Mechanics - Shukla
731
-
Upload
elleguen09 -
Category
Documents
-
view
473 -
download
16
Transcript of Mechanics - Shukla
This pageintentionally left
blank
Copyright © 2006, New Age International (P) Ltd., PublishersPublished by New Age International (P) Ltd., Publishers
All rights reserved.No part of this ebook may be reproduced in any form, by photostat, microfilm,xerography, or any other means, or incorporated into any information retrievalsystem, electronic or mechanical, without the written permission of the publisher.All inquiries should be emailed to [email protected]
PUBLISHING FOR ONE WORLD
NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS4835/24, Ansari Road, Daryaganj, New Delhi - 110002Visit us at www.newagepublishers.com
ISBN (13) : 978-81-224-2709-7
PREFACE
The present book is based on my experience, extending over a period of about fourteenyears of teaching graduate and postgraduate students of the University of Lucknow.
Various universities have continuously been updating postgraduate courses in physics.The courses at senior school level have also been fairly modernized. This book has beenprepared for students of B.Sc. (General) and B.Sc. (Hons.) keeping these trends in mind. Thesubject matter has been selected and developed in such a manner so as to provide a bridgebetween these advanced and introductory level courses.
Mechanics being the vital building-block of physics, the treatment of key concepts hasto be fairly exhaustive. Only then the students would be able to comprehend advanced levelcourses in classical mechanics, quantum mechanics and other areas of physics at postgradu-ate level.
Keeping in view the special requirements of undergraduate students in particular anumber of worked-out examples are distributed throughout the book to elucidate and, insome cases, to extend the various concepts as well as to give the students a good grasp ofthe basic concepts and theories of mechanics.
This book is designed as a textbook according to the updated syllabus primarily for theundergraduate physics and applied physics students of the various Indian universities andcolleges.
This book while being equally useful and inevitable for the students preparing forcompetitive examinations for admission into engineering and medical colleges will continueto be useful also for the first year engineering students.
Some advanced topics like “special theory of relativity” have also been treated in detailin this book.
Some of the problems of mechanics which are expected to help the students gain morepractical knowledge in the subject have been given as follow-up problems at the end of eachchapter.
In other words, the presentation is oriented towards introducing the fundamental con-cepts of mechanics leading to, and associated with, practical applications.
The most distinguishing feature of the book is that the material has been treated andpresented—and the various results derived—systematically and thoroughly, yet to the pointas to make it suitable for self-study by the students.
Suitable diagrams are given in the book to illustrate the basic principles.
Short-Answer Type Questions with Answers have been given in each chapter to helpstudents in the comprehension and appreciation of the finer points of physics.
The book has been written in a very simple and lucid way. Every effort has been madeto make the treatments simple and comprehensive. The difficult topics are explained withthe help of neat and clean diagrams.
Throughout the book emphasis has been laid on the physical concepts.
I am absolutely sure that all readers, particularly students, will be tremendously benefittedby this book.
This book will be extremely useful to teachers and students of various courses—involvingstudies in fundamental sciences—being pursued in various institutions across the globe.
AUTHORS
(vi)
CONTENTS
Preface (v)
1. MEASUREMENT 1
1.1 What is Physics 1
1.2 Scope and Excitement in Physics 1
1.3 Measurement 2
1.4 Physical Quantities 3
1.5 Fundamental Units of S.I. System 3
1.6 Reference Frames 4
1.7 Inertial and Non-inertial Frames 5
1.8 Scalars and Vectors 9
1.9 Addition of Vectors 11
1.10 Resolution of Vectors 16
1.11 Multiplication of Vectors 17
1.12 Vectors and the Laws of Physics 21
1.13 Speed and Velocity 22
1.14 Acceleration 24
1.15 Rectilinear Motion 24
1.16 Acceleration of Gravity 25
1.17 Accuracy and Errors in Measurement 25
SUMMARY 29
2. FORCE AND MOTION 30
2.1 Mechanics 30
2.2 Cause of Motion: Force 30
2.3 Newton’s Laws of Motion 31
2.4 Freely Falling Bodies 33
2.5 Motion in a Vertical Plane 33
2.6 Projectile Motion 34
2.7 Equilibrium of Forces 37
2.8 Frictional Forces 37
2.9 Static Friction and Coefficient of Static Friction 38
2.10 Dynamic Friction and Coefficient of Dynamic Friction 38
PROBLEM 39
3. DYNAMICS OF CIRCULAR MOTION AND THE
GRAVITATIONAL FIELD 86
3.1 Uniform Circular Motion 86
3.2 Centrifugal Force 87
3.3 The Centrifuge 88
3.4 Banking of Curved Roads and Railway Tracks (Banked Track) 88
3.5 Bicycle Motion 89
3.6 Conical Pendulum 90
3.7 Motion in a Vertical Circle 91
3.8 Motion of Planet 92
3.9 Kepler’s Laws of Motion 93
3.10 Derivation of Law of Gravitation 93
3.11 Newton’s Conclusions from Kepler’s Laws 94
3.12 Newton’s Universal Law of Gravitations 95
3.13 Gravity and the Earth 96
3.14 Acceleration due to Gravity 96
3.15 Expression of Acceleration due to Gravity g in Terms ofGravitations Constant G 96
3.16 Difference between Mass and Weight 97
3.17 Inertial Mass and Gravitational Mass 97
3.18 Gravitational Field and Potential 97
3.19 Equipotential Surface 98
3.20 Weightlessness in Satellites 99
3.21 Velocity of Escape 100
3.22 Relation between Orbital Velocity and Escape Velocity 101
3.23 Satellites 101
3.24 Orbital Velocity of Satellite 101
3.25 Orbital Speed and Period of Revolution of a Satellite VeryClose to Earth 103
3.26 Artificial Satellites 104
3.27 Black Holes 105
4. WORK, ENERGY AND MOMENTUM 142
4.1 Work 142
(viii)
4.2 Power 143
4.3 Work in Stretching a Spring 143
4.4 Energy 144
4.5 Kinetic Energy 144
4.6 Potential Energy 145
4.7 Gravitational Potential Energy 145
4.8 Work-Energy Theorem 146
4.9 Significance of the Work-Energy Theorem 147
4.10 Conservative Force : First Definition 147
4.11 A Central force is Conservative 148
4.12 Non-Conservative Force 149
4.13 Relation between Conservative Force and Potential Energy 151
4.14 The Curl of a Conservative Force is Zero 152
4.15 Linear Restoring Force 154
5. LINEAR AND ANGULAR MOMENTUM 222
5.1 Conservation of Linear Momentum 222
5.2 Centre of Mass 223
5.3 Cartesian Components of the Centre of Mass 225
5.4 Centre of Mass of a Solid Body 225
5.5 Position Vector of the Centre of Mass 226
5.6 Velocity of the Centre of Mass 226
5.7 Center of Mass Frame of Reference 227
5.8 Motion of the Center of Mass of a System of Particles
Subject to External Forces 227
5.9 Linear Momentum in Center of Mass Frame of Reference 228
5.10 System of Variable Mass 228
5.11 Motion of a Rocket 230
5.12 Multi Stage Rocket 232
6. COLLISION 241
6.1 Collision 241
6.2 Elastic Collision in One Dimension 241
6.3 The Ballistic Pendulum 245
6.4 Collision in Two Dimension 246
6.5 Value of the Scattering Angle 247
6.6 Scattering Cross-Section 249
(ix)
6.7 Differential Scattering Cross-Section 250
6.8 Total Cross-Section 251
6.9 Impact Parameters 251
6.10 Rutherford Scattering 253
7. ROTATIONAL KINEMATICS 311
7.1 Rigid Body 311
7.2 Moment of a Force or Torque 311
7.3 Angular Acceleration 311
7.4 Relation between Angular Acceleration andLinear Acceleration 312
7.5 Kinetic Energy of Rotation 313
7.6 Angular Momentum 314
7.7 Relation between Torque and Angular Momentum 314
7.8 Conservation of Angular Momentum 315
7.9 Torque Acting on a Particle 315
7.10 Angular Momentum of the Center of Mass of a System ofParticles 316
7.11 Precession 318
7.12 The Top (Precession of a Top Spinning in Earth’s
Gravitational Field) 318
7.13 The Gyrostat 321
7.14 Man with Dumb-bell on a Rotating Table 322
8. MOMENT OF INERTIA 336
8.1 Moment of Inertia 336
8.2 Role of Moment of Inertia in Rotational Motion 337
8.3 Radius of Gyration 337
8.4 Analogous Parameters in Translational and Rotational Motion 338
8.5 General Theorems on Moment of Inertia 339
8.6 Calculation of Moment of Inertia 342
8.7 Moment of Inertia of a Uniform Rod 342
8.8 Moment of Inertia of a Rectangular Lamina (or Bar) 343
8.9 Moment of Inertia of a Thin Circular Ring (or a Hoop) 345
8.10 Moment of Inertia of a Circular Lamina or Disc 345
8.11 Moment of Inertia of an Angular Ring or Disc 346
8.12 Momnt of Inertia of a Solid Cylinder 347
(x)
8.13 Moment of Inertia of a Solid Cone 348
8.14 Moment of Inertia of a Hollow Cylinder 350
8.15 Moment of Inertia of a Spherical Shell 351
8.16 Moment of Inertia of a Solid Sphere 351
8.17 Moment of Inertial of a Hollow Sphere or a Thick Shell 352
8.18 M.I. of a Uniform Triangular Lamina 353
8.19 Kinetic Energy of Rotation 354
8.20 A Body Rolling Down an Inclined Plane(Its Acceleration Along the Plane) 356
8.21 Identification of Hollow and Solid Sphere 358
8.22 Compound Pendulum 358
8.23 Fly Wheels 360
9. MECHANICAL PROPERTIES OF MATTER 393
9.1 Rigid Body 393
9.2 Elasticity 393
9.3 Perfectly Elastic and Perfectly Plastic 393
9.4 Stress 393
9.5 Strain 394
9.6 Hook’s Law 394
9.7 Elastic Limit 394
9.8 Young’s Modulus 395
9.9 Bulk Modulus of Elasticity 396
9.10 Modulus of Rigidity (Shear Modulus) 396
9.11 Poisson’s Ratio 397
9.12 Potential Energy in Stretched Wire 397
9.13 Equivalence of a Shear to a Tensile and a CompressiveStrain at Right Angles to Each Other and Each Equal toHalf the Shear 398
9.14 Equivalence of Compression and Equal PerpendicularExtension to a Shear 399
9.15 Equivalence of a Shearing Stress to an Equal ExtensionalStress Plus an Equal and Perpendicular Compressional Stress 400
9.16 Relations Connecting the Elastic Constants 401
9.17 Theoretical Limiting Values of Poisson’s Ratio 404
9.18 Poisson’s Ratio for an Incompressible Material 405
9.19 Twisting Couple on a Cylinder 406
(xi)
9.20 Determination of the coefficient of Rigidity (η)for the Material of a Wire 408
9.21 Torsional Oscillations 411
9.22 Determination of Modulus of Rigidity of a TorsionalOscillations 411
9.23 Beam 412
9.24 Bending Moment 413
9.25 Cantilever 416
9.26 Beam Supported at its Ends and Loaded in the Middle 418
9.27 Determination of Young’s Modulus by Bending of a Beam 420
9.28 Determination of Elastic Constants by Searle’s Method 421
10. FLUIDS 463
10.1 Molecular Forces 463
10.2 Definition of Surface Tension 464
10.3 Explanation of Surface Tension 465
10.4 Surface Energy 465
10.5 Relation between Surface Tension and Work Done inIncreasing the Surface Area 465
10.6 Shape of Liquid Meniscus in a Glass Tube 466
10.7 Angle of Contact 467
10.8 Capillary Action 468
10.9 Rising of Liquid in a Capillary Tube of Insufficient Length 469
10.10 Effects on Surface Tension 469
10.11 Ideal Liquid 470
10.12 Steady or Stream Line Flow 470
10.13 Equation of Continuity of Flow 471
10.14 Energy of the Fluid 472
10.15 Bernoulli’s Theorem 473
10.16 Velocity of Efflux 475
10.17 Velocity of Efflux of a Gas 476
10.18 Viscosity 477
10.19 Flow of Liquid in a Tube: Critical Velocity 478
10.20 Velocity Gradient and Coefficient of velocity 478
10.21 Poiseuille’s Equation for Liquid-Flow through aNarrow Tube 479
10.22 Poiseuille’s Method for Determining Coefficient of Viscosityof a Liquid 481
(xii)
10.24 Rate of Liquid-Flow through Capillaries in Parallel 483
10.25 Poiseuille’s Formula Extended to Gases 483
10.26 Determination of Viscosity of a Gas 484
10.27 Stoke’s Law of Viscous Force 486
10.28 Stoke’s Formula for the Terminal Velocity of a FallingSphere 487
10.29 Velocity of Rain Drops 487
11. HARMONIC MOTION 523
11.1 Periodic Motion 523
11.2 Simple Harmonic Motion (S.H.M.); As a Projection ofUniform Circular Motion 523
11.3 Displacement Equation of S.H.M. 524
11.4 Conditions for Linear S.H.M. 525
11.5 Equation of Motion of a Simple Harmonic Oscillator 526
11.6 Importance of S.H.M. 527
11.7 Energy of Harmonic Oscillator 527
11.8 Average Values of Kinetic and Potential Energies 529
11.9 Position-Average of Kinetic and Potential Energies 530
11.10 Fractions of Kinetic and Potential Energies 530
11.11 Mass Attached to a Horizontal Spring 531
11.12 Motion of a Body Suspended by a Vertical Spring 532
11.13 Mass Suspended by a Heavy Spring 533
11.14 Oscillations of a Floating Cylinder 535
11.15 Oscillations of a Liquid in a U-tube 536
11.16 Helmholtz Resonator 536
11.17 Composition of Two Simple Harmonic Motions of EqualPeriods in a Straight Line 537
11.18 Composition of Two Rectangular S.H.M. of Equal Periods 538
11.19 Composition of Two Rectangular S.H.M.’s of Time PeriodsNearly Equal 540
11.20 Lissajou’s Figures from Two Rectangular S.H.M. inFrequency Ratio 2:1 542
11.21 Uses of Lissajous Figures 544
12. WAVE MOTION 576
12.1 Wave Motion 576
12.2 Transverse Wave 576
(xiii)
12.3 Longitudinal Wave 576
12.4 General Equation of Wave Motion 577
12.5 Equation of a Plane Progressive Harmonic Wave 578
12.6 Principle of Superposition 580
12.7 Longitudinal Waves in Rods 580
12.8 Longitudinal Waves in Gases (Pressure Variations for PlaneWaves) 581
12.9 Waves in a Linear Bounded Medium 583
12.10 Flow of Energy in Stationary Waves 588
12.11 Characteristics of Stationary Waves 590
12.12 Wave Velocity (or Phase Velocity) 590
12.13 Group Velocity 591
12.14 Relation between Group Velocity and Wave Velocity 592
12.15 Normal and Anomalous Dispersion 593
12.16 Non-dispersive Medium 593
13. STATIONARY WAVES (WAVES IN A LINEAR
BOUNDED MEDIUM) 597
14. DAMPED AND FORCED HARMONIC OSCILLATION 608
14.1 Damping Force 608
14.2 Damped Harmonic Oscillator 609
14.3 Logarithmic Decrement 612
14.4 Power Dissipation in Damped Harmonic Oscillator 613
14.5 Quality Factor Q 614
14.6 Forced (Driven) Harmonic Oscillator 615
14.7 Amplitude Resonance 617
14.8 Sharpness of Resonance 619
14.9 Velocity Resonance 620
14.10 Power Absorption 621
14.11 Driven LCR Circuit 623
15. RELATIVITY 643
15.1 Back Ground of Michelson-Morley Experiment 643
15.2 The Speed of Light Relative to Earth 645
15.3 Michelson Morley Experiment 646
15.4 Explanation of Negative Results: Principle of Constancy ofSpeed of Light 648
(xiv)
15.5 The Relativity of Simultaneity: The Relativistic Concept ofSpace and Time 649
15.6 Postulates of Special Theory of Relativity 651
15.7 Lorentz Transformation Equations 651
15.8 Length Contraction 656
15.9 Time Dilation 659
15.10 An Experimental Verification of Time Dilation 659
15.11 Transformation and Addition of Velocities 663
15.12 Relativistic Doppler’s Effect 666
15.13 Confirmation of Doppler’s Effect 670
15.14 Conservation of Momentum: Variation of Mass with Velocity 671
15.15 Mass-Energy Relation 673
15.16 Relation between Momentum and Energy 680
15.17 Transformation of Momentum and Energy 681
Appendices 685
(xv)
This pageintentionally left
blank
1.1 WHAT IS PHYSICS
Physics is branch of science, which deals with the study of the phenomena of nature.The word ‘physics’ is derived from a Greek word meaning nature. The word ‘Science’ comesfrom the Latin word scientia, which means ‘to know’.
Man has been observing various natural phenomena from time immemorial. He hasalways been curious about nature and the world around him. The motion of the moon andother heavenly bodies in the sky has aroused owe and amazement in him. The regularrepetition of sunrise and sunset and the seasons of the year have fascinated him. Man hasobserved these and other natural phenomena and responded to them in an orderly manner.The experiences gained over a period of time were transmitted from generation to generationand began to be termed as ‘knowledge’. Each generation added new fact to the knowledgeobtained from the previous generation. The systematized knowledge thus gained was termed‘Science’.
Every bit of knowledge is not called science. Only such knowledge as is collected by whatis called the scientific method. The scientific method is the basis of scientific development.The method involves four steps:
1. Observation of the relevant facts,
2. Proposal of a hypothesis or a theory based on these observation,
3. Testing of the proposed theory to see if its consequences or predictions are actuallyobserved in practice, and
4. Modification of the theory, if necessary.
1.2 SCOPE AND EXCITEMENT IN PHYSICS
The various sciences may be divided into two broad classes, physical and biological.Physical sciences deal with nonliving matter and biological sciences with living matter. Abouta hundred years ago, it was possible for one man to master the knowledge of both sciencesand many outstanding workers in physical sciences were also competent doctors and biologists.There was no clear-cut division between the several branches of physical sciences, as weknow them today. In fact, all were included in the term natural philosophy. Aristotle, Archimedesand even Galileo and Newton called themselves natural philosophers. But today the situation
1
MEASUREMENT
1
2 Mechanics
is different. The tremendous upsurge of scientific activity and the accumulation of knowledgein the last century have forced scientists to narrow down their field of activity. Not even agenius could hope to keep up with the developments in the various branches of physicalsciences as we know them today, namely, physics, chemistry, astronomy, biology, medicine,geology, engineering etc.
Those concerned mainly with the application of science to the betterment of human lifeand environment are called engineers. The invention of the steam engine and the electricmotor gave rise to engineering. Physicists and chemists are concerned more with the basicaspects of nonliving matter chemistry deals primarily with molecular changes and therearrangement of the atoms that form molecules. Physics deals with the phenomena of thenon-living world such as mechanics (motion), heat, sound, electricity, magnetism and light.The division of the subject of physics into these and other branches is largely a matter ofconvenience. These branches are inter-related, as you will discover in the course of yourstudy of the subject. Physics may be defined as that branch of knowledge that deals with thephenomena of non-living matter.
In physics, we deal with many physical phenomena and experiences. Merely readingabout the experiences and observation of others is not enough. If students are to understandand enjoy physics, they must have same of these experiences themselves. These experiencesare not only exciting but also very educative. The swinging hanging lamp in a church ledGalileo to a method of measuring time. The fall of an apple and the motion of the moon ledNewton to his famous law of gravitation. The rattling (or dancing) of the lid of a kettle ledto the invention of the steam engine. The flowing of a flute causes vibrations that producesound. The light from stars tells us something about stars and their evolution. The study ofelectricity helps us to design motors and dynamos. The study of semiconductors helps us todesign radios, televisions, calculators and even computers.
1.3 MEASUREMENT
Observation can be subjective or objective. An observation that varies from individual toindividual is subjective. For example, different individuals observing the same thing, e.g. apainting or a flower, feel differently. Physics does not deal with such subjective observations.Physics is a science of objective observation, an observation that is the same for all individuals.An individual observer through his sense of touch or sight, but these senses is not alwaysreliable. To illustrate the inaccuracy of our sense of touch, we consider three pans containingcold, warm and hot water. If you put your finger first in cold water and then in warm water,your sense of touch will tell you that it is hot. But if you put your finger first in hot waterand then in warm water, your sense tells you it is cold. This clearly suggests the necessityof making a measurement to arrive at the truth. It is necessary to measure the degree ofhotness of water in each pan. In other words, it is not enough to describe a phenomenon ina general and qualitative way. A number must be tied to it. Thus, physics is a science ofmeasurement.
Lord Kelvin, a leading physicist of the 19th Century, once said: “When you can measurewhat you are talking about and express it in numbers, you know something about it; butwhen you cannot, your knowledge is of a meagre and unsatisfactory kind; it may be thebeginning of knowledge, but you have scarcely in your thoughts advanced to the stage of ascience”.
Measurement 3
1.4 PHYSICAL QUANTITIES
Physical quantities are often divided into fundamental quantities and derived quantities.Derived quantities are those whose defining operations are based on other physical quantities.Fundamental quantities are not defined in terms of other physical quantities. The number ofquantities regarded as fundamental is the minimum number needed to give a consistent andunambiguous description of all the quantities of physics. All Physical quantities occurring inmechanics can be expressed in the units of ‘length’, ‘mass’ and ‘time’. The units of these threequantities are independent of one another and no one can be changed or related to any otherunit. These quantities are called ‘fundamental quantities’ and their units are called ‘fundamentalunits’.
In the same way, to fix the units of physical quantities occurring in electromagnetism,thermodynamics and optics, electric current, temperature and luminous intensity are takenas fundamental quantities and their units as ‘fundamental units’.
The system of units based on the units of the seven fundamental quantities (length,mass, time, electric current, temperature, luminous intensity and amount of substance) iscalled International system of units. It is abbreviated as SI from the French name Le SystemeInternational d’units. It is based on the following seven fundamental (or Basic) and twosupplementary Units:
Basic physical Quantity Name of the Unit Symbol
1. length meter m
2. mass kilogram kg
3. time second s
4. Electric Current ampere A
5. Temperature Kelvin K
6. Luminous intensity candela Cd
7. Amount of substance mole mol
Supplementary physical Name of the Unit SymbolQuantity
1. Plane angle radian rad
2. Solid angle steradian Sr
1.5 FUNDAMENTAL UNITS OF S.I. SYSTEM
Basic and Supplementary Units of SI SystemThe seven fundamental and two supplementary units of SI system are defined as follows:
1. Metre: On atomic standard meter is defined as to be equal to 1,650,763.73wavelengths in vacuum of the radiation emitted due to transition between thelevels 2p10 and 5d5 of the isotope of krypton having mass number 86. Krypton-86 emits light of several different wavelengths. The light emitted by Krypton-86due to transition between the levels 2p10 5d5 in orange red in colour and haswavelength 6057.8021 Å or 6.0578021 × 10–7 m. The number of these wavelengthsin 1 m can be counted by using an optical interferometer which comes out to be1,650,763.3.
4 Mechanics
2. Kilogram: There is no definition of unit kilogram on atomic standards. Thereforein SI system, kilogram is the mass of a platinum-iridium cylinder kept in theInternational Bureau of weights and measures at Paris. In practice, the mass of1 litre of water at 4°C is 1 kilogram.
3. Second: Unit second can again be defined on atomic standards. One second isdefined to be equal to the duration of 9,192,631,770 vibrations corresponding to thetransition between two hyperfine levels of Caesium –133 atom in the ground state.
4. Kelvin: It was adopted as the unit of temperature. The fraction 1/273.16 of thethermodyamics temperature of triple point of water is called 1K.
5. Ampere: It was adopted as the unit of current. It is defined as the current generatinga force 2 × 10–7 Newton per metre between two straight parallel conductors ofinfinite length and negligible circular cross-section, when placed at a distance of onemeter in vacuum.
6. Candela: It was adopted as the unit of luminous intensity. One candela is theluminous intensity in perpendicular direction of a surface of 1/600,000 meter2 of ablack body at a temperature of freezing platinum (2046.64 Kelvin) and under apressure of 101,325 N/m2. Candela was redefined in 1979 as below:
It is the luminous intensity in a given direction due to a source which emitsmonochromatic radiation of frequency 540 × 1012 Hz and of which the radiantintensity in that direction is 1/683 watt per steradian.
7. Mole: It was adopted as the unit of amount of substance. The amount of asubstance that contains as many elementary entities (molecules or atoms if thesubstance is monoatomic) as there are number of atoms in 0.012 kg of carbon-12is called a mole. This number (number of atoms in 0.012 kg of carbon –12) is calledAvogadro constant and its best value available is 6.022045 × 1023.
8. Radian: It was adopted as the unit of plane angle. It is the plane angle betweenthe two radii of a circle, which cut off from the circumference, an arc equal to thelength of the radius.
Plane angle in Radian = length of arc/radius
9. Steradian: It was adopted as the unit of solid angle with its apex at the centre ofa sphere that cuts out an area on the surface of the sphere equal to the area ofthe square, whose sides are equal to the radius of the sphere.
Solid angle in steradian = area cut out from the surface of sphere/radius2
1.6 REFERENCE FRAMES
The same physical quantity may have different values if it is measured by observers whoare moving with respect to each other.
The velocity of a train has one value if measured by an observer on the ground, adifferent value if measured from a speeding car, and the value zero if measured by anobserver sitting in the train itself. None of these values has any fundamental advantage overany other; each is equally ‘correct’ from the point of view of the observer making themeasurement.
Measurement 5
In general, the measured value of any physical quantity depends on the reference frameof the observer who is making the measurement. To specify a physical quantity, eachobserver may choose a zero of the time scale, an origin in space and an appropriate coordinatesystem. We shall refer to these collectively as a frame of reference. Since the space of ourexperience has three dimensions, we must in general specify three coordinates to fix uniquelythe position of an object. The Cartesian coordinates x, y, z are commonly used in mechanics.Thus, the position and time of any event may be specified with respect to the frame ofreference by three Cartesian coordinates x, y, z and the time t.
1.7 INERTIAL AND NON-INERTIAL FRAMES
A system relative to which the motion of any object is described is called a frame ofreference. The motion of a body has no meaning unless it is described with respect to somewell defined system. There are generally two types of reference systems:
1. The frames with respect to which unaccelerated body is unaccelerated. This alsoincludes the state of rest.
2. The frames with respect to which an unaccelerated body is accelerated.
The frames with respect to which an unaccelerated body is unaccelerated, i.e., is at restor moving with constant linear velocity are called inertial frames i.e., unaccelerated framesare inertial frames.
Let us consider any coordinate system relative to which body in motion has coordinates(x, y, z). If the body, is not acted upon by any external force, then
md xdt
2
2 = 0, md ydt
md zdt
2
2
2
20 0= =,
Hence,
d xdt
2
2 0= , d ydt
2
2 0= , d zdt
2
2 0=
Which gives
dxdt
ux= = constant, dydt
uy= = constant, dzdt
uz= = constant
Where ux, uy and uz are the components of velocity in x, y and z directions respectively.From above equations we see that the components of velocity are constant, i.e., we say thatwithout application of an external force, a body in motion continues its motion with uniformvelocity in a straight line, which is Newton’s first law.
Hence we may say ‘An inertial frame is one in which law of inertia or Newton’s first lawinvalid’.
The frame with respect to which an unaccelerated body is accelerated are called non-inertial frames, i.e., accelerated frames are called non-inertial frames.
A frame of reference moving with constant velocity relative to an inertial frame is alsoinertial. Since acceleration of the body in both the frames is zero, the velocity of the bodyis different but uniform.
Experiments suggest that a frame of reference fixed in stars is an inertial frame. Acoordinate system fixed in earth is not an inertial frame, since earth rotates about its axisand also about the sun.
6 Mechanics
Galilean TransformationGalilean transformations are used to transform the coordinates of a particle from one
inertial frame to another. They relate the observations of position and time made by two ofobservers, located in two different inertial frames.
Let us consider two inertial frames S and S′. S being at rest and S′ moving with aconstant velocity v relative to S. The positions of two observers O and O′ observing an eventat any point P coincide with the origin of the two frames S and S′. Then the problem is totransform the data of the event recorded in the first frame to those recorded in the second.
Case I: When the second frame moves relative to first along positive direction of x-axis.
Let the origins O and O′ of two frames S and S′ coincide initially. Let the event happeningat P be denoted by (x, y, z, t) in frame S and by (x′, y′, z′ and t′) in frames S′, if we counttime from the instant when O and O′ momentarily coincide, then after a time t, the frameS′ is separated from frame S by a distance vt in the direction of x-axis as shown in figure 1.Then the observations of two observers O and O′ of the same event happening at P may beseen to be related by the following equations:
x x vt′ = −
y y′ =
z z′ =
t t′ =
Fig. 1
These equations are called Galilean transformation equations and relate to observationsof position and time made by two sets of observers, located in two different inertial frames.
Case II: When the second frame is moving along a straight line relative to first alongany direction as shown in Fig. 2.
Let the second frame S′ be moving relative to first frame S with a velocity v such that
v→ = $ $ $iv jv kvx y z+ +
Where vx, vy, and vz are the components of v along x, y and z axes respectively.
Let (x, y, z, t) and (x′, y′, z′, t)′ be the coordinates of an event happening at P at anyinstant as observed by the two observers situated at O and O′ of frames S and S′ respectively.If the origins of two systems coincide initially, then after a time t the frame S′ is separatedfrom frame S by a distance vxt, vyt and vzt along x, y and z axes respectively. Then referringto figure 2, we have
x ′ = x – vxt
y ′ = y – vy t
z′ = z – vzt ...(2)
t ′ = t
U
V
|||
W
|||
...( )1
U
V
|||
W
|||
Y Y ′
S ′
O ′ X, X ′
Z ′Z
O
SP
V t
Measurement 7
Y ′ x ′ P
y ′
x ′v ty
O ′
v tx
v tz
z ′z
O
Y
X
Fig. 2
These are the Galilean transformations, relating to the observations of position and timemade by two observers in two different inertial frames.
Case III: When the second frame has uniform angular velocity relative to first. Let usconsider two frames S and S′, the latter moving with uniform angular velocity ω relative toS about z-axis. Let the origins and axes of two frames coincide initially, i.e., at t = t′ = 0. InFig. 3, z-axis is taken perpendicular to the plane of the paper. After time t, x′ and y′ axesare rotated by an angle ωt relative to x and y-axes respectively as shown in Fig. 3.
The observations (x′, y′, z′, t′) taken by observer in S′ are related to those (x, y, z, t)taken by observer in S of the same event of P by the equations.
x ′ = Component of x along x′ + component of y along x′+ component of z along x′
y ′y
ωt
ωtx′
xz , z′
Fig. 3
= x cos ωt + y sin ωt + z cos 90°
or x ′ = x cos ωt + y sin ωt ...(3)
also y ′ = Component of x along y′ + component of y along y′+ component of z along y′
= x cos (90° + ωt) + y cos ωt + z cos 90°
or y ′ = –x sin ωt + y cos ωt ...(4)
and z′ = Component of x along z′ + component of y along z′+ component of z along z′
= x cos 90° + y cos 90° + z cos 0°
or z′ = z ...(5)
8 Mechanics
and t ′ = t ...(6)
Equations (3), (4), (5) and (6) are called time dependent Galilean transformations sincethey are time dependent and were obtained by Galileo.
Galilean Transformation of the Velocity of a Particle: Let us consider two inertialframes S and S′, the frame S′ moving with velocity v relative to S, which is given by
v→ = $ $ $iv jv kvx y z+ +
Let r and r′ be the position vectors of any particle at time t as observed by observer inframe S and S′ respectively. Then from Galilean transformations, we have
r′ = r – vt
and t ′ = t ...(1)
Differentiating eq. (1), keeping v constant, we get
dr′ = dr – vdt
and dt′ = dt ...(2)
\drdt
′′
=drdt
vdtdt
drdt
v′
−′
= − ...(3)
because since drdt
u′′
= ′ = velocity of a particle relative to frame S and drdt
u= = velocity
of particle relative to frame S′Hence, Eq. (3) gives
u′ = u – v ...(4)
Which shows that the velocities measured by the observers in the two frames of referenceare not the same. Alternatively, we can say that velocity of a body is not invariant underGalilean transformations.
The inverse transformation from S′ to S, is obviously given by
u = u′ + v ...(5)
Relation (4) or (5) are known as Galilean law of addition of velocities.
Invariance of Newton’s Second LawNewton’s second law of motion, which also includes the first law, is the real law of
motion. To test it let a force F→
be acting on a mass ‘m’ in frame S, then since force is therate of change of momentum, we have
F→
= ddt
m v md vdt
ma( )→
→→
=′
= ′
Where the mass has been assumed to be independent of velocity and a→
is the accelerationproduced in the mass
Similarly, in frame S′, if F′ be the force acting on m, then
F′→
= ddt
mv mdvdt
ma′
′ = ′′
= ′→
→→
( )
Measurement 9
where a′→
, is the acceleration produced in mass m in frame S′. But since in inertial frames,
a→
= a′→
We have F→
= F′→
This shows that force and hence Newton’s second law of motion is invariant underGalilean transformation.
1.8 SCALARS AND VECTORS
The physical quantities are of two types: Scalars and Vectors.
Scalar Quantities: The quantities which have only magnitude and no direction, arecalled ‘Scalar Quantities’, e.g., mass, distance, time, speed, volume, density, work, charge,electric current, potential, frequency etc.
A scalar quantity can be completely defined by a number and a unit. For example, the‘mass’ of the truck is 200 Kg, the distance of my college is 5 km from my residence. In allthese statements we have given complete information about the quantity. The summation,subtraction, multiplication and division of scalar quantities can be done by ordinary algebra.
Vector QuantitiesThere are certain physical quantities whose complete description not only requires their
magnitude (i.e., a numerical value with appropriate unit) but also their direction in space e.g.,velocity of a train. The magnitude of velocity is represented by a number such as 100 Km/hour. This tells how fast the train is moving. But the description of velocity is complete onlywhen we specify the direction of velocity also. We can represent the magnitude of velocity andthe tip of the arrow represents its direction.
If a particle is subjected to two velocities simultaneously its resultant velocity is differentfrom the two velocities and is obtained by using a special rule. Suppose a particle is movinginside a long tube with a speed of 6 m/sec and the tube itself is moving in the room at a speedof 8 m/sec along a direction perpendicular to its length. Figure 1 represents the position ofthe tube and the particle at initial instant and after a time interval of 1 sec. Geometricalanalysis gives the result that particle has moved a distance of 10 m in a direction θ = 53° fromthe tube. Hence, the resultant velocity of the particle is 10 m/sec along this direction.
8m
Resu ltan tVe loc ity
2nd veloc ity
1st ve loc ity
t = 1 sec
C
BA
6m t = 0
Fig. 4
In figure 4, line AB represents the first velocity with point B as the head. Then we drawanother line BC representing the second velocity with its tail coinciding with the head of thefirst line. Thus the line AC with C as head and A as the tail represents the resultant velocity.The resultant may also be called as the sum of the two velocities. We have added twovelocities AB and BC and have obtained the sum AC. This rule of addition is called as thetriangle law of addition.
10 Mechanics
Thus, the physical quantities which have magnitude and direction and which can beadded according to the triangle rule, are called Vector quantities.
Different Types of Vectors1. Like Vectors: Two vectors are said to be like vectors if they have same direction,
but different magnitude. Fig. 5 show two vectors A→
and B→
which have differentmagnitude but are paralled to each other.
A→
B→
Fig. 5
2. Equal Vectors: Two vectors are said to be equal, if they have the same magnitude
and direction. Fig. 6 shows two vectors A→
and B→
having the same magnitude and
same direction and therefore, A→
= B→
Fig. 6
For two vectors to be equal, it does not matter, whether the two vectors have theirtails at the same point or not. If the scales selected for both the vectors is the same,they are represented by two equal and parallel lines.
3. Unlike Vectors: The vectors having opposite direction and different magnitude,
are called unlike vectors. Fig. 7 shows two such vectors A→
and B→
which havedifferent magnitudes and are antiparallel to each other.
A→
B→
Fig. 7
4. Opposite Vectors: The vectors having same magnitude but opposite direction, are
known as opposite vectors. Fig. 8 shows two such vectors A→
and B→
having the samemagnitude and opposite direction and therefore,
A→
= –B→
Fig. 8
5. Unit Vectors: A vector divided by its magnitude is called a unit vector along thedirection of the vector. Obviously, the unit vector has unit magnitude and directionis the same as that of the given vector.
A unit vector in the direction of same vector A→
is written as $A and is read as ‘Acap’ or ‘A caret’ or ‘A hat’. Therefore, by definition
Measurement 11
or $AAA
=→
or A AA→
= $
Thus, any vector can be expressed as magnitude times the unit vector along its owndirection, unit vectors along x, y and z axes are represented by $, $ $i j k and respectively.
6. Co-initial Vectors: Vectors are said to be co-initial, if they have a common initial point. In
Fig. 9, A and B→ →
starting from the same pointO as their origin are called Co-initial vectors.
B A→ →
7. Co-linear Vectors: Two vectors having equal or unequal magnitude, which eitheract along the same line. Fig. 10(a) or along the parallel lines in the same directionFig. 1-(b) or along the parallel lines in opposite direction. Fig. 10(c), are called co-linear vectors. Like, unlike, equal and opposite vectors are collinear.
A→
B→
A→
B→
B→
A→
( )a ( )b ( )c
Fig. 10
8. Coplanar Vectors: Vectors lying in the same plane are called coplanar vectors.
Fig. 11(a) shows three vectors x→
, y→ and z
→ along mutually ⊥ axis x, y and z
respectively. These vectors are non coplanar but the vectors x→
– y→ , y
→– z
→ and
z→
– x→
are coplanar Fig. 11(b).
y→
x→
z→
x – y→ →
y – x→ →
z – x→ →
x→
y→
z→
( )a ( )b
Fig. 11
9. Null Vector: It is defined as a vector having zero magnitude. It has a direction,which is indeterminate as its magnitude is zero.
1.9 ADDITION OF VECTORS
Since vectors have both magnitude and direction, they cannot be added by ordinary
algebra. In Fig. 12 is shown the method of addition of two vectors A and B→ →
. Fig. 12(a). For
Fig. 9
12 Mechanics
this, we first draw vector A→
. Then starting from the arrow-head of A→
we draw the vector
B→
Fig. 12(b). Finally, we draw a vector R→
starting from the initial point of A→
and ending
at the arrow-head of B→
. Vector R→
would be the sum of A and B→ →
.
R→
= A + B→ →
The magnitude of A→
+B→
can be determined by measuring the length of R→
and the
direction can be expressed by measuring the angle between R→
and A→
( or B→
).
B→
A→
A→
B→R = A + B
→ → →
R = A + B→ → →
( )a ( )b
Fig. 12
Geometrical Method of Vector AdditionThere are three laws of Vectors addition, namely triangle, parallelogram and polygon
laws of vector addition. These laws can be used to add two or more vectors having aninclination with each other.
(i) Triangle law of Vector addition: According to this law if two vectors arerepresented both in magnitude and direction by the two sides of a triangle takenin the same order, then their resultant is represented totally (both in magnitudeand direction) by the third side of the triangle taken in the opposite order.
B→
A→
A→
B→R = A + B
→ → →
D E
F A→
FE
D
R = B + A→ → →
B→
Fig. 13
Figure 13 shows two vectors A and B→ →
. In order to find the sum of these vectors by
using triangle law of vector addition, draw vector DE A→ →
= . Then move vector B→
parallel to
itself until its tail coincides with the tip of vector A→
. Show the arrow head of B→
by point F.
Then vector DF = R→
drawn from the tail of vector A→
to the tip of vector B→
is the sum or
resultant of vector A and B→ →
. Thus
R→
= A B→ →
+
Measurement 13
(ii) Vector addition in Commutative: From Figure 13, it is clear that A B→ →
+ = B A→ →
+ .This property of the vector addition according to which the vector addition isindependent of the order in which the vectors are added is called cumulativeproperty of vector addition. A physical quantity having both magnitude and directionis not a vector, if it does not obey commutative law.
(iii) Parallelogram law of Vector addition: According to this law if two vectors actingsimultaneously at a point can be represented both in magnitude and direction bytwo adjacent sides of a parallelogram, the resultant is represented completely (bothin magnitude and direction) by the diagonal of the parallelogram passing throughthat point.
Suppose we have to find the resultant of two vector A and B→ →
as shown in Fig. 14.
In order to find the resultant, draw the vector DE A−→ →
= . Then move vector B→
parallel to itself, till its tail coincides with the tail of vector A→
. If we represent it
arrow head by point G, then vector DG−→
represents vector B→
. Then complete theparallelogram.
Now EF DG B−→ −→ →
= =
and GF DE A−→ −→ →
= =
Fig. 14
It is clear from figure that diagonal DF→
of the parallelogram represents the resultant
of sum of vectors A and B→ →
.
(iv) Polygon Law of Vector Addition: This law helps us to obtain the resultant ofmore than two vectors and is just the extension of the triangle law of vectoraddition.
B→
C→
A→
D→
A + B + C + D→ → → →
D→ C
→
A→
O
P
Q
R
S
B→
A + B + C→ → →
A + B→→
Fig. 15
According to this law if a number of vectors are represented in magnitude and directionby the sides of a polygon taken in same order, then their resultant is represented in magnitude
B→
A→
A→
B→
D E
F
B→
A→
G
R→
14 Mechanics
and direction by the closing side of polygon taken in the opposite order. Figure 15 represents
the addition of four vectors A B C and D→ → → →
, , by this method.
Analytical Method of Vector Addition
(i) Triangle law of vector addition: Suppose two vectors A and B→ →
represent both
in magnitude and direction the sides PQ and QR−→ −→
of the triangle PQR taken in
same order. Then according to triangle law of vector addition, the resultants R→
isrepresented by the closing side PR taken in the opposite order.
Magnitude of the resultant R→
: Draw a perpendicular RS from the point R, on the sidePQ which meets the line PQ at point S when produced forward. Then, from the ∆PSR , weget
(PR)2 = (PS)2 + (SR)2 = (PQ + QS)2 + (SR)2
= (PQ)2 + (QS)2 + 2 PQ.QS + (SR)2
But (QS)2 + (SR)2 = (QR)2
∴ (PR)2 = (PQ)2 + (QR)2 + 2PQ.QS ...(i)
A→
R→
B→
Q
θ
S
R
P
α
Fig. 16
From right angled triangle QRS, we get
cos θ = QS/QR
or QS = QR cos/QR cos = θTherefore, eq. (i) becomes
(PR)2 = (PQ)2 + (QR)2 + 2 PQ. QR. cos θNow, PR = R, PQ = A, QR = B
\ R2 = A2 + B2 + 2 AB cos θ
or R = [A B AB cos ]2 2+ + 2 θ
Direction of the resultant R→
: Suppose the resultant R→
makes an angle α with the
direction of A→
then from right angled triangle PRS, we get from Fig. 16
tan α =RSPS
RSPQ + QS
=
Now PQ = A, QS = B cos θ, RS = B sin θ
Measurement 15
Hence, tan α =B sin
A + B cosθ
θ
(ii) Parallelogram Law of Vector Addition: Suppose two vectors A and B→ →
inclinedto each other at an angle θ be represented in magnitude and direction both by the
concurrent sides PQ and PT−→ −→
of the parallelogram. PQUT as shown in figure 17.
Then according to parallelogram law, resultant of A and B→ →
is represented both in
magnitude and direction by the diagonal PU−→
of the parallelogram.
A→
B→R
→B→
A→
θα θ
U
SQP
T
Fig. 17
Magnitude of the resultant R→
: Drop a perpendicular from the point U on the line PQwhich meets the line PQ at some point S.
From right handed triangle PSU, we get
PU2 = PS2+ SU2
= (PQ + QS)2 + SU2
= (PQ)2 + (QS)2 + 2 PQ.QS + (SU)2
But (QS)2 + (SU)2 = (QU)2
∴ (PU)2 = (PQ)2 + (QU)2 + 2 PQ.QS
Now PU = R, PQ = A and QU = B
Further QS = B cos θHence, R2 = A2 + B2 + 2 AB cos θ
or R = [A B AB cos ]2 + +2 2 θ
Direction of the resultant R→
: Suppose the resultant vector R→
makes an angle α with
vector A→
.
Then from right angled triangle PUS, we get
tan α =USPS
USPQ + QS
B sinA + B cos
= = θθ
16 Mechanics
1.10 RESOLUTION OF VECTORS
The process of splitting up a vector into two or more vectors is known as resolution ofa vector. The vectors into which a given vector is split are called component vectors. Theresolution of a vector into two mutually perpendicular vectors is called the rectangularresolution of vector in a plane or two dimensions.
Figure 18 shows a vector OR−→ →
= r in the X-Y plane drawn from the origin O. Let thevector makes an angle α with the x-axis and β with the y-axis. This vector is to be resolved
into two component vectors along two mutually perpendicular unit vectors $ $i j and respectively,
where $ $i j and are the unit vectors along x-axis and y-axis respectively as shown in figure 15.From point R, drop perpendiculars RP and RQ on x and y-axis respectively. The length OP
is called the projection of OR−→
on x-axis while length OQ is the projection of OR→
ony-axis. According to parallelogram law of vector addition
r→
= OR OP OQ−→ −→ −→
= +
Thus we have resolved the vector r→
into two parts, one along OX and the other alongOY, the magnitude of the part along OX is OP = rx = r cos α and the magnitude of the part
along OY is OQ = ry = r cos β i.e., in terms of unit vector $ $i j and , we can write
OP→ = $ cos $i r i rx
→=α
and OQ→ = $ sin $j r j ry
→=α
= $ cos $j r j ry
→=β
Thus r→
= $ cos $ cosi r j r→
+α β
= $ $ir jrx y+
If the vector rr
is not in the X-Y plane, it may have non zero projection along x, y andz-axes and we can resolve it into three components i.e., along the x, y and z-axes. If α, βand γ be the angles made by the vector r
→ with respect to x, y and z-axes respectively, thenwe can write
r→ = $ cos $ cos $ cosi r j r krα β γ+ +
r→ = $ $ $i r j r krx y z+ +
Where $, $ $i j k and are the unit vectors along x, y and z axes respectively, the magnitude
(r cos α) is called the component of r→
along x-axis, r cos β is called the component alongy-axis and r cos γ is called the component along z-axis.
j
O
iX
Y
R
α
β
r→
θ
ry
rx
Q
^P
Fig. 18
Measurement 17
Above equation also shows that any vector in three-dimensional can be expressed as a
linear combination of the three unit vectors $, $ $i j k and .
1.11 MULTIPLICATION OF VECTORS
There are three kinds of multiplication operations for vectors
i. multiplication of a vector by a scalar,
ii. multiplication of two vectors in such a way as to yield a scalar, and
iii. multiplication of two vectors in such a way as to yield another vector.
There are still other possibilities, but we shall not consider them here.
The multiplication of a vector by a scalar has a simple meaning. The product of a scalar
k and a vector a→
, written k a→
, is defined to be a new vector whose magnitude is k times the
magnitude of a→
. The new vector has the same direction as a→
if k is positive and the oppositedirection if k is negative. To divide a vector by a scalar we simply multiply the vector by thereciprocal of the scalar.
Scalar Product (Dot Product): The scalar product of two vectors is defined as a scalarquantity having magnitude equal to the product of the magnitude of two vectors and the
cosine of the smaller angle between them. Mathematically, if θ is the angle between vectors
A and B→ →
then
A B→ →
. = AB cosθ
Above equation can also be expressed as below
A B→ →
. = A(B cos θ) = B(A cos θ)
where B cos θ is the magnitude of component of B→
along the direction of vector A→
and A cos θ
is the magnitude of component of A→
along the direction of vector B→
. Therefore, the dotproduct of two vectors can also be interpreted as the product of the magnitude of one vectorand the magnitude of the component of other vector along the direction of first vector.
Dot product of two vectors can be positive or zero or negative depending upon θ is lessthan 90° or equal to 90° or 90° < θ < 180°.
Properties1. Dot product of two vectors is always commutative i.e.,
A→
.B→
= B→
.A→
A→
.B→
= AB cos θ
where θ is the angle between A and B→ →
measured in anti clock wise direction
and B A BA cos( AB cos→ →
= − =. )θ θ
Thus A B B A→ → → →
=. .
Fig. 19
18 Mechanics
B→
A→
θ
B cos θ
B→
A→
A cos θ
0θ
Fig. 20
2. The dot product of a vector with itself gives square of its magnitude i.e.,
A A A A cos 0 A2→ →
= ° =. .
3. The dot product of two mutually perpendicular vectors is zero i.e., if two vectors
A→
and B→
are perpendicular then
A B→ →
. = AB cos 90 = 0°4. The dot product obeys the distributive law i.e.,
A B C→ → →
+. ( ) = A B A C→ → → →
+. .
5. Two vectors are collinear, if their dot product is numerically equal to product oftheir magnitudes i.e.,
When θ = 0° or 180°
A B→ →
. = AB
For example $ . $i i = $ . $j j = k k→ →
. = 1 (as θ = 0°)
$ . $ $. . $i j j k k i= = = °→ →
0 (as = 90 )θ .
6. Dot product of two vectors in term of their rectangular components in threedimensions.
A B A A A B B B→ →
= + + + +. ($ $ $ ) . ($ $ $ )i j k i j kx y z x y z = AxBx + AyBy + AzBz
Examples of some physical quantities which can be expressed as scalar productof two vectors:
(a) Work (W) is defined as the scalar product of force (F→
) and the displacement (S→
) i.e.,
W = F S→ →
.
(b) Power (P) is defined as the scalar product of force (F→
) and the velocity ( v→
) i.e.,
P = F→
. v→
Measurement 19
(c) Magnetic flux (φ) linked with a surface is defined as the dot product of magnetic
induction (B→
) and area vector (A→
) i.e.,
φ = B A→ →
.Vector Product (Cross Product): The vector product of two vectors is defined as a
vector having magnitude equal to the product of the magnitudes of two vectors with the sineof angle between them and direction perpendicular to the plane containing the two vectorsin accordance with right handed screw rule or right hand thumb rule.
If θ is the angle between vectors A and B→ →
, then
A→
× B→
= AB sin θ $n
The direction of vector A→
× B→
is the same as that of unit vector $n. It is decided by anyof the following two rules:
Fig. 21
Right handed screw rule: Rotate a right handed screw from vector A→
to B→
throughthe smaller angle between them, then the direction of motion of screw gives the direction of
vector A→
× B→
.
Right hand thumb rule: Bend the finger of the right hand in such a way that they point
in the direction of rotation from vector A→
to B→
through the smaller angle between them,
then the thumb points in the direction of vector A→
× B→
.
The cross product of two vectors in term of their rectangular components is
A→
× B→
= ($ $ $ ) × ($ $ $ )i j k i j kx y z x y zA A A B B B+ + + +
= (Ay Bz – Az By) $i + (Az Bx – Ax Bz) $j
+ (Ax By – Ay Bx) $k
=
$ $ $i j k
x y z
x y z
A A A
B B B
20 Mechanics
Properties of Vector Product1. The cross product of the two vectors does not obey commutative law.
A B→ →
× = −→ →
( × )B A
i.e., A B→ →
× ≠→ →
( × )B A
2. The cross product follows the distribution law i.e.
A B C→ → →
−× ( ) = A B A C→ → → →
−× ×
3. The cross product of a vector with itself is a NULL vector i.e.,
A A→ →
× = (A) . (A) sin 0° $n = 0
4. The cross product of two vectors represents the area of the parallelogram formedby them.
From figure 22, show a parallelogram PQRS whose adjacent sides PQ and PS are
represented by vectors A→
and B→
respectively
Now, area of parallelogram = QP × SM
= AB sin θ
B→
A
θ
B sin θ
R
MP Q
S
→
Fig. 22
Because, the magnitude of vector A→
× B→
is AB sin θ, hence cross product of twovectors represents the area of parallelogram formed by it. It is worth noting that
area vectors A→
× B→
acts along the perpendicular to the plane of two vectors A and B→ →
.
5. The Cross product of unit vectors are$ × $i i = $ × $ $ × $ ( ) ( ) sin $j j k k n= = ° =1 1 0 0
$ × $i j = (1) (1) sin 90° $ $k k=
Where $k is a unit vector perpendicular to the plane of $ $i j and in a direction in
which a right hand screw will advance, when rotated from $ $.i j to
Also − $ × $j i = (1) (1) sin 90° ( $) $− =k k
Similarly, $ × $j k = − =$ × $ $k j i
and $ × $k i = − =$ × $ $i k j
Measurement 21
Examples of some physical quantities which can be expressed as cross product oftwo vectors:
(a) The instantaneous velocity ( )v→
of a particle is equal to the cross product of its
angular velocity ( )ω→
and the position vector ( )r→
i.e.,
v→
= ω→ →
× r
(b) The tangential acceleration ( )at
→ of a particle is equal to cross product of its angular
acceleration ( )α→
and the position vector ( )r→
i.e.,
at→
= α→ →
× r
(c) The centripetal acceleration (ac
→) of a particle is equal to the cross product of its
angular velocity and the linear velocity ( )v→
i.e.,
ac
→= ω
→ →× v
(d) The force F→
on a charge q moving inside magnetic field is equal to charge times
the cross product of its velocity ( )v→
and magnetic induction (B→
) i.e.,
F→
= q v( × )→ →
B
(e) The torque ( τ→
) of a force (F→
) is equal to cross product of the position vector r→
and
the force (F→
) applied i.e.,
τ→
= r→ →
× F
(f) The angular momentum (L→
) is equal to cross product of position vector ( r→
) and
linear momentum (P→
) of the particle i.e.,
L→
= r→ →
× P
1.12 VECTORS AND THE LAWS OF PHYSICS
Vectors turn out to be very useful in physics. It will be helpful to look a little , more
deeply into why this is true. Suppose that we have three vectors a b r→ → →
, and which have
components ax, ay, az: bx, by, bz and rx, ry, rz respectively in a particular coordinate systemxyz of our reference frame. Let us suppose further that the three vectors are related so that
r→
= a b→ →
+ ...(1)
22 Mechanics
By a simple extension of above equation
rx = ax + bx
ry = ay + by ...(2)
and rz = az + bz
Now consider another coordinate system x′ y′ z′ which has these properties
1. Its origin does not coincide with the origin of the first, or xyz, system and
2. Its three axes are not parallel to the corresponding axes in the first system.
In other words, the second set of coordinates has been both translated and rotated withrespect to the first.
The components of the vectors a b r→ → →
, and in the new system would all prove, in
general, to be different, we may represent them by ax′, ay′, az′, bx′, by′, bz′ and rx′, ry′, rz′respectively. These new components would be found, however, to be related in that.
rx′ = a bx x′ ′+
ry′ = a by y′ ′+ and ...(3)
rz′ = a bz z′ ′+
That is, in the new system we would find once again that
r→
= a b→ →
+
In more formal language: relations among vectors, of which Eq.(1) is only one example,are invariant (that is, are unchanged) with respect to translation or rotation of the coordinates.Now it is a fact of experience that the experiments on which the laws of physics are basedand indeed the laws of physics themselves are similarly unchanged in form when we rotateor translate the reference system. Thus the language of vectors is an ideal one in which toexpress physical laws. If we can express a law in vector form the invariance of the law fortranslation and rotation of the coordinate system is assured by this purely geometrical propertyof vectors.
1.13 SPEED AND VELOCITY
The speed of a moving object is the rate at which it covers distance. It is important todistinguish between average speed and instantaneous speed. The average speed v of somethingthat travels the distance S in the time interval t is
v =st
Average speed = distance traveled/time interval
Thus a car that has gone 150 km in 5 h had a average speed of
v =st
= (150/5) km/h = 30 km/h
Measurement 23
The average speed of the car is only part of the story of its journey, however, becauseknowing v does not tell us whether the car had the same speed for the entire 5 h orsometimes went faster than 30 km/h and sometimes slower.
Even though the car’s speed is changing, at every moment it has a certain definite value(which is what is indicated by its speedometer). To find this instantaneous speed v at aparticular time t, we draw a straight-line tangent to the distance- time curve at that valueof t. The length of the line does not matter. Then we determine v from the tangent line fromthe formula
v = ∆s/∆t
Where ∆s is the distance interval between the ends of the tangent and ∆t is the timeinterval between them. (∆ is the Greek capital letter delta). The instantaneous speed of thecar at t = 40 s is from figure
Total distance, (m) 0 100 200 300 400 500
Elapsed time, (S) 0 28 40 49 57 63
V = ∆s/∆t = 100 m/10 s = 10 m/s
500
400
300
200
100
010 20 30 40 50 60 70
Time (S )
Dis
tanc
e (m
)
Fig. 23
When the instantaneous speed of an object does not change, it is moving at constantspeed.
The speed of a moving object is a scalar quantity. Its unit is meter/second and itsdimension are (LT–1).
VelocityThe object’s velocity, however, includes the direction in which it is moving and is a vector
quantity. If the object undergoes a displacement s→
in the time interval, its average velocity
v→
during this interval is
v→
= s/t
Instantaneous velocity v is the value of ∆s/∆t at a particular moment, and for straight-line motion is found by the same procedure as that used for instantaneous speed v but withthe direction specified as well.
24 Mechanics
1.14 ACCELERATION
If the velocity of a moving object is changing, then its motion is called ‘acceleratedmotion’. The change in velocity may be in magnitude (speed), or in direction or in both. Ifthe object is moving along a straight line, then only the magnitude of velocity (speed) changes.
The time-rate of change of velocity of an object is called the ‘acceleration’ of that object.That is
Acceleration = Change in velocity/time interval
Acceleration is generally represented by ‘ a→
’. Since velocity is a vector quantity, theacceleration is also a vector quantity.
Suppose, the velocity of a moving object is v1 at time t1 and becomes v2 at time t2. Itmeans that in the time-interval (t2 – t1), the change in the velocity of the object is (v2 – v1).Hence, the average acceleration of the object in time-interval t2 – t1 is
a→
=v vt t
vt
2 1
2 1
−−
= ∆∆
If the time interval ∆t is infinitesimally small (∆t → 0) then the above formula givesacceleration ‘at a particular time’. This is called ‘instantaneous acceleration’ and is given by
a→
= Lim∆
∆∆t
vt
dvdt→
=0
The unit of acceleration is meter /second2 and its dimensional formula is (LT–2).
If the velocity of an object undergoes equal changes in equal interval of time, then itsacceleration is said to be ‘uniform’. If the magnitude of the velocity (speed) of an object isincreasing with time, then the acceleration of the object is positive. If the magnitude of thevelocity (speed) is decreasing, then the acceleration is negative and it is called ‘retardation’.
1.15 RECTILINEAR MOTION
When a particle moves along a straight line, its motion is said to be ‘rectilinear’. In thistype of motion the acceleration of the particle is either zero (velocity constant) or arises froma change in the magnitude of the velocity.
Let us consider a particle in rectilinear motion, say along the x-axis, under a constantacceleration a. We can derive relations between the kinematical variable x, v, a and t.
Suppose the particle has velocity v0 at the origin (x = 0) at time t = 0, and moving withconstant acceleration a, acquires a velocity v at time t. From the definition of acceleration,we have
a =∆∆
vt
v vt
= −−
0
0
or v = v0 + at ...(1)
Since the acceleration is constant, the average velocity vav in time-interval t equal to onehalf the sum of the velocities at the beginning and the end of the interval.
Thus vav =v v0
2+
Measurement 25
Hence, the displacement at time t is
x = v tv v
tav × = +0
2Substituting for v from Eq.(1), we get
x =v v at
t0 0
2+ +( )
x = vo t + ½ at2 ...(2)
Eliminating t between (1) and (2), we can get
v2 = v20 + 2ax ... (3)
Eqs. (1), (2) and (3) are equations of rectilinear motion under constant acceleration.
1.16 ACCELERATION OF GRAVITY
When a body is allowed to fall over earth’s surface from a certain point above it, it isseen that
(a) its velocity increases continuously and the acceleration is found to be uniform and
(b) that the acceleration is the same for all bodies.
This uniform acceleration of all the bodies when falling down under the earths action ofgravity is called acceleration due to gravity. It is represented by ‘g’.
If the resistance offered by air to the motion of body is assumed negligible, then all thebodies falling freely under gravity have the same acceleration ‘g’ acting vertically downwards.
The value of g varies from a value 9.781 ms–2 at the equator to the value 9.831 ms–2
at the poles.
The equation derived for rectilinear motion along the x-axis can be used for freely fallingbodies as well. Let us take y-axis as vertical, here the constant acceleration is g directedvertically downwards. Thus after replacing x by y and a by g in the equation of motion, weobtain equations for vertically downward motion:
v = vo + gt
y = vot + ½ gt2
v2 = vo2 + 2gy
In case of bodies moving vertically upward, the acceleration due to gravity g→
acts in theopposite direction i.e., as retardation. Hence the equations of motion would be
v = vo – gt
y = vo t – ½ gt2
v2 = vo2 – 2gy
1.17 ACCURACY AND ERRORS IN MEASUREMENT
In many experiments in the physics laboratory, the aim is to determine the value ofphysical constant. To determine a physical constant, we have to measure the various quantities
26 Mechanics
which are connected with that physical constant by a formula. For example, to determine thedensity (ρ) of a metal block, we have to measure its mass (m) and its volume (V) which arerelated to ρ by the formula:
ρ = m/V
The accuracy in the value of ρ obviously depends upon the accuracy in the measurementsof m and V. Measurements of the quantities in the formula involve errors which are of twotypes: (a) Random errors and (b) Systematic errors
Random ErrorsRandom errors may be due to (i) small changes in the conditions of a measurements and
(ii) the incorrect judgment of the observer in making a measurement. For example, supposeyou are determining the weight of a body with the help of a spring balance. You will usuallymake an error in estimating the coincidence of the pointer with the scale reading or inassessing the correct position of the pointer when it lies between two consecutive graduationsof the scale. This error, which is due to incorrect judgment of the observer, is also an exampleof random errors.
Random errors cannot be traced to any systematic or constant cause of error. They donot obey any well-defined law of action. Their character can be understood or appreciatedfrom the illustration of firing shots at a target using a rifle. The target is usually a bull’s eyewith concentric rings round it. The result of firing a large number of shots at the target iswell known. The target will be marked by a well-grouped arrangement of shots. A largenumber of shots will be nearer a certain point and other shots will be grouped around it onall sides. These shots which are grouped on both sides of the correct point obey the law ofprobability which means that large random errors are less probable to occur than small ones.A study of the target will show that the random shots lie with as many to one side of thecentre as to the other. They will also show that small derivations from the center are morenumerous than large deviation and that a large deviation is very rare.
Method of Minimizing Random ErrorsIf we make a large number of measurements of the same quantity then it is very likely
that the majority of these measurements will have small errors which might be positive ornegative. The error will be positive or negative depending on whether the observedmeasurement is above or below the correct value. Thus random errors can be minimized bytaking the arithmetic mean of a large number of measurements of the same quantity. Thisarithmetic mean will be very close to the correct result. If one or two measurements differwidely from the rest, they should be rejected while finding the mean.
Systematic ErrorsDuring the course of some measurements, certain sources of error operate constantly or
systematically making the measurement. Systematically greater or smaller than the correctreading. These errors, whose cause can be traced, are called systematic errors. All instrumentalerrors belong to this category, such as the zero error in vernier callipers and micrometerscrew, the index error in an optical bench, the end error in a meter bridge, faulty graduationsof a measuring scale, etc.
Elimination of Systematic ErrorsTo eliminate systematic errors, different methods are used in different cases.
Measurement 27
1. In some cases, the errors are determined previously and the measurements correctedaccordingly. For example, the zero error in an instrument is determined before ameasurement is made and each measurement is corrected accordingly.
2. In some cases the error is allowed to occur and finally eliminated with thedata obtained from the measurements. The heat loss due to radiation is takeninto account and corrected for from the record of the temperature at differenttimes.
Order of AccuracyEven after random and systematic errors are minimized, the measurement has a certain
order of accuracy which is determined by the least count of the measuring instruments usedin that measurement.
Suppose we determine the value of a physical quantity u by measuring three quantitiesx, y and z whose true values are related to u by the equation.
u = xα yβ z–γ ...(a)
Let the expected small errors in the measurement of quantities x, y and z be respectively± ± ±δ δ δx y z, and so that the error in u by using these observed quantities is ± δu. Thenumerical values of δx, δy and δz are given by the least count of the instruments used tomeasure them.
Taking logarithm of both sides of Eq. (a), we have
log u = α log x + β log y – γ log z
Partial differentiation of the above equation gives
δuu
= α δ β δ γ δxx
yy
zz
+ − ...(b)
The proportional or relative error in u is δuu
. The values of δx, δy and δz may be
positive or negative and in some case the terms on the right hand side of Eq. (b) maycounteract each other. This effect cannot be relied upon and it is necessary to consider theworst case which is the case when all errors add up giving an error δu given by the equation:
δuu
FHG
IKJmaximum
= α δ β δ γ δxx
yy
zz
+ + ...(c)
Thus to find the maximum proportional error in u, multiply the proportional errors ineach factor (x, y and z) by the numerical value of the power to which each factor is raisedand then add all the terms so obtained.
The sum thus obtained will give the maximum proportional error in the result of u.When the proportional error of a quantity is multiplied by 100, we get the percentage errorof that quantity. It is evident from Eq. (c) that a small error in the measurement of thequantity having the highest power will contribute maximum percentage error in the value ofu. Hence, the quantity having the highest power should be measured with a great precisionas possible. This is illustrated in the following example.
28 Mechanics
Example: In an experiment for determining the density ( ρ ) of a rectangular block of ametal, the dimension of the block are measured with callipers having a least count of 0.01 cmand its mass is measured with beam balance of least count 0.1 g. The measured values are:mass of the block (m) = 39.3 gm, length of block (x) = 5.12 cm, Breadth of block (y) = 2.56 cm,Thickness of block (z) = 0.37 cm.
Error in m = δm 0.1 g= ±
Error in x = δx 0.01 cm= ±
Error in y = δy 0.01 cm= ±
Error in z = δz 0.01 cm= ±
Find the maximum proportional error in the determination of ρ.
Ans: The density of the block is given by
ρ = m/xyz
Calculating ρ omitting errors, ρ = 39 3
5 12 2 56 0 378 1037.
. × . × ..= − g cm 3
Proportional error in ρ is given by the equation
δρρ =
δ δ δ δmm
xx
yy
zz
− − −
The maximum proportional error in ρ is given by the equation
δρρ
FHG
IKJmax
=δ δ δ δmm
xx
yy
zz
+ + +
=0 1
39 30 015 12
0 012 56
0 010 37
..
.
...
.
.+ + +
= 0.0025 + 0.0019 + 0.0039 + 0.0270
= 0.0353
∴ Maximum percentage error = 0.0353 × 100 = 3.53%
Hence the error is 3.53% of 8.1037. Clearly it is absurd to give the result for ρ to fivesignificant figures. The error in ρ is given by
δρ = 0.0353 × ρ = 0.0353 × 8.1037 = 0 286 0 3. .≅
Hence the value of ρ = 8.1037 is not accurate up to the fourth decimal place. In fact, itis accurate only up to the first decimal place. Hence the value of ρ must be rounded off as8.1 and the result of measurements is written as
ρ = ( . . )8 1 0 3 3± − g cm
It is clear that such a large error in the measurement of ρ is due to a large error (=0.027)in the measurement of z, the smallest of the quantities measured. Hence the order ofaccuracy of ρ should be increased by measuring z with an instrument having a least countwhich is smaller than 0.01 cm. Thus a micrometer screw (least count = 0.001 cm), rather thana vernier caliper should be used to measure z.
Measurement 29
SUMMARY
Physics is a branch of science that deals with the study of the phenomena of nature. Thescientific method used in the study of science involves observation, proposal of a theory,testing the consequences of the proposed theory and modification or refinement of the theoryin the light of new facts. The applications of physics have played a very great role in technologyand in our daily lives.
Physics is a science of measurement. All quantities which can be measured either indirectlyor directly such as length, mass, time, force, temperature, light intensity, electric current etc.are called physical quantities of numerous such quantities, length, mass and time are regardedas fundamental quantities. The measurement of these quantities involves the choice of a unit.The internationally accepted units of length, mass and time respectively are metre, kilogramand second. The units of all other mechanical quantities are derived from these three basicunits. These quantities are measured by direct and indirect methods. The measured valuesof these quantities, show a very wide range variation.
Different physical quantities have different dimensions. The dimensions of a physicalquantity are the number of times the fundamental units of mass, length and time appear inthat quantity.
30 Mechanics
2.1 MECHANICS
Mechanics, the oldest of the physical sciences, is the study of the motion of objects. Thecalculation of the path of an artillery shell or of a space probe sent from Earth to mars areamong its problems.
When we describe motion (or trajectories) we are dealing with that part of mechanicscalled kinematics, ignoring the forces producing the motion. When we relate motion to theforces associated with it and to the properties of the moving objects, we are dealing withdynamics.
2.2 CAUSE OF MOTION: FORCE
From daily experience we know that the motion of a body is a direct result of itsinteractions with the other bodies around it (which form its environment). When a batsmanhits a ball, he interacts with the ball and modifies its motion. The motion of a freely fallingbody or a projectile is the result of its interaction with the earth. The motion of an electronaround a nucleus is the result of its interaction with the nucleus.
An interaction is quantitatively expressed in terms of a concept called ‘force’. An intuitivemotion of force is derived in terms of a push or a pull. When we push or pull on a body, weare said to exert a (muscle) force on it. Earth which pulls all bodies towards its centre is saidto exert a (gravitational) force on them. A stretched spring pulling a body attached to its endis said to exert a (elastic) force on the body. A locomotive exerts a force on the train it ispulling or pushing. Thus every force exerted on a body is associated with some other bodyin the environment.
It is not always that an application of force will result in motion or change in motion.For example, we may push a wall, i.e., there is an interaction between us and the wall, andhence there is a force, but the wall may not move at all.
Thus, force may be described as a push or pull, resulting from the interaction betweenbodies, which produces or tends to produce motion or change in motion.
The analysis of the relation between force and the motion of a body is based on threelaws of motion which were first stated by Sir Issac Newton.
FORCE AND MOTION
2
30
Force and Motion 31
2.3 NEWTON’S LAWS OF MOTION
Newton’s laws of motion are the basis of mechanics. Galileo’s version of inertia wasformalized by Newton in a form that has come to be known as Newton’s first law of motion.Newton’s first law of motion: stated in Newton’s words, the first law of motion is:
“Every body continues in its state of rest or of uniform motion in a straight line unlessit is compelled to change that state by forces impressed upon it”.
Newton’s first law is also known as law of inertia and the motion of a body not subjectto the action of other forces is said to be inertial motion. With the help of this law we candefine force as an external cause which changes or tends to change the state of rest or ofuniform motion of a body.
Newton’s first law is really a statement about reference frames. We know that themotion of a body can be described only relative to some other body. Its motion relative to onebody may be very different from that relative to another. A passenger in an aircraft whichis on its take-off run is at rest relative to the aircraft but is in accelerated motion relativeto the earth. Therefore we always choose a set of coordinate axes attached to a specified body,relative to which the motion of a given body is described. Such a set of coordinate axes isknown as a “reference frame”.
The first law tells us that we can find a reference frame relative to which a body remainsat rest or in uniform motion along a straight line (i.e., it has no acceleration) when no netexternal force acts upon it. Such a reference frame is called an ‘inertial frame’. Thus aninertial frame is one in which Newton’s first law correctly describes the motion of a body notacted on by a net force. Such frames are either fixed with respect to the distant stars ormoving at uniform velocity with respect to them.
A reference frame attached to the earth can be considered to be an inertial frame formost practical purposes, although it is not precisely so, due to the axial and orbital motionsof the earth. But if a frame of reference is inertial, then every other frame which is inuniform motion relative to it is also inertial.
Newton’s first law makes no distinction between a body at rest and one moving with aconstant velocity. Both states are “natural” when no net external force or interaction acts onthe body. That this is so becomes clear when a body at rest in one inertial frame is observedfrom a second (inertial) frame moving with constant velocity relative to the first. An observerin the first frame finds the body to be at rest, an observer in the second frame finds the samebody to be moving with uniform velocity. Both observers find the body with no acceleration.Thus both observe that Newton’s first law is being obeyed.
Newton’s Second Law of MotionNewton’s second law tells us what happens to the state of rest or of uniform motion of
a body when a net external force acts on the body i.e., when the body interacts with othersurrounding bodies.
This law states: “The change of motion of an object is proportional to the forceimpressed, and is made in the direction of the straight line in which the force isimpressed”.
By “change of motion” Newton meant the rate of change of momentum (p) with time.So mathematically we have
32 Mechanics
F→
∝→d
dtp( )
or F→
= kddt
p( )→
...(1)
Where F→
is the impressed force and k is a constant of proportionality. The differentialoperator d/dt indicates the rate of change with time. Now, if the mass of the body remainsconstant (i.e., neither the body is gaining in mass like a conveyer belt nor it is disintegratinglike a rocket), then
d pdt
→
=ddt
m v( )→
= md vdt
m a→
→=
Where ad vdt
→→
= = the acceleration of the body. Thus Eq. (1) becomes
F→
= km a→
F = km a
This law provide a quantitative definition of force.
Thus force is equal to mass times acceleration, if the mass is constant. The force hasthe same direction as the acceleration. This is an alternative statement of the second law.
We note that the Newton’s first law is contained in the second law as a special case, if
F then → →
= =0 0, .a In other words, if the net force on a body is zero, the acceleration of thebody is zero. Therefore, in the absence of a net force a body will move with constant velocityor be at rest, which is the first law.
Newton’s Third Law of MotionA force acting on a body arises as a result of its ineraction with another body surrounding
it. Thus any single force is only one aspect of a mutual interaction between two bodies. Wefind that whenever one body exerts a force on a second body, the second body always exertson the first a force which is equal in magnitude but opposite in direction and has the sameline of action. A single isolated force is therefore an impossibility.
The two forces involved in every interaction between the bodies are called an ‘action’ anda ‘reaction’. Either force may be considered the ‘action’ and the other the ‘reaction’.
This property of forces was stated by Newton in his third law of motion: “To everyaction there is always an equal and opposite reaction”.
If a body A exerts a force FAB on a body B, then the body B in turn exerts a force FBAon A, such that
FAB = –FBA
So, we have FAB + FBA = 0
Force and Motion 33
2.4 FREELY FALLING BODIES
It is a matter of common observation that bodies fall towards the ground. Leaves of thetree, fruits from the tree, a body just dropped from the top of a tower, a stone thrown upwardetc. all reach the ground. According to Aristotle it was believed that the time taken by aheavier body to reach the ground as less as compared to a lighter body dropped from the sameheight. This belief continued until the sixteenth century and nobody either challenged it ortried to prove its correctness.
In 1590, Galileo disapproved this idea. He showed from his famous experiments at theleaning tower of Pisa that all bodies dropped from the same height reach the ground at thesame instant irrespective of their masses.
Galileo showed that bodies of different masses when dropped from the top of the leaningtower of Pisa, reached the ground at the same time. A large crowd was present to see thecorrectness of Galileo’s idea. When a stone and a piece of paper were dropped from the towerat the same time, it was found that the stone reached the ground earlier. This difference intime was explained by Galileo to be due to the resistance offered by air on the paper. Heshowed that if a paper and a stone are dropped from the same height in vacuum, both takethe same time to fall through the same height.
Laws of Falling Bodies1. All bodies fall with equal rapidity in vacuum irrespective (Equations of Motion in
Free Fall) of their masses
2. The velocity acquired by a body falling freely from rest is directly proportional tothe time of its fall
v ∝ t
or v = gt
3. The distance moved by a body falling freely from rest is directly proportional to thesquare of the time of fall.
S ∝ t2
S = ½ gt2
2.5 MOTION IN A VERTICAL PLANE
Equations of motion in a vertical plane is given by
v = u – gt
S = ut – (gt2 )/2
v2 = u2 – 2 gs
Components of VectorsIn mechanics and other branches of physics, we often need to find the component of a
vector in a certain direction. The component is the ‘effective part’ of the vector in thatdirection. We can illustrate it by considering a picture held up by two strings OP and OQ eachat an angle of 60° to the vertical. Figure 1(a). If the force or tension in OP is 6 N, its verticalcomponent S acting upwards at P helps to support the weight W of the picture, which actsvertically downwards. The upward component T of the 6 N force in OQ acting in the directionQT, also helps to support W. Figure 1(b) shows how the component value of a vector F can
34 Mechanics
be found in a direction OX. If OR represents F to scale, we draw a rectangle OPRQ whichhas OR as its diagonal. Now F is the sum of the vectors OP and OQ. The vector OQ has noeffect in a direction OX at 90° to itself. So the effective part or component of F in the directionOX is the vector OP.
If θ is the angle between F and the direction OX, then
OP/OR = cos θ or OP = OR cos θ = F cos θSo the component of any vector F in a direction making an angle θ to F is always given by
= F cos θIn a direction OY perpendicular to OX, F has a component F cos (90° – θ) which is F
sin θ. This component is represented by OQ in the figure (b).
P Q
W
60°60°
6N
S TO
90°
θ
F
RY
Q
O P X
F sin θ
( )a ( )b
F cos θ
Fig. 1 Components of vectors.
2.6 PROJECTILE MOTION
A body projected in the horizontal direction witha uniform velocity is under the action of (i) uniformvelocity in the horizontal direction and (ii) uniformacceleration due to gravity in the vertically downwarddirection. Such a body is known as projectile e.g., abomb released from an aeroplane moving with a uniformspeed and a bullet fired from a rifle. The path traversedby the projectile is called its trajectory.
Consider a tower AB of height h. A stone is thrownin the horizontal direction from the top of tower. Assoon as the stone is released, it is under the action of(i) uniform velocity ‘u’ in the horizontal direction and(ii) acceleration due to gravity ‘g’ in the verticallydownward direction. Under the combined effect of the two, the stone does not reach the footof the tower but it falls at some distance away from the foot of the tower. Suppose the stonereaches the point C. The distance BC is the horizontal range.
Suppose the time taken by stone to reach the ground = t
A u
h
B C
Fig. 2
Force and Motion 35
∴ Vertical distance h = ½ gt2 ...(1)
Since the initial velocity in the vertical directionis zero, the horizontal distance
BC = x = u.t
∴ t = x/u ...(2)
From Eqs. (1) and (2)
h = ½ g (x/u)2
or x2 =2 2u
gh
This is the equation of a parabola. Therefore the path traversed by the stone is aparabola.
Take the case of an aeroplane flying at a height h from the ground. It is to drop a bombat C. If the bomb is released just at the moment when the aeroplane is vertically above c,the bomb will not fall at C but it will fall at a point some distance away from C. From theknowledge of the height h, time t can be calculated from the relation:
h = ½ gt2, or t hg
= 2
and BC = ut
Therefore the bomb should be released at a pointA which is vertically above B and t seconds earlier.Then after traversing a parabolic path the bomb willreach the point C. It is to be remembered that theaeroplane will also reach a point vertically above Cjust at the moment when the bomb reaches the pointC. In calculating the distance BC air resistance isneglected.
Show that the horizontal range is maximum when the angle of projection is 45° with thehorizontal.
Suppose a body is projected with a velocity u making an angle θ with the horizontal.Resolve this velocity u into two components.
(i) u cos θ in the horizontal direction,
(ii) u sin θ in the vertically upward direction.
B
θ
CA u cos θ
u s in θ u
Fig. 5
++
CB
A u
h
+
Fig. 3
CB
A
h
+
+
+
Fig. 4
36 Mechanics
The horizontal component remains constant throughout the flight of the body in air,whereas the vertical component of the velocity changes continuously. The acceleration dueto gravity is negative.
Maximum height: Suppose the maximum vertical height reached is h.
From the equation, v2 – u2 = 2gh
v = 0, u = u sin θ and g is negative
∴ 0 – u2 sin2 θ = –2gh
or h =u
g
2 2
2sin θ
...(i)
After time t, the distance travelled in the horizontal direction,
x = (u cos θ)t ...(ii)
Distance travelled in the vertical direction
y = (u sin θ) t – 12
gt2 ...(iii)
From equation (ii), t = x/(u cos θ)
y =u xu
g xu
sincos cos
θθ θ
−FHG
IKJ
12
2
= x g xu
tancos
θθ
−FHG
IKJ2
2
...(iv)
This is the equation of a parabola
Time in air: Suppose the projectile remains for time t in air: It means y = 0
From equation (iii)
0 = (u sin θ) t – 12
gt2
or t =2u
gsin θ
...(v)
Maximum range: The horizontal range
x = (u cos θ) t
Putting value of t from Eq. (v), we get
x = ( cos ). sinu ug
θ θ2
=u
g
2 2sin θ...(vi)
The range x will be maximum
When sin 2θ = 1 or 2θ = 90° i.e., θ = 45°
Hence the horizontal range is maximum when the projectile make an angle of 45° withthe horizontal.
Force and Motion 37
2.7 EQUILIBRIUM OF FORCES
When a particle is in equilibrium, then the resultant of all the forces acting on it is zero.It then follows from Newton’s first law of motion that a particle in equilibrium is either atrest or is moving in a straight line with constant speed. It is found that for a large numberof problems, we have to deal with equilibrium of forces lying in a plane. Therefore, we shallrestrict our discussion to the case when a particle is in equilibrium under the influence of
a number of coplanar forcesF F F1 2 3
→ → →, , ... . The required condition is given by
F F F1 2 3
→ → →+ + … = 0 ...(i)
Since the forces are coplanar, we can resolve them along two mutually perpendiculardirections of x and y-axes, O being the particle. So above equation (i) can be rewritten as
F F F F1 1 2 2x y x yi j i j$ $ $ $ ...+ + + +e j e j = 0
or F F F F1 2 1 2x x y yi j+ + +b g d i$ $ = 0
Σ( ...)F F1 2x x+ + = 0
and Σ ( ...)F F1 2y y+ + = 0
Equation (ii) can be expressed in a conciseform as
Σ Fx = 0,
and Σ Fy = 0
Where Σ denotes summation of the x- ory-components of the forces.
2.8 FRICTIONAL FORCES
When one body moves in contact with another, its motion is opposed by a force whichcomes into play at the plane of contact of the two bodies. This resistive force which tends todestroy the relative motion between the two is called the “force of friction”. Frictional forcesmay occur between the surfaces of contact even when there is no relative motion betweenthe bodies. Work is done in overcoming the friction and consequently friction produces powerlosses in moving machinery whose efficiency falls below hundred percent.
Friction has many useful aspects also. We could not walk without friction and if somehowwe start moving we could not stop. It would have been difficult even to fix nail in the wall.Though annoying, friction is a necessity for us as our everyday activities depend upon friction.
Origin of frictionThe phenomenon of friction occurring between dry surfaces of two sliding bodies is
complicated when it is observed at microscopic level. The force laws for friction do not havesimplicity and accuracy of very high order.
Large number of surface irregularities is the key reason of friction. When a pull isexerted on a body so as to let it slide over the other surface; the relative motion is resistedon account of large number of surface irregularities and an opposing force is developed whichis the force of friction. As the pull is increased, the opposing force also increases and at acertain pull, the body begins sliding.
UVW
( )ii
F 1
F 2
F 4
F3
X
Y
O
Fig. 6
38 Mechanics
2.9 STATIC FRICTION AND COEFFICIENT OF STATIC FRICTION
Suppose a block rests over a horizontal table Fig. 7(a). weight of block is balanced by thenormal reaction. Next a very small horizontal force Fa is applied to the block Fig. 7(b). Theblock still remains stationary. One can say that “frictional force Fs” has come into existence(Fs – Fa). The frictional force preventing a body from sliding over the surface of other bodyis termed static friction. Again if applied force Fa is increased a little, the block still remainsstationary which suggests that frictional force Fs has also increased with Fa. Increasing Fa,a stage is reached when the block begins to move. Till the motion does not start, the Fs isequal and opposite to applied force in all stages. The greatest value of Fs at the plane ofcontact of surfaces of two bodies in the stage when one body is just to slide over the otheris termed “limiting friction”, and is denoted by Fs. Which acts between surfaces at rest withrespect to each other is called ‘force of static friction’. Again
Maximum force of static friction, Fs = Smallest force needed to start motion FL
The coefficient of static friction (µs) is defined as
µs =Magnitude of maximum forces of static friction
Magnitude of normal force
i.e., µs = Fs/N
or Fs = µsN
N
m g
N
m g
F aFs
( )No fo rce app lied
a ( )Applied force ex is ts , bu t no m otion
b
Fig. 7
2.10 DYNAMIC FRICTION AND COEFFICIENT OF DYNAMIC FRICTION
In general a smaller force is needed to maintain uniform motion as compared to startthe motion of a body over the surface. The force acting between surfaces in relative motionis called forces of sliding friction (or kinetic friction or dynamic friction) and is denoted by Fd.
The kinetic friction is measured by the force necessaryto keep the two surfaces in uniform relative motion incontact with one another, coefficient of dynamic friction,(µd) is defined as µd = Magnitude of force of dynamicfriction/magnitude of normal force
i.e., µd = Fd/N
or Fd = µd N
Dependence of Coefficient of FrictionCoefficient of friction (µ) , depends upon (i) finishing of the surface, (ii) nature of material,
(iii) surface films, (iv) contamination, (v) temperature.
N
m g
F o
Fd = µ Nd
Un ifo rm m otion
Fig. 8
Force and Motion 39
The explanation of the two laws of friction lies in surface adhesion theory. In “rolling”’the frictional force is proportional to the velocity of the body; i.e.,
F = –kv
Where k = constant of proportionality
If, motion of the body is so fast that fluid swirls around (aeroplane moving in air) thanfrictional drag is given by
F = –kv2
For still higher velocities above equation does not hold good.
Angle of Friction and Cone of FrictionLet us consider a block A of mass m. HL is a horizontal rough surface. A force P is
applied parallel to surface on the block. Fs is frictional force, N normal reaction force.
If body be at rest, the resultant of Fs and N is a single force R that makes angle θ withnormal to surface, we have
N = R cos θFs = R sin θ
and so tan θ = Fs/N
NP
Rθ
F S
m gF s
LP
Aθ
H
N
Fig. 9 Fig. 10
As P is slowly increased, Fs also increases and thus θ also increases and θ is maximumwhen P is such that block is just to move. The friction in this case is termed “limitingfriction”.
The angle of friction (λ) is defined as the angle which the resultant reaction (R) of thesurface makes with the normal when the friction is limiting. Thus,
(tan θ)limiting = tanF
Nlimitingλ =
( )s
or tan λ =FNL = µs
As FL = (Fs)limiting
Resultant of FL and N lies on the surface of cone havingλ as the semi-vertical angle and direction of N as its axis. Thiscone is termed “cone of friction” as shown in figure 11.
PROBLEM
Q. 1. Two blocks of masses m1 and m2 are connected by a massless spring on a horizontalfrictionless table. Find the ratio of their accelerations a1 and a2 after they are pulled apart andthen released.
Fig. 11
40 Mechanics
Solution. The blocks are pulled apart by equal and opposite forces. On being releasedthey start moving toward each other under equal and opposite (elastic) force exerted by themassless spring. Since force is equals to mass multiplied by acceleration, we have
F = m1 a1 = m2a2
∴aa
1
2=
mm
2
1
Q. 2. A car of mass 1200 kg moving at 22 m/s is brought to rest over a distance of 50m.Find the breaking force and the time required to stop.
Solution. As soon as the brakes are applied, the car is decelerated. If a be the (negative)
acceleration, then from the relationν ν2 2 2= +o ax, we have
0 = (22)2 + 2a (50)
so that a = − = −22 222 50
4 84××
. m/s2
The braking (retarding) force on the car is, therefore,
F = mass × acceleration
= 1200 × (–4.84)
= –5808 nt.
The negative sign signifies retardation.
Let t be the time the car takes to stop. Then, from the relation v = v0 + at, we have
0 = 22 + (–4.84)t
∴ t = 22/4.84 = 4.55 sec.
Q. 3. A car having a mass of 150 kg is moving at 60 km/hour. When the brakes are appliedto produce a constant deceleration, the car stops in 1.2 min. Determine the force applied tothe car. [Ans. –347 nt.]
Q. 4. A gun fires ten 2 gm bullets per second with a speed of 500 m/sec. The bullets arestopped by a rigid wall. (a) What is the momentum of each bullet? (b) What is the kinetic energyof each bullet? (c) What is the average force exerted by the bullets on the wall?
Solution: (a) The momentum of each bullet is
P = mv
= (2 × 10–3 kg ) (500 m/s) = 1 Kg-m/s.
(b) The kinetic energy of each bullet is
K = ½ mv2
= ½ (2 × 10–3 kg) (500 m/s)2 = 250 joule.
(c) The bullets are stopped by a rigid wall. The magnitude of the change of momentumfor each bullet is therefore 1 kg-m/s. Since 10 bullets are fired per second, the rate of changeof momentum is
dP/dt = 10 kg-m/s2
But this must be equal to the average force exerted on the wall. Thus
F = dP/dt = 10 kg-m/s2 = 10 nt.
Force and Motion 41
Q. 5. A body of mass 10 gm falls from a height of 3 meter into a pile of sand. The bodypenetrates the sand a distance of 3 cm before stopping. What force has the sand exerted onthe body?
Solution. Let v be the velocity of the body at the instant it reaches the pile of sand.
Then from the relationv v gyo2 2 2= + , we have
v2 = 0 + 2 × ( 9.8 meter/sec2) × 3 meter.
= 58.8 (meter/sec)2
This velocity is reduced to zero due to the deceleration ‘a’ produced by the sand.
Thus, from the relationv v ayo2 2 2= + , we have
0 = 58.8 + 2a (0.03 meter)
a = − = −58 82 0 03
980.× .
meter/sec2
The mass of the body is 10 gm = 0.01 kg. Hence the (retarding) force exerted by the sandon it is
F = ma
= 0.01 kg × (–980 m/s2)
= –9.8 nt
Q.6. An electron (mass 9.0 × 10–31 kg) is projected horizontally at a speed of 1.2 × 107
m/sec into an electric field which exerts a constant vertical force of 4.5 × 10–15 nt on it.Determine the vertical distance the electron is deflected during the time it has moved forward3.0 cm horizontally.
Solution. The time taken by the electron to move 3.0 cm (= 0.03 meter) horizontallyis given by
t =0 03 2 5 10 9. . × meter
1.2 × 10 meter/sec sec.7 = −
The vertical acceleration of the electron is
a =ForceMass
nt/kg= =−
−4 5 109 0 10
5 0 1015
3115. ×
. ×. ×
The vertical distance moved under this acceleration in time t is
y =12
at2
=12
× (5.0 × 1015) × (2.5 × 10–9)2
= 1.56 × 10–2 meter = 1.56 cm.
Q. 7. A block of mass m = 2.0 kg is pulled along a smooth horizontal surface by ahorizontal force F. Find the normal force exerted on the block by the surface. What mustbe the force F if the block is to gain a horizontal velocity of 4.0 meter/sec in 2.0 secstarting from rest?
42 Mechanics
Solution. Let N be the normal force exerted by the smooth surface on the block. Thenet vertical force on the block is N – mg. Since there is no vertical acceleration in the block,we have by Newton second law (net force = mass × acceleration)
N – mg = 0
or N = mg = 2.0 × 9.8 = 19.6 nt
The horizontal acceleration ‘a’ may be obtained from therelation v = v0 + at :
4.0 = 0 + a (2.0)
∴ a = 2.0 m/sec2
The required horizontal force is therefore, by Newton’s Law, given by
F = ma
= 2.0 × 2.0 = 4.0 nt.
Q. 8. A block of mass M is pulled along a smoothhorizontal surface by a rope of mass m. A force F isapplied to one end of the rope. Find the acceleration ofthe block and the rope. Also deduce the force exerted bythe rope on the block.
[Ans. F/(M + m), MF/M + m)]
Q. 9. Two blocks of masses m1 = 2.0 kg and m2 = 1.0 kg are in contact on a frictionlesstable. A horizontal force F = 3.0 nt is applied to m1. Find the force of contact between m1 andm2. Again, find the force of contact if F is applied to m2.
Solution. Let F′ be the force of contact between m1 and m2. The block m1 exerts a forceF on m2. and by the law of action and reaction, m2 also exerts force F′ on m1, as shown inthe figure.
Fig. 14 Fig. 15
The new force on m1 is F – F′, and that on m2 is F′. Let a be the (common) accelerationproduced. Then, by Newton’s second law, we have
F – F′ = m1a
and F′ = m2a
Solving: F′ =m
m m2
1 2+F
Putting the given values:
F′ =1 0 3 0 1 0. ( . ) .kg
2.0 kg + 1.0 kgnt nt.=
If F is applied to m2., then we would have F ′ = 2.0 nt.
Q. 10. Three blocks of masses m1 = 10 kg, m2 = 20 kg and m3 = 30 kg are connected asshown in the Fig. 16 on a smooth horizontal table and pulled to the right with a force T3 =60 nt. Find the tension T1 and T2. (Lucknow, 1985)
Fig. 12
Fig. 13
Force and Motion 43
Solution. The force T3 is exerted on the entire system. The block m3 exerts a forwardforce T2 on m3; while by Newton’s third law the block m2 exerts a backward force T2 on m3.Similarly, m3 exerts a forward force T1 on m1 and m1 exerts a backward force T1 on m1. Thisis shown below.
Fig. 16
Now, suppose all the blocks acquire an acceleration a. First consider the system as awhole.
Net force on the system = mass × acceleration
T3 = (m3 + m2 + m1)a ...(i)
Again, we consider m2 and m1 together.
Net force on m2 and m1 = mass × acceleration.
T2 = (m2 + m1)a ...(ii)
Finally, we consider m1 only.
T1 = m1 a ...(iii)
Here, T3 = 60 nt; m1 = 10 kg, m2 = 20 kg and m3 = 30 kg. Substitution in eq. (i) gives
60 = (30 + 20 + 10 ) a
a = 60/60 = 1 m/sec2
Substituting the value of a in (ii) and (iii), we get
T2 = 30 nt.
T1 = 10 nt.
Q. 11. An inextensible string connecting blocks m1 = 10 kg and m2 = 5 kg passes over alight frictionless pulley as shown. m1 moves on a frictionless surface while m2 moves vertically.Calculate the acceleration a of the system and the tension T in the string.
Solution: The diagram shows the forces on each block. The hanging block m2 (due togravity) exerts through the tension force T on m2.
Now, the forces on m1 are: its weight mg vertically downward, normal force N exertedby the smooth surface and the tension T. If a be the acceleration of the block in the horizontaldirection then by Newton’s law, we have
T = m1a ...(i)
And N – m1g = 0 ...(ii)
Since the string is inextensible, the acceleration of the block m2 is also a. The new forceon m2 is m2g – T. Hence by Newton’ second law, we have
m2 g – T = m2a ...(iii)
Adding (i) and (iii), we get
m2 g = (m1 + m2)a
44 Mechanics
or a =m
m mg2
1 2+
Substituting the given value of a in (i) we get
T =m m
m mg1 2
1 2+
Substituting the given values: m1 = 10 kg and m2 = 5 kg.
a = (1/3) g = (1/3) × 9.8 = 3.27 nt/kg
and T = (10/3) g = (10/3) × 9.8 = 32.7 nt.
Q. 12. An inextensible string connecting unequal masses m1 = 1.0 kg and m2 = 2.0 kgpasses over a massless and frictionless pulley as shown in figure 18. Find the tension in thestring and the acceleration of the masses.
Solution. As the string is inextensible, both masses have the same acceleration a. Also,the pulley is massless and frictionless, hence the tension T at both ends of the string is thesame. The heavier mass m2 is accelerated downward and the lighter mass m1 is acceleratedupward.
The new (upward) force on m1 is T – m1g, and the net (downward) force on m2 ism2g – T. Therefore, by Newton’s second law, we have
T – m1 g = m1a ...(i)
and m2 g – T = m2a ...(ii)
Adding (i) and (ii), we get
(m2 – m1) g = (m1 + m2)a
or a = m mm m
g2 1
1 2
−+
...(iii)
Substituting for a in eq. (i) and solving , we get
T =2 1 2
1 2
m mm m
g+
Thus, for m1 = 1.0 kg and m2 = 2.0 kg, we have
a = (1/3) × 9.8 = 3.27 m/sec2
and T = (4/3) g = (4/3) × 9.8 = 13.1 nt.
Q. 13. A uniform flexible chain of length l with mass per unitlength λ, passes over a small frictionless, massless pulley. It isreleased from a rest position with a length of chain x hanging fromone side and a length l – x from the other side. (a) Under whatcondition will it accelerate? (b) If this condition is met, find theacceleration a as a function of x. (Lucknow 1997)
[Ans. (a) l ≠ 2x , (b) a = g (1 – 2x/l)]
Q. 14. A block of mass m1 = 3.0 kg on a smooth inclined plane of 30° is connected by acord over a small, frictionless pulley to a second block of mass m2 = 2.0 kg hanging vertically
N
m g1
m g2
T
a
a
T
m 2
m 1
Fig. 17
m g1
m g2
a
T
T
a
Fig. 18
Force and Motion 45
as shown in figure 19. Calculate the acceleration with which the blocks move, and also tensionin the cord. (take g = 10 m/s2).
Solution. Let a be the acceleration of the blocks m1 and m2, and T the tension in thecord, which is uniform throughout as the cord is massless and the pulley is frictionless.
The force acting on the block m1 are: (i) its weight m1g which is the force exerted bythe earth, (ii) the tension T in the string, and (iii) the normal force N exerted by the inclinedsurface (the force is normal because there is no frictional force between the surfaces).
The net force on m1 along the plane is T-m1g sin θ, and that perpendicular to the planeis N – m1g cos θ. The block m1 is accelerated along the incline while the accelerationperpendicular to the plane is zero. Therefore, by Newton’s second law, we have
T – m1 g sinθ = m1a ...(i)
and N – m1g cosθ = 0 ...(ii)
Again, the forces on m2 are : (i) its weight m2g and (ii) the tension T in the string. Thenet vertical (downward) force is m2 g – T. As the block is accelerated downward, Newton’ssecond law gives
m2 g – T = m2a ...(iii)
Solving (i) and (iii), we get
a =m m
m mg2 1
1 2
−+
sinθ
and T =m m
m mg1 2
1 2
1( sin )++
θ
N
mg sin
1
θ
m g2
T
a
a
T
θ
θ
m 1
m 2
mg cos
1
θ
m g1
Fig. 19
Here m1 = 3.0 kg, m2 = 2.0 kg, sin θ = sin 30° = ½ and g = 10 m/s2
a =2 0 3 0 1
23 0 2 0
10 1 0. . ×
. ..
− FHG
IKJ
+= m/s2
and T =( . ) ( . )
. .
3 0 2 0 1 12
3 0 2 010 18
+FHG
IKJ
+= nt.
Q. 15. A car, mass 1000 kg moves uphill along a smooth road inclined 30°. Determine theforce which the engine of the car must produce if the car is to move (a) with uniform motion
46 Mechanics
(b) with an acceleration of 0.2 m/s2. (c) Find also in each case the force exerted on the car bythe road. [Ans. (a) 4900 nt, (b) 5100 nt, (c) 4900 √3 in each case]
Q. 16. A block of mass m = 2.0 kg is kept at rest on a smooth plan, inclined at an angleθ = 30° with the horizontal, by means of a string attached to the vertical wall. Find the tensionin the string and the normal force acting on the block. If the string is cut, find the accelerationof the block. Neglect friction [Ans. 9.8 nt, 17 nt, 4.0 m/s2]
Q. 17. A space traveller, whose mass is 100 kg leaves the earth. Compute his weight onearth (g = 9.8 m/s2) and on Mars (g = 3.8 m/s2).
Solution: Weight = mass × acceleration due to gravity. Therefore, weight on earth
WE = 100 kg × 9.8 m/sec2
= 980 nt,
and weight on Mars
WM = 100 kg × 3.8 m/sec2
= 380 nt.
The weight in interplanetary space would be zero. Of course, the mass will remain 100kg at all locations.
Q. 18. A body has a mass of 100 kg on the Earth. What would be it (i) mass and (ii) weighton the Moon where the acceleration due to gravity is 1.6 m/sec2.
[Ans. (i) 100 kg, (ii) 160 nt.]
Q. 19. A massless string pulls a mass of 50 kg upward against gravity. The string wouldbreak if subjected to a tension greater than 600 Newtons. What is the maximum accelerationwith which the mass can be moved upward? (Lucknow 1990, 1996)
Solution. The force acting upon the mass are (i) its weight mg and (ii) tension T in thestring. The net (upward) force, T – mg, is responsible for the upward acceleration a of themass. By Newton’s second law (net force = mass × acceleration), we have
T – mg = ma
Here (maximum) tension T = 600 nt and m = 50 kg.Therefore; the (maximum) acceleration a is given by
a =T − mg
m
a =600 50 9 8
502 2− =( × . ) . m/sec2
Q. 20. A body of mass 50 kg is hanging by a rope attached to the ceiling of an elevator.Calculate the tension in the rope if the elevator is (i) accelerating upward at 4 m/s2,(ii)downward at 4 m/s2, (iii) falling freely, (iv) moving with uniform velocity of 5 m/s upward ordownward.
Solution. The forces of the body are its weight mg and the tension T in the rope whichis the upward force exerted on the body by the rope. (If the body were suspended by a springbalance, then T would be the reading of the balance. T is also known as the ‘apparent weight’of the body).
Fig. 20
Force and Motion 47
(i) The body is at rest relative, to the elevator, and hence has an upward acceleration‘a’ relative to the earth. Taking the upward direction as positive, the resultant force on thebody is T – mg. Hence from Newton’s law (force = mass × acceleration), we have
T – mg = ma
∴ T = mg + ma
Thus the apparent weight T is greater than the true weight mg, and the body appears“heavier”
Substituting the given values: m = 50 kg, a = 4 m/s2
T = m (g + a) = 50 (9.8 + 4) = 690 nt.
(ii) When the elevator is accelerating downward (‘a’ negative), we may write
T – mg = – ma
∴ T = mg – ma
In this case the apparent weight T is less than true weight mg, and the body appears“lighter”.
Substituting the given values:
T = m(g – a) = 50 (9.8 – 4) = 290 nt
(iii) If the elevator falls freely , a = g, then we have
T = m(g – a) = m (g – g) = 0
In this case the tension (apparent weight) is zero and the body appears “weightless”. Ifthe body were suspended by a spring balance in a freely falling elevator, the balance wouldread zero.
(iv) If the elevator is at rest, or moving vertically (either up or down) with constantvelocity, a = 0. In this case
T – mg = 0
T = mg
The apparent weight equals the true weight.
T = mg = 50 × 9.8 = 490 nt.
a a
T
m g m g
Ta = 0
T
m g
( )a ( )b ( )c
Fig. 21
Q. 21. A body hangs from a spring balance supported from the roof of an elevator. (a) Ifthe elevator has an upward acceleration of 2.45 m/s2 and the balance reads 50 nt, what is thetrue weight of the body? (c) What will the balance read if the elevator cable breaks?
[Ans. (a) 40 nt (b) 2.45 m/s2 downward, (c) zero]
48 Mechanics
Q. 22. An elevator weighing 500 kg is pulled upward by a cable with an acceleration of2.0 m/s2 (a) What is the tension in the cable? (b) What is the tension when the elevator isaccelerating downward at 2.0 m/s2. [Ans. (a) 5900 nt, (b) 3900 nt]
Q. 23. An elevator and its load have a total mass of 800 kg. Find the tension T in thesupporting cable when the elevator, moving downward at 10 m/s, is brought to rest withconstant acceleration in a distance of 25 m.
Solution. Let us take the upward direction as positive. The elevator having an initialdownward velocity of 10 m/s is brought to rest within a distance of 25 m. Let us use therelation
v2 = v02 + 2ax
Here v = 0 , v0 = –10 m/s and x = –25 m. Thus
0 = (–10)2 + 2a (–25)
or a =− −
−( )( )
102 25
2
= 2 m/s2
The acceleration is therefore positive (upward). The resultant upward force on the elevatoris T – mg, so that by Newton’s law
T – mg = maT = mg + ma
= m (g + a)= 800 (9.8 + 2) = 9440 nt.
Q. 24. An elevator is moving vertically upward with an acceleration of 2 m/s2. Find theforce exerted by the feet of a passenger of mass 80 kg on the floor of the elevator. What wouldbe the force if the elevator were accelerating downward?
Solution. The force exerted by the passenger on the floor will always be equal inmagnitude but opposite in direction to the force exerted by the floor on the passenger. Wecan therefore calculate either the force exerted on the floor (action force) or the force exertedon the passenger (reaction force).
Let us take the upward direction as positive, and considerthe forces acting on the passenger. These are the passenger’strue weight mg acting downward, and the force P exerted onhim by the floor which is acting upward. P is the apparentweight of the passenger. The resultant upward force on thepassenger is thus P – mg. If a be the upward acceleration ofthe passenger (and of the elevator), we have by Newton’ssecond law
P – mg = ma
or P = mg + ma
Thus the apparent weight P is greater than the true weight mg. The passenger feelshimself pressing down on the floor with greater force (the floor is pressing upward on himwith greater force) than when he and the elevator are at rest (or moving with uniformvelocity).
Fig. 22
Force and Motion 49
Substituting the given values: m = 80 kg and a = 2 m/s2
P = m (g + a) = 80 (9.8 + 2) = 944 nt.
If the elevator is acelerating downward (a negative),we may write
P = mg – ma.
In this case the apparent weight P is less than the true weight mg. The passenger feelshimself pressing down on the floor with less force than when he and the elevator are at rest(or moving with uniform velocity).
Substituting the given values:
P = m (g – a) = 80 (9.8 – 2) = 624 nt
If the elevator cable breaks, the elevator falls freely (a = g). In the case
P = m(g – a) = m(g – g) = 0
Then the passenger and floor would exert no forces on each other. Then passenger’sapparent weight would be zero.
Q. 25. A 90-kg man is in an elevator. Determine the force exerted on him by the floorwhen: (a) the elevator goes up/down with uniform speed, (b) the elevator accelerates upwardat 3 m/s2, (c) the elevator accelerates downward at 3 m/s2, and (d) the elevator falls freely.
[Ans. (a) 882 nt, (b) 1152 nt, (c) 612 nt, (d) zero]
Q. 26. The total mass of an elevator with a 80-kg man in it is 1000 kg. This elevatormoving upward with a speed of 8 m/s, is brought to rest over a distance of 20 m. Calculate(a) the tension T in the cables supporting the elevator, and (b) the force exerted on the manby the elevator floor.
Solution. (a) Let us take the upward direction as positive. The elevator having an initialupward speed of 8 m/s is brought to rest within a distance of 20 m. Let use the relation
v2 = v02 + 2ax
Here v = 0, v0 = 8 m/s, and x = 20 m. Thus
0 = (8)2 + 2a (20)
a = − = −( )×
.82 20
1 62
m/s2
The acceleration is therefore negative (downward). The resultant upward force on theelevator is T – mg, where mg is the total weight of the elevator. By Newton’s law
T – mg = ma
or T = mg + ma
= m (g + a)
= 1000 (9.8 – 1.6) = 8200 nt.
(b) Let P be the (upward) force exerted on the man by the elevator floor. If m′ be the massof a man, its weight acting downward is m′g. Thus the net upward force on the man is P –m′g. Since a is the acceleration of the man (and of the elevator), we have by Newton’s law
P – m′g = m′aor P = m′g + m′a = m′ (g + a)
= 80 (9.8 – 1.6) = 656 nt.
50 Mechanics
Q. 27. A bob hanging from the ceiling of a car acts as an accelerometer. Derive the generalexpression relating the horizontal acceleration a of the car to the angle θ made by the bob withthe vertical.
Solution. When the car has an acceleration ‘a’ toward the right, the rope carrying theball makes an angle θ with the vertical.
a
T cos θ
T sin θ
θ
T
m g
θ
Fig. 23
Two forces are exerted on the ball: its weight mg and the tension T in the rope (shownseparately). The resultant vertical force is T cos θ – mg, and the resultant horizontal forceis T sin θ. The vertical acceleration of the system is zero, while the horizontal accelerationis a. Hence by Newton’s second law, we have
T cos θ – mg = 0and T sin θ = ma
Eliminating T, we gettan θ = a/g
or a = g tan θThis is the required result
Q. 28. Consider free fall of an object of mass m from rest through the air which exertsa frictional force on it, proportional to its velocity. If the proportionality constant is k : (a) Writedown its equation of motion. (b) Show that it ceases to accelerate when it reaches a terminalvelocity vT = mg/k. (c) Verify that the velocity of the object varies with time asv = vT(1 – e–kt/m).
Solution. (a) The forces acting on the object are (i) the force of gravity mg in thedirection of motion and (ii) the frictional force kv against the motion. The net force in thedirection of motion is therefore mg – kv. This, by Newton’s second law, is equal to mass mmultiplied by the acceleration a of the body. Thus
mg – kv = ma ...(i)
If y represents the (vertical) displacement, then v = dydt
and advdt
d ydt
= =2
2 . The above
equation may therefore be written as
mg – kdydt
=md ydt
2
2
d ydt
2
2 = − +km
dydt
g ...(ii)
This is the equation of motion of the object
(b) Let vT be the ‘terminal’ velocity attained by the object when the acceleration abecomes zero. Making this substitution in eq. (i), we have
Force and Motion 51
mg – kvT = 0
vT =mgk
...(iii)
(c) The equation of motion (ii) can be written as
dvdt
= − +km
v g
But g = (k/m) vT from equation (iii). Therefore
dvdt
= − + = −km
vkm
vkm
v vT T( )
dvv vT − =
km
dt
Integrating, we get
− −log ( )e v vT=
km
t + C (constant)When t = 0 , v = 0. Thus
– loge vT = C
Therefore, the above equation becomes
− −log ( )e v vT =km
t ve− log T
or log( )
ev v
vT
T
− =−km
t
or( )v v
vT
T
−= e
ktm
−
or 1 − vvT
= ektm
−
orvvT
= 1 −−
ektm
or v = v ektm
T 1 −F
HG
I
KJ
−
This is the required expression.
Q. 29. A block weighing 20 nt rests on a rough surface. A horizontal force of 8 nt isrequired before the block starts to slide, while a force of 4 nt keeps the block moving atconstant speed once it has started sliding. Find the coefficients of static and sliding (kinetic)friction.
52 Mechanics
Solution. The smallest (horizontal) force F required to start the motion is equal to themaximum force of static friction fs , i.e.,
F = fsBut we know that (maximum) fs = µs N = µs W,
Where µs is the coefficient of static friction and N is the normal reaction which is equalto the weight W of the block. Thus
F = µs W,
or µs =FW
ntnt
= =820
0 4.
Again; the force F′ required to maintain a uniform motion is equal to the force of kineticfriction, fk . That is
F′ = fk = µk N = µkW
or µk =FW
nt20 nt
′ = =4 0 2.
Q. 30. A block weighing 10 nt is rest on a horizontal table. The coefficient of static frictionbetween block and table is 0.50. (a) What is the magnitude of the horizontal force that will juststart the block moving? (b) What is the magnitude of a force acting upward 60° from thehorizontal that will just start the block moving? (c) If the force acts down at 60° from thehorizontal, how large can it be without causing the block to move?
Solution. (a) As shown in Fig. 24, the horizontal force F that will just start the blockmoving is equal to the maximum force of static friction. Thus
F = fs = µsN
Where µs is the coefficient of static friction and N is the normal reaction. Also
N = W,
So that F = µsW = 0.50 × 10 nt = 5.0 nt.
(b) In this case the forces on the block are shown in Fig. The applied force is inclinedat φ upward from the horizontal. Its horizontal and vertical components are Fx = F cos φ andFy = F sin φ. In this case, we have, for equilibrium
Fx = fs = µs N
and Fy = W – N
Thus Fx = µs (W – Fy)
or F cos φ = µs (W – F sin φ)
or F =µ
φ µ φsW
cos + s sin
Here µs = 0.50, W = 10 nt,
cos φ = cos 60° = 0.50 and sin 60° = 0.866
F =0 50 10
0 50 0 50 0 8665 36. ×
. ( . × . ).
+= nt.
Fig. 24
Force and Motion 53
Note that in this example the normal reaction N is not equal to the weight of the block,but is less than the weight by the vertical component of the force F.
(c) In this case (Fig. 25), the equilibrium equations will be:
Fx = fs = µs N
and Fy = N – W
From these we can see that
F =µ
φ µ φs
s
Wcos − sin
F =0 50 10
0 50 0 5 0 86674 6. ×
. ( . × . ).
−= nt
In this case the normal reaction is greater than the weight of the block.
Q. 31. A block is pulled to the right at constant velocity by a 10-nt force acting 30° abovethe horizontal. The coefficient of sliding friction between the block and the surface is 0.5. Whatis the weight of the block? [Ans. 22.3 nt]
Q. 32. A block of mass 10 kg rests on a horizontal surface. What constant horizontal forceF is needed to give it a velocity of 4 m/s in 2 sec, starting from rest, if the friction force betweenthe block and the surface is constant and is equal to 5 Newton?
Solution. Let a be the acceleration in the block when in motion. From the relation v= v0 + at, we have
a =v v
t− 0
Here v = 4 m/s, v0 = 0 and t = 2s. Thus
a =4 0
2−
= 2 m/s2
If F be the force applied on the block, then the net force producing the above accelerationwould be F – fs , where fs is the static force of friction between the block and the surface.Thus, by Newton’s second law,
F – fs = ma
or F = ma + fs
Here m = 10 kg, a = 2 m/s2 and fs = 5 nt.
∴ F = (10 × 2) + 5 = 25 nt.
Q. 33. A body of mass 2000 kg is being pulled by a tractor with a constant velocity of5 m/s on a rough surface. If the coefficient of kinetic friction is 0.8, what are the magnitudesof the frictional force and of the applied force?
Solution. The force of kinetic friction is given by
fk = µk N,
where µk is the coefficient of kinetic friction and N is the normal reaction which is equal tothe weight W (= mg) of the body. Then
fk = µk mg
Fig. 25
54 Mechanics
= 0.8 × 2000 kg × 9.8 nt/kg
= 15680 nt, in the direction opposite to motion.
In order to pull the body with a constant velocity, the applied force must be just equalto the frictional force, and applied in the direction of motion.
Q. 34. The coefficient of friction between a block of wood of mass 20 kg and a table surfaceis 0.25. (a) What force is needed to give it an acceleration of 0.2 m/s2, (b) What forces are actingwhen it is moving with a constant velocity of 2 m/s, (c) If no external force is applied, howfar will block travel before coming to rest.
[Ans. (a) 53 nt, (b) applied and frictional force each 49 nt. (c) 0.816 m]
Q. 35. A block sliding initially with a speed of 10 m/s on a rough horizontal surface, comesto rest in a distance of 70 meters. What is the coefficient of kinetic friction between the blockand the surface?
Solution. Let us assume that the force of kinetic friction fk between the block and thesurface is constant. Then we have a uniformly decelerated motion. From the relation
v2 = v ax02 2+
with the final velocity v = 0, we obtain
0 = (10 m/s)2 + 2a (70 m)
so that a = − = −10 102 70
0 71××
. m/s2
The negative sign indicates deceleration. The frictional force on the block is, by thesecond law of motion, given by
fk = –ma
where m is the mass of the block. But we know that
fk = µkN
where µk is the coefficient of kinetic friction and N is the normal reaction which is equal tothe weight W (= mg) of the block. Thus
fk = µk mg
µk =f
mgma
mga
gk = − = −
[fk = –ma]
= 0 719 8
0 072..
. .m/sm/s
2
2 =
Q. 36. If the coefficient of static friction between the tyres and the road is 0.5, what isthe shortest distance in which an automobile can be stopped when traveling at 72 km/hour?
Solution. Let fs be the constant force of static friction between the tyres and the road.The automobile is under uniformly decelerated motion. From the relation
v2 = v02 + 2ax
with the final velocity v = 0, we obtain xva
= − 02
2, ...(i)
Force and Motion 55
where the minus sign means that the acceleration a caused by fs is opposite to the directionof motion.
Now, by Newton’s second law, we have
fs = – ma,
Where m is the mass of the automobile. But we know that
fs = µsN = µs mg
where µs is the coefficient of static friction and N (= mg) is the normal reaction.
From the last two expression, we have
–ma = µs mg
or a = –µs g.
Then from eq. (i) the distance of stopping is
x =v
gs
02
2µ
Here v0 = 72 km/hour = 20 m/sec and µs = 0.5
∴ x =(
× . × ( . ).
202 0 5 9 8
40 8m/s)
m/s meter.
2
2 =
Q. 37. Block A weight 100 nt. The coefficient of static friction between the block and tableis 0.30. The block B weights 20 nt and the system is in the equilibrium. Find the friction forceexerted on block A. Also find the maximum weight of block for which the system will be inequilibrium. (Lucknow 1991, 1997)
Solution. The forces acting on the knot O are the tension T, T′ and W (weight of B)in the cords. (The reactions to T and T′ which act on A and on the wall are also shown). Sincethe knot is in equilibrium, we have
T = T′ cos 45°
and T′ sin 45° = W
Fig. 26
56 Mechanics
These equations give
T = W = 20 nt.
Now, the forces on the block A are its weight W, normal reaction N, tangential force ofstatic friction fs and the tension T. Again, since the block is in equilibrium, we have
N = W
and fs = T = 20 nt.
As W (and hence T) increases, the friction force fs also increases until it acquires itsmaximum value, at which
fs = µs N (where µs is coeff. of static friction)
= µs W (∴ N = W)
= 0.30 × 100 = 30 nt.
At this limiting state of equilibrium, W, T and fs all are equal and maximum. Thus
Wmax = 30 nt.
Q. 38. A 4.0 kg block A is put on top of a 5.0 kg block B; which is placed on a smoothtable. It is found that a horizontal force of 12 nt is to be applied on A in order to slip it onB. Find the maximum horizontal force F which can be applied to B so that both A and B movetogether, and the resulting acceleration of the blocks.
Solution. Let m1 and m2 be the masses of the blocks A and B. If µk be the coefficientof sliding friction between A and B, then the frictional force on A exerted by B is
fk = µk N = µk m1 g.
A force of 12 nt maintains a slip of A on B against theforce of friction fk. Thus
12 = fk = µk m1 g
Here m1 = 4.0 kg.
µk = 12 124 0
3
1m g g g= =
. ×B is placed on a smooth table. Hence in order to move A and B together we require a
force F on B which just overcomes the frictional force on B exerted by A and that on A exertedby B. Thus
F = µk m1 g + µk m2 g
= µk (m1 + m2) g
= 3/g (4.0 + 5.0) g = 27 nt
This force is causing both A and B to move on the smooth table. Hence the accelerationproduced in them is
a =F
m/s2
m m1 2
274 0 5 0
3+
=+
=. .
Q. 39. Block A weight 4 nt and block B weight 8 nt. The coefficient of sliding frictionbetween all surfaces is 0.25. Find the force F to slide B at a constant speed when (a) A restson B and moves with it, (b) A is held at rest, (c) A and B are connected by a light cord passingover a smooth pulley. (Lucknow, 1994)
[Ans. (a) 3 nt, (b) 4 nt, (c) 5 nt]
Fig. 27
Force and Motion 57
FA
BF
A
BF
AB
( )a ( )b ( )c
Fig. 28
Q. 40. A block of mass 5 kg resting on a horizontal surface is connected by a cord passingover a light frictionless pulley to a hanging block of mass 5 kg. The coefficient of kinetic frictionbetween the block and the surface is 0.5. Find the tension in the cord and the acceleration ofeach block.
Solution. Let m1 and m2 be the masses of the blocks A and B. The diagram shows theforces on each block. The block B exerts through the tension in the cord of force T on block.A, while A exerts an equal reaction force T on B.
Now, the forces on A are: its weight W1 (= m1 g) vertically downward, normal reactionN exerted by surface, tangential frictional force fk exerted by the surface, and the tension T.If a be the acceleration of the block A in the horizontal direction, then by second law ofmotion, we have
T – fk = m1a
and N – W1 = 0
or N = W1 = m1g.
But fk = µkN = µkm1g
∴ T – µkm1g = m1a ...(i)
Since the cord is inextensible, the acceleration of the block B is also a. The net force onB is W2 – T. Hence by law of motion, we have
W2 – T = m2a or m2g – T = m2a ...(ii)
Substituting the value of T from (i) into (ii), we get
m2g – (m1a + µkm1g) = m2a
or m2g – µk m1g = (m1 + m2)a
or a =( )m m
m mgk2 1
1 2
−+µ
...(iii)
Substituting this value of a in Eq. (ii), we get
m2 g – T = mm mm m
gk2
2 1
1 2
( )−+µ
T = m gm mm m
k2
2
1 21 − −
+LNM
OQP
µ
or T =m m
m mgk1 2
1 2
1( )++
µ...(iv)
58 Mechanics
A T
N
a
T
B a
W (= m g)1 1
W (= m g)2 2
fk
Fig. 29
Substituting the given values in (iii) and (iv); m1 = 5 kg, m2 = 5 kg, µk = 0.5
a =( . × ) ( . )5 0 5 5
5 59 8−
+ = 2.45 m/s2
and T =5 5 1 0 5
5 59 8× ( . ) ( . )+
+ = 36.75 nt.
Q. 41. The masses of A and B in Figure 30 are 10kg and 5 kg. Find the minimum mass of C that willprevent A from sliding, if µs between A and the table is0.20. Compute the acceleration of the system if C isremoved. Take µk between table and A also to be 0.20.
[Ans. 15 kg, g/5]
Q. 42. A 10 kg block is sliding down a plane inclined at an angle of 30° with the horizontal.Find the normal reaction and the acceleration of the block. The coefficient of kinetic frictionis 0.5.
Solution. The forces on the block (Fig. 31) are its weight W and the normal and the(kinetic) frictional components of the force exerted by the plane, namely, N and fk respectively.
The weight W may be resolved into two components, W sin θ parallel to the plane andW cos θ perpendicular to the plane. Thus the net force along the plane is W sin θ – fk andthat perpendicular to the plane is N – W cos θ. If a be the acceleration down the plane, wehave by second law of motion
W sinθ – fk = ma ...(i)
Since there is no motion perpendicular to the plane, we have
N – W cos θ = 0 ...(ii)
Also fk = µk N ...(iii)
From eq. (ii), we get
N = W cos θHere W = mg = 10 × 9.8 = 98 nt and cos θ = cos 30° = 0.866
Fig. 30
Force and Motion 59
Substituting this value of N and µk = 0.5 in (iii), we get
fk = 0.5 × 84.87 = 42.43 nt
Fig. 31
Now, Eq. (i)
a =W sinθ − f
mk
= g sin θ – (fk/m) [∴ W = mg]
= (9.8 × 0.5) – (42.43/10) [sin θ = sin 30° = 0.5]
= 4.9 – 4.243 = 0.657 m/s2
Q. 43. A piece of ice slides down a 45° inclined plane in twice the time it takes to slidedown a frictionless 45° inclined plane. What is the coefficient of kinetic friction between ice andinclined plane? (Lucknow, 1983)
Solution. Let a be the acceleration of the ice piece down the first (rough) plane. Theforces on the piece are shown in Fig. 31. The net force down the plane is W sin θ – fk andthat perpendicular to the plane is N – W cos θ. By second law of motion, we have
W sinθ – fk = ma ...(i)
and N – W cos θ = 0
so that N = W cos θAlso, fk = µk N = µk W cos θPutting this value of fk in (i), we get
W sin θ – µk W cos θ = ma
or a = g sin θ – µk g cos θ, ...(ii)
because W = mg, here θ = 45°, so that sin θ = cos θ = 1/ 2 . Thus
a =g
k2
1( )− µ
Let t be the time taken by the piece to slide down the plane. Then fromx v t at= +021
2,
we have (here v0 = 0)
x =12 2
1 2g tk( )− µ ...(iii)
60 Mechanics
For the frictionless plane (µk = 0), we have
a = g sinθ = g/ 2
The time is now t/2. Thus
x =12 2 2
2g tFHG
IKJ ...(iv)
Comparing (iii) and (iv), we get
1 – µk = ¼
´µk = 1 – ¼ = 3/4 = 0.75.
Q. 44. A block slides down an inclined plane of slope angle θ with constant velocity. Itis then projected up the same plane with an initial speed v0. How far up the incline will it movebefore coming to rest? Will it slide down again? (Lucknow 1986, 1990, 1993)
Solution. Refer to Fig. 31. The net force on the block down the plane is W sin θ – fk whichis zero because the block slides down with constant velocity (zero acceleration). Thus
W sin θ – fk = 0
or fk = W sin θWhen the block is projected up (now fk will be down the plane), the net force down the
plane becomes W sinθ + fk = W sin θ + W sin θ = 2 W sin θ. Hence the acceleration downthe plane would be
a =forcemass
W= =22
sinsin
θ θm
g
Now, using the relation
v2 = v02 + 2ax
with the final velocity v = 0, we have
0 = v g x02 2 2+ ( sin )θ
x = − vg
02
4 sinθ
Hence the distance covered up the plane
–x =v
g02
4 sinθ
As soon as the block stops, the frictional force becomes static, which is larger than thekinetic. Hence the block will not slide down again.
Q. 45. A block of mass 0.2 kg starts up a plane inclined 30° with the horizontal with avelocity of 12 m/s. If the coefficient of sliding friction is 0.16, how far up the plane the blocktravels before stopping? What would be the block’s speed when it is made to return to thebottom of the plane? [Ans. 11.5 m, 9.01 m/s]
Q. 46. A body with a mass of 0.80 kg is on a plane inclined at 30° to the horizontal. Whatforce must be applied on the body so that it slides with an acceleration of 0.10 m/s2 (a) uphill,(b) downhill? The coefficient of sliding friction with the plane is 0.30.
Force and Motion 61
Solution. Let us first consider the body moving uphill (Fig. 32a). The forces acting onthe body are its weight W (= mg) acting vertically downward, the normal force N and thekinetic frictional force fk exerted by the plane, and the force F applied uphill. The frictionalforce fk is always against the motion and hence in this case downhill. The weight W may beresolved into two components, W sin θ parallel to the plane and W cos θ perpendicular to theplane. Thus the net force on the body along the plane directed uphill is F – W sin θ –fk, andthat perpendicular to the plane is N – W cos θ. If a be the acceleration up the plane, thesecond law of motion gives
F – W sin θ – fk = ma ...(i)
Since there is no motion perpendicular to the plane, we have
N – W cos θ = 0
or N = W cos θAlso, we know that
fk = µkN = µk W cos θ.
W sin θ
W cos θfk
θ
W
N
θ
W sin θ
W cos θ
fk
θ
W
N
θ
F
( )b( )a
F
Fig. 32
Where µk is the coefficient of kinetic friction. Substituting this value of fk in eq. (i), weget
F – W sin θ – µk W cos θ=
ma
Replacing W by mg, we get
F – mg (sin θ + µk cos θ) = ma
or F = m[a + g (sin θ + µk cos θ)] ...(ii)
Here m = 0.80 kg, a = 0.10 m/s2, µk = 0.30 and θ = 30°, thus
sin θ = sin 30° = 0.5 and cos θ = cos 30° = 0.866. Whence
F = 0.80 [0.10 + 9.8 (0.50 + 0.30 × 0.866)]
= 6.04 nt.
(b) When the body is moving down hill, the forces are as shown in Fig. (b). We nowassume that the applied force F is downhill. fk is always against the motion and hence nowit is uphill. If a be the acceleration down the plane, then the equation of motion is
62 Mechanics
But fk = µk W cos θ (as before), therefore
F + W sin θ – µk W cos θ = ma
or F = m[a – g (sin θ – µk cos θ)] ...(iv)
Putting the values:
F = 0.80 [0.10 – 9.8 (0.5 – 0.30 × 0.866)] = –1.80 nt.
The negative sign means the force F is uphill instead of downhill as we had assumed.Still the motion is downhill.
Q. 47. Forces required to move a body on a rough inclined plane with a uniform velocityin the upward and downward directions are in the ratio 2:1. Find the inclination of the planeif the coefficient of friction is 3. (Lucknow, 1987)
Solution: Let F1 and F2 be the required forces. Putting a = 0 (because velocity isuniform) in eqs. (ii) and (iv) in the last problem, we get
F1 = mg (sin θ + µk cos θ)
and F2 = –mg (sin θ – µk cos θ)
But F1 = 2 F2. This gives
sin θ + µk cos θ = –2 (sin θ – µk cos θ)
or tan θ =µk
3Now µk = 3
tan θ = 1
or θ = 45°
Q. 48. A block A weighing 100 nt is placed on aninclined plane of slope angle 30° and is connected to asecond hanging block B by a cord passing over a smallsmooth pulley. The coefficient of sliding friction is 0.30.Find the weight of the block B for which the block A movesat constant speed (a) up the plane, (b) down the plane.
[Ans. (a) 76 nt, (b) 24 nt]
Q. 49. Two blocks with masses m1 = 1.65 kg andm2 = 3.30 kg are connected by a string and slide downa plane inclined at an angle θ = 30° with the horizontal.The coefficient of sliding friction between m1 and theplane is 0.226 and that between m2 and the plane is0.113. Calculate the common acceleration of the twoblocks and the tension in the string.
[Ans. 3.62 meter/sec2, 1.06 nt]
Q. 50. A block is at rest on an inclined plane making an angle θ with the horizontal. Asthe angle of incline is increased, slipping just starts at an angle of inclination θs. Find thecoefficient of static friction between block and incline in terms of θs.
Solution. The forces on the block are shown in Fig. 35. W is the weight of the block,N is the normal reaction and fs the tangential force of static friction exerted on the block by
Fig. 33
Fig. 34
Force and Motion 63
the inclined surface. The weight W may be resolved into two components, W sin θ parallelto the inclined plane and W cos θ perpendicular to it. Since the block is in equilibrium, thenet force along the plane as well as perpendicular to the plane is zero. That is
fs – W sin θ = 0
and N – W cos θ = 0
Also fs < µsN.
When we increase θ to θs, slipping just begins. Thus for θ = θs, we can use fs = µs N.Substituting this above, we get
µsN = W sin θs
and N = W cos θs
Dividing, we get
µs = tan θs
a
W sin θ
W cos θ
fs
θ
W
N
θ
Fig. 35
Hence measurement of the angle of inclination at which slipping just starts provides usa method for determining the coefficient of static friction between two surfaces.
In a similar way, we can show that the angle of inclination θk required to maintain aconstant speed of the block sliding down the plane is given by
µk = tan θk
where θk < θs and µk is the coefficient of sliding (kinetic) friction.
Q. 51. A block rests on a plane inclined at θ with the horizontal. The coefficient of slidingfriction is 0.50 and that of static friction is 0.75. (a) As the angle θ is increased, find theminimum angle at which the block starts to slip. (b) At this angle find the acceleration oncethe block has begun to move. (c) How long is required for the block to slip 8 meter along theinclined plane. [Ans. (a) 37°, (b) 1.96 m/s2, (c) 2.86 s]
Q. 52. A particle starting from rest moves in a straight line with acceleration a = (0.5 +0.1 t) m/s2, where t is the time. Calculate its velocity and the distance travelled after 2 seconds.
(Lucknow, 1984, 1989)
Solution. Suppose that initial position of particle is at origin i.e., at t = 0, x = 0, v = 0
Given a = 0.5 + 0.1t m/s2
or dv/dt = 0.5 + 0.1t or; dv = (0.5 + 0.1 t) dt
Integrating we get,
64 Mechanics
v = 0 50 1
2
2.
.;t
tk+ +
where k is constant of integration, determined by boundary condition.
At start; v = 0 and t = 0; so above equation yields; k = 0
So, v = 0 50 1
2
2.
.t
t+
At t = 2 sec; v = 0.5 × 2 + ½ × 0.1 × 4
or v = 1 + 0.2 = 1.2 m/sec
Again rewriting v,
dxdt
= 0 5 0 12
2. . ;t t+ i.e., dx t
tdt= +
FHG
IKJ
0 5 0 12
2. .
integrating, one gets
x =0 5
20 12 3
2 3. .t tk+ +
at start, x = 0, t = 0; so above equation gives; k = 0
so x =0 5
20 12 3
2 3. .t t+
and at t = 2;
x =0 52
4 0 12
83
. × . ×+
= 1 + .133
or x = 1.133 m
Q. 53. Two blocks of masses m1 and m2 are connected by a massless spring on ahorizontal frictionless table. Find the ratio of their acceleration a1 and a2 after they are pulledapart and released. (Lucknow, 1988)
Solution. Situation is shown in the Fig. 36 below.
Here F1 and F2 are forces in opposite directions on m1 and m2.
T1 and T2 are resulting tension in the spring due to F1 and F2.
T1, F1 are one action-reaction pair
T2, F2 are other action-reaction pair
As the system is in equilibrium
F1 + F2 = 0 and T1 + T2 = 0
i.e., m1a1 + m2a2 = 0
i.e.,aa
1
2=
−mm
2
1
Fig. 36
Force and Motion 65
Thus, the ratio of the acceleration is inversely proportional to their masses. The negativesign suggests that acceleration of the block m1 and m2 are oppositely directed.
Q. 54. The displacement (x) of a particle moving in one dimension, under the action of
a constant force is related to time (t) by the equation t x 3,= + where x is in meter and tin sec. Find displacement of particle when its velocity is zero.
Solution. Since t = x 3+x = (t – 3)2 = t2 – 6t + 9
Differentiating
v = dx/dt = 2t – 6
When v = 0, t = 3 sec.
∴ Displacement at time t = 3 sec. Will be
x = (3)2 – 6(3) + 9
= 0
Q. 55. A bullet of mass 20 gm, moving with velocity 16 m/s penetrates a sand bag andcomes to rest in 0.05 sec. Find (i) depth of penetration, (ii) average retarding force of sand.
(Lucknow, 1989)
Solution. Given, u = 16 m/s, v = 0 and t = 0.05 sec.
The retardation in sand is given by
0 = u – at
i.e., a = 16/.05 = 320 m/s2
(i) The depth of penetration is obtained by using equation
v2 = u2 – 2as
or 0 = (16)2 – 2(320)s
or s =16 162 320
0 4××
.= m
(ii) The average retarding force of sand is
F =mv mu
t−
= mv u
tma
−FHG
IKJ =
= (20 × 10–3) × 320
= 6.4 Newton
Q. 56. A light string which passes over a frictionless pulley, and hangs vertically on eachside of it, carries at each end a mass of 240 gm. When a rider of 10 gm is placed over oneof the masses, the system moves from rest a distance of 40 cm in 2 seconds. Calculate g.
Solution. Suppose P and Q are two masses each, equal to 0.240 kg. They are hangingfrom the pulley as shown. when rider R of mass 0.010 kg. is placed on mass Q, the systemmoves through 0.40 m in 2 sec. The final position is shown by dotted lines.
66 Mechanics
Let T be the tension in string and a be the acceleration either P or Q. Since P movesupward we have for its motion;
T – 0.240 g = 0.240 a ...(i)
and as Q moves downwards, therefore
(0.250 g – T) = 0.250 a ...(ii)
from (i) and (ii); eliminating T;
0.250 g – 0.240 g = 0.240 a + 0.250 a
or 0.010 g = 0.490 a
g = 49 a
To find a, we have
s = 0.40 m, t = 2, u = 0, a = ?
Using the formula s = ut + ½ at2
0.40 = 0 + 12
a × 4 = 2a
or a = 0.20 m/sec2
So, g = 49 a = 49 × 0.20 m/sec2 = 9.80 m/sec2
Q. 57. A uniform rope of length D resting on a frictionless horizontal surface is pulled atone end by a force P. What is the tension in the rope at a distance ‘d’ from the end where theforce is applied?
Solution. Suppose the rope to be divided into two segments A and B of length ‘d’ and(D – d). Let T be the tension at the boundary of A and B. If M is total mass of rope and ‘a’,the acceleration produced a = P/M in system,
Now the equation of motion of segment B is
T =MD
D PM
MD
mass /length− FHG
IKJ =F
HGIKJdb g ;
= P DD−L
NMOQP
d
T = P 1D
−LNM
OQP
d
Q. 58. A uniform rod of length L and density ρ is being pulled along a smooth floor witha horizontal acceleration a. Find the magnitude of stress at the transverse cross-sectionthrough the mid-point of rod? (I.I.T. 1993)
Solution. If A is cross sectional area of rod, its mass M = AL.ρ∴ Force acting forward on rod, F = A LραIf p is stress at the mid-cross section of rod, the resultant force on right half length ‘b’
of rod will be= F – P.A
Fig. 37
40 cmPQ
Q
TT
m gP
m gQ
a
aP
Fig. 36(a)
Force and Motion 67
∴ From Newton’s second Law,F – PA = mass of length b × accn
A L ρα – PA = LA2
ρ αFHG
IKJ
i.e., P =12
L ρα
Q. 59. A mass of 500 gm is placed on a smooth table with a string attached to it. Thestring goes over a frictionless pulley and is connected to another mass of 200 gm. At t = 0,the mass of 500 gm is at a distance of 200 cm from the end and moving with a speed of 50cm/sec towards the left (see Fig. 39). what will be its position and speed at t = 1 sec.
Solution. Suppose ‘a’ represent the acceleration of both the block at t = 0
The equation of motion of 200 gm is200 g – T = 200 a ...(i)
The equation of motion of 500 gm isT = 500 a ...(ii)
Adding equation (i) and (ii) we get200 g – 500 a = 200 a ...(iii)
i.e., a = 2/7 gso a = (2/7) × 980 = 280 cm/sec2
for displacement of block of 500 gm
Using, s = ut at+ 12
2
u = –50 cm/sec, a = 280 cm/sec2, t = 1 sec
∴ s = –50 × 1 +12
280 2× × t
= –50 + 140s = +90 cms
Thus, the block of 500 gm has moved 90 cms to the right of its initial position. Thevelocity v of block of 500 gm is then
v = u + at; u = –50 cm/sec, a = 280 cm/sec2, t = 1 sec= –50 + 280 × 1= 230 cm/sec (to the right)
Q. 60. A uniform flexible chain of length L and mass per unit length λ passes over asmooth massless small pulley. At an instant a length x of the chain hangs on one side of thepulley and length (L – x) on the other side. Find the instantaneous acceleration. Also find thetime interval in which x changes from L/4 to 0, if the chain moves with a negligible velocityat x = L/2. (Lucknow, 1987)
Solution. Suppose instantaneous common acceleration is ‘a’. Weight of chain of lengthx = x λ g. Net upward force on chain of length x = T – xλg = xλaor, T = xλa + xλg = xλ(a + g) ...(i)weight of chain of length (L – x) = (L – x) λg
Fig. 38
Fig. 39
68 Mechanics
Net force acting downward on chain of length (L – x) is(L – x) λg – T = (L – x) λa ...(ii)
Putting T from Eq. (i) in Eq. (i), we get(L – x) λg – xλ (a + g) = (L – x) λa
or Lλg – x λg – a xλ – g xλ = (L – x) λaor Lλg – 2xλg = (L – x) λ.a + a.xλor λg (L – 2x) = aλL
or a =λ
λg x g
x( ) ( )L
L LL− = −2 2
or a =g
xL
L( )− 2
Q. 61. Two rough planes, inclined at 30° and 60° to horizontal and of same height areplaced back to back. Two masses of 5 kg and 10 kg are placed on the faces and connected by
a string passing over the top of the planes. If coefficient of friction is 1/ 3 , find the resultantacceleration.
Solution. The 10 kg mass moves down and 5 kg mass moves up the plane.
Let; a = acceleration of system, T = the tension of the string and N1 and N2 be thenormal reaction of the plane.
For 5 kg mass, (ΣF = )ma
T – (µN1 + 5 g sin 30°) = 5a ...(i)
and 5 g cos 30° = N1 ...(ii)
For 10 kg mass
10 g sin 60° – T – µ N2 = 10 a ...(iii)
10 g cos 60° = N2 ...(iv)
(ii) and (i) gives
T – 5 g (sin 30° + µ cos 30°) = 5 a ...(v)
and (iv) in (iii) yields
10g (sin 60° – µ cos 60°) – T = 10 a ...(vi)
So, (v) and (vi) give
10 32
12 3
5 12
12
g g−FHG
IKJ
− +FHG
IKJ = 15 a
or a =2 3 3
9−
g
Q. 62. A block of mass 2 kg slides on an inclined plane which makes an angle of 30° with
the horizontal. The coefficient of friction between the block and surface is 3/2;
(i) What force should be applied to the block so that block moves down without anyacceleration?
(L – x)x
x gλ
T
– Ta
Fig. 40
30°
µN 1
N 1T
5g 10g60°
µN 2
T
N 2
a
Fig. 41
Force and Motion 69
(ii) What force should be applied to the block so that block moves up without anyacceleration.
(iii) Calculate the ratio of power, needed in the above two cases: if the block moves withthe same speed in both the cases.
Solution. (i) The figure 42 shows the situation, in which various forces are depicted,Resolving the forces along and perpendicular to the plane; we get
µN = P1 + mg sin 30° ...(i)
N = mg cos 30° ...(ii)
(ii) and (i) yields
µ mg cos 30° = P1 + mg sin 30°
or P1 = mg (µ cos 30° – sin 30°)
= 2 9 8 32
32
12
× . . −FHG
IKJ
=19 6 3 2
2 219 6 1 586
2 1 414
. . × .× .
−=
e j; P1 = 11 Newton
(ii) The forces acting in this case are shown in figure 43. Resolving along and perpendicularto the rough inclined plane; one gets;
P2 = mg sin 30° + µ N ...(iii)
N = mg cos 30° ...(iv)
(iv) and (iii) gives
P2 = mg (sin 30° + µ cos 30°)
or = 2 9 8 12
32
32
× . .+LNMM
OQPP
= 2 9 8 2 32 2
9 81 414
4 414× . ..
.+LNMM
OQPP
=
P2 = 30.6 Newtons
(iii) As Power = (force × displacement)/time = force × velocity
As the velocity is same in both the cases, the ratio of powers shall be equal to the ratioof forces applies i.e.,
PP
1
2=
FF
1
2= =11
30 60 36
..
63. A block weighing 400 kg rests on a horizontal surface and supports on top of it,another block of weight 100 kg as shown. The block W2 is attached to a vertical wall by a string6 m long. Find the magnitude of the horizontal force P applied to the lower block as shownin figure which shall be necessary so that slipping of W2 occurs.
m g sin 30°P 1
N
µN
3 0
m gm g cos 30°
3 0 °
Fig. 42
m g sin 30°
P 2
µN30°
m gm g cos 30°
30°
N
Fig. 43
70 Mechanics
(µ for each surfaces in contact is ¼ )
Solution. Forces on block W2 are shown below:
It may be seen, that
T cos θ = µN2 ...(i)
N2 + T sin θ = 100g ...(ii)
From figure 44(a) :
sin θ = 11 6 56
11 5/ : cos ; tan /θ θ= =
From (i) T =µ
θN2
cos; then (ii) is,
N2 + µ N2 tan θ = 100 g
or N2 (1 + µ tan θ) = 100 g
or N2 =100
1g
+ µ θtan ...(iii)
90° w 1
w 25m
6m
A
B
P
T sin θ
N 2
T cos θw 2
µN 2
100 g
µN 1
µN 2
N 1
N 2
Pw 1
( )a ( )b ( )c
T
θ
C
400 g
Fig. 44
Forces on block W1 are as shown in Fig. 44(b)As observed for block W1:
P = µN2 + µN1
=µ
µ θµ
µ θ×
tan tan100
1100
1400g g g
++
++
FHG
IKJ
LNMM
OQPP
(as N2 = N1 + 400 g)
P = µg µµ θ µ θ
×tan tan
1001
1001
400g+
++
+FHG
IKJ
LNMM
OQPP
= 9 84
200 2020 11
400. ×+
+LNMM
OQPP = 9 8 1000
20 3 316100.
.++
LNM
OQP
= 142.8 kg wtQ. 64. A block of mass 3 kg rests on a rough horizontal table, the coefficient of friction
between the surface is 0.5. A massless string is tied to the block which passes over a smoothlight pulley at the edge of the table and supports a block of 5 kg. Find the acceleration of thesystem, if any.
Force and Motion 71
Solution. For 3 kg mass, eqn. of motion is,T – µN = 3 a ...(i)
For 5 kg mass,5 g – T = 5 a ...(ii)
and µN = 0.5 × 3 × 9.8= 14.7 Newton ...(iii)
adding (i) and (ii) and using (iii)5 g – µN = 8 a
or 5 × 9.8 – 14.7 = 8 a
or a =49 14 7
834 3
8− =. .
a = 4.29 m/s2
Q. 65. Find the least force required to drag a particle of mass m along a horizontal surface.
Solution. Refer Fig. 46
P = force acting at ∠θ, R = resultant of N and Fs, λ = angle of friction
Resolving the forces horizontally and vertically
P cos θ – R sin λ = 0 ...(i)
P sin θ + R cos λ = mg ...(ii)
from (i) R =P cossin
θλ
...(iii)
then (ii) is;
P sin +P cossin
θ θλ
λ. cos = mg
or, P sin θ + P cos θ cot λ = mg ...(iv)
Differentiating (iv) w.r.t. to θ
sinθθ
θ λ θθ
θdd
dd
PP cos + cot cos
PP sin+ −F
HGIKJ = 0
ddP
P(θ
θ θ λ θ λ θ(sin cos cot ) cos cot sin )+ + − = 0
ddP P (sin
cotθθ λ θθ θ λ
= − −+
cot cos )sin cos
...(v)
for P to be minimum; dP/dθ = 0 i.e.,
sin cot cosθ λ θ− = 0; or; tan θ = tan λ,
i.e., θ = λ ...(vi)
Fig. 45
R
λθ
P
m g
Fs
N
Fig. 46
72 Mechanics
Putting (vi) and (iv)
P sin λ + P cos λ cot λ = mg
Multiplying both sides by sin λ,P sin2 λ + P cos2 λ = mg sin λ
or P = mg sin λQ. 66. A block slides down an inclined plane of slope θ, with a constant velocity. It is then
projected up the same plane with an initial velocity v0 . How far will it move before comingto rest. Will it slide down again ? (Lucknow, 1982, 1986)
Solution. We have, while sliding as shown in Fig. 47(a) mg sin θ = µN, mg cos θ = Ni.e., µ = tan θ ...(i)Now, in upward motion on same plane shown in Fig. 47(b)
Retarding Force = mg sin θ + µ N; (N = mg cos θ)= mg sin θ + µ mg cos θ= mg (sin θ + µ cos θ)
mg s in θ
µN
θ
m gm g cos θ
θ
N
mg s in θµN
θ
m gm g cos θ
NM otion w ith
re
tardation
( )a ( )b
θ
velocity unifo rm
Fig. 47
Hence, retardation = g (sin θ + µ cos θ)
= g (sin θ + tan θ . cos θ) = 2 g sin θusing v2 = u2 + 2as;
with u = v0; a = –2g sin θ, v = 0
0 = v g xs02 4− sinθ
or s =v
g02
4 sinθ
The block will not slide down again since the component of force along the plane downwardswill be just balanced by the force of the motion.
Q. 67. A uniform rod rests in limiting equilibirum in contact with a floor and a verticalwall (rod in a vertical plane). Supposing the wall and floor to be unequally rough, compute theangle between rod and wall.
Solution. Situation is depicted in the figure 48.
LK is rod; MS is wall; θ is angle between rod and the wall. Let, 2l = length of the rod;µ = coeff. of static friction between the rod and the wall; µ′ = coeff. of static friction betweenrod and floor.
Force and Motion 73
For equilibrium in horizontal direction,
µ′N′ = N or N = N′′µ ...(i)
For equilibrium in vertical direction
N′ + µN = W
orN Nµ
µ′
+ = W or, N 1µ
µ′
+FHG
IKJ = W
or N =1 + ′
′FHG
IKJ
=µµµ
W ...(ii)
Now taking moment about point L, we have
W.LD = N.KS + µ N.LS
but as LD = l sin θ; KS = 2l cos θ; and LS = 2l sin θabove equation becomes
W.l sin θ = N.2l cos θ + µN 2l sin θand putting the value of N from (ii)
W.l sin θ =µ
µµθ µ µ
µµθ′
+ ′+ ′
′W W
1 +12 2. cos . . sinl l
or W.l sin θ – 2µµ θ
µµ′
′W sin
1 +l
=2
1µ θ
µµ′
+ ′W cosl
W 1l − ′+ ′
FHG
IKJ
21
µµµµ
θsin =2µ θ
µµ′
′W cos
1 +l
or tan tanθ µµµ
θ µµµ
= ′− ′
= ′− ′
FHG
IKJ
−21
21
1or
Q. 68. Forces required to move a body on a rough inclined plane with a uniform velocityin the upward and downward direction are in the ratio 2:1. Find the inclination of the planeif the coefficient of friction is 0.3. (Lucknow, 1987)
Solution. (i) Uphill uniform motion: Net force in uphill direction is zero because bodymoves with constant velocity; i.e.,
F1 – (mg sin θ + µ N) = 0
or F1 = mg sin θ + µ mg cos θ (or N = mg cos θ)
or F1 = mg (sin θ + µ cos θ)
(ii) Downhill uniform motion: Again, resultant force in downhill direction is zero as thebody moves with constant velocity. If F2 is applied downward force, then
(F2 + mg sin θ) – µN = 0
or F2 = µ mg cos θ – mg sin θ = 0
or F2 = mg (µ cos θ – sin θ)
Kθ
µN
N
l
l
GN ′
Lw = m g
µ N ′′ D S
M
Fig. 48
74 Mechanics
m g sin θµN
θ
m gm g cos θ
θ
N
U niform M otion
( )a ( )b
mg s in θ
µN
θ
m gm g cos θ
N
Uniform motion
θ
F 2
F 1
Fig. 49
According to question
FF
1
2=
21
, i.e., F1 = 2F2
mg (sin θ + µ cos θ) = 2 mg (µ cos θ – sin θ)
or sin θ + µ cos θ = 2 µ cos θ – 2 sin θor 3 sin θ = µ cos θor tan θ = µ /3 = 0.3/3 = 0.1; θ = tan–1 (0.1)
Q. 69. A piece of ice slides down a 45° incline in thrice the time it takes to slide downon a frictionless 45° incline. Find the coefficient of kinetic friction between ice and the incline.
