Lectures Notes in Physics (I) PHY 107

1

Modern University For Technology & Information

Faculty of Engineering

Department of Physics and

Engineering Mathematics

Lectures Notes in

Physics (I)

PHY 107

Prepared By

Dr. A b d e l M a g i e d D I A B

Lecture Notes

in Physics (I)

PHY 107

--------------------------

Prepared By:


Vision

The vision of the Faculty of Engineering at MTI university is to be a

center of excellence in engineering education and scientific research in

national and global regions. The Faculty of Engineering aims to

prepare graduates meet the needs of society and contribute to

sustainable development.

Mission

The Faculty of Engineering, MTI university aims to develop

distinguished graduates that can enhance in the scientific and

professional status, through the various programs which fulfill the

needs of local and regional markets. The Faculty of Engineering hopes

to provide the graduates a highly academic level to keep up the global

developments.

Table of Contents Syllabus of the course: Physics (I) [PHY 107] ......................................................................................... iv

Part ( I ) Discussion .................................................................................................................................... 1

Chapter (I) ................................................................................................................................................... 1

1. Physical Quantity ............................................................................................................................ 1

2. Fundamental and Derived Units .................................................................................................... 1

3. Base dimensions and their SI units ................................................................................................ 5

4. Dimensional analysis ....................................................................................................................... 7

4.1 Uses of dimensional equation ........................................................................................................ 11

5. SI prefixes ...................................................................................................................................... 21

Sheet (I) ................................................................................................................................................. 23

Chapter ( II ).............................................................................................................................................. 27

1. Periodic Motion ............................................................................................................................. 27

2. Some Important definition ........................................................................................................... 28

3. The motion of a spring-mass system .......................................................................................... 30

4. Energy in Simple Harmonic Motion ........................................................................................... 36

4.1. Describe the energy conservation of the system of a mass and a spring ............................... 36

4.2 Explain the concepts of stable and unstable equilibrium points ....................................... 41

Sheet (II) .................................................................................................................................................... 50

Chapter (III) .............................................................................................................................................. 54

1. Newton’s Law of Universal Gravitation .................................................................................... 54

2. Free-Fall Acceleration and the Gravitational Force ................................................................. 61

3. Kepler’s Laws of Planetary Motion ........................................................................................... 64

Sheet (III) .............................................................................................................................................. 72

Chapter (IV) .............................................................................................................................................. 75

1. States of Matter ............................................................................................................................ 75

2. Elastic Properties of Solids .......................................................................................................... 76

3. Stress–strain curve ....................................................................................................................... 77

4. Thermal Expansion of Solids and Liquids ................................................................................. 80

(i) Linear thermal Expansion............................................................................................................ 81

(ii) Superficial (areal) expansion: ...................................................................................................... 82

(iii) Volume (Bulk) expansion: ............................................................................................................ 82

Sheet (IV) ................................................................................................................................................... 89

II) Fluid Mechanics ........................................................................................................................... 93

1. Fluid Statics (Fluid at rest) .......................................................................................................... 93

1.1 Pressure in Fluids .......................................................................................................................... 93

1.2 Variation of Pressure with Depth ................................................................................................ 94

1.3 Buoyant Forces and Archimedes ‘Principle ............................................................................... 96

2. Fluid Dynamics (Fluid at motion) ............................................................................................ 107

2.1 Equation of Continuity ................................................................................................................ 107

2.2 Bernoulli’s Equation .................................................................................................................... 108

2.3 Viscosity ........................................................................................................................................ 110

2.3.1 Newton’s law of Viscous fluid ................................................................................................ 110

2.3.2 Stokes’ law of Viscous fluid .................................................................................................... 111

Sheet (V) ................................................................................................................................................... 122

Chapter (V) ........................................................................................................................................ 126

1. Temperature ................................................................................................................................ 126

2. Zeroth Law of Thermodynamics ............................................................................................... 128

3. Thermodynamic Systems ........................................................................................................... 128

4. Quantity of Heat .......................................................................................................................... 129

4.1 Heat Capacity and Specific Heat ................................................................................................ 130

4.2 Calorimetry and Phase Changes ............................................................................................... 133

5. Work Done During Volume Changes ........................................................................................ 141

6. Internal Energy and the First Law of Thermodynamics ........................................................ 145

7. Kinds of Thermodynamic Processes ......................................................................................... 146

7.1 Adiabatic Process ................................................................................................................ 146

7.2 Isochoric Process ...................................................................................................................... 148

7.3 Isobaric Process ................................................................................................................... 149

7.4 Isothermal Process ....................................................................................................................... 150

Sheet (VI) ................................................................................................................................................. 161

Useful Data ............................................................................................................................................... 164

Bibliography .............................................................................................................................................. 168

Syllabus of the course: Physics (I) [PHY 107]

Assessment

Final

Exam Midterm Assignment

Attendance

(practical) Quizzes Practical Exam

40% 20% 10% 10% 10% 10%

Wee

k

Ch

ap

ter

Chapter title Topic(s)

Lec

ture

Tu

tori

al

Pra

ctic

al

1 1 Physics and

Measurements

Units -system of units -Dimensions-

Dimension analysis 2 h 2 h 1 h

2 2 Oscillatory Motion

Simple Harmonic Motion (SHM) - velocity

and acceleration of SHM – Energy Of SHM-

Simple Pendulum – Spring Pendulum

2 h 2 h 1 h

3 3 Universal

Gravitation

Newton’s law of gravitation - Kepler’s laws -

satellite motion - escape velocity 2 h 2 h 1 h

4 4 Properties of Matter Elasticity- stress and strain- Modulus of

elasticity (Young, Bulk and Shear modulus) ,) 2 h 2 h 1 h

5 Properties of Matter thermal expansion (Linear, surface and bulk

expansion 2 h 2 h 1 h

6 4 Properties of Matter Fluid statics- density- pressure – Buoyancy

and Archimedes’ principle 2 h 2 h 1 h

7 Midterm Exam

8 4 Properties of Matter Fluid dynamics- continuity equation-

Bernoulli's equation- Viscosity of the fluid 2 h 2 h 1 h

9 5 Thermodynamics

Zeroth law of thermodynamics- thermal

equilibrium- Meteorically- heat- specific heat-

heating curve- Work.

2 h 2 h 1 h

10 5 Thermodynamics First Law of Thermodynamics -

thermodynamic processes 2 h 2 h 1 h

11 5 Thermodynamics Modes of heat transfer (conduction Convection-

Radiation)- Thermal conductivity 2 h 2 h 1 h

12 Final Laboratory Exam

13 Final Written Exam

Total hours 20 20 10

Credit hours = 3 h.

Lecture Tutorial Practical

2 2 1

Modern University of Technology and Information

Faculty of Engineering

Department: Physics and Eng. Mathematics

Instructors: Dr. Abdel Magied DIAB

Part ( I )

Discussion

Chapter (I)

Physics and

Measurements

1. Physical Quantity

A quantity which can be measured and expressed in form of laws is called a physical

quantity.

𝑃ℎ𝑦𝑠𝑖𝑐𝑎𝑙 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 (𝑄) = 𝑀𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 × 𝑈𝑛𝑖𝑡 = 𝑛 × 𝑢

Where, n represents the numerical value and u represents the unit. as the unit(u)

changes, the magnitude(n) will also change but product ‘nu’ will remain same.

i. e. n u = constant , or n1u1 = n2u2 = constant;

e.g. Mass of stool = 15 kg

Mass = Physical quantity 15 = Numerical value Kg = Standard unit

2. Fundamental and Derived Units

Any unit of mass, length and time in mechanics is called a fundamental, absolute or

base unit. Other units which can be expressed in terms of fundamental units, are

called derived units.

System of units: A complete set of units, both fundamental and derived for all kinds

of physical quantities is called system of units.

Physics (I) Chapter (I) Physics and measurements

Dr. A b d e l M a g i e d D I A D

2

For measurement of physical quantities, the following systems are commonly used:

-

(i) M.K.S system: In this system, the unit of length is metre, unit of mass is kg

and the unit of time is second.

(ii) C.G.S system: In this system, the unit of length is centimetre, the unit of mass

is gram and the unit of time is second.

(iii) F.P.S system: In this system, the unit of length is foot, the unit of mass is

pound and the unit of time is second.

(iv) S.I System: This system is an improved and extended version of M.K.S

system of units. It is called international system of unit.

Table 1.1: System of units.

(v) S.I. system: It is known as International System of units. There are seven

fundamental quantities in this system.

Fundamental Quantity: The quantity which is independent of other physical

quantities. In mechanics, mass, length and time are called fundamental quantities.

Units of these fundamental physical quantities are called Fundamental units.

MTI University Faculty of Engineering


3

Thus, MKS system was modified with addition of four other fundamental quantities

and two supplementary quantities.

These quantities and their units are given in the following table.

Table 1.2: Fundamental units.

Derived Quantity: The quantity which is derived from the fundamental quantities

e.g. area is a derived quantity. For example: -



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Table 1.3: Derived units.

• Some Units of Length, Mass, and Time



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Some typical lengths in the universe. (f) is a scanning tunneling microscope image

of atoms on a crystal surface; (g) is an artist’s impression.

3. Base dimensions and their SI units

In order to facilitate communication of scientific information, the International

System of units (SI for the french, Systeme International d’unites) was developed.

This allows us to use a well-defined convention for which units to use when

describing quantities. For example, the SI unit for the dimension of length is the

meter and the SI unit for the dimension of time is the second.

In order to simplify the SI unit system, a fundamental (base) set of dimensions was

chosen and the SI units were defined for those dimensions. Any other dimension can

always be re-expressed in terms of the base dimensions shown in Table 1.3 and its

units in terms of the corresponding combination of the base SI units.



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Dimension SI Unit

Length [L] meter [m]

Time [T] second [s]

Mass [M] kilogram [kg]

Temperature [θ] kelvin [K]

Electric Current [I] ampere [A]

Amount of Substance [N] mole [mol]

Luminous Intensity [J] candela [cd]

Dimensionless [l] unitless []

Table 1.3 : Base dimensions and their SI units with abbreviations.

From the base dimensions, one can obtain “derived” dimensions such as “speed”

which is a measure of how fast an object is moving.

The dimension of speed is 𝐿 𝑇−1 (length over time) and the corresponding SI unit is

m/s. Many of the derived dimension have corresponding derived SI units which can

be expressed in terms of the base SI units.

Dimension SI unit SI base units

Speed [L/T] meter per second [m/s] [m/s]

Frequency [1/T] hertz [Hz] [1/s]

Force [𝑀 ⋅ 𝐿 ⋅ 𝑇−2] newton [N] [𝑘𝑔 ⋅ 𝑚 ⋅ 𝑠−2]

Energy [𝑀 ⋅ 𝐿2 ⋅ 𝑇−2] joule [J] [𝑁 ⋅ 𝑚 = 𝑘𝑔 ⋅ 𝑚2 ⋅ 𝑠−2]

Power [𝑀 ⋅ 𝐿2 ⋅ 𝑇−3] watt [W] [𝐽/𝑠 = 𝑘𝑔 ⋅ 𝑚2 ⋅ 𝑠−3]

Electric Charge [I⋅T] coulomb [C] [𝐴 ⋅ 𝑠]

Voltage [𝑀 ⋅ 𝐿^2 ⋅ 𝑇−3 ⋅ 𝐼−1] volt [V] [𝐽/𝐶 = 𝑘𝑔 ⋅ 𝑚2 ⋅ 𝑠−3 ⋅ 𝐴−1]

Table 1.4 : Example of derived dimensions and their SI units with abbreviations.



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Table 1.4. shows a few derived dimensions and their corresponding SI units and

how those SI units are obtained from the base SI units.

By convention, we can indicate the dimension of a quantity, X, by writing it in

square brackets, [X]. For example,[𝑋] = 𝐼, would mean that the quantity X has the

dimension I, so it has the dimension of electric current. Similarly, we can indicate

the SI units of X with SI [X]. Referring to Table 1.3, since X has the dimension of

current, 𝑆𝐼[𝑋] = 𝐴.

4. Dimensional analysis

We call “dimensional analysis” the process of working out the dimensions of a

quantity in terms of the base dimensions and a model prediction for that quantity. A

few simple rules allow us to easily work out the dimensions of a derived quantity.

Suppose that we have two quantities, X and Y, both with dimensions. We then have

the following rules to find the dimension of a quantity that depends on X and Y:

1. Addition/Subtraction: You can only add or subtract two quantities if they have

the same dimension: [𝑋 + 𝑌] = [𝑋] = [𝑌]

2. Multiplication: The dimension of the product, [XY] , is the product of the

dimensions: [𝑋𝑌] = [𝑋] · [𝑌]

3. Division: The dimension of the ratio, [𝑋/𝑌] , is the ratio of the

dimensions: [𝑋/𝑌] = [𝑋]/[𝑌]

4. Numbers, and logarithm, triangle functions are dimensionless.



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Dimensional formula SI unit of Physical Quantities.



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Example 1

Calculate the dimensional formula of energy from the equation 𝑬 =𝟏

𝟐 𝒎 𝒗𝟐

Solution

Dimensionally, E = mass × (velocity)2.

Since 1

2 is a number and has no dimension.

Then [E] = M × (L T−1)2 = M L2T−2

Example 2

Kinetic energy of a particle moving along elliptical trajectory is given by 𝑲 =

𝒔𝟐 where s is the distance travelled by the particle. Determine dimensions of

.

Solution 𝐾 = 𝑠2

[𝛼] =[𝐾]

[𝑆2]=

M L2T−2

𝐿2= M T−2

Example 3

The position of a particle at time t, is given by the equation,

𝒙(𝒕) =𝒗𝟎

(𝟏 – 𝒆−𝜶 𝒕 ),

where 𝒗𝟎 is a constant and > 0. Find the dimensions of 𝒗𝟎 & .

Solution

The dimension of the exponent is dimensionless i.e.

[𝜶] =𝟏

[𝑻]= 𝑻−𝟏



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[𝒙(𝒕)] =[𝒗𝟎]

[] → [𝒗𝟎] = [𝒙(𝒕)] [] = 𝑳 𝑻−𝟏

PRINCIPLE OF HOMOGENEITY

According to this principle ,we can multiply physical quantities with same or

different dimensional formulae at our convenience, however no such rule applies to

addition and subtraction, where only like physical quantities can only be added or

subtracted. e.g. If 𝑷 + 𝑸 𝑷 & 𝑸 both represent same physical quantity.

Example 4

A physical relation must be dimensionally homogeneous, i.e., all the terms on

both sides of the equation must have the same dimensions. In the equation,

𝑺 = 𝒖 𝒕 + ½ 𝒂 𝒕𝟐

The length (S) has been equated to velocity (u) & time (t), which at first seems to be

meaningless, but if this equation is dimensionally homogeneous, i.e., the dimensions

of all the terms on both sides are the same, then it has physical meaning.

Now, dimensions of various quantities in the equation are

Distance [𝑆] = 𝐿 velocity [𝑢] = 𝐿 𝑇−1 time [𝑡] = 𝑇

Acceleration [𝑎] = 𝐿 𝑇−2

1

2 is a constant and has no dimensions.

Thus, the dimensions of the term on L.H.S. [𝑆] = 𝐿

Thus, the dimensions of the term on RH.S.



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[𝒖] [ 𝒕] + [𝒂] [ 𝒕𝟐] = ((𝐿 𝑇−1 )(𝑻)) + ((𝐿 𝑇−2)(𝑻𝟐)) = 𝑳 + 𝑳 = 𝑳

Here, the dimensions of all the terms on both sides of the equation are the same.

Therefore, the equation is dimensionally homogeneous.

4.1 Uses of dimensional equation

The principle of homogeneity & dimensional analysis has put to the following uses:

(i) Checking the correctness of physical equation.

(ii) To convert a physical quantity from one system of units into another.

(iii) To derive relation among various physical quantities

(i) To check the correctness of Physical relations:

According to principle of Homogeneity of dimensions a physical relation or

equation is correct, if the dimensions of all the terms on both sides of the equation

are the same. If the dimensions of even one term differs from those of others, the

equation is not correct.

Example 5

Check the correctness of the following formulae by dimensional analysis.

(i) 𝑭 = 𝒎𝒗𝟐/𝒓 (ii) 𝒕 = 𝟐 𝝅 √𝓵

𝒈

Solution

(i) 𝑭 = 𝒎𝒗𝟐/𝒓

the dimensions of the term on L.H.S. [𝐹] = 𝑀 𝐿 𝑇−2



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the dimensions of the term on R.H.S. [𝑚][𝑣2]

[𝑟]= 𝑀

(𝐿 𝑇−1)2

𝐿= 𝑀 𝐿 𝑇−2

The dimensions of the term on the L.H.S are equal to the dimensions of the term on

R.H.S. Therefore, the relation is correct.

(ii) 𝒕 = 𝟐 𝝅 √𝓵

𝒈

the dimensions of the term on L.H.S. [𝑡] = 𝑇

the dimensions of the term on R.H.S 𝟐 𝝅 √𝓵

𝒈

𝟐 𝝅 is a constant and has no dimensions.

[ℓ]12 [𝑔]−1/2 = 𝐿

12 (𝐿 𝑇−2)−1/2 = 𝐿

12 𝐿

−12 𝑇−2×−

12 = 𝑇



Example 6

Check the correctness of the following equation on the basis of dimensional

analysis, 𝒗 = √𝑷

𝝆 . Here 𝒗 is the velocity of sound, 𝑷 is the pressure and 𝝆 is the

density of the medium.

Solution

the dimensions of the term on L.H.S. [𝒗] = L 𝑇−1



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the dimensions of the term on R.H.S [√𝑷

𝝆 ]

The dimension of the pressure [𝑝] =[𝐹]

[𝐴2]=

𝑀 𝐿 𝑇−2

𝐿2= 𝑀 𝐿−1𝑇−2

The dimension of the density [𝜌] =[𝑚]

[𝑉]=

𝑀

𝐿3= 𝑀 𝐿−3

Then [√𝑷

𝝆 ] = (

𝑀 𝐿−1𝑇−2

𝑀 𝐿−3 )

𝟏/𝟐

= 𝑳 𝑻−𝟏



Example 7: Using Principle of Homogeneity of dimensions, check the

correctness of equations:

(i) 𝑬 = 𝒎𝒈𝒉 + ½ 𝒎𝒗𝟐

(ii) 𝒗𝟑 − 𝒖𝟐 = 𝟐 𝒂 𝒔𝟐

Solution

(i) 𝑬 = 𝒎𝒈𝒉 + ½ 𝒎𝒗𝟐

The dimension of the L.H.S, the energy [E]

[𝐸] = [𝐹𝑜𝑟𝑐𝑒]. [𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡] = 𝑀 𝐿 𝑇−2 × 𝐿 = 𝑀𝐿2 𝑇−2

The dimension of the R.H.S, the term 𝒎𝒈𝒉 + ½ 𝒎𝒗𝟐

[𝒎𝒈𝒉 ] + ½ [𝒎𝒗𝟐] = (𝑴 𝑳𝑻−𝟐 𝑳) + (𝑴 𝑳𝟐𝑻𝟐) = 𝑴𝑳𝟐𝑻−𝟐 + 𝑴𝑳𝟐𝑻−𝟐

= 𝑴𝑳𝟐𝑻−𝟐



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Thus, the dimension of both terms is the same, the dimension of the LHS equals to

the dimension of the RHS. Therefore, the relation is dimensionally correct.

(iii) 𝒗𝟑 − 𝒖𝟐 = 𝟐 𝒂 𝒔𝟐

The dimension of the L.H.S, the term [𝒗𝟑 − 𝒖𝟐 ]

[𝒗𝟑 − 𝒖𝟐] = (𝑳𝑻−𝟏)𝟑 − (𝑳𝑻−𝟏)𝟐 = 𝑳𝟑𝑻−𝟑 − 𝑳𝟐𝑻−𝟏

The dimension of the R.H.S, the term 𝟐 𝒂 𝒔𝟐

[𝟐 𝒂 𝒔𝟐] = 𝑳𝑻−𝟐 × 𝑳𝟐 = 𝑳𝟑𝑻−𝟐

Thus, the dimension of both terms is the different, the dimension of the LHS is not

equal to the dimension of the RHS. Therefore, the relation is dimensionally

incorrect.

Example 8

In the gas equation (𝑷 + 𝒂/𝒗𝟐) (𝒗 – 𝒃) = 𝑹𝑻, where T is the absolute

temperature, R is universal gas constant, P is pressure and v is volume of gas.

What are dimensions of a and b?

Solution

The quantities are added or subtracted, i.e. the dimension of the (𝑷 + 𝒂/𝒗𝟐) is

given as

The dimension of the pressure

𝑃 = [𝐹

𝐴] =

𝑀𝐿𝑇−2

𝐿2= 𝑀 𝐿−1𝑇−2

The relation to be correct, then the term

[𝐹

𝐴] = [

𝒂

𝒗𝟐] = 𝑀 𝐿−1𝑇−2

The dimension of the parameter a



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[𝑎] = 𝑀 𝐿−1𝑇−2 × [𝑣2] = 𝑀 𝐿−1𝑇−2 × (𝐿3)2 = 𝑀 𝐿5 𝑇−2

Then

[𝑎] = 𝑀 𝐿5 𝑇−2

In the same way, the dimension of the (𝒗 – 𝒃) as

The dimension of b is the same dimension of the volume

[𝑣] = [𝑏] = 𝐿3

Then

[𝑏] = 𝐿3

(ii) To convert a physical quantity from one system of units into another.

Physical quantity can be expressed as

𝑄 = 𝑛𝑢

Let 𝑛1𝑢1 represent the numerical value and unit of a physical quantity in one system

and 𝑛2𝑢2 in the other system.

If for a physical quantity 𝑄; 𝑀1𝐿1𝑇1 be the fundamental unit in one system and

𝑄; 𝑀2𝐿2𝑇2 be fundamental unit of the other system and dimensions in mass, length

and time in each system can be respectively a,b,c.

𝑢1 = [ 𝑀1𝑎𝐿1

𝑏𝑇1𝑐]

And

𝑢2 = [ 𝑀2𝑎𝐿2

𝑏𝑇2𝑐]

We are known that

𝑛1𝑢1 = 𝑛2𝑢2

Then 𝑛2 = 𝑛1 × 𝑢1

𝑢2

And

𝑛2 = 𝑛1 × [ 𝑀1

𝑎𝐿1𝑏𝑇1

𝑐

𝑀2𝑎𝐿2

𝑏𝑇2𝑐]



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𝑛2 = 𝑛1 [(𝑀1

𝑀2)

𝑎

(𝐿1

𝐿2)

𝑏

(𝑇

𝑇2)

𝑐

]

While applying the above relations the system of unit as first system in which

numerical value of physical quantity is given and the other as second system.

Thus knowing [𝑀1𝐿1𝑇1], [𝑀2𝐿2𝑇2] a, b, c and 𝑛1, we can calculate 𝑛2.

Example 9. Convert a force of 1 Newton to dyne.

Sol: Here we need to convert work from CGS system to MKS system Thus in the

equation

Example 10. Convert work of 1 erg into Joule.



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Sol. In the same way, the quantity 𝑄 = 𝑛1𝑢1 = 𝑛2𝑢2, therefore

Thus, 𝟏 𝒆𝒓𝒈 = 𝟏𝟎−𝟕 𝑱𝒐𝒖𝒍𝒆.

(iii) To derive relation among various physical quantities

If we know the various factors on which a physical quantity depends, then we can

find a relation among different factors by using principle of homogeneity.



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Example 11.

Let us find an expression for the time period t of a simple pendulum. The time

period t may depend upon (i) mass m of the bob of the pendulum, (ii) length

of pendulum, (iii) acceleration due to gravity g at the place where the pendulum

is suspended.

Solution

Let we can combine all the three factors, we get

𝑡 ∝ 𝑚𝑎 ℓ𝑏 𝑔𝑐 or 𝑡 = 𝑘 𝑚𝑎 ℓ𝑏 𝑔𝑐

Where 𝑘 is a dimensionless constant of proportionality.

Writing down the dimensions on either side of equation

[𝑡] = [𝑚𝑎] [ℓ𝑏][𝑔𝑐]

𝑇 = 𝑀𝑎 𝐿𝑏 (𝐿 𝑇−2)𝑐 = 𝑀𝑎 𝐿𝑏 𝐿𝑐 𝑇−2𝑐

Comparing dimensions 𝑀0𝐿0 𝑇 = 𝑀𝑎 𝐿𝑏+𝑐 𝑇−2𝑐 =

Then 𝑎 = 0, 𝑏 + 𝑐 = 0 𝑎𝑛𝑑 − 2𝑐 = 1

𝑎 = 0 , 𝑏 =1

2 𝑎𝑛𝑑 𝑐 = −

1

2

𝑡 = 𝑘 𝑚0 ℓ1

2 𝑔−1

2 𝑡 = 𝑘 √ℓ

𝑔



19

Example 12.

When a solid sphere moves through a liquid, the liquid opposes the motion with

a force F. The magnitude of F depends on the coefficient of viscosity of the

liquid, the radius r of the sphere and the speed v of the sphere. Assuming that

F is proportional to different powers of these quantities, guess a formula for F

using the method of dimensions.

Solution

Suppose the formula 𝐹 ∝ 𝜂𝑎 𝑟𝑏 𝑣𝑐 or 𝐹 = 𝑘 𝜂𝑎 𝑟𝑏 𝑣𝑐



[𝐹] = [𝜂𝑎] [ 𝑟𝑏] [𝑣𝑐]

𝑀 𝐿 𝑇−2 = (𝑀 𝐿−1𝑇−1)𝑎 𝐿𝑏 (𝐿 𝑇−1)𝑐

𝑀 𝐿 𝑇−2 = 𝑀𝑎 𝐿−𝑎+𝑏+𝑐 𝑇−𝑎−𝑐

Comparing dimensions

𝑎 = 1

−𝑎 + 𝑏 + 𝑐 = 1

−𝑎 − 𝑐 = −2

Then −1 − 𝑐 = −2 → 𝑐 = 1 and

−1 + 𝑏 + 1 = 1 → 𝑏 = 1



20

Thus, the formula for 𝐹 = 𝑘 𝜂 𝑟 𝑣

Example 13.

If P is the pressure of a gas and is its density, then find the dimension of

velocity in terms of P and .

Solution

Suppose the formula v ∝ pa 𝜌𝑏 or 𝑣 = k pa 𝜌𝑏



[𝑣] = [pa ] [𝜌𝑏]

𝑀0𝐿 𝑇−1 = (𝑀 𝐿−1 𝑇−2)𝑎 (𝑀 𝐿−3)𝑏 = 𝑀𝑎+𝑏 𝐿−𝑎−3𝑏 𝑇−2𝑎

Comparing dimensions

𝑎 + 𝑏 = 0

−𝑎 − 3𝑏 = 1

−2𝑎 = −1 → 𝑎 =1

2 and 𝑏 = −

1

2

Thus, the formula for 𝑣 = √𝑝

𝜌



21

Example 14.

Consider a planet of mass (m), revolving round the sun. The time period (T) of

revolution of the planet depends upon the radius of the orbit (r), mass of the

sun (M) and the gravitational constant (G). Using dimensional analysis, verify

Kepler’s third law of planetary motion.

