :
Эрика, 1994. Т. 1, 2.
5. Прокошева, И. И. Основные математические понятия в английском языке :
метод. указания / И. И. Прокошева. Оренбург : ГОУ ВПО ОГУ, 2003. 50с.
6. Сикорская, Н. П. Английский язык для физиков : учеб. / Н. П. Сикорская. 2-е
изд., перераб. и доп. Минск : БГУ, 1981. 301 с.
7. Giancoli, D. C. Physics for scientists and engineers with modern physics /
D. C. Giancoli. 4th ed. Pearson education, 2009. 1172 р.
8. Halliday, D. Fundamentals of Physics Extended / D. Halliday, R. Resnick, J. Walker.
10th ed. John Wiley & Sons, 2014. 1448 p.
9. Konev, V. V. The elements of Mathematics. Textbook / V.V. Konev. 2nd ed. Tomsk :
TPU Press, 2009. 140 p.
10. Lohwater’s, A. J. Russian-English dictionary of the mathematical sciences /
A. J. Lohwater‘s ; ed. R. P. Boas. Aмerican Mathematical Society, 1990.
11. Medical and biological physics for medical students / L. V. Kukharenko [et al.].
Minsk : BSMU, 2015. 260 p.
12. Serway, R. A. Principles of Physics : a calculus-based text / R. A. Serway,
J. W. Jewett. 5th ed. Prime Student, 2014.
13. Weltner, K. Mathematics for Physicists and Engineers / K. Weltner, S. John,
W. J. Weber. Springer, 2014. 588 р.
14. Young, H. D. University Physics with Modern Physics / H. D. Young,
R. A. Freedman. 14th ed. Pearson, 2015. 1600 p.
215
Appendix
Fundamental Constants
Quantity
Speed of light in vacuum
Gravitational constant
Avoqadro‘s number
Gas constant
Symbol
c
G
NA
R
Boltzmann‘s constant
Charge on electron
Stefan–Boltzmann
constant
Permittivity of free space
Permeability of free space
Planck‘s constant
Electron rest mass
k
e
0
Approximate Value
3.00 × 10 m/s
6.67 × 10–11 N·m2/kq2
6.02 × 1023 mol–1
8.315 J/mol·K = 1.99 cal/mol·K =
= 0.082 atm·liter/mol·K
1.38 × 10–23 J/K
1.60 × 10–19 C
5.67 × 10–8 W/m2·K4
8
= (1/c2 0)
0
h
me
Proton rest mass
mp
Neutron rest mass
mn
Atomic mass unit (1 u)
8.85 × 10–12 C2/N·m2
4 × 10–7 T·m/A
6.63 × 10–34 J·s
9.11 × 10–31 kg = 0.000549 u =
= 0.511 MeV/c2
1.6726 × 10–27 kg = 1.00728 u =
= 938.3 MeV/c2
1.6749 × 10–27 kg = 1.008665 u =
= 939.6 MeV/c2
1.6605 × 10–27 kg = 931.5 MeV/c2
Other Useful Data
Joule equivalent (1 cal)
Absolute zero (0 K)
Earth: Mass
Radius (meam)
Moon: Mass
Radius (meam)
Sun: Mass
Radius (meam)
Earth-sun distance (meam)
Earth-moon distance (meam)
4.186 J
–273.15 °С
5.97 × 1024 kg
6.38 × 103 km
7.35 × 1022 kg
1.74 × 103 km
1.99 × 1030 kg
6.96 × 105 km
149.6 × 106 km
384 × 103 km
216
The Greek Alphabet
Alpha
Beta
Gamma
Delta
Epsilon
Zeta
Eta
Theta
Iota
Kappa
Lambda
Mu
A
B
Г
Nu
Xi
Omicron
Pi
Rho
Sigma
Tau
Upsilon
Phi
Chi
Psi
Omega
Е
Z
H
I
N
O
П
P
T
,
Mathematical Signs and Symbols
is proportional to
is equal to
is approximately equal to
is not equal to
is greater than
is much greater than
is less than
is much less than
is less than or equal to
is greater than or equal to
sum of
average value of x
change in x
x approaches zero
n (n – 1) (n – 2) … (1) of
x
x
n!
