The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4
Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on
30.01.2020 and it has been decided to implement it from academic year 2020-21
PHYSICS
Standard XII
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2020
Maharashtra State Bureau of Textbook Production and
Curriculum Research, Pune.
First Edition : © Maharashtra State Bureau of Textbook Production and
2020
Curriculum Research, Pune - 411 004.
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and Curriculum Research reserves all rights relating to
the book. No part of this book should be reproduced
without the written permission of the Director, Maharashtra
State Bureau of Textbook Production and Curriculum
Research, ‘Balbharati’, Senapati Bapat Marg, Pune 411004.
Committee: Illustration
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The Constitution of India
Preamble
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solemnly resolved to constitute India into a
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IN OUR CONSTITUENT ASSEMBLY this
twenty-sixth day of November, 1949, do HEREBY
ADOPT, ENACT AND GIVE TO OURSELVES
THIS CONSTITUTION.
NATIONAL ANTHEM
Preface
Dear Students,
With great pleasure we place this detailed text book on basic physics in the hands of
the young generation. This is not only a textbook of physics for XIIth standard, but contains
material that will be useful for the reader for self study.
This textbook aims to give the student a broad perspective to look into the physics
aspect in various phenomena they experience. The National Curriculum Framework (NCR)
was formulated in the year 2005, followed by the State Curriculum Framework (SCF) in
2010. Based on the given two frameworks, reconstruction of the curriculum and preparation
of a revised syllabus has been undertaken which will be introduced from the academic year
2020-21. The textbook incorporating the revised syllabus has been prepared and designed
by the Maharashtra State Bureau of Textbook Production and Curriculum Research,
(Balbharati), Pune.
The objective of bringing out this book is to prepare students to observe and analyse
various physical phenomena is the world around them and prepare a solid foundation for
those who aspire for admission to professional courses through competitive examinations.
Most of the chapters in this book assume background knowledge of the subject covered by
the text book for XIth Standard, and care has been taken of mentioning this in the appropriate
sections of the book. The book is not in the form of handy notes but embodies a good
historical background and in depth discussion as well. A number of solved examples in
every chapter and exercises at the end of each one of them are included with a view that
students will acquire proficiency and also will get enlightened after solving the exercises.
Physics is a highly conceptual subject. Problem solving will enable students understand the
underlying concepts. For students who want more, boxes entitled ‘Do you know?’ have been
included at a number of places.
If you read the book carefully and solve the exercises in each chapter, you will
be well prepared to face the challenges of this competitive world and pave the way for a
successful career ahead.
The efforts taken to prepare the textbook will prove to be worthwhile if you read the
textbook and understand the subject. We hope it will be a wonderful learning experience for
you and an illuminating text material for teachers too.
Pune (Vivek Gosavi)
Date : 21 February, 2020 Director
Bhartiya Saur : 2 Phalguna, 1941
Maharashtra State Bureau of Texbook
Production and Curriculum Research, Pune
- For Teachers -
Dear Teachers, introduced in a graded manner to facilitate
We are happy to introduce the revised knowledge building.
P 'Error in measurements' is an important
textbook of Physics for XIIth standard. This topic in physics. Please ask the students to
book is a sincere attempt to follow the use this in estimating errors in their
maxims of teaching as well as develop a measurements. This must become an
‘constructivist’ approach to enhance the integral part of laboratory practices.
quality of learning. The demand for more P Major concepts of physics have a scientific
activity based, experiential and innovative base. Encourage group work, learning
learning opportunities is the need of the through each other’s help, etc. Facilitate
hour. The present curriculum has been peer learning as much as possible by
restructured so as to bridge the credibility reorganizing the class structure frequently.
gap that exists between what is taught and P Do not use the boxes titled ‘Do you know?’
what students learn from direct experience or ‘Use your brain power’ for evaluation.
in the outside world. Guidelines provided However, teachers must ensure that students
below will help to enrich the teaching- read this extra information and think about
learning process and achieve the desired the questions posed.
learning outcomes. P For evaluation, equal weightage should be
P To begin with, get familiar with the assigned to all the topics. Use different
textbook yourself, and encourage the combinations of questions. Stereotype
students to read each chapter carefully. questions should be avoided.
P The present book has been prepared for P Use Q.R. Code given in the textbook. Keep
constructivist and activity-based teaching, checking the Q.R. Code for updated
including problem solving exercises. information. Certain important links, websites
P Use teaching aids as required for proper have been given for references. Also a list
understanding of the subject. of reference books is given. Teachers as well
P Do not finish the chapter in short. However, as the students can use these references for
in the view of insufficient lectures, standard extra reading and in-depth understanding of
derivations may be left to the students for the subject.
self study. Problem sloving must be given Best wishes for a wonderful teaching
due importance. experience!
P Follow the order of the chapters strictly as
listed in the contents because the units are
References:
1. Fundamentals of Physics - Halliday, Resnick, Walker; John Wiley (Sixth ed.).
2. Sears and Zeemansky's University Physics - Young and Freedman, Pearson Education (12th ed.)
3. Physics for Scientists and Engineers - Lawrence S. Lerner; Jones and Bartlett Publishers, UK.
Front Cover : Picture shows part of Indus 2, Synchrotron radiation source (electron accelerator) at
RRCAT, Department of Atomic Energy, Govt. of India, Indore. Indus offers several research
opportunities. The photoelectron spectroscopy beamline is also seen.
Picture credit : Director, RRCAT, Indore. The permission to reproduce these pictures by Director,
RRCAT, DAE, Govt. of India is gratefully acknowledged.
Back Cover : Transmission Electron Microscope is based on De Broglie's hypothesis. TEM picture
shows a carbon nanotube filled with water showing the miniscus formed due to surface tension. Other
picture shows crystallites of LaB6 and the electron diffraction pattern (spot pattern) of the crystallite.
Picture credit : Dr. Dilip Joag, Savitribai Phule Pune University. Pune
DISCLAIMER Note : All attempts have been made to contact copy right/s (©) but we have not heard from them. We will be
pleased to acknowledge the copy right holder (s) in our next edition if we learn from them.
Competency Statements :
Standard XII
Area/ Competency Statements
Unit/ After studying the content in Textbook students would be able to....
Lesson
• Distinguish between centrifugal and centripetal forces.
Unit I • Visualize the concepts of moment of inertia of an object.
Ratational Motion and Mechanical • Relate moment of inertia of a body with its angular momentum.
• Differentiate between translational and rotational motions of rolling objects.
Properties of fluids • Relate the pressure of a fluid to the depth below its surface.
• Explain the measurement of atmospheric pressure by using a barometer.
Unit II • Use Pascal's law to explain the working of a hydraulic lift and hydraulic brakes.
Kinetic theory and • Relate the surface energy of a fluid with its surface tension.
Thermodynamics • Distinguish between fluids which show capillary rise and fall.
• Identify processes in daily life where surface tension plays a major role.
Unit III • Explain the role of viscosity in everyday life.
Oscillations and waves • Differentiate between streamline flow and turbulent flow.
• Relate various gas laws to form ideal gas equation.
• Distinguish between ideal gas and a real gas.
• Visualise mean free path as a function of various parameters..
• Obtain degrees of freedom of a diatomic molecule.
• Apply law of equipartition of energy to monatomic and diatomic molecules.
• Compare emission of thermal radiation from a body with black body radiation.
• Apply Stefan’s law of radiation to hot bodies .
• Identify thermodynamic process in every day life.
• Relate mechanical work and thermodynamic work.
• Differentiate between different types of thermodynamic processes.
• Explain the working of heat engine, refrigerator and air conditioner.
• Identify periodic motion and simple harmonic motion.
• Obtain the laws of motion for simple pendulum.
• Visualize damped oscillations.
• Apply wave theory to understand the phenomena of reflection, refraction, interference and
diffraction.
• Visualize polarized and unpolarized light.
• Apply concepts of diffraction to calculate the resolving power.
• Distinguish between the stationary waves in pipes with open and closed ends.
• Verify laws of vibrating string using a sonometer.
• Explain the physics involved in musical instruments.
Unit IV • Use Gaus's law to obtain the electric field for a charge distribution.
Electrostatics and electric current • Relate potential energy to work done to establish a charge distribution.
• Determine the electrostatic potential for a given charge distribution.
• Distingusih between conductors and insulators.
• Visualize polarization of dielectrics.
• Categorize dielectrics based on molecular properties.
• Know the effect of dielectric material used between the plates of a capacitor on its capacitance.
• Apply Kirchhoff’s laws to determine the current in different branches of a circuit.
• Find the value of an unknown resistance by using a meter bridge.
• Find the emf and internal resistance of a cell using potentiometer.
• Convert galvanometer into voltmeter and ammeter by using a suitable resistor.
Unit V • Realize that Lorentz force law is the basis for defining unit of magnetic field.
Magnetism • Visualize cyclotron motion of a charged particle in a magnetic field.
• Analyze and calculate magnetic force on a straight and arbitrarily shaped current carrying
Unit VI
Modern Physics wires and a closed wire circuit.
• Apply the Biot-Savart law to calculate the magnetic field produced by various distributions
of currents.
• Use Ampere’s law to get magnetic fields produced by a current distribution.
• Compare gravitational, magnetic and electrostatic potentials.
• Distinguish between paramagnitic, diamagnetic and ferromagnetic materials.
• Relate the concept of flux to experiments of Faraday and Henry.
• Relate Lenz’s law to the conservation of energy.
• Visualize the concept of eddy currents.
• Determine the mutual inductance of a given pair of coils.
• Apply laws of induction to explain the working of a generator.
• Establish a relation between the power dissipated by an AC current in a resistor and the value
of the rms current.
• Visualize the concept of phases to represent AC current.
• Explain the passage of AC current through circuits having resistors, capacitors and inductors.
• Explain the concept of resonance in LCR circuits.
• Establish validity of particle nature of light from experimental results.
• Determine the necessary wavelength range of radiation to obtain photocurrent from given
metals.
• Visualize the dual nature of matter and dual nature of light.
• Apply the wave nature of electrons to illustrate how better resolution can be obtained with
an electron microscope.
• Check the correctness of different atomic models by comparing results of various experiments.
• Identify the constituents of atomic nuclei.
• Differentiate between electromagnetic and atomic forces.
• Obtain the age of a radioactive sample from its activity.
• Judge the importance of nuclear power.
• Explain use of p-n junction diode as a rectifier.
• Find applications of special purpose diodes for every day use.
• Explain working of solar cell, LED and photodiode.
• Relate the p-n junction diode and special purpose diodes.
• Realize transistor as an important building block of electronic circuits, analyze situations in
which transistor can be used.
CONTENTS Sr. No Title Page No
1 Rotational Dynamics 1-25
2 Mechanical Properties of Fluids 26-55
3 Kinetic Theory of Gases and Radiation 56-74
4 Thermodynamics 75-108
5 Oscillations 109-130
6 Superposition of Waves 131-157
7 158-185
8 Wave Optics 186-213
9 Electrostatics 214-229
10 Current Electricity 230-250
11 Magnetic Fields due to Electric Current 251-264
12 Magnetic Materials 265-287
13 Electromagnetic induction 288-305
14 AC Circuits 306-323
15 Dual Nature of Radiation and Matter 324-343
16 Structure of Atoms and Nuclei 344-364
Semiconductor Devices
1. Rotational Dynamics
Can you recall? of the right hand along the sense of rotation,
1. What is circular motion? with the thumb outstretched. Tohf eωo.utstretched
2. What is the concept of centre of mass? thumb then gives the direction
3. What are kinematical equations of
motion?
4. Do you know real and pseudo forces,
their origin and applications?
1.1 Introduction: Fig. 1.1: Directions of angular velocity.
Circular motion is an essential part of our
If T is period of circular motion or periodic
daily life. Every day we come across several 2S
revolving or rotating (rigid) objects. During time and n is the frequency, Z 2S n ciTrcular
revolution, the object (every particle in the Uniform circular motion: During
object) undergoes circular motion about some
point outside the object or about some other motion if the speed of the particle remains
object, while during rotation the motion is about
an axis of rotation passing through the object. constant, it is called Uniform Circular Motion
1.2 Characteristics of Circular Motion:
1) It is an accelerated motion: As the (UCM). In this case, only the direction of its
direction of velocity changes at every velocity changes at every instant in such a way
instant, it is an accelerated motion.
2) It is a periodic motion: During the motion, that the velocity is always tangential to the
the particle repeats its path along the same
trajectory. Thus, the motion is periodic. path. The acceleration responsible afor r this 2irs
the centripetal or radial acceleration Z
iFt(soaowlroaanUrgdCsZ−Mr2trh,),eihtsvecrn2emcnetarvgceZnailotl.uefddItetcheieisnstarcciliopwrcneautsaytlaals.nr tdmiareonctditoeidnt
1.2.1 Kinematics of Circular Motion:
As seen in XIth Std, in order to describe
a circular motion, we use the quantities
aZngudldaTtr displacement θ , angular DvelocddiZtty
and angular acceleration
which are v =anddastlogaonudsaccteoleradtiisopnlaace=mdednvt t
s , velocity
Fig. 1.2: Directions of linear velocity and
used in translational motion. acceleration.
v ZA ulsro,wthheerteanωgeinsttihael velocity is given by Illustration: Circular motion of any particle
angular velocity. of a fan rotating uniformly.
Here, the position vector r is the radius Non-uniform circular motion: When a fan is
switched ON or OFF, the speeds of particles
vector from the centre of the circular motion. of the fan go on increasing or decreasing
The mDairgenctiitoundeofofωvisisavlw=ayωs r. for some time, however their directions are
along always tangential to their circular trajectories.
the axis of
rotation and is given by the ωrig,hctu-rhlatnhde thumb
rule. To know the direction of fingers
1
During this time, it is a non-uniform circular always change only the direction of ω and
motion. As othrervaedlioaclitaycciselsetrialltitoanngaenr tiiasl,stthilel never its magnitude thereby continuously
centripetal
changing athneacpclaelneeraotifonroatatpioenrp.en(Tdhicius laisr
there. However, for onfoanr-uisninfoortmconcsitracnutl.ar similar to
motion, the magnitude
to velocity v changing only its direction).
The acceleration responsible for changing acceleration α
of rotation, all
the magnitude of velocity is directed along If the angular Tis, cZonasntdaDnt
along the axis
or opposite to the velocity, hence always and
tangential aanTd. is called as tangential will be directed along the axis. This makes it
acceleration velocity v
possible to use scalar notation and write the
As magnitude of tangential kinematical equations of motion analogous to
ims octihoann, gthinegcodrurersinpgondainngoann-ugnuilfaorrmveloccirictyulωar those for translational motion as given in the
table 1 at the end of the topic.
is also changing at every DinstadndZtt. This is due to Example 1: A fan is rotating at 90 rpm.
the angular acceleration
It is then switched OFF. It stops after 21
Though the motion is non-uniform, the revolutions. Calculate the time taken by it
particles are sotfillαinisthsetilslamaleonpglatnhee. Hence, to stop assuming that the frictional torque
the direction axis of
is constant.
irdtoiritsaectoitopionpn.osFoifoterωtionwcthrheaiatlesoidnf ugωrisn.pgeedde,cirteaissianlgonspgetehde, Solution: rad
s
n0 90 rpm 1.5 rps ?Z0 2S n0 3S
The angle through which the blades of
the fan move while stopping is θ = 2πN
= 2π (21) = 42 π rad, ω = 0 (fan stops).