Solution

Suppose the formula T ∝ 𝑟𝑎 Mb 𝐺𝑏 or T = k 𝑟𝑎 Mb 𝐺𝑏



[T] = [𝑟𝑎] [Mb] [𝐺𝑐]

[𝐺] = 𝐹 𝑟2

𝑀2=

𝑀 𝐿 𝑇−2 𝐿2

𝑀2= 𝑀−1 𝐿3 𝑇−2

𝑀0 𝐿0 𝑇 = 𝐿𝑎 𝑀𝑏 (𝑀−1 𝐿3 𝑇−2)𝑐 = 𝑀𝑏−𝑐 𝐿𝑎+3𝑐 𝑇−2𝑐

𝑏 − 𝑐 = 0 𝑎 + 3𝑐 = 0 − 2𝑐 = 1

𝑐 =−1

2 𝑏 = −

1

2 𝑎 = −3𝑐 =

3

2

Thus, the formula for 𝑇 = 𝑘 𝑟3

2 𝑀

−1

2 𝐺−1/2 𝑇2 = 𝑘 𝑟3

𝐺 𝑀

5. SI prefixes

There are 20 prefixes used in SI in order to form decimal multiples and submultiples

of SI units are given in the Table. The goal is to reach a scientific expression for very

large or very small quantities.Obviously, it is no longer usual, to say for example,

1000,000,000 cm or 0,00000001 km.



22

Factor Name Symbol

1024 yotta Y

1021 zetta Z

1018 exa E

1015 peta P

1012 tera T

109 giga G

106 mega M

103 kilo k

102 hecto h

101 deka da

Factor Name Symbol

10-1 deci d

10-2 centi c

10-3 milli m

10-6 micro µ

10-9 nano n

10-12 pico p

10-15 femto f

10-18 atto a

10-21 zepto z

10-24 yocto y

Other units outside the SI that are currently accepted for use with the SI, subject to

further review

Name Symbol Value in SI units

nautical mile 1 nautical mile = 1852 m

knot 1 nautical mile per hour =

(1852/3600) m/s

are a 1 a = 1 dam2 = 102 m2

hectare ha 1 ha = 1 hm2 = 104 m2

bar bar 1 bar = 0.1 MPa = 100 kPa =

1000 hPa = 105 Pa

ångström Å 1 Å = 0.1 nm = 10-10 m

barn b 1 b = 100 fm2 = 10-28 m2

curie Ci 1 Ci = 3.7 x 1010 Bq

roentgen R 1 R = 2.58 x 10-4 C/kg

rad rad 1 rad = 1 cGy = 10-2 Gy

rem rem 1 rem = 1 cSv = 10-2 Sv



23

Sheet (I)

I) Short Answer Questions

1. What do you mean by physical quantity?

2. Differentiate between fundamental and derived unit.

3. Write full form of the following system of unit (i) CGS (ii) FPS (iii) MKS

4. Write definition of Dimensions.

5. What is the suitable unit for measuring distance between sun and earth?

6. Write the dimensional formula of the following physical quantity - (i)

Momentum (ii) Power (iii) Surface Tension (iv) Strain 8. What is the principle

of Homogeneity of Dimensions?

7. . Write the S.I & C.G.S units of the following physical quantities- (a) Force

(b) Work

8. What are the uses of dimensions?

II) Long Answer Questions

1. Check the correctness of the relation 𝜆 = ℎ /𝑚𝑣; where𝜆 is wavelength, h-

Planck’s constant, m is mass of the particle and v - velocity of the particle.

2. Check the correctness of the following relation by using method of dimensions

(i) 𝑣 = 𝑢 + 𝑎𝑡 (ii) 𝐹 = 𝑚𝑣

𝑟2 (iii) 𝑣2 – 𝑢2 = 2𝑎𝑠

3. If Force = (x/density) + C is dimensionally correct, find the dimension of x.

4. If 𝐹 = 𝑎𝑥 + 𝑏 𝑡2 + 𝑐 where F is force, 𝑥 is distance and t is time. Then what

is dimension of 𝑎 𝑥 𝑥

𝑏 𝑡2.

5. The frequency of vibration of a string depends on the length L between the nodes,

the tension F in the string and its mass per unit length m. Guess the expression

for its frequency from dimensional analysis.



24

6. The distance moved by a particle in time from center of ring under the influence

of its gravity is given by 𝑥 = a sin𝑡 where 𝑎 and are constants. If is found

to depend on the radius of the ring (r), its mass (m) and universal gravitation

constant (G), find using dimensional analysis an expression for in terms of

𝑟, 𝑚 𝑎𝑛𝑑 𝐺.

7. Using dimensional analysis, determine if the following equations are

dimensionally correct or incorrect:

(a) 𝑣 = 𝑣0 + 𝑎𝑆

(b) 𝑣2 = 𝑣02 + 2𝑎𝑠

(c)𝑆 = 𝑆0 + 𝑣0𝑡 +1

2𝑎𝑡2

(d) 𝑆 = 𝑆0𝑐𝑜𝑠(𝐾𝑡), where K is a constant that has the dimension of the inverse of

time. Here 𝑣 represents the velocity at any time t, 𝑣0 the initial velocity at t=0 sec.,

a the uniform acceleration and 𝑆 is the displacement.

8. Find the dimension of the quantity on the left-hand side of the equation:

(a)F = m a (b) K.E = 1

2 m v2 (c) p = m v (d) W = m a S

(e) L = m v r

Here F, m, a, v, p, r, S, and L are defined as force, mass, acceleration, velocity,

momentum , radius, displacement, and angular momentum of an object,

respectively.

9- The Boltzmann distribution in thermodynamics involves the factor

𝑒𝑥𝑝[−𝐸 (𝑘𝑇)⁄ ] where E represents energy, T is the temperature and k is

Boltzmann's constant. Find the dimensions of k.

10- Suppose the dimension of the volume [𝑉] = 𝐿3 , density [𝜌] = 𝑀𝐿–3, and time

[t] = T.

(a) What is the dimension of ∫ρ dV? (b) What is the dimension of dV/dt?



25

(c) What is the dimension of ρ(dV/dt)?

11- Hook's law states that the force, F, in a spring extended by a length x is given by

𝐹 = − 𝑘 𝑥. From Newton's second law 𝐹 = 𝑚𝑎, Where m is the mass and a is the

acceleration, estimate the dimension of the spring constant k.

12- If 𝐹 = 𝑎𝑥 + 𝑏𝑡2 + 𝑐, where x, t and F are the displacement, time and force of

an object, respectively. What is the dimension of 𝑎𝑥𝑐 𝑏⁄ 𝑡2 ?

13- Express the dimensional analysis of the viscosity coefficient 𝜂 of a fluid,

𝐹 = −𝜂𝐴𝑣2−𝑣1

𝑥2−𝑥1

Where, 𝐴 is the area, 𝑣2, 𝑣1 are the velocities, 𝑥2, 𝑥1 are the displacements.

14- A simple pendulum has periodic time T given by the relation: 𝑇 = 2𝜋√𝑙

𝑔, show

that this equation is dimensionally correct.

15- A particle moves with a speed v in a circular orbit of radius r, Given that the

magnitude of the acceleration a is proportional to some power of 𝑟𝑚, and some

power of 𝑣𝑛, determine the powers of 𝑟 and 𝑣.

10- The frequency of vibration of a string 𝑓 depends on the length L between the

nodes, the tension F in the string and its mass per unit length 𝜇. Find the expression

for its frequency from dimensional analysis.

16- When a solid sphere moves through a liquid, the liquid opposes the motion with

a force 𝐹. The magnitude of 𝐹 depends on the coefficient of viscosity 𝜂 of the liquid,

the radius 𝑟 of the sphere and the speed 𝑣 of the sphere. Assuming that 𝐹 is

proportional to different powers of these quantities. Express the formula for 𝐹 by

using the method of dimensions.

27

Chapter ( II )

1. Periodic Motion

A motion, which repeat itself over and over again after a regular interval of time is

called a periodic motion and the fixed interval of time after which the motion is

repeated is called period of the motion.

Examples :

(i) Revolution of earth around the sun (period one year)

(ii) Rotation of earth about its polar axis (period one day)

(iv) Motion of hour’s hand of a clock (period 12-hour)

(iv) Motion of minute’s hand of a clock (period 1-hour)

(v) Motion of second’s hand of a clock (period 1-minute)

(vi) Motion of moon around the earth (period 27.3 days)

Figure 2.1

Physics (I) Chapter (II) Oscillatory Motion


28

Oscillatory or vibratory motion is that motion in which a body moves to and from

or back and forth repeatedly about a fixed point in a definite interval of time.

Oscillatory motion is also called as harmonic motion.

Example :

(i) The motion of the pendulum of a wall clock.

(ii) The motion of a load attached to a spring when it is pulled and then

released.

(iii) The motion of liquid contained in U- tube when it is compressed once in

one limb and left to itself.

(iv) A loaded piece of wood floating over the surface of a liquid when pressed

down and then released executes oscillatory motion.

Figure 2.2

2. Some Important definition

(1) Time period (T) : It is the least interval of time after which the periodic motion

of a body rep S.I. units of time period is second.



29

(2) Frequency (𝒇) : It is defined as the number of periodic motions executed by

body per second. SI unit of the frequency is hertz (Hz).

(3) Angular Frequency () : Angular frequency of a body executing periodic

motion is equal frequency of the body with factor 2. Angular frequency = 2 𝑓

S.I. units of is Hz [S.I.] also represents angular velocity. In that case unit will

be 𝑟𝑎𝑑/𝑠𝑒𝑐.

(4) Displacement: In general, the name displacement is given to a physical quantity

which undergoes a change with time in a periodic motion.

Examples:

(i) In an oscillation of a loaded spring, displacement variable is its deviation

from the mean position.

(ii) During the propagation of sound wave in air, the displacement variable is the

local change in pressure

(iii) During the propagation of electromagnetic waves, the displacement

variables are electric and magnetic fields, which vary periodically.

(5) Amplitude: The maximum absolute value of some quantity that varies.

(5) Phase : phase of a vibrating particle at any instant is a physical quantity, which

completely express the position and direction of motion, of the particle at that instant

with respect to its mean position.

In oscillatory motion the phase of a vibrating particle is the argument of sine or

cosine function involved to represent the generalized equation of motion of the

vibrating particle.

𝑦 = 𝑎 𝑠𝑖𝑛 = 𝑎 𝑠𝑖𝑛( 𝑡 +0

) here, = 𝑡 + 0

= 𝑝ℎ𝑎𝑠𝑒 𝑜𝑓 𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑛𝑔 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒.



30

(i) Initial phase or epoch : It is the phase of a vibrating particle at 𝑡 = 0. In

= 𝑡 + 0 , when 𝑡 = 0 ; = 0 here, 0 is the angle of epoch.

(ii) Same phase : Two vibrating particle are said to be in same phase, if the

phase difference between them is an even multiple of or path difference

is an even multiple of (𝜆/ 2) or time interval is an even multiple of (𝑇

/ 2) because 1 time period is equivalent to 2 rad or 1 wave length (𝜆)

(iii) Opposite phase : When the two vibrating particles cross their respective

mean positions at the same time moving in opposite directions, then the

phase difference between the two vibrating particles is 180o Opposite

phase means the phase difference between the particle is an odd multiple

of (say 𝝅, 𝟑𝝅, 𝟓𝝅, 𝟕𝝅 … . . ) or the path difference is an odd multiple of 𝜆

(say 𝜆

2,

3𝜆

2, … . . ) or the time interval is an odd multiple of (T / 2).

(iv) Phase difference: If two particles performs S.H.M and their equation are

𝑦1 = 𝑎 𝑠𝑖𝑛( 𝑡 +1

) 𝑎𝑛𝑑 𝑦2 = 𝑎 𝑠𝑖𝑛( 𝑡 +2

)

then phase difference = ( 𝑡 +2

) − ( 𝑡 +1

) = 2

−1

3. The motion of a spring-mass system

As an example of simple harmonic motion,

we first consider the motion of a block of

mass mm that can slide without friction

along a horizontal surface.

The mass is attached to a spring with spring

constant 𝜅 which is attached to a wall on the

other end.

Figure 2.3 A horizontal spring-mass system

oscillating about the origin with an amplitude A.



31

We introduce a one-dimensional coordinate system to describe the position of the

mass, such that the x axis is co-linear with the motion, the origin is located where

the spring is at rest, and the positive direction corresponds to the spring being

extended. This “spring-mass system” is illustrated in Figure 2.3.

restoring force which is always directed towards the mean position and whose

magnitude at any instant is directly proportional to the displacement of the particle

from the mean position at that instant.

Hooke’s Law

𝑅𝑒𝑠𝑡𝑜𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 ∝ 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑓𝑟𝑜𝑚 𝑚𝑒𝑎𝑛 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛.

𝑭 ∝ −𝒙 → 𝑭 = −𝒌 𝒙

Where 𝑘 is known as force constant. Its S.I. unit is Newton/meter and dimension is

[𝑀𝑇–2 ].

Restoring force: A variable force that gives rise to an equilibrium in a physical

system. If the system is perturbed away from the equilibrium, the restoring force will

tend to bring the system back toward equilibrium. The restoring force is a function

only of position of the mass or particle. It is always directed back toward the

equilibrium position of the system

We modeled the motion of a mass attached to a spring and found that its

position, 𝑥(𝑡), was described by the following differential equation:

According to the Newton’s 2nd law 𝑭 = 𝒎 𝒂 = 𝒎 𝒅𝟐 𝒙

𝒅𝒕𝟐

By equaling between both forces, one can find



32

𝒎 𝒅𝟐 𝒙

𝒅𝒕𝟐= −𝒌 𝒙 →

𝒅𝟐 𝒙

𝒅𝒕𝟐+

𝒌

𝒎 𝒙 = 𝟎

The angular frequency of oscillations is given by

𝜔2 = 𝑘/𝑚

In one complete rotation, OM describes an angle 2π and it takes time T to complete

one rotation. Hence 𝜔 = 2 𝜋/𝑇.

we get an expression for The period of a mass on a spring is given by the equation :

𝑇 = 2𝜋 √𝑚

𝑘

Figure 2.4

Figure 2.4 A block is attached to one end of a spring and placed on a frictionless

table. The other end of the spring is anchored to the wall. The equilibrium position,

where the net force equals zero, is marked as 𝑥 = 0 𝑚. Work is done on the block,



33

pulling it out to 𝑥 = +𝐴, and the block is released from rest. The block oscillates

between 𝑥 = +𝐴 and 𝑥 = −𝐴. The force is also shown as a vector.

A possible solution to that DF was given by:

𝒙(𝒕) = 𝑨 𝑪𝒐𝒔 (𝝎 𝒕 + 𝝓 )

where A = Amplitude, 𝝎 = Angular frequency, 𝑡 = Instantaneous time, and 𝝓 is

the phase shift measured in radians (Figure 2.5). It should be noted that because

sine and cosine functions differ only by a phase shift, this motion could be

modeled using either the cosine or sine function.

Figure 2.5 (a) A cosine function. (b) A cosine function shifted to the left by an

angle ϕ. The angle ϕ is known as the phase shift of the function.

The velocity of the mass on a spring, oscillating in SHM, can be found by taking

the derivative of the position equation:

𝒗(𝒕) =𝒅 𝒙(𝒕)

𝒅𝒕=

𝒅

𝒅𝒕[𝑨 𝑪𝒐𝒔 (𝝎 𝒕 + 𝝓 )] = −𝑨𝝎 𝑺𝒊𝒏(𝝎 𝒕 + 𝝓 )

Another formula to the velocity of SHM 𝑣 = −𝑨𝝎 𝑺𝒊𝒏(𝝎 𝒕 + 𝝓 )

∵ cos2 𝜃 + sin2 𝜃 = 1 𝑣 = −𝐴 𝜔 √𝟏 − 𝑪𝒐𝒔𝟐 (𝝎 𝒕 + 𝝓 )

𝑣 = − 𝜔 √𝑨𝟐 − 𝑨𝟐𝑪𝒐𝒔𝟐 (𝝎 𝒕 + 𝝓 ) = − 𝜔 √𝑨𝟐 − 𝒚𝟐

https://openstax.org/books/university-physics-volume-1/pages/15-1-simple-harmonic-motion#CNX_UPhysics_15_01_OffsetExam



34

𝒗(𝒕) = − 𝝎 √𝑨𝟐 − 𝒚𝟐

Because the sine function oscillates between –1 and +1, the maximum velocity is the

amplitude times the angular frequency,

𝒗𝒎𝒂𝒙 = 𝑨𝝎.

The maximum velocity occurs at the equilibrium position (𝑥 = 0) when the mass is

moving toward 𝑥 = +𝐴. The maximum velocity in the negative direction is attained

at the equilibrium position (𝑥 = 0) when the mass is moving toward 𝑥 = −𝐴 and

is equal to −𝒗𝒎𝒂𝒙.

Figure 2.6



35

The acceleration of the mass on the spring can be found by taking the time

derivative of the velocity:

𝒂(𝒕) =𝒅 𝒗(𝒕)

𝒅𝒕=

𝒅

𝒅𝒕 [ −𝑨𝝎 𝑺𝒊𝒏(𝝎 𝒕 + 𝝓 )] = −𝑨𝝎𝟐 𝑪𝒐𝒔 (𝝎 𝒕 + 𝝓 )

The maximum acceleration is

𝒂𝒎𝒂𝒙 = 𝑨 𝝎𝟐

The maximum acceleration occurs at the position (𝑥 = −𝐴), and the acceleration

at the position (𝑥 = −𝐴) and is equal to −𝑎𝑚𝑎𝑥 .

Example 1

A 𝟐. 𝟎𝟎 − 𝒌𝒈 block is placed on a frictionless surface. A spring with a force

constant of 𝒌 = 𝟑𝟐. 𝟎𝟎 𝑵/𝒎 is attached to the block, and the opposite end of

the spring is attached to the wall. The spring can be compressed or extended.

The equilibrium position is marked as 𝒙 = 𝟎. 𝟎𝟎𝒎.

Work is done on the block, pulling it out 𝒕𝒐 𝒙 = +𝟎. 𝟎𝟐𝒎. The block is released

from rest and oscillate between 𝒙 = +𝟎. 𝟎𝟐𝒎 and. 𝒙 = −𝟎. 𝟎𝟐𝒎. The period of

the motion is 𝟏. 𝟓𝟕 𝒔. Determine the equations of motion.

Solution

We first find the angular frequency. The phase shift is zero,𝜙 = 0.00𝑟𝑎𝑑, because

the block is released from rest at 𝑥 = 𝐴 = +0.02𝑚. Once the angular frequency is

found, we can determine the maximum velocity and maximum acceleration.

The angular frequency can be found and used to find the maximum velocity and

maximum acceleration:



36

• The angular frequency 𝝎 =𝟐𝝅

𝑻=

𝟐𝝅

𝟏.𝟓𝟕= 𝟒. 𝟎𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄

• The maximum velocity 𝒗𝒎𝒂𝒙 = 𝑨𝝎 = 0.02 × 4 = 0.08𝑚

𝑠

• The maximum acceleration 𝒂𝒎𝒂𝒙 = 𝑨 𝝎𝟐 = 𝟎. 𝟎𝟐 × (𝟒)𝟐 = 𝟎. 𝟑𝟐 𝒎/𝒔𝟐

All that is left is to fill in the equations of motion:

𝒙(𝒕) = 𝑨 𝑪𝒐𝒔 (𝝎 𝒕 + 𝝓 ) = 0.02 𝑪𝒐𝒔 (𝟒 𝒕 )

𝒗(𝒕) = −𝒗𝒎𝒂𝒙 𝑺𝒊𝒏(𝝎 𝒕 + 𝝓 ) = 0.08 𝑺𝒊𝒏(𝟒 𝒕)

𝒂(𝒕) = −𝒂𝒎𝒂𝒙 𝑪𝒐𝒔(𝝎 𝒕 + 𝝓 ) = 0.32 𝑪𝒐𝒔(𝟒 𝒕)

4. Energy in Simple Harmonic Motion

By the end of this section, you will be able to:

• Describe the energy conservation of the system of a mass and a spring

• Explain the concepts of stable and unstable equilibrium points

4.1. Describe the energy conservation of the system of a mass and a spring

To produce a deformation in an object, we must do work. That is, whether you pluck

a guitar string or compress a car’s shock absorber, a force must be exerted through

a distance. If the only result is deformation, and no work goes into thermal, sound,

or kinetic energy, then all the work is initially stored in the deformed object as some

form of potential energy.

Consider the example of a block attached to a spring on a frictionless table,

oscillating in SHM. The force of the spring is a conservative force (which you

studied in the chapter on potential energy and conservation of energy), and we can

define a potential energy for it.



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This potential energy is the energy stored in the spring when the spring is extended

or compressed. In this case, the block oscillates in one dimension with the force of

the spring acting parallel to the motion:

𝑼 =𝟏

𝟐 𝒌 𝒙𝟐

In a simple harmonic oscillator, the energy oscillates between kinetic energy of the

mass 𝑲 =𝟏

𝟐𝒎𝒗𝟐 and potential energy 𝑼 =

𝟏

𝟐 𝒌 𝒙𝟐 stored in the spring.

In the SHM of the mass and spring system, there are no dissipative forces, so the

total energy is the sum of the potential energy and kinetic energy.

In this section, we consider the conservation of energy of the system. The concepts

examined are valid for all simple harmonic oscillators, including those where the

gravitational force plays a role.

Consider Figure 2.7, which shows an oscillating block attached to a spring. In the

case of undamped SHM, the energy oscillates back and forth between kinetic and

potential, going completely from one form of energy to the other as the system

oscillates.

So for the simple example of an object on a frictionless surface attached to a spring,

the motion starts with all of the energy stored in the spring as elastic potential energy.

As the object starts to move, the elastic potential energy is converted into kinetic

energy, becoming entirely kinetic energy at the equilibrium position. The energy is

then converted back into elastic potential energy by the spring as it is stretched or

compressed.

The velocity becomes zero when the kinetic energy is completely converted, and this

cycle then repeats. Understanding the conservation of energy in these cycles will

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provide extra insight here and in later applications of SHM, such as alternating

circuits.

Figure 2.7 The transformation of energy in SHM for an object attached to a spring

on a frictionless surface.

(a) When the mass is at the position 𝑥 = +𝐴, all the energy is stored as potential

energy in the spring 𝑼 =𝟏

𝟐 𝒌 𝒙𝟐. The kinetic energy is equal to zero because the

velocity of the mass is zero.

(b) As the mass moves toward 𝑥 = −𝐴, the mass crosses the position 𝑥 = 0. At this

point, the spring is neither extended nor compressed, so the potential energy stored

in the spring is zero. At 𝑥 = 0, the total energy is all kinetic energy where K =

1

2𝑚 𝑣𝑚𝑎𝑥 2.



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(c) The mass continues to move until it reaches 𝑥 = −𝐴 where the mass stops and

starts moving toward 𝑥 = +𝐴. At the position 𝑥 = −𝐴, the total energy is stored as

potential energy in the compressed 𝑼 =𝟏

𝟐 𝒌 𝑨𝟐 and the kinetic energy is zero.

(d) As the mass passes through the position 𝑥 = 0, the kinetic energy is K =

1

2𝑚 𝑣𝑚𝑎𝑥 2 and the potential energy stored in the spring is zero.

(e) The mass returns to the position 𝑥 = +𝐴, where 𝐾 = 0 and 𝑼 =𝟏

𝟐 𝒌 𝑨𝟐 .

Consider Figure 2.7 which shows the energy at specific points on the periodic

motion. While staying constant, the energy oscillates between the kinetic energy of

the block and the potential energy stored in the spring:

𝐸𝑇𝑜𝑡𝑎𝑙 = 𝑈 + 𝐾 =𝟏

𝟐 𝒌 𝒙𝟐 +

𝟏

𝟐𝒎𝒗𝟐

The motion of the block on a spring in SHM is defined by

the position 𝑥(𝑡) = 𝑨 𝑪𝒐𝒔 (𝝎 𝒕 + 𝝓 )

with a velocity of 𝒗(𝒕) = −Aω 𝑺𝒊𝒏(𝝎 𝒕 + 𝝓 )

Using these equations, the trigonometric identity cos2 𝜃 + sin2 𝜃 = 1 and 𝜔 = √𝑘

𝑚,

we can find the total energy of the system:

𝐸𝑇𝑜𝑡𝑎𝑙 = 𝑈 + 𝐾 =𝟏

𝟐 𝒌 𝒙𝟐 +

𝟏

𝟐𝒎𝒗𝟐

𝐸𝑇𝑜𝑡𝑎𝑙 =𝟏

𝟐 𝒌 [𝑨 𝑪𝒐𝒔 (𝝎 𝒕 + 𝝓 )]𝟐 +

𝟏

𝟐𝒎 [−Aω 𝑺𝒊𝒏(𝝎 𝒕 + 𝝓 )]𝟐



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𝐸𝑇𝑜𝑡𝑎𝑙 = 𝟏

𝟐 𝒌 𝑨𝟐 [ 𝑪𝒐𝒔 (𝝎 𝒕 + 𝝓 )]𝟐 +

1

2𝑚 𝑨𝟐 ω2 [𝑺𝒊𝒏(𝝎 𝒕 + 𝝓 )]𝟐


𝟐 𝒌 𝑨𝟐 [ 𝑪𝒐𝒔 (𝝎 𝒕 + 𝝓 )]𝟐 +

1

2𝑚 𝑨𝟐

𝑘

𝑚 [𝑺𝒊𝒏(𝝎 𝒕 + 𝝓 )]𝟐


𝟐 𝒌 𝑨𝟐 [ 𝑪𝒐𝒔 (𝝎 𝒕 + 𝝓 )]𝟐 +

1

2𝐾 𝑨𝟐 [𝑺𝒊𝒏(𝝎 𝒕 + 𝝓 )]𝟐


𝟐 𝒌 𝑨𝟐( [ 𝑪𝒐𝒔 (𝝎 𝒕 + 𝝓 )]𝟐 + [𝑺𝒊𝒏(𝝎 𝒕 + 𝝓 )]𝟐) =

𝟏

𝟐 𝒌 𝑨𝟐


𝟐 𝒌 𝑨𝟐

The total energy of the system of a block and a spring is equal to the sum of the

potential energy stored in the spring plus the kinetic energy of the block and is

proportional to the square of the amplitude 𝐸𝑇𝑜𝑡𝑎𝑙 = 𝟏

𝟐 𝒌 𝑨𝟐. The total energy of

the system is constant.

A closer look at the energy of the system shows that the kinetic energy oscillates like

a sine-squared function, while the potential energy oscillates like a cosine-squared

function. However, the total energy for the system is constant and is proportional to

the amplitude squared. Figure 2.8 shows a plot of the potential, kinetic, and total

energies of the block and spring system as a function of time.

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Figure 2.8 Graph of the kinetic energy, potential energy, and total energy of a block

oscillating on a spring in SHM. Also shown are the graphs of position versus time

and velocity versus time. The total energy remains constant, but the energy oscillates

between kinetic energy and potential energy. When the kinetic energy is maximum,

the potential energy is zero. This occurs when the velocity is maximum and the mass

is at the equilibrium position. The potential energy is maximum when the speed is

zero. The total energy is the sum of the kinetic energy plus the potential energy and

it is constant.