(Lucknow, 1983, 1988)
Solution. Frictionless incline:
Let length of incline be s, initial velocity = 0
downward force = mg sin 45° = ma
a =mg
mgsin 452
° =
Now, if t1 is the time taken to slide down the incline, using
S = ut at+ 12
2; with u = 0, t = t1, a = g/ 2
We have,
s =12 2
12g
tFHG
IKJ ...(i)
m g sin 45°
m gm g cos 45°
45° m g sin 45°
N
µN
m gm g cos 45°
45°( )a ( )b
N
Fig. 50
Force and Motion 75
For incline having friction:
Downward force = mg sin 45° – µN
= mg sin 45° – µ mg cos 45°
ma = mg (sin 45° – µ cos 45°)
= mg 12
12
−FHG
IKJµ
so, downward acceleration; a = g2
1( )− µ
if t2 is the time the body takes to slide down the incline using, s =ut at+ 12
2; with u = 0,
t = t2
and a =g2
1( )− µ
s = 0 12 2
1 22+ −× ( )g tµ ...(ii)
from (i) and (ii)
12 2
12g
tFHG
IKJ =
g t2 2
1 22( )− µ
or t12 = ( )1 2
2− µ t ...(iii)
But according to question, t2 = 3 t1 then (iii) is
t12 = ( ) ( ) /1 3 1 1 91
2− − =µ µt or
or µ = 8/9 = 0.889
Q. 70. Two blocks are kept on a horizontal smooth table as shown. Find out the minimumforce that must be applied on 10 kg block to cause 6 kg block just to slide. Coeff. of staticfriction between two blocks is 0.3. (Lucknow, 1994)
Solution. Let F be the minimum force applied on 10 kgblock, the acceleration produced in system is, α = F/(10 + 6)= F/16. This gives a reactionary force 6 α which will tend toslide the 6 kg block where as frictional force will tend to stopit. Hence for 6 kg not to move relative to 10 kg.
6 α = µs × 6 × g
(6F/16) = 0.3 × 6 × 9.8 or
F = 0.3 × 16 × 9.8 = 47.04 Newton
Q. 71. Determine the frictional force of air on a body of mass 1 kg falling with anacceleration of 8 m/sec2 (g = 10 m/sec2).
Fig. 51
76 Mechanics
W′ = mg′ = mg – Fr
so Fr = mg – mg′ = 10 – 8 = 2 Newton
Q. 72. Block B in the figure 52 has mass 160 kg. The coefficient of static friction betweenblock and table is 0.25, what is the maximum mass of block A for which the system will bein equilibrium ? (Lucknow, 1988)
Solution. Let m be the maximum mass of block A for which system is in equilibrium.Various forces are shown in the figure 52. For equilibrium;
T cos 45° = µN ...(i)
N = Mg ...(ii)
and mg = T sin 45° ...(iii)
(i) and (ii) give, T cos 45° = µ M g
T/ 2 = 0.25 × 160 × 9.8
or T = 0.25 × 160 × 9.8 × 2 ...(iv)
Given M = 160 kg; g = 9.8 m/s2; µ = 0.25
µNB
N
M g
m g
A
Wall
T cos 45°
45°
T sin 45°
T
Fig. 52
Putting (iv) in (iii)
mg = 0.25 × 160 × 9.8 × 2 × /1 2
or m = 0.25 × 160 kg or m = 40 kg.
Q. 73. A body rests upon inclined plane and will just slide down the plane when the slopeof the plane is 30°. Find the acceleration of the body down the plane when the slope is increasedto 60°.
Solution. When body just slides down the inclined plane, the inclination of the plane is30°;
so mg sin 30° = µN
Force and Motion 77
or µ = tan 30° (as N = mg cos 30°)
When slope is 60°; the body moves down the plane.The force acting down is:
mg sin 60° – µN = ma
(with N = mg cos 60°)
so, m(g sin 60° – µ g cos 60°) = ma
where a = accn
or a = g (sin tan cos )60 30 60° − ° °
= 9 8 32
13
12
5 66. × .−FHG
IKJ
= m/s2
Q. 74. A body with a mass of 0.84 kg is on a plane inclined at 30° to the horizontal. Whatforce must be applied on the body so that it moves with uniform acceleration of 0.12 m/sec2.(a) uphill (b) downhill; (The coefficient of sliding friction with the plane is 0.32).
(Lucknow, 1978, 1982)
Solution. (a) Uphill motion: Let F be the applied force so that body moves uphill withan acceleration = 0.12 m/sec2.
F – (µ N + mg sin 30°) = ma
or F = ma + (µ mg cos 30° + mg sin 30°)
= ma + mg (µ cos 30° + sin 30°)
= 0.84 × 0.12 + 0.84 × 9.8 [0.32 × ( / ) ( / )]3 2 1 2+
or F = 0.1008 + 6.397 = 6.498 Newton
mg sin θ
m gm g cos θ
θ
m g sin θ
N
m gm g cos θ
( )a ( )b
θ
θ
θ = 30°
F
µN
Motion
µN
θ
Motion
F
Fig. 54
(b) Downhill motion: Suppose F is the applied force down the hill so that body moves withan acceleration = 0.12 m/s2 then,
F + mg sin θ – µ N = ma
or, F = ma – mg sin 30° + µ mg cos 30°
(as N = mg cos 30°)
= ma – mg (sin 30° – 0.32 cos 30°)
= 0.84 × 0.12 – 0.84 × 9.8 [(1/2) – 0.32 ( / )3 2 ]
= 1.7330 Newtons; i.e., in upward direction.
mg s in 60°
m gm g cos 60°
60°
NµN
6 0 °
Fig. 53
78 Mechanics
Q. 75. A body of 5 kg weight is just prevented from sliding down a rough inclined planeby a force of 2 kg weight acting up the line of greatest slope. When this force is increased to3 kg weight; the body just begins to slide up the plane. Prove that coefficient of friction between
the body and the plane is 3 /15.
Solution. Body is just prevented from sliding down the plane when 2 kg force acts upthe inclined plane; it suggests that friction is limiting and acts up the plane. Resolving theforces on the body, along and perpendicular to the plane;
N = 5 g cos θ; and µN + 2g = 5 g sin θ ...(i)
i.e., µ 5g cos θ + 2g = 5 g sin θor 5 sin θ – 5 µ cos θ = 2 ...(ii)
When 3 kg weight is applied; the body just slides up the plane. The friction is againlimiting but now will act down the inclined plane.
Resolving the forces along and perpendicular to the inclined plane;
3 g=
5 g sin θ + µ N and N = 5 g cos θ...(iii)
i.e., 3 g = 5 g sin θ + µ 5 g cos θor 5 sin θ + µ 5 cos θ = 3 ...(iv)
adding (ii) and (iv)
10 sin θ = 5 or sinθ θ= °12
or = 30
then from (iv)
5 12
5 3 2× × ( / )+ µ = 3
or µ =1
5 30 115= .
Q. 76. A block is released from the rest at the top of a frictionless inclined plane 16 mlong. It reaches the bottom 4.0 sec later. A second block is projected up the plane from thebottom at the instant the first block is released in such a way that is returns to the bottomsimultaneously with the first block.
(i) Find the acceleration of each block on the incline.
(ii) What is the initial velocity of the second block.
(iii) How far up the inclined plane does it travel?
Solution. Taking the displacement down the incline to be positive the acceleration of
both the blocks when they return simultaneously is, a = g sin θ; from s =ut at+ 12
2 with u
= 0, t = 4 sec., s = 16 m.
16 = 012
42+ a
a = (2 × 16)/16 = 2 m/s2
Force and Motion 79
direction of v0 (up the plane) is taken as positive, then acceleration is negative i.e., a = –2m/s2 using t = 4, u = v0
then s = ut at+ 12
2; with a = –2 m/s2, t = 4, u = v0
i.e., 0 = v0 × 4 − −12
2 42× ( ) ×
or v0 = 4 m/s
(iii) Supposing the second block to travel a distance s, up the incline where it comes torest, gives; v2 = u2 + 2as
with v = 0, u = v0 = 4 m/s, a = –2 m/s2
then 0 = 42 + 2 × (–2)s
or s = 4 m
Q. 77. Two blocks of mass 2.9 kg and 1.9 kg are suspended from rigid support by twoextensible wires each 1 meter long. The upper wire has negligible mass and lower wire hasuniform mass of 0.2 kg/m. The whole system has upward acceleration 0.2 m/s2 as shown.
Find (i) Tension at mid point of lower wire
(ii) Tension at mid point of upper wire (I.I.T., 1989)
Solution. Since upper wire has negligible mass, the tension T1 in upper wire will besame all along its length.
If T2 is tension at the upper point of lower wire, for equilibrium of m1.
T1 – T2 – m1 g = m1a ...(i)
and taking m3 as mass of lower wire, for equilibrium of m2,
T2 – m2 g – m3 g = (m2 + m3) a ...(ii)
Adding the two equations
T1 – (m1 + m2 + m3) g = (m1 + m2 + m3) a
i.e., T1 = (m1 + m2 + m3) (g + a)
= (2.9 + 1.9 + 0.2) (9.8 + 0.2)
= 5 × 10 = 50 Newton
This will be the tension at mid point of upper wire.
Putting value of T1 in equation (i)
50 – T2 = 2.9 (9.8 + 0.2) = 29
i.e., T2 = 21 Newton
At mid point tension will be less than at upper point since only half length of lower wirewill contribute to tension due to its own weight.
i.e., Tm + FHG
IKJ +0 2
12
9 8 0 2. × × ( . . ) = 21
or Tm + 0.1 × 10 = 21
i.e., Tm = 20 Newton
T 1
T 2
T 2
m 1
m 2
1.9 kg
0.2 kg /m
2.9 kg
Fig. 55
80 Mechanics
Q. 78. How can a mass of 100 kg be lowered with a massless rope which can support,without breaking, a mass of 80 kg only. (Lucknow, 1994)
Solution. Let, the 100 kg mass be lowered with aceleration ‘a’ then taking T < 80 × 9.8Newton,
the equilibrium of motion for the system will be 100g – T = 100 a
taking maximum possible of T, we get
100 g – 80 g = 100 a
or a = (1/5) g
Hence, the given mass can be lowered with acceleration of g/5 or more than this.
Q. 79. A stone weighing 2 kg and tied at the end of string of length 2.5 meters, revolveswith frequency of 10 revolutions/sec. Find out the force on the stone as measured (i) in inertialframe (ii) in frame rotating with the string.
Solution. (i) The actual force which makes body to rotate, in inertial frame, is the
centripetal force = − = −mvr
mr2
2ω
(–ve sign shows that force is towards center).
So, force on stone, in inertial frame,
= –m ω2 r = –2 (2π × 10)2 × 2.5
= –19719.2 Newton
force being supplied by tension of string
(ii) The frame rotating with the string is non-inertial and in this frame the accelerationof the stone is 0. So total force in this frame = 0.
The equilibrium of the stone in rotating frame is obtained with the help of centrifugalforce (Fictious force).
Q. 80. Two blocks A and B are joined to each other by a string and a spring of forceconstant 1960 N/m which passes over a pulley, attached to another block C as shown. BlockB slides over block C and A moves down along C with same speed. Coefficient of frictionbetween blocks and surface of C is 0.2. Taking mass of block A as 2 kg, calculate.
(a) mass of block B
(b) energy stored in spring.
Solution. When both block B and A at same speed the spring is in condition of maximumextension. Taking T as tension of spring and the string, and ‘m’ as the mass of block B, wehave, from Fig. 56
For motion of B, T – µ mg = 0 ...(i)
For motion of A, T = 2 g ...(ii)
combining (i) and (ii) we get
µ mg = 2 g
or, 0.2 m = 2 so, m = 10 kg
Force and Motion 81
B
µ = 0 .2
C
A
( )a
B
µ = 0 .2
CA
( )b
K = 1960 N /M
M T
Tµm g
2g
Fig. 56
Putting value of ‘m’ in equation (i) or using equation (ii) we get
T = 2 × 9.8 = 19.6 N
The elongation of a string is therefore
x =Tension
Force constant m
−= =19 6
19600 01
..
Hence, energy stored in spring
U =12
12
1960 0 012 2kx = × × ( . )
= 98 × 10–3 = 0.098 joule
Q. 81. A body falling freely from a given height H hits an inclined plane in its path ata height ‘h’. As a resutl of impact the direction of velocity of body becomes horizontal. Forwhat value of (h/H), the body will take maximum time to reach the ground.
Solution. Let t1 be the time of fall through height (H – h) and t2 the time of free fallthrough height ‘h’ then
H – h =12 1
2gt
and h =12 2
2gt (since after impact there is no
initial velocity along vertical)
t = t1 + t2 = 2 2( ) / /H − +h g h g
or t =2 1 2 1 2
gh hH − +b g / /
for time to be maximum the condition is dt/dh = 0
− − +− −1
212
12
12( )H h h = 0
or h = ( )H − h
which yields,
hH
=12
82 Mechanics
Q. 82. A projectile is launched with an initial speed v in direction θ to the horizontalwhich hits an inclined plane of inclination β with horizontal (β < θ) and passing through thelaunching point. Derive an expression for the range R on this inclined plane. When will theprojectile hit this plane?
Solution. Let the projectile hit the inclined plane at M and OM = R (range). At theinstant of hitting the plane, the x and y displacements are ON and NM respectively as shownin Fig. 57.
Now ON = R cos β; NM = R sin β
Clearly time taken to cover horizontal distance ON, tv v
= =ONcos
R cosθ
βθcos ...(i)
For vertical distance NM; we have s = NM; u = v sin θ, a = –g, t = R cos β/v cos θ
Then s = ut at+ 12
2 gives;
Fig. 57
R sin β = vv
gv
sin( cos )( cos )
θ βθ
βθ
R coscos
R− 12
2
2
12
2
2 2gvR2 cos
cosβθ
= R(tan cos sin )θ β β−
or R =2 2 2
2vg
coscos
sin cos cos sin
cosθ
βθ β θ β
θ −
or R =2 2
2v
gcos sin( )
cosθ θ β
β−
Putting above in (i)
t =2 22
2v
g vv
gcos sin( )
coscoscos
sin( )cos
θ θ ββ
βθ
θ ββ
− = −
Q. 83. A block Q (refer figure 58) having mass 0.2 kg. is placed on the top of block P ofmass 0.8 kg. Coefficient of sliding friction between P and the table is 0.2 and that between Pand Q is 0.5. The pulley is light and smooth. What horizontal force A will maintain the motionof P with uniform speed?
84 Mechanics
Q. 85. The water of 0.5 km wide river is flowing with a velocity of 4 km/hr. A boat manstanding on one of the bank of the river wishes to take his boat to a point on the opposite bankexactly infront of his present position. He can row his boat with a velocity of 8 km/hr relativeto water. In which direction should he row his boat. Obtain the time to cross the river.
Solution. Let the boatman start from P and Q be his destination. Then resultantvelocity (relative to earth i.e., bank) should be in direction PQ. So boat should be rowed alongPR. Let resultant velocity be v.
Sin α = RQ/PR = 4/8 i.e., α = 30°
and PQ = 8 4 48 6 92 2− = = .
i.e., v = 6.9 km/hr
So boat man should row in direction 30° with PQ, upstream,Time to cross the river;
t =PQ
Speed-in-direction PQ= 0 5
6 9..
=1
13 860
13 8. .hr min = 4.35 min=
Q. 86. An aeroplane is flying with velocity 70 km/hr in northeast direction. Wind isblowing at 30 km/hr from north to south. What is the resultant displacement of aeroplane in4 hours?
Solution. Let v1 = velocity of aeroplane; v2 = velocity of wind then resultant velocityof aeroplane, v is, v = v1 + v2
Resolving v1 and v2 in x and y direction;
v1 = v0 cos 45° i + v0 sin 45° j
v1 =70
2($ $)i j+
v2 = –30 j
= 35 2 35 2 30$ ( ) $i j+ −v = (49.5 i + 19.5 j) km/hr.
The result and displacement r = (49.5 i + 19.5j) × 4 = (198 i + 78j)
and r→L
NMOQP
= 189 78 204 462 2+ = . km.
if φ = angle made by r with x-axis (i.e., east)
tan φ = y/x = 19.5/49.5 or φ = 23° nearly.
Q. 83 An aeroplane is flying in a horizontal direction with a velocity of 600km per hourand a height of 1960 meters. When it is vertically above the point A on the ground, a bodyis dropped from it. The body strikes the ground at a point B. Calculate the distance AB.
(IIT, 1973)
Ans. When a body is dropped, it is acted upon by
(i) Uniform velocity = 600 km/hour = 6000003600
m/s in the horizontal direction.
α
R4 km /hr
Q
8 km /hrV
P
Fig. 61
Force and Motion 85
(ii) Acceleration due to gravity in the vertical direction.
Here h = 1960 m
Suppose the time to reach the ground = t second
h = ½ gt2; here g = 9.80 m/s2
1960 =9 80
2.
t2
t2 =19604 90.
= 400 second2
t = 20 second
Distance AB = S = ut
S =6000003600
× 20 = 3333.33 m.
SELECTED PROBLEMS
1. What do you understand by a frame of reference. Define inertial frame. Comment on thereference frame attached with earth. (Lucknow, 1994, Agra 1996)
2. What is a coriolis force? Find an expression for it. Exaplain its importance.
(Lucknow, 1995)
3. Explain
(a) Fictitious force, (Agra)
(b) Galilean transformations,
(c) Coriolis force. (IAS)
4. A reference frame ‘a’ rotates with respect to another reference frame ‘b’ with uniform angularvelocity ω. If the position, velocity and acceleration of a particle in frame ‘a’ are representedby ‘R’, va and fa show that acceleration of that particle in frame ‘b’ is given by where:
fB = fa + 2ω × va + ω × (ω × R)
Integrate this equation with reference to the motion of bodies on earth surface.
5. Show that the motion of one projectile as seen from another projectile will always be astraight line motion. (Agra, 1970)
6. A plane flies across the north pole at 400 km/hr and follow a longitude 450 (rotating withearth) all the times. Compute what angle does a freely suspended plumb line in the planemakes as it passes over the pole, make with a plumb line which is situated on the earthsurface at north pole.
7. A thin uniform rod AB of mass m and length 2λ can rotate in a vertical plane about A aspendulum. A particle of mass 2 m is also fixed on the rod at a diatance x from A. Find xso that the periodic time of small swing may be minimum.
8. A person is running towards a bus with a velocity of constant magnitude u→
and alwaysdirected towards the bus. The bus is moving along a straight road with a velocity of constant
magnitude v→
. Initially at t = 0. a→
is perpendicular to v→
and the distance between the personand the bus l. After what time will the person catch the bus?
Ans.ul
v u2 2−
LNM
OQP
86 Mechanics
3
3.1 UNIFORM CIRCULAR MOTION
Centripetal Force: suppose a body is moving on a horizontal circle (radius r) with auniform velocity v. At any point on the circle, the direction of the velocity is directed alongthe tangent to the circle at that point i.e., at A the direction of the velocity AX, at B alongBY, at C along CZ and so on.
If the body is free at any point it will move tangential tothe circle at that point. But as the body is moving along thecircumference of the circle, there must be a force acting towardsthe center of the circular path and this force is called thecentripetal force.
Centripetal force is defined as that force which acts towardsthe center along the radius of a circular path on which thebody is moving with a uniform velocity.
In figure 1 let A and B be two positions of the body afteran interval of time t.
Then AB = velocity. time = vt
Let o′a and o′b be vectors representing the velocities at A and B respectively in figure2(ii). Then ∠ao′b = θ (the angle between the tangents equal to the angle between the radii).ab is the change in velocity from A to B.
Acceleration =Change in Velocity
Time= ab
t
Since A and B are very close to each other, therefore arc AB can be taken as a straightline.
∆OAB and ∆o′ab are similar
∴ABOA
=abo a′
∴vtr
=abv
abt
vr
or =2
DYNAMICS OF CIRCULAR MOTIONAND THE GRAVITATIONAL FIELD
Or
r
r
C
v
z
yx
A
B vv
Fig. 1
86
Dynamics of Circular Motion and the Gravitational Field 87
∴ Acceleration =abt
vr
=2
O
r
vA
B
v
θ
αα
(i)
α
v
θαb V
a
(ii)
r
c
Fig. 2
∴ The acceleration on the body acting towards the center of the circular path = vr
2
∴ Centripetal force = mass × acceleration = mvr
2
F =mv
r
2
Direction of the Centripetal ForceWhen a body moves along a circular path with uniform speed, the magnitude of the
velocity remains the same but its direction changes at every point. It means there is changein velocity whenever there is a change in direction, there must be some acceleration. As abody has mass also, a force must act upon the body. This force must act along such a directionthat the magnitude of the velocity does not change. As a force has no component at rightangles to its direction, the force must act at every point in a direction perpendicular to thedirection of the velocity. As the velocity is tangential to the circle at every point, the forcemust be acting along the radius and towards the centre of the circular path. Due to thisreason it is called centripetal force.
3.2 CENTRIFUGAL FORCE
When a body is rotating on a circular path, it has a tendency to move along a tangent.If a body A leaves the circular path at any instant, for an observer A who is not sharing themotion along the circular path (i.e., a body B standing outside the reference circle), the bodyA appears to fly off tangentially at the point of release. For an observer C, who is sharingthe same circular motion as that of the body A, the body A appears to be at rest before it isreleased. According to C, when A is released, it appears to fly off radially away from thecenter. It appears to the body C as if the body A has been thrown off along the radius awayfrom the center by some force. This inertial force is known as centrifugal force. Its magnitudeis mv2/r. It is not a force of reaction. Centrifugal force is a fictitious force and holds good ina rotating frame of reference.
When a car is on turning round a corner, the persons sitting inside the car experience anoutward force. This is due to the fact that no centripetal force is provided by the passengers.Therefore to avoid this outward force, the passengers are to exert an inward force.
88 Mechanics
3.3 THE CENTRIFUGE
It is an appliance to separate heavier particles from the lighter particles in a liquid. Theliquid is rotated in a cylindrical at a high speed with the help of an electric motor. The heavierparticles move away from the axis of rotation and the lighter particles moves near to the axisof rotation.
Application(a) Sugar crystals are separated from molasses with the help of a centrifuge.
(b) In cream separators, when the vessel containing milk is rotated at high speed, thelighter cream particles collect near the axle while the skimmed milk moves awayfrom the axle.
(c) In drying machines, the wet clothes are rotated at high speed. The water particlesfly off tangentially through the holes in the wall of the outer vessel.
(d) Honey is also separated from bees with the help of a centrifuge.
(e) Precipitates, sediments, bacteria etc., are also separated in a similar way.
In an ultra-centrifuge, the speed of rotation is very high and is of the order of 30 to 40thousand rotations per minute.
3.4 BANKING OF CURVED ROADS AND RAILWAY TRACKS (BANKED TRACK)
A car moving on road, or a train moving on rails, requires a centripetal force while takinga turn. As these vehicles are heavy, the necessary centripetal force may not be provided byfriction, and moreover, the wheels are likely to suffer considerable wear and tear. In this case,the centripetal force is produced by slopping down the road inward at the turns. Similarly, whilelaying the railways tracks, the inner rail is laid slightly lower than the outer rail at turns. Bydoing so, the car, or the train leans inward while taking turn and the necessary centripetal forceis provided. This force is produced by the normal reaction of the earth, or the rail.
In figure 3, is shown a car taking a turn on a road while is given a slope of angle θ, Gis the center of gravity of the car at which the weight mg of the car acts. When the car leansthe total normal reaction R of the road exerted on the wheels makes an angle θ with thevertical. The vertical component R cos θ balances the weight mg of the car, while thehorizontal component R sin θ provides the necessary centripetal force mv2/r. Thus
R cos θ = mg ...(i)
and R sinθ =mv
r
2
...(ii)
Dividing Eq. (ii) by (i), we get
tan θ =vrg
2
From this formula the angle θ can be calculatedfor given values of v and r. Thus the slope θ is properfor a particular speed of the car. Therefore, the driverdrives the car with that particular speed at the turn.Since m does not appear in the formula, hence thisspeed does not depend upon the weight of the car.
R sin θ
R cos θ
CGθ
θm g
m vr
2
R
G
Fig. 3
Dynamics of Circular Motion and the Gravitational Field 89
If a car is moving at a speed higher than the desired speed, it tends to slip outward atthe turn, but then the frictional force acts inward and provides the additional centripetalforce. Similarly, if the car is moving at a speed lower than the desired speed it tends to slipinward at the turn, but now the frictional force acts outward and reduces the centripetalforce.
The hilly roads are made slopping inward throughout so that the vehicles moving onthem lean automatically towards the center of the turn and are acted upon by the centripetalforce. If these road were in level then, at the turns, the outer wheels would be raised up thusoverturning the vehicle.
The flying aeroplane leans to one side while taking a turn in the horizontal plane. In thissituation the vertical component of the force acting on the wings of the aeroplane balancesits weight and the horizontal component provides the necessary centripetal force. In the “wellof death” the driver while driving the motor-cycle fast on the wall of the well, leans inward.The vertical component of the reaction of the wall balances the weight of the motor-cycle andthe driver, while the horizontal component provides the required centripetal force.
3.5 BICYCLE MOTION
As shown in figure 4, a rider taking a turn towards his left hand. Let m be the massof the rider and the bicycle, v the speed of the bicycle and r the radius of the (circular) turn.The centripetal force mv2/r necessary to take a turn is provided by the friction between thetyres and the road. Therefore, whenthe bicycle is turned, a frictional forceF (= mv2/r) towards the center of theturn is exerted at the point A on theroad. Let us imagine two equal andopposite forces F1 and F2, each equaland parallel to F ( F1 = F2 = F), actingat the center of gravity G of the riderand the bicycle. The force F acting at Acan now be replaced by the force F1acting at G, and the anticlockwisecouple formed by F1 and F2 (= F). Theforce F1(= F) is the necessary cen-tripetal force. If the rider remainsstraight while turning, then his weightmg acting vertically downward at G andthe earth’s normal reaction R (= mg)acting vertically upward at A would cancel each other. In this situation the anticlockwisecouple would overturn the bicycle outward. If, however, the rider leans inward towards thecenter of the turn, then his weight mg and the normal reaction R ( = mg) of the earth froma clockwise couple which balances the anticlockwise couple (Fig. 5). Hence the rider moveson the turn without any risk of overturning.
Suppose the rider leans through an angle q from the vertical for balancing the twocouples then:
couple formed by mg and R (= mg) = couple formed by F1 and F2 (= F)
F1 F 2G
m g
A F
R
( )a ( )b
A F
R
θ
m g
F1F1
G
Centre o fTurn
Fig. 4
90 Mechanics
mg × GA sin θ = F × GA cos θ
or tan θ =F
mg
But F =mv
r
2
∴ tan θ =vrg
2
From this formula, the angle θ can be calculated.
Since the centripetal force mv2/r is provided bythe frictional force, the value of mv2/r should be lessthan the limiting frictional force µr otherwise the bicyclewould slip. This is why the rider slow down the speedof the bicycle while taking a turn and follows the path of a larger radius. During rainy seasonthe frictional force decrease appreciably and cannot provide the centripetal force. Henceduring rains bicycle rider usually slip on the roads while taking turn.
3.6 CONICAL PENDULUM
Suppose a small object A of mass m is tied to a string OA of length l and then whirledround in a horizontal circle of radius r, with O fixed directly above the center B of the circle,If the circular speed of A is constant, the string turns at a constant angle θ to the vertical.This is called a conical pendulum.
Since A moves with a constant speed v in a circle of radius r, there must be a centripetal
force mv
r
2
acting towards the center B. The horizontal component, T sin θ , of the tension
T in the string provides this force along AB. So
T sin θ =mv
r
2
...(i)
Also, since the mass does not move vertically, its weight mg must be counter balancedby the vertical component T cos θ of the tension. So
T cos θ = mg ...(ii)
Dividing (i) by (ii), then
tan θ =vrg
2
A similar formula for θ was obtained for the angle ofbanking of a track, which prevented side-slip.
A pendulum suspended from the ceiling of a train doesnot remain vertical while the train goes round a circular track.Its bob moves outwards away from the center and the stringbecomes inclined at an angle θ to the vertical, as shown infigure 6. In this case the centripetal force is provided by thehorizontal component of the tension in the string.
Br
θ
θ
T cos θ
T sin θA
O
T
l
m
m g
Fig. 6
θm g
C.G . Cm v
r2
θR
S
dd θ
F
r
F1
F2
Fig. 5
Dynamics of Circular Motion and the Gravitational Field 91
3.7 MOTION IN A VERTICAL CIRCLE
When a body tied to the end of a string is rotated in a vertical circle, the speed of thebody is different at different points of the circular path. Therefore, the centripetal force onthe body and the tension in the string change continuously.
Let us consider a body of mass ‘m’ tied to the end of a string of length R and whirledin a vertical circle about a fixed point O to which the other end of the string is attached. Themotion is circular but not uniform, since the body speeds up while coming down and slowsdown while going up.
The force acting on the body at any instant are its weight mg directed vertically downward,
and the tension T→
in the string directed radially inward. The weight m g→
can be resolved intotangential component mg sin θ and a radical (normal) component mg cos θ. Thus the body hastangential force mg sin θ and a resultant radial force T – mg cos θ acting on it.
m g cos θ θ
mg s in θ
m g
O T R θ
Fig. 7
The tangential force gives to the body a tangential acceleration, which is responsible forthe variation in its speed. The radial force provides the necessary centripetal force.
Thus T – mg cos θ =mv2
R
The tension in the string is therefore
T = mv
g2
R+
LNM
OQPcosθ ...(i)
Let us consider two special cases of the equation:
(i) At the lower point A of the circle, θ = 0 so cos θ =1
T = TA and v = vA (say).
Equation (i) takes the form
TA = mv
gA2
R+
LNM
OQP
Which means that the tension must be large enough not only to overcome the weight
mg but also provide the centripetal force mvA
2
R.
A vA
m g
O TA
Fig. 8
92 Mechanics
(ii) At the highest point B, θ = 180°, so cos θ = –1, T = TB and v = vB (say)
Then we have Equation (i)
TB = mv
gB2
R−
LNM
OQP
TB is less than TA, because the weight mg providesa part of the centripetal force.
If vB decreases, the tension TB would decrease and vanishat a certain critical speed vC (say). To determine this criticalspeed, we put TB = 0 and vB = vC in the last expression. Then
0 = mv
gC2
R−
LNM
OQP
vc = Rg
This speed of the body is called the “critical speed”. In this state the centripetal force isprovided simply by the weight of the body. If the speed of the body at the highest point B is
less than the critical speed R g , then the required centripetal force would be less than the
weight of the body which will therefore fall down (the string would slack).
If a bucket containing water is rotated fast in a verticalplane, the water does not fall even when the bucket iscompletely inverted. This can be explained in the followingmanner. The bucket rotates in a vertical plane under a changingcentripetal force. The water contained in the rotating bucketexperiences a centrifugal force which is always equal andopposite to the centripetal force. When the bucket is at thehighest point B of this circular path, then the centrifugal force
mvB2
R on the water is directed upward. If the speed of rotation
of the bucket is quite fast, the centrifugal force is greater than
the weight W of the water. The difference mvB
2
RW−
FHG
IKJ keeps
the water pressed to the bottom of the bucket.
3.8 MOTION OF PLANET
Our solar system consists of a sun which is stationary at the center of the Universe andnine planets which revolve around the sun in separate orbits. The name of these planets are:Mercury, Venus, Earth, Mass, Jupiter, Saturn, Uranus, Neptune and Pluto. The planet Mercuryis closest to the Sun and Pluto is farthest.
These are certain celestial bodies which revolve around the planets. These are called‘satellites’. For example, moon revolves around the earth, hence moon is a satellite of theearth. Similarly, Mars has two satellites, Jupiter has twelve satellites, Saturn has ten satellites,and so on.
Fig. 9
B
. W
m vR
B2
Fig. 10
Dynamics of Circular Motion and the Gravitational Field 93
3.9 KEPLER’S LAWS OF MOTION
Kepler found important regularities in the motion of the planets. These regularities areknown as ‘Kepler’s three laws of planetary motion’.
(i) Shape of the Orbit: All planets move around the sun in elliptical orbits havingthe sun at one focus of the orbit. This is the law of orbits.
(ii) Velocity of the orbit: A line joining any planet to the sun sweeps out equal areasin equal times, that is, the areal speed of the planet remains constant. This is thelaw of areas. When the planet is nearest the sun then its speed is maximum andwhen it is farthest from the sun then its speed is minimum. In Figure 11, if a planetmoves from A to B in a given time-interval, and from C to D in the same time-interval, then the areas ASB and CSD will be equal.
B A
C
D
S
Fig. 11
(iii) Time periods of Planets: The square of the period of revolution of any planetaround the sun is directly proportional to the cube of its mean distance from thesun. This is the law of periods.
If the period of a planet around the sun is T and the mean radius of its orbit is r,then
T2 ∝ r3
or T2 = Kr3
where K is a constant. Thus, larger the distance of a planet from the sun, largerwill be its period of revolution around the sun.
3.10 DERIVATION OF LAW OF GRAVITATION
Suppose the mass of the planet A is M1, the radius of its orbit is R1 and time period ofrevolution is T1. It is assumed that the orbit is circular. The force of attraction exerted bythe sun on the planet (centripetal force)
F1 = M R M RT1 1 1 1
1ω π
12
22=
LNM
OQP ...(i)
Similarly for a second planet B of mass M2, Radius R2 and period of revolution round thesun T2
F2 = M R M RT2 2 2 2
2ω π
22
22=
LNM
OQP ...(ii)
94 Mechanics
FF
1
2=
MM
RR
TT
1
2
1
2
2
1
FHG
IKJFHG
IKJFHG
IKJ
2
...(iii)
But according to Kepler’s third law,
TT
2
1
FHG
IKJ
2
=RR
2
1
FHG
IKJ
3
...(iv)
Substituting these values in equation (iii)
FF
1
2=
MM
RR
1
2
2
1
FHG
IKJFHG
IKJ
2
F RM1 1
2
1=
F RM2 2
2
2 = constant
∴ F ∝ M/R2
or (i) F∝ M and
(ii) F∝ 1/R2
Thus, the force of attraction exerted by the sun on a planet is proportional to its massand inversely proportional to the square of its distance from the sun.
3.11 NEWTON’S CONCLUSIONS FROM KEPLER’S LAWS
Newton found that the orbits of most of the planets (except Mercury and Pluto) arenearly circular. According to Kepler’s second law, the areal speed of a planet remains constant.This means that in a circular orbit the linear speed of the planet will be constant. Since theplanet is moving on a circular path; it is being acted upon by a centripetal force directedtowards the center (Sun). This force is given by
F =mv
r
2
Where m is the mass of the planet, v is its linear speed and r is the radius of its circularorbit. If T is the period of revolution of the planet, then
v =linear distance travelled in one revolution
period of revolution= 2πr
T
∴ F =mr
r mr2 42 2π πT T2
FHG
IKJ =
But according to Kepler’s third law, T2 = K r3
∴ F = 4 42
3
2
2π πmr
rmrK K
= FHG
IKJ ...(i)
or F ∝mr2
Dynamics of Circular Motion and the Gravitational Field 95
Thus, on the basis of Kepler’s laws, Newton drew the following conclusions:
(i) A planet is acted upon by a centripetal force which is directed towards the sun.
(ii) This force is inversely proportional to the square of the distance between the planet
and the sun F 1∝FHG
IKJr2 .
(iii) This force is directly proportional to the mass of the planet (F∝ m). Since the forcebetween the planet and the sun is mutual, the force F is also proportional to the
mass M of the sun (F ∝ M). Now, we can replace the constant 4 2πK
in Equation
(i) by GM, where G is another constant. Then, we have
F = G Mmr2
Newton stated that the above formula is not only applied between sun and planets, butalso between any two bodies (or particles) of the universe. If m1 and m2 be the masses of twoparticles, then the force of attraction between them is given by
F = G m mr1 2
2
This is Newton’s Law of Gravitation.
3.12 NEWTON’S UNIVERSAL LAW OF GRAVITATIONS
In 1686, Newton stated that in the Universe each particle of matter attracts every otherparticle. This universal attractive-force is called ‘gravitation’:
The force of attraction between any two material particles is directly proportional to theproduct of the masses of the particles and inversely proportional to the square of the distancebetween them. It acts along the line joining the two particles.
Suppose two particles of masses m1 and m2 are situated at a distance r apart. If the forceof attraction acting between them is F, then according to Newton’s law of gravitation, we have
F ∝m m
r1 2
2
or F = G m mr1 2
2
where G is the constant of proportionality which is called ‘Newton’s gravitation constant’. Itsvalue is same for all pairs of particles. So it is a universal constant.
If m1 = m2 = 1 and r = 1, then F = G
Hence, gravitation constant is equal in magnitude to that force of attraction which actsbetween two particles each of unit mass separated by a unit distance apart.
From the above formula G =Fr
m m
2
1 2
96 Mechanics
Thus, if force (F) be in Newton, distance (r) in meter and mass (m1 & m2) in kg, thenG will be in Newton-meter2/kg2. It dimensional formula is [M–1L3T–2].
In rationalized MKS units G = 6.670 × 10–11 newton-m2/kg2.
3.13 GRAVITY AND THE EARTH
In Newton’s law of gravitation, the gravitation is the force of attraction acting betweenany two bodies. If one of the bodies is earth then the gravitation is called ‘gravity’. Hencegravity is the force by which earth attracts a body towards its center. Clearly gravity is aspecial case of gravitation. It is due to gravity that bodies thrown freely ultimately fall on thesurface of the earth.
3.14 ACCELERATION DUE TO GRAVITY
When a body is dropped down freely from a height, it begins to fall towards the earthunder gravity and its velocity of fall continuously increases. The acceleration developed in itsmotion is called ‘acceleration due to gravity’. Thus, the acceleration due to gravity is the rateof increase of velocity of a body falling freely towards the earth. It is denoted by ‘g’. It doesnot depend upon the shape, size, mass, etc. of the body. If m be the mass of a body then forceof gravity acting on it is mg (weight of the body). Therefore, the acceleration due to gravityis equal in magnitude to the force exerted by the earth on a body of unit mass. The unit ofacceleration due to gravity is meter/second2 or Newton/kg.
The universal constant G is different from g→
. The constant G has the dimension
M–1 L3 T–3 and is a scalar, g→
has the dimension LT–2 and is a vector, and is neither universalnor constant.
3.15 EXPRESSION OF ACCELERATION DUE TO GRA VITY g IN TERMS OF GRAVITATIONS CONSTANT G
Suppose that the mass of the earth is Me, its radius is Re and the whole mass Me isconcentrated at its center. Let a body of mass m be situated at the surface of the earth orat a small height above the surface, this height being negligible compared to the radius ofthe earth. Hence the distance of the body from the center of earth may be taken as Re.According to the law of gravitation, the force of attraction acting on the body due to the earthis given by
F =GM
R2e
e
m...(i)
The acceleration due to gravity g in the body arises due to the force F. According toNewton’s second law of motion, we have
F = mg ...(ii)
From Eqs. (i) and (ii), we have
mg = GMR
e
e
m2
Dynamics of Circular Motion and the Gravitational Field 97
g = GMR2
e
e
This expression is free from m. This means that the value of g does not depend uponthe mass of the body. Hence if two bodies of different masses be allowed to fall freely (in theabsence of air), they will have the same acceleration. If they are allowed to fall from the sameheight, they will reach the earth simultaneously.
In the presence of air, however, the buoyancy effect and the viscous drag will causedifferent accelerations in the bodies. In this case the heavier body will reach the earth earlier.
3.16 DIFFERENCE BETWEEN MASS AND WEIGHT
Mass is the amount of matter contained in a body. This is fixed for a given body. Its valueis the same at all places. Weight is the force with which a body is attracted towards the centerof the earth. It is different at different places. It depends on the value of g. A body has thesame mass at the poles and at the equator whereas its weight will be more at the poles thanat the equator.
With the help of a beam balance, the mass of the body is determined. With the help ofa spring balance the weight of the body is determined. The units of mass are gram or Kg.The units of weight are dyne or Newton.
3.17 INERTIAL MASS AND GRAVITATIONAL MASS
Inertial Mass: According to Newton’s second law of motion, when a force is applied ona body, the body moves with an acceleration
F = ma or m = F/a
This mass is called the inertial mass of the body. If the force is increased, the accelerationalso increases and F/a = constant for a given body. If the same force is applied on two differentbodies and the acceleration produced are equal, then the inertial masses of the two bodiesare equal. If the same force is applied on two different bodies, the inertial mass of that bodyis more in which the acceleration produced is less and vice versa.
Gravitational Mass: According to the law of gravitation, the Gravitational force ofattraction of a body towards the center of the earth is equal to the weight of the body.
Let the weight of the body be W
W = mg
m = W/g
This mass is called the gravitational mass of the body. This is determined with the helpof a beam balance. It will be the same even on the surface of the moon. If the value of g isless at the moon, the weight of the body will also be less but the gravitational mass is thesame.
3.18 GRAVITATIONAL FIELD AND POTENTIAL
The intensity of gravitational field at a point due to a mass is defined as the forceexperienced by unit mass, placed at that point. It is, however, supposed that the introductionof the unit mass at the point does not change the configuration of the field. If m be the massof a gravitating particle, the intensity of gravitational field at a distance r from it is given by
98 Mechanics
E = − Gmr
r2$
where $rrr
=→
is a unit vector along r→ .
Here the negative sign indicates that the direction of force is opposite to $r i.e., towardsthe mass m. Thus, the force per unit mass is a measure of the field intensity.
The gravitational potential at a point in a gravitational field of a body is the amount ofwork, required to be done on a unit mass in bringing it from infinity (i.e., the state of zeropotential) to that point. In other words, the potential at a point is equal to the potential
energy per unit mass. Hence, the potential at a distance r→
from a mass m is given by
V = − = = −→ →
∞ ∞z zE G G.d r m
rdr m
r
r r
2
Therefore, the potential energy of a mass m′, placed at that point, is given by
U = m′V = − ′Gmmr
Similar to the case of an electric field, the expression for the intensity of the gravitationalfield is
E = − ddrV
3.19 EQUIPOTENTIAL SURFACE
An equipotential surface is the surface, at all the points of which the potential is constant.The potential due to a point mass m at a distance r is
V = − Gmr
and it is constant on a spherical surface of radius r. This is the equipotenial surface. Theintensity of the field E at r is given by
E = −∇ −V = Vddr
r$
where $rrr
=→
is along OA
Now if a unit mass moves a small distance AB→
= dr on the surface,
then E→ →
.d r = − =ddr
r drV
$ . 0
because vector $r is perpendicular to the vector dr at the surface. This means that for
B d r E A
m o
Fig. 12
Dynamics of Circular Motion and the Gravitational Field 99
equipotential surface, the intensity of the field E is perpendicular to the surface at any pointA. If a unit mass is moved from one point to the other on an equipotential surface, the
amount of work done is equal to zero (because E V VA
B
B A
→ →FHG
IKJ = − =z d r 0)
If we have a mass m′ at A, then force on the mass is given by
F = −∇ − ∇ − ∇U = V) = V = E(m m m
Thus the force is also directed perpendicular to the equipotenial surface at A. Hence, ifany mass is moved on an equipotential surface, no work will be done.
3.20 WEIGHTLESSNESS IN SATELLITES
The weight of a body is felt due to a reactionary force applied on the body by some otherbody (which is in contact with the first body). For example, when we stand on a plane we feelour weight due to the reaction of the plane on our feet. If under some special circumstancesthe reaction of this plane becomes the zero then we shall feel as our weight has also becomezero. This is called “State of weightlessness”. If the ropes of a descending lift are broken, thenpersons standing in the lift will feel this state.
Weightlessness is also felt by a space-man inside an artificial satellite. Suppose an artificialsatellite of mass m is revolving around the earth (mass Me) with speed v0 in an orbit of radiusr. The necessary centripetal force is provided by the gravitational force.
GMem
r2 =mv
r02
orGMe
r2 =vr02
...(i)
If there is a space-man of mass m′ inside the satellite, he is acted upon by two forces:
(i) gravitational force GMem
r′
2 , (ii) reaction R of the base of the satellite, in the opposite
direction. Thus there is a net force GM
Remr
′ −LNM
OQP2 on the man. It is directed towards the
center of the orbit and is the necessary centripetal force on the man. That is,
GM Remr
′ −2 =m v
r′ 0
2
orGM Rr m2 −
′=
vr02
Substituting the value of v r02 / from equation (i), we get
GM Re
r m2 −′ =
GMe
r2
∴ R = 0
100 Mechanics
Thus, the reactionary force on the man is zero. Hence he feels his weight zero. If hestands on a spring-balance, the balance will read zero. In fact, every body inside the satelliteis in a state of weightlessness. If we suspend a body by a string, no tension will be producedin the string. Space-man cannot take water from a glass because on tilting the glass, watercoming out of it will float in the form of drops. Space-man take food by pressing a tube filledwith food in the form of paste.
Although moon is also a satellite of the earth, but a person on moon does not feelweightlessness. The reason is that the moon has a large mass and exerts a gravitational forceon the person (and this is the weight of person on the moon). On the other hand, the artificialsatellite having a smaller mass does not exert gravitational force on the space-man.
3.21 VELOCITY OF ESCAPE
It is a thing of common experience that when a body is projected in the upward direction,it returns back due to the gravitational pull of the earth on it. Now, if the body is projectedupwards with such a velocity which will just take the body beyond the gravitational field ofthe earth, then it will never come back. This velocity of the body is called the velocity ofescape.
Let us consider a body of mass m lying at a distance r (>R) from the center of the earth(or planet or sun). The force with which the earth attracts the body is given by
F = − GMmr2
where M is the mass of the earth.
If this mass be moved through a distance dr away from the earth, then the small workdone on the body is
dW =GMm
rdr2
In order that the body may not return on the earth, sufficient work must be done onit to import the body so much kinetic energy that it moves to infinity. Therefore the workdone in moving the body from the surface of the earth to infinity is
W = GM GM GMR
R R
mr
dr mr
m2
∞ ∞
z = −LNM
OQP
=
Evidently, a body of K.E. greater than this GMm/R would escape completely from theearth. Hence the minimum velocity v of projection of the body to escape into space is givenby the relation
12
2mv =GM
Rm
or v = 2GMR
...(i)
But g = GM/R2
∴ v = 2gR ...(ii)
Expressions (i) and (ii) are for the escape velocity of a body from earth.
Dynamics of Circular Motion and the Gravitational Field 101
If at the surface of the earth, we take g = 9.8 m/sec2 and its radius R = 6.4 × 106 m,then escape velocity
v = 2 9 8 6 4 106× . × . ×
v = 11.2 × 103 m/sec
v = 11.2 km/sec
Hence, if a body is projected in the upward direction with a velocity 11.2 km/sec or morethan this, it will never return to the earth. Escape velocity is the same for all bodies ofdifferent masses.
3.22 RELATION BETWEEN ORBITAL VELOCITY AND ESCAPE VELOCITY
The orbital velocity of a satellite close to the earth is v g e0 = R , and the escape velocity
for a body thrown from the earth’s surface is v g e0 2= R . Thus
vve
0=
g
ge
e
R
R212
=
∴ ve = 2 0v
If the orbital velocity of a satellite revolving close to the earth happens to increase to 2times, the satellite would escape.
3.23 SATELLITES
In the solar system, different planets revolve round the sun. The radii of the orbits andtheir time periods of revolution are different for different planets. In these cases, the forceof gravitation between the sun and the planet provides the necessary centripetal force.Similarly the moon revolves around the earth and the force of gravitation between the earthand the moon provides the necessary centripetal force for the moon to be in its orbit. Heremoon is the satellite of the earth.
From 1957, many artificial satellites around the earth have been launched. These satellitesare put into orbits with the help of multi-stage rockets.
3.24 ORBITAL VELOCITY OF SATELLITE
When a satellite (such as moon) revolves in a circular orbit around the earth, a centripetalforce acts upon the satellite. This force is the gravitational force exerted by the earth on thesatellite.
In figure, a satellite of mass m is revolving around the earth with a speed v0 in a circular
orbit of radius r. The centripetal force on the satellites is mv
r02
.
Let Me be the mass of the earth. The gravitational force exerted by the earth on the
satellite will be GMem
r2 , where G is gravitation constant. As the gravitational force provides
the required centripetal force, we have
Dynamics of Circular Motion and the Gravitational Field 103
But GMe = g Re2
∴ T = 23
π (R +R 2
e
e
hg
)
From Eq. (iii) or (iv) the period of revolution of the satellite can be calculated. It isevident from these equations that the period of revolution of a satellite depends only uponits height above the earth’s surface. Greater is the distance of a satellite above the earth’ssurface, greater is its period of revolution. This is why the moon, which is at a height of3,80,000 km above earth, completes one revolution of earth in nearly 27 days, while anartificial satellite revolving near the earth’s surface completes 10 to 20 revolution in a day.If the height of an artificial satellite above earth’s surface be such that its period of revolutionis exactly equal to the period of revolution (24 hours) of the axial motion of the earth, thenthe satellite would appear stationary over a point on earth’s equator. It would be synchronouswith earth’s spin. Such a satellite is known as “geostationary satellite”. It is used to reflectT.V. signals and telecast T.V. programs from one part of the world to another.
Now, from Eq. (iv), we have
(Re + h)3 =T R
2
4 22
π× g e
(Re + h) =T R2g e
2
2
1 3
4πLNM
OQP
/
...(v)
Substituting T = 24 h = 24 × 60 × 60 = 86400 s, Re = 6.37 × 106 m and g = 9.8 m/s2 inEq. (v), we get
Re + h = 42,200 Km
This is the orbital radius of a geostationary satellite.
The height of the satellite above the earth’s surface is
h = 42,200 – 6370 = 35,830 km.
Thus, for a satellite to appear stationary, it must be placed in an orbit around the earthat a height of 35,830 km from the earth’s surface. This is often called ‘parking orbit’ of thesatellite. Artificial satellites used for telecasting are placed in parking orbits.
The orbiting speed of the geostationary satellite is given by
v =2π ( )R
Te h+
v =2 3 14 42 200 11042× . × , km
24km/hr
h=
3.25 ORBITAL SPEED AND PERIOD OF REVOLUTION OF A SATELLITE VERY CLOSE TO EARTH
If a satellite is very close to the earth’s surface (h << Re), then h will be negligiblecompared to Re. In this case putting h = 0 in Eq (ii), the orbital speed of the satellite is givenby
104 Mechanics
v0 = RR
Ree
eg
g=
Putting g = 9.8 meter/sec2 and Re = 6.37 × 106 meter, we have
v0 = 9 8 6 37 10 7 9 106 3. × ( . × ) . ×= ≅ m/s 8 km/s
Similarly, putting h = 0 in Eq. (iv), the period of revolution of the satellite is given by
T = 2 2 3 14 6 37 10 9 86π Re
g= × . × ( . × ) / .
= 5063 seconds
≈ 84 minutes.
Thus, the speed of a satellite revolving very close to the earth’s surface is nearly8 km/sec and its period of revolution is nearly 84 minutes.
3.26 ARTIFICIAL SATELLITES
We have seen above that when a satellite revolves around the earth in an orbit near theearth’s surface then its orbital velocity is about 8 km/s. Therefore, if we send a body a fewhundred kilometers above the earth’s surface and give it a horizontal velocity of 8 km/s, thenthe body is placed in an orbit around the earth. Such a body is called an artificial satellite.
Like moon, an artificial satellite also revolves, around the earth under the gravitationalattraction exerted by the earth which acts as the centripetal force. One may ask, how is itthat the satellite continues to revolve in an orbit at a definite height above the earth, insteadof falling towards the centre of the orbit under the centripetal force. In fact, the satellitecontinuously falls towards the centre of the earth, but due to the curvature of the earth itis maintained at the same height. This we can see in Fig. 14. If there were no centripetalforce on the satellite then it would have moved along a straight-line path PQR————. Butdue to the presence of centripetal force it moves along a circular path PST————. Thus itis continuously falling towards the center of the earth through the distances QS, RT——.
An artificial satellite is placed in an orbit by means of a multi-stage rocket. The satelliteis placed on the rocket. On being fired, the rocket moves vertically upward with an increasingvelocity. When the fuel of the first stage of the rocket is exhausted, its casing is detached andthe second stage comes in operation. The velocity of the rocket increases further. Thisprocess continues. When the rocket, after crossing the dense atmosphere of the earth attainsproper height, a special mechanism gives a thrust to the satellite producing a pre-calculatedhorizontal velocity. A satellite carried to a height (<< earth’s radius) and given a horizontalvelocity of 8 km/s is placed, by earth’s gravity, almost in a circular orbit around the earth,It then continues revolving (without using any fuel). Even due to a slight mistake in calculatingthe velocity, the orbit of the satellite would change considerably. The satellite is always givensuch a velocity that its orbit remains outside the earth’s atmosphere; otherwise the frictionof atmosphere would cause so much heat that it will burn.
The satellite revolves around the earth in an orbit with earth as center, or a focus. Ifa packet is released from the satellite, it will not fall to the earth but will remain revolvingin the same orbit with the same speed as the satellite.
Dynamics of Circular Motion and the Gravitational Field 105
Uses of Artificial Satellite(1) They are used to study the upper regions of the atmosphere.
(2) Information about the space of the earth can be obtained.
(3) Weather forecast can be communicated to the earth.
(4) Radiation from the sun and outer space can be studied.
(5) Distant telecasting can be achieved.
(6) Meteorites can be studied.
(7) Space flights are possible due to artificial satellites.
OEarth
Fig. 14
3.27 BLACK HOLES
Gravitational attraction occurs throughout the universe. It is used in theories of galaxiesand stars in addition to explaining the motion of planets and earth satellites as we have done.
Stars appear to have formed after a ‘big bang’ or explosion of matter when the universefirst began. Initially, the hydrogen gas atoms produced moved towards each other undergravitational attraction, the speed and energy of the atoms increased and their temperaturerose. Many atoms then had sufficient energy to fuse together and form helium atoms. Thefusion of hydrogen atoms to form helium atoms produces nuclear energy and the hightemperature rise produces light. A star reaches a stable size or radius when the outwardpressure of the gas is balanced by the inward gravitational attraction between the atoms.
When all the hydrogen in the star is used up in the fusion process, which may occur afterby millions of years, the unbalanced gravitational attraction between the atoms will result inthe star becoming smaller and smaller. Eventually the star will collapses. From the wave-particle duality of light, light can be considered as particles (photons). So as the star collapsesin size and becomes denser, the gravitational attraction on light from its surface increases.This make the light bend more and more towards the inside of the star and soon no lightescapes from the star.
So a collapsed star has such a strong gravitational field that not even light can escapefrom it, which is why it appears black. The region round the star from which no light canescape is called a black hole. Fortunately, our star, the sun, is estimated to continue producingnuclear energy for about 5000 million years. In 1994 the Hubble Space Telescope discovered
106 Mechanics
one of the biggest black holes in the universe, about 500 light years across and in a galaxy52 million light years away from the earth. (1 light year = 9.46 × 1012 km approx.).
PLANETS AND SATELLITES
Q. 1. The maximum and minimum distances of a comet from the sun are 1.6 × 1012 mand 8.0 × 1010 m respectively. If the speed of the comet at the nearest point is 6.0 × 104
m/sec, calculate the speed at the farthest point.
Ans. The speed of a satellite round a planet in elliptical orbit varies constantly. In order
to conserve angular momentum.L→ → →
=( × )r m v at these points is
L→
= mv1r1 = mv2r2
or v1r1 = v2 r2
Putting the given values, we have
v1 × (1.6 × 1012) = (6.0 × 104) × (8.0 × 1010)
∴ v1 =( . × ) × ( . × )
. ×6 0 10 8 0 10
1 6 10
4 10
12
= 3 0 103. × meter /sec Ans.
Q. 2. Earth’s revolution round the sun has radius 1.5 × 1011 meters and period 3.15 ×107 second (one year). If the gravitational constant is 6.67 × 10 –11 newton-m2/kg2, deduce themass of the sun. The formula is to be deduced.
Ans. The period of revolution T of the earth round the sun (mass Ms) in a circular orbitof radius r is given by
T2 =4 2 3π r
sGM
∴ Ms =4 2 3π rGT2
Substituting the given values:
Ms =4 3 14 1 5 10
6 67 10 3 15 10
2 11
11 7 2× ( . ) × ( . ×
( . × / ) ( . × )m)
newton-m kg s
3
2 2−
= 2.0 × 1030 kg Ans.
Q. 3. Determine the mass of the earth from the moon’s revolution around the earth in acircular orbit of radius 3.8 × 105 km with period 27.3 days, G = 6.67 × 10–11 newton-m2/kg2.
Ans. The moon revolves round the earth. If r be the radius of its orbit and T the orbitingperiod, then the mass of the earth is given by
M =4 2 3π rGT2
Dynamics of Circular Motion and the Gravitational Field 107
Here r = 3.8 × 105 km m 3.8 × 108 meter, T = 27.3 days = 27.3 × 24 × 60 × 60 sec.
∴ M =4 3 14 3 8 10
6 67 10 27 3 24 60 60
2 8 3
11 2× ( . ) × ( . × )
( . × ) × ( . × × × )−
= 5 8 1024. × kg.
Q. 4. With what horizontal velocity must a satellite be projected at 800 km above thesurface of the earth so that it will have a circular orbit about the earth. Assume earth’s radius6400 km. What will be the period of rotation? g = 9.8 m/sec2.
Ans. The orbital velocity of a satellite revolving round the earth at a height h fromearth’s surface is given by
v = RR +
gh
FHG
IKJ
Putting the given values R = 6400 km = 6.4 × 106 meter
and R + h = 6400 + 800 = 7200 km
= 7.2 × 106 meter
we get v = ( . × ) ×.
. ×6 4 10
9 87 2 10
66
= 7.5 × 103 m/sec = 7.5 km/sec.
The satellite must be given a horizontal velocity of 7.5 km/sec after taking it to a stableorbit round the earth.
The period of revolution is T =2 2 3 14 7200π (R + km
7.5 km/sech
v) × . ×=
= 6029 sec = 100.5 min Ans.
Q. 5. The period of a satellite in circular orbit of radius 12000 km around a planet is 3hours. Obtain the period of a satellite in circular orbit of radius 48000 km around the sameplanet.
Ans. The period of a satellite is proportional to the32
th power of its orbital radius.
i.e., T ∝ ( ) /r 3 2
Thus, if T′ be the period in an orbit of radius r′, thus
TT′
=rr′
FHG
IKJ
3 2/
Here T = 3 hours, r = 12000 km and r′ = 48000 km. Therefore
3T′
=1200048000
14
18
3 2 3 2FHG
IKJ = F
HGIKJ =
/ /
T ′ = 24 hours Ans.
Dynamics of Circular Motion and the Gravitational Field 109
Q. 8. A satellite moves in a circular orbit around the earth at a height R2
from earth’s
surface, where R is the radius of the earth, calculate its period of revolution (R = 6.38 × 106
meter).
Ans. The period of revolution of a satellite at a height h from earth’s surface is givenby
T =2π ( )R +
RR +h h
gFHG
IKJ
For h = R2
, we have
T =2 1 5 1 5π × . .R
RR
g
= 2 1 5 3 2π Rg
FHG
IKJ ( . ) /
= 2 3 14 6 38 10 1 56
3 2× . × . × × ( . ) / meter9.8 meter/sec2
= 5067 sec × (1.5)3/2 = 9310 sec
= 2 hours 35 min Ans.
Q. 9. Calculate the limiting velocity required by an artificial satellite for orbiting veryclosely round the earth (R = 6.4 × 106 meter, g = 9.8 meter/sec2). What would be the period.
Ans. The orbital velocity of a satellite revolving round the earth (radius R) at a heighth from earth’s surface is given by
v = RR
gh+
For a satellite orbiting very closely the earth’s surface, h << R, so that
v = RR
R
m/sec
= 7.92 km/sec.
g g=
= =9 8 6 4 10 7 92 106 3. × . × . ×
The period is T =2π (R +
RR +h h
g)
For h << R T = 2 2 3 14 6 4 109 8
5 075 106
3π R secg
= =× . × . ×.
. ×
= 84.6 min = 1.41 hours Ans.
110 Mechanics
Q. 10. A satellite is revolving round near the equator of a planet of mean density ρ. Showthat the period T of such an orbit depends only on the density of the planet.
Ans. Let m be the mass and v the velocity of satellite which orbits round the planet justabove its surface. The radius of the orbit is practically equal to the radius R of the planet.Therefore, the centripetal force acting upon the satellite is mv2/R.
If M be the mass of planet, the gravitational force between the planet and the satelliteis GMm/R2 and this supplies the required centripetal force. Thus
mv2
R= G M
R2m
or v =GMR
The period of the satellite is
T =2 R R
GM
3π πv
= 2
Now, the mass M of the planet (mean density suppose ρ) is
M =43
3π ρr
Making the substitution in the last expression, we get
T =3π
ρG
Thus the period depends only on the density of the planet.
Q. 11. A satellite is launched into a circular orbit 1600 km above the surface of the earth.Find the period of revolution if the radius of the earth is R = 6400 km and the accelerationdue to gravity is 9.8 m/sec2. At what height from the ground, should it be launched so thatit may appear stationary over a point on the earth’s equator?
Ans. The orbiting period of a satellite at a height h from earth’s surface is
T =2π ( )R +
RR +h h
g
Here R = 6400 km, h = 1600 km = R4
Then.
T =2π R + R
4R
R + R4
FHG
IKJ
FHG
IKJ
g
= 2 1 14
3 2
π Rg
FHG
IKJ +F
HGIKJ
/
Dynamics of Circular Motion and the Gravitational Field 111
= 2 1 25 3 2π Rg
( . ) /
Putting the given values:
T = 2 3 14 6 4 10 1 254
3 2× . × . × ( . ) /m9.8 m/sec2
= 7092 sec = 1.97 hours.
Now, a satellite will appear stationary in the sky over a point on earth’s equator if itsperiod of revolution round the earth is equal to the period of revolution of the earth roundits own axis which is 24 hours. Let us find the height h of such a satellite above earth’ssurface in terms of earth’s radius. Let it be nR. Then
T =2π (R + R)
RR + Rn n
g
= 2 1 3 2π Rg
n( ) /+
= 2 3 14 6 4 10 16
3 2× . . × ( ) / meter9.8 m / sec2 + n
= (5067 sec) (1 + n)3/2
= (1.41 hours) (1 + n)3/2
For T = 24 hours, we have
(24 hours) = (1.41 hours) (1 + n)3/2
or (1 + n)3/2 =24
1 4117
.=
or 1 + n = (17)2/3 = 6.61
or n = 5.61
The height of the geo-stationary satellite above earth’s surface is
nR = 5.61 × 6400 km = 3.59 × 104 km Ans.
Q. 12. Two satellites of same mass are launched in the same orbit round the earth so asto rotate opposite to each other. They collide in elastically and stick together as wreckage.Obtain the total energy of the system before and just after the collision. Describe the subsequentmotion of the wreckage.
Ans. The potential energy of a satellite in its orbit
is − GMmr
, and the kinetic energy is GMm
r2. The total
energy is
E = K + U = GM GMm
rm
r2−
v
m
→ v
m
→r
M
Fig. 15
112 Mechanics
= − GMmr2
where m is the mass of the satellite, M the attracting mass (earth) and r the orbital radius.When there are two satellites, each of mass m, in the same orbit the energy would be
− −GM GMmr
mr2 2
= − GMmr
Let v→
be the velocity of the wreckage after the collision. Then, by the law of conservationof momentum, we have
m v m v→ →
+ = ( )m m v+ ′→
or mv – mv = (m + m) v′∴ v′ = 0
Therefore, the wreckage of mass 2m has no kinetic energy but only potential energy.Hence the total energy just after the collision would be
− GM (2 )mr
As the velocity of the wreckage is zero, the centripetal force disappears and the wreckagefalls down under gravity.
Q. 13. A small satellite revolves round a planet in an orbit just above planet surface.Taking G = 6.66 × 10–11 MKS units and mean density of planet as 8.00 × 103 MKS units,calculate the time period of the satellite.
Ans. T =3π
ρG
=3 3 14
6 66 10 8 00 1011 3× .
. × × . ×−
= 0.420 × 104 sec = 70 min Ans.
Q. 14. Two earth satellites; A and B, each of mass m are to be launched into circularorbits about earth’s centre. Satellite A is to orbit at an altitude of 6400 km and B at 19200km. The radius of earth is 6400 km.
(a) What is the ratio of the potential energy of B to that of A, in orbit,
(b) Ratio of kinetic energy,
(c) Which one has the greater total energy.
Ans. (a) The potential energy of an earth’s satellite in a circular orbit of radius r is
U(r) = − GMmr
where M is the mass of the earth and m that of the satellite.
For A: r = 6400 + 6400 = 12800 km
Dynamics of Circular Motion and the Gravitational Field 113
For B: r = 19200 + 6400 = 25600 km
∴UU
B
A=
rrA
B= =12800
2560012
.
(b) The kinetic energy of an earth’s satellite moving with velocity v in a circular orbitof radius r is
K =12 2
2mvm
r= GM
∴KK
B
A=
r
rA
B= 1
2
(c) The total energy of the satellite is
E = U + K = − + = −GM GM GMmr
mr
mr2 2
Which is negative. Clearly the farther the satellite is from the earth, the greater (that is, lessnegative) is its total energy E. Hence the satellite B has the greater (less negative) totalenergy.
Q. 15. A stream of α-particles is bombarded on mercury nucleus (z = 80) with a velocity1.0 × 109 cm/sec. If an α-particle is approaching in head-on direction, calculate the distanceof closest approach. The mass of α-particle is 6.4 × 10 –24 gm and electronic charge is4.8 × 10–10 esu.
Ans. The distance of closest approach of an α-particle to a nucleus is given by
ro =2 2ze
kicm
where ki is the initial kinetic energy. Here
ki =12
2mv
=12
6 4 10 24× ( . × − gm) (1.0 × 10 cm/sec)9 2
= 3.2 × 10–6 erg.
∴ ro =2 80 4 8 10
3 2 10
10
6× × ( . ×
. ×
−
− esu)
erg
2
ro = 1.152 × 10–11 cm Ans.
Q. 16. In a double star, two stars (one of mass m and the other of 2m) distant d apartrotate about their common centre of mass. Deduce an expression for the period of revolution.Show the ratio of their angular momenta about the centre of mass is the same as the ratioof their kinetic energies.
Ans. The centre of mass c will be at distances d d3 3
and 2 from the masses. 2m and m
114 Mechanics
respectively. Both the stars rotate round c in their respective orbits with the same angularvelocity ω. The gravitational force acting on each star due to the other supplies the necessary
centripetal force. The gravitational force on either star is G (2 )m m
d2 . If we consider the
rotation of the smaller star, the centripetal force (mrω2) is md23
2FHG
IKJ ω .
Fig. 16
∴G (2 )
2m m
d= m
d23
2FHG
IKJ ω
or ω =3
3G md
Therefore, the period of revolution is given by
T =2 2
3
3πω
π= dmG
The ratio of the angular momenta is
( )( )
II
big
small
ωω
=I
Ibig
small=
FHG
IKJ
FHG
IKJ
=( )2
3
23
12
2
2
m d
m d
Since ω is same for both. The ratio of their kinetic energies is
12
12
2
2
I
I
big
small
ω
ω
FHG
IKJ
FHG
IKJ
=I
Ibig
small= 1
2
which is the same as the ratio of their angular momenta.
Dynamics of Circular Motion and the Gravitational Field 115
Q. 17. Estimate the mass of the sun assuming the orbit of the earth around the sun tobe circular. The distance between sun and earth is 1.49 × 1013 cm, G = 6.66 × 10–8 egs unitsand earth takes 365 days to make one revolution around the sun.
Q. 18. An artificial satellite of the earth moves at an altitude of 640 km along a circularorbit. Find the orbital velocity of the satellite. (Radius of earth = 6400 km).
Q. 19. Show that the time of revolution of a satellite just above the earths surface is 84.4min. Density of earth is 5.51 × 103 kg/meter3 and G = 6.67 × 10–11 nt-m2/kg.
Q. 20. What is the smallest radius of a circle at which a bicyclist can travel if his speedis 7 m/sec and the coefficient of static friction between the tyres and the road in 0.25. Underthese conditions what is the largest angle of inclination to the vertical at which the bicyclistcan without falling?
Ans. The (maximum) force of friction is given by
fs = µsN = µsmg
This must supply the required centripetal force. If R be the smallest radius, the maximumrequired centripetal force would be mv2/R. Thus
µs mg =mv2
R
or R =v
gs
2
µ
here v = 7 m/sec, µs = 0.25 and g = 9.8 m/sec2
∴ R =( )
. × ( . )7
0 25 9 8200
2m / sm/sec
meter2 =
The (largest) angle of inclination θ is given by
tan θ =vg
2 720 9 8
0 25R
m/sec)m × m/sec
2
2= =(.
.
θ = tan–1 (0.25) = 14° Ans.
Q. 21. A cord is tied to a pail of water and the pail is swing in a vertical circle of radius1 meter. What must be the minimum velocity of the pail at the highest point of the circle ifno water is to spill from the pail? (g = 9.8 m/sec2).
Ans. At the highest point of the vertical circle, the centripetal force is supplied by thetension in the cord and also by the weight of the pail plus water. The force must be atleastequal to the weight, otherwise water would spill from the pail under gravity. Hence if vc bethe minimum velocity required at the highest point, we have
mvc2
R= mg
or vc = Rg
Here R = 1 meter
vc = 1 9 8 3 13× . .= m/sec Ans.
116 Mechanics
Q. 22. A circular curve of highway is designed for traffic moving at 15 m/sec. (a) If theradius of the curve is 100 m, what is the correct angle of banking of the road. (b) If the curveis not banked, what is the minimum coefficient of friction between tyres and road that wouldkeep traffic from skidding at this speed?
Ans. (a) Let θ be the correct angle of banking. Then
tan θ =vg
2
RPutting the given values, we have
tan θ =(
( ).
15100
0 23 m/s)
m) (9.8 m/s
2
2 =
θ = tan–1 (0.23) = 13°
(b) If the road is not banked, then the required frictional force f (say) must supply the
entire centripetal force,mv2
R. Thus
f =mv2
R
But f = µN = µmg. Thus
µmg =mv2
R
or µ =vg
20 23
R= . Ans.
Q. 23. The radius of curvature of a railway line at a place is 800 meter and the distancebetween the rails is 1.5 meter. What should be the elevation of the outer rail above the innerone for a safe speed of 20 km/hour?
Ans. Let θ be the correct angle of banking. Then
tan θ =vg
2
R
Here v = 20 km/hour = 5.56 m/sec and R = 800 m
tan θ =5 56 5 56800 9 8
0 004. × .× .
.=
If x be the elevation of the outer rail and l the distance between the rails, then
tan θ =xl
or x = l tan θ= (1.5 m) (0.004) = 0.006 m = 0.6 cm.
Dynamics of Circular Motion and the Gravitational Field 117
Q. 24. A mass m on a frictionless table is attached to a hanging mass M by a cord througha hole in the table. Find the condition (v and r) with which its must spin for M to stay at rest.
Ans. The mass M will stay at rest when its weight Mg is used up in supplying the
required centripetal force mv
r
2 to the mass m, that is when
Mg =mv
r
2
orvr
2=
Mgm
This is the required condition. The equilibrium will,however, be unstable.
Q. 25. A smooth table is placed horizontally and an idealspring of spring-constant k = 1000 nt/m and unextended lengthof 0.5 m has one end fixed to its centre the other end is attached to a mass of 5 kg which ismoving in a circle with constant speed 10 m/s. Find the tension in the spring and the extensionof the spring beyond its normal length.
Ans. The centripetal force required for the mass m to move in a circle is supplied bythe tension T produced in the stretched spring. The streched length of the spring is R, equalto the radius of the circle. If lo be the unextended length of the spring, then the elongationis (R – lo) and the tension produced is given by
T = k (R – lo), ...(1)
where the spring-constant k is the tension per unit elongation. But this is equal to thecentripetal force mv2/R. Therefore,
k(R – lo) =mv2
R
or kR (R – lo) = mv2
Here k = 1000 nt/m, lo = 0.5 m, m = 5 kg, v = 10 m/sec.
∴ 1000 R(R – 0.5) = 5 × 100
or R(R – 0.5) = 0.5
or 2R2 – R – 1 = 0
Solving this quadratic eq. we get R = 1.0 m
The extension of the spring beyond its normal length is
R – lo = 1.0 – 0.5 = 0.5 m
Substituting this value in (1), the tension is
T = 1000 × 0.5 = 500 nt. Ans.
Q. 26. An electron is moving in a circular orbit of radius 5.3 × 10–11 meter around theatomic nucleus at a rate of 6.6 × 1015 rev/sec. Find the acceleration of the electron andcentripetal force acting on it. The mass of the electron is 9.1 × 10–31 kg.
Ans. Let R be the radius of the orbit and f the number of revolutions per second. Thenthe velocity of the electron is given by
Fig. 17
118 Mechanics
v = 2πR f
Hence its acceleration is
a =v
f2
2 24R
R= π
= 4 × (3.14)2 × (5.3 × 10–11) × (6.6 × 1015)2
= 9.1 × 1022 m/sec2, towards the nucleus.
The centripetal force is
Fc = ma
= (9.1 × 10–31) × (9.1 × 1022)
= 8.3 × 10–8 nt towards the nucleus.
Q. 27. A smooth table is placed horizontal and a spring of unstretched length lo and force-constant k has one end fixed to its centre to the other end of the spring is attached a massm which is making f revolutions per second around the centre. Show that the radius R of thisuniform circular motion is klo/(k – 4π2mf 2) and the tension T in the spring in 4π2mklof
2/(k – 4π2mf 2).
Ans. The centripetal force required for the mass m to move in a circle is supplied bythe tension T produced in the stretched spring. The stretched length of the spring is R, equalto the radius of the circle. Thus the elongation in the spring is (R – lo) and the tensionproduced is given by
T = k (R – lo) ...(1)
as k (force-constant) is the tension per unit elongation.
The (linear) velocity of motion is given by
v = 2πRf
Therefore, the required centripetal force is
mv2
R= 4π2Rf2m ...(2)
SinceT =R
mv2, we have by (1) and (2) we get
k l( )R o− = 4π2Rf 2 m
or R =kl
k f mo
− 4 2 2π
Substituting this value of R in Eq. (1) we get
T = kkl
k f mloo( )−
−LNM
OQP4 2 2π
T =4
4
2 2
2 2π
πf ml k
k f mo
− Ans.
Dynamics of Circular Motion and the Gravitational Field 119
Q. 28. An artificial satellite is revolving round the earth at a distance of 620 km. Calculatethe minimum velocity and the period of revolution. Radius of earth is 6380 km and accelerationdue to earth’s gravity at the surface of the earth is 9.8 metres/sec2.
Ans. Radius of earth’s satellite orbit r = Radius of earth + Distance of satellite from earth’s surface
= 6380 + 620 = 7000 km = 7 × 106 m/sec.
Radius of earth R = 6380 × 103 m, g = 9.8 m/sec2
∴ Period of revolution T =2 2 7 10
6380 107 10
9 85775
6
3
6π πr rgR
sec.= =× ××
×.
and orbital velocity v = R gr
= 6380 10 9 87 10
36× .
×
= 7 55 103. × m/sec.
Q. 29. A stone of mass 1 kg is attached to one end of a string 1 m long, of breakingstrength 400 nt, and is whirled in a horizontal circle on a frictionless surface the other endof the string is kept fixed. Find the maximum velocity the stone can attain without breakingthe string.
Ans. The required centripetal force mv2
R is to be supplied by the tension in the string
which should not exceed 400 nt. Thus, if v be the maximum velocity the stone can attain,we have
mv2
R= T = 400 nt
Here m = 1 kg and R = 1 m. Then
v2 = 400
∴ v = 20 m/sec Ans.
Q. 30. A mass of 1 standard kg is placed at sea level on the earth’s equator and is movingwith the earth in a circle of radius 6.40 × 106 meter (earth’s radius) at a constant speed of465 m/s. Determine the centripetal force needed. Also find the force exerted by the mass ona spring balance from which it is suspended at the equator (its weight). Assume that the masswould weight exactly 9.80 nt the equator of the earth did not rotate about its axis.
Ans. The centripetal force is given by
Fc =mv2
61 465
6 40 100 0338
Rnt.= =×
. ×.
The weight w′ = w – Fc
= 9.80 – 0.0338
= 9.77 nt Ans.
120 Mechanics
Q. 31. Escape velocity from solar system. Show that the escape velocity of a body from
solar system, launched from the earth is 2GM /Rs es where Ms is the mass of the sun and Res
is the distance of the earth from the sun. Neglect the earth’s rotation and influence of earth’sgravity. What is the escape velocity, if G = 6.67 × 10–11 S.I. unit, Me = 1.33 × 1030 kg and Res= 1.49 × 1011 m.
Ans. In the gravitational field of the sun, the potential energy of a body of mass msituated on the earth is
v = − GMR
s
es
m
If the body is to be escaped from the gravitational influence of the sun, the body should
be imparted with kinetic energy 12
2mvFHG
IKJ equal in magnitude to GMsm/Res. So that
12
2mv =GM
R or
GMR
s
es
s
es
mv = 2
From the given data, the escape velocity (launched from the earth) from the solar systemis
v =2 6 67 10 1 33 10
1 49 10
11 30
11× . × × . ×
. ×
−
= 4.2 × 104 m/sec = 42 km/sec.
Q. 32. Compare the root mean square velocity of oxygen molecules at 27°C with the escapevelocity from earth’s surface (Boltzmann constant k = 1.4 × 10– 23 joule/k).
Ans. If V is the R.M.S. velocity of the oxygen molecular, then the mean kinetic energyper molecule is given by
12
2mv =12
kT
where m [ = 32 × (mass of a proton) = 32 × 1.7 × 10–27 kg] is the mass of an oxygen moleculeand T the absolute temperature.
∴ V =3 3 1 4 10 30
32 1 7 10
23
27km
T = °−
−× . × ×
× . ×
Q T = 273 + 27 = 300 k
Now, escape velocity from the earth
v = 2 2 9 8 6 4 106g eR = × . × . ×
∴Vv
=3 1 4 10 300
32 1 7 10 2 9 8 6 4 10
23
27 6× . × ×
× . × × × . × . ×
−
−
Dynamics of Circular Motion and the Gravitational Field 121
Vv
≈ 0 04. Ans.
Q. 33. What is the potential energy of a mass of 1 kg on the surface of the earth, referenceto zero potential energy at infinite distance? Calculate also its potential energy at a distanceof 10 5 km from the centre of the earth.
Ans. In case of earth (or solid sphere) the potential energy is given by
v(r) = − GMmr
(i) For the surface of the earth, r = R
∴ v(r) = − = − −GMR
m ( . × ) × . × ×. ×
6 67 10 5 98 10 16 37 10
11 24
6
= –6.23 × 107 joules
(ii) when r = 105 km = 108 m
v(r) = −−( . × ) × . × ×6 67 10 5 98 10 1
10
11 24
8
= –3.98 × 106 joules
Q. 34. If a body is to be projected vertically upwards from earth’s surface to reach a heightof 10R, how much velocity should be given?
Ans. The body should be supplied so much kinetic energy, so as it can reach a height10R from the surface of the earth or 11R from the centre of earth.
If v is the velocity given to the body initially, then its kinetic energy = 12
2mv
Increase in potential energy = − +g mr
g mr
R R2 2
2 1
= − + =g m g mmg
RR
RR
R2 2
111011
∴12
2mv =1011
mgR
or v =2011
2011
9 8 6 4 106gR = × . × . ×
= 10.6 × 103 m/sec.
Q. 35. For the earth-sun system, calculate (i) the kinetic energy and (ii) the work whichwill have to be done in doubling the radius of the orbit of the earth.
Ans. The gravitational potential of the earth due to the sun = –GMs/r
∴ Potential energy of the earth = − GM Ms e
r
122 Mechanics
K.E. of the earth =12
2Mev
In case of earth the necessary centripetal force for its rotation is provided by gravitationalattraction i.e.,
Mevr
2=
GM Ms e
r2
∴12
2Mev =12
GM Ms e
r
∴ Total energy of the earth = − + = −GM M GM M GM Ms e s e s e
r r r12 2
Evidently negative sign indicates that the earth is bound with the earth with this energyi.e.,
The binding energy =GM M
Ms eerv
212
2= = 12
2 2Mer ω
=12
5 98 10 1 5 10 2 3 14365 24 60 60
2411 2
× . × × . × × × .× × ×
FHG
IKJ
= 2.7 × 1033 joules
Q r = =FHG
IKJ
1 5 10 2365 24 60 60
11. ×× × ×
m and rad/secω π
when the radius of the earth’s orbit is doubled, then the final binding energy will be
= GM M GM Ms e s e
r r2 212 2( )
=
=2 72
10 1 35 1033 33. × . ×= joules.
Therefore the amount of work done in doubling the radius of the orbit= 2.7 × 1033 – 1.35 × 1033 = 1.35 × 1033 joules Ans.
Q. 36. Two bodies of masses M1 and M2 are placed distant d apart, show that at theposition, where the gravitational field due to them is zero, the potential is given by
V = − + +Gd
( )M M M M1 2 1 22
Ans. Let the gravitational field be zero at a point distant r from the mass M1. Then, thedistance of the same point from the mass M2 = d – r. At the point, under consideration, thefield due to M1 and M2 is zero, i.e.,
− GM1
r2 = −−
GM2
( )d r 2
ord r
r−
=MM
2
1
Dynamics of Circular Motion and the Gravitational Field 123
Adding 1 on both sides, we have
dr
=M M
M or
M
M M2 1
1
1
1 2
+=
+r
d
∴ d – r = dd d
−+
=+
M
M M
M
M M1
1 2
2
1 2
Therefore, potential at the point is
V = − −−
GM GM1 2
r d r( )
= − + − +GMM
M MGM
MM M1
11 2
2
21 2
d d( ) ( )
= − + + +G M M M M M M1 1 2 2 1 2d e j
V = − + +G M M M M1 2 1 2d( )2 Ans.
Q. 37. If a mass 50 kg is raised to a height 2R from the earth’s surface, calculate thechange in potential energy (g = 9.8 m/sec2; R = 6.4 × 106 m).
Ans. The potential energy of a mass m distance r from a solid sphere of a mass M willbe given by
U(r) = Potential × GMm
mr
= − QV GM= −r
(referring zero potential energy at infinite distance)
But g =GMR2 , where R is the radius of earth,
∴ U(r) = − g mr
R2
Difference of potential energy of mass ‘m’ between the points at r1 and r2 is
U(r2) – U(r1) = − +g mr
g mr
R R2 2
2 1
here r1 = R and r2 = 2R + R = 3R
∴ U(r2) – U(r1) = − +gg gm
RM3
RM = 23
R
=23
9 8 50 6 4 10 2 1 106 9× . × × . × . ×= joules Ans.
Q. 38. (a) A small block of mass ‘m’ slides along the friction less track. Calculate theheight ‘h’ at which mass must be released on the track to be able to go round the track ofradius ‘R’.
Dynamics of Circular Motion and the Gravitational Field 125
so NC =2 2mg
h mgR
R)( − −
=2 5 2mg
mgR
R R)( − −
NC = 5 mg Ans.
Q. 39. A bead can slide without friction on a circular hoop of radius 10 cm in a verticalplane. The hoop rotates at a constant rate of 2 rev/sec, about a vertical diameter.
(a) Find angle θ at which the bead will be in vertical equilibrium,(b) Can bead rise up to the height of the centre of hoop,(c) What will happen if hoop rotates at 1 rev/sec.Ans. For equilibrium along the hoop
mrω2 cos θ = mg sin θ
or (R sin )θ ω θ2 cos = g sin θ
or cos θ =g
Rω π2 29 8
0 1 2 2= .
. ( × )
or θ = cos–1 (0.62)
(b) The bead can not rise up to the height of centre of hoop because, then, there will beno vertical component of mrω2 to balance the weight mg of bead.
(c) If ω = 2π × 1 rev/sec.
cos θ =9 8
0 1 2 19842 2
.. ( × )
,π π
= which is greater than 1.
This means no value of θ is possible. Hence bead will remain stationary at the bottomof hoop.
Q. 40. A hemispherical bowl of radius R = 0.1 m is rotating about its own axis, whichis vertical, with angular velocity ω. A particle of mass m = 10–2 kg, on frictionless inner surfaceof bowl is also rotating with same angular velocity. The particle is at a height ‘h’ from thebottom of bowl.
(a) Obtain a relation between h and ω. What is minimum value of ω needed to have non-zero value of h?
(b) It is desired to measure ‘g’ using this set-up by measuring ‘h’ accurately. Assumingthat R and ω are known and least count in the measurement of ‘h’ is 10–4 m, find minimumpossible error in measurement of g.Ans. (a) The forces acting on rotating particle P are
(1) Weight, mg acting downwards(2) Normal reaction, N(3) Centrifugal force, mrω2
For equilibrium of rotating particleN cos θ = mg ...(1)
and N sinθ – mrω2 = Centripetal force
= mrω2
i.e., N sin θ = 2 mrω2 ...(2)
ω
θm rω2
m g
P
θR-h
C
N
o
r
Fig. 19
Fig. 18
126 Mechanics
From (1) and (2);
tan θ =2 2r
gω
But tan θ =PCCO R
=−r
h
sor
hR − =2 2r
gω
or ω2 =g
h2( )R − ...(3)
i.e., ω2 =9 8
2 0 14 9
0 1.
( . ).
.−=
−h h
which is desired relation. For non-zero value of h, ω2 4 90 1
> ..
. Hence, ω > 7 radian/sec.
(b) Since from Eq. (3), we have
g = 2ω2 (R – h)
Taking log on both side log g = log 2ω2 + log (R – h)
differentiating,∆gg = 0 −
−∆h
hR
taking h → 0, ( )min∆g = gh
. . ×.
. ×∆R
= =−
−9 8100 1
9 8 104
3
Q 41. A thread is passing through a hole at the center of a frictionless table. At the upperend, a block of 0.5 kg is tied and another mass 8.0 kg is tied to the lower end which is freelyhanging. The smaller mass is rotated about an axis passing through hole so as to balance theheavier mass. If hanging mass is changed to 1.0 kg, what is fractional change in radius andthe angular velocity of smaller mass to balance the hanging mass again.
Ans. Let r = radius of circular path traced by smaller mass
and ω = angular velocity of rotation
then tension in string
T = mrω2
Since, the string balances hanging mass M,
Hence T = Mg
For equilibrium of system, mrω2 = Mg
or 0.5 rω2 = 8 g ...(1)
if r′ is the new radius and ω′ the angular velocity when 8.0 kgmass is changed to 1.0 kg
0.5 r′ω′2 = 1 g ...(2)
Fig. 20
Dynamics of Circular Motion and the Gravitational Field 127
Dividing (1) by (2) we get
rr′ ′FHG
IKJ
ωω
2
= 8 ...(3)
Again, since angular momentum of system will be conserved
mr2ω = mr′2ω′
orωω′
=rr′F
HGIKJ
2
putting the value in Eq. (3) we get
rr′
FHG
IKJ = 8 or
δrr
r rr
= ′ − = − =8 11
7
andδωω
=ω ω
ωωω
′ = ′ =~ 1
orδωω
=rr′
FHG
IKJ − = =
2
1 164
1 6364
~
Q. 42. A particle describes a horizontal circle on a smooth surface of inverted cone. Theheight of the plane of circle above the vertex is 9.8 cm. Find speed of the particle.
Ans. The particle while describing circular pathalong smooth surface of inverted cone is acted upon bytwo forces along the inclined surface of cone.
(1) mg cosθ, component of its weight down theinclined surface.
(2) mv
r
2
sin θ, the component of reaction of
centripetal force up the inclined plane.
Hence for circular motion of particle to be possible,without sliding along surface, condition is,
mvr
2sinθ = mg cos θ; where θ is half above the cone angle
v = rg cot θ
From figure r = h tan θ = 9.8 × 10–2 tan θ
Hence v = 9 8 10 9 82. × tan × . cot− θ θ
= 0 98. tan cotθ θ
v = 0.98 m/sec Ans.
θ
θ
9.8
cm
r
mg cos θ
m vr
2
m g
Fig. 21
Dynamics of Circular Motion and the Gravitational Field 129
i.e., vmax = µ = =gR m/sec0 4 9 8 30 10 84. × . × . Ans.
Q. 46. Calculate the altitude of an artificial satellite if it is always above a certain placeon earth’s surface, assuming its orbit to be circular. (Mean radius of earth = 6400 km, g = 9.80m/s2).
Ans. When the period of satellite in its orbit is equal to the period of earth rotation aboutown axis, then the satellite shall always be at a fixed point in reference to earth.
Thus, period of satellite; T = 24 hrs
Since T =2πω
or ω =2 2 3 142
24 60 60π
Trad/s= × .
× ×
For satellite in circular orbit,
GMmr2 =
mvr
mr2
2= ω
or r3 =GMω2 ...(1)
where r = satellite to earth distance,
M = mass of earth,
m = mass of satellite,
ω = angular velocity of satellite.
Again on earth’s surface,
GMR2 = g or GM = gR2
putting this in eq. (1) we get
r3 =gR2
ω2
2 2
29 8 6400000 24 3600
2 3 142= . × ( ) × ( × )
( × . )
or r = 42400 km
height of satellite from earth surface h = r – R
or h = 42400 – 6400
h = 36000 km Ans.
Q. 47. A satellite revolves round a planet in an elliptic orbit. Its maximum and minimumdistances from the planet are 1.5 × 107 meters and 0.5 × 107 meters respectively. If the speedof the satellite at the farthest point be 5 × 103 m/sec. Calculate the speed at the nearest point.
Ans. Let, F = Farthest position of satellite from planet situated at focus at the ellipticalpath
130 Mechanics
Planet
NF
r1 M
m r2
Fig. 22
N = Nearest position of satellite,
m = mass of satellite,
M = mass of planet,
r1 = farthest distance,
r2 = nearest distance,
v1 = velocity at farthest position,
v2 = velocity at nearest position.
Assuming that path of satellite nearly circular at farthest and nearest position, we havefor farthest position,
GMmr1
2 =mv
r12
1
or GMr1
= v12 ...(1)
at nearest position,
GMmr2
2 =mvr
22
2
orGMr2
= v22 ...(2)
Dividing (2) by (1), we get
vv
22
12 =
rr1
2
or v2 = v rr11
2
12F
HGIKJ ...(3)
putting values, we get
v2 = 5 10 1 5 10 0 5 10 5 10 33 7 7 3× ( . × ) / ( . × ) ×=
or v2 = 8.660 × 103 m/s Ans.
Q. 48. Find the magnitude of centripetal acceleration of a particle on the tip of a fan bladeof 0.3 meter diameter, rotating at speed of 1000 rev/min.
132 Mechanics
Ans. Let R = actual distance of earth from the sun and R′ =R3
= supposed distance
between sun and earth.
The gravitational force of attraction providesrequired centripetal force i.e.,
GMR2
m=
mv2
R...(1)
But v = Rω = R2πh
where M = mass of sun,
m = mass of earth,
R = Sun to earth distance,
v = tangential velocity,
T = time taken by earth to complete one revolution.
Since v =2πR
TPutting the value of v in Eq. (1), we get
GMR2
m=
mR
RT
2 2πFHG
IKJ
or T2 =4 2π RGM
3...(2)
Now, if T′ is time of one revolution of earth when distance is R = R3
′ , then by analogy,
we have,
T 2′ =4 2
3
π R3
GM
FHG
IKJ
...(3)
Dividing Eq. (3) by Eq. (2), we get
TT
2
2′
=4
127
23
π
π
R3
GMGM
4 R2 3
FHG
IKJ
=×
or T 2′ =T2
27 but T = 1 year = 365 days
So T′ =T days
3 3365
3 1 73270 3= =
× .. Ans.
Q. 52. A large mass M and a small mass m hang at the two ends of a string that passesover a smooth tube as shown in the figure 24. The mass m moves around a circular path which
Fig. 23
Dynamics of Circular Motion and the Gravitational Field 133
lies in a horizontal plane. The length of the string from the mass m to the top of the tube isl and θ is the angle which the length makes with the vertical, what should be the frequencyof rotation of the mass m so that the mass M remains stationary.
Ans. Various forces on masses m and M are as shown in figure 24.
For equilibrium of mass m, we have
T cos θ = mg ...(1)
and T sin θ =mv
rm r
22= ω ...(2)
For equilibrium of mass M,
T = Mg
From the diagram r = l sin θputting the value of T and r in Eqn. (2)
T cos θ
m vr
2
T s in θ
m g
l
T
M
M g
Fig. 24
Mg sin θ = mω2 l sinθ
or ω =Mgml
and frequency: n =ωπ2
n =1
2πMgml
Ans.
Q. 53. A boy is sitting on the horizontal platform of a joy wheel at a distance of 7m fromthe centre. Wheel begins to rotate and when the angular speed exceeds 10 rev/min, the boy justslips. What is the coefficient of friction between boy and platform. (g = 9.8 m/s2).
Ans. The boy slips when centrifugal force exceeds the limiting force of friction. If ω isthe maximum angular velocity of wheel when boy just slips,
µ mg = mrω2
i.e., µ =r
gω2
134 Mechanics
given, ω =2 10
60 3π π× = rad/sec
and r = 7m
we get µ =π3
79 8
0 782F
HGIKJ =×
..
µ = 0.78 Ans.
Q. 54. A 40 kg. mass hanging at the end of a rope of length l, oscillates in a vertical planewith an angular amplitude θo. What is the tension in the rope when it makes an angle θ withthe vertical. If the breaking strength of the rope is 80 kg, what is the maximum amplitude withwhich the mass can oscillate without breaking the rope?
Ans. The figure shows the position of the oscillating mass at angular amplitudes θ0 andθ. The velocity v at θ (after a desent of distance NM from extreme position) is given by
v2 = 2 gNM
But NM = SM – SN = l cos θ – l cos θ0
then v2 = 2gl (cos cos )θ θ− 0
If T is tension in the rope at position Q. Then
T = mgmv
lcosθ +
2
Putting the value of v2
T = mgml
glcos × (cos cos )θ θ θ+ −2 0
or T = 3 2 0mg mgcos cosθ θ−
Putting m = 40 kg
T = 3 × 40 × g cos θ – 2 × 40 × g × cos θ0
or T = 40 3 2 0g cos cosθ θ−
The maximum tension in the rope occurs when θ = 0 (mean position), i.e.,
Tmax = 40 3 2 0g [ cos ]− θ
But given Tmax = 80 kg
So, 80 g = 40 3 2 0g [ cos ]− θ
i.e., 2 cos θ0 = 1
or cos θ0 =12
i.e., θ0 = 60° Ans.
Q. 55. A body slides down from the top of a hemisphere of radius r. The surfaces of blockand hemisphere are frictionless. Show that the height at which body loses contact with the
surface of the sphere is 23
r.
R
θθ0
Tl
N
M
Pθ
m g cos θm g
S
Q
Fig. 25
Dynamics of Circular Motion and the Gravitational Field 135
Ans. Let height at which body loses contact be ‘h’, then
OR = h and ∠QOR = θIf at position Q, velocity is v, then taking limiting case of equilibrium for circular motion
mg cos θ =mv
r
2...(1)
where cos θ =OROQ
= hr ...(2)
But the velocity ‘v’ at Q, in vertical desentPR = (r – h), is given by
12
2mv = mg (r – h)
or v2 = 2g (r – h) ...(3)
putting the values of cos θ from Eq. (2) and value of v2 from Eq. (3) into Eq. (1), we get forcontact upto point Q
mghr
=m g r h
r× ( )2 −
which gives 3h = 2r
or h =23
r Ans.
Q. 56. The radius of curvature of a railway track at a place is 800 meter and the distancebetween the rails is 1.5 meters. What should be the elevation of the outer rail above the innerone for a safe speed of 20 km per hour?
Ans. Suppose the outer rail is elevated by an angle θ with respect to inner rail i.e., his height to which outer rail is raised vertically, then
tan θ = sin θ θ θ= hd
h or tan =1.5
(as is small)
The banking angle is related to safe speed by
tan θ =vrg
v2
20 where km/hr =10018
m/sec=
putting values, we get
h1 5.
=100 100
18 18 800 9 8×
× × × .
or h = 0.59 cm Ans.
Q. 57. A heavy particle at the end of a tight string (length 20 cm), the other end of whichis fixed is allowed to fall from a horizontal position of the string. When the string is vertical,it encounters an obstruction at its middle point and the particle continues its motion in a circleof 10 cm radius. Find the height which the particle will attain before the string slackens.
θ
θ
O
hm g
P
QR
m g cos θ
r
Fig. 26
Fig. 27
136 Mechanics
Ans. Let OP be the string in horizontal position with end O fixed, and the particle fallsfrom position P. Then the velocity at position Q when string is vertical is given by
v2 = u2 + 2gh where u = 0
or v2 = 2 × 980 × 20 = 39200 ...(1)
Let K be mid point of OQ, QM–circular path whichparticle will describe with K as centre, M–the positionwhere string slackens.
r = KQ = KM
At M, the component of mg along MK providesthe centripetal force, i.e.,
mg cos θ =mv
rv1
2
1; = tangential velocity at M
or g cos θ =vr12
i.e., v1 = rg cosθ
again cos θ =KLKM
QL QKQK
= − = −h rr
we get v1 = rg h rr
( )−
Considering the journey of particle from Q to M, we have
v12 = v2 – 2gh
putting the value of v1 get
rgh r
r−F
HGIKJ = v2 – 2gh
or g(h – r) = v2 – 2gh
which gives 3gh = v2 + rg
or h =v rg
g
2
3+
or h =39200 980 10
3 980+ ××
h = 16.6 cm Ans.
Q. 58. A sphere of mass 200 gm is attached to an inextensible string of length 130 cmwhose upper end is fixed to the ceiling. The sphere is made to describe a horizontal circle ofradius 50 cms.
(1) Calculate the time period of one revolution.
(2) What is the tension in the string.
Ans. Given OP = l = 130 cm, PC = r = 50 cm
r = l sin θThe vertical component of tension balances the
Fig. 28
Dynamics of Circular Motion and the Gravitational Field 137
weight and horizontal component provides thecentripetal force; Thus
mg = T cos θ
andmv
r
2= T sin θ
Dividing we get
tan θ =vrg
2
or v = rg tanθ ...(1)
again tan θ =CPCO OP CP2 2
=−
=−
=r 50
130 50
5122 2( ) ( )
Putting values of r, g, tan θ in Eq. (1) we get
v = 50 980 512
3506
× × = cm/sec.
The time period of one revolution,
T =2 2 2πω
πω
π= =. rr
rv
T = 2 50350 6
2 2π ×/
.= sec. Ans.
(2) Using the condition for vertical equilibrium, T =cosmg
θ
where cos θ =12
25 1441213+
=
then T =200 980
12 132 12 105×
/. ×= dynes Ans.
Q. 59. A nail is locked at a certain distance vertically below the point of suspension of asimple pendulum. The pendulum bob is released from a position where the string makes anangle of 60° with the vertical. Calculate the distance of nail from the point of suspension suchthat the bob will just perform revolution with the nail as centre. Assume the length ofpendulum to be one meter.
Ans. Let SP be the initial position with S as point of suspension then,
SR = PS cos 60 = PS2
° = 12
and RQ = SQ – SR = SP – SR
= 1 12
12
− = m
Fig. 29
138 Mechanics
Now, bob comes from P to Q descending vertically bydistance RQ.
Using v u gd2 2 2= + for this journey with u = 0, g = 9.8
m/sec2, S = 12
m
we get v2 = 0 + 2 × 9.8 ×12
9 8= . ...(1)
We know that velocity of the particle at bottom of avertical circle, so that the particle may successfully describevertical circle, is given by
v = 5rg ; r is the radius of circle.
or v2 = 5 × r × 9.8 ...(2)Equating Eq. (1) and (2), we get
9.8 = 5 × r × 9.8
or r =15
0 2= . meter = NQ
i.e., RN = RQ – NQ= 0.5 – 0.2 = 0.3 m
∴ SN = SR + RN= 0.5 + 0.3 = 0.8 m
SN = 0.8 m Ans.
Q. 60. A particle of mass 0.2 kg is moving inside a smooth vertical circle of radius r =50 cm. If it is projected horizontally with velocity v = 4 m/sec from its lowest position, findthe angle θ at which it will loose contact with the circle.
Ans. Situation is shown in figure 31.Here m = 0.2 kg, r = 50 cm = 0.5 m, g = 9.8 m/sec2
Let h = height from lowest position where particle loses contact with vertical circle.θ = angle with vertical and v = velocity at Q then, component of mg along QO = mg cos θ.
centrifugal force =mv
rmg
2= T + cos θ
But T = 0 at Q.
So, mg cosθ =mv
r
2...(1)
where cosθ =ROOQ
= −h rr
Again, velocity at Q is, from v2 = u2 – 2gs, with u = 4 m/sec
v2 = 16 – 2 × 9.8 × h ...(2)
putting value of cos θ in (1), we get
r
R
N
P
1m
S
60°
Q
Fig. 30
Dynamics of Circular Motion and the Gravitational Field 139
mgh r
r−F
HGIKJ =
mvr
2
or g (h – r) = v2
Substituting the value of v2 from Eq. (2) we getg (h – r) = 16 – 2 × 9.8 × h
or 9.8 (h – 0.5) = 16 – 2 × 9.8 × hor 9.8 h – 4.9 = 16 – 19.6 hwhich gives h = 0.71 m
then cos θ =h r
r− = − =0 71 0 50
0 500 42. .
..
or θ = cos–1 (0.42) Ans.
Q. 61. A smooth table is placed horizontally and an ideal spring of constant k = 1000nt/m and unextended length of 0.5 m has one end fixed to its centre. The other end is attachedto a mass of 5 kg which is moving in a circle with constant speed, 10 m/sec. Find
(1) The tension in the spring.(2) Extension of spring beyond its normal length.Ans. Let l = unextended length of spring
= 0.5 mx = extended length of spring
then, extension in the spring = (x – l)At position P of ball, force on the spring towards centre O, is
F = k × extension = centripetal force
=mv
x
2...(1)
Butmv
x
2=
5 10 5002×x x
= ...(2)
and Spring force F = k (x – 0.5)
= 1000 × (x – 0.5)
then Eq. (2) gives
500x
= 1000 x – 1000 × 0.5
or500
x= 1000 x – 500
or 500 = 1000 x2 – 500 x
or 2x2 – x – 1 = 0
θm g cos θ
m gO
r
QR
v = 4m /s
h
T
Fig. 31
Fig. 32
140 Mechanics
which gives x = 1, x =− 12
; but negative elongation is not possible,
So, x = 1, Hence tension in string
So, F = T = mv
x
2 5 10 101
500= =× × Newton
and Extension (x – l) = (1 – 0.5) = 0.5 meter
F = 500 Newton
Extension = 0.5 meter Ans.
Q. 62. A string with a ball is held horizontally as shown in the figure 33. A nail locatedat a distance ‘d’ vertically below the point of suspension. Show that ‘d’ must be at least 0.61l. (l being the length of the string) if the ball is to swing completely around a circle centeredon the nail.
Ans. Ball falls under gravity from P reaches M (verticaldescent = OM = l) then it has to swing completely aroundthe circle with centre at C (nail) of diameter MN = h.
From v2 = u2 + 2gh, velocity of ball at M is given byv2 = 0 + 2gl ...(1)
Let velocity of ball at N be v1. Considering the ascentMN and using
v2 = u2 + 2hg
where u = v, v = v1, h = MN, g = –g,
we have v12 = v2 – 2gh or v1
2 = 2gl – 2gh ...(2)
For successful completion of circle, condition at N is,
Centrifugal force = weight
i.e.,mvh
12
2/= mg or v
gh12
2= ...(3)
Equating Eqs. (2) and (3) we get
2gl – 2gh =gh
h l h2
4 4 or = −
So MN = h = 45
0 8l
l= .
then O C = d = OM – CM = l – 0.4 l
∴ d = 0.6 l Ans.
UNSOLVED PROBLEMS
1. Show that a centripetal acceleration acts on a particle moving on a circular path.
2. Explain the statement “Centrifugal force is a pseudo force”.
3. Explain the principle of cream separating machine.
4. If a car is suddenly turned to left, a passenger of car strikes with right wall, explain.
Fig. 33
Dynamics of Circular Motion and the Gravitational Field 141
5. A passenger in spaceship uses spring clock-give reason.6. The speed of a body is constant. Can it have a path other than a straight line or
circle.7. Why are tracks banked at sharp turns. An automobile overturns when going too
fast around a curve. Which wheel leaves the ground first.8. The handle of the carpenter’s screw driver is much thicker then the handle of a
watch makers screw-driver. Explain the reason.9. A simple pendulum with a bob of mass m swings with an angular amplitude of 40°.
When its angular displacement is 20°, then the tension in the string is greater thanmg cos 20°. Is the statement correct.
10. As astronaut in spaceship feels weightlessness. Explain why ? If a satellite orbitsearth at an altitude of 100 km. Find its time period of revolution in circular orbit.
11. A particle of mass m rotates in a circle of radius a, with a uniform angular speedω. It is observed from a frame rotating about the z-axis with a uniform angularspeed ω0. Find the centrifugal force on the particle.
12. A smooth circular tube is held firmly in a vertical plane. A particle which can slideinside the tube is slightly displaced from rest at its highest position in the tube.Find the pressure between the tube and the particle in terms of its mass ‘m’ andthe angular displacement θ from its highest position.
13. Obtain expression for the orbiting velocity of artificial satellite.14. A satellite is moving in a circular orbit round the earth at a distance of 5R from
its centre. Calculate its orbital speed and time period.15. Find an expression for the total energy of a satellite of mass ‘m’ moving round the
earth in circular orbit of radius r.16. Write short note on
(a) Satellite launching (b) Escape velocity (c) Geo-stationary satellite.17. In an orbit of radius 1.5 × 1011 meter, earth completes one revolution in 365 days
around the sun. Calculate mass of sun (G = 6.67 × 10–10 nt-m2/kg2).18. A motor cyclist with the motor cycle has a mass of 250 kg and travels round a curve
in a road of 36 metre radius. Snow and ice on the road reduce the coefficient offriction between the tyres and road to 0.2. If the curved road is banked to 15° inmotor cyclist’s favour; Find (a) maximum velocity attainable without slipping (b) theangle, the rider must make with the road surface at this velocity assuming thatrider and motorcycle remain in one plane.
19. A train is travelling at 64 km/h and the diameter of one of the wheels of the engineis 1.5 m. Find the velocities of the two points on this wheel. Which are at a heightof 1.2 m above the ground.
20. A fighter plane flying in the sky dives with a speed of 360 km/hr in a vertical circleof radius 200 m, weight of the pilot sitting in it is 75 kg. Find the value of forcewith which pilot presses his seat when aeroplane is (a) at highest position (b) atlowest position.
21. The moon revolves around the earth in a circle of radius 3.8 × 106 m and requires27.3 days to make a complete revolution. What is the acceleration of the moontowards the earth.
142 Mechanics
4.1 WORK
Whenever a force acting on a body produces a change in the position of the body, workis said to be done by the force. If there is no change in the position of the body, work is notdone. We may exert a large force on a wall, but if the wall remains intact in its presentposition then we have not done any work. If a coolie having a heavy box on his head isstanding at a fixed place, he is not doing any work. Work is said to be done only when thereis a displacement in the direction of the force.
Work is measured by the product of the applied force and the displacement of the bodyin the direction of the force, that is
Work = force × displacement in the direction of the force.If a force F acting on a body produces a displacement ∆s in the body in the direction of
the force (Fig. 1a), then the work done by the force is given byW = F × ∆s
If the force F is making an angle θ with the direction of displacement of the body(Fig. 1b), then the work done is
W = F cos θ × ∆s,because F cos θ is the component of F in the direction of displacement.
∆s
F
F
∆s
θF cos θ
( )a ( )b
Fig. 1
If θ = 90°, then cos θ = 0 and so W = 0.
This means that if the displacement is perpendicular to the force, no work is done.When a satellite revolves around the earth, the direction of the force applied by the earthis always perpendicular to the direction of motion of the satellite. Hence no work is doneon the satellite by the centripetal force.
142
4
Work, Energy and Momentum 143
Force and displacement both are vector quantities but work is a scalar quantity. The unitof work is ‘joule’. If a force of 1 newton produces a displacement of 1 meter in the directionof the force, the work done is called 1 joule:
1 joule = 1 newton × 1 meter
4.2 POWER
The rate of doing work by an agent or a machine is called power:
Power =Worktime
If W is the work done by an agent in t second, then his power P is given by
P =Wt
Since the unit of work is joule, the unit of power will be ‘joule/second’. This is called‘watt’. If 1 joule of work is done in 1 second, the power is 1 watt.
1 watt = 1 joule/second
Another unit of power is ‘horse-power’
1 horse-power = 746 watts
The power of a normal person is from 0.05 to 0.1 horse power. If an agent exerts a forceF in the direction of motion, then
P =W F Ft
rt
v= =
where v = r/t
4.3 WORK IN STRETCHING A SPRING
When a spring is stretched slowly, the stretching force increases steadily as the springelongates, i.e., the force is ‘variable’. Let one end of the spring be attached to a wall its lengthbeing along the x-axis. Let the origin x = 0, coincide with the free end of the spring in itsnormal, unstretched state. Let the spring be stretched through a distance x by applying aforce Fapp at the free end. The spring, on account of its elasticity, will exert a restoring forceF on the stretching agent given by (within elastic limit)
F = –kx; (Hooke’s law)
Where k is the force-constant or stiffness of thespring. The minus sign indicates that the restoringforce is always opposite to the displacement x. Sincethe restoring force F is equal and opposite to the appliedforce Fapp; the latter is given by
Fapp = –F = kx
The work done by the (varying) applied force Fapp.in the displacement from x = 0 to x = x is
W = F F→ →
=
=
=
=
=app
x
x x
app
x
x x
d x dx.
0 0
x
x = 0
F a p p
Fig. 2
144 Mechanics
∴ Angle between F and → →
app d x is zero.
= kx dx kx
kxx x
0
2
0
2
212 =
=
Similarly if the spring is stretched so that its free end moves from x1 to x2, the workdone is
W = Fapp
x x
x x
x
x
x
x
dx k x dx k x
=
=
= =
1
2
1
2
1
22
2
=12
12
122
212
22
12kx kx k x x− = −
4.4 ENERGY
The capacity of doing work is called energy. Energy has various forms such as mechanicalenergy, heat energy, light energy, magnetic energy, sound energy, chemical energy etc.
Mechanical energy has two forms: Kinetic energy and potential energy.
4.5 KINETIC ENERGY
The kinetic energy of a moving body is measured by the amount of work which has beendone in bringing the body from the rest position to its present position, or which the bodycan do in going from its present position to the rest position.
Let a body of mass m be in the rest position. When we apply a constant force F on thebody, it starts moving under an acceleration. If a be the acceleration, then by Newton’s secondlaw, we have
a =Fm
suppose the body acquires a velocity v in moving a distance S. According to equation v2 = u2
+ 2as (u = 0, since the body was initially at rest) we have
v2 = 2 2asm
s= × ×F
F × S = ½ mv2
But F × S is the work which the force F has done on the body in moving it a distance‘S’. It is due to this work that the body has itself acquired the capacity of doing work. Thisis the measure of the kinetic energy of the body. Hence if we represent kinetic energy of abody by K, then
K = F × S = ½ mv2
Thus, the kinetic energy of a moving body is equal to half the product of the mass (m)of the body and the square of its speed (v2). In this formula, v occurs in the second powerand so the speed has a larger effect, compared to mass, on the kinetic energy. It is becauseof this reason that the bullet fired from a gun injures seriously inspite of its very small mass.
Work, Energy and Momentum 145
Now, suppose a body is initially moving with a uniform speed u. When a force is appliedon it then its speed increases from u to v in a distance S. Now, we have
v2 = u2 + 2as
v2 – u2 = 2 × (F/m) × s
F × S = ½ mv2 – ½ mu2
But F × S is the work done W on the body by the force. Hence
W =12
mv mu2 212
−
According to definition, W = 12
mv mu2 212
−
is the increase in the kinetic energy of the
body. Thus when a force acts upon a moving body, then the kinetic energy of the bodyincreases, and the increase is equal to the work done.
4.6 POTENTIAL ENERGY
Bodies can do work also by virtue of their ‘position’ or ‘state of strain’. The energy ina body due to its position or state of strain is called the ‘potential energy’ of the body. Forexample, the water at the top of a water-fall can rotate a turbine when falling on it. Thewater has this capability by virtue of its position (at a height). Similarly, a wound clock-springkeeps the clock running by virtue of its state of strain. Thus water and wound spring bothhave potential energy, the former has ‘gravitational’ potential energy and the later has‘elastic’ potential energy.
A system of electric charges also has potential energy which is called ‘electrostatic’potential energy. If two opposite charges are taken further away from each other, the workdone against their mutual attraction is stored in the form of potential energy, so that thepotential energy of the system increases. In case of similar charges, the potential energyincreases when they are brought closer.
The potential energy of a body is measured by the amount of work which has been donein bringing the body from its zero-position to the present position or which that body can doin going from its present position to the zero-position. Therefore, if a body is taken, underthe action of a force, from one position to the other, then the work done is stored in the bodyin the form of its potential energy (provided there is no loss of work against friction, etc.).
4.7 GRAVITATIONAL POTENTIAL ENERGY
A mass held stationary above the ground has energy, because, when released, it can raiseanother object attached to it by a rope passing over a pulley, for example. A coiled spring alsohas energy, which is released gradually as the spring uncoils. The energy of the weight orspring is called potential energy, because it arises from the position or arrangement of thebody and not from its motion. In the case of the weight, the energy given to it is equal tothe work done by the person or machine which raises it steadily to that position against theforce of attraction of the earth. So this is gravitational potential energy. In the case of thespring, the energy is equal to the work done in displacing the molecules from their normalequilibrium positions against the forces of attraction of the surrounding molecules. So this ismolecular potential energy.
146 Mechanics
Suppose a body of mass m is raised to a height h from the earth’s surface. In this process,work is done against the force of gravity (mg). This work is stored in the body in the formof gravitational potential energy U. Thus
U = work done against force of gravity
= weight of the body × height
= mg × h = mgh
4.8 WORK-ENERGY THEOREM
Let us consider a body of mass m acted upon by a resultant accelerating force F alongthe x-axis. Suppose as the body moves from a position x1 to a position x2 along the x-axis,its velocity increases from v1 to v2. The work done by the force in the displacement is
W = F dxx x
x x
=
=
1
2
By Newton’s second law, we have
F = ma mdvdt
mdvdx
dxdt
mvdvdx
= = =.
v = dx/dt
∴ W = m v dvdx
dx m v dvx x
x x
v v
v v
=
=
=
=
=1
2
1
2
= m v mv mvv
v2
22
12
212
12
1
2 = −
The quantity ½ mv2 is defined as the kinetic energy K of the body. Thus the aboveequation may be written as
W = K2 – K1
Where K2 and K1 are the final and initial kinetic energies of the body. Thus, if ∆Krepresents the change in kinetic energy, ∆K = K2 – K1, then we have
W = ∆K
Thus we conclude that “whenever a body is acted upon by a number of forces such thatthe resultant force is not zero, then the work done by the resultant force (whether constantor variable) is equal to the change in the kinetic energy of the body. This is known as thework-energy theorem.
If the kinetic energy of the body decreases, the work done on it by the resultant forceis negative, or the work is done by the body against the resultant force. Therefore, a movingbody is said to have a store of (kinetic) energy in it, which it loses in doing work. Hence thekinetic energy of a moving body is defined as the work it can do before coming to rest.
The units of kinetic energy and of work are the same. Kinetic energy, like-work, is ascalar quantity.
Work, Energy and Momentum 147
When several forces act upon the body, the work done by the resultant force is thealgebraic sum of the works done by the individual forces i.e.,
W = W1 + W2 + -----------------
Hence the work-energy theorem may also be written as
W1 + W2 + ------------- = ∆K
4.9 SIGNIFICANCE OF THE WORK-ENERGY THEOREM
The work-energy theorem is useful for solving problems in which the work done by theresultant force is easily computed and in which we are interested in finding the particle’sspeed at certain positions of greater significance, perhaps is the fact that the work-energytheorem is the starting point for a sweeping generalization in physics. It has been emphasizedthat the work-energy theorem is valid when W is interpreted as the work done by theresultant force acting on the particle. However, it is helpful in any problems to computeseparately the work done by certain types of force and give special names to the work doneby each type. This leads to the concepts of different types of energy and the principle of theconservation of energy.
4.10 CONSERVATIVE FORCE: FIRST DEFINITION
A force acting on a particle is conservative if the particle, after going through a completeround trip, returns to its initial position with the same kinetic energy as it had initially.
v
A
( )a
v = 0
( )b
Fig. 3
When we throw a ball upward against gravity, the ball reaches a certain height comingmomentaily to rest so that its kinetic energy becomes zero. Then it returns to our handunder gravity with the same kinetic energy with which it was thrown (provided the air-resistance is assumed zero). Thus the force of gravity is conservative. Similarly, when a blockis moved from a position A on a horizontal plane with a velocity v so as to compress on aspring (Figure 3a). It is first brought to rest by the elastic (restoring) force of the spring andloses all its kinetic energy (Figure b). Then the compressed spring re-expands and the blockmoves back under the elastic force gaining kinetic energy. As the block returns to its initialposition A, it has the same velocity v, and hence the samekinetic energy, as it had before (provided the horizontal planeis frictionless and the spring is ideal). Thus the elastic forceexerted by an ideal spring is conservative. The electrostaticforce is also conservative.
A force is termed to be conservative if the work done byit on a particle in moving it from one point to another inspace, depends only on the location or position of the particleand not on the path followed by the particle.
F N
θ A
F
CM
F→
Fig. 4
148 Mechanics
Imagine a particle moving from point M to N under action of force F→
. There may be
three alternative paths, Path MAN, Path MBN, Path MCN. If the force F→
is conservative,then
WMN = F F FM
N
M
N
M
N→ → → → → → = =. . .d r d r d r
Path MAN Path MBN Path MCN
Work done by a conservative force around a closed path
Consider the closed path MANBM. under action of a conservative force F→
, a particlemoves from M to N along path MAN and returns back to M, along the path NBM.
Now, work done by the force F→
in moving the particle from M to N along path 1, is
WMN = F T TM
N
N M
→ → = −.d r
Where TN = K.E. of particle at N and TM = K.E. of particle at M. Similarly the work
done in carrying the particle from N to M, by the force F→
is given by
WNM = F T TN
M
M N
→ → = −.d r
So, total work done by the conservative force F→
, along closed path MANBM, is
F→ → .d r = F F
→ → → → +. .d r d r
Path MANBM Path 1(MAN) Path 2(MBN)
WMANBM = TN – TM + TM – TN = 0
or F→ → .d r = 0
Thus, work done by a conservative force around a closed path is always zero i.e., thechange in kinetic energy of the particle in motion due to a conservative force, over a closedpath is zero.
Examples of conservative forcesPosition dependent forces depend on the instantaneous position of the particle or body
(not on its velocity). Central forces are position dependent and in addition, are directedtowards a fixed center. Position dependent forces are (a) Gravitational force (b) Electrostaticforce (c) Elastic force. Most of the position dependent forces are conservative in nature.
4.11 A CENTRAL FORCE IS CONSERVATIVE
A force is termed central force if it acts on a particle in such a way that it is directedtowards or away from a fixed point and its magnitude depends only on the distance of particlefrom the fixed point.
Fig. 5
Work, Energy and Momentum 149
It may be observed, that central forces are conservative. Consider a central force F→
, bydefinition:
F→
= F Ss
Where Fs is function of position ‘s’ only and S is unit vector along S→
. The above forceis directed away from the fixed point and acts on a particle. Let the particle move from Mto N under action of this force. Now, total work done by the central force in this movementis,
WMN = FM
N → → .d r
Putting the value of force F→
,
WMN = F S FM
N
M
N
s sd r dr . . cos→ = θ
or WMN = F SM
N
s dAs Fs is function of s only, so its integral (say α) is also a function of S, hence
WMN = F SM
N
MN
N Msd = = −α α α
Thus, the work done by the central force depends only on the position of points M andN (not on path followed). This suggests that central force is a conservative force.
Properties of Work done by conservative force(i) It is independent of path.
(ii) It is equal to difference between final and initial values of energy function.
(iii) It is completely recoverable.
4.12 NON-CONSERVATIVE FORCE
A force is non-conservative if the work done by it on the particle moving between twopoints depends on the path followed by the particle. Thus, if a particle moves from M to N
under influence of a non-conservative force F→
, then work done along three possible paths bythe force is not same (i.e., it is path dependent). Thus
WMN = F F FM
N
M
N
M
N→ → → → → → ≠ ≠. . .d r d r d r
Path 1 Path 2 Path 3
Fig. 6
150 Mechanics
Thus, work done by a non-conservative force around aclosed path is not zero i.e.,
WMN = FM
N → → ≠.d r 0
Again, if a conservative and a non-conservative both, forcesact upon the particle and WC be the work done by conservativeforce and WM-C be the work done by non-conservative force,then
WC + WN–C = Change in K.E. of Particle = ∆T
Example of non-conservative force: Frictional forces, viscous forces etc are examplesof path dependent forces.
Potential EnergyPotential energy of a body is defined as the energy stored in body due to its position,
configuration or state of strain. Consider a particle lying in a conservative force field due toforce FC. The force FC that acts on the particle may be balanced by applications of suitableapplied force Fapp. Now the particle may be moved extremely slowly (so that no K.E. develop)by applied force against the conservative force FC. The work done in moving the particle willappear as potential energy of the particle.
Thus, potential energy is measured by the amount of work that it can do when it movesfrom referred position to some standard position. Potential energy is generally denoted byU.
Potential Energy DifferenceThe difference in potential energy between two positions of a particle is defined as the
work done by applied force on the particle in moving it from the first position to secondposition. Thus, if r2 and r1 are the final and initial position, then by definition
U(r2) – U(r1) = F F.→ → → → = −app
r
r
c
r
r
d r d r
1
2
1
2
. (as Fapp = –Fc)
U(r2) – U(r1) = −→ →Fc
r
r
d r.
1
2
If first position is at infinity (i.e., r1 = ∞) having potentialenergy zero there (i.e., U(r1) = 0), then potential energy atposition r (i.e., r2 = r) is given by
0 – U(r) = F→ →
∞ c
r
d r.2
or – U(r) = F→ →
∞ c
r
d r.2
Fig. 7
dr
Fa p p
FC
Fig. 8
Work, Energy and Momentum 151
or U(r) = Fapp→ →
∞ .d rr
Thus potential energy of a particle at a point r may be defined as the work done by theexternally applied force in moving the particle from infinity to that point.
The position at which the conservative force acting on the particle is zero, is the referenceposition with respect to potential energy.
In case of gravitational and electrostatic forces the reference position is infinity.
In case of springs the normal unstretched length is considered as reference position.
In case of earth’s gravitational force field, earth’s surface is the reference position.
4.13 RELATION BETWEEN CONSERVATIVE FORCE AND POTENTIAL ENERGY
By definition of potential energy
U(r) = – F→ →
∞ .d rr
In rectangular Cartesian system one may write
F→ = F i j kx y zF F+ +
and d r→
= i dx j dy kdz+ +Then the potential energy will be given by
U( r→
) = U (x, y, z)
i.e., U(r) = − + + + +∞ ( ).( )i j k i dx j dy kdzx y z
r
F F F
or U( r→
) = − − −∞ ∞ ∞ F F Fx
x
y
y
z
z
dx dy dz
Differentiating the above equation partially
∂∂Ux
= − = − = −F U F U Fx y zy z, ,∂
∂∂∂
In view of these relations, the force F→
can be expressed as
F→
= − − − ix
jy
kz
∂∂
∂∂
∂∂
U U U
or F→
= − + +
ix
jy
kz
∂∂
∂∂
∂∂
U
152 Mechanics
or F→
= −∇→
U
where ∇→
= + + + ix
jy
kz
∂∂
∂∂
∂∂
Vector operator ∇→
is called ‘del’ or ‘nebla’.
Thus, a conservative force is the negative gradient of potential energy U. Only for onedimension
F = − ddxU
or U = F− dx
4.14 THE CURL OF A CONSERVATIVE FORCE IS ZERO
Now let us see what is the value of curl F for a conservative force:
A conservative force may be expressed as negative gradient of potential energy i.e.,
F→
= −∇→
U
The curl of F→
is,
curl F→
= ∇ = ∇ − ∇→ → → →
× × (F U)
= −
i j k
x y z
x y z
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
U U U
= − −
− −
+ −
iy z z y
jx z z x
kx y y x
∂∂ ∂
∂∂ ∂
∂∂ ∂
∂∂ ∂
∂∂ ∂
∂∂ ∂
2 2 2 2 2 2U U U U U U
Since, U is perfect differential, so
∂∂ ∂
2Ux y
=∂∂ ∂
2Uy x
and so on
∴ Curl F→
= − − + = ( ) ( ) ( )i j k0 0 0 0
i.e., ∇→ →
× F = 0
Thus, the curl of a conservative force is zero.
Work, Energy and Momentum 153
The Total Mechanical Energy is conserved in a conservative force field
Consider a particle to be displaced by a conservative force F→
from a position M to N.By work-energy theorem, the work done by force is equal to the increase in K.E. of particle,i.e.,
WMN = F T TM
N
N M
→ → = −.d r
one can express the same work done by decrease in potential energy of the particle, i.e.,
WMN = − = −→ →F U U
M
N
M N.d r
Equating above,
or TN – TM = UM – UN
TN + UN = TM + UM
which suggests that the sum of kinetic and potential energy of a particle under action ofconservative force remains constant, i.e.,
T + U = Total Mechanical Energy
(Constant in conservative force field)
Sometimes it is convenient to call the quantity E = T + U as the energy function. Thisenergy function is invariant with respect to change in time.
Motion of a Body near the Surface of the EarthLet us consider a body of mass m be situated at a height h in the rest position above
the earth’s surface. Suppose that its potential energy is zero at the earth’s surface. Let thex direction be normal to the surface of the earth and directed upwards.
If the body starts to fall down and at any instant its height above the earth’s surface isx, then work done by the gravitational force – mg on the body is
W = ( ) ( ) ( )− = − − = − mg dx mg x h mg h xh
x
Where the direction of force is opposite to x.
This amount of work done on the body will increase its kinetic energy
W = ½ mv2 ∴ initial velocity = 0
∴ ½ mv2 = mg (h – x)
or mgh = ½ mv2 + mgx ...(i)
But a body situated at a height x has the capacity to work mg × x, hence the potentialenergy of the body at this height is mgx. Thus, initially, the body has no kinetic energy, butonly potential energy mgh.
∴ Initial total energy = initial K.E. + initial P.E.
= 0 + mgh = mgh
At any height x, total energy = ½ mv2 + mgx = mgh (from Eq. (i))
154 Mechanics
At earth surface (x = 0), total energy = mgh + 0 = mgh
Hence, the total energy of a freely falling body is the same initially, finally and in anymiddle position.
In other words, the sum of potential and kinetic energies of a freely falling body remainsconstant throughout the motion.
4.15 LINEAR RESTORING FORCE
When a force acting on a particle is directly proportional to the displacement of theparticle from a certain fixed point (called mean position) and is directed oppositely to thedisplacement, it is called linear restoring force. Thus
F→
α − r→
or F→
= −kr r
Where r→
= position vector of particle at any instant = r r
Now r→
= r r = ix jy kz+ +
or F→
= −→
k rLinear restoring force is a conservative force: For a conservative force
∇→ →
× F = 0
and F→
= − = − + +→
k r k ix jy kz( )
so ∇→ →
× F = ∇ −→ →
× ( )k r
=
i j k
x y zkx ky kz
k
i j k
x y zx y z
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
− − −
= − = 0
Thus, linear restoring force is a conservative force.
Massless Horizontal Spring – One End FixedFor small displacement, an ideal compressed or stretched spring produces a linear restoring
force (due to elastic properties of the spring in accordance to Hook’s law). Let there be amassless spring whose one end is rigidly attached to a wall and other end be tied to a blockwhich could slide on a horizontal table having no friction.
Normal PositionNo force is produced in the spring. Let us regard the origin at the normal position of
block.
Work, Energy and Momentum 155
Stretched Position: If an external force stretches the spring slowly, then work will bedone on the system against the restoring force which constitutes the potential energy of thesystem. The linear force produced in the spring is in the direction opposite to displacement;i.e.,
F→
= −kxi
where k = force constant of spring (or spring factor) and x = displacement.
Now, potential energy, by definition, is
U = − = − = − −→ → → F F. . cos ( ).d r dx kx dx
x x x
0 0 0
θ
or U = kxdx kxx
0
212 =
∴ U = ½ kx2
Which gives the potential energy stored in stretched spring suffering extension x. If ‘a’is maximum stretching (or elongation) of spring
Umax =12
2ka ...(1)
Motion after release: If body be stretched up to maximum stretching and then releasedit moves under linear restoring force towards origin.
Initial total energy just at the instant of release = ½ ka2
Total energy at position x during return = 12
2kx + ½ mv2
(where v = velocity in position x)
Using energy conservation principle,
½ ka2 = ½ kx2 + ½ mv2
or ½ mv2 = ½ k (a2 – x2)
or v0 = ± −km
a x( )2 2
the velocity will be maximum in mean position (x = 0), is given by
∴ vmax = ± akm
...(2)
Compressed Position: When spring is compressed then the displacement is –x and,thus, the linear restoring force produced is
F→
= − −k ix.( )
= i kx
156 Mechanics
The spring linear force in compressed spring is in +x direction. The maximum energy(Umax) and maximum velocity (Vmax) gained in mean position will be given by Equation (1)and Equation (2) respectively.
Equation of motion: Restoring force at stretching x of the spring is F = –kx. ByNewton’s law
F = m d xdt
2
2
So equation of motion is
md xdt
2
2 = –kx
ord xdt
km
x2
2 + = 0
Fig. 9
which is the differential equation of motion of spring having time period
T = 2π mk
Vertical spring in uniform gravitational field with mass attached to spring:Figure 10 (a) shows the natural length of an ideal massless, vertical spring of force constantk. When a weight mg is gently attached to the free end of spring, an elongation x0 isproduced. Figure (b) shows the force in the spring (acting upward) balances the weight mg.Thus
mg = kx0
or mg – kx0 = 0
If a further, displacement x is given to the mass, figure 10 (c) then net or effectiverestoring force
= [mg – k(x0 + x)]
= kx0 – kx0 – kx
= –kx
Work, Energy and Momentum 157
From Newton’s law
Force = m d xdt
2
2
thus md xdt
2
2 = –kx
ord xdt
km
x2
2 + = 0
Fig. 10
which is the equation of motion. Note that gravity has no effect in the motion which is simpleharmonic of time period given by
T = 2π mk
or frequency (n) = 1
2πkm
Potential Energy CurveIf the potential energy U of a particle is a function of position (i.e., changes from point
to point), then a graph may be plotted to show the variation of potential energy with theposition of the particle. Such a graph is known as potential energy curve. In case, if theparticle is allowed to move in one dimension only, say along x-axis, the potential energy Uwill be the function of x-coordinate only, and then the force (conservative in nature) on theparticle will be given by
F = −d
dxxU( ) ...(1)
158 Mechanics
U
EU
K
A
QP
C
B
GDY
O x x0X
x
U 0
Fig. 10(a)
Fig. 10(a) shows a possible potential energy curve for one dimensional motion. The slope(dU/dx) of this curve at any point gives the force (F = –dU/dx), acting on the particle placedat that point. The slope (dU/dx) is positive, whenever U increases with the increase in x (e.g.,for the part AB and CD of the curve) and negative whenever U decreases with increase inx (e.g., for the part GA and BC). Therefore the force F is negative or directed to the leftwhenever the potential energy is increasing with x and positive or directed towards rightwhenever the potential energy is decreasing with x. This means that the force acting on theparticle at any point tries to bring the particle to the region of lower potential energy.
At points A, B, C and D, the potential energy U is minimum or maximum, the value ofdU/dx is zero. Therefore, if a particle is placed at such a point with zero velocity, it willexperience no force (F = –dU/dx = 0) and so it will remain at rest. These points of the curveare known as positions of equilibrium.
B, and D are the points corresponding to maximum potential energy. A particle at restat such a point will remain at rest. However, if the particle is displaced even the slighestdistance from this point the force F(x) = –dU/dx will tend to push the particle farther awayfrom the equilibrium position. Hence the points B and D are the positions of unstableequilibrium.
A and C are the points corresponding to minimum potential energy. A particle at rest atsuch a point will remain at rest. If the particle is displaced slightly in either direction bygiving a little energy the force will tend to take it back towards the equilibrium position.Hence the points A and C are the positions of stable equilibrium.
In case, if U is constant in a region, then the slope dU/dx and hence the force actingon a particle placed at any point of that region is zero. In such a region, if a particle isdisplaced from one point to the other, it will remain there without experiencing any restoringforce. So we call the region of constant potential energy as the region of neutral equilibrium.For example, a book placed on a table anywhere remains in equilibrium.
Bounded region or potential well. Now let us discuss about the point A (or C) of stableequilibrium and the region near to it, which is of great interest in physics. A particle, placedat the point A at rest, can be displaced from this point by giving some energy so that the total
Work, Energy and Momentum 159
energy E of the particle becomes more than minimum value of the potential energy U0. Thetotal energy E = K + U of the particle has been indicated.
NUMERICALSQ.1. An object of mass 5 kg falls from rest through a vertical distance of 20 m and reaches
a velocity of 8 m/s. Calculate the work done by push of air exerted on the object.Solution. The forces acting on the body are:(i) the weight, acting downward,
(ii) the air push, acting upward.The body is getting accelerated downward under influence of resultant force (mg – P),
P being the push of air.Now work done by resultant force in 20 m fall
= (mg – P) × 20 Joules
Gain in KE in this fall =12
122
212mv mv−
=12
8 12
0 322m m m( ) ×− = Joules
as, work done = change in K.E.
∴ (mg – P) × 20 = 32 m
or mg – P =3220
m; or P = 5 × 9.8 –
32 520×
or P = 41 Newton
So, work done by push of air (force)
= –41 × 20 = –820 Joule
(–ve sign refers to expense of work).
Q.2. A man pulls a block weighing 10 kg upto a distance of 10 meter on a horizontal roughsurface (coefficient of kinetic function = 0.20) at a constant speed. Calculate the work done bythe man if the pull exerted by him makes an angle of 30° with the horizontal.
Solution. In the figure various forces are shown
where P = pull exerted by man,
mg = weight of block,
µN = frictional force.
Then; work done by the man
W = P→ →
. r = Pr cos 30° (r = displacement)
As block moves with uniform speed; net force is zero
∴ P cos 30° – µN = 0 ...(i)
and also, N + P sin 30° – mg = 0
which gives, N = mg – P sin 30° ...(ii)
Fig. 11
160 Mechanics
Putting this value in (i)
P cos 30° – µ mg + µP sin 30° = 0
or P (cos 30° + µ sin 30°) = µ mg
or P =µ
° + µ °=
+
mgcos sin
. × × .
/ . ×30 300 2 10 9 8
3 2 0 2 12
⇒ P =19 600 966
20 3..
.= Newton
So, work done = Pr cos 30° = 20.3 × 10 3 2/= 175.803 Joule
Q.3. A block initially sliding on a rough horizontal surface with velocity of 10 m/sec stopsafter travelling a distance of 70 m. Find the coefficient of kinetic friction between surface andthe block.
Solution. Initial K.E. =12
12
10 502 2mv m m= =× Joules
Frictional force acting on block = µN = µ mg
Work done against frictional force in moving 70 m
= µ mg × 70 Joule
This work done must be equal to the initial K.E. possessed by the block
i.e., µ mg × 70 = 50 m
or µ =50
7050
70 9 80 073
g= =
× ..
Q.4. A body of mass 40 kg climbs a vertical rope 10 meters in length in 15 sec withconstant velocity. Calculate:
(i) Work done by the body,
(ii) Power output.
Solution. Velocity of the body =1015
0 667= . m/sec
The body has to exert force to support his own weight.
So force exerted = mg = 40 × 9.8 = 392 Newton
Thus, Work = force × displacement
= 392 × 10 = 3920 J
and Power =W Wattt
= =392015
261 33.
Q.5. If a net force of 10 Newton acts on a body, initially at rest, of mass 30 kg; then whatis the
(i) Work done by the force in the third second.
(ii) Instantaneous power exerted by force at the end of third second.
Work, Energy and Momentum 161
Solution. (i) The acceleration of the body, a = ForceMass
i.e., a =1030
m= 13
2/sec
Displacement in 1 sec.
S = ut at+ 12
2
= 0 12
13
1 16
2+ =. ( ) meter
Work done at the end of 1 sec.
W = F × displacement
= 10 16
1 667× .= Joule
Displacement in 2 sec.
S = ut at+ 12
2 = 0 12
13
2 23
2+ =× ( ) meter
Work done at the end of 2 sec = W = F × displacement
= 10 23
6 667× .= Joule
Displacement in 3 sec.
S = 012
13
396
2+ =× × .m
Work done at the end of 3 sec;
W = F × displacement
= 10 96
15× = Joule
∴ Work done in the 3rd sec = Work done in 3 sec – Work done in 2 sec.
= (15 – 6.667) = 8.33 Joules
(ii) Velocity at the end of 3rd sec.
v = u + at
or v = 0 13
3 1+ =× m/sec.
So, instantaneous power,
P = F × v
= 10 × 1 = 10 watt.
Q.6. When a 500 kg stone is pushed over a level horizontal rough surface, by a horizontalforce of 100 Newton, its velocity is found to be uniform. Calculate the work done in followingcases, assuming frictional force to be constant:
162 Mechanics
(a) Velocity remains constant.
(b) Velocity increases from 0 to 1 m/sec.
(c) Velocity decreases from 1 to 0 m/sec. in a distance of 5 meter.
Solution. (a) For uniform velocity over rough surface
Fapplied = Ffrictional
∴ Work done in 5 meter distance against frictional force;
W = 100 × 5 = 500 Joule
(b) Here, the applied force increases the velocity from 0 to 1 m/sec. in addition toovercoming the friction.
Using v2 = u2 + 2aS
= 0 + 2a × 5
So a =1
10 m/sec2
Force required for this acceleration,
F = ma = 500 × 1
10 = 50 Newton
Total force = 100 + 50 = 150 Newton
So, Work done in 5 meter distance
= 150 × 5 = 750 Joule
(c) In this case velocity decreases from 0 to 1 m/sec.
So, v2 = u2 + 2aS
0 = 1 + 2a × 5
a = − 110
m/sec2 (retardation)
So, retarding force = ma = – 500 × 1
10 = –50 Newton
∴ Total force acting = 100 – 50 = 50 Newton
So, Work done = 50 × 5 = 250 Joules
Q.7. A small block of mass m slides along thefrictionless track as shown in the figure 12. Calculatethe height h at which the mass must be released on thetrack to be able to go round the rack of radius R.
If h = 5R, what is the reaction force exerted by thetrack on the block when it is at (i) point A (ii) point B(iii) point C.
Solution. Motion is under gravitational force,which is conservative, so mechanical energy isconserved i.e., loss in P.E. in coming from P to C =Gain in K.E.
Fig. 12
Work, Energy and Momentum 163
∴ mg(h – 2R) =12
2mv ...(i)
[where v = velocity at C]
Now velocity at C should be such that centrifugal force at C = weight of block forsuccessful completion of loop.
i.e.,mv2
R= mg or v2 = Rg ...(ii)
From (i) and (ii)
Rg = 2g (h – 2R)
R = 2h – 4R
h =5R2
if h = 5R, velocity at A is given by
Loss in P.E. = gain in K.E.
i.e., mgh =12
mvA2
or vA2 = 2gh = 2g × 5R = 10gR
At pt. A, centrifugal force,
=mv m g
mgA2
RR
R= =× 10
10
Weight of block at A; acting down wards = mg
So, total outward force (action) = 10 mg + mg = 11 mg
Thus, reaction by track on block at A = 11 mg
Similarly, velocity of block at B will be
mg(h – R) =12
mvB2
or mg(5R – R) =12
mvB2
or vB2 = 8 gR
i.e., Centrifugal force at B;
=mv m g
mgB2
RR
R= =× 8
8
Since mg acts at 90° to centrifugal force at B; so action by block on the track = 8mg
Thus, reaction exerted by track on the block at B = 8 mg
164 Mechanics
The velocity of block at C, is given by
mg(h – 2R) =12
mvC2
or mg(5R – 2R) =12
mvC2
vC2 = 6gR
Then, centrifugal force at C =mv m g mgC
2
RR
R= =6 6
weight mg of block acts in opposite direction to centrifugal force;
So, Action force of block = 6 mg – mg = 5 mg
Thus, reaction exerted by track on the block at C = 5 mg.
Q.8. A particle slides along a track with elevated ends and a flat central part, as shownin the figure. The flat part has a length l = 2.0 meter. The curved parts are frictionless whilefor the flat part the coefficient of kinetic friction is 0.20. The particle is released at a point A,which is at a height h= 1.0 meter above the flat part of the track. Where does the particlefinally come to rest ?
Solution.
A
m g
B C
D
l
Fig. 12
P.E. of particle at A w.r.t. B = mgh
On reaching B, entire energy is kinetic; so
K.E. at B =12
2mv so; 12
2mv = mgh ...(i)
Suppose particle travels distance d in flat part before coming to rest. Thus all K.E. isused against friction, so
12
2mv = µN × distance = µ mgd (as N = mg)
or12
2mv = µmgd ....(ii)
(i) and (ii) give
mgh = µmgd
d = hµ
= =10 2
5.
meters
Work, Energy and Momentum 165
Thus particle travels B to C (2 meter) then climbs towards D (no energy loss, as pathis frictionless), then descends to C, then comes from C to B (2 meters); again climbs towardsA (no energy loss, as track is frictionless); comes back to B then travels half of BC (1 meter)and stops.
Total distance travelled on flat part = 2 + 2 + 1 = 5 meters.
Q.9. A constant force of 5 Newton acts for 10 sec. on a body whose mass is 2 kg. Thebody was initially at rest. Calculate
(a) Work done by the force,
(b) final kinetic energy,
(c) average power of the force.
Solution. (a) Acceleration, a =Forcemass
m/sec2= =52
2 5.
Now displacement is given by
S =ut at+ 12
2
= 0 12
2 5 10 2+ × . × ( )
= 125 meter
So work done by force = 5 × 125 = 625 J.
(b) The final velocity v is given by
v = u + at
v = 0 + 2.5 × 10 = 25 m/sec.
So, final K.E. =12
12
2 25 6252 2mv = =× × ( ) J.
(c) Pavg =Work done
timeWatt.= =625
1062 5.
Q.10. A rod of length 1.0 m and mass 0.5 kg fixed at one end, is initially hangingvertically. The other end is now raised until it makes an angle of 60° with respect to thevertical. How much work is required ?
Solution. Given OP = 1 m
∠POQ = 60°, m = 0.5 kg
Rod is moved aside against gravity, from position P to Q.This makes C.G. of rod to move from C to D. The verticaldisplacement of C.G.
= CD′ = OC – OD′= 0.5 – 0.5 cos 60°
= 0.25 meter.
So, work done in moving the rod against gravity.
= mg (CD′) = mg × 0.25
= 0.5 × 9.8 × 0.25 = 1.225 Joule.
Fig. 13
166 Mechanics
Q.11. A body of mass 0.5 kg starts from rest and slides vertically down a curved trackwhich is in the shape of one quadrant of a circle of radius 1 meter. At the bottom of track,speed of the body was 3 m/sec. What was the work done by the frictional force ?
Solution. The motion is under gravity. Frictionless force resist the motion and in doingso it does work.
Supposing that curved track were frictionless, the velocity of body in vertical descent of1 m is,
v2 = u2 + 2gS
= 0 + 2g × 1 or v = 2 2 9 8g = × .
v = 4.427 m/sec.
and K.E. would have been =12
mv2 = 12
× 0.5 × 19.6 = 4.9 Joules.
but available K.E. at bottom =12
× 0.5 × (3)2
= 2.25 Joules.
The difference of K.E. is 4.9 – 2.25 = 2.65 J has been spent in overcoming friction. Sowork done by frictional force = 2.65 J.
Q.12. Find the force described by the potential energy U = P rr
→ →.3
where P is constant.
Solution. As force is negative gradient of potential energy i.e.,
F = −∇U
So force, F = −∇
= −
→ → →P P. . rr r
rr3 3
δδ
=δδ
θr r
P cos3
= 2 2 23 4P cos × 1
3θr
p rr
p rr
= =→ → →
. .
Q.13. The potential energy of a particle is given by
U = 40 + 6x2 – 7xy + 8y2 + 32 z
where U is in joule and x, y and z are in metre.
Find the force acting on the particle when it is in position (–2, 0, 5).
Solution. Using the relation F = − ddrU
, we can get components of force along x, y and
z directions.
Thus Fx = − = − +ddx
x yU
12 7
Work, Energy and Momentum 167
Fy = − = + +ddy
x yU 7 16
Fz = − = −ddzU 32
In position (–2, 0, 5) we have, Fx = 24, Fy = –14 and Fz = –32. Resultant force on theparticle in given position
F = ( ) ( ) ( )24 14 322 2 2+ − + −= 42.38 Newton.
Q.14. The potential energy function for the force between two atoms in a diatomic moleculeis approximately given by
Ux =a
xbx12 6−
where a and b are the +ve constants and x is distance between atoms, determine
(a) value of x at which Ux is zero.
(b) Value of x at which Ux is minimum.
(c) Force between atoms.
(d) Dissociation energy of molecules.
Solution. (a) Ux =a
xbx12 6 0− =
ora
x12 =bx6
or x =ab
1 6/
(b) For (Ux)min; ddx x( )U = 0
orddx
ax
bx12 6−
= 0 or
− + =12 6 013 7a
xb
x
or2
6a
x=
b1
or xab
xab
61 62 2= = ,
/
or
(c) Force is –ve gradient of potential energy so;
F = − = − −
ddx
ddx
ax
bxx( )U 12 6
= − − +
12 613 7
ax
bx
or F =12 6
13 7a
xb
x−
168 Mechanics
(d) Dissociation energy (D) is changed in potential energy from its minimum value (whichoccurs at equilibrium separation) to zero (which occurs at x equal to infinity). So,
D = ( ) ( )U Uat minx x x= ∞ −
= O −
−
a
ab
bab
2 22
= − + =ba
ba
ba
2 2 2
4 2 4
Thus, dissociation energy, D =ba
2
4.
Q.15. Show that following forces are conservative;
(a) F yz i zx j xy k→
= + +
(b) F ( 2xy z ) i x j 2xz k2 2→
= + + +
Solution. (a) We know that for conservative force, ∇ =→ →
× F 0.
Substituting the value of F→
,
∇→ →
× F =
i j k
x y zyz zx xy
∂∂
∂∂
∂∂
= ( ) ( ) ( ) ( )iy
xyz
zx jz
yzx
xy∂∂
− ∂∂
− ∂∂
− ∂∂
+ ∂∂
− ∂∂
( ) ( )kx
xzy
yz
= i x x j y y k z z− + − + −
= 0
So, given force is conservative.
(b) ∇→ →
× F =
( )
i j k
x y zxy z x xz
∂∂
∂∂
∂∂
+2 22 2
Work, Energy and Momentum 169
= ( ) ( ) ( ) ( )iy
xzz
x jz
xy zx
xz∂∂
− ∂∂
+ ∂∂
+ − ∂∂
2 2 22 2
+ ∂∂
− ∂∂
+
( ) ( )kx
xy
xy z2 22
= i j z z k x x0 0 2 2 2 2− + − + −
= 0
So, force is conservative.
Q.16. Calculate the work done in following cases:
(a) Force F 2i xyj xz k2→
= + + acts on a particle and displaces it from position (2, 3, 1)to (2, 3, 4) parallel to z-axis.
(b) Force F ( 2xy z )i x j 2xzk2 2→
= + + + makes a particle to move from position (0, 1, 2)to (5, 2, 7).
(c) Force F Ax Bx2→
= + acts parallel to x-axis on a particle and moves it from x = 1 tox = 2.
Solution. (a) W = FM
N
M
N→ → = + + + +. ( ) ( )dr i xyj xz k dxi dyj dzk2 2
= 2 02dx xydy xz dz i j i k+ + = =MN
(since . . )
Since motion occurs along z-axis, so no change in x and y occurs i.e., dx = dy = 0 andwe have
W = xz dz x z dz xz2 2
3
3M
N
M
N
M
N
= =
= 23
23
4 1 423
1
43. [ ]
z = − = units .
(b) W = F F F FM
N
M
N→ → = + +. ( ).dr xi y j zk ( )dxi dy j dzk+ +
= F F FM
Nxdx ydy zdz+ +
= (2 )M
Nxy z dx x dy xz dz+ + + 2 2 2
= (2M
Nxy dx x dy z dx xz dz+ + + 2 2 2) ( )
170 Mechanics
= d x y d z x d x y z x( ) ( ) ( )2 2 2 2+ = + M
N
M
N
= x y z x x y z x2 2 2 2
0 1 2
5 2 7+ = +
M
N
( , , )
( , , )
= 295 units.
(c) W = Fdxx
x
=
=
1
2
= ( )A Bx x dx+ 2
1
2
=A Bx x2 3
1
2
2 3+
=A2
B3
( ) ( )2 1 2 12 2 3 3− + −
=32
A + 73
B
Q.17. An ideal massless spring can be compressed 2 meter by a force of 200 N. The springis located at the bottom of a frictionless inclined plane which makes angle of 30° with thehorizontal. A 20 kg mass is released from rest at the top of inclined plane which comes to restafter compressing the spring by 4 metres. Calculate
(a) Distance travelled by mass before coming to rest.
(b) Speed of mass before it reaches the spring.
Solution. (a) The linear restoring force in the spring is given by,
F = –kx
here, k =200
2100= N/m.
Let, l = distance along inclined plane that mass travelsbefore coming to rest.
x = compression of spring = 4 meter.
The initial P.E. of mass will be converted into P.E. of spring during compression of 4meter.
i.e., mgh =12
2kx
or mgl sin θ =12
2kx
or 20 × 9.8 × l × sin 30° =12
100 42× ×
or l = 8.16 meter.
l′
30°
4
l
Fig. 14
Work, Energy and Momentum 171
(b) Let l' be a distance travelled when mass just touches the spring and v be its velocitythen,
l = l' + 4
or l' = l – 4 = 8.16 – 4 = 4.16 meter.
As K.E. of mass at the instant of touching the spring = Initial P.E. of mass
i.e.,12
2mv = mgl' sin θ
or v2 = 2gl' sin θ = 2 × 9.8 × 4.16 × 12
v = 6.385 m/sec.
Q. 18. An ideal spring (Force constant = 8 × 104 dynes/cm) hangs vertically and supports0.8 kg mass at rest. Calculate the distance by which the mass should be pulled down so thatit may pass through, on being released, the equilibrium position with a velocity of 1 m/sec.