Using equations analogous to kinematical
equations of motion
Fig. 1.3: Direction of angular acceleration. D Z Z0 Z 2 Z02
t 2T
Do you know? ? 0 3S 0 3S 2 28 s
If the angular acceleration α is along t 2 42S ?t
any direction other than axial, it will have Remark: One can also use the unit
‘revolution’ for angle and get rid of π
a component perpendicular to otfheω axis.
Thus, it will change the direction also, throughout (for such data). In this case,
Z0 1.5 rps and T 21 rev.
wωhiicshawlwilalycshpaenrgpeenthdeicpullaanr etoofthreotpaltaionne as
of 1.2.2 Dynamics of Circular Motion
rotatiIof n.α is (Centripetal Force and Centrifugal Force):
constant in i) Centripetal force (CPF): As seen above,
magnitude, the acceleration responsible for circular
but always marotio nZi2srth. e centripetal or radial acceleration
The force providing this
tpoerpωen, diitcuwlailrl
acceleration ris the centripetal or radial force,
CPF mZ 2
2
Remember this It must be understood that centrifugal
force is a non-real force, but NOT an
(i) The word centripetal is NOT the name imaginary force. Remember, before the merry-
go-round reaches its uniform speed, you were
or type of that force (like gravitational really experiencing an outward pull (because,
centrifugal force is greater than the resultant
force, nuclear force, etc). It is the force towards the centre). A force measuring
instrument can record it as well.
adjective or property of that force
On reaching the uniform speed, in the
saying that the direction of this force frame of reference of merry-go-round, this
centrifugal force exactly balances the resultant
is along the radius and towards centre of all the real forces. The resultant force in
that frame of reference is thus zero. Thus, only
(centre seeking). in such a frame of reference we can say that
the centrifugal force balances the centripetal
(ii) While performing circular or rotational force. It must be remembered that in this case,
centrifugal force means the ‘net pseudo force’
motion, the resultant of all the real and centripetal force means the ‘resultant of
all the real forces’.
forces acting upon the body is (or, must
There are two ways of writing force
be) towards the centre, hence we call equation for a circular motion:
Resultant force mZ2r or
this resultant force to be centripetal
mZ2r ¦ real forces 0
force. Under the action of this resultant
Activity
force, the direction of the velocity is
Attach a suitable mass to spring balance so
always maintained tangential to the that it stretches by about half is capacity.
Now whirl the spring balance so that the
circular track. mass performs a horizontal motion. You will
The vice versa need not be true, notice that the balance now reads more mass
for the same mass. Can you explain this?
i.e., the resultant force directed towards
1.3 Applications of Uniform Circular Motion:
the centre may not always result into a 1.3.1 Vehicle Along a Horizontal Circular
Track:
circular motion. (In the Chapter 7 you
Figure 1.4 shows vertical section of a car
will know that during an s.h.m. also the on a horizontal circular track of radius r. Plane
of figure is a vertical plane, perpendicular to
force is always directed to the centre of the track but includes only centre C of the
track. Forces acting on the car (considered
the motion). For a motion to be circular, to be a particle) are (i) weight mg, vertically
correspondingly matching tangential downwards, (ii) normal reaction N, vertically
upwards that balances the weight mg and (iii)
velocity is also essential.
(iii) Obviously, this discussion is in an
inertial frame of reference in which
we are observing that the body is
performing a circular motion.
(iv) In magnitude, centripetal force
mrZ 2 mv 2 mvZ
r
ii) Centrifugal force (c.f.f.):
Visualize yourself on a merry-go-round
rotating uniformly. If you close your eyes, you
will not know that you are performing a circular
motion but you will feel that you are at rest. In
order to explain that you are at rest, you need
to consider a force equal in magnitude to the
arewsualytafnrot mreathl efocrecnet,reb.uTt hdiisrefcotrecde,op pomsZite2r,i.e.,
is
the centrifugal (away from the centre) force. It
is a pseudo force arising due to the centripetal
acceleration of the frame of reference.
3
force of static friction fs between road and the but above it. Thus, the frictional force
tyres. This is static friction because it prevents and the centrifugal force result into a
torque which may topple the vehicle
the vehicle from outward slipping or skidding. (even a two wheeler).
(ii) For a two wheeler, it is a must for
This is the resultant force which is centripetal. the rider to incline with respect to the
vertical to prevent toppling.
Use your brain power
Fig. 1.4: Vehicle on a horizontal road. (I) Obtain the condition for not toppling
for a four-wheeler. On what factors
While working in the frame of reference does it depend, and in what way? Think
about the normal reactions – where are
attached to the vehicle, it balances the those and how much are those! What
is the recommendation on loading the
centrifugal force. mrZ 2 mv 2 vehicle for not toppling easily? If a
? mg N and fs r vehicle topples while turning, which
wheels leave the contact? Why? How
? fs rZ 2 v2 does it affect the tyres? What is the
N g rg recommendation for this?
For a given track, radius r is constant. For (II) Determine the angle to be made with
the vertical by a two wheeler rider while
given vehicle, mg = N is constant. Thus, as the turning on a horizontal track.
speed v increases, the force of static friction fs Hint: For both (I) and (II) above, find the
torque that balances the torque due
also increases. However, fs has an upper limit to centrifugal force and torque due to
static friction force.
fs max Ps .N , where µs is the coefficient of
(III)We have mentioned about static friction
static friction between road and tyres of the between road and the tyres. Why is it
static? What about the kinetic friction
vehicle. This imposes an upper limit to the between road and the tyres?
speed v. (IV) What do you do if your vehicle is
trapped on a slippery or a sandy road?
At the maximum possible speed vs, we can What is the physics involved?
write
1.3.2 Well (or Wall) of Death: (मौत का कुआँ):
f s max Ps v2 ? v max P s rg This is a vertical cylindrical wall of radius
N max
rg r inside which a vehicle is driven in horizontal
circles. This can be seen while performing
Do you know? stunts.
(i) In the discussion till now, we had As shown in the Fig. 1.5, the forces acting
assumed the vehicle to be a point. on the vehicle (assumed to be a point) are (i)
In reality, if it is a four wheeler, the Normal reaction N acting horizontally and
resultant normal reaction is due to all
the four tyres. Normal reactions at all
the four tyres are never equal while
undergoing circular motion. Also, the
centrifugal force acts through the centre
of mass, which is not at the ground level,
4
Fig. 1.5: Well of death. due to the weight. What about a four-
wheeler?
towards the centre, (ii) Weight mg acting (iv) In this case, the angle made by the road
surface with the horizontal is 90°, i.e., if
vertically downwards, and (iii) Force of static the road is banked at 90°, it imposes a
lower limit on the turning speed. In the
friction fs acting vertically upwards between previous sub-section we saw that for an
vertical wall and the tyres. It is static friction unbanked (banking angle 0) road there
is an upper limit for the turning speed.
because it has to prevent the downward It means that for any other banking
angle (0 < θ < 90°), the turning speed
slipping. Its magnitude is equal to mg, as this will have the upper as well as the lower
limit.
is the only upward force. Example 2: A motor cyclist (to be treated
Normal reaction N is thus the resultant as a point mass) is to undertake horizontal
centripetal force (or the only force that can circles inside the cylindrical wall of a well
balance the centrifugal force). Thus, in of inner radius 4 m. Coefficient of static
magnitude, mv2 and mg friction between the tyres and the wall is
r
N mrZ 2 fs 0.4. Calculate the minimum speed and
frequency necessary to perform this stunt.
Force of static friction fs is always less than
(Use g = 10 m/s2)
or equal to µs N.
Solution:
§ mv 2 ·
? fs d PsN ? mg d Ps ¨ r ¸ v min rg 4 u10 10 m s 1 and
© ¹ nmin 0.4
Ps
?g d Psv2 ?v2 t rg v min 10 # 0.4 rev s 1
r Ps 2S r 2uS u4
rg
? v min Ps 1.3.3 Vehicle on a Banked Road:
As seen earlier, while taking a turn on
Remember this a horizontal road, the force of static friction
between the tyres of the vehicle and the road
(i) N should always be equal to mv 2 provides the necessary centripetal force (or
r balances the centrifugal force). However, the
mv 2 mg frictional force is having an upper limit. Also,
? N min min Ps its value is usually not constant as the road
surface is not uniform. Thus, in real life, we
r should not depend upon it, as far as possible.
For this purpose, the surfaces of curved roads
(ii) In this case, fs = isN is valid only for the are tilted with the horizontal with some angle
minimum speed as fs should always be θ . This is called banking of a road or the road
equal to mg. is said to be banked.
(iii) During the derivation, the vehicle is Figure 1.6 Shows the vertical section of
a vehicle on a curved road of radius r banked
assumed to be a particle. In reality, it
is not so. During revolutions in such
a well, a two-wheeler rider is never
horizontal, else, the torque due to her/
his weight will topple her/him. Think
of the torque that balances the torque
5
Use your brain power
As a civil engineer, you are given contract
to construct a curved road in a ghat. In order
to obtain the banking angle θ , you need to
decide the speed limit. How will you decide
the values of speed v and radius r ?
Fig 1.6: Vehicle on a banked road. are (i) weight mg acting vertically downwards
and (ii) normal reaction N acting perpendicular
at an angle θ with the horizontal. Considering to the road. As seen above, only at this speed,
the vehicle to be a point and ignoring friction the resultant of these two forces (which is
(not eliminating) and other non-conservative Nsinθ ) is the necessary centripetal force (or
forces like air resistance, there are two forces balances the centrifugal force). In practice,
acting on the vehicle, (i) weight mg, vertically vehicles never travel exactly with this speed.
downwards and (ii) normal reaction N, For speeds other than this, the component of
perpendicular to the surface of the road. As
the motion of the vehicle is along a horizontal force of static friction between road and the
circle, the resultant force must be horizontal
tyres helps us, up to a certain limit.
and directed towards the centre of the track. It
means, the vertical force mg must be balanced.
Thus, we have to resolve the normal reaction
N along the vertical and along the horizontal.
Its vertical component Ncosθ balances weight
mg. Horizontal component Nsinθ being
the resultant force, must be the necessary
centripetal force (or balance the centrifugal
force). Thus, in magnitude, Fig 1.7: Banked road : lower speed limit.
N cosT mg and
mrZ 2 mv 2 v2
r rg --- (1.1)
N sinT ? tan T
(a) Most safe speed: For a particular road, r
and θ are fixed. Thus, this expression gives
us the expression for the most safe speed (not
a minimum or a maximum speed) on this road
as vs rg tanT
(b) Banking angle: While designing
a road, this expression helps us in Fig 1.8: Banked road : upper speed limit.
knowing the angle of banking as For speeds v1 rg tanT , mv12 N sinT
r
T tan 1 § v2 · --- (1.2) (or N sinθ is greater than the centrifugal force
¨ rg ¸ mv12 ). In this case, the direction of force of
© ¹
(c) Speed limits: Figure 1.7 and 1.8 show r
vertical section of a vehicle on a rough static friction fs between road and the tyres
curved road of radius r, banked at an angle is directed along the inclination of the road,
θ . If the vehicle is running exactly at the speed upwards (Fig. 1.7). Its horizontal component
vs rg tanT , the forces acting on the vehicle is parallel and opposite to Nsinθ . These two
6
forces take care of the necessary centripetal Example 3: A racing track of curvature
9.9 m is banked at tan−1 0.5 . Coefficient
force (or balance the centrifugal force). of static friction between the track and
the tyres of a vehicle is 0.2. Determine the
? mg fs sinT N cosT and speed limits with 10 % margin.
Solution:
mv12 N sinT fs cosT
r
For minimum possible speed, fs is
maximum and equal to µsN. Using this in the v min rg § tanT Ps ·
equations above and solving for minimum ¨ 1 Ps tanT ¸
© ¹
possible speed, we get
v1 min vmin rg § tanT Ps · --- (1.3) 9.9u10 § 1 0.5 0.2 ·
¨ 1 Ps tanT ¸ ¨¨© ¸¸¹
© ¹ 0.2u 0.5
For Ps t tanT ,vmin = 0. This is true for most 27 5.196 m / s
of the rough roads, banked at smaller angles.
Allowed vmin should be 10% higher than
mv 2 this.
(d) For speeds v2 ! rg tanT , 2 ! N sinT
r ? v min allowed 5.196 u 110
100
(or N sinθ is less than the centrifugal force
2
mv 2 ). In this case, the direction of force 5.716 m
s
r
of static friction fs between road and the
tyres is directed along the inclination of the v max rg § tanT Ps ·
¨ 1 Ps tanT ¸
road, downwards (Fig. 1.8). Its horizontal © ¹
component is parallel to Nsinθ . These two § 0.5 0.2 ·
©¨¨ ¸¹¸
forces take care of the necessary centripetal 9.9u10 0.2u 0.5
force (or balance the centrifugal force). 1
? mg N cosT fs sinT and 77 8.775 m / s
mv 2 N sinT fs cosT Allowed vmax should be 10% lower than
2 this.
r
For maximum possible speed, fs is
90
maximum and equal to µs N . Using this in the ∴ v max allowed 8.775 u 100 7.896 m / s
equations above, and solving for maximum
possible speed, we get Use your brain power
v2 max vmax rg § tanT Ps · --- (1.4) • If friction is zero, can a vehicle move on
¨ 1 Ps tanT ¸ the road? Why are we not considering
© ¹ the friction in deriving the expression
for the banking angle?
If µs = cot θ , vmax = ∞. But ( µs )max = 1.
Thus, for θ ≥ 45°, vmax = ∞. However, for • What about the kinetic friction between
heavily banked road, minimum limit may the road and the tyres?
be important. Try to relate the concepts used 1.3.4 Conical Pendulum:
A tiny mass (assumed to be a point object
while explaining the well of death.
and called a bob) connected to a long, flexible,
(e) For µs = 0, both the equations 1.3 and 1.4 massless, inextensible string, and suspended
give us v rg tanT which is the safest speed to a rigid support is called a pendulum. If the
on a banked road as we don’t take the help of
friction.
7
string is made to oscillate in a single vertical Z2 g sinT
plane, we call it a simple pendulum (to be r cosT
studied in the Chapter 5).
Radius r of the circular motion is r L sinT .
We can also revolve the string in such a
way that the string moves along the surface of If T is the period of revolution of the bob,
a right circular cone of vertical axis and the
point object performs a (practically) uniform Z 2S g
horizontal circular motion. In such a case the T L cosT
system is called a conical pendulum.