4.2 Explain the concepts of stable and unstable equilibrium points

We have just considered the energy of SHM as a function of time. Another

interesting view of the simple harmonic oscillator is to consider the energy as a

function of position. Figure 2.9 shows a graph of the energy versus position of a

system undergoing SHM.

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Figure 2.9 A graph of the kinetic energy (K), potential energy (U), and total energy

(top straight line) of a simple harmonic oscillator. The force is equal to 𝐹 = −𝑑𝑈

𝑑𝑥.

The equilibrium position is shown as a black dot and is the point where the force is

equal to zero. The force is positive when 𝑥 < 0, negative when 𝑥 > 0, and equal to

zero when 𝑥 = 0.

The potential energy curve in Figure 2.9 resembles a bowl. When a marble is placed

in a bowl, it settles to the equilibrium position at the lowest point of the bowl (x=0).

This happens because a restoring force points toward the equilibrium point. This

equilibrium point is sometimes referred to as a fixed point.

When the marble is disturbed to a different position (𝑥 = +𝐴), the marble oscillates

around the equilibrium position. Looking back at the graph of potential energy, the

force can be found by looking at the slope of the potential energy graph 𝐹 = −𝑑𝑈

𝑑𝑥.

Since the force on either side of the fixed-point points back toward the equilibrium

point, the equilibrium point is called a stable equilibrium point.

stable equilibrium point

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point where the net force on a system is zero, but a small displacement of the

mass will cause a restoring force that points toward the equilibrium point

The points 𝑥 = 𝐴 and 𝑥 = −𝐴 are called the turning points. (See Potential Energy

and Conservation of Energy.)

Stability is an important concept. If an equilibrium point is stable, a slight

disturbance of an object that is initially at the stable equilibrium point will cause the

object to oscillate around that point. The stable equilibrium point occurs because the

force on either side is directed toward it. For an unstable equilibrium point, if the

object is disturbed slightly, it does not return to the equilibrium point.

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The following figure shows that

(a) through (e) Several instants in the simple harmonic motion for a block–spring

system. Energy bar graphs show the distribution of the energy of the system at each

instant. The parameters in the table at the right refer to the block–spring system,

assuming at 𝑡 = 0, 𝑥 = 𝐴; hence, 𝑥 = 𝐴 cos 𝜔𝑡. For these five special instants,

one of the types of energy is zero. (f) An arbitrary point in the motion of the

oscillator. The system possesses both kinetic energy and potential energy at this

instant as shown in the bar graph.

Example 2

A 1.50 Kg mass on a horizontal frictionless surface is attached to a horizontal

spring with spring constant 𝒌 = 𝟐𝟎𝟎 𝑵/𝒎. The mass is in equilibrium at 𝒙 =

𝟎. 𝟎 𝒎. The mass is released when 𝒕 = 𝟎. 𝟎 𝒔𝒆𝒄 at coordinate 𝒙 = 𝟎. 𝟏 𝒎 with

velocity 𝒗𝒙 = 𝟐. 𝟎 𝒎/𝒔.

(i) Calculate the angular frequency 𝜔, periodic time T, phase shift 𝜙 and

Amplitude A.

(ii) Determine the maximum of velocity and the maximum acceleration of the

SHM.

(iii) Determine the maximum force and its total energy of the SHM.

Solution

(i) the angular frequency 𝜔, periodic time T, phase shift 𝜙 and Amplitude A.



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• The angular frequency 𝜔 = √𝐾

𝑚= √

200

1.5= 11.5 𝑟𝑎𝑑/𝑠𝑒𝑐

• periodic time T 𝜔 =2𝜋

𝑇 → 𝑇 =

2𝜋

𝜔=

2𝜋

11.5= 0.546 𝑠𝑒𝑐

• To find the phase shift 𝜙, at 𝑡 = 0.0 𝑠𝑒𝑐 at coordinate 𝑥 = 0.1 𝑚 with

velocity 𝑣𝑥 = 2.0 𝑚/𝑠.

𝑥(𝑡) = 𝐴 𝐶𝑜𝑠 (𝜔 𝑡 + 𝜙 ) → 0.1 = A Cos 𝜙 (1. 𝑎)

and 𝑣(𝑡) = −𝐴 𝜔 𝑆𝑖𝑛(𝜔 𝑡 + 𝜙 ) → 2.0 = −11.5 𝐴 𝑆𝑖𝑛𝜙 (1. 𝑏)

From Eqs. (1.a) and (1.b)

−11.5 𝐴 𝑆𝑖𝑛𝜙

A Cos 𝜙= −11.5 tan 𝜙 =

2.0

0.1

𝜙 = −1.05 𝑟𝑎𝑑

• The Amplitude A, From Eq. (1.a) 0.1 = A Cos 𝜙

0.1 𝑚 = 𝐴 𝐶𝑜𝑠 (−1.05 ×180

𝜋) → 𝐴 = 0.201 𝑚

(ii) the maximum of velocity and the maximum acceleration of the SHM.

• The maximum velocity 𝑣𝑚𝑎𝑥 = 𝐴𝜔 = 0.201 × 11.5 = 2.31𝑚

𝑠

• The maximum acceleration 𝑎𝑚𝑎𝑥 = 𝐴 𝜔2 = 0.201 × (11.5)2 = 26.6𝑚

𝑠2

(iii) the maximum force and its total energy of the SHM.

the maximum force 𝐹𝑚𝑎𝑥 = 𝑚 𝑎𝑚𝑎𝑥 = 1.5 × 26.6 = 39.9 𝑁

the maximum energy. 𝐸 =1

2 𝑘 𝐴2 =

1

2× 200 × (0.201)2 = 4.0401 𝐽



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Example 3

A mass of 100 gm is attached at the end of a light spring which oscillates on a

frictionless horizontal table with an amplitude equal to 0.16 meter and the time

period equal to 2 sec. Initially the mass is released from rest at 𝒕 = 𝟎. 𝟎 𝒔𝒆𝒄

and displacement 𝒙 = – 𝟎. 𝟏𝟔 meter. Find the expression for the displacement

of the mass at any time.

Solution

The displacement of the mass at any time in SHM reads as

𝑥(𝑡) = 𝐴 𝐶𝑜𝑠( 𝜔𝑡 + 𝜙)

Where the angular frequency 𝜔 =2𝜋

𝑇=

2𝜋

2= 𝜋 𝑟𝑎𝑑/𝑠, the amplitude 𝐴 = 0.16 𝑚

and the phase shift 𝜙 is given as

At 𝑡 = 0 𝑆𝑒𝑐, the displacement 𝑥 = – 0.16 meter, then

−0.16 = 0.16 𝐶𝑜𝑠 𝜙

𝜙 = −𝜋

2

Then

𝑥(𝑡) = (2.0 𝑚)𝐶𝑜𝑠 ( 𝜋 −𝜋

2)

Example 4

A particle starts SHM from the mean position. Its amplitude A and time period

T. At the time when it speeds is half of the maximum speed. Find the

displacement of the particle.

Solution



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The velocity 𝑣 =𝑣𝑚𝑎𝑥

2 =

1

2𝐴 𝜔

The displacement of SHM is defined as 𝑣 = −𝜔√𝐴2 − 𝑦2

Then 1

2𝐴 𝜔 = −𝜔√𝐴 − 𝑦2

1

4 𝐴2𝜔2 = 𝜔2(A2 − y2)

1

4 𝐴2 = 𝐴2 − 𝑦2 → 𝑦2 =

3

4 𝐴2 → 𝑦 =

√3

2 𝐴

Example 5

Describe how the kinetic and potential energies of S. H. M. vary periodically

with double the frequency of S. H. M.

Solution

The total Energy 𝐸 = 𝐾. 𝐸 + 𝑃. 𝐸 =1

2𝑚𝑣2 +

1

2𝑘𝑦2

Taken that 𝐶𝑜𝑠2𝑥 =1

2 (1 + 𝐶𝑜𝑠 2𝑥) 𝑆𝑖𝑛2𝑥 =

1

2 (1 − 𝐶𝑜𝑠 2𝑥)

𝐾. 𝐸 =𝟏

𝟐 𝒌 [𝑨 𝑪𝒐𝒔 (𝝎 𝒕 )]𝟐 =

1

4 𝑘 𝐴2 (1 + 𝐶𝑜𝑠 2𝜔𝑡) =

1

2𝐸(1 + 𝐶𝑜𝑠 2𝜔𝑡)

𝑃. 𝐸 = 𝟏

𝟐𝒎 [−Aω 𝑺𝒊𝒏(𝝎 𝒕 )]𝟐 =

1

4 𝑘 𝐴2 (1 − 𝐶𝑜𝑠 2𝜔𝑡) =

1

2𝐸(1 − 𝐶𝑜𝑠 2𝜔𝑡)

The displacement of SHM oscillates with angular frequency 𝜔, while the terms of

energies oscillates with double frequency of SHM 2𝜔.



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From the graph we note that potential energy or kinetic energy completes two

vibrations in a time during which S.H.M. completes one vibration. Thus the

frequency of potential energy or kinetic energy double than that of S.H.M.

Example 6

Two particles execute S.H.M. of same amplitude and frequency along the same

straight line. They pass one another when going in opposite directions. Each

time their displacement is half of their amplitude. Estimate the phase difference

between them.

Solution

Let two simple harmonic motions are 𝑥1 = 𝐴 𝐶𝑜𝑠 𝜔𝑡 and 𝑥2 = 𝐴 𝐶𝑜𝑠 (𝜔𝑡 + 𝜙)

In the first case 𝑥1 =𝐴

2 = 𝐴 𝐶𝑜𝑠 𝜔𝑡 → 𝐶𝑜𝑠 𝜔𝑡 =

1

2 & 𝑆𝑖𝑛 𝜔𝑡 =

√3

2

In the second case

𝑥2 =𝐴

2 = 𝐴 𝐶𝑜𝑠 (𝜔𝑡 + 𝜙) = 𝐴[ 𝐶𝑜𝑠 𝜔𝑡 𝐶𝑜𝑠 𝜙 − 𝑆𝑖𝑛 𝜔𝑡 𝑆𝑖𝑛 𝜙]



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1

2= [

1

2 𝐶𝑜𝑠 𝜙 −

√3

2 𝑆𝑖𝑛 𝜙] → 1 = 𝐶𝑜𝑠 𝜙 − √3 𝑆𝑖𝑛 𝜙

( 𝐶𝑜𝑠 𝜙 − 1)2 = 3 (1 − 𝐶𝑜𝑠2𝜙)

1 + 𝐶𝑜𝑠2𝜙 − 2𝐶𝑜𝑠𝜙 = 3 − 3 𝐶𝑜𝑠2𝜙

2 𝐶𝑜𝑠2𝜙 − 𝐶𝑜𝑠𝜙 − 1 = 0

By using

𝐶𝑜𝑠 𝜙 =1

4(1 ± √1 + 8) =

1

4 (1 ± √9)

𝐶𝑜𝑠 𝜙 = +1 𝑜𝑟 𝐶𝑜𝑠 𝜙 = −1

2

Then 𝜙 = 0 𝑜𝑟 𝜙 = 120.



50

Sheet (II)

1- In an engine, a piston oscillates with simple harmonic motion so that its

position varies according to the expression

𝑥 = 5.00 𝐶𝑜𝑠 ( 2 𝑡 +𝜋

6 ) ,

where x is in centimeters and t is in seconds. At 𝑡 = 0, find (a) the position of

the particle, (b) its velocity, and (c) its acceleration. Find (d) the period and (e)

the amplitude of the motion.

2- The position of a particle is given by the expression

𝑥 = 4.00 𝑐𝑜𝑠 (3.00𝜋𝑡 + 𝜋),

where 𝑥 is in meters and 𝑡 is in seconds. Determine (a) the frequency and (b)

period of the motion, (c) the amplitude of the motion, (d) the phase constant, and

(e) the position of the particle at 𝑡 = 0.250 𝑠.

3- If a simple pendulum oscillates with small amplitude and its length is doubled,

what happens to the frequency of its motion? (a) It doubles. (b) It becomes √2

times as large. (c) It becomes half as large. (d) It becomes 1/√2 times as

large. (e) It remains the same.



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4- A piston in a gasoline engine is in simple harmonic motion. The engine is

running at the rate of 3600 𝑟𝑒𝑣/𝑚𝑖𝑛. Taking the extremes of its position

relative to its center point as ± 5.00 𝑐𝑚, find the magnitudes of the (a)

maximum velocity and (b) maximum acceleration of the piston.

5- A 200-g block is attached to a horizontal spring and executes simple harmonic

motion with a period of 0.250 𝑠. The total energy of the system is 2.00 𝐽. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

6- A 50.0-g object connected to a spring with a force constant of 35.0 N/m

oscillates with an amplitude of 4.00 cm on a frictionless, horizontal surface.

Find (a) the total energy of the system and (b) the speed of the object when its

position is 1.00 cm. Find (c) the kinetic energy and (d) the potential energy

when its position is 3.00 cm.

7- A particle of mass 𝒎 slides without friction inside a hemispherical bowl of

radius R. Show that if the particle starts from rest with a small displacement

from equilibrium, it moves in simple harmonic motion with an angular

frequency equal to that of a simple pendulum of length R. That is, 𝝎 = √𝒈

𝑹.

8- A simple pendulum makes 120 complete oscillations in 3.00 min at a location

where g 5 9.80 m/s2. Find (a) the period of the pendulum and (b) its length.



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9- A 50.0-g object connected to a spring with a force constant of 35.0 N/m

oscillates with an amplitude of 4.00 cm on a frictionless, horizontal surface.

Find (a) the total energy of the system and (b) the speed of the object when its

position is 1.00 cm. Find (c) the kinetic energy and (d) the potential energy

when its position is 3.00 cm.

10- A 200-g block is attached to a horizontal spring and executes simple

harmonic motion with a period of 0.250 s. The total energy of the system is

2.00 J. Find (a) the force constant of the spring and (b) the amplitude of the

motion.

11- A simple harmonic oscillator of amplitude A has a total energy E. Determine

(a) the kinetic energy and (b) the potential energy when the position is one-third

the amplitude. (c) For what values of the position does the kinetic energy equal

one-half the potential energy? (d) Are there any values of the position where the

kinetic energy is greater than the maximum potential energy? Explain.

12- A 1.00-kg glider attached to a spring with a force constant of 25.0 N/m

oscillates on a frictionless, horizontal air track. At 𝑡 = 0, the glider is released

from rest at 𝑥 = − 3.00 𝑐𝑚 (that is, the spring is compressed by 3.00 cm). Find

(a) the period of the glider’s motion, (b) the maximum values of its speed and

acceleration, and (c) the position, velocity, and acceleration as functions of time

53

54

Chapter (III)

Universal Gravitation

1. Newton’s Law of Universal Gravitation

In 1687, Newton published his work on the law of gravity in his

treatise Mathematical Principles of Natural Philosophy.

Newton’s law of universal gravitation states that

Every particle in the Universe attracts every other particle with a force that

is directly proportional to the product of their masses and inversely proportional to

the square of the distance between them

If the particles have masses 𝒎𝟏and 𝒎𝟐 and are separated by a distance 𝑟, the

magnitude of this gravitational force is

𝑭𝒈 = 𝑮𝒎𝟏 𝒎𝟐

𝒓𝟐 (1)

where 𝐺 is a constant, called the universal gravitational

constant. Its value in SI units is 𝐺 = 6.674 × 10−11 𝑁. 𝑚2/

𝑘𝑔2.

The form of the force law given by Equation 1 is often

referred to as an inverse-square law because the magnitude



55

of the force varies as the inverse square of the separation of the particles. The

gravitational force is just one type of force that represents the interaction of a particle

with other particles in its environment.

The Vector Force

Figure 1 represents the gravitational force exerted by two particles on each other,

which form an action–reaction pair according to Newton’s third law.

- The first particle exerts an attractive force �⃗�12

on the second particle along the line joining

the two particles, and similarly

- the second particle exerts a force �⃗�21 on the

first. The forces are oppositely directed and

always equal in magnitude, even though the

two masses may not be equal.

That is, these forces form an action–reaction pair

�⃗�12 = − �⃗�21

Henry Cavendish (1731–1810) measured the universal gravitational constant in an

important 1798 experiment. Cavendish’s apparatus consists of two small spheres,

each of mass m, fixed to the ends of a light, horizontal rod suspended by a fine fiber

or thin metal wire as illustrated in Figure 2. When two large spheres, each of mass

M, are placed near the smaller ones, the attractive force between smaller and larger

spheres causes the rod to rotate and twist the wire suspension to a new equilibrium

orientation. The angle of rotation is measured by the deflection of a light beam

reflected from a mirror attached to the vertical suspension.

Figure 1

Physics (I) Chapter (III) Universal gravitation


56

Figure 1 Cavendish apparatus for measuring G.

Why Gravitational Forces Are Important

The gravitational forces are negligible between ordinary household-sized objects,

but very substantial between objects that are the size of stars. Indeed, gravitation is

the most important force on the scale of planets, stars, and galaxies. It is responsible

for holding our earth together and for keeping the

planets in orbit about the sun. The mutual gravitational

attraction between different parts of the sun compresses

material at the sun’s core to very high densities and

temperatures, making it possible for nuclear reactions

to take place there. These reactions generate the sun’s

energy output, which makes it possible for life to exist

on earth and for you to read these words.

The gravitational force is so important on the cosmic scale because it acts at a

distance, without any direct contact between bodies. Electric and magnetic forces

have this same remarkable property, but they are less important on astronomical

scales because large accumulations of matter are electrically neutral; that is, they



57

contain equal amounts of positive and negative charge. As a result, the electric and

magnetic forces between stars or planets are very small or zero. Their influence is

negligible at distances much greater than the diameter of an atomic nucleus

Example 1

Calculate the magnitude of the gravitational force exerted on a cantaloupe of

mass 𝒎𝒄 = 𝟏. 𝟎𝟎 𝒌𝒈 on the surface of the Earth due to (a) the Earth, (b) the

Moon, (c) the Sun.

Solution (a) The gravitational force on the cantaloupe due to the Earth is simply the

weight of the cantaloupe:

𝐹𝑐𝐸 = 𝑚𝑐𝑔 = (1.00 𝑘𝑔) × (9.8𝑚

𝑠2) = 9.8 𝑁

To find the force due to the Moon, we use Eq. 1

𝑭𝑪𝑴 = 𝑮𝒎𝐜 𝒎𝐄

𝒓𝑴𝟐

=6.67 × 10−11 × 1.00 × 7.36 × 1022

(3.82 × 108)2= 3.36 × 10−5 𝑁

To find the force due to the Sun, we use Eq. 1

𝑭𝑪𝑺 = 𝑮𝒎𝐜 𝒎𝐒

𝒓𝑺𝟐

=6.67 × 10−11 × 1.00 × 1.99 × 1030

(1.50 × 1011)2= 5.90 × 10−3 𝑁

Clearly the Earth is the dominant influence on the behavior of objects on its surface.

Note that the force due to the Sun on an object at the Earth’s surface is much larger

than the force due to the Moon. (However, the tidal effect of the Moon on the Earth’s

oceans is greater than that of the Sun.



58

Example 2

A properly suited astronaut (𝒎𝒂 = 𝟏𝟎𝟓 𝒌𝒈) is drifting through the asteroid

belt on a mining expedition. At a particular instant he is located near two

asteroids of masses 𝒎𝟏 = 𝟑𝟒𝟔 𝒌𝒈 (𝒓𝟏 = 𝟐𝟏𝟓 𝒎) and 𝒎𝟐 = 𝟏𝟖𝟒 𝒌𝒈 (𝒓𝟐 =

𝟏𝟒𝟐 𝒎) as shown in Figure. The lines connecting the astronaut to the two

asteroids form an angle of 𝟏𝟐𝟎𝒐. At that instant, what is the magnitude and

direction of the gravitational force on the astronaut due to these two asteroids?

Assume that the astronaut and the asteroids can be considered as particles.

Solution Equation 1 gives the magnitudes of the two

forces:

𝑭𝒂𝟏 = 𝑮𝒎𝐚 𝒎𝟏

𝒓𝟏𝟐

=6.67 × 10−11 × 105 × 346

(215 )2

= 5.24 × 10−11𝑁 = 52.4 𝑝𝑁

𝑭𝒂𝟐 = 𝑮𝒎𝐚 𝒎𝟐

𝒓𝟐𝟐

=6.67 × 10−11 × 105 × 184

(142 )2= 6.39 × 10−11𝑁 = 63.9 𝑝𝑁



59

Using either the component method or the parallelogram method, we can add these

two vectors to obtain the magnitude of the total force on the astronaut to be

�⃗� = �⃗�𝑎1 + �⃗�𝑎2 = (∑ 𝐹𝑥 ) 𝑖̂ + (∑ 𝐹𝑦 ) 𝑗̂

= (𝐹𝑎1 − 𝐹𝑎2 𝑆𝑖𝑛 30 ) 𝑖̂ + (𝐹𝑎2 𝐶𝑜𝑠 30)𝑗̂

The magnitude of the total force on the astronaut to be

|𝐹| = √𝐹𝑥2 + 𝐹𝑦

2 = 58.9 𝑝𝑁

and its direction is as 𝜙 = 𝑇𝑎𝑛−1 (∑ 𝐹𝑦

∑ 𝐹𝑥) = 69.7𝑜

Example 3

Many stars belong to systems of two or more stars held together by their mutual

gravitational attraction. See figure, shows a three-star system at an instant

when the stars are at the vertices of a right triangle. Find the total gravitational

force exerted on the small star by the two large ones.

Solution

We use the principle of superposition: The total force 𝐹 ⃗on the small star is the

vector sum of the forces �⃗�1 and �⃗�2 due to each large star,



60

The vector sum using components:

𝑭𝟏 = 𝑮𝒎𝐨 𝒎𝟏

𝒓𝟎𝟏𝟐

=6.67 × 10−11 × (8 × 1030 )(1 × 1030)

(2 × 1012)2 + (2 × 1012)2= 6.67 × 1025𝑁

𝑭𝟐 = 𝑮𝒎𝐨 𝒎𝟐

𝒓𝟎𝟐𝟐

=6.67 × 10−11 × (8 × 1030 )(1 × 1030)

(2 × 1012)2= 1.33 × 1026𝑁

Using either the component method or the parallelogram method, we can add these

two vectors to obtain the magnitude of the total force on the astronaut to be

�⃗� = �⃗�1 + �⃗�2 = (∑ 𝐹𝑥 ) 𝑖̂ + (∑ 𝐹𝑦 ) 𝑗̂ = (𝐹2 + 𝐹1𝐶𝑜𝑠𝜃 ) 𝑖̂ + (𝐹1𝑆𝑖𝑛 𝜃)𝑗̂

= 1.81 × 1026 𝑖̂ + 4.72 × 1025 𝑗̂ 𝑁

The magnitude of the total force on the astronaut to be

|𝐹| = √𝐹𝑥2 + 𝐹𝑦

2 = √ (1.81 × 1026)2 + (4.72 × 1025)2 = 1.87 × 1026 𝑁

and its direction is as 𝜙 = 𝑇𝑎𝑛−1 (∑ 𝐹𝑦

∑ 𝐹𝑥) = 14.6𝑜



61

2. Free-Fall Acceleration and the Gravitational Force

We have called the magnitude of the gravitational force on an object near the

Earth’s surface the weight of the object, where the weight is given by 𝑊 = 𝑚𝑔

Equation 1 is another expression for this force. Therefore, we can set

both equal to each other to obtain

𝑊 = 𝑚𝑔 𝑎𝑛𝑑 𝐹𝑔 = 𝐺𝑀𝑝 𝑚

𝑅𝑝2

Then

𝑚𝑔 = 𝐺𝑀𝑝 𝑚

𝑅𝑝2

Therefore

𝑔 = 𝐺 𝑀𝑝

𝑅𝑝2 (2)

Equation 2 relates the free-fall acceleration g to physical parameters of the Earth—

its mass and radius—and explains the origin of the value of

9.80 𝑚/𝑠2 that we have used in earlier chapters. Now consider an

object of mass m located a distance h above the Earth’s surface or

a distance r from the Earth’s center, where 𝑟 = 𝑅 + ℎ. The

magnitude of the gravitational force acting on this object is

𝑊 = 𝑚𝑔′ 𝑎𝑛𝑑 𝐹𝑔 = 𝐺𝑀𝑝 𝑚

𝑅𝑝2

Where 𝑟 = 𝑅𝑝 + ℎ

Then, 𝑔′ = 𝐺𝑀𝑝

𝑅𝑝2



62

𝑔′ = 𝐺𝑀𝑝

(𝑅𝑝 + ℎ)2 = 𝐺

𝑀𝑝

𝑅𝑝 (1 +ℎ

𝑅𝑝)

2 =𝑔

(𝑅𝑝 + ℎ

𝑅𝑝)

2

𝑔′ = 𝑔 (𝑅𝑝

𝑅𝑝+ℎ)

2

(3)

This shows that as you move farther from planet’s center, that is, as 𝑅𝑝 becomes

larger, the acceleration due to gravity is reduced according to this inverse square

relationship. What happens to your weight, 𝑚𝑔, as you move farther and farther from

Earth’s center?

Example 4

The International Space Station operates at an altitude of 350 km. Plans for the

final construction show that material of weight 𝟒. 𝟐𝟐 × 𝟏𝟎𝟔 𝑵, measured at the

Earth’s surface, will have been lifted off the surface by various spacecraft.

What is the weight of the space station when in orbit?

Solution

The mass of the space station is fixed; it is independent of its location. Based on the

discussion in this section, we realize that the value of g will be reduced at the height

of the space station’s orbit. Therefore, its weight will be smaller than that at the

surface of the Earth.

Find the mass of the space station from its weight at the surface of the Earth:

𝑚 =𝑊

𝑔=

𝟒. 𝟐𝟐 × 𝟏𝟎𝟔

9.80 = 𝟒. 𝟑𝟏 × 𝟏𝟎𝟓 𝑲𝒈



63

Using eq. (3)

𝑔′ = 𝑔 (𝑅𝑝

𝑅𝑝 + ℎ)

2

= 9.8 × (6.37 × 106

6.37 × 106 + 350 × 103)

2

= 8.82 𝑚/𝑠2

Use this value of g to find the space station’s weight in orbit:

𝑊′ = 𝑚 𝑔′ = 𝟒. 𝟑𝟏 × 𝟏𝟎𝟓 × 8.82 = 3.8 × 106𝑁

Example 5

Using the known radius of the Earth and that 𝒈 = 𝟗. 𝟖𝟎 𝒎/𝒔𝟐 at the Earth’s

surface, find the average density of the Earth.

Solution

Assume the Earth is a perfect sphere. The density of material in the Earth varies, but

let’s adopt a simplified model in which we assume the density to be uniform

throughout the Earth. The resulting density is the average density of the Earth.