Solution. The linear restoring force produced in the stretched spring is
F = –kx
In stretching the spring, work is done which is stored as P.E. If a vertical downward pullof ‘l’ cm is made,
the work done against the restoring force = potential energy = 12
2kl
On releasing the mass after stretching, it moves upwards and passes the mean positionwith velocity v (say). Then
K.E. at mean position =12
2mv
since entire P.E. at extreme position totally converts into K.E. at mean position,
12
2kl =12
2mv
as m = 0.8 kg, k = 8 × 104 dynes/cm and v = 1 m/sec.
l = mv k2 2 4800 100 8 10 10/ × ( ) /( × )= = cm
Q.19. A block of mass 2.0 kg is dropped from a height of 0.4 m. onto a spring of forceconstant K = 1960 N/m. Find the maximum distance through which the spring will becompressed [Neglect Friction]
Solution. Situation is shown in the diagram
Loss in P.E. of mass = gain in pot. energy by the compressed spring
i.e., mg (d+ y) =12
2ky
given m = 2 kg, d = 0.4 m; k = 1960 Nt/m, g = 9.8 m/sec2; y = ?
So, 2 × 9.8 × (0.4 + y) =12
1960 2× × y
172 Mechanics
or y2 – 0.02 y – 0.008 = 0
or y = 0 02 0 02 4 0 0082
2. ( . ) × .± −
=0 02 0 0004 0 032
2. . .± −
= 0.0999 m (taking +ve value).
Q.20. The scale of spring balance, reading from zero to 100Newton is 20 cm long. A body suspended from the balance isobserved to oscillate vertically at 2 vib/sec. What is weight ofthe body ?
Solution. Force constant of spring = 1000 2
500.
= Nt/m.
Since frequency of oscillation of vertical spring with mass is given by,
n =1
2 2π πK
orK
4 2mm
n; =
with n = 2, K = 500 Nt/m, π = 3.14 kg
m =500
4 3 14 4500
157 752× ( . ) × .= = 3.169 kg
weight mg = 3.169 × 9.8 = 31.06 Newton.
Q.21. An object is attached to a vertical spring and slowly lowered to its equilibriumposition. This stretches the spring by an amount ‘d’. If the same object is attached to the samespring but permitted to fall instead, through what distance does it stretch the spring ?
Solution. Fig. 16(a) shows the normal unextended spring. In Fig. 16(b), by slowly placingthe mass, equilibrium state is shown and elongation is d.
d′
d(a)
(b )
(c)
Fig. 16
2.0 kg
d
Fig. 15
Work, Energy and Momentum 173
So, Restoring force = weight
i.e., Kd = mg or K = mgd
...(i)
In Fig. 16(c); If mass is permitted to fall, the mass will go down to an extreme, will riseup and thus will execute S.H.M.
Suppose in extreme position elongation is d then;
Work done by force mg on the spring
= mg × d' ...(ii)
P.B. corresponding to extreme position = 12
2Kd′ ...(iii)
Equating eqs. (ii) and (iii)
mgd′ =12
2Kd'
Putting for K from (i), mgd′ =12
2mgd
d'.
or d' = 2d
i.e., stretching is double of the previous case.
Q.22. Two block A and B are connected to each other by a string and a spring, the stringpasses over a frictionless pulley as shown.
The block B slides over the horizontal top surface of a stationary block C and the blockA slides along the vertical side of C, both with the same uniform speed. The coefficient offriction between the surfaces of blocks is 0.2. Force constant of spring is 1960 Nt/m. If massof A = 2 kg, calculate the mass of block B and energy stored in the spring.
Solution. Let m = mass of B; T = tension in the spring
For motion of B;
T = µN = µmg = 0.2 mg ...(i)
A
B
C T
T
Fig. 17
for motion of A T = mg = 2g ...(ii)
from (i) and (ii) 2g = 0.2 mg or m = 10 kg
again; T = 2 × 9.8 = 19.6 Newton (from (ii))
if x is the elongation of the spring, then magnitude of linear restoring force;
F = kx (upwards –ve) or x = Fk
174 Mechanics
The linear restoring force acting in the present case is T itself i.e., F = T
So, x =Tk
= 19 61960
....(iii)
Energy stored in the spring =12
2kx
=12
1960 19 61960
0 0982
× × . .
= Joule
Q.23. A body of mass 2 kg is attached to a horizontal spring of force constant 8 Nt/m.A constant force of 6 Nt. is applied along the length of the spring. Find
(a) The speed of the body when it is displaced through 0.5 m
(b) If force is removed, then; how much farther shall the body move before reaching thestate of rest.
Solution. (a) Work done by the constant force = 6 × 0.5 = 3 Joule
If v is velocity, K.E. of body in displaced position will be given by
12
2mv =12
2 2× v
Again, potential energy stored in the spring at displaced position of 0.5 m
12
2kx =12
8 0 5 2× × ( . )
So, Work done in 0.5 m displacement = K.E. + P.E.
i.e., W =12
2 12
8 0 52 2× × × × ( . )v +
or W = v2 + 1
Since W = 3 Joule, 3 = v2 + 1 or v = 1.4 m/sec.
(b) Total energy of system at the time of removal of force = 3 Joule.
Suppose body moves distance ‘l’ still onwards before coming to rest; on removal of force
P.E. at the extreme position = total energy of system
12
8 0 5 2× × ( . )+ l = 3
or (0.5 + l)2 =34
So, l = 0.36 m.
Q.24. A 10 lb block is thrust up a 30° inclined plane with an initial speed 16 ft./sec. Ifit is found to travel 5 ft. along the plane; stop and then slide back again to the bottom.Calculate
(a) The force of friction (regarded as constant) on block.
(b) Speed of block when it reaches bottom of the inclined plane.
Solution. (a) Upward motion: At the top where block momentarily stops; total energy,
Work, Energy and Momentum 175
E = K.E. + P.E.
= mg × 5 sin 30°
= 10 × 5 × 12
= 25 ft lb
At foot of incline,
Ei = (K.E.)i + (P.E.)i
Ei =12
1032
16 02 +( )
Ei = 40 ft-lb
Energy lost = Work done by frictional force in 5 ft. distance
i.e., 25 – 40 = –F × 5 or F = 3 lb
(b) Downwards motion: Suppose block returns to the foot of incline with speed ‘v’.
At bottom where motion ends:
E = K.E. + P.E. = 12
1032
0532
2 2 + =v v ft - lb
At top where motion starts: total energy = 25 ft-lb
again energy lost = work done by frictional force
532
252v − = –15 orv2
322= or v = 8 ft/sec.
Q.25. Figure shows a vertical section of a frictionless track surface. A block of mass 2 kgis released from position A. Compare its K.E. at the positions B, C and D. (g = 9.8 m/sec2).
Solution. Motion occurs under gravitational force which is conservative, so energy isconserved. In coming down block looses its P.E. and corresponding gain in K.E. occurs;
at point B: (K.E.)B = (P.E.)A – (P.E.)B = (mgh)A – (mgh)B
= mg (14 – 5) = 2 × 9.8 × 9
= 176.4 Joules.
at Point C: (K.E.)C = (P.E.)A – (P.E.)C = mghA – mghC
= mg (14 – 7) = 0.2 × 9.8 × 7 = 137.2 Joules
at Point D: (K.E.)D = (P.E.)A – (P.E.)D = mghA – mghD
= mg (14 – 0) = 2 × 9.8 × 14 = 274.4 Joules.
Q.26. A point mass m starts from rest and slides down the surface of a frictionless sphereof radius r. Measure angles from the vertical and P.E. from the top. Find
(a) Change in P.E. of the mass with angle.
(b) K.E. as function of angle.
(c) Radial and tangential acceleration as function of angle.
(d) Angle at which mass flies off the sphere.
176 Mechanics
(e) If friction be present between mass and sphere, shall the mass fly off at greater orlesser angle as compared to situation in (d).
Solution. (a) At top position P (reference point) P.E. = 0
at bottom position at depth, y = 2r ; ∴ P.E. = –mgy
at a general position M;
P.E. = –mgPM' = – mg (PO – M'O)
= – mg r − ′
M OMO
MO.
= – mgr (1 – cos θ)
So change in P.E.; (P.E.)M – (P.E.)P = – mgr (1 – cos θ) – 0
∆(P.E.) = – mgr (1 – cos θ) ...(i)
(b) At top total energy is wholly potential = 0. so from energy conservation
At general position M; K.E. + P.E. = 0
i.e., K.E. + [– mgr (1 – cos θ)] = 0
or K.E. = mgr (1 – cos θ) ...(ii)
(c) At general position M; v = tangential velocity
∴ Radial acceleration =vr
2
But K.E. at same position is 12
2mv = mgr (1 – cos θ)
or v2 = 2gr (1 – cos θ) ...(iii)
So, Radial acceleration =2 1 2 1gr
rg( cos ) ( cos )− = −θ θ
tangential acceleration =dvdt
dvd
ddt
=θ
θ.
∴ as v = rw = rddt
ddt
vr
θ θ∴ =
so,dvdt
=vr
dvdθ
...(iv)
From (iii) ∴ dvdθ
=12
2 1 212gr gr( cos ) . sin− −θ θ
then tangential acceleration is
dvdt
=vr
grv
g. sin sinθ θ=
(d) At position where mass flies off the sphere-surface the condition for such event is
Centrifugal force = Component of weight along the radius
m P
MM ′
Or
B
θ
Fig. 18
M
O
r
N
m g sin θm g
θm g cos θ
Fig. 19
Work, Energy and Momentum 177
i.e.,mv
r
2
= mg cos θ
or v2 = gr cos θ ...(v)
equating (iii) and (v);
2gr (1 – cos θ) = gr cos θ
or 3 cos θ = 2 or θ = cos–1 23
(e) If friction is present, then velocity at any general point M shall be less. v is less meanscos θ is less which suggests θ is large. Therefore mass leaves at greater angle as comparedto previous (frictionless) case.
Q.27. A body of mass 5 kg moves down from state of rest on an inclined plane of length5 meters having inclination of 60° with the horizontal. If coefficient of friction be 0.2 find thespeed of body at the bottom. How far shall it slide on the rough horizontal surface having samecoefficient of friction as the inclined plane.
Solution. At top: P.E. = mgh = mgd sin θ ; and K.E. = O
Now, frictional force = µN = µmg cos θWork done by frictional force = –µmg cos θ × d (–ve sign gives, nature or direction)
Since, change in total energy = work done
i.e.,change in P.E. + Change in K.E. = work done
So; ∆T + (–mgd sin θ) = – µ mgd cos θ
or ∆T = mgd (sin θ – µ cos θ) = 12
mv2
or v2 = 2gd (sin θ – µ cosθ )
Now, g = 9.8 m/sec2; d = 5 meters, θ = 60°, µ = 0.2
v2 = 2 × 9.8 × (sin 60° – 0.2 cos 60°) × 5
which gives, v = 8.664 m/sec
At the start on horizontal rough surface the K.E. possessed by body = 12
mv2,
This becomes zero finally due to expense in doing work against frictional force;
i.e., O – 12
mv2 = Work done against friction
Supposing that l distance is travelled before coming to rest,
then frictional work done = – µ mgl
So, – 12
mv2 = – µ mgl or v2 = 2µ gl
as calculated earlier, v2 = 75.06 ; µ = 0.2, g = 9.8
l =v
g
2
275 06
2 0 2 9 819 15
µ=
× ×=.
. .. .meter
θ
Fig. 20
178 Mechanics
Q.28. A block of mass 1 kg collides with a horizontal weightless spring of force constant2Nt/meter. The block compresses the spring 4.0 meter from its normal position. Calculate thespeed of block at the instant of collision if kinetic friction between block and surface is 0.25.
Solution. Suppose v = speed of block at the instant of collision then its kinetic energy
=12
mv2
This energy is spent in overcoming the friction and compressing the spring, if x iscompression of spring;
12
mv2 =12
2kx + µmgx
given x = 4 meters, k = 2 Nt/meter, m = 1 kg,
g = 9.8 m/sec2, µ = 0.25
∴12
1 2× × v =12
2 4 0 25 1 9 8 42× × ( ) . × × . ×+
or v = 51 6. = 7.18 m/sec.
Q.29. A horizontal force pushes a 10 kg mass up an inclined plane (θ = 30°) from bottom,by a distance of 3 meters on the inclined plane. If the initial and final speeds of the mass are1 m/sec and 3 m/s, calculate the work done by the force.
Solution. Height in the final position,
h = AB sin 30° = 3 × 12
= 1.5 m
∴ Change in P.E., ∆U = mgh = 10 × 9.8 × 1.5 = 147 Joule
Change in K.E. = ∆T = 12
122
212mv mv− =
12 2
212m v v( )−
=12
× 10 × (32 – 12) = 40 J
∴ Change in total energy = (147 + 40) = 187 Joule
which gives the work done.
Q.30. A chain of length d and mass m lies on a frictionless horizontaltable such that one third of its length hangs over the edge. Calculatethe work needed to pull the hanging part back onto table.
Solution. Mass per unit length of chain = m/d
Mass of infinitesimally small length dy of chain = md
. dy work done in lifting this
segment against force of gravity; by a distance y; is
dW =md
dy
.g.y
So, work done in lifting or raising the d3
length
E W
N
S
Work, Energy and Momentum 179
W =mgd
ydyd
.
/
0
3
=mgd
ydymgd
yd
d=
2
0
3
0
3
2
//
=mg
dd mgd
2 9 18
2
= .
Q.31. A 0.5 kg block slides from point A on a horizontal track with initial speed3m/s towards a weightless horizontal spring of length 1m and force constant 2 Nt/m. PartAB of track is frictionless while part BC has coefficient of static and kinetic friction 0.22 and0.2. If AB = 2m and BD = 2.14m, find total distance block moves before coming to rest(g =10m/s2).
Solution. The block will reach B with initial K.E.
V1 =12
0 5 3 2 252× × =. ( ) . J
In going from B to D work will be done against kinetic friction.
∴ W = µN (BD)
= 0.2 × 0.5 × 10 × 2.14
= 2.14 J
Therefore, the K.E. with which block will hit the spring at D
= V1 – W = 0.11 J
If block compresses the spring through x, this energy will be partly used up in doingwork against friction during the compression x and rest will be stored in spring as its P.E.
i.e.,12
2K Nx x+ µ = 0.11
or12
2 0 2 0 5 102× × . × . ×x x+ = 0.11
i.e., x2 + x – 0.11 = 0 which gives x = − ± +
=1 1 0 44
20 1
.. m
It means total distance converted by block before it comes to rest is
D = AB + BD + x
= 2 + 2.14 + 0.1
= 4.24 m.
Q.32. A body of mass m is accelerated uniformly from rest to a speed V in time. Showthat work done on the body as a function of time is
W =12
mVT
. t2
22.
180 Mechanics
Solution. Let ‘a’ be the acceleration in body then, since u = 0
V = 0 + a.T or a = VT
The velocity gained by body in time t will be
v = at = VT
. t
Hence K.E. gained in time t, which gives work done
W =12
VT
mv m t221
2=
.
=12
2m tVT
2
2 .
Q.33. The scale of a spring balance reads from 0 to 250 kg and is 25 cm long. What isthe potential energy of the spring when a 20 kg weight hang from it ?
Solution. As given, 25 cm elongation of spring takes place corresponding to 250 kg. Sothe force constant of spring is given by
250 × 9.8 = K × 0.25
K = 9.8 × 103 N/m
If 20 kg weight hangs from balance, elongation will be obtained by
20 × 9.8 = Kx'
i.e., x' =20 9 8× .
Kmeter
∴ Potential energy stored in spring will be
U = 12
12
9 8 1020 9 89 8 10
2 33
2
K ( ) × . ×× .
. ×x' =
=12
20 9 89 8 10
2
3×( × . )
. ×
= 1.96
Q.34. Calculate the work done by a force F = kx2 acting on a particle at an angle of 60°with x-axis to displace it from x1 to x2 along the x-axis.
Solution. The work done is given by
W = F→ −→ . dr
= Fdx cos k x dxx
x
x
x60
12
2
1
2
1
2° =
= 12 3
16
3
23
13
1
2
k x k x xx
x = − .
Work, Energy and Momentum 181
Q.35. A cord is used to lower vertically a block of mass M a distance d at a constantdownward acceleration of g/4. Find the work done by the cord on the block.
Solution. Let T be the tension in the cord acting vertically upward. The net force onthe block is T-Mg, acting upward, where Mg is the weight of the block. If a is the downwardacceleration of the block, then by Newton’s second law, we have
T – Mg = –Ma
or T = M(g – a)
Here, a = g/4
∴ T = M Mggg−
=
434
Therefore, work done by the tension T is given by
W = tension (T) × displacement in the direction of T
= T × (–d) [ d is downward]
= − 34
Mgd .
Q.36. A block of mass 10.0 kg is pulled up at constant speed from the bottom to the topof a smooth incline 5.00 meter off the ground at the top. Calculate the work done by the appliedforce which is parallel to the incline. (g = 9.8 m/s2).
Solution.
Fig. 21
The force acting on the block at any instant are its weight mg, the normal reaction Nand the applied force F. The weight can be resolved into a component mg sin θ to the inclinedplane and a component mg cos θ perpendicular to it. The net force parallel to the plane isF mg sin θ. Since the block moves on the plane at constant speed, the net force is zero. Thus
F – mg sin θ = 0
or F = mg sin θ
= 10 × 9.8 × 35
= 58.8 Nt
182 Mechanics
The work done is
W = Force × displacement along the force
= 58.8 Nt × 5 m = 294 Joule
Q.37. A boy pulls a 5 kg block 10 meter along ahorizontal surface at a constant speed with a forcedirected 45° above the horizontal. If coefficient of kineticfriction is 0.20, how much work does the boy do onblock ?
Solution. The force acting on the block are itsweight mg, normal reaction N exerted by the surface,sliding frictional force fK against the motion and theapplied force F at 45° above the horizontal. The nethorizontal force is F cos 45° – fK, and net vertical forceis N + F sin 45° – mg.
The block is unaccelerated, so that from Newton’s second law, we obtain
F cos 45° – fK = 0 ...(i)
and N + F sin 45° – mg = 0 ...(ii)
Also, we know fK = µKN ...(iii)
where µK is the coefficient of kinetic friction. Substituting the value of N from Eq. (ii) inEq. (iii), we get
fK = µK (mg – F sin 45°)
Now putting this value of fK in Eq. (i); we get
F cos 45° – µK (mg – F sin 45°)= 0 ...(iv)
or F (cos 45° + µK sin 45°) = µK mg
or F =µK
K
mgcos sin45 45° + µ °
with µK = 0.20, mg = 5 × 9.8 = 49 Nt and cos 45° = sin 45° = 0.707,
we obtain
F = 11.55 Nt.
The block is pulled through a horizontal distance r = 10 meter. Then the work done is
W = Fr cos 45°
W = (11.55) (10) (0.707)
= 81.66 Joule.
Q.38. A block of mass m = 3.57 kg is pulled at constant speed through a distance r = 4.06m along a horizontal surface by a rope exerting a constant force F = 7.68 Nt inclined atθ = 15° to the horizontal. Find (i) the total work done on the block (ii) work done on the blockby the rope and by friction, (iii) coefficient of kinetic friction between block and surface.
Solution. (i) Since the block moves at constant speed, the net force on it is zero. Hencethe total work done on it is zero.
(ii) The force exerted by the rope is inclined at 15° to the horizontal . Hence the workdone on the block by the rope through a distance r = 4.06 m is given by
Fig. 22
Work, Energy and Momentum 183
M = Fr cos 15°´
= 7.68 × 4.06 × 0.966
= 30.1 Joule
Since the total work done on the block is zero, the work done on it by friction is –30.1Joule, i.e., work is done by the block against friction.
(iii) From eq. (iv) of the last problem, we can write
µK = FFcos
sin15
15°
− °mg
=7 68 0 966
3 57 9 8 7 68 0 259. × .
. × . . × .−
=7 41933 00..
= 0.225
Q.39. A running man has half the kinetic energy that a boy of half his mass has. The manspeeds up by 1.0 m/s and then has the same kinetic energy as the boy has. Find speed of manand boy.
Solution. Let m be the mass of man, and m/2 that of boy. Let v1 and v2 be their original
speeds. Then kinetic energy of man is 12 1
2mv and that of boy is 12 2 1
2mv
. Since the energy
of man is half that of boy, we have
12 1
2mv =12
12 2 2
2mv
or v12 =
14 2
2v ...(i)
When the speed of the man becomes (v1 + 1), his kinetic energy equals to that of theboy. That is,
12
112m v( )+ =
12 2 2
2mv
(v1 + 1)2 =12 2
2v ...(ii)
Solving eqs. (i) and (ii) we get
v1 = 2.41 m/s
v2 = 4.82 m/s.
Q.40. A 30 gm bullet initially travelling 500 m/s penetrates 12 cm into a wooden block.What average force does it exert ?
Ans. The kinetic energy of the bullet is given by
K =12
2mv
=12
30 10 5003 2( × ) × ( )− = 3750 Joule
184 Mechanics
It stops after travelling a distance r = 12 cm. = 0.12 m in wooden block. If F Nt be the(retarding) force exerted by the block, then the work done by the bullet against this force is
W = Fr
= F × 0.12 Joule
By work-energy theorem, the kinetic energy equals this work done, that is
3750 = F × 0.12
F =37500 12
31250.
= Nt.
Q.41. Show with the help of work-energy theorem that the minimum stopping distancefor a car of mass m moving with speed v along a level road is v2/2µsg, where µs is thecoefficient of static friction between tyres and road.
Solution. The kinetic energy of the car is
K =12
2mv
It is retarded due to the force of static friction fs (rolling friction neglected), whosemaximum value is given by
fs = µsN = µsmg
N being the normal reaction which equals weight mg of the car on a level road. If x bethe minimum stopping distance, then the work done against the force fs is given by
W = fsx = µs mgx
This, by work-energy theorem, equals the initial kinetic energy K of the car. Thus,
µsmgx =12
2mv
x =v
gs
2
2µ
Q.42. A mass of 0.675 kg is being revolved on a smooth table in a horizontal circle bymeans of a string which passes through a hole in the table at the centre of the circle. If theradius of circle is 0.50 m and the uniform speed is 10.00 m/s, find the tension in string. If theradius of the circle is reduced to 0.30 m by drawing the string down through the hole, thetension is increased by a factor 4.63. Find the work done by the string on the revolving massduring the reduction of the radius.
Solution. The tension T of the string supplies the required centripetal force mv2/R i.e.,
T =mv2 20 675 10 0
0 5135
RNt= =. × ( . )
( . )On reducing the radius, the new tension is
T' = 4.63 T = 4.63 × 135 = 625 Nt
If v' be the new velocity, then we have,
T' =mv′
′
2
R
Work, Energy and Momentum 185
625 =0 675
0 30
2..
v′
v'2 =625 0 3
0 675278 2× .
.( )= m/s
The increase in kinetic energy of the moving mass is therefore
∆K =12
2 2m v v( )′ −
=12
0 675 278 100× ( . ) × ( )−
= 60 Joule
By work energy theorem, this is equal, to the work done on the moving mass.
Hence, W = ∆K = 60 Joule
Q.43. A constant force of 5 Nt acts for 10 seconds on a body whose mass is 2 kg. Thebody was initially at rest. Calculate the work done by the force, the final kinetic energy andthe average power of the force.
Solution. Let a be the (constant) acceleration in the body. Then
a =F m/s2
m= =5
22 5.
The distance x moved in 10 sec is
x =12
12
2 5 102 2at = × . × ( )
= 125 m.
The work done by the force F during this distance is
W = Fx = 5 × 125
= 625 Joule
The initial kinetic energy of the body is zero, By work-energy theorem the final kineticenergy is the same as the work done, i.e., 625 Joule.
The body covers distance x (= 125 m) in 10 sec. Then its average velocity is
v =xt
= =12510
12 5. m/s
The average power of the force is therefore
P→
= F v→
= 5 × 12.5
= 62.5 Watt.
Q.44. A boy whose mass is 51 kg climbs, with constant speed, a vertical rope 6 m longin 10 second. How much work does the boy perform ? What is his power output during theclimb ?
186 Mechanics
Solution. The body does work against his weight in climbing. This is given by
W = force × distance
= (51 × 9.8) × 6
= 3000 Joule
This work is done in 10 sec. Hence the power output is
P =Wt
=300010
= 300 Watt.
l
h
Xθ
dy
m g
A
θ
T
Y
F
O dx
θ0
Fig. 23
potential energy when the bob has been raised to a vertical height h in the x-y plane is
U(x, y) = U(x, h) = mgh ...(i)
At the point A, the kinetic energy is zero i.e., the potential energy is equal to themechanical energy E. Then
E = mgh ....(ii)
If the bob be released at the point A, it returns under the gravitational (restoring) force,and the potential energy begins to convert into the kinetic energy. At any point in its paththe sum of the two energies remains equal to the mechanical energy. Thus if v be the velocityand y the vertical height of the bob at any point, we have by the law of conservation ofmechanical energy.
12
2mv x y+ U( , ) = E ....(iii)
But U(x, y) = mgy
∴12
2mv mgy+ = E ...(iv)
Comparing eq. (ii) and (iv) we have
12
2mv mgy+ = mgh
Work, Energy and Momentum 187
or12
2mv = mg (h – y)
or v = 2g h y( )−
Thus at O (y = 0); v = 2gh (maximum); the energy is all kinetic. At A (y = h); v = 0
(minimum); the energy is all potential.
Thus v varies between 2gh and O or 12
2mv varies between mgh and O. This means
that from equation (iii), U(x, y) can never be greater than E i.e., U(x, y) /> mgh. In other wordsthe bob cannot rise higher than h, its release point A.
(iv) Magnetic Potential Energy. Let τ be the (restoring) couple acting upon a magnetof moment M when its axis makes an angle θ with a magnetic field H. Then we have
τ = –MH sinθ
Let U(0) be the potential energy in the θ = 0 position. The work in turning the magnetfrom θ = 0 to θ position is the potential energy further acquired by the magnet. Thus if U(θ)be the potential energy at position θ, then
U(θ) – U(0) = − τ θθ
d0
= MH sin θ θθ
d0
= MH (1 – cos θ)
If U(0) be assumed to be zero, then
U(θ) = MH (1 – cos θ)
This is the required expression.
If the magnet is released from the position θ, its potential energy is converted into thekinetic energy of rotation and in position θ = 0 the entire energy becomes kinetic. It I be themoment of inertia of the magnet about the axis of suspension and ω the angular speed
attained at θ = 0 position, then the kinetic energy is 12
2Iω . Therefore,
12
2Iω = MH (1 – cos θ)
ω =2MH (1 cos
I− θ)
Q.45. If a force F ( 2xy z ) i x j 2xzk,2 2→
= + + + then show that it a conservative force.
Determine its potential function.
Solution. Let us find the curl of given force.
188 Mechanics
curl F→
=
i j k
x y z
i j k
x y zxy z x xzx y z
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
F F F
=
+2 22 2
= ( ) ( ) ( )i j z z k x x0 0 2 2 2 2− − − + −= 0
As curl F→
is zero, the force F→
is conservative.
The potential energy function of F→
is given by.
U = − = − + +→ −→ F F F F. ( )dr dx dy dzx y z
= − + + + ( )2 22 2xy z dx x dy xz dz
= − + + + ( ) ( )2 22 2xy dx x dy z dx xzdz
= − + = − + d x y d z x d x y z x( ) ( ) ( )2 2 2 2
= − +( )x y z x2 2
Q.46. Show that the force represented by F yz i zx j xy k→
= + + is conservative.
Q.47. The position of a moving particle at an instant is given by r A cos i A sin j→
= +θ θ .Show that the force acting on the particle is conservative.
Solution. The position vector of the particle at an instant t is given by
r→
= A A Acos sin (cos sin )θ θ ω ωi j t i t j+ = +
where ω is the angular velocity of the particle and ωt = θThe linear velocity and acceleration at t are given by
v→
=drdt
t i t j→
= − +Aω ω ω( sin cos )
and a→
= d vdt
A t i t j r→
→= − + = −ω ω ω ω2 2(cos sin )
If m be the mass of the particle, then the force acting upon it is given by
F→
= m a m r→ →= − ω2
Since curl r→
= 0, we have
curl F→
= 0
and so the force is conservative.
Work, Energy and Momentum 189
Q.48. What is the potential energy of an 800 kg elevator at the top of a building 380 mabove street level ? Assume the potential energy at street level to be zero. What would happento this energy if the elevator comes down to the street level ?
Solution. The (gravitational) potential energy of the elevator at a height h when itsvalue at street level is zero, is given by
U(h) = mgh
Here, m = 800 kg and h = 380 m
∴ U(h) = 800 × 9.8 × 380
= 2.98 × 106 Joule.
When the elevator comes down to the street level, the potential energy is first convertedinto kinetic energy and then into heat when the elevator has come to a stop.
Q.49. A rod of length 1.0 meter and mass 0.5 kg is fixed at one end and is initially havingvertically. The other end is now raised until it makes an angle of 60° with the vertical. Howmuch work is required ?
Fig. 24
Solution. The weight mg of the rod acts at its centre of gravity G. As the lower end ofthe rod is rotated through 60°, G moves to G', i.e., it is raised through a height h, where,
h = CG = OG – OC
= OG – OG' cos 60°
= OG (1 – cos 60°)
because CG = OG'
Here OG = 0.5 meter and cos 60° = 0.5
∴ h = 0.5 (1 – 0.5) = 0.25 meter
Thus; the gain in the potential energy of the rod is
U = mgh
= 0.5 × 9.8 × 0.25
Since the work done W has been stored as potential energy, we have
W = U = 1.225 Joule
190 Mechanics
Q.50. A uniform chain is held on a frictionless table with one-fifth of its length hangingover the edge. If the chain has a length l and a mass m, how much work is required to pullthe hanging part back on the table ?
Solution.l/5
Fig. 25
The mass of the hanging part of the chain is m/5. The weight mg/5 of this part of thechain acts at its centre of gravity which is distance l/10 below the surface of the table. Hencethe gain in potential energy in pulling the hanging part on the table is
U =mg l mgl5 10 50
=
This is also the work done in pulling the chain. Thus
W = U = mgl/50
Q.51. The potential energy function for the force between two atoms in a diatomic moleculecan be expressed as follows:
U(x) =a
xbx12 6−
where a and b are positive constants and x is the distance between the two atoms. Derive anexpression for the force between the two atoms and show that the two atoms repel each otherfor x less than x0 and attract each other for x greater than x0. What is the value of x0 ?
Solution. U(x) =a
xbx12 6−
We know that the force is the negative gradient of the potential energy. Therefore theforce between the two atoms is given by
F = − d xdxU( )
= − −
ddx
ax
bx12 6
=12 6
13 7a
xb
x−
Let the force F be zero, when x = x0, then we have
12 6
013
07
ax
bx
− = 0
x06 =
2ab
x0 =2 1 6ab
/
Work, Energy and Momentum 191
This is the equilibrium point (a point of stable equilibrium). Now, we may write
F =12 6
13 7a
xb
x−
= 6 21
617 6 7
06b
xa bx
bx
xx
/ −
= −
Clearly, when x < x0 then F is positive and the atoms repel each other. When x > x0 thenF is negative and the atoms attract each other.
Q.52. The potential energy function for the force between two atoms in a diatomic moleculemay be expressed as follows:
U(x) =a
xbx
;10 5−
where a and b are positive constants and x is the distance between the atoms. (a) Calculatethe distance x at which P.E. is minimum. (b) Assume that one of the atoms remains at restand that the other moves along x. Describe the possible motion. (c) The energy needed to breakup the molecule into separated atoms (x = ∞) is called the dissociation energy. What is thedissociation energy of the molecule ?
Solution. (a) The value of x at which U(x) is a minimum is found from ddx
xU( ) = 0
so thatddx
ax
bx10 5−
= 0
or − +10 511 6a
xb
x= 0
or x5 =2ab
or x =2 1 5ab
/
(b) The force between the two atom is
F = − = −ddx
x ax
bx
U( ) 10 511 6
The force is zero when x = 2 1 5ab
/
. When x is less than (2a/b)1/5, the force is positive
and the atoms repel each other. When x is greater than (2a/b)1/5, then force is negative andthe atoms attract each other. If one of the atoms is fixed, then the other atom would oscillateabout the equilibrium separation (2a/b)1/5.
(c) The dissociation energy D is equal to the change in potential energy from its minimumvalue at equilibrium separation x = (2a/b)1/5 to the zero value at x = ∞ .
192 Mechanics
∴ D = U(x = –∞) – Ux = (2a/b)1/5
= D – a
a bba b( / ) /2 25 −
= b2/4a
If the kinetic energy at the equilibrium position is equal to or greater than its value, themolecule will dissociate.
Q.53. The potential energy function of the force between two atoms in a diatomic moleculecan be expressed approximately as:
U(x) =a
xbx12 6−
where a and b are positive constants and x is the distance between atoms (i) At what valuesof x is U(x) equal to zero and U(x) a minimum ? (ii) Determine the force between atoms (iii)Calculate the dissociation energy of the molecule.
Ans. ( or ,2
iab
ab
iia
xb
xiii b a) ; ( ) , ( ) /
/ / ∞
−
1 6 1 6
3 7212 6
4
Q.10. The potential energy between the protons and neutrons inside a nucleus is given by
U(r) = − −rr
U e00
r/r0
Find the corresponding expression for the force of attraction and compute the ratio of thisforce at r0 , 2r0 , 4r0 and 10r0 to the force at r = r0. What conclusion you draw from theresults?
Solution. The force is the negative gradient of potential energy. Thus
F(r) = − =
−ddr
xddt
rr
e r rU( U0) /0 0
= rr
er
er
r r r r0 2
0
1 1 10 0U0 −
+
−
− −/ /
=− +
−rr
er r
r r0
0
01 1U0 /
This is the expression for the force of ‘attraction’. At r = r0 the force is
F0r = −
−U0 er
1
0
2
Similarly, F02r = − −U0 e
r2
0
34
F4r0= − −U
5160
4
0e
r
Work, Energy and Momentum 193
F10 0r = − −U0 er
10
0
11100
Therefore we have
F
F2r
r
0
0
=38
1e− ~ 0.14 ~ 1.4 × 10–1 [ . ] e = 2 718
F
F4 0
0
r
r=
532
3e− ~ 0.0078 ~ 7.8 × 10–3 [ . ] e3 20 09=
F
F10 0
0
r
r=
11200
9e− ~ 0.0000067 ~ 6.7 × 10–6 [ ] e9 8108=
The result shows that the force is short range.
Q.54. The potential energy of a body is given by
U = 40 + 6x2 – 7xy + 8y2 + 32z
where U is the joule and x, y, z in meter. Deduce the x, y, z components of the force on thebody when it is in position (–2, 0, +5).
Solution. U = 40 + 6x2 – 7xy + 8y2 + 32z
The negative gradient of the potential energy with respect to the potential variablesgives the intrinsic force. Therefore, the (x, y, z) components of the force at the position(–2, 0, +5) are given by
Fx =− = − + = +∂∂
UNt
xx y12 7 24 [putting x = –2, y = 0]
Fy =− = − = −∂∂
U Nty
x y7 16 14 [putting x = –2, y = 0]
Fz =− = −∂∂
UNt
z32
Q.55. If the potential in a plane is given by U = 4x2 – 10xy + 2y2, deduce the x and ycomponents of field at the point x = 2, y = 4.
Solution. U = 4x2 – 10xy + 2y2
The x and y components of the field are the negative gradients of U with respect to xand y respectively. Thus
Fx =− = − +∂∂
Ux
x y8 10
and Fy =− = −∂∂
Uy
x y10 4
at x = 2, y = 4 we have
Fx = 24 and Fy = 4.
Q.56. The electric potential in a region of space is given by V = 5x – 7x2y + 8y2 + 16yz – 4zvolt.
194 Mechanics
Deduce an expression for the electric field E→
. Calculate the y-component of the field at thepoint (2, 4, –3).
Solution. V = 5x – 7x2y + 8y2 + 16yz – 4z volt ...(i)
The electric field E→
may be written in terms of its cartesian components as
E→
= E E Ex y zi j k + +Since the electric field in a direction is the negative potential gradient in that direction,
we may write
E→
= − − −∂∂
∂∂
∂∂
V V Vx
iy
jz
k ...(ii)
By partial differential of eq. (ii) we get
∂∂Vx
= 5 – 14 xy, ∂∂
= − + + ∂∂
= −V and Vy
x y zz
y7 16 16 16 42
Putting these values in eq. (ii), we obtain
E→
= ( ) ( ) ( ) − + + − − + − +5 14 7 16 16 16 42x xy i x y z j y kThis is the required expression. The y-component at the point (2, 4, –3) is
Ey = 7x2 – 16y – 16z = 12 volt/meter.
Q.57. The electric potential in a system is given by
V(x, y, z) = 20 + 6x2 – 5xy + 4y2 + 3z2 joule/coulomb.
where x, y, z are in meter. Deduce in , , i j k notation the force on a 2 × 10–15 coulomb charge
placed at position (2, 0, –3) meter.
Solution. The electric field is the negative derivative of the potential and is given by
E→
=− − −∂
∂∂∂
∂∂
VZx
i Vy
j V k
= − − − − + −( ) ( ) 12 5 5 8 6x y i x y j zk Nt/Coul.
At position (2, 0, –3) meter, the field is therefore
E→
= − − − − + − −( ) ( ) ( ) 24 0 10 0 18i j k
or E→
= − + +24 10 18 i j k Newton/Coulomb
The force on a charge q = 2 × 10–15 coulomb at this point is
F→
= q i j kE Newton→
−= − + −( ) ×48 20 36 10 15
This is the required expression.
Work, Energy and Momentum 195
Q.58. A scalar potential field is given by
V = 6x + 8y – 12xy2 + 7yz2 – 5y2 joule/coul.
where x, y, z are in meters. Calculate (a) the work done on a 5 coulomb charge in movingit from position (2, 0, 0) to (2, 5, 0). (b) the force on a 4-coulomb charge placed at the origin(0, 0, 0).
Solution. (a) The potential at positions (2, 0, 0) and (2, 5, 0) are obtained by the givenexpression as
V2, 0, 0 = –12 joule/coulomb and V2, 5, 0 = –697 joule/coulomb
Therefore the work done in carrying a charge q = 5 coulomb from a point V2, 0, 0 toV2, 5, 0 is given by (work = charge × potential difference)
W = q (V2, 0, 0 ~ V2, 5, 0)
= 5 × 685 = 3425 Joule
(b) The expression for the electric field is
E→
= − ∂∂
− ∂∂
− ∂∂
V V Vx
iy
jz
k
= − − − − + − −( ) ( ) 6 12 8 24 7 10 142 2y i xy z y j yz k
At (0, 0, 0) we have
E0
→= –6 8 i j− Newton/coulomb.
The force on a charge q = 4 coulomb is therefore
F→
= q i jE Newton.0
→= − −24 32
The scalar magnitude of this force is
F = ( ) ( )24 322 2+ −= 40 Newton.
Q.59. Show that for the same initial speedv0 , the speed v of a projectile will be the sameat all points at the same elevation regardless ofthe angle of projection.
Solution. In the absence of air resistance,the only force on a projectile is its weight, andthe mechanical energy of the projectile remainsconstant. Fig. 26 shows two trajectories of aprojectile with the same initial speed (and hencethe same total energy) but with different anglesof departure.
Now, at all points at the same elevationthe potential energy is the same; hence thekinetic energy is the same and so the speed isthe same.
X
X
v0 v0
Fig. 26
196 Mechanics
Q.60. The bob of a 2m long pendulum has mass m= 0.5 kg. It is pulled to a side to make θ = 30° anglewith the vertical. Calculate the change in potentialenergy, the work done and the speed of the bob whenit passes the lowest point after being released.
Solution. As the bob of mass m is moved from Ato B, it rises through the height h, where
h = CA = OA – OC
= l – l cos θ = l (1 – cos θ)
Thus the gain in potential energy is
U = mgh
= mgl (1 – cos θ)
Here m = 0.5 kg, l = 2 meter and cos θ = cos 30° = 0.866
∴ U = 0.5 × 9.8 × 2 (1 – 0.866)
= 1.31 Joule
All the work done is stored as potential energy. Therefore
Work done = 1.31 Joule
On being released at B, when bob passes the lowest point A, it has lost all its potential
energy which appears as kinetic energy 12
2mv , where v is the velocity at A. No work is done
by the tension in the string because it is always perpendicular to the displacement. Henceby the conservation of mechanical energy, we have
mgh =12
2mv ,
or v = 2gh
Here h = l (1 – cos θ) = 2 (1 – 0.866) = 0.268 meter
∴ v = 2 9 8 0 268 2 3× . × . .= m/s
Q.61. (a) The bob of a simple pendulum of lengthl is released from a point in the same horizontal line asthe point suspension and at a distance l from it. Calculatethe velocity of the bob and the tension in the string atthe lowest point of its swing.
(b) If the string of the pendulum is catched by anail located vertically below the point of suspensionand the bob just swings around a complete circle aroundthe nail, find the distance of the nail from the point ofsuspension.
(c) If the string of the pendulum is made of rubber then show that it will stretch by 3 mg/K (where K is the force constant) on reaching the bob at the lowest point.
CBm
l cos θ
h
l
l
θ
O
m
A
Fig. 27
Fig. 28
Work, Energy and Momentum 197
Solution. (a) Let S be the point of suspension and A the point from which the bob ofmass m is released (SA = l). The point A is at a vertical height l above the lowest point Bof the swing. Hence is the potential energy mgl. On being released, the bob swings along the
dotted arc, reaching the lowest point B where its entire energy is kinetic, 12
2mv (v being the
velocity at B). By the conservation of mechanical energy, we have
mgl =12
2mv
∴ v = 2gl
This is the velocity at the lowest point.
As the bob moves toward B, the tension in the string increases and becomes maximumat B. If T be this (maximum) tension, the net vertically upward force on the bob is T-mg, andthis provides the required centripetal force mv2/l to the swinging bob. Thus
T – mg =mv
l
2
But at B, v2 = 2gl (proved above).
T = mg + mv
l
2
T = 3mg
(b) Now, suppose a nail N is located at distance d vertically below S. As the bob reachesthe point B, the swing of the pendulum is caught by the nail and the bob swings around acomplete circle of radius r(say). C is the highest point of this circular swing. Clearly the bob
will do so provided its velocity at C, say vc, is such that the required centripetal force mv
rc2
(downward) is provided by its entire weight mg. That is
mvr
c2
= mg
or vc2 = gr
The decrease in kinetic energy as the bob goes from B to C is 12
2 2m v vc( ).− This appears
as gravitational potential energy mg(2r) of the bob at the point C. Thus
12
2 2m v vc( )− = mg (2r)
Substituting the values of v2 and vc2 from above, we get
12
2m gl gr( )− = mg (2r)
2gl – gr = 4gr
198 Mechanics
2gl = 5gr
r =25
l
∴ The distance of the nail from the point of suspension is
d = l – r = l – 25
35
l l=
(c) If the string is made of rubber, then on reaching the bob at B the string is stretchedby an amount ∆l (say). The string would therefore experience an upward elastic force K(∆l).This minus the (downward) weight mg would provide the centripetal force mv2/l at B. Thus
K(∆l) – mg =mv
l
2
at B; v2 = 2gl
∴ K(∆l) – mg =m gl
lmg
( )2 2=
∆l = 3mg/K.
Q.62. A light meter stick, pivoted about a horizontal axis through its centre, has a 2kgbody attached to one end and a 1 kg body to the other. The system is released from rest withthe stick horizontal. What is the velocity of each body as the stick swings through a verticalposition ?
Solution.
Fig. 29
Initially the bodies are at A and B, having gravitational potential energies mgh only,where h is the vertical height from the lowest point A'(h = 0.5 meter). The mechanical(potential) energy of the system is
= 2 × 9.8 × 0.5 + 1 × 9.8 × 0.5= 3 × 9.8 × 0.5 Joules
On being released, the heavier body comes down in the position A' and the lighter one
goes upto B'. Let v be the velocity of each. Now the body at A' has kinetic energy 12
2mv
Work, Energy and Momentum 199
alone, but the body at B' has both the kinetic energy 12
2mv
and the potential energy (mgh);
h being now 1 meter. Thus the mechanical energy of the system
=12
212
1 1 9 8 12 2× × × × × . ×v v+ +
=32
9 82v + .
By the law of conservation of mechanical energy, we have
3 × 9.8 × 0.5 =32
9 82v + .
or32
2v = 3 × 9.8 × 0.5 – 9.8 = 0.5 × 9.8
∴ v =2 0 5 9 8
31 81
× . × ..= m/s
Q.63. A light rod of length l and with a mass m attached to its end is suspendedvertically. It is turned through 180° and then released. Calculate the velocity of the mass andthe tension in the rod when the mass reaches at lowest point.
If the system be released with the rod horizontal, at what angle from the vertical thetension in the rod would be equal to the weight of the body ?
Solution. Let S be the point of suspension and A thepoint from which the mass m is released. The point A is at a
vertical height 2l above the lowest point B of the swing. Henceat A the mass has gravitational potential energy mg(2l).
On being released, the mass swing along the dottedsemicircle, reaching the lowest point B where its energy is
entirely kinetic, 12
mv2 (v being the velocity at B). By energy
conservation, we have mg (2l) = 12
mv2
∴ v = 2 gl
As the mass is moving in a circle or radius l, it must be acted upon by a centripetal force.At the point B, the net upward force on the mass is T-mg, where T is the vertical tensionin the string. This provides the required centripetal force. Thus
T – mg =mv
l
2
But v2 = 4gl
∴ T = mgmgll
mg+ =4 5
Now, suppose the mass is released from the point A'. On reaching the point B', itdescends vertically through a distance l cos θ, thus losing potential energy by mgl cos θ. It,
however, gains kinetic energy 12
2mv′ , where v' is the velocity at B'.
A
S
T
B
l
v
m g
Fig. 30
200 Mechanics
Thus
mgl cos θ =12
2mv′
∴ v′ 2 = 2gl cos θAt the point B', the net radically inward force on the
mass is T – mg cosθ, which provides the required centripetalforce mv′2/l. Thus
T – mg cosθ =mv
l′2
or T = mg cos θ + m ( cos ) cos2 3gl
lmg
θ θ=
Let the angle θ be such that T = mg, then we have mg = 3mg cos θ
or cos θ =13
∴ θ = cos− = °1 1
371
Q.64. A ball is tied to a cord and set in rotation in avertical circle prove that the tension in the cord at the lowestpoint exceed that at the highest point by six times the weightof the ball.
Solution. Let v1 be the velocity of the ball as, it passesthe highest point. The force acting on it are its weight mg andthe tension T1 in the cord both acting downward. The resultantforce is thus T1 + mg, which provides the centripetal forcemv1
2
R. Thus
T1 + mg =mv1
2
RSimilarly, at the lowest point we shall have
T2 – mg =mv2
2
R
T2 – T1 =m
v v mgR
( )22
21 2− + ...(1)
At the highest point, the ball has gravitational potential energy mg (2R) and the kinetic
energy 12
21mv . At the lowest point it has entirely kinetic energy
12
22mv . By the law of
conservation of energy, we have
mg (2R) + 12
21mv =
12
22mv
R
T2
T1m g
v2
v1
m g
O
Fig. 32
θ
B
l cos θ
lA ′ S
B ′θ
m g cos θ m g
T
Fig. 31
Work, Energy and Momentum 201
orm
v vR
( )22
21− = 4 mg
Substituting this value in eq. (1) we have
T2 – T1 = 4 mg + 2 mg = 6 mg.
Q.65. A body slides down a curvedfrictionless track which is one quadrant of acircle of radius R. Find its speed at the bottomof the track. Point out the usefulness of theenergy method of solving dynamical problems.Will the conclusion regarding the speed at thebottom hold if friction were present ?
Ans. Motion on a frictionless inclinedplane: Let m be the mass of the body whichstarting from rest at the point 1, slides down africtionless curved track to the point 2. As there is no friction the forces acting on the bodyare (i) its weight mg, and (ii) the normal force N exerted on it by the track. Since the forceN is always normal to the direction of motion of the body, it does no work. In other words,only the gravitational force mg does work. Since the force is conservative, the mechanicalenergy of the body is conserved. Thus we can write.
E = K1 + U1 = K2 + U2
12
mv12 + mgh1 =
12
22 2mv mgh+
Here v1 = 0, h1 = R, h2 = 0, v2 = ?
O + mgR =12
022mv +
v = 2gR
The speed is therefore the same as if the body had fallen vertically through a height R.This proves the fact that the work done by a conservative force in moving a body througha distance is independent of the path chosen.
Here the resultant force acting on the body, and hence the acceleration depends on theslope of the track at each point, i.e., it varies from point to point. Thus the acceleration isvariable. As such we cannot use the Newton’s equation of motion which holds for constantacceleration. Hence to solve the problem by other methods we shall have to determine theacceleration at each point and then applying integration. All this is avoided in the energymethod.
We have seen above that the speed at the bottom is independent of the shape of thesurface. This conclusion would not hold if friction were present. This is so because the workof the friction force does depend on the path; the longer the path, the greater the work.
Q.66. A body of mass 0.5 kg. starts from rest and slides vertically down a curved trackwhich is in the shape of one quadrant of a circle of radius 1 met. At the bottom of the trackthe speed of the body is 3 m/s. What is the work done by the frictional force?
Solution. Let m be the mass and R the radius of the circle. When the mass is at reston the top of the track, its energy is entirely potential being equal to mg R. On reaching at
1
2Reference
level
N
m g
m
Fig. 33
202 Mechanics
the bottom the energy is entirely kinetic, being equal to 12
2mv . Let the work done against
the frictional force be Wf. Then, by the conservation of total energy. We have
mgR =12
2mv f+ W
Wf = mg mvR − 12
2
= (0.5 × 9.8 × 1) – 12
× 0.5 × (3)2
= 2.65 Joules
Q.67. A 0.5 kg block is released from rest at a point on a track which is one quadrantof a circle of radius 4m. It slides down the track and reaches its bottom with a velocity of6 m/s. Then it further slides a distance of 9 m on alevel surface and stops. How much work is done againstfriction in sliding on the circular track and what is thecoefficient of sliding friction on the horizontal surface?
[Ans. 10.6 Joule, 0.20]
Q.68. A body slides along a track with elevatedends and horizontal flat central part. The flat part hasa length l = 3met. The curved portions of the track arefrictionless. For the flat part the coefficient of kinetic friction is µk = 0.20. The body is releasedat point A which is at a height h = 1.5 meter above the flat part of the track. Where does theparticle finally comes to rest.
Solution. Suppose the body is releasedform rest at A where it has gravitationalpotential energy. It reaches B with kineticenergy which then carries it to C againstfriction. From C it rises up to D (say) wherethe kinetic energy in it at C is once againconverted into potential energy. It thenagain descends to C, goes to B rises up toa point E and this process continue untilfinally it comes to rest some where on thehorizontal part of the track.
The initial (potential) energy of the body at the point of release A is mgh. Since thecurved parts are frictionless energy is not spent up in moving the body on them. Energy isspent up in doing work against the friction only when the body moves on the horizontal part.The friction force against the direction of motion is
fk = µN = µmg
Suppose whole of the initial energy, mgh, is consumed in moving a distance d on thehorizontal part, the work done being.
W = fk × d = µ mgd
4m
B C9m
4m
A
Fig. 34
Fig. 35
Work, Energy and Momentum 203
By the law of energy conservation, we write
mgh = W = µmgd
∴ d =hµ
Here h = 1.5 meter and µ = 0.20
d =1 50 2
7 5..
.= meter
Thus the body will cover a distance of 7.5 m on the horizontal part which is 3.0 meterlong. This means that it will come to rest at a point F such that
BC + CB + BF = 7.5
3 + 3 + BF = 7.5
BF = 1.5 meter
Thus the point F is the middle point of the horizontal part where the body would cometo rest.
Q.69. A 3000 kg automobile at rest at the top of an incline 30 m high and 300 m longis released and rolls down the hill. What is its speed at the bottom of the incline if the averageretarding force due to friction is fk = 200 g ?
Solution. Let m be the mass of the automobile, h the height of the incline, x the lengthand v the velocity at bottom. The potential energy of the automobile when at rest at the top
of incline is mgh. The kinetic energy at the bottom is 12
2mv , and the energy dissipated
against the friction is fkx.
By conservation of total energy, we have
mgh =12
2mv + fkx
v2 = 2gh – 2xm
fk
= (2 × 9.8 × 30) – 2 300
3000200 9 8
×× × .
= 196
v = 14 m/s.
Q.27. A block of mass m slides down at 30° inclined plane of length l and coefficient offriction µ. With what speed it will reach the bottom. If it further slides on a similar horizontalsurface, how far will it go before coming to rest ?
204 Mechanics
Solution.
µNN
l
µN
N
l sin θ
θ
θ
m g
m g
Fig. 36
The weight of the block, mg, acts vertically downwards. Let N be the normal reactionexerted by the plane of the block.
Then N = mg cos θThe friction force opposing its sliding down will be
fs = µN = µmg cos θLet v be the velocity of the body on reaching the bottom of the plane. At the moment
of start the vertical height of the block is l sinθ. Its potential energy is, therefore, mg (l sin θ).
This is used up in providing the kinetic energy 12
2mv to the block and in doing work against
the friction fs. Thus, by the conservation of total energy. We have
mgl sin θ =12
12
2 2mv f mv mgls+ = + µ θcos
or 2gl sin θ – 2 µgl cos θ = v2
or v = [ (sin cos )].2gl θ µ θ−
On reaching the bottom, the block has kinetic energy 12
2mv . Suppose now it moves a
distance d on a horizontal surface before coming to rest. Thus the kinetic energy is used upin doing work against this friction. Now the friction force is µN = µmg and the work doneis µmgd. Then
12
2mv = µmgd
or12
2m gl[ (sin cos )]θ θ− µ = µmgd
or d =l (sin cos )θ µ θ−
µ
Work, Energy and Momentum 205
30°
2m
h = 2 s in 30°
Fig. 37
Q.71. A 0.2 kg block is thrust up a plane inclined at an angle of 30° to the horizontal withan initial velocity of 5 m/sec. It goes up 2m along the plane and then slides back to the bottom.Calculate the force of friction and the velocity of the block with which it reaches the bottomof the plane.
Solution. Let us first consider the upward motion.Let f be the force of friction. At the bottom where theblock starts moving, the energy is wholly kinetic (K =1/2 mv2). At the top where the block momentarily stops,the energy is wholly potential (U = mgh). The workdone against the friction during this motion is f × x,where x is the distance travelled. By the general lawof conservation of energy, we have
K = U + fx
or12
2mv = mgh + fx
Here, m = 0.2 kg, v = 5 m/s, x = 2 m, h = 2 sin 30° = 1.
∴12
0 2 5 2× . ( ) = 0.2 × 9.8 × 1 + f × 2
or 2.5 = 1.96 + 2f
f =2 5 1 96
2. .−
= 0.27 nt
Let us now consider the downward motion. Let v be the velocity of the block with whichit reaches the bottom. Now the potential energy (U = mgh) of the block at the top is used
up in giving kinetic energy (K = 1/2 mv′2) to the block on its reaching the bottom and in doing
work against the friction (f × x). Thus
U = K + fx
mgh =12
2mv′ + fx
0.2 × 9.8 × 1 =12
× 0.2 × v'2 + 0.27 × 2
or 1.96 = 0.1 v'2 + 0.54
or v'2 =1 96 0 54
0 114 2
. ..
.− =
or v'2 = 3.8 m/sec.
Q.72. A 12 kg block is pushed 20 m up the sloping surface of a plane inclined at an angle37 to the horizontal by a constant force of 120 nt acting parallel to the plane. The coefficientof friction between the block and the plane is 0.25. Calculate (i) work done by F (ii) increasein potential energy of the block (iii) work done against the friction and (iv) increase in kineticenergy. What becomes the work done ?
Solution. (i) The force F (= 120 nt) pushes to block through a distance x (= 20m). Hencethe work done
206 Mechanics
W = Fx = 120 × 20 = 2400 Joule.
(ii) initially, the potential energy is zero, finally itbecome mg. Here m = 12 kg and h = 20 sin 37° = 20× 0.6 = 12 m. Hence the increase in P.E.
∆U = mgh
= 12 × 9.8 × 12
= 1411 Joule.
(iii) The force of friction is µN, where µ is thecoeff. of friction and N the normal force of reactionexerted by the plane on the block. But N = mg cos 37°(see Fig.). Thus the force of friction is µmg cos 37°. Thework done against the force
W = force × displacement
= µmg cos 37° × x
= 0.25 × 12 × 9.8 × 0.8 × 20 = 470.4 joule
This work is converted into heat.
(iv) Initially, the kinetic energy is zero. Let ∆K be the change in kinetic energy. By thelaw of conservation of energy
W = ∆U + ∆K + w
∆K = W – ∆U – w
= 2400 – 1411 – 470 = 519 Joule
Q.73. Figure 39 shows the vertical section of a frictionless surface. A block of mass 2 kgis released from position A. Compute its kinetic energy as it reaches positions B, C, D. (givegravitational field = 9.8 jm–1 kg–1).
Solution. The (gravitational) force is conservative so that the mechanical energy isconserved. The loss in potential energy as the block comes down to B, C and D from A equalsthe corresponding gain in the kinetic energy
∴ K.E. at B = loss in P.E. between A and B (UA – UB)
= mg (hA – hB)
= 2 kg × 9.8 jm–1 kg–1 × (14 – 5) m
= 176.4 Joule
K.E. at C = UA – UC
= mg (hA – hC)
= 2 × 9.8 × 7.0 = 137.2 Joule
K.E. at D = UA
= mghA
= 2.0 × 9.8 × 14 = 274.4 Joule
NF
µN
20 m
37°
m g
37°
h =
20
sin
37°
Fig. 38
Work, Energy and Momentum 207
Fig. 39
Q.74. A small block of mass m slids along a frictionless track as shown in figure 40. Ifit starts from rest at P, what is the reaction exerted by the track on the block when it is atA, B and C. At what height from the bottom A of the loop should the block be released so thatthe force it exerts against the track at the top C is equal to its weight.
5R m g
C
N C
N Bm gN A
R
Am g
O
P
B
Fig. 40
Solution. Let us first calculate the velocities at A, B and C. At P the entire energy is
potential being equal to mg (5R) and at A it is entirely kinetic, being equal to 12
mvA2 . By
conservation of mechanical energy, we have
mg (5R) =12
mvA2
∴ vA2 = 10gR ...(i)
At B, the energy is partly kinetic 12
mvB2 and partly potential mg(R). By energy conservation.
208 Mechanics
mg(5R) =12
mvB2 + mg(R)
∴ vB2 = 8gR. ...(ii)
Similarly, at C we have
mg(5R) =12
mvC2 + mg(2R)
vC2 = 6gR ...(iii)
Now, at any point on the track, the force acting on the block are its weight mg actingvertically downward and the reaction force N exerted by the track acting radially inward. Theresultant radial force supplies the required centripetal force at that point.
NA – mg =mvA
2
R
=mR
(10gR) [from eq. (i)]
NA = 10mg + mg = 11mg.
For the point B, we have
NB =mv2
B
R
=m
gR
R( )8 = 6mg [from eq. (ii)]
= 8 mgAgain for the point C, we write
NC + mg =mv c
2
R...(iv)
=m
gR
R( )6 = 6 mg [from eq. (iii)]
∴ NC = 6mg – mg = 5mg.
Now, let h be the height from which the block must be released so that NC equals mg.Then from eq. (iv), the velocity vc must be given by
mg + mg =mvC
2
R
or vC2 = 2gR
By energy conservation at C, we have
mgh =12
mvC2 + mg(2R)
=12
m(2gR) + mg(2R)
= 3mgR
h = 3R
Work, Energy and Momentum 209
Q.75. A body is allowed to slide on an inclined frictionless track from rest position underearth’s gravity. The track ends in a circular loop of radius R. Show that the minimum height
h of the body, so that it may successfully complete the loop is given by h = 52
R.
h m g
C
O
P
N
v
B
A
R
Fig. 41
Solution. Let us first calculate the velocity v of the body which it acquires at the highestpoint C of the loop, when released from a height h. For this we use conservation of mechanicalenergy.
At point P, the body is at rest so that its mechanical energy is entirely potential and isequal to mgh At point C it has a velocity v and a height 2R, so that it has kinetic energy equal
to 12
2mv as well as potential energy equal to mg(2R). By energy conservation the mechanical
energy at P must be the same as at C because the entire track is frictionless. That is,
mgh =12
2mv + mg(2R)
or mg(h – 2R) =12
2mv
or v2 = 2g(h – 2R) ...(i)
Now, the forces acting on the body at C are its weight mg acting vertically downwardand the reaction force N exerted by the track which is directed toward the centre O of thecircle. In fact, at the point C both are directed toward O. Thus the resultant radial force atC is N + mg, and this supplies the required centripetal force mv2/R. That is
N + mg =mv2
RSince N cannot be negative, the minimum velocity of the body at C if it is to describe
the circle must correspond to N = 0 so that
mg =mvmin
2
Ror v2
min = gR ...(ii)
or vmin = gR
If the velocity is less than gR , the downward pull of the weight will be larger than therequired centripetal force and the body will lose contact from the loop. (The velocity canhowever be grater than gR ).
210 Mechanics
Now from eq. (i)
v2min = 2g (hmin – 2R).
Putting this value in eq. (ii) we get
gR = 2g(hmin – 2R)
or hmin =R2
+ 2R = 52
R
Q.76. A small body of mass m slideswithout friction around the loop the loopapparatus, starting at a height 3R above thebottom of the loop, where R, is the radius ofthe loop. Compute its radial acceleration at theend of a horizontal diameter of the loop. Fromwhat minimum height above the bottom of theloop it should start so that it may loop the loop?
Ans. 4 52
g, R
Q.77. A body is moving on a vertical circular frictionless track. Calculate the minimumvelocity which it should have at the lowest point of its path in order to go completely roundthe track.
Solution. Let v1 be the velocity at the highest point of the path, and v2 the correspondingvelocity at the lowest point. In order that the body does not leave the track at the highestpoint (where it is most likely to do so), the magnitude of V1 must be atleast such that thecentripetal force mV1
2/R at that point utilises the entire weight mg of the body. That is
mVR
12
= mg
or V12 = gR
Fig. 43
As the body slides down from the top to the bottom of the track, it loses potential energy
by an amount mg(2R) but gains an equivalent amount of kinetic energy 12
m (V22 – V1
2). Thus
mg(2R) =12
m (V22 – V1
2)
or V22 = V1
2 + 4gR
Fig. 42
Work, Energy and Momentum 211
But V12 = gR (proved above)
V22 = gR + 4gR = 5gR
or V2 = ( )5gR
Q.78. A particle of mass m = 0.2kg is moving inside a smooth vertical circle of radius R= 50 cm. If it is projected horizontally with velocity v0 = 4m/s from its lowest position findthe angle θ at which it will lose contact with the circle.
Solution. As the particle moves up the circular track, its velocity decreases. Let v bethe velocity at the point P. In this rise the particle gains gravitational potential energy by an
amount, mg (R + R sinθ), while it loses an equivalent amount of kinetic energy given by 12
m
(v02 – v2). By energy conservation, we have
mg (R + R sinθ) =12
m (v02 – v2)
or v2 = v02 – 2gR (1 + sin θ) ...(i)
Let us now consider the forces acting on the particle at the point P. These are particle'sweight mg acting vertically downward and the normal reaction N exerted by the track actingradially inward. The net force along the radius is N + mg sinθ which supplies the necessarycentripetal force mv2/R. That is,
N + mg sin θ = mv2/R.
The particle will loose contact for that value of θ for which N = 0. Thus
mg sin θ =mv2
R
Putting the value of v2 from eq. (i) we get
g sin θ =v g0
2 2− R (1 + sin )R
θ
=v0
2
R – 2g – 2g sin θ
or 3g sin θ =v0
2
R – 2g
or sin θ =v0
2 233gR
−
Here v0 = 4 m/s and R = 0.5 m
sin θ =( )
( . ) ( . )4
3 9 8 0 523
2
2m/s
m/s m−
= 1.09 – 0.67 = 0.42
or θ = sin–1 (0.42) = 25°.
Q.79. A small mass m starts from rest and slides down the smooth surface of a solidsphere of radius R. Assume zero potential energy at the top. Find (a) the change in potential
R
m g sin θN
m gR s in θ
O Rθ
θmg cos θv
P
R
Fig. 44
212 Mechanics
energy of the mass with angle (b) the kinetic energy as a function of angle (c) the angle at whichthe mass flies off the sphere. If there is friction between the mass and the sphere, does the massfly off at a greater or lesser angle ?
Solution. (a) The potential energy of the mass m at A is zero. As it slides down to P,it descends vertically through. AC. Hence the loss in its potential energy is given by
A
P
B
θ
m
A
P
R
θm gO
C
mg cos θ
m g sin θ
N
B
R
θ
Fig. 45
U = –mg (AC)
= –mg (AO – CO)
= –mg (R – R cos θ)
= –mg R (1 – cos θ)
(b) This lost energy appears as gain in kinetic energy K, as the surface is smooththerefore
K = + mgR (1 – cos θ)
(c) The net radially inward force on the mass m at P is mg cos θ – N, where N is thenormal reaction of the surface. This supplies the centripetal force, so that
mg cos θ – N =mv2
R
where v is the velocity at P the mass will fly off the surface at angle θ for which N is zero.i.e.,
mg cos θ =mv2
R
The kinetic energy at P is 12
2mv , which is mgR (1 – cos θ), as shown above. Substituting
this in the last expression, we get
mg cos θ = 2mg (1 – cos θ)
or 3mg cos θ = 2mg
or cos θ =23
or θ = cos–1 23 .
Work, Energy and Momentum 213
In presence of fiction, the velocity at P will be smaller and hence the mass will fly offthe surface at a large angle.
Q.80. What will be the spring constant if it stretches 10 cm where it has a potential energyof 5600 joule ?
Solution. The potential energy of a spring stretched through a distance is given by
U =12
2Kx
Hence the spring constant is given by
K =2
2U
x
Here U = 5600 joule and x = 10 cm = 0.1 m.
K =2 5600
0 1 2
×( . )
= 1.12 × 106 nt/m
Q.81. A spring-gun has a spring constant of 80 Nt/cm. The spring is compressed 12 cmby a ball of mass 15 gm. How much the potential energy of spring ? If the trigger is pulled,what will be the velocity of the ball be ?
Ans. The spring (elastic) force is conservative and hence the mechanical energy of thesystem is conserved. Before the trigger is pulled the spring has an elastic potential energyof compression given by
U =12
2Kx
where K is spring-constant and x is the distance of compression.
Here K = 80 nt/cm = 8 × 103 nt/m and x = 12 cm = 0.12 m
∴ U =12
× (8 × 103) (0.12)2 = 57.6 joule
This energy, when the trigger is pulled is converted into the kinetic energy 12
2mv of the
ball, where m is the mass and v the velocity of the ball. Thus
12
2mv = 57.6
Here m = 15 gm = 0.015 kg
∴ v2 =2 57 6 2 57 6
0 0157680
× . × ..m
= =
∴ v = 7680 = 87.6 m/s
Q.82. A mass of one kg suspended by a spring of force constant 8 × 104 dyne/cm. Findthe distance through which it should be pulled so that on releasing it passes through itsequilibrium position with a velocity of 1.0 m/s. [Ans. 11.2 cm]
214 Mechanics
Q.83. A block of mass 1 kg is forced against a horizontal spring of force constantK = 100 nt/m. Which is compressed by x1 = 0.2 m. When released, the block moves on a levelsurface a distance x2 = 1.0 m before coming to rest. Obtain the coefficient of friction µk betweenthe block and the surface.
Fig. 46
Solution. When the block is released, it receives the (elastic) potential energy 12 1
2Kx
stored in the compressed spring. This energy is used up in doing work against the frictionforce µKN= µKmg, in moving the body a distance x2. Thus
12 1
2Kx = µKmg × x2
or12
× 100 × (0.2)2 = µK × 1 × 9.8 × 1.0
∴ µK = 0.20
Q.84. A 2kg body moving on a level surface collides and compresses a horizontal springof force constant k = 2 nt/m through 2 m. Compute the velocity of the body while colliding.(Between block and surface µK = 0.25).
Fig. 47
Solution. The kinetic energy of the block, 12
2mv supplies the elastic potential energy,
12
2Kx , to the spring and does work against the friction force µKN = µKmg, as the block moves
the distance x. Thus, by the general law of energy conservation, we have
12
2mv =12
2Kx + µKmgx (g = 9.8 m/s2)
12
× 2 × v2 =12
× 2 × (2)2 + 0.25 × 2 × 9.8 × 2
v2 = 4 + 9.8 = 13.8
v = 13 8.v = 3.7 meter/sec.
Q.85. A 10 kg block slides from the top of a 30° inclined smooth plane and compressesa spring placed at the bottom of the plane through 2.0 meter before coming to rest. Calculate
Work, Energy and Momentum 215
the distance through which the block has slide before coming to rest and its speed on reachingthe bottom. The spring can be compressed 1.0 meter by a force of 100 Newton.
Ans. When a force F compresses a spring by x, we have
F = kx
where k is the force constant of the spring.
30° 30°
s s in 30°
S
Fig. 48
The given spring can be compressed by 1.0 meter by a force of 100 newton. That is
100 = K × 1
∴ K = 100 Nt/m
Let s the distance through which the block has slide before coming to rest. The verticaldistance through which it descends is s sin 30°. Now, at the moment of start, the block hasgravitational potential energy, the kinetic energy being zero. At the moment the maximumcompression of the spring occurs, again there is no kinetic energy. Hence the loss of gravitationalpotential energy of the block in sliding down, mg s sin 30°, equal to the gain of (elastic)
potential energy of the spring, 12
2Kx . That is
mg s sin 30° =12
2Kx
Here m = 10 kg, g = 9.8 m/s2, K = 100 Nt/m and x = 2.0 m
∴ 10 × 9.8 × s2
=12
× 100 × 4
s =100 410 9 8
×× . = 4.1 meter.
This distance of 4.1 meter includes the distance of 2.0 meter travelled while compressingthe spring before coming to rest. Thus the distance travelled on reaching the spring (butbefore compressing) is 2.1 meter. Up to this moment, the loss in potential energy of the block
is mg × 2.1 sin 30° which appears as kinetic energy of the block 12
2mv . Thus
mg × 2.1 sin 30° =12
2mv
9.8 ×2.1 × 12
=12
2v
v = 9 8 2 1. × .
v = 4.5 meter/sec.
216 Mechanics
Q.86. A 20 kg block of mass, initially at rest is droppedfrom a height of 0.40 meter on to a spring whose force constantis 1960 Nt/meter. Find the maximum distance that the springwill be compressed.
Ans. At the moment of release, the block has onlygravitational potential energy. At the moment of maximumcompression of the spring, the block loses some gravitationalpotential energy which is converted into the elastic potentialenergy of the spring.
Let m be the mass of the block, and K the force constantof the spring. Let y be the distance through which the spring
is compressed. Then the elastic potential energy in the spring is 12
2Ky . The total vertical fall
of the block is (h + y), so that the loss in its gravitational potential energy is mg(h + y). Bythe law of conservation of mechanical energy, we have
12
2Ky = mg(h + y) ...(i)
or y2 – 2mg
yK
– 2mgh
K= 0
∴ y =12
2 2 82mg mg mghK K K
±
+
...(ii)
The block of mass 20.0 kg and is dropped from a height of 0.40 meter. The force-constantK = 1960 newton/meter. On solving for y, we get
y = 0.4 meter
Q.87. A 20 kg body is released from rest so as to slidein between vertical rails and compress a vertical spring offorce constant K = 1920 Nt/m, placed at a distance h = 1.0meter from the starting position of the body. The rails offera friction force of f = 36 Nt opposing the motion of thebody. Find (i) the velocity v of the body just before strikingwith the spring (ii) the distance y through which the springis compressed, and (iii) the distance h through which thebody is rebounded off.
Ans. (i) The body slides a distance h against the friction force f just before striking thespring. Hence, the loss of gravitational potential energy of the body equals the gain of kineticenergy plus the work done against the friction of f. That is,
mgh =12
2mv fh+
20 × 9.8 ×1.0 =12
× 20 × v2 + 36 × 1.0
Fig. 49
Fig. 50
Work, Energy and Momentum 217
v2 =196 36
10−
= 16.0
∴ v = 4.0 m/s
(ii) At the moment when maximum compression y of the spring occurs, there is nokinetic energy. Hence, the loss of gravitational potential energy of the body equals the gainof elastic potential energy of the spring plus the work done against the friction force f. Nowthe total fall of the body is (h + y). Thus
mg (h + y) =12
2Ky + f (h + y)
20 × 9.8 × (1.0 + y) =12
× 1920 × y2 + 36 (1.0 + y)
160 (1.0 + y) = 960 y2 or 6y2 – y – 1 = 0
or y =1 1 24
12± +
or y = 0.5 (– sign is inadmissible).
(iii) The compressed spring with elastic energy 12
2Ky rebounds the body to a height h'.
Therefore the loss of elastic energy equals the gain of gravitational energy of the body plusthe work done against the friction force f. That is.
12
2Ky = mgh' + fh'
12
× 1920 × (0.5)2 = 20 × 9.8 × h′ + 36h'
240 = 232 h'
h' =240232
= 1.03 m.
Q.88. A body of mass 4kg slides on a horizontal frictionless table with a speed of 2 m/s. It is brought to rest in compressing a spring in its path. By how much is the springcompressed if the force constant of the spring is 64 kg/m.
Ans. Kinetic energy of the body
=12
2mv = 12
× 4 × 22 = 8 Joule.
Let the spring be compressed through a distance x so that the body loses its all kineticenergy and comes to rest. Obviously, the lost kinetic energy will be stored in the form of
potential energy 12
Cx2 of the system. Thus,
12
Cx2 =12
2mv = 8 Joule
x2 =16 16
6414C
= =
whence, x = 0.5 meter.
218 Mechanics
Q.89. A body of mass 2 kg is attached to a horizontal spring of force constant 8 N/m andthen a constant force of 6 newtons is applied on the body along the length of the spring. Findthe speed of the body when it has been displaced through 0.5 met. Now if the force is removed,how much farther the body will move before coming to rest?
Ans. Work done by constant force 6 Newton in displacing the mass through 0.5 meter,W = 6 × 0.5 = 3 Joule. The potential energy of the system at this displacement
U =12
Cx2 = 12
× 8 × (0.5)2 = 1 Joule
The work done W on the system increases the potential energy and kinetic energy ofthe system. If the speed acquired by the body is v m/sec, then
W =12
Cx2 + 12
2mv
3 = 1 + 12
× 2 × v2
v = 2 = 1.4 m/sec.
At the displacement 0.5 meters, the total energy of the system is 3 joule. Now, if theconstant force is removed, only 3 joule energy will remain with the system.
Now if the body is further displaced through a distance α against the elastic force of thespring so that it comes to rest, the total energy will become in the form of potential energyi.e.,
12
C (0.5 + α)2 = 3 or 82
(0.5 + α)2 = 3
or 0.5 + α = 3 2/ = 0.866
or α = 0.366 meter.
Q.90. A body of mass m is suspended from a spring. It comes to rest after a downworddisplacement x0. If a linear force is acting, then prove that
(i) the spring (force constant) is mg/x0.
(ii) the gravitational energy lost is mgx0.
(iii) the elastic potential energy gained is 12
mgx0.
What happened to the remaining energy ?
Ans. In the equilibrium position, the downward force mg on the spring will be balancedby the restoring force in the spring. So that
mg = –Restoring force = –(–Cx0) = Cx0
whence C = mg/x0 ...(i)
Loss in gravitational potential energy = mgx0 ...(ii)
Elastic potential energy = Cx dxx
0
0 = 12
C 02x =
12
mgx0 ...(iii)
Work, Energy and Momentum 219
Remaining energy = mgx0 – 12
mgx0 = 12
mgx0 ...(iv)
This energy is first changed to kinetic energy and finally to heat, when initial oscillationsstop.
Q.91. A car carries a framework ABCD shown below, in which a 200 gm mass P issupported between two spring of force constant 5 N/m each. Side AB is kept horizontal andalong the length of the car. And the pointer attached to P reads zero when the car is at rest.What will the pointer read when the car has:
(i) uniform speed 20 m/s on a straight road.
(ii) uniform speed 10 m/s on a circular road of radius 20 m
(iii) uniform acceleration 0.5 m/s2 on a straight road.
(iv) uniform acceleration –1.0 m/s2 on a straight road.
Fig. 51
Ans. (i) When the car is moving with constant speed on a straight road, then pointerwill read zero.
(ii) When the car is moving on a circular path, the force will be perpendicular to thespring and hence again the pointer will read zero.
(iii) In the case when car has forward acceleration 0.5 m/s2 the mass P will experiencea force F = ma = –0.2 × 0.5 = –0.1 newton. As the both spring exerts the force on the massP, the force at the displacement x is
F = – (C1 + C2)x i.e.,
0.1 = (5 + 5)x or x = 0 110.
= 1 cm = 10 mm
Hence, the pointer will read + 10 mm back to the zero position of the scale.
(iv) In this case, F = 0.2 × 1 = –(5 + 5)x
or x = − 0 210.
m = –2cm = – 20 mm
Hence the pointer will read –20mm forward to the zero of the scale.
Q.92. A body of mass 1 kg, initially at rest is dropped from a height of 2 meters on toa vertical spring having force constant 490 N/m. Calculate the maximum compression.
Ans. Loss in gravitational potential energy = Gain in potential energy by the spring
i.e., mg(h + x) =12
2Cx
220 Mechanics
Here, m = 1 kg, g = 9.8 m/s2, h = 2m, C = 490 N/m.
∴ 1 × 9.8 (2 + x) =4902
2x
or 100x2 = 4x + 8
or 25x2 – x – 2 = 0
whence x =1 1 200
5018 177
50± +
= .
= 0.3035 m
(–ve value is inadmissible)
SELECTED PROBLEMS
1. Explain clearly the work-energy theorem.
2. A force of 5 Nt acts on a body of 10 kg mass initially at rest. Compute the workdone by the force in the third second and also the instantaneous power exerted bythe force at the end of zero second.
3. Define conservative force (a) show that for a conservative force, ∇ =→ →
× F 0.
(b) Central force is conservative.
4. A body of mass 5 kg is released from a position of rest on a frictionless sphericalsurface. It then moves on a horizontal surface CD whose coefficient of kineticfriction is 0.2. An elastic spring with the force constant K = 900 Nt/m is placed atC. Find the maximum compression of the spring ; (g = 9.8 m/s2).
5. Prove that the rate of change of K.E. of a body is equal to the power exerted bythe force acting on it.
6. An electric field is given by E = ( ) ( ).i x y j x y2 3 5 4+ + + Find the scalar point.
7. What is a conservative force ? Show that for a conservative force, the work donearound (a) closed path is zero.
(b) Show that a conservative force can be expressed as negative gradient of potentialenergy.
(c) Prove that curl of a conservative force is zero.
8. (i) A light and heavy body have equal momentum, which one has greater kineticenergy ?
(ii) A body of mass m is moved to a height h equal to the radius of earth. Whatis the increase in potential energy ?
(iii) Two particles whose masses are in the ratio 1 : 4 have equal momentum. Whatis ratio of their kinetic energies ?
(iv) A 2000 kg motor car was running at a speed of 10 m/s. It was brought to a stopsuddenly by breaking. What is the amount of heat produced at the breakers ?
(v) A ball is dropped from rest at a height of 12 m, if it loses 25% of its kinetic
Work, Energy and Momentum 221
energy on striking the ground, what is the height to which it bounces ? Howdo you account for the loss in K.E. ?
9. A ball falls under gravity from a height of 10 m with an initial downward velocityv0. It collides with the ground and loses 50% of its energy in the collision and thenrises back to same height. Find (i) initial velocity v0 (ii) the height to which theball would rise after collision if the initial velocity was directed upward instead ofdownwards.
10. A bullet of mass m moving with a horizontal velocity ‘v’ strikes a stationary blockof mass M suspended by a string of length L.
222 Mechanics
5.1 CONSERVATION OF LINEAR MOMENTUM
If a force F→
is acting on a particle of mass m, then according to Newton’s second lawof motion, we have
F→
= d pdt
ddt
m v→
→= ( )
Where p m v→ →
= is the linear momentum of the particle. Momentum is a vector quantity.If the external force acting on the particle is zero, then
F→
=d pdt
→
= 0 or p m v→ →
= = a constant
Thus, in absence of an external force, the linear momentum of a particle remainsconstant. The law of conservation of momentum for a system of two particles, which areinteracting mutually, in absence of external forces is given by
p p1 2→ →
+ = constant
or m v m v1 1 2 2→ →
+ = constant
Now, let us consider a system of n particles whose masses are m1, m2 ........, mn. Thesystem can be a rigid body in which the particles are in fixed positions with respect to oneanother, or it can be a collection of particles in which there may be all kinds of internalmotion.
Suppose that the particles of the system are interacting with each other and are also
acted by external forces. If p m v p m v p m vn n n1 1 1 2 2 2→ → → → → →= = =, , ..., are the momenta of the
particles of masses m1, m2, ..., mn respectively, then the total momentum P→ of the system
is the vector sum of the momenta of individual particles i.e.,
222
5
Linear and Angular Momentum 223
P→
= p p p pn1 2 3
→ → → →+ + + +...
p1
→= m v m v m v m vn n1 1 2 2 3 3
→ → → →+ + + +...
Differentiating it with respect to time t, we have
ddtP
−→
= dpdt
dpdt
dpdt
dpdt
n1 2 3
−→ −→ −→ −→
+ + + +...
ordpdt
−→
= F F F→ → →
+ + +1 2 ... n
Where F F F→ → →
1 2, , ... n represent the forces acting on the particles of masses m1, m2, m3... mn respectively.
These forces include external and internal forces both. But according to Newton’s thirdlaw, the internal forces exist in pairs of equal and opposite forces, they balance each otherand so do not contribute any thing to the total force. Hence the right hand side in above
equation represents the resultant force F→
ext only due to the external forces acting on all theparticles of the system. The internal forces cannot change the total momentum of the system,because being equal and opposite they produce equal and opposite changes in the momentum.Hence, if we want to change the total momentum of the system of the particles, it isnecessary to apply external force on that system. Then the sum of external forces is
F→
ext = dpdt
ddt
p p pn
−→→ → →
= + + +( ... )1 2
If the resultant external force is zero, then
dpdt
−→
= 0 or P→
= a constant
Thus, if the resultant external force acting on a system of particles is zero, the totallinear momentum of the system remains constant. This simple but quite general result iscalled the law of conservation of linear momentum for a system of particles. It is to be notedthat the momenta of individual particles may change, but their sum i.e., the total linearmomentum remains unaltered in the absence of external forces.
The law of conservation of momentum is fundamental and exact law of nature. Noviolation of it has ever been found and it has been thoroughly checked by all kinds ofexperiments.
5.2 CENTRE OF MASS
Every physical system has associated with it a certain point whose motion characterizesthat of the system as a whole. When the system moves under an external force then thispoint moves in the same way as a single particle would move under the same external force.This point is called the “center of mass” of the system. The motion of the system can bedescribed in terms of the motion of its center of mass.
224 Mechanics
Let us consider a system of n particles of masses m1, m2, m3, ... mn with position vectors
r r r rn1 2 3
→ → → →, , , ..., relative to a fixed origin O. The position vector r
→cm of the center of mass of
this system is defined by
r→
cm=
m r m r m rm m m
n n
n
1 1 2 2
1 2
→ → →+ ++ + +
......
=
m r
m
m ri i
i
n
ii
n i ii
n
→
=
=
→
=
∑
∑∑=1
1
1
1M
...(i)
Fig. 1
where M =
=∑mii
n
1
is the total mass of the system. This is the expression for the position
vector of the center of mass of a system of particles.
Qualitatively, r→
cm represents a geometric point located at the “average” position of theparticle weighted in proportion to their masses. From the above definition, we draw twoconclusions:
(i) If all the n particles have the same mass m, then
r→
cm=
mr
nri
i
n
ii
n
M
→
=
→
=∑ ∑=
1 1
1
That is, the center of mass coincides with the geometric center of the system.
(ii) If the origin O is at the center of mass ( ),r→
=cm 0 then
m ri ii
n →
=∑
1
= 0
Linear and Angular Momentum 225
That is, the sum of the moments of the masses of the system about the center of massis zero.
5.3 CARTESIAN COMPONENTS OF THE CENTRE OF MASS
The position vectors r ri→ →
cm and are related to their Cartesian components by
r→
cm = x i y j z kcm cm cm + +
and ri→
= ix jy kzi i i+ +
Making these substitutions in Equation (i), the Cartesian components of r→
cm are givenby
xcm =1 1 1
1 1 1M M Mcm cmm x y m y z m zi i
i
n
i ii
n
i ii
n
= = =∑ ∑ ∑= =, , ,
where xi, yi, zi are the Cartesian co-ordinates of the ith particle
5.4 CENTRE OF MASS OF A SOLID BODY
Let us now consider a body with continuous distribution of mass. If the surface may besub divided into n small elements, the ith element being of mass ∆mi and located approximatelyat the point (xi, yi, zi). The Co-ordinates of the center of mass are then approximately givenby
xcm =
∆
∆
∆
∆
∆
∆
m x
m
y
m y
m
z
m z
m
i ii
n
ii
n
i ii
n
ii
n
i ii
n
ii
n=
=
=
=
=
=
∑
∑
∑
∑
∑
∑= =1
1
1
1
1
1
, ,cm cm
An n → ∞, the co-ordinates are defined precisely, by
xcm =x dm
dm
But dm = M (mass of the body).
∴ xcm =1M
x dm
ycm =1M
y dm
and zcm =1M
z dm
226 Mechanics
The vector expression corresponding to these three scalar expressions is
r→
cm=
1M
r dm→
The center of mass of a “homogeneous body” (having a uniform distribution of mass)
must coincide with the geometric center of the body. Thus if the homogeneous body has apoint, a line, or a plane of symmetry, its center of mass must lie at its point, line or planeof symmetry. This fact can be easily understood. From symmetry, the first moment of mass
r dm→
of the “homogeneous” body must be zero with respect to its geometric center:
r dm→
= 0
r→
cm= 0
i.e., the center of mass coincides with the geometric center. From this we see that the centerof mass of a body does not necessarily lie within the body. For example, in a homogeneous,ring the center of mass would be at its geometric center which does not lie within thematerial of the ring.
5.5 POSITION VECTOR OF THE CENTRE OF MASS
Let us consider a system of n particles of masses m1, m2, m3, ... mn with position vectors
r r r rn1 2 3
→ → → →, , , ..., relative to a fixed origin. The position vector r
→cm of the center of mass of this
system is defined by
r→
cm =m r m r m r
m m mn n
n
1 1 2 2
1 2
→ → →+ + ++ + +
......
=
m r
m
i ii
n
ii
n
→
=
=
∑
∑1
1
=1
1M
m ri ii
n →
=∑
where M =
=∑mii
n
1 is the total mass of the system.
5.6 VELOCITY OF THE CENTRE OF MASS
Let us write the expression for the position vector of a system of particles:
r→
cm =1
1 1 2 2M( ... )m r m r m rn n
→ → →+ + +
Linear and Angular Momentum 227
Differentiating it with respect to time, we obtain
d rdt
→cm =
11
12
2
Mm
drdt
md rdt
md rdtn
n
→ → →
+ + +
.....
or v→
cm=
11 1 2 2M
( ... )m v m v m vn n
→ → →+ + +
v→
cm=
1
1M
m vi ii
n →
=∑
where v→
cm is the velocity of the center of mass and vi
→ the velocity of the ith particle. This
is the required expression.
5.7 CENTER OF MASS FRAME OF REFERENCE
It is a reference frame attached with the center of mass of a system of particles (or abody). It is also known as C-frame of reference. In this frame the velocity of the center of
mass is zero by definition ( v→
cm = 0).
5.8 MOTION OF THE CENTER OF MASS OF A SYSTEM OF PARTICLES SUBJECTTO EXTERNAL FORCES
Let us consider a system of n particles of masses m1, m2, m3, ... mn with position vectors
r r r rn1 2 3
→ → → →, , , ..., with respect to a fixed origin, and subjected to external forces F F F1 2
→ → →, , ..., n
exerted by the surroundings.
The position vector of the center of mass of the system is defined by
r→
cm =m r m r m r
m m mn n
n
1 1 2 2
1 2
→ → →+ + ++ + +
......
r→
cm=
11 1 2 2M
( ... )m r m r m rn n
→ → →+ + +
where M (= m1+ m2 + m3 + ... + mn) is the total mass of the system. Differentiating it withrespect to time, we obtain
d rdt
→cm =
11
12
2
Mm
drdt
mdrdt
mdrdtn
n
→ → →
+ + +
...
or v→
cm=
11 1 2 2M
( ... )m v m v m vn n
→ → →+ + +
228 Mechanics
Where v→
cm is the velocity of the center of mass and vi
→ is the velocity of the particle m1,
and so on.
Differentiating again, we get
d vdt
→cm =
11
12
2
Mm
dvdt
mdvdt
mdvdtn
n
→ → →
+ + +
...
or M cma→
= m a m a m an n1 1 2 2→ → →
+ + +...
where a→
cm is the acceleration of the center of mass, a1
→ is the acceleration of the particle m1
and so on.
Now, from Newton’s second law, the external force F1
→ acting on the first particle is given
by F1
→ →= m a1 1 and so on. Thus,
M cma→ = F F F1 2
→ → →+ + +... n
or M cma→ = Fext
→
Where Fext→
is the sum of the external force acting on all the particles. Thus the productof the total mass of a system of particles and the acceleration of its center of mass is equalto the sum of the ‘external’ force acting on the particles. This means that the center of massof a system of particles moves as if all the mass of the system were concentrated at it andall the ‘external’ force were applied at it. This result holds whether the system is a rigid bodywith particles in fixed position or a system of particles with internal motions.
5.9 LINEAR MOMENTUM IN CENTER OF MASS FRAME OF REFERENCE
In the reference frame attached with the center of mass (C-frame) the velocity of the
center of mass, v→
cm is zero by definition. Therefore, the total linear momentum of a systemof particles is also zero:
P→
= M cmv→
= 0
Whatever may be the velocities of the constituent particles. Thus the center-of masscoordinate system is also referred to as the center of-momentum coordinate system, to stressthe fact that in such a system the total momentum is zero.
The C-frame is important because many experiments that we perform in our laboratoryor L-frame of reference can be more simply analyzed in the C-frame of reference (which
moves with a velocity vcm→
relative to the L-frame).
5.10 SYSTEM OF VARIABLE MASS
There are certain systems whose mass does not remain constant during their motion,but varies with time. A rocket is propelled by ejecting burnt fuel which causes the total mass
Linear and Angular Momentum 229
of the rocket to decrease as the rocket accelerates. A falling rain drop collapses with smallerdrops which increase its mass. Let us see what form Newton’s law takes for such system of‘variable mass’.
Let a system of mass M be moving at a velocity v→
at any instant t in a particularreference frame. Let at a later instant t + δt, a part δM separated from the system be moving
with a velocity u→
and the remaining system M-δM with new velocity v v→ →
+ δ . If we still treatboth parts as forming one and the same system, then for the finite time interval δt, we canwrite:
Fext→ = P P M M) ( M. M
→ → → → → →− = − + − −f i
tv v u v
tδδ δ δ
δ[( ) ] [ ]
= MM M Mδ
δδδ
δ δδ
δδ
vt
vt
vt
ut
→→ → →
− − +
Mv δM
u
M - Mδ
At t + tδv + vδ→ →
At t
→
→
Fig. 2
Now if δt → 0 (i.e., the mass of the system is continuously decreasing), then
δδvt
→
=d vdt t
ddt
v
→→
= − =; ;δδ
δM M0
so that Fext→ = M
M MM
Md vdt
vddt
uddt
ddt
v uddt
→→ → → →
+ − = −( )
This expresses Newton’s second law as applied to a body of variable mass. We note thatddt
v( )M→
is not equal to the external force acting on the system (unless the ejected mass
comes out with zero speed).
We can write the above equation as
Md vdt
→
= FM
ext→ → →
+ −( )u vddt
or Md vdt
→
= F Mext rel
→ →+ v d
dt
Where v→
rel is the velocity of the ejected mass relative to the main body. The last termin the above equation is the rate of change of momentum of the system due to the massleaving it. It can be taken as the reaction force exerted on the system by the leaving mass.Thus we can write
Md vdt
→
= F Fext reaction
→ →+
230 Mechanics
5.11 MOTION OF A ROCKET
The rocket is the most interesting example of a system of variable mass. Its motion canbe explained on the basis of Newton’s third law of motion and the momentum principle. Itconsists of a combustion chamber in which liquid or solid fuel is burnt. The heat of combustionraises the pressure very high inside the chamber. Therefore, hot gases (produced by combustion)are expelled from the tail of the rocket in the form of a jet with a very high exhaust velocity.Consequently, the rocket rushes in the forward direction.
The rocket exerts an action force on the gas-jet in the backward direction, while the gas-jet exerts a reaction force on the rocket in the forward direction. These are the internalforces in the (rocket + gas) system. In the absence of external forces, the total momentumof the system (rocket + gas) is constant. The gas-jet acquires momentum in the back warddirection and the rocket acquires an equal momentum in the forward direction.
Let us now derive an expression for the final velocity of the rocket. Let M (a variable)
be the mass of the (rocket + unburnt fuel) at an instant t and v→
its velocity in a fixed(laboratory) frame of reference. Suppose in a time-interval dt an amount of Mass dM isejected from the rocket in the form of gas-jet.
If u→
be the velocity of the gas-jet in the laboratory reference frame, then its velocity
relative to the rocket v→
rel , would be given by
v→
rel= u v
→ →−
v→
rel is known as “exhaust velocity”. Now, according to the Newton’s second law asapplied to a system of variable mass, we have
Md vdt
→
= FM
ext rel→ →
+ vddt
v0→
Y
O
v→
vR e s→ dM/dt
Thrust
x
M
Fig. 3
Where Fext→
is the ‘external’ force acting on the system and vddt
→rel
M in the reaction force
exerted on the system by the leaving mass. In the case of a rocket, this term is called the
‘thrust’ exerted on the rocket by the ejecting gas-jet. The external force Fext→
is the force ofgravity on the rocket and the air resistance.
To solve the above equation, let us assume that the exhaust velocity v→
rel is constant.Also, neglecting air resistance and the variation of the gravity with altitude, we may write
Fext→
= M g→
, so that the last equation becomes:
Linear and Angular Momentum 231
Md vdt
→
= MM
relg vddt
→ →+
Now, suppose the motion of the rocket is vertical. Then v→
is directed upward and v→
rel
and g→ downward. The last equation may now be written as
Md vdt
→
= − −→
M Mrelg v d
dt
or dv = − −gdt vd
relM
M
Integrating from the beginning of the motion (t = 0), when the velocity is v0 and the massof the rocket + fuel is M0, upto an arbitrary time t, we have
dvv
v
0
= − − g dt v dt
0 0
rel
M
MM
M
or v – v0 = − −gt v erel0
MM
log
or v – v0 = − +gt v erel0M
Mlog
or v = v v gte0 + −rel0M
Mlog
If t is the time required for burning all the fuel, then M is the final mass and v is themaximum velocity attained by the rocket.
If the force of gravity is ignored, then the above expression reduces to
v = v v e0 + rel0M
Mlog
Further, if the initial velocity v0 of the rocket is zero, then
v = v erel0M
Mlog
Limitation of One-stage RocketIn the absence of gravity and air resistance, the ultimate velocity of the rocket at burn
out (when all the fuel has been used up) is v erel0M
Mlog . Thus it can be increased by increasing
the exhaust velocity relv and the ratio MM
0
. However, the exhaust velocity cannot be more
than 2.5 km/sec with the conventional chemical fuels. Further, the fact that the rocket shell
232 Mechanics
must be strong enough to hold the fuel sets an upper limit on the ratio MM
0
also. In actual
practice, this ratio is not greater than 4. Therefore, the maximum velocity of the rocket atburn out is
v = 2.5 loge 4 = 2.5 × 1.39 = 3.5 km/sec
This velocity is much less than the escape velocity (11 km/sec) or the orbital velocity nearthe surface of the earth (8 km/sec). Thus a single-stage rocket is incapable to put spacesatellites in orbits or escape the earth’s gravitational field.
5.12 MULTI STAGE ROCKET
To attain higher velocities, the rocket is designed in stages. For example, a two-stagerocket means that one rocket is placed on the top of another rocket. When the fuel of thefirst-stage (lower) is exhausted, its rocket casing is detached and drops off. The velocityattained so far becomes the initial velocity of the second stage which is now ignited . Theremoval of the surplus mass contained in the first stage considerably helps in attaining stillhigher velocity.
Suppose that the initial mass of each of the first-and second stage rocket (plus fuel) isM0/2, and the mass of each rocket casing is M0/2. The total initial mass is thus M0. Thevelocity v′ attained when the first stage is detached is given by
v′ = v ve erel0
relM
Mlog
/log
22=
This is now the initial velocity for the second stage. The final velocity v attained, whenthe fuel of the second stage is exhausted, is given by
v = v v e′ + rel0M
Mlog
//22
v = v ve erel rel0M
Mlog log2 +
Taking vrel = 2.5 km/sec and M0/M = 4, then we have
v = 2.5 km/sec (loge 2 + loge 4)
= 2.5 km/sec × loge 8
= 2.5 × 2.08 km/sec
= 5.2 km/sec
Thus the final velocity attained by a two-stage rocket is greater than that attained bya single-stage rocket of the same weight and fuel supply. The velocity can be further increasedby adding more stages.
The rocket of equal stages is not generally the optimum construction. In fact the firststage should be made much larger than the second in order to obtain a high final speed.
NUMERICALS: SYSTEMS OF VARIABLE MASS: ROCKET
Q. 1. Solve the last problem is M = 200 kg, m = 10 gm, n = 10/sec and vrel = 500 m/s.
Linear and Angular Momentum 233
Solution. The magnitude of the acceleration of the trolly at the instant the mass is M,
is given by a = vmn
rel M
=500 10 10
2000 25
2× ×.
−= m/s2
The magnitude of the average thrust on the system is given byFreaction = vrel mn
= 500 × 10–2 × 10= 50 nt.
Q. 2 (a). A man of mass m is standing on a trolly of mass M which is moving with a
velocity v→
on frictionless horizontal rails. If the man starts running opposite to the direction
of motion of the trolly and his velocity relative to the trolly is v→
rel just before he jumps off,find the change in velocity of the trolly.
(b) Let us now assume that there are n men, each of mass, m, on the trolly. Should theyall run and jumps off together or should they do so one by one in order to give a greatervelocity to the trolly.
Solution. (a) The initial mass of the (trolly + man) system is (M + m) which is moving
with a velocity v→
in a fixed frame of reference. As the man jumps, he acquires a backwardmomentum and the trolly acquires forward momentum. The Newton’s second law, in theabsence of external forces, gives
( )M +V
mddt
→
= VM +
rel→ d m t
dt( )
( )M V+→
m d = V (Mrel→
+d m) ...(i)
(M V+→
m) ∆ = V M +rel→
∆ ( )m
Here ∆(M + m) = –m (the man of mass m jumps off). Therefore
(M V+→
m) ∆ = −→Vrel m
∴ Change in velocity
∆ V→
= −+
→V
Mrel
mm
This equation shows that the speed of the trolly will increase by mvrel/(M + m) in its
initial direction of motion (i.e., opposite to v→
rel )(b) Let us now suppose that n men are standing on that trolly. If they all jumps off
together, the change in velocity would be
∆ V→
= −+
→V
Mrel
mnmn
...(ii)
234 Mechanics
But if they jump one by one, the mass of system will go on changing till the last manjumps. In this case we can write eqn. (i) as
d V→
= VM)
( M)rel
→ ++
d mnmn(
∆ V→
= V MM
V Mrel
M +
M
rel MM→ →
++
+= + d mn
mnmn
mn
e mn( )
( )log ( )
= VM
M +V
M +M
rel rel→ →
= −log loge emnmn
= − +
→V
Mrel loge
mn1
= −→V
Mrel
mn...(iii)
A comparison of eqns. (ii) and (iii) shows that the men would give a greater velocity tothe trolly by running and jumping off one by one.
Q. 3. Fine particles of sand are being dropped continuously from a fixed container on toa moving belt. Find out the force necessary to have the belt moving at a constant speed andshow that the power supplied by this force is twice the rate of increase of the K.E. of thesystem.
Solution. Suppose the belt is moving with a constant velocity V→
in a certain referenceframe. The container is fixed in this reference frame. As sand particles are falling on the belt,the mass of the system (belt + material on it) is continuously increasing. Let M be the massof the system at any instant, dM/dt the rate at which the material is falling on the belt. Thenthe Newton’s second law as applied to a system of varying mass, gives
MVd
dt
→
= F VM
ext rel→ →
+ ddt
...(i)
Where Fext→
is the external force acting on the system and Vrel→
is the relative horizontalvelocity of the falling mass relative to the belt.
Here ddtV→
= 0 (as V→
is constant)
Vrel→
= −→V
and ddtM
is positive (the mass of the system is increasing with time).
Therefore the equation (i) becomes
0 = F V Mext
→ →− d
dt
Linear and Angular Momentum 235
Fext→
= VM→ d
dt...(ii)
X
v→
F
Z
Y
Fig. 4This is the external force acting on the system under which is moving at a constant rate.Alternative Method. We can also determine the force by applying momentum principle.
Let M′ be the mass of the belt and M that of the material on the belt at any instant. Themomentum of the system is
P→
= (M + M) V′→
Now the rate of change of momentum gives the force. Thus
F→
= ddt
ddt
PM + M) V
→→
= ′(
= VM
as V and M are constants)→ →
′ddt
, (
where ddtM
is the rate at which the material is falling on the belt i.e., the rate of increase
of mass of the system.
Now, the power P supplied by the force Fext→
is the rate of doing work. In vector rotation,it is given by the dot product of the force and velocity. That is
P→
= F Vext→ →
.
= VM
V VM→ →
=ddt
ddt
. 2 V V V→ →
=
. 2
=ddt
MV2 ( V is constant)
= 212
22ddt
ddt
( ) ,MVK=
236 Mechanics
where K = 12
MV2
is the K.E. Thus the power supplied by the external force acting on the
system is twice the rate of increase of K.E. of ‘whole’ system. This is an example in whichthe mechanical energy is not conserved.
Q. 4. A wagon filled with sand has a hole so that sand leaks through the bottom at a
constant rate − =dmdt
λ. A force F→
acts on the wagon in the direction of its motion. If its
instantaneous velocity is V→
, write the equation of motion.
Solution. Let m be the instantaneous mass and V→
the instantaneous velocity of thesystem (wagon + sand). The instantaneous momentum is
P→
= mV→
By Newton’s law, the rate of change of momentum is the force F→
acting on the system.That is
F→
= ddt
ddt
mP
V
→→
= ( ) = mddt
dmdt
VV
→→
+
The rate of loss of mass is − =dmdt
λ, so that
F→
= mddtV
V
→→
− λ
If m0 be the initial mass, then m = m0 – λt, so that
F→
= ( )m tddt0V
V− −→
→λ λ .
This is the required equation.Q. 5. A rocket having initial mass of 240 kg ejects fuel at the rate of 6 kg/s with a velocity
of 2 km/s vertically downward relative to itself. Calculate its velocity 25 seconds after start,taking initial velocity to be zero and neglecting gravity.
Solution. Neglecting gravity, the velocity of a rocket at any time t is given by
v = v v e0 rel0M
M+ log .
where M0 is the initial (at t = 0) mass of the rocket plus fuel and M is the mass remainingat time t. If the initial velocity v0 is zero, then
v = v erel0M
Mlog
Linear and Angular Momentum 237
Here vrel = 2 km/s, M0 = 240 kg. As the fuel is ejected at the rate of 6 kg/sec, the massconsumed in 25 sec is 150 kg. Therefore, the mass remaining after 25 sec is M = 240 – 150= 90 kg. Thus
v = 2 24090
loge
= 2 2 3 2 6710× . log ( . )
= 2 × 2.3 × 0.4265= 1.96 km/sec.
Q. 6. An empty rocket weighs 5000 kg and contains 40,000 kg of fuel. If the exhaustvelocity of the fuel is 2.0 km/s, find the maximum velocity gained by the rocket. (loge10 = 2.3,log10 3 = .4771).
Solution. Ignoring gravity effect, the velocity of a rocket at any time t is given by
v = v v e0 rel0M
M+ log
When Mo is the initial (at t = 0) mass of the rocket plus fuel and M is the mass remainingat time t. The velocity v attains maximum value when all the fuel is burnt. M is then themass of the empty rocket.
Here, initial velocity v0 = 0, M0 = 5000 + 40,000 = 45,000 kg, M = 5000 kg andVrel = 2 km/s.
∴ vmax = 2450005000
× loge
= 2 × loge(3)2
= 2 × 2 loge 3 = 2 × 2 × 2.3 × 0.4771= 4.4 km/sec.
Q. 7. From the nozzle of a rocket 100 kg of gases are exhausted per second with a velocityof 1000 m/s. What force (thrust) does the gas exert on the rocket.
Solution. The thrust ( )Freaction→
exerted by the escaping gas on the rocket is given by.
Freaction→
= VM
rel→ d
dt
Here vrel = 1000 m/s and ddtM kg / sec= 100
Freaction→
= 1000 × 100= 105 nt.
Q. 8. The first and second stages of a two stage rocket have weights 100 kg and (10 kgand carry 800 kg and 90 kg of fuel supply. The velocity of ejected gases relative to the rocketis 1.5 km/s. Find the final velocity attained by the rocket (loge10 = 2.3).
Solution. Ignoring gravity, the velocity of a rocket at any time t is given by
v = v v e0 rel0M
M+ log
238 Mechanics
where v0 is the initial velocity, M0 the initial mass of the rocket plus fuel, and M the massof the rocket plus unburnt fuel at time t.
For the operation of the first stage. We have; v0 = 0, M0 = 100 + 10 + 800 + 90 = 1000kg and M = 100 + 10 + 90 = 200 kg.
∴ v = 0 + 1.5 loge 5This is the initial velocity for the second stage. Now, the first stage rocket of mass 100
kg drops off. Thus for the second stage, we havev0 = 1.5 loge 5 km/s, Mo = 10 + 90 = 100 kg and M = 10 kgv = 1.5 loge 5 + 1.5 loge 10
= 1.5 loge 50= 1.5 × 2.3 × log10 50= 1.5 × 2.3 × 1.6990 = 5.86 km/s.
If we would have a single stage rocket of the same total mass 110 kg carrying sameamount of fuel of 890 kg, then for its operation we would have
v0 = 0, Mo = 110 + 890 = 1000 kg and M = 110 kg
∴ v = 0 1 5 1000110
1 5 9 1+ =. log . log .e e
= 1.5 × 2.3 × log10 9.1= 1.5 × 2.3 × 0.959 = 3.31 km/sec.
Thus the final velocity attained by a multistage rocket is much greater than that attainedby a single stage rocket of the same total mass and fuel supply.
Q. 9. A 6000 kg rocket is set for vertical firing. If the gas exhaust speed is 1000 m/s, howmuch gas must be ejected each second to supply the thrust needed (a) to overcome the weightof the rocket, (b) to give the rocket an initial upward acceleration of 20 m/s2.
Solution. (a) The equation for the vertical motion of the rocket is
M dvdt
= − −M Mrelg v
ddt
...(i)
where M is the mass of the rocket + fuel at any instant t. The term M dvdt
represents the
instantaneous net upward force acting on the rocket, and vddtrelM
is the thrust. In order to
just overcome the weight of the rocket, the thrust need not give any net upward force i.e.,
M dvdt
= 0. Then eqn. (i) becomes
− −M Mrelg v
ddt
= 0
ddtM
= −M
rel
gv
= − = −6000 9 81000
58 8× . . kg/sec.
Hence the gas should be ejected at the rate of 58.8 kg/sec.(b) If the rocket is to be given an initial upward acceleration of 20 m/s2, then plutting
dvdt
= 20 in eqn. (i) we get
Linear and Angular Momentum 239
20M = − −M Mrelg v
ddt
vddtrelM
= − −M Mg 20
ddtM
=− +M
relvg( )20 = − +6000
10009 8 20( . )
= –178.8 kg/sec.Q. 10. A 8000 kg rocket is set for vertical firing. If the exhaust speed is 800 m/s, how much
gas must be ejected/second to supply the thrust needed (i) to overcome the weight of therocket, (ii) to give the rocket an initial upward acceleration of 3g.
[Ans. (i) –98 kg/s, (ii) –392 kg/s]Q. 11. A rocket of mass 20 kg has 100 kg of fuel. The exhaust velocity of the fuel is 1.6
km/s. Calculate the minimum rate of consumption of fuel so that the rocket may rise fromthe ground. Also calculate the final vertical velocity gained by the rocket when the rate ofconsumption of the fuel is (i) 2.0 kg/s, (ii) 20 kg/s.
Solution. With the minimum rate of fuel consumption, the thrust, vddtrelM, supplied to
the rocket will just overcome its initial weight Mog and raise it from the ground. Thus
vddtrelM
= M0g
ddtM
=M
kg/sec.0
rel
gv
= =200 9 81 6 10
1 2253× .
. ×.
Now, the velocity attained by the rocket at any time t is
v = v v gte0 + −rel0M
Mlog
where M is the mass of the rocket plus unburnt fuel at time t. If T is the time in which theentire fuel has burnt, then v is the final velocity attained by the rocket. Thus, if the initialvelocity v0 = 0, we have
vmax = v gerel0M
MTlog −
(i) vrel = 1.6 km/s = 1.6 × 103 m/s, log log log .e e eMM
0 = = =20020
10 2 3 and T = 180
290= sec
vmax = (1.6 × 103 × 2.3) – (9.8 × 90) = 3680 – 882 = 2798 m/sec∼ 2.8 km/s
(ii) T = 18020
sec= 9
vmax = 3680 – 88.2 = 3591.8 m/sec~ 3.6 km/sec.
240 Mechanics
Q. 12. A 103 kg rocket is set vertically on its launching pad. The propellant is expelledat the rate of 2 kg/s. Find the minimum velocity of the exhaust gases so that the rocket justbegins to rise. Also find the rocket’s velocity 10 sec after ignition, assuming the minimumexhaust velocity.
Solution. With the minimum velocity of exhaust gases, the thrust, vddtrelM , supplied to
the rocket will just overcome its initial weight M0g and raise it from the ground. Thus
vddtrelM
= M0g
vrel =MM
0gd dt/
× .= 10 9 82
3
= 4.9 × 103 m/sNow, the velocity attained by a rocket at any time t is given by
v = v v gte0 reloM
M+ −log ,
where M is the mass of the rocket plus unburnt fuel at time t. If the initial velocity vo is zero,then
v = v gterel0M
Mlog −
Here vrel = 4.9 × 103 m/s, M0 = 103 kg, and t = 10 sec. As the fuel is expelled at the rateof 2 kg/sec, the mass consumed in 10 sec is 20 kg. Therefore the mass remaining is M = 1000– 20 = 980 kg. Thus
v = ( . × ) log ( . ) ( )4 9 10 1000980
9 8 103e −
= ( . × ) × . log log4 9 10 2 3 1000 980 98310 10− −
= 4.9 × 103 × 2.3 × (3.0000 – 2.9912) – 98
= 99 – 98 = 1.0 m/s
6
6.1 COLLISION
It is defined as the phenomenon of the change in the velocities of bodies during the verysmall time interval of their contact. In the event two bodies collide with one another theyare deformed. The kinetic energy before collision of the system transforms completely orpartially into potential energy of elastic deformation and into internal energy of the bodies.
In collision process it is not at all assumed that the particles actually come in contactas contrary to every day life’s sense of collision. The significance of collision studies lies inthe fact that they yield the information regarding the forces which act between particles.
Elastic CollisionIf the force of interaction between the colliding bodies are conservative, the kinetic
energy remains conserved in the collision and the collision is said to be elastic. Collisionbetween atomic, nuclear and fundamental particles are usually elastic. Collisions betweenivory and glass balls are approximately elastic.
Inelastic Collision: When the kinetic energy is changed in the collision, the collisionis said to be inelastic (the momentum as well as the total energy is still conserved). Collisionsbetween gross bodies are always inelastic to some extent. When two bodies stick togetherafter collision, the collision is said to be completely inelastic. When a bullet hitting a targetremains embedded in the target, the collision is completely inelastic.
6.2 ELASTIC COLLISION IN ONE DIMENSION
Suppose two particles of following specification undergo elastic head-on collision.
Let m1 = mass of particle 1, m2 = mass of particle 2, u1 = velocity of mass 1 beforecollision, u2 = velocity of mass 2 before collision, v1 = velocity of mass 1 after collision,v2 = velocity of mass 2 after collision.
By the law of conservation of energy
K.E. before collision = K.E. after collision
∴12
121 1
22 2
2m u m u+ =12
121 1
22 2
2m v m v+
241
242 Mechanics
or m u v1 12
12( )− = m v u2 2
222( )− ...(i)
1 2 1 2
mu
1
1
mu
2
2
mv
1
1
mv
2
2
Befo re co llis ion After co llis ion
Fig. 1
From law of conservation of Momentum
m1u1 + m2u2 = m1v1 + m2v2
or m1 (u1 – v1) = m2 (v2 – u2) ...(ii)
Dividing (i) by (ii),
m u v
m u v1 1
212
1 1 1
−
− =
m v u
m v u2 2
222
2 2 2
−
−
or (u1 + v1) = (v2 + u2)
or (u1 – u2) = – (v1 – v2) ....(iii)
Which suggests that relative velocity of the particles after elastic collision is equal andopposite to the relative velocity before collision.
Putting Equation (iii), in Eq. (ii), we obtain
m1(u1 – v1) = m2 (u1 + v1 – u2 – u2)
or u1(m1 – m2) + 2m2u2 = v1 (m1 + m2)
or v1 =m mm m
um
m mu1 2
1 21
2
1 22
2−+
+
+
...(iv)
Similarly
v2 =m mm m
um
m mu2 1
1 22
1
1 21
2−+
+
+
...(v)
General Case: Newton experimentally showed that two smooth spherical balls, whenapproach, collide and separate, then they obey the rule.
v1 – v2 = – e (u1 – u2)
v vu u
1 2
1 2
−− = – e
That is, ratio of velocities of separation and approach are in the ratio e:1. Where ‘e’ has+ve value and is called coefficient of Restitution, which depends on (i) the material of theobject (ii) shape and size of colliding objects. Its value lies between 0 and 1. For perfectlyinelastic collision, e = 0 and for perfectly elastic collision, e = 1.
Special cases:
(1) when m1 = m2 then the equation (iv) and (v)
Collision 243
v1 = u2 and v2 = u1
i.e., colliding bodies, simply exchange velocities as a result of collision.
(2) when u2 = 0, i.e., second body is initially at rest, then
v1 =m mm m
u1 2
1 21
−+
...(vi)
and v2 =2 1
1 21
mm m
u+
...(vii)
(3) When u2 = 0 and m2 >> m1 then from Equations (vi) and (vii) becomes
v1 = –u1 and v2 = 0 (nearly)
Thus, when a lighter body collides with too heavy body at rest, the lighter bodyreverses the course of motion with no change in the rest state of heavier body.
(4) When m2 << m1, in addition to u2 = 0, then from Equation (vi) and (vii)
v1 = u1 and v2 = 2v1
Thus, when a very heavy body collides with a lighter body at rest, then velocity ofheavier body remains nearly unchanged and lighter body moves with nearly twicethe velocity of heavy body.
Fraction of K.E. Transferred in Elastic Head on CollisionSuppose a particle 1 collide with particle 2 which is initially at rest. After collision,
particle 2 acquire velocity v2 and has K.E. = ½ m2 v22. Before collision, it has K.E. = 0. So
kinetic energy transferred = ½ m2v22 .
Initial total K.E. = ½ m1 u12
Thus, fractional K.E. transferred = ½½
m vm u
2 22
1 12
Putting the value of v2 from equation (vii)
v2 =2 1
1 21
mm m
u+
We get, Fractional K.E. transferred =
12
2
12
42
1
1 21
2
1 12
1 2
1 22
m mm m
u
m u
m mm m
+
=+( )
If m1 = m2, Fractional K.E. transferred = 1.
Thus, when the mass ratio is unity, the whole of the kinetic energy of the moving ballis transferred to the ball initially at rest.
Inelastic Collision (Particle Stick After collision)(i) Description in Inertial frame: Let a particle of mass m2 be at rest at origin in an
inertial frame. Suppose a particle of mass m1 moving with velocity u1 along x-direction
244 Mechanics
collides with m2, both stick and then travel together with velocity v after collision in the samedirection.
From law of conservation of momentum
m1u1 = (m1 + m2)v
final velocity of composite particle v, after collision
v =m
m mu1
1 21( )+ ...(viii)
Ratio of K.E. after collision to K.E. before collision
K.E. before collision Ki = ½ m1 u12
K.E. after collision Kf = ½ (m1 + m2)v2
∴ ratio of final to initial K.E. of system
KK
f
i=
12
12
1 22
1 12
( )m m v
m u
+
Fig. 2
=
12
12
1 2
1 12
1 1
1 2
2( ).
m m
m u
m um m
+
+
using Equation (viii)
orKK
f
i=
mm m
1
1 2+
It shows Kf < Ki which means during inelastic collision, there is a loss in K.E. of system.
If a moving particle of mass m1 suffers inelastic collision with a stationary particle ofmass m2, the loss in kinetic energy of system is given by
δ.E =12
121 1
21 2
2m u m m v− +( )
Since, v = mm m
u1
1 21( )+
from Eq. (viii)
∴ δ.E = 12
1 2
1 212m m
m mu
( )+
Collision 245
Description in center of mass frame of reference (C-Frame): velocity of center ofmass vcm is given by
vcm =m u m
m m1 1 2
1 2
0++
.
or vcm =m u
m m1 1
1 2+
Velocity of 2nd particle, before collision, in C-frame
u′2 = u vm u
m m21 1
1 20− = −
+cm
i.e., u′2 = −+
m um m
1 1
1 2
Velocity of 1st particle, before collision, in C-frame
u′1 = u v1 − cm
= um u
m m11 1
1 2−
+
=m u m u m u
m m1 1 2 1 1 1
1 2
+ −+
or u′1 = ++
m um m
2 1
1 2
Velocity of combined particle after collision in C-frame
v′ = v vm u
m mm u
m m− =
+−
+=cm
1 1
1 2
1 1
1 20
v′ = 0
Thus combined particle is at rest in C-frame.
6.3 THE BALLISTIC PENDULUM
It is a device for measuring the speeds of bullets. It consists of a large wooden block massM hanging vertically by two cords. A bullet of mass m, moving with a speed v, strikes theblock and remains embedded in it. If the collision time is very small compared to the timeof swing of the pendulum, the supporting cords remain approximately vertical during thecollision. Hence no external horizontal force acts on the system during the collision, and thehorizontal momentum is conserved.
Let v′ be the velocity of the block-bullet combination immediately after the collision.The initial momentum of the system is mv, and the momentum just after the collision is(m + M)v′, so that
mv = (m + M)v′ ...(i)
246 Mechanics
The kinetic energy of the system immediately after the collision is 12
2(m v+ ′M) . The
block-bullet combination now swings up to a height y at which its kinetic energy is convertedinto gravitational potential energy. From the conservation of mechanical energy for this partof the motion, we obtain
12
2(m v+ ′M) = (m + M) gy
or v′ = 2 gy
Substituting this of v′ in Equation (i) and solvingfor v, we get
v =m
mgy
+ M2
By measuring m, M and y, we can compute the initial speed v of the bullet. The kineticenergy is not conserved in the collision. The ratio of the kinetic energy of block-bulletcombination just after the collision to the initial kinetic energy of the bullet is
12
12
2
2
(m v
mv
+ ′M)=
mm + M
As m << M, a very small fraction of the original kinetic energy remains, most of theenergy is converted into heat, etc.
6.4 COLLISION IN TWO DIMENSION
(1) Description in Laboratory Frame: Suppose a particle of mass m1 collides withother particle of mass m2 which is at rest in the laboratory frame of reference. Suppose initialvelocity of m1 is u1 in this frame and after collision it has velocity v1 in a direction that makesangle φ1 with the initial direction of travel. Here φ1 is the angle by which colliding particleis deflected and is called scattering angle. Let v2 be the velocity of m2, in direction φ2 withinitial line. If direction of u1 is along x-axis and u1 and v1 are contained in x-y plane thenthere is no z-component of v2.
According to principle of conservative of linear momentum:
x-component
m1u1 = m1v1 cos φ1 + m2v2 cos φ2 ...(i)
Fig. 4
yMv
m
Fig. 3
Collision 247
y-component
0 = m1v1 sin φ1 – m2v2 sin φ2 ...(ii)Since the collision is elastic, K.E. is also conserved
∴12 1 1
2m u =12
121 1
22 2
2m v m v+ ...(iii)
If we solve Equations (i), (ii) and (iii), we can know the three unknown quantities ofinterest. However, the velocity of center of mass in laboratory frame, vcm is given by
(m1 + m2) vcm = m1 u1 + m2 . 0
vcm =m u
m m1 1
1 2+(2) Description in C-frame: Centre of mass frame of reference moves with velocity vcm
in lab frame i.e., center of mass remains at rest in c-frame.In this frame:
Initial velocity of m1 is, u′1 = u1 – vcm
Initial velocity of m2 is, u′2 = u2 – vcm = 0 – vcm
Final velocity of m1 is, v′1 = v1 – vcm
Final velocity of m2 is, v′2 = v2 – vcm
vcm is always zero in this frame, hence total momentum before and after collision is zero i.e.momenta of two particles are always equal and opposite as observed in C-frame. So the angleof scattering of both the particles should be same and, thus, they move along a linear pathin opposite direction.
Thus m1u′1 = m2u′2 and m1v′1 = m2v′2Again K.E. conservation yields
12 1 1
2m u′ =12 1 1
2m v′ or u v′ = ′1 1
and12 2 2
2m u′ =12 2 2
2m v′
or u′2 = v′2Thus, in C-frame, the magnitude of velocities of the particles in an elastic collision do
not alter.
6.5 VALUE OF THE SCATTERING ANGLE
(i) In the center-of-mass frame of referenceIn this frame of reference, there are absolutely no limitations on the value of the
scattering angle θ, so that it can have all possible values.
(ii) In the laboratory frame of referenceIn this frame of reference, there are some restriction on the value of the scattering
angle, θ1, as will be clear from the following:
we have,
tan θ1 =vv
1 1
1 1
sincos
θθ
248 Mechanics
Now, the y-component of the final velocity of the first particle being the same in thelaboratory reference frame as well as the center-of-mass frame, we have v1 sin θ1 = v′1 sinθ. And, since the x-components of this velocity differ by V in the two frames of reference, wehave v1 cos θ1 = v′1 cos θ + V. Substituting these values in the relation above, we, therefore,have
tan θ1 =v
vv
′′ +
=+
′
1
1
1
sincos
sin
cos
θθ
θ
θV V
As we known,
V =m u
m mm u
m m1 1
1 2
1 1
1 2+= ′ +
+( V)
=m u
m mm
m m1 1
1 2
1
1 2
′+
++V
or V 1 1
1 2−
+
mm m =
m um m
1 1
1 2
′+
or Vm
m m2
1 2+
=
m um m
1 1
1 2
′+
Whence V =mm
umm
v1
21
1
21′ = ′ ∴ u v′ = ′1 1
Substituting this value of V in Equation (i) above, we have
tan θ1 =sin
cos
θ
θ +
mm
1
2
This shows that
(a) If m1 > m2, so that m1/m2 a little >1, the denominator can never be zero and hencetan θ1 can never be ∞. And therefore, θ1 must be less than 90°.
(b) If m1 = m2, so that m1/m2 = 1, the denominator can be zero for cos θ = –1 and,therefore, tan θ1, can be ∞. Thus, in this case , θ1 can have any value up to thelimiting value of 90°.
(c) If m1< m2 , so that m1/m2 < 1, the value of tan θ1 can also be negative. In this casealone therefore, all values of θ1 can be possible.
It follows from case (a) that in an elastic collision, if a massive particle collides againsta lighter one at rest, it can never bounce back along its original path. On the other hand,it follows from case (c) that if a lighter particle collides against a massive one at rest, it maywell bounce back along its original path. And, it follows from case (b) that in an elasticcollision between two particles of equal mass, one of which is initially at rest, the twoparticles move at right angles to each other after the collision.
Collision 249
6.6 SCATTERING CROSS-SECTION
We can consider three distinct stages in the entire scattering process. We show thesethree stages of the collision process in below figure. The first stage shown in figure (a)corresponds to a time long before the interaction of the colliding particles. At this stage eachparticle is effectively free, i.e., its energy is positive. As the particles approach each other(figure b), interaction forces much larger than any other force acting on them come into play.Finally, long after the interaction (figure C), the emerging particles are again free and movealong straight lines with new velocities in new directions. The emerging particles may ormay not be the same as the original particles.
v1
m 1 v2
m 2
m 1
m 2
m 1
m 2
v ′1
v ′2
( )a ( )b ( )c
Fig. 5 Scattering of two particles
In a typical scattering experiment, a parallel beam of particles, also called projectiles, ofgiven energy and momentum is incident upon a target (below figure). The particles interactwith the target for a short time, which deflects or scatters them in various directions.Eventually these particles are detected at large distances from the target. The scatteredparticles may or may not have the same energies and momenta.
An experimenter may be interested in knowing the velocities, linear momenta andenergies of the particles before and after scattering. Then the changes brought about in thesequantities can be determined.
Fig. 6 A typical scattering process
The probability of scattering in a given direction is found by determining the scatteringcross-sections. Let us now define the scattering cross-section for a typical scattering process.
Let us suppose that a uniform parallel beam of n particles, all of the same mass andenergy, is incident upon a target containing N number of identical particles or scatteringcenters. Such scattering centers might, for example, be the positive nuclei of atoms in a thinmetal foil which could be bombarded by α-particles. Let us assume that the particles in thebeam do not interact with each other and the scattering centers in the target are sufficientlyfar apart. With these assumptions we can regard the incident particles and target particles
250 Mechanics
to be sufficiently far apart. Then we can think of this scattering event as if at a given timeonly one projectile was being scattered by one target particle, without being affected by thepresence of other particles. So, effectively at any instant we deal with a two-body collisionprocess. For convenience, we choose the origin of the coordinate system at the position of thetarget and one of the axes, say z-axis, in the direction of the incidents beam. The directionof scattering is given by the angles (θ, φ) as shown in figure (a). The angle θ, called the angleof scattering, is the angle between the scattered and the incident directions. These twodirections define the plane of scattering. The angle φ specifies the orientation of this planewith respect to some reference plane containing the z-axis. The shaded plane in figure is areference plane. The probability of the scattering of a particle in a given direction (θ, φ) ismeasured in terms of the differential cross-section.
θφ
Scatte red ParticleScatte ring p lane
O ZInciden tpartic le Target
( )ay
z
x
dΩ
dADetecto r
rInciden tpartic le O ta rge t
(θ, φ)
( )b
Fig. 7
6.7 DIFFERENTIAL SCATTERING CROSS-SECTION
Let F be the number of projectile incident per unit area per unit time on the target. Frepresents the incident flux. Let ∆n be the number of particles scattered into a small solidangle dΩ about the angle (θ, φ) in time ∆t. Then the number of scattered particles by a singletarget particle in time ∆t, must be proportional to the incident flux F, the duration ∆t andalso the solid angle in which they are scattered, i.e.,
∆n ∝ F (d tΩ ∆) ( )
The constant of proportionality is defined as the differential scattering cross-section and
is denoted by the symbol dd
σΩ
. So that
∆n =dd
d tσΩ
Ω ∆
F ( ) ( )
ordd
σΩ
=∆
∆ Ωn
t dF
Thus, we can also express the differential scattering cross-section as the following ratio:
dd i e
σ θ φΩ
Ω= The number of particles scattered per unit time in a solid angle d in the direction ( , )Incident flux . ., the number of particles incident on the target per unit area per unit time
The differential scattering cross-section gives a probability. In fact, it is a measure ofthe probability that an incident particle will be scattered in solid angle dΩ in the direction
Collision 251
(θ, φ). The dimension of dd
σΩ is area. This explains the use of the term “cross-section”.
dd
σΩ
is equal to the cross-sectional area of the incident beam that contains the number of particles
scattered into the solid angle dΩ by a single target particle. The unit of dd
σΩ is m2 sr–1. The
dd
σΩ depends only on the parameters of the incident particle, nature of the target and the
nature of the interaction between the two.
For the N scattering centers the number of particles scattered will be just N times thenumber scattered by a single scattering center. Thus for N scattering centers, the numberof particles scattered is
∆n′ =dd
d tσΩ
Ω∆NF
Above equation is valid only when the target scattering centers are far enough apart sothat the same particle is not scattered by two of them.
6.8 TOTAL CROSS-SECTION
Let us place the detector at all possible values of (θ, φ) and count the total number ofscattered particles entering all the corresponding solid angles. Then we will get the totalscattering cross-section. It is denoted by σ. It can also be calculated from the differentialscattering cross-sections by integrating over all possible values of dΩ. Thus
σ =dd
dσΩ
Ω
So the total scattering cross-section represents the number of particles scattered in alldirection per unit flux of incident particles. It has the dimension of area. So its unit is m2.Now, we also define the solid angle subtended by an area to be dΩ = sinθ dθ dφ, where thelimits of θ and φ are 0 to π and 0 to 2π, respectively. Using these relations we get,
σ = dd
d dσ θ θ φππ
Ω
sin . .
0
2
0
We can show that for the cases in which the force is central and its magnitude depends
only on r, dd
σΩ is independent of φ. We can integrate over φ so that
σ = 20
2
π σ θ θπ
dd
dΩ
sin .
6.9 IMPACT PARAMETERS
Let us suppose that the projectile does not make a head-on collision with the target.Instead, it travels along a path, which if continued in a straight line, would pass target at adistance b (as shown in figure 8). The distance b is known as the impact parameter. b is theperpendicular distance between the projectiles initial path and the target.
252 Mechanics
(a) The impact parameter b (b) The particle having impact parametersbetween b and b + db are scattered into anglesbetween θcm and θcm + dθcm.
Fig. 8
Let us now express the differential scattering cross-sections in terms of the impactparameter. We will study the scattering process in the c.m. frame of reference with θcm asthe angle of scattering (figure b). Let us consider a circular ring having radii between b andb + db. The area of the ring is 2πbdb for infinitesimal values of db. If the incident flux isF then,
The number of incident particles having an impact parameter between b and (b + db)
= F(∆t) (2πbdb)
Let us suppose that these particles are scattered into angles between θcm and θcm + dθcm.The particles with larger b will be scattered through smaller angles as shown in figure (c).This happens because larger b means lesser interaction, i.e., less scattering. For very largeb, scattering will be minimal and the particles will go almost undeflected in a straight line.Now in the c.m frame of reference the number of particles scattered in the solid angle dΩin time ∆t is given as
n =dd
tσΩ
Ω ∆
cm
cmF (d )
Otarge t
θL
dΩ
b
(c) The scattering angle decreases with increasing impact parameter.
Fig. 8
This is the same as the number of incident particles in time ∆t having impact parametersbetween b and b + db; given by
F(∆t)2π.b.db =dd
d tσΩ
Ω ∆
cm
cmF( )
or 2π.b.db = −
dd
dσ π θ θΩ cm
cm cm2 sin
Collision 253
Here we have assumed that dd
σΩ is independent of φ. Taking into account all values of
φ in dΩ, we have dΩ = 2π sinθ.dθ. The negative sign expresses the fact that as b increases,θcm decreases, i.e., db and dθcm have opposite signs. From above Equation we get,
dd
σΩ
cm
=b db
dsinθ θcm cm
We have not written the negative sign in above equation because dd
σΩ
cm
has the
dimension of area and its magnitude has to be positive. So, if we know b as a function ofscattering angle θcm, we can calculate the differential scattering cross-section using aboveequation.
6.10 RUTHERFORD SCATTERING
The Rutherford scattering experiment was an important milestone in understanding thestructure of the atom. Until the early twentieth century Thomson’s plum pudding model ofthe atom was believed to be valid. J.J. Thomson had proposed, in 1898, that atoms wereuniform spheres of positively charged matter in which electrons were embedded. It wasalmost 13 years later that a definite experimental test of this model was made. Now, themost direct way to find out what is inside a plum pudding is to plunge a finger into it. Asimilar technique was used in the classic experiment performed in 1911, by Geiger andMarsden who were working with Lord Rutherford. They bombarded thin foils of variousmaterials with α-particles (helium nuclei) and recorded the angular distribution of the scatteredα-particles.
It was found that most of the α-particles pass through the foil (i.e., scattering angleθ < 90°). However, about 1 in 6.17 × 106 alpha particles was scattered backward, i.e., deflectedthrough an angle greater than 90°. This result was unexpected according to Thomson’s model.It was anticipated that the alpha particles would go right through the foil with only slightdeflections. This follows from the Thomson model. If this model were correct, only weekelectric forces would be exerted on alpha particles passing through a thin metal foil. In sucha case their initial momenta should be enough to make them go through with only slightdeflections. It would indeed need strong forces to cause such considerable deflections inα-particles as were observed.
In order to explain these results Rutherford proposed a nuclear model of the atom.Using this model he calculated the differential scattering cross-section. In doing so, hereasoned that the backward scattering could not be caused by electrons in the atom. Thealpha particles are so much more massive than electrons that they would hardly be scatteredby them. He assumed that the positive charge in the atom was concentrated in a very smallvolume, which he termed the nucleus, rather than being spread out over the volume of theatom. So the scattering of alpha particles was due to the atomic nucleus.
Let us consider the scattering of a particle carrying charge q by the atomic nuclei havingcharge q′. For this scattering process Rutherford derived the relation between the impactparameter b and the angle of scattering θcm to be
b =r02 2
cot θcm
254 Mechanics
Where rqq
004
= ′πε Ecm
, Ecm is the total mechanical energy of the projectile and the target
in the c.m. system.
ε0 is known as the permittivity of free space. Its value is 8.8 × 10–12 C2N–1 m–2.
Fig. 9
The differential scattering cross-section in the c.m. system for Rutherford scattering isthen given by
dd
σΩ
cm
=b db
dsinθ θcm cm
=r r0
022 2
12 2
cot
sin. cos
θ
θθ
cm
cm
2 cmec
=r0
2
2
162 2
2
cot
sin .cos
θ
θ θθ
cm
cm cm
2 cmcosec
ordd
σΩ
cm
=r0
2
16 2cosec2 cmθ
where r0 =qq′
4 0πε Ecm
This is the Rutherford scattering cross-section. For scattering of an α-particle by anucleus of atomic number z, qq′ = (2e) (Ze) = 2Ze2 where e is the electronic charge. TheRutherford scattering cross-section is strongly dependent on both the energy of the incomingparticle and the scattering angle. The number of particles scattered to increase as Z2 withincreasing atomic number.
NUMERICALS
Q.1. Assuming that all collisions are completely elastic, find the value of m for which bothblocks move with the same velocity after m has collided once with M and once with the wall.(The wall has effectively infinite mass).
Collision 255
Ans. Suppose after the collision,
v1 = speed of mass M towards right
v2 = speed of mass m towards left
Hence, momentum conservation requires;
Fig. 10
momentum before collision = momentum after collision
mu2 = Mv1 – mv2 ...(i)
The mass m rebounds elastically from the wall and its speed is reversed after thecollision with the wall. The mass m has the same speed as that of mass M after its collisionwith the wall. (v2 = v1 as given); so (i) is
mu2 = (M – m) v1 ...(ii)
The collision is elastic, so,
12 2
2mu =12
12 1
2M 12v mv+
or, mu22 = (M + m) v1
2 ...(iii)
Substituting the value of v1 from eqn. (ii) in eqn. (iii) we get,
mu22 = ( ) ( )
( )M +
Mm mu
m2
2
2−
or (M – m)2 = (M + m) (m)
or M2 + m2 – 2Mm = Mm + m2
or M2 = 3Mm
or M = 3m
or m =M3
kg.= =1003
33 33.
Q. 2. A ball moving with a speed of 8m/sec strikes an identical ball at rest such that afterthe collision the direction of each ball makes an angle of 30° with the original line of motion.Find the speeds of two balls after the collision. Is the K.E. conserved in this collision?
Solution. Momentum before collision = m × 8 + m × 0 = 8 m ...(i)
where m is the mass of each ball.
Suppose after collision their velocities are v1 and v2 respectively. Now the final momentumof the balls after the collision along the same line
256 Mechanics
= mv1 cos 30° + mv2 cos 30°
=mv mv1 23
23
2+ ...(ii)
From the law of conservation of momentum
8 m =mv mv1 23
23
2+
8 23
×= v1 + v2 ...(iii)
3030
V1
V2
u = 8 m /sec m Ball at rest
m
mm
Fig. 11
The initial momentum of the balls along perpendicular direction is zero and final momentumof the balls along perpendicular direction
= mv1 sin 30° – mv2 sin 30° = mv v
2 −( )1 2Then momentum conservation gives:
0 =m
v v2 −( )1 2
∴ v1 – v2 = 0 ...(iv)
from eqns. (iii) and (iv), we get
v1 =83
83
m/s and m/s2v =
Since, Energy before collision = Energy after collision
∴12
121
222mu mu+ =
12
12
mv mv12
22+
or12
0m (8)2 + = 12
83
2 2
m 83
+
Collision 257
or64
2m
=1282 3
m×
Since, L.H.S. ≠ R.H.S.
Hence, energy is not conserved in this collision so, collision is inelastic.
Q. 3. A plastic ball is dropped from a height of 1 metre and rebounds several times fromthe floor. If 1.3 seconds elapse from the moment it is dropped to the second impact with thefloor. What is the coefficient of restitution.
Ans. Suppose the plastic ball makes first strike on the floor with a velocity V, then ballrebounds with a velocity eV. Now the ball rises to a certain height where its velocity becomeszero and then it retraces its path. The ball strikes the floor with velocity eV making thesecond impact.
After the second impact, it rebounds with a velocity e2V. Time taken by the ball beforefirst impact is given by.
V = gt1
(Initial velocity u = 0, Final velocity V, and acceleration = g)
t1 = Vg
...(i)
Time interval between first and second impact t2 = 2 × time taken for the velocity tochange from eV to O under gravity
t2 = 2 × egV
...(ii)
∴ t1 + t2 =V V Vg
eg g
e+ = +2 1 2( ) ...(iii)
According to question;
Vg
e( )1 2+ = 1.3 ...(iv)
Again, V2 = 0 + 2g.1
V = 2g ...(v)
Using (v) in (iv)
21 2
gg
e( )+ = 1.3
or 2 1 2/ ( )g e+ = 1.3; or 2 9 8 1 2 1 3/ . ( ) .+ =e
or (1 + 2e) = 1 3 4 9 2 8777. . .=
1+2e = 2.8777
which gives e = 0.94
258 Mechanics
Q. 4. A meteor of M = 2 kg mass moving with velocity u = 12 km/s explodes above theatmosphere into two pieces which travel along the same path with 15 km/s and 11 km/s.Calculate the mass of each piece and energy set free by the explosion.
Solution. Due to the absence of external forces;
Linear momentum before explosion = After explosion
i.e., 2 × 12 = 15m1 + 11m2 ...(i)
and m1 + m2 = 2 ...(ii)
From (i) and (ii);
m1 =24
= 0.5 kg
m2 = 2 – 0.5 = 1.5 kg
Energy released ∆E =12
12
121 2
2m m muV V12
22+
−
= 3 × 106 J
Q. 5. A sand bag of mass 10 kg is suspended with a 3 meters long weightless string. Abullet of mass 200 gm is fired with speed 20 m/s into the bag and stays in bag. Calculate
(i) Speed acquired by the bag.
(ii) Maximum displacement of bag.
(iii) The energy converted to heat in collision.
Solution. (i) Collision is inelastic so the momentum is conserved not the K.E. if m =mass of bullet, M = mass of bag, u = velocity of bullet, V = velocity of bag after hit.
then mu = (m + M) V
0.2 × 20 = (0.2 + 10) V
V =0 2 20
10 24
10 20 39. ×
. ..= = m/s
(ii) Bag shall oscillate like pendulum after it is hit. If x is the maximum displacement,
Restoring force at maximum displacement; F = –(M + m) g sinθ
i.e., F = − = −( )M + Km gxl
x ...(i)
or K =10 2 9 8
3. × .
...(ii)
At maximum displacement, x, K.E. of system will be changed into potential energy
Ko
x dxx
=12
12
12
10 2 0 392 2K V2x m= = . ( . )
or, x2 =10 2 0 39 0 39 10 2 0 39 0 39
10 2 9 83. × . × . . × . × .
. × .×
K=
Collision 259
or, x = ( . × . × )/( . ) . / . .0 39 0 39 3 9 8 0 39 3 9 8 0 21= =
max. displacement, x = 0.21 metre
(iii) Energy converted into heat =12
12
2mu m− +( M)V2
=12
0 2 20 12
10 2 0 392 2× . × ( ) × . × ( . )−
= 40 0 8 38 2− =. . J
Q. 6. A steel ball weighing 1 lb is fastened to a cord 27 inches long and is released whenthe cord is horizontal. At the bottom of its path, the ball strikes a block weighing 5.0 lb whichis initially at rest on a frictionless surface as shown.
The collision is elastic. Find the speed of the ball and the speed of the block just aftercollision.
Solution. Let, m1, m2 = masses of the ball andblock respectively u1 = velocity obtained by the ballafter falling through 27′′, the velocity by which ballhits the block; V2 = velocity obtained by the block; V1= velocity of ball after impact;
Conservation of momentum gives.
m1u1 + 0 = m1V1 + m2V2
K.E. obtained in 27”(= 9/4 ft) descent
12 1
2m u1 = m1gh = m × ×32 94
or u1 = 12 ft/sec.
Since collision is head on, we have
V2 =2
12 12
1 5 1246
41
2 1
um m+
=+
= =/
×/
ft /sec.
Then from (i); 1 × 12 = 1 × V1 + 5 × 4
or V1 = – 8 ft/sec.
thus, ball recoils in opposite direction with a velocity = 8 ft/sec.
Q. 7. A moving particle of mass m makes head on collision with a particle of mass 2m whichis initially at rest. Show that the colliding particle looses (8/9)th of its energy after collision.
Ans. Let initial velocity of colliding particle be ‘u’ and final velocity ‘ V ’, after collision.If V1 is velocity of the other particle after collision, from law of conservation of momentum.
mu = mV + 2mV1
which gives, V1 =u − V
2Again from law of conservation of energy
12
2mu =12
12
22m mV V12+ . .
27′′
m 1
m 2
Fig. 12
260 Mechanics
or u2 = V V2
2 + −
2
2u
or u2 = VV
V22
+ + −uu
2
2 2
or 3V2 = u u2 2+ V
or 3V2 – 2uV – u2 = 0
which gives final velocity of colliding particle, which is
V = 2 4 126
2 2u u u± +
= uuor −3
Since +ve value of V will not satisfy the law of conservation of momentum.
∴ V = − u3
Therefore, loss in K.E. of colliding particle
∆U =12
12 3
22
mu mu− −
=12
1 19
89
2mu − = initial K.E.
Q. 8. A bomb moving with velocity 40 i 50 j 25 k m/s + − exploded into pieces of mass
ratio 1 : 4. The small piece goes out with velocity 200i 70 j 15 k m/s. + + Deduce the velocityof larger piece after explosion.
Ans. Force of explosion is the internal force acting on bomb. In absence of externalforces, thus, total momentum of the system remains constant.
∴ Momentum before explosion = momentum after explosion
i.e., 5 50 25M (40 )i j k+ − = M (200 MV2 )i j k+ + +70 15 4
or 4V2 = 200 200 i j k i j k+ − − − −250 125 70 15
or 4V2 = 180 140 j k−
or V2 = 45 35 j k−
Q. 9. A body at rest explodes and breaks up into 3 pieces. Two pieces having equal mass,fly off perpendicular to each other with the same speed of 30 m/sec. The 3rd piece has 3 timesthe mass of each of the other piece. Find the magnitude and direction of its velocity immediatelyafter the explosion.
Collision 261
Ans. Mass ratio is 1:1:3 in explosion. Since only internal force acts; so that momentumis conserved;
Momentum before explosion = after explosion
i.e., 5M(O) = M × 30 M × 30 MV i j+ +→
3 3
or 3 3V→ = − −30 30 i j
or V→
3= − − = − +10 10 10 10 ( )i j i j
or |V3| = 10 10 10 22 2+ = m/sThe direction is given by
or tan θ =− = −1010
1 ∴ θ = 90 + 45 = 135°
Q. 10. A U238 nucleus emits an α-particle and is converted into Th234. If the velocity ofthe α-particle be 1.4 × 107 m/s and the kinetic energy be 4.1 MeV, calculate the velocity of therecoil and the K.E. of the residual nucleus (Th234).
Ans. Only internal force act; so total momentum is same before and after thefragmentation (or the nuclear reaction)
i.e., momentum before reaction = after reaction
or O × [MU] = MTh VTh + MαVα
or O = 234VTh + 4 × 1.4 × 107 m/s
or VTh = –2.4 × 105 m/s
So, magnitude = 2.4 × 105 m/s and direction will be opposite to motionof α particle.
Kinetic Energy of Th234 nucleus:
Again;(K.E.)K.E.
Th
( )α=
1212
2344
2 4 101 4 10
5
7
2M V
M V
Th Th2
2α α
=
. ×
. ×
(.K.E.)
MeVTh
4 1 = 58 5 14449
10 171 19 104 4. × × . ×− −=
(K.E.)Th = 171.19 × 4.1 × 10–4 = .0704
K.E. of thorium nucleus = 0.0704 MeV
Q. 11. A 5 kg object with a speed of 30 m/s strikes a steel plant at an angle 45° andrebounds at the same speed and same angle. What is the change (magnitude and direction)in the linear momentum of the object?
Ans. Linear momentum before striking; is as below,
P1
→= mu i mu jx y
−
262 Mechanics
Linear momentum after striking is given as;
P2
→= mu i mu jx y
−
Change in linear momentum:
∆ P→
= ( )P P→ →
−2 1
= mu i mu j mu i mu jx y x y − − +
∆ P→
= 2mu jx
But, ux = u cos 45° = 30 12
uy = u sin 45° = 30 12
So, ∆ P→
= 2 5 302
3002
22
× × × j j=
or ∆ P→
= 150 2 j
So, magnitude, | |∆ P Kg m/s→
= 150 2 , in direction perpendicular to steel plate.
Q. 12. A particle of mass M, moving with a velocity u, makes a head on collision witha particle of mass m initially at rest so that their final velocities V and v are along the same
line. Assuming an elastic collision, prove that v = 2u
1 m/M+. If the particle coalesce on
colliding, find the final common velocity and the loss in K.E.
Ans. (i) Law of conservation of momentum gives;
Mu = MV + mv ...(i)
and law of conservation of energy gives;
12
2Mu =12
12
2MV2 + mv
u2 = V M)2 + ( /m v2 ...(ii)
Putting for V from (i), we obtain,
u2 =M
M Mu mv m
v−
+
22
u2 = um
v um
vm
v22
2 22+ − +
M M M
Collision 263
2um
vM
= vm m2
2
M M +
2u = vm
vumM
orM
+
=+
1 21 /
(ii) When particles stick together (inelastic collision); conservation of momentum gives
Mu = (m + M)v
v = M( M)m
u+
.
Loss of energy =12
12
2 2M M +u m v− ( )
=12
12
2
2M M + )M
M2
2u m
um
−+
(( )
=12
12
22
MM
M
2u
um
−+( )
=12
12
2 2 2 2 2Mu M) MM)
M M MM
2 2 2((m u
mmu u u
m+ −
+
= + −
+
∴ Loss of energy, ∆E =12
2MM)
mum( +
Q. 13. A radioactive nucleus initially at rest, decays by emitting an electron and aneutrino at right angles to one another. The momentum of the electron is 1.2 × 10–22 kg-m/sand that of the neutrino is 6.4 × 10–23 kg m/s. Find the direction and magnitude of the recoilnucleus. If its mass is 5.8 × 10–26 kg, deduce K.E. of recoil.
Ans. The nucleus is at rest, before decay. Let Pe =momentum of electron; Pn = momentum of neutrino; PN =momentum of recoil nucleus. Before decay; total momentumof nucleus is zero. In decay process, no external forces act(only internal one) so momentum remains constant before andafter decay.
x-component of momentum
O = Pe – PN cos αPN cos α = Pe ...(i)
y-component of momentum
O = Pn – PN sin αPN sin α = Pn ...(ii)
(i) and (ii) give; PN = P P2 2e n+
Fig. 13
264 Mechanics
= ( . × ) ( . × )1 2 10 6 4 1022 2 23 2− −+
= 1 44 10 40 96 1044 46. × . ×− −+
So, magnitude of momentum of recoil nucleus = 1 85 10 22. × − Kg-m/s
Its direction will be given by
tan α =PP
n
e= = =
−
−6 4 101 2 10
0 641 2
0 5323
22. ×. ×
..
.
α = tan–1 0.53 = 28° (nearly).
From the given diagram,
θ = 180 – α = 180 – 28 = 152°
recoil nucleus will move making angle 152° from electron motion or 242° from neutrino’spath.
K.E. of recoil nucleus;
K.E. =12 2
M VPMN N
2 N2
N=
=( . × )
× . ×. ×. ×
1 85 102 5 8 10
1 85 1011 6 10
22 2
26
44
26
−
−
−
−=
=1 8511 6
10 1018 19..
× ×− −J =18.511.6
J
=18 511 6
.
.eV
So, K.E. of recoil nucleus = 1.58 eV (nearly)
Q. 14. A body of 3kg makes elastic collision with another body at rest and afterwardscontinues to move in the original direction but with one half of its original speed. What is themass of the struck body?
Ans. Let, mass of struck body = M
Velocity of struck body = V
Velocity of moving body which collides = v
From law of conservation of momentum,
Total momentum before collision = Total momentum after collision
3 × v = 32
× v v+ MV or, V = 32M
Law of conservation of energy gives
12
3 2× v =12
32
12
2
×v + MV2
94
2v = MV2
Collision 265
Substituting the value of V we get
94 2
22
v v=
M 3
M or M = 1 kg.
Q. 15. The empty stages of a two stage, rocket separately weigh 6000 and 500 kg andcontain 40000 and 3500 kg fuel respectively. If the exhaust velocity is 1.8 km/s. Find the finalvelocity. (loge 10 = 2.3; log10 2 = 0.30).