SI Derived Units and Their Abbreviations
Quantity
Force
Energy and work
Power
Pressure
Frequency
Electric charge
Electric potential
Electric resistance
Capacitance
Magnetic field
Magnetic flux
Inductance
Unit
newton
joule
watt
pascal
hertz
coulomb
volt
ohm
farad
tesla
weber
henry
Abbreviation
N
J
W
Pa
Hz
C
V
F
T
Wb
H
217
In Terms of Base Units
kg·m/s2
kg·m2/s2
kg·m2/s3
kg/(m·s2)
S–1
A·s
kg·m2/(A·s3)
kg·m2/(A2·s3)
A2·s4/(kg·m2)
kg/(A·s2)
kg·m2/(A·s2)
kg·m2/(s2·A2)
Metric (SI) Multipliers
Prefix
exa
peta
tera
giga
mega
kilo
hecto
deka
deci
centi
milli
micro
nano
pico
femto
atto
Abbreviation
E
P
T
G
M
k
h
da
d
c
m
n
p
f
a
Value
1018
1015
1012
109
106
103
102
101
10–1
10–2
10–3
10–6
10–9
10–12
10–15
10–18
Elastic moduli
Material
Young’s modulus, E (N/m2)
Solids
100 × 109
200 × 109
100 × 109
70 × 109
20 × 109
14 × 109
50 × 109
45 × 109
Iron, cast
Steel
Brass
Aluminum
Concrete
Brick
Marble
Granite
Wood (pine)
(parallel to grain)
(perpendicular to grain)
Nylon
Bone (limb)
10 × 109
1 × 109
5 × 109
15 × 109
218
Shear modulus, G (N/m2)
40 × 109
80 × 109
35 × 109
25 × 109
80 × 109
Densities of Substances*
Density, p (kg/m3)
Substance
Solids
2.70 × 103
7.8 × 103
8.9 × 103
11.3 × 103
19.3 × 103
2.3 × 103
2.7 × 103
0.3–0.9 × 103
2.4–2.8 × 103
0.917 × 103
1.7–2.0 × 103
Aluminum
Iron and steel
Copper
Lead
Gold
Concrete
Granite
Wood (typical)
Glass, common
Ice
Bone
Liquids
1.00 × 103
1.025 × 103
1.03 × 103
1.05 × 103
13.6 × 103
0.79 × 103
0.68 × 103
Water (4 °C)
Sea water
Blood, plasma
Blood, whole
Mercury
Alcohol, ethyl
Gasoline
Gases
Air
Helium
Carbon dioxide
Water (steam) (100 °C)
1.29
0.179
1.98
0.598
Densities are given at 0 °C and 1 atm pressure unless otherwise specified.
Latent Heats (at 1 atm)
Substance
Water
Lead
Silver
Iron
Tungsten
Melting
point
(°C)
0
327
961
1808
3410
Heat of Fusion
kcal/kg*
79.7
5.9
21
69.1
44
J/kg
3.33 × 105
0.25 × 105
0.88 × 105
2.89 × 105
1.84 × 105
Boiling
Point (°C)
100
1750
2193
3023
5900
Numerical values in kcal/kg are the same in cal/g.
219
Heat of
Vaporization
kcal/kg*
J/kg
539
22.6 × 105
208
8.7 × 105
558
23 × 105
1520
63.4 × 105
1150
48 × 105
Resistivity and Temperature Coefficients (at 20 °C)
Resistivity, ρ (Ω·m)
Material
Conductors
1.59 × 10–8
1.68 × 10–8
2.44 × 10–8
2.65 × 10–8
5.6 × 10–8
9.71 × 10–8
10.6 × 10–8
98 × 10–8
100 × 10–8
Semiconductors
(3–60) × 10–5
(1–500) × 10–3
0,1–60
Insulators
109–1012
1013–1015
Silver
Copper
Gold
Aluminum
Tungsten
Iron
Platinum
Mercury
Nichrome (alloy of Ni, Fe, Cr)
Carbon (graphite)
Germanium
Silicon
Glass
Hard rubber
Temperature
Coefficients, α (°C)–1
0.0061
0.0068
0.0034
0.00429
0.0045
0.00651
0.003927
0.0009
0.0004
–0,0005
–0,05
–0,07
Values depend strongly on presence of even slight amounts of impurities.
Indices of refraction*
Material
Vacuum
Air (at STP)
Water
Ethyl alcohol
Glass
Fused quartz
Crown glass
Light flint
Lucite or Plexiglas
Sodium chloride
Diamond
1.0000
1.0003
1.33
1.36
1.46
1.52
1.58
1.51
1.53
2.42
λ = 589 nm.
220
Rest Masses in Kilograms, Unified Atomic Mass Units, and MeV/c2
Object
Mass
u
0.00054858
1.007276
1.007825
1.008665
kg
9.1094 × 10–31
1.67262 × 10–27
1.67353 × 10–27
1.67493 × 10–27
Electron
Proton
atom
Neutron
The Types of Radioactive Decay
α decay:
A
ZN
A 4
Z 2N
4
2 He
β decay:
A
ZN
A
Z 1N
e
v
A
ZN
A
Z 1N
e
v
A
ZN
A **
ZN
e
A
Z 1N
A
ZN
e
v [EC]*
γ
* Electron capture; ** indicates the excited state of a nucleus.