The gravitational acceleration at the surface of the Earth

𝑔 = 𝐺 𝑀𝑝

𝑅𝑝2 → 𝑀𝐸 = 𝑔

𝑅𝐸2

𝐺

Substitute this mass into the definition 𝜌𝐸 =𝑀𝐸

𝑉𝐸=

𝑀𝐸4

3 𝜋 𝑅𝐸

3 =

3

4

𝑔𝑅𝐸

2

𝐺

𝜋 𝑅𝐸3 =

3

4

𝑔

𝜋𝐺 𝑅𝐸

Then 𝜌𝐸 = 3

4

𝑔

𝜋𝐺 𝑅𝐸 =

3

4 ×

9.80

𝜋× 6.67×10−11 ×6.37×106 = 5.51 ×

103𝐾𝑔/𝑚3



64

What if you were told that a typical density of granite at the Earth’s surface is

𝟐. 𝟕𝟓 × 𝟏𝟎𝟑𝑲𝒈/𝒎𝟑. What would you conclude about the density of the

material in the Earth’s interior?

Answer

Because this value is about half the density we calculated as an average for the entire

Earth, we would conclude that the inner core of the Earth has a density much higher

than the average value.

3. Kepler’s Laws of Planetary Motion

Johannes Kepler, a 29-year-old German, became

one of Brahe’s assistants when he moved to Prague.

Upon his death in 1601, Kepler inherited 30 years’

worth of Brahe’s observations. He studied Brahe’s

data and was convinced that geometry and

mathematics could be used to explain the number,

distance, and motion of the planets. Kepler

believed that the Sun exerted a force on the planets

and placed the Sun at the center of the system. After

several years of careful analysis of Brahe’s data on

Mars, Kepler discovered the laws that describe the

motion of every planet and satellite.

Kepler’s three laws describe the motion of the planets

KEPLER’S LAWS OF PLANETARY MOTION

First Law (Law of orbits): Each planet travels in an elliptical orbit around the sun,

and the sun is at one of the focal points.



65

Second Law (Law of Areas): An imaginary line drawn from the sun to any planet

sweeps out equal areas in equal time intervals.

Third Law (Law of periods) : The square of a planet’s orbital period (𝑇2) is

proportional to the cube of the average distance (𝑟3) between the planet and the sun,

or 𝑇2 ∝ 𝑟3.

Kepler did not know why the planets moved in this way. Three generations later,

when Newton turned his attention to the motion of the planets, he discovered that

each of Kepler’s laws can be derived; they are consequences of Newton’s laws of

motion and the law of gravitation. Let’s see how each of Kepler’s laws arises.

Kepler’s first law (Law of orbits) states that the paths of the planets are ellipses,

with the Sun at one focus. An ellipse has two foci, as shown in Figure 4. Like

planets and stars, comets also orbit the Sun in elliptical orbits. Comets are

divided into two groups—long-period comets and short-period comets—

based on orbital periods, each of which is the time it takes the comet to

complete one revolution.

Figure 4



66

Kepler found that the planets move faster when they are closer to the Sun and slower

when they are farther away from the Sun. Thus, Kepler’s second law.

The distance of each focus from the center of the ellipse is 𝑒𝑎, where e is a

dimensionless number between 0 and 1 called the eccentricity. If the ellipse is a

circle. The actual orbits of the planets are fairly circular; their eccentricities range

from 0.007 for Venus to 0.206 for Mercury.

The terms perihelion and aphelion describe different points in the Earth’s orbit of

the Sun.

Remember that the Earth orbits the Sun in an elliptical path—which is oval, not

circular. This means that the Earth is about 3 million miles nearer to the Sun in

January at its nearest point than in July at its farthest point.

• Aphelion is the point of the Earth’s orbit that is farthest away from the Sun.

• Perihelion is the point of the Earth’s orbit that is nearest to the Sun.

The words come from Ancient Greek, in which helios means “Sun,” apo means

“far,” and peri means “close.”



67

Kepler’s second law (Law of Areas) states that an imaginary line from the Sun to

a planet sweeps out equal areas in equal time intervals, as illustrated in

Figure 5

Universal Gravitation and Kepler’s Third Law

Newton stated his law of universal gravitation in terms that applied to the motion of

planets about the Sun. This agreed with Kepler’s third law and confirmed that

Newton’s law fit the best observations of the day.

Consider a planet orbiting the Sun, as shown in Figure 6.

Newton’s second law of motion, 𝐹𝑛𝑒𝑡 = 𝑚 𝑎, can be written

as𝐹𝑛𝑒𝑡 = 𝑚𝑝 𝑎𝑐, where F is the gravitational force, 𝑚𝑝 is the

mass of the planet, and 𝑎𝑐 is the centripetal acceleration of the

planet.

the centripetal force of the planet reads as 𝐹𝑐 = 𝑚 𝑎𝑐 = 𝑚𝑣2

𝑟

the centripetal acceleration of the planet. 𝑎𝑐 =𝑣2

𝑟 ; 𝑣 =

𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

𝑡𝑖𝑚𝑒



68

𝑎𝑐 =𝑣2

𝑟=

(2 𝜋 𝑟

𝑇)

2

𝑟 =

4 𝜋2 𝑟

𝑇2

In this equation, T is the time required for the planet to make one complete revolution

about the Sun.

If you set the right side of this equation equal to the right side of the law of universal

gravitation, you arrive at the following result:

𝐹𝑐 = 𝐹𝑔 → 𝑚 𝑎𝑐 = 𝐺 𝑀 𝑚

𝑟2

4 𝜋2 𝑟

𝑇2 = 𝐺

𝑀

𝑟2

𝑇2 =4 𝜋2

𝐺 𝑀 𝑟3

- Period of a Planet Orbiting the Sun

𝑻 = 𝟐𝝅 √𝒓𝟑

𝑮𝑴

- The velocity of the planet around the sun (orbital velocity )

𝒗𝒐 =𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕

𝒑𝒆𝒊𝒐𝒅𝒊𝒄 𝒕𝒊𝒎𝒆=

𝟐 𝝅 𝒓

𝑻=

𝟐 𝝅 𝒓

𝟐𝝅 √𝒓𝟑

𝑮𝑴

= √𝑮𝑴

𝒓

𝒗𝒐 = √𝑮𝑴

𝒓



69

Kepler’s third law relates the orbital period and mean distance for two orbiting

planets as follows

𝑇12

𝑇22 =

𝑟13

𝑟23

This law also applies to satellites orbiting Earth, including our moon. In that case, r

is the distance between the orbiting satellite and Earth. The proportionality constant

depends on the mass of the central object.

Example 6

The color-enhanced image of Venus shown here was compiled from data taken

by Magellan, the first planetary spacecraft to be launched from a space shuttle.

During the spacecraft’s fifth orbit around Venus, Magellan traveled at a mean

altitude of 361 km. If the orbit had been circular, what would Magellan’s period

and speed have been?



70

Solution

Find r by adding the distance between the spacecraft

and Venus’s surface (𝑟1) to Venus’s radius (𝑟2).

𝑟 = 𝑟1 + 𝑟2 = (3.61 × 105) + (6.05 × 106)

= 6.41 × 106 𝑚

Period of a Planet Orbiting the Sun

𝑇 = 2𝜋 √𝑟3

𝐺𝑀= 2𝜋 √

(6.41 × 106)3

6.67 × 10−11 × 4.87 × 1024 = 5.66 × 103 Sec

= 94 min

The velocity of the planet around the sun (orbital velocity )

𝒗𝟎 = √𝑮𝑴

𝒓= √

6.67 × 10−11 × 4.87 × 1024

6.41 × 106= 7.12 × 103 m/s

Example 7

Assume that a satellite orbits Earth 225 km above its surface. Given that the

mass of Earth is 𝟓. 𝟗𝟕 × 𝟏𝟎𝟐𝟒 𝒌𝒈 and the radius of Earth is 𝟔. 𝟑𝟖 × 𝟏𝟎𝟔 𝒎,

what are the satellite’s orbital speed and period?

Solution

Find r by adding the distance between the satellite and

Earth’s surface (ℎ) to Venus’s radius (𝑟𝐸).



71

𝑟 = ℎ + 𝑟𝐸 = (0.225 × 106) + ( 𝟔. 𝟑𝟖 × 𝟏𝟎𝟔) = 6.61 × 106 𝑚

Period of a Planet Orbiting the Sun

𝑇 = 2𝜋 √𝑟3

𝐺𝑀= 2𝜋 √

(6.61 × 106)3

6.67 × 10−11 × 𝟓. 𝟗𝟕 × 𝟏𝟎𝟐𝟒 = 5.35 × 103 Sec

= 89 𝑚𝑖𝑛

The velocity of the planet around the sun (orbital velocity )

𝒗𝟎 = √𝑮𝑴

𝒓= √

6.67 × 10−11 × 𝟓. 𝟗𝟕 × 𝟏𝟎𝟐𝟒

6.61 × 106= 7.76 × 103 m/s



72

Sheet (III)

1. The center-to-center distance between Earth and Moon is 384 400 km. The Moon

completes an orbit in 27.3 days. Determine the Moon’s orbital speed.

2. Titan, the largest moon of Saturn, has a mean orbital radius of 1.22𝑥109 m. The

orbital period of Titan is 15.95 days. Hyperion, another moon of Saturn, orbits at a

mean radius of 1.48𝑥109 m. Use Kepler’s third law of planetary motion to predict

the orbital period of Hyperion in days.

3. The planet Mercury travels around the Sun with a mean orbital radius of

5.8𝑥1010m. The mass of the Sun is 1.99𝑥1030 kg. Use Kepler’s third law to

determine how long it takes Mercury to orbit the Sun. Give your answer in Earth

days.

4. An artificial satellite circles the Earth in a circular orbit at a location where the

acceleration due to gravity is 9.00 m/s2. Determine the orbital period of the satellite.

5. A 20 kg satellite has a circular orbit with a period of 2.4 h and a radius of

8.0 × 106m around a planet of unknown mass. If the magnitude of the gravitational

acceleration on the surface of the planet is 8.0 𝑚 𝑠2⁄ , what is the radius of the planet?

6. A satellite of mass 200 kg is placed into Earth orbit at a height of 200 km above

the surface. (a) Assuming a circular orbit, how long does the satellite take to

complete one orbit? (b) What is the satellite’s speed? (c) Starting from the satellite

on the Earth’s surface, what is the minimum energy input necessary to place this

satellite in orbit?

7. A satellite orbits a planet of unknown mass in a circle of radius 2.0 × 107m. The

magnitude of the gravitational force on the satellite from the planet is F = 80 N. (a)

what is the kinetic energy of the satellite in this orbit? (b) What would F be if the

orbit radius were increased to 3.0 × 107m?

8. A 1000-kg satellite orbits the Earth at a constant altitude of 100 km. (a) How much

energy must be added to the system to move the satellite into a circular orbit with



73

altitude 200 km? What are the changes in the system’s (b) kinetic energy and (c)

potential energy?

9. Calculate the scape speed from the earth for a 500 kg space craft, and

determine the kinetic energy it must have at the earth surface in order to escape from

the earth gravitational field

10. Four particles of masses m, 2m, 3m and 4m

are kept in sequence at the corners of a square of

side a. Find the magnitude of gravitational force

acting on a particle of mass m placed at the center

of the square.

11. Plaskett’s binary system consists of two stars that revolve

in a circular orbit about a center of mass midway between

them. This means that the masses of the two stars are equal

(Fig. ). Assume the orbital speed of each star is 220 km/s and

the orbital period of each is 14.4 days. Find the mass M of

each star. (For comparison, the mass of our Sun is 1.99𝑥1030

kg.)

12. The moon revolves around the earth 13 times per year. If the ratio of the distance

of the earth from the sun to the distance of the moon from the earth is 392, find the

ratio of mass of the sun to the mass of the earth.

13. Comet Halley (Figure ) approaches the Sun to

within 0.570 AU, and its orbital period is 75.6

years. (AU is the symbol for astronomical unit,

where 1 AU equals 1.50𝑥1011 m is the mean

Earth–Sun distance.) How far from the Sun will

Halley’s comet travel before it starts its return

journey?

74

75

Chapter (IV)

Properties of

Matter

1. States of Matter

The three states of matter differ from each other due to the following two factors.

(1) The different magnitudes of the interatomic and intermolecular forces.

(2) The extent of random thermal motion of atoms and molecules of a substance

(which depends upon temperature).

Physics (I) Chapter (IV) Properties of Matter


76

Note: ❑ The fourth state of matter in which the medium is in the form of positive

and negative ions, is known as plasma. Plasma occurs in the atmosphere of stars

(including the sun) and in discharge tubes.

2. Elastic Properties of Solids

In solid, the molecules are very closely packed together and are held in relatively

fixed position by strong intermuscular forces. As a result of this, it is difficult for an

external force that is applied on the solid to displace the molecules of the solid due

to the resistance of the intermuscular forces between the

molecules

Elasticity is the ability of solid materials to return to their

original shape and size after the forces deforming them have

been removed.

Elastic materials are those materials that regains their

original shape and size after the force that caused the

deformation is removed.

Plastic Material The bodies which remain in deformed

state even after removed of the deforming force are defined

as plastic bodies.

We shall discuss the deformation of solids in terms of the concepts of stress and

strain.

Stress is a quantity that is proportional to the force causing a deformation;

more specifically, stress is the external force acting on an object per unit cross-

sectional area.



77

𝑆𝑡𝑟𝑒𝑠𝑠 =𝐹

𝐴 (𝑁/𝑚2)

The result of a stress is strain, which is a measure of the degree of deformation.

𝑆𝑡𝑟𝑎𝑖𝑛 =change in size of the body

original size of the body (Unitless)

for sufficiently small stresses, stress is proportional to strain; the constant of

proportionality depends on the material being deformed and on the nature of the

deformation. We call this proportionality constant the elastic modulus. The elastic

modulus is therefore defined as the ratio of the stress to the resulting strain:

𝐸𝑙𝑎𝑠𝑡𝑖𝑐 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 =𝑆𝑡𝑟𝑒𝑠𝑠

𝑆𝑡𝑟𝑎𝑖𝑛 (𝑁/𝑚2)

3. Stress–strain curve

Proportion Limit :

The limit in which Hook's law is valid and stress is directly proportional to strain is

called proportion limit.

𝑆𝑡𝑟𝑒𝑠𝑠 𝑆𝑡𝑟𝑎𝑖𝑛

𝑆𝑙𝑜𝑝𝑒 = 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 =𝑆𝑡𝑟𝑒𝑠𝑠

𝑆𝑡𝑟𝑎𝑖𝑛



78

Elastic limit : That maximum stress which on removing the deforming force makes

the body to recover completely its original state.

Yield Point :The point beyond elastic limit, at which the length of wire starts

increasing without increasing stress, is defined as the yield point

Breaking Point :The position when the strain becomes so large that the wire breaks

down at last, is called breaking point. At this position the stress acting in that wire is

called breaking stress and strain is called breaking strain.

Creep: If a small force is applied for a long time then it causes breaking of metal.

We consider three types of deformation and define an elastic modulus for each:

1. Young’s modulus (a) measures the resistance of a solid to a change

in its length.

2. Shear modulus (b, c) measures the resistance to motion of the planes

within a solid parallel to each other.

3. Bulk modulus (d) measures the resistance of solids or liquids to

changes in their volume.



79

Exchange Types Diagram Stress Strain E. Moduli

Change

occurs in

length

Young’s

modulus

(Tensile)

𝝈 =𝑭⊥

𝑨

=𝒎 𝒈

𝝅𝒓𝟐

𝜺 =𝚫𝓵

𝓵𝟎 𝒀 =

𝝈

𝜺=

𝑭⊥ 𝚫𝓵

𝑨𝓵𝟎

Change

occurs in

angle

Shear

Modulus

𝝈 =𝑭∥

𝑨

𝜺 = 𝑻𝒂𝒏 𝜽

=𝐱

𝒉

𝑺 = 𝝈

𝜺

Change

occurs in

volume

Bulk

Modulus

𝝈 = 𝚫𝑷 𝜺 = −𝚫𝑽

𝑽𝟎

𝑩 =−𝚫𝑷

𝚫𝑽𝑽𝟎

Compressibility

𝜿 = 𝟏/𝚩

Change

occurs in

area

Poisson

ratio

𝑷. 𝑹 =

−𝚫𝒓𝒓𝟎

𝚫𝒍𝒍



80

4. Thermal Expansion of Solids and Liquids

Most bodies expand as their temperatures increase. This

phenomenon plays an important role in numerous engineering

applications, such as the joints in buildings, highways, railroad

tracks, bridges . . . etc. Such thermal expansion is not always

desirable.

Microscopically, thermal expansion arises from the change in

the separation between the constituent atoms or molecules of the solid. To

understand this, we consider a crystalline solid of a regular array of atoms or

molecules held together by electrical forces.

A mechanical model can be used to imagine the

electrical interaction between the atoms or

molecules, as shown in Fig. 1. At an ordinary

temperature, the average spacing between the atoms

is of the order of 10−10 𝑚, and they vibrate about

their equilibrium positions with an amplitude of about 10−11 𝑚 and a frequency of

about 1013 𝐻𝑧. As the temperature increases, the atoms vibrate with larger

amplitudes and the average separation between them increases.

Thermal expansion is the tendency of matter to change its shape, area, volume,

and density in response to a change in temperature, usually not including phase

transitions. Solids can expand in one dimension (Linear expansion), two dimension

https://www.wikiwand.com/en/Shape

https://www.wikiwand.com/en/Area

https://www.wikiwand.com/en/Volume

https://www.wikiwand.com/en/Density

https://www.wikiwand.com/en/Temperature

https://www.wikiwand.com/en/Phase_transitions

https://www.wikiwand.com/en/Phase_transitions



81

(Superficial expansion) and three dimension (Volume expansion) while liquids and

gases usually suffers change in volume only.

(i) Linear thermal Expansion

If a rod of length 𝐿 and temperature 𝑇 experiences a small change in temperature

Δ𝑇, its length changes by an amount L, see Fig. 2.

For a sufficiently small change Δ𝑇, experiments show that Δ𝐿 is proportional to both

L and Δ𝑇. We introduce a proportionally coefficient 𝛼 for the solid and write:

Δ 𝐿 = 𝛼 𝐿 Δ𝑇

𝐿𝑛𝑒𝑤 = 𝐿 (1 + 𝛼 Δ𝑇)

where the proportionality constant α is called the coefficient of linear expansion for

a given material. Thus the unit of 𝛼 is (𝑑𝑒𝑔𝑟𝑒𝑒−1)

Note that it describes an expansion when T is positive and a contraction when T is

negative.



82

(ii) Superficial (areal) expansion:

For a sufficiently small change Δ𝑇,

Δ𝐴 = 𝛿 𝐴0 Δ𝑇

where the new area

𝐴 = 𝐴0 (1 + 𝛿 Δ𝑇)

ℓ2 = ℓ02 (1 + 𝛿 Δ𝑇) → (1)

but ℓ2 = ℓ02 (1 + 𝛼 Δ𝑇)2 = ℓ0

2 (1 + 2𝛼 Δ𝑇) By using Binomial expansion

𝛿 = 𝟐𝜶

(iii) Volume (Bulk) expansion:

The change in volume V at a constant pressure is

proportional to the original volume V and to the

change in temperature T according to the

following relation: Δ𝑉 = 𝛽 𝑉0 Δ𝑇

𝑉 = 𝑉0 ( 1 + 𝛽 Δ𝑇 )

In the same way, one can find 𝜷 = 𝟑 𝜶



83

Example 1. A cube of aluminum of sides 0.1 m is subjected to a shearing force

of 100 N. The top face of the cube is displaced through 0.02 cm with respect to

the bottom face. Find (a) The shearing Stress (shearing) shearing strain and (c)

the shearing modulus.

ℓ = 0.1 m F∥= 100N Δ x = 0.02cm

S. stress =?? S. Strain=??? , and S. Modulus

= ???

S. stress =F∥

A=

100N

(0.1 × 0.1 )m2 = 104N/m2

S. Strain = tan ϕ =Δx

ℓ=

0.02 × 10−2 m

0.1 m= 0.002

S. Modulus = S. stress

S. strain = =

104

0.002= 5 × 106 N/m2

Example 2 . A 5m long aluminum wire (𝐘 = 𝟕 × 𝟏𝟎𝟏𝟎𝐍 𝐦−𝟐) of diameter 3mm

supports a 40kg mass. In order to have the same elongation in the copper

wire (𝐘 = 𝟏𝟐 × 𝟏𝟎𝟏𝟎𝐍 𝐦−𝟐) of the same length under the same weight, the

diameter should now be (in mm).

YAl = 7 × 1010N m−2 and Ycu = 12 × 1010N m−2

DAl = 3 mm and m = 40Kg Δ lAl = Δ lCu ℓAl =

ℓCu

DCu = ? ?

The Young’s modulus is given as Y = F ℓ

A Δ l =

mg ℓ

π r2 Δ l

Therefore YAl

YCu =

m g ℓAl

π rAl2 Δ lAl

× π rCu

2 Δℓcu

m g ℓCu =

rCu 2

rAl2 ⇒

YAl

YCu =

rCu 2

rAl2

The Diameter of copper wire



84

DCu = 2 rCu = 2 ( rAl × √YAl

YCu ) = 2 ( (1.5 × 10−3) × √

7 × 1010

12 × 1010 )

= 2.3 mm

Example 3. The force required to stretch a steel wire of 𝟏 𝐜𝐦𝟐 cross-section

to 1.1 times its length would be (𝐘 = 𝟐 × 𝟏𝟎𝟏𝟏𝐍 𝐦−𝟐)

ℓ2 = 1.1 ℓ1 A = 1cm2 Y = F ℓ

A Δ ℓ ⇒ F =

Y A Δ ℓ

ℓ

Δℓ = ℓ2 − ℓ1 = 1.1 ℓ1 − ℓ1 = 0.1 ℓ1

F = Y A Δ ℓ

ℓ =

(2×1011N m−2) (1×10−4) (0.1 ℓ1)

ℓ1 = 2 × 106 N

Example 4. Two wires of equal length and cross-section are suspended as

shown. Their Young’s moduli are 𝐘𝟏 and 𝐘𝟐 respectively. The equivalent

Young’s modulus will be

ℓ1 = ℓ2 = ℓ and A1 = A2 = A

Solution:

The relation between the force constant (K) and Young’s modulus Y = F ℓ

A Δ ℓ =

Kℓ

A.

Note that the Force constant (K) in series combinations of wires 1

Keq =

1

K1 +

1

K2 +

⋯ and for parallel combinations of wires Keq = K1 + K2 + ⋯

Therefore, for parallel combinations of wires Keq = K1 + K2



85

Then Yeq Aeq

ℓ =

Y1 A1

ℓ1 +

Y2 A2

ℓ2

Note that Aeq = Awire1 + Awire2 = 2A ℓ1 = ℓ2 = ℓ and A1 =

A2 = A

Yeq 2A

ℓ =

Y1 A

ℓ +

Y2 A

ℓ ⇒ 2 Yeq = Y1 + Y2 ⇒

Yeq =Y1+Y2

2

Example 5. When a pressure of 100 atmosphere is applied on a spherical ball

of rubber, then its volume reduces to 0.01%. Find the bulk modulus of the

material of the rubber in 𝐝𝐲𝐧𝐞/𝐜𝐦𝟐 .

Solution 1 𝑎𝑡𝑚 ∼ 105 𝑝𝑎 𝑃 = 100 × 105 =

107 𝑝𝑎

Δ𝑉

𝑉0= − 0.01% = −

0.01

100= − 0.0001

𝛽 = −𝑃

Δ𝑉/𝑉0 =

107

0.0001= 1011𝑁/𝑚2

𝑁 = 𝐾𝑔 𝑚

𝑠2 =

(103 𝑔𝑚 ) (102𝑐𝑚)

𝑠2= 105

𝑔𝑚 𝑐𝑚

𝑠2= 105𝑑𝑦𝑛𝑒

𝑚2 = 104 𝑐𝑚2

∴ 𝛽 = 1011 𝑁

𝑚2 = 1011 ×

105𝑑𝑦𝑛𝑒

104 𝑐𝑚2 = 1012 𝑑𝑦𝑛𝑒/𝑐𝑚2



86

Example 6: A cylindrical steel rod has a length of 2 m and a radius of 0.5 cm. A

force of magnitude 𝟐 × 𝟏𝟎𝟒 𝑵 is acting normally on each of its ends. Find the

change in its length and radius, if the Young’s modulus Y is 𝟐𝟎𝟎 × 𝟏𝟎𝟗 𝑵/𝒎𝟐

and the Poisson’s ratio 𝝁 is 0.25.

Solution

The Young’s modulus reads as Y = F ℓ

A Δ ℓ

The Change in its length

𝚫𝓵 =𝑭 𝓵

𝒀 𝑨 =

𝟐 × 𝟏𝟎𝟒 × 𝟐

𝟐𝟎𝟎 ×𝟏𝟎𝟗 × (𝝅×(𝟎.𝟓 ×𝟏𝟎−𝟐)𝟐) = 𝟐. 𝟓 × 𝟏𝟎−𝟑𝒎 = 𝟎. 𝟐𝟓 𝒄𝒎

The Change in its radius, the Poisson’s ratio 𝝁 = −𝚫𝒓

𝒓𝚫𝓵

𝓵

= − 𝚫𝒓 𝓵

𝒓 𝚫 𝓵

𝚫𝒓 = −𝝁 𝒓 𝚫𝓵

𝓵 = −

𝟎.𝟐𝟓 ×𝟎.𝟓 𝒄𝒎 ×𝟎.𝟐𝟓𝒄𝒎

𝟐𝟎𝟎 𝒄𝒎 = −𝟏. 𝟓𝟔 × 𝟏𝟎−𝟒 𝒄𝒎

Example 7 A steel rod has a length L = 8 m and radius r = 1.5 cm when the

temperature is 𝟐𝟎𝒐𝑪. Take 𝜶 = 𝟏𝟏 × 𝟏𝟎−𝟔 (𝑪)−𝟏 and Young’s modulus of

the rod to be 𝒀 = 𝟐𝟎𝟎 × 𝟏𝟎𝟗 𝑵/𝒎𝟐 (a) What is its length on a hot day when

the temperature is 𝟓𝟎𝒐𝑪? (b) If the rod’s ends were originally fixed, then find

the compression force on the rod?

Solution

As result of heating the steel rod from 𝟐𝟎𝒐𝑪 𝒕𝒐 𝟓𝟎𝒐𝑪, the elongation is

Δ𝐿 = 𝛼 𝐿 Δ𝑇 = = 𝟏𝟏 × 𝟏𝟎−𝟔 × 𝟖 × (𝟓𝟎 − 𝟐𝟎 ) = 2.64 × 𝟏𝟎−𝟑 𝑚

Therefore, the rod’s new length at 𝟓𝟎𝒐𝑪 is 8.00264 m.