Ans. Total mass of rocket = (6000 + 500 + 40000 + 3500) kg = 50,000 kg
Mass at the end of I stage = 6000 + 3500 + 500 = 10,000 kg
exhaust velocity = 1.8 km/s
Thus, velocity at the end of I stage, V1; is
V1 = VMM0
0+
= +
v e elog . log ,
,0 1 8 50 000
10 000
= 1 8 5 1 8 102
1 8 10 2. log . log . [log log ]e e e e= = −
= 1.8 × [2.3 – 2.3 × 0.3] = 1.8 × [2.3 – 0.69]
= 1.8 × 1.61 = 2.898
V1 = 2.898 km/s, which gives initial velocity of second stage
For II stage, V0 = 2.9 (nearly) M = 500 kg
M0 = 3500 + 500 = 4000
V2 = V0 +
= +1 8 4000
5002 9 1 8 23. log . . loge e
= 2.9 + 1.8 × 3 × 2.3 log10 2 = 2.9 + 1.8 × 3 × 2.3 × 0.3
= 2.9 + 3.726 = 6.626 km/sec
∴ Final velocity at the end of II stage = 6.626 km/sec.
Q. 16. A 5000 kg rocket is set for vertical firing. If the exhaust velocity is 500 m/s, howmuch gas must be ejected per second to supply the thrust needed to
(i) Overcome the weight of rocket.
(ii) Give rocket an initial upward acceleration 19.6 m/sec2
Ans. The net upward force on rocket at an instant is given by
F = M. V V M Mddt
ddt
g= − −
To just overcome the weight of rocket, net force is zero, i.e.,
0 = − −V M Mddt
g,
or ddtM
=MV
kg/secg = =5000 9 8500
98× .
266 Mechanics
i.e., rate of gas ejection should be 98 kg/sec.
(ii) If ‘a’ is the upward acceleration of rocket, we have
F = − − =V M M Mddt
g a
or,ddtM
= − − +1 1500
5000 9 8 19 6V
M ( ) =g + a . ( . . )
= –294 kg/sec.
So, gas ejection should be at the rate of 294 kg/sec.
Q. 17. A rocket starts vertically upwards with speed u0, show that its speed v at a height
h is given by u V 2gh
1 hR
02 2− =
+.
where R is the radius of earth, and g is acceleration due to gravity at earth’s surface. Hencededuce an expression for maximum height reached by a rocket fired with speed 90% of escapevelocity.
Ans. The velocity of rocket at height h is
V2 = u g dho
h
02 2− ′ ...(i)
[As rocket rises up, h and g change]
At earth surface g =GMR2
and g′ at height h is given by
g′ =GM
(R + )GM
R R) R2 2 2h hgh
=+
=+( / ( / )1 1 2
putting the value of g′ in eqn. (i)
V2 = ugh
dhh
02
20
21
−+ ( / )
.R
V2 = u gh
h
02
0
2− −
R1 + R/
= u gh0
2 2+ −
R1 + R
R/
V2 = u gh
hu
ghh0
2022
21
+ − −
= −+
R R1 + R R/ /
Collision 267
V2 – u02 = −
+− =
+2
12
102gh
hu gh
h r/ /R or V2
Since the escape velocity = 2gR, the escape velocity in present case = 0.9 2gR . Athighest point attained by rocket V = 0, hence the above equation gives
0 – 0.81 × 2gR =2gh
hR +R.
0.81 =h
hh h
R + or 0.81R + .81 =0
h – 0.81h = 0.81 R
h =8119
R = 4.26 R
Q. 18. The weight of an empty rocket is 5000 kg and it contains 40,000 kg fuel. If theexhaust velocity of the fuel is 2 km/s, find the maximum velocity attained by the rocket.(loge 10 = 2.3 and log10 3 = 0.4171).
Ans. When gravity effect is neglected, the velocity of the rocket at any instant is givenby;
V1 = V VMM0
0+ loge
where, M0 = initial mass; M = mass at any instant, V = exhaust velocity; V0 = initial velocity,V1 is velocity of the rocket at any instant.
Maximum speed is obtained when all the fuel is exhausted.
Now, V0 = 0, m0 = 5000 + 40,000 = 45000 kg,
M = 5000 Kg, V = 2 km/sec.
So, Vmax. = 0 + 2 loge (45000/5000) = 2 loge 32 = 4 loge 3
= 4 × 2.3 × log10 3 = 4 × 2.3 × 0.4771
= 9.2 × 0.4771 km/s = 4.4 km/s
Q. 19. A rocket of mass 30 kg has 200 kg of fuel and the exhaust velocity of fuel is 1.6km/sec. Calculate the minimum rate of consumption of fuel, so that the rocket may rise fromthe ground. Also calculate the final vertical velocity gained by the rocket when fuel consumptionrate is (i) 2 kg/sec. (ii) 20 kg/sec.
Ans. Rocket rises from the ground when fuel consumption occurs at minimum rate suchthat the thrust, balances the initial weight of rocket.
∴ Net force on rocket = 0
V Mddt
= Mg ...(i)
Now, M = 200 + 30 = 230 kg
V = 1.6 km/s = 1.6 × 103 m/s
268 Mechanics
from (i);ddtM
=MV
kg/sg = =230 9 8
1 6 1014093
× .. ×
Final velocity, V = V0 + V loge (M0/M) – gt
Case I: V = 1.6 km/s = 1.6 × 103 m/s
t =200
2kg
kg/sec = 100 sec.
given, g = 9.8 m/s2
v0 = 0
M0 = 230, M = 30
Vmax = 1.6 × 103 × loge (230/30) – 9.8 × 100
= 1.6 × 103 × 2.3 log10 (23/3) – 980.0
= 2.284 km/s
Case II: All the quantities will be same as in case I:
but, t =200 10 kg20 kg/s
= sec
∴ Vmax = 3.16 km/s
Q. 20. A body moving in straight line suddenly explodes into two parts. If one of theseis twice as heavy as the other one and the two parts move in opposite direction with samespeed, show that mechanical energy released in explosion is at least 8 times the initial kineticenergy.
Ans. Let, V1 and V2 be the speeds of smaller and larger parts. Then using law ofconservation of momentum
–mV1 + 2mV2 = 3mV (V-speed of original body)
But as given, V2 = –V1 = V′ (say)
–mV′ + 2mV′ = 3 mV
V′ = 3 V
This means each part moves with thrice the initial speed of original body in oppositedirection. So, from law of conservation of energy, the energy released in explosion.
∆U =12
3 12
32
2 32 2( ) ( ) × ( )m m mV V V2 − + 1
=12
3 1 3 6 8( ) ( )m V U2 − + = −
where U is initial K.E. This proves that released energy is 8 times the initial energy.
Q. 21. A bullet of mass m moving with horizontal velocity v strikes a stationary block ofmass M suspended by a string of length L. The bullet gets embeded in the block. What is themaximum angle made by the string after impact.
Ans. Suppose block and bullet system moves horizontally with speed V, after bullet getsembeded. Then from law of conservation of momentum:
Collision 269
mv = (M + m) V
V =mv
mM +
The K.E. gained by block will be converted into P.E. when string reaches final deflectedposition. Hence if θ = is the maximum angle string makes after impact,
12
( )M + V2m = (M + m) gh
12
V2 = g (L L cos )− θ
V = 2 4 2g gL (1 L sin2− =cos ) /θ θ
Putting the value of V
mvmM +
= 4 2g L sin2 θ/
sin θ2
=mv
m g2 ( )M + L
θ = 2 11sin /−
mm
gV
2 (M + )L
Q. 22. The maximum and minimum distances of a comet from the sun are 1.4 × 1012 mand 7 × 1010 m. If its velocity nearest to the sun is 6 × 104 m/s, what is its velocity when itis in farthest position. Assume in both position that the comet is moving in circular path.
Ans. From law of conservation of angular momentum;
m1V1r1 = m2V2r2 (for circular orbit)
here, m1 = m2 = m
V1r1 = V2r2
V2 =V1rr
1
2
4 10
126 10 7 10
1 4 10= × × ×
. ×
V2 = 3000 m/s
Q. 23. A particle of mass m performs motion along a path which is given by equation
r i a cos wt j b sin wt .→
= +
Calculate the angular momentum and torque about the origin.
Ans. r→ = cos sini a t j b tω ω+
∴ V→
= d rdt
→
= ω ω ω( sin cos )i a t j b t+
270 Mechanics
The angular momentum is given by
J→
= r r m→ → → →
=× ×P V
= ( cos sin ) × ( sin cos )i a t j b t m i a t j b tω ω ω ω ω+ +
= ab m k t t (cos sin )ω ω ω2 2+
= m ab kω
by definition, torque is,
τ→
= ddt
J→
∴ τ→
=ddt
m abk( )ω
= 0 (as all quantities are constant)
Q. 24. A particle of mass 20 gm moving in a circle of 4 c.m. radius with constant speedof 10 cm/s. What is its angular momentum (1) about the centre of the circle (2) a point on theaxis of the circle and at 3 cm distance from its centre.
Ans. (1) J = mvr = 20 × 10 × 4 = 800 erg. sec.
indirection perpendicular to plane of circle.
(2) QP = distance of particle from point of reference
= 3 4 52 2+ = cm
So, J = mvr = 20 × 10 × 5 = 1000 erg. sec.
Its direction will also be perpendicular to plain containing the radius vector (QP) and
instantaneous velocity; [Note: The direction of J→
change in this case in First case direction
of J→
does not change].
Q. 25. A particle of mass 1/2 gm starts from the point (1, 0, 2) at a time t = 0. If force
acting on it is 2 1 6 ( ) i t j+ + dynes. Calculate the torque and angular momentum about thepoint (2, 2, 4) at time t = 1 sec.
Ans. Force F→
= md rdt
2
2
→
; or d rdt m
2
2
→ →
= F
d rdt
2
2
→
= 2 2 1 6 ( ) i t j+ + Integration yields;
d rdt
→
= 2 2 6 22ti t t j ( / ) + + + K ...(1)
(K = constant of integration)
Collision 271
as velocity, V = 0, at t = 0; (1) then yields, K = 0
So (1) is d rdt
→
= 4 2 6 2ti t t j ( ) + + ...(2)
Further integration gives
r→
= 2 22 2 3t i t t j k ( ) + + + ′ ...(3)
Given at t = 0; particle is at (1, 0, 2)
ro→
= 1 0 2 2 i j k i k k+ + = + = ′
thus, r→
= 2 2 22 2 3. ( ) t i t t j i k+ + + + ...(4)
So, at t = 1; r→
1 = 3 3 2 i j k+ + ...(5)
Distance R→
from given point of reference; ( )2 2 4i j k+ + is
∴ R→
= 3 3 2 2 2 4 i j k i j k+ + − − − = + − i j k2
Thus at time t = 1, (refer eqns. (5), (2) and (1))
R→
= i j k+ − 2
V→
= 4 8 i j+ ...(6)
F→
= 2 7 i j+
So, torque; τ→
= R F→ →
×
=
i j k1 1 22 7 0
−
= ( ) ( ) ( )i j k0 14 0 4 7 2+ − + + −
τ→
= 14 4 5 i j k− +
and the angular momentum at t = 1, is = R V→ →
× m
=12
1 1 24 8 0
i j k−
=12
16 0 8 8 4 ( ) ( ) ( )i j k− + + −
= 8 4 2 i j k− +
272 Mechanics
Q. 26. A particle of mass m moving in a circular orbit of radius r has angular momentum
L→
about the centre. Calculate the K.E. of the particle and the centripetal force acting on it.Ans. Angular momentum of a particle moving in circular path
L = mr2ω; or, rmr
ω = 1
K.E. =12
12
2 2mv m r= ( )ω
So, K.E. =12
1 12
2
2mmr mr
= L2
Centripetal force =mv
rmr
r mr mr
22
2
= =
.( ) .ω L
=L2
mr3
Q. 27. A meter stick of mass M lies on a smooth horizontal table. A fall of mass m movingwith velocity V collides elastically and normally with stick at a distance d as shown in figure.Find the velocity of ball after collision.
Ans. The linear momentum, angular momentum as well as K.E. will be conserved inelastic collision.
Let, v′ = velocity of ball after collision,
V = velocity of centre of mass of stick,
ω = the angular velocity of stick after collision,
and I =M × l2
12, the moment of inertia of stick about centre of mass.
From law of conservation of linear momentum
mV = MV + mV′
therefore V =m ( )V V
M− ′
...(i)
Law of conservation of angular momentum gives
mVd = mV′d + Iw
m(V – V′)d =Mwl2
12
ω = 122
m dl
(V V )M
− ′ ...(ii)
Also law of conservation of K.E. yields
12
mV2 =12
12
12
2mV I MV2 2′ + +ω
Collision 273
m ( )V V2 2− ′ =M
MV2w l2 2
12+ ...(iii)
Putting the values of V and ω in eqn. (iii)
m ( )V V2 2− ′ =12 2 2 2
4
2m d ll
m( (V V )M
V V )M
2 2− ′ + − ′
which gives V + V′ =m
ld l( ( )V V )
M− ′ +2
2 212
V′ =m d l
l( )V V ) (12
MV
− ′ + −2 2
2
Q. 28. Determine the impact parameters of 2 MeV, α-particle whose distance of closestapproach to a gold nucleus (Z = 79) is 2 × 10–11 cm.
Ans. Initial K.E. = 12
2 10 1 6 102 6 12mu = −
−
[ × × . × ]
]
ergs
= [3.2 × 10 6
Angular momentum is conserved during scattering
At G; Angular momentum of α++ particle about N
= mαPu ...(i)
Angular momentum of α++ particle about N: (when approach is closest)
= mα.S.V ...(ii)
or mα Pu = mα.S.V
V =PSu
...(iii)
Energy conservation gives, 12
2mu = 12
2 12
222 2
m e e muV ZS
PS
ZeS2
2+ = +.( )
12
12
22 2
mumu− P
S2 =2 Z
S
2e
12
122
mu −
PS2 =
2 ZS
2e
12
2− Ps
=2 2
2Z 2es mu
×
12
− PS2 =
2 79 4 8 102 10
13 2 10
0 5710 2
11 6× × ( . × )
×.
. ×.
−
− − =
PS
2
2 = 1 – 0.57 = 0.43
P = S 0.43 = −2 10 0 4311× .
P = 1.3 × 10–11 cm
274 Mechanics
Q. 27. Calculate the angular momentum of 1 MeV neutron about a nucleus of impactparameter 10 –10 cm. (m = 1.7 × 10–24 gm).
Ans. Initial K.E. =12
1 102 6mu = × eV
= 1.6 × 106 × 10–12 ergs
= 1.6 × 10–6 ergs.
So, u =2 1 6 10
1 7 10
6
24
1 2× . ×. ×
/−
−
cm/s
Neutron being a light neutral particle experiences no force and goes straight way (P =impact parameter). Its angular momentum.
= m × P × u
= 1 7 10 102 1 6 10
1 7 1024 10
6
24
1 2
. × × ×× . ×. ×
/− −
−
−
= 1.93 × 10–25 gm-cm/sec.
Q. 30. A light weight man holds heavy dumb-bells in straightened arms when standingon a turn table. The turn table then start rotating at rate of 1 rev/min. The man then pullsthe dumb-bells inside toward chest. If initially dumb-bells are 60 cms from his axis of rotationand are pulled to 10 cms from rotation axis, find the frequency of revolution of turn table.Neglect the angular momentum of man in comparison to that of dumb-bells.
Ans. Initial angular momentum of system (neglecting the angular momentum of man)= 2Mω1 r1
2; where M is mass of dumb-bells, ω is initial rotational frequency, r1 is distanceof dumb-bells from axis.
Final angular momentum when arms are stretched; = 2Mω2 r22, where ω2 and r2 are
changed frequency and distance due to arm stretching. Since no external torque is applied,angular momentum will be conserved.
i.e., 2Mω1 r12 = 2Mω2 r2
2
2πf1 r12 = 2πf2 r2
2
f2 =f rr1 1
2
22
2
21 60
103600100
= =× ( )( )
f2 = 36 rev/min.
Q. 31. Compute the orbital angular momentum and total energy of the electron in hydrogenatom, assuming the path to be circular.
Ans. Hydrogen atom is one electron and one proton system. The required centripetalforce for circular motion is provided by Coulombian attraction force;
∴ mrV2
=Z
2e.er
er
=2
2 ...(i)
mV2r2 = e2r
(mVr)2 = me2r
Collision 275
putting the standard values,
m = 9 × 10–28 gm; e = 4.8 × 10–10 esu; r = 0.5 × 10–8 cms:
The angular momentum of revolving electron is
J = mVr = me r2
J = 9 10 4 8 10 0 5 1028 10 2 8× × ( . × ) × ( . × )− − −
= 4 8 10 3 0 5 14 4 10 0 528 28. × × . . × .− −=
= 1.02 × 10–27 g cm2/sec.
from Eqn. (i) K.E. =12 2
2m
er
V2 =
also P.E. = − er
2
∴ Total energy =er
er
er
2 2 2 10 2
82 24 8 10
2 0 5 10− = − = −
−( . × )× . ×
= –23.05 × 10–12 ergs
=− = −
−
−23 05 101 6 10
14 412
12. ×
. ×. eV
Q. 32. A nucleus of mass m emits a gamma ray photon of frequency νo. Show that the
loss of internal energy by the nucleus is not hνo but is h 1 h2mco
o2ν ν+.
Ans. The initial momentum of nucleus is zero. Let its momentum after emission ofphoton be P.
The momentum of photon =h
coν
The conservation of linear momentum suggests,
0 = P + or P =o ohc
hc
ν ν;
−
Then the internal energy which has converted into K.E. of nucleus;
K.E. = P2o
2 2
2
2mhmc
= −( )ν
Internal energy which has converted into radiant energy of photon = hνo
So total loss in the internal energy of nucleus
=( )h
mch h
hmc
ν ν ν νoo o
o2
2 221
2+ = +
276 Mechanics
Q. 33. Locate the centre of mass of a system of three particles of masses 1 kg, 2 kg and3 kg placed at the corners of an equilateral triangle of 1 meter.
Ans. Choosing the co-ordinate system as shown in figure:
Position of m1 = 0, Position of m2 = 1 i
Position of m3 = 12
60 12
32
sin i j i j+ = +
Then position of C.M. R→
=mr m r m r
m m m1 2 2 3 3
1 2 3
→ → →+ +
+ +
=1 0 2 1 3 1
23
1 2 3
× × ( ) × ( )+ + +
+ +
i i j
=2 3
23 3
26
i i j+ +
=712
34
i j+
1 kg(m )1
2 kg(m )2
( i )^
(m )3 kg
3( j )^
Y
X
( k )^
Z
60O
Fig. 14
So co-ordinates of C.M. is, 712
34
0, ,
meters.
Q. 34. A particle moves in a force field given by F f r r→
= ( ) , where r is the unit vector
along position vector of particle. Prove that the angular momentum of the particle is conserved.
Ans. Torque acting on particle is
τ→
= r r f r r f r r r→ → → → →
= = =× × ( ) ( ) ( × ),F τ 0
Collision 277
Since external torque on the particle is zero, so according to theorem of conservation ofangular momentum, its angular momentum shall be conserved.
Q. 35. A stream of α-particles is bombarded on a mercury nucleus (z = 80) with velocity1 × 107 m/s. If an α-particle is approaching the nucleus in head on direction. Calculate thedistance of closest approach.
Ans. Distance of closest approach (d) for α-particles is given, in head on approach, by
d =1
44
π ∈o
2
o2
ZVe
m
=9 10 4 80 1 6 10
6 4 10 1 10
9 19 2
27 7 2× × × × ( . × )
( . × ) × ( × )
−
−
= 1.15 × 10–13 m
Q. 36. Express in term of angular momentum (J), the kinetic, potential and total energyof a satellite of mass m in a circular orbit of radius r.
Ans. Angular momentum of the satellite in circular orbit is;
(a) J = mVr if the K.E. be T; then
T =12 2
22m
mrV J2
= ...(1)
(b) Potential energy of satellite is: U = − GMmr
, where M = mass of the earth. But for
a satellite moving in circular orbit we have;
GMmr2 =
mrV2
, orGM V2m
rm=
then, U = –mV2 and with J = mVr
U = − J2
mr2 ...(2)
(c) total energy (E) = K.E. + P.E. = 12
12
m m mV V V2 2 2− = −
then in view of Eqn. (2): E = − J2
2
mr2
Q. 37. “The ratio of maximum to minimum velocity of a planet in its circular orbit equalsthe inverse of the ratio of the radii of the orbits” – Establish the statement.
Ans. Let V1 and V2 be the maximum and minimum velocity of the planet with mass mand r1 and r2 be the radii of the circular orbits. The principle of conservation of angularmomentum suggests;
mV1r1 = mV2r2
orVV
1
2=
rr2
1 hence the statement.
278 Mechanics
Q. 38. Calculate the momentum of an electron accelerated by a potential of 100 volts.
Ans. K.E. of electron, T =12
12 2
2mm
mm
V V)P2
2= =(
where P = mV = linear momentum of electron.
But T = eV; so we have
eV =P
or, P = 2 V2
2mme,
here, m = 9 × 10–28 gm, e = 4.8 × 10–10 esu, V = 100 volts = 100300
stat volts.
P = 2 9 10 4 8 10 100 30028 10× × × . × × /− −
= 5.37 × 10–19 gm cm/sec.
Q. 39. A meteorite burns in the atmosphere before it reaches the earth surface. Whathappens to its momentum.
Ans. Taking the earth, atmosphere and meteorite to form a closed system, the momentumlost by meteorite is taken up by the combustion products, molecules of the atmosphere (i.e.,air) which interact with the meteorite and the earth. The momentum lost and gained remainsame and so the total momentum of system is conserved.
Q. 40. A 20 gm bullet passes through a plate of mass M1 = 1 kg and then comes to restinside a second plate of mass M2 = 2.98 kg. It is found that the two plates initially at rest nowmove with equal velocities. Find the percentage loss in the initial velocity of the bullet whenit is between M1 and M2. Neglect any loss of material of the plates due to action of bullet.
Ans. Let V1 = initial velocity of bullet; V2 = bullet velocity after emergence from firstplate M1. Also let m be the mass of the bullet and V be the velocity of plates after impact.Collision is inelastic, so linear momentum is conserved i.e., for first plate.
mV1 = mV2 + M1V ...(i)
For second plate when bullet enters it with initial velocity V2, the conservation ofmomentum gives;
mV2 = (m + M2)V ...(ii)
from (i); M1V = m(V1 – V2)
or V =mM
V V1
1 2( )− ...(iii)
Using this value in eqn. (ii)
mV2 = ( ) ( )mm+ −MM
V V21
1 2
M1V2 = (m + M2) (V1 – V2) = mV1 – mV2 + M2V1 – M2V2
or V2(m + M1 + M2) = (m + M2)V1
or V2 =( )
( )( . . )
( . . ).
mm
++ +
= ++ +
=M VM M
VV2 1
1 2
11
0 02 2 980 02 1 2 98
0 75
Collision 279
So loss in velocity after emergence from first plate
= V1 – V2 = V1 – 0.75V1 = 0.25V1
hence % loss =0 25
100 25.
× % V
V1
1=
Q. 41. A .5 kg block slides down from the point A as shown in figure on a horizontal trackwith an initial speed of 3 m/s towards weightless horizontal spring of length 1 m and forceconstant 2 Nt/meter. The part AB of the track is frictionless and the part BC has the coefficientof static and kinetic friction as 0.22 and 0.2 respectively. If the distance AB and BD are 2mand 2.14 m respectively, find total distance through which block moves before it comes to restcompletely.
Ans. Speed of block when it comes at B is same as that at A because path AB isfrictionless. Then K.E. of block at position B is
=12
12
0 5 3 94
2mV Joules.2 = =× . ×
Fig. 15
Suppose the block after travelling frictionful path BD, further travels distance x in whichit comes to rest after compressing the spring. Total frictionful path traversed in this journeyis (BD + x) i.e., (2.14 + x). Work done in this journey against force of friction is;
W(BD + x) = µk.mg × (2.14 + x) J
= 0.2 × 0.5 × 10 × (2.14 + x) J
= (2.14 + x) J.
Work done in compressing the spring by x;
12
K 2x =12
2 2 2× x x= Joule
So total work done = (2.14 + x + x2) J
then loss of K.E. of block = work done
12
0mV2 − = (2.14 + x2 + x)
12
0 5 32× . × = 2.14 + x + x2 + x)
x2 + x – 0.11 = 0; or; x = − ± + =12
1 1 4 11 0 1× . . m
Now, the compressed spring forces back the block with a force.
280 Mechanics
F = Kx = 2 × 0.1 = 0.2N
But force of friction on the block = µs mg = 0.22 × 0.5 × 10 = 1.1 N
It is evident that block will not move back as the pushing back force is smaller than thefrictional force. Then total distance moved by the block.
= AB + BD + x = 2 + 2.14 + 0.1 = 4.24 m
Q. 42. Calculate the intrinsic angular momentum of (i) photon, (ii) electron and (iii) π-meson.
Ans. (i) Photon: P S (S + 1)sh= .
2π, S = 1, h = 6.626 × 10–27 erg-sec.
So, Ps = 1 1 16 626 10
2 3 14
27( ) ×
. ×× .
+−
= 1.414 × 1.053 × 10–27 = 1.49 × 10–27 gm cm2/sec.
(ii) Electron: Again Ps = S (S + 1). ;h2π
here S = 12
So, Ps =12
1 2 1 6 62 102 3 14
27( / ) × . ×
× .+
−
=1 732
21 054 10 9 127 1027 28. × . × . × /− −= gm cm s2
(iii) π-meson Ps = S (S + 1) × ;h2π
here S = 0
So PS = 0
Q. 43. A cart of mass is M moving with a constant velocity v on a horizontal track. Amonkey of mass m jumps from a tree on to cart just from above. Find the velocity of cart afterthe event.
Ans. Considering cart and monkey as a single system, it is noted that there is noexternal horizontal force on the system.
Supposing that cart is moving in the +x distance towards right, it is found, initialhorizontal momentum of monkey = 0.
Initial horizontal momentum of cart = Mv
The final horizontal momentum of monkey–cart system if after the event system moveswith velocity V.
= (M + m) V
The horizontal momentum of the system shall be conserved in the absence of externalforce, thus
O + Mv = (M + m) V
So; V = MM+
vm
Collision 281
SELECTED PROBLEMS
1. State and explain the law of conservation of linear momentum, Explain the following
(a) A meteorite burns in the atmosphere before it reaches earth. What happensto its linear momentum ?
(b) When a ball is thrown up, its momentum first decreases, then increases. Doesthis violate the principle of conservation of momentum.
(c) Show that conservation of linear momentum is equivalent to Newton’s thirdlaw.
2. What do you mean by “Centre of mass” of a system of particles/show that inabsence of any external force the velocity of the centre of mass remains constant.
3. Show that in head on elastic collision between two particles, the transfer of energyis maximum when their mass ratio is unity.
4. Two bodies of mass M and m (at rest) collide and stick. Describe the collision incentre of mass frame. What is the ratio of kinetic energy transferred.
5. Describe principle of rocket. Establish the relation for the motion of rocket in
horizontal motion, V = VMM0
0+ log ,e where the symbols have their usual meanings.
6. (a) What is multistage rocket. Discuss its motion when it is moving in a field feespace with frictional forces are present as also when it moves in a regionwhere gravitational force is present.
(b) What is multistage rocket? What is the advantage of a two stage rocket on asingle stage rocket.
7. Show that the rocket speed is equal to the exhaust speed when the ratio MM
0 = e2.
Show that the rocket speed is twice the exhaust speed when MM
0 = e2.
CENTRE OF MASS : CONSERVATION OF LINEAR MOMENTUM
Q. 1. Show that centre of mass of two particles must lie on the line joining them, andthe ratio of the distances of the two particles from the centre of mass is the inverse ratio oftheir masses.
Solution. Let there be two particles of masses m1 and m2 whose position vectors are
r→
1 and r→
2 with respect to a fixed origin O. Let c be the centre of mass whose position vectors
with respect to m1 and m2 are d d1 2
→ → and .
The position vector of C with respect to the origin O is defined by
r→
cm=
m r m rm m
1 1 2 2
1 2
→ →++
Now by the law of vector addition, we have
r d1 1
→ →+ = r
→
cm and r d r
→ → →+ =cm 2 2
282 Mechanics
O X
Y
Z
m 1d1→ C
d2
m 2r1→ rc m
→
r2→
Fig. 16
These give
d1
→= r r
m r m rm m
rm r r
m m
→ →→ →
→→ →
− = ++
− = −+cm 1
1 1 2 2
1 21
2 2 1
1 2
( )
d2
→= r r r
m r m rm m
mm m
r r2 21 1 2 2
1 2
1
1 22 1
→ → →→ →
→ →− = − +
+=
+−cm ( )
∴ d1
→=
mm
d2
12
→
This shows that the vectors d d1 2
→ → and are collinear i.e., the centre of mass O lies on the
line joining m1 and m2.
Further, from the last equation, we have
d
d
1
2
→
→ =mm
2
1
Hence the ratio of the distance of the particles from the centre of mass is the inverseratio of their masses. This also shows that the position of the centre of mass is independentof the origin or the reference of frame chosen.
Q. 2. The mass of the moon is about 0.013 times the mass of the earth and the distancefrom the centre of the moon to the centre of the earth is about 60 times the radius of the earth.Find the distance of the centre of the mass of the earth moon system from the centre of theearth. Take the radius of the earth 6400 km.
Solution. Let d be the distance between earth and moon. Then if x be the distance ofthe centre of mass from earth, its distance from the moon will be d – x. The ratio of x tod – x will be equal to the inverse ratio of the masses of earth and moon. That is,
Collision 283
xd x−
=mass of moonmass of earth
= 0 013.
or x = (d – x) 0.013
or x (1 + 0.013) = 0.013 d
or x =0 013
1 0 0130 0128.
..
+=d d .
Here d = 60 × 6400 km
∴ x = 0.0128 × (60 × 6400 km)
= 4928 km.
Q. 3. The distance between the centres of carbon and oxygen atoms in the carbon monoxidegas molecule is 1.130 × 10–10 meter. Locate the centre of mass of the molecule relative to thecarbon atom.
Solution. Let the molecule be along the x-axis, the carbon atom being at the origin. Thecentre of mass relative to the carbon atom is given by
xcm =m x m x
m m1 1 2 2
1 2
++
where x1 and x2 are the distance relative to the carbon atom.
xcm =( × ) ( × .12 0 16 1 130+ Å)
12 + 16= 0.6457 Å Along the line of symmetry.
Q. 4. Locate the centre of mass of a system of three particles of masses 1.0 kg, 2.0 kg,and 3.0 kg placed at the corners of an equilateral triangle of 1 meter side.
Solution. Let the triangle lie in the x – y plane with its corner m1 (1.0 kg) at the originand side m1, m2 along the x-axis. Let (xcm, ycm) be the coordinates of the centre of mass C.By the definition of the centre of mass, we have
Y
m (3 gm )3
Cyc m
xc m (2 gm )(1 gm )Om 1
3
m 2X
x(c m )
1
Y(c
m)
Fig. 17
284 Mechanics
xcm =m x m x m x
m m m1 1 2 2 3 3
1 2 3
+ ++ +
=( . ) ( ) ( . ) ( ) ( . ) ( . )
( . . . )1 0 0 2 0 1 3 0 0 5
1 0 2 0 3 0kg kg meter kg meter
kg+ +
+ +
=7
12 meter
ycm =m y m y m y
m m m1 1 2 2 3 3
1 2 3
+ ++ +
=( .
( . . . )
1 02
1 0 2 0 3 0
kg) (0) + (2.0 kg) (0) + (3.0 kg) 3 meter
kg
+ +
= 34
meter
The co-ordinates of the centre of mass are 712
34
,
meter.
Q. 5. Two masses 6 and 2 units are at positions 6 i 7 j − and 2i 10 j 8k + − respectively.Deduce the position of their centre of mass.
Solution. The position vectors of the masses m1 = 6 and m2 = 2 are r i j1 6 7→
= − and
r i j k2 2 10 8→
= + − .
The position vector of the centre of mass is defined by
r→
cm=
m r m rm m1 1 2 2
1 2
→ →++
=18
6 6 7 2 2 10 8( ) ( )i j i j k− + + −
= 5 2 75 2 . i j k− −
The coordinates of the centre of mass are (5, –2.75, –2).
Q. 6. Calculate the position of the C.M. of a uniform semi-circular plane lamina.
Ans. ACB is a semi-circular lamina of radius a.
From symmetry we can conclude that the C.M. lies on OC, the perpendicular bisectorof the base AB of the lamina.
Let mass of the lamina be M. Let the C.M. be at a distance R from O on the straightline OC.
Collision 285
|R| = OR =1M
r dmo
a
...(1)
Here r is the radius of a semicircular element xy with centre at O and breadth ‘dr’ andmass ‘dm’.
Area of entire lamina =πa2
2 (semicircular)
∴ Mass perunit area =M M
2 2π πa a2
2= ...(2)
Area of semicircular element, xy = length × breadth dt
= πr dr
∴ Mass of xy element, dm = ( ) .ππ
r dra a
rdr2 2
2M M
2 = ...(3)
Substituting in (1), we get,
OR =1 2 1 2 2
MM
MM
2 2ra
r dra
r dro
a
o
a
=
=2
323
232
3 3
2ar a
aa
o
a = =
x y
dr
a
r
C
BA0
Fig. 18
The centre of mass of a plane semicircular lamina of radius ‘a’ will be at a height 2a/3 from the centre and lies on the perpendicular bisector of the base of the lamina.
Q. 7. A system consists of masses 7, 4 and 10 gm located at (1, 5, –3), (2, 5, 7) (3, 3,–1) respectively. Find the position of its centre of mass.
Solution. The coordinates of the centre of mass (xcm, ycm, zcm) are given by
xcm = m x m x m xm m m
1 1 2 2 3 3
1 2 3
7 1 4 2 10 37 4 10
4521
+ ++ +
= + ++ +
=( ) ( ) ( )
ycm =m y m y m y
m m m1 1 2 2 3 3
1 2 3
7 5 4 5 10 37 4 10
8521
+ ++ +
= + ++ +
=( ) ( ) ( )
286 Mechanics
zcm =m z m z m z
m m m1 1 2 2 3 3
1 2 3
7 3 4 7 10 17 4 10
321
+ ++ +
= − + + −+ +
= −( ) ( ) ( )
The centre of mass is located at 4521
8521
321
, ,−
.
Q. 8. If the centre of mass of three particles of masses 2, 4 and 6 gm be at the point(1, 1, 1) then where should the fourth particle of mass 8 gm be placed so that the position ofthe centre of the new system is (3, 3, 3).
Solution. Let (x1, y1, z1), (x2, y2, z2), (x3, y3, z3) be the positions of the particles ofmasses 2, 4, and 6 gm respectively then
xcm =m x m x m x
m m mx x x1 1 2 2 3 3
1 2 3
1 2 32 4 62 4 6
+ ++ +
= + ++ +
=16
2 31 2 3( )x x x+ +
It is given that (xcm, ycm, zcm) is (1, 1, 1). Thus
1 =16
2 31 2 3( )x x x+ +
or x1 + 2x2 + 3x3 = 6 ...(i)
Let (x4, y4, z4) be the position of the fourth particle of mass 8 gm, so that the new centreof mass is (3, 3, 3). Then we have
3 = 2 4 6 82 4 6 8
1 2 3 4x x x x+ + ++ + +
x1 + 2x2 + 3x3 + 4x4 = 30 ...(ii)
Subtracting eqn. (i) from (ii), we get x4 = 6
In the same way we get y4 = 6, z4 = 6
Hence 8 gm mass should be placed at (6, 6, 6).
Q. 9. Three particles of masses m1, m2, m3 are at positions r r r1 2 3
→ → →, , and are moving with
velocities V V V1 2 3
→ → →, , respectively (i) what is position vector of the centre of mass? (ii) what is
the velocity of the centre of mass ?
Solution. (i) The position vector of the centre of mass is given by
r→
cm=
m r m r m rm m m
1 1 2 2 3 3
1 2 3
→ → →+ +
+ +
(ii) The velocity of the centre of mass is given by vm v m v m v
m m m
→→ → →
= + ++ +
cm1 1 2 2 3 3
1 2 3
Q. 10. The velocity of two particles of masses m1 and m2 relative to an inertial observer
are v1
→ and v2
→ . Determine the velocity of the centre of mass relative to the observer and thevelocity of each particle relative of the centre of mass.
Collision 287
Solution. The velocity of the centre of mass relative to the observer is given by
v→
cm = m v m vm m1 1 2 2
1 2
→ →++
The velocity of each particle relative to the centre of mass is, by Galilean transformationof velocities, given by
v1
→= v v v m v m v
m m1 11 1 2 2
1 2
→ → →→ →
− = − ++
cm
=m v v
m m2 1 2
1 2
( )→ →
−+
and v2
→= v v v m v m v
m m2 21 1 2 2
1 2
→ → →→ →
− = − ++
cm
=m v v
m m1 2 1
1 2
( )→ →
−+
Thus, in c-frame, the two particles appear to be moving in opposite directions.
Q. 11. Two bodies of masses 10 kg and 2 kg are moving with velocities 2i 7 j 3k − + and
– 10 35 i j 3k+ − meter/sec respectively. Find the velocity of the centre of mass.
Solution. The velocity of the centre of mass in the laboratory frame of reference (L-frame) is defined by
v→
cm =1
1m
m vi ii
n →
=∑
in the case of two bodies only, we have
v→
cm=
1
1 21 1 2 2m m
m v m v+
+
→ →
=1
10 210 2 7 3 2 10 35 3
+− + + − + −( ) ( )i j k i j k
=112
20 70 30 20 70 6 i j k i j k− + − + −
= 2 k meter/sec.
Q. 12. In a system comprising two particles of masses 2.0 kg and 5.0 kg the positions of
the particles at t = 0 are 4i 3 j + and 6i 7 j 7k − + respectively, and the velocities are 10 6 i k−
and 3 6 i j+ respectively. Deduce the velocity of the centre of mass and the position of thecentre of mass at t = 0 and t = 4 units.
Solution. v→
cm =1
1 21 1 2 2m m
m v m v+
+
→ →
288 Mechanics
=1
2 0 5 02 0 10 6 5 0 3 6
. .. ( ) . ( )
+− + +i k i j
=1
7 020 12 15 30
.( )i k i j− + +
=1
7 035 30 12
.[ ]i j k+ −
The position vector of the centre of mass at t = 0.
r→
cm=
1
1 21 1 2 2m m
m r m r+
+
→ →
=1
2 0 5 02 0 4 3 5 0 6 7 7
. .. ( ) . ( )
++ + − +i j i j k
=1
7 08 0 6 0 30 35 35
.. . i j i j k+ + − +
=1
7 038 29 35
. i j k− +
The displacement of the centre of mass in 4 sec is
∆ Scm→ = 4
47 0
35 30 12v i j k→
= + −.
∴ position vector at t = 4 sec.
=1
7 038 29 35 4
7 035 30 12
.
. i j k i j k− + + + −
=1
7 0178 91 13
. i j k+ −
Q. 13. Two particles of masses 100 and 300 gm have position vectors 2i 5 j 13k + + and
− + −6i j 2k 4 and velocity (10i 7 j 3k ) and (7i 9 j 6k ) − − − + cm/sec respectively. Deducethe instantaneous position of the centre of mass, and the velocity of the second particle in aframe of reference travelling with the centre of mass.
Solution. The position of the centre of mass in the laboratory frame (L-frame) is givenby
r→
cm=
1
1 21 1 2 2m m
m r m r+
+
→ →
=1
100 300100 2 5 13 300 6 4 2
++ + + − + −( ) ( )i j k i j k
=14
16 17 7( )− + +i j k cm.
290 Mechanics
Solution. F→
= − + +6 8 12 i j i
= 6 8 i j+ newton
We know that the product of the mass of the system M, and acceleration of its centre
of mass a→
cm is equal to the resultant (external) force acting on the system. Thus
a→
cm=
FM
newton(3 + 4 + 5) kg
→
= +( )6 8i j
=12
23
i j+ meter/sec2
its magnitude is
a =12
23
0 832 2
2 +
= −. ms
its direction with the x-axis is
θ = tan−
1 aa
y
x
= tan //
−
1 2 31 2
= tan−
1 43 = 53°6'
Q. 17. Two particles P and Q of masses m1 = 0.10 kg and m2 = 0.30 kg respectively areinitially at rest 1.0 meter apart they attract each other with a constant force of 1.0 × 10–2 nt.No external forces act on the system. Describe the motion of the centre of mass. At whatdistance from the original position of P do the particles collide?
Solution. Initially the system is at rest so that its centre of mass is also at rest (zero
velocity). If M be the total mass of the system and a→
cm the acceleration of the centre of mass,then
M cma→ = Fext
→
Since no external forces act on the system (Fext→
= 0), we have
a→
cm = ddtVcm→
= 0
Vcm→ = constant ...(i)
Hence the centre of mass remains at rest.
Collision 291
Let the particle m1 lie at the origin of co-ordinates, and the particle m2 at a distance rfrom it along the x-axis. The distance of the centre of mass of the system from the originwould be given by
xcm =( × ) ( × )m o m r
m mm
m mr1 2
1 2
2
1 2
++
=+ ...(ii)
Each particle is acted upon by a constant force F. Therefore the accelerations of theseparticles would be
a1 =Fm1
and am2
2= F
Suppose the particles collide after time t at a distance x from the origin. This means thatin time t, the particle m1 travels a distance x and particle m2 travels a distance r – x. Thus,
using the formula x at= 12
2, we have
x =12 1
2a t
r – x =12 2
2a t
r xx−
=aa
mm
2
1
1
2= (From above)
rx
− 1 =mm
1
2
rx
=mm
m mm
1
2
1 2
21+ = +
x =m
m mr2
1 2+
Here m1 = 0.10 kg, m2 = 0.30 kg, r = 1.0 m
x =0 30
0 10 0 301 0 0 75.
. .( . ) .
+= meter.
Infact, the particle collides at the centre of mass.
Q. 18. A bullet of mass m and velocity v passes through a pendulum bob of mass M and
emerges with a velocity v2
. The pendulum bob is at the end of a string length l. What is the
minimum value of v such that the pendulum bob will swing through a complete circle.
Solution. In order that the pendulum bob may describes a complete circle, its velocity
at the lowest point must be 5 gl . This must be the velocity after collision.
From the law of conservation of momentum.
292 Mechanics
mv = Mxmv+2
where x is the velocity of the bob.
Mx =mv2
x =mv
gl2
5M
=
v =2 5Mm
gl.
Q. 19. A weightless string is passing over a light frictionless pulley of diameter 5 cm, anda tray of sand weiging 50 gm is tied at each of its two ends. The system is in equilibrium andat rest. Find its centre of mass. Now, 5 gm of sand is transferred from one tray to anotherbut the system is not allowed to move. Find the new centre of mass. If the system is now setfree to move, what would be acceleration of the centre of mass.
Solution. In the first case, by symmetry the centre of mass C will be at the mid pointof the line joining the centres of the two trays. In the second case, let x be the distance ofthe first tray (45 gm) from some arbitrary origin on the left and on the same line as joiningthe centres of the two trays. The distance of the second tray (55 gm) from the same originwould be x + 5. Then the distance of the centre of mass C from the same origin
=45 55 5
45 552 75× ( ) .x x x+ +
+= +
C
50 gm 50 gm
5 cm
C
45 gm 45 gm
5 cm
( )a ( )b
Fig. 20
Thus the centre of mass lies at a distance of 2.75 cm from the lighter tray on the linejoining the centres.
When the system is released, the heavier tray goes down and the lighter one goes up. Weknow that the product of the total mass of the system and the acceleration of its centre of massis equal to the vector sum of the external forces acting on all parts of the system. That is
M cma→
= F F1 2
→ →+
(45 + 55) acm = –45g + 55 g = 10 g
acm =10
10g g
100(down)=
Collision 293
Q. 20. A 10 kg boy standing in a 40 kg boat floating on water is 20 meter from the shoreof the river. If he moves 8 meter on the boat toward the shore then how far is he now fromthe shore ? Assume no friction between boat and water.
C 1C 2C
x
C 1 C 2C
( )a
( )b
(X + 8 )md
20m
Fig. 21
Solution. Let C1 be the centre of mass of the boy, C2 that of the boat and C that of thesystem (boy + boat), as shown in (a). Let x be the distance between C1 and C2. The distanceof C1 (10 kg) from the shore is 20 meter and that of C2 (40 kg) is 20 + x meter. Therefore,the distance of the centre of mass C of the system is
rcm =(10 × 20) + 40 × (20 + x)
10 40+
=1000 40
50+ x meter
when the boy moves 8 meter towards the shore, the distance between C1 and C2 becomes(x + 8) meter. Let d be the new distance of the boy from the shore. The distance of centreof mass from the shore is given by
r′cm =( × ) × ( )10 40 8
10 40
d d x+ + ++
=50 40 320
50d x+ + meter
As no external forces are acting on the system, the velocity of the centre of mass C ofthe system remains constant. Initially the centre of mass was at rest, So it must continueto remain at rest i.e., its position will still remain unchanged with respect to shore. Thus
rcm = r′cm
1000 + 40x = 50d + 40x + 320
d = 13.6 meter
Q. 21. (a) What is the momentum of a 10000 kg truck whose velocity is 20 m/s? (b) Whatvelocity must a 5000 kg-truck attain in order to have the same momentum (c) the same K.E.
Solution. (a) The momentum of the truck of mass m, moving with velocity v is givenby
P = mv = 10000 × 20
= 2 × 105 kg m/s
(b) m1v1 = m2v2
10000 × 20 = 5000 × v2
v2 = 40 m/s
294 Mechanics
(c)12 1 1
2m v =12 2 2
2m v
10000 × (20)2 = (5000) v22
v2 = 20 2 28 2= . m/s.
Q. 22. Two masses, one n times heavier than other, are dropped from the same height.How do their momenta compare just before they hit the ground. If these masses have same
K.E. Then how do their momenta compare. [Ans. n : 1, n : ]1
Q. 23. A rifleman, who together with his rifle has a mass of 100 kg stands on a smoothsurface and fires 10 shots horizontally. Each bullet has a mass of 10 gm and a muzzle velocityof 800 m/s (a) What velocity does the rifleman acquire at the end of 10 shots. (b) If the shotswere fired in 10 sec. What was the average force exerted on him. (c) compare his K.E. withthat of the 10 bullets.
Solution. (a) Let m1 and m2 be the masses and v1 and v2 be the velocities of therifleman and the bullet respectively after the first shot. The initial momentum of the systemis zero, and so must be the final momentum. That is,
m1v1 + m2v2 = 0
v1 = − mm
v2
12
Here m1 = 100 kg, m2 = 10 gm = 10–2 kg and v2 = 800 m/s
v1 = − = −−10
100800 0 08
2( ) . m/s
∴ velocity acquired after 10 shots = 10 v1 = – 0.8 m/s
The negative sign signifies that this velocity is opposite the velocity of the bullets.
(b) The momentum acquired by rifleman is,
P = mass × velocity acquired
= 100 × 0.8 = 80 kg m/s
This momentum is acquired in 10 sec. Hence the rate of change of momentum, whichis equal to the average force exerted, is
F =ddtP
nt.= =8010
8
(c) Kinetic energy of the rifleman,
K1 =12
101 12m v( )
=12
100 0 8 322× × ( . ) = joule
Kinetic energy of 10 bullets, K2 =12
102 22m v ×
Collision 295
=12
10 800 10 320002 2× × ( ) ×− = Joule
∴KK
1
2=
3232000
= 11000
Q. 24. A machine gun fires 50 gm bullets at a speed of 1000 m/s. The gunner holding themachine gun in his hands, can exert an average force of 180 nt against the gun. Determinethe maximum number of bullets he can fire per minute.
Solution. The initial momentum of the bullet is zero, while the momentum acquired onfire is
P = mv = 50 × 10–3 × 1000
= 50 kg-m/s
If n be the number of bullets fired per sec, then the rate of change of momentum is
ddtP
= 50n kg-m/s2
But this must be equal to the average force (180 nt) exerted. Thus
50 n = 180
n =18050
per sec
=18050
60× per minute = 216 per minute.
Q. 25. Find the average recoil force on a machine gun firing 120 shots per minute. Themass of each bullet is 10 gm and the muzzle velocity is 800 m/s. [Ans. 16 nt]
Q. 26. A rocket of mass 20 kg with a capsule of mass 10 kg attached to it, is moving witha velocity of 25000 m/s. A compressed spring when allowed to re-expand, separates the capsulefrom the rocket which now moves with a relative velocity of 3000 m/s along the same line asthe rocket. Calculate the new velocities of the rocket and the capsule. What happens to thekinetic energy of the system?
Solution. Let mr and mc be the masses and vr and vc the respective velocities of therocket and the capsule (after separation). As capsule is lighter, its velocity would be greater.Therefore vc – vr = 3000 m/s. Now, by the conservation of momentum, we have
Initial momentum = Final momentum
(mr + mc) × v = mrvr + mcvc
30 × 25000 = 20 × vr + 10 × (3000 + vr)
750000 = 30vr + 30000
vr =720000
3024 000= , m/sec
and vc = 3000 + vr = 27,000 m/sec.
But the rocket and the capsule move forward.
296 Mechanics
Now, initial kinetic energy =12
2( )m m vr c+
=12
30 25000 2× × ( )
= 9.38 × 109 joule
Final kinetic energy =12
12
2 2m v m vr r c c+
=12
20 24000 12
10 270002 2× × ( ) × × ( )+
= 9.40 × 109 joule
This energy comes from the compressed spring.
Q. 27. A bomb moving with velocity ( )40i 50 j 25k m/s+ − explodes into two pieces of
mass ratio 1:4. The smaller piece goes out with velocity 200i 70 j 15k m/s + + . Deduce thevelocity of the larger piece after the explosion.
Also, calculate the velocities of the piece in the centre-of-mass reference frame.
Solution. The forces of explosion exerted by a part of the system (bomb) on other partsof the system are all internal forces which may the momenta of all the individual fragmentsbut they cannot change the total momentum of the system.
Before the explosion the bomb can be thought of as consisting of two pieces of respectivemasses M/5 and 4M/5, where M is the total mass of the bomb. Now using the given data,applying the conservation of linear momentum, we have
Initial momentum = Final momentum
M ( 04 50 25 )i j k+ − =M5
M5
( )200 70 154
i j k v+ + +→
Thus the velocity v→
of the larger piece is
v→
=5 40 50 25 200 70 15
4( ) ( )i j k j j k+ − − + +
= 200 250 125 200 70 154
i j k i j k+ − − − −
=180 140
445 35
j kj k
− = − meter/sec
As there is no external force, the centre of mass moves after the explosion with same
velocity 40 50 25 i j k+ − as before the explosion. Hence the velocity of the smaller piece inthe centre of mass reference frame is
( ) ( )200 70 15 40 50 25i j k i j k+ + − + − = 160 20 40 i j k+ + meter/sec.
Collision 297
and that of the larger piece is
( ) ( )45 35 40 50 25j k i j k− − + − = − − −40 5 10 i j k meter/sec.
Example 28. A U238 nucleus emits an α-particle and is converted into Th234. If thevelocity of the α particle be 1.4 × 107 meter/sec and the kinetic energy be 4.1 MeV, calculatethe velocity of the recoil and the kinetic energy of the remaining Th234 nucleus.
Solution. Initially the U238 nuclues is at rest so that its momentum is zero. No externalforce acts upon it, the radio-active emission being purely an internal process. Hence, afteremission, the sum of the momenta of the α particle and the remaining Th234 nucleus is alsozero. That is
m v m va a th th→ →
+ = 0
vth→ = −
→m vm
a a
th
The α-particle is 2He4 nucleus. Therefore mm
a
th= 4
234; and va = 1.4 × 107 meter/sec.
Therefore
vth = − 4234
1 4 107× ( . × )
= −2 4 105. × meter /sec
Thus the Th234 moves with a velocity of 2.4 × 107 m/s opposite to the motion of the α-particle.
The ratio of the kinetic energy of Th234 nucleus and α-particle would be equal to theinverse ratio of their masses. That is
KK
th
a=
mm
a
th
Kth =mm
a
tha× × . .K MeV= =4
2344 1 0 07
Q. 29. A nucleus of mass 190 amu emits an α-particle (mass 4 amu) with velocity
( 4i 10 j ) 10 m/s6 − in laboratory frame. Deduce the velocity of recoil of the residual nucleus,as also the fraction of total energy of disintegration shared by the recoil.
Solution. We think of the system (α particle + residual nucleus) as initially bound andforming the original nucleus. The system then splits into two separate parts. The momentumof the system before splitting is zero. In the absence of external forces, the total (vector)momentum of the parts is also zero. Hence
Initial momentum = Final momentum
0 = m v m va a n n→ →
+
= 4 4 10 10 1866( )i j vn− +→
298 Mechanics
The velocity of the residual nucleus is therefore
vn→
= − −4186
4 10 106( )i j m/sec.
The minus sign indicates that the residual nucleus recoils in a direction exactly oppositeto the motion of the α-particle, so as to give a resultant vector momentum of zero.
The ratio of the kinetic energies acquired by the recoiling nucleus and the α-particle is
kk
n
α=
1212
2
2
2
2
m v
m v
m v mm v m
n n
a a
n n
a a n
= ( )( )
α
But mnvn = –mava
kk
n
a=
mm
a
n= =4
186293
The source of the K.E. of the products is the so called binding energy of the originalnucleus.
Q. 30. A shell of mass 2 kg and moving at a rate of 4 m/s suddenly explodes into twoequal fragments. The fragments go in directions inclined with the original line of motion withequal velocities. If the explosion imparts 48 joules of translational K.E. to the fragmentsdivided equally find the velocity and direction of each fragment.
Solution. A moving shell has a small amount ofK.E., but a large amount of hidden chemical energy,which is released when it explodes. Some of this energyis converted into heat, but a large part goes to theK.E. of the fragments of the explosion.
As no external force acts on the shell (explosion ispurely an internal affair), the vector sum of themomenta of the fragments is equal to the momentumof the unexploded shell. Further, the centre of mass ofthe fragments continue to move along the original lineof motion. Since, here the two fragments are of equalmass, they move in directions equally inclined to theoriginal line of motion.
Now, by conservation of momentum, we have
initial horizontal momentum = final horizontal momentum
2 × 4 = 1 × v cos θ + 1 × v cos θv cos θ = 4 ...(i)
The initial K.E. of the, shell is 12
2 4 162× × ( ) = joule. An amount of 48 joule is imparted
by explosion. Thus the total energy of the fragments is 64 joule i.e., each fragment has 32joule of K.E. If v be the velocity of each fragment, then
12
2mv = 32
Fig. 22
Collision 299
12
1 2× × v = 32
v2 = 64
v = 8 m/s
substituting this value of v in equation (i), we get
8 cos θ = 4
cos θ =12
θ = 60°
Q. 31. A body of mass 8.0 kg moving at 2.0 m/s without any external force on it. Aninternal explosion suddenly breaks the body into equal parts and 16 joules of translationalkinetic energy is imparted to the system by the explosion. Neither piece leaves the line of theoriginal motion. Determine the speed and direction of each piece after the explosion.
[Ans. One piece comes to rest the other moves ahead with a speed of 4.0 m/s].
Q. 32. A body at rest explodes and breaks up into three pieces. Two pieces having equalmass, fly off perpendicular to one another with the same speed of 30 m/s. The third piece hasthree times the mass of each of the other pieces. Find the magnitude and direction of itsvelocity immediately after the explosion.
Solution. Let v be the velocity, and θ the direction as shown of the third piece. Beforeexplosion, the momentum of the body is zero. Hence by conservation of momentum, themomentum of the system (all the three pieces) after explosion must be zero.
Equating the momentum of the system along OA and OB to zero, we get
m × 30 – 3m × v cos θ = 0 ...(i)
and m × 30 – 3m × v sin θ = 0 ...(ii)
these give 3 mv cos θ = 3 mv sin θor cos θ = sin θ
∴ θ = 45°
∴ ∠AOX = ∠BOX = 180° – 45° = 135°
Putting the value of θ in eq. (i), we get
30 m = 3 mv cos 45° = 312
mv
∴ v = 10 2 meter/sec.
The third piece will go with a velocity of 10 2 meter/sec in a direction making an angleof 135° with either piece.
Q. 33. A radioactive nucleus, initially at rest, decays by emitting an electron and aneutrino at right angles to each other. The momentum of the electron is 1.2 × 10–22 kg-m/sec,and that of the neutrino 6.4 × 10–23 kg-m/sec. Find the direction and magnitude of themomentum of the recoil nucleus. If its mass, is 5.8 × 10–26 kg, deduce its kinetic energy ofrecoil.
Bm
m
A
3m
v
θ
0
300 Mechanics
Solution. Let P P and PNe n
→ → →, be the momenta of the electron, neutrino and the recoiling
nucleus respectively, with directions as shown. Before decay, the momentum of the nucleusis zero. As no external force is acting on the system, the vector sum of the momenta of e,n and N must also be zero.
θ
eP e
Pn
P N
N
O
n
Fig. 23
Equating the sum of the components of the various momenta along two chosen directionsto zero, we have
Pe – PN cos θ = 0
Pn – PN sin θ = 0
Pn = P P2e n+ 2
= ( . × ) ( . × )1 2 10 6 4 1022 2 23 2− −+
= ( . ) ( . ) ×1 2 0 64 102 2 22+ −
= 1.36 × 10–22 kg-m/sec
Also, tan θ =PP
n
e= =0 64
1 20 53.
..
θ = tan–1 (0.53) = 28°
The direction of recoil is (180 – 28) = 152° from the electron track and 118° from theneutrino track.
The kinetic energy of the recoiling nucleus is
KN =12
MN N2v
Collision 301
=PN
2
N21 36 10
2 5 8 10
22 2
26m=
−
−( . × )× ( . × )
= 0.16 × 10–18 joule.
Now 1 eV = 1.6 × 10–19 joule
KN =0 16 101 6 10
1 018
19. ×. ×
.−
− = eV
Q. 34. A cue striking a billiard ball at rest, exerts an average force of 50 nt over a timeof 10 millisecond. The mass of the ball is 0.20 kg. Find the speed acquired by the ball.
Solution. The impulse of the force exerted on the ball is
F∆t = 50 nt (10 × 10–3 sec)
= 0.50 nt-sec.
This is equal to the change in momentum of the ball. The ball was initially at rest. Itacquires a velocity v after the impact, the change in momentum is mv = 0.20 v. Thus
0.20 v = 0.50
∴ v =0 500 20
2 5..
.= m/sec.
Q. 35. A 1.0 kg ball falls vertically onto the floor with a speed of 25 m/sec. It reboundswith an initially speed of 10 m/sec. Find the coefficient of restitution. What impulse acts onthe ball during contact. If the ball is in contact for 0.020 sec, what is the average force exertedon the floor.
Solution. The coefficient of restitution e is defined as the ratio between the relativevelocity of separation after the collision and the relative velocity of approach before thecollision. Thus, in this case,
e =10 0 5m/sec20 m/sec
= .
The initial velocity of the ball is v1 = 25 m/sec and the final velocity is v2 = –10 m/s. Thus
m (v1 – v2) = 1.0 (25 – (–10) = 35 kg-m/sec
This is also the impulse, F∆t, acting on the ball.
Now, F∆t = 35 nt-sec
∴ F =35 35 1750∆t
= =nt - sec0.020 sec
nt Ans.
Q. 36. The scale of a balance is adjusted to read zero. Particles falling from a height of5 meters collide with a pan of the balance, the collisions are elastic. If each particle has a massof 1/60 kg and collisions occur at the rate of 32 particles per second, what is the scale readingin kg? (g = 10 m/sec2).
Solution. The velocity of the particle before colliding in
v = 2 2 10 5 10gh = =× × m/sec.
302 Mechanics
and so its momentum is mv =160
10 16
× = kg m/sec.
The collision is elastic, i.e., the particle rebounds upward with the same speed. Hencethe change in momentum due to a collision is
2mv =18
kg m/sec
This is also momentum acquired by the pan in one collision. Since collision occur at therate of 32 per second, the average force exerted on the pan.
= rate of change of momentum
=18
32 4× = kg m/sec2
Hence scale reading in kg =410
0 4kg m/sec m/sec
kg.2
2 = . Ans.
Q. 37. A 6.0 kg cart is moving on a smooth surface at a speed of 9.0 m/sec when a 12 kgpackage is dropped into it vertically. Describe the subsequent motion of the cart.
Solution. The initial horizontal momentum of the cart is mv while that of the packageis zero. After the package is dropped into the cart, the two move as a single mass. Thus theimpact is inelastic, but the horizontal momentum is conserved.
mv = (m + M)v′where v′ is the velocity of the cart-package system. Putting the given values:
6.0 × 9.0 = (6.0 + 12)v′
or v′ =6 0 9 0
183 0. × . .= m/sec
The speed of the cart slow down to 3.0 m/sec. Ans.
Q. 38. A body of mass of 3.0 kg collides elastically with another body at rest and thencontinues to move in the original direction with one-half of its original speed. What is the massof the struck body.
Solution. Let m1 and m2 be the masses of the two bodies and v1i the initial velocity ofm1. Let v1f and v2f be the velocities of m1 and m2 after the collision. By the conservation ofmomentum, we have
m1v1i = m1v1f + m2v2f
or 3.0 v1i = 3 021
2 2.v
m vif
+ v vf i1 2= /
or v2f =1 3
221m
v i
...(1)
By the conservation of kinetic energy, we have
12 1 1
2m v i =12
121 1
22 2
2m v m vf f+
Collision 303
or12
3 12( ) v i =
12
3 02
12
12
2 22( . )
vm vi
f +
or32 1
2v i =38
121
22 2
2v m vi f+
Substituting the value of v2f from Eq. (1), we get
32
381
212v vi i− =
12
1 942
22 1
2mm
v i
or98 1
2v i =9
8 212
mv i
or m2 = 1.0 kg Ans.
Q. 39. A bullet of mass 10 gm moving horizontally with speed of 500 m/sec passes througha block of wood of mass 1 kg, initially at rest on frictionless surface. The bullet comes out ofthe block with a speed of 200 m/sec. Calculate the final speed of the block.
Solution. Let m1 and m2 be the masses of the bullet and the block and v1i the initialvelocity of the bullet. Let v1f and v2f be the velocities of the bullet and the block after thecollision. By the conservation of momentum, we have
m1v1i = m1v1f + m2v2f
(0.01 kg) (500 m/sec) = (0.01 kg) (200 m/sec) + (1 kg) (v2f)
∴ v2f = 3 m/sec Ans.
Q. 40. An open-topped freight car of mass 10,000 kg is coasting without friction along alevel track. It is raining very hard, with the rain falling vertically downward. The car isoriginally empty and moving with a velocity of 1 m/sec. What is the velocity of the car afterit has travelled to collect 1000 kg of rain water.
Solution. The initial horizontal momentum of the car is (10,000 kg) (1 m/sec), while thatof the rain water is zero (as it is falling vertically). Finally, the water sticks to the car. Now,the mass of the car plus water is 11,000 kg. If v′ be the velocity, then by the conservationof momentum, we have
(10,000 kg) (1 m/sec) = (11,000 kg) v′
∴ v′ =10 00011 000
1 0 91,,
× .= m/sec Ans.
Q. 41. An empty freight car of mass m1 = 10,000 kg rolls at v1 = 2 m/sec on a level roadand collides with a loaded car of mass m2 = 20,000 kg standing at rest. If the cars coupletogether find their speed v′ after the collision, and also the loss in kinetic energy. What shouldbe the speed of the loaded car toward the empty car, in order that both shall be brought torest by the collision.
Solution. By the conservation of momentum, we havem1v1 = (m1 + m2) v′
so that v′ =m
m mv1
1 21+
304 Mechanics
=10 000
10 000 20 0002 2
30 66,
, ,( ) .
+
= = m/sec
Loss in kinetic energy =12
121 1
21 2
2m v m m v− + ′( )
=12
10 000 2 30 00023
22
, ( ) ( , )−
=40 000
313 333 33, , .= Joule.
Let v2 be the speed of the loaded car toward the empty car, in order that v′ = 0. Thus
m1v1 – m2v2 = 0
or v2 =mm
v1
22
=10 00020 000
2,,
×
v2 = 1 m/sec Ans.
Q. 42. A slow-moving electron collides elastically with a hydrogen atom at rest. The initialand final motions are along the same straight line. What fraction of the electron’s kineticenergy is transferred to the hydrogen atom? The mass of hydrogen atom is 1840 times themass of the electron.
Solution. Let m1 be the mass of the electron and m2 that of the hydrogen atom. Letv1i and v1f be the initial and final velocities of the electron. Then the initial and final kinetic
energies of the electron are ki = 12 1 1
2m v i and k m vf f= 12 1 1
2 respectively. The fractional decrease
in kinetic energy is
k k
ki f
i
−=
v v
v
v
vi f
i
f
i
12
12
12
12
121
−= −
But, for such a collision, we have
v1f =m mm m
v i1 2
1 21
−+
so thatk k
ki f
i
−= 1 1 2
1 2
2
− −+
m mm m
=4 4
1
1 2
1 22
2 1
2
1
2m m
m mm m
mm
( )( / )
+=
+
Collision 305
Heremm
2
1= 1840
∴k k
ki f
i
−=
4 18401841
0 002172×
( ).= or 0.217%
Q. 43. A bullet of mass 5 gm is shot through a 1 kg wood block suspended on a string2 m long. The centre of gravity of the block rises a distance of 0.50 cm. Find the speed of thebullet as it energies from the block if the initial speed is 300 m/sec.
Solution. Let v be the velocity of the emerging bullet and v′ the velocity acquired bythe block just after the collision. From the conservation of momentum, we have
initial momentum of bullet = final momentum of bullet + momentum of block
(0.005 kg) (300 m/sec) = (0.005 kg) v + (1 kg) v′ ...(1)
The kinetic energy 12
2mv′ acquired by the block is converted into gravitational potential
energy mgy. so that
v′ = 2gy
= [ × . × ( . × .2 9 8 0 50 10 0 31252m/s m)] m/sec.2 − =
Substituting this value of v′ in Eq. (1), we get
1.5 = 0.005 v + 0.3125
or v =1 5 0 3125
0 005. .
.−
v = 237.5 m/sec. Ans.
Q. 44. A moving particle of mass m collides head-on with a particle of mass 2m whichis initially at rest. Show that the particle m will loose 8/9th part of its initial kinetic energyafter the collision.
Solution. Let v1i be the initial velocity of the mass m1 and v1f and v2f the velocities afterthe collision of mass m and 2m respectively. By conservation of momentum, we have
mv1i = mv mvf f1 22+
or v vi f1 1− = 2 2v f ...(1)
Again, by the conservation of kinetic energy, we have
12 1
2mv i =12
12
212
22mv m vf f+ ( )
or v vi f12
12− = 2 2
2v f
or ( ) ( )v v v vi f i f1 1 1 1− + = 2 22v f ...(2)
Dividing Eq. (2) by Eq. (1), we get
v vi f1 1+ = v f2
306 Mechanics
Substituting the value of v f2 in Eq. (1), we get
v vi f1 1− = 2 1 1( )v vi f+
or v f1 = − 13 1v i ...(3)
Hence, the initial and final kinetic energies of m are
Ki =12 1
2mv i
and Kf =12
12
191
212mv m vf i=
∴ Fraction lost, k k
ki f
i
−=
12
1 19
12
12
12
mv
mv
i
i
−
=89
Ans.
Q. 45. Two blocks of masses m1 = 2.0 kg and m2 = 5.0 kg are moving in the same directionalong a frictionless surface with speeds 10 m/sec and 3.0 m/sec respectively, m2 being aheadof m1. An ideal spring with k = 1120 nt/m is attached to the backside of m2. Find themaximum compression of the spring when the blocks collide.
v1
m 1
v2
m 2
k
Fig. 24
Solution. Let v′ be the speed of the combination after collision. By the conservation ofmomentum, we have
m v m v1 1 2 2+ = ( )m m v1 2+ ′
or (2.0) (10) + (5.0) (3.0) = (2.0 + 5.0) v′
∴ v′ =( . ) ( ) ( . ) ( . )
( . . )2 0 10 5 0 3 0
2 0 5 05+
+= m/sec
Now, by the conservation of mechanical energy, we have
12
121 1
22 2
2m v m v+ =12
121 2
2 2( )m m v kx+ ′ +
or 200 + 45 = 175 + 1120 x2
or x2 =200 45 175
11201
16+ − =
∴ x =14
0 25= . meter Ans.
Collision 307
Q. 46. A 5-kg object with a speed of 30 m/sec strikes a steel plate at an angle of 45° andrebounds at the same speed and same angle. What is the change in magnitude and directionof the linear momentum of the object.
Solution. The vector OA−→
represents the momentum of the object before the collision,
and the vector OB−→
that after the collision. The vector AB−→
represents the change in the
momentum of the object ∆P→
Fig. 25
As the magnitude of OA and OB−→ −→
are equal, the components of OA and OB−→ −→
along theplate are equal and in the same direction, while those perpendicular to the plate are equaland opposite. Thus the change in momentum is due only to the change in direction as theperpendicular components. Hence
∆P = OB sin 45° – (–OA sin 45°)= mv sin 45° – (–mv sin 45°)= 2mv sin 45°= 2 (5 kg) (30 m/sec) (sin 45°)= 212 kg m/sec.
∆P→
is directed perpendicular and away from the plate.Q. 47. A ball of mass m is projected with speed v into the barrel of a spring-guest of mass
M initially at rest on a frictionless surface. The mass m sticks in the barrel at the point ofmaximum compression of the spring. What fraction of the initial kinetic energy of the ball isstored in the spring? Neglect friction.
Solution. Let v′ be the velocity of the system after the mass m sticks in the barrel.From the law of conservation of linear momentum, we have
Mm
v
Fig. 26
mv = (m v+ ′M) ...(1)
308 Mechanics
The initial kinetic energy of the ball is 12
2mv . It is converted into the elastic potential
energy 12
2kx of the spring, and the kinetic energy 12
2(m v+ ′M) of the system. That is
12
2mv =12
12
2 2kx m M v+ + ′( )
where k is the force-constant of the spring and x is its maximum compression. Dividing the
last Equation by 12
2mv throughout, we obtain
1 =
1212
2
2
2
2
kx
mv
m vmv
+ + ′( M)
or
1212
2
2
kx
mv= 1
2
2− + ′(m vmv
M)
Putting the value of vv′22 from Eq. (1) we get
1212
2
2
kx
mv= 1
2− +
+( .
(m
mm
mM)
M)2
= 1 −+
=+
mm m( M)
MM
12
2kx is the energy stored in the spring, while 12
2mv is the initial kinetic energy of the
ball. Thus 12
12
2 2kx mv/ is fraction of the initial energy which is stored in the spring. This
fraction is M
Mm +.
Q. 48. Two balls A and B, having different but unknown masses, collide, A is initially at
rest while B has a speed v. After collision B has a speed v2
and moves at right angles to its
original motion. Find the direction in which ball A moves after collision. Can you determinethe speed of A from the given information?
Solution. Let v′ be the speed of A after the collision by the conservation of momentumalong the original direction of motion, we have
Collision 309
θ
B
A
Bv
v2
v′
Fig. 27
mBv = mAv′cos θ ...(1)
Again, by the conservation of momentum of right angles to the original motion, we have
0 = mv
m vB A2− ′sin θ
or mv
B 2= m vA ′sin θ ...(2)
Dividing Eq. (2) by Eq. (1) we get
tan θ =12
∴ θ = tan–1 (0.5) = 27°
Thus the ball A moves at 27° + 90° = 117° from the final direction of B. We cannotdetermine v′ unless we know mA and mB or their ratio.
Q. 49. A ball moving at a speed of 2.2 m/sec strikes an identical stationary ball. Aftercollision one ball moves at 1.1 m/sec at 60° angle with the original line of motion. Find thevelocity of the other ball.
Solution. Let m be the mass of each ball, and v1i the initial velocity of the first ball.Let v1f and v2f be the final velocities of the balls after the collision in directions as shown infigure. By the conservation of momentum along the original line of motion, we have
Fig. 28
mv1f = mv mvf f1 260cos cos° + θ ( cos .60 0 5° = )
or 2.2 = 1.1 (0.5) + v2f cos θ
or v f2 cosθ = 2.2 – 0.55 = 1.65 ...(1)
By conservation of momentum perpendicular to the original line of motion, we have
0 = mv mvf f1 260sin sin° − θ
310 Mechanics
0 = 1.1 (0.866) – v f2 sinθ ( sin .60 0 866° = )
or v f2 sinθ = 1 1 0 866 0 953. ( . ) .= ...(2)
Squaring and adding Eq. (1) and Eq. (2), we get
v2f = ( . ) ( . ) .1 65 0 953 1 92 2+ = m / sec
Dividing Eq. (2) by Eq. (1), we get
tan θ =0 9531 65
0 577..
.=
or θ = tan–1 (0.577) = 30° Ans.
PROBLEM
1. A ball of mass m and speed v strikes a wall perpendicularly and rebounds with undiminishedspeed. If the time of collision is t, what is the average force exerted by the ball on the wall.
2. A 10 gm bullet moving horizontally at 100 m/sec strikes and is embedded in a 25-kg blockplaced on a rough surface. The block slides 3m before coming to rest. Find the frictional forceretarding the motion of the block on the surface.
3. Find the fractional decrease in the kinetic energy of neutron when it collides elastically head-on with (1) lead nucleus (2) hydrogen nucleus, initially at rest. The ratio of nuclear massto neutron mass is 206 for lead and 1 for hydrogen.
7
7.1 RIGID BODY
If an external force applied to a body does not produce any displacement of the particlesof the body relative to each other, then the body is called a rigid body. No real body isperfectly rigid. However, in solid bodies (leaving rubber, etc.) the relative displacement by theexternal force is so small that it can be neglected. Hence ordinarily, when we speak of a body,we mean a rigid body.
7.2 MOMENT OF A FORCE OR TORQUE
When an external force acting on a body has a tendency to rotate the body about an axis,then the force is said to exert a ‘torque’ upon the body about that axis. The moment of aforce, or the torque, about an axis of rotation is equal to the product of the magnitude of theforce and the perpendicular distance of the line of action of the force from the axis of rotation.
In figure 1 shown a body, which is freeto rotate about an axis passing through apoint O and perpendicular to the plane ofthe paper. When a force F is applied on thebody in the plane of the paper, the bodyrotates about this axis. If r be theperpendicular distance of the line of actionof the force from the point O, then themoment of the force F, that is, the torqueabout the axis of rotation is
τ = F × r
If the torque tends to rotate the bodyanticlockwise then it is taken as positive, ifclockwise then negative. The unit of torque is‘newton-metre’.
7.3 ANGULAR ACCELERATION
When a body rotates about a fixed axis, the rotation is called a ‘rotatory motion’ or‘angular motion’ and the axis is called the ‘axis of rotation’. In rotatory motion, every particle
Or
F
L ine o f action
Fig. 1
311
312 Mechanics
of the body moves in a circle and the centers of all these circles lie at the axis of rotation.The rotating blades of an electric fan and the motion of a top are examples of rotatory motion.
If the angular velocity of a rotating body about an axis is changing with time then itsmotion is ‘accelerated rotatory motion’. The rate of change of angular velocity of a body aboutan axis is called the angular ‘acceleration’ of the body about that axis. It is denoted by α.If the angular velocity of a body about an axis changes from ω1 to ω2 in t second, then theangular acceleration of the body about that axis is
α =change in angular velocity
time-interval= −ω ω2 1
tThe unit of angular acceleration is radian/sec2 and its dimensional formula is [T–2].
7.4 RELATION BETWEEN ANGULAR ACCELERATION AND LINEAR ACCELERATION
In figure 2, a body suspended from a point O and free to rotate about an axis passingthrough O. When a torque is applied to the body, the body rotates with accelerated motionabout the axis of rotation. Suppose, the body rotates through an angle θ in t second. It meansthat the angular displacement of the body in t sec is θ. Therefore, the angular velocity(=angular displacement/time-interval) of the body is given by
ω =θt
Different particles of the body cover different distances in t second, hence their linearvelocities are different. But all the particles rotate through the same angle θ in t sec, hencethe angular velocity of each particle at any instant is same.
Suppose a particle P of the body is at a distance r from the point O. As the body rotates,the particle P rotates through angle θ in t second and reaches P′. Thus the particle movesfrom P to P′ in t second. Therefore, its linear velocity is
v =PP′ =
trtθ
arc = radius × angle
But θt
= ω
∴ v = r × ω
or ω =vr
...(1)
Suppose, at any instant the angular velocityof the body is ω1 and after t second it becomesω2. Then the angular acceleration of the bodyis
α =ω ω2 1−
t
P P ′
r
Axis o f ro ta tion
O
θ
Fig. 2
Rotational Kinematics 313
If at the above instants the linear velocities of the particle P be v1 and v2 respectively,
then according to equation (1), ω ω11 2= =vr
vr
and 2 . Therefore,
α =
vr
vr
tv v
rt
2 1
2 1
−
= − ...(2)
v2 – v1 is the change in the linear velocity of the particle P in t sec. Therefore(v2 – v1)/t is the rate of change of linear velocity of the particle, that is, it is the linear
acceleration a of the particle P. Substituting at
vv =− 12 in equation (2), we get
α =ar
or a = r × αThus, the linear acceleration of a particle of a body is equal to the product of the angular
acceleration of the body and the distance of the particle from the axis of rotation.
7.5 KINETIC ENERGY OF ROTATION
Suppose a body is rotating about an axis with a uniform angular velocity ω. The angularvelocity of all the particles of the body will be the same but their linear velocities will bedifferent. Suppose a particle of the body whose mass m1 is at a distance r1 from the axis ofrotation. Let v1 be the linear velocity of this particle. As the linear velocity of a particle isequal to the product of its angular velocity and its distance from the axis of rotation, we have
v1 = r1 ωThe kinetic energy of this particle is
12 1 1
2m v =12
121 1
21 1
2 2m r m r( )ω ω=
In the same way, if the masses of the other particles be m2, m3, ....... and their respectivedistances from the axis of rotation be r2, r3, ....... then their kinetic energies will be ½ m2(r2)2 ω2, ½ m3 (r3)2 ω3, ....... respectively. The kinetic energy K of the whole body will beequal to the sum of the kinetic energies of all the particles:
K =12
12
121 1
2 22 2
2 23 3
2 2m r m r m rω ω ω+ + + .......
K =12
121 1
22 2
23 3
2 2 2 2m r m r m r mr+ + + =...... ω ωΣ
But Σmr2 is the moment of inertia I of the body about the axis of rotation. Therefore
K =12
I 2ω
This is the formula for the kinetic energy of uniform rotation.
314 Mechanics
7.6 ANGULAR MOMENTUM
When a body moves along a straight line, then the product of the mass ‘m’ and the linear
velocity v of the body is called the ‘linear momentum’ p→
of the body (p = mv). If a body isrotating about an axis then the sum of the moments of the linear momenta of all the particlesabout the given axis is called the ‘angular momentum’ of the body about that axis. It is
denoted by j→
.
Let a body is rotating about an axis with an angular velocity ω. All the particles of thebody will have the same angular velocity, but their linear velocities will be different. Let aparticle be at a distance r1 from the axis of rotation. The linear velocity of this particle isgiven by
v1 = r1 ωIf the mass of the particle be m1, then its linear momentum = m1 v1
The moment of this momentum about the axis of rotation
= momentum × distance
= m1 v1 × r1
= m1 (r1 ω) × r1
= m1 r12 ω
Similarly, if the masses of other particles be m2, m3, ........ and their respective distancesfrom the axis of rotation be r2, r3, ........, then the moments of their linear momenta aboutthe axis of rotation will be m2 r2
2 ω, m3 r32 ω, ........ respectively. The sum of the moments
of linear momenta of all the particles, that is, the angular momentum of the body is givenby
J = m1 r12 ω + m2 r2
2 ω + m3 r32 ω + ........
or J = (m1 r12 + m2 r2
2 + m3 r32) ω, = (Σmr2) ω
But Σmr2 is the moment of inertia I of the body about the axis of rotation. Hence theangular momentum of the body about the axis of rotation is
J = I × ωThe unit of angular momentum is ‘kg-meter2/sec’ or ‘joule-sec’. Its dimensional formula
is [M L2 T–1].
7.7 RELATION BETWEEN TORQUE AND ANGULAR MOMENTUM
Suppose a body is rotating about an axis under the action of a torque, τ and its angularacceleration is α.
Then τ = IαWhere I is the moment of inertia of the body about the axis of rotation. Now,
α =∆∆ωt
(rate of change of angular velocity)
∴ τ = I ∆∆ωt
...(i)
Rotational Kinematics 315
The angular momentum of the body about the axis of rotation is J = IωThe rate of change of angular momentum is given by
∆∆Jt
= I ∆∆ωt
[∴ I is constant]
Hence, from equation (i) we have
∆∆Jt
= τ
Thus, the rate of change of angular momentum of a body is equal to the external torqueacting upon the body.
7.8 CONSERVATION OF ANGULAR MOMENTUM
If no external torque is acting upon a body rotating about an axis, then the angularmomentum of the body remains constant, that is,
J = Iω = constant
This is called the ‘law of conservation of angular momentum’. If I decreases, ω increases,and vice-versa.
7.9 TORQUE ACTING ON A PARTICLE
A force is required to produce linear acceleration in a particle. In a similar way, a torque(or moment of force) is required to produce angular acceleration in a particle about some axis.Like force, the torque is a vector quantity. It is defined in the following way:
τ→Z
O
r s in θ
r→ F→
θP
Y
X
Fig. 3
Let a force F→
acts on a particle P whose position vector is r→
with respect to the origin
O of an inertial reference frame. Then the ‘torque’ τ→
acting on the particle about on the
particle about the origin O is defined as the vector product of r→
and F→
. That is,
τ→
= r→
× F→
Its scalar magnitude is
τ = rF sin θ
316 Mechanics
where θ is the angle between r→
and F→
, and is along a direction perpendicular to the plane
containing r→
and F→
. In the figure, r→
and F→
lie in the x-y plane and the torque τ→
is alongthe positive z axis.
It is clear that torque is maximum when the force is applied at right angles to r→
(i.e.,
θ = 90°). This is why in opening or closing a heavy revolving door the force is applied (byhand) at right angles to the door at its outer edge. Besides this, the torque depends also onthe position of the point relative to origin at which the force is applied a force applied at the
origin 0 (i.e., r→
is zero) produces zero torque about 0. This is why we cannot open or closea door by applying force at the hinge.
7.10 ANGULAR MOMENTUM OF THE CENTER OF MASS OF A SYSTEM OF PARTICLES
The total angular momentum of a system of particles about a fixed point can also be
expressed in terms of the angular momentum of the center of mass about that point. Let r→
cm
be the position vector of the center of mass c of the system with respect to the origin, and
r i′→ the position vector of the particle i (mass mi) with respect to the center of mass. Then,
by the rule of vector sum, we have
r i
→ = r ri→ →
′ + cm
rc m→ ri
→
ri′→
X
Z
Y
O
m iC
Fig. 4
Similarly, if v→
cm be the velocity of the center of mass and v i
→′ the velocity of the particle
i with respect to the center of mass, then the velocity of the particle with respect to the originis
vi
→= v vi
→ →′ + cm
and the linear momentum with respect to the origin is
Rotational Kinematics 317
pi
→= m v m v m vi i i i
→ → →= ′ + cm
= p m vi i
→ →′ + cm
where pi
→′ is the linear momentum with respect to the center of mass. Therefore, the total
angular momentum of the system about the origin is
L→
= r pi ii
n → →
=∑ ×
1
or L→
= r r p m vi i i
→ → → →′ +
′ +
∑ cm cm×
or L→
= r p ri i′ ′ +→ →
∑ × ×cm
p r m v r m vi i i i
→ → → → →′ + ′ + ∑∑∑ . ×cm cm cm
or L→
= r p r p r r m vi i i i i
→ → → → → → →′ + ′ + ′ +
∑∑∑ × . ×cm cm cm
or L→
= r p r p r r m vi i i i i
→ → → → → → →′ ′ + ′ + ′ +∑∑∑ × . ( ) ×cm cm cm
Now, r pi i
→ →′ =∑ × ,Lcm angular momentum of the system about the center of mass, and
p i′ =→
∑ 0 (we know that the total linear momentum of a system of particles about the center
of mass is zero).
L→
= Lcm cm
→ → → →+ ∑ ( ) ×m r vi i since r r ri i
→ → →= ′ = cm
By the definition of the center of mass, rm ri i→
→
= ∑cm
M, where M is the total mass of the
system.
∴ L→
= L Mcm cm cm→ → →
+ r v×
or L→
= L Mcm cm cm→ → →
+ r v×
But M cmv→
known to the total linear momentum P→
of the system. Thus
318 Mechanics
L→
= L Pcm
→ → →+ ri ×
The term r→ →
cm P× is the angular momentum of the center of mass about the origin.
7.11 PRECESSION
When a body rotates about a fixed axis, under the action of a torque, the axial componentof the torque alone is effective in producing rotation, its component perpendicular to the axisbeing neutralized by an equal and opposite torque due to the constraining forces applied tothe axis to maintain its original position, i.e., to keep it fixed.
In the absence of any such constraining forces, obviously, the axis of rotation, and hencealso the plane of rotation, of the body will turn from its initial position under the action ofthe component of the torque perpendicular to the axis, which we may denote by the symbolτ1.
Since this torque τ1 (i.e., the axis of this torque) is perpendicular to the axis of rotationof the body, the magnitude of the angular velocity, and hence also that of the angularmomentum of the body remains unaltered, with only its direction changed in exactly thesame manner that in a linear motion, a constant force (viz, the centripetal force), acting ona body in a direction perpendicular to that of its velocity, simply changes its direction and notits magnitude. So that if the torque τ1 be a constant one, the rotation-axis and the plane ofrotation of the rotating body continue to change their direction at a constant rate, with theaxis of the torque always perpendicular to the axis of rotation of the body.
This change in the direction of the rotation-axis, or in the direction of the plane ofrotation, of the rotating body under the action of a constant torque perpendicular to the axisof rotation is called precession.
The torque τ1 which brings this about is, therefore, called the precessional torque andthe rate of rotation of the axis of rotation or the plane of rotation of the body is called therate of precession, usually denoted by the symbol φ.
7.12 THE TOP (PRECESSION OF A TOP SPINNING IN EARTH’S GRAVITATIONAL FIELD)
A symmetrical body rotating (or spinning) about an axis, one point of which is fixed, iscalled a ‘top’.
Figure shows a top spinning with angular velocity ω about its own axis of symmetry, thefixed point O being at the origin of an inertial reference frame. Its angular momentum is
represented by the vector L→
pointing upward along the axis of rotation (right-hand rule). Theaxis is inclined at an angle θ with the vertical.
Let C be the center of mass of the top, with position vector r→
with respect to O. The
weight of the top, m g→
, acting at C exerts a (external) torque about the fixed point O whichis given by
τ→
= r m g→ →
×
Rotational Kinematics 319
Its scalar magnitude is
rmg sin (180° – θ) = rmg sin θ ...(i)
and its direction is perpendicular to the plane containing r→
and m g→
as shown in the figure.
Thus the torque τ is perpendicular to L→
i.e., perpendicular to the axis of rotation of the top.
The torque τ→
changes the angular momentum L→
of the top, the change taking place in the
direction of the torque i.e., perpendicular to L→
. Let ∆L→
be the change produced in a time ∆t.We know that the time-rate of change of angular momentum of a system is equal to thetorque, that is
τ→ = ∆
∆L→
t
The angular momentum L L→ →
+ ∆ , after a time ∆t, is the vector sum of L→
and ∆L→
. These
three vectors are shown clearly in figure. Since ∆L→
is perpendicular to L→
(or paralled to τ→
)
and is very small, the new angular momentum vector L L→ →
+ ∆ has the same magnitude as
the initial angular momentum vector L→
, but a different direction. This means that theangular momentum remains constant in magnitude but varies in direction. As time goes on,
the tip of the angular momentum vector L→
moves on a horizontal circle. But the vector L→
always lies along the axis of rotation of the top, the axis of rotation itself rotates about avertical axis (z-axis) and sweeps out a cone whose vector is the fixed point O. This is theprecessional motion of the top.
The angle ∆φ through which the vector L→
turns in time ∆t, is given by
∆φ =∆L
L.sin θ [ ∴ ≤∆L L]
∴ the angular velocity of precession, ωp is given by
ωp =∆∆
∆∆
φθt t
= LL sin ( )
But∆∆Lt
= τ
∴ ωp =τ
θL sin ...(ii)
320 Mechanics
x
z
y
180° θ
c
rθ
m g
τ→
L→
ω
Fig. 5
Putting the value of τ from equation (i), we get
ωp =mgr
L...(iii)
Thus the rate of precession is independent of the inclination of the axis of rotation withthe vertical, and inversely proportional to the angular momentum of the spinning top. Smallerthe angular momentum, larger the precessional velocity. As the spinning top slows down, saydue to friction, its angular momentum L (= Iω) decreases and the precession correspondinglybecomes faster.
The precessional velocity would be represented by a vector ω→
p along the z-axis directedupward. Now, from Equation (ii), we have
τ = ω θp L sin
We note in figure (6) that ω→
p and L→
are vectors
with θ as the angle between them, and τ→
is a vector
perpendicular to the plane formed by ω→
p and L→
. Thus
ω θp L sin is the vector product of ω→
p and L→
. Hencethe above expression can be written in the followingvector form.
τ→
= ω→ →
p × L
We obtain an important result from this. An angular momentum vector L→
of constant
magnitude but rotating about a fixed axis with angular velocity vector ω→
p is always associated
τ→
x
z
y
O
θ
L ∆L
ωP→∆p
ωP
L + L∆→ →
Fig. 6
Rotational Kinematics 321
with a torque τ→
which acts perpendicular to the plane formed by ω→
p and L→
and is directedaccording to the right-hand rule.
7.13 THE GYROSTAT
A gyrostat, in actual practice, usually takes the form of a heavy circular disc, free torotate at a high speed about an axle passing through its center of mass and perpendicular toits plane, i.e., about its axis of symmetry, and is so mounded that the disc, along with theaxle can turn freely about any one of the three mutually perpendicular axes.
Thus the disc or the gyrostat G is free to rotate about the axle, coincident with the axisof x and is carried by a circular ring R1 which is itself free to turn about the axis of y insideanother ring R2, which is free to rotate about the axis of z, within a rigid frame work F. Thethree axes being obviously perpendicular to each other, intersect in the same point O andtherefore, for any given position of F, the disc or the gyrostat G can turn freely about anyone of them.
XX ′
Y ′
Y
Z
Z ′
R 2
R 1
G
F
Fig. 7
When the disc rotating fast then (i) any rotation of the frame work leaves the positionand direction of the axis of rotation of the disc unaffected with respect to it and (ii) any torqueapplied to the axis of rotation of the disc results in its precession.
Since the rate of precession is inversely proportional to the angular momentum Iω of thedisc, the greater its M.I. and the higher its angular velocity about the axis of rotation, thesmaller the precession of the latter. Obviously, therefore, a massive disc rotating at a highspeed makes for a greater stability of its axis of rotation. Hence the use of gyrostats forsteadying of motions and ensuring stability of direction.
In view of the property of stability, the gyroscopes are used as stabilizers in ships, boatsand aeroplanes. For examples, a gyroscope may be used to reduce the roll of a boat. For this,the gyroscope wheel, kept rotating at high speed by a motor, is mounted with its axis verticalin such a way that its axis may be tilted forward and backward but not sideways. When the
322 Mechanics
boat rolls, say to the right, a control-gyro closes contacts of a motor which tilts the axisforward, thereby producing a torque opposing the roll. On the other hand, if the roll is tothe left, the motor tilts the axis of the gyroscope backward, again opposing the roll. Theinherent stability of the gyroscope suggests its use as a compass, and a gyro-compass ispreferable to the magnetic compass in many respects. It is so constructed that it always setsitself with its axis parallel to the axis of the earth. If it is in any other direction, there isa torque which causes a precession and brings it parallel to the axis of the earth.
Another important application of the directional stability of a rapidly spinning (rotating)body is the rifling of the barrels of the rifles. Spiral grooves are cut in the barrels (the processis known as rifling) so that the bullet is forced to move along them before it emerges out inair, and thus acquires a rapid spin about an axis in the direction of motion. This spin motionprevents the deflection of the bullet from its path due to air and gravity effects, and causesonly very little precession. Thus the uniformity of flight of the bulled is increased.
The rolling of hoops and the riding of bicycles (which are statically unstable since bothof them cannot remain in equilibrium when at rest) are possible because of the gyroscopiceffect. This effect produces a movement of the plane of rotation, tending to counter balancethe disturbing action of gravity.
Many modern aircraft instruments such as automatic pilot, bomb sights, artificial horizon,turn and back indicators, etc. have been developed on gyroscope-controlled principles.
7.14 MAN WITH DUMB-BELL ON A ROTATING TABLE
Let I0 be the total moment of inertia of the man (with outstretched arm), the dumb-bells,and the table about the axis of rotation and ω0 the initial angular speed. The angularmomentum is I0ω0.
As the friction is assumed to be zero, it exerts no torque about the vertical axis ofrotation. The only external force is gravity acting through the center of mass, and that exertsno torque about the axis of rotation. Hence the angular momentum about the axis of rotationis conserved.
ω0→
I0
ω0
( )a
ω0→
ω
( )b
Fig. 8
When the man pulls in his arms figure (b), the moment of inertia of the system decreasesto I. Therefore, in order that the angular momentum remains conserved, the angular speedmust increases to ω, such that
I0 ω0 = Iω
Rotational Kinematics 323
The table now rotates faster.
The kinetic energy of the system, however, does not remain conserved as the man pullsin his arms. In fact, the kinetic energy increase. This increases comes from the work doneby the internal forces of the system when the dumb-bells were pulled in. In other words,the kinetic energy increases at the expense of the chemical energy stored in the man’s body.The total energy is thus conserved.
NUMERICALS
Q. 1. A top has a mass of 0.50 kg and moment of inertia of 5.0 × 10–4 kg-meter2. Its centreof mass is 4.0 cm from the pivot point. The top spins at a rate of 1800 rev/min with its axismaking an angle of 30° with the vertical. Find the angular velocity of precession. If the rotation(spin) of the top is clockwise as one looks downward along the axis, is the precession clockwiseor counter-clockwise ?
Solution. ωp =mgr mgr
L I=
ωHere m = 0.50 kg, g = 9.8 meter/sec2, r = 4.0 cm = 0.04 meter
x
z
y
180° θ
cr θ
m g
τ→
L→ ω
O
→
Fig. 9
I = 5.0 × 10–4 kg-meter2 and ω = 2π × 30 rad/sec
∴ ωp =0 50 9 8 0 04
5 0 10 2 3 14 302 084
. × . × .. × × ( × . × )
.− = rad/sec
A consideration of the direction of the torque vector and angular momentum vector andthe right-hand rule would show that the precession will be clockwise as seen from above.
Q. 2. A gyroscope wheel weighing 1.5 kg and having a radius of gyration 30 cm is spinningabout its axle at a rate of 240 rev/min. The axle is held horizontal and is supported by a pivotat one end which is 10 cm from the centre of mass of the wheel. Calculate the angular velocityof the precession of the wheel.
Ans. The gyroscope is shown in figure.
The angular velocity of precession is given by
ωp =mgr mgr
L I=
ωwhere m is the mass of the wheel and L (= Iω) is the angular momentum about the axle.
324 Mechanics
r
ω
Fig. 10
Here m = 1.5 kg, r = 10 cm = 0.1 m, I = mk2
= (1.5 kg) (0.3 m)2 = 0.135 kg-m2
and ω = 240 rev/min = 4 rev/sec = 4 × 2π rad/sec
∴ ωp =1 5 9 8 0 1
0 135 4 2 3 140 43. × . × .
. × × × ..= rad/sec
Q. 3. A torpedo boat with propeller making 4.5 rev/sec makes a complete turn in 84second. The moment of inertia of the propeller is 94.5 kg-m2. Compute the processional torqueon the propeller.
Solution. ωp =mgrIω
∴ torque mgr = Iω (ωp)
= 94.5 × (4.5 × 2π) × .284
199 66π = nt - m.
Q. 4. Use the Bohr’s postulate of the quantisation of angular momentum to obtain (i) thepossible orbits for the rotation of electron about the proton in the hydrogen atom, and (ii) thepossible energy levels of a diatomic molecule behaving like a rigid rotator.
Ans. (i) Orbits for the Electron Motion in Hydrogen Atom: In the hydrogen atom, anelectron revolves in a circular orbit about the proton (nucleus). Then electron carries anegative charge e and the proton an equal positive charge.
The electrical force of attraction between them is 1
4 0
2
2π ∈er
, where r is the radius of the
orbit. This force supplies the necessary centripetal force to the electron which is mv
r
2, where
m is the mass and v linear velocity of the electron. Clearly
14 0
2
2π ∈er
=mv
r
2
or r =1
4 0
2
2π ∈e
mv...(1)
If I be the moment of inertia and ω the angular velocity of the electron about the proton,then the angular momentum of the electron would be given by
Rotational Kinematics 325
L = Iω
= mrvr
2 I = and mr
vr
2 ω =
Now, according to Bohr’s postulate of quantisation, the angular momentum of the electron,
L, can only have values that are integral multiples of h
2π, where h is Planck’s constant. That
is
L =nh2π
where n is 1, 2, 3, ...
or mvr =nh2π
or r =nhmv2π
...(2)
Let us now eliminate v between Eq. (1) and (2). For this we divide the square of Eq. (1)by Eq. (2), when we obtain
rr
2=
n h m ve mv
2 2 2 2 2
20
24
4/
/π
π ∈
or r =n h
me
2 20
2∈
π
Putting n = 1, 2, 3, .... we obtain the radii of the orbits allowed to the electron.
(ii) Rotational Energy Levels for Diatomic Molecules: A diatomic molecule is a dumb-bellmodel, consisting of two atoms (point-masses) joined by a weightless rigid rod. It rotates aboutan axis passing through the centre of the joining rod (internuclear axis) and at right angleto it. If I be the moment of inertia of the molecule and ω the angular velocity of rotation,then its kinetic energy of rotation is given by
K =12
2Iω ...(1)
The angular momentum of the molecule is given by
L = Iω ...(2)
The two equations give
K =LI
2
2...(3)
Now, according to Bohr’s postulate of quantisation, the angular momentum L of the
molecule can only have values which are integral multiple of h
2π. That is,
L = J h2π
where J = 0, 1, 2, ...
326 Mechanics
Substituting this value of L in Eq. (3) we have
K = JI
2 h2
28π
Thus the kinetic energy also can only have a set of discrete values obtained by puttingJ = 0, 1, 2, 3,... in this expression. (In other words, the energy is also quantised). These valuesare
08
48
98
2
2
2
2
2
2, , , ...h h hπ π πI I I
Figure represents an energy level diagram which indicates how the spacing between theenergy levels varies as J increases.
We can also obtain the possible angular velocities with which the molecule would rotate.
J = 6
J = 5
J = 4
J = 3J = 2J = 1J = 0
K
Fig. 11
For this we put L =Jh2π
in Eq. (2), we get
Jh2π
= Iω
or ω = JI
h2π
; J = 0, 1, 2, ....
= 02
22
32
, , , ,...h h hπ π πI I I
Q. 5. The moon revolves about the earth so we always see the same face of the moon (a)How are the spin angular momentum and orbital angular momentum of the moon with respectto the earth related? (b) By how much its spin angular momentum must change so that wecould see the entire moon’s surface during a month?
Solution. (a) Let M be the mass and R the radius of the moon. Its spin angularmomentum about its own axis is
LS = IωS
where I is the moment of inertia of the moon and ωS its angular velocity about its own axis.
Treating it as a solid sphere, I = 25
MR2, so that
LS =25
MR2Sω
Rotational Kinematics 327
The moon not only spins about its axis, but also revolves in a orbit around the earth.The orbital angular momentum of the moon with respect to the earth is given by
L0 = I0ω0
where I0 is the moment of inertia of the moon (now treated as a particle) about the earthand ω0 is its orbital velocity. If r be the distance between earth and moon, then I0 = Mr2, sothat
L0 = Mr2ω0 ...(2)
Since we always see the same face of the moon, the moon makes one rotation about itsaxis in the same time as it makes one revolution around the earth. That is
ωS = ω0
Then Eqs. (1) and (2) giveLL
S
0=
25
Rr
2
(b) LS must increase or decrease by one-half of its present value if we have to see theentire moon’s surface during a month.
Q.6. The earth has a mass of 6 × 1024 kg and a radius of 6.4 × 106 meter. It spins onits own axis and also revolves in an orbit of radius 1.5 × 1011 meter around the sun. Computeits spin angular momentum about its own axis and its orbital angular momentum about thesun.
Ans. The earth makes one rotation (turns through 2π radian) about its own axis in 24hours. Treating it as a solid sphere, its spin angular momentum is given by
LS = Iωwhere I is its moment of inertia about its own axis. Now
I =25
MR2
=25
6 1024( × kg) (6.4 × 10 m)6 2
= 98.3 × 1036 kg-m2
and ωS =2 3 14
24 60 6072 6 10 6× .
× ×. ×= − rad / sec.
∴ LS = (98.3 × 1036) × (72.6 × 10–6)
= 7.13 × 1033 kg-m2/sec
The earth revolves around the sun once in an year. In order to find the angular momentumof the earth’s motion about the sun, we may treat the earth as a point mass. Then, its orbitalangular momentum is
L0 = I0ω0
where I0 is its moment of inertia about the sun and ω0 is its orbital velocity. If r be thedistance between sun and earth, then
I0 = Mr2
= (6 × 1024 kg) (1.5 × 1011 m)2
328 Mechanics
= 13.5 × 1046 kg-m2
and ω0 =2 3 14
365 24 60 601 99 10 7× .
× × ×. ×= − rad / sec
∴ L0 = (13.5 × 1046) × (1.99 × 10–7)
= 2.7 × 1040 kg-m2/sec
Q. 7. A body of mass M = 30 kg is running with a velocity of 3.0 meter/sec on groundjust tangentially to a merry go round which is at rest. The boy suddenly jumps on the merry-go-round. Calculate the angular velocity acquired by the system. The merry-go-round has aradius of r = 2.0 meter and a mass of m = 120 kg and its radius of gyration is 1.0 meter.
Solution. The merry-go-round rotates about the central axis. The moment of inertia ofthe body about the axis of rotation is
I1 = Mr2 = 30 × (2.0)2 = 120 kg-meter2
and that of the merry-go-round is
I2 = mk2 = 120 × (1.0)2 = 120 kg-meter2
The body is running with an angular velocity about the axis of rotation which is given by
ω1 =vr = =3 0
2 01 5.
.. rad/sec.
while the merry-go-round is at rest (ω2 = 0). Let ω be the angular velocity of the system whenthe boy jumps on the merry-go-round.
As no external torque is acting on the system (friction being ignored), we have
initial angular momentum = final angular momentum
I1ω1 + I2ω2 = (I1 + I2)ω(120 × 1.50) + (120 × 0) = (120 + 120)ω
∴ ω =120 1 5
2400 75× . .= rad/sec. Ans.
Q. 8. A meter stick lies on a frictionless horizontal table. It has a mass M and is free tomove in any way on the table. A small body of mass m moving with speed v as shown in figure,collides elastically with the stick. What must be the value of m if it is to remain at rest afterthe collision?
Ans. There is no external force (or torque) actingupon the system, and the collision is elastic. Therefore,the linear momentum, the angular momentum abouta point say, the center of mass C of the stick and themechanical energy are all conserved.
The initial linear momentum of the system is thatof the body mv, angular momentum about C is
Iω = md2(v/d), and the kinetic energy is 12
2mv . Let V
be the linear velocity acquired by the stick after collisionand ω the angular velocity about C. The bodyconservation of linear momentum, we have
l
m v
d
M
C
Fig. 12
Rotational Kinematics 329
mv = MV ...(1)
By the conservation of angular momentum, we have
mdvd
2 =
MI12
2ω ...(2)
where MI2/12 is the moment of inertia of the stick about C. By the conservation of mechanicalenergy, we have
12
2mv =12
12
MV MI12
22
+
ω ...(3)
From Eq. (1) V =Mmv , from Eq. (2), ω = mvd 12
MI2
Substituting these values of V and ω in eq. (3), we get
mv2 = MM
MI12 MI
mvmvd
+
2 2
2
212
×
or 1 =m
mdM MI2+
2 12
=m dM I
1 12 2
2+
∴ m =MI
I
2
2 212+ d
Q. 9. A point-mass is tied to one end of a cordwhose other end passes through a vertical hollow tube,caught in one hand, and is caught in the other hand.The point-mass is being rotated in a horizontal circle ofradius 2 meter with a speed of 4 meter/sec. The cordis then pulled down so that the radius of the circlereduces to 1 meter. Compute the new linear and angularvelocities of the point-mass and compare the kineticenergies under the initial and final states.
Ans. Let m be the point-mass v1 its linear velocityand ω1 the angular velocity of radius r1. Let v2 and ω2be the linear and angular velocities respectively whenthe circle is shortened to a radius r2.
The force on the point-mass due to the cord isradial, so that the torque about the centre of rotationis zero. Therefore the angular momentum must remainconstant as the cord is shortened. Thus,
v1
ω1r1
r2
v2ω2
Fig. 13
330 Mechanics
initial angular momentum = final angular momentum
I1ω1 = I2ω2
or mr vr1
2 1
1= mr v
r22 2
2
or r1v1 = r2v2
∴ v2 =rr
v1
21 =
21
4 8× = m/sec
and ω2 =vr2
2 =
81
8= rad/sec
Final K.E.Initial K.E.
=
12
I
12
I
2
1
ω
ω
22
12
22 2
2
2
12 1
1
2=
mr vr
mr vr
=vv
22
12
2
284
41
= =( )( )
we see that the kinetic energy increases. The increase comes from the work done in pullingin the cord against the centrifugal force.
Q. 10. (a) The time-integral of a torque is called the ‘angular impulse’. Show that theangular impulse about a fixed axis is equal to the change in angular momentum about thataxis.
(b) Show that in order to make a billiard ball roll smoothly, without sliding on top or back
spin, the cue must hit the ball at a height 75
R above the table, where R is the radius of the
ball.
Ans. (a) Angular Impulse: When a large torque acts for a short time, the product ofthe torque and the time for which it acts is called ‘impulse’ of the ‘torque’ or ‘angularimpulse’.
We know that the time-rate of change of the total angular momentum of a system aboutan axis is equal to the resultant external torque acting on the system about that axis. Thatis
τ =ddtL
or τ dt = dL
If the torque acts for a short time ∆t, and the angular momentum of the system increasesfrom L1 to L2 during this time, then
Rotational Kinematics 331
τ dtt
0
∆
= dL L LL
L
2 1
1
2
= −
The quantity τ dtt
0
∆
is the ‘angular impulse’. Thus the impulse of a torque about an axis
is equal to the change in angular momentum about that axis.
(b) Suppose the billiard-ball is initially at rest, and is hit by the cue at a height h abovethe table. Let the cue exert a force F for a short interval of the time ∆t.
h
C
R
ωR
(h – R )
F Cue
ω
Fig. 14
The moment of the force (torque) about the axis perpendicular to the paper through thecentre C of the ball in F(h – R). If ω is the angular velocity of the ball after impact, then byconsideration of angular momentum which increases from 0 to Iω, we have
F(h – R) ∆t = Iωwhere I is the moment of inertia of the sphere.
But I =25
MR2
∴ F(h – R) ∆t =25
MR2 ω ...(1)
The original linear velocity of the ball is zero, and if there is to be no slipping at the pointof contact with the table after impact, the velocity of the ball will be ωR. By consideration oflinear momentum, we have
F∆t = MωR ...(2)
Dividing Eq. (1) by Eq. (2), we get
h – R =25
R
∴ h =25
R + R
∴ h =75
R Ans.
Q. 11. A block of mass M is moving with velocity v1 on a frictionless surface as shown.It passes over to a cylinder of radius R and moment of inertia I which has a fixed axis andis initially at rest. When it first makes contact with the cylinder, it slips on the cylinder, but
332 Mechanics
the friction is large enough so that slipping ceases before it loses contact with the cylinder.Finally, it goes to the dotted position with velocity v2. Compute v2 in terms of v1, M, I andR.
Ans. The initially angular momentum of M about C when it first makes contact with thecylinder, is
L1 = Moment of inertia about C × angular velocity
= (MR)R
M R2 vv11
=
The final angular momentum when it loses contact with the cylinder is
L2 = Mv2R
The loss in angular momentum is
∆L = L1 – L2 = MR (v1 – v2)
This must be equal to the angular impulse τ∆t on the cylinder about C. Thus
MR (v1 – v2) = τ∆t ...(1)
Now, if α be the angular acceleration produced in the cylinder, then
= Iα = I∆∆ωt
...( 2)
CR
v1
M
v2
Fig. 15
As the cylinder is initially at rest and finally moves with a velocity v2, we have
∆ω =v2
RSo that from Eq. (2),
τ =I
R∆tv2
Putting this value in (1), we get
MR (v1 – v2) = IRv2
or v1 – v2 =I
MR2
v2
or v2 =v1
1 + IMR2
Ans.
Rotational Kinematics 333
Q. 12. As cockroach, mass m, runs round the rim of a disc of radius R with speed vcounter clockwise. The disc has rotational inertia I about its axis and it rotates clockwise withangular speed ω0. If the cockroach stops, compute the angular speed of the disc and also thechange in kinetic energy.
Ans. As no external torque is acting on the system, the angular momentum is conserved.Now
Initial angular momentum = I R)R
2ω0 − (m
v
= Iω0 – mvR
(the negative sign is used because the angular momentum of the insect of oppositelydirected to that of the disc).
Let ω be the final angular velocity of the system when the cockroach stops. Thus
Final angular momentum = (I + mR2)ωBy the conservation of angular momentum, we have
Iω0 – mvR = (I + mR2)ω
∴ ω =I R
I + R0
2ω − mv
m
Though angular momentum is conserved, but the angular velocity has changed. Hencethe energy is also changed. The change can be easily computed.
Q. 13. If the earth suddenly contracts to half its radius, what would be the length of theday.
Ans. Let R1 be the present radius of earth and T1 its period of revolution about its axis,which is the length of the day (T1 = 24 hours). Let R2 be the new value of radius and period.By conservation of angular momentum, we have
I1ω1 = I2ω2
or25
2MRT1
2
1
π=
25
2MRT2
2
2
π
or T2 = TRR
hours) ×(R
R122
12
1
12= (/ )
242 2
= 6 hours Ans.
Q. 14. A man stands at the center of a rotating platform holdings his arms extendedhorizontally with a weight in each hand, the platform making 1 revolution per second. Themoment of inertia of the whole system is 6 kg-meter2. When the man pulls in his hands tohis sides, the moment of inertia becomes one-third of its initial value. What is the new angularvelocity of the platform. Has the kinetic energy been conserved, If not, explain it. Assume thatthere is no friction in the platform.
Ans. If friction in the platform is neglected, no external torque acts about the verticalaxis through the center of mass, and hence the angular momentum about the axis is constant.(the force of gravity acts through the center of mass and hence exerts on torque the axis ofrotation). Thus
334 Mechanics
I1ω1 = I2ω2
where I1 and ω1 are the initial and I2 and ω2 the final moment of inertia and angular velocityabout the axis of rotation.
I1 = 6 kg-m2, ω1 = 1 rev/sec = 2π rad/sec
I2 = 2 kg-m2, ω2 = ?
∴ 6 × 2π = 2 × ω2
or ω2 = 6π rad/sec = 3 rev/sec
Initial kinetic energy12 1
2I1 ω =12
6 2 122 2× × ( )π π= joule
Final kinetic enegy12 2
2I2 ω =12
2 6 362 2× × ( )π π= joule
Clearly, the kinetic energy has not been conserved. It has increased.
Increase in kinetic energy = 36 π2 – 12 π2
= 24 π2 = 24 × (3.14)2 = 236.6 Joule
This increase in energy is supplied by the man who does work when he pulls in his handsto his sides.
Q. 15. A uniform disc of mass M and radius R is rotating about its central axis whichis horizontal with angular velocity w0
(a) write down its kinetic energy and angular momentum.
(b) suddenly a piece of mass m breaks off the disc and rises vertically. To what height thepiece would rise.
(c) what are the final energy and momentum of the broken disc?
Ans. (a)Initial kinetic energy Ki =12
I 02ω
=12
12
140
202MR MR2 2
=ω ω
Initial angular momentum Li = I MR02ω ω= 1
2 0
(b) Thus moment of inertia of the mass m, which later on breaks off the disc is mR2.
Its rotational energy is therefore 12 0
2( .mR )2 ω On breaking off, the mass goes up and its
kinetic energy is converted into gravitational potential energy mgh, where h is the maximumheight to which it goes. Thus,
12 0
2mR2 ω = mgh
or h =R2 ω0
2
2g(c) Let us consider the disc to be made up of the piece of mass m and the remainder
(broken) disc. Now, by the conservation of energy, we have
Rotational Kinematics 335
ki = k k mf f( )of broken disc) + (of mass ′
or14 0
2MR2 ω = k mf + 12 0
2( )Rω
∴ kf =12
12 0
2M R2−
m ω
Similarly, by the conservation of angular momentum, we have
Li = Lf (of broken disc) + Lf (of mass m)
or12 0MR2 ω = Lf + mR2 ω0
∴ Lf =12 0M R2−
m ω Ans.
PROBLEM
1. A wheel of moment of inertia 8 kg-m2 is momented on a shaft and revolving at a rate of600 rev/min. When second wheel, initially at rest, is coupled to the same shaft, the combinationrevolves at a rate of 400 rev/min. What is the moment of inertia of the second wheel. Howmuch energy is lost in the process ? (Neglect all bearing friction).
[Ans. 4 kg-m2, 5258 joule]
2. A certain mass tied to one of a light string of length 100 cm is making 4 rev/sec in ahorizontal plane about the other end. How many rev/sec does it make if the string lengthis changed to 50 cm ? How much were kinetic energy does the revolving mass have if it weighs0.5 kg ? From where does this additional energy come ? [Ans. 16, 48π2 joule]
3. An ice-skater spinning at the rate 1 rev/sec changes posture in such a way that his momentof inertia is reduced to half its initial value. What becomes his speed of spin ?
4. A turn table rotates about a fixed vertical axis, making one revolution in 10 sec. The momentof inertia of the turn table about the axis is 1200 kg-m2. An 80 kg man, initially standingat the center on the turn table, moves along a radius. What is the angular velocity of theturn table when the mass is 2 meter from the center.
336 Mechanics
8
8.1 MOMENT OF INERTIA
The inability of a body to change by itself its state of rest or uniform motion along astraight line (Newton’s first law of motion) is an inherent property of matter and is calledinertia. The greater the mass of the body, the greater the resistance offered by it to anychange in its state of rest or linear motion. So that, mass is taken to be a measure of inertiafor linear or translatory motion.
In the same manner, a body free to rotate about an axis opposes any change in its stateof rest or uniform rotation. In other words, it possess inertia for rotational motion, i.e., itopposes the torque or the moment of the couple applied to it to change its state of rotation.It must also be, therefore, of the nature of the moment of a couple, for only a couple canoppose another couple. Hence the name moment of inertial given to it.
It will thus be seen at once that the moment of inertia (usually written as M.I.) abouta given axis plays the same part in rotational motion about that axis as the mass of a bodydoes in translational motion. In other words the moment of inertia, in rotational motion, isthe analogue of mass in linear or translational motion.
If a particle of mass m is at a distance r from an axis of rotation, its moment of inertiaI about the axis is given by the product of the mass and the square of its distance from theaxis.
Thus
I = mr2
In an extended body, each particle ofmatter in the body add to the moment of inertiaand amount mr2. Figure represents such a bodyrotating about an axis through O andperpendicular to the plane of the body. If m1,m2, m3, ... be the masses of the particlescomposing the body and r1, r2, r3, ... theirrespective distances from the axis of rotation,the moment of inertia I of the body about thataxis is given by
O
r5 r4
m 5m 4
m 3
r3
r2
m 2
m 1
r1
Fig. 1
336
Moment of Inertia 337
I = m r m r m r1 12
2 22
3 32+ + + ...
= Σ mr2
For a body having a continuous distribution of matter, we put
I = r dm2where dm is the mass of a infinitesimally small element of the body taken at a distance rfrom the axis of rotation.
Hence the moment of inertia of a body about a given axis is the sum of the productsof the masses of its particles by the squares of their respective distance from the axis ofrotation. It thus depends not only on the mass of the body but also upon the manner in whichthe mass is distributed around the axis of rotation.
8.2 ROLE OF MOMENT OF INERTIA IN ROTATIONAL MOTION
According to the Newton’s first law of motion, every body maintains its state of rest oruniform transnational motion unless some external force is applied on it to change that state.In other words, a force is necessary to change the motion of a body that is, to produce a linearacceleration in it. This property of bodies is called ‘inertia’. The greater is the mass of a body,the greater is the force required to produce a given linear acceleration in it. Thus the massof a body is a measure of its inertia.
Similarly, if the rotational motion of a boy about an axis is to be changed, that is, if anangular acceleration is to be produced, a torque (moment of force) about that axis must beapplied. The body is, therefore, said to possess a ‘moment of inertia’ about that axis. Thegreater is the moment of inertia, the greater is the torque required to produce a givenangular acceleration in the body.
Thus, moment of inertia plays the same role in the rotational motion as mass plays intransnational motion.
8.3 RADIUS OF GYRATION
The radius of gyration, K, of a body about its axis of rotation, which we have called theeffective distance of its particles from the axis, is actually that distance from the axis at whichif its center of mass (M) be supposed to be concentrated, its moment of inertia about the givenaxis would be the same as with its actual distribution of mass.
If a change in the position or inclination of the axis of rotation of a body will bring abouta change in the relative distances of its particles and hence in their effective distance or theradius of gyration of the body about the axis. And so will a change in the distribution of massabout the axis, e.g., the transference of a portion of matter (or mass) of the body from onepart to another, despite the fact that the total mass of the body remains unaltered.
Thus, whereas the mass of a body remains the same irrespective of the location orinclination of its axis of rotation, its radius of gyration about the axis depends upon (i) theposition and inclination of the axis and (ii) the distribution of mass of the body about the axis.
We usually put it as
I = Σ mr2 2= MK
Where M is the total mass of the body (e.g. equal to m1 + m2 + m3 + ..... etc) and K,the effective distance of its particles from the axis, called its radius of gyration about the axisof rotation.
338 Mechanics
8.4 ANALOGOUS PARAMETERS IN TRANSLATIONAL AND ROTATIONAL MOTION
Translational Motion1. Mass (or translational inertia), m or M
2. Distance covered (or linear displacement), S
3. Linear velocity, vdsdt
=
4. Linear velocity, advdt
d sdt
= =2
2
5. Linear momentum, p = mv
6. Force (or rate of change of linear momentum), F = =ddt
maφ
7. Work done by force, W = F S FSd =
8. Power, P = Fv
9. Translational Kinetic Energy, ½ mv2
10. Equations of translational motion:
(i) v = u + at
(ii) S = ut + ½ at2
(iii) v2 – u2 = 2as
Potential Motion1. Moment of Inertia, I = MK2 (or rotational inertia)
2. Angular described, θ (radius or angular displacement)
3. Angular velocity, ω θ= ddt
(radian/sec)
4. Angular acceleration, α ω θ= =ddt
ddt
2
2
5. Angular momentum, J = Iω6. Torque or moment of couple (or rate of change of angular momentum), τ or
C = J Iddt
ddt
= ω
7. Work done by torque or couple, W or = τ θ θd cd or τθ or Cθ
8. Power, P = τω or Cω
9. Rotational kinetic energy = 12
I 2ω
10. Equation of rotational motion:
(i) ω ω α ω ω2 1 1= + = +
tddt
t
Moment of Inertia 339
(ii) θ ω α ω ω= + = +
1
21
212
12
t t tddt
t
(iii) ω ω αθ ω θ22
12 2 2− = = d
dt
8.5 GENERAL THEOREMS ON MOMENT OF INERTIA
There are two important theorems on moment of inertia which, in some cases, enablethe moment of inertia of a body to be determined about an axis, if its moment of inertia aboutsome other axis be known.
(i) The theorem of perpendicular axes: The moment of inertia of a uniform planelamina about an axis perpendicular to its plane is equal to the sum of its moments of inertiaabout any two mutually perpendicular axes in its plane intersecting on the first axis. Thisis the theorem of perpendicular axis for a laminar body.
Thus, if Ix and Iy be the moments of inertia of a plane lamina, about the perpendicularaxis OX and OY respectively, which lie in the plane of the lamina and intersect each otherat O, its moment of inertia (I) about an axis passing through O and perpendicular to its planeis given by
i.e., I = Ix + Iy
For, considering a particle of mass m at P, distance r fromO and x from y from the axes OY and OX respectively, wehave
I = Σmr2, Ix = Σmy2 and Iy = Σmr2
so that Ix +Iy = Σmy2 + Σmx2
= Σm (y2 + x2)
= Σmr2 [y2 + x2 = r2]
i.e., Ix + Iy = I
For a three-dimensional body: In this case, the theorem demands that the sum of themoments of inertia of a three-dimensional body about its three mutually perpendicular axesis equal to twice the summation Σmr2 about the origin.
Suppose, we have a three-dimensional or a cubical body with OX, OY and OZ as its threemutually perpendicular axes, representing its length, breadth and height respectively.
Consider a small element of mass ‘m’ of the body at a point P somewhere inside it. Dropa perpendicular PM from P on the x-y plane to meet it in M. Join OM and OP and from Mdraw MQ parallel to the x-axis and MN parallel to the y-axis. Also, from P draw PR parallelto OM. Then , clearly, the coordinates of the point P are x = ON = QM, y = OQ = MN andz = MP = OR.
Q
O
rN
MY
Z
X
R m P
y
x
Fig. 3
O
y
Px
r
X
Y
m
Fig. 2
340 Mechanics
Since plane x – y is perpendicular to the z-axis, any straight line drawn on this planeis also perpendicular to it. So that, both OM and PR are perpendicular to the z-axis (PR beingparallel to OM). ∠OMP is a right angle, [OM || PR and PM || OR.]
We, therefore, have
OM2 + MP2 = OP2, or OM2 + z2 = r2 ...(i) ∴ OP = r
But OM2 = QM2 + OQ2 = x2 + y2
∴ OQM is a right angle, being the angle between the axes of x and y.
Substituting the value of OM2 in relation (i), we have
x2 + y2 + z2 = r2 ...(ii)
Join PN and PQ. Then, PN and PQ are the respective normals to the axes of x and y.For ∠PMN being a right angle (the angle between the axes of y and z), we have PN2 = MN2
+ PM2 = y2+ z2 and, therefore x2 + PN2 = x2 + y2 + z2 = r2 or ON2 + PN2 = r2 (x = ON).Angle PNO is thus a right angle and, therefore, PN perpendicular to the x-axis.
Similarly, in the right-angled triangle ∆PMQ, we have
PQ2 = MQ2 + PM2 = x2 + z2
Again, because x2 + y2 + z2 = r2, we have PQ2 + y2 = r2
or PQ2 + OQ2 = OP2 ∴ y = OQ and r = OP
Angle PQO is a right angle and, therefore, PQ perpendicular to the y-axis. Now, momentof inertia of the element at P about the z-axis = m × PR2 = m × OM2, because PR = OMis the perpendicular distance of the mass from the axis.
∴ moment of inertia of the whole body about the z-axis, i.e.,
Iz = Σm.OM2 = Σm (x2 + y2)
Similarly, moment of inertia of the body about the y-axis, i.e.,
Iy = Σm.PQ2 = Σm (x2 + z2)because PQ is the perpendicular distance of mass m from this axis. And , moment of inertiaof the body about the x-axis, i.e.,
Ix = Σm.PN2 = Σm (y2 + z2)because PN is the perpendicular distance of mass m from the x-axis,
∴ adding up the moments of inertia of the body about the three axes, we haveIx + Iy + Iz = Σm (y2 + z2) + Σm (x2 + z2) + Σm (x2 + y2)
= 2Σm (x2 + y2 + z2)= 2Σm r2
The Theorem of parallel axes: The moment ofinertia I of a body about any axis is equal to its momentof inertia Icm about a parallel axis through its centerof mass, plus the product of the mass M of the bodyand the square of the perpendicular distance h betweenthe two axes. That is
I = Icm + Mh2
This is the theorem of parallel axes.
(a) For a plane laminar body: Let AB be theaxis in the plane of the paper about which the moment
rO
m
P
A
B D
C
x
Fig. 4
Moment of Inertia 341
of inertia (I) of the plane lamina is to be determined and CD, an axis parallel to AB, throughthe center of mass O of the lamina, at a distance r from AB.
Considering a small element of mass of the lamina at the point P, distance x from CD,we have moment of inertia of the element about the axis AB
= m (x + r)2
and; therefore, moment of inertia of the whole lamina about the axis AB, i.e.,
I = Σm (x + r)2
or I = Σmx2 + Σm2 + 2Σmxr
Σmx2 = Icm the moment of inertia of the lamina about the axis CD, through its centreof mass. So that,
I = Icm + Σmr2 + 2Σmxr
Now, Σmr2 = r2 Σm (r being constant) = Mr2, where M is the mass of the whole lamina,and Σmx = the sum of the moments of all the particles of the lamina about the axis CD,passing through its center of mass and, therefore, equal to zero, as we know, a body alwaysbalances about its c.m., showing that the algebraic sum of the moments of all its particlesabout the c.m. is zero, therefore, we have
I = Icm + Mr2
i.e., the moment of inertia of the lamina about the axis AB = its moment of inertia about a parallelaxis CD through its center of mass + mass of the lamina × (distance between he two axes)2
(b) For a three-dimensional body: Let AB be the axis about which the moment ofinertia of a cubical or a three-dimensional body, is to be determined.
Draw a parallel axes CD through the center of mass O of the body at a distance r from it.
Imagine an element of the body, of mass m, at a point P outside the plane of the axesAB and CD and let PK ad PL be perpendicular drawn from P to AB and CD respectively andPT, the perpendicular dropped from P on to KL produced.
Put PL = d, LK = r, LT = x and ∠PLK = θ.
Then, if I be the moment of inertia of the body about the axis AB and Icm, its momentof inertia about the axis CD (through its center of mass O), we have
r
r
dO
DB
K
A C T
P
θ
x
L
Fig. 5
I = Σm.PK2 and Icm = Σm.PL2 = Σm d2
Now, from the geometry of the figure, we have
PK2 = PL2 + LK2 – 2 PL.LK cos PLK = d2 + r2 – 2 dr cos θ. ...(1)
And, in the right-angled triangle PTL, we have cos PLT = LT/PL, where, ∠PLT = 180°–∠PLK = 180° – θ . So that cos (180° – θ) = x/d or –cos θ = x/d, where d cos θ = –x
342 Mechanics
Substituting this value of d cos θ in the expression for PK2 above Eq. (1), we have
PK2 = d2 + r2 + 2rx, and therefore,
I = ΣmPK2 = Σm (d2 + r2 + 2rx)
= Σmd2 + Σmr2 + 2r Σmx
or I = Icm + Mr2 + 2r Σmx
Σm = M, mass of the body and r is constant.
clearly Σmx = 0, being the total moment about an axis passing through the center of massof the body. We, therefore, have
I = Icm + Mr2
the same result as for a plane laminar body.
8.6 CALCULATION OF MOMENT OF INERTIA
In the case of a continuous homogeneous body of a definite geometrical shape, itsmoment of inertia is calculated by (i) first obtaining an expression for the moment of inertiaof an infinitesimal element (of the same shape) of mass dm about the given axis, i.e., dm.r2,where r is th distance of the infinitesimal element from the axis and then (ii) integrating thisexpression over appropriate limits so as to cover the entire body, making full use of thetheorems of perpendicular and parallel axes, where necessary.
8.7 MOMENT OF INERTIA OF A UNIFORM ROD
(i) About an axis through its centre and perpendicular to its length: Let AB bea thin uniform rod, of length l and mass M, free to rotate about an axis YOY' passing throughits center O and perpendicular to its length. Since the rod is uniform, its mass per unit lengthM/l.
Considering a small element of the rod, oflength dx at a distance x from the axis throughO, we have mass of the element = (M/l).dx andtherefore, its M.I. about the axis (YOY') throughO
= (M/l).dx.x2
The moment of inertia (I) of the whole rodabout the axis YOY' is thus given by the integralof the above expression between the limits x =– l/2 and x = + l/2 or by twice its integralbetween the limits x = 0 and x = l/2, i.e.,
I = 22
32
24 122
0
2 3
0
2 3 2M M M Ml
x dxl
xl
l ll
l/
. =
= =
(ii) About an axis through one end of the rod and perpendicular to its length: Themoment of inertia of the rod about the axis; passing through one end A of the rod, byintegrating the expression for the M.I. of the element dx of the rod, between the limitsx = 0 at A and x = l at B, i.e.,
I =M M Ml
x dxl
x ll l
2
0
3
0
2
3 3 =
=
Odx
B
l2
A
Y
Y ′
l2
x
Fig. 6
Moment of Inertia 343
x
Odx
B
l2
A
Y
Y ′
Fig. 7
8.8 MOMENT OF INERTIA OF A RECTANGULAR LAMINA (OR BAR):
(i) About an axis through its center and parallel to one side: Let ABCD be a rectangularlamina of length l, breadth b and mass M and let YOY' be the axis through itscenter O and parallel to the side AD or BC about which its moment of inertia isto be determined.
Consider an element, or a small rectangular strip of the lamina, parallel to YY',and at a distance x from, the axis. The area of this strip or element = dx.b. And,since the mass per unit area of the lamina = M/(l × b), we have mass of the strip
or element = M M
l bdx b
ldx
×. . . ,= and, therefore, M.I. of the element about the axis
YOY'
=Ml
dx x. 2
O
dxB
l2
A
Y
Y ′
l2
xD C
b
Fig. 8
The moment of inertia (I) of the whole rectangular lamina is thus given by twice theintegral of the above expression between the limits x = 0 and x = l/2.
I = 222 2
0
2
0
2M Ml
x dxl
x dxll
= //
=2
32
24 12
3
0
2 3 2M M Ml
xl
l ll
= =.
344 Mechanics
This result show that if b small, the rectangular lamina becomes a rod of length l whoseM.I. about the axis YOY’ passing through its center and perpendicular to its length would beMl2/12.
(ii) About One Side: In this case, since the axis coincides with AD or BC, we integratethe expression for the M.I. of the element of length dx and distance x from the axis, i.e.,(M/l) dx.x2 between the limits x = 0 at AD and x = l at BC. So that M.I. of the lamina aboutside AD or BC is given by
I =M M Ml
x dxl
x ll l
2
0
3
0
3
3 3 =
=
(iii) About an axis passing through its center and perpendicular to its plane: Thisis obtained by an application of the principle of perpendicular axes to case (i), according towhich, M.I. of the lamina about an axis through O and perpendicular to its plane = M.I. ofthe lamina about an axis through O parallel to b + M.I. of the lamina about an axis throughO parallel to l, i.e.,
I =M M M
12l b
l b2 2
2 2
12 12+ = +( )
(iv) About an axis passing through the mid-point of one side and perpendicularto its plane: In this case the axis passes through the mid-point of side AD or BC, andperpendicular to the plane of the lamina, so that, it is parallel to the axis through O (the c.m.of the lamina) in case (iii).
In accordance with the principle of parallel axes, therefore, the M.I. of the lamina aboutthis axis is given by
I = M Ml b l2 2 2
12 2+
+
= M Ml b l l b2 2 2 2 2
12 4 3 12+ +
= +
And if the axis passes through the mid-point of AB or DC, then
I = M Ml b b2 2 2
12 2+
+
= M Ml b l b2 2 2 2
12 12 3+
= +
(v) About an axis passing through one of its corners and perpendicular to itsplane: Let the axis pass through the corner D of the lamina and it is perpendicular to theplane of the lamina, it is parallel to the axis through its center of mass O in case (iii). Again,by the principle of parallel axes, we have moment of inertia of the rectangular lamina aboutthis axis through D is given by
I = M M 2l br
2 2
12+
+ ,
where r is the distance between the two axes.
Moment of Inertia 345
We have, r2 =l b l b2 2 4
2 2 2 2
+
= +
So that, I = M Ml b l b2 2 2 2
12 4+
+ +
= Ml b l b2 2 2 23 3
12+ + +
I = Ml b2 2
3+
8.9 MOMENT OF INERTIA OF A THIN CIRCULAR RING (OR A HOOP)
(i) About an axis through its center ad perpendicular to its plane: Let the radiusof the hoop or the thin circular ring be R and its mass M.
Consider a particle of mass m of the hoop or the ring, its M.I. about an axis passingthrough the center O of the hoop or the ring and perpendicular to its plane = mR2
∴ M.I. of the entire hoop or ring about this axis passing through its center andperpendicular to its plane,
I = ΣmR2 = MR2
Where åm = M, the mass of the hoop or ring.
(ii) About its diameter: Due to symmetry, the M.I. of thehoop or the ring will be the same about one diameter as aboutanother. Thus, if I be its M.I. about the diameter XOX', it willalso be I about the diameter YOY' perpendicular to XOX'.
By the principle of perpendicular axes, therefore, the M.I. of the hoop or the ring aboutthe axis through its center O and perpendicular to its plane is equal to the sum of itsmoments of inertia about the perpendicular axes XOX' and YOY' in its own plane andintersecting at O, i.e.,
I + I = MR2
or 2I = MR2
or I =MR2
2
8.10 MOMENT OF INERTIA OF A CIRCULAR LAMINA OR DISC:
(i) About an axis through its center and perpendicular to its plane: Let M be the
mass of the disc and R its radius, so that its mass per unit area is equal to MR2π
.
Considering a ring of the disc, of width dx, and distant x from the axis passing throughO and perpendicular to the plane of the disc, we have
Area of the ring = circumference × width
X ′ X
Y
Y ′
O
Fig. 9
346 Mechanics
= 2π x dx and hence its
Mass =MR
M.R2 2π
π
=22
. .x dxxdx
And, therefore, its M.I. about the perpendicular axis through O
=2 22
3M.R
M.R2 2
x dxx
x dx. .=
Since the whole disc may be supposed to be made up of such concentric rings of radiiranging from O to R, the M.I. of the whole disc about the axis through O and perpendicularto its plane, i.e., I , is obtained by integrating the above expression for the M.I. of the ring,between the limits x = 0 and x = R. Thus,
I =2 2
43
0
4
0
MR
MR2
R
2
R
x dxx =
=2
4 2M
RR MR
2
4 2=
(ii) About a diameter: Due to symmetry, the M.I. of the disc about one diameter is thesame as about another. So that, if I be the M.I. of the disc about each of the perpendiculardiameters XOX' and YOY', then by the principle of perpendicular axes,
I + I = M.I. of the disc about an axisthrough O and perpendicularto its plane.
i.e. 2I =MR2
2
or I =MR2
4
8.11 MOMENT OF INERTIA OF AN ANGULAR RING OR DISC
(i) An angular disc is just an ordinary disc, with a smaller coaxial disc removed from itleaving a concentric circular hole in it.
If R and r be the outer and inner radii of the disc and M, its mass, then mass per unitarea of the disc
= massarea
M(R2=
−π r2)Now, the disc may be imagined to be made up of a number
of circular rings, with their radii ranging from r to R. So that,considering one such ring of radius x and width dx, we have
Face area of the ring = 2πxdx and
∴ its mass = 2πxdx [M/p (R2 – r2)]
=2
2M
R2xr
dx( )−
xR
Fig. 10
X ′ X
Y
Y ′
O
Fig. 11
xRr2
O
dx
Fig. 12
Moment of Inertia 347
and hence its M.I. about the axis through O and perpendicular to its plane
= 2 22
23
2M
RM
R2 2xr
dx xxr
dx( )
.( )−
=−
The M.I. of the whole annular disc i.e., I, is therefore, given by the integral of the aboveexpression between the limits x = r and x = R
or I =2 2 2
4
3
2 2 23
2
4MR
M(R
M(R2
R R
2
Rxr
dxr
x dxr
x
r r r( ) )
.)−
=−
=−
I =2
4 22
4 2M(R
R M (R2
4 2
−−
= +
rr r
))
(ii) About a diameter: Due to symmetry, the M.I. of the annular disc about onediameter is the same as about another, say, I. Then, in accordance with the principle ofperpendicular axes, the sum of its moments of inertia about two perpendicular diametersmust be equal to its M.I. about the axis passing through its center (where the two diametersintersect) and perpendicular to its plane, i.e., I + I = M (R2 + r2)/2
2I = M(R2 + r2
2)
or I =M (R2 + r2
4)
8.12 MOMENT OF INERTIA OF A SOLID CYLINDER
(i) About its own axis of cylindrical symmetry: A solid cylinder is just a thick circulardisc or a number of thin circular disc (all of the same radius) piled upon over the other, sothat its axis of cylindrical symmetry is the same as the axis passing through the centre ofthe thick disc and perpendicular to its plane.
∴ M.I. of the solid cylinder about its axis ofcylindrical symmetry, i.e., I = M.I. of the thick disc (orthe pile of thin discs) of the same mass and radiusabout the axis through its center and perpendicular toits plane.
I = MR2
2case 8.10(i)
(ii) About the axis through its center and perpendicular to its axis of cylindricalsymmetry: If R be the radius, l, the length and M, the mass of the solid cylinder, supposedto be uniform and of a homogeneous composition, We have its mass per unit length = M/l.
Now, imagining the cylinder to be made up of a number of discs each of radius R, placedadjacent to each other, and considering one such disc of thickness dx and at a distance x fromthe center O of the cylinder, we have
mass of the disc = (M/l) dx and radius = R
∴ M.I. of the disc about its diameter AB =M R2
ldx.
4
X ′ X
Y
Y′
O
l2 dx
R
lB
Ax
Fig. 13
348 Mechanics
and its M.I. about the parallel axis YOY', passing through the center O of the cylinder andperpendicular to its axis of cylindrical symmetry (or its length), in accordance with theprinciple of parallel axes,
=M R M2
ldx
ldx x
42+ .
Hence, M.I. of the whole cylinder about this axis, i.e.,
I = twice the integral of the above expression between the limit x = 0 and x = l/2, i.e.,
I = 24
24
2 2
0
2
0
2M R M M R2 2
ldx
ldx x
ldx x dx
ll
+
= +
. .//
=2
4 3
3
0
2M R2
lx x
l
+
/
or I = 24 2 8 3
3M R2
ll l
..
+
=2
8 24 4 12
3 2M R M R2 2
ll l l+
= +
8.13 MOMENT OF INERTIA OF A SOLID CONE
(i) About its vertical axis: Let M be the mass of the solid cone, h its vertical heightand R, the radius of its base.
Volume of the cone = (1/3)π R2h and if ρ be the density of its material, its mass M
M =13
π ρR2h
or ρ =3MR2π h
The cone may be imagined to consist of anumber of disc of progressively decreasing radii,from R to O, piled up one over the other.
Considering one such disc of thickness dx and at a distance x from the vertex (A) of thecone, we have
radius of the disc r = x tan a, where a is the semi-vertical angle of the cone, and, therefore,its volume = πr2dx = πx2 tan2 α dx and its mass
= πx2 tan2 α ρ dx
Hence, M.I. of the disc about the vertical axis AO of the cone (i.e., an axis passingthrough its center and perpendicular to its plane) = mass × radius2/2
= πx2 tan2 αρ dx. r2/2
= πx2 tan2 αρ dx. x2 tan2 α/2
= (πρ tan4 a/2) x4 dx
h
X ′ X
αr
dx
O R
x
A
Fig. 14
Moment of Inertia 349
and M.I. of the entire cone about its vertical axis AO is given by
I =πρ αtan
.4
4
02
x dxh
=πρ α πρ αtan . tan4
4
0
4 5
02 2 5
x dxx
h h
=
=πρ α πρtan
. .4 5
4
5
2 5 2 5h
hh= R4
Substituting R/h for tan α or substituting the value of ρ obtained above, we have
I =ππ
32 5
3104
5MR
R MR2
4 2
h hh
. . =
(ii) About an axis passing through the vertex and parallel to its base: Consideringthe disc at a distance x from the vertex of the cone, we have its M.I. about its diameter =mass × (radius)2/4
= πx2 tan2 αρ dx. r2/4
= π x2 tan2 αρ dx.x2 tan2 α/4 = (πρ tan4 α/4)x4 dx
∴ Its M.I. about the parallel axis XX', distant x from it.
= (πρ tan4 α/4) x4 dx + π x2 tan2 αρ . x2 dx
= (πρ tan4 α/4) x4 dx + π tan2 αρ.x4 dx
Hence M.I. of the entire cone about the axis XX', parallel to its base is given by
I =πρ α πρ αtan
. tan .4
0
4
0
2 4
04
h h h
x dx x dx +
=πρ α πρ αtan
. tan4
4 2
0
4
04
x dx x dxh h
+
=πρ πρ4 5 54
5
02
5
0
.R R4 2
hx
hx
h h
+
=πρ πρR R4 2
4 5 54
5
2
5
hh
hh. .+
or, substituting the value of ρ, we have M.I. of the cone about the axis XX', i.e.,
I =ππ
ππ
35
354
5
2
5M4 R
R MR
R2
4
2
2
h hh
h hh. . . .+
or I =3
203
5
2MR M2+ h
350 Mechanics
8.14 MOMENT OF INERTIA OF A HOLLOW CYLINDER
(i) About its axis of cylindrical symmetry: A hollow cylinder may be considered to bea thick annular disc or a combination of thin annular discs, each of the same external andinternal radii, placed adjacent to each other, the axis of the cylinder (i.e., its axis of cylindricalsymmetry) being the same as the axis passing through the centre of the thick annular disc(or the combination of thin annular discs) and perpendicular to its plane.
The M.I. of the hollow cylinder about its own axis is, therefore, the same as that of athick annular disc (or a combination of thin annular discs) of the same mass M and externaland internal radii R ad r respectively about the axis passing through its center and perpendicularto its plane, i.e.,
I = M(R2 + r2)/2
(ii) About an axis passing through its centre and perpendicular to its own axis:If R and r be the external and internal radii respectively of the hollow cylinder, l, its lengthand M, its mass, we have mass per unit volume of the cylinder = M/π (R2 – r2) l.
Imagining the hollow cylinder to be made up of a large number of annular discs, ofexternal and internal radii R and r respectively, placed adjacent to each other, and consideringone such disc at a distance x from the axis YOY' passing through the center O of the cylinderand perpendicular to its own axis, we have
Surface area of the disc = p(R2 – r2), its volume (R2 – r2) dx and its mass
= ππ
( ) .)
RM
(RM.2
2−−
=r dxr l
dxl
22
So that, M.I. of the disc about its diameter (AB) = Mdx (R2 + r2)/4l
And, its M.I. about the parallel axis YOY' distant x from it, in accordance with theprinciple of parallel axes,
=M R M2
ldx
rl
dx x+
+
22
4.
Hence, M.I. of entire hollow cylinder about the axis YOY' is equal to twice the integralof the above expression between the limits x = 0 and x = l/2 i.e.,
I = 24
22
0
2M (R M2 + +
r
ldx
lx dx
l) .
/
=2
4
22
0
2M R2
lr dx x dx
l( ) .
/+ +
=2
4 32
4 2 8 3
2 3
0
2 2 3M (R M (R2 2
lr x x
lr l l
l+ +
= + +
) )× ×
/
I = M R2 + +
r l2 2
4 12
x
l2
Y
Y ′
O
dxA
Bl
Fig. 15
Moment of Inertia 351
8.15 MOMENT OF INERTIA OF A SPHERICAL SHELL
(i) About its diameter: Let ACBD be the section through the center O, of a sphericalshell of radius R and mass M, whose moment of inertia is to be determined about a diameterAB, its value being obviously the same about any other diameter.
Surface area of the shell = 4π r2 and, therefore, mass per unit area of the shell =M/4πR2.
Consider a thin slice of the shell, lying between two planesEF and GH, perpendicular to the diameter AB at distances xand x + dx respectively from its center O. This slice is, a ringof radius PE and width EG (and not PQ which is equal to dx,the distance between the two planes).
Area of the ring = circumference × width
= 2 πPE × EG and hence its mass
= 2 πPE × EG × M/4πR2
Join OE and OG and let angle COE be equal to θ and angle EOG = dθ.
Then, PE = OE cos OEP = R cos θ, because OE = R and ∆OEP = alternate ∆COE = θAnd OP = OE sin OEP, x = R sin θ and ∴ dx/dθ = R cos θ, OP = x and OE = R.
dx = R cos θ dθ = PE. dθ because R cos θ = PE and EG = OE dθ = R dθ∴ mass of the ring = 2 π PE × R dθ × (M/4θR2) = Mdx/2R, PE dθ = dx
Hence, M.I. of the ring about diameter AB of the shell (i.e., an axis passing through thecenter of the ring and perpendicular to its plane = mass × (radius)2 = (Mdx/2R) (R2 – x2)because PE2 = OE2 – OP2 = R2 – x2
M.I. of the whole spherical shell about the diameter (AB) is equal to twice the integralof this expression between the limits x = 0 and x = R, i.e.,
I = 2 2
0
M2R
R2R
( )− x dx
=MR
R2R
( )− x dx2
0
=MR
RMR
RR2
R3
3x
x−
= −
3
03 3
M =MR
R MR3 2. 23
23
=
8.16 MOMENT OF INERTIA OF A SOLID SPHERE
(iv) About a diameter: Figure shows a section, through the center, of a solid sphere ofradius R and mass M, whose moment of inertia is to be determined about a diameter AB,its value being the same about any other diameter.
Since the value of sphere = 4πR3/3, its mass per unit volume (or density)
Rθ
C E
D FH
BA P Q
xO
dx
dθG
Fig. 16
352 Mechanics
=M
43
R
M4 R3 3
π π= 3
A thin circular slice of the sphere at a distance x from its center O and of thickness dx,
we have surface area of the slice (which is a disc of radius R2 − =x p2 ) (R2 – x2) and itsvolume = area × thickness = π (R2 – x2) dx and hence its mass = volume × density
= ππ
( ) . )R M4 R
M (RR
23
23− = −x dx x dx2 23 3
4
M.I. of the whole sphere about its diameter (AB) is equal to twice the integral of theabove expression between the limits x = 0 and x = R, i.e.,
I = 23
8
2
0
M (RR
2
3
R− x
dx)
= 2 3 2 2
0
.( )
M8R
R32
R
− x dx
= 34
2 2 4
0
MR
R R34 2
R
( )− + x x dx
or I =3
23 5
3 5
0
4M
4 RR R3
4 2xx x− +
=34
23
15
MR
R R R35 5 5− +
=3 8
1525
M4R
R MR3
52=
(v) About a tangent: A tangent drawn to the sphere at any point will be parallel to oneor the other diameter of it (i.e., an axis passing through its center or center of mass) andat a distance equal to the radius of the sphere, R, from it. By the principle of parallel axes.Moment of inertia of the sphere about a tangent, i.e.,
I = (2/5) MR2 + MR2
= (7/5) MR2
8.17 MOMENT OF INERTIAL OF A HOLLOW SPHERE OR A THICK SHELL
(i) About its diameter: A hollow sphere is just a solid sphere from the inside of whicha small concentric sphere has been removed.
Since the M.I. about a given axis is a scalar quantity. The moment of inertia of thehollow sphere about a diameter= moment of inertia of the solid sphere minus moment ofinertia of the smaller solid sphere removed from it, both about the same diameter.
R
BAxO
dx
R – x2 2
Fig. 17
Moment of Inertia 353
Let R and r be the external and internal radii of the hollow sphere, i.e., the radius ofthe bigger solid sphere and the smaller solid sphere (removed from it) respectively. Then,if r be the density of the material of the given hollow sphere, we have
mass of the bigger sphere =43
π ρR3 and
mass of the smaller sphere =43
π ρr3
so that mass of the hollow sphere, M = 43
3π ρ( ) ,R3 − r and therefore
ρ =3
3M
4 (R3π − r )
Moment of inertia of the bigger and the smaller spheres about a given diameter are
respectively 25
43
43
3 2π ρ π ρR R and 25
3 2
r r
Therefore, M.I. of the hollow sphere about the same diameter is given by
I =25
43
25
43
25
43
3 2 5π ρ π ρ πρR R R3 2 5
−
= −r r r( )
or, substituting the value of r obtained above, we have
I =25
43
3 253
55
3.)
)ππ
M4 (R
(R M RR3
55
3−− = −
−
rr
rr
(ii) About a tangent: A tangent to the sphere at any point being parallel to one diameteror the other (i.e., the axis passing through the center of mass) of the sphere and at a distanceequal to its external radius R from it, we have, by the principle of parallel axes,
M.I. of hollow sphere about a tangent, i.e.,
I =25
5
3M RR
MR5
32−
−
+r
r
8.18 M. I. OF A UNIFORM TRIANGULAR LAMINA
M.I. about base of a triangular lamina: Let the massof the triangular lamina ABC of mass M, the length of ANis equal to P. Divide the lamina by strips parallel to BC.Let PQ be one such strip of breadth δx at a distance x fromA. Let AN be perpendicular to BC such that AN = P. LetBC = a. From similar ∆’s APQ and ABC, we have
PQa
=xP
so that PQ =Pxa
Now mass per unit area =M
12
P
MPa a
= 2
Pdx
M
NB C
A
x
PQ
a
Fig. 18
354 Mechanics
∴ Mass of the strip PQ =2 2 2
2MP
PQ MP P
MPa
xa
ax x x x. . .δ δ δ= =
∴ M.I. of the strip about BC =2
22M
PPx x xδ ( )−
∴ M.I. of the triangle about BC
=2 2
222
02
2 2
0
MP
PM
PP P
P Px
x dx x x x dx( ) ( )− = − +
=2
22
3 42
2 2 3 4
0
MP
P PP
x x x− +
=2
22
3 42
4 4 4MP
P P P− +
= MP2
6
8.19 KINETIC ENERGY OF ROTATION
In pure rotation, in which the center of mass of the rotating body has zero linearvelocity, e.g., the rotation of a body about an axis through its center of mass, and a rotationin rolling of a body along a plane or an inclined surface.
(i) Kinetic energy of a body rotating about anaxis through its center of mass: Consider a body ofmass M, rotating with angular velocity ω about an axisAB, passing through its center of mass, O, so that thecenter of mass has zero linear velocity. It is thus acase of pure rotation.
The body possesses kinetic energy in virtue of itsmotion of rotation which is, therefore, its kinetic energyof rotation. Let us obtain an expression for it.
The body is made up of a large number of particlesof masses m1, m2, m3 .…etc, at respective distances r1,r2, r3 ... etc. from the axis AB through O. Since theirangular velocity is the same (ù), their linear velocitiesare respectively r1ω = v1, r2ω = v2 and r3ω = v3 ... etc.and hence their respective kinetic energies equal to½ m1v1
2 = ½ m1r12ω2, ½ m2v2
2 = ½ m2r22ω2, ½ m3v3
2
= ½ m3r32ω2 ... etc.