? Period T 2S L cosT --- (1.7)
Fig. 1.9 (a): In an inertial frame g
Frequency of revolution,
n 1 1 g --- (1.8)
T 2S L cosT
In the frame of reference attached to the
bob, the centrifugal force should balance the
resultant of all the real forces (which we call
CPF) for the bob to be at rest.
∴ T0 sinθ = mrω 2 --- (in magnitude). This is
the same as the Eq. (1.5)
Do you know?
(i) For a given set up, L and g are constant.
Thus, both period and frequency
Fig. 1.9 (b): In a non- inertial frame depend upon θ .
Figure 1.9 shows the vertical section of (ii) During revolutions, the string can
a conical pendulum having bob (point mass) NEVER become horizontal. This can
of mass m and string of length L. In a given be explained in two different ways.
position B, the forces acting on the bob are (i) (a) If the string becomes horizontal,
the force due to tension will also be
its weight mg directed vertically downwards horizontal. Its vertical component will
and (ii) the force T0 due to the tension in the then be zero. In this case, nothing will
string, directed along the string, towards
be there to balance mg.
the support A. As the motion of the bob is a (b) For horizontal string, i = 90°. This will
horizontal circular motion, the resultant force indicate the frequency to be infinite
must be horizontal and directed towards the and the period to be zero, which are
centre C of the circular motion. For this, all impossible. Also, in this case, the tension
mg
the vertical forces must cancel. Hence, we T0 cosT in the string and the kinetic
shall resolve the force T due to the tension. energy 1 mv2 1 mr 2Z 2 of the bob
0 2 2
If θ is the angle made by the string with the
vertical, at any position (semi-vertical angle will be infinite.
of the cone), the vertical component T0 cos Activity
θ balances the weight mg. The horizontal
A stone is tied to a string and whirled
component T0 sinθ then becomes the resultant such that the stone performs horizontal
force which is centripetal. circular motion. It can be seen that the string
is NEVER horizontal.
?T0 sinT centripetal force mrZ2 --- (1.5)
Also, T0 cosθ = mg --- (1.6)
Dividing eq (1.5) by Eq. (1.6), we get,
8
Example 1.4: A merry-go-round usually Solution: N sinT mg and N cosT mv 2
consists of a central vertical pillar. At the r
top of it there are horizontal rods which can rg v2 tanT
rotate about vertical axis. At the end of this ? tanT v2 ? r g
horizontal rod there is a vertical rod fitted
like an elbow joint. At the lower end of ? rmax v 2 tanT 0.3 m
each vertical rod, there is a horse on which max
the rider can sit. As the merry-go-round is
set into rotation, these vertical rods move g
away from the axle by making some angle v rZ 2S rn
with the vertical.
If we go for the lower
limit of the speed (while
rotating),
v o 0?r o 0 , but the
frequency n increases.
Hence a specific upper
limit is not possible in the case of frequency.
Thus, the practical limit on the frequency of
rotation is its lower limit. It will be possible
The figure above shows vertical section for r = rmax
of a merry-go-round in which the ‘initially ? nmin v max 1 # 1rev / s
2S rmax 0.3S
vertical’ rods are inclined with the vertical at
370, during rotation. Calculate the frequency
of revolution of the merry-go-round.
(Use g = π2 m/s2 and sin 37° = 0.6) Activity
Solution: Length of the horizontal rod, Using a funnel and a marble or a ball bearing
try to work out the situation in the above
H = 2.1 m question. Try to realize that as the marble
goes towards the brim, its linear speed
Length of the ‘initially vertical’ rod, increases but its angular speed decreases.
When nearing the base, it is the other way.
V = 1.5 m, θ = 37°
1.4 Vertical Circular Motion:
∴ Radius of the horizontal circular motion Two types of vertical circular motions are
of the rider = H + V sin 37° = 3.0 m commonly observed in practice:
(a) A controlled vertical circular motion such
If T is the tension along the inclined rod,
as a giant wheel or similar games. In this
T cos θ = mg and T sinθ = mrω 2 = 4π2 mrn2 case the speed is either kept constant or
NOT totally controlled by gravity.
? tanT 4S 2rn2 (b) Vertical circular motion controlled only
g by gravity. In this case, we initially
supply the necessary energy (mostly) at
?n tanT 1 rev s 1 }as g S2 the lowest point. Then onwards, the entire
4r 4 kinetics is governed by the gravitational
force. During the motion, there is
Example 1.5: Semi-vertical angle of the interconversion of kinetic energy and
conical section of a funnel is 370. There is a gravitational potential energy.
small ball kept inside the funnel. On rotating
the funnel, the maximum speed that the ball
can have in order to remain in the funnel is 2
m/s. Calculate inner radius of the brim of the
funnel. Is there any limit upon the frequency
of rotation? How much is it? Is it lower or
upper limit? Give a logical reasoning. (Use
g = 10 m/s2 and sin 370 = 0.6)
9
1.4.1 Point Mass Undergoing Vertical realized with minimum possible energy),
Circular Motion Under Gravity:
Case I: Mass tied to a string: TA = 0 ? vA min rg --- (1.10)
The figure 1.10 shows a bob (treated as Lowermost position (B): Force due to the
a point mass) tied to a (practically) massless,
inextensible and flexible string. It is whirled tension, TB is vertically upwards, i.e., towards
along a vertical circle so that the bob performs the centre, and opposite to mg. In this case also
a vertical circular motion and the string rotates
in a vertical plane. At any position of the bob, their resultant is the centripetal force. If vB is
there are only two forces acting on the bob: the speed at the lowermost point, we get,
A TB mg mv 2 --- (1.11)
B
r
While coming down from the uppermost to
the lowermost point, the vertical displacement
is 2r and the motion is governed only by
gravity. Hence the corresponding decrease in
the gravitational potential energy is converted
into the kinetic energy.
1 1
?mg 2r 2 mv 2 2 mv 2
B A
? v 2 v 2 4rg --- (1.12)
B A
Using this in the eq (1.11), and usin?g vA mminin
B rg
from Eq. (1.10) we get,
vB min 5rg --- (1.13)
Fig 1.10: Vertical circular motion. Subtracting eq (1.9) from eq (1.11) , we can
(a) its weight mg, vertically downwards, which
write, mv 2 v 2 --- (1.14)
is constant and (b) the force due to the tension TB TA 2mg B A
r
along the string, directed along the string and
Using eq (1.12) and rearranging, we get,
towards the centre. Its magnitude changes
TB TA 6mg --- (1.15)
periodically with time and location.
Positions when the string is horizontal (C
As the motion is non uniform, the resultant
and D): Force due to the tension is the only
of these two forces is not directed towards
force towards the centre as weight mg is
the center except at the uppermost and the
perpendicular to the tension. Thus, force due
lowermost positions of the bob. At all the other to the tension is the centripetal force used to
positions, part of the resultant is tangential and change the direction of the velocity and weight
is used to change the speed. mg is used only to change the speed.
Uppermost position (A): Both, weight mg and Using similar mathematics, it can be shown
force due to tension TA are downwards, i.e., that
towards the centre. In this case, their resultant TC TA TD TA 3mg and
vC min
is used only as the centripetal force. Thus, if v D min 3rg
vA is the speed at the uppermost point, we get, Arbitrary positions: Force due to the tension
2
mg TA mv A --- (1.9) and weight are neither along the same line,
r nor perpendicular. Tangential component of
Radius r of the circular motion is the weight is used to change the speed. It decreases
length of the string. For minimum possible the speed while going up and increases it while
speed at this point (or if the motion is to be coming down.
10
Remember this 1.4.2 Sphere of Death (मतृ ्यु गोल):
This is a popular show in a circus. During
1. Equation (1.15) is independent of v and r.
this, two-wheeler rider (or riders) undergo
2. T can never be exactly equal to zero in rounds inside a hollow sphere. Starting with
A small horizontal circles, they eventually
the case of a string, else, the string will perform revolutions along vertical circles. The
slack. ∴ TB > 6 mg. dynamics of this vertical circular motion is
the same as that of the point mass tied to the
3. None of the parameters (including the string, except that the force due to tension T is
linear and angular accelerations) are replaced by the normal reaction force N.
constant during such a motion. Obviously,
kinematical equations given in the table1 If you have seen this show, try to visualize
are not applicable. that initially there are nearly horizontal circles.
The linear speed is more for larger circles but
4. We can determine the position vector or angular speed (frequency) is more for smaller
velocity at any instant using the energy circles (while starting or stopping). This is as
conservation. But as the function of the per the theory of conical pendulum.
radius vector is not integrable (definite 1.4.3 Vehicle at the Top of a Convex Over-
integration is not possible), theoretically Bridge:
it is not possible to determine the period
or frequency. However, experimentally
the period can be measured.
5. Equations (1.10) and (1.13) give only
the respective minimum speeds at the
uppermost and the lowermost points. Any
higher speeds obeying the equation (1.14)
are allowed.
6. In reality, we have to continuously Fig. 1.11: Vehicle on a convex over-bridge.
supply some energy to overcome the air
resistance.
Case II: Mass tied to a rod: Consider a bob Figure shows a vehicle at the top of a
(point mass) tied to a (practically massless and convex over bridge, during its motion (part
rigid) rod and whirled along a vertical circle. of vertical circular motion). Forces acting on
The basic difference between the rod and the the vehicle are (a) Weight mg and (b) Normal
string is that the string needs some tension at reaction force N, both along the vertical line
all the points, including the uppermost point. (topmost position). The resultant of these two
Thus, a certain minimum speed, Eq. (1.10), is must provide the necessary centripetal force
necessary at the uppermost point in the case (vertically downwards) if the vehicle is at the
of a string. In the case of a rod, as the rod is uppermost position. Thus, if v is the speed at
rigid, such a condition is not necessary. Thus the uppermost point,
(practically) zero speed is possible at the mg N mv 2
r
uppermost point.
Using similar mathematics, it is left to the As the speed is increased, N goes on
readers to show that decreasing. Normal reaction is an indication
vlowermost min 4rg 2 rg of contact. Thus, for just maintaining contact,
vmin at the rod horizontal position = 2rg N = 0. This imposes an upper limit on the speed
as vmax = rg
T Tlowermostuppermost 6mg
11
Do you know? ∴ 1 0.02 8 2 K.E.min 0.02 10 1.8
2
Roller coaster is a common event in the 1 ? K.E.min 0.28 J
amusement parks. During this ride, all 2
the parts of the vertical circular motion K.E.max 0.02 8 2 0.02 10 1.8
described above can be experienced. The ? K.E.max 1 mvm2 ax 1J
major force that we experience during this is 2
the normal reaction force. Those who have =2(K . E .)max 10 m s-1
m
experienced this, should try to recall the =∴ vmax
changes in the normal reaction experienced
at the lowermost position, for which θ = 0.
by us during various parts of the track.
mv 2
T mg cosT r --- at any angle θ ,
Use your brain power where the speed is v.
• What is expected to happen if one travels Thus, if T = mg, we get, mv 2
fast over a speed breaker? Why? r
mg mg cosT
• How does the normal force on a concave
suspension bridge change when a vehicle ?rg 1 cosT v2 --- (A)
is travelling on it with constant speed?
Vertical displacement at the angular
Example 1.6: A tiny stone of mass 20 g is
tied to a practically massless, inextensible, position θ is r 1 cosT . Thus, the energy
flexible string and whirled along vertical
circles. Speed of the stone is 8 m/s when equation at this position can be written as
the centripetal force is exactly equal to the 1 1
force due to the tension. 2 m 10 2 2 mv 2 mg ª¬r 1 cosT º¼
Calculate minimum and maximum kinetic
energies of the stone during the entire circle. By using Eq. A, we get
Let θ = 0 be the angular position of the 50 1 rg 1 cosT rg 1 cosT
string, when the stone is at the lowermost 2
position. Determine the angular position of 3
? 50 2 rg 1 cosT
? cos T 23 ?T 1480 25'
27
the string when the force due to tension is
numerically equal to weight of the stone. 1.5 Moment of Inertia as an Analogous
Use g = 10 m/s2 and length of the string = Quantity for Mass:
1.8 m
Solution: When the string is horizontal, the In XIth Std. we saw that angular
force due to the tension is the centripetal displacement, angular velocity and
force. Thus, vertical displacements of the angular acceleration respectively replace
bob for minimum and maximum energy displacement, velocity and acceleration for
positions are radius r each. various kinematical equations. Also, torque is
If K.E.max and K.E.min are the respective an analogous quantity for force. Expressions of
kinetic energies at the uppermost and the linear momentum, force (for a fixed mass) and
kinetic energy include mass as a common term.
lowermost points, 1 In order to have their rotational analogues, we
2 need a replacement for mass.
K.E.max m 8 2 mgr and
If we open a door (with hinges), we give a
1 m 8 2 K.E.min mgr certain angular displacement to it. The efforts
2
12
needed for this depend not only upon the mass ¦If I miri2 replaces mass m and angular
of the door, but also upon the (perpendicular)
distance from the axis of rotation, where we speed ω replaces linear speed v, rotational
apply the force. Thus, the quantity analogous
to mass includes not only the mass, but also K.E. 1 IZ 2 is analogous to translational
takes care of the distance wise distribution of 2
the mass around the axis of rotation. To know 1
K.E. = 2 mv 2 . Thus, I is defined to be the
rotational inertia or moment of inertia (M.I.)
the exact relation, let us derive an expression of the object about the given axis of rotation.
for the rotational kinetic energy which is the It is clear that the moment of inertia of an
sum of the translational kinetic energies of all object depends upon (i) individual masses and
the individual particles. (ii) the distribution of these masses about the
given axis of rotation. For a different axis, it
will again depend upon the mass distribution
around that axis and will be different if there is
no symmetry.
During this discussion, for simplicity, we
assumed the object to be consisting of a finite
number of particles. In practice, usually, it
Fig. 1.12: A body of N particles. is not so. For a homogeneous rigid object of
Figure 1.12 shows a rigid object rotating mathematically integrable mass distribution,
with a constant angular speed ω about an the moment of inertia is to be obtained by
axis perpendicular to the plane of paper.
For theoretical simplification let us consider integration as I ³r2dm . If integrable mass
distribution is not known, it is not possible to
the object to be consisting of N particles obtain the moment of inertia theoretically, but
of masses m1, m2, …..mN at respective it can be determined experimentally.
perpendicular distances r1, r2, …..rN from the
axis of rotation. As the object rotates, all these
particles perform UCM with the same angular
speed ω , but with different linear speeds
v1 r1Z, v2 r2Z, }} vN rNZ .
Translational K.E. of the first particle is
1 1
K.E.1 2 m1v12 2 m1r12Z 2 Fig. 1.13: Moment of Inertia of a ring.