87

(b) If the rod is not allowed to expand, we then calculate what force would be

required to compress the rod by the amount 2.64 × 𝟏𝟎−𝟑 𝑚

The compression force can be found from Young’s modulus

F = Y A Δ ℓ

ℓ=

200 × 109 × (𝜋 (1.5 × 10−2)2 × 2.64 × 𝟏𝟎−𝟑

(8)

= 4.7 × 104𝑁

Example 8: On a hot day, when the temperature was 𝑻𝒊 = 𝟒𝟓𝒐𝑪, an oil trucker

fully loaded his truck from an oil station with 10,000 gal of gasoline

(𝟏 𝒈𝒂𝒍𝒍𝒐𝒏 ∼ 𝟑. 𝟖 𝒍𝒊𝒕𝒆𝒓). On his way to a delivery city, he encountered cold

weather, where the temperature went down to 𝑻𝒇 = 𝟐𝟎𝒐𝑪, see Figure. The

coefficients of volume expansion of gasoline and steel are 𝜷 = 𝟗. 𝟔 ×

𝟏𝟎−𝟒 (𝑪)−𝟏 and 𝜷 = 𝟏𝟏 × 𝟏𝟎−𝟔 (𝑪)−𝟏, respectively. How many gallons did

the trucker deliver?

Solution: The change in temperature from the production city to the delivery city is

Δ 𝑇 = 20 − 45 = −25𝑜𝐶

we can find the change in the gasoline volume V as follows:

Δ𝑉𝐺 = 𝛽𝐺 𝑉0 Δ𝑇 = 𝟗. 𝟔 × 𝟏𝟎−𝟒 × (𝟏𝟎𝟎𝟎𝟎) × (−𝟐𝟓) = −240 𝑔𝑎𝑙𝑙𝑜𝑛



88

Thus, the amount of gasoline delivered was: 𝑉𝑓 = 𝑉0 + Δ 𝑉 = 10000 − 240 =

9760 𝑔𝑎𝑙𝑙𝑜𝑛

The thermal expansion of the volume of the steel tank can also be calculated as

follows:

Δ𝑉𝑆 = 𝛽𝑆 𝑉0 Δ𝑇 = 𝟏𝟏 × 𝟏𝟎−𝟔 × (𝟏𝟎𝟎𝟎𝟎) × (−𝟐𝟓) = −2.75 𝑔𝑎𝑙𝑙𝑜𝑛

This change is very small and has nothing to do with the problem, since the decrease

in the gasoline volume is much bigger than that of the steel. Question: Who paid for

the missing gasoline?

Example 9: A second's pendulum clock has a steel wire. The clock is calibrated

at 200 C. How much time does the clock lose or gain in one week when the

temperature is increased to 𝟑𝟎𝒐𝑪? ( 𝜶𝒔 = 𝟏. 𝟐 × 𝟏𝟎−𝟓 𝑪−𝟏 ).

Solution

The time period of second's pendulum is 𝑡 = 2 second. As the temperature increases

length time period increases. Clock becomes slow and it loses the time. 𝑡 ∝ √ℓ

The change in time period is

Δ 𝑡 = 1

2𝛼 𝑡 Δ𝑇 =

1

2× 𝟏. 𝟐 × 𝟏𝟎−𝟓 × 2 × (30 − 20) = 𝟏. 𝟐 × 10−4𝒔

The new period time is

𝒕 = 𝒕𝟎 + 𝚫 𝒕 = 𝟐 𝒔𝒆𝒄 + 𝟏. 𝟐 × 𝟏𝟎−𝟒𝒔 = 𝟐. 𝟎𝟎𝟎𝟏𝟐 𝒔𝒆𝒄

Time lost in one week, we have

𝚫 𝒕

𝒕=

𝒕𝒊𝒎𝒆 𝒍𝒐𝒔𝒕

𝒐𝒏𝒆 𝒘𝒆𝒆𝒌 𝒕𝒊𝒎𝒆

𝒕𝒊𝒎𝒆 𝒍𝒐𝒔𝒕 = 𝚫 𝒕

𝒕 × 𝑜𝑛𝑒 𝑤𝑒𝑒𝑘 = (

𝟏. 𝟐 × 10−4

𝟐. 𝟎𝟎𝟎𝟏𝟐) × (7 × 24 × 3600)

= 36.28 𝑠𝑒𝑐.



89

Sheet (IV)

1. A mass of 5 kg is suspended from the end of a copper wire that has a diameter

of 1 mm. Find the tensile stress on the wire?

2. A 4 m long structural steel rod with a cross-sectional area of 0.5 𝑐𝑚2

stretches 1 mm when a mass of 250 kg is hung from its lower end. Find the

value of Young’s modulus for this steel.

3. An iron rod 10 m long and 0.5 𝑐𝑚2 in cross section, stretches 2.5 mm when

a mass of 300 kg is hung from its lower end. Find Young’s modulus for the

iron rod.

4. A wire has a length L = 3 m and a radius r = 0.75 cm, see Fig.

A force acting normally on each of its ends has a magnitude 𝐹⊥ =

9 × 104 𝑁. Find the change in the wire’s

length and radius, when its Young’s modulus

Y is 190 × 109 𝑁/𝑚2 and its Poisson’s ratio

𝜇 is 0.25.

5. A block of gelatin resting on a rough dish has

a length L = 60 cm, width d = 40 cm, and height h = 20 cm, see the vertical

cross section abcd in Fig. . A force F = 0.6 N is applied tangentially to the

upper surface, leading to a new shape abef and hence a displacement x = 5

mm for the upper surface relative to the lower one. Find: (a) the shearing

stress, (b) the shearing strain, and (c) the shear modulus.



90

6. Two parallel but opposite forces, each having a magnitude 𝐹 = 4 × 103 𝑁,

are applied tangentially to the upper and lower faces of a cubical metal block

of side 𝑎 = 25 𝑐𝑚 and shear modulus 𝑆 = 80 × 109 𝑁/𝑚2, see Fig. Find

the displacement Δ𝑥 of the upper surface relative to the lower one, and hence

find the angle of shear 𝜃.

7. The pressure of the atmosphere around a metal block is reduced to almost zero

when the block is placed in vacuum. Find the fractional change in volume if the bulk

modulus of the metal is 𝐵 = 150 × 109 𝑁/𝑚2.

8. The pressure around a cube of copper of side 40 mm is changed by 𝑃 =

2 × 1010 𝑁/𝑚2. Find the change in volume if the bulk modulus for copper is 𝐵 =

125 × 109 𝑁/𝑚2.

9. A copper rod is 8 m long at 20 ◦C and has a coefficient of linear expansion 𝛼 =

17 × 10−6 (𝐶)−1. What is the increase in the rod’s length when it is heated to 40

◦C?



91

10. A road is built from concrete slabs, each of 10 m long when

formed at 10 ◦C, see Fig. . How wide should the expansion

cracks between the slabs be at 10 ◦C to prevent road buckling if

the range of temperature changes from −5 ◦C in winter to +40 ◦C

in summer? The coefficient of linear expansion for concrete is

𝛼 = 12 × 10−6 (𝐶)−11.

11. An iron steam pipe is 100 m long at 0 ◦C and has a

coefficient of linear

expansion 𝛼 = 10 × 10−6 (𝐶)−1. What will be its length when heated to

100 ◦C?

12. An ordinary glass window has a coefficient of

linear expansion 𝛼 = 9 × 10−6 (𝐶)−1. At

20 C the sides a and b have the values 1 m and

0.8 m respectively, see Fig.. By how much

does the area increase when its temperature

rises to 40 ◦C?

13. A bar of length L = 4 m and linear expansion 𝛼 = 25 × 10 − 6 (𝐶)−1 has

a crack at its center. The ends of the bars are fixed as shown in Fig.. As a result

of a temperature rise of 40 C◦, the bar

buckles upwards. Find the vertical rise d

of the bar’s center.

14. A composite rod of length L is made from two different rods of lengths 𝐿1 and

𝐿2 with linear expansion coefficients of 𝛼1 and 𝛼2, respectively, see Fig.



92

(a) Show that the coefficient of linear expansion 𝛼 for this composite rod is

given by 𝛼 =𝛼1𝐿1 + 𝛼2 𝐿2

𝐿.

(b) Using the linear expansion coefficients of steel and brass given in Table

find 𝐿1and 𝐿2 in the case where L = 0.8 m and 𝛼 = 14 × 10−6 (𝐶)−1.

15. A flask is completely filled with mercury at 20 ◦C and is sealed off. Ignore

the expansion of the glass and assume that the bulk modulus of mercury is

𝐵 = 2.5 × 109 𝑁/𝑚2 and its coefficient of volume expansion is 𝛽 =

1.82 × 10−4 (𝐶)−1. Find the change in pressure inside the flask when it is

heated to 100 ◦C.

16. A glass flask of volume 200 𝑐𝑚3 is filled with mercury when the temperature

is T = 20 C, see Fig. The coefficient of volume expansion of the glass and

mercury are 𝛽 = 1.2 × 10−5 (𝐶)−1 and 𝛽 = 18 × 10−5 (𝐶)−1

respectively. How much mercury will overflow when the temperature of the

flask is raised to 100 ◦C?



93

II) Fluid Mechanics In this section, you will learn about buoyant force, fluid pressure, and the basic

equations that govern the behavior of fluids. This section will also introduce moving

fluids and the continuity equation and Bernoulli’s equation. Furthermore, we give

short notes for the viscous fluid. Liquids and gases are fluids.

1. Fluid Statics (Fluid at rest)

If you have ever dived deep into a swimming pool or lake, you know that your body,

especially your ears, is sensitive to changes in pressure. You may have noticed that

the pressure you felt on your ears did not depend on whether your head was upright

or tilted, but that if you swam deeper, the pressure increased.

1.1 Pressure in Fluids

Deep-sea explorers wear atmospheric diving suits like the one

shown in Figure to resist the forces exerted by water in the

depths of the ocean. You experience the effects of similar forces

on your ears when you dive to the bottom of a swimming pool,

drive up a mountain, or ride in an airplane.

Pressure is a measure of how much force is applied over a given area. It can be

written as follows: 𝑷 =𝑭

𝑨 (𝑵/𝒎𝟐) = 𝑷𝒂𝒔𝒄𝒂𝒍

The SI unit of pressure is the pascal (Pa), which is equal to 𝑵/𝒎𝟐. The pascal is a

small unit of pressure. The pressure of the atmosphere at sea level is about

1.013 × 105 𝑃𝑎. This amount of air pressure under normal conditions is the basis

for another unit, the atmosphere (atm).



94

1.2 Variation of Pressure with Depth

As indicated in the previous section, all points at the

same depth from a liquid surface have the same value of

pressure. The variation of pressure 𝑃 with depth h in a

liquid of density 𝝆 open to the atmosphere can be found

by considering a small horizontal area 𝑑𝐴 at that depth,

as shown figure

Accordingly, we have the pressure at depth h gives

𝑷 = 𝑷𝒂 + 𝝆 𝒈 𝒉

This relation verifies that the pressure is the same at all

points having the same depth from a liquid surface.

Moreover, the pressure is not affected by the shape of

the container, see Fig.

Pascal’s principle Blaise Pascal, a French physician, noted that the pressure in a

fluid depends upon the depth of the fluid and has nothing to do with the shape of the

fluid’s container.

He also discovered that any change in pressure applied at any point on a confined

fluid is transmitted undiminished throughout the fluid, a fact that is now known as

Pascal’s principle.



95

An important application of Pascal’s law is the hydraulic lever illustrated in Fig. Let

an external input force of magnitude 𝐹1 be exerted downwards on a small piston of

area 𝐴1. The pressure will be transmitted through an incompressible fluid which then

exerts an output force 𝐹2 on a larger piston of area 𝐴2, balancing the load.

The pressure on both leveled pistons is the same. That is:

𝑷𝟏 = 𝑷𝟐 𝑭𝟏

𝑨𝟐=

𝑭𝟐

𝑨𝟐

Thus, the force 𝐹2 is larger than 𝐹1 by the multiplying factor 𝐴2/𝐴1. Hydraulic

brakes, car lifts, etc make use of this principle.

When we move the input piston downwards a distance 𝑑1, the output piston moves

upwards a distance 𝑑2, such that the same volume V of the incompressible liquid is

displaced at both pistons. Then, we get:

𝑽𝟏 = 𝑽𝟐 𝑨𝟏𝒅𝟏 = 𝑨𝟐 𝒅𝟐

Thus, for 𝐴2 > 𝐴1, the output piston moves a smaller distance than the input piston.

On the other hand, for small values of 𝑑1, we can find the following input/output

work: 𝑊2 = 𝐹2 𝑑2 = (𝐴1

𝐴2) 𝐹1 ×

𝐴1𝑑1

𝐴2 = 𝐹1 𝑑1 = 𝑊1



96

which shows that the work 𝑊1done on the input piston by the applied force equals

the work 𝑊2done by the output piston in lifting the load.

Application: Gauge Pressure of Blood Blood-pressure

readings, such as 130/80, give the maximum and

minimum gauge pressures in the arteries, measured in

mm Hg or torr. Blood pressure varies with vertical

position within the body; the standard reference point

is the upper arm, level with the heart.

1.3 Buoyant Forces and Archimedes ‘Principle

In a swimming pool, you may have noticed that it is relatively easy to carry an object

that is totally or partially immersed in the water. This is because you must support

only part of the object’s weight, while the buoyant force supports the remainder.

This important property of fluids in hydrostatic equilibrium is summarized by

Archimedes’ principle, which can be stated as follows:

Archimedes’ Principle:

A body fully or partially immersed in a fluid is buoyed up by a force equal to the

weight of the fluid displaced by the body.

Let us show that the buoyant force is equal in magnitude to the weight of the

displaced fluid. We can do this by considering a cube of fluid of height h (and hence

volume 𝑉𝑓 = ℎ3) as in Fig. a.



97

(a) External forces acting on a cube of fluid (colored blue). Under equilibrium, the fluid’s weight 𝑾𝒇

is equal to the buoyant force 𝑭𝑩. (b) A cube of weight Wo is buoyed by a force 𝑭𝑩 = 𝑾𝒇.

The cube of this fluid is in equilibrium under the action of the forces on it. One

of the forces is its own weight �⃗⃗⃗⃗�𝑓 . Apparently, the rest of the fluid in the container

is buoying up the cube and holding it in equilibrium. Therefore, the magnitude of

this buoyant force, 𝑭𝑩, must be exactly equal in magnitude to the weight of the fluid.

That is: 𝐹𝐵 = 𝑊𝑓

Now, imagine we replace the cube of fluid by a cubical object of the same

dimensions. The fluid surrounding the cube will behave the same way, regardless

whether the cube is a fluid or a solid. Therefore, the buoyant force acting on an

object of any density will be equal to the weight of the fluid displaced by the object,

i.e. 𝐹𝐵 = 𝑊𝑓.

To show this result explicitly, we notice in Fig. b that the pressure at the bottom of

the object is greater than the pressure at the top by Δ𝑃 = 𝜌𝑓𝑔ℎ , where 𝜌𝑓 is the

density of the fluid. Since the pressure difference Δ𝑃 equals the buoyant force per

unit area, then Δ𝑃 = 𝐹𝐵/𝐴, where 𝐴 = ℎ2 is the area of one of the cube’s faces.

Therefore:



98

𝑭𝑩 = Δ𝑃 𝐴 = 𝜌𝑓𝑔ℎ × ℎ2 = 𝜌𝑓𝑔 ℎ3 = 𝜌𝑓 𝑉 𝑔 = 𝑊𝑓

Consider the object of Fig. b to be of weight 𝑊𝑜 = 𝜌𝑜𝑉𝑜 𝑔,

where 𝜌𝑜 and 𝑉𝑜 are its density and volume, respectively.

If the object is totally immersed in a fluid of density 𝜌𝑓, the buoyant force will be

𝑭𝑩 = 𝑾𝒇 = 𝝆𝒇 𝑽𝒇𝒈, 𝒘𝒉𝒆𝒓𝒆 𝑽𝒇 = 𝑽𝒐.

Example 1

The U shaped tube shown in Fig. contains oil in the right arm and

water in the left arm. In static equilibrium, the measurements

give h = 18 cm and d = 2 cm. What is the value of the density of

the oil 𝝆𝒐?

Solution

The pressure 𝑃𝐴 at the oil–water interface of the right arm must be equal to the

pressure 𝑃𝐵 in the left arm at the same level. 𝑃𝐴 = 𝑃𝐵

In the right arm, we use

𝑷𝑨 = 𝑷𝒂 + 𝝆𝒐 𝒈 (𝒉 + 𝒅)

In the left arm, we use the same

𝑷𝑩 = 𝑷𝒂 + 𝝆𝒘 𝒈 𝒉

𝝆𝟎 (𝒉 + 𝒅) = 𝝆𝒘 𝒉

Then

𝝆𝟎 =𝒉

(𝒉 + 𝒅) 𝝆𝒘 =

𝟏𝟖

𝟐𝟎 × 𝟏𝟎𝟑 = 𝟗𝟎𝟎 𝑲𝒈/𝒎𝟑

Note that the answer does not depend on the atmospheric pressure.



99

Example 2

The special manometer shown in Fig., uses mercury

of density 𝝆 = 𝟏𝟑. 𝟔 × 𝟏𝟎𝟑 𝒌𝒈/𝒎𝟑 If the

atmospheric pressure is 100 kPa and the height of

mercury above the surface of separation at point B

is 𝒉 = 𝟏𝟎 𝒄𝒎, what is the pressure of the gas tank?

Solution

The pressures at the same level are the same. Then 𝑃𝐴 = 𝑃𝐵

𝑃 = 𝑃𝑎 + 𝜌𝑚 𝑔 ℎ = 105 + (13.6 × 103 × 9.81 × 10 × 10−2) = 1.133 × 105 𝑃𝑎

Example 3

For the car lift shown in the Fig., the pistons on

the left and right have areas

𝟐𝟓 𝒄𝒎𝟐 𝒂𝒏𝒅 𝟕𝟓𝟎 𝒄𝒎𝟐 respectively. The car

and the right piston have a total weight of

𝟏𝟓, 𝟎𝟎𝟎 𝑵. What force must be applied on the

left piston (if it has negligible weight)? What

pressure will produce this force?

Solution

The pressure on both leveled pistons is the same. That is:

𝑷𝟏 = 𝑷𝟐 𝑭𝟏

𝑨𝟐=

𝑭𝟐

𝑨𝟐 𝑭𝟏 = 𝑭𝟐

𝑨𝟏

𝑨𝟐 = 𝟏𝟓𝟎𝟎𝟎 ×

𝟐𝟓

𝟕𝟓𝟎 = 𝟓𝟎𝟎 𝑵

The pressure that produce this force is given by:



100

𝑃 = 𝐹

𝐴 =

500

25 × 10−4 = 2 × 105 𝑃𝑎

Example 4

The areas of the car lift pistons, shown in Fig. ,

are 𝑨𝟏 = 𝟐𝟓 𝒄𝒎𝟐 𝒂𝒏𝒅 𝑨𝟐 = 𝟓𝟎𝟎 𝒄𝒎𝟐 . The

car and the right piston have a total weight of

𝟏𝟎𝟒 𝑵, while the left piston has a negligible

weight and is at a height 𝒉 = 𝟏𝟎 𝒄𝒎 with

respect to the right one. The apparatus is filled

with oil of density ρ = 800 kg/m3. What is the

value of the force 𝑭𝟏 needed to keep the system

in equilibrium?

Soln. The pressures at the same level are the same.

𝑃𝑎 + 𝜌oil 𝑔 ℎ +𝐹1

𝐴1 = 𝑃𝑎 +

𝐹2

𝐴2

𝐹1

𝐴1=

𝐹2

𝐴2− 𝜌oil 𝑔 ℎ ⟹ 𝐹1 = 𝐴1 [

𝐹2

𝐴2− 𝜌oil 𝑔 ℎ ]

𝐹1 = (25 × 10−4 ) [104

500 × 10−4 − (800 × 9.8 × 10 × 10−2)] = 498 𝑁

Example 5

A person dives to a depth 𝒉 = 𝟓𝟎 𝒄𝒎 below the water surface without inhaling

first. (a) Find the pressure on his body and on his lungs. (b) Repeat part (a)

when he dives to a depth 𝒉 = 𝟓 𝒎. When the diver ignores diving rules and

foolishly uses a snorkel tube at that depth, find the pressure on his lungs? Why

he is in danger? Assume that 𝑷𝒂 = 𝟏. 𝟎𝟏 × 𝟏𝟎𝟓 𝑷𝒂, 𝝆 = 𝟏𝟎𝟑 𝒌𝒈/𝒎𝟑, and

𝒈 = 𝟗. 𝟖 𝒎/𝒔𝟐.

Solution

(a) The external pressure on the diver’s body will be:

𝑃𝑏𝑜𝑑𝑦 = 𝑃𝑎 + 𝜌 𝑔 ℎ = (𝟏. 𝟎𝟏 × 𝟏𝟎𝟓) + (𝟏𝟎𝟑 × 𝟗. 𝟖 × 𝟎. 𝟓)

= 1.059 × 105 𝑃𝑎 = 𝟏. 𝟎𝟓 𝒂𝒕𝒎



101

The diver’s body adjusts to that pressure by a very slight contraction until the

internal pressure is in equilibrium with the external pressure. Consequently,

his average blood pressure increases, and the average air pressure in his lungs

increases until it balances the external pressure. Thus, his lung pressure will

be at:

Plungs = 𝟏. 𝟎𝟓 𝒂𝒕𝒎

(b) When h = 5 m, the external pressure on the diver’s body will be:

𝑃𝑏𝑜𝑑𝑦 = 𝑃𝑎 + 𝜌 𝑔 ℎ = (𝟏. 𝟎𝟏 × 𝟏𝟎𝟓) + (𝟏𝟎𝟑 × 𝟗. 𝟖 × 𝟓)

= 1.5 × 105 𝑃𝑎 = 𝟏. 𝟓 𝒂𝒕𝒎

Again, as in part (a), the pressure inside his lungs will be:

Plungs = 𝟏. 𝟓 𝒂𝒕𝒎

If the diver foolishly uses a 5 m snorkel tube, the pressurized air in his lungs

will be expelled upwards through the tube to the atmosphere. Consequently,

the air pressure in his lungs will drop rapidly from 1.5 to 1 atm. This 0.5 atm

pressure difference is sufficient to collapse his lungs and force his still-

pressurized blood into them.

Example 6

A cubic decimeter, 𝟏. 𝟎𝟎 × 𝟏𝟎−𝟑 𝒎𝟑, of a granite building block is

submerged in water. The density of granite is 𝟐. 𝟕𝟎 × 𝟏𝟎𝟑 𝒌𝒈/𝒎𝟑. a.

What is the magnitude of the buoyant force acting on the block? b. What

is the apparent weight of the block?



102

Solution

(a) the buoyant force on the granite block.

𝐹𝐵 = 𝜌𝑤 𝑉 𝑔 = 103 × 1.00 × 10−3 × 9.8

= 9.8 𝑁

(b) the granite’s weight and then find its apparent weight.

𝑊 = 𝑚 𝑔 = 𝜌granite 𝑉 𝑔 = 2.70 × 103 × 10−3 × 9.8 = 26.5 𝑁

𝐹𝑎𝑝𝑝 = 𝑊 − 𝐹𝐵 = 26.5 − 9.8 = 16.7 𝑁

Example 7

A bargain hunter purchases a “gold” crown at a flea

market. After she gets home, she hangs the crown

from a scale and finds its weight to be 7.84 N. She then

weighs the crown while it is immersed in water, and

the scale reads 6.86 N. Is the crown made of pure

gold? Explain.

Solution

𝑊 = 7.84 𝑁

𝐹𝑎𝑝𝑝 = 6.86 𝑁

The total mass of the gold crown in air

𝑚𝐺 = 𝑊

𝑔 =

7.84

9.8= 0.8 𝐾𝑔

The apparent weight of the crown in

water



103

𝐹𝑎𝑝𝑝 + 𝐹𝐵 = 𝑊

𝐹𝐵 = 𝐹𝑎𝑝𝑝 − 𝑊 = 7.84 − 6.86 = 0.98 𝑁

The buoyant force 𝐹𝐵 = 𝜌𝑤𝑉𝑤 𝑔 = 103 × 𝑉𝑤 × 9.8 = 0.98 𝑁

Then 𝑉𝑤 = 10−4 𝑚3

For total immersed 𝑉𝑤 = 𝑉𝐺

The density of the Gold 𝜌𝐺 = 𝑚𝐺

𝑉𝐺 =

0.8

10−4 = 8 × 103 𝐾𝑔/𝑚3

Form the table the density of the pure gold is 19.3 × 103 𝐾𝑔/𝑚3 .Therefore, the

density of the crown 8 × 103 𝐾𝑔

𝑚3< 19.3 × 103

𝐾𝑔

𝑚3. 𝑇ℎ𝑒𝑛 the crown cannot be pure

gold.

Example 8

A piece of steel has a mass 𝒎𝒔 = 𝟎. 𝟓 𝒌𝒈 and a

density 𝝆𝒔 = 𝟕. 𝟖 × 𝟏𝟎𝟑 𝒌𝒈/𝒎𝟑. The steel is

suspended in air by a string attached to a scale, see

Fig. After that, the steel is immersed in a container

filled with water of density 𝝆𝒘 = 𝟏𝟎𝟑 𝒌𝒈/𝒎𝟑.

Find the tension in the string before and after the

steel is immersed.

Solution

When the piece of steel is suspended in air, the tension in the string Ta equals the

weight 𝑚𝑠 𝑔 of that piece of steel. That is: 𝑇𝐴 = 𝑚𝑠 𝑔 = 0.5 × 9.8 = 4.9 𝑁

When the steel is immersed in water, it experiences an upward buoyant force FB.

Thus, the tension in the string will be reduced to a new value 𝑇𝑤 , see the figure.

Equilibrium in this case gives:



104

𝑇𝑤 + 𝐹𝐵 = 𝑊𝑠

The Buoyant force 𝐹𝐵 = 𝜌𝑤 𝑉𝑤 𝑔 to calculate the buoyant force, we need to find

the volume of water displaced by the piece as

𝜌𝑠 =𝑚𝑠

𝑉𝑠 → 𝟕. 𝟖 × 𝟏𝟎𝟑 =

0.5

𝑉𝑠 𝑉𝑠 = 6.4 × 10−5𝑚3

The piece is total immersed in the water, This volume equals the volume of the

displaced water. That is: 𝑉𝑤 = 𝑉𝑠 = 6.4 × 10−5 𝑚3

Since the buoyant force equals the weight of the displaced water, then:

𝐹𝐵 = 𝜌𝑤 𝑉𝑤 𝑔 = 103 × 6.4 × 10−5 × 9.8 = 0.63 𝑁

Therefore, the tension in the string Tw (the apparent weight) will be:

𝑇𝑤 = 𝑊 − 𝐹𝐵 = 4.9 𝑁 − 0.63 𝑁 = 4.3 𝑁

Example 9

The approximate density of ice 𝒊𝒔 𝝆𝒊 = 𝟗𝟏𝟖 𝒌𝒈/𝒎𝟑 and

the approximate density of sea water in which an iceberg

floats is 𝝆𝒘 = 𝟏, 𝟎𝟐𝟎 𝒌𝒈/𝒎𝟑, see Fig. What fraction of

the iceberg is beneath the water surface?