∴ Total K.E. of all the particles = 12
12
121 1
2 22 2
2 23 3
2 2m r m r m rω ω ω+ + + ....
= 12
21 1
22 2
23 3
2ω ( ...)m r m r m r+ + +
A
ω
O
B
Fig. 19
Moment of Inertia 355
=12
12
2 2 2ω ωmr∑ = MK2
where Σmr2 = MK2, with, M, as the mass of the body and K, its radius of gyration about theaxis of rotation AB.
Since MK2 = I, the moment of inertia of the body about the axis AB, we have kineticenergy of rotation of the body about the axis (AB) through its center of mass = ½ Iω2.
(ii) Kinetic energy of a rotating body whose center of mass has also a linearvelocity
(a) A body rolling along a plane surface: Let us consider a body, like a circulardisc, a cylinder, a sphere etc. (i.e., a body with a circular symmetry), of mass M,radius R and with its centre of mass at O, rolling, without slipping, along a planeor a level surface, such that it rotate clockwise and moves along the x direction.
P
Q 2v
R
O
X
Fig. 20
At any given instant, the point P, where the body touches the surface, is at rest, so thatan axis through P, perpendicular to the plane of the paper is its instantaneous axis of rotationand the linear velocities of its various particles are perpendicular to the lines joining themwith the point of contact P, their magnitudes being proportional to the lengths of these lines.Thus, if the linear velocity of the center of mass O (where PO = R) be v, that of the particleat Q (where PQ = 2R) is 2v.
This shows that the particles have all the same angular velocity with respect to the pointP or that the body is rotating about the fixed axis through P with an angular velocity ω, say,given by v/R, where v is the linear velocity of the centre of mass.
The motion of the body is thus equivalent to one of pure rotation about the axis throughP, with an angular velocity ω. The whole of the kinetic energy of the body is therefore, thesame as its kinetic energy of rotation about this axis and hence equal to Ip ω2 where Ip isthe M.I. of the body about the axis through P.
If Icm be the moment of inertia of the body about a parallel axis through its center ofmass, then by the principle of parallel axes, IP = Icm + MR2.
So that, K.E. of the rolling body = ½ (Icm + MR2) ω2
= ½ Icmω2 + ½ MR2 ω2 = ½ Icm ω2 + ½ Mv2 ...(1)
where v is the linear speed of its center of mass with respect to P.
Now Icm = MK2, where K is the radius of gyration of the body about the axis throughits center of mass, and ω = v/R. So that
K.E. of the rolling body =12
12
12
12
2 22
MKR
M M KR
22 2
vv v+ = +
...(2)
In equation (1) the first term gives its K.E. of pure rotation about the center of mass,i.e., its K.E. when it is simply rotating with angular velocity ù about the axis through its
356 Mechanics
center of mass, without executing any translational motion (i.e., with the linear velocity ofits center of mass zero). And the second term gives its K.E. of pure translational, i.e., its K.E.when it is simply moving with linear velocity v without performing any rotational motion(i.e., with its angular velocity zero).
8.20 A BODY ROLLING DOWN AN INCLINED PLANE (ITS ACCELERATION ALONG THE PLANE)
Let a body of circular symmetry (e.g., a disc, sphere, cylinder etc.) of mass M roll freelydown a plane, inclined to the horizontal at an angle q and rough enough to prevent slippingand hence any work done by friction.
If v be the linear velocity acquired by the body on covering a distance S along the plane,its vertical distance of descent = S sin θ.
∴ Potential energy lost by the body = Mg.S sin θ ...(1)
This is equal to the kinetic energy gained by the body, i.e., equal to its K.E. of rotationplus its K.E. of translation.
Now, K.E. of rotation of the body = ½ Iω2 where ω is its angular velocity about aperpendicular axis through its center of mass, and its K.E. of translation = ½ Mv2, becauseits center of mass has a linear velocity v.
∴ Total K.E. gained by the body =12
12
12
12
22
2I M MKR
M2 22ω + = +v
vv.
Because I = MK2, where K is the radius of gyration of the body above the axis throughits center of mass, and ω = v/R.
∴ Total K.E. gained by the body = 12
12M KR
2
2v +
...(2)
From conservation of energy Eq. (1) are equal to Eq. (2)
12
12M KR
2
2v +
= MgS sin θ
v2 ( )K RR
2 2
2+
= 2g sin θS
or v2 = 2 RK R
S2
2 2( ).sin .
+g θ
Comparing this with the kinetic relation v2 = 2 a.s for a body starting from rest, we haveacceleration of the body along the plane, i.e.,
a =R
K R KR
2
2 2 2
2( )
.sinsin
+=
+g
gθ θ
1
i.e., the acceleration is proportional to R
K R
2
2 2( )+ for a given angle of inclination (θ) of the
planes.
v
R
ω
θ
s
s s in θ
Fig. 21
Moment of Inertia 357
Let us consider the case when the body is a solid sphere, cylinder, spherical shell andring. M, K and R will be the mass, radius of gyration and the radius of the rolling sectionrespectively in each case.
(i) Solid Sphere: The moment of inertia of a solid sphere about its diameter is givenby
I = MK MR2 2= 25
orKR
2
2 =25
∴ a =g
gsin θ θ
1 25
57+
= sin
(ii) Cylinder (or Disc): The moment of inertia of a cylinder or disc about the axispassing through its center and perpendicular to its plane (which is the axis ofrotation) is given by
I = MK MR2 2= 12
orKR
2
2 =12
∴ a =g
gsinθ θ
1 12
23+
= sin
(iii) Spherical Shell: Its moment of inertia about the diameter (axis of rotation) isgiven by
I = MK2 = 23
MR2
orKR
2
2 =23
∴ a =g
gsinθ θ
1 23
35+
= sin
(iv) Ring: It moment of inertia about the axis passing through its center andperpendicular to its plane (axis of rotation) is given by
I = MK2 = MR2
orKR
2
2 = 1
∴ a =g gsin sinθ θ1 1
12+
=
358 Mechanics
We see that the accelerators for a solid sphere, disc, shell and ring are in a decreasingorder. Therefore, if all of them start rolling at the same instant, the sphere will reach downthe plane first, then the disc, then the shell, and then the ring.
8.21 IDENTIFICATION OF HOLLOW AND SOLID SPHERE
The given spheres are taken at the top of an inclined plane and allowed to start rollingat the same instant. The one which reaches the bottom of the plane earlier is the solidsphere. The other which reaches afterwards is ‘hollow’.
Reason: The acceleration with which the sphere rolls is given by
a =g gsin sinθ θ
1 1+=
+KR
IMR
2
2
2
2
∴ I = MK2
Now, the moment of inertia of a hollow sphere about the axis of rotation (diameter) isgreater than that of a solid sphere of the same mass and radius, because in case of hollowsphere the material particle are, on the average, at a greater distance from the axis. Thismean that the hollow sphere will roll down the plane with a smaller acceleration and hencereach the bottom after the solid sphere does so.
8.22 COMPOUND PENDULUM
Any rigid body which is so mounted that it can swing in a vertical plane about ahorizontal axis passing through it, is called a ‘physical’ or ‘compound’ pendulum.
Figure represents the vertical section of an irregular rigid body pivoted at a point S init. In the equilibrium position, the center of mass C of the body lies vertically below S. Whenthe body is displaced to one side and then released, it oscillates in the vertical plane containingS and C about a horizontal axis perpendicular to this plane and passing through S. It is nowacting as a compound pendulum.
Motion of Compound Pendulum: Let m be the mass of the body and l the distanceof C from S. Suppose at some instant during the oscillation, the body makes an angle θ withthe vertical. At this instant the moment of its weight = mg (l sin θ). This is the restoringtorque τ at this instant tending to bring the body to its equilibrium position. Thus
τ = –mgl sin θ–ve sign is used since the torque acts in the direction of θ decreasing. If the angle θ is small,then sin θ = θ (radian). Therefore,
τ = –mglθ ...(i)
If I be the moment of inertia of the body about the axis through S and ddt
2
2θ
the
instantaneous angular acceleration, then the torque acting on the body at this instant mustbe given by n
τ = Iddt
2
2θ
...(ii)
From Eqs. (i) and (ii), we get
Iddt
2
2θ
= –mgl θ
Moment of Inertia 359
orddt
2
2θ
= − = −mglI
θ ω θ2
where ω2 =mgl
I
∴ddt
2
2θ
∝ θ
that is, the angular acceleration is proportional to theangular displacement. The motion is, therefore, simpleharmonic and its time-period is given by
T =2
2π
ωπ= I
mgl
Let K be the radius of gyration of the pendulum about a plane of oscillation. Then itsmoment of inertia about this axis will be mK2. Therefore, the moment of inertia I about theaxis through S will be
I = mK2 + ml2 (Theorem of parallel axes)
T = 2 22 2
2
π πm mlmgl
ll
gK
K+ =
+
This is the expression for the time-period. Comparing it with the period of a simplependulum.
T = 2π Lg
We see that the period of a compound pendulum is the same as that of a simple
pendulum of length L = K2
ll+
. Thus
K2
ll+
is the length of the equivalent simple
pendulum.
Let us suppose that the whole mass of the compound pendulum is concentrated at a
point O on SC produced such that SO = K2
ll+
. Thus it would be like a simple pendulum
of the same period. The point O which is at a distance equal to the length of the equivalentsimple pendulum from S is called the ‘centre of oscillation’ corresponding to the center ofsuspension S.
Interchangeability of the centres of Suspension and Oscillation: When the body is suspendedfrom the centre of suspension S, its period of oscillation is given by
T = 2
2
π
Kl
l
g
+
where l is the distance of S from C.
l
k2
l+ lS
θ
Ol s in θ
m g
Fig. 22
360 Mechanics
Let the body be now inverted and suspended from the center of oscillation O. Thedistance of O from C is K2/l. The new period of oscillation T′ is therefore obtained bysubstituting K2/l for l in the about expression i.e.,
T′ = 2 2
2
2
2
π π
KK
K K2
/ l lg
ll
g
+=
+
Hence T = T'
Thus the periods of oscillation about S and O are equal i.e., the center of suspension andoscillation are interchangeable.
8.23 FLY WHEELS
It is a heavy wheel whose M.I. about its axis of rotation is large. Most of the mass isconcentrated on rim (to increase its M.I.). These have wide application in stationary enginesand various instruments of every day use.
Determination of M.I. of Fly wheel: Acord is wound round the horizontal axle of flywheel and a mass is attached to the free endof the cord. Releasing the mass, the fall of themass is noted and number of turns in the notedtime are counted when fly wheel comes to rest.
Let N is the number of revolutions of fly wheel till mass detaches from wheel, I ismoment of inertia of fly wheel, v is velocity of the mass acquired, m is mass attached, h isdistance descended, t is time taken to come to rest after mass is detached. n is the totalrevolutions till wheel comes to rest, after mass is detached. W is the work done/revolutionsagainst friction.
From Energy conservation gives:
mgh = ½ mv2 + ½ Iω2 + nW
where nW is the total work done against friction in n revolutions.
or 2 mgh = mv2 + Iω2 + 2 nW ...(1)
Now work done in n revolutions = nW
K.E. of rotation of wheel = ½ Iω2
i.e., nW = ½ Iω2
or W =12
I 2ωn
Substituting this value of W in Eq. (1) we get
2 mgh = mv2 + I ω2 + 2N. 12
I 2ωn
or 2mgh – mv2 = Iω2 (1+ N/n)
or I =2
1
2
1
2
2
2 2
2
mgh mv
n
mgh mr
n
−
+
= −
+
ω
ω
ωN N
MM g
Fig. 23
Moment of Inertia 361
or I =
2
1
22mgh mr
n
ω−
+
N...(2)
Average angular velocity of wheel = (ω + 0)/2 = ω/2
If n rotation of wheel, angle described = 2ω n/t
∴ Average angular velocity = 2π n/t = ω/2
or ω =4πn
tSubstituting this value of ù in Eq. (2) we get
I =
216
1 1 8
2
2 22
2
2 22
mghtn
mr
n
m
n
ghtn
rππ
−
+
=+
−
N N
or I = mn
nght
nr
+
−
N
2
2 22
8πHence, the value of I, the moment of inertia of the fly wheel about its axis of rotation,
may be obtained.
NUMERICALS
Q.1. The torque τ acting on a rotating body of moment of inertia I about the axis ofrotation is given by
τI
= a bt ct ,2+ +
where a, b, c are constants. Express angular displacement of the body (starting from rest att = 0) as a function of τ.
Solution. The torque τ and the angular acceleration a are related by
τ = Iα = I2d
dtθ2
Therefore, from the given expression, we have
ddt
2θ2 = a + bt + ct2
Integrating:ddt
θ= at b
tc
t+ + +2 3
2 3A
At t = 0, ddtθ = 0 so that constant A = 0.
Again integrating: θ = at
bt
ct2 3 4
2 6 12+ + + ′A
Again at τ = 0, θ = 0 so that A' = 0.
362 Mechanics
∴ θ =t
abt ct2 2
2 3 6+ +
Ans.
Q.2. A flywheel of mass 10 kg and radius of gyration 50 cm is being acted on by a torqueof 108 dyne-cm. Calculate the angular acceleration produced.
Solution. In usual rotation :
τ = Ia
or α =τI
Here τ = 108 dyne-cm, I = MK2 = (10,000 gm) (50 cm)2 = 2.5 × 107 gm-cm2.
∴ α =10
2 5 104
8
7. ×= rad/sec2 Ans.
Q.3. An automobile engine develops 100 kilowatt when rotating at a speed of 1800 rev/min. What torque does it deliver ?
Solution. The power developed by the torque τ exerted on a rotating body is given by
P = τ ω
∴ τ =Pω
Hee P = 100 kilowatt = 100,000 watt = 100,000 Joule/sec and ω = 1800
602× π = 60π
rad/sec.
∴ τ =100 000 531, Joule/sec60 × 3.14 rad/sec
Joule= Ans.
Q.4. Show that the rotational inertial of a dumb-bell is twice as great about an axisthrough one end as it is about an axis through the centre. Compute these values if the dumb-bell is a light rad of length one meter carrying 5 kg sphere, one at each end.
Solution. Let M be the mass of each sphere, and l the length of the rod. Let us treatspheres as point masses. The moment of inertia of the system about the axis through l is
A C
l2
l2
BM M
Fig. 24
IC = M Ml l2 2
2 2
+
=12
2Ml
Here M = 5 kg and l = 1 meter
Moment of Inertia 363
∴ IC = 2.5 kg-meter2
By the theorem of parallel axes, the moment of inertial of the system about an axisthrough A is
IA = IC + (2M) l l l l2
12
12
22 2
= + =M M M 2
= 5 × (1)2 = 5.0 kg-meter2.
Here we see that IA is twice IC.
Q.5. Deduce the moment of inertial of HCl molecule about an axis passing through itsC.M. and perpendicular to the bond. Given : internuclear distance 1.3 Å, proton mass = 1.7× 10–27 kg, atomic weight of chlorine = 35.
Solution. The mass of H atom (proton) is 1.7 × 10–27 kg and that of Cl atom is 35 ×(1.7 × 10–27) = 59.5 × 10–27 reduced mass of HCl molecule is therefore
µ =m m
m mH Cl
H Cl+=
+
− −
− −( . × ) ( . × )
( . × ) ( . × )1 7 10 59 5 10
1 7 10 59 5 10
27 27
27 27
= 1.65 × 10–27 kg.
The internuclear distance is r = 1.3 Å = 1.3 × 10–10 meter.
The moment of inertial about the centre of mass is therefore given by
I = µr2
= (1.65 × 10–27 kg) (1.3 × 10–10 m)2
= 2.79 × 10–47 kg/m2.
Q.6. A uniform thin bar of mass 3 kg and length0.9 is bent to make an equilateral triangle. Calculatethe moment of inertia I about an axis passing throughthe centre of mass and perpendicular to the plane ofthe triangle.
Solution. Each side of the triangle, such as PQhas a mass m = 1 kg and a length l = 0.3 m. Itsmoment of inertia about a perpendicular axis throughits middle point O is
ml2
12=
1 0 312
0 00752×( . )
.= Kg-m2
The distance OC is d = PO tan 30° = l2
tan 30° = 0 32
0 577 0 0865. × . .= m.
The moment of inertial about C, by the theorem of parallel axes, is
mlmd
22
12+ = 0 0075 1 0 0865 0 0152. ( ) ( . ) .+ = kg-m2
By symmetry, the moment of inertial of each side about C will be same. Hence for thewhole triangle
I = 3 × 0.015 = 0.045 kg-m2. Ans.
30°
P
RQ
O
C
Fig. 25
364 Mechanics
Q.7. The inter-nuclear distance between the two hydrogen atoms in a hydrogen moleculeis 0.71 Å. Calculate its moment of inertia and the first two rotational energy levels. Mass ofthe proton is 1.6 × 10–24 gm and Planck’s constant is 6.6 × 10–27 erg-sec. Given that angular
momentum J→
is quantised by the rule | | ( )J j jh→
= +22
214π
with j = 0, 1, 2, ... .
Solution. The reduced mass of the H2 molecule is given by µ = m m
m mmH H
H HH+
= 12
= 12
1 6 10 0 8 1024 24× ( . × ) . ×− −=gm gm.
The internuclear distance is r = 0.71 Å = 0.71 × 10–8 cm. Hence the moment of inertiaof the molecule about an axis passing through the centre of mass and perpendicular to theline joining the two atoms is
I = µr2
= (0.8 × 10–24 gm) (0.71 × 10–8 cm)2
The rotational energy of the system is E =
12
I 2ω , and angular momentum is J (=Iω).
∴ E =JI
2
2
Here J2 = j jh
( )+ H2
24π
∴ E = j jh
( )+ 18
2
2π IFor the first two levels j = 1, 2.
∴ E1 =28
2 6 6 108 3 14 4 0 10
2 8 102
2
27 2
2 4114h
π Ierg.= =
−
−−× ( . × )
× ( . ) . ×. ×
But 1.6 × 10–12 erg = 1 eV
∴ E1 =2 8 101 6 10
1 7 1024
122. ×
. ×. ×
−
−−= eV
E2 =68
3 5 1 102
22h
π IE eV.1= = −. ×
Q.8. Four particles each of mass m are symmetricallyplaced on the sim of a uniform disc of mass M and radiusR. What is the moment of inertia of the system about anaxis passing through one of the particles and perpendicularto the plane of the disc ?
Solution. The arrangement is shown in figure 26.The moment of inertial of the disc about an axis through
its centre O and perpendicular to its plane is 12
MR2. Hence,
by the theorem of parallel axes, its M.I. about an axisthrough B and perpendicular to its plane
mA Cm
m
B
OR
m
R2R
Fig. 26
Moment of Inertia 365
=12
32
MR MR MR2 2 2+ =
Now, the M.I. of the particle at B about the same axis is zero, while that of the particle
at A, C and D is m m m[ , [ ] .2 22 2R] and (2R)2
Hence the M.I. of the entire system
=32
2 2 2MR R] R] R)2 2 2 2+ + +m m m[ [ (
=32
M + 8 R2m
Q.9. A circular hole of diameter R is cut from a disc of radius R, such that thecircumference of the hole passes through the centre of the disc. Find the moment of inertiaof the remaining disc about the straight line joining the centres of the disc and the hole interms of the remaining mass.
Solution. Let O be the centre of the disc and O'that of the hole. Let M be the mass of the remainingdisc.
Area of the original disc = πR2
Area of the removed material disc = π πR2
R2
=2 1
4
Area of the removing disc = π π πR R R2 2 2− =14
34
∴ Mass per unit area, σ =M
34
R2π
Mass of the original disc = π σR M2 = 43
Mass of the removed material =14
13
π σR M2 =
Moment of inertial of the original disc about OO' (diameter)
=14
14
43
13
(mass) R M R MR2 2 2=
=
Moment of inertia of the removed material about OO'
=14
14
13
148
2 2
(mass) R2
M R2
MR2
=
=
∴ Moment of inertia of the remaining disc about OO'
=13
148
1548
MR MR MR2 2 2− = .
O
R
O ′P
Fig. 27
366 Mechanics
Q.10. A circular hole of diameter R is cut from a uniform disc of radius R and mass Msuch that the perimeter of the hole passes through the centre of the disc. Find the momentof inertia of the remaining disc about (i) an axis passing through the centre of the disc andperpendicular to the plane of the disc (ii) a tangent in the plane of the disc and touching thehole.
Solution. (i) Refer to the above figure. M is now the mass of the original disc. The massper unit area is
σ =MR2π
Mass of the removed material =14
14
π σR M2 =
Moment of the inertia of the original disc about an axis through O and perpendicular tothe plane.
=12
MR2
Moment of inertia of the removed material about an axis through O' and perpendicularto the plane
=12
14
132
2
M R2
MR2
=
By the theorem of parallel axes, the moment of inertia of the removed material aboutthe parallel axis through O
=132
14
332
2
MR M × R2
MR2 2+
= .
∴ moment of inertia of the remaining disc about an axis through O and perpendicularto the plane
=12
332
1332
MR MR MR2 2 2− = .
(ii) Moment of inertia of the original disc about a tangent in the plane of the disc andpassing through P
=54
MR2,
and that of the removed material.
=54
14
564
2
=R2
MR2
∴ moment of inertia of the remaining disc
=54
564
7564
MR MR MR2 2 2− =
Q.11. Find the M.I. of a rod 4 cm in diameter, 2m long and of mass of 8 kg (i) about anaxis normal to its lengths and passing through its centre (ii) about an axis passing throughone end and normal to its length (iii) about a longitudinal axis through the centre of the rod.
Moment of Inertia 367
Solution. (i) The M.I. about an axis normal to the rod and passing through its centreis given by
I = MR
kg(2m)2 2l2 2
12 48
120 02
4+
= +
( . ) = 2.67 kg-m2
(ii) By the theorem of parallel axes, the MI about an axis normal to the rod and passingthrough an end is given by
I = MR2 2l l
12 4 2
2
+
+
M
= 2.67 kg-m2 + (8 kg) 2 10 67
2m2
kg-m2
= . .
(iii) The M.I. about the central axis of the rod is given by
I =12
12
8 0 0016MR kg) × (0.02 m) kg-m2 2 2= =× ( . .
Q.12. Four spheres each of radius ‘r’ and mass ‘m’ are placed with their centres on fourcorners of a square of side ‘l’. Calculate the moment of inertia of the arrangement about any(i) diagonal of the square (ii) any side of the square.
Solution. (i) The moment of inertia of each of the spheres A and C about the diagonal
AC of the square (an axis passing through their centres) is 25
2mr .
Similarly, the moment of inertia of each of the spheres B and D about an axis through
its centre and parallel to AC is 25
2mr . The distance between this axis and AC is l/ ,2 where
l is the side of the square. Therefore, the moment of inertial of each of the spheres B and
D about AC is, by the theorem of parallel axes, 25 2
22
mrml+ .
D
A B
C
O
l/ 2
Fig. 28
Hence the moment of inertia of the all the four spheres about AC.
= 225
25 2
2 22
mr mrml
+ +
368 Mechanics
=15
4 4 52 2 2m r r l( )+ +
=m
r l5
8 52 2( )+
(ii) The moment of inertia of all the spheres about a side of the square would be
mr l
58 102 2( )+ =
25
4 52 2mr l( ).+
Q.13. The flat surface of a hemisphere of radius R is cemented to one flat surface of acylinder of radius R and length l and made of the same material. If the total mass be M, showthat the moment of inertial of the combination about the axis of the cylinder is
MR l2
4R15
l 2R3
.2 +
+
Solution. Let m1 and m2 be masses of the cylinderand the hemisphere respectively and ρ the density oftheir material. The moment of inertia of the cylinder
about its own axis is 12 1m R2, and that of the hemisphere
about its own diameter is 25 2m R2.
(The moment of inertia of a hemisphere about adiameter is same as that of a sphere of same mass andsame radius). Therefore, the moment of inertial of thecombination about this common axis is
I =12
251 2m mR R2 2+
=12
25
23
( )π ρ π ρR R R R2 2 2 2l +
= ( )πR R R2 2 3l l12
415
+
...(i)
Now M = m1 + m2 = π ρ π ρ π ρR23
R RR3
2 3 2l l+ = +
2
so that π ρR2 = M R3
l +
2
Substituting this value of π ρR2 is eq. (i) we get
I =MR
R3
4R15
2
l
l
+
+
2 2
l
R R
Fig. 29
Moment of Inertia 369
Q.14. What will be the radius of gyration for a solid sphere about a diameter whoseradius is half meter ?
Solution. Let K be the radius of gyration to be determined. The moment of inertia ofthe solid sphere about a diameter is given by
I = MK2 = 25
MR2
∴ K =25
0 5 0 316
=R =25
m m.× . .
Q.15. A circular metal disc of mass 4 kg and diameter 0.4 meter makes 10 revolution persecond about its centre, the axis of rotation being normal to the plane of the disc (i) what isthe moment of inertial about this axis ? (ii) What is the angular momentum about the sameaxis (iii) calculate the torque which will increase the angular momentum by 20% is 10 seconds.
Solution. (i) The M.I. about the given axis is
I =12
12
4 0 08MR kg × (0.2 m) kg - m2 2 2= =× .
(ii) The angular momentum about the same axis is
L = Iω = (0.08 kg-m2) (2 × π × 10 rad/sec)
= 5.02 kg-m2/sec.
(iii) The torque is rate of change of angular momentum.
τ =ddtL
In 10 seconds, L increases by 20% i.e.,
dL = 5.0 × 20
1001 0= .
∴ τ =1 010
0 10. .= nt - m.
Q.16. A solid cylinder of mass 16 kg and radius 0.4 m is rotating about its axis. If itsrotational speed increases in 10 second from 500 to 2500 revolution per minute, Calculate (a)its angular acceleration assuming to be constant (b) the torque which must be applied to thecylinder.
Solution. (a) The moment of inertia of the cylinder about its axis is
I =12
12
16 1 28MR kg × (0.4) kg - m2 2 2= =× .
The initial and final rotational speeds are
ω1 = 2π × no. of rev. per sec
= 2 50060
503
π π× = rad/sec.
and ω2 = 2 250060
2503
π π× = rad/sec.
370 Mechanics
This increase takes place in 10 seconds. Therefore, the angular acceleration is
α = ω ωπ π
π2 1
2503
503
10200
3021− =
−= =
∆trad/sec2.
(b) The torque to be applied isτ = Iα = (1.28 kg-m2) (21 rad/sec2) = 27 nt-m.
Q.17. A flywheel in the form of a solid cylinder of mass 5000 kg and diameter 2 metersis rotating, making 120 revolutions per minute. Compute the kinetic energy and the angularimpulse, if the flywheel is brought to rest is 2 seconds.
Solution. The M.I. of the cylinder about its axis is
I =12
12
5000 2500MR kg) (1m) kg-m2 2 2= =( .
If n is number of revolution per second, then the angular velocity is
ω = 2 2 3 14 12060
12 56πn = =× . × . rad/sec.
Therefore, the rotational kinetic energy is
K =12
12
2500 12 56 1 97 102 5I Joule.2ω = =× × ( . ) . ×
The angular impulse is the time integral of torque, that is
τdtt
0 = I I I (
1
2ddt
dt dω ω ω ωω
ω
= = −2
1)
provided I is constant. Thus the angular impulse for the fly-wheel coming to rest.
= moment of inertial × change in angular velocity
= 2500 × 12.56 = 3.14 × 104 kg-m2/sec2.
Q.18. A thin plane disc of mass M and radius R, initially at rest, is set into rotation withangular velocity ω. Calculate the work done if the axis of rotation is (i) through its centre andperpendicular to its plane, (ii) through a point on its edge and perpendicular to its plane (iii)a diameter, (iv) a tangent parallel to a diameter.
Solution. The work done appears as the kinetic energy of rotation, 12
I 2ω , is each case.
(i) W =12
I 2ω
=12
12
14
MR MR2 2 2 2
=ω ω
(ii) W = Itω2
=12
12
12
34
2 2 2( )I + MR MR MR MR2 2 2 2ω ω ω= +
=
Moment of Inertia 371
(iii) W =12
2Idω
=12
14
18
2 2MR MR2 2
=ω ω
(iv) W =12
2Itω
= 12
12
14
58
2 2 2( ) .I MR MR MR MR2 2 2 2d + = +
=ω ω ω
Q.19. A circular disc of radius 0.1 m and mass 1.0 kg is rotating at the rate of 10revolution a second about its axis. Find the work that must be done to increase the rate ofrevolution to 20 per second.
Solution. The M.I. of the disc about its axis is
I =12
12
1 0 5 10 3MR kg) (0.1) kg - m2 2 2= = −( . × .
Initial and final angular velocity are
ω1 = 2π × number of revolution per sec
= 2π × 10 rad/sec.
and ω2 = 2π × 20 rad/sec.
The work done is in increasing the rate of revolutions
= final kinetic energy – initial kinetic energy
=12
12
I I22
12ω ω−
=12
5 10 40 203 2 2 2× ( × ) ( ) ( ) / sec− −kg-m2 π π
=12
5 10 1200 29 63 2× × × × .− =π joule.
Q.20. Calculate the angular momentum of a body whose rotational kinetic energy is 10joule, if the angular momentum vector coincides with the axis of rotation and its moment ofinertia about this axis is 8 gm-cm2.
Solution. The angular momentum and rotational kinetic energy are given by
L = Iω and K = 12
I 2ω
∴ L = 2KIHere K = 10 joule and I = 8 gm-cm2 = 8 × 10–7 kg-m2.
∴ L = 2 10 8 10 4 107 3× × × × .− −= kg-m /sec2
Q.21. Calculate the angular momentum and rotational kinetic energy of earth about itsown axis. How long could this amount of energy supply 1 kilowatt power to each of the 3.5× 109 persons on earth ? (mass of earth = 6.0 × 1024 kg, radius = 6.4 × 103 km).
372 Mechanics
Solution. Let us assume the earth to be a solid sphere. Its moment of inertia of aboutits axis is
I =25
25
6 0 1024MR kg (6.4 × 10 meter)2 6 2= × . ×
= 9 8 1037. × kg-meter2
The earth makes one revolution (traces an angle of 2π radian) is one day (24 × 60 × 60sec.) Therefore its angular velocity
ω =2
24 60 607 27 10 5π
× ×. ×= − rad/sec.
Hence its angular momentum
Iω = ( . × ) × ( . × ) . × /sec.9 8 10 7 27 10 7 1 1037 5 33− = kg-m2
Its rotational energy 12
I 2ω =12
9 8 10 7 27 10 2 6 1037 5 2 29( . × ) ( . × ) . ×− = Joule
The power supplied by this energy
P =energytime
Joule/sec (or watt)= 2 6 1029. ×t
=2 6 10
1000
29. ×t
kilowatt.
We require 1 kilowatt power to each of the 3.5 × 109 persons, that is 3.5 × 109 kilowatt.
∴ 2 6 101000
29. ×t
= 3.5 × 109
or t =2 6 10
1000 3 5 10
29
9. ×× . ×
sec
=2 6 10
1000 3 5 10 365 24 60 60
29
9. ×
× . × × ( × × × )
= 2 35 109. × years.
Q.22. A thin rod of length l and mass M is suspended freely at its end. It is pulled asideand swung about a horizontal axis passing through its lowest position with an angular speedω. How high does its centre of mass rise above its lowest position ? Neglect friction.
Solution. Let M be the mass of the rod. Suppose its centre of mass rises to a heighth above the lowest position. The potential energy is then Mgh. As the rod passes its mean
(lowest) position, the energy becomes entirely kinetic (rotational), 12
I 2ω . If air resistance is
neglected, then
Mgh =12
I 2ω
Moment of Inertia 373
where I is the moment of inertia of the rod about the horizontal axis through its end. Weknow that
I =M 2l
3
Thus Mgh =12 3
22Ml ω
or h =( )Iω 2
6g
Q.23. A rod of mass 6.40 kg and length 1.20 meter carries a sphere of mass 1.06 kg atits each end. It is rotated with an initial angular velocity of 39.0 rev/sec in a horizontal planeabout a vertical axle through its centre and fixed in ball-bearings. It is found that the systemcomes to rest in 32.0 sec on account of friction in bearings,calculate (i) the angular acceleration (ii) frictional torque(iii) work done by frictional torque and (iv) the numberof revolutions made before coming to rest.
Solution. The arrangement is shown in figure.
The moment of inertia of the rod is Ml2
12, so that the
moment of inertia of the whole system about thevertical axis through the centre C is
I = 1 06 0 60 1 06 0 60 6 40 1 2012
2 22
. × ( . ) . × ( . ) . × ( . )+ +
= 1.53 kg-meter2.
(i) The initial angular velocity is
ω = 39.0 rev/sec. = 39.0 × 2π = 245 rad/sec.
The final angular velocity is zero. Therefore the angular acceleration is
α =∆∆ωt
= − = −0 25432 0
7 66.
. (rad/sec retarding)2
(ii) The frictional torque is τ = Iα = 1.53 × (–7.66) = –11.7 newton-meter
(iii) Work done by the frictional torque
= initial kinetic energy of the system
=12
12
1 53 245 4 59 102 4I joule.2ω = =× . × ( ) . ×
(iv) Let n be the number of revolution. Then
Work = torque × angle traced before, coming to rest
i.e., 4.59 × 104 = 11.7 × 2πn
∴ n =4 59 10
11 7 2 3 14624
4. ×. × × .
= rev.
1.2 m eter
(1 .06 kg)
(1 .06 kg)
C
Fig. 30
374 Mechanics
Q.24. A meter stick is held vertically with one end on the floor and is then allowed tofall. Find the velocity of the other end when it hits the floor. Assuming that the end on thefloor does not slip.
Solution. Let M be the mass and l the length of the stick (l = 1 meter). When it is held
vertically, its centre of mass is at a height 12
l from the floor. So that the potential energy
in the stick is (Mgl/2). On releasing, the stick falls i.e., it rotates about the end on the floor
and the potential energy is converted into potential kinetic energy 12
I 2ω , where I is the
moment of inertia of the rod about the lower end and ω the angular velocity when it hits thefloor. Thus
Mg l2
=12
I 2ω
But I = Ml2
3
∴ Mg l2
=12 3
22Ml ω
or ω =3gl
If v be the linear velocity of the end hitting the floor, then
v = lω = 3gl
Now g = 9.8 meter/sec2 and l = 1 meter
∴ v = 3 9 8 1 5 42× . × .= meter/sec.
Q.25. A uniform disc of radius R = 0.5 m and mass M = 20 kg can rotate without frictionaround a fixed horizontal shaft through its centre. A light cord is wound around the rim ofthe disc and a steady downward pull T = 9.8 nt is exerted on the cord. Find the angularacceleration of the disc, and the tangential acceleration of a point on the rim. Also find theangular velocity after 2 second.
Solution. The steady downward pull T exerts a torque τ about the central axis ofrotation, which is given
τ = TR ...(i)
But we know that
τ = Iα ...(ii)
where I is that moment of inertia of the disc and α its angularacceleration about the axis of rotation.
From eq. (i) and (ii), we get
TR = Iα
or α =TRI
R
T
Fig. 31
Moment of Inertia 375
Now, for the disc I =12
2MR , so that
α =2TMR
Given T = 9.8 nt, M = 20kg and R = 0.5 m.
∴ α =2 9 820 0 5
1 96× .× .
.= rad/sec2
The tangential acceleration of a point on the rim is given by a = Rα =0.5 × 1.96 = 0.98m/sec2. If the disc started from rest, then its angular velocity after 2 sec is given by
ω = αt = (1.96 rad/sec2) (2 sec) = 3.92 rad/sec.
Q.26. Suppose we hang a body of mass m = 1.0 kg from the cord in the previous problem.Find the angular acceleration of the disc (M = 20 kg, R = 0.5m) and the tangential accelerationof a point on the rim of the disc. Also compute the tension in the cord.
Solution. Let T be the tension in the cord. The forces onthe downward suspended body m are its weight mg actingdownward and the tension T in the cord acting upward. Thenet downward force on it is mg-T.
If this force produces a linear acceleration a in thesuspended body (which is same as tangential acceleration of apoint on the rim of the disc), then from Newton's second law,we have
mg – T = ma
or T = m (g – a)
Let us now consider the disc. The tension T in the cordexerts a torque τ on the disc which is given by
τ = TR
But we know that τ = Iαwhere I is the moment of inertia of the disc and α its angular acceleration about the axisof rotation. From the last two equations, we get
ΤR = IαBut T = m(g – a) = m(g – Rα), because we know that a = Rα. Thus,
m(g – Rα) R = Iαor α(I + mR2) = mgR
or α =mg
mR
I + R2
But for the disc I =12
MR2, so that
α =mg
m
mg
m
mm
R
MR R MR + R M + 2gR21
212
22+
= =
R
TT
m g
am
Fig. 32
376 Mechanics
Given m = 1.0 kg, M = 20 kg and R = 0.5 meter
∴ α =2 1 0
20 2 1 09 80 5
1 78× .× .
.
..
+
= rad/sec2
The tangential acceleration is a = Rα = 0.5 × 1.78 = 0.89 meter/sec2
We note that the acceleration are less for a suspended 1.0 kg body (equivalent to 9.8 ntweight) then they were for a steady 9.8 nt pull on the card (as in previous example). Thisis because the tension in the cord exerting the torque on the disc in now smaller (less than9.8 nt). Infect, the tension in the string must be less than the weight of the suspended bodyif the body is to accelerate downward.
Let us compute the tension
T = m(g – a) = 1.0 (9.8 – 0.89) = 8.91 nt,
which is less than 9.8 nt. it must be.
Q.27. A uniform sphere, a spherical shell and a cylinder are released from rest at the topof an inclined plane. If the spherical shell reaches the bottom with speed 42 cm/sec. Calculatethe speeds with which the sphere and the cylinder reach the bottom.
Solution. A body of radius R and radius of gyration K rolls down a θ-inclined plane withan acceleration a given by
α =gsin θ
1 2+ KR
2
If the body is released from rest, and the length of the plane is S, the velocity of reachingthe in given by
v2 = 22
1as
gs=+
sin θKR
2
2
For the spherical shall KR
2
2 = 23
and v = 42 cm/sec. Thus.
(42)2 =2
1 23
65
gs sin θ
+= gs sin θ ....(i)
For the sphereKR
2
2 =25
. Thus
v2 =2
1 25
107
gsgs
sin sinθ θ+
= ...(ii)
Dividing (ii) by (i), we get
v2
242( )=
107
56
×
Moment of Inertia 377
or v2 =107
56
42 42 2100× × × =
or v2 = 2100 45 8= . cm/sec
For the cylinder KR
2
2 = 12
. Thus
v2 =2
1 12
43
gsgs
sin sinθ θ+
= ...(iii)
Dividing (iii) by (i), we get
v2
242( )=
43
56
×
or v2 =43
56
42 42 1960× × × =
or v = 44.3 cm/sec.
Q.28. A disc of 10 cm diameter pivoted at its rim and made to swing like a compoundpendulum. Find its time period.
Solution. Let m be the mass and r the radius of the disc. When it swings like acompound pendulum, the time-period is
T = 2π Imgl
where I is the moment of inertial of the disc about the axis of suspension of l is the distanceof the point of suspension (pivot) from its centre of mass. Here l = r. Now, the moment of
inertial of a disc about an axis perpendicular to its plane through its centre is 12
2mr , and,
thus by the theorem of parallel axes, its moment of inertia about a parallel axis through thepivot is
I =12
32
2 2 2mr mr mr+ =
∴ T = 2
32 2
32
2
π πmr
mgr
r
g=
The period is the same as that of a simple pendulum of length l = 32
r .
Now r = 10 cm and g = 980 cm/sec2.
∴ T = 2 3 14
32
10
9800 78× . ×
×.= sec.
378 Mechanics
Q.29. A wire, bent into a circular ring of radius 40 cm hangs over a horizontal knife-edge.Assuming that there is no slipping on the knife-edge, find the periodic-time of oscillation andthe frequency of the ring.
Solution. Let m be the mass and r the radius of the ring. The ring is swinging like acompound pendulum with period
T = 2π Imgl
where I is the moment of inertia of the ring about the axis of suspension and l is the distanceof the point of suspension (knife-edge) from the centre. Hence l = r.
Now the moment of inertia of a ring about an axis perpendicular to its plane throughits centre is mr2, and thus, by the theorem of parallel axes, its moment of inertia about aparallel axis through the knife edge is
I = mr2 + mr2 = 2mr2
∴ T = 22
222
π πmrmgr
rg
=
Now r = 40 cm and g = 980 cm/sec2
∴ T = 2 3 142 40980
1 8× . ××
.= sec
and Frequency, n =1 1
1 80 55
Tsec 1= = −
.. .
Q.30. Show that a cylinder will slip on an inclined plane of inclination angle θ if the
coefficient of static friction between plane and cylinder is less than 13
tan θ.
Solution. Let C be the centre of mass of the cylinder. The translational motion of, bodycan be obtained by assuming that all the external forces act at the centre of mass. Now, theexternal forces acting on C are the weight mg of the cylinder acting vertically downward, andthe normal force N and the tangential (static frictional) force f exerted by the plane. Let abe the (linear) acceleration of the centre of mass. Resolving forces parallel and perpendicularto the plane, we have
mg sin θ – f = ma ...(i)
and N – mg cos θ = 0 ...(ii)
N
mg sin
1
θ
a
T 1
T 1 R
am 2
m g2
T 2T 2
m g cos θ1
m g1θ
Fig. 33
Moment of Inertia 379
N
mg sin θ
CR
m g cos θm gθ
θ
f
Fig. 34
As the cylinder rolls down, it is acted on by a torque τ exerted by the frictional force f,such that
τ = f R
where R is the radius of the cylinder. But τ is equal to be product of the moment of inertiaI of the cylinder about the axis of rotation and the angular acceleration α produced in it. Thatis
τ = IαThe last two expression give
fR = Iα = IRa
α = aR
or f =IR2a
...(iii)
Substituting then value of f in Eq. (i), we get
mg sin θ =IR2a ma=
or a =mg
msinI / R)
θ+ (
Now I = 12
mR2. Therefore
a = mg
m msin sinθ θ+
=
2
23
Thus the acceleration of the centre of mass of the rolling cylinder is 23
gsin ,θ which is
less than the acceleration of the centre of mass of a cylinder sliding down the plane (g sin θ).
Now, substituting the above values of a and I is Eq. (iii), we get
f =
12
23 1
3
m gmg
R
R
2
2
=
sinsin
θθ
This is the minimum force of static friction needed for rolling. If it is less, the cylinderwould slip down the plane.
380 Mechanics
We known that, for just equilibrium
f = µs smgN = µ θcos
or13
mg sin θ = µ θsmg cos
or µs =13
tan θ
This is the minimum value of µs needed for rolling. If µs is less than 13
tan ,θ the cylinder
would slip.
Q.31. A solid sphere of mass 0.5 kg and diameter 1 m rolls without slipping with aconstant velocity of 5 m/sec along a smooth straight line. Calculate its total energy.
Solution. When the sphere rolls about its diameter, it has translator y as well asrotatory motion. Hence its total kinetic energy is
Ktotal = Ktrans + Krot
=12
12
2 2M Iv + ω
For sphere I = 25MR2 about diameter
and ω =vR
, where R is the radius of sphere
∴ Ktotal =12
12
25
22
M MRR
2v v+
=12
15
2 2M Mv v+
=7
102Mv
=7
100 5× ( . kg) (5 m / sec)2
= 8.75 joule.
Q.32. A block of mass m = 5 kg slides down a surface inclined 37° to the horizontal. Thecoefficient of sliding friction is µk = 0.25. A cord attached to the block is wrapped around aflywheel on a fixed axis at O, as shown. The flywheel has a mass M = 20 kg, an outer radiusR = 0.2 meter, and a radius of gyration with respect to the axis of rotation, k = 0.1 meter.Find the acceleration of the block down the plane and the tension in the cord.
Solution. Let a be the acceleration down the plane. The force acting on the block areits weight mg, the normal force N and the tangential (frictional) force fk (opposite to themotion) exerted by the plane, and the tension T in the cord. mg can be resolved into acomponent mg sin θ along the plane, and a component mg cos θ perpendicular to the plane.The net force on the block down the plane is mg sin θ – T – fk, and that perpendicular to
Moment of Inertia 381
the plane is N – mg cos θ. The block is accelerated down the plane, while its accelerationperpendicular to the plane is zero. Therefore, by Newton’s second law, we have
mg sin θ – T – fk = ma ...(i)
and N – mg cos θ = 0
or N = mg cos θ ...(ii)
Now fk = µkN = µkmg cos θ, so that Eq. (i) becomes
mg sin θ – T – µk mg cos θ = ma ...(iii)
N
mg sin θ
a
TT
R
m g cos θ
m gθ
O
θ
Fig. 35
Let us now consider the flywheel. The tension in the cord events a torque τ = τR on it.If I be the moment of inertia of the flywheel about the axis of rotation and α the angularacceleration produced in it, then
τ = Iα
or τR = IRa,
where a is the linear acceleration. This gives
T =IR2a
...(iv)
Substituting this value of τ in Eq. (iii), we get
mg sin θ – IR2a mgk− µ θcos = mg
or mg k(sin cos )θ µ θ− = a m +
IR2
or a =mg
mk(sin cos )θ µ θ−
+ I / R2
Here m = 5 kg, sin θ = sin 37° = 0.6, µk = 0.25, cos θ = cos 37° = 0.8, I = MK2 = (20) (0.1)2
= 0.2 kg-m2 and R = 0.2 m
∴ a =( ) ( . ) [ . ( . ) ( . )]
. / ( . ).
5 9 8 0 6 0 25 0 85 0 2 0 2
1 962−
+= m/sec2
Substituting the value of a in Eq. (iv), we get
T =IR
nt.2a = =0 2 1 96
0 29 82
. × .( . )
.
382 Mechanics
Q.33. In an Atwood’s machine the pulley mounted in horizontal frictionless bearings hasa radius R = 0.05 meter. The cord passing over the pulley carries a block of mass m1 = 0.50kg at one end and a block of mass m2 = 0.46 kg at the other. When set free from rest, theheavier block is observed to fall a distance of 0.75 meter in 5 sec. Compute the moment ofinertia I of the pulley.
Solution. Let a be the linear acceleration of the system.The heavier block moves a distance 0.75 meter in 5 sec.
Using x =12
2at , we have
0.75 =12
5 2a( )
∴ a = 0.06 meter/sec2
If v be the linear velocity at the end of 5 second, then
v = at = 0.06 × 5 = 0.3 meter/sec.
The heavier block descends 0.75 meter, thus losing apotential energy of (0.50 × g × 0.75) joules, while the lighterone ascends the same distance and gains energy of (0.46 × g× 0.75) joule. Thus net loss is P.E. of the system.
∆U = (0.50 × g × 0.75) – (0.46 × g × 0.75) = 0.294 Joule
If there is no friction in the pulley, this must be equal to the gain in the kinetic energyof the blocks and the pulley which is
∆K =12
121 2
2( )m m v+ + I 2ω
=12
0 50 0 46 0 3 12
22
( . . ) ( . )+ +
I 0.30.05
= 0.0432 + 18 I
Equating the loss in P.E. to the gain in K.E., we have
0.294 = 0.0432 + 18 I
∴ I =0 294 0 0432
180 0139. . . .− = kg-meter2
Q.34. Suppose the disc (M = 20 kg, R = 0.5 m) in the problem 26 is initially at rest. Asthe body (m = 1.0 kg) is hanged from the cord, the disc begins to rotate. Calculate the workdone by the applied torque on the disc in 2.0 second. Compute also the increase in rotationalkinetic energy of the disc.
Ans. The angular acceleration produced in the disc is
α =mg
mm
mgR
I + R M + 2 R2 =
2 I = 1
2MR2
= 2 1 020 2 1 0
9 80 5
1 78× .( × . )
.
.. .
+
= rad/sec2
M
R
a
am 2
0.46 kg
m 1
0.50 kg
Fig. 36
Moment of Inertia 383
The torque τ exerted on the disc is therefore
τ = Iα =12
MR2α
=12
20 1 782( ) ( . )kg) (0.5 m rad/sec2
= 4.45 joule
Let us now calculate the angle θ through which the disc has rotated in 2.0 second. Weuse
θ = ω α021
2t t+ .
Here ω0 = 0 (starts from rest), α = 1.78 rad/sec2 t = 2.0 sec
∴ θ =12
1 78 2 0 3 562( . ) ( . ) .= rad.
Hence, the work done by the torque on the disc is
ω = τθ = (4.45 joule) (3.56 rad) = 15.8 joule.
As there is no friction, this work is used up in giving rotational kinetic energy to thedisc. Let us compute it
Knet =12
I 2ω
=12
12
2MR2
( )αt
=14
20 0 5 1 78 2 02 2( ) ( . ) ( . × . )
= 15.8 joule.
Q.35. Two mass m1 = 12 kg and m2 = 8 kg are attachedto the ends of a cord which passes over the pulley of an Atwood’smachine. The mass of the pulley is M = 10 kg and its radiusis R = 0.1 m. Calculate the tension in the cord and theacceleration a of the system.
Solution. Since the pulley has a finite mass M, the twotension T1 and T2 are not equal. The net force on the largermass m1 is m1g – T1 acting downward, and that on the smallermass m2 is T2 – m2g acting upward. If a bet the (linear)acceleration of the system, we have by Newton’s second law
m1g – T1 = m1a ...(i)
and T2 – m2g = m2a ...(ii)
Let us now consider the pulley which may be assumed asa solid disc. The tension in the cord exert torques on thepulley, the net torque is (T1 – T2)R, which is clockwise.
M
R
a
a
m 2
m g2
m 1
m g1
T 2
T 2
T 1
T 1
Fig. 37
384 Mechanics
If α be the angular acceleration produced in the pulley, whose moment of inertia (I) than,we have
torque = moment of inertia × ang. accel.
i.e., (T1 – T2)R = Iα
But I = 12
MR2 and α = aR
, so that
T1 – T2 =12
Ma ...(iii)
Adding (i) and (ii), we get
m g m g1 2− + −T T1 2 = m a m a1 2+
or ( )m m g1 2− = ( ) ( )T T1 2− + +m m a1 2
Putting the value of T1 – T2 from Eq. (iii), we have
( )m m g1 2− =12 1 2Ma m m a+ +( )
∴ a =( )m m g
m m
1 2
1 212
−
+ +
M...(iv)
Given m1 = 12 kg, m2 = 8 kg and M = 10 kg
∴ a =( ) × .( )
.12 8 9 812 8 5
1 57−+ +
= m/sec2
Now from Eq. (i), we have
T1 = m1 (g – a)
= 12 (9.8 – 1.57)
= 99 nt.
Again, from Eq. (ii), we have
T2 = m2 (g + a)
= 8 (9.8 + 1.57)
= 91 nt.
Q.36. A sphere rolls up an inclined plan of inclination 30°. At the bottom of the inclinethe centre of mass of the sphere has a translational speed of 7m/sec. How for does the spheretravel up the plane. How long does it take to return to the bottom.
Solution. We know the acceleration of a body rolling down an inclined plane is givenby
a =gsinθ
12
2+ KR
For a sphere I = MK2 = 25
2MR (about a diameter), so that KR
2
225
= . His given
Moment of Inertia 385
a =57
sin θ
Lets be the distance travelled by the sphere up the plane, after which its velocity isreduced to zero. We apply the translational equation of motion
v2 = v02 + 2as
Here v = 0, v0 = 7 meter/sec, a = − = − ° = −57
57
305
14g g gsin sin .θ
∴ 0 = (7)2 – 1014
gs
or s =7 7 1410 9 6
7× ×
× .= meter
After travelling the distance s up the plane, the sphere return. Let t be the time it takesin reaching the bottom. Thus, let us use
s = v0t + 12
2at
We have 7 = 012
514
2+
g t
∴ t =7 285 9 8
2 0×× .
. sec.=
Q.37. A sphere/spherical shell rolling don on inclined plane without shipping, which islarger, the rotational kinetic energy of the translations kinetic energy find the ratio.
Ans. In usual nattiness, the rotational end translational kinetic energies are given by
Krot =12
12
22
2I MKR
2cm
vω = ?( )
and Ktrans = 12
2Mv
ThusK
Koat
trans=
KR
2
2
For a solid sphere KR
2
2 = 25
. Which for < thin spherical shell KR
2
2 = 23
. Thus in each care
the translational energy in larger.
Q.38. A ball starts for rest and rolls down a 30° inclined plane. How much time wouldit take to cover 7 meter?
Solution. Let M be the mass and R the radius of the ball. Let v be the velocity of thecentre of mass of the ball after it covers a distance of the plane. In doing so, it loses potentialenergy Mg (s sinθ) which appears as kinetic energy. Thus
386 Mechanics
Mg (s sinθ) =12
12
2 2I Mw v+
=12
25
12
2
22MR
RM2
+vv
=7
102Mv
∴ v2 =107
gs sinθ
If a be the acceleration of the centre of men, then
v2 = 2as
or107
gs sin θ = a2s
∴ a =57
gsinθ
Here θ = 30° and g = 9.8 m/sec
∴ a =57
9 8 12
3 5( . ) .
= m/sec2
Let t be the time taken in covering 7 meter, than from the relation
S =12
2at , we have
7 =12
3 5 2( . )t
or t2 =7 23 5
4×.
=
∴ t = 2 sec
Q.39. A circular disc of man 50 gm and radius 10 cm rolls down a plan inclined at 30°to the horizontal. It rolls down form rest through 100 cm in 10 sec. Find the kinetic energyof the disc at the end of 10 sec and the moment of inertia of the disc about its areas.
Solution. Let S be the distance on the inclined plane through which the disc rolls down.In doing so it comes down a vertical distance of S sin 30°, and thus loses potential energyequal to mg S sin θ which appears as kinetic energy K at the end of the journey. Thus
K = Mg S sin 30°
= (0.05 kg) (9.8 nt/kg) (cm) 12 = 0.245 joule
The moment of inertia of the disc about its axis is given by
I =12
12
2MR = (0.05 kg) (0.1m)2 = 2.5 × 10–4 kg–m2.
θ
s
s s in θ
Fig. 38
Moment of Inertia 387
Q.40. A small solid ball rolls without shipping along the track shown. The radius of thecircular part of the track is r. If the ball starts from rest at a height of 6 r above the bottom,what are the horizontal and vertical force acting on it at the point Q.
Solution. Let m be the mass and R radius of theball. The ball on reaching the point Q dexends througha vertical distance of (6r – r) = 5, thus loosinggravitational potential energy, mg (5r). This appearsas kinetic energy at Q.
If v be the speed at Q, then the total kineticenergy is
Ktotal = Krot + Ktrans
=12
12
2 2Iω + mv
=12
25
2 12
710
22
2 2mv
mv mvRR
+ =
Equating it to the loss in potential energy, we get
710
2mv = mg(5r)
or v2 =50
7gr
The velocity v of the ball at the point, q is tangentially directed. The horizontal forceF acting on the ball is the counterplot force directed toward the centre of the circular track.That is
F =mv
vmg2 507
=
The vertical force on the ball is its weight mg.
Q.41. One and of a horizontal spring of force constant 6.0 nt/meter is tied to a fixed walland the other end to a solid cylinder which can roll without shipping on the horizontalground. The cylinder is pulled s distance of 0.1 meter and released. Calculate the kineticenergies of the cylinder when passing through equilibrium position.
Solution. When the cylinder is pulled by 0.1 meter, the spring is stretched and acquirespotential energy U, where
U = 12
12
6 0 0 1 0 032 2kx = =× . ×( . ) . joule
This entire energy is converted into the kinetic energy of rotation and the kinetic energyof translation when the cylinder is passing through its equilibrium position. Thus
Krot + Ktrans = 0.03 joule
Now, Krot = 12
12
12
14
2 22
22w
vv=
=MRR
M
and Ktrans =12
2Mv
6r
2r Q
Fig. 39
388 Mechanics
∴14
12
2 2M Mv v+ = 0.03
or Mv2 = 0.03 × 43
0 04× . Joule
Krot =14
14
0 04 0 012Mv = =( . ) . Joule
and Ktrans = 12
12
0 04 0 022Mv = =( . ) . Joule.
Q.42. A 3.0 kg horizontal cylinder has threads wrapped round it near each end. The twothreats are held vertical with their force ends tied to hooks on the ceiling. When the cylinderis released it falls under gravity and the threads are unwind. Find out the linear accelerationof the falling cylinder and the tension in the thread.
Ans. The cylinder rotating under gravity has botha motion of translation as well as of rotation. Let Mbe the mass and I the moment of inertia of the cylinderabout the axis of radiation. Let v be the linear velocityof its centre of mass and ω its angular velocity aboutthe axis of rotation when it has fallen through a verticaldistance h. Then
Loss in potential energy = mgh
gain in kinetic energy of translation = 12
2Mv
and gain in kinetic energy of rotation = 12
12
12
14
2 22
2I MRR
Mω =
2
=vv
where R in the radius of the cylinder.
By the conservation of energy, we have
Mgh = 12
14
2 2M Mv v+
or v2 =43
gh
If a be the acceleration with which the cylinder falls, we have from the formula
v2 = 2ah
43
gh = 2ah
or a =23
g ...(i)
Let T be the tension in each thread acting vertically up ward. The weight Mg of thecylinder acts down ward. Then the resulting force acting downward on the falling cylinder inMg – 2T. But this should be equal to Ma, where a is the accelerate of the falling cylinder.Then
hT T
vω
Fig. 40
Moment of Inertia 389
Mg – 2T = Ma
or T =12
M (g – a)
Substituting the value of a from Eq. (i) we have
T =12
M (g – 23
g) = 16
Mg
Here M = 3.0 kg and g = 9.8 ht/kg.
T =16
× 3.0 × 9.8 = 4.9 nt
Q.43. A uniform circular slab 5.0 m and mass 100 kg rolls along a horizontal at a speedof 4 m/sec. How much work has to be done to stop it.
Solution. The total kinetic energy of a circular slab of mass M and radius R rollingwithout shipping is the sum of its kinetic energy of translation and that of ration, that is
K = Ktrans + Krot
=12
12
2 2M Iv + ω
=12
12
2 22
2M MRR
vv+ ( )
= Mv2 ∴ For disc, I = MR2
Here M = 100 kg and v = 4m/sec
∴ K = 100 × (4)2 = 1600 Joule
By work-energy theorem the work need to stop it is
W = K = 1600 joule
Q.44. The oxygen molecule has a mass of 5.30 × 10–26 kg and a moment of inertia of 1.94× 10–46 kg-m2 about an axis perpendicular to the inter-nuclear axis and passing through thecentre. The molecule in a gas has a mean speed of 500 m/sec, and its rotational kinetic energyis two-third of its translational kinetic energy. Find its average angular velocity.
Solution. Ktrans = 12
12
5 30 102 26M kgv = −( . × ) (500 m/sec)2
= 6.625 × 10–21 joule
Krot =12
12
2Iω = (1.94 × 10–46 kg-m2)
2
= 0.97 × 10–46 2 Joule
But Krot =23
Ktrans (given). Therefore
0.97 × 10–46 2
=23
(6.625 × 10–21)
or 2 =
2 6 625 103 0 97 10
21
46
× . ×× . ×
−
− = 45.5 × 1024
∴ = 6.75 × 1012 rad-sec.
390 Mechanics
Q.45. A body of radius R and mass m is rolling horizontally without slipping with speed
v. It then rolls up a hill to a maximum height h. If h = 34
2vg
, (a) what is the moment of inertia
of the body (b) what might the shape of the body be?
Solution. (a) The total kinetic energy of a rolling body is
Ktotal = Krot + Ktrans
=12
12
2 2Iω + mv
=12
12
2
22I
Rv
mv+ = 12
22v m
IR
+
when it rolls up a hill to a maximum height h =
349
2v. The entire kinetic energy is converted
into gravitational potential energy mgh. Thus
12
22v m
IR
+
= mgh = mg
34
2vg
orI
R2 + m =32
m
or I =12
2mR
(b) The body may be solid circular cylinder or disc.
Q.46. A small ball of mass m and radius r, starting from rest, rolls down on the innersurface of a large hemisphere of radius R. What is its kinetic energy when it reaches thebottom of the hemisphere? What forcing of it is rotational and what translatable.
Solution. Initially the ball is at the top of thehemisphere. Its centre of mass is at vertical height(R – r) from the bottom of the hemisphere, so that ithas gravitational potential energy mg (R – r). Onreaching the bottom, this energy is totally convertedinto kinetic energy. Hence total kinetic energy
= mg (R – r)
We know that for a sphere of mass m and radius r rolling with linear velocity v andangular velocity w;
rotational kinetic energy =12
12
25
15
2 22
22Iω =
=mrvr
mv
and transactional kinetic energy =12
2mv
∴ total energy =15
12
710
2 2 2mv mv mv+ =
r
R
Fig. 41
Moment of Inertia 391
Now, fractional rotational energy =
157
10
27
2
2
mv
mv=
and fractional translational energy =
127
10
57
2
2
mv
mv=
Q.47. A uniform this bar of mass 4 kg and length 2 meter is bent to made a square.Calculate its moment of inertia about an axis passing through the centre of mass andperpendicular to the plane of the square
Ans. 12
kg meter2−
Q.48. Three masses each 2 kg are placed at the vertices of an equilateral triangle of side10 cm. Calculate the M.I. and radius of gyration of the system about an axis passing through(i) a centre (ii) mid point of one side and (iii) centre of man. In all cases the axis is perpendicularto the plane of the triangle.
[Ans. (i) 0.04 kg-m2, 0.082 m, (ii) 0.025 kg-m2, 0.064 m, (iii) 0.02 kg-m2, 0.058 m]
Q.49. Find the radius of gyration of a disc of mass 100 gm and radius 5 cm about anaxis perpendicular to its plane and passing through its centre of gravity. [Ans. 3.5 cm.]
Q.50. A square frame ABCD of side 40 cm and negligible mass, has four masses, 2, 5,6 and 3 kg placed at the corners and a mass 10 kg at the centre X. It is be rotated about anaxle passing through and perpendicular to the frame. Compute the torque needed to producean angular acceleration of 20 per sec2.
Q.51. A fly wheel of mass 10 kg and radius of gyration 20 cm is rotating with anglervelocity 10 rotations per minute. Calculate its anguler momentum. [Ans. 0.42 kg-m2/sec.]
Q.52. A solid cylindrical fly-wheel of 16 kg man and 20 cm radius is making 5.0 rev/sec.Find the kinetic and momentum.
Q.53. A light cord is wrapped around the rim of a disc of mass M and radius R, whichrotates without friction about a fixed horizontal axis. The free end of the cord is tied to a massm, which is released from rest a distance h above the floor show that the mass m strikes the
floor with a linear speed given by v = [ 2gh/1 ( M/2m)]+ .
Q.54. A cord is wrapped around the rim of a fly wheel 0.5 m is radius and steady downword pull of 50 at is exerted on the cord. The wheel is mounted in fraction less bearings ona horizontal shaft through its centre. The moment of inertia of the wheat is 4 kg-m2. (a)compute the angular acceleration of the wheel (b) find the work done in unwinding 5 meterof cord (c) If a mass having a weight of 50 mt from the cord, compute the annular accelerationof the wheel. Why is this not the same as in part (a).
[Ans. (a) 6.25 rad/sec2 (b) 250 Joule (c) 4.74 rad/sec2, it is smaller than that in part(a) because the pull on the wheel is now smaller.]
Q.55. A block of mass m1 = 3 kg is put on a plane inclined at 0 = 30° to the horizontaland is attache by a cord parallel to the plane over a pulley at the top to a hanging block of
392 Mechanics
mass m2 = 9kg. The pulley has a mass m = 1kg and has a radius R = 0.1 meter. The coefficientof kinetic friction between block and plane is µk = 0.10. Find the acceleration a of the hangingblock and the tension in the cord on each side of the pulley. Assume the pulley to be a uniformdisc. [Ans. a = 5.7 meter/sec2, T2 = 37 nt. T1 = 34 nt.]
Q.56. A thin hollow cylinder of 25 kg-wt, open at both ends (a) slides with a speed of 2meter/sec (b) rolls with the same speed without slipping. Compare the kind energies of thecylinder in the two cases.
Q.57. The moment of inertia of a reel of thread about its axis is MK2. If the loose endof the thread is held in the hand and the real is allowed to unroll itself while falling undergravity, show that it falls down with an acceleration of gR2 (R2 + K2) where R is the radiusof the real.
Q.58. A said cylinder (a) rolls, (b) slides from rest down an inclined plane. Neglet friction.Show that the ratio of the velocities of the centre of mass in the two cases, when the cylinder
reaches the bottom of the incline is 3/2 .
393
9.1 RIGID BODY
A body is said to be rigid, if the distance between any two particles of which remainsunaltered whatever the external forces applied to it and in whatever manner they may vary,so that it remains un-deformed, i.e., its size, shape and volume remain unaffected. In actualpractice, however, we come across no such body, for all material bodies are found to getdeformed to a greater or a smaller extent under suitably applied external forces, tending torecover their original state. (i.e., length, volume or shape) on the removal of these forces.
9.2 ELASTICITY
A body remains in natural equilibrium under the internal molecular forces of attractionwhose magnitude depends upon the spacing between the molecules. When external forces areapplied on the body, new internal forces are developed which cause a change in the relativespacing between the molecules. Hence, the body changes in size, or shape or both and is saidto be ‘deformed’. When the external forces are removed, the new internal forces bring thebody to its normal state. The property of the body by virtue of which it recovers its originalsize and shape when the external forces are removed is called the ‘elasticity of the body.
9.3 PERFECTLY ELASTIC AND PERFECTLY PLASTIC
If the deformation of a body under a given deforming force, at a given temperature,remains unchanged (i.e., neither increases nor decreases) by the prolonged application of thatforce and which completely regains its original state on the removal of that force, it is saidto be perfectly elastic.
On the other hand, if the body remains deformed and shows no tendency to recover itsoriginal condition on the removal of the deforming force, it is said to be perfectly plastic. Thenearest approach to the former is a quartz fibre and to the latter, ordinary putty. All otherbodies lie between these two extremes.
9.4 STRESS
A body is equilibrium under the influence of its internal forces is in its natural state.But when external or deforming forces are applied to it, there is a relative displacement ofits particles and this gives rise to internal forces of reaction tending to oppose and balancethe deforming forces, until the elastic limit is reached and the body gets permanently deformed.The body is then said to be stressed or under stress.
9
394 Mechanics
If this opposing or recovering force be uniform, i.e., proportional to area, it is clearly adistributed force like fluid pressure and is measured in the same manner, as force per unitarea, and termed as stress. Stress is equal in magnitude; though opposite in direction, to theapplied or deforming force per unit area.
Thus, if F be the deforming force applied uniformly over an area A, then stress = F/A.
The stress is not uniformly distributed over a surface, let a force δF act normally on anelemental area δA of the surface, when stress at a point on the surface = δF/δA which in thelimit δA → 0 is equal to dF/dA.
If the deforming force were inclined to the surface, its components perpendicular andalong the surface are respectively called normal and tangential (or shearing) stress. Thestress is, however, always normal in the case of a change of length or volume and tangentialin the case of a change of shape of a body.
9.5 STRAIN
When a body suffers a change in its size, or shape, under the action of external forces,it is said to be ‘deformed’ and the corresponding measured per unit dimension i.e., per unitlength, per unit volume or the angular deformations produced in it, is called strain and isreferred to as linear, volume and shearing strain (or shear) in the three cases respectively.
9.6 HOOK’S LAW
This is the fundamental law of elasticity and was given by Robert Hook in 1679 in theform ‘Ut tension sic vis’, i.e., ‘As the tension, so the strain’. It may given by:
‘Provided the strain is small, the stress is proportional to the strain. It follows, therefore,that if the strain be small, the ratio between stress and strain is a constant, called modulusof elasticity (a name given to it by Thomas Young) or Coefficient of elasticity (E). Thus:
E =StressStrain
The dimensions and units of the modulus or coefficient of elasticity are the same asthose of stress or pressure.
9.7 ELASTIC LIMIT
In the case of a solid, if the stress be gradually increased, the strain too increases withit in accordance with Hook’s law until a point is reached at which the linear relationshipbetween the two just ceases and beyond which the strain increases much more rapidly thanis warranted by the law. This value of the stress for which Hook’s law just ceases to beobeyed is called the elastic limit of the material of the body for the type of stress in question.
Behaviour of a Wire Under an Increasing LoadLet us consider a wire suspended vertically from a rigid support. When a load is applied
at its lower end, its length increases. The load divided by its area of cross-section gives the‘stress’ while the increases in length divided by the initial length gives the ‘strain’. When theload is increased step by step, and a graph is drawn between the stress and the correspondingstrain a curve is obtained.
According to Hook’s law, the stress in a body is proportional to the corresponding strainproduced, provided the strain is small. Now, up to the point A, the curve is a straight line,
Mechanical Properties of Matter 395
indicating that the length of the wire increases in proportion to the load applied. Further, itis found that the wire recovers its original length when the load is removed. The point A iscalled the ‘limit of proportionality’.
Stress
Strain
A
BC
D
Fig. 1
When the load is further increased, the extension is greater than that allowed byproportionality. However, up to a small range beyond A the wire continues to return to itsoriginal length when the load is removed, but after that it refuses to do so and a permanentincrease in length is produced. The stress at which the wire first refuses to return to itsoriginal length is called ‘elastic limit’.
On further increasing the load, a stage is reached when the extension increases veryrapidly for very slowly increasing load i.e., the material on the wire flows like a viscous liquid.This happens at point B, which is called the ‘yield point’.
When the load is increased still further, a point C is reached when the stress takes amaximum possible value. This is called the ‘breaking stress’. The wire begins to thin downso that its cross-section does not remain uniform, and breaks with a release of stress. Thispoint where the wire actually breaks is D.
9.8 YOUNG’S MODULUS
When a body whose length is very large as compared to its breadth and thickness (awire) is acted upon by two equal and opposite forces in the direction of its length, the lengthof the body is changed. The change in length per unit length of the body is called thelongitudinal strain and the force applied per unit area of cross-section of the body is calledthe longitudinal stress. When strain is small, the ratio of the longitudinal stress to thecorresponding longitudinal strain is called the ‘Young’s Modulus’ of the material of the body.It is denoted by ‘Y’.
Let there be a wire of length L and radius r. Its one endis clamped to a rigid support and a weight Mg is applied at theother end. Suppose within the elastic limit, its length isincreased by l. Then
Longitudinal stress =Force (weight suspended)
Area of cross - section
=Mg
rπ 2
Longitudinal strain =Increase in length
Original length L= l
l
L
M g
Fig. 2
396 Mechanics
∴ Young’s modulus of the material of the body is
Y =Longitudinal stressLongitudinal strain
= ML
M L2
g rl
gr l
//
ππ
2=
‘Y’ can be determined only for solids and it is the characteristic of the material of a solid.
9.9 BULK MODULUS OF ELASTICITY
When a uniform pressure (normal force) is applied all over the surface of a body, thevolume of the body changes, but its shape remains unchanged. Such a strain may appear inall the three states of matter, solid, liquid and gas. The change in volume per unit volumeof the body is called the ‘Volume strain’ and the normal force acting per unit area of thesurface (pressure) is called the ‘normal stress’. When strain is small, the ratio of the normalstress to the volume strain is called the ‘bulk modulus’ of the material of the body. It isdenoted by B.
Let the initial volume of a body be V, which changes, by v when a pressure p is applied,when the pressure increases the volume decreases and vice-versa. Then
Normal stress=
p
Volume strain =vV
∴ Bulk modulus of the material of the body is
B =p
vpv/VV=
Compressibility: The reciprocal of the Bulk modulus of the material of a body is called‘Compressibility’ of that material.
9.10 MODULUS OF RIGIDITY (SHEAR MODULUS)
When a body is acted upon by an external force tangential to a surface of the body, theopposite surface being kept fixed, it suffers a change in shape, its volume remaining unchanged.Then the body is said to be ‘sheared’. The ratio of the displacement of a layer in the directionof the tangential force and the distance of that layer from the fixed surface is called the‘shearing strain’ and the tangential force acting per unit area of the surface is called the‘shearing stress’. For small strain, the ratio of the shearing stress to the shearing strain iscalled the ‘modulus of rigidity’ of the ‘material of the body’. It is denoted by η.
Let us consider a cube ABCDHIJK, whose lower surface is fixed. When a tangential forceF is applied at its upper surface ABIH, then all the layers parallel to ABIH are displaced inthe direction of the force. The displacement of a layer is proportional to its distance from thefixed surface. The cube thus takes the new form A′B′CDH′I′JK. The angle θ through whichthe line AD or BC, initially perpendicular to the fixed surface, is turned is called the ‘angleof shear’ or ‘shearing strain’. If θ is small, then
Mechanical Properties of Matter 397
θ = tan θ = ′BBBC
=displacement of the upper surface
distance of the upper surface from the fixed surface
If A be the area of the upper surface ABIH, then
Shearing stress = F/A
∴ Modulus of rigidity of the material of the cube is
D
A
θ
A ′
H H ′ I I′
J
C
K
θ
BF
Fixed
B ′
Fig. 3
η =F A F
Aθ θ=
The unit of η is Newton/meter2 and its dimensional formula is [ML–1T–2]
9.11 POISSON’S RATIO
When two equal and opposite forces are applied to a body along a certain direction, thebody extends along that direction. At the same time, it also contracts along the perpendiculardirection. The fractional change in the direction along which the forces have been applied iscalled the ‘longitudinal strain’, while the fractional change in a perpendicular direction iscalled the ‘lateral strain’. The ratio of the lateral strain to the longitudinal strain is calledthe ‘Poisson’s ratio’. It is a constant for the material of the body. In figure a wire of originallength l and diameter D is subjected to equal and opposite forces F, F along its length. If thelength increases to l + ∆l and the diameter decreases to D – ∆D, then
Longitudinal strain =∆ll
and Lateral strain =∆DD
The Poisson’s ratio, σ, of the material of the wire is
σ =∆∆D/D
/l l
9.12 POTENTIAL ENERGY IN STRETCHED WIRE
When a wire is stretched, work is done against the inter-atomic forces. This work isstored in the wire in the form of elastic potential energy. Suppose on applying a force F on
D D– D∆ ll l+ ∆
F
F
Fig. 4
398 Mechanics
a wire of length L, the increase in length is l. Initially the internal force in the wire was zero,but when the length is increased by l, the internal force increases from zero to F (equal tothe applied force). Thus for an increase l in length, the average internal force in the wireis
O + F2
=F2
Hence the work done on the wire is
W = average force × increase in length = ½ F × l
This is stored as elastic potential energy U in the wire
U = ½ F.l
Let A be the area of cross-section of the wire. The last expression may be written as
U = ½ (F/A) (l/L) LA
= ½ stress × strain × volumes of the wire
∴ Elastic potential energy per unit volume of the wire is
u = ½ stress × strain
But Stress = Young’s modulus × strain
∴ u = ½ Young’s modulus × strain2
9.13 EQUIVALENCE OF A SHEAR TO A TENSILE AND A COMPRESSIVE STRAINAT RIGHT ANGLES TO EACH OTHER AND EACH EQUAL TO HALF THE SHEAR
Let ABCD be the section of a cube fixed at the lowersurface. Let a tangential force F be applied to its upper surface,so that, it is sheared through a small angle θ and takes theshape A′B′CD. In this process the diagonal DB is extended toDB′ while AC is compressed to A′C.
Let arcs BM and A′N be drawn with centres D and C, andradii DB and CA′ respectively. These may be taken asperpendiculars to DB′ and CA respectively, since θ is small.Again, since θ is small, ∠BB′M is very nearly equal to 45°.
Let us consider the triangle BMB′.MB′ = BB′ cos BB′M = BB′ cos 45° = BB′/√2
Now, let us consider the triangle, DBC.
DB =BC
cos DBCBC
cos 45BC 2=
°=
Extensional strain along DB =Extension in DB
Initial length
=DB DB
DBDB DM
DB′ − = ′ −
= MB′/DB
θ
M
FB ′BA ′A
D C
θ
N
Force
Fig. 5
Mechanical Properties of Matter 399
Putting the value of MB′ and DB from above, we get
Extensional strain =
BB2
BC 2BBBC
′
= ′12
=12
12
BBBC
tan ′ = θ
When θ is small, we can put tan θ = θExtensional strain = θ/2
Similarly, we can show that the compressional strain along AC is θ/2. Thus, a shear θis equivalent to an extensional strain and a compressional strain at right angles to each otherand each of value θ/2.
9.14 EQUIVALENCE OF COMPRESSION AND EQUAL PERPENDICULAREXTENSION TO A SHEAR
Let ABCD be the section of a cube of side l. Let it be compressed along the diagonal ACand extended along the perpendicular diagonal BD so that the new diagonals are A′C′ andB′D′. Let AA′ = CC′ = BB′ = DD′ = y, where y is small.
Let ∠ ANA′ = ∠C′MC = ϕAO = AB Cos BAO
= l cos 45° = l/√2
∴ A′O = AO – AA′ = l/√2 – y
Similarly, B′O = BO + BB′ = l/√2 + y
∴ A′B′ = ( (A O) B O)2 2′ + ′
=l
yl
y2 2
2 2
−
+ +
= l y2 22+
But 2y2 << l2
∴ A′B′ = l = AB
That is, the sides of the cube remain almost unchanged in length. Let us imagine thatthe deformed cube A′B′C′D′ has been rotated clockwise through the angle φ, and D′C′ hasbeen made to coincide with DC. In this position A'D' makes with AD on anlge θ which isequal to 2φ. This position represents the case of shear where the plane DC is fixed and theplanes above DC suffer parallel displacement. The angle of Shear is θ.
Let A′E Fig. 7 be perpendicular to AD. Now
θ = 2φ = 2 sin φ (where φ is small, then sin φ = φ)
= 2 A EA N
′′
...(i)
Let us now consider the traingle AA′E
θ
B ′BA ′A
D C
θ
D ′ C ′
Fig. 6
400 Mechanics
A′E = AA′ sin EAA′ = y sin 45° = y/√2
Also A′N =l2
Putting these values of A′E and A′N in (i) we obtain
θ = 2 22
2 2yl
yl
//
= ...(ii)
Let α be the compressional strain along the diagonal AC.Then
α =compressioninitial length
=AAAO
′ = =yl
yl/ 22
But from (ii)2yl
=θ2
∴ α =θ2
Thus, the compressional strain along AC is one-half the angle of shear. Similarly theextensional strain along the other diagonal BD is one half the angle of shear. Thus, compressionθ/2 and a perpendicular extension θ/2 are equivalent to a shear θ.
9.15 EQUIVALENCE OF A SHEARING STRESS TO AN EQUAL EXTENSIONALSTRESS PLUS AN EQUAL AND PERPENDICULAR COMPRESSIONAL STRESS
Let ABCDHIJK be a cube of side l having its base CDKJ fixed. Let a tangential force Fact at its upper surface ABIH. The tangential (shearing) stress P is given by
P =tangential force
surface areaF2=
l
This force must have a tendency of moving the cube as a whole. But actually the baseof the cube is fixed. Hence an equal and opposite force of reaction F is developed in the baseCDKJ. These two forces form a couple which tends to rotate the cube clockwise. Again, sincethe cube is not free to rotate, the base gives rise to an equal but anticlockwise couple byexerting forces F, F, on the faces BCJI and DAHK.
Let us imagine a diagonal plane ACHJ, cutting the cubeinto two halves and consider the equilibrium of the upper half.The forces F, F acting on the faces. ABIH and BCJI give aresultant force F√2 parallel to DB acting normally over theplane ACJH. Since this upper half is in equilibrium, theresultant force F√2 must be balanced by an equal and oppositeforce. This force is provided by the lower half of the cube.Thus we have equal and opposite forces F√2, F√2 actingnormally over the plane ACJH and tending to extend thematerial of the cube parallel to the diagonal DB. The area ofthe plane ACJH is
M
B ′
BA ′
A
DC
θ
N
D ′
C ′
φ
O
Fig. 7
D
A
H I
J
C
K
B
F
F
F F
Fig. 8
Mechanical Properties of Matter 401
AC × AH = l √2 × l = l2 √2
Therefore the extensional stress parallel to DB
=ForceArea
=F 2l2 2
= F P2l
=
Similarly, if we cut the cube into two halves by an imaginary plane BDKI, we can provethat a compressional stress P is acting parallel to the diagonal AC. Thus a shearing stressP is equivalent to an extensional stress and a compressional stress at right angles to eachother and each of value P.
9.16 RELATIONS CONNECTING THE ELASTIC CONSTANTS
The elastic constant Y, K, n and the poission’s ratio σ, are all interconnected.
Let us consider a cube whose sides each of length l, are parallel to the three coordinateaxes x, y and z. Let a uniform normal stress P be applied over its surface. Then each sideof the cube is under an extensional stress P.
The stress P acting parallel to the x-axis will produce extension parallel to the x-axis,and compressions parallel to y and z axis. These will be as follows:
Extension parallel to x-axis = longitudinal strain × initial length.
=stress
Young,s modulusinitial length×
=PY
× l
Compression parallel to y-axis = lateral strain × initial length
= Poission’s ratio × longitudinal strain × initial length
= σ PY
× l
Compression parallel to z-axis = σ PY
× l
Similarly, the stress P acting parallel to y-axis produces.
Extension parallel to y-axis =PY
× l
Compression parallel to x-axis = σ PY
× l
and Compression parallel to z-axis = σ PY
× lP
P
x
P
P
Py
Pz
O
Fig. 9
402 Mechanics
Also, the stress P acting parallel to z-axis produces
Extension parallel to z-axis =PY
× l
Compression parallel to x-axis = σ PY
× l
and compression parallel to y-axes = σ PY
× l
Hence, net extension parallel to x-axis
=PY
PY
PY
PY
l l l l− − = −σ σ σ( )1 2
Net extension parallel to y-axis = − + − = −σ σ σPY
PY
PY
PY
l l l l( )1 2
and net extension parallel to z-axis = − + − = −σ σ σPY
PY
PY
PY
l l l l( )1 2
Thus each side of the cube increases from l to l l+ −PY
( )1 2σ
The new volume is therefore,
l l+ −
PY
( )1 23
σ = l l33
31 1 2 1 3 1 2+ −
= + −
PY
PY approx
( ) ( )σ σ
Since the initial volume was l3, the change in volume
= l l l3 3 313
1 23
1 2+ −
− = −PY
PY
( ) ( )σ σ
Hence, Volume strain =Change in volume
Initial volume
=
3PY P
Y
( )( )
1 2 3 1 2
3
3
−= −
σσ
l
l
Now, considering the cube as a whole, its surface being under a uniform normal stress
P undergoes the volume strain3PY
( ).1 2− σ Its bulk modulus is therefore
K =normal stressvolume strain
P3PY
Y3(1
=−
=−( ) )1 2 2σ σ
Y = 3K (1 – 2σ) ...(i)
Let us now consider a cube of side l, upon which an extensional stress P parallel tox-axes, and a compressional stress P parallel to y-axes have been applied. There is no stressalong the z-axes
Mechanical Properties of Matter 403
The extensional stress P produces
Extension along x-axis =PY
× l
Compression along y-axis = σ PY
× l
and Compression along z-axis = σ PY
× l
The compressional stress P produces
Compression along to y-axis =PY
× l
Extension along to x-axis = σ PY
× l
and Extension along to z-axis = σ PY
× l
Thus, net extension parallel to x-axis = PY
PY
PY
l l l+ = +σ σ( )1
and net compression along y-axis = σ σPY
PY
PY
l l l+ = +( )1
There are equal extension and compression along the z-axis. Hence there is no changealong z-axis.
Since, each side of the cube is of initial length l, the extensional and compressionalstrains along the x-and y-axes are each equal to
=
PY P
Y
( )( )
11
+= +
σσ
l
l
Now, simultaneous equal extensional and compressional strains at right angles to eachother, are equivalent to a shear θ
∴PY
PY
( ) ( )1 1+ + +σ σ = θ
or2
1P
Y( )+ σ = θ
orPθ
=Y
2 (1 + )σ
Further, the extensional stress P parallel to the x-axis, and the compressional stress Pparallel to the y-axis will be equivalent to a shearing stress P. Hence, the modulus of rigidityof the material of the cube is
η =Shearing stress
ShearP=θ
P
x
P
P
y
P
z
Fig. 10
404 Mechanics
Putting the value of P/θ from above, we have
η =Y
2 (1 + )σ
or Y = 2η (1 + σ) ...(ii)
orY2η
= 1 + σ
or σ =Y2η
− 1
Let us put this value of σ in Eq. (i). Then
Y = 3 2 1K 1 Y2
− −
η
= 3 3 9 3K Y K KY−
= −η η
Y + 3KYη
= 9 K
Y =9η
ηK
+ 3K ...(iii)
Equations (i) and (ii) give
Y = 3K (1 – 2σ) = 2η (1 + σ)
3K – 2η = σ (6K + 2η)
or σ =3 26 2
KK
−+
ηη ...(iv)
Equations (i), (ii), (iii) and (iv) are the various relations among the elastic constants.
9.17 THEORETICAL LIMITING VALUES OF POISSON’S RATIO
The relation between Poission’s ratio σ to bulk modulus K and modulus of rigidity η is
3K (1 – 2σ) = 2η (1 + σ)
If σ is positive, right-hand side is positive. Therefore left-hand side must also be positive.This is so when
2σ < 1
or σ < (1/2)
or σ < 0.5
Now, if σ is negative, the left-hand-side is positive and hence right-hand side must alsobe positive. This is so when
1 + σ > 0
or σ > –1
Mechanical Properties of Matter 405
Thus theoretically σ must lie between –1 and 12
σ is, however, never negative. Hence, in practice, σ lies between 0 and 12
.
9.18 POISSON’S RATIO FOR AN INCOMPRESSIBLE MATERIAL
For a material whose volume remains unchanged by changing pressure, we have
K = ∞An infinite value of bulk modulus means that the body is incompressible.
Putting K = ∞ in the relation
Y3K
= 1 − 2σ, we get
1 – 2σ = 0
or σ = ½
Determination of Poisson’s Ratio (σσσσσ)(i) For rubber: About a meter long rubber tube R (like
a cycle tube) is closely fitted with rubber bungs(smeared with glue) and metal caps (A and B) at itstwo ends. The upper cap A is clamped tightly in awall bracket or a massive stand and a small weightplaced in the scale pan suspended from a hook inthe lower cap B, so that the tube hangs verticallywith no bends or kinks in it.
A glass tube G, about half a meter long and of radius r is just inserted into the rubbertube which is then filled with air free water until some of it rises in tube G.
When conditions become steady, the positions of the water meniscus in G and a pointor pin P carried by hook H are noted with the help of two separate travelling microscope.A suitable weight (W) is then placed in the scale pan so that the length of the rubber tube,as also its internal volume increases. Again, when conditions becomes steady, the positionsof the water meniscus and the point or pin P are noted.
Several such readings are taken by increasing the weight in equal steps of W each andthe mean increase in length, say dL, of the tube for a stress P = W/A as also the mean falldh in the level of the water meniscus in G obtained.
If L be the original length of the rubber tube, longitudinal strain = dL/L and if Y be theYoung’s modulus for the material of the tube,
Y =PL
Ld
And since the fall in the level of the water meniscus in G corresponds to increase involume of the rubber tube R, we have increase in volume of the rubber tube = (πr2)dh = dV.And, therefore, Bulk modulus for the material of the tube is given by
Water
B
PH
R
A
G
W
Fig. 11
406 Mechanics
K = P3 ( V / V)d
, where P is the stress applied along the length of the tube, i.e., only in
one direction (so that the increase in volume of the tube is one-third of what it would be withan equal stress (P) in all the three directions, and V, the initial volume of the tube (obtainedby measuring the volume of the water filling it completely).
Now, we have the relation K = Y
3 (1 − 2σ)Substituting for K and Y, therefore, we have
dVV
=dLL
( )1 2− σ
or σ =12
1 −
dd
VV
LL
(ii) For Glass: We proceed here too in the same fashion as in case (i) for rubber. Since,however, the change in volume in this case is very much smaller, we use acapillary tube in place of tube G above.
9.19 TWISTING COUPLE ON A CYLINDER
(i) Case of a Solid Cylinder or WireLet a solid cylinder (or wire) of length L and radius R be fixed at its upper end and let
a couple be applied to its lower end in a plane perpendicular to its length (with its axiscoinciding with that of the cylinder) such that it is twisted through an angle θ.
This will naturally bring into play a resisting couple tending to oppose the twistingcouple applied, the two balancing each other in the position of equilibrium.
To obtain the value of this couple, let us imagine the cylinder to consist of a largernumber of hollow, coaxial cylinder, one inside the other and consider one such cylinder ofradius x and thickness dx as shown in figure (i).
It is seen from figure (ii) that each radius of the base of the cylinder will turn throughthe same angle θ but the displacement (BB′) will be the maximum at the rim, progressivelydecreasing to zero at the center (O), indicating that the stress is not uniform all over.
dx
B
B ′R
O xθ
( )i
φ
θ
xO
BB ′
L
A
O ′
( )ii
φ φ
2πxA D
B B ′ C ′C
L
( )iii
Fig. 12
Mechanical Properties of Matter 407
Thus, a straight line AB, initially parallel to the axis OO′ of the cylinder will take upthe position AB′ or the angle of shear (or shear) = ∠BAB′ = φ. This may be easily visualizedif we imagine the hollow cylinder to be cut along AB and spread out when it will initially havea rectangular shape ABCD figure (iii) and will acquire the shape of parallelogram AB′C′D afterit has been twisted, so that angle of shear = BAB′ = φ.
Now BB′ = xθ = Lφ, whence, the shear φ = xθ/L and will obviously have the maximumvalue when x = R, i.e., at the outermost part of the cylinder and the least at the innermost.
If η be the coefficient of rigidity of the material of the cylinder, we have
η =Shearing stress
Shearor Shearing stress = η × Shear
Shearing stress = ηφ = η θx
L
∴ Shearing force on face area of the hollow cylinder = η θxL
× face area of the cylinder
=η θ π πηθx
xdx x dxL L
=
× .2
2 2
And the moment of this force about the axes OO′ of the cylinder
=2 22 3πηθ πηθ
L L
=
x dx x x dx. .
∴ Twisting couple on the whole cylinder = 2
23πηθ πη θ
LRL
4
O
R
x dx = Or twisting couple per unit twist of the cylinder or wire, also called torsional rigidity of
its material, is given by
C =πηR2L
4
(ii) Case of a Hollow CylinderIf the cylinder be a hollow one, of inner and outer radii R1 and R2 respectively, we have
Twisting couple on the cylinder =2
23πηθ πη θ
L LR R
R
R
24
14
1
2
x dx = −( )
∴ Twisting couple per unit twist, C′ = πη2L
R R24
14( )− .
Now, if we consider two cylinders of the same material, of density ρ, and of the samemass M and length L, but one solid, of radius R and the other hollow of inner and outer radiiR1 and R2 respectively, we have
408 Mechanics
CC
′= R R
R
R R R R
R24
14
4
22
12
22
12
4− =
+ −
Since M = π ρ π ρR R L = R L ,22
12 2− we have R R R2
212 2− =
orCC
′=
R R R
R
R R
R22
12 2
4
22
12
2
+=
+
Again, because R R22
12− = R2, We have R R R2
2 212= +
And therefore, R R22
12+ = R R R R R2
12
12 2
12+ + = + 2 i e. .,
R R22
12+ > R2
Clearly, therefore, CC
′ > 1; C′ > C
or, the twisting couple per unit twist is greater for a hollow cylinder than for a solid one ofthe same material, mass and length.
This explain at once the use of hollow shafts, in preference to solid ones, for transmittinglarge torques in a rotating machinery.
9.20 DETERMINATION OF THE COEFFICIENT OF RIGIDITY (ηηηηη) FOR THE MATERIAL OF A WIRE
1. Statical Method (Barton’s Statical Method of Determining Modulus of Rigidity)The torsional rigidity for a rod of length l and radius r, which is fixed at the upper end
and twisted at the lower end is,
C =πηr
l
4
2
Where η is coefficient of rigidity of the material. This expression has already given.
The given wire or rod is hung vertically with its upper end rigidly clamped at A. To itslower end a brass cylinder C is attached. Two flexible threads are wound round the cylinderC such that they leave it tangentially at the opposite ends of a diameter and then pass overto two frictionless pulleys P1 and P2. The ends of these threads carry pans. A pointer isattached to the wire at a known distance from the upper end. When the wire is twisted, thepointer moves over a degree scale S.
Theory: Suppose equal weights mg, mg are placed on the pans. So that the threadsexperience equal and opposite parallel forces mg, mg. These forces form a couple of momentmgD, where D is the diameter of the cylinder C. This is a twisting couple applied to the lowerend of the wire. It causes each cross-section of the wire to twist through an angle proportionalto the distance of the cross-section from the upper fixed end. In the equilibrium state, thetwisting couple is equal and opposite to the elastic restoring couple set up in the wire.
Let r be the radius of the wire and φ radian the twist in it at a distance l from the fixedend. The elastic restoring couple set up in the wire is
Mechanical Properties of Matter 409
Cφ =πη φr
l
4
2
This, at equilibrium, is equal to the twisting couple mg D.That is
πη φrl
4
2= mgD
η =2
4mg l
rD
π φ
Method: Equal weights are placed on the pans and thetwist is noted by reading both ends of the pointer on the scale(to eliminate the error due to eccentricity of the wire withrespect to the scale). The weights are then increased step bystep and the corresponding twists are noted. The procedure isrepeated by decreasing the weights in the same steps. A graph is then plotted between theweights mg and the corresponding mean twists φ. This is a straight line. Its slope givesmg/φ. The length l is measured directly by a metre scale. To find the radius r, the diameterof the wire is measured by a screw gauge at a number of places in two mutually perpendiculardirection and the mean value is calculated. The diameter D is measured by vernier callipers.η is calculated from the above expression.
φO
m g
Fig. 14
On the circular scale φ is read in degree. Hence its value is multiplied by π/180 to convertit into radian before substituting in the above expression.
2. Dynamical method (Maxwell’s vibrating needle method)In this method, the time period of torsional vibration of the suspended body is observed
directly, its total mass kept the same throughout and yet its distribution ingeniously alteredabout the axis of suspension so as to bring about a known change in its moment of inertiaabout the axis and hence also a change in its time-period.
The so called vibrating needle is a hollow tube of length a, into which can be fitted twohollow and two solid cylinders of equal length (a/4), one pair on either side, such that, placedend to end, they just completely fill the tube. The tube itself is suspended from a torsion headH by means of the wire under test, of length L and radius R and a small piece of mirror Mattached to it to enable the torsional vibration of the tube to be observed by the telescopeand scale method.
The solid cylinders are first placed in the inner position, as shown in figure (i) and thetime-period T1 of the loaded tube determined. The position of the solid and hollow cylindersare then interchanged (figure ii), and time-period T2 of the tube determined again.
l
A
S
C
P2
P 1 D
m gm g
Fig. 13
410 Mechanics
Since the total mass suspended from the wire remains the same, there is no question
of any possible change in its torsional rigidity (or restoring couple per unit twist) C =RL
4πη2
.
If I1 and I2 be the moments of inertia of the loaded tube about the wire as axis in the twocases respectively, we have
T1 = 2 2π πIC
and TIC
12
2=
whence T T22
12− =
4 2πC
I I2 1
− ...(1)
H HS S
( )i
M
H
HS S
( )ii
M
H
H
Fig. 15
To obtain the value of (I2 – I1), we note that the c.g. of either inner cylinder lies at a
distance a / ,4
2 or
a8
and that of each outer cylinder at a distance3 4
2a /
or38a
from the axis.
So that, if m1 and m2 be the masses of a hollow and a solid cylinder respectively, the changefrom position (i) to (ii) merely means the shifting of a mass (m2 – m1) from a distance a/8to a distance 3.a/8 from the axis on either side. Therefore, in accordance with the principleof parallel axes, we have
I2 = I1 + − −
238 82 1
2 2
( )m ma a
or I2 – I1 = m ma
2 1
2
4−
Substituting this value of I2 – I1 in expression (1) above, we get
T T22
12− =
44
2
2 1
2πC
−m ma
Substituting the value of C = πηR
L
4
2 we get
T T22
12− =
4 24
2
2 1
2ππη
× LR4 m m
a−
which gives η =2 2 1πL
T T R
2
22
12 4
a m m( )( )
−−
Mechanical Properties of Matter 411
R must be measured most accurately, in view of its 4th power being involved in theexpression for η.
9.21 TORSIONAL OSCILLATIONS
Let a body, say a disc, be hung by a long and thin vertical wire whose upper end is rigidlyclamped.
Let the disc be given a slight rotation in the horizontal plane by applying a couple (byhand). The wire is twisted and an equal and opposite elastic restoring couple is set up in it,when the twisting couple is withdrawn, the restoring couple is unbalanced and produces anangular acceleration in the disc in a direction opposite to that of the twist. The disc thereforereturns to its mean position, crosses it (due to inertia) and rotates further in the oppositedirection. The wire is now twisted in the opposite direction. A restoring couple is again setup in it which arrests its motion and makes it return. The whole phenomenon is thenrepeated. Thus the disc oscillates in the horizontal plane about the wire as axis. Theseoscillations are called ‘torsional oscillations’ and the system is called a “torsion pendulum”.
To find the periodic time of a torsion pendulum by considering a disc, when the disc isoscillating the twist at the lower end of the wire at an instant t is φ, the restoring couple atthus instant will be – C φ, where C is the restoring couple per unit twist. The minus sign isput as the couple is directed opposite to the twist φ.
Now, it I be the moment of inertia of the disc about the axis of oscillation, the coupleacting upon it at this instant must be I(d2φ/dt2), where (d2φ/dt2) is the angular accelerationat this instant.
∴ –Cφ = Iddt
2
2φ
orddt
2
2φ
= − = −CI
φ ω φ2
where ω2 =CI
constant
∴ddt
2
2φ
∝ φ
Thus the angular acceleration is proportional to the twist or the angular displacement.Hence the motion is simple harmonic, so that its period is given by
T =2πω
or T = 2π IC
9.22 DETERMINATION OF MODULUS OF RIGIDITY OF A TORSIONALOSCILLATIONS
The upper end of the given wire is clamped to a rigid support, and a disc is attached toits lower end. The disc is slightly rotated by hand and left. The system begins to perform
r
l
Fig. 16
412 Mechanics
torsional oscillations. The period of these oscillations is obtained by timing 50 (say) oscillations.The period is given by
T = 2π IC
...(i)
Where I is the moment of inertia of the disc about the axis of oscillation and C is thetorsional constant of the wire.
Now, a regular auxiliary body is placed centrally upon the disc and the period of thesystem is again obtained. Let it be T1. Then,
T1 = 2π I + IC
1 ...(ii)
Where I1 is the moment of inertia of auxiliary body about the axis of oscillation. Squaringand subtracting equation (i) from equation (ii) we get
T T12 2− =
4 2πC
I1
ButC =4πηr
l2, where l is the length, r is the radius and η is the modulus of rigidity of
the material of the wire,
∴ T T12 2− = 4 22
4ππη
( )lr
I1
or η =8
4
πI
T T1
12 2
l
r− ...(iii)
The length l of the wire is measured by a meter scale. To obtain r, the diameter of thewire is measured with a screw gauge at several points of the wire in two perpendiculardirections and the mean value is calculated. I1 is calculated from the mass and dimensionsof the auxiliary body. Finally, η is calculated from the last expression.
Defects: (a) The expression (iii) is obtained on the assumption that C remains constant.Actually it suffers a small change when the auxiliary body is placed upto the disc. (b) Incalculating the moment of inertia of the auxiliary body from its mass and dimensions, it issupposed to be of uniform density throughout, which is not always the case.
9.23 BEAM
A rod or a bar of a circular or rectangular cross section, with its length very muchgreater than its thickness (so that there are no shearing stresses over any section of it) iscalled a beam.
Now, a beam may just rest on a support, like a knife-edge or have a small part of itfirmly clamped or built into a wall at either end. In the former case, it is called a supportedbeam and in the latter, a built-in or on oncastre beam or usually, simply, a fixed beam.
In a supported beam, obviously, the support can merely exert a force on the beam butin a fixed beam, it can also exert a couple on it.
If the beam be fixed only at one end and loaded at the other, it is called cantilever.
Mechanical Properties of Matter 413
Longitudinal Filament: A rectangular beam may be supposed to be made up of anumber of thin layers placed in contact and parallel to one another. Similarly, a cylindricalbeam may be supposed to be made up of thin cylindrical layers placed in contact and coaxialto one another. Further, each layer may be considered as a collection of thin fibres lyingparallel to the length of the beam. These fibres are called ‘longitudinal filaments’ of the beam.
Neutral Surface: When a beam is clamped horizontally at one end and loaded at theother, it undergoes bending. The filament (or fibres) of outward side are lengthened andsubjected to tension, while those of the inner side are shortened and compressed. In betweenthese two portions, there is a layer or surface in which the filaments are neither elongatednor shortened. Such a surface is called neutral surface.
Neutral axis: The line of intersection of the plane of bending with the neutral surface(both are perpendicular to each other) is called the neutral axis.
9.24 BENDING MOMENT
When a horizontal beam is fixed at one end and loaded at the other, a bending isproduced due to the moment of the load. In this position the beam remains in equilibrium,provided the limit of elasticity has not been exceeded.
A
D
N
W
E
FB
N ′C
W
Fig. 17
Let ABCD represent a section of the beam fixed rigidly in the wall at AD with the otherend BC, loaded with a weight W, as shown in figure. Such a beam is called cantilever.
Let us consider the equilibrium of a portion BCFE of the beam, cut by a transverse planeEF across it. A force W acts downwards at the end BC, hence an equal and opposite reactionforce equal to it is acting vertically towards along EF. These two equal and opposite forcesconstitute a couple. This clockwise couple which bends the beam, is called bending couple andthe moment of this couple is called the bending moment.
But the beam is in equilibrium, therefore, there must also be an equal and oppositecouple to balance it. The only other forces which act on EBCF are obviously the forces dueto stretching and compression of the filaments of the beam passing across EF. The filamentsin the upper portion of the beam are elongated and thus, they must be in tension, the portionof the filaments to the left of EF exerts a pulling force on the portion to the right of EF,similarly a portion of the filaments in the lower half of the beam, to the left of EF, exertsa pushing force on the portion of the right of EF, because the filaments below the neutralsurface are shortened. The distribution of these forces at the section EF has been shown inabove figure and their resultant is a couple which balances the bending couple.
414 Mechanics
The moment of this balancing couple (due to tensile and compressive stresses) is calledthe moment of the resistance to bending. Since in the position of equilibrium it is equal andopposite to the bending couple, it is referred as bending moment of the beam.
Expression for bending momentLet us consider a small portion of the beam bounded by two transverse sections AB and
CD close to each other. After bending as in 18(b), AC is elongated to A′C′ and BD is shortenedto B′D′. Line EF represents the neutral surface, which is neither stretched nor shortened.
Let the small portion, under consideration, be bent in the form of a circular arc, subtendingan angle θ at the center of curvature O. Let R be the radius of curvature of the neutralsurface E′F′.
Consider a filament GH distant z from EF in the unbent position of the beam. Afterbending its position is shown by G′H′. Now
G′H′ = (R + Z)θBefore bending GH = EF
E′F′ = EF (for neutral surface)
GH = E′F′But E′F′ = Rθ
A C
B D
G H
E FZ
( )a
A ′G ′
C ′H ′Z
E ′ F ′
D ′B ′
O( )b
θ R
Fig. 18
Change in the length of the filament = G′H′ – GH
= (R + Z)θ – Rθ = Zθ
Strain = Change in lengthInitial length =
G H GHGH
ZR
ZR
′ ′ − = =θθ
If PQRS represents a section of the beam at right angles to its length and the plane ofbending. In figure 19 then, clearly, the forces acting on the filaments are perpendicular tothis section. The line MN lies on the neutral surface.
Let the breadth of the section be PQ = b and its depth, QR = d.
The forces, producing elongations and contractions in filaments, act perpendicular to theupper and lower halves, PQMN and MNRS respectively, of the section PQRS and theirdirections being opposite to each other.
Now, consider a small area δa of the section PQRS about a point A, distant Z from theneutral surface. The strain produced in a filament passing through this area will be Z/R, asshown above.
Mechanical Properties of Matter 415
Now
Y = Stress/strain
∴ Stress = Y × strain
Hence, stress about the point A = Y × Z/R
Where Y is Young’s modulus for the material of the beam,
b
d
P Q
RS
M N
AZ
δaO
Fig. 19
Force on the area δa = Y (Z/R) δa
Moment of this force about the line MN = Y.(z/R) . δa.z = Y.
Rδa z. 2
But, the moments of the moments acting on both the upper and the lower halves of thesection are in the same direction, therefore, the total amount of the moments acting on thefilaments in the section PQRS is given by
Y.R
δa z. 2
∑ =YR
YIR
δa z. 2 =∑Where I = Σδa.Z2 and is called the second moment (or geometrical moment of inertia)
of the sectional area about MN and therefore, equal to aK2, where a is the whole area of thesurface PQRS and K its radius of gyration about MN.
This quantity YI/R is the restoring couple or the bending moment of the beam, beingequal and opposite to the moment of bending couple due to the load in the position ofequilibrium. Thus
Bending moment =YIR
The quantity YI is known as flexural rigidity.
In case of a beam of rectangular cross-section of breadth b and depth d, the area of cross-section is bd and (radius of gyration)2 in equal to d2/12. Hence
I = area of cross-section . d2/12 = bd.d2/12 = bd3/12
Similarly, in the case of cylindrical rod of radius r the area of cross-section is πr2 andthe square of radius of gyration of the cross-section about an axis through the center of thesection and perpendicular to the plane of bending is r2/4. Thus
I = πr2.r2/4 = πr4/4
In case, the rod is hollow, having r1 and r2 as the internal and external radii, then
I =π4
r r24
14−
416 Mechanics
9.25 CANTILEVER
A horizontal beam, fixed at one end and loaded at the free end, is called a cantilever.
Let a beam of length l be fixed at its one end A and loaded with a load W at B. Let theend B be deflected to the position B′ under the action of the load W. Suppose the axis of Xlies horizontally in the direction of the unbent beam and the axis of Y, vertically downwards,consider a section of the beam, as at P, at a distance x from the end A. Then the momentof the bending couple due to the load W.
= W (l – x)
The cantilever is in equilibrium in the bent position, hence the bending couple must bebalanced at the section under consideration by the restoring couple or bending moment.
The bending moment =YIR
Where Y is the young’s modulus of the material of the beam, I geometrical moment ofinertia of the section and R the radius of curvature of the neutral axis at the section.
Thus,YIR
= W (l – x) ...(i)
Ax ( –x)l
l
y
Pδ
B ′
B
W
Fig. 20
The radius of curvature R is given by the equation
R =
1 + dydx
d ydx
232
2
2
Where y is the depression of the beam at a distance x from the fixed end. Since y is
small,dydx
2
can be neglected in comparison to l. Hence, we get
R =1
2 2d y dx/
Substituting for R in equation (i) above, we have
YI.d ydx
2
2 = W(l – x)
Mechanical Properties of Matter 417
ord ydx
2
2 =WYI
( )l x− ...(ii)
This equation may now be solved to get the value of the depression of the free end i.e.,at x = l.
Integrating equation (ii), we get
dydx
=WYI
C1lxx−
+2
2 ...(iii)
Where C1 is constant of integration, which can be known from the conditions of theproblem. One such condition is at x = 0,
dydx
= 0
Applying this condition in equation (iii), we have C1 = 0
and thusdydx
=WYI
lxx−
2
2 ...(iv)
It is to be pointed out that at P, the slope of the curve tan θ = dθ/dx, θ being the anglebetween x-axis and tangent at P of the bent portion APB′.
Thusdydx
= tanθ = −
WYI
lxx2
2Integrating once again
y =WYI
C2lx x2 3
2 6−
+ ...(v)
Where C2 is another constant of integration.
But at x = 0 , y = 0, hence substituting these values in Eq. (v), we get
C2 = 0
Thus, the expression for the depression y at a distance x from the fixed end is
y =WYI
lx x2 3
2 6−
At the free end x = l and let y = δ, then
δ =WYI
l l3 3
2 6−
or δ =W3YI
l3
For a beam of rectangular cross-section of breadth b and depth d,
I = bd3/12
Therefore, δ =4 3
3W
Yl
bd
418 Mechanics
For a beam of circular cross-section of radius r, I = πr4/4
δ =4 3
4W
3 Yl
rπ
9.26 BEAM SUPPORTED AT ITS ENDS AND LOADED IN THE MIDDLE
Consider a beam supported on two knifes edges K1 and K2 at a distance l apart in ahorizontal plane. Let it be loaded at its middle point O. The reaction of each knife-edge isobviously equal to W/2 acting upwards. The beam bends as shown in the figure and thedepression is maximum in the middle.
The middle part of the beam is almost horizontal. Hence the beam may be consideredas equivalent to two-inverted cantilever OA and OB. The length of each cantilever will bel/2, clamped at the end O and carrying an upward load W/2 at the other. Therefore, thedepression of O below the knife edges in the elevation of the loaded end of such a cantileverfrom the lowest position O.
Let us consider a vertical section at P, distant x from O and consider the part PB asshown in Fig. 22. The moment of the deflecting couple is
=W2
PB =W2
.l
x2
−
In the equilibrium position this deflection couple must be balanced at the section underconsideration by the restoring couple or bending moment, established inside the beam, i.e.,
W2
lx
2−
=
YIR
...(i)
If y is the elevation of the section P above x-axis, then radius of curvature of the neutralaxis at this section is given by
R =1
d ydx
2
2
l
K1 K 2
W
O
A B
W2
W2
Fig. 21
Hence Eq. (i) is
W2
lx
2−
= YI
d ydx
2
2
Mechanical Properties of Matter 419
ord ydx
2
2 =W
2YIl
x2
−
Integrating this expression, we have
dydx
=W
2YIC1
lx
x2 2
2−
+
Where C1 is the constant of integration.
At O, x = 0 and dy/dx = 0, hence 0 = 0 + C1, or C1 = 0
∴dydx
=W
2YIl
xx
2 2
2−
Integrating further, we have
y = W2YI
C2l x x2 2 6
2 3−
+
Where C2 is the constant of integration.
Again, at 0 , x = 0 and y = 0 ∴ C2 = 0
∴ y =W
2YIl x x2 2 6
2 3−
Now, at the free end, x = l/2 and y = δ (say). Therefore
δ =W
2YIl l l2 8 48
2 3−
or δ =W
YIl3
48
If ‘b’ and ‘d’ be the breadth and thickness of beam respectively, then
I = bd3/12
δ =WY
lbd
3
34...(ii)
If the cross-section of the beam be a circle of radius r, then
δ = WY 4
lr
3
12 πThis is the same as the depression at the middle point O. If M be the mass with which
the beam has been loaded in the middle, then Eq. (ii) is
δ =WY
MY
lbd
glbd
3
3
3
34 4=
The use of the equation is made in the construction is girders.
Oy
x
l2
P
B
δ
W2Y
W2
Fig. 22
420 Mechanics
The depression of a beam at the middle is directly proportional as the cube of its lengthand inversely as the width and cube of the depth or thickness. This, in order that the girdermay not bend appreciably, its span, i.e., length is made small and its breadth and depth large.At the same time Young’s modulus for the material of the beam must be larger.
When a girder is supported at its ends, its filaments above the neutral surface arecompressed and those below the neutral surface (or in the lower half) are elongated. Thecompression or extension of the filaments is maximum at the upper or lower faces of thegirder respectively; hence the stresses are maximum at these faces and they decrease, as wemove towards the neutral surface from either side. Consequently, the upper and lower facesof the beam should be much stronger than its middle position. In other words, the middlepositions may be made of much smaller breadth than the upper and the lower faces. This isdone by manufacturing the girders with their section in the form of I so that breadth at theupper and lower faces may be sufficiently larger than that of the inner parts. This saves quitea good amount of the material without sacrificing the strength of the girder.
Stiffness of a beam: The stiffness of a beam is defined as the ratio of maximumdepression of the beam to its span.
9.27 DETERMINATION OF YOUNG’S MODULUS BY BENDING OF A BEAM
In the last section it is shown that the depression of a beam of rectangular cross-section,supported at the ends and loaded in the middle, is given by
δ =MY
glbd
3
34
or Y =Mglbd
3
34 δ
S
K2
K 1
G
W
Batte ry
M
δ
Y
X
Fig. 23
Thus, to determine the young’s modulus of the material of a beam, the beam is supportedhorizontally and symmetrically on two knife edges, K1 and K2, fixed at a distance l apart, asshown in figure. A hanger, supported by a knife edge is then placed on the middle of the beamand the load is applied by placing the weight on it. The depression δ of the beam, so produced,is determined by a micrometer screw. At the moment, when micrometer screw just touchesthe beam, a current starts flowing in the galvanometer G, from the battery, causing to deflectthe needle of the former and the reading is noted. A series of loads are applied and observations
Mechanical Properties of Matter 421
are taken for load increasing and decreasing. A graph is then plotted between the load ofmass M as abscissa and the corresponding mean depression as the ordinate. The slope of thestraight line, so obtained, gives the values δ/M.
Now Y = M Mglbd
glbd
3
3
3
34 4δ δ= .
Thus, knowing the values of l, b and d, Young’s Modulus of the material of the beammay be determined.
9.28 DETERMINATION OF ELASTIC CONSTANTS BY SEARLE’S METHOD
The elastic constants of the material of a thin short wire can be determined by thismethod.
The apparatus consists of two exactly equal metal rods AB and CD of square or circularcross-section. They are connected together at their middle points by the specimen wire andsuspended by silk threads from a rigid support such that the rods and wire lie in a horizontalplane.
When the rods, so suspended, are brought near to each other through equal distancesand then released, they execute torsional vibrations in a horizontal plane, due to the coupleexerted by the specimen wire on the two rods.
Young’s Modulus: In the Fig. 25 let θ be the angle through which each rod is turnedfrom its normal position. Then the angle subtended by the bent wire at the centre ofcurvature O is 2θ and obviously
l = R.2θWhere R is the radius of curvature of the bend wire.
From the theory of bending of beam, moment of the internal couple or bending moment
=YI
Rg
Where Ig is the geometrical moment of inertia of the cross-section of the wire and Y theYoung’s Modulus of the material of the wire.
Moment of the couple = YIg 2θ/l
For a wire of radius r,
Ig =πr4
4
∴ Moment of the couple =π θr
l
4Y2
.
Now this couple produces an angular accelerationd2θ/dt2 in each rod. If I is the moment of inertia ofeither rod AB or CD about the thread from which it issuspended, then the moment of the couple is Id2θ/dt2.
Hence, the equation of motion is
I Y2
ddt
rl
2
2
4θ π θ+ . = 0
B D
CA
Fig. 24
422 Mechanics
orddt
rl
2
2
4θ π θ+ Y2 I
. = 0
BD
CA θ
R
O
Fig. 25
Therefore, the above equation represents a simple harmonic motion of period.
T1 = 22
4ππ
lr
IY
whence Y =8 IT1
2πlr4 ….(i)
Modulus of rigidity: Now the suspension threads are removed. One rod say CD, isclamped horizontally to a rigid support and the other is given a slight rotations so as to twistthe specimen wire. When released, it executes torsional vibrations about the wire as axis,in a horizontal plane. Its time period T2 is determined.
Then T2 = 2 224
4π π
ηπηπI
CI.2 , C == l
rrl
∴ η =8
4πlrI
T22 ….(ii)
Poisson’s Ratio: From Eqs. (i) and (ii), we have
Yη =
TT
22
12
But Poisson’s ratio σ =Y2η
− 1 ∴ σ = TT22
122
– 1 ...(iii)
Thus, by substituting the values of various experimentally determined quantities inEqs. (i), (ii) and (iii), we can find the values of elastic constants Y, η and σ.
This method has the following advantages:
1. This method requires only a small specimen of the material in the form of a wire.
2. Young’s Modulus of the material of a wire is measured by measuring the periodictime and not by observing any deflection.
3. Here, for determining the Poisson’s ratio, the radius of the wire is not required themeasurement of the radius of the wire always introduces some error.
Mechanical Properties of Matter 423
NUMERICALS
Q. 1. A copper rod of length 1 m and cross-section 3 cm2 is fastened end to end to a steelrod of length L and cross-section 1 cm2. The compound rod is subjected to equal and oppositepulls of magnitude 3kg-wt at its ends.
(a) Find the length L of the steel rod if the elongations of the two rods are equal.
(b) What is the stress in each rod ?
(c) What is the strain in each rod ?
Ycopper = 1 × 1012 dyne/cm2, Ysteel = 2 × 1012 dyne/cm2.
Solution. (a) By Hooke’s law Ystressstrain
we have,=
strain =stress
Y
In usual notations,
lL
=F /
Ya
or l =FLYa
The elongation, l, are equal and the pulls, F, are equal so that
L
Ycopper
copper copper× a= L
Ysteel
steel steel× a
Putting the given values, we get
13
m(1 × 10 dyne / cm cm12 2 2) ( ) =
L dyne / cm cm
steel2 2( × ) ( )2 10 112
or Lsteel =23
m.
(b) The stress in the copper rod is
F
coppera =3 9 8
3 109 8 104
4× .×
. ×nt m
nt / m22
− =
and in steel rod isF
steela =3 9 8
29 4 104× .. ×
nt1 × 10 m
nt / m4 22
− = .
(c) The strain in the steel rod is
stressYsteel
=29 4 10
2 1014 7 10
4
117. ×
×. ×nt / m
nt / m
2
2 = −
and the strain in the copper rod is
424 Mechanics
stressYcopper
= 9 8 101 10
9 8 104
117. ×
×. ×
nt / mnt / m
2
2 = − .
Q. 2. (a) Show that the potential energy per unit volume of a strained wire is 12
(stress
× strain).
(b) Calculate the work done in stretching a uniform wire of cross-section 10–6 m2 andlength 1.5 m through 4 × 10–3 m. The Young’s modulus is 2 × 1011 nt/m2.
Solution. (a) Energy in the strained wire. Suppose a wire of length l and area of cross-section A is stretched through a distance x by applying a force F along its length. Then
tensile stress =FA
and tensile strain= xl
Therefore the Young’s modulus of its material is
Y =stressstrain
F/A=x l/
Hence the force F required to stretch it through x is
F =YAl
x
If the wire is now stretched a further small distance dx, the work done will be
dW = F YAdx
lx=
Hence the total work done to stretch the wire from its original length to l + x (i.e., fromx = 0 to x = x) will be
W =YA YAl
x dxl
xx x
0
2
02 =
=12
12
2YA A
Yxl
lx
lxl
=
( )
=12
× volume stress × strain.×
Hence the work per unit volume is 12
(stress × strain). This work is stored as the (potential)
strain energy in the wire.
(b) For the given wire, we have
l = 1.5 m, A = 10–6 m2, x = 4 × 10–3 m and Y = 2 × 1011 nt/m2
From this,
volume of the wire = A l = 10–6 × 1.5 m3
tensile strain =xl
=−4 10
1 5
3×.
Mechanical Properties of Matter 425
and tensile stress = Y strain = (2 × 10 nt / m11 2× ) × ×.
4 101 5
3−
Hence, the work done
W =12
× volume × stress × strain.
=12
1 5 2 10 4 101 5
4 101 5
611 3 3
(10 m nt / m3 2−− −
× . ) × × × ×.
× ×.
= 1.07 Joule.
Q. 3. Young’s modulus Y for brass is 9 × 1010 nt/m2. Determine the energy stored in abrass rod of 2 cm2 cross-section, 0.01 meter in length if compressed by a force of 10 Newton.
Solution. The work done in compressing the rod is given by
W =12
× volume × stress × strain
Hence, volume of rod = length × cross-section
= 0.01 m × 2 × 10–4 m2 = 2 × 10–6 m3
stress =compression force
cross- section
=10 nt
(2 × 10 m nt / m4 2
2− =
)×5 104
and strain =stress
Y= 5 10
9 10
4
10××
=59
10 6× −
∴ W =12
2 10 5 1059
106 4 6× ( × ) × ( × ) × ×− −
m nt /m3 2
= 2.78 × 10–3 Joule.
Q. 4. A steel wire of 2mm diameter is just stretched between two fixed points at atemperature of 20°C. Determine the tension when the temperature falls to 10°C coefficient oflinear expansion of steel is 0.000011 per°C and Young’s Modulus for steel is 2.1 × 1011 newton/meter2.
Solution. Let l be the length of the wire and a its area of cross-section when thetemperature falls, the wire contracts. The contraction (decrease in length) is
dl = l α dθwhere α is the coefficient of linear expansion and dθ is the fall in temperature. The strainin the wire is
dll
= α dθ
426 Mechanics
If Y be the young’s modulus of the material of wire the stress is given by
stress = young’s modulus × strain
= Y α dθThe force (tension developed) in the wire is
F = stress × area of cross-section
= Y α dθ × a
Substituting the given values:
F = (2.1 × 1011 nt/m2) × (0.000011/°C) × (10°C)
× 3.14 × (1 × 10–3 m2)
= 72.5 nt.
Q. 5. A material has Poisson’s ratio 0.20. If a uniform rod of it suffers longitudinal strain4.0 × 10–3, deduce the percentage change in its volume.
Solution. Poissons ratio σ =lateral strain (
longitudinal strain (∆
∆r r
l l/ )
/ )
Here,∆ ll
= 4.0 × 10–3
and∆ rr
= σ × . × ( . × ) . ×∆ ll
= =− −0 20 4 0 10 0 80 103 3
Let V be the initial volume, and V + ∆V the final volume. Then
V = πr2l ...(i)
and V + ∆V = π (r – ∆r)2 (l + ∆l) ...(ii)
substracting Eq. (i) from (ii) and neglecting smaller terms, we get
∆V = πr2∆l – 2πrl∆r
∴∆VV
=∆ ∆ll
rr
− 2
substituting the values:
∆VV
= 4.0 × 10–3 – 2(0.80 × 10–3) = 2.4 × 10–3
∴ percentage change in volume
∆VV
× 100 = (2.4 × 10–3) × 100
= 0.24
Q. 6. Calculate Y for rubber if a rubber tube 0.4 meter long, whose external and internaldiameters are 0.01 meter and 0.004 meter respectively, extends 0.0006 meter when stretchedby a force of 5 kgm-wt.
Solution. If r1 and r2 be the initial and external radii of the tube, then the area of cross-section of the tube is
Mechanical Properties of Matter 427
A = π ( )r r22
12−
= 3 14 0 005 0 002 2. ( . ( . )m) m2 −
= 3.14 × 0.000021 m2
= 6.6 × 10–5 m2
The stretching force is
F = 5 kg-wt = 5 × 9.8 = 49 nt
∴ Stress = FA
nt /m2= =−49
6 6 107 4 105
5
. ×. ×
and Strain =δll
= =0 00060 4
0 0015..
. m
The youngs modulus of rubber is therefore
Y =stressstrain
nt/m2= =7 4 100 0015
4 9 105
8. ×.
. ×
Q. 7. An iron wire of length 1 mts and radius 0.5 mm elongates by 0.32 mm whenstretched by a force of 5 kg-wt, and twists through 0.4 radian when equal and opposite torquesof 3 × 104 dyne-cm are applied at its ends. Calculate elastic constants for iron.
Solution. Let a wire of length l and radius r be elongated by x with force F.
By Hooke’s law, the young’s modulus of the material is given by
Y =stressstrain
=F / Fπ
πr
x ll
r x
2
2/=
Here F = 5000 × 981 dyne, r = 0.5 mm = 0.05 cm, x = 0.32 mm = 0.032 cm
l = 1 mts = 100 cm
∴ Y =5000 981 100
3 14 0 05 0 03219 5 102
11× ×. × ( . ) × ( . )
. ×= dyne/cm2
Now the torque necessary to produce a twist of φ radian in the wire is given by
τ =π ηr
l
4
2φ
⇒ η =τπ φ
( )24l
r
Here τ = 3 × 104 dyne-cm, l = 100 cm, r = 0.05 cm and φ = 0.4 radian
∴ η =3 10 200
3 14 0 05 0 4
4
2× ×
. × ( . ) × .
= 7.64 × 1011 dyne/cm2.
428 Mechanics
Now, the Poisson’s ratio σ, and Y andη are related as
σ =Y2
1η
−
=19 5 10
2 7 64 101
11
11. ×
× . ×−
= 0.276
Q. 8. Calculate Poisson’s ratio for silver: Given its Young’s modulus = 7.25 × 1010
nt/m2 and bulk modulus = 11 × 1010 nt/m2.
Solution. Y = 3K ( 1 – 2σ)
or σ =12
13
−
YK
Putting the given values:
σ =12
1 7 25 103 11 10
10
10−
. ×× ×
= 0.39
Q. 9. A wire of length 1 meter and diameter 2mm is clamped at one of its ends. Calculatethe couple required to twist the other end by 45° (η = 5 × 1011 dyne/cm2).
Solution. The couple required to twist the free end of a clamped wire of length lthrough φ radian is
τ =πη φr
l
4
2
For φ = 45° = π τ π η4 8
4radian, we have =
2 rl
Here l = 1 meter = 100 cm, r = 1.0 mm = 0.1 cm and η = 5 × 1011 dyne/cm2
∴ τ =( . ) × ( × ) × ( . )
×3 14 5 10 0 1
8 100
2 11 4
= 6.2 × 105 dyne-cm.
Q. 10. A wire of radius 1 mm and length 2 meter is twisted through 90°. Calculate theangle of shear at the surface, at the axis of the wire and at a point mid-way. If the modulusof rigidity of material is 5.0 × 1011 dyne/cm2, what is the torsional couple ?
Solution. Let l be the length of the wire, r the radius and φ the twist at the free end.Then the angle of the shear at a radial distance x from the axis of the wire is
θ =xlφ
At the axis of the wire, x = 0, so that θ = 0.
At the surface of the wire, x = r, so that
Mechanical Properties of Matter 429
θ =rlφ
Here, r = 1mm = 0.1 cm, l = 200 cm and φ = 90°
∴ θ =0 1 90
2000 045. × .° = °
At a point mid-way, x = r/2
∴ θ =12
0 045 0 0225× . .° = °
The torsional couple is
τ =πη φr
l
4
2
For φ = 90° = π/2 radian, we have
τ =π η2 4
4rl
Substituting the given values:
τ =( . ) × ( . × ) × ( . )
×3 14 5 0 10 0 1
4 200
2 11 4
= 6.2 × 105 dyne-cm
Q. 11. Two wires A and B of the same material and equal lengths but of radii r and 2rare soldered coaxially. The free end of B is twisted by an angle φ. Find the ratio of the relativeangles of the twist between the ends of the two wires, the twist at the junction, and thetorsional rigidity of the complete wire.
Solution. Let τ be the couple applied at the free end of B and φ′ the twist at the junction.The couple produce a relative twist φ′ between the ends A and a relative twist(φ − φ′) between the ends of B. Therefore, if η be modulus of rigidity of the material of thewires and l the length of each wire, we have
τ =πη φr
l
4
2′ ...(1)
τ =πη φ φ( ) ( )2
2
4rl
− ′ ...(2)
and comparing Eqs. (1) and (2) we get
φ′ = 16 (φ – φ′)
⇒φ
φ φ′
− ′ = 16
Also, solving we get, φ′ =1617
φ
Now, substituting this value of φ′ in the expression τ πη φ= ′rl
4
2
430 Mechanics
we get, τ =πη φr
l
4
21617
=8
17
4πη φrl
∴ The torsional rigidity of the complete wire is
c =τφ
πη= 817
4rl
Q. 12. The solid cylindrical rods of same material having lengths l and 2l and radii 2rand r respectively are soldered coaxially. The free end is twisted through 11°, the other endbeing clamped. Calculate the twist of the thicker rod.
Solution. (1/3°).
Q. 13. Two solid cylinders of the same material having equal lengths l and 2l and radiir and 2r are joined coaxially. Under a couple applied between the free ends the shorter cylindershows a twist of 30°. Calculate the twist of the longer cylinder.
Solution. Let τ be the couple, and let it produced a twist φ in the shorter cylinder andtwist φ′ in the longer cylinder. Then,
τ =πη φ πη φr
lrl
4 4
22
2 2= ( )
( )
φ′ =φ8
308
3 75= ° = °.
Q. 14. (a) Find an expression for the elastic (potential) energy stored (or work done) ina cylinder whose one end is fixed and the other end is twisted through an angle φ.
(b) Find the work done in twisting a steel of radius 1 mm and length 0.25 mts throughan angle of 45°. The modulus of rigidity of steel is 8 × 1010 nt/m2.
Solution. (a) Let c be the couple per unit twist in the wire. Thus couple for a twist φis cφ. The work done for additional small twist dφ is cφ dφ. Hence the work done for producinga total twist φ will be
W = c d cφ φ φφ
0
212 =
Now, c =πηr
l
4
2
∴ W =πη φr
l
4 2
4
This is also the elastic potential energy stored in strained wire.
(b) For the given wire, we have,
η = 8 × 1010 nt/m2, r = 1 mm = 10–3 m, l = 0.25 m and φ = 45° = π4
radians
Mechanical Properties of Matter 431
∴ W =3 14 8 10 10
4 0 253 14
4
10 3 4 2. × ( × ) × ( )× .
× .−
= 0.155 Joule.
Q. 15. A uniform tube of outer radius 5 units and inner radius 4 units is twisted by angle20° under a couple c. If a solid cylindrical rod of same material and same length, but of radius3 units, is placed under the same couple c what would be the angle of twist ?
Solution. The torsional rigidities c and c′ of the solid and hollow cylinders are given by
c =πηr
l
4
2...(i)
and c =πη ( )r r
l24
14
2−
where r → radius of solid cylinder
r1 and r2 → external and internal radii of the hollow cylinder
But r2 = 2r1 (given), so that
c′ =πη ( )15
214r
l...(ii)
From eqs. (i) and (ii), we have
cc′
=rr
4
1415 ...(iii)
The masses of two shafts are equal, so that
π ρr l2 = π ρ( )r r l22
12−
= π ρ( )3 12r l
where ρ is the density of the material of the cylinders. This gives
r2 = 3r12
r = 3 1r ...(iv)
Making this substitution in eq. (iii) we get
cc′
=rr
4
1415
35
=
Now if φ be the twist produced at the free end of a solid cylinder under twisting torqueτ, then
τ =πη φr
l
4
2
If θ be the (maximum) shearing strain at the surface, then
θ =rlφ
432 Mechanics
⇒ φ =lrθ
Thus we have
τ =πη θ πη θr
llr
r4 3
2 2 = ...(v)
If θ′ be the maximum strain on the hollow cylinder for the same torque τ, then
τ =πη θ( )r r
llr
24
14
22− ′
=πη θ( )15
2 214
1
rl
lr′
=πη θ( )15
413r ′
...(vi)
Comparing (v) and (vi) we obtain,
πη θr3
2=
πη θ( )154
13r ′
orθθ′
=152
13
3
rr
But r = 3 1r from eq. (iv)
∴θθ′
=15
2 3 313
13
rr×
=5
2 35
2 1 732=
× .
=1 44
1.
Q. 17. A cylinder of radius r and length l is recast into a pipe of same length. If the pipepossesses torsional rigidity which is 19 times greater than that of the original cylinder,calculate the inner radius of the pipe.
Solution. Let c and c′ be the torsional rigidity of the solid and hollow cylinders respectively.Then,
c = πη πηrl
cr r
l
424
14
2 2 and ′ = −( )
where r is the radius of solid cylinder and r1 and r2 the internal and external radii of the
hollow cylinder. Sincecc′ = 19, we have
r r24
14− = 19r4
Mechanical Properties of Matter 433
or ( ) ( )r r r r12
22
22
12+ − = 19r4 ...(i)
since the length and mass remain unchanged, we have
π ρ( )r r l22
12− = π ρr l2
or r r22
12− = r2 ...(ii)'
Making this substitution in eq. (i), we get
r r22
12+ = 19r2 ...(iii)
substracting (ii) from (iii) and solving, we get
r1 = 3r.
Q. 18. A thin-walled circular tube of mean radius 10 cm and thickness 0.05 cm is meltedup and recast into a solid rod of same length. Compare the torsional couples in the two cases.
Solution. Internal radius of tube, r1 = 10 12
0 05 9 975− =× . . cm
and external radius of tube, r2 = 10 12
0 05 10 025+ =× . . cm
Let r be the radius of the recast solid rod of the same length, Since the mass remainsunchanged, we have,
π ρ( )r r l22
12− = π ρr l2
where ρ is the density of the material. Thus
r2 = r r22
12 2 210 025 9 975 1 000− = − =( . ) ( . ) .
r = 1.0 cm
The torsional couple per unit twist for the hollow tube is πη ( )r r
l24
14
2−
and that for solid
rod is πηr
l
4
2 Therefore, their ratio is
r rr
24
14
4
−=
( . ) ( . )( . )
10 025 9 9751 0
4 4
4−
= 200 : 1
Q. 19. A solid cylinder of radius 3 cm is melted and recart into 0 hollow cylinder of outerradius 5 cm, while length remain unchanged. By what ratio torsional rigidity change ?
[Ans. It shall increase 4.5 times.]
Q. 20. The restoring couple per unit twist in a solid cylinder of radius 5.0 cm is 1 × 108
dyne-cm. Find the restoring couple per unit twist in a hollow cylinder of the same material,mass and length but of internal radius 12.0 cm.
Solution. The restoring couple per unit twist in a solid cylinder of length l and radiusr is
434 Mechanics
c =πηr
l
4
2Here r = 5.0 cm and c = 1 × 108 dyne-cm
∴πη2l
=cr4
8
41 10
5 0= ×
( . )If r1 and r2 be internal and external radii of the hollow cylinder of the same length,
material and mass m, then
m = π ρ π ρ( )r r l r l22
12 2− = [ρ is density]
or r r22
12− = r2
Here r1 = 12.0 cm and r = 5.0 cm
∴ r22 = r r1
2 2 2 2 25 0 12 0 13 0+ = + =( . ) ( . ) ( . )
⇒ r2 = 13.0 cm
The restoring couple per unit twist for this hollow-cylinder is
c′ =πη ( )r r
l24
14
2−
=1 10
5 013 0 12 0
8
44 4×
( . )( . ) ( . )−
= 1.25 × 109 dyne-cm.
Q. 21. An inertia table is a uniform disc of mass M and radius r. Its period of oscillationis T0. Now a thin ring of mass 10 M and radius 0.8 r is placed symmetrically on the table.Deduce the new period of oscillation. How would the period further change if the length ofsuspension wire is reduced to one-third ?
Solution. The period of oscillation of the inertia table is given by
To = 2π Ic
where I is the moment of inertia of the table about the axis of suspension and c the torsionalcontant of the suspension wire. Let I′ be the moment of inertia of the ring placed symmetricallyon the table. The new period is
T1 = 2π I + I I + II
T0′ = ′
c
Now I = 12
M and I M) (0.8 ) M2 2r r r2 10 6 4′ = =( .
∴ T1 =0 5 6 4
0 5
2 2
2. .
.M M
Mr r
r+
; T T0 0=
695
Now let c′ be the torsional constant of the new wire. The period of oscillation becomes
T2 = 2π I + I T1′
′=
′c
cc
Mechanical Properties of Matter 435
Now crl
cr
lcc
= ′ =′
=πη πη4 4
2 2 313
and so that ( / )
∴ T2 =13
6915
=T T1 0
Q. 22. A steel wire of 1.0 mm radius is bent in the form of a circular arc of radius 50cm. Calculate the bending moment and the maximum stress. (Y = 2.0 × 1012 dyne/cm2)
Solution. The bending moment is YIR
where I is the geometrical moment of inertia of the cross-section of the wire and R is theradius of circular arc
For a wire of radius r, I =π ρr4
43 14 0 1
4= . × ( . cm)
and = 50 cm4
∴ Bending moment, YIR
=( . × ) × . × ( . )
×2 0 10 3 14 0 1
4 50
12 4
= 3.14 × 106 dyne-cm
In bending of wire, the longitudinal stress on the cross-section of a filament at a distanceZ from the neutral surface is
YR
Z
This is maximum when Z is maximum i.e., when Z = r
∴ Maximum stress =YR
dyne/cm cm)50 cm
2r = ( . × ) × ( .2 0 10 0 112
= 4.0 × 106 dyne/cm2
Q. 23. A cantilever of length 0.5 meter has a depression of 15 mm at its free end.Calculate the depression at a distance of 0.3 meter from the fixed end.
Solution. The depression of the cantilever at a distance x from its fixed end is givenby
Y =W
Iylx x2 3
2 6−
=W
Ixy
lx2
2 3− ...(i)
and that at the free end (x = l) is
δ =W
YIl3
3...(ii)
436 Mechanics
where l is the length of the cantilever, W is the load and Y is young’s modulus of thematerial. I is the geometrical moment of inertia. From eqs. (i) and (ii) we write
yδ
=x l x
l
2
32 3
3/ ( / )
/−
or y =3 3
2
2
3x l x
l( / )− δ
Here l = 0.5 meter, x = 0.3 meter and δ = 15 mm = 15 × 10–3 meter
∴ y =3 0 3 0 5 0 3
315 10
2 0 5
2 3
3
× ( . ) × . . × ( × )
× ( . )
−
−
= 6.48 × 10–3 meter
Q. 24. One end of a 1.0 mts rod is clamped rigiditly. When a 0.5 kg load is suspendedat the other end, it is depressed by 0.5 cm. Calculate the depression at 40 cm distance fromthe fixed end when a 5 kg load is suspended at
(i) free end,
(ii) 40 cm from the free end.
Solution. The depression of the rod at a distance x from its fixed end is
y =W
YIx
l x2
23( / )− ...(i)
where W is the load suspended at the free end and l is the length of the rod.
At the free end (x = l), the depression is
δ =W
YIl3
3...(ii)
Here l = 100 cm, W = 0.5 kg-wt = 500 g dyne, δ = 0.5 cm
0.5 =500 100
3
3gYI× ( )
...(iii)
(i) Now, when a 5 kg-wt is suspended at the free end (W = 500 g), the depressions ata distance of 40 cm (x = 40 cm) is, by eq. (i) given by
y =5000
2100 40
3g × (40)YI
2× −
=5000 × (40) × 260/3
2YI
2g
Dividing this eq. by eq. (iii), we get
y0 5.
= 5000 2602 2000
g × (40)g × (100)
2
3×
×
= 2.08
Mechanical Properties of Matter 437
∴ y = 2.08 × 0.5
= 1.04 cm
(ii) When 5 kg-wt is suspended at 40 cm from the fixed end, then
x = l = 40 cm
Then by eq. (ii) we have,
y =5000 40
3
3gYI× ( )
Dividing by eq. (iii) we get
y0 5.
=5000500
0 64 g × (40)
g × (100) cm
3
3 = .
∴ y = 0.64 × 0.5 = 0.32 cm.
Q. 25. The end of the given strip cantilever depresses 10 mm under a certain load.Calculate the depression under the same load for another cantilever of same material, 2 timesin length, 2 times in width and 3 times in thickness (vertical).
Solution. The depression δ at the end of a cantilever is
δ =W
YIl3
3
For a rectangular cantilever, I = bd3/12
∴ δ =4 3
3W
Yl
bd
The depression for another cantilever of the same material (Y same) but length 2l, width2d and thickness 3d would be
δ′ =4
2 3
3
3W (2
Yl
b d)
( ) ( )
=427
427
10δ = × ( mm)
= 1.48 mm
Q. 26. The rectangular cross-section of a long rod (cantilever) has sides in the ratio1:2. Calculate the ratio of depressions under the same load when
(i) the smaller side is vertical,
(ii) the longer side is vertical.
Solution. The depression at the free end of a cantilever of rectangular cross-sectionunder a load is
δ =4 3
3W
Yl
bd
Where l is length, b is width and d is thickness of the cantilever rod. Let a and 2a bethe sides (ratio 1:2) of the cross-section.
438 Mechanics
(i) when the smaller side is vertical, b = 2a, d = a
∴ δ1 = 42
12
43
3
3
4W
YW
Yl
a al
a( ) ( )=
(ii) when the longer side is vertical, b = a, d = 2a
∴ δ2 = 42
18
43
3
3
4W
YW
Yl
a al
a( ) ( )=
Thus,δδ
1
2=
1 / 21 / 8
= 41
or δ1 : δ2 = 4 : 1
Q. 27. If a cross-section of a beam of rectangular with breadth b and thickness d and ifthe depression of the free end for the given load are δ1 and δ2 respectively when b and d are
vertical, show that δδ
1
2
2
2db
.=
Q. 28. For the same cross-section areas, show that beam of a square cross-section is stifferthan one of circular cross-section of the same length and material, show also that for a givenload the depressions are in the ratio 3:π .
Solution. The depression δ at the loaded end of the beam whose other end is clampedhorizontally is given by
δ =W
YIl3
3
Where W is load, l is length of beam, I is geometrical moment of inertia of its section.
For a rectangular beam,
I =bd3
12where b and d are breadth and depth of its section. If the beam has a square section b = dthen
I =b4
12Therefore, if δ1 be the depression for this beam, we have
δ1 =W
Y
WY
l
b
lb
3
4
3
4
312
4
= ...(i)
For a beam of circular cross-section of radius r, we have
I =πr4
4
Therefore, if δ2 be the depression for this beam, we have
Mechanical Properties of Matter 439
δ2 =W
Y
WY
l
r
lr
3
4
3
4
34
43π π
= ...(ii)
Dividing eq. (i) and (ii) we get,
δδ
1
2=
3 4
4πrb
But the cross-section areas of two beam are equal
∴ b2 = πr2
∴δδ
1
2=
3 34
2 2π
π πr
r( )=
which is less than 1
∴ δ1 < δ2.
Thus the depression at the loaded end of the beam of square cross-section is smallerthan that of the beam of circular cross-section. Hence the former is stiffer than the later.
Q. 29. Compare loads required to produce equal depressions for two beams of samematerial, length and weight when one has a square cross-section and the other has a circularcross-section. [Ans. π:3]
Q. 30. A solid cylindrical rod of radius 6 mm bends by 8 mm under a certain load.Calculate the bending if the rod is replaced by a hollow cylindrical rod of the same material,mass and length but of internal radius 8 mm.
Solution. If r1 and r2 be the internal and external radii of a same hollow cylinder ofthe same material (density ρ), same length l and same mass m as of a solid cylinder of radiusr, then we have,
m = π ρ π ρ( )r r l r l22
12 2− =
or r r22
12− = r2
Here r = 6 mm, r1 = 8 mm so that
r22 = r r2
12+
= (6)2 + (8)2 = 100
∴ r2 = 10 mm
Let δ1 and δ2 be the depression for the solid and hollow cylinders. Then
δ1 =W
YIW
Y
l l
r
3 3
433
4
=
π
and δ2 =W
Y
l
r r
3
24
143
4π ( )−
440 Mechanics
∴δδ
2
1=
rr r r r
4
22
12
22
12( ) ( )+ −
Here r = 6 mm, r1 = 8 mm and r2 = 10 mm
δδ
2
1=
(6)4
( ) ( ) ( ) ( )10 8 10 82 2 2 2+ −
=36 36
164 36941
××
=
But δ1 = 8 mm
∴ δ2 =941
mm.× .8 1 76=
This shows the longer stiffness of the hollow rod.
Q. 31. A uniform bar ABCD is supported horizontally at B and C, here AB12
BC = CD = l.=
If two equal masses m, m are suspended at A and D, the bar rises by height h in the middle.Deduce an expression for h in term of relevant quantities.
Solution. The bar ABCD is placed horizontally upon two knife-edges at B and C. Onsuspending weigths mg, mg at A and D, the bar bends. The bent position is shown in thefigure. The reaction at each knife edge is also mg. The couple cutting at each knife edge isalso mg.
hm g m g
CB
m g m g
A D
O
E
F
Fig. 26
The couple acting at each end is
T = mgl where l = AB = CD.
The bar bend into a circular arc of radius R (say). Let us complete the circle. EF is theelevation h. Now by the property of circle, we have
EF × FG = BF2
that is h (2ρ – h) = (2l)2 [ BF = 1/2 BC = 2l]
solving and neglecting smaller terms, we have
h =2 2lR
...(i)
Mechanical Properties of Matter 441
Now when the bar is bent, a restoring couple YI/R is developed where Y is Young’smodulus and I is geometrical moment of inertia of the cross-section. This is equal andopposite to the external couple τ. Thus,
YIR
= τ = mgl or R YI=mgl
putting this value in eq. (i), we obtain
h =2 3mgl
YI
This is required expression.
Q. 32. A 10 cm wide and 0.2 mm thick metal steel is bent to form a cylinder of 10 cmlength and 50 cm radius. If the young’s modulus of the metal is 1.5 × 1012 dyne/cm2, calculate
(i) the stress and strain on the convex surface,
(ii) bending moment.
Solution. The cylindrical metal sheet is equivalent to a beam for which breadth b = 10cm, thickness d = 0.02 cm and radius R = 50 cm.
(i) When a beam is bent, the strain in a filament at a distance Z from the neutral surfaceis Z/R. For the convex surface,
Z =d2
0 01= . cm.
∴ strain =ZR
= = −0 0150
2 10 4. ×
Now stress = Y × strain
= ( . × ) × ( × )1 5 10 2 1012 4−
= 3 × 108 dyne/cm2
(ii) The bending moment is YIR
where I is geometrical moment of inertia which is bd3
12
bending moment =YR
bd3
12
=1 5 10
5010 0 02
12
12 3. ××
× ( . )
= 2 × 105 dyne-cm.
Q. 33. A rectangular bar is 2 cm in breadth and 1 cm in depth and 100 cm in length issupported at its ends and a load of 2 kg is applied at its mid-point. Calculate the depressionif the young’s modulus of the material of the bar is 20 × 1011 dyne/cm2.
Solution. The depression at middle point of the beam supported horizontally ontwo-knife-edges distant l apart and loaded in the middle by a weight W is given by
δ =W
YIl3
48
442 Mechanics
where I is geometrical moment of inertia of cross-section. For a rectangular beam of widthb and thickness d, we have
I =bd3
12
∴ δ =WY
lbd
3
34
Here W = 2000 × 980 dyne, l = 100 cm, b = 2 cm, d = 1 cm and Y = 20 × 1011 dyne/cm2
∴ δ =( × ) ( )
( × ) ( ) ( )2000 980 100
4 20 10 2 1
3
11 3
= 0.1225 cm.
Q. 34. A bar 1 meter long, 5 mm square in sections, supported horizontally at its endsloaded at the middle is depressed 1.96 mm by a load of 100 gm. Calculate young’s modulusfor the material of the bar (g = 980 cm/sec2). [Ans. 2.0 × 1012 dyne/cm2]
Q. 35. A wire of length 2.0 meter and cross-section 1 × 10–6 m2 is stretched in horizontalposition between two supports with force 200 N. If mid point of the wire is pulled through 5mm, find change in the tension. (Y = 2.2 × 1011 N/m2).
Solution. On being pulled, new length of wire = 2 (AC)
= 2 1 5 102 3 2+ −( × )
= 2 1 12
25 10 6+
−× × binomially
= 2 + 25 × 10–6 m.
∴ Change in length of wire AB, ∆l = 25 × 10–6 m
If ∆F is extra force exerted in clamps,
Y =∆∆F/A
/l l
or 2.2 × 1011 =∆F
10 6− −××2
25 10 6
or ∆F =2 2 10 25 10 10
2
11 6 6. × × × ×− −
= 2.75 Newton.
Q. 36. Find the greatest length of steel wire that can hang vertically without breaking.Breaking stress for steel 7.8 × 109 dyne/cm2.
Density of steel = 7.9 gm/sec
Solution. Let the maximum length be ‘l’ and cross-sectional area of wire be A, then itsweight,
Mg = lA dg, ∴ =stress = MgA
l dg
Mechanical Properties of Matter 443
For wire not to break, ldg = Breaking stress
∴ l =7 8 107 9 981
1 01 109
6. ×. ×
. ×= cm
Q. 37. A light rod 2.0 meter long is suspended from ceiling horizontally by means of twovertical wires of equal length tied to ends. One is of steel having cross-section 10–3 m2 andother is of brass with cross-section 2 × 10–3 m2. Find the position along the rod at which aweight may be hung to produce
(i) Equal stress in both wires,
(ii) Equal strain in both wires.
Given Ysteel = 2 × 1011 N/m2, Ybrass = 1011 N/m2
Solution. (I) If T1 and T2 are tensions in steel and brass wires. When weight W is hungat distance x from steel wire, for equal stress
Stee l Brass
x
T1 T2
Fig. 27
T1
10 3− =T2
2 10 3× −
i.e.,TT
1
2=
12
...(i)
also for equilibrium of rod,
T1x = T2 (2 – x)
orTT
1
2=
2 − xx
...(ii)
From (i) and (ii),12
=2 − x
xor x = 1.33 m
(II) If equal strain in two wire is desired
TY
1
steel
/ 10 3−
=T
Y2
brass
/ ×2 10 3−
444 Mechanics
orT1
2 1011× =T2
2 1011× i.e., TT
1
2= 1
∴ For equilibrium of rod if weight is suspended at distance y from steel wire
T1y = T2 (2 – y)
or 1 =2 − y
y
or y = 1.0 m.
Q. 38. By how much a rubber string of length 10 m increases in length under its ownweight ?
Density of rubber = 1.5 × 103 kg/m3
Y for rubber = 5 × 108 N/m2.
Solution. If A is cross-sectional area, weight of rubber string (i.e., elongating) force,
F = Aldg
∴ Normal stress =FA
= lgd
If ∆l is elongation in rubber string, this elongation will take places in only upper halflength since the weight of string acts from C.G. of string.
∴ Tensile strain =∆ll/2
Hence, Y =F/A∆ll/2
=ldg
ll l dg
l∆ ∆.
2 2
2=
∴ ∆ l =l dg2 3
8210 10 1 5 10 9 8
2 5 10Y= × × . × × .
× ×
= 1.47 mm
Q. 39. A gold wire 0.32 mm in diameter elongates by 1mm when subjected to force 330gm-wt and twists through/radian when a couple 145 dyne × cm is applied. Find Poisson’s ratiofor gold.
Solution. We have Y =M L L
× (0.016)2 2gr lπ π
= 330 9800 1
× ×× .
Also η =2
4L.τ
π θr as =
L
4τ πη θr
2
= 2L × 145(0.016)
L4π π× ( . )1
2900 016 4=
Mechanical Properties of Matter 445
Using the relation, Y = 2η (1 + σ), i.e., ση
= −Y2
1
we get σ =330 980
0 10 016
2 2901
4×× .
× ( . )×
L(0.016) L2π
π
−
= 1.428 – 1 = 0.428
Q. 40. A wire 1.0 m long and 0.5 mm in radius elongates 0.32 mm when 5 kg load isapplied. It is twisted through 0.4 radian under a torque 3 × 10–3 Nm. Determine elasticcoefficient for material of wire.