221
MeV/c2
0.51100
938.27
938.78
939.57
CONTENTS
PREFACE .................................................................................................................. 3
THE BASICS OF ELEMENTARY MATHEMATICS
AND DIFFERENTIAL CALCULUS ........................................................................ 4
1. THE BASIC MATHEMATICAL CONCEPTS AND FORMULAS .................... 4
1.1. Fraction. Operations with fractions. Exponents and radicals.
Factoring and expanding ..................................................................................... 4
1.2. Functional dependence. Basic functions and their graphs .................................. 6
1.2.1. Linear function and its graph .................................................................... 7
1.2.2. Inverse proportionality function and its graph.......................................... 9
1.2.3. Quadratic function and its graph ............................................................... 9
1.2.4. Quadratic equations. Quadratic formula ................................................... 11
1.2.5. Cubic function and its graph ..................................................................... 11
1.2.6. The exponential function and its graph .................................................... 13
1.2.7. Logarithm. Common and natural logarithms. The properties
of logarithms. Logarithmic function and its graph ................................... 14
1.2.8. Trigonometric functions and their graphs.
Properties of trigonometric functions ....................................................... 15
1.2.9. Main trigonometric formulas .................................................................... 18
1.3. Vectors................................................................................................................. 19
1.3.1. Vector addition .......................................................................................... 20
1.3.2. Vector subtraction ..................................................................................... 21
1.3.3. Vector multiplication (Scalar multiplication) ........................................... 21
1.3.4. Vector decomposition ............................................................................... 22
1.3.5. Projection of vector on a coordinate axis .................................................. 22
1.4. Elementary geometry figures and formulas ........................................................ 24
1.5. Limit of a function ............................................................................................... 26
1.5.1. Limits of special interest ........................................................................... 26
1.6. Derivatives and integrals ..................................................................................... 27
1.6.1. Derivatives. General rules ......................................................................... 27
1.6.2. Differentiation rules .................................................................................. 29
1.6.3. Maxima and minima of functions ............................................................. 31
1.6.4. Differential of a function ........................................................................... 33
1.6.5. Indefinite integrals. General rules ............................................................. 33
1.6.6. Definite integral ......................................................................................... 35
THE BASICS OF PHYSICS...................................................................................... 39
2. KINEMATICS ....................................................................................................... 39
2.1. Mechanical motion characteristics ...................................................................... 40
222
2.2. Uniform linear motion ......................................................................................... 41
2.3. Non-uniform linear motion.................................................................................. 43
2.3.1. Average and instantaneous velocity and speed ......................................... 43
2.3.2. Average and instantaneous acceleration .................................................... 46
2.4. Uniformly accelerated linear motion ................................................................... 47
2.5. Freely falling objects ........................................................................................... 51
2.6. Uniform circular motion ...................................................................................... 53
3. DYNAMICS ........................................................................................................... 59
3.1. Newton‘s laws of motion..................................................................................... 59
3.2. Main forces .......................................................................................................... 61
3.2.1. The Gravitational force.............................................................................. 61
3.2.2. Gravity near the Earth surface ................................................................... 61
3.2.3. The force of elasticity and Hooke‘s law .................................................... 62
3.2.4. Normal force .............................................................................................. 63
3.2.5. Tension ...................................................................................................... 64
3.2.6. Weight ....................................................................................................... 64
3.2.7. Friction....................................................................................................... 64
3.2.8. Problem solving ......................................................................................... 65
3.3. Conservation of momentum ................................................................................ 68
3.3.1. Linear momentum and impulse equation .................................................. 68
3.3.2. The law of conservation of momentum ..................................................... 69
4. WORK. POWER. ENERGY .................................................................................. 74
4.1. Work .................................................................................................................... 74
4.2. Power ................................................................................................................... 75
4.3. Energy.................................................................................................................. 76
4.3.1. Kinetic energy ........................................................................................... 76
4.3.2. Potential energy ......................................................................................... 77
5. MECHANICAL OSCILLATIONS AND WAVES ............................................... 82
5.1. Mechanical oscillations ....................................................................................... 82
5.1.1. Characteristics of oscillations .................................................................... 82
5.1.2. Simple harmonic motion ........................................................................... 83
5.1.3. Examples of mechanical oscillations......................................................... 84
5.2. Mechanical waves ............................................................................................... 85
6. STATICS ................................................................................................................ 89
6.1. Conditions for equilibrium .................................................................................. 90
6.2. Types of equilibrium ........................................................................................... 92
6.3. Center of mass and center of gravitation ............................................................. 92
223
7. FLUID MECHANICS............................................................................................ 97