Solution

The iceberg floats, as shown in the figure, due to the effect of an upward buoyant

force and the weight of the iceberg is given by:

𝐹𝐵 = 𝑊𝑖

𝜌𝑤 𝑉𝑤 𝑔 = 𝜌𝑖 𝑉𝑖 𝑔

Therefore



105

𝑉𝑖

𝑉𝑤=

𝜌𝑤

𝜌𝑖 → 𝑉𝑖 =

𝜌𝑤

𝜌𝑖 × 𝑉𝑤

𝑉𝑖

𝑉𝑤 =

𝜌𝑤

𝜌𝑖 =

1000

1020 = 0.90 𝑜𝑟 90 %

Thus, 90% of the iceberg lies below water level. This means that, only 10% of an

iceberg—its tip is above the surface of the water.

Example 10

A helium-filled balloon has a volume 𝑽𝒃 = 𝟖 × 𝟏𝟎𝟑 𝒎𝟑

and balloon mass 𝒎𝒃 = 𝟐𝟎𝟎 𝒌𝒈, see Fig. What is the

maximum mass m of a load that keeps the balloon in

equilibrium? Neglect the air displaced by the load. Take

𝝆𝑯𝒆 = 𝟎. 𝟏𝟖 𝒌𝒈/𝒎𝟑 to be the density of helium and

𝝆𝒂𝒊𝒓 = 𝟏. 𝟐𝟖 𝒌𝒈/𝒎𝟑 to be the density of air.

Solution

the displaced air equals the balloon’s volume, i.e. 𝑉𝑎𝑖𝑟 = 𝑉𝑏. Since the balloon’s

volume is approximately equal to the volume of helium, i.e. 𝑉𝐻𝑒 = 𝑉𝑏,

𝑉𝐻𝑒 = 𝑉𝑏 = 𝑉𝑎𝑖𝑟 = 𝑽𝒃 = 𝟖 × 𝟏𝟎𝟑 𝒎𝟑

According to Archimedes’ principle, the buoyant force is the weight of the displaced

air. At equilibrium

𝐹𝐵 = ( 𝑚𝑏 + 𝑀𝐻𝑒 + 𝑚𝑙𝑜𝑎𝑑)𝑔

𝜌𝑎𝑖𝑟 𝑉𝑎𝑖𝑟 𝑔 = ( 𝑚𝑏 + 𝜌𝐻𝑒 𝑉𝐻𝑒 + 𝑚𝑙𝑜𝑎𝑑)𝑔

𝜌𝑎𝑖𝑟 𝑉𝑎𝑖𝑟 = ( 𝑚𝑏 + 𝜌𝐻𝑒 𝑉𝐻𝑒 + 𝑚𝑙𝑜𝑎𝑑)

Then

1.28 × 8 × 103 = ( 200 + (0.18 × 8 × 103) + 𝑚𝑙𝑜𝑎𝑑)

Then



106

𝑚𝑙𝑜𝑎𝑑 = 8600 𝐾𝑔

Example 11

A piece of wood has a mass 𝒎 = 𝟎. 𝟐𝟓 𝒌𝒈 and a density 𝝆 = 𝟕𝟓𝟎 𝒌𝒈/𝒎𝟑. The

wood is tied by a string to the bottom of a container of water in order to have

the wooden piece fully immersed, see Fig. Take the water

density to be 𝝆𝒘 = 𝟏. 𝟎 × 𝟏𝟎𝟑 𝒌𝒈/𝒎𝟑𝟑. (a) What is the

magnitude of the buoyant force 𝑭𝑩 on the wood? (b) What is

the magnitude of the tension T in the string?

Solution

a) The buoyant force for the water 𝑭𝑩 = 𝝆𝒘 𝑽𝒘 𝒈

the volume of the displaced water 𝑽𝒘 will equal be the volume of the

piece of the wood 𝑉𝑠

𝑽𝒘 = 𝑽𝒔 =𝒎

𝝆=

𝟎. 𝟐𝟓

𝟕𝟓𝟎=

𝟏

𝟑𝟎𝟎𝟎 𝒎𝟑

Then, the buoyant force for the water 𝑭𝑩 = 𝝆𝒘 𝑽𝒘 𝒈 = 𝟏𝟎𝟑 ×𝟏

𝟑𝟎𝟎𝟎×

𝟗. 𝟖 = 𝟑. 𝟐𝟕 𝐍

b) At Equilibrium, as shown in the fig.

𝑭𝑩 = 𝑻 + 𝑾

Therefore, the tension reads as

𝑻 = 𝑭𝑩 − 𝑾 = 𝑭𝑩 − 𝒎𝒈

𝑻 = 𝟑. 𝟐𝟕 − (𝟎. 𝟐𝟓 × 𝟗. 𝟖 ) = 𝟎. 𝟖 𝑵



107

2. Fluid Dynamics (Fluid at motion)

Ideal Fluids

The motion of a real fluid is very complicated. Instead, we shall discuss the motion

of an ideal fluid that will obey the following four assumptions:

1. Steady flow: The velocity of the fluid at any specific point does not change with

time. However, in general the velocity might vary from one point to another.

2. Incompressible flow: The density of the fluid does not change with time. That is,

the density has a constant uniform value.

3. Non-viscous flow: A tiny object can move through the fluid without experiencing

a viscous drag force; that is, there is no resistive force due to viscosity.

4. Irrotational flow: A tiny object can move through the fluid without rotating about

an axis passing through its center of mass.

2.1 Equation of Continuity

Figure below shows two cross-sectional areas 𝑨𝟏 and

𝑨𝟐 in a thin stream tube of fluid of varying cross-

sectional areas. The fluid particles are moving steadily

through this stream tube.

In a small time interval t, the fluid at the area 𝐴1

moves a small distance Δ 𝑥1 = 𝑣1 Δ𝑡. Assuming



108

uniform density over the area 𝐴1, then the mass in the colored segment of Fig. is:

𝑚1 = 𝜌1(𝐴1 Δ𝑥1) = 𝜌1(𝐴1 𝑣1 Δ𝑡) = 𝑀𝑎𝑠𝑠 𝑖𝑛𝑡𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡

Similarly, the fluid that moves through the area 𝐴2 in the same time interval will

be:

𝑚2 = 𝜌2(𝐴2 Δ𝑥2) = 𝜌2(𝐴2 𝑣2 Δ𝑡) = 𝑀𝑎𝑠𝑠 𝑖𝑛𝑡𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡

Since mass is conserved and because the flow is steady, the mass that crosses 𝐴1in

time interval t must be equal to the mass that crosses 𝐴2 in the same time interval.

Then, 𝑚1 = 𝑚2 and we get:

• For Steady flow 𝜌1 𝐴1 𝑣1 = 𝜌2 𝐴2 𝑣2

• Incompressible Steady flow 𝐴1 𝑣1 = 𝐴2 𝑣2

This is the equation of continuity.

• The mass flow rate 𝑅𝑚 =𝑚𝑎𝑠𝑠

𝑡𝑖𝑚𝑒 = 𝜌 𝐴 𝑣

• the volume flow rate 𝑅𝑉 =𝑉𝑜𝑙𝑢𝑚𝑒

𝑡𝑖𝑚𝑒 = 𝐴 𝑣

2.2 Bernoulli’s Equation

Bernoulli's equation is mathematical expression of the law of mechanical energy

conservation in fluid dynamics. Bernoulli’s theorem is applied to the ideal fluids.

𝑻𝒉𝒆 𝒕𝒐𝒕𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒇𝒍𝒖𝒊𝒅 𝒆𝒏𝒕𝒆𝒓𝒊𝒏𝒈

= 𝑻𝒉𝒆 𝒕𝒐𝒕𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒇𝒍𝒖𝒊𝒅 𝑳𝒆𝒂𝒗𝒊𝒏𝒈

Every point in an ideal fluid flow is associated with three kinds of energies:

➢ Kinetic Energy

If a liquid of mass (m) and volume (V) is flowing with velocity (v) then

𝐾. 𝐸

𝑉𝑜𝑙=

1

2

𝑚

𝑉 𝑣2 =

1

2 𝜌 𝑣2

➢ Potential energy



109

If a liquid of mass (m) and volume (V) is at height (h) from the surface of the

earth then

𝑃. 𝐸

𝑉𝑜𝑙 =

𝑚 𝑔 ℎ

𝑉 = 𝜌 𝑔 ℎ

➢ Pressure Energy

If P is the pressure on area A of a liquid and the liquid moves through a

distance () due to this pressure then

𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝒇𝒐𝒓𝒄𝒆 𝒙 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕

= 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒙 𝒂𝒓𝒆𝒂 𝒙 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕

𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒆𝒏𝒆𝒓𝒈𝒚

𝑽𝒐𝒍 =

𝑭 . 𝒅

𝑽=

𝑷 𝑨 𝒅

𝑽=

𝑷 𝑨 𝒅

𝑨 𝒅 = 𝑷

According to Bernoulli's Theorem , in case of steady flow of incompressible

and non–viscous fluid through a tube of non–uniform cross–section, the sum of the

pressure, the potential energy per unit volume and the kinetic energy per unit volume

is same at every point in the tube

𝑷 + 𝟏

𝟐 𝝆 𝒗𝟐 + 𝝆 𝒈 𝒉 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕

Or 𝑷𝟏 + 𝟏

𝟐 𝝆 𝒗𝟏

𝟐 + 𝝆 𝒈 𝒉𝟏 = 𝑷𝟐 + 𝟏

𝟐 𝝆 𝒗𝟐

𝟐 + 𝝆 𝒈 𝒉𝟐

This equation is the Bernoulli's equation and represents conservation of

mechanical energy in case of moving fluids. This tells us that if the speed of



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a fluid increases as it travels horizontally, then its pressure must decrease, and

vice versa.

2.3 Viscosity

Fluids cannot withstand a shearing stress. However, fluids show some degree

of resistance to shearing motion, and this resistance is called viscosity.

The degree of viscosity can be understood by considering a fluid between two

sheets of glass where the lower one is kept fixed; see the sketch in Fig. below.

It is easier to slide the upper glass if the fluid is oil as opposed to tar because

tar has a higher viscosity than oil.

We can think of fluids as a set of adjacent layers. Thus, a reasonable shearing

stress produces smooth relative displacement of adjacent layers in fluids,

called laminar flow. When we apply a force of magnitude 𝑭 to the upper glass

of area 𝐴, it will move to the right with a speed 𝑣. As a result of this motion,

a portion of the fluid with shape abcd will take a new shape abef after a short

time interval Δ𝑡.

2.3.1 Newton’s law of Viscous fluid

we can define the shearing stress and the shearing strain on the fluid of Fig.

as follows:



111

𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑒𝑒𝑠 =𝐹

𝐴 𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑎𝑖𝑛 =

Δ𝑥

𝑑

Since the upper sheet is moving with speed 𝑣, the fluid just beneath it will

move with the same speed. Thus, in time Δ𝑡, the fluid just beneath the upper

sheet moves a distance Δ 𝑥 = 𝑣 Δ𝑡. Accordingly, we define the coefficient

of viscosity or simply the viscosity: at a given temperature, the ratio of the

shear stress to the rate of shear strain.

𝜂 =shear stress

𝑡𝑖𝑚𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑎𝑖𝑛

𝜂 =𝐹/𝐴

𝑣/𝑑 =

𝐹 𝑑

𝐴 𝑣

The SI unit of viscosity is 𝑵. 𝒔/𝒎𝟐 = 𝑷𝒂. 𝑆 which is called the poiseuille

(abbreviated by 𝑷𝒍) while in cgs it is 𝒅𝒚𝒏𝒆. 𝒔/𝒄𝒎𝟐 which is called the poise

(abbreviated by P).

Thus, s 𝑵. 𝒔/𝒎𝟐 = 𝑷𝒂. 𝑆 = 1 𝑃𝑙 = 10 𝑃 = 103 𝑐𝑃.

2.3.2 Stokes’ law of Viscous fluid



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When the small object is a sphere of radius 𝑟 and it moves with a terminal

speed 𝑣 through a viscous medium with viscosity 𝜂, it experiences a drag

force 𝐹𝑣𝑖𝑠which by Stokes’ law has a magnitude:

𝐹𝑣𝑖𝑠 = 6 𝜋 𝜂 𝑟 𝑣

As an application to Stokes’ formula, Fig. below displays the fall of a small

metallic spherical ball of volume 𝑉𝑠 =4

3 𝜋𝑟𝑠

3, density ρs, and mass 𝑚𝑠 =

𝑉𝑠𝜌𝑠 in a viscous liquid of density 𝜌. The forces that act on the sphere when

it reaches its terminal (constant) speed 𝑣 will be:

1. The sphere’s weight 𝑊 = 𝑚𝑠 𝑔 = 𝜌𝑠 𝑉𝑠 𝑔 (downwards)

2. The buoyant force 𝐹𝐵 = 𝜌 𝑉 𝑔 (upwards)

3.The viscous force 𝐹𝑣𝑖𝑠 = 6 𝜋 𝜂 𝑟 𝑣

(upwards)

From this equation we get the following

relation for the viscosity η:

𝜼 =𝟐

𝟗

(𝝆𝒔 − 𝝆𝒍) 𝒓𝟐 𝒈

𝒗



113

Example 12

A fluid flows in a cylindrical pipe of radius 𝒓𝟏 with a speed 𝒗𝟏. (a) What would

be the speed of this fluid at a point where the fluid is confined to a cylindrical

part of radius 𝒓𝟐 = 𝒓𝟏 /𝟓, see the top part of Fig. 1. (b) What is the effect of

elevating the constriction in the pipe by h = 10 m, see the lower part of Fig.

Soln.

(a)By using the continuity eq. 𝑨𝟏𝒗𝟏 = 𝑨𝟐 𝒗𝟐

where 𝑨 = 𝝅 𝒓𝟐

∴ 𝒗𝟐

𝒗𝟏 =

𝑨𝟏

𝑨𝟐= (

𝒓𝟏

𝒓𝟐)

𝟐= (

𝟓 𝒓𝟏

𝒓𝟏)

𝟐= 𝟐𝟓 → → 𝒗𝟐 =

𝟐𝟓 𝒗𝟏

(b)The volume flow rate of the pipe should be

constant. i.e. 𝑹𝒗 =𝚫𝑽

𝚫𝒕= 𝑨𝒗 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕

∴ No effect, because the continuity equation does not depend on altitude.

Example 13

Water with a density of 𝟏𝟎𝟑 𝒌𝒈/𝒎𝟑 is flowing steadily through a closed-pipe

system via the motor M, see Fig. 2 . At height 𝒚𝟏, the water speed and pressure

are 𝒗𝟏 = 𝟒 𝒎/𝒔 𝒂𝒏𝒅 𝑷𝟏 = 𝟑𝟎 𝒌𝑷𝒂, respectively. At height 𝒚𝟐, which is a

point higher than by a height h = 2

m, the water speed is 𝒗𝟐 = 𝟔 𝒎/𝒔.

(a) What is the pressure at 𝒚𝟐 ?

Soln.

𝒗𝟏 = 𝟒𝒎

𝒔 𝒂𝒏𝒅 𝑷𝟏 = 𝟑𝟎 𝒌𝑷𝒂

h = 2 m 𝒗𝟐 = 𝟔𝒎

𝒔.



114

a) 𝒚𝟐 = 𝒉 + 𝒚𝟏 = 𝟐 + 𝒚𝟏

By using the Bernoulli Eq.

𝑷𝟏 +𝟏

𝟐 𝝆 𝒗𝟏

𝟐 + 𝝆 𝒈 𝒚𝟏 = 𝑷𝟐 +𝟏

𝟐 𝝆 𝒗𝟐

𝟐 + 𝝆 𝒈 𝒚𝟐

(𝟑𝟎 × 𝟏𝟎𝟑 ) + (𝟏

𝟐× 𝟏𝟎𝟑 × 𝟏𝟔) + (𝟏𝟎𝟑 × 𝟗. 𝟖 × 𝒚𝟏)

= 𝑷𝟐 + (𝟏

𝟐× 𝟏𝟎𝟑 × 𝟑𝟔) + (𝟏𝟎𝟑 × 𝟗. 𝟖 × (𝟐 + 𝒚𝟏))

𝑷𝟐 = 𝟒𝟎𝟎 𝑷𝒂

Example 14

Water enters a house (Figure) through a pipe with an inside diameter of 2.0 cm

at an absolute pressure of 𝟒 × 𝟏𝟎𝟓 𝑷𝒂 (about 4 atm). A 1.0-cm-diameter pipe

leads to the second-floor bathroom 5.0 m above. When the flow speed at the

inlet pipe is 𝟏. 𝟓 𝒎/𝒔 find the flow speed, pressure, and volume flow rate in the

bathroom.



115

We assume that the water flows at a steady rate. Water is effectively incompressible,

so we can use the continuity equation. It’s reasonable to ignore internal friction

because the pipe has a relatively large diameter, so we can also use Bernoulli’s

equation. Let points 1 and 2 be at the inlet pipe and at the bathroom, respectively.

We are given the pipe diameters at points 1 and 2, from which we calculate the areas

𝐴1and 𝐴2, as well as the speed 𝑣1 = 1.5 𝑚/𝑠 and pressure 𝒑𝟏 = 𝟒 × 𝟏𝟎𝟓 𝑷𝒂 at the

inlet pipe. We take 𝑦1 = 0 and 𝑦2 = 5 𝑚 . We find the speed 𝑣2using the continuity

equation and the pressure 𝑝2 using Bernoulli’s equation. Knowing we calculate the

volume flow rate 𝐴 𝑣.

using the continuity equation 𝐴1 𝑣2 = 𝐴2 𝑣2

𝑣2 =𝐴1

𝐴2 𝑣1 =

𝜋 (1 𝑐𝑚)2

𝜋 (0.5 𝑐𝑚)2 (1.5

𝑚

𝑠) = 6 𝑚/𝑠

From Bernoulli’s equation

𝑷𝟏 +𝟏

𝟐 𝝆 𝒗𝟏

𝟐 + 𝝆 𝒈 𝒚𝟏 = 𝑷𝟐 +𝟏

𝟐 𝝆 𝒗𝟐

𝟐 + 𝝆 𝒈 𝒚𝟐

Then

𝑷𝟐 = 𝑷𝟏 −𝟏

𝟐 𝝆 (𝒗𝟐

𝟐 − 𝒗𝟏𝟐 ) − 𝝆 𝒈 (𝒚𝟐 − 𝒚𝟏)

𝑷𝟐 = (𝟒 × 𝟏𝟎𝟓 ) − ( 𝟏

𝟐 × 𝟏𝟎𝟑 × (𝟑𝟔 − 𝟐𝟐. 𝟓)) − (𝟏𝟎𝟑 × 𝟗. 𝟖𝟎 × (𝟓 − 𝟎))

𝑷𝟐 = 𝟑. 𝟑 × 𝟏𝟎𝟓 𝑷𝒂 = 𝟑. 𝟑 𝒂𝒕𝒎

The volume flow rate is

𝑹𝒗 =𝒅𝑽

𝒅𝒕= 𝑨𝟐 𝒗𝟐 = (𝝅 (𝟎. 𝟓 × 𝟏𝟎𝟐)𝟐) × (𝟔) = 𝟒. 𝟕 × 𝟏𝟎−𝟒 𝒎𝟑/𝒔 = 𝟎. 𝟒𝟕 𝐋/𝐬



116

Example 15

A large tank is kept full at a height of h = 4 m as shown in Fig. 3 . Take the

water density to 𝒃𝒆 𝝆 = 𝟏𝟎𝟑 𝒌𝒈/𝒎𝟑. (a) Find the speed 𝒗𝟐 of the jet of water

emerging from a small pipe at the bottom of the tank. (b) If 𝒚𝟐 = 𝟐 𝒎, then

find the horizontal distance 𝒙 (from the base of the tank) that the water stream

travels before striking the floor.

Soln.


𝑷𝟏 +𝟏

𝟐 𝝆 𝒗𝟏

𝟐 + 𝝆 𝒈 𝒚𝟏 = 𝑷𝟐 +𝟏

𝟐 𝝆 𝒗𝟐

𝟐 + 𝝆 𝒈 𝒚𝟐

𝑷𝟏 = 𝑷𝟐 = 𝑷𝒂

The water is steady flow, i. e. 𝒗𝟏 = 𝟎 𝒎/𝒔 and 𝒉 = 𝒚𝟏 − 𝒚𝟐 = 𝟒 𝒎

a) By simplification 𝒗𝟐 = √𝟐𝒈𝒉 = √𝟐 ∗ 𝟗. 𝟖 ∗ 𝟒 = 𝟖. 𝟖𝟓𝟒 𝒎/𝒔

b) If 𝒚𝟐 = 𝟐 𝒎 𝒙 = ⋯

By using Newton’s law of kinematics

i- The vertical distance 𝒚 = 𝒗𝒚 𝒕 +𝟏

𝟐 𝒈 𝒕𝟐 = (𝒗 𝒔𝒊𝒏 𝜽 ×

𝒕 ) +𝟏

𝟐 𝒂 𝒕𝟐



117

ii- The Horizontal distance 𝒙 = 𝒗𝒙 𝒕 = 𝒗 𝑪𝒐𝒔 𝜽 × 𝒕

𝜽 = 𝒊𝒔 𝒗𝒆𝒓𝒚 𝒔𝒎𝒂𝒍𝒍 𝜽 ∼ 𝟎

Then 𝟐 = (𝒗 𝒔𝒊𝒏 𝟎 × 𝒕 ) + ( 𝟏

𝟐× 𝟗. 𝟖 × 𝒕𝟐) Or 𝟎 − 𝟐 = 𝟎 −

(𝟏

𝟐 × 𝟗. 𝟖 × 𝒕𝟐)

𝒕 = √𝟒

𝟗.𝟖 = 𝟎. 𝟔𝟑𝟖𝟗 𝑺𝒆𝒄

𝒙 = 𝒗𝒙 𝒕 = 𝒗 𝑪𝒐𝒔 𝜽 × 𝒕 = 𝒗 × 𝒕 = 𝟖. 𝟖𝟓𝟒 ∗ 𝟎. 𝟔𝟑𝟖𝟗 = 𝟓. 𝟔𝟓𝟕 𝒎

Example 16

Assume the faucet in the previous exercise is closed. In terms of the cross-

sectional areas 𝑨𝟐 and 𝑨𝟏 of the small pipe and the tank, respectively, show that

𝒗𝟐 and 𝒙 depend on the variable height h as follows:

Soln.


𝐏𝟏 +𝟏

𝟐 𝛒 𝐯𝟏

𝟐 + 𝛒 𝐠 𝐲𝟏 = 𝐏𝟐 +𝟏

𝟐 𝛒 𝐯𝟐

𝟐 + 𝛒 𝐠 𝐲𝟐

Where 𝐏𝟏 = 𝐏𝟐 = 𝐏𝐚 and 𝐡 = 𝐲𝟏 − 𝐲𝟐 = 𝟒 𝐦 and

𝑨𝟏 𝒗𝟏 = 𝑨𝟐𝒗𝟐

𝟏

𝟐 𝐯𝟏

𝟐 + 𝐠( 𝐲𝟏 − 𝒚𝟐) =𝟏

𝟐 𝐯𝟐

𝟐

𝒗𝟏 = 𝒗𝟐 (𝑨𝟐

𝑨𝟏)



118

𝟏

𝟐 𝒗𝟐

𝟐 (𝑨𝟐

𝑨𝟏)

𝟐

+ 𝐠 𝐡 =𝟏

𝟐 𝐯𝟐

𝟐 → 𝐯𝟐𝟐 [𝟏 − (

𝑨𝟐

𝑨𝟏)

𝟐

] = 𝟐 𝐠𝐡 → 𝐯𝟐

= √

𝟐 𝒈𝒉

[𝟏 − (𝑨𝟐𝑨𝟏

)𝟐

]

b) The horizontal distance

𝒙 = 𝒗𝒙 𝒕 = 𝒗𝟐 𝑪𝒐𝒔 𝜽 × 𝒕

𝜽 = 𝒊𝒔 𝒗𝒆𝒓𝒚 𝒔𝒎𝒂𝒍𝒍 𝜽 ∼ 𝟎

𝒙 = 𝒗𝟐 × 𝒕

To find the time t, the vertical distance 𝒚𝟐 = 𝒗𝒚 𝒕 +𝟏

𝟐 𝒂 𝒕𝟐 =

𝟏

𝟐 𝒈 𝒕𝟐

Then 𝒕 = √𝟐 𝒚𝟐

𝒈 Therefore 𝒙 = 𝒗𝟐 × 𝒕 =

√

𝟐 𝒈𝒉

[𝟏− (𝑨𝟐𝑨𝟏

)𝟐

]

× √𝟐 𝒚𝟐

𝒈

𝒙 = 𝟐√

𝟐 𝒚𝟐 𝒉

[𝟏 − (𝑨𝟐𝑨𝟏

)𝟐

]



119

Example 17

A steel plate of area 𝑨 = 𝟎. 𝟐 𝒎𝟐 is placed over a thin film of lubricant of thickness 𝒅 =

𝟎. 𝟒 𝒎𝒎 sprayed over the flat horizontal surface of a table, see Fig. When connected via a

cord that passes over a massless and frictionless pulley to a mass 𝒎 = 𝟏𝟎 𝒈, the steel plate

is found to move with a constant speed 𝒗 = 𝟎. 𝟎𝟓 𝒎/𝒔. Find the viscosity of the lubricant

oil.



120

Example 18

A spherical ball is moving with terminal velocity inside a liquid. Determine the relationship

of rate of heat loss with the radius of ball.

Soln. The rate of heat loss =𝐸𝑛𝑒𝑟𝑔𝑦

𝑇𝑖𝑚𝑒= 𝐹.

𝐷

𝑡= 𝐹. 𝑣

By using Stoke’s law of viscosity 𝐹𝑣𝑖𝑠 = 6 𝜋 𝜂 𝑟 𝑣 where 𝑣 =2

9 𝑟2 (𝜌𝑠−𝜌𝑙)𝑔

𝜂

The rate of heat loss = 𝐹. 𝑣 = 6 𝜋 𝜂 𝑟 𝑣2 = 6 𝜋 𝜂 𝑟 [2

9 𝑟2 (𝜌𝑠−𝜌𝑙)𝑔

𝜂]

2

= 𝐶𝑜𝑛𝑠𝑡. × 𝑟5 .

Example 18

A cubical block (of side 2m) of mass 20 kg slides on inclined plane

lubricated with the oil of viscosity = 𝟏𝟎–𝟏 poise with constant

velocity of 10 m/s. Find out the thickness of layer of liquid. (g = 10

m/s2)

Soln. = 10–1 poise =10–1

10 = 0.01 𝑁. 𝑠/𝑚2

The viscosity coefficient 𝜂 =𝐹𝐷 𝑑

𝐴 𝑣 where d is the thickness of layer of liquid

In Equilibrium

𝐹𝐷 = 𝑊 → 𝜂 𝐴𝑣𝑑

= 𝑚𝑔co s 60

0.01 × 4 ×10

𝑑= 20 × 10 𝐶𝑜𝑠 60

𝑑 = 4 × 10−3 = 4 𝑚𝑚

Example 19

Water is moving with speed 5 m/s through a pipe with a cross sectional area of 4 𝑐𝑚2.

The water gradually descends 10 m as the pipe increases in area to 8 𝑐𝑚2. (a) What is

the speed of flow at the lower level? (b) If the speed at the upper level is 1.5 × 105 𝑃𝑎

what is the pressure at the lower level? (Hint: water density is1000 𝑘𝑔/𝑚3.)