Solution. Y =M Lg
r lπ 2
=5 9 81 1
0 5 10 0 32 103 2 3× . ×
( . × ) . ×π − −
= 49 05 103 14 0 25 32
2 109
11. ×. × . ×
×= N/m2
Also τ =πη θr4
2L
∴ η =2 1 3 10
3 14 0 5 10 0 4
3
3 4× × ×
. × ( . × ) × .
−
−
= 7.6 × 1010 N/m2
Using relation, Y = 2 12
1η σ ση
( )+ ⇒ = −Y
or σ =2 10
2 7 6 101 0 31
11
10×
× . ×.− =
Since Y = 3K (1 – 2σ)
or K =2 10
3 1 2 0 311 75 10
1111×
( × . ). ×
−= N/m2
∴ K = 1.75 × 1011 N/m2.
Q. 41. One end of a wire 50 cm in length and 1 mm in radius is twisted through 45° atone end relative to the other. Calculate the angle of shear on the surface.
Solution. If θ is the angle of twist at free end of a rod of length l and radius r, the angleof shear is given by
φ =rlθ
Putting the given values
φ = 0 1 4550
0 09. ×
.° = °
446 Mechanics
Q. 42. If a solid cylinder of length l and radius r is clamped at its middle section, whatwill be the couple acting on the clamp due to the cylinder, when the two ends are twistedaround the axis through equal angles
(i) In same direction.
(ii) In opposite directions.
Solution. For any twist of the couple acting on each half length of cylinder will be
τ =πη θ πη θr
lrl
4 4
2 2( / )=
(i) When the two ends are twisted in same direction, the couple in the two halves willhave additive effect on the clamps and so total couple exerted at the clamp will be
= 22 4
τ π η θ= rl
(ii) When the two ends of cylinder are twisted in opposite directions, the two will haveneutralizing effect at the clamp and so net couple exerted at the clamp will be 0.
Q. 43. Two cylinders of length L and 2L and radii 2R and R respectively are weldedcoaxially. One end being fixed, the other is twisted by 110°. Calculate the twist of the thickerrod.
Solution. Let θ be the twist at welded point as shown in Fig. 28. Since in equilibriumcondition, torque on the two cylinders will be same.
2LL
θ
110°
Fig. 28
110 – θ =τπη
. 2LR4 (for thinner cylinder)
and θ =τ
πηLR)4(2
(for thicker)
Dividing we get
110 − θθ
= 32
i.e., θ = 3.33°.
Q. 44. A solid cylinder of radius 3 cm is melted and recast into hollow cylinder of radius5 cm while length remains unchanged. By what ratio torsional rigidity changes in doing so?
Solution. If r is inner radius of hollow cylinder but mass and length remaining same
π ρ× ( )3 2 L = π ρ( )52 2− r L.
i.e., r2 = 52 – 32 or r = 4 cm.
Now for solid cylinder, torsional rigidity, CS = πη πη( )
.3
281
2
4
l l=
Mechanical Properties of Matter 447
and for hollow cylinder, torsional rigidity CH = πη πη( )5 4
2369
2
4 4− =l l
∴CC
H
S=
36981
= 419
i.e., torsional rigidity of hollow cylinder will be419
times that of solid.
Q. 45. A thin-walled tube of mean radius 10 cm and thickness 0.05 cm is melted to forma rod of same length. Compare torsional rigidity in the two cases.
Solution. Since mean radius of tube is 10 cm and thickness of wall is 0.05 cm, the outer
radius will be r2 =10 0 052
10 025+ =. . and inner radius r1 =10 0 052
9 475− =. . cm. If tube is
melted to form a rod of radius r, we have
πr ld2 = π r r ld22
12−
i.e., r2 = ( . ) ( . )10 025 9 9752 2−
or r = 20 × 0.05 = 1 cm
If CH and CS are torsional rigidity
CC
H
S=
πη
πη
r rl
r l
24
14
4
2
2
−
( / )
=( . ) ( . )
( )10 025 9 975
1
4 4
4−
=2001
Q. 46. Two cylinderical steel rods have the same mass per unit length. One is solid whileother which is hollow and has external radius twice the internal. Compare their torsionalrigidity and show that maximum strains produced by equal twisting couple also bear the sameratio.
Solution. Let rs be radius of solid rod and 2r, r be external and internal radii of hollowrod. Then mass per unit length being same,
πrs2d = π ( ) ( )2 2 2r r d−
i.e., rS2 = 3r2
The ratio of torsional rigidities is
CC
S
H=
πηπη
r l
r r l
rr
S4
S4/
( ) ( )
2
2 2 159
154 4 4−= =
448 Mechanics
Also for equal torques
πηθ
rlS4
S2=
πηθ
( ) ( )2
2
4 4r r
l
−H
i.e., rS4
Sθ = 15 4r θH
orθθ
H
S=
rr
S4
159
154 =
Since for any given cylinder φ α θ ∴ φφ
H
S= 9
15
Q. 47. Calculate the work done in twisting a steel wire 0.25 m long and 2 mm thickthrough 45° (η = 8 × 1010 N/m2).
Solution. The work done in twisting the wire through of radian is given by
W =12
12
2
4
2cr
lθ
πηθ2 =
As given, η = 8 × 1010 N/m2
r = 1 × 10–3 m, l = 0.25 m
θ = π/4 radian
Substituting W =12
3 14 8 10 102 0 25
3 144
10 3 4 2
× . × × × ( )× .
.−
= 0.154 joule.
Q. 48. A steel bar of length 50 cm, breadth 2 cm and thickness 1 cm is bent into an arcof radius 2.0 meter. Determine,
(i) Stress and strain at convex surface,
(ii) Bending moment (Y = 2 × 1012 dyne/cm2).
Solution. (i) The tensile strain at distance Z from neutral surface
=ZR
= = −1 2200
2 5 10 3/ . ×
and so stress = Y × strain
= 2 × 1012 × 2.5 × 10–3 = 5 × 109 dyne/cm2
(ii) Bending moment =YI
RY
Rg bd= =( / ) × × ×
×
3 12 312 2 10 2 112 200
=16
10 1 66 1010 9× . ×= dyne × cm
Q. 49. A circular bar is supported horizontally on two horizontal knife edges 0.4 cm apartand projects by equal amounts beyond knife edges. The total length of bar is 0.6 meter andits radius is 0.4 cm. When a load 5 kg is suspended from each end, the centre rises by 0.5cm. Find the value of Y for material of bar.
Mechanical Properties of Matter 449
Solution. The situation is same as if beam of length, l = 0.4 cm supported horizontallyis loaded in the middle by 10 kg, producing depression δ = 0.5 cm.
Hence we must have
δ = MY Igl
g
3
48 .
Putting the values, Mg = 10 × 9.8 N, l = 0.4 m
Ig =π δ( . × )0 4 10
4
2 4−− and = 0.5 × 10 m2
we get Y =10 9 8 0 4 4
48 0 4 10 0 5 10
3
2 4 2× . × ( . ) ×
× ( . × ) × . ×π − − = 13 × 1010 N/m2.
Q. 50. A rod AD consisting of three segments AB, BC and CD hangs vertically from afixed support at A as shown in figure 29. Cross section of rod is uniform equal to 10 – 4 m2
and a weight 10 kg is hung from D. Calculate displacements of points B, C and D given,
YAB = 2.5 × 1010 N/m2
YBC = 4 × 1010 N/m2
YCD = 1.0 × 1010 N/m2
Solution. The elongation of a wire given by I = MAY
gl
Since the same load 10 kg will be effective on all thethree section of rod,
IAB =10 9 8 0 1
10 2 5 103 92 104 10
6× . × .× . ×
. ×−−= m
IBC =10 9 8 0 210 4 10
4 9 104 106× . × .
× ×. ×−
−= m
ICD =10 9 8 0 15
10 1 0 1014 7 104 10
6× . × .× . ×
. ×−−= m
Obviously the displacement of point B is = 3.92 × 10–6 m
the displacement of point C is = 3.92 × 10–6 + 4.9 × 10–6 m
= 8.82 × 10–6 m
the displacement of point D is = 14.7 × 10–6 + 8.82 × 10–6 m
= 23.52 × 10–6 m
Q. 51. Two different rod of length l1 and l2 but of same cross-section A are placed betweentwo massive walls. The first rod has young’s modulus Y1 and coeff. of linear expansion α1.The corresponding values for second rod are Y2 and α2. If temperature of both rods isincreased by T°, find the force with which rods act on each other.
Solution. The thermal expansion in first rod due to increase in temp T° will be
∆l1 = l1 α1 T
A
B
C
D
10 kg
0 .15
0 .2
0 .1
Fig. 29
450 Mechanics
and increase in length of second rod
∆l2 = l2 α2 T
∴ Total expansion in length of composite rods
= ∆l1 + ∆l2
= T (l1 α1 + l2α2)
Since expansion cannot take place hence it is neutralized byelastic compressive forces, so we can write
T (l l1 1 2 2α α+ ) =FY A
FY A
l l1
1
2
2+
(where F is force exerted by one rod at other)
i.e., F ×Y A Y Al l1
1
2
2+
= T l l1 1 2 2α α+
or F =A T
Y Yl l
l l1 1 2 2
1 1 2 2
α α++/ /
Q. 52. Two rods of equal cross-section A, one of copper and other of steel, are joined toform a composite rod of length 2.0 m at 20°C. The length of copper rod is 0.5 m. When tempis raised to 120°C, the composite rod is increases to 2.002 m. If the composite rod is fixedbetween two rigid walls and not allowed to expand, the length of two components do notchange. Find young’s modulus and coeff. of linear expansion for steel, given YC = 1.3 × 1013
N/m2 αC = 1.6 × 10–5/°C.
Solution. The increase in length of Cu rod is
(∆l)cm = l tCu Cu. .α ∆= 0.5 × 1.6 × 10–5 × (120 – 20)
= 0.8 × 10–3 m
If αS is expansion coeff. for steel
(∆l)S = 1.5 × αS × (120 – 20) = 150 αS
But total increase in length of the composite rod is 0.002 m
∴ 150 αS + 8 × 10–4 = 20 × 10–4
or αS = 0.8 × 10–5/°C
Since, stress in steel rod = Stress in Cu rod
∴ YS S.α ∆t = YCu Cu. .α ∆t
∴ YS = YCu Cu
S
.αα
=1 3 10 1 6 10
0 8 102 6 10
13 5
513. × × . ×
. ×. ×
−
− = N/m2
Q. 53. Two rods of different materials have same length and cross-section and are joinedto make a rod of length 2.0 m. The metal of one rod has young’s modulus 3 × 1010 N/m2 andcoeff. of thermal expansion 10–5/°C. The corresponding values for other metal are
Mechanical Properties of Matter 451
1010 N/m2 and 2 × 10–5/°C. How much pressure must be applied to prevent the expansion ofrod when temperature is raised by 100°C ?
Solution. The increase in length of two rods due to increase in temperature will be
∆l1 = l t1 15 31 10 100 10α ∆ = =− −× × m
∆l2 = l t2 25 31 2 10 100 2 10α ∆ = =− −× × × × m
∆l =FYA
l ∴ F = YA ∆l
lIt means Tension developed in two rods would be
F1 =3 10 1010 3× × − A
1 newton
and F2 =10 2 1010 3× × − A
1 newton
Hence total pressure to be applied to prevent expansion must be
F = F1 + F2
= 5 × 107 A Newton
∴ Pressure =Force
A= 5 × 107 N/m2.
Q. 54. A solid sphere of radius R made of a material of bulk modulus B is surroundedby a liquid in a cylindrical container. A massless piston of area A floats on the surface of liquid.Find the fractional change in radius of the sphere, when a mass M is placed on the piston tocompress the liquid.
Solution. The volume of spherical body is
v =43
R3π
∴ dv = 4πR2 dR
Hencedvv
= 3 dRR
...(i)
The Bulk Modulus to given by B = − ∴ =vddv
dvv
dP PB
Since dpg= M
A, so M
A.Bdvv
g=
Substituting this value in (i) we get,
dRR
=MA.B
g3
Q. 55. A body of mass 3.14 kg is suspended from one end of a wire of length 10 meter.The radius of wire changes from 9.8 × 10–14 m at one end to 5 × 10–4 m at the other. Find
452 Mechanics
the change in length of wire. What will be the change in length if ends are interchanged.(Y = 2 × 10–11 N/m2)
Solution. Consider an element of length dx atdistance x from fixed end of wire from definition, thechange in length of this element will be
dl =F.YA
L =dxdx( )∴
Hence, A = πr2
= π (a + x tan θ)2
So, dl =F
Ydxxπ α θ( tan )+ 2
The total change in length of wire is therefore;
∆l = FY
L
π θ.( tan )
dxa xa + 2
Putting a x+ tanθ = t, dx =dt
tanθ
∴ ∆L = (tan
at dt
a+ L tan )
F.Y
Lθ
π θ
2
= 1 1 1− = −
F
Y tanF
Yπ θ π θ. tant a ba
b
from fig. tan θ =b a−
L
So, ∆L =F.L
Yπ ab ...(i)
Substituting the values given
∆L =3 14 9 8 10
9 8 10 2 10 5 10104 11 4
3. × . ×( . × ) ( × ) ( × )π − −
−= m
If ends of wire are interchanged, values of a and b will replaced each other. Hence, fromeq. (i), ∆l will remain the same.
Q. 56. A gold wire, 0.32 mm in diameter, elongates by 1 mm when stretched by a forceof 330 gm.wt and twist through 1 radian, when equal and opposite torques of 145 × 10–7 n.mare applied at its ends. Find the value of Poisson’s ratio for gold.
Solution. When a wire of radius r and length L is elongated by l with a force F, theyoung’s modulus of its material is given by
Y =F L
πr l2 .
dxL
α
b
x
Fig. 30
Mechanical Properties of Matter 453
Here, F = mg = 0.33 × 9.81 newton, r = 1.6 × 10–4 m, l = 10–3 m
∴ Y =0 33 9 81
102 3. × . ×
) ×L
× (1.6 × 10n/m4
2
π − −
The torque applied to twist the wire by 1 radian, is given by
C = ηπ r4 2/ L = 145 × 10 n.m7−
where η =145 10 27
4 4× ×
)
−
−L
× (1.6 × 10π
∴ Pisson’s ratio σ =Y2
1η
−
=0 330 9 81
1 6 10 10 2 145 21
4
4 2 3. × . × )× ( . × ) × × × ×
L × × (1.6 × 10L
4ππ
−
− − −
= 1.429 – 1 = 0.429
Q. 57. A circular rod of young’s modulus 2.04 × 106 kg. wt/sq cm. Poisson’s ratio 0.4,length 1 meter, and cross-section 0.95 sq.mm is stretched by a weight of 10 kg. Find theextension and dimension in cross-section.
Solution. When a wire of length L and cross-section A is elongated through l by a forceF, the young’s modulus of its material is given by
Y =F × LA × l
⇒ l =F × LY A×
...(i)
=10 9 81 1
2 04 10 9 81 10 0 95 106 4 6× . ×
. × × . × × . × −
= 5.16 × 10–4 m
Poisson’s ratio is given by σ =Lateral strain
Longitudinal strain L= δr r
l/
/
where r is radius of wire and δr the diminuation produced in it
∴ δr = σ × ×lr
L...(ii)
cross-section area of wire A = πr2 = 0.95 × 10–6 sq.m
∴ r = 0 95 10 6. × /− π= 5.499 × 10–4 m
Poisson’s ratio for material σ = 0.4.
Substituting the values of various quantities in (ii), we get
δr =0 4 5 16 10
15 499 10
44. × . ×
× . ×−
−
454 Mechanics
⇒ δr = 11.35 × 10–8 m
A = πr2 ∴ δA = 2π r dr
where δA is the diminuation in cross-section A,
∴ δA = 2 5 499 10 11 35 104 8× × . × × . ×π − −
= 3.92 × 10–10 sq/m.
Q. 58. A circular bar one meter long and 8 mm diameter is rigidly clamped at one endin a vertical position. A couple of magnitude 2.5 n.m is applied at the other end. As a resulta mirror fixed at this end deflects a spot of light by 15 cm on a scale one meter away. Calculatethe modulus of rigidity of the bar.
Solution. When the mirrors turn through an angle θ, the reflected ray turns throughan angle 2θ. As the spot of light is reflected through 15 cm on a scale 100 cm away, then
tan 2θ = 0.15/1
⇒ 2θ = 0.15 if θ is small
∴ θ = 0.075 radian.
The mirror is fixed at the lower end of the bar, therefore its lower end is deflected by0.075 radian. Now, to twist the bar by 0.075 radian, the couple required is
Cθ =ηπr
l
4
20 075× . n/m
Here l = 1 m, r = 4 × 10–3 m and Cθ = 2.5 n/m
∴ηπ × ( × ) × .
×4 10 0 075
2
3 4−
l= 2.5
or η =2 5 2 1
3 14 4 10 0 0758 29 103 4
10. × ×. × ( × ) × .
. × .− = n/m2
Q. 59. An iron wire of 1 mm radius 100 cm length is twisted through 18°. Where isshearing stress is maximum ? Find the maximum angle of shear. If modulus of rigidity of ironis 8 × 1010 n/m2, what is torsional couple?
Solution. η = shearing stressshearing strain
∴ shearing stress = ηφHence the shearing stress is maximum when φ is maximum. But φ is maximum at the
surface of the wire, the reason is given below:
The strain at distance x (radially) will be φ = xθ/l. It has maximum values, when x = r,the radius of the wire, i.e.,
φmax = rθ/l, at the surface of wire
Here r = 10–3, θ = 18° and l = 1 m
∴ φmax = 10–3 × 18°/1 = 0.018°
Torsional couple Cθ =ηπr
l
4
2
Mechanical Properties of Matter 455
=8 10 3 14 10
2 118
180
10 3 4× × . × ( )×
×−
= π
= 3.8 × 10–2 n.m.
Q. 60. Two cylinders A and B of radius r and 2r are soldered coaxially. The free end ofA is clamped and the the free end of B is twisted by an angle θ. Calculate twist at the junction,taking the material of the cylinders to be the same and lengths equal.
Solution. Let τ be the couple applied at the free end of B and θ′ the twist at thejunction. The couple T produces a relative twist θ′ between the ends of the cylinder A andrelative twist( )θ θ− ′ between the ends of the cylinder B. Thus, if η be the modulus of rigidityof the cylinder and length l of each cylinder, we have
τ =πη θ πη θ θr
lr
l
4 4
22
2′ = − ′( ) ( )
∴ θ′ = 16 (θ – θ′)
⇒ θ′ =1617
θ .
Q. 61. Two solid cylinders of same materials having length l, 2l and radii r, 2r respectivelyare joined coaxially. Under a couple applied between the free ends the shorter cylinder showsa twist of 30°. Calculate the twist of longer cylinder.
Solution. Torque, τ =πηθ πη θ1
4 4
22
2 2r
lrl
= ( )( )
or θ1 = 8θ
∴ θ =θ1
8308
3 75= = °.
Q. 62. A thin walled cylinder tube of mean radius 10 cm and thickness 0.05 cm is meltedup and recast into solid rod of the same length. Compare the torsional rigidities in two cases.
Solution. Mean radius of the hollow tube = 10 × 10–2 m
Its thickness = 0.05 × 10–2 m
External radius r1 = 10 + 0.05/2 cm = 10.025 × 10–2 m
Internal radius r2 = 10 – 0.05/2 cm = 9.975 × 10–2 m
Now if r be the radius of recast solid rod having the same length and same mass as thehollow one, then
π ρ( )r r l12
22− = π ρr l2
where l is length of rod and ρ is density of material
∴ r r12
22− = r2
or r2 = ( . × ) ( . × )10 025 10 9 975 102 2 2 2− −−
= 0.05 × 20 × 10–4 = 10–4
⇒ r = 10–2 m
456 Mechanics
Torsional rigidity of hollow rod C1 = ηπ ( )/r r l14
24 2−
Torsional rigidity of solid rod C2 = ηπr l4 2/
∴CC
1
2=
r rr
14
24
4
2 2
210 025 9 975
1−
= +( . ) ( . )( )
∴CC
1
2=
2001
Q. 63. A body suspended symmetrically from the lower end of a wire, 100 cm long and1.22 mm in diameter, oscillator about the wire as axis with a period of 1.25 sec. If the modulusof rigidity of the material of the wire 8.0 × 1010 n/m2. Calculate the M.I. of the body aboutthe axis of rotation.
Solution. T = 24 2π
πI / C I = CT2
∴ ...(i)
Where I = moment of inertia of body about the axis of rotation and C = ηπr l4 2/
Here, η = 8 × 1010 n/m2, l = 1m, r = 0.61 × 10–2 m
∴ C =8 10 3 14 0 61 1010 2 4× × . × ( . × )−
2 × 1n.m
Time period T = 1.25 sec.
Substituting the values of C and T in (i), we get,
I =8 10 3 14 0 61 10
2 11 25
4 3 14
10 4 4 2
2× × . × ( . × )
×× ( . )
× ( . )
−
= 8.889 × 10–4 kg-m2.
Q. 64. A wire of 4 mts long, 0.3 mm in diameter is stretched by a force of 8000 gm wt.If the extension in the length amounts to 1.5 mm, calculate the energy stored in the wire.
Solution. The work done in stretching the wire through under a longitudinal force Fis given by
W = F.AYL
dlldl
o
l
o
l =
=AY
LFl l
2
212
=
Here F = 8000 gm-wt = 8 kg wt = 8 × 9.8 n; l = 1.5 × 10–3
∴ work done in the wire =12
9 8 8 1 5 10 3× . × × . × −
=5.88 × 10–3 Joules.
Q. 65. Find the amount of work done in twisting a steel wire of radius 1 mm and length
Mechanical Properties of Matter 457
Solution. The work done in twisting a wire from zero to θ angle is given by
W = C C 2θ θ θθ
do = 1
2
Here C =πηr
l
4 10 3 4
28 10 3 14 10
2 0 25=
−× × . × ( )× .
and θ = 45 × π/180 = π/4 radian.
∴ Work done W =12
C 2θ
=12
8 10 3 14 102 0 25 4
10 3 4 2
× × × . × ( )× .
×−
π
= 0.1547 Joule.
Q. 66. A steel wire of 1 mm radius is bent to form circle of 10 cm radius. What is thebending moment and the maximum stress, if Y = 2 × 1011 n-m–2.
Solution. Bending moment =YIR
Here I = πr4 3 44 3 14 10 4/ . × ( ) /= −
∴ Bending moment = 2 10 3 14 104 0 1
1 5711 3 4× × . × ( )
× ..
−= n-m
stress =YR
Z
For maximum stress the value of Z should be maximum, i.e., Z = r
∴ Maximum stress =YR
n-m 2r = =−
−2 10 100 1
2 1011 3
9× × ( ).
×
Q. 67. A cantilever of length 50 cm is depressed by 15 mm at the loaded end. Calculatethe depression at distance 30 cm from the fixed end.
Solution. Depression y =M
Ygx
ll x
2
23( / )− ...(i)
where x = 0.5 m, Y = 0.015 m
0.015 = MY
gl
( . ).
.0 52
0 50 53
2−
...(ii)
when x = 0.3 m, y = ?
y = MY
gl
( . ).
.0 32
0 50 33
2−
Dividing eq. (ii) by eq. (i) we get,
458 Mechanics
Y0 015.
=( . )( . )
×..
0 30 5
1 21 0
2
2
⇒ Y = 6 48 10 3. × − m
Q. 68. Cantilever of length l and uniform cross-section shows a depression of 2 cm at theloaded end. What will be the depression at distances l/4, l/2 and 3l/4 from the fixed end?
Solution. y =M
YIg x
lx2
2 3−
when x = l, y = 0.02 m
0.02 =M
YIg l
ll2
2 3− =
MYIgl3
3
When x = l/4, y1 =MYIg l l l
2 4 12
2 −
=M
YIM
YIgl gl3 3
321112
11128 3
. .=
=11
1280 02 1 72 10 3× . . ×= − m
When x = l/2, y2 = MYIg l l l
2 2 6
2 −
=M
YIM
YIgl gl3 3
856
516 3
. ×=
=5
160 02 6 2 10 3× . . ×= − m
when x = 3l/4, y3 =MYIg l l l
234 4
2 −
=M
YIgl3
22764
. = M
YIgl3
381
128.
=81
1280 02 1 27 10 2× . . ×= − m.
Q. 69. A 10 cm wide and 0.2 mm thick metal sheet is bent to form a cylinder of 10 cmlength and 50 cm radius. If the young’s modulus of the metal is 1.5 × 1011 n-m–2, calculate
(i) the stress and strain on the convex surface and,
(ii) the bending moment.
Solution. (i) Strain =ZR
= =−
−2 102
10 5
2 104
4××
.×
Mechanical Properties of Matter 459
( for convex surface Z = half thickness of the sheet)
stress = Y × strain
= 1 5 10 2 1011 4. × × × −
= 3.0 × 107 n-m–2
(ii) Bending moment =YIR
YR
= bd3
12
=1 5 10
0 50 1 2 10
12
11 4 3. ×.
×. × ( × )−
= 2 × 10–2 n-m
Q. 70. A light metal rod of length 60 cm and of radius 1 cm is clamped at one end loadedat the free end with 5.5 kg. Calculate the depression of free end assuming Y = 9 × 1011 dyneand g = 980 cm/sec2.
Solution. When a beam of length l is fixed at one end and loaded by weight W at thefree end, the depression is given by
δ =W
YIl3
3For a circular cross-section of radius r, I = πr4/4
∴ δ =43
3
4WY
lrπ
Here W = 5.5 × 103 × 980 dynes, l = 60 cm, Y = 9 × 1011 dyne/cm2 and r = 1 cm
Therefore, δ =4 5 5 10 980 603 9 10 3 14 1
3 3
11 4× . × × ×( )× × × . × ( )
= 0.5495 cm.
Q. 71. A vertical rod of circular cross section of radius 1 cm is rigidily fixed in the earth.Its upper end is 3mts from the earth. A thick string which can stand a maximum tension of2 kg is tied at the upper end of the rod and pulled horizontally. Find how much the rod willbe deflected before the string snaps. (Y for steel = 2 × 1012 C.G.S. units, g = 1000 cm/sec2).
Solution. The lower end of a vertical rod is fixedand the upper end is pulled by a horizontal force, hencethis is the case of cantiliver.
The deflection of the upper end, is given by
δ =W
YIl3
3Here the horizontal force, W = 2 × 1000 × 1000 dynes/cm2,
length of the rod l = 300 cm; value of Y = 2 × 1012 dyne/cm2;
radius of the rod r = 1 cm
For circular cross-section I =π π πr4 4
41
4 4= =× ( )
∴ δ =2 1000 1000 300 4
3 2 10 3 1411 46
3
12× × × ( ) ×
× × × ..= cm
BW
A
Fig. 31
460 Mechanics
Q. 72. Compare loads required to produce equal depression for two beams, made of thesame material and having the same length and weight with the only difference that while onehas circular cross-section, the cross-section of the other is square.
Solution. The depression δ at the middle point of a beam of length l of circular cross-section of radius r, is given by
δ =WY1l
r
3
412. π...(i)
where W1 is the weight loaded.
In the case of the rectangular cross-section beam the weight loaded by W2, then thedepressions δ′, is given by
δ′ =W
Y2l
bd
3
34
Two beams have same length and are made of same material
But for square cross-section, b = d
∴ δ′ =W
Y2l
b
3
44...(ii)
The two beam have equal weights and made of same material,
∴ Their masses and hence volumes must be equal, i.e.,
πr2l = b2l or b2 = πr2
substituting the value of b2 in (ii), we get
δ′ = WY
2 lr
3
2 44π...(iii)
But depression is same in both cases, i.e., δ = δ′,∴ equating (i) and (iii) we get,
WY1 l
r
3
412 π= W
Y2lr
3
2 44π
orWW
1
2=
3π
which is the required ratio.
Q. 73. A cantilever of length l is loaded at a point x1, calculate the depression of the tipof the cantilever. If the same load is shifted to the tip, what is the depression of the point x1?
Solution. The expression for the depression y at distance x from fixed end, is given by
y = WYI
lx x2 3
2 6−
Also at x,dydx
= tanθ = −
WYI Z
bxx2
Mechanical Properties of Matter 461
In the first case the cantilever is loaded at OP = x1. Hence the beam will not bendbeyond x1 by the load W1 but the portion P′y will be target at P′ and the total depress δ =xy will be given by
δ = xx x y′ + ′P
= PP P′ + ′ ′x tanθ
Hence PP =W
YI′ x1
3
3, P′x′ = l – x1
and tan θ =WYI
WYI
xx x
12 1
212
2 2−
=
∴ δ =W
YIW
YIx
l xx1
3
112
3 2+ −( )
=WYI
lx x12
13
2 6−
=W
YIx
lxl1
2
2 3−
This is the depression of tip of the cantiliver.
In the 2nd case, the same load is shifted to the tip, then the depression at any point x1is given by
δ′ =WYI YI
WYI
lx x xl
x12
13
12
1
6 2 3−
= − −
We see that in two cases the depression is same.
Q. 74. A lath of width 2 cm and thickness 3 mm supported horizontally on two knifeedges 80 cm apart is loaded with the weight of 10 gm, hang its ends, which project 15 cmbeyond the knife edges. If the centre of the lath is thereby elevated 2 mm, calculate young’smodulus for material.
Solution. Let the beam be placed on two knife edges P and Q. It takes the shape asshown in fig., when loaded at both ends by 0.01 g newton. Therefore, according to staticallaws, the reactions at each knife edge will be 0.01 g of newton.
∴ Couple acting at P = 0.01 g × 0.15 n-m = YIR
Here I =bd3
12
=0 02 3 10
124 5 10
3 310. × ( × ) . ×
−−=
Consider the lath is bent into a circle of radius ofcurvature R, then from the geometry of the wide,
PO.OQ = (2R – OO′) OO′
θ
Px
hs
l
O
Fig. 32
C
RR
O 0.4m
0.01g0.01g
0.01g 0.01g
O ′
0.4mQP
462 Mechanics
or 2 × 10–3 × 2R = 0.4 × 0.4[ 2R >> OO′]Whence R = 40 m
Now,YIR
=Y × . ×
. × . × .4 5 10
400 01 9 81 0 15
11−=
⇒ Y =4 0 0 01 9 81 0 15
4 5 101 308 1010
10. × . × . × .. ×
. ×− = n - m2
Q. 75. A steel strip is clamped horizontally from one end. On applying a 500 gm load atthe free end the bending in equilibrium state is 5 cm. Calculate (i) the potential energy of strip,(ii) the frequency of vibration, if the load is disturbed from equilibrium (neglect mass of stripitself).
Solution. (i) The bending is 5 cm by the load 500 gm = 0.05 × 9.81 newton
Hence force per unit bending or force constant is, C = 0 5 9 81
0 0598 1 1. × .
..= −n - m
∴ Potential energy = C Cx dx x0
212
π =
=12
9 81 0 05 0 122× . × ( . ) .= Joules.
(ii) Frequency n =1
2πCM
=1
29 810 5
2 23π
..
.= −sec (nearly)1
Q. 76. A solid cylinder rod of radius 6 mm bends by 8 mm under a certain load. Deducethe bending when all other things remaining the same the beam is replaced by a hollowcylindrical one with external radius 10 mm and internal radius 8 mm.
Solution. If δ and δ′ are the depressions of two beams, then,
δ =M
YI and =
MYI
II
gl gl3 3
3 3δ δ δ′
′′ =
′;
For solid beam I = πr4 4/
For hollow beam I′ =π ( )r r2
414
4−
∴ δ′ =6
10 88 10 1 76 10
4
4 43 3
−=− −× × . × m.
10.1 MOLECULAR FORCES
There are two kinds of molecular forces:
(i) Force of adhesion or adhesive force and
(ii) Force of Cohesion or Cohesive force.
The force of attraction between the molecules of different substances is called force ofadhesion, while that between molecules of the same substance is called force of cohesion.
The force of cohesion is entirely different from ordinary gravitational force and does notobey the ordinary inverse square law. Probably, this force varies inversely as the eight powerof the distance between two molecules and therefore falls of rapidly with distance and becomesalmost negligible at a distance of 10–7 cm.
The maximum distance upon which the force of cohesion between two molecules can actis called the molecular range. It is of the order of 10–7 cm being different for differentsubstances. The sphere, drawn with the molecule as center, having radius equal to molecularrange, is called sphere of influence.
Free Surface of a Liquid Tends to Contract to a Smallest Possible AreaThe free surface of a liquid always behaves like a stretched elastic membrane and hence
has a natural tendency to assume as small an area as possible. This peculiar property of aliquid can be illustrated by the following experiments:
(a) Let us tie loosely a closed loop of cotton thread between two points of a wire frameand dip it in a soap solution. A thin film of soap solutions is formed on the frameand the loop of thread is placed on the film in an irregular manner Fig (i). If nowthe film inside the loop is broken by a needle, the thread immediately assumes acircular form Fig. (ii). This shows that as soon as the film inside the loop is broken,the film outside the loop contracts, thus pulling the thread into a circle. We knowthat for a given perimeter the circle has the largest area. This means that the filmoutside the loop contracts to a smallest area.
(b) A freely suspended drop of water, formed at the end of a tap, assumes sphericalshape. We know that for a given volume, sphere has the least surface area. Therefore,to acquire minimum possible surface area, the drop takes the spherical shape.
(c) If we place a greased needle on a piece of blotting paper and put the paper lightlyon the surface of water, the blotting paper will soon sink to the bottom, but the
463
10
464 Mechanics
needle is found to float on the surface although it is much heavier than water. Onexamining the surface of water below the needle, it is seen to be slightly depressedas if the liquid surface were acting as a stretched membrane, so that the weightof the needle is balanced by the inclined upward tension forces in the membrane.If one end of the needle is pressed to pierce a hole in the surface of water, it goesslantingly downward.
T T
Need le
m g
Fig. 1
The above experiments clearly show that the surface of a liquid behaves as though itwere covered with stretched elastic membrane, having a natural tendency to contract. Butthere is one important difference between the two that in an elastic membrane the tensionincreases with stretching or its surface area is increased, in accordance with Hook’s law,whereas the tension in the liquid surface is quite independent of stretching or of the areaof its surface, unless the stretching force reduces the thickness of the film below 5 × 10–6 mmafter which the tension diminishes rapidly.
( )i ( )ii
Fig. 2
10.2 DEFINITION OF SURFACE TENSION
Let an imaginary line AB be drawn in any direction ina liquid surface. The surface on either side of this line exertsa pulling force on the surface on the other side. This forcelies in the plane of the surface and is at right angles to theline AB. The magnitude of this force per unit length of ABis taken as measure of the surface tension of the liquid.Thus if F be the total force acting on either side of the lineAB of length l, then the surface tension is given by
T =Fl
If l = 1 then T = F. Hence the surface tension of a liquid is defined as the force per unitlength in the plane of the liquid surface, acting at right angles on either side of an imaginaryline drawn in that surface. Its unit is ‘newton/meter’.
The value of the surface tension of a liquid depends on the temperature of the liquid,as well as on the medium on the other side of the surface. It decreases with rise in temperatureand becomes zero at the critical temperature.
B
A
Fig. 3
Fluids 465
10.3 EXPLANATION OF SURFACE TENSION
The tendency of a liquid surface to contract to a minimum possible area has suggestedthat it is in a state of tension, which we call as surface tension. This property of liquid surfacehas been explained on the basis of molecular theory given by Laplace.
Consider three molecules A, B and C of a liquid, withtheir spheres of influence drawn around them. Since the sphereof influence of A lies completely inside the liquid, it is equallyattracted in all directions by molecules within its sphere ofinfluence. Hence there is no resultant cohesive force acting onit.
In the case of molecule B, the sphere of influencelies partly outside the liquid. The number of molecules in theupper half, attracting the molecule B in the upward direction is less than the number ofmolecules in the lower half attracting it in the downward direction. Hence a resultantdownward force is acting on it.
The sphere of influence of molecule C is exactly half outside and half inside the liquid.Hence this molecule is attracted in the downward direction with maximum force, perpendicularto the surface.
Now, if we draw a plane RS parallel to free surface PQ of the liquid at a distance equalto the molecular range, hence the layer of the liquid between the planes PQ and RS is termedas the surface film. Clearly, what is true for molecules B and C is true for all other moleculesin the surface film. Hence all the molecules in the surface film are pulled downwards dueto the resultant cohesive force, the magnitude of which increases from surface RS to freesurface PQ.
Now, if a molecule from the interior of the liquid is brought up to the surface film, workhas to be done against the downward cohesive force on it and its potential energy increase.Therefore the potential energy of the molecules in the surface film is greater than thepotential energy of the molecules lying below it. A system is equilibrium has a tendency toacquire lowest possible potential energy. Thus to decrease the potential energy of the moleculesin the surface film, the film tries to have minimum number of molecules in it. This can bedone only by decreasing the surface area of the film, because its thickness is already fixed(being equal to molecular range). This is the reason, why the free surface of a liquid alwaystends to have minimum possible area.
10.4 SURFACE ENERGYWhen the surface area of a liquid is increased, the molecules from the interior rise to
the surface. This requires work against the force of attraction of the molecules just belowthe surface. This work is stored in the form of potential energy in the newly formed surface.Besides this, there is cooling due to the increase in the surface area. Therefore, heat flowsinto the surface from the surroundings to keep its temperature constant and is added to itsenergy. Thus the molecules in the surface have some additional energy due to their position.This additional energy per unit area of the surface is called ‘surface energy’.
10.5 RELATION BETWEEN SURFACE TENSION AND WORK DONE ININCREASING THE SURFACE AREA
Let a liquid film be formed between a bent wire ABC and a straight wire PQ which canslide on the bent wire without friction. As the film surface tends to contract, the wire PQ
R SQP
BC
A
Fig. 4
466 Mechanics
moves upward. To keep PQ in equilibrium, a uniform force F (which includes the weight ofthe wire) has to be applied in the downward direction. It is found that the force F is directlyproportional to the length l of the film in contact with the wire PQ. Since there are two freesurfaces of the film, we have
F ∝ 2l
or F = T × 2l
Where T is a constant called ‘surface tension’ of the liquid.Now, suppose the wire is moved downward through a smalldistance ∆x. This results in an increase in the surface areaof the film. The work done by the force F (= Force ×distance) is given by
W = F × ∆x = T × 2l × ∆x
But 2l × ∆x is the total increase in area of both the surfaces of the film. Let it be ∆A.Then
W = T × ∆A
or T = W/∆A
If ∆A = 1, then T = W. Thus the work done in increasing the surface area by unity willbe equal to the surface tension T. Hence, the surface tension of a liquid is equal to the workrequired to increase the surface area of the liquid film by unity at constant temperature.Therefore, surface tension may also be expressed in ‘joule/meter2’.
10.6 SHAPE OF LIQUID MENISCUS IN A GLASS TUBE
When a liquid is brought in contact with a solid surface, the surface of the liquid becomescurved near the place of contact. The nature of the curvature (concave or convex) dependsupon the relative magnitudes of the cohesive force between the liquid molecules and theadhesive force between the molecules of the liquid and those of the solid.
In figure (i), water is shown to be in contact with the wall of a glass tube. Let us considera molecule A on the water surface near the glass. This molecule is acted upon by two forcesof attraction:
(i) The resultant adhesive force P, which acts on A due to the attraction of glassmolecules near A. Its direction is perpendicular to the surface of the glass.
(ii) The resultant cohesive force Q, which acts on A due to the attraction of neighboringwater molecules. It acts towards the interior of water.
The adhesive force between water molecules and glass molecules is greater than thecohesive force between the molecule of water. Hence, the force P is greater than the forceQ, their resultant R will be directed outward from water (figure i).
In figure (ii), mercury is shown to be incontact with the wall of a glass tube. Thecohesive force between the molecules ofmercury is far greater than the adhesive forcebetween the mercury molecules and the glassmolecules. Hence, in this case, the force Q willbe much greater than the force P and theirresultant R will be directed towards the interiorof mercury.
A
R
P
Water
G lass( )i
A
R
P
G lass( )ii
M ercury
Q
Fig. 6
2Tl
B
P Q
CFA
∆x
Fig. 5
Fluids 467
The resultant force R acts on all the molecules on the surface of water or mercury. Forthe molecules more and more away from the wall, the adhesive force P goes on decreasingwhile the cohesive force Q becomes more and more vertical. Consequently, the resultant Ralso becomes more and more vertical. In the middle of the surface, P becomes zero and Qbecomes vertical. Hence the resultant R becomes exactly vertical (Figure iii & iv).
If the surface of the liquid is in equilibrium, the resultant force acting on any moleculein the surface must be perpendicular to the surface. Hence the liquid surface sets itselfperpendicular to the resultant force everywhere. This is why the water surface assumes aconcave shape while the mercury surface assumes a convex shape in a glass tube. In eithercase the resultant force in the middle is vertical and the surface there is horizontal.
10.7 ANGLE OF CONTACT
When the free surface of a liquid comes in contact of a solid, it becomes curved near theplace of contact. The angle inside the liquid between the tangent to the solid surface and thetangent to the liquid surface at the point of contact is called the ‘angle of contact’ for thatpair of solid and liquid.
G lass
M ercury
G lass
θθ
Water
Water M ercury
( )i( )ii
( )iii ( )iv
Fig. 7
The angle of contact for those liquids which wet the solid is acute. It is zero for purewater and clean glass for ordinary water and glass it is about 8o. The liquid which do not wetthe solid have obtuse angle of contact. For mercury and glass the angle of contact is 135o.The angle of contact for water and silver is 90o. Hence in a silver vessel the surface of waterat the edges also remains horizontal.
When a liquid is kept in container, a liquid molecule like A, touching the wall experiencesthree forces:
(i) Weight mg, acting downwards.
468 Mechanics
(ii) Cohesive force Q due to liquid molecules, acting inwards at 45°.
(iii) Adhesive force P between liquid & solid molecule, acting horizontally outwards.
If cohesive force is less than adhesive force, net forces (P–Q cos 45°) and (mg + Q sin45°) act on liquid molecule A. So that resultant force R is directed outwards as shown in figure(a). The meniscus then takes concave shape as in the case of water.
If cohesive force is greater than the adhesive force, (Q cos 45° – P) and (mg + Q sin 45°)act as shown in figure (b) so that the resultant of these will be directed inwards giving convexshape to the meniscus as shown in figure (b). This is the case of mercury.
Water M ercury
45°Q
θH
P
m g + Q sin 45°
A
(P–Q cos 45°)
m g + Q sin 45°
P
Q
( )a ( )b
Fig. 8
10.8 CAPILLARY ACTION
When a capillary is dipped a little in liquid theliquid rises or falls down a little. This action is calledcapillarity. Liquids having concave meniscus are foundto rise and those like mercury which have convexmeniscus are found to be depressed. This is because inconcave meniscus, the vertical component T cos θ ofsurface tension force acts vertically upward per unitlength, all along the circumference, which makes liquidto rise in capillary till it is balanced by the weight ofliquid column in capillary. Thus for equilibrium,
2πr T cos θ = ( )π πr h dg r dg2 313
+
= πr hr
dg2
3+
or T =rdg
hr
2 3cosθ+
.
Which gives the rise as,
h =2
3Tcos θrdg
r−
where d is the density of water.
When meniscus is convex, the vertical component T cos θ acts vertically downwards andliquid is depressed inside capillary through depth given by above formula.
θθ T c
os θ
T c
os θ
θθT sin θ T sin θ
h
r
Fig. 9
Fluids 469
10.9 RISING OF LIQUID IN A CAPILLARY TUBE OF INSUFFICIENT LENGTH
If radius of capillary is negligibly small the rise of liquid is given by
h =2T cos θ
rdg ; neglecting r/3
which gives h × r =2T cos θ
dg .
The quantities on right hand side are constants for a liquid-solid contact and soh × r = h′ × r′ = ...... constant. It suggests that if capillary has sufficient length, liquid willrise up to the required height and radius of meniscus will be same as that of capillary. Butif length of capillary is insufficient, liquid rises upto available height h and radius of meniscusis increased so as to keep product (h′ × r′) same as (h × r). In this condition liquid does notoverflow.
h
r
r
h h′<
r′′
Fig. 10
10.10 EFFECTS ON SURFACE TENSION
Surface tension is affected by the following factors:
(i) Effect of contamination: If the water surface has dust, grease or oil, the surfacetension of water reduces.
(ii) Effect of Solute: If the solute is very soluble then the surface tension of liquidincreases, as by dissolving salt in water the surface tension increases. If the soluteis less soluble then the surface tension decreases, as by pouring soap or phenol thesurface tension of water decreases.
(iii) Effect of Temperature: Rise of temperature, decreases inter molecular, force(called cohesive force) and so the surface tension in general, decreases astemperature is increased. It becomes zero at about 6o to 8o below critical temperatureof liquids.
The most reliable relation showing variation of surface tension is the Etvos-Ramseyshield formula.
TpV23 = To( )θ θc d− −
where
T = Surface tension at θoC,
To = Surface tension at 0oC,
θc = Critical temperature of liquid,
470 Mechanics
V = Specific volume,
d = Density of liquid.
It is obvious that at temperature θ = θc – d, then the surface tension T will be zero.
10.11 IDEAL LIQUIDAn ideal liquid is one which has the following two properties:(i) Zero Compressibility: An ideal liquid is incompressible, that is, on pressing the
liquid there is no change in its volume (or density). Most of the liquids may beconsidered approximately incompressible, because on pressing them the change intheir volume is negligible. For example, on pressing water by one atmosphericpressure its volume changes only by a fraction of 0.000048.
(ii) Zero Viscosity: An ideal liquid is non-viscous, that is, when there is a relativemotion between different layers of the liquid then there is no tangential frictionalforce in between the layers. In actual practice, however, there is some viscosity inall liquids (and gases). It is less in gases, but larger in liquids.
10.12 STEADY OR STREAM LINE FLOWThe flow of a fluid is said to be steady. Orderly, streamline or laminar, if the velocity
at every point in the fluid remains constant (in magnitude as well as direction), the energyneeded to drive the fluid being used up in overcoming the ‘Viscous drag’ between its layers.In other words, in a steady or streamline flow, each infinitesimally small volume element ofthe fluid, which we may call a particle of the fluid, follows exactly the same path and hasexactly the same velocity as its predecessor, i.e., its velocity does not change with time.
Thus, if a fluid flows along a path ABC, where A, B and C are points inside the fluid,each new particle arriving at A will always have the same velocity VA in a direction tangentialto the curve ABC at A, and the pressure and density will always be the same there. Similarly,the particle arriving at B will always have the same velocity VB which may or may not bethe same as VA. And the same is true for the point C.
This line ABC along which the particles of the fluid move, one after the other, with theirvelocities constant at the various points on it and directed along the tangents to these points,is called a stream line.
A
B
CvC
vB
vA
Fig. 11
The fluid-flow, however, remains steady or stream line if its velocity does not exceed alimiting value, called its criticial velocity, beyond which the flow loses all steadiness ororderlines and becomes zigzag and sinuous, acquiring what is called turbulence.
X Y
Fig. 12
Fluids 471
Stream-lined TubeIf we imagine a tube made up of a large number of stream-lines, then this imaginary tube
is called stream-lined tube. In figure 12 a stream-lined tube XY is shown. The boundary wallsof the stream-lined tube are everywhere parallel to the direction of flow of the liquid. Therefore,the liquid can neither go out nor come in through the walls of the tube. The liquid that entersat one end X leaves at the other end Y, as if it were flowing through an actual tube.
10.13 EQUATION OF CONTINUITY OF FLOWThe equation of continuity is a fundamental equation of fluid-flow and is a special case
of the general physics law of conservation of matter. For an incompressible fluid it may bededuced as follows:
Imagine the fluid to be flowing through a pipe AB, with a1 and a2 as its cross-sectionalareas at sections A and B.
Then, although the streamlines are all parallel to the axis of the pipe and perpendicularto its cross-section at every point, the velocity of flow is not necessarily the same at allsections of pipe. Let it be v1 at section A and v2 at section B.
Consider an infinitely narrow tube of flow, shown dotted in the figure 13 of cross-sectional areas da1 and da2 at sections A and B respectively. Then, if the fluid coversdistances dS1 and dS2 in time δt at the two ends respectively and if ρ1 and ρ2 be the densitiesof the fluid at A and B, we have
mass of fluid entering the tube of flow per unit time at section A = da1ds1P1/δt = da1v1P1and mass of fluid leaving the tube of flow per unit time at section B = da2ds2P2/δt = da2v2P2it being assumed that there is no appreciable change in the cross-sections of the tube atsections A and B over the small distances dS1 and dS2 respectively.
∴ Mass of fluid entering the whole section A of the pipe per unit time i.e.,
mass rate of fluid-flow at A = da v a va
1 1 1 1
0
1
P P1 1=and mass of fluid leaving the whole section B of the pipe per unit time;
or mass rate of fluid-flow at B = da v a va
2 2 2 2
0
2
P P2 2=Since the fluid is incompressible, ρ1 = ρ2 = ρ, say, and since there can be no flow across
the sides of a tube of flow and there is no ‘source’ or ‘sink’ in between the sections A andB to add more fluid or to drain away some of it on its way from A to B, we have, from thelaw of conservation of matter,
Mass of fluid entering section A per second = mass of fluid leaving section B per second.
or a1v1ρ = a2v2ρi.e., a1v1 = a2v2
v1
da 1
a 1 dS1
A
dS2
B
da2
v2
a 2
Fig. 13
472 Mechanics
This is called the equation of continuity and clearly means that the quantity av remainsconstant throughout the fluid-flow if it be steady or stream line. av is the volume of the fluidflowing across any cross-section of the pipe in unit time. It is therefore, called the volumerate of flow, or, simply, the rate of flow or the rate of discharge (or volume flux) and maybe denoted by the letter V or Q, the latter being the more appropriate symbol, used inEngineering practice.
Thus, the equation of continuity tells us that the volume rate of flow of an incompressiblefluid (i.e., a liquid) remains constant throughout the flow, and so does it mass rate of flow (a vρ).
Since a v = constant, it follows that v α 1/a i.e., the velocity of the fluid-flow at anysection of the pipe is inversely proportional to the cross-section of the pipe at the section.
10.14 ENERGY OF THE FLUID
A fluid in steady or streamline flow may posses any or all of the three types of energy,viz., (i) kinetic energy because of its inertia, (ii) potential energy because of its positionrelative to the earth’s surface and (iii) Pressure energy because of its pressure; for, if workbe done on it against its pressure, it can do the same amount of work back for us and thusacquires energy.
Let us obtain the value of each type of energy per unit mass and per unit volume of thefluid.
(i) Kinetic EnergyIf m be the mass of a fluid, of density ρ, flowing with velocity v, its kinetic energy is
clearly ½ mv2.
Kinetic energy per unit mass of the fluid = v2/2
and, since mass of unit volume of the fluid = its density ρ, we have
Kinetic energy per unit volume of the fluid = 12
v2 ρ.
(ii) Potential energyIf we have a mass m of the fluid at a height h above the earth’s surface, its potential
energy = mgh and, therefore,
potential energy per unit mass of the fluid = gh
and potential energy per unit volume of the fluid = gh ρ = ρ gh
(iii) Pressure EnergySuppose we have a tube of uniform area of cross-section a, containing an incompressible
non-viscous fluid, with its hydrostatic pressure equal to p. If we introduce an additional massδm of the fluid into the tube against this pressure so gradually as not to impart any velocity,and hence any kinetic energy, to it and if the mass occupies a length δl of the tube, clearly,
Work done on the mass = force × distance = (pa) × δl.
Since the work done on mass δm of the fluid forms its pressure energy, we have
pressure energy per unit mass of the fluid
= ( ) ×palmδ
δ
Fluids 473
δm = Volume × density = aδl × ρSo that, pressure energy per unit mass of the fluid
= pal
a lpδ
δ ρ ρ=
∴ Pressure energy per unit volume of the fluid
=p
pρ
ρ× ,= its pressure
10.15 BERNOULLI’S THEOREM
This theorem states that the total energy of an incompressible, non-viscous fluid insteady flow remains constant throughout the flow. The theorem may be easily deduced asfollows:
Consider an infinitesimally small portion AB of a tube of flow of length dl, and in viewof its small length, let its area of cross-section be assumed to be uniform and equal to da.
If ρ be the density of the fluid (supposed to be incompressible); its mass in portion AB= m = dadlρ.
Its weight mg = dadlρg, therefore, acts vertically downwards at its centre of gravity O,making an angle θ with the direction of flow, as shown. So that its component mg cos θ =dadlρg cos θ acts in the direction of flow, with its other component mg sin θ = dadlg sin θacting perpendicularly to the wall of the tube, its effect being offset by the lateral thrust dueto the adjoining tube of flow.
θ
A
BO
0, v
p pl d l
+ δδ
mg s in θm g = dad l ρg
dl
da m g cos θp
Fig. 14
If p be the pressure on face A of the tube, that on face B will be ppl
dl+ δδ
, in the
direction shown. And, therefore force acting on face A in the forward direction = pda and force
acting on face B in the backward direction = ppl
dl da+
δδ
.
Hence, net force acting on mass dadlρ of the fluid in the tube, say
F = pda ppl
dl da da dl g− +
+δ
δρ θ. . cos
or F = − +δδ
ρ θpl
dl da dadl g( ) . cos ...(i)
474 Mechanics
If v be the velocity of the fluid as it enters the tube at A, we have its acceleration =dv/dt, where dt is the time taken by the fluid to cover the length dl of the tube. So that,force acting on mass dadlρ of the fluid in the tube is also
= mass × acceleration
= da dldvdt
. . ×ρ ...(ii)
Now, taking the most general case where the velocity, being a function of both distanceand time, may change from point to point or from moment to moment, we have
dv =δδ
δδ
vl
dlvt
dt+ and
∴dvdt
=δδ
δδ
δδ
δδ
vl
dldt
vt
vvl
vt
+ = +
Substituting this value of dv/dt in expression (ii), above, we have
F = dadlρδδ
δδ
vvl
vt
+
and cos ,θ δδ
= − hl
where h is the vertical height of tube AB from a choosen datum plane.
Relation (i) above thus becomes
da dl vvl
vt
. .ρ δδ
δδ
+
= − −δ
δρ δ
δpl
dl da da dlhl
g. . .
or ρ δδ
δδ
vvl
vt
+
= − −δ
δρ δ
δpl
ghl
...(iii)
This is an important equation, applicable to both steady and unsteady fluid-flow and isreferred to as Euler’s equation.
In the case of steady or streamline flow, δδvt
= 0 and the other partial derivatives all
become total derivatives. So that, relation (iii) becomes
ρvdv + dp + ρgdh = 0
or vdv + (dp/ρ) + gdh = 0
Integrating along the streamline, therefore, we have
12
2vdp
gh+ + ρ= constant.
This is called Bernoulli’s equation which, in the case of the fluid being incompressible,and, therefore, ρ, a constant, takes the familiar form
12
2vp
gh+ +ρ = Constant ...(iv)
or, if we consider unit volume of the fluid, i.e., a mass ρ of the fluid (because mass of unitvolume is the density ρ), we have on multiplying relation (i) by ρ,
Fluids 475
½ ρv2 + p + ρgh = Constant ...(v)
This is another form of Bernoulli’s equation in terms of pressure, every term on the lefthand side has the dimensions of pressure.
Now, the term (p + ρ gh) represents the pressure of the fluid even if it be at rest. Itis, therefore, called static pressure of the fluid. And, the term ½ ρv2 represents the pressureof the fluid in virtue of its velocity v, and is, therefore, called dynamic pressure of the fluid.Thus we have,
Static pressure + dynamic pressure = constant.
Again, on dividing relation (iv) by g, we have
12
2v gpg
h/ + +ρ
= Constant ...(vi)
Which is yet another form of Bernoulli’s equation in terms of lengths or heads, as theyare called. For, as can be easily seen, each term on the left hand side has the dimensionsof length or a height and hence called a ‘head’. Thus ½ v2/g, is the velocity head, p/ρg, thepressure head, h, the gravitational head. So that, we have
Velocity head + pressure head + gravitational head = total head = constant.
10.16 VELOCITY OF EFFLUX
Let a vessel be filled with a liquid upto a height H and let there be an orifice at a depthh below the free surface of the liquid. The pressure at the free surface of the liquid and alsoat the orifice in atmosphere, and so there will be no effect of atmospheric pressure on theflow of liquid from the orifice. The liquid on the surface has no kinetic energy, but onlypotential energy, while the liquid coming out of the orifice has both the kinetic and potentialenergies.
Applying Bernoulli’s theorem, we have total energy per unit mass of the liquid at thesurface = K.E. + P.E.+ pressure energy
= 0 + gh + 0 = gh
and total energy per unit mass of the liquid at the orifice
= ½ v2 + 0 + 0 = ½ v2
H
A
B
h
h ′
O
R
Fig. 15
Since total energy of the liquid must remain constant in steady flow, in accordance withBernoulli’s theorem, we have
½ v2 = gh whence v2 = 2gh or
476 Mechanics
velocity of efflux = 2gh
This formula was first established in 1644 by Torricelli and is called ‘Torricellip theorem’.
If a body is dropped freely (u = 0) from a height h, then from the third equation of motionv2 = 2gh, we have
v = 2gh
Clearly, the velocity of the liquid falling from a height h is 2gh . Hence the velocity ofefflux of a liquid from an orifice is equal to that velocity which the liquid acquires in fallingfreely from the free surface of the liquid upto the orifice.
No liquid being free from internal friction or viscosity, however, this ideal velocity isnever attained in practice. The observed velocity is always less than v and is equal toCv 2gh , where Cv is called the coefficient of velocity, its value lying between 0.95 to 0.99 inthe case of water, depending upon the ‘head’ of water (i.e., depth of the orifice from the watersurface) and the shape of the orifice.
After emerging from the orifice the liquid adopts a parabolic path.
If it takes t second in falling through a vertical distance (H-h) then according to equationS = ½ at2, we have
H – h = ½ gt2
∴ t =2( )H − h
g
Since there is no acceleration in the horizontal direction, the horizontal velocity remainsconstant. The horizontal distance covered by the liquid is
x = horizontal velocity × time
= vt
= 2 2gh hg
( )H −
= 2 h h( )H −
This formula shows that whether the orifice in the vessel is at a depth h or at a depth(H-h) from the free surface of the liquid, the emerging liquid will fall at the same distancei.e., the range x of the liquid will remain the same.
Now, h (H – h) will be maximum when h = H – h i.e., h = H/2. Hence the maximumrange of the liquid is given by
xmax = 2 H2
H H2
H −
=× .
Therefore, when the orifice is exactly in the middle of the wall of the vessel, the streamof the liquid will fall at a maximum distance (equal to the height of the liquid in the vessel).
10.17 VELOCITY OF EFFLUX OF A GAS
Since a gas issues out of an orifice in a reservoir under adiabatic conditions, we have,from Bernoulli’s equation for adiabatic flow of a gas,
Fluids 477
12 1
2vp+
−γ
γ ρ =12 10
2 0
0v
p+−γ
γ ρ
where v0, p0 and ρ0 are the velocity, pressure and density of the gas inside the reservoir andv, p and ρ their respective values as the gas issues out of the orifice in the reservoir, so thatv is the velocity of efflux of the gas.
Since, obviously, v0 = 0, we have
12
2v =γ
γ ρ ρ−−
1
0
0
p p
or v2 =2
10
0
γγ ρ ρ−
−
p p
Because under adiabatic conditions, p p0
0ρ ργ γ− , we have
v2 =2
110
0 0
1
γγ ρ ρ
γγ
−−
−
p p
Whence, v the velocity of efflux of the gas, may be easily obtained.
10.18 VISCOSITY
When a solid body slides over another solid body, a frictional force begins to act betweenthem. This force opposes the relative motion of the bodies. Similarly, when a layer of a liquidslides over another layer of the same liquid, a frictional force acts between them whichopposes the relative motion between the layers. This force is called internal frictional-force.
A BZero ve locity Horizon ta l p lane
M axim um ve loc ity
c
b
a
Fig. 16
Suppose a liquid is flowing in stream lined motion on a fixed horizontal surface AB. Thelayer of the liquid which is in contact with the surface is at rest, while the velocity of otherlayers increases with distance from the fixed surface. In the figure 16, the lengths of thearrows represent the increasing velocity of the layers. Thus there is a relative motionbetween adjacent layers of the liquid. Let us consider three parallel layers a, b and c. Theirvelocities are in the increasing order. The layer a tends to retard the layer b, while b tendsto retard c. Thus each layer tends to decrease velocity of the layer above it. Similarly, each
478 Mechanics
layer tends to increase the velocity to the layer below it. This means that in between anytwo layers of the liquid, internal tangential forces act which try to destroy the relative motionbetween the layers. These forces are called ‘viscous forces’. If the flow of the liquid is to bemaintained, an external force must be applied to overcome the dragging viscous forces. Inthe absence of the external force, the viscous forces would soon bring the liquid to rest. Theproperty of the liquid by virtue of which it opposes the relative motion between its adjacentlayers is known as ‘viscosity’.
10.19 FLOW OF LIQUID IN A TUBE: CRITICAL VELOCITY
When a liquid flows in a tube, the viscous forces oppose the flow of the liquid. Hencea pressure difference is applied between the ends of the tube which maintains the flow of theliquid. If all particles of the liquid passing through a particular point in the tube move alongthe same path, the flow of the liquid is called ‘stream-lined flow’. This occurs only when thevelocity of flow of the liquid is below a certain limiting value called critical velocity’. Whenthe velocity of flow exceeds the critical velocity, the flow is no longer stream lined butbecomes turbulent. In this type of flow, the motion of the liquid becomes zig-zag and eddy-currents are developed in it.
Reynold proved that the critical velocity for a liquid flowing in a tube is vc = kη/ρa,Where ρ is density and η is viscosity of the liquid, a is radius of the tube and k is Reynoldnumber (whose value for a narrow tube and for water is about 1000).
When the velocity of flow of the liquid is less than the critical velocity, then the flow ofthe liquid is controlled by the viscosity, the density having no effect on it. But when thevelocity of flow is larger than the critical velocity, then the flow is mainly governed by thedensity, the effect of viscosity becoming less important. It is because of this reason that whena volcano erupts, then the lava coming out of it flows speedly inspite of being very thick (oflarge viscosity).
10.20 VELOCITY GRADIENT AND COEFFICIENT OF VELOCITY
Suppose a liquid is flowing in stream-lined motion on a horizontal surface OX. The liquidlayer in contact with the surface is at rest while the velocity of other layers increases withincreasing distance from the surface OX. The highest layer flows with maximum velocity. Letus consider two parallel layers PQ and RS at distance Z and Z + ∆Z from OX. Let vx and vx+ ∆vx be their velocities in the direction OX. Thus the change in velocity in a perpendiculardistance ∆Z is ∆vx. That is, the rate of change of velocity with distance perpendicular to the
direction of flow is ∆∆vx
Z. This is called velocity gradient.
Now let us consider a liquid layer of area A at a height Z above OX. The layer of theliquid immediately above it tends to accelerate it with a tangential viscous force F, while thelayer immediately below it tends to retard it backward with the same tangential viscous forceF. According to Newton, the viscous force F acting between two layers of a liquid flowing instream-lined motion depends upon two factors:
(i) It is directly proportional to the contact-area A of the layers.
(ii) It is directly proportional to the velocity-gradient ∆∆vx
Z between the layers.
Fluids 479
S v + v→ ∆x x
Q v→ x
∆z
z
O X
F
F
A
z
( )a ( )b
P
R
O X
Fig. 17
Combining both these laws, we have
F α AZ
∆∆vx
F = ±ηA∆∆vx
Z
Where η is a constant called ‘coefficient of viscosity’ of the liquid.
If A = 1 and ∆∆vx
Z = 1, then η = ± F. Thus, the coefficient of viscosity of a liquid is defined
as the viscous force per unit area of contact between two layers having a unit velocitygradient between them.
In the above formula, ± sign indicate that the force F between two layers is a mutualinteraction force.
10.21 POISEUILLE’S EQUATION FOR LIQUID-FLOW THROUGH A NARROWTUBE
In deducing an expression for the rate of flow of a liquid through a narrow tube,poiseuille made the following assumptions:
(i) The liquid-flow is steady or streamline, with the streamlines parallel to the axis ofthe tube.
(ii) Since there is no radial flow, the pressure, in accordance with Bernoulli’s theorem,is constant over any given cross-section of the tube.
(iii) The liquid in contact with the wall of the tube is stationary.
All these assumption are found to be quite valid if the tube be narrow and the velocityof liquid-flow really small.
Let a liquid of coefficient of viscosity η be flowing through a narrow horizontal tube ofradius r and length l and when the condition become steady, let the velocity of flow at allpoints on an imaginary coaxial cylindrical shell of the liquid, of radius x, be v and, thereforethe velocity gradient, dv/dx.
Since the velocity of the liquid in contact with the walls of the tube is zero and goes onincreasing as the axis is approached, where it is the maximum, it is clear that the liquid layerjust inside the imaginary shell is moving faster, and the one just outside it, slower than it.So that, in accordance with Newton’s law of viscous flow, the backward dragging force on theimaginary liquid shell is given by
F = η π ηA Addx
xldvdx
= 2
480 Mechanics
Where A is the surface area of the shell equal to 2πxl. And, if the pressure differenceacross the two ends of the tube be P, the force on the liquid shell, accelerating it forwards= P × πx2.
Where πx2 is the area of cross-section of the shell.
For the liquid-flow to be steady, therefore, we must have driving force equal to backwarddragging force, i.e., P × πx2 = – 2 πxlη(dv/dx), the –ve sign of the dragging force indicatingthat it acts in a direction opposite to that of the driving force. We, therefore, have
dv = –P Pπ
π η ηx dxxl
xdxl
2
2 2= − , which, on integration, gives
v = –P P
2C12 2
2
η ηlxdx
lx = − +.
where C1 is a constant of integration.
Since at x = r, v = 0, we have
0 = –Pr
C2
14ηl+
or C1 =Pr2
4ηl.
∴ Velocity of flow at distance x from the axis of the tube, i.e.,
v = –P P P2 2x
lr
l lr x
4 4 42 2
η η η+ = −( )
Which, incidentally shows at once that the profile or the velocity-distribution curve ofthe advancing liquid is a parabola, the velocity increasing from zero at the walls of the tubeto a maximum at its axis.
Now, if we imagine another coaxial cylindrical shell of the liquid of radius. x + dx,enclosing the shell of radius x, the cross-sectional area between the two is clearly 2πxdx and,therefore, volume of the liquid flowing per second through this area is, say, dQ = 2πxvdx.
Imagining the whole of the liquid inside the tube to consist of such coaxial cylindricalshell, the volume of the liquid flowing through all of them per second i.e., the rate of flowthrough the tube, as a whole, say, Q, is obtained by integrating the expression for dQbetween the limits x = 0 and x = r. We thus have rate of liquid-flow through the arrow tube,i.e.,
Q = 2 20
2 2
0
π πη
xvdx xl
r x dxr r
= −P4
( )
=πη
πη
πη
P2
P2
P2l
x r xl
r rl
rr2 2 4
0
4 4 4
2 4 2 4 4−
= −
= .
or Q =π
ηPr4
8 l
or η = πPrQ
4
8 l.
Fluids 481
10.22 POISEUILLE’S METHOD FOR DETERMINING COEFFICIENT OF VISCOSITY OF A LIQUID
The method is suitable only for liquids having a low viscosity, like water, consist incollecting in a weighed beaker the liquid flowing out in a trickle, in a given time t (in second),from a tall cylinder C through a horizontal capillary tube T of radius r and a known lengthl fitted near its bottom, the liquid head h being kept constant by means of an overflow tubeO, as shown in figure.
C
Liqu id
O
T
To s ink
h
Fig. 18
Dividing the mass of the liquid collected in the beaker by its density ρ and the time t,
its rate of flow Q is obtained. Then, from the relation η π= PQ
4rl8
, where, P = hρg, the value
of η for the liquid can be easily calculated.
There are two important sources of error in above experiments.
(i) The liquid-flow all along the length of the capillary flow-tube (T) is not uniform orstreamline, as assumed. For, the motion of the liquid where it enters the flow-tubeis clearly accelerated and does not become uniform or streamline until after coveringa good length of it. This error is eliminated if the effective length of the flow-tubeis taken to be (l + α) instead of l, where α is almost invariably found to be equalto 1.64 r. The correction (α) is thus quite independent of the length of the tube butdepends upon its radius.
(ii) Part of the pressure difference P between the two ends of the flow-tube is used upin imparting kinetic energy to the liquid flowing through the tube, so that, only theremainder, say P′, is really responsible for overcoming the viscous resistance of theliquid. The effective pressure difference for our purpose is P′, which should,therefore, be substituted for P in the relation for η.
Analogy between Liquid-flow and Current FlowThe rate of flow of a liquid through a capillary tube is given by Poiseuillis equation
Q =π
ηPr4
8 l
If we put 84
ηπ
lr
= R and call it effective viscous resistance, then
Q = P/R
482 Mechanics
Which is clearly a relation similar to I = V/R for flow of elastic current in accordancewith ohm’s law.
A liquid flow through a capillary tube is thus analogous to the flow of elastic currentthrough a conductor, with the rate of liquid-flow, Q, corresponding to current or rate of flowof charge, I, the pressure difference P across the two ends of the capillary, to potential
difference V across the two ends of the conductor and the effective viscous resistance, 8
4η
πl
rto electrical resistance R.
10.23 RATE OF LIQUID-FLOW THROUGH CAPILLARIES IN SERIES
Let two capillaries A and B, of lengths l1 and l2 and radii r1 and r2 respectively beconnected in series, as shown in figure 19 and let a liquid of coefficient of viscosity η flowthrough them is steady or stream line motion.
Since liquids are incompressible, the same volume of liquid that passes through capillaryA, in a given time, also passes through capillary B in the same time.
The rate of liquid-flow through each capillary as also through the combination, as awhole, is the same (like the flow of current through conductors in series). Let it be Q.
Let the pressure difference across the capillaries A and B be p1 and p2 and that acrossthe combination, as a whole, P. So that
P = p1 + p2 ...(i)
If, therefore, R1 and R2 be the effective viscous resistances for capillaries A and B andR, for the combination, as a whole, we have
Rate of liquid flow Q =PR R R1 2
= =p p1 2
Or P = RQ, p1 = R1Q and p2 = R2Q
AB
Fig. 19
Substituting these values in relation (i) above, we have
RQ = R1Q + R2Q
Or R = R1 + R2
i.e., the total effective viscous resistance of the combination is equal to the sum of theeffective viscous resistances for the individual capillaries. Therefore, rate of liquid-flow throughthe combination, i.e.,
Q = PR
PR R1 2
=+
Since R1 = 8 81
14
2
24
ηπ
ηπ
lr
lr
and R2 =
∴ Q = πηP
8lr
lr
1
14
2
24
1
+
−
Fluids 483
10.24 RATE OF LIQUID-FLOW THROUGH CAPILLARIES IN PARALLEL
Let A and B be two capillaries, of lengths l1 and l2 and radii r1 and r2 respectively lyingin the same horizontal plane and in parallel with each other as shown in figure 20. Let aliquid of coefficient of viscosity η be flowing through them in steady or stream-line motion.
The pressure difference across each capillary, as also across the combination as a wholewill be the same, say, P. The rate of liquid-flow through each capillary will, however, bedifferent. Let it be Q1 and Q2 through the two capillaries respectively. Then if Q be the rateof flow through the combination, we have
Q = Q1 + Q2
Now, Q = P/R, Q1 =P/R1 and Q2= P/R2.
where R is the total effective viscous resistance for the combination and R1 and R2, theeffective viscous resistances for capillaries A and B respectively. Substituting these values inrelation (i) we have
PR
=PR
PR1 2
+
or1R
=1
R1
R1 2+
A
l1
B
l2
Fig. 20
Which corresponds to the law of electrical resistances in parallel.
Thus, the reciprocal of the total effective viscous resistance for the combinations is equalto the sum of the reciprocals of the individual effective viscous resistances for the twocapillaries. The same will, of course, be true for any number of them.
Rate of liquid-flow through the combination is given by
Q =PR
P 1R R1 2
= +
1
or since R and R1 2= =8 81
14
2
24
ηπ
ηπ
lr
lr
, we have
∴ Q = Pr P
814π
ηπη
πη8 81
24
2
14
1
24
2lrl
rl
rl
+
= +
.
10.25 POISEUILLE’S FORMULA EXTENDED TO GASES
According to Poiseuille’s formula, the volume of a liquid flowing per second through acapillary is given by
Q =π
ηpr
l
4
8
484 Mechanics
where l and r are the length and radius of the capillary tube and p the difference of pressureacross its ends. In deriving this formula it has been assumed that the volume per secondcrossing each section of the tube is constant. This is true for an incompressible substance andhence fairly true for a liquid. When a gas flows along a tube, however, the volume per secondincreases as the pressure decreases along the tube, and hence the above formula must bemodified to take this into account.
Let us imagine a short length dl of the tube for which the volume per second can beconsidered constant. Let p be the average pressure and dp the difference in pressure betweenits ends. The pressure gradient is –dp/dl, the minus sign indicating that the pressure diminishesas l increases. From Poiseuille’s formula, the volume of the gas flowing per second throughthis length is given by
Q = –π
ηr dp
dl
4
8.
Now, if p1 is the pressure at the inlet end and Q1 is the volume entering the tube persecond then, from Boyle’s Law
p1Q1 = pr
pdpdl
Q = − πη
4
8.
so that p1Q1dl = − πηr
p dp4
8.
Integrating this expression for the entire length of the tube, we get
p dll
1
0
Q1 = − πηr p dp
p
p4
81
2
.
Where p2 is the pressure at the outlet end of the tube. Hence
p1Q1l = − −πηr p p4
22
12
8 2( )
or p1Q1 =π
ηr p p
l
412
22
16( )
.−
This is Poiseuille’s formula for the flow of a gas through a capillary tube.
10.26 DETERMINATION OF VISCOSITY OF A GAS
One method for determining the viscosity of a gas is due to Rankine. The apparatusconsists of a uniform tube ABDE bent as shown and provided with taps T1 and T2. A pelletof clean mercury is introduced into the tube. The tube is then joined to a capillary tube Cby rubber tubes at A and E. The whole arrangement is then mounted on a board. Two fixedmarks H and L are made on the tube such that the volume of the portion ABH = volumeof the portion LDE = v (say).
A quantity of the experimental gas is introduced into the system by opening the taps T1and T2 and connecting them to the gas supply. The taps are then closed. Now, the board isheld vertical causing mercury pellet to descent in BD under its own weight, thus forcing the
Fluids 485
gas below it through the capillary C. The time-interval t between the top surface of the pelletpassing H and the bottom surface reaching L is measured. This is repeated by inverting theboard several times.
Theory: Let V be the total volume ABDE less the volume of the pellet. If the apparatusbe placed horizontally, the pressure of the gas will be uniform throughout. Let it be P. Letthe density of the gas at this pressure be ρP, where ρ is the density per unit pressure. Themass of the gas enclosed in this ρPV. With the tube vertical and the mercury pellet inposition M, the pressure in ABH will be different than that in HDE. Let them be p and p1respectively. The corresponding densities are ρp and ρp1 respectively. The volume ABH andHDE are v and (V – v) respectively. Since the mass of the gas is constant. We have
ρPV = ρpv + ρp1 (V – v)
or PV = pv + p1 (V – v)
But p1 = p + (mg/a), where m is the mass of the pellet and a is the cross-sectional areaof the tube ABDE.
∴ PV = pv pmga
v+ +
−( )V
or p = PV
− +mga
mga
v
and p1 = P +V
mga
v
At the end of the interval t, when the pellet reaches theposition M', these pressures becomes
p' = P (VV
− + −mga
mga
v)
and p'1 = P (VV
+ −mga
v).
Now, with the mercury pellet at M, the mass of the gas below the pellet is ρp1(V – v),and with the pellet at M', the mass below it is ρp′1v. Hence the mass of the gas which haspassed through the capillary during the interval t
= ρp1(V – v) –ρp′1v
= ρ ρP +V
V P + VV
mga
vv
mga
vv
− − −
( ) ( )
= ρP (V – 2v)
Thus the average rate of flow is
ρP Vt
v( )− 2 ...(A)
Now, when the pellet is at M, the volume per second (Q1 say) entering the capillary is,from Poiseuille’s formula for the flow of a gas, given by
p1Q1 = πη
rp p
l4 1
222
16( )−
AH M
B
L M ′
DE
T2
T1
C
Fig. 21
486 Mechanics
Thus the mass entering per second is
p1Qρ =π ρ
ηr
lp p
4
12
22
16( )−
=π ρ
ηr
lp p p p
4
1 2 1 216( ) ( )− +
=π ρ
ηr
lmga
mga
v mga
4
162P +
2V
−
Similarly, when the pellet reaches at M′, the mass per second entering is π ρ
ηr
lmga
4
16
2P +2 V
Vmga
v mga
− −
.
Hence the average rate of flow through the capillary is
π ρη
rl
mga
4
162P
Equating this value in that obtained in (A), we obtain
π ρη
rl
mga
4
162P =
ρR Vt
v( )− 2
η =πr mgt
al v
4
8 2( )V −Where (V – 2v) is the volume between the two marks H and L less the volume of the
pellet. Hence η for the gas can be calculated.
10.27 STOKE’S LAW OF VISCOUS FORCE
When a solid body falls under gravity through a liquid (or gas), the layer of the liquidin contact with the body moves with the velocity of the body, while the liquid at a greatdistance from it is at rest. Thus, the body produces a relative motion between the layers ofthe liquid. This is opposed by the viscous force in the liquid. The viscous force increases withthe velocity of the body and ultimately becomes equal to the force driving the body. The bodythen falls with a constant velocity, which is called the ‘terminal velocity’.
Stokes showed that when a small sphere is moving slowly with a constant velocitythrough a perfectly homogeneous viscous fluid of infinite extension; the resistive forceexperienced by the sphere is
F = 6πηrv
Where r is the radius of the sphere, v its terminal velocity, and η the viscosity of thefluid. This is stoke’s law. The conditions for its validity are
(i) The sphere should be small in size,
(ii) It should move slowly with a constant velocity,
(iii) The fluid should be perfectly homogeneous and of infinite extension.
Fluids 487
10.28 STOKE’S FORMULA FOR THE TERMINAL VELOCITY OF A FALLINGSPHERE
Let us suppose that a small sphere of radius r and density ρ is falling freely from restunder gravity through a fluid of density σ and coefficient of viscosity η, when it acquires theterminal velocity v, the various forces acting upon it are
(i) downward force due to gravity = (4/3) πr3ρg
(ii) upward thrust due to buoyancy = (4/3) πr3σg
(iii) resistive viscous force = 6 πηrv
Thus the resultant downward driving force is (4/3)πr3 (ρ − σ)g.
Since the sphere has attained a constant velocity, the resultant driving force must beequal to the resistive force. That is,
(4/3) πr3 (ρ − σ)g = 6 πηrv
or v =29
2r g( )ρ ση−
This is the required formula.
10.29 VELOCITY OF RAIN DROPS
Rain drops are formed by the condensation of water vapour on dust particles. When theyfall under gravity, their motion is opposed by the viscous drag in air. As the rain drop fallswith increasing velocity, the viscous drag increases and finally becomes equal to the effectiveforce of gravity. The drop then falls with a constant terminal velocity v. Let r be the radiusof the drop and ρ its density. Let σ be the density and η the viscosity of air. Now, the variousforce acting upon the drop are:
(i) its weight acting downward = (4/3) πr3ρg
(ii) upthrust of the air = (4/3) πr3σg
(iii) viscous drag = 6 πηrv
Since the drop is falling with a constant velocity, we must have
6 πηrv = (4/3) πr3 (ρ − σ)g
v =29
2r g( )ρ ση−
or v α r2
The radii of the rain drops are very small. Hence their terminal velocities, beingproportional to the square of radii, are also very small. Therefore, to an observer on earth,the drops appear as cloud particles floating in the sky.
For the same reason, a larger dust particle (larger r) acquires a larger velocity v andhence falls faster than a smaller one in air.
NUMERICALS
Q.1. Calculate the work done in blowing a soap bubble from a radius of 10 cm to 20cm.(S.T. = 25 × 10 –3N/m).
Solution. Total surface area of two surfaces or soap bubble before blowing
= 4πr12 + 4πr1
2
488 Mechanics
⇒ 8rπ (0.1)2 m2
Similarly total surface area of bubble after blowing
= 4πr22 + 4πr2
2
= 8π (0.2)2 m2
i.e., increase in surface area
∆A = 8π ( . ) ( . )0 2 0 12 2−
= 0.24π m2.
The work done is therefore
∆W = T.∆A
= 25 × 10–3 × 0.24 π = 6π × 10–3 joules
Q.2. Calculate the work done in spraying a spherical mercury drop of 1mm radius intomillion identical droplets, the surface tension of mercury being 550 dyne/cm.
Solution. Initially the surface area of mercury drop
= 4π (1 × 10–1)2 = 0.4π cm2
When its sprayed into million droplets, let ‘r’ be the radius of one droplet then
10 43
6 3× πr =43
1 10 1 3π ( × )−
(102r)3 = (1 × 10–1)3
r = 10–3cm.
Final surface area of million droplets
= 10 4 10 46 3 2 2× ( )π π− = cm
Increase in surface area
∆A = 4π – 0.04 π = 3.96 π∴ Work done against S.T. in spraying
∆W = T.∆A
= 550 × 3.96 × 3.14 ⇒ 6839 ergs
Q.3. Eight air bubble each of radius ‘a’ coalesce to form a bigger bubble of radius ‘R’. If
P is atmospheric pressure, T the surface tension and ∈ =Tap << 1, Show that
R = 2a 1 +23 ∈
Solution. Taking isothermal conditions, we must have for 8 air bubbles to coalesce
PV + PV +..... = p'v'
8PV = p'v' ... (i)
Since ‘a’ is radius of original bubble
p = P +T2a
Fluids 489
v =43
πa3
Similarly ‘R’ is the radius of bigger bubble and we have for it
p' = P +2TR
v' =43
3πR .
Substitute these values in eqn. (i) gives
843
3P +2Ta
a
π = P+
2TR
R
43
3π
8Pa3 + 16Ta2 = PR3 + 2TR2
Dividing both sides by 8Pa3, we get
1 2+ TPa
=R TR
P
3 2
38 43a a+
But TPa
= ∈
1 2+ ε =R2
R2
a a +
3
24ε
R2a
3
= 1 2 1+ −
ε R8
2
2a
R = 2 1 + 2 1R2
aa
ε −
8 2
1 3/
R = 2 1 + 13
R2a
a. ....2 1
8 2ε −
+
= 2 1 23
1aa
+ −
+
ε R8
2
2 ....
Which concludes that
R = 2 1 + 23
a
ε
Q.4. A number of soap bubbles coalesce to form one spherical soap bubble under isothermalconditions. If ‘V’ represents change in volume ‘S’ the change in surface area, ‘P’ is atmosphericpressure and T is S.T. of soap solution, show that
3PV + 4ST = 0
490 Mechanics
Solution. Let ‘n’ soap bubbles coalesce to form one soap bubble, under isothermalconditions. Then from Boyles law product PV for system will be constant.
Hence n(pv) = p'v'
Suppose ‘r’ is radius of small bubbles and ‘R’ be the radius of single bubble formed.
Then p = P +4Tr
and v =
43
3πr
Similarly for bigger bubble
p' = P +4TR
and v' =
43
3πR
Putting these value
nr
rP +4T
43
3π = P +4TR
R
43
3π
or, P 4 R3 3π πrn
34
3−
=
44 4 2T
3R2( . )π π− n r
Now, nr
. 43
43
3π π−
R3
= V, the change in volume
and ( )4 42π πr n − R2 = S the change in surface area
PV = − 4T3
S
which gives 3PV + 4ST = 0
Q.5. Two soap bubbles of radii ‘a’ and ‘b’ combine to form a single bubble of radius ‘c’.If external pressure is P. Show that surface tension S is given by
S =P c a b
a b c( )( )
3 3 3
2 2 24− −+ −
Solution. If bubbles coalesce under isothermal conditions, product of pressure andvolume of enclosed air must remain same for the system.
i.e., p1v1 + p2v2 = pv
or, P +4S
P +4S
aa
bb
+
43
43
3 3π π = P +4Sc
c
43
3π
or, P(a3 + b3 – c3) = 4S (c2 – b2 – a2)
S =P(a b c
c b a
3 3 3
2 2 24+ −− −
)( )
S =P(c a b
a b c
3 3 3
2 2 24− −+ −
)( )
.
Fluids 491
Q.6. Two air bubbles of radii 0.002 m and 0.004 m formed in same liquid of surfacetension 0.07 N/m come together to form a double bubble. Find the radius of surface commonto both the bubbles.
Solution. The pressure inside air bubble formed in liquid of surface tension T is given by
p = P +2Tr
Hence, pressure inside smaller bubble
p1 = P +2Tr1
and pressure inside bigger bubble
p2 = P +2Tr2
Since pressure inside smaller bubble will be greater, when the two bubbles come together,the common surface will have concave side towards smaller and convex side towards largerbubble. The difference of pressure on the two sides of common surface will be
p = P +2T
P +2T
r r1 2
−
= 21
2T
1
1r r−
If R is the radius of curvature of common surface the excess pressure on its concave side
must be p = 2TR
So,2TR
= 2T1r r1 2
1−
i.e.,1R
= 1 1
1 2r r−
R =r r
r r1 2
2 1−
Putting r1 = 0.002 m and r2 = 0.004 m we get
R =0 002 0 0040 004 0 002
0 004. × .( . . )
.−
⇒ m Ans.
Q.7. A plate of 100 cm2 is placed on upper surface of 2 mm thick oil having coeff. ofviscosity η = 15.5 Poise. Calculate the horizontal force needed to move the plate with velocity3 cm/sec.
Solution. The viscous force due to viscosity which acts to oppose relative motion is
492 Mechanics
F = −ηA dvdx
As given η = 15.5 C.G.S. units A = 100 cm2
and velocity gradient dvdx
=30 2cm/sec
cm. = 15 sec–1
Putting these values the viscous force is
F = – 15.5 × 100 × 15 dynes
= –23250 dynes
Thus, the forward force needed to overcome the viscous force and move the plate willbe
F = 2.325 × 104 dynes.
Q.8. A horizontal capillary of diameter 2 mm and length 20 cm is connected to a watertank and 0.2 cc. water flow per sec. Calculate the rate of flow, if another capillary of length10 cm and diameter 1 mm is formed in series with the first capillary.
Solution. The rate of flow of liquid through a capillary is given by
V = πη η
π
P P8
rl l
r
4
48
=.
...(i)
If two capillaries are joined in series the combined viscous resistance will
R =8 81
14
2
24
ηπ
ηπ
lr
lr
+
=8 1
14
2
24
ηπ
lr
lr
+
Hence for same pressure difference the new rate of flow will be
V =P
8 1
14
2
24
ηπ
lr
lr
+
As given in first case
V = 0.2 c.c., l1 = 20 cm, r1 = 1 mm = 0.1 cm
0.2 =P
8 200 1 4
ηπ ( . )
...(ii)
In second case, when another capillary is also used having.
l2 = 10 cm and r2 = 0.5 mm = 0.05 cm
V′ =P
8 200 1
100 054 4
ηπ ( . ) ( . )
+
...(iii)
Fluids 493
Dividing (ii) and (iii)
0 2.V′
=
200 1
100 05
200 1
4 4
4
( . ) ( . )
( . )
+
= 11020
0 10 05
4
+
.. = [1 + 8]
V' =0 29
0 022. .= cc/sec
Q.9. A capillary of radius 0.2 mm and length 12cm is fixed horizontally at the bottomof a large reservoir filled upto 25 cm with alcohol (density = 0.8 gm/c.c., η = 0.006 poise). Findvelocity of liquid flowing along axis of capillary.
Solution. If pressure difference P is applied across a capillary of length l and radius r,the velocity of flow of layer lying at distance x from axis is given by
V =P
4ηlr x( )2 2−
Putting x = 0, We get the velocity of liquid flowing along axis of tube
Vaxis =P4
2rlη
As given P = hdg =25 × 0.8 × 9.8
r = 0.02cm, l = 12cm and h = 0.006 Poise
Substuting the values,
Vaxis =25 × 0.8 × 980 × (0.02)2
4 0 006 12× . ×
=25 8 98 4 10
4 6 12 10
5
3× × × ×
× × ×
−
−
=25 98 10
927 2
1× ×. /sec.
−= cm
Q.10. Three capillaries of same length but radii in the ratio 3:4:5 are connected in seriesand liquid flows through them. If pressure across third capillary is 8.1 mm, deduce thepressure across the first capillary.
Solution. Let l be the length of each capillary and radii 3r, 4r, 5r. Since they are inseries rate of flow must be same through the three
πη
p rl
1
8(3 )4
=π
ηπ
ηp r
lp r
l2
43
448
58
( ) ( )=
494 Mechanics
p1(3)4 = p2 (4)4 = p3 (5)4
p1 =8 1 5
3
4
4. × ( )
( ) =
8 1 62581
. × = 62.5 mm.
Q.11. Two tubes A and B of length 100 cm and 59 cm have radii 0.1 mm and 0.2 mmrespectively. A liquid passing through both enters in tube A at pressure 80 cm of Hg and leavestube B at 76 cm of Hg. Find the pressure at junction of two tubes.
Solution. Let P be the pressure at the junction of tubes A and B. Then for tube A,p1 = (80 – P), l1 = 100 cm, r1 = 0.1 mm and for tube B, p1 = (P – 76), l2 = 59 cm, r2 = 0.2 mm.
Since same quantity of liquid passes through both tube per sec,
V =π
ηπ
ηp r
lp r
l1 1
4
1
2 24
28 8=
(80100
− P)(0.1)4
=(P 76) (0.2)4−
59
80− P100
=P 76
59− × 16
59 (80 – P) = 1600 (P – 76)
126320 = 1659 P
P = 76.051 cm
Q.12. A cylindrical vessel of radius 5 cm is completely filled with water and carries ahorizontal capillary of length 20 cm and internal diameter 0.8 mm. What time it will take forvessel to become half empty ? (η = 0.01 poise)
Solution. Let A be cross-sectional area of vessel. If height falls by ‘dh’ in vessel due toescape of liquid through capillary in time dt,
rate of flow − Adhdt
=π
ηpr
l
4
8
− Adhdt
=π
η( )hdg r
l
4
8
dt = − 8ηπ
ldgr
dhh
A4
For vessel to become half empty, level will fall from h to h2
. Hence total time will be
given by
t = − 82
ηπ
ldgr
dhh
h
hA
4
/
=8
2η
πl
dgr eA
4 log
Fluids 495
As given A = π(5)2, l = 20 cm, r = 0.08 cm, η = 0.01 poise, d = 1 gm/cc, g = 980 cm/sec2
t =8 0 01 20 5
1 980 0 0899 6
2
4× . × × ( )× × × ( . )
. secππ
=
Q.13. A horizontal pipe of non-uniform bore has water flowing through it such that thevelocity of flow is 40 cm/sec at a point where the pressure is 2 cm of mercury column. Whatis the pressure at a point where the velocity of flow is 60 cm/sec ? (Take g = 980 cm/sec2 anddensity of water = 1 gm/cc).
Solution. Here p1 = 2 cm of mercury column = 2 × 13.6 × 980 dyne/cm2
ρ1 = ρ2 = ρ = 1 gm/cc, v1 = 40 cm/sec, h1 = h2 = h
In accordance to Bernoulli’s theorem
pgh v1
121
2ρ+ + =
pgh v2
221
2ρ+ +
Where p2 is the pressure at the second point
or,12 2
212( )v v− =
p pp p1 2
1 2ρ ρ−
= −
so that,12
60 402 2( )− = 2 × 13.6 × 980 – p2
or 1000 = 26650 – p2
p2 = 26650 – 1000 = 25650 dyne/cm2
= 25650/13.6 × 980
= 1.925 cm of mercury column.
Q.14. Calculate the velocity of efflux of kerosene oil from a tank in which the pressureis 50 lb wt/square inch above the atmospheric pressure. The density of Kerosene is 48 lb/c.ft.
Solution. We know the velocity of efflux v = 2gh , where h is the height of the liquid
surface from the axis of the orifice.
Now, pressure due to kerosene at the level of the axis of the orifice
= hρg poundals/ft2 = hρ lb wt/ft2.
But this is given to be 50 lb wt/(inch)2 = 50 × 144 lb wt/ft2
hρ = 50 × 144 or, h = 50 × 144/ρ = 50 × 144/48 ft
Hence, velocity of efflux, v = 2gh
= 2 32 50 144 48× × × ×
= 97.97 or 98 ft/sec.
Q.15. A water main of 20 cm diameter has pilot tube fixed into it and the pressuredifference indicated by the gauge is 5 cm of water column. Calculate the rate of flow of waterthrough the main. (g = 980 cm/sec2 and ρ for water = 1 gm/cc).
496 Mechanics
Solution. Radius (r) of the main = 20/2 = 10 cm and therefore its area of cross-sectionα = πr2 = π(10)2 = 100 π sq cm
Since loss of kinetic energy per unit mass on stoppage of flow =12
2v = pressure energy
per unit mass = p/ρ = p/1 = p = 5 × 1 × 980
v2 = 10 × 980
v = 9800 99 0= . cm/sec.
Rate of flow of water through the main = velocity of flow × area of cross-section of the main
= 99 × 100 π = 31100 cc/sec. or 31.1 liters/sec.
Q.16. The diameters of a water main where a venturimeter is connected to it are 20 cmand 10 cm. What is the rate of water flow. If the water levels in the two piezometer tubes differby 5 cm ? (g = 980 cm/sec2).
Solution. The rate of flow of water through the main is given by the relation
Q = a v a a hg a a1 1 1 2 12
222= −/ ( )
a r a r1 12 2
2 22 220 2 100 12 2= = = = =π π π π π( / ) ( / )sqcm, = 36 π sq cm and h = 5 cm, we have
Rate of flow of water through the main i.e.,
Q = 100 36 2 5 980100 362 2π π
π π× × ×
( ) ( )−
= 360098008704
11920π = cc/sec or 11.92 or 12 liters/sec.
Q.17. Calculate the mass of water flowing in 10 minutes through a tube 0.1 cm indiameter, 40 cm long, if there is a constant pressure head of 20 cm of water. The coefficientof viscosity of water is 0.0089 C.G.S. units.
Solution. From Poisueille’s Eqn. volume rate of water given by
Q = πη
πPcc/sec.
rl
4 4
820 1 981 0 058 0 0089 40
0 1353= =× × × × ( . )× . ×
.
Volume of water flowing out in 10 minutes
= 0.1353 × 10 × 60, say
V = 81.18 cc
mass of water flowing out in 10 minutes
= V × P
= 81.18 × 1 = 81.18 gm Ans.
Q.18. A cylindrical vessel of radius 7 cm is filled with water to a height of 50 cm. It hasa capillary tube 10 cm long, 0.2 mm radius protruding horizontally at its bottom. If thevelocity of water is 0.01 C.G.S. units and g = 980 cm/sec2. Find the time in which the levelwill fall to a height of 25 cm.
Fluids 497
Solution. Let h be the height of the water column in the vessel at any given instantand dh, the fall in its height in a small interval of time dt. Then if A be the area of crosssection of the vessel. We have
rate of flow through the capillary tube, i.e.,
Q = –Adh/dt
the –1 ve sign indicating that h decreases as t increases.
But, as we know, the rate of flow of water through a capillary tube is given by poiseuille’seqn., Q = πPr4/8ηl. Where P = hρg (ρ = 1 gm/cc) we have
−A dhdt
=π
ηη
πhgr
ldt
lgr
dhh
4
488or A= −
And ∴ dtt
0 = −
h
hl
grdhh
1
28ηπ
A4
. , whence tl
grhhe= 8 1
2
ηπ
A4 log
Or, putting the values of A = π × 72, r = 0.02 cm, h1 = 50 cm and h2 = 25 cm
We have t = 8 0 01 10 7
9 8 0 022 3026 50
25
2
4 10× . × × ×
× . × ( . )× . logπ
π
= 1.734 × 105sec = 48.16 hrs.
The water level in the vessel will thus fall to 25 cm in 48.16 hrs.
Q.19. Write down Poiseuille's formula for the rate of flow of a liquid through a capillarytube. From this show that if two capillaries of radii a1 and a2 having length l1 and l2respectively are set in series the rate of flow Q* is given by
Q =π
ηP
8la
la
1
14
2
24
1
+
−
Where P is the pressure across the arrangement and η , the coefficient of viscosity of theliquid.
Solution. Let P1 be the pressure across the first and P2 across the second, capillary.So that, P = P1 + P2, and, therefore P2 = P – P1.
Obviously, in accordance with the eqn. of continuity, the rate of flow through eithercapillary will be the same, say Q and therefore, from Poiseuille’s eqn., we have
Q = πP1a14/8ηl1 = πP2a2
4 /8ηl2 = π(P – P1) a24 /8ηl2,
whence P1a l14
1/ = P P24
1 24a l a l/ /2 2−
or, P1al
al
14
1
24
2+
=
P 24a
l2
P1 = P P2
41a l
a l a l
l
a l a l a
/
/ / / /2
14
1 24
2 14
2 24
1 14 +
=+
498 Mechanics
=P 1
14l
ala
la
2
24
1
14
1
+
−
Substituting this value of P1 in expression Q = πP1a14 /8ηl1, we have
Q =πηP
8la
la
1
14
2
24
1
+
−
Q.20(a). Three capillaries of length 8L, 0.2L and 2L with their radii r, 0.2r and 0.5rrespectively are connected in series. If the total pressure across the system in an experimentis p deduce the pressure across the shortest capillary.
(b) Fig. 22 below shows two wide tubes P and Q connected by three capillaries A, B, Cwhose relative lengths and radii are indicated. If a pressure p is maintained across A, deduce(i) the ratio of liquid flowing through A and B (ii) the pressure across B and across C.
A ( r)l,
CP Q
B
( /2, 2 r)l ( /2, r/2 )l
Fig. 22
Solution. (a) Obviously the rate of flow of liquid across each capillary is the same. Sothat, if p1, p2 and p3 be the pressure across the three capillaries respectively, we have, inaccordance with Poiseuille’s eqn.,
Q =πηp r1
4
8 8( )L =
πη
πη
p r p r24
340 2
8 0 20 5
8 2( . )( .
( . )(L) L)
=
Whencep1
64=
p p2 3
1000 256=
p1 =P2 641000
× =
8125 1000
25632
1252 32
2p pp
pand = =×
Now p = p1 + p2 + p3 = 8125
32125
165125
22
22
pp
pp+ + =
p2 = 125165
0 7575p p= . .
(b) (i) Clearly, rate of flow liquid through capillary A is Q = π ηpr l4 8/
and rate of liquid flow through B and C is, say
Q′ = π
ηπ
ηp r
lp r
l1
42
428 2
28 2
( )( / )
( / )( / )
=
Whence (2r)4 p1 = (r/2)4p2 or p2 = 256 p1
Fluids 499
Since p = p1 + p2, we have p = p1 + 256 p1 = 257 p1
∴ Ratio of liquid flowing through A and B = QQ
= π ηπ η
pr lp r l
4
14
82 8 2
/( ) / ( / )
=257 832 8
257 3214
14
π ηπ η
p r lp r l
//
:=
(ii) Since p = 257 p1, we have
pressure across capillary B, i.e., p1 = p/257 and
pressure across capillary C, i.e., p2 = p – p1 = p – p/257 = 256257
p Ans.
Q.21. A gas bubble of diameter 2 cm rises steadily through a solution of density1.75 gm/c.c. at the rate of 0.35 cm/sec. Calculate the coefficient of viscosity of the solution.(Neglect density of the gas).
Solution. We know, the coefficient of viscosity of the solution is given by the relation
η =29
2ρ σ−
v
r g
Here radius of the bubble r = 2/2 = 1 cm density of the solution σ = 1.75 gm/c.c., densityof bubble ρ = 0 and velocity of the bubble v = – 0.35 cm/sec (because it is directed upward).We have
η =29
6 29
1 75 1 9810 35
2 2r gv
= ×. × ×
.= 1.09 × 103 Poise.
Q.22. A glass bulb of volume 500 c.c. has a capillary tube of length 40 cm and radius0.020 cm leading from it. The bulb is filled with hydrogen at an initial pressure of 86 cm ofmercury density 13.6 gm/c.c. and it is found that if the volume of the gas remaining in thevessel is kept constant, the pressure falls to 80 cm of mercury in 254 sec. If the height of thebarometer is 76 cm and g = 981 cm/sec2, find the viscosity of hydrogen.
Solution. This is obviously a straight application of Searle’s method for determining thecoefficient of viscosity of a gas and we have, therefore
η =π ρr gt
l hh
hh
4
1
2
2H
8 V × 2.3026 log H +2H +10
2
1.
H = 76 cm, ρ = 13.6 gm/c.c., g = 981 cm/sec2, t = 25.4 sec, l = 40 cm, V = 500 c.c.h1 = 86 – 76 = 10 cm and h2 = 80 – 76 = 4 cm. We have coefficient of viscosity of hydrogen.
η =π ( . ) × × . × × .
× × × . log .
0 02 76 13 6 981 25 4
8 40 500 2 3026 104
152 4152 10
4
10++
=π ( . ) × × . × × .
× × × . × .0 02 76 13 6 981 25 4
8 40 500 2 3026 0 3815
4
= 9 204 10 5. × − Poise.
500 Mechanics
Q.23. Two horizontal capillary tubes A and B are connected together in series so that asteady stream of fluid flows through them. A is 0.4 mm is internal radius and 256 cm long.B is 0.3 mm in internal radius and 40.5 cm long. The pressure of the fluid at the entranceis 3 inches of mercury above the atmosphere. At the exit end of B, it is atmospheric (30 inchesof mercury). What is the pressure at the function of A and B if the fluid is (i) a liquid (ii) agas ?
Solution. Case (i): When the fluid is a liquid—Let the pressure at the junction ofcapillary tubes A and B be h inches of mercury, so that pressure across capillary A is say, P1= (33 – h) inches of mercury, and pressure across capillary B is, say, P2 = (h – 30) inches ofmercury. The two tubes being connected in series. The volume rate of flow of the liquid isthe same through both and we have
Q =π
ηπ
ηp r
lp r
lpp
ll
rr
1 14
1
2 24
2
1
2
1
2
2
1
4
8 8= =
, whence
Where l1, l2 and r1, r2 are the length and the radii of the two tubes respectively.Substituting the values of p1 and p2 from above, we have
3330
−−
hh =
25640 5
0 30 4
25640 5
81256
24
... .
×
= =
or, 33 – h = 2(h – 30) = 2h – 60, or 3h = 93, h = 31 inches
i.e., the pressure at the junction of the two capillary tubes is 31 inches of mercury.
Case (ii): When the fluid is gas—The mass rate of flow of the gas through eithercapillary tube is the same and we thus have
ηη
r hl
14 2 2
1
3316( )−
=η
ηr h
l24 2 2
2
3016( )−
Where h, as before, is the pressure in inches at the junction of the two tubes
we have33
30
2 2
2 2−
−h
h=
rr2
1
4
2
=
332 – h2 = 2h2 – 2 × 302, or, h2 = (2 × 302 + 332/3)
= 2889/3 = 963
h = 963 31 03= . inches
The pressure at the junction of the two capillary tubes is, in this case, equal to 31.03inches of mercury.
Q.24. A plate of area 100 cm2 is placed on the upper surface of caster oil 2 mm thick.Taking the coefficient of viscosity to be 15.5 Poise. Calculate the horizontal force necessary tomove the plate with a velocity 3 cm/sec.
Solution. The (horizontal) viscous force is, by Newton’s hypothesis, given by
F = ηAdvdy
Fluids 501
For the present case, we can put
F = ηA vy
η = 15.5 poise, A = 100 cm2, v = 3 cm/sec and y = 2 mm = 0.2 cm
F = 15 5 100 30 2
2 325 104. × ×.
. ×= dynes.
Q.35. Water is conveyed through a horizontal tube 8 cm in diameter and 4 km in lengthat a rate of 20 litre per second. Assuming only viscous resistance. Calculate the pressurerequired to maintain this flow. Viscosity of water may be taken as 0.01 cgs units.
Solution. The volume of water flowing per second through a capillary of length l andradius a under a pressure difference p is given by
Q =π
ηpa
l
4
8
the pressure difference required is, therefore
p = Q 8ηπ
la4
Q = 20 liter/sec = 20 × 103 cm3/sec,η = 0.01 cgs unit, l = 4 km = 4 × 105 and a = 4 cm
p = 20 108 0 01 4 10
3 14 43
5
4× ×× . × ( × )
. × ( )
= 7.96 × 105 dynes/cm2 Ans.
Q.26. Calculate the rate at which water flows through a capillary tube of length 0.5 meterwith an internal diameter of 1 mm. Coefficient of viscosity is 1.3 × 10–3 kg/m-sec. The pressurehead is 20 cm of water.
Solution. By Poiseuille’s formula, the rate of water flow is given by
Q =π
ηpa
l
4
8
p = 20 cm of water = 0.20 m × (1 × 103 kg/m3) × 9.8 nt/kg = 1.96 × 103 nt/m2
a = 0.5 mm = 0.5 × 10–3 m,η = 1.3 × 10–3 kg/m-sec and l = 0.5 m
Q =3 14 1 96 10 0 5 10
8 1 3 10
3 3
3. × ( . × ) × ( . ×
× ( . ×nt m m)kg/m - sec) × (0.5 m)
2 4/ −
−
= 7.4 × 10–8 m3/sec.
Q.27. A capillary tube, 1.0 mm in diameter and 20 cm in length is fitted horizontally toa vessel kept full of alcohol of density 0.8 gm/c.c. The depth of the centre of the capillary tubebelow the surface of alcohol is 30 cm. The viscosity of alcohol is 0.012 C.G.S. unit. Find theamount of alcohol that will flow in 5 minutes.
Solution. The volume of liquid flowing per second through a capillary of length l andradius a under a pressure difference p is given by
Q =π
ηpa
l
4
8
502 Mechanics
p = hρg = 30 × 0.8 × 980 dyne/cm2, a = 0.5 mm = 0.05 cm.
η = 0.012 Poise and l = 20 cm
Q =3 14 30 0 8 980 0 05
8 0 012 200 24
4. × × . × × ( . )× . ×
. /sec= cm2
Volume flowing in 5 minutes
= 0.24 cm3/sec × (5 × 60) sec = 72 cm3
The density of alcohol is 0.8 gm/cm3. Therefore, mass of alcohol flowing in 5 minutes
= 72 cm3 × 0.8 gm/cm3 = 57.6 gm
Q.28. Calculate the volume of water flowing in 10 minutes through a tube 0.1 cm indiameter and 40 cm long if there is a constant pressure head of 20 cm of water. The coefficientof viscosity is 0.0089 cgs units.
Q.29. Find the coefficient of viscosity from the given data: volume of water coming outthrough the capillary in 10 minutes = 80.7 cc, head of water = 39.1 cm, length of capillary= 60.24 cm radius = 0.0514 cm, g= 980 cm/sec.
Solution. By Poiseuille’s capillary flow formula
η =πpa
l
4
8Q
p = 39.1 cm of water = 39.1 × 1 × 980 dynes/cm2, l = 60.24 cm, a = 0.0514 cm
Q =80 7600
0 1345. . /= cm sec3
η =3 14 39 1 1 980 0 0514
8 0 1345 60 24
4. × ( . × × ) × ( . )× . × .
= 0.01296 gm/cm-sec
= 1.296 × 10–2 Poise.
Q.30. Deduce the velocity profile for stream line flow of liquid through a capillary ofcircular cross-section. Deduce the fraction of liquid which flows through the section uptodistance a/2 from the axis, where a is the radius of capillary.
Solution. The velocity of liquid flow at a distance r from the axis of the capillary tubeis given by
v =p
la r
82 2
η( )−
Now, the volume of the liquid flowing per second through a thin cylindrical shell of radiir and r + dr is given by
dQ = v(2πrdr) = πηpl
a r rdr2
2 2( )−
the volume of the liquid Q, flowing per second through the section of the tube upto a distancea/2 from the axis is obtained by integrating this expression between the limits r = 0 and r= a/2
Fluids 503
Q' =πη
πη
pl
a r r dr pal
a
27
16 82 3
4
0
2( )
/− =
the volume Q flowing through the entire tube is π
ηpa
l
4
8
′=
716
.
Q.31. A liquid with coefficient of viscosity 0.50 Poise flows through a capillary tube havingan internal diameter of 1 mm and length 25 cm due to pressure difference of 10 cm of mercuryat its two ends. Find the volume of the liquid flowing out per minute and also its velocity inthe capillary tube along its axis.
Solution. The volume of liquid flowing per second through a capillary
Q =π
ηpa
l
4
8
p = 10 × 13.6 × 980 dyne/cm2, a = 0.5 mm = 0.05 cm,η = 0.50 Poise, l = 25 cm
Q =3 14 10 13 6 980 0 05
8 0 50 250 0261
43. × × . × ( . )
× . ×. /sec= cm
Volume flowing per minute = 0.0261 × 60 = 1.57 cm3/min.
Now, the velocity of liquid flow at a distance r from the axis of the capillary tube is givenby
v =p
la r
42 2
η( )−
At the axis of the tube r = 0
v =pa
l
2 2
410 13 6 980 0 05
4 0 50 25η= × . × × ( . )
× . ×
= 6.7 cm/sec.
Q.32. A capillary of radius 0.2 mm and length 12 cm is fixed horizontally at the bottomof a large reservoir filled upto 25 cm with alcohol of density 0.8 gm/c.c. and viscosity 0.006Poise. Find the velocity of the liquid following along the axis of the capillary.
Q.33. Water is flowing through a capillary tube 40 cm long and of 1 mm internal radiusunder a constant pressure head of 15 cm of water. Calculate the maximum velocity of waterin the tube and verify that the flow is stream-lined. Given: for water viscosity = 0.0098 Poise.Reynold’s number = 1000 and g = 980 cm/sec2.
Solution. The velocity of liquid flow at a distance r from the axis of the capillary tube(length l, radius a) under pressure p is given by
v =p
la r
42 2
η( )−
It is maximum at the axis (where r = 0) thus
vmax =pa
l
2
4η
504 Mechanics
p =15 cm of water = 15 × 1 × 980 dynes/cm2, a = 1mm = 0.1 cm, η = 0.0098 poise andl = 40 cm
vmax = 15 1 980 0 14 0 0098 40
93 752× × × ( . )
× . ×.= cm/sec
Now, the critical velocity of flow through the tube is
vc =Kηρa
Where K is Reynold’s number and ρ is density of water
vc =1000 × 0.0098
1 ×0.1cm/sec= 98
Thus vmax < vc. Hence the flow is streamlined.
Q.34. 0.0314 c.c. of a liquid is flowing out per second through a capillary tube of 1 mmradius. Calculate the velocity of the liquid at a point (i) on the axis of the capillary, (ii) adjacentto the wall, (iii) at 0.5 mm distance from the axis.
Solution. The velocity profile of a viscous liquid flowing in a capillary tube of length land radius a is given by
v =P
4ηla r( )2 2−
where v is the velocity of flow at a distance r from the axis of the tube. By Poiseuille’sformula, the volume of the liquid flowing per second is given by
Q =π
ηpa
l
4
8Eliminatingη between the above two formula: we get
v =2 2 2Q
4ηaa r( )−
Q = 0.00314 cm3/sec and a = 0.1 cm
v =2 0 0314
3 14 0 12004
2 2 2 2× .. × ( . )
( ) ( )a r a r− = −
(i) On the axis of the capillary, r = 0
v = 200a2 = 200 (0.1)2 = 2 cm/sec
(ii) At the wall, r = a
v = 200 (a2 ˆ– a2) = 0
(iii) At 0.05 cm from the axis, r = 0.05 cm
∴ v = 200 [(0.1)2 – (0.05)2] = 1.5 cm/sec Ans.
Q.35. Water at 20 °c is flowing through a capillary tube of length 1 meter and of diameter2 mm under the constant pressure head of 10 cm water. Calculate the velocity of water at apoint (i) on the axis, (ii) adjacent to wall of the tube, (iii) at 0.5 mm distance from the axis.Coefficient of viscosity for water at 20°C = 0.01 poise.
[Ans. [(i) 24.5 cm/sec, (ii) zero, (iii) 18.375 cm/sec]
Fluids 505
Q.36. Water flows in a streamline through a horizontal pipe of cross-sectional radius afrom a vertical tank of cross-section. Find the time when the height of the level in the tankbecomes half.
Solution. Let h be the height of liquid in the tank above the horizontal tube at anyinstant t. The pressure difference across the ends of the tube at that instant is p = ρgh,Where ρ is the density of liquid. If the height of liquid falls through dh is a small timeinterval dt, then the volume of the liquid flowed through the tube in time dt is A(dh). WhereA is the area of cross-section of the tank. Therefore the rate of flow of the liquid at theinstant t is
− A(dhdt
)
The minus sign is put because h decreases with increase in t. But from Poiseuille’sformula, the rate of flow
=π
ηπ ρ
ηpa
lgh a
l
4 4
8 8= ( )
∴ – A ( )dh
dt=
π ρη
( )gh al
4
8
dt = – 8ηπρ
lga
dhh
A4
The initial height of the liquid is H (say). Let t be the time required by the level to fallto H/2. Then, integrating the last expression between proper limits, we have
dtt
0 = − = −8 84
lga
dhh
lga
heη
πρη
πρA A
4 HH/2
H
H/2log
t = − −84
lga e eη
πρA
H/2 H][log log
=8
4lga
lae e e
ηπρ
ηπρ
AH log H/2] =
8 AH4[log log−
This is the required expression.
Q.37. A cylindrical vessel of radius 0.05 meter has at its bottom a horizontal capillarytube of length 0.2 meter and internal radius 0.4 mm. The vessel is completely filled with waterand then the water is allowed to flow out of the capillary. How long it will take for the waterlevel to fall to half its initial height in the vessel. The viscosity of water is 0.001 mks unit(g = 9.8 mtr/sec2, density of water = 1 × 103 kg/m/r3, loge 2 = 0.693).
Solution. The time required for the water level to fall to half its initial height is givenby
t =8
24lga eη
πρA
log
t =8 0 2 0 001 0 051 10 9 8 0 4 10
0 6933 3( . ) ( . ( .
( × ) ( . ) ( . ×× .
m kg/m - sec) × m)kg /m m/sec m)
2
3 2 4π
π −
= 1.1 × 104 sec ≈ 3 hours.
506 Mechanics
Q.38. A horizontal capillary tube of length 10 cm and internal radius 0.5 mm is attachedto the bottom of a vessel of cross-section 20 cm2. The vessel is initially filled with water toa height of 20 cm above the capillary tube. Find the time taken by the vessel to empty one-half of its contents coefficient of viscosity of water is 0.01 Poise. [Ans. 576.5 sec]
Q.39. Prove that if the level of water in a reservoir falls from mark A to B in 120 sec whenit flows through a capillary and in case of a liquid of specific gravity 0.9, the level falls fromA to B in 100 sec, then the ratio of viscosity of the liquid with that of water is 3:4.
Q.40. A horizontal capillary of diameter 2 mm and length 20 cm is connected to a tankof water and 0.2 c.c. of water flows out per second. What will be the rate of flow of water ifanother capillary of length 10 cm and diameter 1 mm is joined in series with the firstcapillary ?
Solution. The rate of flow through the first capillary (length l1, radius a1) along underpressure p is
Q =π
ηpa
l14
18
πηp
8=
Qla
1
14
Let Q' be the rate of flow when the second capillary (length l2, radius a2) is joined tothe first in series. The pressure difference across the composite capillaries is same, p as wasacross the first capillary alone. Now for two capillaries in series, we have
Q' =π
ηp l
ala8
1
14
2
24
1
+
−
=Qla
la
la
1
14
1
14
2
24+
=Q
1 +la
al
2
24
14
1
Q = 0.2 cm3/sec, a1 = 0.1 cm, l1 = 20 cm, a2 = 0.05 cm and l3 = 10 cm
∴ Q' =0 2
1 10 0 10 05 20
0 21 8
0 0224
4
.× ( . )
( . ) ×
.. /
+=
+= cm sec.3
Q.41. A tube of radius R and length L is connected in series with another of radiusR/2 and length L/4. If the pressure across the two tubes taken together is p, deduce thepressure across the tubes separately.
Solution. Let p1 and p2 be the pressure across the first and second tubes respectively.Then
p = p1 + p2 ...(i)
Fluids 507
the rate of flow of a liquid (viscosity η) throughout the system is the same. Thus
πη
p1
8RL
4
=π
ηp2
8((R/2)L/4)
4
p1 = p2/4 ...(ii)
Solving eqn. (i) and eqn. (ii), we obtain
p1 =p
pp
5and 2
45
= .
Q.42. Two capillaries A and B of lengths 16 cm and 4 cm have radii 0.2 cm and 0.1 cmrespectively and are joined in series. In Poisuille’s experiment a pressure difference of 3 cm of wateris employed across the ends of the composite tube. Find pressure difference across tube B.
[Ans. 2.4 cm of water]
Q.43. Two capillaries AB and BC of same length and joined end to end at B have radiir and 2r respectively. A is connected to a vessel of water giving a constant head of 0.1 meterand C is open to air. Calculate the pressure at B.
Ans Let P be the pressure at B, then the pressure difference across AB is (0.17 – P) andthat across BC is P relative to the atmosphere. Since the rate of liquid flow, Q, must be thesame through both the capillaries we have
Q =π
ηπ
η( . )0 17
8 8
4 4− =P) P(2rl
rl
(0.17 – P) = 16P
P = 0.01 meter
Q.44. Two capillaries A and B of diameters 0.4 mm and 0.3 mm and lengths 40 cm and30 cm respectively are placed horizontally in series and water flows slowly through them fromA to B. If the open end of B (water outlet) is at atmospheric pressure and the open end of A(water inlet) is at pressure 10 cm of water above the atmospheric pressure, calculate thepressure at the function of A and B relative to the atmospheric pressure.
Ans. 64091
cm of water
Q.45. Two capillaries of length 70 cm and 30 cm and radii 0.2 mm and 0.1 mm respectivelyare connected in series. If a liquid enters the first capillary at a pressure of 80 cm of Hg andleaves the second capillary at 70 cm of Hg, find the pressure at the junction of the twocapillaries.
[Hint. Let P be the pressure at the junction. Then the pressure difference across thefirst capillary is (80 – P) and that cross the second is (P – 70).]
Q.46. Three capillaries of the same length and radii 2r, 3r, and 4r are connected in seriesand a liquid is flowing through them under streamlined conditions. If the pressure differenceacross the entire-system is 6.3 cm of mercury, deduce the pressure across the first tube.
[Ans. 5 cm of mercury]
Q.47. Let P1 and P2 be the pressure across the first and second capillaries. The rate ofliquid flow is same in all of them. Thus,
508 Mechanics
Q =π
ηπ
ηπ
ηP P mm21
4 4 438
48
8 1 58
( ) ( ) ( . )( )rl
rl
rl
= =
81P1 = 256 P2 = 625 (8.1 mm)
P1 =625 8 1
81× .
= 62.5 mm.
Q.47. A cylindrical vessel of height h is kept maintained full of water. Three capillarytubes of same radius a and same length l are fitted horizontally in the wall of the cylinderat distances h/2, 3h/4 and h measured from the top. The water can flow through these tubes.Show that the length of a single outflow tube of the same radius which can replace the threetubes, when fitted at the bottom is 4/9.
Solution. Let P1, P2, P3 be the pressure heads at the three tubes and Q1, Q2, Q3 thecorresponding rates of flow. Then, by Poiseuille’s formula, we have
Q1 =π
ηP1a
l
4
8, Q2 =
πη
πη
PQ
P24
33
4
8 8al
al
, =
Q = Q1 + Q2 + Q3 = π
ηa
l
4
8( )P P P1 2 3+ + ...(i)
Let l be the length of the single tube fitted at the bottom (at pressure head P3) havingrate of flow Q. Then
Q =π
ηP3
4
8al
...(ii)
Equating (i) and (ii)
PL
3 =P + P P1 2 + 3
l
Now P1 =h h
h2
342, P and P3= =
ThenhL
=h h h
lh
l/ / /2 3 4 9 4+ + =
L =49
l.
Q.49. Three capillary tubes of the same radius r but of lengths l1, l2, l3 are fittedhorizontally to the bottom of a long cylinder containing a liquid at a constant head which canflow through these tubes. Show that the length L of a single tube of the same radius r whichcan replace the three capillaries is given by
1L
=1l
1l
1l1 2 3
+ +
Q.50. Calculate the maximum velocity of an oil drop radius 10–4 cm falling in air.(Density of oil = 0.9 gm/cc, viscosity of air = 1.8 × 10–4 gm. (cm-sec), g = 980 cm/sec2). Neglectair buoyancy.
Fluids 509
Solution. The drop (radius r1 density ρ) falling in air (density σ) acquires maximum(terminal) velocity v when the resultant driving (gravity) force equals the retarding viscousforce i.e., when
43
3π ρ σr g( )− = 6 π ηrv
Where η is the viscosity of air. If air buoyancy is ignored (σ = 0) then
43
3π ρr g = 6 π ηrv
v =29
2r gρη
v =2 10 0 9 980
9 1 8 10
4 2
4× ( ) × ( . ) ×
× ( . × )
−
− = 1.1 × 10–2 cm/sec.
Q.51. Determine the radius of the drop of water falling through air, if the terminalvelocity of the drop is 1.2 cm/sec. Assume the coefficient of viscosity for air = 1.8 × 10–4 Cgsunits and the density of air = 1.21 × 10–3 gm/sec. [Ans. 0.001 cm/sec.]
Hint. r 92( P )g
=−
ηθσ
Q.52. Two equal drops of water are falling through air with a steady velocity of 5 cm/sec.If the drops coalesce, what will be the new velocity?
Solution. Let r be the radius of each drop. The (steady) terminal velocity v of a dropof radius r and density ρ, falling through air is given by
v =29
2r g( )ρ ση−
Where σ is the density and η the viscosity of air. Thus
v∝ r2 = Kr2
Now, volume of each drop =43
3πr
∴Volume of the coalesced drop = 2 43
43
3× π πr r= [(2) ]1/3 3
∴ Radius of the coalesced drop = 21/3 r.
Hence the new terminal velocity of the coalesced drop.
v1 = K(2)1/3r2
= (2)2/3 Kr2 = (2)2/3 v
= (2)2/3 × 5 = 7.94 cm/sec.
Q.53. A gas bubble of diameter 2.0 cm rises steadily through a solution of density 1.75gm/cc at the rate of 0.35 cm/sec. Calculate the coefficient of viscosity of the solution. Neglectthe density of the gas.
510 Mechanics
Solution. The weight of the bubble is negligible. The forces acting upon it are: (i) the
up thrust of solution, 43
3π σr g, where r is the radius of bubble and σ the density of solution,
and (ii) the viscous drag 6 πηrv. Since the bubble has attained a steady velocity, we have
6 πηrv =43
3π σr g
η =23
2r gvσ
η =29
1 1 75 9800 35
×× . ×
.= 1.09 × 108 gm/cm-sec = 1.09 × 103 Poise.
Q.54. An air bubble of 1 cm radius rises through a long column of liquid of density1.47 × 103 kg-mtr3 with a steady velocity of 0.21 cm/sec. Find the viscosity of the liquid.Neglect the density of air (g = 9.8 m/sec2). [Ans. 1.52 × 102 kg/m-sec.]
Q.55. With what terminal velocity will an air bubble 1.0 mm in diameter rise in a liquidof viscosity 150 centipoise and density 0.90 gm/cm3 ?
Hint. vr g=
2
9
2ση
, η = 150 centipoise = 1.5 Poise.
= 0.33 cm/sec.]
Q.56. A glass plate of length 10 cm, breadth 1.54 cm and thickness 0.20 cm weights 8.2gmin air. It is held vertically with the long side horizontal and the lower half under water. Findthe apparent weight of the plate. Surface tension of water = 73 dynes/cm, g = 980 cm/sec2.
Solution. Volume of the portion of the plate immersed in water is
10 ×12
(1.54) × 0.2 = 1.54 cm3.
Therefore, if the density of water is taken as 1 then upthrust.
= wt. of the water displaced
= 1.54 × 1 × 980 = 1509.2 dynes
Now, the total length of the plate in contact with the water surface is
2(10 + 0.2) = 20.4 cm.
∴ Downward pull upon the plate due to surface Tension = 20.4 × 73 = 1489.2 dynes
∴ resultant upthrust = 1509.2 – 1489.2
= 20.0 dynes
=20 0980
0 0204. .= −gm tω
apparent weight of the plate in water = weight of the plate in air – resultant upthrust
⇒ 8.2 – 0.0204 = 8.1796 gm.
Q.57. A glass tube of circular cross – section is closed at one end. This end is weightedand the tube floats vertically in water, heavy end down. How far below the water surface is
Fluids 511
the end of the tube? Given radius of tube 0.14 cm, mass of weighted tube 0.2 gm, surfacetension of water 73 dyne/cm and g = 980 cm/sec2.
Solution. Let l be the length of the tube inside water. The forces acting on the tubeare:
(i) upthrust of water acting upward = πr2l × 1 × 980
=227
(0.14)2l × 980 = 60.368 l dynes.
(ii) Weight of the system acting downward = mg
= 0.2 × 980 = 196 dyne.
(iii) Forces of surface tension acting downward
= 2πrT = 2227
× × 0.14 × 73 = 64.24 dyne.
Since the tube is in equilibrium, the upward force is balanced by the downward forces.That is
60.368 l = 196 + 64.24 = 260.24
l =260 2460 368
4 31.
..= cm
Q.58. Calculate the work done in blowing a soap bubble of 5 cm radius. The surfacetension of soap solution is 25 dyne/cm.
Solution. When a soap bubble of radius R is blown, the total surface are formed is =2 × 4πR2 = 8 πR2
As the bubble has two surfaces. If T be the surface tension of soap solution, the workdone in blowing is
W = T × 8πR2
= 25 × 8 × 3.14 × (5)2 = 1.57 × 104 erg.
Q.59. Calculate the work done against surface tension in blowing a soap bubble from aradius of 10 cm to 20 cm. If the surface tension of soap solution is 25 dynes/cm.
Solution. Original area of both the surfaces of the bubble
= 2 × 4π (10)2 = 800 π cm2
Final area of the surfaces of the bubble
= 2 × 4π (20)2 = 3200 π cm2
Increase in surface area
= 3200π – 800π = 2400π cm2
The work done against surface tension is
W = S.T. × increase in area
= 25 × 2400 × 3.14
= 1.88 × 105 erg.
Q.60. A soap bubble of surface tension 25 dyne/cm has a radius of 5 cm. Find the workdone in blowing the bubble to the radius of 10 cm. [Ans. 4.7 × 104 erg.]
512 Mechanics
Q.61. Find the amount of work done in blowing a soap bubble of surface tension 30dyne/cm from 10 cm radius to 15 cm radius. [Ans. 9.42 × 104 erg.]
Q.62. Obtain an expression for the work necessary to break up a liquid drop of radius Rinto n equal smaller drops.
Solution. When a liquid drop is broken into a number of smaller drops, the surface areaincreases. Hence work is to be done against the surface tension. Let us that compute theincrease in surface area. Let r be the radius of each of the n smaller drops. Since the totalvolume of all the n smaller drops is the same as the volume of the original drop of radiusR, we have
n × 43
3πr =43
3πR
r =R
( ) /n 1 3
Thus, increase in surface area
= n × 4πr2 – 4πR2 = 4π (nr2 – R2)
= 4π nnR
R22
2 3( ) / −
= 4πR2 ( ) ./n 1 3 − 1
W = S.T. × increase in surface area
= T × 4πR2 ( ) ./n 1 3 − 1
= 4πR2T ( ) ./n 1 3 − 1
Q.63. Calculate the work done in spraying a spherical drop of mercury of radius 1mminto a million identical droplets. The surface tension of mercury is 550 dynes/cm.
Solution. The work done in spraying a liquid drop of radius R into n droplets is givenby
W = 4πR2T ( ) ./n 1 3 − 1
R = 1 mm = 0.1 cm, T = 550 dynes/cm and n = 106
W = 4 × 3.14 × (0.1)2 × 550(106)1/3 – 1
= 4 × 3.14 × (0.1)2 × 550 × 99
= 6839 ergs.
Q.64. Calculate the amount of energy needed to break a drop of water 1mm in radius into106 droplets of equal size, taking surface tension of water = 72 dynes/cm. [Ans. 895 ergs.]
Q.65. Find the increase in surface energy in breaking a water drop of radius 0.4 cm into125 smaller drops. (S.T. of water = 72 dynes/cm) [Ans. 579 ergs.]
Q.66. Calculate the amount of energy needed to break a drop of water 2 mm in diameterinto 109 droplets of equal size. (S.T. of water is 73 × 10–3 nt/m).
Fluids 513
Solution. When a drop is sprayed into a number of droplets, the surface area and hencethe surface energy increases. This increases in the surface energy is equal to the energy ofthe newly formed area.
Now, the radius of the drop is R = 1 mm = 10–3 m. Let r be the radius of each of the109 droplets. The total volume of all the 109 droplets is the same as that of the original drop.Thus
109 × 43
3πr =43
10 3 3π ( )−
103r = 10–3
r = 10–6 m.
The increase in surface area = Surface area of 109 droplets – Surface area of thedrop
= 109 × 4πr2 – 4πR2
= 4π 109 × (10–6)2 – (10–3)2
= 4π (10–3 – 10–6)
= 4π 10–3 m2.
The work done W in spraying the drop is stored as the energy of the newly formed areaand is equal to increase in area × S.T.
W = 4 × 3.14 × 10–3 × 73 × 10–3
= 9.17 × 10–4 joule.
Q.67. Calculate the loss of energy is 1000 drops of water each of diameter 2 mm coalesceto form one large drop. The S.T. of water is 72 dynes/cm.
Solution. Let R cm be the radius of the large drop formed, when 1000 small drops, eachof radius 1 mm coalesce. Since volume remains the same, we have
43
3πR = 1000 × 43
π (0.1)3
R = (1000)1/3 (0.1) = 1 cm.
Now, the surface area of the 1000 small drops = 1000 × 4π (0.1)2 = 40 π cm2 and thatof the large drop = 4π(1) = 4π cm2
Thus, decrease in surface area = 40π – 4π = 36 π cm2
Energy liberated = S.T. × decrease in surface area
= 72 × 36 × 3.14
= 8139 erg.
Q.68. Two drops of a liquid, each of radius r, coalesce to form a larger drop. Derive anexpression for the rise in temperature.
Solution. When two or more small drops of a liquid coalesce to form a single large drop,the surface area decreases, and hence the surface energy also decreases. The lost energyappears as heat, resulting in the rise in temperature of the drop.
Let R be the radius of the larger drop formed. Its volume is the same as that of thesmaller drops, each of radius r. Thus
514 Mechanics
43
3πR = 2 × 43
3πr
R = (2)1/3 r.
The decrease in surface area = surface area of 2 small drops – surface area oflarger drop.
= 2 × 4πr2 – 4πR2
= 4π (2r2 – R2) = 4π [2r2 – (2)2/3 r2].
= 8πr2 [1 – (2)–1/3]
If W be the energy released, we have
W = S.T. × decrease in surface area
= T × 8πr2 [1–(2)–1/3] ...(i)
This energy is spent in increasing the temperature of the liquid Let t be the rise intemperature. If ρ be the density and s the specific heat of the liquid, the heat H absorbedby the drop is
H = mass × Sp. heat × temp. rise
=43
3π ρR × s × t = 43
2 3π ρ( ) .r st ...(ii)
Substituting for W form eq. (i) and for H from eq. (ii) in Joule’s relation
W = JH.
T × 8πr2 1– (2)–1/3 = J ×43
(2r3) ρst
t =3
1 2 1 3TJ Sr ρ
[ ( ) ]./− −
Q.69. Two water drops each of radius 0.00002 cm coalesce. What rise in temp. will result ?Surface tension of water is 74 dynes/cm and j = 4.2 × 107 erg./cal. [Ans. 0.055°C]
[Hint. 1 – (2)–1/3 = 0.2063]
Q.70. If a number of little droplets of water, each of radius r1 coalesce to form a singledrop R, show that the rise in temp. will be given by
3 1 1SJ Rr
−
where S is the surface tension of water and J the mechanical equivalent of heat.
Solution. When a number of droplets coalesce to form a single drop the surface areadecreases and the surface energy also decreases. The decrease in surface energy is equal tothe energy of the decreased surface. Now suppose n droplets coalesce. Then
Surface area of n droplets = n × 4πr2
Surface area of the combined drop = 4πR2
decrease in surface area = n × 4πr2 – 4πR2
= 4π (nr2 – R2)
Fluids 515
If W be the energy released, we have
W = decrease in surface area × S.T.
= 4π (nr2 – R2) × S ...(i)
This energy is spent in increasing the temp. of the water drop. Let t be the rise in temp.Then if both the specific heat and density of water are 1, the heat H absorbed by the dropis
H = mass × sp.heat × t = 43
1 13πR × × × t
=43
3πR t ...(ii)
Substituting for W form Eq. (i) and for H from eq. (ii) in the Joules relation W = JH,we have
4π (nr2 – R2) × S = J × 43
3πR t
t =3 2 2
3SJ
RR
nr −
=
3 12
3SJ R R
nr −
...(i)
Since the total volume of n droplets is equal to that of the combined drop, we have
n r× 43
3π =43
πR3
nr3 = R3
nr2
3R =1r
Putting this value in Eq. (i) we get
t =3 1 1SJ Rr
−
.
Q.71. Calculate the pressure inside a small air bubble of radius 0.01 mm situated at adepth of 20 cm below the free surface of a liquid of density 0.8 gm/cm3 and S.T. 75 dyne/cmtaking the atmospheric pressure to be 76 cm of mercury.
Solution. The excess pressure inside an air bubble of radius R is
P =2TR
Where T is surface tension of liquid in which the bubble is formed. Here R = 0.01 mm= 0.001 cm and T = 75 dyne/cm.
P =2 750 001
×.
= 0.15 × 106 dyne/cm2
The pressure outside the bubble is atmospheric plus that due to 20 cm long column ofliquid of density 0.8 gm/cm3 i.e.,
(76 × 13.6 × 980) + (20 × 0.8 × 980) + (0.15 × 106)
= 1.18 × 106 dyne/cm2.
516 Mechanics
Q.72. Calculate the excess of pressure inside a soap bubble of radius 3 mm. Surfacetension of soap solution is 20 × 10–3 nt/mtr. Also calculate the surface energy of the bubble.
Solution. The excess pressure inside a soap bubble over that outside is
P =4TR
Where T is the surface tension of soap solution and R the radius of the bubble
Here T = 20 × 10–3 nt/mtr and R = 3 mm = 3 × 10–3 mtr.
P =4 20 10
3 10
3
3
× ( × )×
−
− = 26.7 nt/mtr2
The soap bubble has two surfaces so its surface area is
2 × 4πR2 = 2 × 4 × 3.14 × (3 × 10–3)2 = 2.26 × 10–4 mtr.2
Now the surface energy of the bubble = S.T. × surface area
= 20 × 10–3 nt/mtr × 2.26 × 10–4 mtr.2
= 4.52 × 10–6 joule.
Q.73. The pressure of air in a soap bubble of 7 mm diameter is 8 mm of water above theatmospheric pressure. Calculate the surface tension of the soap solution (g = 980 cm/sec2).
Solution. The excess pressure inside a soap bubble over that outside is
P =4TR
Where T is the surface tension of soap solution and R is the radius of the bubble
T =PR4
Here, the excess pressure inside the bubble over the outside atmospheric pressure is8mm of water column.
P = (0.8) × 1 × 980
= 784 dyne/cm2
R =72
mm = 0.35 cm.
T =784 0 35
4× .
= 68.6 dyne/cm.
Q.74. Two spherical soap bubbles. A and B of radii 3 cm and 5 cm coalesce so as to havea portion of their surface in common. Calculate the radius of curvature of this commonsurface.
Solution. Let r be radius of curvature ofthe common surface of the soap bubbles A and B.
The excess pressure in A above atmos-pheric is 4T/3 while that in B is 4T/5. Thusthere is an excess pressure on the A-side ofthe common surface above that on the B-side
of 43
45
T T− . But this excess pressure must be
equal to 4T/r. Hence
3cm 5cm
A B
r
Fluids 517
4Tr
=43
45
T T−
1r
=13
15
−
r = 7.5 cm.
Q.75. A capillary tube of 0.8 mm bore stands vertically in a wide vessel containing a liquidof surface tension 30 dyne/cm. The liquid wets the tube and has a density of 0.8 gm/cm3.Calculate the rise of the liquid in the tube. What would be the rise if the capillary is inclinedat 45° with the vertical?
Solution. The surface tension T of a liquid is related to the rise h in a capillary tubeof radius r by
T =r h r g+
3
2
ρ
θcos
Where ρ is the density of liquid. If the liquid wets the tube the angle of contact θ = 0.
Further, r3
can be neglected as compared to h. Then
T =rh gρ
2
h =2Tr gρ
T = 30 dyne/cm, r = 0 82. mm = 0.04 cm and ρ = 0.8 gm/cm2
h =2 30
0 04 0 8 980×
. × . × = 1.9 cm
When the capillary is inclined at 45° with the vertical, the vertical height of liquid in itwill still be h = 1.9 cm. The length of water column in the column would be
hcos45° = 1.9 × 2 = 2.7 cm.
Q.76. Calculate the diameter of a capillary tube in which a liquid of specific gravity 0.85and surface tension 35.0 dynes/cm rises 2.50 cm. [Ans. 0.067 cm.]
Q.77. A liquid of specific gravity 1.5 is observed to rise 3.0 cm in a capillary tube ofdiameter 0.50 mm and the liquid wets the surface of the tube. Calculate the excess pressureinside a-spherical bubble of 1.0 cm diameter blown from the same liquid.
Ans. The surface tension of the liquid is
T =rh gρ
2 =
0 025 3 1 5 9802
. × × . × = 55 dyne/cm
Hence excess pressure inside a spherical bubble
P =4TR
=4 55
0 5×.
= 440 dyne/cm2.
518 Mechanics
Q.78. By how much will the surface of a liquid be depressed in a glass tube of radius 0.02cm if the angle of contact of the liquid is 135° and its surface tension is 547 dyne/cm. Assumethe density of the liquid to be 13.5 gm/cm3 and acceleration due to gravity 981 cm/sec2.
Solution. The surface tension T of liquid is related to its elevation or depression h ina capillary tube of radius r by
T =r h
rg+
3
2
ρ
θcos
Where ρ is the density of liquid and θ the angle of contact. The factor r3
is small
compared to h. Therefore, neglecting it
T =rh gρ
θ2 cos
h =2Tcosθ
ρr g
r = 0.02cm, θ = 135° so that Cos 135° = –1
2 = –0.707
T = 547 dyne/cm, ρ = 13.5 gm/cm3, g = 981 cm/sec2.
h =2 547 0 7070 02 13 5 981× × ( . ). × . ×
− = –2.92 cm
The negative sign indicates that the liquid is depressed.
Q.79. Water rises to a height of 10.0 cm in a certain capillary tube. In the same tube thelevel of mercury surface is depressed by 3.42 cm. Compare the surface tensions of water andmercury. Specific gravity of mercury is 13.6, the angle of contact for water is zero and thatfor mercury is 135°.
Solution. The surface tension T of a liquid is related to its elevation or depression hin a capillary tube of radius r by
T =rh gρ
θ2cos
where ρ is the density of the liquid and θ the angle of contact. Let Tω and Tm be the surfacetensions of water and mercury respectively
Then Tω =r g× . × ×
cos10 0 12 0° = 5rg.
and Tm =r g× ( . ) × . ×
cos−
°3 42 13 62 135 =
r g× ( . ) × . ×× ( . )
−−
3 42 13 62 0 707
=3 42 13 62 0 707. × .
× .rg = 33rg.
TT
ω
m
= 533
rgrg
= 0.15
Fluids 519
Q.80. Ripples are produced on the surface of liquid with the help of a tuning fork offrequency 60 and the wavelength observed is 0.5 cm. Calculate the surface tension of theliquid. Density of liquid is 1.0 gm/cm3 and g = 980 cm/sec.
Solution. If n be frequency and λ the wavelength of ripples formed on a liquid of densityρ, then the surface tension of the liquid is given by
T =n g2 3 2
22 4λ ρπ
λ ρπ
− =
λ ρπ
λπ
22
2 2n g−
Substituting the given values,
T =( . ) ×
× .( ) ( . )
× .0 5 12 3 14
60 0 5 9802 3 14
22 −
= 0.04 × [1800 – 156] = 0.04 × 1644 = 65.8 dyne/cm.
Q.81. Water is flowing through two horizontal pipes of different diameters which areconnected together. In the first pipe the speed of water is 4m/s and the pressure is 2.0 × 104
N/m2. Calculate the speed and pressure of water in the second pipe. The diameter of the pipesare 3cm and 6cm respectively.
Solution. If A is the area of cross-section of a pipe at a point and v is the velocity offlow of water at the point, then by the principle of continuity, we have
Av = Constant
A1v1 = A2v2
(πr12)v1 = (πr2
2)2 v2
v2 =rr
v1
2
2
1
Here r1 = 1.5 cm = 1.5 × 10–2 m, r2 = 3 cm = 3 × 10–2 m and v1 = 4 m/s
v2 =1 5 103 10
2
2
2. ×
×
−
−
× 4 = 1 m/s
By Bernoulli’s theorem:
P1 121
2+ ρv = P2 2
212
+ ρv
P2 = P (1 12
221
2+ −ρ v v )
For water ρ = 1 × 103 kg/m3, v1 = 4 m/s, v2 = 1 m/s and P1 = 2.0 × 104 N/m2
P2 = 2.0 × 104 + 12
× 1 × 103 × (42 – 12)
= 2.0 × 104 + 0.75 × 104
= 2.75 × 104 N/m2.
Q.82. The pressure difference between two points along a horizontal pipe, through whichwater is flowing, is 1.4 cm of mercury. If, due to non-uniform cross-section, the speed of flowof water at the point of greater cross-section is 60 cm/sec, Calculate the speed at the otherpoint.
520 Mechanics
Solution. By Bernoulli’s theorem
P1 +12
ρv12 = P2 +
12
ρv22
v22 – v1
2 =2ρ
( )P P1 2−
The speed of water will be greater at the place when the cross-section is smaller
∴ v22 =
212
ρ( )P P1 2− + v
Here P1 – P2 = 1.4 cm of mercury
= 1.4 × 10–2 × 13.6 × 103 × 9.8
= 1.866 × 103 N/m2,
P = 1 × 103 kg/m2, v1 = 60 cm/sec = 0.6 m/s
v22 =
21 103× × 1.866 × 103 + (0.6)2 = 4.092
v2 = 2 m/s (approx.).
Q.83. Water flows into a horizontal pipe whose one end is closed with a valve and thereading of a pressure gauge attached to the pipe is 3 × 105 N/m2. This reading of the pressuregauge falls to 1 × 105 N/m2 when the valve is opened. Calculate the speed of water flowinginto the pipe.
Solution. According to Bernoulli’s theorem
P1 + 12 1
2ρv = P2 + 12 2
2ρv
12
ρ( )22
12v v− = P1 – P2
Here v1 = 0 (the valve is initially closed and so the velocity of water is zero)
∴ v22 =
2ρ (P1 – P2)
=2
1 103× × [3 × 105 – 1 × 105] = 2 × 2 × 102 = 400
v2 = 20 m/s.
Q.84. A horizontal tube has different cross-sectional areas at points A and B. The diameterof A is 4 cm and that of B is 2 cm. Two manometer limbs are attached at A and B. Whena liquid of density 0.8 gm/cm3 flows through the tube, the pressure difference between thelimbs of the manometer is 8 cm. Calculate the rate of flow of the liquid in the tube.
Solution. The rate of liquid flow in non-uniform horizontal tube is given by
Q = A1A2 2
12
22
gh
A A−
Fluids 521
A1 = πr12 = π ×
42
2 = 4π cm2, A2 = πr2
2 = π22
2 = π cm2, g = 980 cm/s2, h = 8cm.
Q1 = 4π × π 2 980 84 2 2× ×
( ) ( )π π−
= 42 980 8
15π × ×
= 4 × 3.14 × 32.3 = 406 cm3/s.
Q.85. Water tank has a hole in its wall at a distance of 10 m below the free surface ofwater. The diameter of the hole is 2 mm. Compute the velocity of efflux of water from the holeand the rate of flow of water.
Solution. The velocity of efflux of water is
v = 2gh = 2 9 8 10× . × = 14 m/s
The rate of flow of water is
A × v = πr2 × v
= 3.14 (1 × 10–3)2 × 14
= 4.4 × 10–5 m3/s.
Q.86. The relative velocity between two layers of water is 8.0 cm/s. If the perpendiculardistance between the layers is 0.1 cm, find the velocity gradient.
Solution. Velocity gradient = ∆∆vx
Z
Relative velocity between layers, ∆vx = 8.0 cm/s and distance between the layers ∆Z =0.1 cm.
∴ Velocity Gradient =8
0 1. = 80 per/sec.
Q.87. There is a 1 mm thick layer of glycerine between a flat plate of area 100 cm2 anda big plate. If the coefficient of viscosity of glycerine is 10 kg/m-s. Then how much force isrequired to move the plate with a velocity of 7 cm/s.
Solution. To move the plate with a constant velocity, the necessary force will be equalto the viscous force F (say). Now
F = ηA∆∆vzx
η = 1.0 Kg/m-s, A = 100 cm2 = 10–2 m2, ∆vx = 7 × 10–2 m/s and ∆z = 1 mm = 10–3 m
F =1 0 10 7 10
10
2 2
3. × × ×− −
− = 0.7 N.
Q.88. A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosityof air is 1.8 × 10–5 Kg/m-s. What will be the terminal velocity of the drop? (Density of water= 1.0 × 103 kg/m3 and g = 9.8 N/kg) Density of air can be neglected.
Ans. By Stokes’ law, the terminal velocity of a water drop of radius r is given by
v =29
2r g( )ρ ση−
522 Mechanics
Where ρ is the density of water, σ is the density of air and η the coefficient of viscosityof air. Here σ is negligible and r = 0.0015 mm = 1.5 × 10–3 mm = 1.5 × 10–6m.
v =29
1 5 10 1 0 10 9 81 8 10
6 2 3
5× ( . × ) × ( . × ) × .. ×
−
−
= 2.72 × 10–4 m/s.
Q.89. A metallic sphere of radius 1.0 × 10–3 m and density 1.0 × 104 kg/m3 enters a tankof water, after a free fall through a distance of h in the earth's gravitational field. If its velocityremains unchanged after entering water, determine the valve of h. Given coefficient of viscosityof water = 1.0 × 10–3 N-S/m2, g = 10 m/s2 and density of water = 1.0 × 103 kg/m3.
Solution. The velocity attained by the sphere in falling freely from a height h is
v = 2gh ...(i)
This is the terminal velocity of the sphere in water. Hence by Stokes’ law.
we have v =29
2r p g( )− ση
where r is the radius of the sphere, ρ is the density of the material of the sphere, σ(= 1.0 × 103 kg/m3) is the density of water and η is coefficient of viscosity of water
∴ v =2 1 0 10 1 0 10 1 0 10 10
9 1 0 10
3 2 4 3
3
× ( . × ) × ( . × . × ) ×× . ×
− +
−−
= 20 m/s
From Eq. (i) we have
h =vg2
2 =20 202 10
×× = 20 m.
Q.90. An air bubble of radius 1 cm rises up in a liquid column with terminal velocity of0.21 cm/s. If the density of liquid be 1.47 × 103 kg/m3, then Calculate the coefficient of viscosityof the liquid. Density of air is negligible. (g = 9.8 m/s2)
Solution. The weight of the air bubble is negligible two forces act upon it: (i) upthrust
of the liquid 43
3πr σg, where r is the radius of the bubble and σ is the density of the liquid
and (ii) viscous force 6πηrv. Since the bubble has acquired terminal velocity, therefore
6πηrv =43
3π σr g
η =29
2r gvσ
η = 29
1 10 1 47 10 9 80 21 10
2 2 3
2( × ) × ( . × ) × .
. ×
−
−
= 1.52 × 102 kg/m-s = 1.52 × 103 Poise.
11.1 PERIODIC MOTION
When a body repeats its motion continuously on a definite path in a definite interval oftime then its motion is called ‘periodic motion’, and the interval of time is called ‘time-period’,T. The earth completes one rotation around the sun is 1 year. This motion of the earth isperiodic motion whose time-period is 1 year.
If a body is periodic motion moves along the same path to and fro about a definite point(equilibrium position), then the motion of the body is a ‘vibratory motion’ or ‘Oscillatorymotion’. As the body goes to one side of its equilibrium position, comes back to that position,goes to the other side, and again returns to the same position; it is said to complete ‘onevibration’ or ‘one oscillation’.
If figure, drawn the time-displacement curve of a body oscillating to and fro about itsequilibrium position. The displacement y is measured from the equilibrium position and thetime t is measured from the instant when the body passed through its equilibrium position.One oscillation is completed from the point O to the point A and the time T taken for oneoscillation is the periodic-time of the body.
T
AO
t
Y
Fig. 1
11.2 SIMPLE HARMONIC MOTION (S.H.M.); AS A PROJECTION OF UNIFORMCIRCULAR MOTION
Simple harmonic motion is the simplest form of vibratory or oscillatory motion. Supposea particle P is moving with uniform speed along the circumferences of a circle of radius aand center O. When the particle is at the point P, then the foot of the perpendicular drawn
11
523
524 Mechanics
from the particle on the diameter AA' of the circle is at the point N. When the particle wasat B, the foot of the perpendicular was at the point O. When the particle moving along thecircumference reaches the point A, the foot of the perpendicular moving along the diameteralso reaches A. When the particle moving along the circumference reaches the point B', thefoot moving along the diameter reaches from A to O. When the particle goes along thecircumference from B' to A', the foot moves along the diameter from O to A', and when theparticle moves from A' to B, the foot moves from A' to O. Thus the foot N moves in a straightline to and fro about the point O. This straight-line motion of N is simple harmonic motion.The motion of N from O to A, from A to A' and from A' to O is called 1 vibration.
Thus, if a particle is moving with uniform speed along the circumference of a circle, thenthe straight-line motion of the foot of the perpendicular drawn from the particle on thediameter of the circle is called ‘simple harmonic motion’. The circle is called the ‘referencecircle’ of the simple harmonic motion. This is the kinematical definition of simple harmonicmotion.