7.1. Density and pressure ........................................................................................... 98
7.2. Pascal‘s principle................................................................................................. 98
7.3. Archimedes‘ principle and buoyancy.................................................................. 100
7.3.1. Archimedes‘ principle ............................................................................... 100
7.3.2. Condition for flotation ............................................................................... 101
8. FUNDAMENTALS OF KINETIC THEORY OF GASES ................................... 103
8.1. Assumptions of kinetic molecular theory of gases ............................................. 103
8.2. Amount of substance, molar mass....................................................................... 103
8.3. Ideal gas. Gas pressure ........................................................................................ 104
8.4. Temperature as a measure of kinetic energy of molecules ................................. 105
8.5. Ideal gas law ........................................................................................................ 106
8.6. Isometric processes ............................................................................................. 107
9. THERMAL PHENOMENA. BASICS OF THERMODYNAMICS ..................... 111
9.1. Internal energy. Work of gas. First law of thermodynamics............................... 111
9.2. First law of thermodynamics at different processes ............................................ 112
9.3. Heat transfer, types of heat transfer .................................................................... 112
9.4. Amount of heat. Specific heat ............................................................................. 113
9.5. Phase changes ...................................................................................................... 113
9.6. The heat balance equation ................................................................................... 115
9.7. Thermal expansion. Coefficient of linear and volume expansion ...................... 116
10. ELECTRICITY .................................................................................................... 118
10.1. Electric charge ................................................................................................... 118
10.2. Law of conservation of electric charge ............................................................. 118
10.3. Coulomb‘s law .................................................................................................. 118
10.4. The electric field. The electric field strength .................................................... 120
10.5. Electric potential and potential difference ........................................................ 126
10.6. Capacitors. Capacitance. Electric energy storage ............................................. 132
10.7. The equivalent capacitance ............................................................................... 134
10.8. Energy stored in a capacitor .............................................................................. 135
10.9. An electric current ............................................................................................. 138
10.10. Direct current. Ohm‘s law. Resistance ............................................................ 138
10.11. Resistors in series and in parallel .................................................................... 143
10.12. Electric energy and electric power .................................................................. 146
10.13. Electromotive force. Ohm‘s law for a complete circuit .................................. 148
10.14. Alternating current .......................................................................................... 150
224
11. MAGNETIC FIELD ............................................................................................. 156
11.1. The magnetic field produced by electric current ............................................... 156
11.2. Force on an electric current in a magnetic field (Ampere‘s force) ................... 158
11.3. Force on an electric charge moving in a magnetic field (Lorentz‘s force) ....... 160
11.4. Electromagnetic induction and Faraday‘s law .................................................. 164
11.5. Faraday‘s law of induction ................................................................................ 165
11.6. Self-inductance .................................................................................................. 170
11.7. Mutual induction ............................................................................................... 172
11.8. Energy stored in a magnetic field ...................................................................... 175
11.9. LC circuit and electromagnetic oscillations ...................................................... 177
12. GEOMETRICAL OPTICS ................................................................................... 182
12.1. The ray model of light ....................................................................................... 182
12.2. Image formation by a flat mirror ....................................................................... 183
12.3. Refraction. Snell‘s law ...................................................................................... 184
12.4. Phenomenon of total internal reflection ............................................................ 184
12.5. Thin lenses. Ray tracing .................................................................................... 185
12.6. The thin lens equation. Magnification ............................................................... 186
13. THE WAVE NATURE OF LIGHT ..................................................................... 190
13.1. Electromagnetic waves spectrum ...................................................................... 190
13.2. Light interference .............................................................................................. 190
13.3. Diffraction of light ............................................................................................. 193
13.4. Dispersion of light ............................................................................................. 195
13.5. Quantum properties of light. Photoelectric effect ............................................. 197
14. ATOMIC PHYSICS ............................................................................................. 201
14.1. Rutherford‘s model of atom .............................................................................. 201
14.2. Bohr‘s theory of the hydrogen atom .................................................................. 201
14.3. Energy states of a hydrogen atom ..................................................................... 203
15. PHYSICS OF ATOMIC NUCLEUS ................................................................... 206
15.1. Binding energy .................................................................................................. 206
15.2. Nuclear reactions ............................................................................................... 207
15.3. Radioactivity. Alpha, beta, and gamma radiation ............................................. 208
15.4. Radioactive decay law ....................................................................................... 209
15.5. Nuclear fission ................................................................................................... 210
15.6. Elementary particles .......................................................................................... 211
REFERENCES ........................................................................................................... 214
APPENDIX ................................................................................................................ 215
225
Was this manual useful for you? yes no
Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Download PDF

advertisement