1.5.1 Moment of Inertia of a Uniform Ring:
Similar will be the case of all the other
An object is called a uniform ring if
particles. Rotational K.E. of the object, is its mass is (practically) situated uniformly
on the circumference of a circle (Fig 1.13).
the sum of individual translational kinetic Obviously, it is a two dimensional object of
negligible thickness. If it is rotating about its
energies. Thus, rotational K.E. own axis (line perpendicular to its plane and
passing through its centre), its entire mass M
1 m1r12Z 2 1 m2r22Z 2 }} 1 mN rN2Z 2 is practically at a distance equal to its radius
2 2 2 R form the axis. Hence, the expression for the
moment of inertia of a uniform ring of mass M
? Rotational K.E. and radius R is I = MR2.
1 1
2
2
m1r12 m2r22 } mN rN2 Z2 IZ 2
N
¦Where I m1r12 m2r22 }} mN rN2 mi ri2
i1
13
1.5.2 Moment of Inertia of a Uniform Disc: of inertias of objects of several integrable
geometrical shapes can be derived. Some of
Disc is a two dimensional circular object those are given in the Table 3 at the end of the
topic.
of negligible thickness. It is said to be uniform 1.6 Radius of Gyration:
if its mass per unit area and its composition is As stated earlier, theoretical calculation
m mass of moment of inertia is possible only for
the same throughout. The ratio V A area mathematically integrable geometrical shapes.
is called the surface density. However, experimentally we can determine
the moment of inertia of any object. It depends
Consider a uniform disc of mass M and upon mass of that object and how that mass
is distributed from or around the given axis
radius R rotating about its own axis, which is of rotation. If we are interested in knowing
only the mass distribution around the axis of
the line perpendicular to its plane and passing rotation, we can express moment of inertia
of any object as I = MK 2 , where M is mass
through its centre ?V M of that object. It means that the mass of that
S R2 . object is effectively at a distance K from the
As it is a uniform circular object, it can given axis of rotation. In this case, K is defined
as the radius of gyration of the object about
be considered to be consisting of a number the given axis of rotation. In other words, if K
is radius of gyration for an object, I = MK 2 is
of concentric rings of radii increasing from the moment of inertia of that object. Larger the
value of K, farther is the mass from the axis.
(practically) zero to R. One of such rings of
mass dm is shown by shaded portion in the
Fig. 1.14.
Fig .1.14: Moment of Inertia of a disk. Consider a uniform ring and a uniform
Width of this ring is dr, which is so small disc, both of the same mass M and same
that the entire ring can be considered to be radius R. Let Ir and Id be their respective
moment of inertias.
of average radius r. (In practical sense, dr is
less than the least count of the instrument that If Kr and Kd are their respective radii
measures r, so that r is constant for that ring). of gyration, we can write,
dm
Area of this ring is A = 2πr.dr ?V 2S r.dr Ir = MR2 = MK 2 ∴Kr = R and
1 r = R
2
∴ dm = 2πσr.dr. Id = 2 MR2 = MK 2 ∴ Kd ∴ Kd <Kr
d
As it is a ring, this entire mass is at a It shows mathematically that K is
distance r from the axis of rotation. Thus, the decided by the distribution of mass. In a
moment of inertia of this ring is Ir = dm (r2) ring the entire mass is distributed at the
Moment of inertia (I) of the disc can now
distance R, while for a disc, its mass is
be obtained by integrating Ir from r = 0 to distributed between 0 and R. Among any
r = R.R R R R objects of same mass and radius, ring has
? I ³Ir ³dm r2 ³2SV r dr r2 2SV ³r3 dr the largest radius of gyration and hence
00 0 0 maximum M.I.
?I 2SV § R4 · 2S § M ·§ R4 · 1 MR2 1.7 Theorem of Parallel Axes and Theorem
¨ 4 ¸ ¨© S R2 ¸¹ ¨ 4 ¸ 2 of Perpendicular Axes:
© ¹ © ¹
Expressions of moment of inertias of
Using similar method, expressions for moment
14
regular geometrical shapes given in the table 3 ³ >DC@2 2NC.h h2 dm
are about their axes of symmetry. These are
derived by integration. However, every time ³(DC)2 dm 2h³NC.dm h2 ³dm
the axis need not be the axis of symmetry. In Now, ³ DC 2 dm IC and ³dm M .
simple transformations it may be parallel or
perpendicular to the symmetrical axis. For NC is the distance of a point from the
example, if a rod is rotated about one of its centre of mass. Any mass distribution is
ends, the axis is parallel to its axis of symmetry. symmetric about the centre of mass. Thus,
If a disc or a ring is rotated about its diameter, from the definition of the centre of mass,
the axis is perpendicular to the central axis.
In such cases, simple transformations are ³NC.dm 0 .
possible in the expressions of moment of
inertias. These are called theorem of parallel ? IO IC M .h2
axes and theorem of perpendicular axes. This is the mathematical form of the
1.7.1 Theorem of Parallel Axes: theorem of parallel axes.
It states that, “The moment of inertia (IO)
In order to apply this theorem to any of an object about any axis is the sum of its
object, we need two axes parallel to each other moment of inertia (IC) about an axis parallel to
with one of them passing through the centre of the given axis, and passing through the centre
mass of the object. of mass and the product of the mass of the
object and the square of the distance between
the two axes (Mh2).”
Use your brain power
In Fig. 1.15, the point D is chosen such that
we have to extend OC for the perpendicular
DN to fall on it. What will happen to the final
expression of I0, if point D is so chosen that
the perpendicular DN falls directly on OC?
Fig. 1.15: Theorem of parallel axes. 1.7.2 Theorem of Perpendicular Axes:
This theorem relates the moment of
Figure 1.15 shows an object of mass M.
Axis MOP is any axis passing through point inertias of a laminar object about three
O. Axis ACB is passing through the centre mutually perpendicular and concurrent axes,
of mass C of the object, parallel to the axis two of them in the plane of the object and
MOP, and at a distance h from it (∴ h = CO). the third perpendicular to the object. Laminar
Consider a mass element dm located at point object is like a leaf, or any two dimensional
D. Perpendicular on OC (produced) from point object, e.g., a ring, a disc, any plane sheet, etc.
D is DN. Moment of inertia of the object about
N
³the axis ACB is IC (DC)2 dm , and about M
the axis MOP it is IO ³ DO 2 dm .
? IO ³ DO 2 dm ³ >DN@2 >NO@2 dm
³ >DN@2 [NC]2 2.NC.CO >CO@2 dm Fig. 1.16: Theorem of perpendicular axes.
15
Figure 1.16 shows a rigid laminar object able (II) Consider any two mutually perpendicular
to rotate about three mutually perpendicular
axes x, y and z. Axes x and y are in the plane diameters x and y of the flywheel. If the
of the object while the z axis is perpendicular
to it, and all are concurrent at O. Consider a flywheel rotates about these diameters, these
mass element dm located at any point P. PM
= y and PN = x are the perpendiculars drown three axes (own axis and two diameters) will
from P respectively on the x and y axes. The
respective perpendicular distances of point be mutually perpendicular and concurrent.
M from x, y and z axes will then be y, x and
Thus, perpendicular axes theorem is
y 2 + x2 . If Ix, Iy and Iz are the respective
moment of inertias of the body about x, y and applicable. Let Id be the moment of inertia
z axes, we can write, of the flywheel, when rotating about its
³ ³? I x y 2dm, I y x2dm and diameter. ? Id I x I y
Thus, according to the theorem of
³Iz y2 x2 dm
perpendicular axes,
1 MR2
Iz 2 Ix Iy 2Id
? Id 1 MR2
4
³ ³? I z y 2dm x2dm I x I y
This is the mathematical form of the
theorem of perpendicular axes. As the diameter passes through the centre
It states that, “The moment of inertia
of mass of the (uniform) disc, Id = IC
(Iz) of a laminar object about an axis (z) Consider a tangent in the plane of the disc
perpendicular to its plane is the sum of its
moment of inertias about two mutually
perpendicular axes (x and y) in its plane, all
the three axes being concurrent”.
Example 1.7: A flywheel is a mechanical and parallel to this diameter. It is at the
device specifically designed to efficiently distance h = R from the diameter. Thus,
store rotational energy. For a particular parallel axes theorem is applicable about
machine it is in the form of a uniform 20 kg these two axes.
disc of diameter 50 cm, able to rotate about ∴ IT, parallel = Io = Ic + Mh2 = Id + MR2
its own axis. Calculate its kinetic energy = 1 MR2 + MR2 = 5 MR2
when rotating at 1200 rpm. Use S 2 10. 4 4
Calculate its moment of inertia, in case it is 5 5
4 4
rotated about a tangent in its plane. ∴ IT, parallel = MR2 = 20 × 0.252
Solution: (I) As the flywheel is in the form = 1.5625 kg m2
of a uniform disc rotating about its own 1.8 Angular Momentum or Moment of
axis, Iz = 1 MR 2 Linear Momentum:
2
∴ Rotational kinetic energy The quantity in rotational mechanics,
1 IZ2 1 § 1 MR 2 · 4S 2n2 analogous to linear momentum is angular
2 2 ©¨ 2 ¹¸
momentum or moment of linear momentum. It
∴ Rotational kinetic energy is sipmiislatrhteoitnhsetatonrtqauneeobuesinlginmeaormmenotmoefnatufomrcoef.
If
MS 2 Rn 2 20 u10 u 0.25 u 20 2 5000 J
a particle undertaking a circular motion, its
16
aLngurlaur pm,owmeerneturmiasttthheatpionssittaionnceviescgtoivr efnrobmy axis of rotation. The expression for angular
momentum L = Iω is analogous to the
the axis of rotation. expression p = mv of linear momentum, if the
moment of inertia I replaces mass, which is
In magnitude, it is the product of linear its physical significance.
1.9 Expression for Torque in Terms of
momentum and its perpendicular distance from Moment of Inertia:
the axis of rotation. ∴L = P × r sinθ , where θ
is the sm aller angle between the directions of
P and r .
1.8.1 Expression for Angular Momentum in
Terms of Moment of Inertia:
Figure 1.12 in the section 1.5 shows a
rigid object rotating with a constant angular
speed ω about an axis perpendicular to the
plane of paper. For theoretical simplification
let us consider the object to be consisting of
N number of particles of masses m1, m2, ….. Fig 1.17: Expression for torque.
m at respective perpendicular distances r , r ,
Figure 1.17 shows a rigid object rotating
N 12 with a constant angular acceleration α
…..rN from the axis of rotation. As the object about an axis perpendicular to the plane
rotates, all these particles perform UCM with
same angular speed ω , but with different linear of paper. For theoretical simplification let
speeds v = r ω , v = r ω , ..... v = r ω . us consider the object to be consisting of N
1 1 2 2 N N
Directions of individual velocities v1,
v2 , etc., are along the tangents to their number of particles of masses m1, m2, ….. mN
at respective perpendicular distances r1, r2,
respective tracks. Linear momentum of the …..rN from the axis of rotation. As the object
rotates, all these particles perform circular
first particle is of magnitude p1 = m1v1 = m1r1 ω.
Its direction is along that of v1. motion with same angular acceleration α, but
Its angular momentum is thus of with different linear (tangential) accelerations
a1 r1D , a2 r2D ,}}, a N rND , etc.
magnitude L = pr m1r12Z
1 11 Force experienced by the first particle is
f1 m1a1 m1r1D
Similarly, L2 m2r22Z , L3 m3r32Z , …….
LN mN rN2Z As these forces are tangential, their
For a rigid body with a fixed axis of
rotation, all these angular momenta are directed respective perpendicular distances from the
along the axis of rotation, and this direction axis are r1, r2, …..rN.
Thus, the torque experienced by the first
can be obtained by using right hand thumb
particle is of magnitude W1 f1r1 m1r12D
rule. As all of them have the same direction, Similarly, W 2 m2r22D , W 3 m3r32D ……..
their magnitudes can be algebraically added. W N mN rN2D
Thus, magnitude of angular momentum of the If the rotation is restricted to a single
body is given by plane, directions of all these torques are the
L m1r12Z m2r22Z } mN rN2Z
same, and along the axis. Magnitude of the
m1r12 m2r22 } mN rN2 Z IZ
resultant torque is then given by
Where, I m1r12 m2r22 } mN rN2 is the W W1 W2 } WN ID
moment of inertia of the body about the given
m1r12 m2r22 } mN rN2 D
17
where, I m1r12 m2r22 } mN rN2 is the Examples of conservation of angular
momentum: During some shows of ballet
moment of inertia of the object about the given dance, acrobat in a circus, sports like ice
skating, diving in a swimming pool, etc., the
axis of rotation. principle of conservation of angular momentum
is realized. In all these applications the product
The relation W ID is analogous to
f = ma for the translational motion if the L IZ I 2S n is constant (once the players
moment of inertia I replaces mass, which is acquire a certain speed). Thus, if the moment
of inertia I is increased, the angular speed and
its physical significance. hence the frequency of revolution n decreases.
Also, if the moment of inertia is decreased, the
1.10 Conservation of Angular Momentum: frequency increases.
(i) Ballet dancers: During ice ballet, the
In the article 4.7 of XIth Std. we have dancers have to undertake rounds of smaller
and larger radii. The dancers come together
seen the conservation of linear momentum while taking the rounds of smaller radius (near
the centre). In this case, the moment of inertia
which says that linear momentum of an of their system becomes minimum and the
frequency increases, to make it thrilling. While
isolated system is conserved in the absence outer rounds, the dancers outstretch their legs
and arms. This increases their moment of
of an external unbalanced force. As seen inertia that reduces the angular speed and
hence the linear speed. This is essential to
earlier, torque and angular momentum are prevent slipping.
(ii) Diving in a swimming pool (during
the respective analogous quantities to force competition): While on the diving board, the
divers stretch their body so as to increase the
and linear momentum in rotational dynamics. moment of inertia. Immediately after leaving
the board, they fold their body. This reduces
With suitable changes this can be transformed the moment inertia considerably. As a result,
the frequency increases and they can complete
into the conservation of angular momentum. more rounds in air to make the show attractive.
Again, while entering into water they stretch
As seen in the section 1.8, angular their body into a streamline shape. This allows
them a smooth entry into the water.
romwofhotaaemtrsieeoynnsrttueaimmnsdtoihsreptghpiievosesmtinhtoiebomylnienvnLeetacortformlriunofrempoaemrnmttuhomema.exnitsuomf
?DNddioLtfddwfLte,reddddntrtt riar=utuivFnpg awmnidtrhvudrdueddtpsvptp=e cFdtdrtt.outipme, we get,
?BNuoddtwLtr vuv 0
?W
Thus, r u F
×dLF is the moment of force or torque τ .
constant.
Example 1.8: A spherical water balloon
idf tW 0, dL 0 or L is revolving at 60 rpm. In the course of
dt
time, 48.8 % of its water leaks out. With
tHheenacbes,eanncgeuolaf rexmteormnaelnutunmbal Laniscecdontosreqruveedτi.n what frequency will the remaining balloon
revolve now? Neglect all non-conservative
forces. m1 V1 § R1 ·3 1
Solution: m2 V2 ¨ R2 ¸
This is the principle of conservation of angular © ¹ ? R1 § m1 ·3
R2 ¨ m2 ¸
momentum, analogous to the conservation of © ¹
linear momentum.