A ′
A
P
B B′ O
a
θ = ωt
θN
y
Fig. 2
11.3 DISPLACEMENT EQUATION OF S.H.M.
Suppose the particle P starts from the point B and rotates through an angle of θ radianin t second. If the angular velocity of the particle be ω, then ω = θ/t or θ = ω t. In t secondthe displacement of N from O is ON. If this displacement is written as y, then
y = ON = OP sin NPO
But OP = a and ∠NPO = ∠POB = θ = ω∴ y = a sin ωωωωωt
This is the displacement equation of the simple harmonic motion.
Amplitude: The maximum value of sin ωt is 1. Hence, by above equation, the maximumvalue of the displacement y will be a. This maximum displacement is called the ‘amplitude’of motion. It is equal to the radius of the reference circle.
Periodic Time: The time taken by N to complete one vibration is called the ‘periodic-time’T of N. The time during which N completes one vibration, the particle P completes one roundof the circle of reference, that is, it rotates through an angle of 2π radian. Thus the time ofone rotation of P is 2π/ω. Therefore, the periodic-time of N is given by
T = 2πω
Hormonic Motion 525
Frequency: The number of vibrations completed by N in one second is called the‘frequency’ (n) of N. The frequency is reciprocal of the periodic-time,
n =1
2T= ω
πPhase: When a particle vibrates, its position and direction of motion vary with time. The
phase of a vibrating particle at any instant expresses the position and direction of motion ofthe particle at that instant.
A ′
A
P
B B′ O
αN
y
φωt
P0
Fig. 3
The phase is expressed either in terms of the angle θ or in terms of the periodic-time T.For example, in figure, when the particle P reaches the point A, then its phase is π/2 or T/4.
If at any instant two vibrating particles are passing simultaneously through theirequilibrium positions in the same direction, then at that instant they are in the same phase,and if they are passing in opposite directions then they are in opposite phase.
Suppose the time is measured from the instant when the particle moving on the referencecircle was at a point different from B, say at P0, where ∠PoOB = φ. Then the displacementy (=ON) of N from the equilibrium position O after a time t will be given by
y = ON = OP sin NOP = OP sin POB
y = a sin (ωωωωωt + φφφφφ)
ωt is the angle through which the particle P has rotated in moving from Po to P in timet. Thus at the instant t the phase of N will be measured by the total angle (ωt + φ). φ is calledthe ‘initial phase’ or ‘epoch’ of N. The above equation is the general equation of simpleharmonic motion.
11.4 CONDITIONS FOR LINEAR S.H.M.
The following three conditions should be fulfilled for the linear S.H.M. of a particle:
1. The motion of the particle should be in a straight line to and fro about a fixed point.
2. The restoring force (or acceleration) acting on the particle should always beproportional to the displacement of the particle from that point.
3. The force (or acceleration) should always be directed towards that point.
The S.H.M. is Defined on the Basis of These ConditionsWhen a particle moves in a straight line to and fro about its equilibrium position in such
a way that the force (or acceleration) acting upon it is always directly proportional to its
526 Mechanics
displacement and directed towards the equilibrium position, then the motion of the particleis called ‘simple harmonic motion’.
11.5 EQUATION OF MOTION OF A SIMPLE HARMONIC OSCILLATOR
Let us consider a particle P of mass m executing S.H.M. about an equilibrium positionO. By definition, the force under which particle is oscillating is proportional to the displacementof P from O and is directed toward O. Thus to the displacement of P from O at any instantt, the instantaneous force F acting upon it is given by
F = – kx
Where k is the proportionality factor, the minus sign shows that the force F is oppositeto the displacement x.
According to the Newton’s second law, the force acting on the particle is equal to theproduct of the mass and the acceleration. The instantaneous acceleration of the particle is
d xdt
2
2 . Hence
–kx = m d xdt
2
2
ord xdt
2
2 = – (k/m) x
Let us put the constant, k/m = ω2. Thus we have
ord xdt
2
2 + ω2x = 0 ...(i)
This is the differential equation of a simple harmonic oscillator.
Let a solution to this equation be
x = Ceαt
Where C and α are arbitrary constant. On differentiating with respect to t, we get
dxdt
= Ceαt.α
andd xdt
2
2 = Ceαt.α2
Substituting these values of d xdt
2
2 and x in Eqn. (i), we get
C Ce et tα αα ω2 2+ = 0
or C )e tα α ω( 2 2+ = 0
or (α ω2 2+ ) = 0
or α = ± − = ±( )ω ω2 j
Hormonic Motion 527
Where j = −1. Thus, there are two possible solutions from Eqn. (i), which are
x = C +e j tω
and x = Ce j t− ω
Hence the general solution to Eqn. (i) will be the sum of these two solutions, i.e.,
x = C C1 2e ej t j t+ −+ω ω
Where C1 and C2 are arbitrary constants. This gives
x = C1 (cos ωt + j sin ωt) + C2 (cos ωt – j sin ωt)
= (C1 + C2) cos ωt + j (C1 – C2) sin ωt
Let us make a change in arbitrary constants by putting
C1 + C2 = a sin φand j(C1 – C2) = a cos φ
Where a and φ are now constants. This gives
x = a sin φ cos ωt + a cos φ sin ωt
or x = a sin (ωt + φ)
This is, in fact, a solution of the equation of S.H.M. The values of the constants a andφ depends upon how the motion is started.
11.6 IMPORTANCE OF S.H.M.
The study of simple harmonic motion is important for two reasons.
(i) There are a great many physical problem in mechanics, acoustic, optics, electricityand in atomic and molecular Physics, in which the force on a system is approximatelyproportional to its displacement from some equilibrium position. In such cases theresulting motion may be represented approximately be the simple harmonic model.
(ii) Even the complicated periodic motion occurring in physical problem can berepresented as a combination of a number of simple harmonic motion havingfrequencies which are multiple of that of the complicated motion. The vibratingstrings and membranes, the vibrations of atoms in solid, the electrical and acousticaloscillations in a cavity can be treated in this manner.
11.7 ENERGY OF HARMONIC OSCILLATOR
A harmonic oscillator possess two types of energy:
(i) Potential energy, which is due to its displacement from the mean position.
(ii) Kinetic energy, which is due to its velocity.
Hence, at any instant the total energy of the oscillator will be the sum of these twoenergies. As the system is conservative (i.e., no dissipative or frictional forces are acting), thetotal mechanical energy E (=K + U) must be conserved.
Let the displacement of the harmonic oscillator at any instant t be given by
x = a sin (ωt + φ)
Its velocity v =dxdt
a t a x= + = −ω ω φ ωcos ( ) 2 2
528 Mechanics
Accelerationd xdt
2
2 = − + = −ω ω φ ω2 2a t xsin ( )
∴ Force = –m ω2x = –Cx, ω2 = C/m
If the oscillator is displaced through dx distance, then work done on the oscillator is
dW = Cx dx
If it is displaced from x = 0 to x = x, then work done
W = C Cxdx xx
0
212 = .
This work done on the oscillator becomes its potential energy U, i.e.,
U = ½ Cx2
Kinetic energy of the oscillator at the displacement x is
K = ½ mv2 = ½ mω2(a2 – x2) = ½ C (a2 – x2)
∴ Total energy E = U + K = ½ Cx2 + ½ C (a2 – x2) = ½ Ca2
= ½ mω2 a2 = ½ m (2π/T)2 a2
= 2m π2 a2 / T2
or E = 2mπππππ2a2n2
Thus, we see that the total mechanical energy is constant, as we expect and has thevalue ½ Ca2, i.e., the total energy is proportional to the square of the amplitude. Hence, ifthe system is once oscillated the motion will continue for indefinite period with t any decreasein amplitude, provided no damping (frictional) forces are acting on the system.
E (To ta l E nergy)
Energy
P.E. CurveU = ½ Cx2
K.E. CurveK = ½ C(a – x22 )
–a +aO
x
Fig. 4
If we plot a graph between the potential energy U and displacement x, we get a parabolahaving vertex at x = 0. The curve for the total energy E (constant) is the horizontal line. Theparticle cannot go beyond the points where this line intersects the potential energy curvebecause U can never be larger than E. These points are turning points of the motion andcorrespond to the maximum displacement. At these positions (i.e., x = ±a) the total energyof the oscillator is wholly potential (i.e., E = U = ½ Ca2) but the kinetic energy is zero andso the amplitude of motion is a = ± 2E /C. At the equilibrium position, the potential energyis zero, but the kinetic energy has the maximum values K = ½ mv2
max = ½ Ca2 = E with
maximum velocity vmmax .= 2E
While at other intermediate points the energy is partly
kinetic and partly potential but their sum (total energy) is always ½ Ca2.
Hormonic Motion 529
11.8 AVERAGE VALUES OF KINETIC AND POTENTIAL ENERGIES
The instantaneous displacement of particle of mass m executing S.H.M. under a force-constantk is
x = a sin (ωt + φ) where ω2 = k/m
The instantaneous kinetic energy is
K =12 2
2 22 2m dx
dtm a t
= +ω ω φcos ( )
The time-average of the kinetic energy over a period T of the motion is
Kav =
kdt m a t dt0
2 2 2
0
2
1 2
2
T
T
=
+/ cos ( )
/
/
ω ω φ
π ω
π ω
, T = 2π/ω
=m a t dtω
πω φ
π ω3 2
0
2
41 2+ + cos ( )
/
=m a dt t dtω
πω φ
π ωπ ω3 2
0
2
0
2
42+ +
cos ( )
//
=m a
m aω
ππ
ωω
3 22
42
012
+
= 2
Kav = ½ ka2
The potential energy of the particle is
U = ½ kx2 = ½ ka2 sin2 (ωt + φ)
The time-average of the potential energy over a period T of the motion is
Uav =
U
T
T
dt ka t dt0
2 2
0
2
1 2
2
=
+( / ) sin ( )
/
/
ω φ
π ω
π ω
=k a t dtω
πω φ
π ω2
0
2
41 2− + cos ( )
/
= k a dt t dtωπ
ω φπ ωπ ω2
0
2
0
2
42− +
cos ( )
//
=k aω
ππ
ω
2
42
0−
Uav = ½ ka2
530 Mechanics
Thus the time-average kinetic energy is equal to the time-average potential energy andeach is equal to ½ ka2, which is half the total energy. This is a characteristic property of theharmonic oscillator and is not true for anharmonic oscillators.
11.9 POSITION-AVERAGE OF KINETIC AND POTENTIAL ENERGIES
Let us now find out the position-average of the kinetic and potential energies. Thekinetic energy is
K = ½ m (dx/dt)2 = ½ k (a2 – x2)
The position-average over the displacement from x = 0 to x = a is
K =
Kdx
a
k a x dx
a
a a
0
2 2
0
1 2 =
−/ ( )
=12 3
12 3
13
23
0
33
2ka
a xx k
aa
aka
a
−
= −
=
The potential energy is
U =12
kx2
Its position-average is
U =
Udx
a
k x dx
aka
x ka
a a
a0
2
03
0
2
1 212 3
16
= =
=
/
Thus the position-average kinetic energy (1/3 ka2) is not equal to the position averagepotential energy (1/6 ka2).
11.10 FRACTIONS OF KINETIC AND POTENTIAL ENERGIES
The equation for S.H.M. is
x = a sin (ωt + φ) where ω2 = k/m
So that
u =dxdt
a t km
a a t= + = ± − +ω ω φ ω φcos ( ) ( sin ( ))2 2 2
= ± −km
a x( )2 2
The total energy of the particle in S.H.M. is
E = K + U
=12
mv2 + 12
kx2
=12
k (a2 – x2) + 12
kx2
=12
ka2
Hormonic Motion 531
Now the displacement is one-half the amplitude i.e., at x = a/2, we have
(U)a/2 = 12
kx2 = 12
k (a/2)2 = (1/8) ka2 = 14
E
and (K)a/2 = 12
k (a2 – x2) = 12
k (a2 – a2/4) = (3/8) ka2 = 34
E
Thus the potential energy is one-fourth and the kinetic-energy is three-fourth of thetotal energy.
11.11 MASS ATTACHED TO A HORIZONTAL SPRING
Suppose a mass m attached to one end of a massless spring is free to move on africtionless horizontal surface. Let x = 0 be the equilibrium position of the mass at which thespring is unscratched. When the mass is displaced toward right through a distance x, thenthe spring exerts an elastic restoring force F on the mass. By Hooke’s Law,
F = – kx
Where k is the force-constant of the spring. The negative sign indicates that F is directedtowards left (opposite to displacement). If we displace the mass toward left, then F is directedtoward right as in Figure (c). When the displaced mass is released, it oscillates to and froabout the equilibrium position.
m( )a
x
( )b
( )c
x = 0F = 0
F
x
F
Fig. 5
At any instant during oscillation the force F is equal to the mass m multiplied by the
instantaneous acceleration d xdt
2
2 . That is
F = –kx = md xdt
2
2
Where ω2 = k/m. This is the differential equation of the simple harmonic motion of themass. Its solution is
x = a sin (ωt + φ)
Where ω and φ are constants.
If t is replace by t + 2π/ω, the displacement x will be the same. Hence the period of themotion
T =2 2πω
π= mk
532 Mechanics
Now, suppose the spring has a finite mass ms. If ms<< m, then the spring will stretchuniformly along its length, that is, the displacement of the different elements of the springwill be proportional to their respective distances from the fixed end.
Let l be the length of the spring, then its mass per unit length is ms/l. Let ds be a smallelement of length at a distance S from the fixed end. Its mass is (ms/l) ds.
When the displacement of the mass m is x, the displacement of the element ds will be(s/l)x and the velocity will be (s/l)(dx/dt). Hence the instantaneous energy of the element
=12
2ml
ds sl
dxdt
s
=ml
dxdt
s dss
2 3
22
The instantaneous kinetic energy of the whole spring is
ml
dxdt
s dssl
2 3
22
0
=
16
2
mdxdts
The kinetic energy of the system (mass + spring) is
K =12
16
2 2
m dxdt
m dxdts
+
=12 3
2
mm dx
dts+
This shows that the effective mass of the system is (m + ms/3). Hence the new time-periodis
T = 2 3πm m
k
s+
11.12 MOTION OF A BODY SUSPENDED BY A VERTICAL SPRING
Suppose a light spring, whose normal length is l,is hanging from a rigid support. When a body of massm is suspended from its lower end, then due to theweight of the body, the length of the spring is extendedsay by xo. The spring exerts an elastic (restoring)upward force F on the mass m. By Hooke’s law,
F = – kx0
Where k is the force-constant of the spring. Theother force acting on the mass is its weight, mg. Thusthe total force acting on m is –kx0 + mg. Since the masshas acceleration, the total force on it is zero, that is,
–kx0 + mg = 0
or x0 = mg/k ...(i)
m
( )am
( )b
( )c
x0
xm g
m g
F
F ′
l
Fig. 6
Hormonic Motion 533
This is the static equilibrium position of the mass m.
Now, suppose during oscillation, the total extension in the spring at any instant is x0+x(figure c). The (upward) force exerted on m by the spring is then
F' = –k (x0 + x)
= –k (mg/k + x)
= –mg – kx
The other force acting on the mass m is its weight, +mg (downward). Therefore, the netforce acting on the mass is
(–mg – kx) + mg = –kx
This, by Newton’s second law, must be m times the instantaneous acceleration d xdt
2
2 of
the mass. That is
–kx = m d xdt
2
2
ord xdt
2
2 = –(k/m) x = –ω2k
where ω2 = k/m. This is the equation of the S.H.M. Thus the mass m executes S.H.M. aboutthe position x0 = mg/k. The time period is
T =2
2π
ωπ= m
k...(ii)
we have k = mg/x0. Therefore
T = 2 0π xg
...(iii)
Eqn. (ii) and (iii) both can be used to determine the period of oscillations of themass-springsystem. Eqn. (iii) has the advantage that we can find the period of oscillation simply bymeasuring the extension x0 of the spring without knowing the suspended mass and the force-constant of the spring.
Spring cut in two halves: For a given mass hung from the spring, the extension x0 isproportional to the unstretched length l of the spring. Therefore, for half length (l/2) of thespring the extension will be x0/2. Hence the new time-period will be
T = 220π ( / )x
g= T/ 2
on the new frequency is √2 times the odd frequency.
11.13 MASS SUSPENDED BY A HEAVY SPRING
Suppose a mass m is suspended from the end of a spring of mass ms. If ms << m, thenthe spring will stretch uniformly along its length. Let l be the length of the spring. Then itsmass per unit length is ms/l. Let us consider an element of length ds at a distance S fromthe fixed end of the spring. The mass of this element is (ms/l) dS.
534 Mechanics
Let x be the instantaneous displacement of the mass m from its position of rest. Sincethe displacement is proportional to the distance from the fixed end, the displacement of the
element considered will be (S/l) x, and its instantaneous velocity is Sl
dxdt
. Hence the
instantaneous kinetic energy of the element.
=12
2ml
ds sl
dxdt
s
=ml
dxdt
s dss
2 3
22
If the spring is uniform, the total kinetic energy of the suspended mass and the springis
K =12 2
2
3
22
0
mdxdt
ml
dxdt
s dssl
+
=12 2 3
2
3
2 3m dx
dtml
dxdt
ls +
=12 3
2
mm dx
dts+
.
Now, when the mass is at a distance x from its position of rest, the elastic potentialenergy of the system is
U =12
kx2
Where k is the force-constant of the spring. Therefore, the total energy of the system is
E = K + U = 12 3
12
22m
m dxdt
kxs+
+
Since E remains constant; dE/dt = 0. that is
=12 3
212
2 02
2mm dx
dtd xdt
k xdxdt
s+
+ =( )
mm d x
dtkxs+
+
3
2
2 = 0
ord xdt
2
2 = –k
m mx x
s+
= −
3
2. ω
where ω2 =k
m ms+
3
Hormonic Motion 535
Thus the motion is simple harmonic and its period is
T = 2 2 3πω
π=+
m m
k
s
The frequency is
n =1 1
23
T=
+
πk
m ms
If the spring be massless, the period is T = 2π mk
. Thus when the mass of the spring
is taken into account, the period is increased or the frequency is decreased.
11.14 OSCILLATIONS OF A FLOATING CYLINDER
Let us consider a cylinder of mass ‘m’ and cross-sectional area A floating vertically ina liquid of density ρ. Let l be the immersed length of the cylinder. At equilibrium, the weightof the cylinder is equal to the upthrust of the displaced liquid (mg = Alρg). When the cylinderis slightly pressed down and released, it oscillates up and down.
Let x be the instantaneous displacement of the cylinder. Then the mass of the furtherdisplaced liquid will be Axρ and the corresponding upthrust will be Axρg.
This is the restoring force F on the cylinder, that is,
F = –Axρg
The force is directed to the displacement x.
By Newton’s second law, the force F equals and
product of mass m and acceleration d xdt
2
2 of the cylinder.
Thus
–Axρg = m d xdt
2
2
ord xdt
2
2 = − = −Aρ ωgm
x x2
where ω2 = Aρg/m. This represents a S.H.M. of period.
T =2 2πω
πρ
= mgA
But m = Alρ
∴ T = 2π lg
m
ml
x
Fig. 7
536 Mechanics
11.15 OSCILLATIONS OF A LIQUID IN A U-TUBE
Suppose a U-tube of uniform cross-section A contains a liquid. Let ρ be density of theliquid and l be the total length of the liquid in the tube. Suppose, in the equilibrium position,the levels of the liquid are at B and C which are in the same horizontal plane. If the levelat C is depressed downwards to a distance x, the level at B will rise to the same distancex. On releasing the liquid column oscillates up and down in the tube.
Let x be the instantaneous displacement of the liquid level below C. Then the length ofthe extra column of the left hand limb will be 2x. The weight of this extra column is 2xAρgand this is the restoring force F acting on the liquid. That is
F = –2x Aρg
By Newton’s law, this equals the product of the mass m of the liquid and its acceleration
d xdt
2. Thus
–2xAρg = m d xdt
2
or d xdt
2
2= − = −2Aρ ωg
mx x2
Where ω2 =2Aρg
m. This represents a S.H.M. of period
T =2 2πω
πρ
= mg2A
But m = lAρ
T = 2π gl2
11.16 HELMHOLTZ RESONATOR
It consists of a large spherical glass vessel having a small cylindrical neck A and anarrow stem B just opposite to A as shown in figure 9. When the neck A is directed towardsa sound source, the air in the vessel vibrates with a natural frequency depending upon thevolume of the vessel and the dimension of its neck.
Let l be the length of the neck; A its area of cross-section and V the volume of the vessel.The mass of the air within the neck is
m = lAρWhere ρ is the density of air. When sound waves enter the neck, the air in it acts like
a piston, alternately compressing and rarifying the air inside the vessel. If the air within theneck moves inward through a small distance x then the air inside the vessel is compressedby an amount v is given by
v = xA
The resulting increase in pressure, ρ, given by Hooke’s law as
x
CB
l
x
Area
Fig. 8
Hormonic Motion 537
K =ρ
v / V(Bulk modulus =
Volume stressVolume strain
)
or p = −KVv
Therefore the (restoring) force on the air in the neck
F = pA = − = −K AV
K AV
2v x
(Since v = xA)
But by Newton’s second law
F = md xdt
2
2
where m (= l Aρ) is the mass of the air in the neck. Hence
− K AV
2x= m
d xdt
ld xdt
2
2
2
2= Aρ
ord xdt
2
2 = − = −KAVl
x xρ
ω2
where ω2 = KA/(lρV). This is the equation of S.H.M. whose frequency is given by
n =ωπ π ρ2
12
= KAVl
Now Kρ
= v, the velocity of sound in air.
∴ n =v
l2πAV
This is the natural frequency of the air of the resonator. If a sound wave is incident onthe neck A, then the air will, resonate only if natural frequency is equal to the frequency ofthe incident wave. To detect the resonance easily, the stem B is pressed into the ear.
11.17 COMPOSITION OF TWO SIMPLE HARMONIC MOTIONS OF EQUAL PERIODS IN A STRAIGHT LINE
If two simple harmonic motions are acting on a particle simultaneously, it shall performresultant vibration, which is obtained by compounding the component vibrations. There aretwo methods of compounding two S.H.Ms. the geometrical methods and the analytical method.
We will discuss here only analytical method.
Let two S.H.Ms. of equal periods (= 2 π/ω) but of different amplitudes and phases besimultaneously acted on a particle in the same straight line. Let the two component vibrationsbe represented by
x1 = a1 sin (ωt – φ1) ...(i)
and x2 = a2 sin (ωt – φ2) ...(ii)
A
B
Fig. 9
538 Mechanics
where a1 and a2 are the amplitudes and φ1 and φ2 the epochs or initial phases of thecomponent vibrations. The phase difference between them is (φ1 – φ2). The resultantdisplacement from the principle of superposition, is given by
x = x1 + x2
= a1 sin (ωt – φ1) + a2 sin (ωt – φ2)
= a1[sin ωt cos φ1 – cos ωt sin φ1]+ a2 [sin ω t cos φ2 – cos ωt sin φ2)
= sin ωt [a1 cos φ1 + a2 cos φ2 ] – cos ωt [a1 sin φ1+ a2 sin φ2]
Now, let
r cos θ = a1 cos φ1 + a2 cos φ2 ...(iii)
and r sin θ = a1 sin φ1 + a2 sin φ2 ...(iv)
∴ x = r [sin ωt cos θ – cos ωt sin θ]
or x = r sin (ωt – θ) ...(v)
where r is the amplitude and (ωt – θ) the phase of the resultant vibrations. The above relationshows that the resultant motion is simple harmonic of the same period (2π/ω) but of differentamplitude r and epoch θ.
Squaring Eqns. (iii) and (iv) and adding, we get,
r2 = (a1)2 + (a2)2 + 2 a1 a2 cos (φ1 – φ2) ...(vi)
Dividing Eqn. (iv) by (iii), we have
tan θ =a aa a
1 1 2 2
1 1 2 2
sin sincos cos
φ φφ φ
++ ...(vii)
Let us consider some special cases:
(a) If φ1 – φ2 = 0 or 2nπ, where n is an integer, the two vibrations are in same phaseand the resultants amplitude has the maximum value, i.e.,
r = a a a a a a12
22
1 2 1 22+ + = +The negative root has been omitted, as the amplitude is essentially positive.
(b) If φ1 – φ2 = π or (2n + 1)π, where n is any integer, the two vibration are in theopposite phase and the resultant amplitude has the minimum values i.e.,
r = a a a a a a12
22
1 2 1 22+ − = −If in addition φ1 = φ2; the amplitude r = 0 i.e., the particle remains in rest position.
11.18 COMPOSITION OF TWO RECTANGULAR S.H.M. OF EQUAL PERIODS
Lissajous Figures: When a particle is vibrating, simultaneously under two simpleharmonic motions at right angles to each other, the resultant motion of the particle is calledLissajous figures. The nature of the motion, or the curve traced out, depends upon theamplitudes, frequencies and the phase difference of the two component motions. We shallconsider below some simple cases of the composition of two rectangular vibrations.
Let the two simple harmonic motions of equal periods be along the axis of X and axisof Y. If their displacement at any time t be x and y, then they can be represented by
Hormonic Motion 539
x = a sin (sin ωt + φ) ...(i)
y = b sin ωt ...(ii)
where a and b are the amplitudes of the component vibrations having each period 2p/w andthe first vibration is ahead of the second by a phase angle φ, i.e., the phase difference betweenthe two vibrations is φ.
The resultant motion can be obtained by eliminating t from Eqs. (i) and (ii). ExpandingEq. (i), we get
x/a = sin ωt cos φ + cos ωt sin φBut from Eq. (ii)
sin ωt = y/b and cos ωt = 12
2− yb
xa
=yb
yb
cos sinφ φ+ −12
2
orxa
yb
−
cosφ
2
= sin22
21φ −
yb
orxa
yb
xyab
2
2
2
22+ − cosφ = sin2φ ...(iii)
This expression represents the resultant motion, which is in general an ellipse inclinedto the axes of co-ordinates. There are, however, a number of special cases:
(a) when φ = 0 , the relation (iii), becomes
xa
yb
xyab
2
2
2
22+ − = 0
orxa
yb
−
2
= 0
± −
xa
yb = 0
This represent a pair of coincident straight line y = bx/a, passing through the originand lying in the first and third quadrants (Fig. a)
(b) when φ = π/2, the expression (iii) is
xa
yb
2
2
2
2+ = 1
This equation represents an ellipse, whose axes are coincident with the co-ordinateaxes (Fig. b).
Now, if a = b, the ellipse degenerates in the circle.
x2 + y2 = a2
540 Mechanics
Hence, two simple harmonic motions at right angles to each other of equal amplitudesbut with phase, differing by π/2, are equivalent to a uniform circular motion, theradius of the circle being equal to the amplitude of either simple harmonic motion.
(c) When φ = π, Eq. (iii) is
xa
yb
xyab
2
2
2
22+ + = 0 or
xa
yb
+
=
2
0
which represents a pair of coincident straight lines y = − bxa
, passing through the
origin and lying in the second and fourth quadrats as BOB' (Fig. c).
For any other values of φ, except discusses above, the path will be an ellipse,inclined to the co-ordinate axes.
O
Y
A2a
2b
Y′
XX ′O
Y2a
2b
Y ′
XX ′
O
Y
B2a
Y ′
XX ′
( ) = 0a φ ( ) = /2b φ π ( ) = 2c φ π
Fig. 10
11.19 COMPOSITION OF TWO RECTANGULAR S.H.M.’S OF TIME PERIODS NEARLY EQUAL
If the two component vibrations are exactly of equal periods, the elliptical paths, referredto above, remain perfectly steady. But, when the two time periods differ slightly from eachother, there comes about a gradual but progressive change in the relative phase (φ) of the twovibrations and consequently the shape of the ellipse slowly undergoes a change. All theseelliptic path lie within a rectangle of sides 2a and 2b.
Now, starting with the phase in agreement, i.e., φ =0, the ellipse coincides with the
diagonal xa
yb
−
= 0 of the rectangle.
As φ increases from o to π/2, the ellipse opens out to the form xa
yb
2
2
2
2 1+ = passing
through intermediate oblique positions.
When φ increase from π/2 to π, the ellipse closes up again and finally coincides with the
other diagonal xa
yb
+
= 0 of the rectangle.
As the phase difference changes from π to 2π , the reverse process takes place, until theellipse again coincides with the first diagonal. All these changes have been represented asfollows:
Hormonic Motion 541
It should be noted that the frequency of the complete cycle is equal to the difference ofthe frequencies of the two component simple harmonic vibrations.
Phase difference0
2π
π/4
7π/4
π/2
3π/2
3π/4
5π/4
π
Phase difference
Fig. 11
COMPOSITION OF TWO RECTANGULAR S.H.M. OF PERIODS IN THE RATIO 1:2
Let the component vibrations be represented by
x = a sin (2 ωt + φ) ...(i)
and y = b sin ωt ...(ii)
where φ is the phase angle by which the first motion is ahead of the second.
The resultant motion, which can be obtained by eliminating ωt from the equation (i) and(ii), in general will be a curve, having two loops for any value of phase difference andamplitudes.
Phase difference 1:2
Phased ifference1:2
0
2π 7 /4π
π/4
3 /2π
π/2
5 /4π
3π/4 π
Phase difference
Fig. 12
Here, we will consider only two special cases, given below:
(a) When φ = 0, we get
x = a sin 2 ωt = 2a sin ωt cos ωt
or x/a = 2 sin ωt cos ωt
orxa
=2 1
2
2y
byb
−
orxa
2
2 =4 1 4 1
2
2
2
2
2
2
2
2y
byb
yb
yb
−
= − −
orxa
yb
yb
2
2
2
2
2
24 1+ −
= 0
This equation represent the figure of ‘8’.
542 Mechanics
(b) When φ = π/2, we have
x = a sin (2ωt + π/2)
= a cos 2 ωt = a (1 – 2 sin2ωt), sin2ωt = y2/b2
xa
= 12 2
2− yb
or2 2
2y
b= 1 − = − −
x ax a
a/ ∴ y b
ax a2
2
2= − −( )
This is the equation of a parabola with vertex (a, 0).
If the time periods of the two rectangular vibrations depart slightly from the ratio 1:2,the form of the curve slowly varies as φ changes.
11.20 LISSAJOU’S FIGURES FROM TWO RECTANGULAR S.H.M. IN FREQUENCY RATIO 2:1
Let a particle be subjected to two mutually perpendicular simple harmonic vibrationsalong the x and y axes and having frequencies in the ratio 2:1. They may be represented by
x = a sin (2 ωt + φ) ...(i)
and y = b sin ωt ...(ii)
where a is the amplitude of the x-vibration whose angular frequencies is 2 ω, and b is theamplitude of the y-vibration whose frequency is ω. The phase difference is φ.
The equation to the curve of resultant motion can be obtained by eliminating t from Eqs.(i) and (ii). Expanding Eq. (i), we get
x/a = sin 2 ωt cos φ + cos 2 ωt sin φ= 2 sin ωt cos ωt cos φ + (1 – 2 sin2 ωt) sin φ
But from Eq. (ii) sin ωt = y/b and cos sinω ωt t yb
= − = −1 122
2
∴xa
=2 1 1 22
2
2
2y
byb
yb
− + −
cos sinφ φ
orxa
yb
− −
1 2 2
2 sinφ =2
12
2y
byb
−
cosφ
squaring xa
yb
xa
yb
yb
yb
2
2
2
2
22
2
2
2
2
2
221
2 21
2 41+ −
− −
= −
sin sin cosφ φ φ
xa
yb
yb
xa
xyab
yb
yb
2
22
4
42
2
22
2
2
2
22
4
424 4 2 4 4 4+ + − − + = −sin sin sin sin sin cos cosφ φ φ φ φ φ φ
xa
xa
yb
yb
xyab
2
22
4
42 2
2
22 2
2
22 4 4 4
0+ − + + − + + =sin sin (sin cos ) (sin cos ) sinφ φ φ φ φ φ φ
Hormonic Motion 543
or xa
yb
yb
yab
x−
+ − +sin sinφ φ
2 4
4
2
2 24 4 4 = 0
xa
yb
yb
xa
−
+ − +
sin sinφ φ2 2
2
2
24 1 = 0 ...(iii)
This is the general equation for a curve having two loops, which is the general path ofthe particle. Let us consider special cases:
(a) When φ = 0, sin φ = 0 and the Eq. (iii) reduces to
xa
yb
yb
2
2
2
2
2
24 1+ −
= 0
This represents a curve symmetrical about both theaxis, like the figure of ‘8’ as shown in figure (a)
(b) when φ = π/2, sin φ = 1 and the Eq. (iii) reduces to
xa
yb
yb
xa
−
+ − +
1 4 12 2
2
2
2 = 0
orxa
yb
yb
xa
−
+ + −
1 4 4 1
2 4
4
2
2 = 0
orxa
yb
−
+
1
2 2
2
2
= 0
This represents two coincident parabolas symmetrical about the x-axis each being
yba
x a22
2= − −( ), as shown in figure (b)
(c) when φ = π, sin φ = 0. Hence the path of the particle is again the figure of ‘8’.
(d) when φ = 3π/2, sin φ = –1 and the Eq. (iii) reduces to
xa
yb
yb
xa
+
+ − −
1 4 12 2
2
2
2 = 0
orxa
yb
yb
xa
+
+ − +
1 4 4 1
2 4
4
2
2 = 0
orxa
yb
+
−
1
2 2
2
2
= 0
This again represents two coincident parabolas, each being y2 = b2/2a (x + a) whichis the appropriate Lissajous figure (c).
X ′ X
Y
Y ′
a
b
Fig. 13(a)
544 Mechanics
Y
Y ′
XX ′
b
a
Y
Y ′
XX ′
b
aO O
( )b ( )c
Fig. 13
11.21 USES OF LISSAJOUS FIGURES
Lissajous figures are used in obtaining beautiful design for printing in cloth industry.They find useful applications in the acoustical measurement, e.g., determination of unknownfrequency. If the frequencies of vibration of the two rectangular vibrations are not exactlyequal in the ratio 2:1, the form of the Lissajous figure undergoes a gradual progressivechange. The time, taken by the curve to undergo a complete cycle of changes, enables us tofind the frequency of one vibration, if the frequency of the other is known. Let the frequencyof one vibration be n and that of the other be slightly different n′, the frequencies beingnearly in the ratio 1:1. If t be the time in which the cycle undergoes a complete change, thenthe frequency of the other will be
n' = nt
± 1
If the frequencies are nearly in the ratio 2:1, then
n' = 21
nt
±
To find whether n′ is greater or less than n or 2n, a small piece of wax is attached tothe unknown (e.g., the prong of the fork) and again the time of one cycle of changes isdetermined. If this time increases, the unknown frequency is higher and if it decreases, theunknown frequency is lower that n or 2n.
One practical application of Lissajous figures is that they enable us to know by inspectionthe period ratio of its constituent vibrations. It is evident that if the horizontal and verticallines are drawn on a Lissajous figure and if they cut it m and n times respectively, therequired period ratio of the two vibration is m:n.
NUMERICALS
Q.1. A particle is executing S.H.M. along the x-axis. Obtain expressions for its positionvelocity as function of time, assuming the initial conditions. t = 0, x = 0, v = v0.
Solution. Let x be the displacement of the particle from its equilibrium position at anyinstant t. By definition of S.H.M., the instantaneous force acting upon the particle is givenby
F = –Kx
Hormonic Motion 545
where K is proportionality factor. But by Newton’s second law
F =md xdt
2
2
where m is the mass of the particle and d2x/dt2 is the instantaneous acceleration. Thus,
md xdt
2
2 = –Kx
d xdt
2
2 = –ω2x
where ω2 = K/m. Multiplying both sides of this eqn. by 2dx/dt. We have
2 2
2dxdt
d xdt
× = −
ω2 2x
dxdt
Integrating w.r.t. ‘t’ we get
dxdt
2
= − +ω2 2x c (constant) .
When x = 0, v = dxdt
v= 0 . Therefore
c = v02
dxdt
2
= v x02 2 2− ω
dxdt
= v x02 2 2− ω
dx
v x02 2 2/ω −
= ωdt
Integrating, we get
sin/
−1
0
xv ω
= ωt + φ (constant)
ωxv0
= sin (ωt + φ)
orωxv0
= sin (ωt + φ)
Again at t = 0, x = 0 ∴ φ = 0. Thus
ωxv0
= sin ωt
546 Mechanics
x =v
t0
ωωsin
Diff. w.r.t. t we get
dxdt
= ωω
ωvt0 × cos
v = v t0 cos ω Ans.
Q.2. If the displacement of a moving particle at any time is given by x = a cos ωt +b sin ωt. Show that the motion is simple harmonic. If a = 3, b = 4, ω = 2, find the period,maximum velocity and maximum acceleration.
Solution. The displacement is given by
x = a cos ωt + b sin ωt ...(i)
⇒dxdt
= –ωa sin ωt + ωb cos ωt ...(ii)
⇒d xdt
2
2 = –ω2acos ωt – ω2b sin ωt
= –ω2(a cos ωt + b sin ωt)
= –ω2x
This acceleration d xdt
2
2 is proportional to displacement x and directed opp. to it. Hence
the motion is simple harmonic of amplitude A = a b2 2 2 23 4 5+ = + =( ) cm.
The period of motion is
T =2 2 3 14 3 14πω
= =× . . sec.rad2 rad/sec
The maximum velocity is
ωA = 2 × 5 = 10 cm/sec
The maximum acceleration is
ω2A = 22 × 5 = 20 cm/sec2 Ans.
Q.3. A body on the end of spring oscillates with an amplitude of 5 cm at a frequency of1 Hz. At t = 0 the body is at its equilibrium position x = 0. Write the equation describing theposition of the body given the numerical values of A, ω and α in the form x = A cos (ωt + α).What will be its velocity at t = 8/3 sec.
Solution. The required form of the motion is
x = A cos (ωt + α)
Here A = 5 cm and ω = 2 πn = 2 π × 1 = 2π rad/sec. Further at t = 0, x = 0 so that
0 = A cos α.
or α = π/2
Hence the equation of the motion is
x = 5 cos (2πt + π/2) cm
Hormonic Motion 547
or x = 5 cos 212
t +
π cm
The instantaneous velocity of the body is
dxdt
= − +
10 2
12
π πsin t
At t = 8/3 sec we have
dxdt
= −10 350
π πsin
= − −
−
= −
10 66
66 6
π π π π π πsin sin sin
= 106 6
12
π π πsin sin =
= 5π cm. Ans.
Q.4. A particle of mass 10 gm is placed in the potential field given by v = (50x2 + 100)erg/gm. Calculate the frequency of oscillation.
Solution. The potential energy of the 10 gm particle is
U = 10 (50x2 + 100) erg
The force acting upon the particle is given by
F = − = −ddx
xU dyne1000
But F =md xdt
2
2 , by Newton’s law
⇒ m d xdt
×2
2 = –1000x
⇒d xdt
2
2 =− = − = −1000 1000
10100x
mx
x
⇒d xdt
2
2 = –ω2x
Where ω2 = 100. Since the acceleration is proportional to the displacement, the particleis executing S.H.M. The frequency of oscillation is
n = ω/2π = 10/2π = 1.58/sec or Hz.
Q.5. Distinguish between two oscillation given by:
y1 =A cos ωt and y2 = A cos (ωt + π)
Solution. The oscillation have an initial phase difference of π. At start (t = 0), the firstparticle is at one extreme end of its motion (y1 = A) while the second particle is at the opp.end of its motion (y2 = –A).
548 Mechanics
Q.6. Write the equation of S.H.M. if (i) initial phase is zero (ii) initial phase is π/2. Theamplitude is 5 cm and period 8 sec.
Solution. The general solution of S.H.M. is
x = a sin (ωt + φ)
Here a = 5 cm, ω = 2π/T = 2 3 14 0 785× . .rad
8 secrad / sec.=
(ii) When initial phase φ = 0, we have
x = 5 0 7852
sin . t +
π
= 5 cos (0.785 t)
Q.7. A particle executes S.H.M. of period 31.4 sec and amplitude 5 cm. Calculate itsmaximum velocity and maximum acceleration.
Solution. The eqn. of S.H.M. is
x = a sin (ωt + φ)
The velocity is
u =dxdt
a t= +ω ω φcos ( )
The maximum value of cos (ωt + φ) is 1. Therefore
umax = ωa
Here ω = 2π/T = 2 3 14
31 40 2 5× .
..= =rad/sec and cm.a
umax = 0.2 rad/sec × 5 cm = 1 cm/sec
The acceleration is given by
f =d xdt
a t2
22= +ω ω φsin ( )
The maximum value of sin (ωt + φ) is 1. Therefore
(fmax) = ω2a
= (0.2 rad/sec)2 × 5 cm
= 0.2 cm/sec2
Q.8. Show that when a particle is moving in S.H.M. it’s velocity at a distance 3/2 ofits amplitude from the central position is half its velocity in the central position.
Solution. The displacement of particle in S.H.M. is
x = a sin (ωt + φ)
The velocity is
u =dxdt
a t= +ω ω φcos ( )
= ω ωa xa
a x12
22 2− = −
Hormonic Motion 549
At xa= 3
2, the velocity is
u = ω ωa a a223
412
− = ,
which is half the velocity (ωa) in the central position (x = 0).
Q.9. The equation of motion of a particle is x = 2 sint
2 4cm
π π+
. Find the period and
the maximum velocity of the particle.
Solution. The given eqn. is
x = 22 4
sinπ πt +
cm
Comparing it with the general equation of S.H.M.
x = a sin (ωt + φ)
we have,
a = 2 cm, ω = π/2 rad/sec.
The period is
T = 2π/ω = 22
4ππ
radrad/sec/
sec.=
The maximum velocity of the particle is
umax = ωa
=π2
2rad / se cmc ×
= 3.14 cm/sec.
Q.10. The period of S.H.M. is 10 sec and amplitudes is 10–1m . Write its equation. Whatare the phase and displacement at a time 5 sec after a passage of the particle through itsextreme positive elongation? What is the maximum velocity.
Solution. The general equation of S.H.M. is
x = a sin (ωt + ϕ )
Here a =10–1 m = 10 cm and ω = 2 2
105π π π
Trad/sec.= =× /
∴ x = 10 sin π φt5
+
cm.
If the motion starts from of the extreme positive elongation, that is,
x = 10 cm at t = 0
then we have
10 = 10 sin φφ = π/2
550 Mechanics
the equation of motion is
x = 10 sin π πt5 2
+
Now at t = 5, the displacement is
x = 105
5 2sin
×π π+
= 10 sin 32
10π = − cm.
the phase at t = 5 sec is 3π/2.
The velocity is differentiating (i) given by
u =dxdt
t= +
π π π5
105 2
× cos
umax =π π5
10 2 6 28 = =× . cm/sec.
Q.11. Two particle 1 and 2 start vibrating together in S.H.M. along the same straight line.If their periods are 40 and 60 sec resp. Find their phase difference (a) after 20 sec from startand (b) when one particle is at the end of its path while the other is at middle.
Solution. The displacement of particle in S.H.M. about x = 0 is given by
x = a tsin 2πT
(a) for particle 1 and 2, T = 40 sec respectively. Thus
x1 = a sin πt20
and x2 = a sin πt30
at t = 20 sec, we have
x1 = a sin π
and x2 = a sin 23 3π π π= −
a sin
these two eqn. show that the phase diff between particle 1 and 2 is π/3 i.e., 60°.
(b) Particle 1 is at the end of its path and particle 2 is at the middle i.e., x1 = +a andx2 = 0, thus at this instant, we can write
+a = a sin ω1t
0 = a sin ω2t
These give ω1t = π/2 or 3π/2 and ω2t = 0
∴ phase diff.
ω1t ~ ω2t = π/2 or 3π/2
Hormonic Motion 551
Q.12. A particle of mass 5 gm is executing S.H.M. has amplitude of 8 cm. If it makes 16vibration/sec find maximum velocity and energy at mean position.
Solution. The displacement of particle in S.H.M. is
x = a sin (ωt + φ)
and its velocity
u =dxdt
a t= +ω ω φcos ( )
It is maximum when cos (ωt + ϕ ) = 1 i.e.,
umax = ωa
= 2πna
= 2 × 3.14 × 16 × 8 = 802.8 cm/sec.
The energy at mean position is entirely kinetic and is given by
E = Kmax = 12 mumax
2
=12
5 80 8 1 6 102 6× ( . ) . ×= erg.
Q.13. A particle in S.H.M. has velocities u1 and u2 when its displacements from the meanposition are x1 and x2 respectively. Calculate the period, amplitude and maximum speed of theparticle.
Solution. The speed of a particle in S.H.M. at a displacement x is given by
u = ω a x2 2−
where ω is angular speed and a is amplitude
Here u1 = ω a x212− ...(i)
u2 = ω a x222− ...(ii)
u u12
22− = ω2
22
12( )x x−
or ω =u ux x
12
22
22
12
−−
The period is therefore
T = 2 2 22
12
12
22π ω π/ = −
−x xu u
from eqn. (i) and (ii), we have.
uu
12
22 =
a xa x
212
22
22
−−
or a u x u212
22
12− = a u x u2
22
12
22−
552 Mechanics
a =x u x u
u u22
12
12
22
12
22
−−
The maximum particle velocity in S.H.M. is
umax = ωa
=u ux x
x u x uu u
12
22
22
12
22
12
12
22
12
22
−−
−−
×
=x u x u
x x22
12
12
22
22
12
−−
Q.14. A particle is moving with S.H.M. along a straight line. If velocity when passingthrough the points 3 cm and 4 cm from the centre of its path is 16 cm/sec and 12 cm/secrespectively. Find the amplitude and period of motion and the maximum velocity of theparticle.
Solution. The displacement of a particle in S.H.M. is
x = a sin (ωt + ϕ )
where a is amplitude. It’s velocity
dxdt
= ω ω φa tcos ( )+
= ωa 1 2− +sin ( )ω φt
= ωa 12
2− xa
= ω a x2 2−
dxdt
2
= ω2 (a2 – x2)
Now when x = 3,dxdt
= 16 cm/sec
when x = 4,dxdt
= 12 cm/sec
Substituting in eqn. (i) we get
256 = ω2 (a2 –9)
144 = ω2 (a2 – 16)
Dividing we get169
=aa
2
29
16−−
16 (a2 – 16) = 9 (a2 – 9)
7a2 = 156 – 81 = 175
Hormonic Motion 553
a2 =1757
= 25
a = 5 cm.
Substituting in Eqn. (iii) we get
144 = ω2 (25 – 16)
ω2 =144
9
ω = 12/3 = 4
∴ The period is
T =2 2 3 14
41 57π
ω= =× . . sec
Maximum velocity of the particle
dxdtmax
= ω a.
Q.15. A particle executes linear S.H.M. about the point x = 0. At time t = 0, it hasdisplacement x = 2 cm and zero velocity. If the frequency of motion is 0.25/sec find (i) period(ii) circular frequency (iii) Amplitude, (iv) maximum speed, (v) displacement and velocity at t= 3 sec. Calculate the time required for the particle to come half-way in toward the centre ofmotion from its initial position.
Solution. (i) Frequency, n = 0.25/sec
∴ period, T =1n
= 1/0.25 = 4 sec.
(ii) Circular frequency ω = 2π/T = 2π/4 = π/2 rad/sec.
(iii) x = a sin (ωt + φ) ...(i)
So that u =dxdt
= ωa cos (ωt + φ) ...(ii)
At t = 0, x = 2cm and u = 0 (given). Therefore from eqn. (iii) we get
a = + 2 cm ( sin / , sin / ) π π2 1 3 2 1= = −
(iv) From eqn. (ii) we have
umax = ωa = π/2 × 2 = 3.14 cm/sec.
(v) Putting t = 3 sec, ω = π/2, a = 2 cm and φ = π/2 in eqn. (i) and (ii) we get
x = 2 sin 2π = 0
u = 3.14 cos 2π = 3.14 cm/sec.
Let us now compute the time required for the particle to come halfway in toward thecentre of motion from its initial Putting a = 2 cm, ω = π/2 and φ = π/2 in Eqn. (i) we get
554 Mechanics
x = 2 sin π π πt t2 2
22
+
= cos
At half way x =a2
= 1 cm. Hence
1 = 2 cos πt2
cos πt2
= 1/2
πt2
= cos /− =1 1
23π
t =23
sec
Q.16. The equaiton of motion of an oscillating body is x = 6 cos ( 3 t 3) mπ π+ / . What isthe period, frequency and phase constant of motion. Also find the displacement, velocity andacceleration at time t = 2 sec.
Solution. The given eqn.
x = 6 cos 33
π πt +
meters.
represents a S.H.M. Let us compare it with the general eqn. of S.H.M. in the cosine fromwhich is
x = a cos (ωt + φ)
We see that for the given oscillating body
a = 4 meter, ω = 2π and φ = π/3
period T = 2 2
323
πω
ππ
= = sec
frequency, n =1 3
2Tvib / sec=
and phase constant φ =π3
60rad = °
Now, the displacement is
x = 6 cos 33
π πt m+
Af t = 2 sec, x = 6 cos π3
6 60 3= ° =cos meter
The velocity is u = dxdt
t= − +
18 3
3π π π
sin m / sec
At t = 2 sec u = − +
18 6
3π π π
sin
Hormonic Motion 555
= − = −183
18 60π π πsin × sin
= − = −183
249π × m/sec
The acceleration is
f =d xdt
t2
22 254 3
3= − +
π π πcos m/sec
At t = 2 sec f = − +
54 6
32π π π
cos
= − = − = −54 3 54 12
1662 2 2π π πcos / × m/sec
Q.17. A mass of 150 gm hangs from a vertical spring of elastic constant 2 N/m. Calculatethe frequency of small vertical oscillations, assuming that the damping is small.
Solution. m = 250 gm = 0.25 kg and K = 2 N/m.
n =1
2πKm
=1
2 3 142
0 250 45 1
× . .. sec= −
Q.18. A spring is hung vertically and loaded with a mass of 400 gm and made to oscillate.Calculate (i) the time-period and frequency of oscillation. When the spring is loaded with 100gm, it extends by 5 cm.
Solution. If an applied force Fapp extends a spring by x, then by Hooks’s law
Fapp = Kx
where K is force constant of the spring. Here Fapp = 100 gm = 100 × 980 dynes and x = 5 cm.
K =Fapp
x =
100 9805
1 96 104× . ×= dynes/cm
Now, the period of a mass hung from a light spring is given by
T = 2π mK
Here m = 400 gm and K = 1.96 × 104 dyne/cm
T = 2 4001 96 104π. ×
T = 2 14
0 9πa
= . sec
The frequency is n = 1 1
0 91 1 1
T= = −
. sec. sec
556 Mechanics
Q.19. A mass M of 2000 gm hangs from a vertical spring. When a mass m of 600 gm isgently added to 2000 gm, the spring is further stretched by 4 cm. Now m is removed and Mis set in oscillation. Calculate the period.
Sol. The eqn. of S.H.M. of a body of mass m is
T = 2π mK
= 2 3 142000
1470000 73× . × . sec.=
Q.20. A body of mass 0.1 kg hanging from a spring is oscillating with a period of 0.2 secand an amplitude of 1 m. Find the maximum value of the force acting on the body and thevalue of force constant of the spring.
Sol. The equation of S.H.M. of a body of mass m is
x = a sin (ωt + φ) ...(i)
where a is the amplitude and ω = Km
the time period is
T =2 2πω
π= Km
Here T = 0.2 sec
ω =Km
= 10π ...(ii)
The acceleration of the body is, differentiating (i) twice
f =d xdt
a t2
22= − +ω ω φsin ( )
Then fmax = ω2a
Here ω = 10π and a = 1 m (given)
∴ fmax = (10π)2 × 1 = 986 m/sec2
and max. force = mass × max. acceleration
= (0.1 kg) × 986 m/sec2
= 98.6 N
Now from eqn. (ii) we have
Km
= (10π)2
K = 100 2π m
= 100 × (3.14)2 × 0.1
= 98.6 N/m.
Q.21. The potential energy of a harmonic oscillator in its rest position is 5 joule and theavg. kinetic energy is 2 joule. What is the total energy? If the amplitude is 1 meter, what isthe force constant? If the mass is 2 kg. what is the period?
Hormonic Motion 557
Solution. The potential energy in rest position is 5 joule the avg. Kinetic energy(2 joule) is half the energy of oscillation so that the energy of oscillation is 4 joule. Hencetotal energy.
5 + 4 = 9 Joule
At the maximum displacement a, the gain in potential energy is 12
Ka2, where K is force
constant thus.
12
2Ka = 4 Joule
Here a = 1 meter
K = 8/a2 = 8 N/m
The period of oscillation is
T = 2 2 3 14π πmK
2kg8N/m
= = . sec.
Q.22. A simple harmonic oscillator of period 5 sec has 5 joule potential energy when itsdisplacement is 2 cm. (Energy in the rest position may be taken as zero). Calculate (i) forceconstant, (ii) avg. kinetic energy if the amplitude is 4 cm, (iii) Kinetic and potential energywhen the displacement is 1 cm, (iv) oscillation frequency if mass is reduced to 1/100th of it’sinitial value.
Solution. (i) The gain in Potential energy when the displacement is x is 12
Kx2 where
K is force-constant. Here at x = 2 cm = 0.02 m, the gain in potential energy is 5 joule. Thus,
12
0 02 2K ( . ) = 5
K = 2.5 × 104 N/m
(ii) The avg. Kinetic energy is
Kav =12
2K a
Here a = 4 cm = 4 × 10–2 m
Kav =14
2 5 10 4 104 2 2× ( . × ) × ( × )N/m m−
= 10 Joule
(iii) The kinetic energy when the displacement is x is
K =12
2 2K ( )a x−
Here a = 4 × 10–2 m and x = 1 × 10–2 m
K =12
2 5 10 15 104 4× ( . × ) × ( × )−
= 18.75 Joule.
558 Mechanics
The potential energy is
U =12
2Kx
=12
2 5 10 10 14 2 2( . × ) × ( × )N/m m−
= 1.25 Joule.
(iv) The frequency is inversely proportional to the square root of mass. The initialfrequency is
ni =15
0 2= . /sec
If mi and mf be the initial and the final masses, the final frequency is given by
n
nf
c=
mm
i
f
Here mf = 1
100mi
∴mf
0 2.= 100 10=
nf = 0.2 × 10 = 2/sec
Q.23. A body of mass 0.5 kg is projected with an energy 4 joules. If a restoring force of2 N/m acts on it. What maximum displacement will be body acquire? What will be its periodictime?
Solution. At the maximum displacement a (say), the energy would be entirely potentialenergy. That is
12
K 2a = 4 Joule.
Here K = 2 nt/m
a =2 4
2×
= 2 meter.
T = 2π mK
= 2π π0.52
= sec.
Q.24. The amplitude of harmonic oscillations of a point mass of 10 gm is 5 cm and theenergy of oscillations is 3.1 × 10–5 joule. Write the eqn. of motion if the initial phase is 60°
Solution. The energy of oscillation is given by
E = 2 12
22 2 2 2 2π πmn a n ma= ( )
Hormonic Motion 559
=12
2 2ω ma
ω =1 2a m
E
Here E = 3.1 × 10–5 Joule = 310 erg, a = 5 cm, and m = 10 gm.
ω =15
2 31010
625
1 57×
.= = rad/sec
Now, the Eqn. of motion is
x = a sin (ωt + φ)
Putting a = 5 cm, ω = 1.57 rad/sec, φ = 60 = π/3 rad, we get
x = 5 sin 1 573
. t +
πcm.
Q.25. A mass M is controlled by two massless springs between two rigit support asshown. If the mass is 50 gm and the force-constants of springs are 3000 and 2000 dyne/cmfind (i) frequency of oscillations of the mass, (ii) energy of vibration of amplitude 0.4 cm,(iii) velocity of mass when passing through mean position.
Solution. (i) When the mass is displaced through a distance x, the restoring forcesdeveloped in the spring are
F1 = − = −K and F K1 2 2x x
and act in the same direction. The total restoring force is
F = − +K K1 2 xSo that the force constant of the system is
K = K K 3000 2000 5000 dyne/cm1 2+ = + =
Therefore, the frequency of oscillation is
n =1
21
2500050
5π π π
Km
= =
=5
3 141 6
.. /sec.=
(ii) The energy of oscillation is the same as the maximum potential energy.
Thus E = K 12
Kmax2= a
E =12
5000 0 4 4002× ×( . ) = erg.
(iii) The velocity, and hence the kinetic energy, is maximum when the mass passes through,its mean position. Again, the energy of oscillation is same as the maximum kinetic energy.
E = K 12
( )max max2= =m u 400
umax =800m
= 80050
4= ± cm/sec
560 Mechanics
Q.26. A riffle bullet weighing 10 gm moving with a velocity of 16000 cm/sec strikes andembeds itself in a 790 gm block which rests on a horizontal frictionless surface and is attachedto a spring of force constant 8 × 104 dyne/cm. Compute the amplitude of the resulting simpleharmonic oscillation of the block. What happens to the K.E. of the bullet ?
Solution. Let v' be the velocity acquired by the block M when the bullet m strikes itand comes to rest in it. By conservation of momentum, we have
(M + m) v' = mv
800 + v' = 10 × 16000
∴ v' =10 16000
800200× = cm/sec .
The block is set in oscillation about its mean position. If a be the amplitude of motion,the eqn. of motion may be written as
x = a sin ωt
The velocity is dxdt
a t= ω ωcos
In the mean position, the velocity is a maximum and equal to ωa. This we have obtainedas 200 cm/sec. Thus
ωa = 200
a =200ω
If T be the period, then T = 2 2 /Tπω
ω πi.e., =
a =200T
2π
But T = 2M
K, where Mπ + +m
m is the mass of the block + bullet and K is the force
constant of the spring
a = 200M
K+ m
Here M + m = 800 gm and K = 8 × 104 dyne/cm.
a = 200 8008 10
204××
= cm.
The kinetic energy of the moving bullet was
K =12
2mv
=12
10 16000 128 102 7× × ( ) ×= erg.
Hormonic Motion 561
The maximum potential energy stored in the spring is
U =12
2Ka
=12
8 10 20 1 6 104 2 7× ( × ) × ( ) . ×= erg.
∴ The percentage of energy stored in the spring
=1 6 101 28 10
100 1 257
7. ×. ×
× . %=
Thus a very small fraction of the K.E. of a bullet is stored in the spring and the restappears as heat.
Q.27. A block rests on a horizontal surface at the rate of 2 oscillations. The coefficientof static friction µ between block and plane is 0.5. How large can the amplitude be withoutslipping block and surface.
Solution. Let m be the mass of the block and a the largest possible amplitude withoutslipping between block and surface. Then the maximum horizontal force acting on the blockmust be equal to the frictional force the maximum horizontal force on the block is mω2abecause ω2a is the maximum acceleration where ω is the angular frequency. The frictionalforce is µmg. Thus
mω2a = µmg
a =µω
g2
But ω = 2π n, where n is the frequency
a =µπ
gn4 2 2
=0.5 × 980
9 × (3.14)cm2 ×
.2
3 12 =
Q.28. A simple pendulum of length 2m and amplitude 3.03 meter has energy 0.10 joule.What will be its energy (i) if its amplitude is changed to 0.06 meter and length remainsunchanged, (ii) amplitude remains 0.03 meter but length is changed to 1 meter.
Solution. The energy of oscillation of amplitude a is
E =12
2m aω 2
In case of simple pendulum ω = 2πT
= gl
E =mga
l
2
2
Originally l = 2 meter a = 0.03 meter, E = 0.10 joule.
0.10 =mg × ( . )
×0 06
2 2
2
...(i)
562 Mechanics
(i) E =mg × .
×0 06
2 2 ...(ii)
From eqns. (i) and (ii)
E =0 060 03
0 10 0 40..
× . .= joule
(ii) E =mg × ( . )
×0 03
2 1
2
...(iii)
From eqns. (iii) and (i) we get
E =12
0 10 0 20× . .= joule
Q.29. One end of a horizontal spring of force constant 6.0 N/m is tied to a fixed wall andthe other end to a solid cylinder which can roll without slipping on the horizontal ground.The cylinder is pulled a distance of 0.1 meter and released. Calculate the K.E. of the cylinderwhen passing through the equilibrium position.
Solution. When the Cylinder is pulled 0.1 m, the spring is stretched and has a potentialenergy given by
U =12
12
6 0 0 1 0 032 2K joulex = =× . × ( . ) .
Whole of this energy is converted into the K.E. of rotation and the K.E. of translationwhen the cylinder is passing through the equilibrium position. That is
Krot + Ktrans = 0.03 joule.
Krot =12
I 2ω = 12
12
14
22
22× mr v
rmv
=
Ktrans =12
2mv
∴14
12
2 2mv mv+ = 0.03 Joule
or mv2 = 0 0343
0 04. × .= Joule
Krot =14
14
0 04 0 012mv = =( . ) . Joule
Ktrans =12
12
0 04 0 022mv = =( . ) . Joule.
Q.30. Two tunning forks produce Lissajous figures of the shape of st. line, ellipse andfinally to the same straight line in 2 sec. If one of the tunning forks has a frequency of 100vib/sec, find the possible frequency of the other.
Hormonic Motion 563
Solution. The Lissajous’ figures are straight line and ellipse. Hence the frequency ofthe two must nearly be in the ratio 1 : 1
As the cycle of figures is traced in 2 sec, the difference in frequencies must be 12
/ sec.
Therefore, the other fork must have one of the following frequencies.
100 12
± = 100.5 , 99.5
Q.31. The Lissajous’ figure in the case of two tunning forks is a parabola. Prove that thefrequencies are in the ratio 1 : 2.
Solution. Let us draw co-ordinate axes across the given parabola. Let one fork offrequencies n1 be vibrating along the X-axis and the other of frequency n2 be vibrating alongthe Y-axis. Then if the parabola cuts the Y-axis Py times and the X-axis Px, times
nn
1
2=
PP
y
x
Here Py = 2 and Px = 1. Therefore
nn
1
2=
21
.
Q.32. The tunning forks of approximately equal frequencies produce Lissajous’ figuresthat go through a cycle of changes in 15 second. When one fork is loaded with wax, the cycleof changes of Lissajous’ figure takes 10 seconds. If the second turning fork has a frequencyof 400, find the frequency of the first fork before and after loading.
Solution. Let n1 and n2 be the frequencies of two forks. Before loading, the cycle ofLissajous’ figure takes 15 seconds. This shows that in 15 seconds one forks makes onecomplete vibration more (or less) than the other. Thus,
X
Y
Y ′
X ′
Fig. 14
n1 × 15 ~ n2 × 15 = 1
n1 ~ n2 = 1/15
But n2 = 400
n1 = 400 + 1/15
= 400.067 or 399.933
On loading the first fork with wax, the time of cycle decreases to 10 sec. Hence thefrequency of first fork after loading
n′1 = n2 1 10± /
564 Mechanics
= 400 + 0.1
= 400.1 or 399.9
But on loading, the frequency will decrease. Hence its frequency before loading 400.1 andafter loading 399.9.
Q.33. Two tuning forks A and B of nearly equal frequencies are employed to produceLissajous’s figures. On slightly loading A, the cycle of change of figures slows down from 15sec to 20 sec. If the frequency of B is 256, find the frequency of A before and after loading.
Ans. 256115
256120
+ +
,
Q.34. A particle of mass 10 gm lies in a potential field V = 50x2 + 100 erg/gm. Deducethe frequency of oscillation.
Solution. Potential energy of 10 gm.
U = 10 50 1003 2− +( ) .x ergs
Force = – dU/dx = –1000x, ∴ = −10 10002
2d xdt
x
ord xdt
x2
2 100+ = 0
This is the eqn. of S.H.M. whose frequency is
n = 1002
5 1 6π π
= = . oscillations/sec.
Q.34. A particle moves in the potential energy field U = U P Qxx02− + . Find the expression
for the force. Calculate the force constant and time period. At what point does the forcevanish ? Is this a point of stable equillibrium?
Solution. Force F =− = + −ddx
nU P Q2
Evidently, this is a linear restoring force and the force constant is 2Q. Eqn. of motionis
m d xdt
2
2 = P Q or Q P− + =2 22
2x d xdt m
xm
∴ Periodic time = 2 2π m / .Q
The force vanishes, whereddxU
= 0 i.e., P – 2Qx = 0
∴ x = P/2Q
Nowddx
i e2
2 2U Q= + , . ., if Q is positive at x = P/2Q, there is the minima of potential
energy curve and hence this point is in stable equilibrium.
Hormonic Motion 565
Q.35. A particle is moving with S.H.M. in a st. line, when the distance of the particle fromthe equilibrium position has the value x1 and x2 the corresponding values of velocity are u1and u2. Show that the period is
T = 2x x
U U22
12
12
22
1/2
π −−
Solution. Let the equilibrium position of the particle be 0.8 if P, Q are the two positionsof the particle at distances x1 and x2, from O respectively, then the velocity u1 of the particleat the position P is given by
u1 = ω a x212− ...(i)
The velocity u2 at position Q is
u2 = ω a x222− ...(ii)
Squaring (i) and (ii) and substracting we get
u u12
22− = ω2
22
12( )x x−
ω =u ux x
12
22
22
12
−−
.
Hence, time period T =2 2 2
212
12
22
πω
π= −−
x xu u
.
Q.36. A particle of mass 2 gm moves along the x–axis and is attracted towards origin bya force 8 × 10–3x newton. If it is initially at rest at x = 10 cm, find (i) then differential eqn.of motion, (ii) the position of particle at any time, (iii) the velocity of particle at any time,(iv) amplitude and frequency of vibration.
Solution. (i) Frommd xdt
2
2 = –8 × 10 8 1032
23− − −= −x d x
dtxor 2 × 10 3 ×
ord xdt
x2
2 4+ = 0
This is the differential eqn. of motion.
(ii) Let the solution of the equation be
x = a sin (2t + φ)
when t = 0 x = 0.1 m ∴ 0.1 = a sin φ
when t = 0 dx/dt = 0 ∴ 0 = 2a cosφ [dx/dt = 2a cos (2t + φ)]
a = 0 or φ = π/2 but a ≠ 0, ∴φ = π/2
Therefore 0.1 = a sin π/2 or a = 0.1 m.
x = 0.1 sin (2t + π/2) = 0.1 cos 2t
This equation gives the position of the particle at any time t.
566 Mechanics
(iii) Now, v = dx/dt = – 0.2 sin 2t
(iv) Amplitude = 0.1 m
Frequency n = ω/2π = 2/2π = 1π
= 0.3 Hz.
Q.37. A body having a mass of 4 gm executes S.H.M. The force acting on the body, whendisplacement is 8 cm, is 24 gm ωt. Find the period. If the maximum velocity is 500 cm/sec.Find the amplitude and maximum acceleration. (g = 9.8 m/sec.)
Solution. Mass of the body = 4 gm = 4 × 10–3 kg
Force acting on the body at 8 cm
Displacement = mω2x
= 4 × 10–3 × ω2 × 8 × 10–2 = 24 × 10–3 × 9.8
whence ω = 6 9818 27 12× .=
∴ Period T =2 2 3 14
27 120 2315π
ω= =× .
.. sec.
The maximum velocity of the body v = ωa = 0.5 m/sec.
∴ a =vω
= 0 5
27 15..
= 0.1844 m.
The maximum acceleration = ω2a = 735.75 × 0.1844 = 135.67 m/sec2.
Q.38 A vertical U tube of uniform cross-section contains water upto height 30 cm. Showthat if the water on one side is depressed and then released its motion up and down the twosides of the tube is simple harmonic and Calculate its time period.
Solution. Let PQ be the initial level of water in the U tube upto a height h = 0.3 m.
When the level of water in tube A is depressedthrough a distance y, it rises at the same time by thesame amount in the tube B. Now the difference betweenthe two levels P′ and Q′ = 2y.
The weight of this column of length 2y will nowact on the mass of the water in the tube, as a resultit will oscillate up and down.
The weight of the column of length 2y = 2y × A× ρg newton.
Where A is cross-section of the tube and ρ thedensity of the liquid.
Or force acting on the mass of water
= 2y Aρg newton ...(i)
Total mass of the water m = 2hAρ = 2 × 0.3 × Aρ kg
∴ Force acting on the mass of water
= m(d2y/dt2) = 2hAρ(d2y/dt2) ...(ii)
2hAρ (d2y/dt2) = –2yAρg or d ydt
yh
y2
2 + × = 0
B
h
P
P ′
Q′
BA
yQ
Fig. 15
Hormonic Motion 567
Hence the motion of water column is simple harmonic, whose time period is given by
T = 2π hg
= 2 0 309 81
π ..
= 1.098 sec
Q.39. Show that the time period for the swing of a magnet in the earth’s field is given
by T = 2I
MHπ where M is the magnetic moment of the magnet, I its moment of inertia
about the axis of suspension and H the earth’s field.
Solution: Let a magnet Ns be suspended freely in the earth’s magnetic field H. Obviouslytwo opposite forces each equal to mH will constitute a couple equal to mH × NS sin θ, whereθ is the angle making the magnet with the direction of the earth’s magnetic field.
But magnetic moment M = m × Ns
Couple = –MH sin θ = –MHθ (as θ is small)
(Since couple is opposite to θ increasing)
The couple develops an angular accelerationd2θ/dt2 in the magnet. If I is the moment of inertia ofthe magnet about the suspension thread, then thecouple can be depressed as I (d2θ/dt2) Hence theequation of motion of the magnet is
IMH
ddt
2
2θ θ+ = 0 or
ddt
2
2θ θ+ MH
I = 0
Therefore, the motion of the magnet is simpleharmonic, whose period is
t = 2π IMH
Q.40. If the earth were a homogeneous sphere of radius R and a straight hole were boredin it through its centre. Show that a particle dropped into the hole will execute a S.H.M. andfind out its time period.
Solution. If the mass of the earth is M and R its radius then the acceleration due togravity at the surface of the earthy is given by
g = GM/R2
where G is gravitational constant.
If the density of the earth is ρ, the M = 43
3R ρ
∴ g =43
Rρ ...(i)
If g' be the value at depth h below the surface of the earth, then
g' =43
( ).R G− h ρ ...(ii)
P
H m H
Nθ
θ
O
S
m H
Fig. 16
568 Mechanics
Dividing Eq. (ii) by Eq. (i) we have
g'g
=R
R− h
or g′ = g hR
R( )−
If we measure the distance from the centre of the earth, putting R–h = x, accelerationof the particle
d xdt
2
2 = gg
x′ = −R
Negative sign shows that the acceleration acts towards the centre of the earth i.e.,opposite to x increasing.
∴d xdt
gx
2
2 +R
= 0
Thus, the particle, dropped into the hole, will execute S.H.M. Its periodic time is givenby
T = 2π Rg
Q.41. Two cities on the surface of the earth are joined by a straight smooth under groundtunnel of length 640 km. A body is released into the tunnel from one city. How much timewill it take to reach the other city? Derive the formula used. G = 6.67 × 10–11 S.I. Units andρ = 5.520 kg/m3. Calculate the velocity when the body would be nearest to the centre of theearth in its journey.
Solution. Suppose that A and B are the two cities,which are connected by a straight tunnel of length AB= 640 km. Let O be the centre of the earth and R itsRadius. OC is the perpendicular, drawn from the centre0 on AB, where C is the midpoint of the tunnel AB. Leta body of mass m be released into the tunnel from itsend A and let it be at P, distant x from c at any instantt. If M is the mass of the earth, then on its surface theweight (mg) of the body is given by mg = GMm/R2
The weight of the body at P, distant OP (= r) from the centre of the earth is
mg' = GM R
RGM
R R′ = = =m
r
G r m
r
G mr m r2
3
2
3
3 2
43
43
π ρ π ρ.
m r′ =
343
3π ρ π ρand M = 43
R
The direction of this force is PO, whose component along PC
=GM
Rm r
R2 . cos OPC = GM
R Rm r x
r2 . . = GM
Rm
x3
C O
P
A
R
B
Fig. 17
Hormonic Motion 569
Hence the eq. of motion of mass m is
md xdt
mx
2
2 3+ GMR
= 0 or d xdt
x2
2 3+ GMR
= 0
The equation represent a S.H.M. of period.
T = 23
π RGM
i.e., this is the time in which the body will reach from A to B and again form B to A. Hence,time taken by the body to pass through the tunnel is
T2
= π π π ρ πρ
RGM
G.
R
RG
3= =
3
343
34 .
=3 3 14
4 5520 6 62 10 11× .
× × . × − = 2528 sec
velocity at c1(v) = ωa = GMR
AB G AB3 2
43 2
. . .= πρ
= 6 67 1043
3 14 55206402
1011 3. × × × . × ×−
= 3.972 × 102 m/sec.
Q.42. The total energy of a particle executing a S.H.M. of period 2π second is 10.24 ×
10–4 Joule. The displacement of the particle at π/4sec is 8 2 cm. Calculate the amplitude ofthe motion and the mass of the particle.
Solution. The displacement of a particle executing S.H.M. is given by
x = a sin ωt = a sin 2πt/T [ ω = 2π/T]
∴ 8 2 10 2× − = a sin 22 4
ππ
π× = a sin π4 2
= a
∴ Amplitude of vibration a = 0.16 m.
If m be the mass of the particle, then its total energy is given by
E =2 2 2
2m aπ
T = 10.24 × 10–4 or m =
10 24 102
4 2
2 2. × ×
×
− Tπ a
Here T = 2π sec and a = 0.16 m
m =10 24 10 2 2
2 0 16 0 16
4
2. × × ×× × . × .
− π ππ = 0.18 kg
Q.43. If the potential energy of a harmonic oscillator in its resting position is 5 joules andthe total energy is 9 Joules. When the amplitude is 1 m, what is the force constant? If its massis 2 kg what is the period?
570 Mechanics
Solution. Potential energy in rest position = 5 joules. At the maximum displacement(equal to amplitude), the potential energy = 9 Joules.
∴ Gain in P.E. = 9 – 5 = 4 Joule.
The gain in energy at the maximum displacement should be equal to 12
Cx2 max = 12
C.
(1)2 = C/2 Joules.
Thus C/2 = 4 or C = 8 Joule/m.
Period T = 2 2π πmC
28
= = 3.14 sec.
Q.44. Using conservation of energy, show that the angular speed dθ/dt of a simplependulum is given by
ddtθ
= 2ml
E mgl(1 cos ) 2
12
− −
θ
where E is total energy of oscillation. L and m are length and mass of the pendulum and θis angular displacement from the vertical.
Solution. The potential energy of the simplependulum at any position P relative to the lowestposition O of the bob is obviously given by
U = mg(l – l cos θ)
because the bob is displaced through a verticalheight (l – l cos θ)
The kinetic energy of the pendulum at the positionP is
K =12
2mv = 12
2m ld dt( / )θ V( )= =
rlddt
ω θ
∴ Total energy E = K + U = 12
ml2 (dθ/dt)2 + mgl(1 – cos θ) which should be a constant,
according to the law of conservation of energy
∴ddt
θ= 2
12
12
mlE mgl( ( cos )− −
θ
Q.45. A simple pendulum of length l and mass m is oscillating with a maximum angulardisplacement θ0 radians. Show that the tension in the string at the angular displacement θis mg (3 cosθ – 2 cos θ0).
Solution. Let the bob P of mass m be oscillating about the point S. At the angulardisplacement θ of the pendulum the weight mg can be resolved into two components mg cosθand mg sin θ. The latter component provides the restoring force for S.H.M. and the formercomponent reduces the Tension T of the string. So that the necessary centripetal force forthe rotation of the pendulum in a circular path of radius l at the angular displacement θ is.
T – mg cos θ = mv2/l ...(i)
where v is the vocally to the bob at P.
Ol s in θ
θS
P
l
m g
Fig. 18
Hormonic Motion 571
θ
N
mg sin θθ m
g co s θ
P0PN
m g
θ0θ
l
S
Fig. 19
Total energy of the particle = mg ON (at Po relative to 0)
= mgl(1–cos θ0) = mgl (1 – cos θ) + 12
2mv p( )at
or 2mg cos θ – 2mg cos θ0 = mv2/l ...(ii)
∴From Eq. (i) and Eq. (ii), we get
T–mg cos θ = 2mg cos θ – 2mg cos θ0
T = 3mg cos θ – 2mg cos θ0
= mg (3cos θ – 2cos θ0)
Q.46. Derive an expression for the small angleoscillations of a simple pendulum of length L comparablewith the radius of earth R. What is the period of apendulum of an infinite length on the surface of theearth?
Solution. As shown in fig the restoring force onthe bob is
md xdt
2
2 = – mg sin (θ + α )
where θ andα both are small. So that sin (θ + α ) = θ
+α and θ = xL
andα = xR
therefore
d xdt
g x2
21 1+ +
L R
= 0
xP
C
O
S
α m g
θ + α
O
Fig. 20
572 Mechanics
Thus the motion of the pendulum is simple harmonic of Period.
T = 2
1 12π π
gL R
LRg(L + R)
+
=
When 1 1L R
<< or practically L is tending to infinite length
T =2
1 12π π
∞+
=
R
R
gg
Q.47. A 1.6 kg wt extends a spring 8 cm from its unstretched position. The mass is replaceby a body of 50 gm. The mass is pulled and then released. Find the period of oscillation.
Solution. F = –Cx or C = –F/x.
∴ C =1 60 08
1 6 9 80 08
..
. × ..
g = = 196 n/m
∴ Period T = 2 2 50 10196
3π πm
c=
−×
= 0.11 sec.
Q.48. (a) Figure below shows a mass M restingon a smooth table between two firm support. A,Band controlled by two massless springs. If the massM is 30 gm, and the force constants of the twosprings are 2 and 1 n/m deduce (i) the frequency ofsmall oscillations of M, (ii) the energy of oscillationsfor amplitude 0.5 cm.
(b) If the two springs are connected as shown is fig. 21 find the period for oscillation.
Solution. (a) If the mass M is displaced through a distance x towards left or right, therestoring forces in the two springs are
F1 = –C1x and F2 = – C2x
acting on the mass along the same direction. So that, the total restoring force on the massM is F = –(C1 + C2)x. Thus the force constant of the system is
C = C1 + C2 = 2 + 1 = 3 n/m.
(i) Mass M = 30 gm = 0.03 kg
∴ Frequency n =1
21
25
π π πCM
30.03
= =
= 5/3.14 = 1.6 (nearly)
(ii) Energy E = Max. P.E. = 12
2cx (max)
=12
3 0 5 10 2 2× × ( . × )− = 3.75 × 10–5 J.
MC 1 C 2
A B
Fig. 21
Hormonic Motion 573
(b) When the springs of force constant C1 and C2 are connected as shown in Fig. 21 thedisplacement x of mass M produces the same restoring force F in the two springs. If the twosprings are extend through x1 and x2 then
F = –Cx1 = –Cx2
and therefore
x = x1 + x2 = − −FC
FC1 2
= − +
F
C CC C1 2
1 2
F =−
+C C
C C1 2
1 2x
Thus the force constant of the system is
C =C C
C C1 2
1 2+ =2 12 1
×+ =
23
∴ Period T = 2 2π πMC
0.03 × 32
= = 1.33 sec.
Q.49. Show that the frequency of oscillation of a body of mass from M suspended fromuniform spring of force constant C and mass m is given by
n = 12
CM m/3π +
Solution. If l is the length of the spring, then mass per unitlength = m/l.
Now, let us consider a length ds of the spring at a distance smeasured along the length below the fixed upper end. The mass ofthis small element is m/l ds.
If the velocity of the lowest point of the spring is v, then evidentlythe velocity of the length ds at a distance s from rigid support will be(s/l)v
∴Kinetic energy of the element=12
2ml
dsvsl
∴ Kinetic energy of the whole spring
=12
2
32
0
mvl
s dsl
=
12 3
2
3
3mvl
l. = mv2
6Kinetic energy of the mass M at the lower end of the spring
=12
2Mv
Total kinetic energy of the system
=12 6 2 3
32 2
M Mvmv v m+ = +( )
PB
ds
A
M
Fig. 22
574 Mechanics
If x is the displacement of the mass from its mean position, then
v = dx/dt and total K.E. = (M + m/3) (dx/dt)2
Potential energy of the spring
= C Cxdx xx
= 12
2
0
From the law of conservation of energy 12
(M + m/3) (dx/dt)2 + 12
2Cx = a constant
Differenting it with respect to time and dividing by dx/dt, we have
M +3m d x
dt
2
2 + Cx = 0 or d xdt
Cm
x2
2 +( )M + /3 = 0
This is the equation of simple harmonic motion, whose frequency is
n =1
2 3πC
M( / )+ m
Q.50. A solid sphere of mass 4 kg and radius 0.05 m is suspended from a wire. Find theperiod of oscillation, if the torque required to twist the wire is 4 × 10–3 newton–m/radiation.
Solution. Moment of inertia of the Sphere about any diameter is
I =23
MR2 =23
4 0 05 0 05× × . × . = 0.004 kg × m2
∴ Period of oscillation T = 2 2π πIC
0.0044 × 10 3= −
= 2 × 3.14 = 6.25 sec.
Q.51. A body is executing torsional oscillations about a wire of torsional rigidity C. If thetotal energy of oscillation is E and the moment of inertia of the body about the axis of rotationis I. Show that the angular velocity (dθ/dt) at the displacement θ is given by
ddtθ
=CI
2EC
− θ
Solution. Let C be couple per unit twist. If the wire is twisted through an angle θ, thenthe restoring couple is –Cθ. Now if the wire is twisted to angle θ + dθ, then the amount ofwork done is
dW = Cθ dθHence the total amount of work done in twistig the wire through an angle θ is
W = Cθ θθ
d0 =
12
C 2θ
At the angular displacement θ, this work done becomes the potential energy of the
system i.e., U = 12
C 2θ
Hormonic Motion 575
The instantaneous angular displacement θ of a body, executing torsional oscillation isgiven by
θ = θ0 sin (ω t + φ), where ω = CI
So thatddtθ
= θ0 ω cos (ω t + φ )
K.E. at the angular displacement is
K =12
2
I ddt
θ =
12
20
2 2Iθ ω ω φcos ( )t +
=12
120
22
I02θ ω θ
θ−
=
12
202I 2ω θ θ−
=12
2( )θ θ02 − C
∴ Total energy E = K + U = 12
12
120
2 2 202C C C( )θ θ θ θ− + =
∴ Angular amplitude θ0 = 2EC
Now,ddtθ θ θ
θ= −0 1
2
02 =
CI
2θ θ02 − =
CI
2EC
− θ2
Q.52. A sphere of moment of inertia 3355 gm.cm2 when suspended by a thin wire executestorsional oscillations of period 4.3 sec. Calculate the maximum angular velocity of the sphereand it’s average potential energy when the amplitude is π/2 radians.
Solution. Let the equation for the angular displacement of the sphere be θ = θ0 sin(ω t + φ).
∴ Angular velocity ddtθ
= θ0 ω cos (ω t + φ )
Its maximum value = θ0 ω = 2πnθ0
Here θ0 = π/2 radians and frequency n = 1
4 3. per sec.
∴ Max. angular velocity = 2 × 3.14 ×1
4 33 14
2.× .
= 2.3 rad/sec
Average potential energy =14
12
202 2 2I in place ofω θ ωm a
=14
3355 10 24 3 2
72 2
× × ×.
×−
π π
= 4.45 × 10–4 J.
12.1 WAVE MOTION
The concept of waves is very fundamental in nature and is one of the most importantmeans of transferring of energy.
Let us suppose a number of particles in an elastic medium (such as air) so that one ofthem cannot move without disturbing its neighboring particles. Thus any how if a particleis displaced then a disturbance will be produced in the medium which will be handed on fromparticle to particle till all of them have suffered greater or less displacements. Now, if theparticle performs periodic motion, then other particles of the medium perform the same typeof motion with the phase differing regularly from one particle to next. Such a disturbance iscalled wave motion. Thus, a wave motion may be regarded as a disturbance, which is causedby the particles of an elastic medium executing definite periodic vibrations about their meanpositions, the phase varying particle to particle and which travels with a definite velocity,depending upon the density and the elasticity of the medium. If the disturbance or state ofmotion is continuously transmitted along the same direction, then it is called a progressivewave.
It may be noted that in a wave motion there is no bodily transfer of the medium throughwhich the wave propagates. What travels forward in the medium is a state of motion whichis passed on from one particle to the next as the wave advances. Since during the propagationof the wave, the particles of the medium lying in its path are thrown into vibration insuccession, obviously the waves pass on energy from one particle to next along their line oftravel. Hence in a progressive wave motion there is always a transmission of energy alongthe direction of propagation of the wave. In stationary waves, the energy does not transferfrom one place to another.
12.2 TRANSVERSE WAVE
A transverse wave is one in which the direction of oscillation of the particles of themedium is at right angles to the direction of propagation of the wave. For example, the wavesgenerated in ropes and stretched strings and the light (electro-magnetic) waves are transversewaves.
12.3 LONGITUDINAL WAVE
A longitudinal wave is one in which the direction of oscillation of the particles of themedium is the same as the direction of propagation of the wave. For example, if a spring fixed
12
576
at one end is pulled at the other and then let free, a longitudinal wave is generated in it.Sound waves are also longitudinal waves.
12.4 GENERAL EQUATION OF WAVE MOTION
Let a wave train move along the direction of x-axis with velocity v. If we assume thesource of wave or any point of x-axis as the origin (x = 0), then at any instant t thedisplacement of a particle, situated at the origin, can be represented by the relation.
y = f (t)
Where f(t) is any function of time. The wave will reach a point P, distant x from O, inx/v second. Hence at any instant t the displacement y of the particle at P must be the sameas the displacement at the origin x/v seconds earlier i.e.,
y = f (t – x/v) ...(1)
This is the equation for wave, travelling along the positive direction of x-axis withconstant velocity v. The function f determines the shape and size of the wave. Similarly, theequation of a wave moving along the negative direction of x-axis, will be given by
y = f (t + x/v) .…(2)
Thus, the general equation of a wave, moving along x-axis, may be expressed as
y = f txv
g txv
−
+ +
...(3)
Where f and g are any two functions.
In Eq. (1) or (2) y is the function of time and distance both. Now partially differentiatingEq. (1), we get
∂∂yt
= f txv
′ −
.…(4a)
∂∂yx
= − ′ −
1v
f txv ...(4b)
∂∂
2
2y
t= ′′ −
f txv ...(5a)
∂∂
2
2y
x=
12v
f txv
′′ −
...(5b)
Dis
plac
emen
ty
Y
Ox
p
y
X
Fig. 1
578 Mechanics
where f ′ and f ′′ are the first and second differentials of f. From Eq. (4), we have
∂∂yt
= −vyx
∂∂
...(6)
This ∂∂yt
represents the particle velocity at the time t and distant x from the origin and
∂∂yx
is the slope of the y-x curve at the same instant and at the same point. Hence, for those
waves, traveling along +ve direction of x-axis, we have
particle velocity = –wave velocity × slope of y-x curve
From Eq. (5) we have
∂∂
2
2y
x=
12
2
2vy
t∂∂
or∂∂
2
2y
t= v
yx
22
2∂∂
…. (7)
This is called one dimensional differential equation of wave position.
This is very important equation. The striking feature of this equation is that the coefficientof derivative on the right hand side represents the square of the wave velocity so that weneed not to solve the equation to obtain the velocity of propagation of the wave.
12.5 EQUATION OF A PLANE PROGRESSIVE HARMONIC WAVE
The simplest and most important type of wave is the simple harmonic wave in whichthe displacement of a particle is a simple harmonic or sinusoidal function of time and distanceboth. Such waves are produced by vibrating bodies, which execute simple harmonic motion.
Let the source of the wave be situated at the origin 0 (x = 0) and let a regular train ofplane waves travel in the positive direction of x-axis. Since the source of waves performingsimple harmonic motion at any instant t the displacement of a particle at O will be given by
yo = a sin ωt ...(1)
Where a is the amplitude of the vibration and ω/2π is the frequency. If the wave velocityis v, then the disturbance; produced at O, will reach at the point P distance x in x/v seconds.Hence the displacement of the particle P at any instant t will be given by
y = a sin ω(t – x/v) ...(2)
This equation represents the displacement of a particle at any time t and is of the sameform as Eq. (1) in previous section, and it is one of the solutions of general differential onedimensional wave equation. Thus it is the equation of a simple harmonic wave traveling withvelocity v along the positive direction of x-axis.
Similarly the equation y = a sin ω(t + x/v) will represent a wave traveling in the negativedirection of x-axis with velocity v.
The frequency of the wave is equal to the frequency of the oscillating body, hence thefrequency of the wave n = ω\2π = 1/T, where T is the period.
In one vibration (i.e., in period T) the wave moves through a distance equal to wavelengthλ, hence v = λ/T = nλ.
Wave Motion 579
Now the various alternative forms of the equation of wave motion can be written as
y = a n txv
sin 2π −
y = a txv
at x
vsin sin
22
π πT T T
−
= −
y = at x
sin 2πλT
−
y = av
txv
sin2π
λ−
y = a vt xsin ( )2πλ
−
D istan ce x
Y
y
O PX
Fig. 2
The quantity 2
2π
λπ= =v k is called the propagation constant, where v = 1
λ is the wave
number.
y = a sin (ωt – kx)
If φ is the phase constant, then
y = a sin (ωt – kx + φ)
Three-dimensional waveIf a wave is moving in any direction (such a wave is called three dimensional wave), the
wave equation can be written as
∂∂
∂∂
∂∂
2
2
2
2
2
2Ψ Ψ Ψx y z
+ + =12
2
2v t∂∂
Ψ
∇2Ψ =12
2
2v t∂∂
Ψ
One of the solutions of this equation is a three dimensional plane harmonic wave, in
which the displacement (or disturbance) at time t and at a point r xi yj zk= + + is given by
ψ = A sin (ωt – k1x – k2y – k3z + φ)
580 Mechanics
or ψ = A sin (ω φt k x− +→ →
)
where k ik jk kkx y z= + + is called the propagation vector. The direction of this vector is the
direction of wave motion and its magnitude is 2π/λ, the propagation constant.
12.6 PRINCIPLE OF SUPERPOSITION
A very important characteristic of all the waves is the principle of superposition. Thisprinciple states that when a number of waves are simultaneously propagated through amedium, the resultant physical disturbance (e.g., displacement) at any point is the sum of thedisturbance due to separate waves, i.e.,
ψ = ψ ψ ψ1 2 3+ + + ...
Where ψ is the resultant disturbance at a point due to different waves, having ψ1, ψ2,ψ3 ... etc. disturbances separately at the same point.
In case of a one dimensional wave, if y1, y2, ... etc. are the displacements at a point dueto separate waves, then the resultant displacement due to all waves is
y = y1 + y2 + y3 + ...
12.7 LONGITUDINAL WAVES IN RODS
When longitudinal vibrations are produced in a solid rod, its every section vibrates toand fro about its mean position along the axis of the rod, provided that the rod has thesufficient length in comparison to its diameter.
Let us suppose that longitudinal waves are traveling along an elastic solid rod. Now,consider two close planes A and B, perpendicular to the rod at distance x and x + δx respectivelyfrom some arbitrary origin.
Suppose that at any instant A′ and B′ are the positions of the planes A and B respectively.Then, the displacement of x is y (= AA′) and at x + δx is y + δy (=BB′). Hence the initial lengthAB between the two planes = δx
Change in this length = A′B′ – AB = (δx + δy) – δx = δy
A′B′ = (x + δx) + (y + δy) – ( x + y) = δx + δy
∴ Longitudinal strain =δδyx
dydx
=
y + yδ
x
x + xδ
y
A ′ B ′A B
Fig. 3
Wave Motion 581
The change in length AB of the rod occurs due to the difference of force acting on thetwo planes. The force acting on the plane A is given by
Fa = Stress × Area of cross-section of the rod
If α is the area of cross-section of the rod, the Young’s modulus of its material is givenby
Y =StressStrain
or Stress = Y or F
Ydydx
dydx
a
α=
∴ Fa = αY dydx
Similarly the force acting on the plane B is given by
Fb = F F F Fa a a addx
x+ = +δ δ( ) ( )
= α α δY Ydydx
d ydx
x+2
2
∴ Resultant force on AB = F F Yb ad ydx
x− = α δ2
2
If ρ is the density of material of the rod, then the mass of the element AB = ραδx.
If the acceleration of this small element AB is d2y/dt2, then according to Newton’ssecond law of motion, we have
ραδx d ydt
2
2 = α δY x d ydx
2
2
ord ydt
2
2 = Y d ydx
2
2 /ρ
This is the differential equation for the longitudinal waves in a rod.
Comparing this with the differential equation of wave motion.
d ydt
v d ydx
2
22
2
2= , the velocity of the longitudinal waves in the solid rod is given by
v2 =Yρ or v = Y
ρ
Thus, the velocity of longitudinal waves in a rod depends upon the density and Young’smodulus of its material.
12.8 LONGITUDINAL WAVES IN GASES (PRESSURE VARIATIONS FOR PLANEWAVES)
When a plane progressive longitudinal wave moves through a gas, the gaseous particlesexecute simple harmonic motion along the direction of propagation of the wave. The phasesof these particles vary regularly. The distance between the particles so changes that at anyinstant particles are alternately crowded and spreaded out. Hence pressure varies fromparticle to particle inside the gas.
582 Mechanics
Let a progressive wave be moving along the positive direction of x-axis. Now, imaginecylindrical gas column of cross-sectional area α along x-axis. Let us consider in equilibriumposition, two right planes A and B (being very close to each other) of the gas column, situatedat x and x + δx distances respectively from the origin. At some later instant, when the wavepropagates, the particles of the plane A are displaced by an amount y to the position A′ andthose of the plane B are displaced by an amount y + δy to the position B′. As dy/dx is therate of change of displacement with distance.
∴ y y+ δ = ydydx
x+ δ
Initial volume of the gas column between the plane A and B
= α(AB) = αδx
its final volume = α(A′B′) = α(δx + δy) = α δ δxdydx
x+
∴ Change in volume = α δdydx
x
Thus, Volume Strain =α δ
α δ
dydx
x
xdydx
=
If E is the Bulk modulus of the gas, then
K =Volume StressVolume Strain
decrease in pressureVolume Strain
= = − pdy dx/
∴ p = −K dydx
Which represents the variation of pressure in case of a longitudinal progressive wavetravelling through a gas. Here the negative sign shows that an increase in volume is causedby a decrease in pressure. If there is a decrease in volume, p will be positive but negativesign will still be there.
If the equation of the plane waves is taken to be
y = a vt xsin ( )2πλ
−
thendydx
= − −2 2πλ
πλ
avt xcos ( )
∴ p = − = − = −K K 2dydx
avt x p vt xo
πλ
πλ
πλ
cos ( ) cos ( )2 2
where v is the wave velocity and ‘a’ the wave amplitude. The quantity pa
o = 2πλ
K is called
the pressure amplitude.
Wave Motion 583
If the rate of change of pressure is dp/dx, then the pressure difference across the smallelement AB is
δp =dpdx
xδ
Hence the force on this element = δ α αδpdpdx
x× =
= αδ αδxddx
dydx
xd ydx
−
= −K K2
2
If ρ is the density of the gas, then the mass of the element = αδxρAs the force on the element is in the direction B'A', hence the acceleration due to this
force will be –d2y/dt2. Now according to Newton’s law of motion
−αδ ρx d ydt
2
2 = −αδx d ydx
K2
2
d ydt
2
2 =Kρ
d ydx
2
2
This is the differential equation for the waves traveling in a gas.
Comparing it with the standard differential equation of wave motion, we get
v2 =K or Kρ ρ
v =
Thus the velocity of longitudinal waves in a gas depends upon the elasticity and thedensity of the medium.
The formula v = Kρ
for the velocity of sound in a gas was first of all deduced by Newton
and so it is called Newton’s formula. Newton assumed that during alternate compressionsand rarefactions occurring when a sound wave travels through a gaseous medium, thetemperature remains constant. Therefore he put the isothermal value of the elasticity K,which is equal to the pressure P of the gas. Thus
v =Pρ
12.9 WAVES IN A LINEAR BOUNDED MEDIUM
If a wave travels along a fixed line in a medium, then we call such a medium as linearone. Transverse waves traveling on a stretched string longitudinal waves in a solid rod andsound waves traveling through a pipe are such type of examples. If a wave travels along aline, then it must satisfy the one dimensional wave equation, given by
d ydx
2
2 =12
2
2vd ydt
...(1)
when y is the longitudinal or transverse displacement.
584 Mechanics
The solution of Eq. (1) must be some function of a linear combination of t and x. Lety = f (t + αx) be the solution, where α is a constant. So that on substitution in Eq. (1) wehave
α α2 ′′ +f k x( ) =12v
f t x′′ +( )α
where a = ± 1/v and f may be any function. Therefore, the general solution of this equationwill be given by
y = f txv
g txv
−
+ +
...(2)
This means that the total displacement at a point is in general due to two waves, oneof them is moving along the positive direction of x-axis and the other along the negativedirection of x-axis. In the case of a linear medium of infinite length, the functions f and gdepend only on the initial disturbance produced by the external cause.
When the length of a medium is not infinite, it is called a linear bounded medium. Insuch a case, considerable restrictions are applied on the wave functions f and g. Theserestrictions depend on the boundary conditions.
(a) The boundary may be rigid: In such a case, the displacement of the particle atthe boundary is zero at all the times i.e., particles remain permanently at rest.Such a boundary is formed by a fixed rigid body. If the medium tries to vibrate thisbody, then this rigid body exerts a large pressure on the medium. Thus, a boundaryis said to be rigid when it does not yield under the pressure of sound waves, forexample a rigid wall, closed end of a pipe etc.
(b) The boundary may be absolutely Free: A boundary is said to be free when ithas no inertia, for example the open end of an organ pipe. On such a boundary,
the particles will have displacement and the change of pressure i.e., dydx
pdydx
= −
K
will be zero at all the times.
In a linear bounded medium, generally a wave is incident on the boundary and onreflection we get a second wave. By superposition of these two similar waves, traveling inopposite directions, we get stationary waves. If the incident wave is a simple harmonic wavethen Eq. (2) will have the following form:
y = a vt x a vt xsin ( ) sin ( )2 2πλ
πλ
+ + ′ − ...(3)
incident wave reflected wave
Where the incident wave is traveling along negative x-axis and the reflected wave ismoving along positive x-axis. In general, a is the amplitude of incident wave and a' that ofreflected wave.
This equation gives the resultant displacement due to the two harmonic waves.
Case 1. When the boundary of the linear medium is rigid: Let us consider a rigidboundary at x = 0, having linear medium on the positive side. As the displacement y = 0 atthe point x = 0 hence from Eq. (3), we have
Wave Motion 585
0 = avt
avtsin sin2 2π
λπλ
+ ′ or a′ = – a
This means that the harmonic wave, traveling along the +ve direction of x-axis, isreflected back with an equal negative amplitude, i.e., with a phase change of π. Thus, onreflection from a rigid surface, a reversal of phase occurs. Now, the total displacement at anypoint of the medium is given by
y = a vt x a vt xsin ( ) sin ( )2 2πλ
πλ
+ − −
y = 2 2 2a
x vtsin cosπλ
πλ
But 2πv/λ = 2πn = ω
∴ y = 22
ax
tsin cosπλ
ω
...(4)
If the other end of the medium is not bounded, i.e., the medium extends to infinity, thenfrequency (n = ω/2π) can have any value. The amplitude at any point x is 2a sin 2π/λ, whichis zero when
sin 2πλ
x= 0
or2πλ
x= rπ
or x =rλ2
where r = 0, 1, 2, ... etc.
Such points at which the value of the amplitude is zero, are called nodes. First node exitsat the rigid boundary and the other successing nodes are separated from each other by thedistance λ/2. The maximum value of the amplitude is ±2a, when
sin 2πλ
x= ±1
or2πλ
x= ( )2 1
2r + π
or x =rλ λ2 4
+ where r = 0, 1, 2, ...etc.
Such point at which the value of the amplitude is maximum are called antinodes. Firstantinode occurs at a distance λ/4 from the rigid boundary and the adjacent antinodes areseparated by a distance λ/2. This means that one antinodes is situated in between two nodes.Here, the resultant wave is stationary in which the net flow of energy is zero because equalamounts of energy are transmitted in opposite directions by the two component waves.
Now, differentiating Eq. (4), with respect to x, we get
dydx
=4 2a x
tπ
λπλ
ωcos cos
586 Mechanics
Therefore,
p = − = −K K 4dydx
a xt
πλ
πλ
ωcos cos2...(5)
Hence the variation of pressure is proportional to cos 2πx/λ. Its value is maximum, when
cos 2πλ
x= ±1
or2πλ
x= rπ
or x =rλ2
where r = 0, 1, 2, ...etc.
Hence at the points where the value of the amplitude is zero, the variation of pressureis maximum. Similarly at the points of maximum amplitude, the variation of pressure is zero.If we differentiate Eq. (4) with respect to time t, we get the particle velocity, i.e.,
dydt
= −2 2a
xtω π
λωsin sin ...(6)
Hence the velocity of any particle of the wave is proportional to sin 2πx/λ. This meansthat the particle velocity is zero at the nodes and maximum at the antinodes.
For any (particular) value of x, we will discuss the variation of the displacement withrespect to the time t. It is clear from Eq. (4), (5) and (6) that if cos ωt = ±1 or ωt = 0, thenthe value of displacement y and variation of pressure p will be maximum and the particlevelocity will be zero, i.e., when
ωt = mπ or 2π t/T = mπ
∴ t =mT2
, where m = 0, 1, 2, ...
or t = 0, T/2, T, 3T/3 ...
i.e., at these instants the value of y and p is maximum and that of velocity is minimum.These instants occur twice in every period.
when cos ωt = 0 or sin ωt = ±1
ωt = (2p + 1) π/2 where p = 0, 1, 2, ...
or t = (2p + 1) T/4
or t = T/4, 3T/4, 5T/4, ...
Hence it is clear from Eqs. (4), (5) and (6) that at these instants the values of y and pare zero and that of dy/dt is maximum. Such instants occur twice in every period.
Now, if the linear medium is also rigidly bounded at the other end, then in the case ofthe medium of length l, substituting y = 0 at x = l in Eq. (4), we get
0 = 2 2a
ltsin cosπ
λω
As this relation holds at all values of t,
∴2πλ
l= rπ, where r is any integer.
Wave Motion 587
∴ λ =2lr
and frequency nv rv
l= =
λ 2Substituting this value in Eq. (4), we get
y = 2ar xl
r vtl
sin cosπ π
Thus, we see that when a linear medium has rigid boundaries at its both ends, onlyvibrations of certain discrete frequencies are possible (n = v/2l, v/l, 3v/2l, ...). When r = 1,the value of frequency (n = v/2l) is minimum and in such a case the mode of vibration of themedium is called fundamental. When r = 2, 3, ... etc. the medium can vibrate with othermodes. The integral multiples (v/2l, v/l, 3v/2l, ...) of the fundamental frequency are calledharmonics.
Case 2. When the boundary of the linear medium is free: Let us now consider afree boundary at x = 0 in place of a rigid boundary. As the boundary is free, the pressurevariation at x = 0 must be zero, i.e., at x = 0, dy/dx = 0. Differentiating Eq. (3) with respectto x, we get
dydx
=2 2 2 2π
λπ
λπλ
πλ
avt x
avt xcos ( ) cos ( )+ − ′ −
when x = 0, dy/dx = o
∴ 0 =2 2 2 2π
λπλ
πλ
πλ
a vt a vtcos cos− ′
This equation holds for all values of time t.
∴ a′ = a
This means that in this case the phase of the reflected wave does not change. Hence thedisplacement at any point of the medium is given by
y = a vt x a vt xsin ( ) sin ( )2 2πλ
πλ
+ + −
incident wave reflected wave
or y = 2 2 2a
x vtcos sinπλ
πλ
or y = 2 2a
xtcos sinπ
λω ...(7)
∴ Straindydx
= − 4 2πλ
πλ
ωa xtsin sin ...(8)
anddydt
= 2 2a
xtω π
λωcos cos ...(9)
From these equation it is evident that x = 0, displacement y and particle velocitydy/dt are maximum and variation in pressure p (p = –K dy/dx) is zero. Hence, in this casean antinodes is present at the free boundary. The other antinodes will be situated at x =λ/2, λ, 3λ/2,... . Here, the nodes will be at x = λ/4, 3λ/4, ... with the help of these equations,the changes with respect to time t of y, dy/dt and p can be described similar to the first case.
588 Mechanics
Now, if another free boundary is present at x = 1 (e.g. in open end organ pipes, in a solidrod clamped in the middle point etc.), then substituting dy/dx = 0 at x = 1 in Eq. (8), we have
0 =4 2π
λπλ
ωa ltsin sin
2πλ
l= rπ, where r = 0, 1, 2, ...
λ =2lr
and frequency nv rv
l= =
λ 2
or n =vl
vl
vl2
32
, , , ...
Thus, if both the boundaries of a linear bounded medium are free, only vibrations ofcertain discrete frequencies are possible. In the present case, the frequency of the fundamentalis v/2l and the frequencies of harmonics are proportional to the natural numbers 1, 2, 3,4, ... .
Now, if there is a rigid boundary (as in the case of closed organ pipe whose one end isopen and other is rigid) at x = l, we have there y = 0, i.e.,
0 = 2 2a
ltcos sinπ
λω (from Eq. 7)
∴2πλ
l= ( ) ,2 1
2r + π
where r = 0, 1, 2, 3, ...
Where frequency n =v v
lr
λ= +
42 1( )
or n =vl
vl
vl4
34
54
, , , ...
Now, if we put r = 0, 1, 2, ...etc. the frequencies of the harmonics are proportional tothe odd numbers 1, 3, 5, 7, ... . These are odd harmonics.
Thus, we see that in linear bounded medium (i) when both the boundaries are free, allthe harmonics odd and even are present and (ii) when one boundary is free and the otheris rigid, only odd harmonic are present.
12.10 FLOW OF ENERGY IN STATIONARY WAVES
In a linear bounded medium, the incident wave is always reflected back. In general, thedisplacement at a point of the medium is given by
y = a vt x a vt xsin ( ) sin ( )2 2πλ
πλ
φ− + + +
where first and second component waves are incident and reflected ones respectively. Now,
y = 22
22
2a vt
xsin cos
πλ
φ πλ
φ+
+
∴dydx
= − +
+
4 22
22
πλ
πλ
φ πλ
φa x vtsin sin
Wave Motion 589
∴ variations in pressure p = –K (dy/dx), where K is the elasticity of the medium
or p = E4π
λπλ
φ πλ
φa x vtsin sin
22
22
+
+
= pvt
x sin ;2
2πλ
φ+
where p
a xx = +
K 4πλ
πλ
φsin 2
2
Particle velocity U =dydx
av vt x= − +
+
4 22
22
πλ
πλ
φ πλ
φcos cos
or U = uvt
x cos2
2πλ
φ+
where uav x
x = − +
4 22
πλ
πλ
φcos
Now, at any point the energy is flowing across α area. Force on this area
F = pαIf in infinitely small time dt, the medium at that point is displaced through dy distance,
then the work done is
= pαdy = pα(dy/dt)dt = pUαdt
∴ Work done or transferred energy in a period = p dtUT
α0
The rate of transfer of energy across unit area per second is called energy current
∴ Energy current =1 1
0 0α
αT
UT
UT T
p dt p dt =
= 1 22
22
0T
T
p uvt vt
dtx x sin cos +
+
πλ
φ πλ
φ
=p u vt
dtx x
o2
4T
T
sinπλ
φ+
=p u
vvtx x
o2 44
T
T
− +
λπ
πλ
φcos
=p u
vtx x
o2 44
T T
T
− +
λπ
π φcos ∴ vλ
= 1T
=p u
vx x
2T−
+ −λπ
π φ φ4
4cos cos
= 0
590 Mechanics
Thus, we see that in a linear bounded medium, the rate of transference of energy iszero.
12.11 CHARACTERISTICS OF STATIONARY WAVES
The above theory tells us the following characteristics in stationary waves:
(i) All particles of the medium oscillate at the same time and with the same frequency.
(ii) The amplitude of oscillation varies from zero at the nodes to a maximum at theantinodes.
(iii) Consecutive nodes, as well as consecutive antinodes, are λ/2 apart. Nodes andantinodes occur alternately.
(iv) The nodes remain permanently at rest but (in case of longitudinal stationarywaves) suffer maximum pressure-variation and are alternately the center ofcompression and the center of rarefaction.
(v) The antinodes have always maximum displacement and maximum velocity to otherpoints but (in case of longitudinal stationary waves) do not suffer pressure-variation.
(vi) All the points between any pair of nodes are in same phase, so that they all reachtheir maximum displacement at the same instant, but the phase suddenly reversesat each node.
(vii) Twice in one period all the particles simultaneously pass through their meanpositions. The direction of motion is reversed after each half-period.
(viii) Twice in one period all the particles simultaneously have their maximumdisplacements and are momentarily at rest. The displacements are reversed aftereach half-period.
(ix) There is no net flow of energy in any direction.
12.12 WAVE VELOCITY (OR PHASE VELOCITY)
When a monochromatic wave (wave of a single frequency or wavelength) travels througha medium, its velocity of advancement in the medium is called ‘wave velocity’.
The equation of a plane progressive harmonic wave, traveling along x-axis is given by
y = a sin (ωt – kx)
Where a is the amplitude, ω (=2πn) is the angular frequency and k (=2π/λ) is the propagationconstant of the wave.
For a wave, the ratio of ω and k is called the phase velocity or wave velocity i.e.,
vp =ωk
Evidently, this is the velocity with which a plain wavefront, i.e., a plane of constantphase, given by (ωt – kx) = constant, propagates in the medium, because then
Differentiating with respect to time, we get
ω − kdxdt
= 0
dxdt
=ωk
Wave Motion 591
Which is the wave velocity v. Thus the wave velocity is the velocity with which planesof constant phase advance through the medium. Hence the wave velocity is also called as‘phase velocity’.
There may be cases when ω and k are not linearly proportional, but the ratio ω/k iscertainly the speed of the propagation at which the phase of a single wave would move along.
12.13 GROUP VELOCITY
In practice, we come across pulses rather than monochromatic waves. A pulse consistsof a number of waves differing slightly from one another in frequency. A superposition ofthese waves is called a ‘wave packet’ or a ‘wave group’. When such a group travels in adispersive medium, the phase velocities of its different components are different. The observedvelocity is, however, the velocity with which the maximum amplitude of the group advancesin the medium. This is called the ‘group velocity’. Thus the group velocity with which theenergy in the group is transmitted.
The individual waves travel ‘inside’ the group with their phase velocities.
In figure (a) are shown two plane harmonic waves of equal amplitudes but slightlydifferent frequencies traveling from left to right. The dotted curve represents the wave oflower frequency and is travelling faster. At a certain instant the two waves are in phase atthe point P. Therefore, at this instant, the maximum amplitude of the group formed by themalso lies at P (figure b): At a later instant the maximum will be build up a little to the leftof P. That is the maximum will move to the left with time relative to the waves. As a result,the group velocity will be lower than the wave velocities.
Expression for Group Velocity: Let us consider a wave group consisting of twocomponents of equal amplitudes a but slightly different angular frequencies ω1 and ω2, andpropagation constants k1 and k2. Their separate displacements are given by
y1 = a sin (ω1t – k1x) ...(1)
and y2 = a sin (ω2t – k2x) ...(2)
P
( )a
P
( )b
Fig. 4
592 Mechanics
Their superposition gives
y = y1 + y2
y = a[sin (ω1t – k1x) + sin (ω2t – k2x)]
Using the trigonometric relation,
sin A + sin B = 22
sin cosA + B2
A B−
We get
y = 22 2 2 2
1 2 1 2 1 2 1 2at k k x t k k x
sin( ) ( )
cos( ) ( )ω ω ω ω+ − +
− − −
y = 22 2 2 2
1 2 1 2 1 2 1 2at k k x t k k x
cos( ) ( )
sin( ) ( )ω ω ω ω− − −
+ − +
This represents a wave-system with a frequency (ω1 + ω2)/2 which is very close to thefrequency of either component, but with an amplitude given by
A = 22 2
1 2 1 2at k k x
cos( ) ( )ω ω− − −
Thus the amplitude of wave group is modulated both in space and time by a very-slowlyvarying envelope of frequency (ω1 – ω2)/2 and propagation constant (k1 – k2)/2; and has amaximum value of 2a. This envelope is represented by the dotted curve in Fig. (b). Thevelocity with which this envelope moves, that is, the velocity of the maximum amplitude ofthe group is given by
vg =ω ω ω1 2
1 2
−−
=k k k
∆∆
If a group contains a number of frequency-components in an infinitely small frequency-interval, then the expression for the group velocity may be written as
vg =ddkω
12.14 RELATION BETWEEN GROUP VELOCITY AND WAVE VELOCITY
Since ω = kv, where v is wave (phase) velocity, the group velocity is given by
vg =ddk
ddk
kv v kdvdk
ω = = +( )
Now, k = 2π/λ, where λ is wavelength. Therefore
vg = vdvdk
vdv
d+ = +2 2
2π
λπ
λ πλ
vg = v dv
d+
11λλ
Wave Motion 593
But d 1λ = − 1
2λλd .
Therefore, vg = vdvd
− λλ
This is the relation between group velocity vg and wave velocity v in a dispersivemedium (A dispersive medium is one in which the wave velocity is frequency dependent.ω/k = constant).
12.15 NORMAL AND ANOMALOUS DISPERSION
Usually dv/dλ is positive, so that the group velocity vg is smaller than the wave velocityvo. This is called ‘normal dispersion’. But ‘anomalous dispersion’ can arose when dv/dλ isnegative, so that vg is greater than v. An electrical conductor is anomalously dispersive toelectromagnetic waves.
12.16 NON-DISPERSIVE MEDIUM
In a non-dispersive medium (v = ω/k = constant), we have dv/dλ = 0, so that
vg = v
that is, the two velocities are identical. For light waves in free space the group velocity issame as the wave velocity.
Sound waves in gases are also non-dispersive. This is a fortunate circumstance. If soundsof different frequencies travelled at different speeds through the air it would result in chaosand aural anguish.
NUMERICALS
Q.1. A simple harmonic wave train is travelling in the positive x direction with velocity100 m/sec. The amplitude of the wave is 2 cm and frequency 100 Hz. Calculate (a) thedisplacement y, (b) the particle velocity and (c) the particle acceleration at x = 2m from theorigin at t = 5 sec.
Solution. The equation of a simple harmonic wave is given by
y = a vt xsin ( )2πλ
−
= a n txv
sin 2π −
( )v n= λ
where y is the displacement of a particle at a distance x from the arbitrary origin x = 0.
Here, a = 0.02 m, n = 100 Hz, v = 100 m/sec and x = 2 m
∴ y = 0.02 sin 2π × 100 52
100−
= 0.02 sin (2π × 496) = 0
Particle velocity U =∂∂
π πyt
na n txv
= −
2 2cos
Here, sin2πn txv
−
= 0 or cos 2 1πn txv
−
=
U = 2πna = 2 × 3.14 × 100 × 0.02
594 Mechanics
= 12.56 m/sec.
Particle acceleration∂∂
2
2y
t= − −
=4 2 02 2π πn n txv
sin
Q.2. The equation of a progressive wave is
y = 8 sin 2t
0.050.5x 2π −
+
Find the amplitude, frequency, velocity, wavelength and wave number.
Solution. The general equation of a harmonic wave is given by
y = a vt xsin ( )2πλ
φ− +
where φ is a constant
Here y = 8 2 0 50 05
2sin × ..
π tx−
+
∴ Amplitude a = 8 cm
∴2πλ
= 2π × 0.5,
∴ Wavelength λ = 2 cm
velocity v =1
0 025400
.= cm/sec.
Frequency n =vλ
= =4002
200 Hz
Wave number =1 1
20 5
λ= = . cm.
Q.3. Which of the following are the solutions to the one dimensional wave equation
(i) y = 2 sinx cos vt, (ii) y = 5 sin 2x cos vt,
(iii) y = x2 – v2t2, (iv) y = 2x – 5t.
Solution. One dimensional wave equation is
∂∂
2
2y
t= v
yx
22
2∂∂
(i) Differentiate (i) twice partially with respect to t and x separately i.e.,
∂∂
2
2y
t= − = −v y
yx
y22 and
2∂∂
∴ ∂∂
2
2y
t= v
yx
22
2∂∂
Wave Motion 595
Hence (i) is the solution to the one dimensional wave equation.
(ii)∂∂
2
2y
t= − = −v y
yt
y22
2 4 and ∂∂
∴ ∂∂
2
2y
t=
v yx
2 2
24∂∂
Hence (ii) is not the solution of one dimensional wave equation. In the same way do for(iii) and (iv) Eq. (iii) is not the solution and eq. (iv) is the solution of the wave equation.
Q.4. Find the frequency 1 period and wave number for a light of wave length 6000 A.U.
Solution. Velocity of light c = vλ where v is the frequency of light wave and λ thewavelength.
∴ v =cλ
= =−3 10
6000 105 10
8
1014×
×× Hz
Period T =1 1
5 102 1014
15
v= = −
×× sec.
Wave number V =1 1
6000 101 7 1010
6 1
λ= =−
−
×. × m
Q.5. A train of sound waves is propagated along a wide pipe and it is reflected from anopen end. If the amplitude of the waves is 0.002 cm, the frequency 1000 Hz and the wavelength40 cm, find the amplitude of the vibration at a point 10 cm from the open end inside the pipe.
Solution. The equations of incident wave and its reflected wave from open end aregiven by
[Incident wave] y1 = a vt xsin ( )2πλ
−
[Reflected wave] y2 = a vt xsin ( )2πλ
+
By the principal of superposition, the resulting stationary wave is given by
y = y y1 2+ = a vt x a vt xsin ( ) sin ( )2 2πλ
πλ
− + +
= 2 2 2avt xsin cosπλ
πλ
= 2 2 2ax vtcos sinπλ
πλ
The amplitude of the resulting wave is A = 2 2axcos πλ
Amplitude at x = 0.1 m
A = 2 × 0.002 × 10–2 cos × ..
2 0 10 4
0π =
Q.6. A stationary wave produced in a certain rod is expressed as y = 0.01 sin (13x) sin(200t).
596 Mechanics
Calculate the maximum particle velocity and tensile stress at the point x = 4 cm if theyoung’s modulus of the material of the rod is 1011 n.m–2.
Solution. Particle velocity
dydt
= 0.01 × 200 sin (13 x) cos 200t
The maximum value of particle velocity occurs when cos (200t) = 1, i.e.,
maximum particle velocity = 0.01 × 200 sin (13x)
= 2 sin (13x)
But at x = 0.04 m, 13x = 13 × 0.04 = 0.52 radian
= 0 52 1803 14
. ×.
degree = 30°
∴ sin(13x) = sin30 12
° =
Maximum particle velocity = 2 12
1 1× = − ms
Tensile stressdydx
= 10 0 01 13 13 20011 × . × cos ( ) sin ( )x t
Therefore, at 0.04 m the maximum Tensile stress
= 10 0 01 13 3 211 × . × × / cos cos13 30x = °
= 11.26 × 109 n/m2
PROBLEMS
Q.1. A string vibrates according to the equation
y = 5 sinx3
cos 40 tπ π ,
where x and y are in cm and t is second. Find out the amplitude and velocity of the twocomponent waves superposition can give rise to this vibration. What is the distance betweenadjacent nodes? What is the velocity of a particle of the string at x = 1.5 cm at t = 9/8 second.
Solution. The given equation
y = 53
40sin cosπ πxt ...(i)
represents a stationary wave. It may be written as
y =52
403
52
403
sin sinπ π π πt
xt
x+
− −
The two sine terms represent two harmonic waves travelling in opposite directions.These two waves constitute the given stationary wave. The component waves may be writtenas
y1 =52
26
120sin ( )πt x+
and y2 = − −52
26
120sin ( )πt x
comparing the component wave equations with the standard equation
y = ± ±a vt xsin 2πλ
( ), we see that
amplitude, a = 52
cm, wavelength λ = 6 cm, velocity v = 120 cm/sec.
13
597
598 Mechanics
and frequency, n =vλ
= −20 1sec
The distance between adjacent node is
λ2
= 3 cm
Differentiating eq. (i) with respect to t, we get
∂∂yt
= −5 403
40× sin sinπ π πxt
At x = 1.5 cm and t = 98
sec, we have
∂∂yt
= −2002
45π π πsin sin
= 0 sin 45 0π =
Q. 2. The equation of a stationary wave is
x = 6 cosx
3sin 60 t cm
π π
.
Find the (i) equations of waves constituting the stationary wave, (ii) wave length andfrequency of these waves, (iii) distance between successive nodes.
Hint. sin A B = A + B2
A B2
− −
sin sin cos2
Ans. ( ) sin , sin , ( ) ,i x tx
x tx
ii iii1 213 60
33 60
36 30= +
= −
−π π π π cm, sec ( ) 3 cm
Q.3. A longitudinal stationary wave in a rod is expressed as
y = 0.002 sin (0.15x) sin (4.8 × 104 t)
Calculate the maximum particle velocity and maximum tensile stress at the point x =3.5 cm. The young’s modulus of the material of the rod is 8.0 × 1011 dyne/cm2.
Solution. The particle velocity at position x and time t is obtained by differentiating thegiven equation with respect to t.
∂∂yt
= 4 8 10 0 002 0 15 4 8 104 4. × ( . ) sin ( . ) cos ( . × )x t
The maximum velocity at x occurs when cos (4.8 × 104 t) = 1. Thus
∂∂yt
max
= 4.8 × 104 (0.002) sin (0.15x) = 96 sin (0.15 x)
At x = 3.5 cm, we have
0.15 x = 0.15 × 3.5 = 0.525 radian
= °0 525 1803 14
. ×.
= 30° π rad = 180°
Stationary Waves (Waves in a Linear Bounded Medium) 599
∴∂∂yt
max
= 96 sin (30°) = 48 cm/sec.
In longitudinal waves the tensile strain is given by ∂∂yx
. So the tensile stress is young’s
modulus x tensile strain, that is, Y ∂∂yx
. Now, differentiating the given equation. With respect
to x, we get
∂∂yt
= 0 15 0 002 0 15 4 8 104. ( . ) cos( . ) sin( . × )x t
∴ tensile stress = Y ∂∂yx
= ( . × ) ( × )8 0 10 3 1011 4dyne / cm2 −
cos ( . ) sin ( . × )0 15 4 8 104x t
The tensile stress at x is maximum when sin ( . × )4 8 10 14 t =
Thus max. tensile stress = ( . × ) ( × ) cos ( . )8 0 10 3 10 0 1511 4 dyne / cm2 − x
= 24 × 107 cos (0.15x) dyne/cm2
At x = 3.5 cm, we have
max. tensile stress = 24 × 107 cos 30° dyne/cm2
= 24 × 107 × 0.866
= 2.1 × 108 dyne/cm2 Ans.
Q.4. A stationary wave produced in a rod is expressed as
y = 0.01 sin (0.13x) sin (200t) cm.
Calculate the maximum particle velocity and tensile stress at the point x = 4 cm. Theyoung modulus of the material of the rod is 1012 dyne/cm2.
[Ans. 1.0 cm/sec, 1.1 × 109 dyne/cm2]
Q.5. The refractive indices of a medium for two spectral lines of wave length 400Å and5000Å are 1.540 and 1.530 respectively. Calculate the value of the group velocity. Velocity oflight in vacuum is 3 × 1010 cm/sec.
Solution. The group velocity vg is defined by
vg =∆∆
ω ω ωk k k
= −−
2 1
2 1...(i)
where ω is angular frequency and k is propagation constant.
Now ω = 2πn
where n is frequency, which is given by
n =c
λ(vac)
600 Mechanics
where λvac is the wave length of light in vacuum. If λ be the wavelength in a medium ofrefractive index µ, then
λ = λµvac
and so n = cµλ
ω = 2πµλc
Thus k = 2πλ
from Eq. (i), we have
vg =2 1 1
2 1 12 2 1 1
2 1
πµ λ µ λ
πλ λ
c −
−
=c ( )
( )µ λ µ λ
µ µ λ λ1 1 2 2
1 2 1 2
−−
Given: c = 3 × 1010 cm/sec, µ1 = 1.540, µ2 = 1.530, λ1 = 4000Å, λ2 = 5000Å
∴ vg =3 10 1 540 4000 1 530 5000
1 540 1 530 4000 5000
8× ( . × . × ). × . ( )
−−
= 1.897 × 1010 cm/sec
Q.6. A wave packet of mean wavelength 3.6 × 10–5 cm is travelling with phase velocity1.8 × 1010 cm/sec in a dispersive medium of carbon-di-sulphide. From the dispersion of the
medium, dvdλ
is computed to be 3.8 × 1013 cm per sec per cm. Calculate the group velocity in
carbon-di-sulphide.
Solution. The group velocity vg is related to the phase velocity by
vg = vdvd
− λλ
Substituting the given values, we get
vg = (1.8 × 1010 cm/sec) – (3.6 × 10–5 cm)
× (3.8 × 1013/sec)
= 1.8 × 1010 cm/sec – 13.68 × 108 cm/sec
= 1.8 × 1010 cm/sec – 0.1368 × 1010 cm/sec
= 1.66 × 1010 cm/sec
Stationary Waves (Waves in a Linear Bounded Medium) 601
Q.7. (i) Calculate the wave number for gamma rays of wavelength 1Å. (ii) The angularfrequency is five times of the propagation constant. Calculate phase and group velocities.
Solution. The wave number v (say) is defined by
v =1λ
For the given gamma rays, λ = 1Å = 10–8 cm
v =1 1
10108
8
λ= =− cm
per cm. Ans.
(ii) The angular frequency ω is 5 times the propagation constant that is
ωk
= 5.
ωk
is the phase velocity v. Thus
v = 5 cm/sec. Ans.
If k is in per cm.
Since here ωk
is constant, that is
v = constant
dvdλ
= 0
Now, the group velocity is given by
vg = vdvd
v− =λλ
because dvdλ
= 0 Thus
vg = 5 cm/sec.
Q.8. The intensity of a plane wave is reduced to half after travelling one meter in anabsorbing medium. What will be its intensity after travelling three meters? What will happenif the medium is non-absorbing.
Solution. Let I0 be the initial intensity of the wave and I the intensity remaining afterthe wave travels a distance x in an absorbing medium. Then we have
I = I0 e–βx
where β is a constant
when x = 1 m, I = 12
I0. Thus
12
= e–β
602 Mechanics
or β = loge 2 = 0.693
Let I′ be the intensity for x = 3m. Thus
I′ = I0 e–3β
But 3β = 3 × 0.693 = 2.08
I′ = I0e–2.08 = I0
e 2 08.
we can see that e2.08 = 8
I′ =18
I0 .
The given wave is a “plane wave”. Hence while travelling in an ‘non-absorbing’ mediumthe intensity will remain undiminished.
Q.9. The intensity I0 of a plane wave is reduced by 30% after passing through 1 meterin an absorbing medium. What distance should it travel so that the intensity becomesapproximately I0/2 ?
Solution. For a plane wave, the variation of intensity with distance in an absorbingmedium is given by an exponential law of the form
I = I0 e–βx
where I0 is the initial intensity, I is the intensity after travelling a distance x and β is aconstant.
when x = 1 meter, I = 0.7 I0. Thus
0.70 I0 = I0 e–β
10 70.
= eβ
β = loge (10/7) = loge (1.428)
For I = 12
I0 we have
12
I0 = I0 e–βx
2 = eβx
βx = loge 2
x =log log
log .e e
e
2 21 428β
=
=log
log ...
10
10
21 428
0 30100 1549
2= = meter.
Q.10. Which of the following are solution of the one-dimensional wave equation? (i) y = 4sin x cos vt, (ii) y = 9 sin 2x cos vt, (iii) y = x2 – v2t2, (iv) y = 2x – 5t.
(Lucknow, 1992)
Stationary Waves (Waves in a Linear Bounded Medium) 603
Solution. The general differential equations for a progressive wave
y = y vt x( )± are:
∂∂yt
= − v ydx∂
...(1)
and∂∂
2
2y
t= v
ydx
22
2∂
...(2)
where v is the constant wave-velocity. Let us examine the given equations:
(i) y = 4 sin x cos vt ...(i)
differentiating it twice partially with respect to t and x separately, we get
∂∂
2
2y
t= 4 sin x (–v2cos vt) = –v2y
and∂∂
2
2y
x= –4 sin x cos vt = –y
∴ ∂∂
2
2y
t= v
yx
22
2∂∂
This is same as Eq. (2) above. Hence the given eq. (i) is a solution of wave equation.
(ii) y = 9 sin 2x cos vt ...(ii)
∂∂
2
2y
t= 9 2 2 2sin ( cos )x v vt v y− = −
and∂∂
2
2y
x= − = −36 2 4sin cosx vt y
∴ ∂∂
2
2y
t=
v yx
2 2
24.∂∂
This is different from Eq. (2). Therefore eq. (ii) is not a solution of wave equation
(iii) y = x2 – v2t2 ...(iii)
∂∂
2
2y
t= –2v2
∂∂
2
2y
x= 2
∂∂
2
2y
t= −v y
x2
2
2∂∂
Clearly, Eq. (iii) also is not a solution of wave equation
(iv) y = 2x – 5t ...(iv)
This equation can be differentiated only once
∂∂yt
= –5
604 Mechanics
∂∂yx
= 2
∴∂∂yt
=−52
∂∂yx
This is similar to Eq. (i) if v = 52
. Hence the given Eq. (iv) is a solution of wave motion.
Q.11. What is meant by a propagation vector in a general plane harmonic wave? If thepropagation constant of a wave is 132 per cm and its velocity is 300 m/sec. Which must beits wave number, wavelength and frequency?
Solution. The equation of a plane harmonic wave moving along +x-axis, in terms of thepropagation constant k is
y = a t kxsin ( )ω −
A physical displacement y may not be associated with all waves. The common featureof all waves is all transmission of some sort of disturbance with a velocity characteristic ofthe medium. For example, y may correspond to pressure – change in sound waves or electric(or magnetic) field in electromagnetic waves. The disturbance may be represented by ageneral symbol ψ, so that we may write
y = a sin ( )ωt kx−
The equation for a three-dimensional wave should be
ψ = a t k x k y k zx y zsin ( )ω − − −
or ψ = a t k rsin ( . )ω −→ →
where the vector k→
( )= + +k i k j k kx y z is called the ‘propagation vector.’ The direction of thisvector is the direction of wave motion and its magnitude is 2π/λ, the propagation constant k.
Now, propagation constant k =2 132πλ
= per cm.
∴ Wave number V→
= 12
132
2 227
21λ π
= =
=k
× per cm.
Wave length λ =1 1
210 0476
v= = . cm.
Frequency, n =v
vλ
= = =→V
cmsec cm
/ sec.300021
6 3 105× . ×
Q.12. Write down the equation of simple harmonic progressive waves having followingcharacteristics: (i) amplitude 0.6 cm, period 0.5 sec, wavelength 36 cm, travelling along positivex-axis (ii) amplitude 0.003 cm, frequency 660 Hertz, velocity 330 m/sec travelling along negativex-axis. Assume the displacement y = 0 at x = 0 at t = 0.
Stationary Waves (Waves in a Linear Bounded Medium) 605
Solution. (i) The equation of a simple harmonic progressive wave along + x-axis, interms of amplitude a, period T and wavelength λ is
when y = 0 at x = 0 at t = 0
y = at x
sin 2πλT
−
Putting a = 0.6 cm, T = 0.5 sec and λ = 36 cm, we get
y = 0 6 2 236
. sin π tx−
cm.
(ii) Again, the equation of a simple harmonic wave travelling along x-axis, in terms offrequency n and velocity v is
y = a txv
sinω +
y = a n txv
sin 2π +
Putting a = 0.003 cm = 3 × 10–5 m, n = 660 sec–1 and v = 330 m/sec we get:
y = 3 10 2 660660330
5× sin− +
π tx
or y = 3 10 2 660 25× sin ( )− +π t x meter. Ans.
Q.13. Two particles of a medium disturbed by the wave propagation are atx1 = 0 and x2 = 1 cm. The displacement of the particles can be given by the equations:
y1 = 2 sin 3πt
y2 = 2 sin 38
π πt −
Determine the frequency, wavelength, wave-velocity and the displacement of the particleat t = 1 sec and x = 4 cm.
Solution. The particle motion is simple harmonic (y = a sin ωt). This shows that a =2 cm and ω = 3π.
The frequency is
n =ωπ
ππ2
32
1 5= = . Hz
We see that two particles at a distance 1 cm apart have a phase difference of π/8.Therefore, the wavelength, i.e., the distance between two particles having a phase differenceof 2π is
λ =8 2 16π
π× = cm
The wave velocity is
v = ηλ = 1.5 sec–1 × 16 cm = 24 cm/sec
606 Mechanics
The wave equation
y = a vt xsin ( )2πλ
−
becomes y = 2 216
24sin ( )πt x− cm
At t = 1 sec and x = 4 cm, we have
y = 2 sin 2.5 π cm
= 2 22
sin π π+
cm
= 22
2sin π = cm
Q.14. The equation of a wave is given by
y = 10 sin 2t
0.04x
200,π −
where y and x are in cm and t in second. Find the amplitude, wavelength, velocity and thefrequency of the wave. Find the velocity of a point x = 400 cm at t = 0.08 sec.
Solution. Let us compare the given equation with the standard form of the waveequation
y = at x
sin2πλT
−
we find
amplitude, a = 10 cm,
wave length, λ = 200 cm
frequency, n =1 1
0 0425 1
T sec= = −
.and velocity v = nλ = 25 × 200 = 5000 cm/sec
The particle velocity at a point x at time t is
∂∂yt
= 102
0 042
0 04 200π π
.cos
.
−
t x
Putting x = 400 cm and t = 0.08 sec, we get
∂∂yt
= 102
0 040
π.
cos
= 500 π = 500 × 3.14 = 1570 cm/sec.
Q.15. The equation of a transverse wave travelling along a rope is given by y = 5 sin π(0.02x – 4.0 t) cm.
Find the amplitude, frequency, velocity and wavelength. Calculate the maximum transversevelocity of any particle in the rope.
Stationary Waves (Waves in a Linear Bounded Medium) 607
Solution. y = 5 sin π (0.02 x – 4.0 t) cm
= 5 0 024 00 02
sin ...
π xt−
cm
= 5 2100
200sin ( )πx t− cm
= 5 2100
200sin ( )π πt x− − cm
Comparing it with the standard wave equation
y = a vt xsin ( )2πλ
− ,
we get:
amplitude, a = 5 cm, wavelength, λ = 100 cm velocity, v = 200 cm/sec, frequency,
n = vλ
= =200100
2 Hz.
The particle velocity is obtained by partially differentiating Eq. (i) that is
∂∂yt
=2 2π
λπ
λva
vt xcos ( )−
The maximum value of the cos term is 1.
∂∂yt
max
=2π
λva
=2 3 14 200 5
10062 8× . × × .= cm/sec.
608 Mechanics
14.1 DAMPING FORCE
The frictional force, acting on a body opposite to the direction of its motion, is calleddamping force. Such a force reduces the velocity and the kinetic energy of the moving bodyand so it is also called a retarding or dissipative force. These forces usually arise due to theviscosity or friction of the medium and are non-conservative in nature. When the velocitiesare not sufficiently high, the damping force is found to be proportional to the velocity (v) ofthe particle and may be expressed as
F = –γv = –γ (dx/dt) ...(1)
Where γ is a positive constant, called thedamping coefficient.
If it is the only external force acting onthe moving particle, then according to Newton’ssecond law, we have the equation of motion as
mdvdt
= – γv –γv
ordvdt m
v+ γ= 0 ...(2)
It is sometimes useful to define a constantτ = m/γ, called the relaxation time, so that theequation of motion becomes
dvdt
v+τ
= 0 ...(3)
This is a very important differential equation and may be written as
dvv
=−dt
τ
ordvv
v
v
0
=− 1
0τ
dtt
...(4)
608
14
Y
1.0
v/v0 .5
0e – t/τ
O τt
Fig. 1
Damped and Forced Harmonic Oscillation 609
where v0 is the velocity at t = 0. On integrating Equation (4), we have
logev – loge v0 = − tτ
or logevv0
= − tτ
So that v = v0 e–t/τ ... (5)
Therefore, the velocity decreases exponentially with time and we say that the velocityhas been damped with time constant τ.
When t = τ, v = v0/e. Hence the relaxation time or time constant may be defined as thetime in which the velocity becomes 1/e times the initial velocity. Sometimes, 1/τ or γ/m isput equal to 2k, where k is called damping constant.
The instantaneous kinetic energy of the particle is
K = 12
12
202 2
0
2mv mv e k e
t t
= =− −
τ τ ...(6)
Thus, the kinetic energy also decreases exponentially with a relaxation time τ/2. Putting
vdxdt
= in Eq. (5) and then
x = v e dttt
0
0
− τ , where at t = 0, x = 0
∴ x = v e v et t t
0
0
0 1τ ττ τ−
= −
− −...(7)
The example, for representing eq. (1) is furnished by an ohmic resistance. If I is thecurrent at any instant, then voltage drop across R is IR and the induced E.M.F. acrossinductance L is –LdI/dt. There is no external E.M.F., so that
RI = −L Iddt
or L I RI =ddt
+ 0
Which is exactly of the form (2) or (3) with x = L/R.
The current will decrease in an undriven LR circuit as
I = I0 e–(R/L)t
14.2 DAMPED HARMONIC OSCILLATOR
When the effect of frictional force on the oscillator is not considered then a pendulumor a tuning fork, if vibrated once, will oscillate indefinitely with a constant amplitude i.e.,without any loss of energy. In actual practice, the amplitude of oscillation gradually decreasesto zero as a result of frictional forces, arising due to the viscosity of the medium in whichthe oscillator is moving. The motion of the oscillator is said to be damped by friction andthe vibrating system is called damped harmonic oscillator. If damping is to be taken intoaccount, then a harmonic oscillator experiences.
I
L
R
Fig. 2
610 Mechanics
(i) a restoring force proportional to displacement (–x) and
(ii) a damping force proportional to the velocity dxdt but opposite to it −
γ dx
dt
where x is the displacement of the oscillating system and dxdt
is its velocity at this displacement.
Hence the equation of motion of the damped harmonic oscillator is
md xdt
2
2 = − −γ dxdt
xC
or md xdt
dxdt
x2
2 + +γ C = 0
ord xdt
kdxdt
x2
2 022+ + ω = 0 ...(1)
where 2k = γ/m = 1/τ and ω0 = C/m is the natural angular frequency in absence ofdamping forces. This τ is the relaxation time and k is called damping constant.
Equation (1) is the differential equation of damped harmonic oscillator and is applicableto any system for which the equation of motion have this form.
Now, we have to solve equation (1) of damped harmonic motion. Let its possible solutionbe
x = Aeat ...(2)
Substituting the values of x, dx/dt and d2x/dt2 in Equation (1), we have
α α ω2022+ +k eat = 0 or α α ω2
022 0+ + =k
∴ α = − ± −k k202ω
Hence the solutions of the equation (1) are
x = A1
202
ek k t− + −
ω
and x ek k t
= A2
202− − −
ω
Thus the most general solution of equation (1) is
x = A A1 2
202 2
02
e ek k t k k t− + −
− − −
+
ω ω
i.e. x = e e ekt k t k t− −
− −
+
A A1 2
202 2
02ω ω
...(3)
where A1 and A2 are constants, depending upon the initial conditions of motion.
The quantity k202− ω is imaginary, real or zero, it depends on the relative values of
k and ω0. If k < ω0, then k202− ω is imaginary and it is called underdamped case. If k > ω0
or k = ω0, the situations are known as overdamped and critically damped respectively.
Damped and Forced Harmonic Oscillation 611
(i) Underdamped Case: If the damping is so low that If k < ω0, then k202− ω
= − − − = − =ω ω ω02 2
02 2k i k i where i = −1 and ω = ω0
2 2− k , which is a real
quantity. Now, from Equation (3), we have
x = e e ekt i t i t− −+A A1 2ω ω
x = e t i t t i tkt− + + −A A1 2(cos sin ) (cos sin )ω ω ω ω
x = e t i tkt− + + −(A A A A1 2 1 2)cos ( )sinω ω
As x is a real quantity, (A1 + A2) and i(A1 – A2) must be real quantities.
Clearly, A1 and A2 are complex quantities. If A1 + A2 = A0 sin φ and i (A1 – A2) = A0cos φ , then
x = A0e tkt− +sin( )ω φ ...(4)
This equation represents a damped harmonic motion. This motion is oscillatory orballistic whose periodic time is given by
T =2 2
02 2
πω
π
ω=
− k...(5)
Thus, the effect of damping is to increase the periodic time. If k = 0, i.e.; no damping
then T = 2
0
πω
. But in actual cases, the effect of damping on periodic time is usually
negligible except few extreme cases.
The amplitude of the oscillatory motion is given by
A = A A0 0e ektt
− −= 2
τ
Time
A = A e0– t/2τ
Y
Dis
plac
emen
t
A = –A e0– t/2τ
X
Fig. 3
612 Mechanics
Where A0 is the amplitude in the absence of damping.
In the presence of damping, the amplitude decreases exponentially with time. Infigure, time displacement curve is shown for damped harmonic motion. As themaximum value of sin (ωt + φ) is alternately +1 and –1, obviously the timedisplacement curve of the vibrating body lies entirely between the curvesA = A0e
–kt and A = –A0e–kt shown by the dotted lines.
14.3 LOGARITHMIC DECREMENT
The time interval between the successive maximum displacements (i.e., amplitudes) ofleft and right hand sides is T/2, hence if An and An+1 are the successive amplitudes, then
An = A0e–kt and An+1 = A0
T
ek t− +
2
∴A
An
n+1=
e
e
e dkt
k t
k−
− +
= =T2
T2 , a constant for motion ...(6)
Where d is called the decrement, indicating for the reduction in amplitude.
Now loge d =kT2
= λ ...(7)
This quantity λ is called the logarithmic decrement and is equal to the ratio of thenatural logarithmic of the ratio of two successive amplitudes of vibration.
(ii) Overdamped Case: If the damping is so high k > ω0, then k202− ω , say β, is a real
quantity and then from Equation (3), We have
x = A A1 2e ek t k t− − − ++( )β β ...(8)
As k > β, both quantities of right hand side decrease exponentially with time and themotion is non-oscillatory. Such a motion is called dead beat or aperiodic and its mainapplication is in dead beat galvanometers.
(iii) Critically Damped Case: If k = ω0, then from Equation (3), we have
x = A A C1 2+ =− − e ekt kt , where C = A1 + A2
In this equation, there is only one constant, hence it does not provide us thesolution of differential equation (1) of second order.
Now, suppose k h202− =ω , which is a small quantity. Hence from Equation (3), we
have
x = e e ekt ht ht− −+A A1 2
x = e ht htkt− + + + − +A B(11( ...) ...)1
Neglecting the small terms, containing h2 and higher powers of h, we get
x = e h tkt− + + −A A A A1 2 1 2
Damped and Forced Harmonic Oscillation 613
or x = e tkt− ( )P + Q ...(9)
Where P = A1 + A2 and Q = h(A1 – A2)
Also,dxdt
= e e k tkt kt− − −Q + P + Q( ) ( )
If initially at t = 0 the displacement of the particle is x = x0 and there the velocityis v0, then
x0 = P and v0 = Q – kP = Q – kx0 or Q = v0 + kx0
∴ x = x v kx t e kt0 0 0+ + −( ) ...(10)
This equation represents that initially the displacement increases due to the factor
x v kx t0 0 0+ +( ) . But as time elapses, the exponential term becomes relatively moreimportant and the displacement returns continuously from the maximum value to zero andthe oscillatory motion does not just occur. Such a motion is called critically damped or justaperiodic.
Time (t)Dis
plac
emen
t (x)
Heavy Dam ping
Critical Dam ping
L igh t Dam ping
Fig. 4
14.4 POWER DISSIPATION IN DAMPED HARMONIC OSCILLATOR
If a particle oscillates in a medium, then due to the viscosity of the medium dampingforces act on the particle in a direction opposite to its movement. In this process work isdone by the particle in overcoming the resistance forces. Consequently, the mechanicalenergy of the vibrating particle continuously decreases so that the amplitude of oscillationbecomes less and less. Here, we want to find a relation for power dissipation (i.e., rate ofdissipation of energy).
At any instant t, the displacement of a damped harmonic oscillator is given by
x = A0e tkt− +sin( )ω φ
∴ Velocity of the particle
dxdt
= A0e k t tkt− − + + +sin( ) cos( )ω φ ω ω φ
614 Mechanics
∴ Kinetic energy of vibration
K = 12
22 2 2 2 2m e k t t k t tktA02 − + + + − + +sin ( ) cos ( ) sin( ) cos ( )ω φ ω ω φ ω ω φ ω φ
Potential energy U =12
12
202 2
0
C U = Cx m x xdxx
= ω ,
U =12 0
2 202 2m e tktA − +ω ω φsin ( )
The average total energy for a period will be the sum of time average of kinetic andpotential energy. If the amplitude of oscillation does not change much in one cycle of motion,then the factor e–2kt may be taken as constant. Now, we are left with the average of sin2
(ωt + φ), cos2 (ωt + φ) and 2 sin (ωt + φ) cos (ωt + φ), whose average values for a periodare ½ , ½ and 0 respectively.
Hence the average kinetic energy =12
12
12
140
2 2 2 2 2 2m e k m ekt ktA A02− −+
=ω ω.
for damping is low so k2 << ω02 and average potential energy =
12 0
2 2 2m e ktA − ω for low damping
ω ≈ ω0
∴ Average total energy E = 12 0
2 2 2 2m e e ekt ktt
A E or E0 0− − −
=ω τ
∴ Average power dissipation P = − = =−ddt
m ke kktE A E or E02ω
τ2 2 2
This dissipation of energy is caused by the damping force γ dxdt
.
Hence, if we calculate the negative value of the average rate of doing work by this
force, i.e., <γ dxdt
2
>, then we will get exactly the same expression for power dissipation, as
obtained above. This loss of energy generally appears in the form of heat in oscillatingsystem.
14.5 QUALITY FACTOR Q
To represent the efficiency of an oscillating system in a term known as quality factorsQ, is widely used. The quality factor Q of an oscillating system is a measure of damping,or the rate of energy decay, of the system. It is defined as 2π times the ratio of the energystored in the system to the energy lost per period. That is
Q = 2π Energy storedEnergy loss per period
Q = 2π ωEPT
EP
= , 2πT
= ω
Damped and Forced Harmonic Oscillation 615
∴ Q = ωτ, P = E/τEvidently Q is a dimensionless quantity.
In the case of low damping Q = ω0τ
But ω0 =Cm
and τ =mγ
,
∴ Q =mcγ
Thus high value of Q means that the damping of the oscillating system is low.
This Q is a measure of the extent, to which oscillator is free from damping.
For an undamped oscillator, γ = 0, so that Q is infinite.
The energy of the oscillator ( )E = E0et−τ decreases to e–1 of its initial value in the time
τ and in this time the oscillator performs ω τπ π0 2
= Q2
oscillations.
The Q of an excited atom is about 108, while for a violin string it is only about 103 dueto a much larger damping in the string.
14.6 FORCED (DRIVEN) HARMONIC OSCILLATOR
If an external periodic force is applied on a damped harmonic oscillator, then theoscillating system is called driven or forced harmonic oscillator and its oscillations are calledforced (or driven) vibrations.
Let us consider system oscillating about an equilibrium position under and externalperiodic force. Let x be its displacement from the equilibrium position at an instant duringthe oscillation. Its instantaneous velocity is dx/dt. The forces acting upon the system at theinstant are:
(i) A restoring force proportional to the displacement but acting in the opposite direction,this may be written as –Cx, where C is the force constant.
(ii) A frictional force proportional to the velocity but acting in the opposite direction.
This may be written as −γ dxdt
(iii) An external periodic force represented by F0 sin pt, where F0 is the maximum valueof this force and p is its angular frequency. Thus the total force F acting uponsystem is
F = − − +C F0xdxdt
ptγ sin
616 Mechanics
By Newton’s second law this must be equal to the product of the mass ‘m’ of the system
and the instantaneous acceleration d xdt
2
2 . That is
md xdt
2
2 = − − +C F0xdxdt
ptγ sin
d xdt m
dxdt m
x2
2 + +γ C=
F0
mptsin
ord xdt
k dxdt
x2
2 022+ + ω = f pt0 sin ...(i)
where 2k = γ/m, ω0 =Cm
= 2πn0 = the natural angular frequency of the system in the
absence of damping and driven forces, and
f0 =F0
m
The natural frequency of the oscillating system may be different to the frequency(p/2π) of the applied force. When a periodic force acts on a body, it delivers periodic impulsesto the body so that the loss of energy in doing work against the dissipative forces isrecovered. The result of this is that the body is thrown into continuous vibrations. In theinitial stages the body tends to vibrate with its natural frequency whilst the impressed forcetries to impose its own frequency upon it. But soon the free vibrations of the body die outand ultimately the body vibrates with a constant amplitude and with the same frequency asthat of the impressed force. Such vibrations of constant amplitude, performed by a bodyunder the influence of a impressed periodic force, with a frequency equal to that of the force,are known as forced vibrations and the oscillating system is itself called driven or forcedharmonic oscillator. The impressed periodic force is called the driver and the body, executingforced oscillations, is called the driven.
Now let us suppose that steady state solution of Equation (i) is
x = A sin (pt – θ) ...(ii)
Because the amplitude (A) of the forced oscillations remains constant and the frequencyof vibrations is equal to the (p/2π) of the force. Hence θ represents the phase differencebetween the force and the resultant displacement of the system. Now from Equation (ii), wehave
dxdt
= p ptAcos ( − θ)
andd xdt
2
2 = –Ap2 sin (pt – θ)
Substituting these values in Equation (i), we get
− − + − + −p pt kp pt pt2022A sin( Acos ( Asin(θ θ ω θ) ) ) = f pt0 sin ( )− +θ θ
Damped and Forced Harmonic Oscillation 617
A( A cos (02ω θ θ− − + −p pt kp pt2 2) sin( ) ) = f pt f pt0 0cos sin ( ) sin cos ( )θ θ θ θ− + −
If this equation is to satisfied for all values of t, then the coefficients of sin (pt – θ) andcos (pt – θ) on the two sides must be equal. Equating them, we obtain.
A 02ω − p2 = f0 cos θ ...(iii)
and 2kpA = f0 sin θ ...(iv)
Squaring Eqs. (iii) and (iv) and on adding, we get
∴ A =f
p k p
0
02 2 2 2 24ω − +
...(v)
Dividing Equation (iii) by Equation (iv), we get
tan θ =2 2
02 2
1
02 2
kpp
kppω
θω− −
−or = tan ...(v a)
Substituting these values in Equation (ii), we get
x =f
p k ppt kp
p0
02 2 2 2 2
1
02 2
4
2
ω ω− +−
−
−
sin tan ...(vi)
Here, we will assume that the damping is low. Now we will consider the different cases,when ω0 >> p, ω0 = p and ω0<< p.