Solution

d

𝐹𝐷

v

𝑊 = 𝑚𝑔 𝑚𝑔 𝐶𝑜𝑠 60



121

From the continuity equation

𝐴1𝑣1 = 𝐴2𝑣2

𝑣2 =4 × 10−4 × 5

8 × 10−4= 2.5 𝑚/𝑠

From Bernoulli’s equation

𝑃1 + 1

2 𝜌𝑣1

2 + 𝜌𝑔 ℎ1 = 𝑃2 + 1

2 𝜌𝑣2

2 + 𝜌𝑔 ℎ2

𝑃2 = 𝑃1 + 1

2 𝜌(𝑣2

2 − 𝑣12) + 𝜌𝑔 (ℎ2 − ℎ1)

𝑃2 = 1.5 × 105 +1

2 × 1000[52 − 2.52] + 1000 × 9.8 × 10 = 2.57 𝑃𝑎

Example 20

What is the velocity of flow of liquid from a hole of diameter 0.5 cm in the bottom of

the wide tank at depth 20 cm if the acceleration of gravity is 10 𝑚/𝑠2. Calculate the

volume flowing after 2 min.

Solution

𝑣 = √2 𝑔 ℎ = √2 × 10 × 0.2 = 2𝑚/𝑠

𝑄 = 𝑉

𝑡

𝑉 = 𝑄 𝑡 = 𝑣 𝐴 𝑡 = 2 × 𝜋 × (0.0025)2 × 120 = 5.59 × 10−3𝑚3



122

Sheet (V)

1. You purchase a rectangular piece of metal that has dimensions 5 × 15 ×

30 𝑚𝑚 and mass 0.0158 𝑘𝑔. The seller tells you that the metal is gold. To

check this, you compute the average density of the piece. What value do you

get? Were you cheated?

2. A uniform lead sphere and a uniform aluminum sphere have the same mass.

What is the ratio of the radius of the aluminum sphere to the radius of the lead

sphere?

3. Black smokers are hot volcanic vents that emit smoke deep in the ocean floor.

Many of them teem with exotic creatures, and some biologists think that life

on earth may have begun around such vents. The vents range in depth from

about 1500 m to 3200 m below the surface. What is the gauge pressure at a

3200-m deep vent, assuming that the density of water does not vary? Express

your answer in pascals and atmospheres.

4. An electrical short cuts off all power to a submersible diving vehicle when it

is 30 m below the surface of the ocean. The crew must push out a hatch of

area and weight 300 N on the bottom to escape. If the pressure inside is 1.0

atm, what downward force must the crew exert on the hatch to open it?

5. A 950-kg cylindrical can buoy floats vertically in salt water. The diameter of

the buoy is 0.900 m. Calculate the additional distance the buoy will sink when

a 70.0-kg man stands on top of it.

6. A slab of ice floats on a freshwater lake. What minimum volume must the slab

have for a 45.0-kg woman to be able to stand on it without getting her feet

wet?

7. A hollow plastic sphere is held below the surface of a freshwater lake by a

cord anchored to the bottom of the lake. The sphere has a volume 0.650 𝑚3of

and the tension in the cord is 900 N. (a) Calculate the buoyant force exerted



123

by the water on the sphere. (b) What is the mass of the sphere? (c) The cord

breaks and the sphere rises to the surface. When the sphere comes to rest, what

fraction of its volume will be submerged?

8. A solid aluminum ingot weighs 89 N in air. (a) What is its volume? (b) The

ingot is suspended from a rope and totally immersed in water. What is the

tension in the rope (the apparent weight of the ingot in water)?

9. Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 𝑚3/𝑠

(a) How fast will it shoot out of a hole 4.50 cm in diameter? (b) At what speed

will it shoot out if the diameter of the hole is three times as large?

10. A shower head has 20 circular openings, each with radius 1.0 mm. The shower

head is connected to a pipe with radius 0.80 cm. If the speed of water in the

pipe is 3 𝑚/𝑠 what is its speed as it exits the shower-head openings?

11. A sealed tank containing seawater to a height of 11.0 m also contains air above

the water at a gauge pressure of 3.00 atm. Water flows out from the bottom

through a small hole. How fast is this water moving?

12. A small circular hole 6.00 mm in diameter is cut in the side of a large water

tank, 14.0 m below the water level in the tank. The top of the tank is open to

the air. Find (a) the speed of efflux of the water and (b) the volume discharged

per second.

13. Water stands at a depth H in a large, open tank whose side walls are vertical

(Fig.). A hole is made in one of the

walls at a depth h below the water surface. (a) At what distance R from the

foot of the wall does the emerging stream strike the floor? (b) How far above

the bottom of the tank could a second hole be cut so that the stream emerging

from it could have the same range as for the first hole?



124

14. Water flows steadily from an open tank as in Fig. . The elevation of point 1

is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional

area at point 2 is 0.048 𝑚2 at point 3 it is 0.016 𝑚2. The area of the tank is

very large compared with the cross-sectional area of the pipe. Assuming that

Bernoulli’s equation applies, compute (a) the discharge rate in cubic meters

per second and (b) the gauge pressure at point 2.

125

126

Chapter (V)

Thermodynamics

1. Temperature

The concept of temperature is rooted in

qualitative ideas of “hot” and “cold” based on

our sense of touch. A body that feels hot usually

has a higher temperature than a similar body that

feels cold. That’s pretty vague, and the senses

can be deceived. But many properties of matter

that we can measure depend on temperature.

The length of a metal rod, steam pressure in a boiler, the ability of a wire to conduct

an electric current, and the color of a very hot glowing object—all these depend on

temperature.

Temperature is also related to the kinetic energies of the molecules of a material. In

general this relationship is fairly complex, so it’s not a good place to start

in defining temperature. To use temperature as a measure of hotness or coldness, we

need to construct a temperature scale. To do this, we can use any measurable

property of a system that varies with its “hotness” or “coldness.”

To measure the temperature of a body, you place the thermometer in contact with

the body. If you want to know the temperature of a cup of hot coffee, you stick the

thermometer in the coffee; as the two interact, the thermometer becomes hotter and



127

the coffee cools off a little. After the thermometer settles down to a steady value,

you read the temperature. The system has reached an equilibrium condition, in which

the interaction between the thermometer and the coffee causes no further change in

the system. We call this a state of thermal equilibrium.

temperature is a pointer for the direction of energy transfer as heat

It follows that temperature is a property two systems have in common when they are in

thermal equilibrium with each other.

Thermal Equilibrium: 𝑇𝐴 = 𝑇𝐵 and Δ𝑄 = 0

Physics (I) Chapter (V) Thermodynamics


128

Temperature Scales

The Celsius, Kelvin, and Fahrenheit temperature scales are shown in relation to the phase

change temperatures of water. The Kelvin scale is called absolute temperature and the

Kelvin is the SI unit for temperature.

2. Zeroth Law of Thermodynamics

The "zeroth law" states that, if two systems are at the same time in thermal

equilibrium with a third system, they are in thermal equilibrium with each other

If C is initially in thermal equilibrium with both A and B, then A

and B are also in thermal equilibrium with each other. This result

is called the zeroth law of thermodynamics.

State and Equilibrium The state of a system is its condition as

described by a set of relevant energy related properties.

Thermodynamic equilibrium refers to a condition of equilibrium with respect to all possible

changes in a given system.

Two systems are in thermal equilibrium if and only if they have the same temperature.

3. Thermodynamic Systems

We always talk about energy transfer to or from some specific system. The system might be a

mechanical device, a biological organism, or a specified quantity of material, such as the

refrigerant in an air conditioner or steam expanding in a turbine. In general, a thermodynamic

system is any collection of objects that is convenient to regard as a unit, and that may have the

potential to exchange energy with its surroundings.

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html#c1



129

Thermodynamic system The system which can be represented in of pressure (P),

volume (V) and temperature (T), is known thermodynamic system. A specified

portion of matter consisting of one or more substances on which the effects of

variables such as temperature, volume and pressure are to be studied, is called a

system. e.g. A gas enclosed in a cylinder fitted with a piston is a system.

Surroundings Anything outside the system, which exchanges energy with the

system and which tends to change the properties of the system is called its

surroundings.

Heterogeneous System A system which is not uniform throughout is said to be

heterogeneous. e.g. A system consisting of two or more immiscible liquids.

Homogeneous System A system is said to be homogeneous if it is completely

uniform throughout. e.g. Pure solid or liquid.

Isolated System A system in which there can be no exchange of matter and energy

with the surroundings is said to be an isolated system.

Universe The system and its surroundings are together known as the universe.

4. Quantity of Heat

When you put a cold spoon into a cup of hot coffee, the spoon warms up and the

coffee cools down as they approach thermal equilibrium. The interaction that causes

these temperature changes is fundamentally a transfer of energy from one substance

to another. Energy transfer that takes place solely because of a temperature

difference is called heat flow or heat transfer, and energy transferred in this way is

called heat.



130

When a hot body is put in contact with a cold one, the former gets colder and the

latter warmer. From this observation it is natural to conclude that a certain quantity

of heat has passed from the hot body to the cold one.

Heat is a form of energy. Heat is felt by its effects. Some of the effects of heat are :

(a) Change in the degree of hotness

(b) Expansion in length, surface area and volume

(c) Change in state of a substance

(d) Change in the resistance of a conductor

(e) Thermo e.m.f. effect.

We can define a unit of quantity of heat based on temperature changes of some

specific material. The calorie (abbreviated cal) is defined as the amount of heat

required to raise the temperature of 1 gram of water from 14.5°C to 15.5°C.

Because heat is energy in transit, there must be a definite relationship between these

units and the familiar mechanical energy units such as the joule. Experiments similar

in concept to Joule’s have shown that

𝟏 𝑪𝒂𝒍. = 𝟒. 𝟏𝟖𝟔 𝑱𝒐𝒖𝒍𝒆

The calorie is not a fundamental SI unit. The International Committee on Weights

and Measures recommends using the joule as the basic unit of energy in all forms,

including heat.

4.1 Heat Capacity and Specific Heat

The quantity of heat energy 𝑸 required to raise the temperature of an object by some

amount Δ𝑇 varies from one substance to another.

The heat capacity C of an object is defined as: The heat capacity C of an object of

a particular material is defined as the amount of heat energy needed to raise the

object’s temperature by one degree Celsius.



131

Accordingly, if 𝑄 units of heat energy are required to change the temperature by

Δ𝑇 = 𝑇𝑓 − 𝑇𝑖, where 𝑇𝑖 and 𝑇𝑓 are the initial and final temperatures of the object,

then:

𝑄 = 𝐶 Δ𝑇,

Heat capacity C has the unit 𝐽/𝐶𝑜(≡ 𝐽/𝐾).

The heat capacity for any object is proportional to its mass m. For this reason, we

define the “heat capacity per unit mass” or the specific heat c which refers to a unit

mass of the material of which the object is made. Thus, with 𝐶 = 𝑚 𝑐, therefore

𝑄 = 𝑚 𝑐 Δ𝑇,

Specific heat c has the unit 𝐽/𝐾𝑔. 𝐶𝑜(≡ 𝐽/𝐾𝑔 𝐾).

Note that, when heat energy is added to objects, 𝑄 and Δ𝑇 are both positive, i.e. the

temperature increases. Likewise, when heat is removed from objects, 𝑄 and Δ𝑇 are

both negative, i.e. the temperature decreases.

Example 1

During a bout with the flu an 80-kg man ran a fever of 39.0°C instead of the

normal body temperature of 37.0°C Assuming that the human body is mostly

water, how much heat is required to raise his temperature by that amount?

Solution :

Given that 𝑚 = 80 𝑘𝑔 , the specific heat of water 𝑐 = 4190 𝐽/𝐾𝑔𝐾 and the

change of temperature Δ 𝑇 = 39 − 37 = 20𝑜𝐶 = 20 𝐾.

The heat quantity

𝑄 = 𝑚 𝑐 Δ 𝑇 = 80 × 4190 × 20 = 6.7 × 105 𝐽



132

Specific heat c of some substances at atmospheric pressure and room temperature (20 ◦C) with few exceptions

Example 2

A steel metal object of mass 0.05 kg is heated to 225 ◦C and then dropped into

a vessel containing 0.55 kg of water initially at 18 ◦C. When equilibrium is

reached, the temperature of the mixture is 20 ◦C. Find the specific heat of the

metal.

Solution

Given that

𝑚𝑠 = 0.05 𝐾𝑔 𝑎𝑛𝑑 𝑇𝑠 = 225𝑜𝐶

𝑚𝑤 = 0.55 𝐾𝑔 𝑎𝑛𝑑 𝑇𝑤 = 18𝑜 𝐶 𝑎𝑛𝑑 𝑇𝑚𝑖𝑥 = 20𝑜𝐶

The amount of heat gained by water = the amount of hat lost by steel



133

𝑄𝑔𝑎𝑖𝑛 = −𝑄𝑙𝑜𝑠𝑡

𝑚𝑤 𝑐𝑤 (Tmix − Tw) = −𝑚𝑠 𝑐𝑠 (Tmix − Ts)

0.55 × 4186 × (20 − 18) = − 0.05 × 𝑐𝑠 × (20 − 225)

𝑐𝑠 = 449 𝐽/𝐾𝑔. 𝐾

Example 3:

A 0.05 kg of metal is heated to 200.0°C and then dropped into a beaker

containing 0.400 kg of water initially at 20.0°C. If the final equilibrium

temperature of the mixed system is 22.4°C, find the specific heat of the metal.(

water specific heat 4186 J/kg °C)

Solution

The amount of heat gained by water = the amount of hat lost by steel

𝑄𝑔𝑎𝑖𝑛 = −𝑄𝑙𝑜𝑠𝑡

𝑚𝑤 𝑐𝑤 (Tmix − Tw) = −𝑚𝑠 𝑐𝑠 (Tmix − Ts)

0.400 × 4186 × (22.4 − 20) = − 0.05 × 𝑐𝑠 × (22.4 − 200)

𝑐𝑠 = 453 𝐽/𝐾𝑔. 𝐾

4.2 Calorimetry and Phase Changes

Calorimetry means “measuring heat.” We have discussed the energy transfer (heat)

involved in temperature changes. Heat is also involved in phase changes, such as

the melting of ice or boiling of water. Once we understand these additional heat

relationships, we can analyze a variety of problems involving quantity of heat.



134

Phase Changes

We use the term phase to describe a specific state of matter, such as a solid, liquid,

or gas. A transition from one phase to another is called a phase change or phase

transition.

We consider the following two main common phase changes:

1. A phase change from solid to liquid (as ice melting) and from liquid to gas (as

water boiling), where heat energy is absorbed while the temperature remains

constant.

2. A phase change from gas to liquid (as steam condensing) and from liquid to

solid (as water freezing), where heat energy is released while the temperature

remains constant.

The amount of heat energy per unit mass, L, that must be transferred when a

substance completely undergoes a phase change without changing temperature is

alled the latent heat (literally, the “hidden” heat). If a quantity Q of heat energy

transfer is required to change the phase of a pure substance of a mass m, then 𝐿 =

𝑄/𝑚 characterizes an important thermal property of that substance. That is:

𝑄 = ±𝑚 𝐿

A positive sign is used in this equation when energy enters the system, causing

melting or vaporization of the substance, while a negative sign corresponds to energy

leaving the system such that the substance condenses or solidifies.

When a substance experiences a phase change from solid to liquid by absorbing heat,

the heat of transformation is called the latent heat of fusion 𝐿𝐹, see Fig. below.

When the substance releases heat and experiences a phase change from liquid back



135

to solid, the heat of transformation is called the latent heat of solidification and is

numerically equal to the latent heat of fusion, see Fig. below.

When a substance experiences a phase change from liquid to gas by absorbing heat,

the heat of transformation is called the latent heat of vaporization 𝐿𝑉, see Fig.

below. When the gas releases heat and experiences a phase change from gas back to

liquid, the heat of transformation is called the latent heat of condensation and is

numerically equal to the latent heat of vaporization



136

Now, we consider the liquid to gas phase change. The attractive forces between

molecules in liquid form are stronger than in gas form because the average distance

between molecules is smaller in the liquid state. As described in the solid-to-liquid

phase transition, work must be done against these attractive forces. The latent heat

of vaporization is the amount of energy added to the molecules in liquid form to

accomplish this.

To understand the role of latent heat in phase changes, we calculate the energy

required to convert 1 g of ice at −50𝑜𝐶 into steam at 150𝑜𝐶.

Figure shows the results obtained when energy is added gradually to 1 g of ice. The

red curve of the figure is divided into the following five stages:

Stage A : Changing the temperature of ice from −50 to 0 ◦C: With a specific heat

of ice ci = 2,220 J/kg.C◦, the amount of heat added QA is:

𝑄𝐴 = 𝑚 𝑐𝑖 Δ𝑇 = (1 × 10−3) (2220) (0 − (−50)) = 111 𝐽

Stage B: Ice-water mixture remains at 0𝑜𝐶 (even heat is added): With a latent heat

of fusion 𝐿𝐹 = 3.33 × 105 𝐽/𝑘𝑔, the amount of heat added 𝑄𝐵until all of the ice

melts is:

𝑄𝐵 = 𝑚 𝐿𝐹 = (1 × 10−3) (3.33 × 105) = 3.3 × 102 𝐽



137

Stage C: Changing the temperature of water from 0 to 100 ◦C: With a specific heat

of water 𝑐𝑤 = 4186 𝐽/𝑘𝑔. 𝐶◦, the amount of heat added 𝑄𝑐is:

𝑄𝑐 = 𝑚 𝑐𝑤 Δ𝑇 = (1 × 10−3) (4186) (100 − 0) = 419 𝐽

Stage D: Water-steam mixture remains at 100◦C (even heat is added):

With a latent heat of vaporization 𝐿𝑉 = 2.26 × 106 𝐽/𝑘𝑔, the amount of

heat added 𝑄𝐷until all of the water evaporates is:

𝑄𝐷 = 𝑚 𝐿𝑉 = (1 × 10−3) (2.26 × 106) = 2.26 × 103 𝐽

Stage E: Changing the temperature of steam from 100 to 150◦C:

With a specific heat of steam 𝑐𝑠 = 2010 𝐽/𝑘𝑔. 𝐶, the amount of heat

added QE is:

𝑄𝐸 = 𝑚 𝑐𝑠 Δ𝑇 = (1 × 10−3) (2010) (150 − 100) = 101 𝐽

The total heat added to change 1gm of ice at −50◦C to steam at 150◦C is 𝑄𝑡𝑜𝑡 =

3224 𝐽. That is, if we cool 1 g of steam at 150◦C until we have ice at −50◦C, we

must remove 3224 𝐽 of heat.

Example 4

Find the quantity of heat required to convert 40 g of ice at –20°C into water at

20°C. Given 𝑳𝒊𝒄𝒆 = 𝟎. 𝟑𝟑𝟔 × 𝟏𝟎𝟔 𝑱/𝒌𝒈. Specific heat of ice = 2100 J/kg–K,

specific heat of water = 4200 J/kg–K



138

Solution:

Q1 Q2 Q3

T=-20 C T= 0C T=0 C T= 20C

The total energy = 𝑸𝒈𝒂𝒊𝒏 = 𝒎 𝒄𝒊𝒄𝒆 𝚫 𝑻 + 𝒎 𝑳𝒊𝒄𝒆 + 𝒎𝒄𝒘𝚫 𝑻 =

𝑄𝑔𝑎𝑖𝑛 = (40 × 10−3 × 2100 × (0 − (−20)) )

+ (40 × 10−3 × 0.336 × 106 )

+ (40 × 10−3 × 4200 × (20 − 0)) = 18480 J

Example 5

A calorimeter of heat capacity 100 J/K is at room temperature of 30°C. 100 g

of water at 40°C of specific heat 4200 J/kg–K is poured into the calorimeter.

What is the temperature of water in calorimeter?

Soln.

𝑄𝑔𝑎𝑖𝑛 (calorimeter ) = 𝑄𝑙𝑜𝑠𝑠(𝑤𝑎𝑡𝑒𝑟)

𝑀𝑐 𝑐𝑐𝑎𝑙𝑜 Δ𝑇 = 𝑀𝑤 𝑐𝑤 Δ𝑇

Heat Capacity of calorimeter C = 𝑀𝑐 𝑐𝑐𝑎𝑙𝑜 ( 𝐶 =𝑐

𝑀 )

𝐶𝑐𝑎𝑙𝑜 Δ𝑇 = 𝑀𝑤 𝑐𝑤 Δ𝑇

100 ∗ (𝑇𝑓 − 30) = 0.1 ∗ 4200 ∗ (40 − 𝑇𝑓)

100 𝑇𝑓 − 3000 = 16800 − 420𝑇𝑓

19800 = 520 𝑇𝑓

𝑻𝒇 = 𝟑𝟖. 𝟎𝟕°𝐂

Ice Ice

wate

r

water water

+20



139

Example 6:

5g ice at 0°C is mixed with 5g of steam at 100°C.

(a) What is the final temperature? (b) Find the mass of the total

Mass of water.

Hint: - Specific heat of water 𝒄 = 𝟏. 𝟎𝟎𝑪𝒂𝒍 . 𝒈⁄ .𝟎 𝑪

Laten heat of fusion for water 𝑳𝑭 = 𝟖𝟎 𝑪𝒂𝒍 𝒈⁄

Laten heat of vaporization for water 𝑳𝑽 = 𝟓𝟒𝟎 𝑪𝒂𝒍 𝒈⁄

𝑚 𝐿𝐹 𝑚 𝑐 Δ 𝑇

T= 0 C T= 0 C T=100 C

• The amount of heat energy required to convert 5 gm ice at 0 C to water at 100 C is

𝑄𝑔𝑎𝑖𝑛 (𝐼𝑐𝑒) = 𝑚 𝐿𝐹 + 𝑚 𝑐𝑤 Δ 𝑇 = (5 ∗ 80 ) + (5 ∗ 1 ∗ (100 − 0))

= 900 𝐶𝑎𝑙.

• But the amount of heat energy required to convert 5 gm steam at 100 C to water at

100 C is

𝑄𝑙𝑜𝑠𝑠(𝑠𝑡𝑒𝑎𝑚) = 𝑚 𝐿𝑉 = 5 ∗ 540 = 2700 𝐶𝑎𝑙.

We find that 𝑄𝑙𝑜𝑠𝑠(𝑠𝑡𝑒𝑎𝑚) ≫ 𝑄𝑔𝑎𝑖𝑛 (𝐼𝑐𝑒)

T=100 C 𝑚 𝐿𝑉 T=100 C

a) Therefore, the final phase will be (water + steam at T = 100 C)

b) To find the mass of steam that converts into water

Ice Ice+

water

water Steam

water



140

𝑄 = 𝑚 𝐿𝑉 𝑚 =𝑄

𝐿𝑉 =

900

540 = 1.6 𝑔.

Therefore, the total mass of water = 5 𝑔 + 1.6 𝑔 = 6.6 𝑔.

Example 7:

Steam at 100°C is passed into 1.1 kg of water contained in a calorimeter of water

equivalent 0.02 kg at 15°C till the temperature of the calorimeter and its

contents rises to 80°C. What is the mass of steam condensed? Latent heat of

steam = 536 cal/g.

𝑐𝑤 = 𝑐𝑐𝑎𝑙𝑜 = 1 𝐶𝑎𝑙/𝑔 𝑐 cice = 1 𝐶𝑎𝑙/𝑔 𝑐

𝑄𝑔𝑎𝑖𝑛 (calorimeter + Water ) = 𝑄𝑙𝑜𝑠𝑠(𝑆𝑡𝑒𝑎𝑚)

(𝑀𝑐𝑎𝑙𝑜 𝑐𝑐𝑎𝑙𝑜 + 𝑀𝑤 𝑐𝑤 ) Δ 𝑇 = 𝑀𝑖𝑐𝑒 𝐿𝑣 + 𝑀𝑖𝑐𝑒 𝑐𝑖𝑐𝑒Δ𝑇

(0.02 ∗ 1 + 1.1 ∗ 1 )(80 − 15 ) = 𝑀𝑖𝑐𝑒 [536 + (1 ∗ (100 − 80 )) ]

72.8 = 556 𝑀𝑖𝑐𝑒

𝑀𝑖𝑐𝑒 = 0.13 𝑘𝑔.

Signs for Heat and Work in Thermodynamics

We describe the energy relationships in any thermodynamic process in terms of:-

the quantity of heat Q added to the system and the work W done by the system.

Both Q and W may be positive, negative, or zero.

A positive value of Q represents heat flow into the system, with a corresponding

input of energy to it; negative Q represents heat flow out of the system.



141

A positive value of W represents work done by the

system against its surroundings, such as work

done by an expanding gas, and hence corresponds

to energy leaving the system. Negative W, such as

work done during compression of a gas in which

work is done on the gas by its surroundings,

represents energy entering the system.

5. Work Done During Volume Changes

assume we reduce the load from the piston in such a way that the piston will move

upward through a differential displacement 𝑑𝑠 with almost constant upward force

�⃗� as shown in Fig. below. From the definition of pressure, we have 𝐹 = 𝑃𝐴,

where A is the area of the piston. The differential work 𝑑𝑊 done by the gas during

the displacement is:

𝒅 𝑾 = �⃗⃗⃗� ⋅ 𝒅�⃗⃗� = 𝑭 𝒅𝒔 = 𝑷 𝑨 𝒅𝒔 = 𝑷 𝒅𝑽 = 𝑷 (𝑽𝒇 − 𝑽𝒊)

A confined gas in a cylinder at pressure P does work dW on a free piston as the gas expands from volume V to

volume V + dV because of a decreased load



142

If the gas expands, as in Fig. below, then 𝑑𝑉 is positive and the work done by the

gas is positive, whereas if the gas is compressed, 𝑑𝑉 is negative, indicating that the

work done by the gas is negative (which can be interpreted as work done on the gas).

When we remove an appreciable amount of load from the piston, the volume of the

gas changes from 𝑉𝑖 𝑡𝑜 𝑉𝑓, and the total work done by the gas is:

𝑾 = ∫ 𝒅 𝑾 = ∫ 𝑷 𝒅𝑽 𝑽𝒇

𝑽𝒊

𝟐

𝟏

During the change in volume of the gas, the pressure and temperature of the gas may

also change. To evaluate the integral in the last equation, we need to know how the

pressure varies with volume. For example, Fig. below indicates that the work done

by the gas is represented by the area under the PV diagram of the figure.

Hence the work done by a gas is equal to the area under P–V graph.

Following different cases are possible.

(i) Volume is constant



143

(ii) Volume increasing

(iii) Volume decreasing

(iv) Cyclic process



144

Example 8:

A gas is compressed at a constant pressure of 0.800 atm from 9.00 L to 2.00 L.

In the process, 400 J of energy leaves the gas by heat. (a) What is the work done

on the gas? (b) What is the change in its internal energy?