18
Also, m1 100 100 1 Accordingly, the object possesses two
m2 100 48.8 51.2 0.512
types of kinetic energies, rotational and
translational. Sum of these two is its total
1 kinetic energy.
§ m1 ·3 1 1.25 Consider an object of moment of inertia
?¨ m2 ¸ 0.8
¹ I, rolling uniformly. Following quantities can
=n1 ©
6=0 rpm 1rps, n2 = ? be related.
Being sphere, moment of inertia v = Linear speed of the centre of mass
·2 5 R = Radius of the body
¸
I 2 mR 2 ? I1 § m1 ·§ R1 ¹ § m1 ·3 Z Angular speed of rotation of the body
5 I2 ¨ m2 ¸¨ R2 ¨ m2 ¸
© ¹© © ¹ ?Z v for any particle
R
According to principle of conservation of
angular momentum, I1Z1 I2Z2 M = Mass of the body
5 K = Radius of gyration of the body ? I MK 2
? I1 2S n1 I2 2S n2 ? n2 § I1 · n1 § m1 ·3 Total kinetic energy of rolling = Translational
¨ I2 ¸ ¨ m2 ¸
© ¹ © ¹ K.E. + Rotational K.E.
n1 1.25 5 u1 3.052 rps ?E 1 Mv 2 1 IZ 2
2 2
Example 1.9: A ceiling fan having moment
of inertia 2 kg-m2 attains its maximum 1Mv 2 1 MK 2 § v ·2
frequency of 60 rpm in ‘2π’ seconds. 2 ¨© R ¸¹
Calculate its power rating. 2
Solution: 1 Mv 2 ¨§ 1 K2 · --- (1.18)
2 © R2 ¸
Z0 0, Z 2S n 2S u 2 4S rad / s ¹
It must be remembered that static friction
Z Z0 4S 0 2 rad/s2
D t 2S is essential for a purely rolling motion. In this
? P W Z ID Z 2u 2u 4S case, it prevents the sliding motion. You might
16S watt # 50 watt have noticed that many a times while rolling
down, the motion is initially a purely rolling
1.11 Rolling Motion: motion that later on turns out to be a sliding
The objects like a cylinder, sphere,
motion. Similarly, if you push a sphere-like
wheels, etc. are quite often seen to perform
rolling motion. In the case of pure rolling, object along a horizontal surface, initially it
two motions are undertaking simultaneously;
circular motion and linear motion. Individual slips for some distance and then starts rolling.
motion of the particles (except the one at the
centre of mass) is too difficult to describe. 1.11.1 Linear Acceleration and Speed While
However, for theory considerations we can
consider the actual motion to be the result of Pure Rolling Down an Inclined Plane:
(i) circular motion of the body as a whole,
about its own symmetric axis and Figure 1.18 shows a rigid object of mass
(ii) linear motion of the body assuming it to
be concentrated at its centre of mass. In other m and radius R, rolling down an inclined plane,
words, the centre of mass performs purely
translational motion. without slipping. Inclination of the plane with
the horizontal is θ .
Fig. 1.18: Rolling along an incline.
19
As the objects starts rolling down, its K2 = 1 for a uniform disc or a solid cylinder
R2 2
gravitational P.E. is converted into K.E. of
rolling. Starting from rest, let v be the speed of K2 = 2 for a thin walled hollow sphere
R2 3
the centre of mass as the object comes down
through a vertical distance h. (II) When a rod rolls, it is actually a cylinder
From Eq. (1.18), that is rolling.
E 1 Mv 2 1 IZ 2 1 Mv 2 § 1 K2 · (III) While rolling, the ratio ‘Translational
2 2 2 ¨ R2 ¸
© ¹ K.E.: Rotational K.E.: Total K.E.’ is
?E mgh 1 Mv 2 ¨§ 1 K2 · 1: K2 : § 1 K2 ·
2 © R2 ¸ R2 ¨ R2 ¸
¹ © ¹
K2 2
?v 2gh For example, for a hollow sphere, R2 = 3
Thus, for a rolling hollow sphere,
¨§ 1 K2 · --- (1.19)
© R2 ¸
¹ Translational K.E.: Rotational K.E.: Total
Linear distance travelled along the plane 2
h 3 2
is s DsuirninTg K.E. = 1: : §©¨ 1 3 · 3: 2: 5
¸¹
this distance, the linear velocity Percentage wise, 60% of its kinetic energy
has increased from zero to v. If a is the linear is translational and 40% is rotational.
acceleration along the plane, Table 1 : Analogous kinematical equations
2as v2 u2? 2a § h · 2gh 0 (ω0 is the initial angular velocity)
¨© sinT ¸¹ Equation for Analogous equation
§¨ 1 K2 ·
© R2 ¸ translational for rotational
¹
g sinT motion motion
?a
§ K2 · u v
¨ 1 R2 ¸ --- (1.20) vav 2 Zav Z0 Z
¹ 2
©
For pure sliding, without friction, the
a dv v u dZ Z Z0
acceleration is g sinθ and final velocity is dt t D dt t
2gh . Thus, during pure rolling, the factor ? v u at ?Z Z0 Dt
§ 1 K2 · is effective for both the expressions. s vav t T Zav t
¨ R2 ¸
© ¹
§ u v ·
Remarks : ¨© 2 ¹¸ t § Z0 Z ·
I) For a rolling object, if the expression for ¨© 2 ¹¸ t
moment of inertia is of the form n (MR2), the
ut 1 at 2 Z0t Dt 2
2
numerical factor n gives the value of K2 for
that object. R2 v2 u2 2as Z 2 Z02 2DT
For example, for a uniform solid sphere,
2
I 2 MR2 MK 2 ? K 2 2 Internet my friend
5 R 5 http://hyperphysics.phy-astr.gsu.edu/hbase/
Similarly, mi.html
https://issacphysics.org
K2 = 1, for a ring or a hollow cylinder https://www.engineeringtoolbox.com
R2 https://opentextbc.ca/physicstextbook
20
Table 2: Analogous quantities between translational motion and rotational motion:
Translational motion Rotational motion
Quantity Symbol/ Quantity Symbol/ Inter-relation,
expression
Angular expression if possible
displacement
Linear s u r
s θ T
displacement
Linear velocity = Angular velocity Z dT v Z u r
v ds dt
dt
Linear = Angular D dZ D a u r
acceleration a dv acceleration dt
dt
Rotational inertia ³ ¦ I r2dm
Inertia or mass m or moment of I mi ri2
inertia
Linear p = mv Angular L IZ L r u p
momentum momentum W r u f
Force f = dp Torque W dL
dt Work dt
Work W f s
W W ------
Power T
P dW v Power P dW W Z ------
dt f dt
Table 3: Expressions for moment of inertias for some symmetric objects:
Object Axis Expression of Figure
moment of inertia
Thin ring or Central I = MR2
hollow cylinder
Thin ring Diameter I = 1 MR 2
2
Annular ring or
thick walled I 1M
Central 2 r22 r12
hollow cylinder
21
Uniform disc or Central I = 1 MR 2
solid cylinder 2
Uniform disc Diameter I = 1 MR 2
4
Thin walled Central I = 2 MR 2
hollow sphere 3
Solid sphere Central I = 2 MR 2
5
Uniform symmetric Central I 2 M r25 r15
spherical shell 5 r23 r13
Thin uniform rod or Perpendicular to I = 1 ML2
rectangular plate length and passing 12
through centre
Thin uniform rod or Perpendicular to I = 1 MR 2
rectangular plate length and about 3
one end
Uniform plate Central I 1 M ( L2 b2 )
or rectangular 12
parallelepiped
22
Uniform solid Central I = 3 MR 2
right circular cone 10
Uniform hollow Central I = 1 MR 2
right circular cone 2
Exercises
Use g = 10 m/s2, unless, otherwise stated. formula (expression) of moment of
1. Choose the correct option. inertia (M.I.) in terms of mass M of
i) When seen from below, the blades of the object and some of its distance
parameter/s, such as R, L, etc.
a ceiling fan are seen to be revolving (A) Different objects must have different
anticlockwise and their speed is
decreasing. Select correct statement expressions for their M.I.
about the directions of its angular (B) When rotating about their central
velocity and angular acceleration.
(A) Angular velocity upwards, angular axis, a hollow right circular cone and
acceleration downwards. a disc have the same expression for
(B) Angular velocity downwards, the M.I.
angular acceleration upwards. (C) Expression for the M.I. for a
(C) Both, angular velocity and angular parallelepiped rotating about the
acceleration, upwards. transverse axis passing through its
(D) Both, angular velocity and centre includes its depth.
angular acceleration, downwards. (D) Expression for M.I. of a rod and
ii) A particle of mass 1 kg, tied to a 1.2 m that of a plane sheet is the same
long string is whirled to perform vertical about a transverse axis.
circular motion, under gravity. Minimum iv) In a certain unit, the radius of gyration
speed of a particle is 5 m/s. Consider of a uniform disc about its central and
following statements. transverse axis is 2.5 . Its radius of
P) Maximum speed must be 5 5 m/s. gyration about a tangent in its plane (in
Q) Difference between maximum and
minimum tensions along the string is 60 N. the same unit) must be
Select correct option. (A) 5 (B) 2.5
(A) Only the statement P is correct.
(B) Only the statement Q is correct. (C) 2 2.5 (D) 12.5
v) Consider following cases:
(C) Both the statements are correct. (P) A planet revolving in an elliptical orbit.
(D) Both the statements are incorrect. (Q) A planet revolving in a circular orbit.
iii) Select correct statement about the
Principle of conservation of angular
momentum comes in force in which of
these?
23
(A) Only for (P) What can you infer if a uniform ring and
a uniform disc have the same radius of
(B) Only for (Q) gyration?
6. State the conditions under which
(C) For both, (P) and (Q) the theorems of parallel axes and
perpendicular axes are applicable. State
(D) Neither for (P), nor for (Q) the respective mathematical expressions.
7. Derive an expression that relates angular
X) A thin walled hollow cylinder is rolling momentum with the angular velocity of
a rigid body.
down an incline, without slipping. At 8. Obtain an expression relating the torque
with angular acceleration for a rigid
any instant, the ratio ”Rotational K.E.: body.
9. State and explain the principle of
Translational K.E.: Total K.E.” is conservation of angular momentum. Use
a suitable illustration. Do we use it in our
(A) 1:1:2 (B) 1:2:3 daily life? When?
10. Discuss the interlink between
(C) 1:1:1 (D) 2:1:3 translational, rotational and total kinetic
energies of a rigid object that rolls
2. Answer in brief. without slipping.
11. A rigid object is rolling down an
i) Why are curved roads banked? inclined plane. Derive expressions for
the acceleration along the track and
ii) Do we need a banked road for a two- the speed after falling through a certain
vertical distance.
wheeler? Explain. 12. Somehow, an ant is stuck to the rim of a
bicycle wheel of diameter 1 m. While the
iii) On what factors does the frequency bicycle is on a central stand, the wheel
is set into rotation and it attains the
of a conical pendulum depends? Is it frequency of 2 rev/s in 10 seconds, with
uniform angular acceleration. Calculate
independent of some factors? (i) Number of revolutions completed by
the ant in these 10 seconds. (ii) Time
iv) Why is it useful to define radius of taken by it for first complete revolution
and the last complete revolution.
gyration? [Ans:10=rev., tfirst =10 s, tlast 0.5132 s ]
13. Coefficient of static friction between
v) A uniform disc and a hollow right a coin and a gramophone disc is 0.5.
Radius of the disc is 8 cm. Initially the
circular cone have the same formula centre of the coin is π cm away from
the centre of the disc. At what minimum
for their M.I., when rotating about their frequency will it start slipping from
central axes. Why is it so?
3. While driving along an unbanked
circular road, a two-wheeler rider has
to lean with the vertical. Why is it so?
With what angle the rider has to lean?
Derive the relevant expression. Why
such a leaning is not necessary for a four
wheeler?
4. Using the energy conservation, derive
the expressions for the minimum speeds
at different locations along a vertical
circular motion controlled by gravity.
Is zero speed possible at the uppermost
point? Under what condition/s? Also
prove that the difference between the
extreme tensions (or normal forces)
depends only upon the weight of the
object.
5. Discuss the necessity of radius of
gyration. Define it. On what factors does
it depend and it does not depend? Can
you locate some similarity between the
centre of mass and radius of gyration?
24
there? By what factor will the answer motorcycle varies between 6 m/s and 10
change if the coin is almost at the rim? m/s. Calculate diameter of the sphere of
(use g = π2 m/s2) 1 death. How much minimum values are
[Ans: 2.5 rev/s, n2 = 2 n1 ] possible for these two speeds?
14. Part of a racing track is to be designed [Ans: Diameter = 3.2 m,
for a curvature of 72 m. We are not (v1)min = 4 m/s, (v2)min = 4 5 m / s ]
19. A metallic ring of mass 1 kg has moment
recommending the vehicles to drive
faster than 216 kmph. With what angle of inertia 1 kg m2 when rotating about
should the road be tilted? By what height one of its diameters. It is molten and
will its outer edge be, with respect to the remoulded into a thin uniform disc of the
inner edge if the track is 10 m wide? same radius. How much will its moment
[Ans: T tan 1(5) 78.69o ,h 9.8 m ] of inertia be, when rotated about its own
15. The road in the question 14 above is axis.
constructed as per the requirements. The [Ans: 1 kg m2]
coefficient of static friction between the 20. A big dumb-bell is prepared by using a
tyres of a vehicle on this road is 0.8, will uniform rod of mass 60 g and length 20
there be any lower speed limit? By how cm. Two identical solid spheres of mass
much can the upper speed limit exceed in 50 g and radius 10 cm each are at the
this case? two ends of the rod. Calculate moment
[Ans: vmin ≅ 88 kmph , no upper limit as of inertia of the dumb-bell when rotated
the road is banked for T ! 45o ]
about an axis passing through its centre
16. During a stunt, a cyclist (considered to and perpendicular to the length.
be a particle) is undertaking horizontal [Ans: 24000 g cm-2]
circles inside a cylindrical well of 21. A flywheel used to prepare earthenware
radius 6.05 m. If the necessary friction pots is set into rotation at 100 rpm. It is
coefficient is 0.5, how much minimum in the form of a disc of mass 10 kg and
speed should the stunt artist maintain? radius 0.4 m. A lump of clay (to be taken
Mass of the artist is 50 kg. If she/he equivalent to a particle) of mass 1.6 kg
increases the speed by 20%, how much falls on it and adheres to it at a certain
will the force of friction be? distance x from the centre. Calculate x if
[Ans: vmin = 11 m/s, =f s m=g 500 N ]
17. A pendulum consisting of a massless the wheel now rotates at 80 rpm.
=1 m
string of length 20 cm and a tiny bob [An=s: x 8 0.35 m ]
of mass 100 g is set up as a conical 22. Starting from rest, an object rolls down
pendulum. Its bob now performs 75 rpm. along an incline that rises by 3 in every
Calculate kinetic energy and increase in 5 (along it). The object gains a speed of
the gravitational potential energy of the 10 m/s as it travels a distance of 5
bob. (Use S 2 10 ) 3
[Ans: cos T 0.8, K.E. = 0.45 J,
m along the incline. What can be the
' P.E 0.04 J]
possible shape/s of the object?
18. A motorcyclist (as a particle) is [Ans: K2 = 1. Thus, a ring or
R2
undergoing vertical circles inside
a sphere of death. The speed of the a hollow cylinder]
25
2. Mechanical Properties of Fluids
Can you recall? A fluid flows under the action of a force
or a pressure gradient. Behaviour of a fluid
1. How important are fluids in our life? in motion is normally complicated. We can
2. What is atmospheric pressure? understand fluids by making some simple
3. Do you feel excess pressure while assumptions. We introduce the concept of an
ideal fluid to understand its behaviour. An
swimming under water? Why? ideal fluid has the following properties:
1. It is incompressible: its density is constant.
2.1 Introduction: 2. Its flow is irrotational: its flow is smooth,
In XIth Std. we discussed the behaviour of
there are no turbulences in the flow.
solids under the action of a force. Among three 3. It is nonviscous: there is no internal friction
states of matter, i.e., solid, liquid and gas, a
solid nearly maintains its fixed shape and in the flow, i.e., the fluid has no viscosity.
volume even if a large force is applied to it. (viscosity is discussed in section 2.6.1)
Liquids and gases do not have their own shape 4. Its flow is steady: its velocity at each point
and they take the shape of the containing is constant in time.
vessel. Due to this, liquids and gases flow under It is important to understand the difference
the action of external force. A fluid means a between a solid and a fluid. Solids can be
substance that can flow. Therefore, liquids and subjected to shear stress (tangential stress) as
gases, collectively, are called fluids. A fluid shown in Fig. 2.1 and normal stress, as shown
either has no rigidity or its rigidity is very low. in Fig.2.2.
In our daily life, we often experience Fig. 2.1: Shear stress.
the pressure exerted by a fluid at rest and in
motion. Viscosity and surface tension play
an important role in nature. We will try to
understand such properties in this chapter.
2.2 Fluid:
Any substance that can flow is a
fluid. A fluid is a substance that deforms
continually under the action of an external
force. Fluid is a phase of matter that
includes liquids, gases and plasmas.
Do you know?
Plasma is one of the four fundamental states (a) Compressive (b) Tensile
of matter. It consists of a gas of ions, free
electrons and neutral atoms. Fig. 2.2: Normal stress.
We shall discuss mechanical properties Solids oppose the shear stress either by
of only liquids and gases in this Chapter. The
shear modulus of a fluid is zero. In simpler developing a restoring force, which means that
words, fluids are substances which cannot
resist any shear force applied to them. Air, the deformations are reversible, or they require
water, flour dough, toothpaste, etc., are some
common examples of fluids. Molten lava is a certain initial stress before they deform and
also a fluid.
start flowing. (We have studied this behavior
of solids (elastic behaviour) in XIth Std).
Ideal fluids, on the other hand, can only
be subjected to normal, compressive stress
(called pressure). Most fluids offer a very
26
weak resistance to deformation. Real fluids Figure 2.4 shows a fluid exerting normal
display viscosity and so are capable of being force on a vertical surface and Fig. 2.5 shows
subjected to low levels of shear stress. fluid exerting normal force on a horizontal
surface.
Fig. 2.3: Forces acting on a small surface dA Fig. 2.4: Fluid exert force on vertical surface.
within a fluid at rest. Fig. 2.5: Fluid exert force on horizontal surface.
The Fig. 2.3 shows a small surface of Thus, an object having small weight
area dA at rest within a fluid. The surface does can exert high pressure if its weight acts on a
not accelerate, so the surrounding fluid exerts small surface area. For example, a force of
equal normal forces dF on both sides of it. 10 N acting on 1 cm2 results in a pressure of
Properties of Fluids: 105 N m-2. On the other hand, the same force of
1. They do not oppose deformation, they get 10 N while acting on an area of 1 m2, exerts a
permanently deformed. pressure of only 10 N m-2.
2. They have ability to flow.
3. They have ability to take the shape of the Remember this
container. 1 N weight is about 100 g mass, if
A fluid exhibits these properties because g =10 m s-2.
it cannot oppose a shear stress when in The SI unit of pressure is N/m2. Also,
static equilibrium. 1 N/m2 = 1 Pascal (Pa). The dimension of
pressure is [L-1M1T-2]. Pressure is a scalar
Remember this quantity. Other common units of measuring
pressure of a gas are bar and torr.
The term fluid includes both the liquid and gas 1 bar = 105 Nm-2
phases. It is commonly used, as a synonym 1 hectapascal (hPa) = 100 Pa
for liquid only, without any reference to gas.
For example, ‘brake fluid’ is hydraulic oil Can you tell?
and will not perform its required function if Why does a knife have a sharp edge,
there is gas in it! This colloquial use of the and a needle has a sharp tip?
term is also common in the fields of medicine
and nutrition, e.g., “take plenty of fluids”. Use your brain power
2.2.1 Fluids at Rest: A student of mass 50 kg is standing on both
feet. Estimate the pressure exerted by the
The branch of physics which deals student on the Earth. Assume reasonable
value to any other quantity you need. Justify
with the properties of fluids at rest is called your assumption. You may use g = 10 m s-2
By what factor will it change if the student
hydrostatics. In the next few sections we will lies on back?
consider some of the properties of fluids at rest.
2.3 Pressure:
A fluid at rest exerts a force on the surface
of contact. The surface may be a wall or the
bottom of an open container of the fluid. The
normal force (F) exerted by a fluid at rest per
unit surface area (A) of contact is called the
pressure (p) of the fluid.
p = FA
--- (2.1)
27
Remember this Remember this
The concept of pressure is useful in dealing 1. As p = hρg, the pressure exerted by a
with fluids, i.e., liquids and gases. As fluids fluid at rest is independent of the shape
do not have definite shape and volume, it
is convenient to use the quantities pressure and size of the container.
and density rather than force and mass when
studying hydrostatics and hydrodynamics. 2. p = hρg is true for liquids as well as for
gases.
2.3.1 Pressure Due to a Liquid Column:
Example 2.1: Two different liquids of
density ρ1 and ρ2 exert the same pressure at
a certain point. What will be the ratio of the
heights of the respective liquid columns?
Fig. 2.6: Pressure due to a liquid column. Solution: Let h1 be the height of the liquid of
density ρ1. Then the pressure exerted by the
liquid of density ρ1 is p1 = h1ρ1g. Similarly,
let h2 be the height of the liquid of density
ρ2. Then the pressure exerted by the liquid
of density ρ2 is p2 = h2ρ2g.
Both liquids exert the same pressure,
A vessel is filled with a liquid. Let us therefore we write,
calculate the pressure exerted by an imaginary p1 = p2 or, h1 U2
∴ h1ρ1g = h2ρ2g h2 U1
cylinder of cross sectional area A inside the Alternate method:
container. Let the density of the fluid be ρ, and For a given value of p = hρg = constant,
the height of the imaginary cylinder be h as as g is constant. So the hight is inversely
shown in the Fig. 2.6. The liquid column exerts proportional to the density of the fluid ρ. In
a force F = mg, which is its weight, on the this case, since pressure is constant, height
bottom of the cylinder. This force acts in the is inversely proportional to density of the
downward direction. Therefore, the pressure p liquid.
exerted by the liquid column on the bottom of Example 2.2: A swimmer is swimming in
cylinder is, a swimming pool at 6 m below the surface
p = FA
∴ p = mAg of the water. Calculate the pressure on the
Now, m = (volume of cylinder) x (density of swimmer due to water above. (Density of
water = 1000 kg/m3, g = 9.8 m/s2)
Solution: Given,
liquid) h = 6 m, ρ = 1000 kg/m3, g = 9.8 m/s2
= (Ah) × ρ = Ahρ p = hρg = 6 × 1000 × 9.8 = 5.88 × 105 N/m2
? p AhU g (Which is nearly 6 times the atmospheric
A pressure!)
p = hρg --- (2.2) 2.3.2 Atmospheric Pressure:
Earth's atmosphere is made up of a fluid,
Thus, the pressure p due to a liquid of
namely, air. It exerts a downward force due
density ρ at rest, and at a depth h below the to its weight. The pressure due to this force
is called atmospheric pressure. Thus, at any
free surface is hρg. point, the atmospheric pressure is the weight
of a column of air of unit cross section starting
Note that the pressure dose not depend
on the area of the imaginary cylinder used to
derive the expression.
28
from that point and extending to the top of the Substituting Eq. (2.4) and Eq. (2.5) in Eq. (2.3)
atmosphere. Clearly, the atmospheric pressure we get,
is highest at the surface of the Earth, i.e., at the p2A = p1A + ρAg (x1- x2)
sea level, and decreases as we go above the p2 = p1 + ρg (x1- x2) --- (2.6)
surface as the height of the column of air above This equation can be used to find the
decreases. The atmospheric pressure at sea pressure inside a liquid (as a function of
level is called normal atmospheric pressure. depth below the liquid surface) and also the
The density of air in the atmosphere decreases atmospheric pressure (as a function of altitude
with increase in height and becomes negligible or height above the sea level).
beyond a height of about 8 km so that the
height of air column producing atmospheric
pressure at sea level can be taken to be 8 km.
The region where gas pressure is less than
the atmospheric pressure is called vacuum.
Perfect or absolute vacuum is when no matter,
i.e., no atoms or molecules are present.
Usually, vacuum refers to conditions when the
gas pressure is considerably smaller than the
atmospheric pressure.
2.3.3 Absolute Pressure and Gauge Pressure: Fig. 2.7: Pressure due to an imaginary cylinder
Consider a tank filled with water as shown of fluid.
in Fig. 2.7. Assume an imaginary cylinder of To find the pressure p at a depth h below
horizontal base area A and height x1- x2 = h. the liquid surface, let the top of an imaginary
x1 and x2 being the heights measured from a
reference point, height increasing upwards: cylinder be at the surface of the liquid. Let
x1 > x2. The vertical forces acting on the this level be x1. Let x2 be some point at depth h
cylinder are: below the surface as shown in Fig. 2.8. Let p0
be the atmospheric pressure at the surface, i.e.,
1. Force cFy1 laincdtserd,oawnndwisardduseattoththeetowpesiugrhfat coef at x1. Then, substituting x1 = 0, p1 = p0, x2 = -h,
of the
and p2 = p in Eq. (2.6) we get,
the water column above the cylinder. p = p0+ hρg --- (2.7)
2. Force Fo2f acts upwards at the bottom The above equation gives the total
surface the cylinder, and is due to the
pressure, or the absolute pressure p, at a depth
water below the cylinder.
h below the surface of the liquid. The total
3. The gravitational force on the water
pressure p, at the depth h is the sum of:
enclosed in the cylinder is mg, where m is
1. p0, the pressure due to the atmosphere,
the mass of the water in the cylinder. As which acts on the surface of the
the water is in static equilibrium, the forces liquid, and
on the cylinder are balanced. The balance 2. hρg, the pressure due to the liquid at depth
of these forces in magnitude is written as, h.
F2= F1+ mg --- (2.3)
p1and p2 are the pressures at the top and
bottom surfaces of the cylinder respectively
due to the fluid. Using Eq. (2.1) we can write
F1 = p1A, and F2 = p2A --- (2.4)
Also, the mass m of the water in the cylinder
can be written as,
m = density × volume = ρV Fig. 2.8. Pressure at a depth h below the surface
of a liquid.
∴m = ρA(x1-x2) --- (2.5)
29
In general, the difference between the base of the vessel B and the liquid from vessel
C would rise into the vessel B. However,
absolute pressure and the atmospheric pressure it is never observed. Equation 2.2 tells that
the pressure at a point depends only on the
is called the gauge pressure. Using Eq. (2.7), height of the liquid column above it. It does
not depend on the shape of the vessel. In this
gauge pressure at depth h below the liquid case, height of the liquid column is the same
for all the vessels. Therefore, the pressure of
surface can be written as, liquid column in each vessel is the same and
the system is in equilibrium. That means the
p - p0 = hρg --- (2.8) liquid in vessel C does not rise in to vessel B.
Eq. (2.8) is also applicable to levels above
the liquid surface. It gives the pressure at a
given height above a liquid surface, in terms
of the atmospheric pressure p0 (assuming that
the atmospheric density is uniform up to that
height).
To find the atmospheric pressure at a
distance d above the liquid surface as shown
in Fig. 2.9, we substitute x1 = d, p1 = p, x2 = 0,
p2 = p0 and ρ = ρair in Eq. (2.6) we get,
p = po - dρair g --- (2.9) (a) (b)
Fig. 2.10: Hydrostatic paradox.
Fig. 2.9: Change of atmospheric pressure Consider Fig. 2.10 (b). The arrows indicate
with height. the forces exerted against the liquid by the walls
of the vessel. These forces are perpendicular to
Can you tell? walls of the vessel at each point. These forces
can be resolved into vertical and horizontal
The figures show three containers filled components. The vertical components act in
with the same oil. How will the pressures at the upward direction. Weight of the liquid in
the reference line compare? section B is not balanced and contibutes the
pressure at the base. Thus, it is no longer a
paradox!
2.3.5 Pascal’s Law:
Pascal’s law states that the pressure
applied at any point of an enclosed fluid at
rest is transmitted equally and undiminished to
every point of the fluid and also on the walls of
the container, provided the effect of gravity is
neglected.
(a) (b) (c) Experimental proof of Pascal’s principle.
Consider a vessel with four arms A, B, C,
2.3.4 Hydrostatic Paradox: and D fitted with frictionless, water tight
pistons and filled with incompressible fluid
Consider the inter connected vessels as shown in the figure given. Let the area of
cross sections of A, B, C, and D be a, 2a, 3a,
as shown in Fig. 2.10 (a). When a liquid is and a/2 respectively. If a force F is applied
on the piston A, the pressure exerted on the
poured in any one of the vessels, it is noticed liquid is p = F/a. It is observed that the other
three pistons B, C, and D move outward.
that the level of liquids in all the vessels is the In order to keep these three pistons B, C,
same. This observation is somewhat puzzling.
It was called 'hydrostatics paradox' before
the principle of hydrostatics were completely
understood.
One can feel that the pressure of the base
of the vessel C would be more than that at the
30
and D in their original positions, forces 2F, transmitted undiminished to the bigger piston
3F, and F/2 respectively are required to be
applied on the pistons. Therefore, pressure S2. A force F2 = pA2 will be exerted upwards
on the pistons B, C, and D is: on it.