Solution

(a) The work done on the gas

𝑾 = −𝑷 𝚫𝑽 = −𝑷 (𝑽𝒇 − 𝑽𝒊) = −(𝟎. 𝟖 × 𝟏𝟎𝟓) (𝟐 − 𝟗) = +𝟓𝟔𝟕 𝑱

(b) The internal energy

𝚫 𝑼 = 𝑸 + 𝑾 = −𝟒𝟎𝟎 + 𝟓𝟔𝟕 = 𝟏𝟔𝟕 𝑱

Example 9:

Determine the work done on a fluid that expands from i to f as indicated in

Figure. (b) What If? How much work is performed on the fluid if it is

compressed from f to i along the same path?

Solution



145

6. Internal Energy and the First Law of Thermodynamics

Internal energy is one of the most important concepts in thermodynamics. Matter

consists of atoms and molecules, and these are made up of particles having kinetic

and potential energies. We tentatively define the internal energy of a system as the

sum of the kinetic energies of all of its constituent particles, plus the sum of all the

potential energies of interaction among these particles.

We use the symbol 𝑈 for internal energy. During a change of state of the system,

the internal energy may change from an initial value to a final value We denote the

change in internal energy as Δ𝑈 = 𝑈2 − 𝑈1.

When we add a quantity of heat Q to a system and the system does no work during

the process (so 𝑊 = 0 ), the internal energy increases by an amount equal to Q;

that is Δ 𝑈 = 𝑄, When a system does work W by expanding against its

surroundings and no heat is added during the process, energy leaves the system and

the internal energy decreases: W is positive, Q is zero, and Δ𝑈 = −𝑊. When both

heat transfer and work occur, the total change in internal energy is

𝚫𝑼 = 𝑼𝟐 − 𝑼𝟏 = 𝑸 − 𝑾

This is the first law of thermodynamics. It is a generalization of the principle of

conservation of energy to include energy transfer through heat as well as mechanical

work. As you will see in later chapters, this principle can be extended to ever-broader

classes of phenomena by identifying additional forms of energy and energy transfer.

In every situation in which it seems that the total energy in all known forms is not

conserved, it has been possible to identify a new form of energy such that the total

energy, including the new form, is conserved. There is energy associated with



146

electric fields, with magnetic fields, and, according to the theory of relativity, even

with mass itself.

7. Kinds of Thermodynamic Processes

In this section we describe four specific kinds of thermodynamic processes that

occur often in practical situations. These can be summarized briefly as “no heat

transfer” or adiabatic, “constant volume” or isochoric, “constant pressure” or

isobaric, and “constant temperature” or isothermal. For some of these processes we

can use a simplified form of the first law of thermodynamics.

7.1 Adiabatic Process

An adiabatic process (pronounced “ay-dee-ah-bat-ic”) is defined as one with no heat

transfer into or out of a system 𝑄 = 0; We can prevent heat flow either by

surrounding the system with thermally insulating material or by carrying out the

process so quickly that there is not enough time for appreciable heat flow. From the

first law we find that for every adiabatic process,

Q = 0 and Δ𝑈 = −𝑊

When a system expands adiabatically, W is positive (the system does work on its

surroundings), so is negative and the internal energy decreases. When a system is

compressed adiabatically, W is negative (work is done on the system by its

surroundings) and U increases. The compression stroke in an internal-combustion

engine is an approximately adiabatic process. The temperature rises as the air–fuel

mixture in the cylinder is compressed. The expansion of the burned fuel during the

power stroke is also an approximately adiabatic expansion with a drop in

temperature.

An example of a PV diagram and an Energy-Interaction diagram is shown below.



147

Adiabatic processes typically occur very quickly, such that the system has not

time to exchange heat with its environment. Every time you open a carbonated

beverage, the pressured gas quickly expands causing the temperature to drop.

This adiabatic cooling cause the water vapor in the gas to condense, creating a small

cloud. Another example in nature of an adiabatic process is the cooling of air in the

mountains. As wind brings air over a mountain range, the lower pressure cause the

air to expand, and thus cooling it. The cooling air often results in water vapor

condensation creating clouds, causing it to rain. If there is a prevailing wind

direction, this explains why on some mountain ranges there is more vegetation on

only one side. By the time the cloud gets to the other side of the mountain, it has

used up all of its moisture. Although, no processes are truly adiabatic (since they

depend of an either perfectly insulated system or a process of a very quick time

scale), often, it is a good approximation.



148

7.2 Isochoric Process

An isochoric process (pronounced “eye-so-kor-ic”) is a constant-volume process.

When the volume of a thermodynamic system is constant, it does no work on its

surroundings. Then and

𝑾 = 𝟎 𝚫𝐔 = 𝑸

When the volume of a system remains constant during a thermodynamic process,

the process is called isochoric. Consider a sealed container with a gas at equilibrium.

If the sealed container is then heated, the gas particles will start moving around

faster, exerting a greater pressure on the wall of the container. Setting the initial and

According to Ideal Gas law, when the initial and final volumes are the same, we get

the following relationship:

𝑷𝒊

𝑻𝒊=

𝑷𝒇

𝑻𝒇

Assume the piston of Fig. below is clamped to a fixed position to ensure an

isovolumetric process. In such a process, the value of the work done by the gas is

zero, i.e. W = 0, because the volume does not change. Thus, with this restriction and

the application of the first law of thermodynamics to an isovolumetric process.



149

Since the volume stays constant, no work is being done and only the heat entering

the system contributes to the change in internal energy. A physical example of this

process is illustrated below, in addition to the PV diagram and the Energy-

Interaction diagram that describes this particular process.

7.3 Isobaric Process

In isobaric process (pronounced “eye-so-bear-ic”) is a constant-pressure process.

In general, none of the three quantities Q, and W are all non-zero.in an isobaric

process, but calculating W is easy nonetheless.

𝑾 = 𝑷 𝚫 𝑽 = 𝑷 (𝑽𝒇 − 𝑽𝒊)

Assume the piston of Fig. below is free to move in such a way that it is always

in equilibrium under the effect of the net force from a gas pushing upwards and the

weight of the piston plus the force due to atmospheric pressure pushing downwards.

Then, an isobaric process could be established by transferring heat energy

Q to or from the gas by any mechanism. This transfer

causes the gas to expand or contract depending on the

sign of Q.



150

According to Ideal Gas law, when the initial and final pressures are the same, we get

the following relationship:

𝑽𝒊

𝑻𝒊=

𝑽𝒇

𝑻𝒇

7.4 Isothermal Process

An isothermal process is a constant-temperature process 𝑻 = 𝑪𝒐𝒏𝒔𝒕.. For a

process to be isothermal, any heat flow into or out of the system must occur slowly

enough that thermal equilibrium is maintained. In general, none of the quantities Δ𝑈,

Q, or W is nonzero in an isothermal process.

In some special cases the internal energy of a system depends only on its

temperature, not on its pressure or volume. The most familiar system having this

special property is an ideal gas, as we’ll discuss in the next section. For such systems,

if the temperature is constant, the internal energy is also constant;

𝚫𝑼 = 𝟎 and 𝑸 = 𝑾

That is, any energy entering the system as heat Q must leave it again as work W

done by the system.



151

Suppose that an ideal gas is allowed to expand at

constant temperature as described by the PV

diagram in Fig. below, According to Ideal Gas

law,, the curve is a hyperbola with the equation

𝑷𝑽 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕

𝑃𝑖 𝑉𝑖 = 𝑃𝑓 𝑉𝑓

Let us calculate the work done by the gas in the isothermal expansion from state 𝑖 to

state 𝑓, Because the gas is ideal and the process is quasistatic, we can use the

expression 𝑷𝑽 = 𝒏 𝑹𝑻 for each point on the path. Therefore, we have:

𝑊 = ∫ 𝑃 𝑑𝑉 𝑉𝑓

𝑉𝑖

= ∫ 𝑛 𝑅 𝑇

𝑉 𝑑𝑉

𝑉𝑓

𝑉𝑖

= 𝑛 𝑅 𝑇 ∫𝑑𝑉

𝑉

𝑉𝑓

𝑉𝑖

𝑊 = 𝑛 𝑅 𝑇 ln 𝑉 |𝑉𝑖

𝑉𝑓

𝑾 = 𝒏 𝑹 𝑻 𝐥𝐧 𝑽𝒇

𝑽𝒊 = 𝒏 𝑹 𝑻 𝐥𝐧

𝑷𝒊

𝑷𝒇

If the gas expands, the work W equals the positive of the shaded area under the PV

curve shown in Fig. above; this is because 𝑙𝑛(𝑉𝑓/𝑉𝑖) > 0. If the gas is compressed,

𝑉𝑓 < 𝑉𝑖, then 𝑙𝑛(𝑉𝑓/𝑉𝑖) < 0 and the work done is the negative of the area under

the 𝑃𝑉 curve.



152

Four different processes for a constant amount of an ideal gas, all starting at state a.

For the adiabatic process, 𝑄 = 0. for the isochoric process 𝑊 = 0 , and for the

isothermal process Δ𝑈 = 0 , the temperature increases only during the isobaric

expansion.



153

Example 10

The pV-diagram of Figure shows a series of thermodynamic processes. In

process ab, 150 J of heat is added to the system; in process bd, 600 J of heat is

added. Find (a) the internal energy change in process ab; (b) the internal energy

change in process abd (shown in light blue); and (c) the total heat added in

process acd (shown in dark blue).

Solution

𝑄𝑎𝑏 = 150 𝐽 𝑄𝑏𝑑 = 600 𝐽

(a) internal energy change in process ab

No volume change occurs during process ab, so the

system does no work:

Isochoric process 𝑊𝑎𝑏 = 0

Therefore Δ𝑈𝑎𝑏 = 𝑄𝑎𝑏 = 150 𝐽

(b) The internal energy change in process abd

Δ 𝑈𝑎𝑏𝑑 = Δ𝑈𝑎𝑏 + Δ𝑈𝑏𝑑

Δ 𝑈𝑎𝑏𝑑 = (𝑄𝑎𝑏 − 𝑊𝑎𝑏 ) + (𝑄𝑏𝑑 − 𝑊𝑏𝑑 )

To find Δ𝑈𝑎𝑏 = 𝑄𝑎𝑏 − 𝑊𝑎𝑏 = 150 − 0 = 150 𝐽

To find Δ𝑈𝑏𝑑

Isobaric process, the work done bd path

𝑊𝑏𝑑 = 𝑃 Δ𝑉 = 𝑃 (𝑉𝑑 − 𝑉𝑏) = 8 × 104 (5 − 2 ) × 10−3 = 240 𝐽

While the heat energy 𝑄𝑏𝑑 = 600 𝐽, then the change in the internal energy

Δ𝑈𝑏𝑑 = 𝑄𝑏𝑑 − 𝑊𝑏𝑑 = 600 − 240 = 360 𝐽

Therefore, the internal energy change in process abd

Δ 𝑈𝑎𝑏𝑑 = Δ𝑈𝑎𝑏 + Δ𝑈𝑏𝑑 = 150 + 360 = 510 𝐽



154

(c) the total heat added in process acd 𝑸𝒂𝒄𝒅 = 𝑾𝒂𝒄𝒅 + 𝚫𝑼𝒂𝒄𝒅

The internal energy 𝚫𝑼𝒂𝒄𝒅 = 𝚫𝑼𝒂𝒃𝒅 = 𝟓𝟏𝟎 𝑱. To find the work done 𝑾𝒂𝒄𝒅,

𝑾𝒂𝒄𝒅 = 𝑾𝒂𝒄 + 𝑾𝒄𝒅

For path ac (Isobaric Process) 𝑃 = 𝐶𝑜𝑛𝑠𝑡., the work done by system in the path

ac

𝑊𝑎𝑐 = 𝑃 Δ𝑉 = 𝑃 (𝑉𝑐 − 𝑉𝑎 ) = 3 × 104 × (5 − 2) × 10−3 = 90 𝐽

For path cd (Isochoric Process) 𝑉 = 𝐶𝑜𝑛𝑠𝑡., the work done by system in the path

cd is zero 𝑊𝑐𝑑 = 0 𝐽. Then

𝑾𝒂𝒄𝒅 = 𝑾𝒂𝒄 + 𝑾𝒄𝒅 = 𝟗𝟎 + 𝟎 = 𝟗𝟎 𝑱

Therefore, the total heat added in process acd 𝑸𝒂𝒄𝒅 = 𝑾𝒂𝒄𝒅 + 𝚫𝑼𝒂𝒄𝒅

𝑸𝒂𝒄𝒅 = 𝟗𝟎 + 𝟓𝟏𝟎 = 𝟔𝟎𝟎 𝐉

Example 11

At a constant pressure of 1 atm and a temperature of 𝟎𝒐𝑪, the heat fusion of

ice is 𝑳𝑭 = 𝟑. 𝟑𝟑 × 𝟏𝟎𝟓 𝑱/𝒌𝒈, the density of ice is 𝝆𝒊 = 𝟗𝟐𝟎 𝒌𝒈/𝒎𝟑, and the

density of liquid water is 𝝆𝒘 = 𝟏, 𝟎𝟎𝟎 𝒌𝒈/𝒎𝟑. (a) Find the work W done by

𝟏 𝒌𝒈 of ice that melts completely to water. (b) Find the change in internal

energy of this process.

Solution

Given that :

Isobaric process 𝑃 = 𝑐𝑜𝑛𝑠𝑡. = 1.013 × 105 𝑃𝑎 𝑇 = 0𝑜 𝐶 = 273 𝐾



155

the heat fusion of ice is 𝑳𝑭 = 𝟑. 𝟑𝟑 × 𝟏𝟎𝟓 𝑱/𝒌𝒈, the density of ice is 𝝆𝒊 =

𝟗𝟐𝟎 𝒌𝒈/𝒎𝟑 and 𝝆𝒘 = 𝟏, 𝟎𝟎𝟎 𝒌𝒈/𝒎𝟑.

(a) the work W done by 𝟏 𝒌𝒈 of ice that melts completely to water

𝑊 = 𝑃 Δ𝑉 = 𝑃 ( 𝑉𝑤 − 𝑉𝑖)

The initial volume of ice is:

𝑉𝑖𝑐𝑒 =𝑚

𝜌𝑖=

1

920 = 1.087 × 10−3 𝑚3

The final volume of water is:

𝑉𝑤 =𝑚

𝜌𝑤=

1

103 = 1.0 × 10−3 𝑚3

Then, the work W done by 𝟏 𝒌𝒈 of ice that melts completely to water

𝑊 = 𝑃 Δ𝑉 = 𝑃 ( 𝑉𝑤 − 𝑉𝑖) = 1.013 × 105 × (1.0 × 10−3 − 1.087 × 10−3)

= −8.8 𝐽

The minus sign appears because ice contracts when it melts.

b) The heat energy transferred to change the phase of 1 kg of ice to water is:

𝑄 = 𝑚 𝐿𝐹 = 1 𝐾𝑔 × 3.33 × 105 𝐽

𝑘𝑔= 3.33 × 105 J

Thus, from the first law of thermodynamics, we can find the change in internal

energy of this process as follows:

Δ𝑈 = 𝑄 − 𝑊 = 3.33 × 105 − ( −8.8 ) = 3.330088 × 105 𝐽

We see from parts (a) and (b) that |W| is less than 0.003% of Q in this process, i.e.

|𝑊| ≪ 𝑄. That is, the mechanical energy is negligible in comparison to the heat of

fusion. So, all the added heat of fusion shows up as an increase in the internal

energy.



156

Example 12

At a constant pressure of 1 atm, a movable piston encloses 1 kg of water with a

volume of 𝟏𝟎−𝟑 𝒎𝟑 and a temperature of 𝟏𝟎𝟎𝒐𝑪,. Heat is added from a

reservoir until the liquid water changes completely into steam of volume

𝟏. 𝟔𝟕𝟏 𝒎𝟑, see the figure. (a) How much work is done by the system (water +

steam) during the boiling process? (b) How much heat energy is added to the

system? (c) What is the change in the internal energy of the system?

𝑉𝑖 = 10−3𝑚3 , 𝑇𝑖 = 100𝑜𝐶 𝑉𝑓 = 1.671 𝑚3

Solution: (a) The work done by 1 kg of water that is converted completely into steam

under a constant pressure of 1 atm (1.01 × 105 𝑃𝑎) and a constant temperature of

100𝑜𝐶, is:

𝑊 = 𝑃 Δ𝑉 = 𝑃 ( 𝑉𝑓 − 𝑉𝑖) = (1.01 × 105 ) × (1.671 − 10−3) = 169 𝐾 𝐽

Since the heat of vaporization of water at atmospheric pressure is 𝐿𝑉 = 2.26 × 106

J/kg, the heat energy required to change the phase of 1 kg of water to steam will be:

𝑄 = 𝑚 𝐿𝑉 = 1 𝑘𝑔 × 2.26 × 106 = 2260 K J



157

(c ) the internal energy of the system

Δ𝑈 = 𝑄 − 𝑊 = 2260 − 169 = 2091 𝐾 𝐽

We see that about 92.5% of the heat energy goes into internal energy while the

remaining 7.5% goes into external work.

Example 13

The pressure in monoatomic gas increases linearly from 𝟒 ×

𝟏𝟎𝟓 𝑵𝒎–𝟐 𝒕𝒐 𝟖 × 𝟏𝟎𝟓 𝑵𝒎–𝟐 when its volume increases from

𝟎. 𝟐 𝒎𝟑 𝒕𝒐 𝟎. 𝟓 𝒎𝟑. Calculate. (i) Work done by the gas, (ii) Increase in the

internal energy, (iii) Amount of heat supplied,

Solution

(i) Work done by the gas = area under the

curve

𝑊 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐴𝐶𝐷𝐸 + 𝐴𝑟𝑒𝑎 𝐴𝐵𝐶

𝑊 = ((0.5 − 0.2 ) × (4 − 0) × 105 ) + (1

2 (0.5 −

0.2 ) × (8 − 4 ) × 105) = 1.8 × 105 𝐽

(ii) Increase in the internal energy

Δ𝑈 = 𝑛 𝐶𝑉 Δ T



158

For monoatomic gas 𝐶𝑉 =3

2 𝑅

Known that 𝑃𝑉 = 𝑛 𝑅 𝑇

Δ𝑈 = 𝑛 (3

2 𝑅) Δ T = 𝑛 (

3

2 𝑅) (T2 − T1) =

3

2 n (RT2 − RT1)

=3

2 (P2V2 − P1V1) =

3

2 ((0.5 × 8 × 105) − (0.2 × 4 × 105) )

= 4.8 × 105J

(iii) Amount of heat supplied,

𝑄 = Δ 𝑈 + 𝑊 = 4.8 × 105 + 1.8 × 105 = 6.6 × 105 𝐽

Example 14

When a system is taken from state a to state b, in figure along the path

𝒂 → 𝒄 → 𝒃, 𝟔𝟎 𝑱 of heat flow into the system, and 𝟑𝟎 𝑱 of work is done :

(i) How much heat flows into the system along the path 𝒂 → 𝒅 → 𝒃 if the work

is 𝟏𝟎 𝑱.

(ii) When the system is returned from b to a along the curved path, the work

done by the system is –20 J. Does the system absorb or liberate heat, and how

much?

(iii) If 𝑼𝒂 = 𝟎 𝒂𝒏𝒅 𝑼𝒅 = 𝟐𝟐 𝑱, find the heat absorbed in the process

𝒂 → 𝒅 𝒂𝒏𝒅 𝒅 → 𝒃.



159

Solution

For path 𝒂 → 𝒄 → 𝒃 𝑸 = 𝟔𝟎 𝑱 𝒂𝒏𝒅 𝑾 = 𝟑𝟎 𝑱

Δ𝑈𝑎𝑐𝑑 = 𝑄 − 𝑊 = 60 − 30 = 30 𝐽

Δ𝑈𝑎𝑐𝑑 = Δ𝑈𝑎𝑑𝑏 = 30 𝐽

(i) The heat flows into the system along the path 𝑎 → 𝑑 → 𝑏 if the work is

10 𝐽.

𝑄 = Δ𝑈𝑎𝑑𝑏 = +𝑊𝑎𝑑𝑏 = 30 + 10 = 40 𝐽

(ii) Along the curved path b, a 𝑊𝑏𝑎 = – 𝟐𝟎 𝐉.

𝑄 = Δ𝑈𝑏𝑎 + 𝑊𝑏𝑎 = (𝑈𝑎 − 𝑈𝑏) + 𝑊𝑏𝑎 = (0 − 30 ) − 20 = −50 𝐽

heat flows out the system

(iii) 𝑄𝑎𝑑 = ⋯

𝑄𝑎𝑑 = 𝑊𝑎𝑑 + Δ𝑈𝑎𝑑 = 𝑊𝑎𝑑 + (𝑈𝑑 − 𝑈𝑎) = 10 + (22 − 0 ) = 32 𝐽

𝑄𝑑𝑏 = ⋯

𝑈𝑏 = 𝑈𝑎𝑏 − 𝑈𝑎 = 30 − 0 = 30 𝐽

𝑄𝑑𝑏 = Δ 𝑈𝑑𝑏 + 𝑊𝑑𝑏 = (𝑈𝑏 − 𝑈𝑑) + 𝑊𝑑𝑏 = (30 − 22) − 0 = 8 𝐽



160

Example 14 Figure shows a process ABCA performed on an ideal gas. Find

the net work given to the system during the process.

Solution

Since the process is cyclic, hence the change in internal energy is zero.

The heat given to the system is then equal to the work done by it. The work done in

part AB is 𝑊1 = 0 (the volume remains constant).

The part BC represents an isothermal process so that the work done by the gas

during this part is

𝑊2 = 𝑛 𝑅 𝑇2 ln𝑉2

𝑉1

During the part CA 𝑉 𝑇 So, 𝑉/𝑇 is constant and hence 𝑃 =𝑛 𝑅 𝑇

𝑉 = 𝐶𝑜𝑛𝑠𝑡.

The work done by the gas during the part CA

𝑊3 = 𝑃 (𝑉1 − 𝑉2 ) = 𝑃1 𝑉1 − 𝑃2𝑉2 = 𝑛 𝑅 (𝑇1 − 𝑇2)

The net work done by the gas in the process ABCA is

𝑊 = 𝑊1 + 𝑊2 + 𝑊3 = 0 + 𝑛𝑅 𝑇2 ln𝑉2

𝑉1 + 𝑛 𝑅 (𝑇1 − 𝑇2)

𝑊 = 𝑛 𝑅 ( 𝑇2 ln𝑉2

𝑉1 − (𝑇1 − 𝑇2 )).



161

Sheet (VI)

1- (a) An ideal gas is taken from an initial state 𝑖 to a final state 𝑓, as shown in

Fig. below. Find the work done by the gas along the three paths 𝑖𝑎𝑓, if, and

𝑖𝑏𝑓. (b) Answer part (a) if the gas is taken from 𝑓 to 𝑖.

2- An ideal gas of 2 𝑘𝑚𝑜𝑙 is carried around the thermodynamic cycle as shown

in Fig. . The cycle consists of three parts; the isothermal expansion 𝑎𝑏 at 𝑇 =

300 𝐾 an isobaric compression 𝑏𝑐, and an isovolumetric increase in pressure

𝑐𝑎. (a) When 𝑃𝑎 = 4 𝑎𝑡𝑚 𝑎𝑛𝑑 𝑃𝑏 = 1 atm, then find the work done by the

gas per cycle. (b) Answer part (a) when the direction of the cycle is reversed.

3- An amount of work of 100 J is done on a system, and 100 cal of heat are

extracted from it. In light of the first law of thermodynamics, what are the

values (including algebraic signs) of: (a) W, (b) Q, and (c) Δ𝑈 ?

4- An ideal helium gas of 1 kmol is carried around the thermodynamic cycle as

shown in Fig.. The path ab is isothermal, with 𝑃𝑎 = 2 𝑎𝑡𝑚, 𝑃𝑏 = 1 𝑎𝑡𝑚,



162

and 𝑉𝑎 = 22.4 𝑚3. (a) What are the values of 𝑇𝑎, 𝑉𝑏, 𝑎𝑛𝑑 𝑇𝑐? (b) How much

work is done by the gas in this cycle?

5- An ideal gas of one kmol does 4,000 J of work as it expands isothermally to a

volume of 12 × 10−3 𝑚3 that has a final pressure of 2 atm. (a) What is the

temperature of the gas? (b) What is the initial volume of the gas?

6- A fluid is carried through the cycle 𝑎𝑏𝑐𝑑 as shown in Fig. . How much work

(in kilojoules) is done by the fluid during: (a) the isobaric expansion ab, (b)

the isovolumetric process bc, and (c) the isobaric compression cd? (d) What

is the net amount of heat transferred to work during the cycle abcd?

7- An ideal gas has an initial temperature of 27 ◦C and an initial volume of 1 𝑚3.

An isobaric expansion of the gas to a new volume of 3 𝑚3 is achieved by

adding 9,500 J of heat at a constant pressure of 3,000 Pa. (a) Determine the



163

work done by the gas during expansion. (b) What is the change in the internal

energy of the gas? (c) Find the final temperature of the gas.

8- An ideal gas is taken from a to c along the curved path in Fig. . Along this

path, the work done by the gas is 𝑊𝑎𝑐 = 15 𝐽 and the heat added to the gas

is 𝑄𝑎𝑐 = 43 𝐽. In addition, the work done along path 𝑎𝑏𝑐 is 𝑊𝑎𝑏𝑐 = 34 𝐽.

(a) What is the change in internal energy of the gas Δ𝑈𝑎𝑐for path ac? (b) What

is 𝑄𝑎𝑏𝑐 for path 𝑎𝑏𝑐? (c) What is 𝑊𝑐𝑑𝑎 for path cda? (d) What is 𝑄𝑐𝑑𝑎 for path

cda?

9- A 100 g block of ice at 0 ◦C is added to 400 g of water at 30 ◦C. Assuming we

have a perfectly insulated calorimeter for this mixture, what will be its final

temperature when all of the ice has melted?

10- A copper calorimeter has a mass 𝑚𝑐 = 100 𝑔. The calorimeter contains

water of mass 𝑚𝑤 = 500 𝑔 at a temperature of 20𝑜𝐶. How much steam must

be condensed into water if the final temperature of the mixture is to reach 50 ◦C?

Assume the specific heat of copper is 𝑐𝑐 = 840 𝐽/𝑘𝑔. 𝐶◦, the specific heat of

water is 𝑐𝑤 = 4,186 𝐽/𝑘𝑔. 𝐶, and the latent heat of condensation of steam is

𝐿𝑉 = 2.26 × 106 𝐽/𝑘𝑔.



164

Useful Data

165


166


167


168

Bibliography 1. University Physics with Modern Physics Paperback – International Edition,

Francis Weston Sears Hugh D. Young, Roger A. Freedman, A. Lewis Ford, 15

edition (2019).

2.College Physics 11th Edition by Raymond A. Serway , Chris Vuille, Global

edition, (2021).

3. Physics: Principles with Applications, Global Edition, Douglas C. Giancoli, ,

7th edition, (2015).

4. Principles of Physics For Scientists and Engineers by Radi, Hafez A.,

Rasmussen, John O. ISSN 2192-4791.

5. Holt Physics STUDENT EDITION Unstated by Raymond A.

Serway , Jerry S. Faughn, 2006 Edition.

Principles of Physics ( I )

By Dr. Abdel Magied DIAB

Physics ( I )

Documents

Lectures Notes in Physics (I) PHY 107