F2 F1 § A2 · --- (2.10)
¨ A1 ¸
© ¹
Thus, F2 is much larger than F1. A heavy
load can be placed on S2 and can be lifted up
or moved down by applying a small force on
S1. This is the principle of a hydraulic lift.
Observe and discuss
on B, =pB 22=Fa F Blow air in to a flat balloon using a cycle
a pump. Discuss how Pascal’s principle is
applicable here.
on C, p=C 33=Fa F and
a ii) Hydraulic brakes: Hydraulic brakes are
used to slow down or stop vehicles in motion.
on D, =pD Fa=// 22 F It is based on the same principle as that of a
a hydraulic lift.
i.e. pB = pC = pD = p, this indicates that the Figure 2.12 shows schematic diagram
pressure applied on piston A is transmitted of a hydraulic brake system. By pressing the
equally and undiminished to all parts of the brake pedal, the piston of the master cylinder
fluid and the walls of the vessel. is pushed in forward direction. As a result,
the piston in the slave cylinder which has a
Applications of Pascal’s Law: much larger area of cross section as compared
i) Hydraulic lift: Hydraulic lift is used to lift a to that of the master cylinder, also moves in
heavy object using a small force. The working forward direction so as to maintain the volume
of this machine is based on Pascal’s law. of the oil constant. The slave piston pushes the
friction pads against the rotating disc, which
is connected to the wheel. Thus, causing a
moving vehicle to slow down or stop.
Fig. 2.11 Hydraulic Lift.
As shown in Fig. 2.11, a tank containing
a fluid is fitted with two pistons S1 and S2 . S1 Fig. 2.12 Hydraulic brake system (schematic).
has a smaller area of cross section, A1 while
S2 has a much larger area of cross section, The master cylinder has a smaller area of
A2 (A2 >> A1). If we apply a force F1 on the cross section A1 compared to the area A2 of the
smaller piston S1 in the downward direction it slave cylinder. By applying a small force F1
will generate pressure p = (F1/A1) which will be
31
to the master cylinder, we generate pressure i) Mercury Barometer: An instrument that
p = (F1/A1). This pressure is transmitted measures atmospheric pressure is called a
undiminished throughout the system. The force barometer. One of the first barometers was by
Italian scientist Torricelli. The barometer is
F2 on slave cylinder is then, A2 in the form of a glass tube completely filled
F1 A1 with mercury and placed upside down in a
F2 pA2 A1 u A2 F1 § · small dish containing mercury. Its schematic
¨ ¸ diagram is shown in Fig. 2.13.
© ¹
Torricelli’s vacuum
This is similar to the principle used in
Fig. 2.13: Mercury
hydraulic lift. Since area A2 is greater than A1, barometer.
F2 is also greater than F1. Thus, a small force
applied on the brake pedal gets converted into 1. A glass tube of about 1 meter length and
a diameter of about 1 cm is filled with
large force and slows down or stops a moving mercury up to its brim. It is then quickly
inverted into a small dish containing
vehicle. mercury. The level of mercury in the glass
tube lowers as some mercury spills in the
Example 2.3: A hydraulic brake system dish. A gap is created between the surface
of a car of mass 1000 kg having speed of of mercury in the glass tube and the closed
50 km/h, has a cylindrical piston of radius end of the glass tube. The gap does not
of 0.5 cm. The slave cylinder has a radius contain any air and it is called Torricelli’s
of 2.5 cm. If a constant force of 100 N is vacuum. It dose contain some mercury
applied on the brake what distance the car vapors.
will travel before coming to stop? 2. Thus, the pressure at the upper end of the
mercury column inside the tube is zero, i.e.
Solution: Given, pressure at point such as A is pA= zero.
F1 = 100 N, A1 = π (0.5 × 10-2)2 m2, 3. Let us consider a point C on the mercury
A2 = π (2.5 × 10-2)2 m2, F2 = ? surface in the dish and another point B
By Pascal’s Principle, inside the tube at the same horizontal level
F2 F1 as that of the point C.
A2 = A1
4. The pressure at C is equal to the
atmospheric pressure p0 because it is open
100u S (2.5u10 2 )2 to atmosphere. As points B and C are at the
F2 S (0.5u10 2 )2 2500 N same horizontal level, the pressure at B is
also equal to the atmospheric pressure po,
Acceleration of the car = i.e. pB= po.
a = F2 /m = 2500/1000 = 2.5 m/s2. Using 5. Suppose the point B is at a depth h below
Newton's equation of motion, the point A and ρ is the density of mercury
then,
v2 = u2 -2as where final velocity v = 0, PB= PA + hρg --- (2.11)
u = 50 km/h
s § 50 u1000 ·2 u ( 2 1 38.58 m
©¨ 3600 ¸¹ u 2.5)
2 .3.6 Measurement of Pressure:
Instruments used to measure pressure are
called pressure meters or pressure gauges or
vacuum gauges. Below we will describe two
instruments which are commonly used to
measure pressure.
Caution:
Use of mercury is not advised in a
laboratory because mercury vapours are
hazardous for life and for environment.
32
pA = 0 (there is vacuum above point A) and pressure at C is the same as at D, i.e., inside
pB = p0, therefore, p0 = hρg, where h is the length the chamber. Therefore, the pressure p in the
of mercury column in the mercury barometer. container is,
p = pC
Remember this Using Eq. (2.12) and Eq. (2.13) we can write,
p = p0 + hρg --- (2.14)
The atmospheric pressure is generally
As the manometer measures the gauge
expressed as the length of mercury column pressure of the gas in the container D, we can
write the gauge pressure in the container D as
in a mercury barometer. p - p0 = hρg
patm = 76 cm of Hg = 760 mm of Hg is Can you recall?
Another unit for measuring pressure
torr. One torr = 1 mm of Hg
Can you tell? 1. You must have blown soap bubbles in
your childhood. What is their shape?
What will be the normal atmospheric
pressure in bar and also in torr? 2. Why does a greased razor blade float
ii) Open tube manometer: A manometer on the surface of water?
consists of a U – shaped tube partly filled with
a low density liquid such as water or kerosene. 3. Why can a water spider walk
This helps in having a larger level difference comfortably on the surface of still
between the level of liquid in the two arms of water?
the manometer. Figure 2.14 shows an open
tube manometer. One arm of the manometer 4. Why are free liquid drops and bubbles
is open to the atmosphere and the other is always spherical in shape?
connected to the container D of which the
pressure p is to be measured. 2.4 Surface Tension:
Fig. 2.14 Open tube A liquid at rest shows a very interesting
manometer. property called surface tension. We have
seen that water spider walks on the surface
of steady water, greased needle floats on the
steady surface of water, rain drops and soap
bubbles always take spherical shape, etc.
All these phenomena arise due to surface
tension. Surface tension is one of the important
properties of liquids.
Do you know?
The pressure at point A is atmospheric 1. When we write on paper, the ink sticks
to the paper.
partmesosuspreherpe0. because this arm is open to
To find the pressure at point C, 2. When teacher writes on a board, chalk
particles stick to the board.
which is exposed to the pressure of the gas
3. Mercury in a glass container does not
in the container, we consider a point B in the wet its surface, while water in a glass
container wets it.
open arm of the manometer at the same level
2.4.1 Molecular Theory of Surface Tension:
as point C. The pressure at the points B and C All the above observations can be
is the same, i.e., explained on the basis of different types of
forces coming into play in all these situations.
prepssCu=repaB t poin t B is, --- (2.12) We will try to understand the effect of these
The
pB = ptoh+e hρg --- (2.13)
where, ρ is density of the liquid in the
manometer, h is the height of the liquid column
above point B, and g is the acceleration due forces and their relation to the surface tension
to gravity. According to Pascal’s principle, in liquids.
33
To understand surface tension, we need (a)
to know some terms in molecular theory (b)
that explain the behaviour of liquids at their
surface. Fig. 2.15: (a) sphere of influence and
a) Intermolecular force: Matter is made up (b) surface film.
of molecules. Any two molecules attract each
other. This force between molecules is called (e) Free surface of a liquid: It is the surface
intermolecular force. There are two types of of a fluid which does not experience any shear
intermolecular forces - i) Cohesive force and stress. For example, the interface between
ii) Adhesive force. liquid water and the air above. In Fig. 2.15 (b),
i) Cohesive force: The force of attraction XY is the free surface of the liquid.
between the molecules of the same Remember this
substance is called cohesive force or
force of cohesion. The force of attraction While studying pressure, we considered both
between two air molecules or that between liquids and gases. But as gases do not have
two water molecules is a cohesive force. a free surface, they do not exhibit surface
Cohesive force is strongest in solids and tension.
weakest in gases. This is the reason why
solids have a definite shape and gases do (f) Surface tension on the basis of molecular
not. Small droplets of liquid coalesce into theory: As shown in Fig. 2.15 (b), XY is the
one and form a drop due to this force. free surface of liquid and X′Y′ is the inner
ii) Adhesive force: The force of attraction layer parallel to XY at distance equal to the
between the molecules of different range of molecular force. Hence, the section
substances is called adhesive force or XX′-Y′Y near the surface of the liquid acts as
force of adhesion. The force of attraction the surface film. Consider three molecules A,
between glass and water molecule is a B, and C such that molecule A is deep inside
force of adhesion. the liquid, molecule B within surface film and
b) Range of molecular force: The maximum molecule C on the surface of the liquid.
distance from a molecule up to which the
molecular force is effective is called the range As molecule A is deep inside the liquid, its
of molecular force. Intermolecular forces sphere of influence is also completely inside
are effective up to a distance of the order of the liquid. As a result, molecule A is acted
few nanometer (10-9 m) in solids and liquids. upon by equal cohesive forces in all directions.
Therefore, they are short range forces. Thus, the net cohesive force acting on molecule
c) Sphere of influence: An imaginary sphere A is zero.
with a molecule at its center and radius equal
to the molecular range is called the sphere of Molecule B lies within the surface layer
influence of the molecule. The spheres around and below the free surface of the liquid. A
molecules A, B or C are shown in Fig. 2.15 (a) larger part of its sphere of influence is inside
and (b). The intermolecular force is effective the liquid and a smaller part is in air. Due to
only within the sphere of influence. this, a strong downward cohesive force acts on
d) Surface film: The surface layer of a the liquid molecule. The adhesive force acting
liquid with thickness equal to the range of on molecule B due to air molecules above it
intermolecular force is called the surface film. and within its sphere of influence is weak. It
This is the layer shown between XY and X′Y′ points upwards. As a result, the molecule B
in Fig. 2.15 (b). gets attracted inside the liquid.
34
The same holds for molecule C which lies This force per unit length is the surface
exactly on the free surface of the liquid. Half tension. Surface tension T is defined as, the
of the sphere of influence is in air and half in tangential force acting per unit length on both
the liquid. The number of air molecules within sides of an imaginary line drawn on the free
the sphere of influence of the molecule C, surface of liquid.
above the free surface of the liquid is much less T = FL --- (2.15)
than the number of liquid molecules within the SI unit of surface tension is N/m. Its Dimension
sphere of influence that lies within the liquid. are, [L0M1T-2].
This is because, the density of air is less than
that of a liquid. The adhesive force trying to Use your brain power
pull the molecule above the liquid surface is
much weaker than the cohesive force that tries Prove that, equivalent S.I. unit of surface
to pull the molecule inside the liquid surface. tension is J/m2.
As a result, the molecule C also gets attracted
inside the liquid. Example 2.4: A beaker of radius 10 cm
Thus, all molecules in the surface film is filled with water. Calculate the force
are acted upon by an unbalanced net cohesive
force directed into the liquid. Therefore, the of surface tension on any diametrical line
molecules in the surface film are pulled inside
the liquid. This minimizes the total number of on its surface. Surface tension of water is
molecules in the surface film. As a result, the
surface film remains under tension. The surface 0.075 N/m.
film of a liquid behaves like a stretched elastic
membrane. This tension is known as surface Solution: Given,
tension. The force due to surface tension acts
tangential to the free surface of a liquid. L = 2 × 10 = 20 cm = 0.2 m
2.4.2 Surface Tension and Surface Energy:
a) Surface Tension: As seen previously, the T = 0.075 N/m
free surface of a liquid in a container acts as
a stretched membrane and all molecules on We have, = FL
the surface film experience a stretching force.
Imagine a line PQ of length L drawn tangential T
to the free surface of the liquid, as shown in
Fig. 2.16.
∴F = TL = 0.075 × 0.2 = 0.015
= 1.5 × 10-2 N
Table 2.1 – Surface tension of some liquids at 20oC.
Sr. Liquid S.T. S.T.
No. (N/m) (dyne/cm)
0.0727
1 Water 0.4355 72.7
0.025 435.5
2 Mercury
25
3 Soap
solution
4 Glycerin 0.0632 63.2
Fig. 2.16: Force of surface tension. b) Surface Energy: We have seen that a
molecule inside the volume of a liquid (like
All the molecules on this line experience molecule A in Fig 2.15) experiences more
equal and opposite forces tangential to surface cohesive force than a molecule like molecules
as if they are tearing the surface apart due B and C in the surface film of the liquid in
to the cohesive forces of molecules lying on that figure. Thus, work has to be done to bring
either side. any molecule from inside the liquid into the
surface film. Clearly, the surface molecules
possess extra potential energy as compared
to the molecules inside the liquid. The extra
35
energy of the molecules in the surface layer opposite to F) applied isothermally (gradually
is called the surface energy of the liquid. As and at constant temperature), to the arm
any system always tries to attain a state of QR, so that it pulls the arm away and tries
minimum potential energy, the liquid tries to to increase the surface area of the film. The
reduce the area of its surface film. Energy has arm QR moves to Q′R′ through a distance dx.
to be spent in order to increase the surface area
of a liquid. Therefore, the work done against F, the force
Remember this due to surface tension, is given by
1) Molecules on the liquid surface dw = F ′dx
experience net inward pull. In spite of
this if they remain at the surface, they Using Eq. (2.16),
possess higher potential energy. As a dw = T (2Ldx)
universal property, any system tries to
minimize its potential energy. Hence But, 2Ldx = dA, increase in area of the two
liquid surface tries to minimize its surfaces of the film. Therefore, dw = T(dA).
surface area.
This work done in stretching the film is
2) When a number of droplets coalesce stored in the area dA of the film as its potential
and form a drop, there is reduction energy. This energy is called surface energy.
in the total surface area. In this case,
energy is released to the surrounding. ∴ Surface energy = T (dA) --- (2.17)
Thus, surface tension is also equal to the
c) Relation between the surface energy and
surface tension: Consider a rectangular frame surface energy per unit area.
of wire P′PSS′. It is fitted with a movable arm
QR as shown in Fig. 2.17. This frame is dipped Example 2.5: Calculate the work done
in a soap solution and then taken out. A film
of soap solution will be formed within the in blowing a soap bubble to a radius of
boundaries PQRS of the frame.
1 cm. The surface tension of soap solution is
Fig. 2.17: Surface energy of a liquid
Each arm of the frame experiences an 2.5 × 10-2 N/m.
inward force due to the film. Under the action
of this force, the movable arm QR moves Solution: Given
towards side PS so as to decrease the area
of the film. If the length of QR is L, then this T = 2.5 × 10-2 N/m
inward force F acting on it is given by
F = (T) × (2L) --- (2.16) Initial radius of bubble = 0 cm
Since the film has two surfaces, the upper
surface and the lower surface, the total length Final radius of bubble, r = 1 cm = 0.01 m
over which surface tension acts on QR is
2L. Imagine an external force F′ (equal and Initial surface area of soap bubble = 0
(A soap bubble has two surfaces, outer
surface and inner surface).
Final surface area of soap bubble is,
A = 2 × (4πr 2) = 8πr 2
∴change in area = dA = A – 0 = 8πr2
= 0.00251 m2
∴ work done = T × dA
= 2.5 × 10-2 × 0.251 × 10-2
= 6.275 × 10-5 J
Try this
Take a ring of about 5 cm in diameter. Tie
a thin thread along the diameter of the ring.
Keep the thread slightly loose. Dip the ring
in a soap solution and take it out. A soap
film is formed on either side of thread.
Break the film on any one side of the thread.
Discuss the result.
36
Remember this Do you know?
The work done, under isothermal condition, • when we observe the level of water in a
against the force of surface tension to capillary, we note down the level of the
change the surface area of a liquid is stored tangent to the meniscus inside the water.
as surface energy of liquid.
• When we observe the level of mercury
2.4.3 Angle of Contact: in a capillary we note down the level of
When a liquid surface comes in contact the tangent to the meniscus above the
mercury column.
with a solid surface, it forms a meniscus,
which can be either convex (mercury-glass) or a) Shape of meniscus:
concave (water glass), as shown in Fig. 2.18. i) Concave meniscus - acute angle of contact:
The angle of contact, θ , between a liquid and
a solid surface is defined as the angle between
the tangents drawn to the free surface of the
liquid and surface of the solid at the point of
contact, measured within the liquid.
Fig. 2.18 (a): Concave meniscus due to liquids Fig. 2.19 (a): Acute angle of contact.
which partially wet a solid surface.
Figure 2.19 (a) shows the acute angle of
When the angle of contact is acute, the contact between a liquid surface (e.g., kerosene
liquid forms a concave meniscus Fig. 2.18 (a) in a glass bottle). Consider a molecule such as
at the point of contact. When the angle of A on the surface of the liquid near the wall of
contact is obtuse, it forms a convex meniscus the container. The molecule experiences both
Fig. 2.18 (b). For example, water-glass interface cohesive as well as adhesive forces. In this case,
forms a concave meniscus and mercury-glass since the wall is vertical, the net adhesive force
interface forms a convex meniscus. ( AP ) acting on the molecule A is horizontal,
Net cohesive force ( AC ) acting on molecule is
Fig. 2.18 (b): Convex meniscus due to liquids directed at nearly 45o to either of the surfaces.
which do not wet a solid surface. Magnitude of adhesive force is so large that
the net force ( AR ) is directed inside the solid.
This difference between the shapes of
menisci is due to the net effect of the cohesive For equilibrium or stability of a liquid
forces between liquid molecules and adhesive surface, the net force ( AR ) acting on the
forces between liquid and solid molecules as molecule A must be normal to the liquid surface
discussed below.
at all points. For the resultant force AR to
be normal to the tangent, the liquid near the
wall should pile up against the solid boundary
so that the tangent AT to the liquid surface
is perpendicular to AR. Thus, this makes the
meniscus concave. Obviously, such liquid
wets that solid surface.
ii) Convex meniscus - obtuse angle of
contact:
Figure 2.19 (b) shows the obtuse angle
of contact between a liquid and a solid
(e.g., mercury in a glass bottle). Consider a
molecule such as A on the surface of the liquid
37
near the wall of the container. The molecule iv) Angle of contact 900 and conditions for
experiences both cohesive as well as adhesive convexity and concavity:
forces. In this case also, the net adhesive force
( AP ) acting on the molecule A is horizontal
since the wall is vertical. Magnitude of
cohesive force is so large that the net force
( AR ) is directed inside the liquid.
Fig. 2.19 (d): Acute angle equal to 900.
Consider a hypothetical liquid having
angle of contact 900 with a given solid
container, as shown in the Fig. 2.19 (d). In this
case, the net cohesive force AC is exactly at
Fig. 2.19 (b): Obtuse angle of contact. 450 with either of the surfaces and the resultant
For equilibrium or stability of a liquid force AR is exactly vertical (along the solid
surface, the net force ( AR ) acting on all
molecules similar to molecule A must be surface).
normal to the liquid surface at all points. The
liquid near the wall should, therefore, creep For this to occur, AP = AC where, AR is
inside against the solid boundary. This makes 2
the meniscus convex so that its tangent AT is
normal to AR. Obviously, such liquid does not the magnitude of the net force. From this we
wet that solid surface.
iii) Zero angle of contact : can write the conditions for acute and obtuse
angles of contact: AC
2
For acute angle of contact, AP > , and for
obtuse angle of contact, AP < AC .
2
Can you tell?
How does a water proofing agent work?
b) Shape of liquid drops on a solid surface:
When a small amount of a liquid is
dropped on a plane solid surface, the liquid
Fig. 2.19 (c): Angle of contact equal to zero. will either spread on the surface or will form
Figure 2.19 (c) shows the angle of droplets on the surface. Which phenomenon
contact between a liquid (e.g. highly pure
water) which completely wets a solid will occur depends on the surface tension of
(e.g. clean glass) surface. The angle
of contact in this case is almost zero (i.e., the liquid and the angle of contact between
θ → 00). In this case, the liquid molecules near
the contact region, are so less in number that the liquid and the solid surface. The surface
the cohesive force is negligible, i.e., AC = 0
and the net adhesive force itself is the resultant tension between the liquid and air as well as
force, i.e., AP = AR . Therefore, the tangent
AT is along the wall within the liquid and the that between solid and air will also have to be
angle of contact is zero.
taken in to account.
Let θ be the angle of contact for the given
solid-liquid pair.
T1 = Force due to surface tension at the liquid-
solid interface,
T2 = Force due to surface tension at the air-
solid interface,
38
T3 = Force due to surface tension at the air- Table 2.2 – Angle of contact for pair of
liquid interface. liquid - solid in contact.
As the force due to surface tension is Sr. Liquid - solid in contact Angle of
tangential to the surfaces in contact, directions No. contact
of T1, T2 and T3 are as shown in the Fig. 2.20. 1 Pure water and clean glass 0°
For equilibrium of the drop,
2 Chloroform with clean 00
T2 T1 T3cosT , cosT T2 T1 --- (2.18)
T3 glass
From this equation we get the following cases: 3 Organic liquids with clean 00
1) aIfnTd2 >thTe1 aanngdle(To2-fT1c)o<ntTa3c,tcoθs θ is positive glass
is acute as
4 Ether with clean glass 160
shown in Fig. 2.20 (a). 5 Kerosene with clean glass 260
6 Water with paraffin 1070
7 Mercury with clean glass 1400
Solid 2.4.4 Effect of impurity and temperature on
surface tension:
Fig. 2.20 (a): Acute angle. a) Effect of impurities:
i) When soluble substance such as common
2) If T2 < T1 and (T1 – T2) < T3, cosθ is
negative, and the angle of contact θ is salt (i.e., sodium chloride) is dissolved
obtuse as shown in Fig. 2.20(b). in water, the surface tension of water
increases.
Fig. 2.20 (b): Obtuse angle. ii) When a sparingly soluble substance such
as phenol or a detergent is mixed with
3) If (T2 – T1) = T3, cosθ = 1 and θ is nearly water, surface tension of water decreases.
equal to zero. For example, a detergent powder is mixed
with water to wash clothes. Due to this,
4) If (T2 – T1) >T3 or T2> (T1 + T3), cosθ > 1 the surface tension of water decreases and
which is impossible. The liquid spreads water makes good contact with the fabric
over the solid surface and drop will not be and is able to remove tough stains.
formed. iii) When insoluble impurity is added
into water, surface tension of water
c) Factors affecting the angle of contact: decreases. When impurity gets added
The value of the angle of contact depends on to any liquid, the cohesive force of that
the following factors, liquid decreases which affects the angle
i) The nature of the liquid and the solid in of contact and hence the shape of the
meniscus. If mercury gathers dust then
contact. its surface tension is reduced. It does not
ii) Impurity : Impurities present in the liquid form spherical droplets unless the dust is
completely removed.
change the angle of contact. b) Effect of temperature: In most liquids,
iii) Temperature of the liquid : Any increase as temperature increases surface tension
decreases. For example, it is suggested that
in the temperature of a liquid decreases its new cotton fabric should be washed in cold
angle of contact. For a given solid-liquid water. In this case, water does not make good
surface, the angle of contact is constant at contact with the fabric due to its higher surface
a given temperature. tension. The fabric does not lose its colour
because of this.
39
Hot water is used to remove tough stains Fig. 2.21 (b) : Convex surface.
on fabric because of its lower surface tension. downward force f A . This develops greater
pressure at point B, which is inside the liquid
In the case of molten copper or molten and on the concave side of the meniscus. Thus,
cadmium, the surface tension increases with the pressure on the concave side i.e., inside the
increase in its temperature. liquid is greater than that on the convex side
i.e., outside the liquid.
The surface tension of a liquid becomes c) Concave liquid surface:
zero at critical temperature.
2.4.5 Excess pressure across the free surface Fig. 2.21 (c): Concave Surface.
of a liquid: Surface of the liquid in the Fig. 2.21 (c)
is upper concave (concave, when seen from
Every molecule on a liquid surface above). In this case, the force due to surface
experiences forces due to surface tension tension fT , on the molecule at B is vertically
which are tangential to the liquid surface at upwards. The force f A due to atmospheric
rest. The direction of the resultant force of pressure acts downwords. Forces f A and fT
surface tension acting on a molecule on the thus, act in opposite direction. Therefore,
liquid surface depends upon the shape of that the net downward force responsible for the
liquid surface. This force also contributes in pressure at B is less than f A . This develops a
deciding the pressure at a point just below the lesser pressure at point B, which is inside the
surface of a liquid. liquid and on the convex side of the meniscus.
Thus, the pressure on the concave side i.e.,
Figures 2.21 (a), (b) and (c) show surfaces outside the liquid, is greater than that on the
of three liquids with different shapes and their convex side, i.e., inside the liquid.
menisci. Let f A be the downward force due to 2.4.6 Explanation of formation of drops and
the atmospheric pressure. All the three figures bubbles:
show two molecules A and B. The molecule A Liquid drops and small bubbles are
is just above, and the molecule B is just below spherical in shape because the forces of surface
it (inside the liquid). Level difference between tension dominate the gravitational force.
A and B is almost zero, so that it does not These force always try to minimize the surface
contribute anything to the pressure difference. area of the liquid. A bubble or drop does not
In all the three figures, the pressure at the point collapse because the resultant of the external
A is the atmospheric pressure p. pressure and the force of surface tension is
a) Plane liquid surface: smaller than the pressure inside a bubble or
inside a liquid drop.
Figure 2.21 (a) shows planar free surface
of the liquid. In this case, the resultant force
idsuzeetroo.suTrhfeacfeortceensifoAni,tsfeTlfodnecthideems tohleecpurleessautrBe
and the pressure at A and B is the same.
Fig. 2.21 (a): Plane surface.
b) Convex liquid surface:
Surface of the liquid in the Fig.2.21 (b)
is upper convex. (Convex, when seen from
above). In this case, the resultant force due
to surface tension, fT on the molecule at B
is vertically downwards and adds up to the
40
Consider a spherical drop as shown in surface and the outer surface. For a bubble,
Fig. 2.22. Let pi be the pressure inside the drop
and p0 be the pressure out side it. As the drop is Eq. (2.19) charges to dA = 2(8πr∆r). Hence, total
spherical in shape, the pressure, pi, inside the
drop is greater than p0, the pressure outside. increase in the surface area of a soap bubble,
Therefore, the excess pressure inside the drop
is pi- p0. while increasing its radius by ∆r, is 2(8πr∆r)
The work done by this excess pressure is
dW = (pi – p0) 4πr 2∆r = T(16πr∆r)
4T
∴ (pi – p0) = r --- (2.24)
Remember this
Fig. 2.22. Excess pressure inside a liquid drop. The gravitational force acting on a
Let the radius of the drop increase from molecule, which is its weight, is also one
of the forces acting within the sphere of
r to r + ∆r, where ∆r is very small, so that influence near the contact region. However,
within the sphere of influence, the cohesive
the pressure inside the drop remains almost and adhesive forces are so strong that the
gravitational force can be neglected in the
constant. above explanation.
Let the initial surface area of the drop be Brain teaser:
1. Can you suggest any method to measure
∴∴Adr1oAA=p22 A==b4esπ44∆Arππ r22(r,r2=i2sa++4nv8πde2πrr(ytrr∆h∆+serrm∆+f+rai)n∆l24l.ar,πl ∆2∆)srru 2 2rcfaancebearneeagloefcttehde,
∴ A2 = 4πr 2 + 8πr∆r the surface tension of a soap solution?
Thus, increase in the surface area of the drop is Will this method have any commercial
application?
dA = A2 – A1 = 8πr∆r --- (2.19) 2. What happens to surface tension under
Work done in increasing the surface area different gravity (e.g. Space station or
lunar surface)?
Example 2.6: What should be the diameter
by dA is stored as excess surface energy. of a water drop so that the excess pressure
∴dW = TdA= T (8πr∆r) --- (2.20) inside it is 80 N/m2? (Surface tension of
This work done is also equal to the product water = 7.27 × 10-2 N/m)
of the force F which causes increase in the area Solution: Given
of the bubble and the displacement ∆r which is pi – po = 80 N/m2
T = 7.27 × 10-2 N/m
the increase in the radius of the bubble.
∴ dW = F∆r --- (2.21) We have, 2T
(pi – po) = r
The excess force is given by,
(Excess pressure) × (Surface area) ∴r= 2T 2u7.27u10 2 1.8 u10 3 m
pi po 80
∴F = (pi – p0) 4πr 2 --- (2.22)
Equating Eq. (2.20) and Eq. (2.21), we get, ∴d = 2r = 3.6 mm
T(8πr∆r) = (pi – p0) 4πr 2∆r
2T
∴(pi – p0) = r --- (2.23) 2.4.7 Capillary Action:
A tube having a very fine bore ( ~ 1 mm)
This equation gives the excess pressure
and open at both ends is called a capillary
inside a drop. This is called Laplace’s law of a tube. If one end of a capillary tube is dipped in
a liquid which partially or completely wets the
spherical membrane. surface of the capillary (like water in glass)
the level of liquid in the capillary rises. On the
In case of a soap bubble there are two
free surfaces in contact with air, the inner
41
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ENGLISH MEDIUM
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