For low driving frequency, i.e., when ω0 >> p, tan θ → 0 or θ → 0, that is, the drivingforce and the displacement are in the same phase. Hence the amplitude of the oscillationfrom Equation (v) is given by
A =f m
m0
02ω
= =FC /
FC
0 0/
Thus, in this case the response (amplitude) does not depend on the mass of the oscillatingsystem or damping, but it depends only on the force constant C.
For very high driving frequency i.e., when ω0<< p, if the driving frequency is increased,the value of the amplitude A increases.
A ≈fp mp02 2≈ F0
Which shows that the amplitude, which now depends upon the mass, continuouslydecreases as the driving frequency p is further increased.
14.7 AMPLITUDE RESONANCE
Equation (vi) show that the amplitude of the forced oscillations depends upon ( )ω02 2− p
which in turn depends upon the difference between the driving frequency p and the naturalfrequency ω0 of the oscillator. Smaller this difference, larger the amplitude.
618 Mechanics
To find the condition for maximum amplitude (Resonance) Equation (v) can be arrangedas follows:
A = f
k p k k
0
02 2 2 2 2
02 42 4 4ω ω− − + −
...(vi)
Therefore, to have the maximum value of A,
ω02 2 22− −k p = 0
or p = ω02 22− =k pr (say)
∴ Amax =f
k k0
02 22 ω − ...(vii)
Hence for a certain value of driven frequency (pr) the amplitude of the oscillatingsystem is maximum. This phenomenon in which the amplitude of the driven oscillatorbecomes maximum at a particular driven frequency, is called amplitude resonance and thisfrequency is known as the resonant frequency.
For a forced harmonic oscillator, the resonant frequency is p/2π, where pr is the angularresonant frequency of that oscillator. The value of this angular frequency is less than ω0 or
ω ω= −
0
2 2k , which is the frequency of the natural damped oscillations of the body. If the
damping is low, then ω0 = pr and the amplitude.
Amax =f
kf0
0
0
02 ωτ
ωor ...(viii)
Hence in the case of zero damping the amplitude should be infinity. But it is notpossible because the friction on the oscillating system is never zero. If the damping is low,then the ratio of the response at resonance to the response at zero driven frequency is givenby
=f k
f k/
/2
20
02
00
ωω
ω ω τ= = = Q
Hence at resonance, this ratio is equal to the quality factor Q of the system. Thus, wesee that the damping controls the response of resonance.
When p = ω0 from Equation (va) and (v), we have
tan θ = ∞ or θ = π/2
And A =f
kf0
0
0
02 ωτ
ω= , which is less than the amplitude as given by Equation (viii).
In case of low damping, the resonant frequency is equal to the natural frequency of theoscillator and then the amplitude is maximum.
For high driving frequency or when ω0 << p, tan θ = –2k/p tan θ → –θ or θ → p and
amplitude A = fp02 nearly.
Damped and Forced Harmonic Oscillation 619
So on increasing the driving frequency (p), the response or amplitude decreases.
ω0
P
XO
Am
plitu
de
H igh Dam ping
Zero Dam ping
Low D am ping
Y
Fig. 5
The variation of the amplitude A with the frequency p of the periodic force is shown infigure. Initially, when the angular frequency p of the force is increased the amplitude Acontinuously increases and at a certain value (which is nearly equal to ω0 for low damping)of p, the value of the amplitude becomes maximum. This is the condition of amplituderesonance. Now, on further increasing p, the response decreases gradually. The x-coordinatescorresponding to the peaks of these curves, as shown in figure, represent the resonantfrequencies. When the damping is greater, the resonant frequency (pr) is less than ω0 butin case of low damping the resonant frequency of the force is nearly equal to the naturalfrequency (ω0) of the oscillator. It is also clear from these curves that for low damping theheight of the peaks become greater and when the damping is zero (in the ideal case) theheight of the peaks (i.e., the maximum amplitude) rises to infinity.
14.8 SHARPNESS OF RESONANCE
At resonance, the amplitude of the oscillating system becomes maximum. It decreasesfrom this maximum value with the change (decrease or increase) of the frequency of theimpressed force. In above figure, for different values of damping curves have been drawnbetween the driven frequency and the amplitude.
The term sharpness of resonance refers to the rate of fall in amplitude with the changeof forcing frequency on each side of the resonant frequency. When the damping is low, theresponse (amplitude) falls of very rapidly on other side of resonant frequency and we say thatthe resonance is sharp. On the other hand, for high damping the response falls off veryslowly on either side of resonant frequency and the resonance is said to be flat. Thus, theresonance is flat or sharp according to the damping for the oscillating system is large orsmall. Familar examples of a flat or sharp resonance are the resonance of an air columnand a sonometer wire with a tuning fork respectively. Due to its large damping the aircolumn responds with tuning fork over fairly wide range in the neighborhood of resonance.Thus in this case it is actually difficult to get an exact point of resonance and resonance issaid to be flat. But in case of sonometer wire, the damping is small so that the resonanceis sharp and hence to have the resonant points, the length of the wire is adjusted veryprecisely.
620 Mechanics
Half width of resonance Curve: If ph is the value of angular frequency when theamplitude falls to half the value at resonance, then the change in p is called the half widthof the resonance curve i.e.,
Half width ∆p = | |p ph r−
At resonance, Amax =f
k k
f
k k0
02 2
02
02 42 4 4ω ω−
=−
...(1)
At the frequency ph the amplitude is Amax/2
∴Amax
2=
f
k p k kh
0
02 2 2 2 2
02 42 4 4ω ω− − + −
...(2)
From Eqs. (1) and (2), we have
ω ω02 2 2 2 2
02 42 4 4− − + −k p k kh = 4 4 42
02 4( )k kω −
but pr2 = ω0
2 22− k
( )p pr h2 2 2− = 3 4 4 3 4 42
02 4 2 2 4k k k p krω − = +
If damping is small, 4k2 is negligible and then we have
( )p pr h2 2− = ± 3 2( )kpr
or ph2 = p kpr r
2 3 2 ( )
or ph = p kp
p kp
p krr
rr
r1 3 2 1 3 3
12
=
=
∴ Half width ∆p = | |p p kh r− = 3
14.9 VELOCITY RESONANCE
The velocity of the driven oscillator is
v =dxdt
f p
p k ppt=
− +−0
02 2 2 2 24ω
θ
cos( )
= v pt0 2sin( / )− +θ π
where vf p
p k p
f
pp
k
00
02 2 2 2 2
0
02
24 4
=− +
=
−
+ω ω
is the velocity amplitude and
θ = tan−
−
1
02 22kp
pω
Damped and Forced Harmonic Oscillation 621
We see that the velocity amplitude v0 varies with the driven frequency p. When p = 0,
v0 = 0 and when p = ω0, v0 attains the maximum value. Hence at the frequency pm
= =ω0C
of the impressed force, the velocity has the maximum value and we call it velocity resonance.
At velocity resonance
θ = tan−
=1 2
0 2kp π
∴ Velocity phase constant = − + = − + =θ π π π2 2 2
0
ω0
Low dam pin g
H igh dam ping
Y
v
OX
p
Fig. 6
This means that the velocity of the oscillator is in phase with the applied force. Thisis the most favourable situation for transfer of energy from the applied force to the oscillator,because the rate of work done on the oscillator by the impressed force is Fv, which is alwayspositive for F and v in phase.
When p < ω0, the velocity amplitude is smaller than at p = ω0.
14.10 POWER ABSORPTION
When an oscillator executes vibration in presence of damping forces, it loses energy indoing work against these forces. This loss of energy is supplied by the periodic impressedforce so as to continue the oscillations. In a small time interval dt the energy supplied bythe driving force F(= F0 sin pt) will be equal to the work done on the oscillating system bythis force, i.e.,
dE = F.dx = F (dx/dt) dt
orddtE
= Fdxdt
∴ F = mf0 sin pt
anddxdt
=pf pt
p k p
0
02 2 2 2 24
cos( )−
− +
θ
ω
Hence at any time the energy absorbed by the oscillating system is
pddt
dxdt
= =
E F =mf pt
p k ppf pt0
02 2 2 2 2
0
4
sin [cos ( )]ω
θ− +
−
622 Mechanics
Hence the average power absorbed is
P Favav
dxdt
= =
mf p
p k ppt pt av
02
02 2 2 2 24ω
θ− +
−
[sin cos ( )]
Now the average of sin pt cos (pt – θ) for one period T or 2π/ω
= 1
0T
T
sin cos ( )pt pt dt− θ
=1 1
22
12
0T
T
sin( ) sin sinpt dt− − = θ θ θ
=12
2
402 2 2 2 2
kp
p k p( )ω − +
Hence Pav = 12
2
402
2
02 2 2 2 2
mfkp
p k pω − + ...(1)
Replacing f p
p k p
02 2
02 2 2 2 24ω − +
by v02, we have
Pav = mkvmv v
02 0
202
2 2=
τγ
or
from equation (1), we see that when p = ω0, the absorbed average power is maximum and
is equal to 14
02mf
k or
12 0
2mf τ . Thus, the power absorbed is maximum at the frequency of
velocity resonance and net at the frequency of amplitude resonance.
When steady state is attained the average power dissipated should be the same as theaverage power supplied by the driving force.
From equation (1), we see that the power drops to half its maximum value in thecondition
ω0
Y
1.0
P0.5
a v
X
p
Fu ll w id th
Fig. 6
Damped and Forced Harmonic Oscillation 623
12
( )P Resav = (Pav)half power frequency
12 4
02mf
k= 1
22
402
2
02 2 2 2 2
mfkp
p k pω − +
or 8k2p2 = ω02 2 2 2 24− +p k p
or ω02 2− p = +2kp
or ω0 − p = ±+
= ±+
≈ ±2 2
10 0
pkp
k
p
kω ω
or12τ
taking ω0
p = 1 approx.
∴ Half-width of the power verses frequency curve
∆ p = ωτ0 − =p k or 1
2
Hence Q = ω0 τ = ω0
2∆p= Frequency at resonance
Full width at half maximum power
Near resonance, when p = ω0, we see that Q = ω0τ, for low damping Q will be high.Hence Q measures the sharpness of resonance.
14.11 DRIVEN LCR CIRCUIT
LCR circuit is an example of the driven harmonic oscillator in which the oscillations aresustained by an alternating electromotive force.
R L C
EM F
Fig. 7
Let us consider a circuit containing the inductance L, resistance R and condensor C inseries. An alternating E.M.F., E = E0 sin pt is applied to this circuit.
If at any instant t, Q is the charge on the condenser C, then the potential differenceacross its plates is Q/C. If the current in the circuit at this instant is I, then the self-induced
E.M.F. in the inductance is −L Iddt
. According to ohm’s law, the potential difference between
the ends of the resistance is RI. Hence total Potential difference
= RI + QC
total E.M.F. = E L I= − ddt
624 Mechanics
i.e., L I RI + QC
ddt
+ = E
orddt
ddt
2
2Q R
LQ Q
LC+ +. =
EL
0 sin pt ...(1) ∴ I = dQ/dt
This equation is similar to the equation of driven harmonic oscillator only with thedifference that here we have R/L in place of 2k, 1/LC for ω0
2 and E0/L for f0. Hence the steadystate solution of the above equation is
Q =E
LCR
L
0 /sin( )
L
p ppt
1 22 2
−
+
− θ ...(2)
where θ = tan−
−
1
2
p
p
R / L1
LC
...(3)
Differentiating Equation (2) with respect to t, we can find the current I = dQ/dt in thecircuit. Practically, we require the knowledge of the current in the circuit.
∴ I =E
LCL R
0p L
p
pt/
cos( )1 2
2−
+
− θ
or I =E
LC
R
0
pp
pt
−
+
−1
22
sin( )θ ...(4)
where φ = θ π− =−
−
21tan
ppc
L 1
R...(5)
tan φ = − =−
=−
cot φp
p
pp
2 1 1
LCR
L
LC
R
which represents that the current I lags in phase φ from the E.M.F. E.
The quantity R L2 + −
p
pc1
2
is called the impedance of the circuit and is represented
by z. Its unit is ohm. Hence
I =EZ
0 sin ( )pt − θ ...(6)
or I = I0 sin (pt – θ)
Damped and Forced Harmonic Oscillation 625
where IEZ0
0= is the peak value of the current.
The quantity (pL – 1/pC) is called the reactance of the circuit and is denoted by X.Therefore,
Impedance Z = R X Resistance) Reactance)2 2 2 2+ = +( (
Now, the current amplitude I EZ
E
R X0
0 02 2
= =+
is smaller for the larger value of Z.
The expression for Z = R L2 + −
p
pc1
2
depends upon the frequency (p) of the applied
alternating E.M.F.
We will consider the cases, when the driven frequency is increased.
Case I: p << ω0 the term –1/pc in impedance is dominant and is very large and
Z = R L2 + −
≈p
pc pc1 1
2
Therefore, the current amplitudes is (when p → 0)
I0 =E0
1 / pc
and φ =π2
Thus, for low values of p, the current amplitude has a small value. On increasing the
driven frequency (p) gradually, the magnitude of reactance. ppc
L −
1 and hence the
impedance Z decrease resulting in the increase of the current amplitude.
Case II: When p = =ω01LC
, the reactance
X = pp
LC
LLC
LCC
LC
LC
− = − = − =1 0
and so the impedance Z is minimum and is equal to R. In consequence the current amplitudeI0 = E0/Z = E0/R becomes maximum. This is the case of electrical resonance. If R = 0, thenthe current amplitude at resonance becomes infinite. The resonant frequency of the circuitis given by
Pr =p
2 21
20
πω
π π= =
LC
626 Mechanics
In such a case φ = 0 and I = (E0/R) sin pt
Thus at resonance the current and the applied E.M.F. are in the same phase.
Case III: When p > ω0, the term LC in the expression for Z becomes dominant and verylarge and then
I0 =E
L0
pand φ π=
2
π/2
O
−π/2
P
Low R
H igh R
ω0
R = 0
Y
ω0
R = 0
H igh R
Low R
I0
p
Fig. 8
The variation of I0 with ω0 is shown in figure for different values of R. The smaller valueor R, the sharper is the resonance. The variation of phase difference φ between the currentand the E.M.F. is shown in figure. The current lags in phase for p < ω0 and exceeds in phasefor p > ω0.
NUMERICALS
Q.1. The differential equation of an oscillating system is
d xdt
2rdxdt
x2
22+ + ω = 0
If ω >> r, then calculate the time in which (i) amplitude becomes 1/e of its initial value,(ii) energy becomes 1/e of the initial value (iii) energy becomes 1/e4 of its initial value.
Solution. The given differential equation with the condition ω>>r, is the equation ofmotion of an under-damped harmonic oscillator whose solution is
x = ae–rt sin (ω′t + φ),
where a and φ are arbitrary contants and ω′ = ( ) .ω2 2− r Thus the amplitude of oscillationis ae–rt
(i) The damped amplitude at any instant is ae–rt. Let a0 be the amplitude at t = 0 anda0/e at t. Then we have.
a0 = a
andaeo = ae a ert rt− −= 0
Damped and Forced Harmonic Oscillation 627
from this, we have
e−1 = e rt−
–1 = –rt
t =1r
sec (mean life time)
(ii) The instantaneous energy of the damped oscillations is given by
E = E0e–2rt
when the energy falls to E0/e, we have
E0
e= E0e
–2rt
e–1 = e rt−2
–1 = –2rt
t =12r
sec . (relaxation time)
(iii) When the energy falls to E0/e4, we have
E04e
= E0e–2rt
e–4 = e rt−2
–4 = –2rt
t =2r
sec.
Q.2 . The equation of motion of a system is given by
d ydt
2 dydt
y2
22+ +
τω = 0
Explain the significance of the various terms of the equation and show that τ has thedimensions of time. Solve the equation and discuss the motion of the system when ω2τ2 > 1.
Solution. The given equation
d ydt
dydt
y2
222+ +
τω = 0 ...(i)
represent the motion of damped harmonic oscillator along the y-axis let us write it as
d ydt
2
2 =− −2 2
τωdy
dty
The term d y dt2 2/ is the instantaneous acceleration of the oscillator which is the sum
of the acceleration due to damping force, 2τ
dydt
and the acceleration due to restoring force,
ω2y .
628 Mechanics
Since2τ
dydt
is acceleration anddydt
is velocity,τ has the dimensions of time. Let us write
the solution of equation (i) in the form
y = Ceαt ...(ii)
dydt
= Cαeαt
d ydt
2
2 = Cα2eαt
Putting these values in equation (i), we get
C C Cατ
α ωα α α2 22e e et t t+ + = 0
ατ
α ω2 22+ + = 0
α =− ± −
1 12
2
τ τω
Hence the general solution of equation (i) is
y = C e C et t
1
1 1
2
1 12
22
2− + −
− − −
+
τ τω
τ τω
...(iii)
when ω τ2 2 1> , the term12
2
τω−
is imaginary. Then
12
2
τω−
= j jω
τω2
21−
= ′
where ω′ = ωτ
−
12 . Now equation (iii) becomes
y = C C1
1
2
1
e ej t j t− + ′
− − ′
+τ
ωτ
ω
= e e et j t j t− ′ − ′+/τ ω ωC C1 2
= e t j tt− ′ + ′/ cosτ ω ωC sin1
+ ′ − ′C cos sin2 ω ωt j t
= e t j tt− + ′ + − ′/ cosτ ω ωC C C C sin1 2 1 2 .
Damped and Forced Harmonic Oscillation 629
Let us put C1 + C2 = a j asin ) cosφ φand (C C1 2− = where a and φ are constants. Thenwe have
y = e a t a tt− ′ + ′/ ( sin cos cos sin )τ φ ω φ ω
y = ae tt− ′ +/ sin ( )τ ω φ
This is the solution of the given equation under the condition ω τ2 2 1> . This shows thatthe amplitude of motion, ae–t/τ decreases with time exponentially, and the period is
T =2 2
122
πω
π
ωτ
′=
−
The amplitude of motion at t = 0 is a and at t = τ is a/e. That is, τ is the time-intervalin which the amplitude decreases to 1/e of its initial value. Thus here τ is the mean life-timeof the (damped) oscillator.
Q.3. A particle of mass 40 gm experiences only a damping force proportional to itsvelocity. If its velocity is decreased from 100 cm/sec to 10 cm/sec in 23.0 second, calculate(i) relaxation time, (ii) the damping force when its velocity is 50 cm/sec, (iii) the time in whichits kinetic energy is reduced to one-tenth of its initial value, (iv) the total distance travelledif its initial velocity is 100 cm/sec. (Given loge10 = 2.30)
Solution. Let v be the velocity of the particle at any instant t. Then the instantaneousdamping force is
–bv
where b is a constant. If there is no other force acting, then by Newton’s law (force = mass× acceleration), we have
mdvdt
= –bv
dvdt
=− = −bm
v rv2 ,
where b/m = 2r. The expression may be written as
dvv
= –2rdt
Integrating loge v = –2rt + A (constant)
If at t = 0, we have v = v0, then A = loge v0. Then we have
loge v = –2rt + loge v0
logevv0
= –2rt
vv0
= e–2rt ...(i)
v = v0 e–2rt
This is the expression for the (damped) velocity, which decreases exponentially withtime.
630 Mechanics
Now, it is given that the velocity decreases from 100 cm/sec. to 10 cm/sec. in 23.0
seconds. Hence putting vv0
10100
110
= = and t = 23.0 sec in equation (i), we have
110
= e–2r (23.0)
10 = e2r (23.0)
2r (23.0) = loge10 = 2.30
2r =2 3023 0
110
..
=
Equation (i) therefore becomes
v = v e t0
10− / ...(ii)
(i) Let τ be the relaxation time (the time-interval in which the velocity reduces to 1/e
of its initial value). Thus substituting vv e0
1= and t = τ in equation (ii), we get
1e
= e−τ /10
e–1 = e−τ /10
τ10
= 1
τ = 10 sec.
(ii) The damping force is
Fd = – bv = –2rmv.
Here 2r =1
1040 501sec ,− = =m vgm, cm/sec
Fd = − = −110
40 50 200× × dyne.
(iii) The instantaneous kinetic energy is
K =12
12
202 5mv mv e t= − / (by equation (ii))
If K = K0 at t = 0, then
K = K0 e t− /5
Let t′ be the time in which the kinetic energy is reduced to 1/10 of its initial value.
Putting KK0
= 110 and t = t′ in the last expression, we get
110
= e t− ′ /5
Damped and Forced Harmonic Oscillation 631
t ′5
= loge10 = 2.30
t ′ = 5 × 2.30 = 11.5 sec
(iv) Putting v =dxdt
in equation (ii) and integrating, we get
x = v e t0
10 10− − +/ ( ) C (constant)
At t = 0, we have x = 0, so that C = 10 v0.
∴ x = − + = −− −10 10 10 1010
0 010v e v v et t/ /( )
At t → ∞, e t− →/ ,10 0 and x v→ 10 0
Therefore, distance travelled if its initial velocity is 100 cm/sec is x = 10v0 = 10 × 100= 1000 cm.
Q.4. A 3 s gm particle is subjected to an elastic force of 48 dynes/cm and a frictional forceof 12 dynes/(cm-sec–1). If it is displaced through 2 cm and then released, find whether theresulting motion is oscillatory or not. If so find its period.
Solution. The equation of motion of a particle of mass m subjected to an elastic forceand a frictional force is
md xdt
bdxdt
x2
2 + + K = 0
where K is the elastic force per unit displacement and b the frictional force per unit velocity.
Putting bm
r= 2 andKm
= ω2 we get
d xdt
rdxdt
x2
22+ + ω = 0
This represents an oscillatory motion provided r < ω, and the time period is
T =22 2
π
ω( )− r
Here m = 3 gm, K = 48 dynes/cm and b = 12 dynes/(cm-sec–1)
∴ r =bm2
122 3
2= =×
and ω =Km
= = =483
16 4
Thus r < ω hence the motion is oscillatory.
The time period is
T =2 2 3 14
16 42 3 143 464
1 812 2
π
ω −=
−= =
r
× .( )
× ..
. sec
632 Mechanics
Q.5. A body of mass 0.2 kg is hang from a spring of constant 80 nt/m. The body issubject to a resistive force given by bv, where v is the velocity in m/s. Calculate the value
of undamped frequency, and the value of τ if the damped frequency is 3/2 of the undampedfrequency.
Solution. The undamped frequency of a mass m suspended by a spring of force-constant K is given by
n = 12π
Km
Here m = 0.2 kg and K = 80 nt/m.
∴ n = 12
800 2
103 18 1
π π.. sec= = −
The damped frequency n′ (say) is given by
n′ =1
22 2
πω( )− r
where ω = K /m and r = b/2 m.
∴ n′ = 12
2
πKm
r−
But n n′ = ( / ) .3 2
∴1
22
πKm
r−
=
32
10×
π
Km
r− 2 = 300
HereKm
=800 2
400.
=
∴ 400 – r2 = 300
r2 = 100
r = 10
Therefore, the relaxation time is
τ = 12
120
0 05r
= = . sec .
Q.6. If the relaxation time of a damped harmonic oscillator is 50 sec. Find the time inwhich (i) the amplitude falls to 1/e the initial value, (ii) energy of the system falls to 1/e timesthe initial value, (iii) energy falls to 1/e4 of the initial value.
Solution. The equation of a damped harmonic oscillator is
x = ae trt− ′ +sin( )ω φ
Damped and Forced Harmonic Oscillation 633
where r is a damping constant. The relaxation time is
τ =12
50r
= sec
so that r =1
1001sec−
(i) The damped amplitude at any instant is ae–rt. Let a0 be the amplitude at t = 0 anda0/e at t. Then we can write
a0 = a
andae0 = ae–rt = a0e–rt
From this, we have
e–1 = e–rt
t =1r
= 100 sec.
(ii) The energy (or velocity) falls to 1/e times the initial value in the relaxation time.Thus in this case
t = τ = 50 sec.
(iii) The energy of a damped harmonic oscillator is given by
E = E0e–2rt
where E0 is the energy at t = 0. Let t′ be the time during which the energy falls toE0/e
4. Then we have
E0
e4 = E0e rt− ′2
e–4 = e–2rt′
4 = 2rt′
t ′ =2 2 100 200r
= =× sec.
Q.7. The amplitude of an oscillator of frequency 200 cycles/sec falls to 1/10 of its initialvalue after 2000 cycles. Calculate (i) its relaxation time, (ii) its quality factor, (iii) time inwhich its energy falls to 1/10 of its initial value, (iv) damping constant.
Solution. The oscillators having a frequency of 200 cycle/sec, completes 2000 cycles in10 sec. The instantaneous amplitude of the (damped) oscillator is ae–rt where r is a dampingconstant. If a0 be its initial amplitude (at t = 0) and a0/10 after 10 sec, then we have
a0 = a
a0
10= ae–10r = a0e–10r
from this we have
10 = e10r
634 Mechanics
Taking logarithm
loge 10 = 10r
10r = 2.3 log1010 = 2.3
r =2 310
0 23. .=
(i) No, the relaxation time is
τ =12
12 0 23
2 174r
= =× .
. sec.
(ii) The quality factor is
Q = ωτ = 2πnτ= 2 × 3.14 × 200 × 2.174 = 2730.5
(iii) The energy of the oscillator is
E = E0e rt−2
Let t′ be the time after which the energy falls to E0/10. Then
E0
10= E0e rt− ′2
10 = e2rt′
2rt′ = loge 10 = 2.3
t ′ =2 32
2 32 0 23
5 0. .× .
. sec.r
= =
(iii) The damping constant
r = 0.23 (as determined above)
Q.8. The quality factor Q of a tuning fork is 5.0 × 104. Calculate the time interval afterwhich its energy becomes 1/10 of its initial value. The frequency of the fork is 300 sec–1
(loge10 = 2.3).
Solution. The quality factor of a damped oscillator is given by
Q = ωτ,where ω is angular frequency and τ is relaxation time. Therefore
τ =Qω
Here Q = 5.0 × 104 and ω = 2πn = 600 π sec–1
∴ τ =5 0 10
600
4. ×sec
πNow, the energy of the damped system is given by
E = E E0 0e ert t− −=2 /τ τ =
12r
Let t′ be the time after which the energy becomes 1
10 of the initial value. Then putting
E/E0 = 1
10 and t = t′ in the last expression, we get
Damped and Forced Harmonic Oscillation 635
110
= e t− ′ /τ
10 = et′ /τ
loge10 =t ′τ
t ′ = τ loge10
=5 0 10
600 3 142 3 61
4. ×× .
× . sec.=
Q.9. Q of a sonometer wire is 2 ×103. On plucking, it executes 240 vibrations per second.Calculate the time in which the amplitude decreases to 1/e2 of the initial value.
Solution. The quality factor is
Q = ωτHere Q = 2 × 103 and ω = 2πn = 2 × 3.14 × 240 sec–1
τ =Qω
= =2 102 3 14 240
1 3273×
× . ×. sec
Now, if a0 be the initial amplitude (at t = 0) and a0 /e2 the amplitude after time t, we
have,
ae
02 = a e rt
0−
e–2 = e–rt
2 = rt
t =2 4r
= τ τ =
1r
= 4 × 1.327 = 5.3 sec
Q.10. The oscillator of a tuning fork of frequency 200 Cps die away of 1/e time theiramplitude in 1 second. Show that the reduction in frequency due to air damping is exceedingsmall.
Solution. The damped amplitude at any time t is ae–rt where a is a constant. If a0 bethe amplitude at t = 0 and a0/e at time t, then we have
ae0 = a e rt
0−
This gives t =1r
Here t = 1 sec
∴ r = 1 sec–1
The angular frequency in presence of damping is
( )ω2 2− r = ω2 1−
636 Mechanics
Now ω = 2πn = 2 × 3.14 × 200 = 1256, so that ω2 = 1.6 × 106
∴ damped frequency = ( . × ) ( . × )1 6 10 1 1 6 106 6− ≅
≅ ω2 ~ ω,
Which is almost same as damped frequency. Thus the effect of damping on frequencyis exceedingly small.
Q.11. A damped vibrating system starting from rest has an initial amplitude of 20 cmwhich reduces to 2 cm after 100 complete oscillations each of period 2.3 sec. Find thelogarithmic decrement of the system.
Solution. The logarithmic decrement λ is the logarithm of the ratio of two amplitudeof oscillations which are separated by one period. Here the amplitude ratio of oscillations
separated by 100 periods is 202
. Thus
λ =1
100202
loge = 1
10010loge
=1
1002 3× . = 0.023.
Q.12. A damped vibrating system starting from rest reaches a first amplitude of 500 mm,which reduces to 50 mm after 100 oscillations, each of period 2.3 second. Find the dampingconstant, relaxation time correction for the first displacement for damping.
Solution. The amplitude of damped vibrations is given by
a = a e t0
−K
Time period T = 2.3 sec.
First amplitude of the system will be at t = T/4 and the amplitude after 100 oscillationsi.e, 20th amplitude will occur at (100 T + T/4) time, hence
a1 = a e0−KT/4 and a a e201 0= −K(100T + T/4)
∴aa201
1= e e− −=100 100 2 3KT Kor
50500
× .
10–1 = e–230K or loge10 = 230 K
∴ K = 2.3/230 = 0.01 sec–1
∴ Relaxation time τ =1
21
2 0 0150
K= =
× .sec
∴ a1 = a e a e0 00 001 2 3 4500− −∴ =KT/4 . ( . / )
a0 = 500e0.023/4 or log10a0 = log10500 + 0 023
41
2 3.
.log10a0 = 2.699 + 0.0025 = 2.7015
hence a0 = 502.9 mm.
Damped and Forced Harmonic Oscillation 637
Thus in absence of damping, the amplitude of motion would have been 502.9 mm.
Q.13. The differential equation for a certain system is
d xdt
Kdxdt
x02
2
2 2+ + ω = 0
if ω0 >> K, find the time in which
(i) amplitude falls to 1/e times the initial value,
(ii) energy of the system falls to 1/e times the initial value,
(iii) energy falls to 1/e4 times the initial value.
Solution. (i) The given differential equation represent the equation of motion of dampedharmonic, oscillator, whose amplitude of vibration is given by
a = a0e–Kt
when amplitude falls to 1/e times the initial value a
a0e–Kt = a0/e or eKt = e'
Kt = 1, ∴ t = 1K
sec
(ii) Energy Et = E E0K
0e et− =2 /
∴ e2Kt = e or 2Kt = 1 or t = 1/2K sec.
(iii) E0–2Kt = E0/e
4 or 2Kt = 4 or t = 2K
sec.
Q.14. The frequency of a tuning fork is 300 Hz if its quality factor Q is 5 × 104, find thetime after which its energy becomes 1/10 of its initial value.
Solution. Energy E = E E0K
0e et t r− −=2 /
when energy becomes 1/10 of its initial value,
E0/10 = E0e t r− / or 10 = et/r
tτ
= loge10 = 2.3, t = 2.3τ
But τ =Q Qω π
= =2
5 102 3 14 300
4
n×
× . ×
∴ t =2 3 5 10
2 3 14 30061
4. × ×× . ×
= sec
Q.15. A box of 100 gm is attached to one end of a spring whose other end is fixed toa rigid support. When a mass of 900 gm is placed inside the box the system performs4 vibration per second and the amplitude falls from 2 cm to 1 cm in 15 seconds. Calculate(i) the force constant, (ii) the relaxation time, (iii) Q of the system and (iv) Q when only 100gm (empty box) is with the spring (loge10 = 2.3).
Solution. (i) Frequency of vibration of the system n = 4 sec–1
∴ ω0 = 2πn = 2 × 3.14 × 4 = 25 rad/sec
638 Mechanics
(if the damping is small)
But ω0 = C / m or C = mw02 = 1 × 252 = 625 nm.
(ii) At any constant t, the amplitude of the system is given by
a = a0e–Kt or 0.01 = 0.02 e–15K
∴ loge(2 – 15 K) = 0, or K = log
sece215
2 3 0 315
0 046 1= = −. × ..
∴ Relaxation time τ =1
21
2 0 04611
K= =
× .sec
(iii) Q = ω0τ = 25 × 11 = 275
Damping force = γdx/dt, where γ = 2 km = 2 × 0.046 × 1 = 0.092
(iv) when the box is empty, i.e., the mass is 100 gm the damping force γdx/dt andrestoring force Cx remain unchanged. Now
ω0 = C rad. /secm′
= =625
0 179
12
.
τ =1
20 1
2 0 046 11 1
K= ′ = =m
γ.
× . ×. sec
∴ Q = ω0τ = 79 × 1.1 = 87 (approx.)
Example 5. If the suspension of a galvanometer coil exerts a restoring couple 5 × 10–5
n-m/rad. the period is 6.28 sec and the amplitude decreases to 1/10 of its original value in92 seconds, find the value of I, γ and Q factor.
Solution. Period T =2 6 28 1
02 2
2 2π
ωω
−= − =
Ko. or K ...(i)
Now,aao
= e ete
− −∴ = =K Khere 110
or log K, .92 10 92
∴ K = 2.3/92 = 1/40
Substituting this value of K in relation (i), we have
ω02 21 40− ( / ) = 1 or ω0 = 1 nearly
Now, ωo2 = C/I, ∴ = ∴ =− −1 5 10 5 105 5 2× / , × ×I I kg m
γI
= 2 2140
2 5 10 6K or 5 × 10
sec.5γ γ−
−= ∴ =× , . × nm
∴ Q =ω0
21
2 1 4020
K= =
× /
Q.17. A condenser of 2 microfarad capacity is discharged through a 1 ohm resistanceand 2 henry inductance. Calculate the frequency and quality factor of LC circuit and find thetime in which the amplitude of oscillations is reduced to 1%.
Damped and Forced Harmonic Oscillation 639
Solution. n =1
2π1
LCR4L
2
2−
Here, C = 2 × 10–6 farad, L = 2 henry and R = 1 ohm
n =1
21
2 2 101
1610
2 279 56
3
π π× × ×.− − = = Hz
The resistance is low, so
Q =LRω π
π= =2 2
110
2 210
33×
××
The amplitude Q = θ0e− R
2L a a e t= −0
K
As amplitude decreases to 1% of its initial value θ0, we have
θ0
100= θ0
4 1 4 100e tt− =/ /or
Taking logarithm of both sides
t4
= log . × . sec.e t100 2 3 2 18 4= =or
Q.18. In Case of a forced harmonic oscillator, the amplitude of vibrations increases from0.02 mm at very low frequencies to a value 5 mm at the frequency 100 Hz. Find (i) Q factorof the system, (ii) damping constant K and relaxation time and (iii) half width of resonancecurve.
Solution. (i) Q = ω τ ω0
0
2= =
KA
A at zero or very low frequencymax
= 5/0.02 = 250
(ii) Resonant frequency = 100 Hz, ω0 = 2π × 100
τ =Q
secω π0
2502 100
0 4= =×
.
K =12
12 0 4
1 25τ
= =× .
. per sec.
(iii) Half width ∆p = 3 1 25 2 2× . .= rad/sec.Q.19. A damped vibrating system starting from rest reaches a first amplitude of 500 mm,
which reduces to 50 mm after 100 oscillations, each of period 2.3 seconds. Find the dampingconstant and correction for the first displacement for damping.
Solution. The amplitude of damped vibrations as we know, is given by
y = y0e–Kt ...(1)
Here, time period for 1 oscillation, T = 2.3 sec
First amplitude of the system will be at T/4 and the amplitude after 100 oscillations i.e.,201th amplitude will occur at (100T + T/4) time, hence
640 Mechanics
a1 = a0e–KT/4 and a201 = a0e
–K(100T + T/4)
∴aa201
1= e–100KT or
50100
100 2 3= −e × . K
10–1 = e–230K or loge10 = 230 K
∴ K =2 3230
0 01. .=
∴ a1 = a0e–KT/4 ∴ 500 = a0e–0.01(2.314)
a0 = 500e0.023/4 or log10a0 = log10500 + 0 023
41
2 3. .
.log10 = 2.699 + 0.0025 = 2.7015
whence, a0 = 502.9 mm.
Thus in absence of the damping the amplitude of motion would have been 502.9 mm.
Q.20. The time period of pendulum is 2 sec. If the angular retardation due to air is 0.04times the angular velocity of the pendulum and the original amplitude is 1°. Calculate theamplitude after 10 oscillations.
Solution. The amplitude of damped oscillations at any time is given by
y = y0e–Kt ...(i)
where y0 is the initial amplitude and K, the damping constant and 2K, the frictional retardationper unit mass per unit velocity.
Here, y0 = 1°, 2K = .04 and t = 2 sec.
∴ time required for 10° oscillations = 10 × 2 = 20 sec.
Substituting these values in relation (i), we get
y = 1 × e–.02 × 20
∴ log10y = 0 0 42 3
00 1739 1 8261− = =..
. .
y = 0.67° = 0.67 × 60 = 40.2
Q.21. The amplitude of forced vibrations is given by
A =f
P P02 2 2 2( ) /ω τ− +2
If θ = 40, calculate the value of A/Amax when dP/ω0 = 0.99.
Solution. Q = ω0τ = 50 or τ = 50/ω0
and A = f f
( P ) P / P P02 2 2 2 2 2 2ω τ
ω ω
− +=
−
+1 125000
2
2
02
For small damping at resonance.
Amax =11 50
50
02
02ω ω× /
= f
Damped and Forced Harmonic Oscillation 641
When P/ω0 = 0.99
AAmax
=150
f
( . ) ( . ).
1 0 99 12500
0 990 71
2 2− +=
Q.22. If a constant torque of 8 × 10–5 n-m produces an angular displacement of 2° in atorsion pendulum whose moment of inertia is 10–3 kg × m2, Calculate the frequency of theperiodic external torque which will produce resonance.
Solution. Angular displacement = 2° = 2 × π/180 radian.
Applied couple = 8 × 10–5 n-m
Restoring couple for unit displacement
=8 10
2180
5××
−
πn-m/rad .
At resonance, frequency of the force = approx. natural frequency
=1
21
28 10 180
2 100 24
5
31
π π πCI
= =−
−−× ×
×. sec .
Q.23. Calculate the average energy stored in a 20 gm mass attached to a spring andvibrating with an amplitude 1 cm in resonances with a periodic force whose frequency is 20Hz. If the quality of the oscillator be 160, how much energy being dissipated per second.
Solution. During resonance average energy stored
= maximum potential energy
=12
120
202
02Cx m x= ω
= 12
0 02 2 3 14 20 0 01 0 0162 2× . × ( × . × ) × ( . ) .= Joules
Now, Q =2π average energy stored
energy dissipated per cycle
∴ Energy dissipated in 1/20 sec. (i.e., time for one vibration)
=2πQ
average energy stored×
=2 3 14
1600 016× . × .
∴ Energy dissipated per second =20 3 14 2
1600 016 0 0125× . × × . .= Joule.
Q.24. A circuit contains a resistance of 4 ohms and an inductance of 0.68 henry andalternating effective e.m.f. of 500 volts at a frequency of 120 Hz applied to it. Find the valueof the effective current in the circuit and power factor.
Solution. Impedence Z = R P L R L2 2 2 2 2+ = + ( )2 2πn
= 4 4 3 1416 120 0 682 2 2 2+ ( . ) × ( ) × ( . )
642 Mechanics
= 16 262600 512+ = ohms.
∴ Effective current =Effective e.m.f .
Zamp.= =504
5120 98.
Power factor =RZ
R
R P L2 2 2=
+= =4
5121
128
Q.25. An A.C. supply of frequency 10000 and 110 volts is connected across a circuitcontaining a resistance of 10 ohms, an inductance of 10–2 henry and a capacity of 1µF. Findthe value of the current what must be the value of the capacity in order that the current maybe maximum.
Solution. Impedance Z = R PL PC)2 2+ −( /1
= R L C)2 2+ −( /2 1 2π πn n
= 10 2 10000 101
2 10000 102 3
6
2
+ −
−−π
π× ×
× ×
= 48 ohms approx.
∴ Current = e.m.f./Z = 110/48 = 2.293 amp.
Maximum current will flow when the total resistance is zero and resonance is produced.Hence the maximum current will be when PL – 1/PC = 0 i.e., when
C =1 1 1
10 4 102 3 2 4 2LP L × 42 2= = −π πn × × × ( )
= 0.253 × 10–6 farad.
Q.26. An A.C. potential of 1 × 105 sec–1 frequency and 1.0 volt amplitude is applied toa series LCR circuit, if R = 2 ohm, L = 0.50 mili henry, Calculate the value of C to secureresonance. Also calculate the rms value of current and peak potential difference across thecondenser.
Solution. Capacity, required to produce resonance, is given by
C =1
41
4 3 14 10 0 5 102 2 2 5 2 3π n L= −× ( . ) × ( ) × . ×
= 5 × 109 farad
Irms =E
RE
R 2amp.rms 0= = 1
2 2[at resonance Z = R]
Peak potential diff. across the condenser
= Peak current × reactance across condenser
=ER PC
0 × ×× ×
1 12
110 5 105 9= −
= 103 volt.
643
15.1 BACK GROUND OF MICHELSON-MORLEY EXPERIMENT
To send a signal through free space from one point to another as fast as possible, abeam of light or some other electromagnetic radiation such as radiowave is used. No fastermethod of sending the signal has ever been discovered than light. This experimental factsuggests that the speed of light in free space denoted by C ( = 3.00 × 108 meters/sec) is anappropriate limiting reference speed to which other speeds, such as speeds of particles ormechanical waves, can be compared.
The presently accepted value of speed of light is given by
C = (2.997925 ± 0.000003) × 108 meters/sec.
This one of the most fundamental importance due to the following facts:
(1) The speed of light in free space is independent of the reference frame from whichit is observed.
(2) All electromagnetic radiations from longest radio waves to shortest wave travel withspeed of light C independent of the frequency of radiation.
(3) It is impossible to transmit any signal with a velocity greater than the speed oflight, that is, C is an upper limit for the velocity with which a particle can move.
(4) Maxwell’s equations of electromagnetic field in C.G.S. system involve this constantC, i.e.,
div D = 4πρ ...(a)
div B = 0 ...(b)
curl E = − 1c t
∂∂B
...(c) ...(1)
curl H =4 1π ∂
∂cj
c+ D
t...(d)
where D, B, j, E, H and ρ are electric displacement vector, magnetic inductionvector, current density, electric field intensity, magnetic field intensity and chargedensity respectively.
(5) Lorentz force equation in C.G.S. system also involves the speed of light.
644 Mechanics
F = qqc
vE + B× ...(2)
This is Lorentz force equation. F is the total force on a uniformly moving particleof charge q. This force is the sum of electrostatic and magnetic (Lorentz) forces.
(6) Very important relations in relativity involve this constant as
m = m
vc
0
2
21 −, showing variation of mass with velocity. ...(3)
and E = mc2, showing that mass and energy are equivalent and the factor connectingthem is square of C.
(7) The two important units in physics, i.e., electromagnetic units and electrostaticunits of any electrical quantity are related to each other by this constant. Forexample,
1 e.m.u. of charge = c e.s.u. of charge
1 e.m.u. of capacity = c2 e.s.u. of capacity
1 e.m.u. of potential difference =1c
e.s.u. of potential difference.
(8) The reciprocal of the fine structure constant also involves the speed of light givenby
hce2 2π
= 1370.4
where h is the Planck’s constant and e the elementary charge.
In the macroscopic world of our ordinary experience the speed u of moving objects ormechanical waves with respect to any observer is always less than c. For example, soundwaves in air at room temperature move with speed 332 m/sec. So that u/c = 0.0000010. Anartificial satellite circling the earth may move with speed 30000 kilometers/hour. So that
there uc
= 0 000028. .
Such limiting examples of macroscopic world are the basis of formulation of our ideasabout space and time in which Newton developed his system of mechanics.
In the microscopic world it is readily possible to find particles whose speeds are quiteclose to that of light. An electron accelerated through a potential difference of 10 millionvolts has speed 0.9988 c. The validity of Newtonian mechanics can be extended from the
ordinary region of low speeds uc
<<
1 in which it was developed to this high speed region
uc
→
1 ; because experiments show that Newtonian mechanics does not predict the correct
answers when it is applied to such fast moving particles. In fact in principle there is no upperlimit to the speed attainable by a particle in Newtonian mechanics, so that there should beno special role of the speed of light c at all. And yet if the energy of a electron of 10 MeVis increased four times (to 40 MeV), the experimental observation is that the speed is not
Relativity 645
doubled to 1.9976c, as one might expect from Newtonain mechanics relation T =12
mu2
,
but remains below c, it increases only from 0.9988 c to 0.9999 c, a change of 0.11 percentor if an electron of energy 10 MeV moves at right angles to a magnetic field of 2 weber/m2,the observed radius of curvature of its path is not 0.53 cm (as may be calculated from
classical relation r = m uqB
e) but instead to 1.18 cm. Hence no matter how well Newtonian
mechanics may work at low speeds, but fails badly as uc
→ 1. i.e., when the speeds can not
be neglected in comparison with the speed of light.
In 1905 Albert Einstein published his special theory of relativity extending andgeneralising Newtonian mechanics as well. He correctly predicted the results of mechanical
experiments over the complete range of speeds from uc
= 0 to uc
→ 1. Newtonian mechanics
was revealed to be an important special case of Einsteins theory. While developing thetheory of relativity, Einstein critically examined the procedures used to measure length andtime intervals. These procedures require the use of light signals and, in fact an assumptionabout the way light is propagated is one of the two fundamental assumptions upon whichhis theory is based. His theory ruled out the Newton’s concept of the space and time andresulted in a completely new view. The connection between mechanics and electromagnetismis not surprising because light is an electromagnetic phenomenon and plays a basic role inmaking the fundamental space and time measurements that underlie mechanics. Howeverour low speed Newtonian environment is so much a part of our daily life that almost everyone has some conceptual difficulty in understanding Einstein ideas of space and time whenhe first studies them. A careful analysis of basic assumptions of Einstein and of Newtonmakes it clear that the assumptions of Einstein are really much more reasonable than thoseof Newton.
We shall now develop the experimental basis for the ideas of Einsteins theory of relativity.
15.2 THE SPEED OF LIGHT RELATIVE TO EARTH
At the beginning of the present century when the modern techniques were used tomeasure the speed of light in refined manner; it was questioned what was the referencesystem of co-ordinate frame with respect to which the speed of light was measured ? Wasit relative to earth or stars of any other medium? It was to choose the medium for light asair for sound.
We know that air is the medium through which sound travels. Therefore the speed ofsound is measured with respect to air. We also know that velocity of sound relative to theearth can be calculated be imposing the velocity of air with respect to earth to the velocityof sound with respect to air according to Galilean transformations. If the velocity of air withrespect to the earth is u in the direction opposite to that of sound, then the velocity of soundrelative the earth is the velocity of sound relative to air plus velocity of air relative to theearth i.e., u. Then it was assumed as already described, that the preferred medium for lightwas ether which filled all space uniformly. This ether is perfectly transparent to light andthe material bodies may pass through it without any resistance, and since the earth has
646 Mechanics
orbital motion about 3 × 104 m/sec. It was, therefore supposed that acording to Galileantransformations the speed of light should depend upon the direction of motion of light withrespect to the earth’s motion through the ether. Also since the ether was assumed to be atrest and it offers no resistance to material bodies moving through it, i.e., ether remains fixedin space, therefore, it was considered to be possible to find the absolute velocity of any bodymoving through it, but experiments were conducted to find the absolute velocity of the earththrough ether by a number of scientists namely Fizeau, Michelson and Morely, Trouton andNoble and many others, but they could not get success in their attempts. Later Hertzproposed that ether was not at rest, but it was carried completely by bodies moving throughit, that is, the velocity of ether is the same as that of the body moving through it. By thisassumption the phenomenon of aberration could not be explained and hence the assumptionwas rejected.
15.3 MICHELSON MORLEY EXPERIMENT
According to Newtonian mechanics or classical mechanics there existed absolute spaceor luminiferous ether relative to which the measurements are taken. The ether is filleduniformly through all space and penetrates all matter and it was considered that lightpropagates through ether as the sound propagates through air and c is a constant independentof the direction of propagation of light.
The ether was assumed to be at rest. The light passes through ether with fixed velocity3 × 108 m/sec and the earth passes through this stationary ether with velocity 3 × 104
m/sec. It was attempted to determine the velocity of light relative to the earth, for thepurpose of determination of velocity of the earth relative to medium, i.e., ether, throughwhich the velocity of light was measured, was essential. For this purpose the best and mostimportant experiment was first performed by A.A. Michelson in 1881 and then in 1887 incollaboration with E.W. Morley. Michelson-Morley Experiments laid the experimentalfoundations of theory of relativity.
The principle of the experiments lies in noting the shift in fringes in the Michelsoninterferometer due to difference in time taken by light to travel along and opposite thedirection of motion of the earth; for, the time taken by a beam of light to travel along thedirection of motion of the earth is greater than that to travel distance opposite to thedirection of motion of the earth. For his invention of the interferometer and many opticalexperiments, Michelson was awarded the Nobel Prize in Physics in 1907, the first Americanto be so honoured.
The Michelson-Morley experimental arrangement consists of two excellent opticallyplane mirrors M1 and M2 highly silvered on their front surfaces to avoid multiple internalreflections and two plane glass plates P and Q of equal thickness and of same material.These plates are mounted vertically with an inclination of 45° to the interferometer armsas shown in figure.
The plate P is semi-silvered glass plate so that a beam of light from a monochromaticextended source S is partly reflected and partly transmitted when it falls on the plate P. Thereflected beam travels perpendicular to the direction of initial beam and falls normally onmirror M2 due to which it is reflected back to P. The transmitted portion travelling alongthe initial direction of the beam falls normally on mirror M1 at point A and is reflected backto P. The two rays returned to P superimpose to form interference pattern which is observedby telescope T. we see that the reflected ray passes two times through the glass plate P
Relativity 647
while the optical path of the transmitted beam lies wholly in air in the absence of plate Q.Due to this a path difference = 2(µ – 1)t is introduced in the reflected beam. Now if we placea plate Q of same material and of equal thickness parallel to plate P in the path of thetransmitted beam, the extra optical path 2(µ – 1)t is compensated; due to this reason theplate Q is called compensating plate.
If the apparatus is at rest in ether, the two waves, reflected and transmitted, would takeequal time to return to P. But actually the whole apparatus is moving with earth. Considerthe direction of motion of the earth in the direction of the initial beam of light. Due tomotion of apparatus with earth, the positions of reflections of the mirrors and the paths ofthe rays are shown by dotted lines in figure. In this case the time taken by the two beamsto return to P would not be same. This time difference may be calculated as follows:
SP
O
Q
P ′A
M 1
A ′
T Telescope
M 2
B B
Fig. 1 Michelson-Morley Experiment
Let the two mirrors be at equal distances from the plate P, i.e., PA = PB = l (ray) andc and v be the velocities of light and apparatus (earth) respectively. Now the ray reflectedfrom P strikes the mirror at B′ instead of B due to the motion of the earth.
Total path traversed by reflected beam
= PB′P′ = PB′ + B′P′ = 2PB′ since PB′ = B′P′Also PB′2 = PO2 + OB′2 ...(1)
If t is the time taken by the beam to reach from P to mirror M2, then
PB′ = ct and BB′ = vt
∴ Eq. (1) gives c2t2 = v2t2 + l2 since PB = OB′
∴ t =l
c v( )2 212−
If t1 is time taken by the reflected ray to travel total path, then
648 Mechanics
t1 = 2t = 2
2 212
l
c v( )−
= 2 1 2 1
2
2
2
12 2
2l
cvc
lc
vc
−
= +
−
...(2)
The transmitted ray has velocity (c – v) relative to apparatus from P to B and (c + v)from A′ to P′; since from P to A the transmitted ray and apparatus are moving in the samedirection while from A′ to P′ they are moving in opposite directions.
∴ Total time taken by transmitted beam to travel the total path, i.e., from P to A and
then from A′ to P′ = t2 = l
c vl
c v−+
+ ′ since PA = P′A′ = l
∴ t2 =2 2
12 2 2
2
2
1lc
c vlc
cvc−
= −
−
= 2 1
2
2l
cvc
+
The time difference ∆t = t2 – t1 =2 1 2 1
2
2
2
2
2l
cvc
lc
vc
+
− +
=
22
2
2
2
3l
cvc
lvc
. =
∴ The path difference = c∆t = clvc
lvc
.2
3
2
2=
∴ The shift in terms of number of fringes = lvc
2
2λIn Michelson-Morley experiment PA and PB were not exactly equal to get straight
fringes by proper inclination of the mirrors. In performing the experiment the transmittedbeam travelled along the direction of motion of apparatus. Then the whole apparatus wasturned through 90°. In this case the difference of path will be in opposite direction. Theobserved shift after 90° rotation
= 2lv2/c2λ.
If l = 10 meters, v = 3 × 104 meters/sec and c = 3 × 108 metres/sec and the visible lightof wavelength λ ≈ 6 × 10–7 metres, the path difference
= 2 2 10 3 10
3 10 6 10
2
2
4 2
8 2 7lv
c λ= −
× × ( × )( × ) × ×
=13
wavelength (in order).
In spite of all the necessary precautions, the experiment showed no shift. The experimentwas performed by a number of scientists at the different times of year when the directionsof earth’s orbital velocity were different, but no shift was observed.
The negative results observed by the experiment suggest that the space or medium inwhich light propagates is not moving relative to earth. We may say that the existence ofstationary medium carrying light (or absolute space) is disproved.
15.4 EXPLANATION OF NEGATIVE RESULTS: PRINCIPLE OF CONSTANCY OF SPEED OF LIGHT
One way of explaining the negative results of Michelson-Morley experiment is to concludesimply that the measured speed of light is the same (i.e., c) for all directions in every inertial
Relativity 649
frame. This statement is called the principle of constancy of the speed of light and is one ofthe two fundamental postulates of Einstein’s special theory of relativity. This principle wouldlead to no shift in fringes in Michelson-Morley experiment: Since now the speed of light is crather than |(c + v)| in any frame. Hence according to principle of constancy of speed of lightno shift in fringes should be observed. Thus the principle of constancy of speed of light iscapable of explaining negative results of Michelson-Morley experiment. The principle ofconstancy of light being incompatible with Galilean velocity transformations seemed to be toodrastic philosophically at that time. If the observed speed of light did not depend on the motionof the observer, all inertial frames would be equivalent for propagation of light and there couldbe no experimental evidence to indicate the existence of an absolute ether frame.
15.5 THE RELATIVITY OF SIMULTANEITY: THE RELATIVISTIC CONCEPT OFSPACE AND TIME
Michelson-Morley, Trouton Nobel and other carefully performed experiments could notobserve the velocity of earth relative to ether. Lorentz and Fitzgerald tried to explain thenegative results by introducing the hypothesis.
It is necessary in every inertial frame to use a special time (Lorentz hypothesis) andspecial space co-ordinates (Fitzgerald hypothesis) which are different from the time andspace co-ordinates in the absolute ether system.
Lorentz and Fitzgerald, thus, suspected a new concept of space and time along with theconcept of absolute ether frame. Einstein, in 1905, keeping in mind the negative results ofMichelson-Morley and other experiments performed to determine the velocity of earth throughether and the scientific tradition not to postulate things that are by their nature unobservabletook a bold step and stated that the ether does not exist and the motion relative to materialbodies has physical significance while motion through ether in meaningless concept. In otherwords there is no absolute frame and all frames are equally suitable for description ofmotion, although one may be more convenient to use a specific frame in a specific case. Hethus ruled out the absolute concept of space.
To explain the following two contradictory statements:
(1) According to classical mechanics the velocity of any motion has different values forobservers moving relative to each other.
(2) According to experimental observations the velocity of light is not affected by themotion the frame of reference.
Einstein decided that the conflict between them must be due to an imperfection of theclassical ideas about measuring space and time. He ruled out the concept of absolute timeby attacking our pre-conceived ideas about simultaneity.
The nature of his argument may be considered as follows:
Let us see whether the statement “The two events x and y take place simultaneouslywithout reference of any co-ordinate system” can have any meaning. Let us consider a lightsignal which travels in a straight line from one point x to another point y in a given inertialframe. If we imagine the time of emission t1 to be read on a clock placed at x and the timeof arrival t2 to be read on a clock placed at y, the difference (t2 – t1) obtained in this waywill give only the real time taken by the light to travel from x to y if the clocks at x andy are put right. This obviously requires that the hands of both clocks simultaneoutly are inthe same position, i.e., the clocks placed at x and y are synchronized.
650 Mechanics
Now how can we make sure that two events occurring in two different places aresimultaneous ? Whether the two events simultaneous in one frame will also be simutaneousin any other frame ? Let us consider two events occurring at two points x and y which arefixed in inertial frame S. Since the velocity of light is c in all directions, the criterion forthese events to be simultaneous relative to system S is obviously that the two light signalsemitted from x and y at the moments when the events occur shall meet in the centre O ofthe line connecting x and y. A similar criterion for simultaneity is also true relative tosystem S′, having a constant velocity v relative to system S. Now let us consider that thetwo events are simultaneous relative to system S and also the line connecting x and y isparallel to the direction of velocity v of system S′ relative to system S. Then consider twopoints x′ and y′ in system S′ which, at the moment the events occur, coincide with the pointsx and y. Then simultaneously the centre O′ between x′ and y′ will coincide with O. Sincenow O′, just as x′ and y′ moves together with system S′ with a velocity v relative to systemS, O′ will not coincide with O at the moment light signals from x and y meet in O. The lightsignals will thus not meet in O′ and hence, according to above mentioned criterion, the twoevents are not simultaneous relative to system S′.
Thus simultaneity is a relative concept, not an absolute one. Hence the concept ofsimultaneity between two events in different space points has an exact meaning only inrelation to a given inertial system. In other words each frame of reference has its ownparticular time. Therefore there is no meaning in declaring the time of an event, unless werefer the reference system to which the statement of time refers. On account of the absoluteconcept of simultaneity being ruled out, the absolute concept of space is automatically ruledout. To measure the length of an object means to locate its end points simultaneously. Assimultaneity depends upon the frame of reference, the length measurements will also dependon the frame of reference. Thus the length i.e., the space is relative concept, not an absoluteone.
As a result of Einstein’s relativistic concepts of space and time it is possible to reconcileourselves to the experimental fact that observers who are moving relative to each othermeasure the speed of light to be the same. Using these new concepts of space and time wehave to replace Galilean transformation equations by a new type of transformation equationswhich will be based on the invariant character of the speed of light.
The theory of relativity having new concepts of space and time is applicable not onlyto mechanical phenomenon, but also to all optical and electromagnetic phenomenon, and isdivided into two parts.
(1) Special or restricted theory of relativity.
(2) General theory of relativity.
The special theory of relativity deals with systems known as inertial system that is thesystems, which move in uniform rectilinear motion relative to one another. According to this“All systems of co-ordinates are equally suitable for description of physical phenomenon”. Ifwe extend this principle to accelerated system, i.e., the systems moving with acceleratedvelocity relative to one another, the theory of relativity is called “general theory of relativity”.The general theory of relativity is applicable to the laws of gravitation and explains it in amore refined manner than given by Newton.
Relativity 651
15.6 POSTULATES OF SPECIAL THEORY OF RELATIVITY
(1) The fundamental laws of physics have the same form for all inertial systems, i.e.,for reference systems at rest or moving with constant linear velocity relative to oneanother.
(2) The velocity of light in vacuum is independent of the relative motion of the sourceand the observer.
These are the two fundamental postulates used in the special theory of relativity. Thefirst postulate is the extension of the conclusion drawn from Newtonian mechanics, sincevelocity is not absolute, but relative, which is a fact drawn from the failure of the experimentsto determine the velocity of earth relative to ether.
We know that the speed of light is not constant under Galilean transformations and thefirst postulate is the conclusion from Newtonian mechanics; thus second postulate is nottrue according to Galilean transformations. Actually this is true since the velocity of lightcalculated by any means is a constant. Thus the second postulate is very important and onlythis, postulate is responsible to differentiate the classical theory and Einstein’s theory ofrelativity. And according to Einstein the theory of relativity is applicable to laws of optics.Thus for the constancy of velocity of light we have to introduce the new transformationequations which fulfill the following requirements:
1. The speed of light c must have the same value in every inertial frame.
2. The transformations must be linearand for low speeds (i.e., v << c), theyshould approach the Galilean trans-formations.
3. They should not be based on “absolutetime” and “absolute space”.
15.7 LORENTZ TRANSFORMATION EQUATIONS
The above requirements were fulfilled byH. A. Lorentz by introducing transformationequations relating the observations of positionand time made by two observers in two differentinertial frames and are known as “LorentzTransformation Equations”.
Let S and S′ be two inertial frames of reference. S′ having uniform velocity v relativeto S.
Let two observers at O and O′, observe any event P from systems S and S′ respectively.For convenience let us consider that the x axes of two systems coincides permanently andthe velocity is parallel to x-axis. The event P is a light signal and is produced when botht and t′ are zero and when origins of the two frames coincide. The event P is determinedby co-ordinates (x, y, z, t) and (x′, y′, z′, t′) by observers O and O′ respectively.
The light pulse produced at t = 0 will spread out as a growing sphere and the radiusof wavefront produced in this way will grow with speed c, since (x, y, z, t) are co-ordinatesof the event relative to observer in system S at rest therefore the equation of sphericalsurface whose radius grows at the speed c, is
Y Y ′
P
X, X ′O ′O
Z Z ′
v→
S S ′
Fig. 2
652 Mechanics
x y z2 2 2+ + = c t2 2
or x y z c t2 2 2 2 2+ + − = 0 ...(1)
Similarly for observer O′ in system S′ having the co-ordinate of P as (x′, y′, z′, t′), theequation of spherical surface is
x y z′ + ′ + ′2 2 2 = c t2 2′
or x y z c t′ + ′ + ′ − ′2 2 2 2 2 = 0 ...(2)
c is not primed since according to 1st postulate of special theory it is a constant for allinertial frames.
As velocity of S′ is only along x-axis, thus from symmetry.
y = y ′and z = z′ ...(3)
Then from Eq. (1) and Eq. (2), we have
x c t2 2 2− = λ λ( );x c t′ − ′2 2 2 being any, constant ...(4)
Now for the transformation equation relating to x and x′, let us put
x ′ = γ ( ),x vt− ...(5)
γ being independent of x and t.
The reasons for trying the above relation are
(1) The transformations must reduce to Galilean transformation for low speed i.e., forspeed v/c → 0.
(2) The transformation must be linear and simplest. We see that above relation issufficiently general to satisfy the required conditions.
Since motion is relative, we may assume that S is moving relative to S′ with velocity–v along (+ve) direction of x-axis, therefore,
x = γ ( )x vt′ + ′ ...(6)
γ is not primed done to first postulate.
Now putting the value of x′ from (5) in (6), we have
x = γ γ ( )x vt vt− + ′
orxγ = γ γx vt vt− + ′
∴ vt′ =x
vt xx
vt xγ
γ γ γγ
+ − = + −
2
∴ t ′ = γγ
tx
vxv
+ −
2
Relativity 653
= γγ
t xv
− −
1 12 ...(7)
Now substituting in Eq. (4) the value of x′ from Eq. (5) and for t′ from Eq. (7) we have
x c t2 2 2− = λ γ γγ
2 2 2 22
2
11
( )x vt c txv
− − − −
or x c t x vxt v t c txv
2 2 2 2 2 2 2 2 22
2
2 11− − − + + − −
λγ λ γγ
( ) = 0
or x c t x vxt v t c t xtv
xv
2 2 2 2 2 2 2 2 2 22
2
2 2
2
2 2 1 1 1 1− − − + + − −
+ −
λγ λ γγ γ
( ) = 0 ...(8)
Since this equation is an identity, the coefficients of various powers of x and t mustvanish separately.
Equating coefficients of xt to zero, we have
λγ λ γγ
2 2 222 2 1 1. v c
v+ − −
= 0
or 2 2 1 122 2
2γ γγ
vcv
− −
= 0
or ( )v c c2 2 2 2− +γ = 0
this given, γ =1
12
2−
vc
...(9)
Putting coefficients of t2 equal to zero in Eq. (8), we have
− − +c v c2 2 2 2 2λ γ λ γ = 0
λ γ( )v c c2 2 2 2− + = 0 ...(10)
Substituting value of γ from Eq. (9), we get λ = 1
Further, from Eq. (5) we have
x ′ = γ ( )x vt− ...(11)
From Eq. (7) we have
t ′ = γγ
t xv
− −
1 12
654 Mechanics
= γ txv
vc
−
2
2 from Eq. (9)
∴ t ′ = γ tvxc
−
2 ...(12)
Substituting value of γ from Eq. (9) in Eq. (11) and Eq. (12), we have
x ′ =x vt
vc
−
−
1
2
2
...(13)
t ′ =t vx c
vc
−
−
/ 2
2
21
thus combining Eq. (3) and (13) we have
x ′ =x vt
vc
−
−12
2
, y′ = y, z′ = z
and t ′ =t vx c
vc
−
−
/ 2
2
21
...(14)
These are called Lorentz transformation.
If we exchange our frames of reference or consider the given space time co-ordinatesof the event to be those observed in system S′ rather than in system S, the only changeallowed by the relativity principle is the physical one of a change in relative velocity fromv to –v, i.e., the system S moves relative to system S′ with velocity v along negativedirection of x-axis where as system S′ moves relative to system S along positive directionof x-axis. when we solve equations (14) for x, y, z and t in terms of the primed co-ordinatesx′, y′, z′ and t′ we obtain
x =x vt
vc
′ + ′
−
1
2
2
y = y ′ ...(15)
z = z′
Relativity 655
and t =t vx
c
vc
′ + ′
−
2
2
21
These transformation equations are identical in form with equation (11), except that, asrequired, v changes to –v and are called inverse Lorentz transformation equations.
Example 1. Show that direction application of Lorentz transformations
x y z c t2 2 2 2 2+ + − is invariant.
Solution. We have only to prove by the help of Lorentz transformations.
x y z c t2 2 2 2 2+ + − = x y z c t′ + ′ + ′ − ′2 2 2 2 2
where (x, y, z, t) and (x′, y′, z′, t′) are the co-ordinate of the same event observed by twoobservers in systems S and S′ while S′ is moving with a velocity v relative to S.
Let us consider the expression
x y z c t′ + ′ + ′ − ′2 2 2 2 2 = ( )x vtvc
y z ct vx
cvc
−
−+ + −
−
−
2
2
2
2 2 22
2
2
21 1
Putting values of x′, y′, z′ and t′ from Lorentz transformations.
∴ x y z c t′ + ′ + ′ − ′2 2 2 2 2 =c
c vx vt y z c
c vc t vx
c
2
2 22 2 2
2
2 22
2
2
−− + + −
−−
( ) .
=c
c vx vt c t
vxc
y z2
2 22 2
2
22 2
−− − −
+ +( )
=c
c vx vxt v t c t vxt
cv x
cy z
2
2 22 2 2 2 2
2
2 2
42 22 2
−− + − − +
+ +
=c
c vx vxt v t c t vxt
v xc
y z2
2 22 2 2 2 2
2 2
22 22 2
−− + − + −
+ +
=c
c vc x c v t c t v x
cy z
2
2 2
2 2 2 2 2 4 2 2 2
22 2
−+ − −
+ +
=1
2 22 2 2 2 2 2 2 2 2 2
c vc x c t v x c t y z
−− − − + +( ) ( )
=1
2 22 2 2 2 2 2 2
c vc v x c t y z
−− − + +( ) ( )
= x c t y z x y z c t2 2 2 2 2 2 2 2 2 2− + + = + + −
656 Mechanics
Thus we have proved that
x y z c t′ + ′ + ′ − ′2 2 2 2 2 = x y z c t2 2 2 2 2+ + −
That is, the expression x y z c t2 2 2 2 2+ + − is invariant under Lorentz transformations.
Example 2. Show that for low values of v, Lorentz transformations approach to Galilean.
Solution. We have Lorentz transformation equations
x ′ = x vt
vc
y y z z−
−′ = ′ =
12
2
, ,
and t ′ =t vx c
vc
−
−
/ 2
2
21
...(1)
If v is small i.e., v < c so that v/c → 0, then we have from Eq. (1)
x ′ = x vt y y z z t t− ′ = ′ = ′ =, , ,
which are Galilean transformations.
15.8 LENGTH CONTRACTION
Consider two co-ordinate system S and S′, the latter moving with velocity v relative toformer along (+ve) direction of x-axis. Let a rod of proper length (i.e., length at rest) lo beat rest in frame S′ along the x-axis. If x′1, and x′2 are the co-ordinates of abscissae of theends of the rod, in S′ at the same time t′, then
lo = x x′ − ′2 1 ...(1)
since rod is at rest in frame S′.If l is the length of the rod in frame S and the abscissae of the ends of the rod in this
system are x1 and x2, at the same time t, then
l = x2 – x1 ...(2)
According to Lorentz transformations
x′2 =x vt
vc
2
2
21
−
−...(3)
x′1 =x vt
vc
1
2
21
−
−...(4)
Subtracting Eq. (4) from Eq. (3), we have
x x′ − ′2 1 =x x
vc
2 1
2
21
−
−
Relativity 657
or lo = +
−
l
vc
12
2
;
∴ l = lvc
o 12
2−
...(5)
From Eq. (5) we see l < l0.
Thus the length of the rod is reduced in the ratio vc
2
2 :1 as measured by moving
observer with velocity v relative to rod along its length. We may also state that the length
of the object moving with velocity v relative to observer is contracted by a factor 12
2− vc
in the direction of motion while the length at right angles to the direction of motion remainsunchanged. This is also known as “Lorentz-Fitzgerald contraction.”
Example 3. How, fast would a rocket have to go relative to an observer for its lengthto be contracted to 99% of its length at rest.
Solution. According to length contraction
l = l vc
o 12
2−
Here l l= 99100 o i.e.,
llo
=99
100
∴99
100= 1
2
2− vc
i.e., 99
1001
2 2
2
= − v
c
vc
2
2 = 1 99100
2
− =
( ) ( )( ) ( )
100 99100
199100
2 2
2 2− =
v = 199100
0 1416c = . C Ans.
Example 4. A vector in system S′ is represented by 8i + 6j. How can the vector berepresented in system S while S′ is moving with velocity. 0.8ci with respect to S. i and j areunit vectors along the respective directions.
Solution. S′ is moving relative to S with velocity 0.8c along x-axis, therefore the lengthof the vector will only change along x-axis while its length along y-axis will remain the same.
We know l = l vco 1
2
2−
Herevc
=0 8 0 8. .c
c=
658 Mechanics
∴ lx = 8 1 0 64− . = 8 0 36 8 0 6 4 8. × . .= =
Therefore in system S the vector may be represented by 4.8i + 6j.
Example 5. Show that if l03 is the rest volume of a cube, then l v
c03
2
21 − is the volume
viewed from a reference frame moving with uniform velocity v parallel to an edge of the cube.
Solution. Suppose two systems S and S′, S′ moving with velocity v relative to S along+ve direction of x-axis.
Now the volume of the cube in system S = lo3. Hence one edge of the cube = lo. Let
one edge of the cube be along axis of x; then its length along x-axis as observed by an
observer in S ,′ = −l l vcx 0
2
21 . The lengths along y and z-axis remain unchanged. Therefore
volume of the cube as observed from S'
= lx ly lz = lvc
l lo o o12
2−
× × = l vco
3 12
2− .
Example 6. Calculate the percentage contraction of a rod moving with a velocity 0.8cin a direction inclined at 60° to its own length.
Solution. Let lo be the length of the rod at rest.
If 0.8c is the velocity of the rod along the axis of x, then we have
lo = l i l jo ocos sin60 60° + °
where i and j are unit vectors along the x and y-axes respectively.
As the contraction takes place in the direction of motion, the contraction will only takeplace along the axis of x, while the component of length along the axis of y will remain thesame.
If lx′ is the component of the length of the rod, when in motion, along the axis of x,then we have
lx′ = lvc
x 12
2− = lvco cos 60 1
2
2° −
= l cco cos ( . )60 1 0 8 2
2° − = lo2
0 6× . = 0.3 lo
The component of the length of the moving rod along y-axis is given by
ly′ = l lo osin 603
2° =
∴ The length of the moving rod = l lx y′ ′+2 2 = ( . ) .0 312
3 0 9122
l l lo o o+ =
Relativity 659
∴ decrease in length of the rod due to its motion = lo – 0.91 lo = 0.09 l0.
Therefore the percentage contraction = 0 09 100 9. × %lloo
= Ans.
15.9 TIME DILATION
Consider two systems S and S′ , S′ moving with velocity v relative to S along positivedirection of x-axis. Let the clock be situated at x in frame S and gives signals at intervals.
∆t = t t2 1− ...(1)
Then the interval observed by observer in system S′ will be
∆t′ = t t′ − ′2 1 ...(2)
From Lorentz transformation we have
t ′1 =t vx c
v c1
2
2 21
−
−
/
/...(3)
and t ′2 =t vx c
v c2
2
2 21
−
−
/
/...(4)
Substracting Eq. (3) from Eq. (4), we get
t′2 – t ′1 =t t
v c2 1
2 21
−
− /
∴ ∆t′ =∆t
v c1 2 2− /
or ∆t′ > ∆t.
Thus the time interval ∆t appears to be moving observer to the dilated or lengthenedbecause the time interval in system S′ is greater than that in system S. We may state “Aclock will be found to run more and more slowly if the relative velocity between the clockand the observer is increased more and more.
The time interval measured by a clock at rest relative to observer is called the propertime interval.
15.10 AN EXPERIMENTAL VERIFICATION OF TIME DILATION
To observe time dilation in the laboratory the following conditions must be satisfied.
1. The bodies must move with relativistic speed v.
2. The events occuring in bodies should be independent of v.
3. The time interval between events should be sufficiently short so that the eventsmay occur within a reasonably short travel of bodies, otherwise the laboratorynecessary will be inconveniently long.
660 Mechanics
Some nuclear and elementary particles satisfy these three conditions, they undergosome transformation with a proper life time τ which is independent of their speed and theycan be produced with large speeds so that their non-proper time τ′ (measured in laboratoryframe) may differ appreciably from τ.
The nuclear particles called π+ mesons are produced with speed 0.99c when high energyparticle generated by a syncho-cyclotron strike a target. These mesons decay (break into µmesons and neutrinos) in such a way that in every 1.8 × 10–8 sec half of them die out. Theflux of π+ mesons was measured at two places separated by 30 meters. The laboratory timeinterval ∆t′ for travelling the distance was given by
∆t =30
10 10 8m0.99 × 3 × 10 m/sec
sec8 ≈ −×
This is about 5.6 times of 1.8 × 10–8 sec. If T is half life time of particles then after timet, the fractional flux N/N0 is given by
NN0
=12
T =
= ≈−
t / .. %
12
2 25 6
5 6
Hence the flux of π mesons should decrease to approx, 2% of the original flux intravelling 30 metres. But the actual flux at the second place was approx 60% of that at firstplace.
This discrepancy is explained by calculating the proper time ∆t by the relation
∆t′ =∆t
vc
12
2−
or ∆t = ∆t vc
′ −12
2
= 10 10 10 993
2
×.− −
cc
= 1.4 × 10–8 sec.
This is 0.78 times of 1.8 × 10–8 sec. Hence in this much time the flux should fall to
2 600 78− ≈. %. This is exactly what is observed.
From this experiment it is clear that in the laboratory measurements time taken byπ+ mesons for 30 m travel is 10 × 10–8 sec. while π+ mesons themselves measure the timeas 1.4 × 10–8 sec. Thus a 7-fold time dilation has occurred in this case.
Example 7. The half life of a particular particle as measured in the laboratory comesout to be 4.0 × 10–8 sec, when its speed is 0.8c and 3.0 × 10–8 sec, when its speed is 0.6c.Explain this.
Solution. This can be explained on the basis of relativistic time dilation. The timeinterval in motion is given by
∆t′ = ∆t
vc
12
2−
where ∆t is the proper time interval.
Relativity 661
The proper half life of the given particle is
∆t = ∆tvc
′ −
12
2
In first case given ∆t′ = 4.0 × 10–8 sec and v = 0.8 c.
∴ ∆t = 4 10 10 88
2
×.− −
c
c
= 4 10 0 6 2 4 108 8× × . . ×− −= sec.
As proper half life is independent of velocity, therefore half life of the particle whenspeed is 0.6c must be given by
∆t′ =∆t
vc
12
2− =
2 4 10
10 6
8
2
. ×
.
−
−
cc
=2 4 10
0 8
8. ×.
− = 3 × 10–8 sec
which is actual observation. Thus the variation of half life of the given particle is due torelativistic time dilation.
Example 8. A beam of particles of half life 2 × 10–6 sec. travels in the laboratory withspeed 0.96 times the speed of light. How much distance does the beam travel before the flux
fall to 12
times the initial flux.
Solution. The observed half life ∆t′ is given by
∆t′ =∆t
vc
12
2− =
2 10
1 0 96
6
2
×
.
−
−
c
c
=2 10
0 28
6×.
− = 7.14 × 10–16 sec.
According to definition of half life in this observed time the flux falls to 12
times the
initial value. Thus in this time the distance traversed by the beam is
∆t′ = 0.96 × 3 × 10–8 × 7.14 × 10–6
= 2.1 × 103 metres.
Example 9. A particle with a mean proper life of 1 micro-sec (µ-sec) moves through thelaboratory at 2.7 × 108 m/sec.
662 Mechanics
(i) what will be its life time as measured by observer in the laboratory.
(ii) what will be the distance transversed by it before disintegrating.
(iii) find the distance traversed without taking relativity into account.
Solution. (i) By the result of time dilation, we have
∆t′ =∆t
vc
12
2−
Given ∆t = 1 µ sec = 1 × 10–6 sec.
v = 2.7 × 108 m/sec.
Therefore ∆t′ =1 10
1 2 7 103 10
6
8
8
2
×
. ××
−
−
=3 10
3 2 72 3 10
6
2 2
6×
( . ). ×
−−
−=
sec
= 2.3 µ sec.
(ii) Distance transversed by the particle
= v∆t′ = 2.7 × 108 × 2.3 × 10–6 metres
= 620 metres
(iii) Distance transversed without relativistic effects.
v∆t = 2.7 × 108 × 1 × 10–6
= 270 meters.
Example 10. The proper life of π+ mesons is 2.5 × 10–8 sec. If a beam of these mesonsof velocity 0.8c is produced, calculate the distance, the beam can travel before the flux of themeoson beam is reduced to 1/e2 times the initial flux.
Solution. According to time-dilation, we have
∆t′ =∆t
vc
12
2− =
2 5 10
10 8
8
2
. ×
.
−
−
sec
cc
=2 5 10
0 6
8. ×.
− = 4.16 × 10–8 sec.
If No is the initial flux and N is the flux after time τ, then we have
N = No e t− / ,τ τ being mean life
Here N =12e
No
Relativity 663
∴12e
No = No e t− /τ
or t = 2τ = 2∆t′ (since here τ = ∆t′ = 4.16 × 10–8 sec).
∴ 2∆t′ = 2 × 4.16 × 10–8 sec = 8.32 × 10–8 sec.
∴ The distance traversed by the beam before the flux of meson beam is reduced to 12e
times the initial flux.
= 2 0 8∆t c′ × .
= 8.32 × 10–3 × 0.8 × 3 × 108
= 19.96 meters.
Example 11. At what speed should a clock be moved so that it may appear to lose 1minute in each hour ?
Solution. If the clock is to lose 1 minute, in one hour, then the clock must record 59minutes for each hour recorded by clocks stationary with respect to the observer. If v is therequired speed of the clock, then according to Einstein’s apparent retardation of clocks, wemust have
59 = 60 12
2− vc
orvc
= 15960
2
−
or ≈ 0 18.
or v = 0.18 × c= 5.4 × 107 metres/sec Ans.
15.11 TRANSFORMATION AND ADDITION OF VELOCITIES
Let there be two co-ordinate system S and S′ the latter moving with velocity v relativeto former along the +ve direction of x-axis. Let a particle be moving with velocities u andu′ relative to frames S and S′ respectively; then
u = i u j u k ux y z+ +
and u′ = i u j u k ux y z′ + ′ + ′ ...(1)
where ux, uy and uz are components of u along the three axes, , i j k and being unitvectors along x, y and z-axis respectively.
u u ux y z′ ′ ′, and are components of u′ along respective axes. From the definition ofvelocity, we have
ux = dxdt
udydt
udzdty z, ,= = , u x′ =
dxdt
udydt
udzdty z
′ ′ = ′ ′ = ′, and ...(2)
664 Mechanics
Lorentz inverse transformations are given by
x =x vt
vc
y y z z tt vx c
vc
′ + ′
−
= ′ = ′ = ′ + ′
−1 12
2
2
2
2
, ,/
and ...(3)
Differentiating these transformation equations, we get
dx = dx vdt
vc
dy dy dz dz′ + ′
−= ′ = ′
12
2
, , and dt = dt v dx c
vc
′ + ′
−
/ 2
2
21...(4)
Now from these expressions, we have
ux =dxdt
= dx vdt
dt v dxc
dxdt
v
vc
dxdt
′ + ′
′ + ′ =
′′
+
+ ′′2 21
∴ ux =u v
vc
u
x
x
′ +
+ ′1 2
...(5)
uy =dydt
= dy v
c
dt v dxc
dydt
vc
vc
dxdt
′ −
′ + ′ =
′′
−
+ ′′
1 1
1
2
2
2
2
2
2
∴ uy =
u vc
vc
u
y
x
′ −
+ ′
1
1
2
2
2
...(6)
similarly uz =u v c
vc
u
z
x
′ −
+ ′
1
1
2 2
2
/...(7)
Eqs. (5), (6) and (7) represent transformation of velocity components from one inertialsystem to another.
If we consider the particle to be a photon moving with velocity c in frame S′ and thesystem S′ to be moving with velocity c relative to system S along +ve direction of x-axis,then from eq. (5), we have
ux =c c
cc
c+
+=
12
2
Relativity 665
i.e., the speed of light is same for all inertial frames whatever their relative speed maybe and no particle can obtain a speed greater than the speed of light.
Example 12. Two particles, are travelling in opposite directions with speed 0.9c relativeto the laboratory. What is their relative speed.
Solution. To solve the problem consider the particle moving with velocity –0.9c to beat rest in system S. Then it may be considered that S′ is moving with velocity v = 0.9crelative to S, so that the particle in S′ has velocity u cx′ = 0 9. . Then the velocity in S is givenby
ux =u v
vc
u
x
x
′ +
+ ′1 2
= 0 9 0 9
1 0 91 81 81
0 9952
2
. .( . )
.
..
c cc
c
c+
+= = c
Example 13. An electron is moving with a speed of 0.85c in a direction opposite to thatof a moving photon. Calculate the relative velocity of the electron with respect to photon.
Solution. The speed of the photon = c
The speed of the electron = 0.85c
Let the electron be moving along –ve direction of x-axis while the photon along +vedirection of x-axis. Consider that the electron moving with velocity –0.85c is at rest insystem S. Then it may be assumed that the system S′ (laboratory) is moving with velocity.85c relative to system S (electron). Then we have
v = 0.85c, u′ = c;
∴ u =u v
vuc
′ +
+1 2
= c cc c
c
c c c+
+= +
+=0 85
1 0 850 85
1 0 852
.( . ) ( )
..
Example 14. An particle has a velocity u 3i 4 j 12k′ = + + m/sec in a co-ordinate systemmoving with velocity 0.8c relative to laboratory along +ve direction of x-axis. Find u in thelaboratory frame.
Solution. Given u′ = 3 4 12 i j k+ +
or u x′ = 3, u uy z′ = ′ =4 12, meter/sec.
and v = 0.8c
Therefore components of u are given by
ux =u v
vc
u
x
x
′ +
+ ′1 2
= 3 0 8
1 0 8 33 0 8
1 0 8 32
+
+= +
+
.( . ) ( )
.( . ) ( )
ccc
c
c
=( . )
( . ).3 0 8
2 40 8+
+≈c c
cc = 2.4 × 108 metres/sec.
uy =u v
cvc
u
y
x
′ −
+ ′
1
1
2
2
2
666 Mechanics
=4 1 0 8
1 0 8 34 0 6
2 4
2
2
−
+=
+
.
( . ) ( )( × . )
.
cc
cc
cc
= 2 42 4
2 4..
.cc +
≈ m/sec.
and uz =u v c
vc
u
z
x
′ −
+ ′
1
1
2 2
2
/
=12 1 0 8
2 47 2
2 47 2
2
−
+=
+≈
.
..
..
cc
cc
c m/sec.
Therefore u in the laboratory frame is given by
u = i u ju kux y z+ +
= ( . × . . )2 4 10 2 4 7 28 i j k+ + m/sec.
Example 15. A space-ship moving away from earth with velocity of 0.4c fires a rocketwhose velocity relative to space-ship is 0.6c, away from the earth. What will be the velocityof the rocket as observed from the earth.
Solution. We have ux =u v
u vc
x
x
′ +
+ ′1 2
Here ux is velocity of rocket relative to earth.
v is the velocity of space-ship relative to earth.
u′x is the velocity of rocket relative to space-ship.
Here v = 0.4c, u′x = 0.6c
∴ ux = 0 6 0 4
1 0 6 0 42
. .( . ) ( . )
c cc cc
+
+
=c c
1 0 24 1 24+=
. .ux = 0.8 c Ans.
15.12 RELATIVISTIC DOPPLER’S EFFECT
Consider two systems S and S′, the latter moving with velocity v relative to formeralong positive x-axis.
Relativity 667
Let the transmitter and receiver be situated at origins O and O′ of frames S and S′respectively. Let two light signals or pulses be transmitted at time t = 0 and t = T, T beingtrue period of light pulses. Let ∆t′ be the interval between the reception of these pulses byreceiver in S′. Obviously ∆t′ is the proper time interval T′ between these pulses (sinceobserver or receiver is at rest in frame S′).
Since observer continues to be at O′ all the time the distance ∆x′ covered by him inframe S′ during the reception of two pulses is zero.
X ′′
X ′
X
Y Y ′′Y ′
S ′
S
O ′′
O ′
O
∆x
Fig. 3
From inverse Lorentz transformation
x =x vt
vc
′ + ′
−12
2
we have ∆x =∆ ∆x v t
vc
′ + ′
−12
2
=v t
vc
∆ ′
−12
2
since here ∆x′ = 0
or ∆x =v
vc
T
1
′
−2
2
since ∆t′ = T′ ...(1)
This equation shows that the second pulse has to travel this much distance ∆x morethan the first pulse in frame S along x-axis to be able to reach at origin O′ in moving frameS′.
Also from inverse Lorentz transformations
t =t vx
c
vc
′ + ′
−
2
2
21
668 Mechanics
we have ∆t =∆ ∆
∆t v x
c
vc
t
vc
′ + ′
−
= ′
−
2
2
2
2
21 1
since ∆ x′ = 0
or ∆t =T
1
′
− vc
2
2
...(2)
since ∆t′ = T′
This relation includes both the actual time period T of pulses and the time taken ∆ xc
by the second pulse to cover the extra distance ∆ x in frame S. That is
∆t = T + ∆xc
...(3)
Substituting values of ∆ x and ∆t from Eqs. (1) and (2) in Eq. (3), we get
T
1
′
− vc
2
2
= T +Tv
c vc
′
−12
2
on solving
T =T 1
T′ −
−
= ′−
+
vc
vc
vcvc1
1
12
2
...(4)
If ν and ν′ be the actual and observed frequencies of light pulses respectively, we have
ν =1 1T
and T
ν′ =′
∴ Equation (4) gives
1ν
= 11
1ν′
−
+
vcvc
...(5)
i.e., ν′ = ν1
1
−
+
vcvc
Relativity 669
This result is known as Doppler’s relativistic formula for light waves in vacuum.
In terms of actual and apparent wavelengths λ and λ′ given by
λ =cν
and λν
′ =′
c
we getc
λ′=
cvcvc
λ
1
1
−
+
i.e., λ′ = λ1
1
+
−
vcvc
...(6)
In both relations (5) and (6), v is positive or negative according as the observer isreceding form or approaching the source of light, so that during recession the apparentfrequency of pulse ν′ decreases and its apparent wavelength λ′ increases and during approachthe apparent frequency ν′ increases and the apparent wavelength λ′ decreases. The Doppler’seffect so far considered is the longitudinal Doppler’s effect since the observations are madealong the direction of travel of light source.
If the observations are made at right angles to the direction of travel of light source,the Doppler’s effect observed is known as transverse Doppler’s effect. This is found usuallyin the case of atoms. Unlike the classical theory, the theory of relativity predicts transverseDoppler’s effect.
For transverse Doppler’s effect the second signal does not take on extra time since herey = y′ and z = z′ due to no relative motion along y and z axes.
Therefore, from the results of time dilation, the real and observed periods are relatedby
T ′ =T
1 − vc
2
2
...(7)
If v and v′ are the actual and observed frequencies in S and S′ respectively, then
T ′ =1ν ν′
and T = 1
So that equation (7) then gives
ν′ = ν 12
2− vc
...(8)
This expression represents transverse Doppler’s Effect. If v << c, the longitudinal andtransverse Doppler’s Effect take the form
670 Mechanics
ν′ = ν 1 − vc
(longitudinal Doppler’s Effect)
ν′ = ν (transverse Doppler’s Effect)
Then according to the classical theory, there is no transverse Doppler Effect; but accordingto theory of relativity transverse Doppler’s change in frequency is given by Eq. (8).
15.13 CONFIRMATION OF DOPPLER’S EFFECT
The longitudinal Doppler Effect was confirmed by H.E. Lves and G.R. Stilwell in 1941spectroscopically using beams of hydrogen atoms in excited electronic states. Atomic hydrogenwas produced by the result of breaking up accelerated molecular hydrogen ions using electricfield.
They observed the shift in the average wavelength of a particular spectral line ofhydrogen atoms. The shift was observed by taking the average over forward and backwarddirections relative to the line of flight of atoms.
The longitudinal Doppler’s Effect is expressed as
λ′ = λ1
1
+
−
vcvc
...(9)
If v is +ve in forward direction, it is –ve in backward direction, thus if λ1 is wavelengthobserved in forward direction λ0 the wavelength emitted by atom at rest, then from Eq. (9)
λ1 = λ0
1
1
+
−
vcvc
If λ2 is wavelength observed in the backward direction, we have
λ2 = λ0
1
1
−
+
vcvc
Therefore the average observed wavelength is given by
λ =λ λ1 2
2+
= λ0
2
1
1
1
1
+
−+
−
+
vcvc
vcvc
=λ0
2
21 − vc
...(10)
Relativity 671
The shift observed by them was 0.074 × 10–8 cm, while shift calculated from Eq. (10)
taking value of vc
= 0 005. is 0.072 × 10–8 cms.
Thus observed and calculated values very nearly agree, thereby verifying longitudinalDoppler’s Effect.
Example 16. Calculate the wavelength shift in the relativistic Doppler Effect for the Hα(6563 Å) line emitted by a star receding from the earth with a relative velocity 0.1c. Is theclassical (first order) result a good approximation?
Solution. The new wavelength λ′ is given by
λ′ = λ 11
+−
v cv c
//
= 6563 1 0 11 0 1
+−
.
. since
vc
= 0 1.
= 65631 10 9
7256..
= Å
∴ The wavelength shift λ′ – λ = 7256 – 6563 = 693Å
Now looking for the classical result, we have
λ′ = λ 11
+−
v cv c
// = λ 1 1
12
12
+ −
−vc
vc
= λ 1 +
vc
in classical limit
= λ (1 + 0.1)
or λ λ′ − = 0.1 λ = 0.1 × 6563
= 656.3 Å
Thus the classical result, in this case, differs with relativistic result by
693 656 3693
100−
. × % = 5.3%
This percentage depends on the value of vc .
15.14 CONSERVATION OF MOMENTUM: VARIATION OF MASS WITH VELOCITY
According to theory of velocity the laws of mechanics as well as those of electromagnetic,must remain unaltered when a transformation from one inertial frame to another frame isaffected.
According to principle of conservation of momentum. If the net force acting on the bodyis zero, its linear momentum must remain constant.
F =ddt
mv( ) = 0 or mv = constant.
672 Mechanics
We shall see that if the laws of conservation of momentum and energy are to beretained under Lorentz transformations, the mass m must vary with velocity.
Now we shall apply the laws of conservation of momentum and energy to simplehypothetical experiment first revised by Tolman.
Consider two systems S and S′, the latter moving with velocity v relative to formeralong +ve direction of x-axis as shown in figure.
Consider two exactly, similar elastic balls A and B placed in system S and S′ respectively.The two balls are thrown with equal and opposite velocities ± u by two observers situatedon Y and Y′ axes of systems S and S′ respectively as shown in figure.
The masses of the two balls are equal when measured in the same system. Nowconsider the balls A and B collide due to motion of S′ at the instant when the line of centreshas the direction of Y-axis, so that the x-components of velocities do not change as a resultof impact. Then according to law of conservation of energy, their total kinetic energy beforeimpact and after impact must be equal.
YB u
uS ′ S
Y
Xv→
A
Fig. 4
Now before impact the components of velocity are given according to rule for thetransformation of velocities.
As observed from system S before impact.
uax = 0, uay = u, uay = v and u u vcby = − −1
2
2
where uax is the velocity of ball A along x-axis uay is velocity of ball A along y-axis and soon.
The total momentum of the two balls along y-axis before impact as observed from S is
m u m ua ay b by− = m u m uvca b− −
12
2
After impact each ball reverses its y-component of velocity; therefore the componentsof velocity of the two balls along y-axis after impact as observed from system S are givenby
uay = − = −
u u uvc
by and 12
2 ...(1)
Relativity 673
Therefore momentum of the two balls along y-axis after impact as observed fromsystem S is given by
m u m ua ay b by+ = − + −
m u m uvc
a b 12
2...(2)
According to principle of conservation of momentum,
momentum before impact = moment after impact
m u m u vc
a b− −12
2 = − + −m u m u vc
a b 12
2
or 2mau = 2 12
2m u vc
b −
ormm
b
a=
1
12
2− vc
According to this relation mass must vary with velocity.
Now if u → 0, ma → mo which is known as rest mass and mb → m; the above equationtakes the form
m =m
vc
0
2
21 −...(3)
Thus the mass of a body increases with increase of velocity and m is known as apparentmass or momental mass.
The momentum of a body of mass m moving with velocity v can be written as
p = mvm v
vc
=−
0
2
21
...(4)
15.15 MASS-ENERGY RELATION
Force is defined as rate of change of momentum, i.e.,
F =ddt
mv( ) ...(1)
From the principle of conservation of energy, kinetic energy of a moving body is equalto work done by the force that imports the velocity to the body from rest. Work is definedas force multiplied by distance; therefore we have
K.E. = F ds ddt
mv dsv
v v v
=
=
=0 0
( )
674 Mechanics
=ddt
mvdsdt
dt vddt
mv dtv v
( ) ( )0 0 =
= vd mvv
( )0 ,
According to theory of relativity
m =m
vc
0
2
21 −
∴ K.E. = T = vd m v
vc
v
v v0
2
20 1 −
=
=
= m vvc
dv v c
vc
dvv
02
2
2 2
2
2
3 20
1
1 1−+
−
//
= m
vc
vc
vc
vdvv
0
2
2
2
2
2
2
3 20
1
1
−
+
−
/ .
= m vdv
vc
m cvc
v
v
02
2
3 20
02
2
2
3 2
0
1
1
1−
=
−
/ /
∴ K.E. T = m cvc
02
2
2
1
1
1
−−
=m
vc
m c m m c0
2
2
02
02
1 −
= −. ( ) ...(2)
Relativity 675
This is relativistic equation for kinetic energy.
Thus the K.E. of a moving body is equal to gain in mass multiplied by c2. So m0c2 may
be considered as rest energy of a body of rest mass m0. This rest energy may be consideredas internal store of energy in the body. Then total energy E of a body is the sum of restenergy and kinetic energy of a body.
∴ E = rest energy + kinetic energy
= m0c2 + (m – m0)c
2
= m0c2 + mc2 – m0c
2
∴ E = mc2 ...(3)
which is well know Einstein mass-energy relation which states a universal equivalencebetween mass and energy.
This principle has been confirmed a number of times in nuclear physics.
Expanding 1st term in Eq. (2) by Binomial theorem, we have
T = m cvc
vc0
22
2
4
41 12
38
1+ + + −
... ...(3)
If v << c, then kinetic energy, T = 12
m v02 as in classical mechanics.
Hence for non-relativistic cases we may use kinetic energy expression 12
2mv but for
relativistic cases we must use the expression
T = ( )m m c− 02
Example 17. The rest mass of an electron is 9.0 × 10–28 gm. What will be its mass if
it were moving with 45
th the speed of light.
Solution. The mass of the electron if it were moving with speed v is given by
m = m
vc
0
2
21 −, where m0 is the rest mass of the electron
Here v =45
0 8c c= .
and m0 = 9 × 10–28 gm = 9 × 10–31 kg.
∴ m =9 10
10 8
31
2
×
.
−
−
cc
=9 101 0 64
9 100 36
31 31×.
×.
− −
−=
676 Mechanics
=9 10
0 615 10
3131×
.×
−−=
= 1.5 × 10–30 kg. Ans.
Example 18. For what value of vc
= β will be relativistic mass of a particle exceed its
rest mass by a given fraction f.
Solution. We have
f =m m
mmm
− = −0
0 01
Butmm0
=1
12
2− vc
∴ f =1
1
12
2−−
vc
Solving forvc
=f f
f
( )2
1
++
Ans.
Example 19. A certain young lady decides on her twenty fifth birth day that it is timeto slendrize. She weights 100 kg. She has heard that if she moves fast enough, she will appearthinner to her stationary friends.
(i) How fast must she move to appear sleadrized by a factor of 50%.
(ii) At this speed what will her mass appear to be to her stationary friends.
(iii) If she maintains her speed until the day she calls her twenty ninth birth day, howold will her stationary friends claim she is according to their measurements?
Solution. (i) According to Lorentz-Fitzgerald contraction
l = l vc
o 12
2−
Given that the dimensions of lady’s thickness are reduced by 50% i.e.,
l =50
100lo
∴50
100lo = l v
co 1
2
2−
or 12
2− vc
=12
...(1)
Relativity 677
∴ 12
2− vc
=14
34
2
2 or vc
=
∴ v =3
20 866c = . c
(ii) The mass of a body moving with velocity v is given by
m =m
vc
o
12
2−
In this case mo = 100 kg
12
2− vc
=12
from Eq. (1)
∴ m =1001 2
200/
= kg
(iii) According to time-dilation
∆t′ =∆t
vc
12
2− ; Here ∆t = 4 years, 1 1
2
2
2− =vc
∴ ∆t′ =4
1 28
/= years
∴ The lady will appear to be (25 + 8) = 33 years old.
Example 20. Calculate the energy equivalent to 1 atomic mass unit in million elec-tron volt. Given Avogadro Number = 6 × 1023/gm mol.
Solution. We know E = mc2
Here m = 1 a.m.u. = 1
6 1023×gm = 1
6 × 10kg26
c = 3 × 108 m/sec.
∴ E =1
6 103 1026
8
×× ( ×kg m / sec)2
=9 106 10
16
26××
joules
=9 10
6 10 1 6 10
16
26 19×
× × . × − ev
= 937 × 106 ev = 937 Mev.
Example 21. Calculate the fractional change in the mass of the hydrogen atom whenit is ionized from the following data:
678 Mechanics
The binding energy of hydrogen atom = 13.58 eV
The rest hydrogen atom = 1.00797 a.m.u.
Solution. The change in rest mass ∆m0 = ∆Ec2
= 13 58 13 58 1 6 10 19. . × . ×ev(3 × 10 m / sec)
Joules(3 × 10 m / sec)8 2 8 2=
−
=13 58 1 6 10
3 10
19
8 2. × . ×
( × )
−kg
=13.58 × 1.6 × 10
a.m.u.19−
−( × ) × . ×3 10 1 66 108 2 27
= 1.46 × 10–8 a.m.u.
So fractional change in the mass of the hydrogen atom,
∆mm
o
o=
1 46 101 00797
8. ×.
− = 1.45 × 10–8 = 1.45 × 10–6%
Example 22. How much mass is lost when 1 kg. of water at 0°C turns to ice at 0°C.
Solution. The heat energy lost by water,
= (1 kg) × (80 kilo cal/kg)
= 80 kilo cal = 80 × 4.2 × 103 joules
∴ Equivalent loss in mass
∆m =∆Ec2 = 80 4 2 10
3 103 73 10
3
8 212× . ×
( × ). ×kg. kg. = − Ans.
Example 23. A body whose specific heat is 0.2 kilo cal/kg°C is heated through 100°C.Find the percentage increase in its mass.
Solution. The heat energy gained by body
∆E = ms ∆Q = m × (0.2) × 100
= 20 m kilocal = 20 m × 4.2 × 103 Joule
∴ Equivalent gain in mass, ∆m =∆Ec2
=20 4 2 10
3 10
3
8 2m × . ×( × )
∴ Percentage increase in mass =∆mm
× 100
=20 4 2 10
3 10100 9 3 10
3
811m × . ×
( × ) ×× . × %
m= −
Example 24. Find the velocity that an electron must be given so that its momentumis 10 times its rest mass times the speed of light. What is the energy at this speed.
Relativity 679
Solution. The momentum of an electron of rest mass mo moving with velocity v isgiven by
p =m v
vc
o
12
2−
Givenm v
vc
o
12
2−= 10 moc
orvc
= 10 12
2− vc
Squaring it, we getvc
2
2 = 100 1 100 1002
2
2
2−
= −vc
vc
or ( )100 12
2+ vc
= 100
orvc
2
2 =100101
orvc
=100101
0 995= . ...(1)
∴ v = 0.995 c = 0.995 × 3 × 108 m/sec
= 2.985 × 108 m/sec.
The mass of the electron at this speed is given by
m =m
vc
0
2
21 −
where m = rest mass of electron = 9 × 10–31 kg
∴ mo =9 10
1 100101
9 101
101
31 31× ×− −
−=
= 9 10 101 9 10 10 0431 31× × × .− −=
= 90.36 × 10–31 kg = 9.036 × 10–30 kg
∴ Energy of the electron at speed 0.995 c is
E = mc2 = (9.036 × 10–30 kg) (3 × 108 m/sec)2
= 8.13 × 10–13 Joules.
680 Mechanics
Example 25. An electron and a positron practically at rest come together and annihilateeach other, producing two photons of equal energy. Find the energy and equivalent mass ofeach photon.
Solution. The rest mass of electron = 9 × 10–28 gm = 9 × 10–31 kg
= rest mass of positron
∴ The energy of each photon = moc2
= 9 × 10–31 × (3 × 108)2 Joules
= 81 × 10–15 Joules
=81 101 6 10
15
19×
. ×
−
− eV = 5 × 10 eV5
The equivalent mass of each photon = 9 × 10–3 kg.
Example 26. Calculate the velocity of electrons accelerated by a potential of 1 million-volts.
Solution. We know, the kinetic energy
T = ( )m m c− 02
=1
1
12
2
2
−−
v
c
m co
Given T = 1 mev = 106 eV
= 106 × 1.6 × 10–19 Joules.
mo = 9 × 10–28 gm = 9 × 10–31 kg.
∴ 106 × 1.6 × 10–19 =1
1
1 9 10 3 102
2
31 8 2
−−
−
vc
× × × ( × )
Solving we get v = 2 23
2 23
3 108c = × × m/sec
= 2.8 × 108 m/sec.
15.16 RELATION BETWEEN MOMENTUM AND ENERGY
Relativistic momentum is written as
p = mvm
vc
vo=−1
2
2
Relativity 681
and relativistic energy E = mcm
vc
co2
2
2
2
1
=
−
Squaring E2 =m c
vc
o2 4
2
21 − and p m
vc
vo22
2
2
2
1=
−
∴ E2 − p c2 2 =m c
vc
m v cvc
o o2 4
2
2
2 2 2
2
21 1−−
− =
m cvc
c vo2 2
2
2
2 2
1 −−( )
=m c v
cvc
o2 4
2
2
2
2
1
1
−
−
or E2 – p2 c2 = mo2c4.
We see that right hand side is invariant, therefore we may say that E2 – p2c2 is invariantunder Lorentz transformations.
15.17 TRANSFORMATION OF MOMENTUM AND ENERGY
Consider two systems S and S′, S′ moving with velocity v relative to S along +vedirection of x-axis. Now consider in system S a photon on the light sphere having momentump and energy E.
We know p = mc and E = mc2
∴ p2 = m2c2 and E2 = m2c4 = m2c2.c2
or E2 = p2c2
or p2 =E2
c2 ...(1)
If px, py and pz are components of momentum in system S along the three axes, we have
p2 = px2 + py
2 + pz2 ...(2)
Now from Eq. (1) becomes
px2 + py
2 + pz2 =
E2
c2 ...(3)
If E′ and p′ are energy and momentum of photon as viewed from system S′, then wehave
p p px y z′ ′ ′+ +2 2 2 =E 2′c2 ...(4)
682 Mechanics
The equation is similar to x y z c t2 2 2 2 2+ + = ...(5)
Therefore energy and momentum transform in the same way as space and time.
Here x has been replaced by px, y by py , z by pz and t by E/c2. Lorentz transformationfor space and time are
x ′ =x vt
vc
y y z z tt vx
cvc
−
−′ = ′ = ′ =
−
−1 12
2
2
2
2
, , and
Therefore Lorentz transformations for momentum and energy are given by
px′ =p v
cvc
p p p px
y y z z
−
−′ = ′ =
E2
2
21
, , ...(6)
andE′c2 =
Ec
vpcvc
x2 2
2
21
−
−
or E′ =E −
−
vp
vc
x
12
2
...(7)
Inverse Lorentz transformations for momentum and energy are
px =p v
cvc
p p p px
y y z z
′ + ′
−
= ′ = ′
E2
2
21
, ,
and E =E +′ ′
−
vp
vc
x
12
2
...(8)
Example 27. A proton has velocity 0.999c in the laboratory frame. Find the energymomentum as observed in a frame travelling in the same direction with a velocity 0.99crelative to laboratory (mass of the proton = 1.67 × 10–27 kg).
Solution. Let the frame S′, in which momentum and energy are required, be movingrelative to lab frame S along positive direction of x-axis. Also let the direction of motion ofthe proton in the lab frame be x-axis.
If mo is the rest mass of the proton, the momentum of the proton in the lab. frame is,
Relativity 683
p = px = m u
uc
o x
x12
2−
=m c
cc
m coo
( . )
..
0 999
1 0 99922 4
2
−
≈
The energy of the proton in lab. frame is
E = mc2 = m c
uc
m co
x
o
2
2
2
2
1
22 4
−
≈ .
Using transformation equations the energy and momentum of the proton in system S′are
p′x =p uE
cvc
x −
−
2
2
21 =
22 4 0 99 22 4
1 0 99
2
2
2
. . × .
.
m c c m cc
cc
oo−
−
≈ =7 09 0 224 1 59 0. × . .m c m co
= 1.59 × 1.67 × 10–27 × 3 × 108
= 7.966 × 10–19 kg.m/sec.
and E′ =E −
−
vp
vc
x
12
2
= 22 4 0 99 22 4
1 0 990
20
2
. ( . ) ( . )
( . )
m c c m c−
−
= =7 09 22 4 0 01 1 592 2. × . × . .m c m co o
= 1.59 × (1.67 × 10–27 kg) (3 × 108 m/sec)2
= 2.39 × 10–10 joules
=2 39 101 6 10
10
19. ×. × − ev
= 1.5 × 109 ev = 1.5 BeV Ans.
Example 28. Write down the expression for the momentum of a photon. How much isthe rest mass of a photon.
Solution. The momentum of the photon of mass m is given by
p = mc ...(1)
since the velocity of the photon is equal to c.
The energy of the photon is given by
684 Mechanics
E = mc2 = hv
∴ m =hvc2
∴ Substituting the value of m in Eq. (1), we get
p = hvc
c hvc
h2 . = =
λ...(2)
But the momentum is given by
p =m v
vc
o
12
2−
mo =p v
cv
12
2−
But for a photon v = c, therefore the rest mass of the photon is given by
mo =p c
cv
10
2
2−=
Example 29. Calculate the momentum of a photon whose energy is 1.00 × 10–29 joules.
Solution. The momentum of photon is
p =Ec
= 1 00 10
3 103 33 10
19
828. ×
×. ×
−−= kg.m / sec Ans.
685
APPENDICES
APPENDIX ADEFINITION OF STANDARDS AND EQUIVALENTS
Standard Abbreviation Equivalent
Meter m 1,650,763.73 wavelengths in vacuo of the unperturbedtransition 2p10—5d5 in Kr86
Kilogram kg mass of the international kilogram at Sevres, France
Second sec 9,192,631,770 vibrations of the unperturbed hyperfinetransition 4,0—3,0 of the fundamental state 2S1/2 in Cs132 a
Degree Kelvin °K defined in the thermodynamic scale by assigning 273.16 °Kto the triple point of water
Unified atomic mass unit amu112
the mass of an atom of the C12 nuclide
Mole mol amount of substance containing the same number of atomsas 12 gm (exactly) of pure C12
Standard acceleration of gn 9.80665 meter/sec2
free fall
Normal atmospheric atm 101,325 nt/meter2
pressure
Thermochemical calorie cal 4.1840 joules
Liter l 0.001,000,028 meter3
Inch in. 0.0254 meter
Pound (avdp.) lb 0.453,592,37 kg
a There is no measurable difference between this and the previous standard of time,1/31,556,925.9747 of the tropical year at 12h ET, 0 January 1900. For this reason and because evenmore accurate maser standards may soon be available, the Cs standard was adopted provisionallyrather than “permanently”.
686 Mechanics
APPENDIX BFUNDAMENTAL AND DERIVED CONSTANTS
Name Symbol Computational Value Best Experimental
Valueb
Avogadro constant N0 6.02 × 1012/mole 6.02252 ± 0.00028Elementary charge e 1.60 × 10– 19 coul 1.60210 ± 0.00007Permeability constant µ0 1.26 × 10– 6 henry/meter 4π × 10– 7 exactlyPermittivity constant ∈0 8.85 × 10– 12 farad/meter 8.85418 ± 0.00002Speed of light c 3.00 × 108 metres/sec 2.997925 ± 0.000003
Electron rest mass me 9.11 × 10– 31 kg 9.1091 ± 0.0004Faraday constant F 9.65 × 104 coul/mole 9.64870 ± 0.00016Neutron rest mass mn 1.67 × 10– 27 kg 1.67482 ± 0.00008Planck constant h 6.63 × 10– 34 joule sec 6.6256 ± 0.0005Proton rest mass mp 1.67 × 10– 27 kg 1.67252 ± 0.00008
Electron charge/mass ratio e/me 1.76 × 1011 coul/kg 1.758796 ± 0.000019Electron Compton wavelength λC 2.43 × 10– 12 meter 2.42621 ± 0.00006Fine structure constant α 7.30 × 10– 2 7.29720 ± 0.00010
Proton Compton wavelength λCp1.32 × 10– 15 meter 1.32140 ± 0.00004
Quantum/charge ratio h/e 4.14 × 10– 15 joule sec/coul 4.13556 ± 0.00012
Bohr magneton µB 9.27 × 10– 24 joule/teslaa 9.2732 ± 0.0006Bohr radius a0 5.29 × 10– 11 meter 5.29167 ± 0.00007Nuclear magneton µN 5.05 × 10– 27 joule/teslaa 5.0505 ± 0.0004Proton magnetic moment µp 1.41 × 10– 26 joule/teslaa 1.41049 ± 0.00013Rydberg constant R∞ 1.10 × 107/meter 1.0973731 ± 0.0000003
Boltzmann constant k 1.38 × 10– 23 joule/°K 1.38054 ± 0.00018First radiation constant π 2 hc2 c1 3.74 × 10– 16 watt/meter2 3.7405 ± 0.0003Second radiation constant hc/k c2 1.44 × 10– 2 meter °K 1.43879 ± 0.00019Standard volume of ideal gas — 2.24 × 10– 2 meter3/mole 2.24136 ± 0.00030Universal gas constant R 8.31 joule/°K mole 8.3143 ± 0.0012
Gravitational constant G 6.67 × 10– 11 nt meter2/kg2 6.670 ± 0.015Stefan-Boltzmann constant σ 5.67 × 10– 8 watt/meter2 °K4 5.6697 ± 0.0029Wien displacement constant b 2.90 × 10– 3 meter °K 2.8978 ± 0.0004
a Tesla = weber/meter2.b Same units and power of ten as the computational value.
Appendices 687
APPENDIX CMISCELLANEOUS TERRESTRIAL DATA
Standard atmosphere 1.013 × 105 nt/meter2
14.70 lb/in2
760.0 mm-Hg
Density of dry air at STPa 1.293 kg/meter3
2.458 × 10– 3 slug/ft3
Speed of sound in dry air at STP 331.4 meter/sec1089 ft/sec742.5 miles/hr
Acceleration of gravity, g (standard value)b 9.80665 meters/sec2
32.1740 ft/sec2
Solar constantc 1340 watts/m2
1.92 cal/cm2-min
Mean total solar radiation 3.92 × 1026 watts
Equatorial radius of earth 6.378 × 106 meters3963 miles
Polar radius of earth 6.357 × 106 meters3950 miles
Volume of earth 1.087 × 1021 meter3
3.838 × 1022 ft3
Radius of sphere having same volume 6.371 × 106 meters3959 miles2.090 × 107 ft
Mean density of earth 5522 kg/meter3
Mass of earth 5.983 × 1024 kg
Mean orbital speed of earth 29,770 meters/sec18.50 miles/sec
Mean angular speed of rotation of earth 7.29 × 10– 5 radians/sec
Earth’s magnetic field, B (at Washington, DC) 5.7 × 10– 5 teslad
Earth’s magnetic dipole moment 6.4 × 1021 amp-m2
a STP = standard temperature and pressure = 0°C and 1 atm.b This value, used for barometer corrections, legal weights, etc., was adopted by the International
Committee on Weights and Measures in 1901. It approximates 45° latitude at sea level.c The solar constant is the solar energy falling per unit time at normal incidence on unit area
of the earth’s surface.d Tesla ≡ weber/meter2.
688M
echanics
APPENDIX D
THE SOLAR SYSTEMa
Planet Mercury Venus Earth , Mars Jupiter Saturn Uranus , Neptune Pluto
Meandiameter km 5,000 12,400 12,742 6,870 139,760 115,100 51,000 50,000 12,700?
Earthdiameters 0.39 0.973 1.000 0.532 10.97 9.03 4.00 3.90 0.46
Volume (earthvolumes) 0.06 0.92 1.00 0.15 1,318 736 64 39 0.10
Mass (earthmasses) 0.04 0.82 1.00 0.11 318.3 95.3 14.7 17.3 1.0?
Density (earthdensities) 0.69 0.89 1.00 0.70 0.24 0.13 0.23 0.29 ?
Mean density
gm/cm3 3.8 4.86 5.52 3.96 1.33 0.71 1.26 1.6 ?
Surface gravity
(earth’s) 0.27 0.86 1.00 0.37 2.64 1.17 0.92 1.44 ?
Velocity of
escape, km/sec 3.6 10.2 11.2 5.0 60 36 21 23 11?
Length of day
(earth days) 58.6d 30d? 1d 1d37m23s 9h55m 10h38m 10.7h 15.8h ?
Earth Uranus
Appen
dices
689
Planet Mercury Venus Earth , Mars Jupiter Saturn b Uranus , Neptune Pluto
Period,siderial days 87.97 224.70 365.26 686.98 4,332.59 10,759.20 30,685.93 60,187.64 90,885
Inclination ofequator to orbit — 0° ? 23°27′ 25°12′ 3°7′ 26°45′ 98.0° 29° ?
Oblateness 0.00 0.00 1/296 1/192 1/15.4 1/9.5 1/14 1/45 ?
Atmosphere,main con-stituents none N2, CO2, N2, O2 N2, CO2, CH4, NH3 CH4, NH3 CH4, NH3 CH4, NH3 none
A H2O
Maximum sur-face tempera-ture, °K 700 700 350 320 153 138 110? 90? 80?
Distance fromSun, 106 km 58 108 149 228 778 1426 2869 4495 5900
The Sun 329,390 earth masses, mean density 1.42, mean diameter 1,390,600 km, surface gravity 28 (earth’s).
The Moon 0.01228 earth masses, mean density 3.36, mean diameter 3,476 km, surface gravity 0.17 (earth’s), distance from earth 38 × 104 km.
aAdapted from Payne-Gaposchkin and Handbook of Chemistry and Physics.
Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Nepture Pluto
690M
echanics
APPENDIX EPERIODIC TABLE OF THE ELEMENTS
Atomic weights are expressed in atomic mass units (amu), one atom of the isotope C12 being defined to have a mass of (exactly)12 amu. For unstable elements the mass number of the most stable or best known isotope is given in brackets.
Group →Period Series I II III IV V VI VII VIII 0
1 11
1 00797H
.2
4 0026He
.
2 236 939
Li.
49 0122
Be.
510 811
B.
612 01115
C.
714 0067
N.
815 9994
O.
918 9984
F.
1020 183
Ne.
3 31122 9898
Na.
1224 312
Mg.
1326 9815
Al.
1428 086
Si.
1530 9738
P.
1632 064
S
.17
35 453Cl
.18
39 948A
.
44
1939 102
K.
2040 08
Ca.
2144 956
Sc.
2247 90
Ti.
2350 942
V.
2451 996
Cr.
2554 9380
Mn.
2655 847
Fe
. 2758 9332
Co.
2858 71
Ni.
52963 54
Cu.
3065 37
Zn.
3169 72
Ga.
3272 59
Ge.
3374 9216
As.
3478 96
Se.
3579 909
Br.
3683 80
Kr.
56
3785 47
Rb.
3887 62
Sr.
3988 905
Y.
4091 22
Zr.
4192 906
Nb.
4295 94
Mo.
4399
Tc 44101 07
Ru.
45102 905
Rh.
46106 4
Pd.
747
107 870Ag
.48
112 40Cd.
49114 82
In.
50118 69
Sn.
51121 75
Sb.
52127 60
Te.
53126 9044
I.
54131 30
Xe.
68
55 Cs132.905
56137 34
Ba.
57 71−Lanthanideseriesa
72178 49
Hf.
73180 948
Ta.
74183 85
W.
75186 2
Re.
76190 2
Os.
77192 2
Ir.
78195 09
Pt.
979
196 967Au
.80
200 59Hg.
81204 37
Tl.
82207 19
Pb.
83208 980
Bi.
84210
Po 85210
At 86222
Rn
7 1087223
Fr 88226
Ra89Actinideseriesb
aLanthanideseries :
57138 91
La.
58140 12
Ce.
59140 907
Pr.
60144 24
Nd.
61145
Pm 62150 35
Sm.
63151 96
Eu.
64157 25
Gd
.65
158 924Tb.
66162 50
Dy.
67164 930
Ho.
68167 26
Er.
69168 934
Tm.
70173 04
Yb.
71174 97
Lu.
bAseries:
ctinide 89227
Ac 90232 038
Th.
91231
Pa 92238 04
U.
93237
Np 94242
Pu 95243
Am 96245
Cm 97249
Bk 98249
Cf 99254
Es 100252
Fm 101256
Md 102254
103257
Lw
Appen
dices
691
(Cont...)
APPENDIX FTHE PARTICLES OF PHYSICSa
Family Particle Symbol Mass Spin Strange- Charge Antiparticlec No. of Averaged Typicalname name nessb particles lifetime, decay
seconds products
Photon γ (gamma 0 1 0 0 Same 1 Infinite —
ray) particle
Electron Electron e– 1 — – e e+ 2 Infinite —
family
Electron’s ve 0 — 0 2 Infinite —
neutrino
Muon Muon µ– 206.77 — – e µ+ 2 2.212 × 10– 6
family
Muon’s vµ 0(?) — 0 2 Infinite —
neutrino
Mesons Pion π+ 273.2 0 0 + e π– 3 2.55 × 10– 8 µ+ + vµ
π0 264.2 0 0 0 π0 1.9 × 10– 16 γ + γ
Kaon K+ 966.6 0 + 1 + e 4 1.22 × 10– 8 π+ + π0
K0 974 0 + 1 0 1.00 × 10– 10 π+ + π–
and 6 × 10– 8
e– + ve + vµ
692M
echanics
Family Particle Symbol Mass Spin Strange- Charge Antiparticlec No. of Averaged Typicalname name nessb particles lifetime, decay
seconds products
Baryons Nucleon p+ (proton) 1836.1212
0 + e p+ 4 Infinite
n0 (neutron) 1838.6512
0 0 n0 1013 p e ve+ +−
Lambda Λ0 2182.812
– 1 0 Λ0 2 2.51 × 10– 10 p + π–
particle
Sigma Σ+ 2327.712
– 1 + e Σ+ 8.1 ×10– 11 n + π+
particle Σ– 2340.512
– 1 – e Σ− 6 1.6 × 10– 10 n + π–
Σ0 233212
– 1 0 Σ0 about 10– 20 Λ0 + γ
Xi particle Ξ– 258012
– 2 – e Ξ− 4 1.3 × 10– 10 Λ0 + π–
Ξ0 257012
– 2 0 Ξ0 about 10– 10 Λ0 + π0
aAdapted and modified from The World of Elementary Particles, by Kenneth W. Ford.bThis is a “quantum number” whose assignment permits an understanding of the inter-relationships of the particles.cAntiparticles have the same mass and spin as the particles but their charges and strangeness numbers are opposite in sign.dThe K0 meson has two different Lifetimes; all other particles have only one.
Appendices 693
APPENDIX GSYMBOLS, DIMENSIONS, AND UNITS FOR PHYSICAL QUANTITIESAll units and dimensions are in the mksq (rationalized) system. The primary units can
be found by reading kilograms for M, meters for L, seconds for T, and coulombs for Q. Thesymbols are those used in the text.
In practice, Q is defined in terms of M, L and T. However, the addition of Q to thetraditional M, L, and T enables us to avoid the use of fractional exponents in dimensionalconsiderations. The term ‘rationalized’ simply means that a factor 1/4π is separated out ofCoulomb’s law in order to remove the factor 4π that would otherwise appear in many otherformulas in electricity.
Quantity Symbol Dimensions Derived Units
Acceleration a LT – 2 meters/sec2
Angular acceleration ααααα T– 2 radians/sec2
Angular displacement θ — radian
Angular frequency and speed ω T – 1 radians/sec
Angular momentum L ML2T – 1 kg-m2/sec
Angular velocity ω T – 1 radians/sec
Area A, S L2 meter2
Displacement r, d L meter
Energy, total E ML2T– 2 joule
kinetic K ML2T – 2 joule
potential U ML2T – 2 joule
Force F MLT – 2 newton
Frequency v T – 1 hertz ≡ cycles/sec
Gravitational field strength g LT – 2 nt/kg
Gravitational potential V L2T – 2 joules/kg
Length l L meter
Mass m M kilogram
Mass density ρ ML– 3 kg/m3
Momentum p MLT – 1 kg-m/sec2
Period T T second
Power P ML2T–3 watt
Pressure p ML– 1T – 2 nt/m2
Rotational inertia I ML2 kg-m2
Time t T second
Torque τττττ ML2T – 2 nt-m
Velocity v LT – 1 meters/sec
Volume V L3 meter3
Wavelength λ L meter
(Cont...)
694 Mechanics
Work W ML2T – 2 joule
Entropy S ML2T – 2 joules/K°
Internal energy U ML2T – 2 joule
Heat Q ML2T – 2 joule
Temperature T — Kelvin degree
Capacitance C M – 1L– 2T 2 Q2 farad
Charge q Q coulomb
Conductivity σ M– 1L– 3TQ2 (ohm-meter)– 1
Current i T– 1Q ampere
Current density j L– 2T – 1Q amp/meter2
Electric dipole moment p LQ coul-meter
Electric displacement D M– 2Q coul/meter2
Electric polarization P M– 2Q coul-meter2
Electric field strength E MLT – 2Q–1 volts/meter
Electric flux Φ E ML3T – 2Q– 1 volt-meter
Electric potential V ML2T – 2Q– 1 volt
Electromotive force ε ML2T – 2Q– 1 volts
Inductance L ML2Q– 2 henry
Magnetic dipole moment µµµµµ L2T – 1Q amp-meter2
Magnetic field strength H MT – 1Q amp-meter
magnetic flux Φ B ML2T – 1Q– 1 weber = volt-sec
Magnetic induction B MT– 1Q– 1 tesla ≡ webers/meter2
Magnetization M L – 1T – 1Q amp/meter
Permeability µ MLQ– 2 henrys/meter
Permittivity ∈ M– 1L– 3T 2Q2 farads/meter
Resistance R ML2T – 1Q– 2 ohm
Resistivity ρ ML3T – 1Q– 2 ohm-meter
Voltage V ML2T – 2Q– 1 volt
Appendices 695
APPENDIX HCONVERSION FACTORSa
Conversion factors for common and not-so-common units may be read off directly fromthe tables below. For example, 1 degree = 2.778 × 10– 3 revolutions, so 16.7° = 16.7 × 2.778× 10– 3 rev. The mks quantities are capitalized in each table.
PLANE ANGLE
° ′ ″ radian rev
1 degree = 1 60 3600 1.745 × 10– 2 2.778 × 10– 3
1 minute = 1.667 × 10– 2 1 60 2.909 × 10– 4 4.630 × 10– 5
1 second = 2.778 × 10– 4 1.667 × 10– 2 1 4.848 × 10– 6 7.716 × 10– 7
1 radian = 57.30 3438 2.063 × 105 1 0.1592
1 revolution = 360 2.16 × 104 1.296 × 106 6.283 1
1 rev = 2π radians = 360° 1° = 60′ = 3600″SOLID ANGLE
1 sphere = 4 steradians = 12.57 steradiansπa Adapted in part from G. Shortley and D. Williams, Elements of Physics, Prentice-Hall, EnglewoodCliffs, New Jersey, Second Edition, 1955.
LENGTH
cm meter km in. ft mile
1 centimeter = 1 10– 2 10– 5 0.3937 3.281 6.214
× 10– 2 × 10– 6
1 Meter = 100 1 10– 3 39.37 3.281 6.214
× 10– 4
1 kilometer = 105 1000 1 3.937 3281 0.6214
× 104
1 inch = 2.540 2.540 2.540 1 8.333 1.578
× 10– 2 × 10– 5 × 10– 2 × 10– 5
1 foot = 30.48 0.3048 3.048 12 1 1.894
× 10– 4 × 10– 4
1 statute mile = 1.609 1609 1.609 6.336 5280 1
× 105 × 104
1 angstrom (A) = 10– 10 meter 1 light-year = 9.4600 × 1012 km 1 yard = 3 ft
1 X-unit 10– 13 meter 1 parsec = 3.084 × 1013 km 1 rod = 16.5 ft
1 micron = 10– 6 meter 1 fathom = 6 ft 1 mil = 10– 3 in.
1 millicron (mµ) = 10– 9 meter
1 nautical mile = 1852 meters = 1.1508 statute miles = 6076.10 ft.
696 Mechanics
AREA
meter2 cm2 ft2 in.2 circ mil
1 square meter = 1 104 10.76 1550 1.974 × 109
1 square centimeter = 10– 4 1 1.076 × 10– 3 0.1550 1.974 × 105
1 square foot = 9.290 × 10– 2 929.0 1 144 1.833 × 108
1 square inch = 6.452 × 10– 4 6.452 6.944 × 10– 3 1 1.273 × 106
1 circular mill = 5.067 × 10– 10 5.067 × 10– 6 5.454 × 10– 9 7.854 × 10– 7 1
1 square mile = 27,878,400 ft2 = 640 acres 1 acre = 43,560 ft2
1 barn = 10– 28 meter2
VOLUME
Meter3 cm3 1 ft3 in.3
1 cubic meter = 1 106 1000 35.31 6.102 × 104
1 cubic centimeter = 10– 6 1 1.000 × 10– 3 3.531 × 10– 5 6.102 × 10– 2
1 liter = 1.000 × 10– 3 1000 1 3.531 × 10– 2 61.02
1 cubic foot = 2.832 × 10– 2 2.832 × 104 28.32 1 1728
1 cubic inch = 1.639 × 10– 5 16.39 1.639 × 10– 2 5.787 × 10– 4 1
1 U.S. fluid gallon = 4 U.S. fluid quarts = 8 U.S. pints = 128 U.S. fluid ounces = 231 in.3
1 British imperial gallon = the volume of 10 lb of water at 62 ° F = 277.42 in.3
1 liter = the volume of 1 kg of water at its maximum density = 1000.028 cm3
MASS
Note: Those quantities to the right of and below the heavy lines are not mass units at all but areoften used as such. When we write, for example,
1 kg “ = ” 2.205 lb
this means that a kilogram is a mass that weighs 2.205 pounds. Clearly this “equivalence” isapproximate (depending on the value of g) and is meaningful only for terrestrial measurements. Thus,care must be employed when using the factors in the shaded portion of the table.
gm kg slug amu oz lb ton
1 gram = 1 0.001 6.852 6.024 3.527 2.205 1.102× 10– 5 × 1023 × 10– 2 × 10– 3 × 10– 5
1 kilogram = 1000 1 6.852 6.024 35.27 2.205 1.102× 10– 2 × 1026 × 10– 3
1 slug = 1.459 14.59 1 8.789 514.8 32.17 1.609× 104 × 1027 × 10– 2
1 amu = 1.660 1.660 1.137 1 5.855 3.660 1.829× 10– 24 × 10– 27 × 10– 28 × 10– 26 × 10– 27 × 10– 30
1 ounce 28.35 2.835 1.943 1.708 1 6.250 3.125(avoirdupois) = × 10– 2 × 10– 3 × 1025 × 10– 2 × 10– 5
1 pound 453.6 0.4536 3.108 2.732 16 1 0.0005(avoirdupois) = × 10– 2 × 1026
1 ton = 9.072 907.2 62.16 5.465 3.2 2000 1× 10+ 5 × 1029 × 104
Appendices 697
DENSITY
Note: Those quantities to the right or below the heavy line are weight densities and, as such, aredimensionally different from mass densities. Care must be used. (See note for mass table.)
slug/ft3 kg/meter3 gm/cm3 lb/ft3 lb/in.3
1 slug per ft3 = 1 515.4 0.5154 32.17 1.862 × 10– 2
1 kilogram per
meter3 = 1.940 × 10– 3 1 0.001 6.243 × 10– 2 3.613 × 10– 5
1 gram per cm3 = 1.940 1000 1 62.43 3.613 × 10– 2
1 pound per ft3 = 3.108 × 10– 2 16.02 1.602 × 10– 2 1 5.787 × 10– 4
1 pound per in.3 = 53.71 2.768 × 104 27.68 1728 1
TIME
yr day hr min sec
1 year = 1 365.2 8.766 × 103 5.259 × 105 3.156 × 107
1 day = 2.738 × 10– 3 1 24 1440 8.640 × 104
1 hour = 1.141 × 10– 4 4.167 × 10– 2 1 60 3600
1 minute = 1.901 × 10– 6 6.944 × 10– 4 1.667 × 10– 2 1 60
1 SECOND = 3.169 × 10– 8 1.157 × 10– 5 2.778 × 10– 4 1.667 × 10– 2 1
SPEED
ft/sec km/hr meter/sec miles/hr cm/sec knot
1 foot per
second = 1 1.097 0.3048 0.6818 30.48 0.5925
1 kilometer per
hour = 0.9113 1 0.2778 0.6214 27.78 0.5400
1 meter per
second = 3.281 3.6 1 2.237 100 1.944
1 mile per hour = 1.467 1.609 0.4470 1 44.70 0.8689
1 centimeter per
second = 3.281 × 10– 2 3.6 × 10– 2 0.01 2.237 × 10– 2 1 1.944 × 10– 2
1 knot = 1.688 1.852 0.5144 1.151 51.44 1
1 knot = 1 nautical mile/hr 1 mile/min = 88 ft/sec = 60 miles/hr
FORCE
Note: Those quantities to the right of and below the heavy lines are not force units at all but areoften used as such, especially in chemistry. For instance, if we write
1 gram-force = 980.7 dynes,
we mean that a gram-mass experiences a force of 980.7 dynes in the earth’s gravitational field. Thus,care must be employed when using the factors in the shaded portion of the table.
dyne nt lb pdl gf kgf
698 Mechanics
dyne nt lb pdl gf kgf
1 dyne = 1 10– 5 2.248 7.233 1.020 1.020
× 10– 6 × 10– 5 × 10– 3 × 10– 6
1 newton = 105 1 0.2248 7.233 102.0 0.1020
1 pound = 4.448 4.448 1 32.17 453.6 0.4536
× 105
1 poundal = 1.383 0.1383 3.108 1 14.10 1.410
× 104 × 10– 2 × 10– 2
1 gram-force = 980.7 9.807 2.205 7.093 1 0.001
× 10– 3 × 10– 3 × 10– 2
1 kilogram-force = 9.807 9.807 2.205 70.93 1000 1
× 105
1 kgf = 9.80665 nt 1 lb = 32.17398 pdl
PRESSURE
atm dyne/cm2 inch of cm Hg nt/ lb/in.2 lb/ft2
water meter2
1 atmosphere = 1 1.013 406.8 .76 1.013 14.70 2116
× 106 × 105
1 dyne per cm2 = 9.869 1 4.015 7.501 0.1 1.450 2.089
× 10– 7 × 10– 4 × 10– 5 × 10– 5 × 10– 3
1 inch of water at 2.458 2491 1 0.1868 249.1 3.613 5.202
4°Ca = × 10– 3 × 10– 2
1 centimeter of 1.316 1.333 5.353 1 1333 0.1934 27.85
mercury at 0°Ca = × 10– 2 × 104
1 newton per 9.869 10 4.015 7.501 1 1.450 2.089
meter2 = × 10– 6 × 10– 3 × 10– 4 × 10– 4 × 10– 2
1 pound per in.2 = 6.805 6.895 27.68 5.171 6.895 1 144
× 10– 2 × 104 × 103
1 pound per ft2 = 4.725 478.8 0.1922 3.591 47.88 6.944 1
× 10– 4 × 10– 2 × 10– 3
Ap
pen
dices
69
9
ENERGY, WORK, HEAT
The electron volt (ev) is the kinetic energy an electron gains from being accelerated through the potential difference of one volt
in an electric field. The Mev is the kinetic energy it gains from being accelerated through a million-volt potential difference.
The last two items in this table are not properly energy units but are included for convenience. They arise from the relativistic
mass-energy equivalence formula E = mc2 and represent the energy released if a kilogram or atomic mass unit (amu) is destroyed
completely.
Again, care should be used when employing this table.
Btu erg ft-lb hp-hr JOULES cal kw-hr ev Mev kg amu
1 British 1 1.055 777.9 3.929 1055 252.0 2.930 6.585 6.585 1,174 7.074
thermal unit = × 1010 × 10– 4 × 10– 4 × 1021 × 1015 × 10– 14 × 1012
1 erg = 9.481 1 7.376 3.725 10– 7 2.389 2.778 6.242 6.242 1.113 670.5
× 10– 11 × 10– 8 × 10– 14 × 10– 8 × 10– 14 × 1011 × 105 × 10– 24
1 foot-pound = 1.285 1.356 1 5.051 1.356 0.3239 3.766 8.464 8.464 1.509 9.092
× 10– 3 × 107 × 10– 7 × 10– 7 × 1018 × 1012 × 10– 17 × 109
1 horsepower- 2545 2.685 1.980 1 2.685 6.414 0.7457 1.676 1.676 2.988 1.800
hour = × 1013 × 106 × 106 × 105 × 1025 × 1019 × 10– 11 × 1016
1 Joule = 9.481 107 0.7376 3.725 1 0.2389 2.778 6.242 6.242 1.113 6.705
× 10– 4 × 10– 7 × 10– 7 × 1018 × 1012 × 10– 17 × 109
1 calorie = 3.968 4.186 3.087 1.559 4.186 1 1.163 2.613 2.613 4.659 2.807
× 10– 3 × 107 × 10– 6 × 10– 6 × 1019 × 1013 × 10– 17 × 1010
1 kilowatt- 3413 3.6 2.655 1.341 3.6 8.601 1 2.247 2.270 4.007 2.414
hour = × 1013 × 106 × 106 × 105 × 1025 × 1019 × 10– 11 × 1016
1 electron volt = 1.519 1.602 1.182 5.967 1.602 3.827 4.450 1 10– 6 1.783 1.074
× 10– 22 × 10– 12 × 10– 19 × 10– 26 × 10– 19 × 10– 20 × 10– 26 × 10– 36 × 10– 9
1 million 1.519 1.602 1.182 5.967 1.602 3.827 4.450 106 1 1.783 1.074
electron volts = × 10– 16 × 10– 6 × 10– 13 × 10– 20 × 10– 13 × 10– 14 × 10– 20 × 10– 30 × 10– 3
1 kilogram = 8.521 8.987 6.629 3.348 8.987 2.147 2.497 5.610 5.610 1 6.025
× 1013 × 1023 × 1016 × 1010 × 1016 × 1016 × 1010 × 1035 × 1029 × 1026
1 atomic mass 1.415 1.492 1.100 5.558 1.492 3.564 4.145 9.31 931.0 1.660 1
unit = × 10– 13 × 10– 3 × 10– 10 × 10– 17 × 10– 10 × 10– 11 × 10– 17 × 108 × 10– 27
1 m-kgf = 9.807 joules 1 watt-sec = 1 joule = 1 nt-m 1 cm-dyne = 1 erg
700 Mechanics
POWER
Btu/hr ft-lb/min ft-lb/sec hp cal/sec kw WATTS
1 British thermal 1 12.97 0.2161 3.929 7.000 2.930 0.2930
unit per hour = × 10– 4 × 10– 2 × 10– 4
1 foot-pound per 7.713 1 1.667 3.030 5.399 2.260 2.260
minute = × 10– 2 × 10– 2 × 10– 5 × 10– 3 × 10– 5 × 10– 2
1 foot-pound per 4.628 60 1 1.818 0.3239 1.356 1.356
second = × 10– 3 × 10– 3
1 horsepower = 2545 3.3 × 104 550 1 178.2 0.7457 745.7
1 calorie per 14.29 1.852 3.087 5.613 1 4.186 4.186
second = × 102 × 10– 3 × 10– 3
1 kilowatt = 3413 4.425 737.6 1.341 238.9 1 1000
× 104
1 WATT = 3.413 44.25 0.7376 1.341 0.2389 0.001 1
× 10– 3
ELECTRIC CHARGE
abcoul amp-hr coul faraday statcoul
1 abcoulomb
(1 emu) = 1 2.778 × 10– 3 10 1.036 × 10– 4 2.998 × 1010
1 ampere-hour = 360 1 3600 3.730 × 10– 2 1.079 × 1013
1 coulomb = 0.1 2.778 × 10– 4 1 1.036 × 10– 5 2.998 × 109
1 faraday = 9652 26.81 9.652 × 104 1 2.893 × 1014
1 statcoulomb
(1 esu) = 3.336 × 10– 11 9.266 × 10– 14 3.336 × 10– 10 3.456 × 10– 15 1
1 electronic charge = 1.602 × 10– 19 coulomb
ELECTRIC CURRENT
abamp amp statamp
1 abampere (1 emu) = 1 10 2.998 × 1010
1 ampere = 0.1 1 2.998 × 109
1 statampere (1 esu) = 3.336 × 10– 11 3.336 × 10– 10 1
ELECTRIC POTENTIAL, ELECTROMOTIVE FORCE
abv volts statv
1 abvolt (1 emu) = 1 10– 8 3.336 × 10– 11
1 volt = 108 1 3.336 × 10– 3
1 statvolt (1 esu) = 2.998 × 1010 299.8 1
Appendices 701
ELECTRIC RESISTANCE
abohm ohms statohm
1 abohm (1 emu) = 1 10– 9 1.113 × 10– 21
1 ohm = 109 1 1.113 × 10– 12
1 statohm (1 esu) = 8.987 × 1020 8.987 × 1011 1
ELECTRIC RESISTIVITY
abohm-cm µohm-cm ohm-cm statohm-cm ohm-m ohm-circ
mil/ft
1 abohm-centi- 1 0.001 10– 9 1.113 10– 11 6.015
meter (1 emu) = × 10– 21 × 10– 3
1 micro-ohm- 1000 1 10– 6 1.113 10– 8 6.015
centimeter = × 10– 18
1 ohm-centi- 109 106 1 1.113 0.01 6.015
meter = × 10– 12 × 106
1 statohm-centi- 8.987 8.987 8.987 1 8.987 5.406
meter (1 esu) = × 1020 × 1017 × 1011 × 109 × 1018
1 ohm-meter = 1011 108 100 1.113 1 6.015
× 10– 10 × 108
1 ohm-circular 166.2 0.1662 1.662 1.850 1.662 1
mil per foot = × 10– 7 × 10– 19 × 10– 9
CAPACITANCE
abf farads µf a statf
1 abfarad (1 emu) = 1 109 1015 8.987 × 1020
1 farad = 10– 9 1 106 8.987 × 1011
1 microfarad = 10– 15 10– 6 1 8.987 × 105
1 statfarad (1 esu) = 1.113 × 10– 21 1.113 × 10– 12 1.113 × 10– 6 1
aThis unit is frequently abbreviated as mf.
INDUCTANCE
abhenry henrys µh mh stathenry
1 abhenry (1 emu) = 1 10– 9 0.001 10– 6 1.113 × 10– 21
1 henry = 109 1 106 1000 1.113 × 10– 12
1 microhenry = 1000 10– 6 1 0.001 1.113 × 10– 18
1 millihenry = 106 0.001 1000 1 1.113 × 10– 15
1 stathenry (1 esu) = 8.987 × 1020 8.987 × 1011 8.987 × 1017 8.987 × 1014 1
702 Mechanics
MAGNETIC FLUX
maxwell kiloline weber
1 maxwell (1 line or 1 emu) = 1 0.001 10– 8
1 kiloline = 1000 1 10– 5
1 weber = 108 105 1
1 esu = 299.8 webers
MAGNETIC INDUCTION B
gauss kiloline/in.2 weber/ milligauss
meter2
= tesla
1 gauss (line per
cm2) = 1 6.452 × 10– 3 10– 4 1000 105
1 kiloline per in.2 = 155.0 1 1.550 × 10– 2 1.550 × 105 1.550 × 107
1 weber per
meter2 =
1 tesla = 104 64.52 1 107 109
1 milligauss = 0.001 6.452 × 10– 6 10– 7 1 100
1 gamma = 10– 5 6.452 × 10– 8 10– 9 0.01 1
1 esu = 2.998 × 106 webers/meter2
MAGNETOMOTIVE FORCE
abamp-turn amp-turn gilbert
1 abampere-turn = 1 10 12.57
1 ampere-turn = 0.1 1 1.257
1 gilbert = 7.958 × 10– 2 0.7958 1
1 pragilbert = 4π amp-turn 1 esu = 2.655 × 10– 11 amp-turn
MAGNETIC FIELD STRENGTH, H
abamp-turn/cm amp- amp-turn/in. amp- oersted
turn/cm turn/
meter
1 abampere-turn per
centimeter = 1 10 25.40 1000 12.57
1 ampere-turn per
centimeter = 0.1 1 2.540 100 1.257
1 ampere-turn per inch = 3.937 × 10– 2 0.3937 1 39.37 0.4947
1 ampere-turn per
meter = 0.001 0.01 2.540 × 10– 2 1 1.257 × 10– 2
1 oersted = 7.958 × 10– 2 0.7958 2.021 79.58 1
1 oersted = 1 gilbert 1 esu = 2.655 × 10– 9 amp-turn meter
1 praoersted = 4π amp-turn/meter
Appendices 703
MATHEMATICAL SYMBOLS AND THE GREEK ALPHABET
Mathematical signs and symbols= equals
≅ equals approximately
≠ is not equal to
≡ is identical to, is defined as
> is greater than (>> is much greater than)
< is less than (<< is much less than)
≥ is more than or equal to (or, is no less than)
≤ is less than or equal to (or, is no more than)
± plus or minus e.g., 4 2= ± ∝ is proportional to (e.g., Hooke’s law : F ∝ x, or F = –kx)
Σ the sum of
x the average value of x
The Greek alphabetAlpha Α α Nu Ν νBeta Β β Xi Ξ ξGamma Γ γ Omicron Ο οDelta ∆ δ Pi Π πEpsilon Ε ε Rho Ρ ρZeta Ζ ζ Sigma Σ σEta Η η Tau Τ τ
Theta θ θ ϑ, Upsilon ϒ υ
Iota Ι ι Phi Φ φ, ϕKappa Κ κ Chi Χ χLambda Λ λ Psi Ψ ψMu Μ µ Omega Ω ω
APPENDIX I
704 Mechanics
Y
X0
θ
r
x
y
Fig. App. I
APPENDIX JMATHEMATICAL FORMULAS
Quadratic formula
If ax2 + bx + c = 0, then x =− ± −b b ac
a
2 42
.
Trigonometric functions of angle θθθθθ
sin θ = yr
cos θ = xr
tan θ = yx
cot θ = xy
sec θ = rx
csc θ = ry
Pythogorean theoremx2 + y2 = r2
Trigonometric identities
sin2 θ + cos2 θ = 1 sec2 θ – tan2 θ = 1 csc2 θ – cot2 θ = 1
sin (α ± β) = sin α cos β ± cos α sin βcos (α ± β) = cos α cos β – sin α sin β
+
tan (α ± β) =tan tan
tan tanα β
α β±
1
sin 2 θ = 2 sin θ cos θcos 2 θ = cos2 θ – sin2 θ = 2 cos2 θ – 1 = 1 – 2 sin2 θ
sin θ =e e
iq
e eiq iq iq iq− = +− −
2cos
2
e±iθ = cos θ ± i sin θ
Taylor’s series
f(x0 + x) = f(x0) + f '(x0)x + f "(x0) x
!f "'( x )
x!
....2
0
2
2 3+ +
Series expansions (these expansions converge for – 1 < x < 1,
11 + x = 1 – x + x2 – x3 + ....
1 + x = 12 8 16
2 3+ − + −x x x
....
Appendices 705
1
1 + x = 12
38
516
2 3− + − +x x x
....
ex = 12 6
2 3+ + + +x
x x.... (– ∞ < x < ∞)
sin x = xx x− + −
3 5
6 120.... (– ∞ < x < ∞)
cos x = 12 24
2 4− + −x x
.... (– ∞ < x < ∞)
tan x = xx x+ + +
3 5
3215
.... (– π/2 < x < π/2)
(x + y)n = xn
x yn n
x yn n n+ + − +− −1
12
1 2 2
!( )
!.... (x2 > y2)
Derivatives and indefinite integralsIn what follows, the letters u and v stand for any functions of x, and a and m are constants. To
each of the integrals should be added an arbitrary constant of integration. A Short Table of Integralsby Peirce and Foster (Ginn and Co.) gives a more extensive tabulation.
1.dxdx
= 1 1. dx = x
2.ddx
au( ) = adudx
2. au dx = a u dx3.
ddx
u v( )+ = dudx
dvdx
+ 3. (u v)dx+ = u dx v dx +
4.ddx
xm = mxm – 1 4. x dxm = xm
m +
+
1
1 (m ≠ – 1)
5.ddx
xln = 1x
5.dxx = ln x
6.ddx
uv( ) = udvdx
vdudx
+ 6. udvdx
dx = uv vdudx
dx− 7.
ddx
ex = ex 7. e dxx = ex
8.ddx
xsinh = cosh x 8. cosh x dx = sinh x
9.ddx
xcosh = sinh x 9. sinh x dx = cosh x
10.ddx
xarctan = 1
1 2+ x10.
dx
x1 2+ = arctan x
x in radians
706 Mechanics
11.ddx
xarcsin = 1
1 2− x11.
dx
x1 2− = arcsin x
12.ddx
xarcsec = 1
12x x −12.
dx
x x2 1− = arcsec x
13.ddx
xcos = – sin x 13. sin x dx = – cos x
14.ddx
xsin = cos x 14. cos x dx = sin x
15.ddx
xtan = sec2 x 15. tan x dx = ln sec x
16.ddx
xcot = – csc2 x 16. cot x dx = ln sin x
17.ddx
xsec = tan x sec x 17. sec x dx = ln sec tanx x+
18.ddx
xcsc = – cot x csc x 18. csc x dx = ln csc cotx x−
Vector productsLet i, j, k be unit vectors in the x, y, z directions. Then
i . i = j . j = k . k = 1, i . j = j . k = k . i = 0,
i × i = j × j = k × k = 0,
i × j = k, j × k = i, k × i = j.
Any vector a with components ax, ay, az, along the x, y, z axes can be written
a = axi + ayj + azk.
Let a, b, c be arbitrary vectors with magnitudes a, b, c. Then
a × (b + c) = a × b + a × c
(sa) × b = a × (sb) = s(a × b) (s a scalar).
Let θ be the smaller of the two angles between a and b. Then
a . b = b . a = axbx + ayby + azbz = ab cos θ
a × b = – b × a = i j k
a a ab b b
x y z
x y z
= (aybz – byaz)i + (azbx – bzax)j + (axby – bxay)k
a b× = ab sin θ
a . (b × c) = b . (c × a) = c . (a × b)
a × (b × c) = (a . c)b – (a . b)c
Appendices 707
(Cont...)
APPENDIX KVALUES OF TRIGONOMETRIC FUNCTIONS
TRIGONOMETRIC FUNCTIONS
Radians Degrees Sines Cosines Tangents Cotangents
.0000 0 .0000 1.0000 .0000 ∞ 90 1.5708
.0175 1 .0175 .9998 .0175 57.29 89 1.5533
.0349 2 .0349 .9994 .0349 28.64 88 1.5359
.0524 3 .0523 .9986 .0524 19.08 87 1.5184
.0698 4 .0698 .9976 .0699 14.30 86 1.5010
.0873 5 .0872 .9962 .0875 11.430 85 1.4835
.1047 6 .1045 .9945 .1051 9.514 84 1.4661
.1222 7 .1219 .9925 .1228 8.144 83 1.4486
.1396 8 .1392 .9903 .1405 7.115 82 1.4312
.1571 9 .1564 .9877 .1584 6.314 81 1.4137
.1745 10 .1736 .9848 .1763 5.671 80 1.3963
.1920 11 .1908 .9816 .1944 5.145 79 1.3788
.2094 12 .2079 .9781 .2126 4.705 78 1.3614
.2269 13 .2250 .9744 .2309 4.332 77 1.3439
.2443 14 .2419 .9703 .2493 4.011 76 1.3285
.2618 15 .2588 .9659 .2679 3.732 75 1.3090
.2793 16 .2756 .9613 .2867 3.487 74 1.2915
.2967 17 .2924 .9563 .3057 3.271 73 1.2741
.3142 18 .3090 .9511 .3249 3.078 72 1.2866
.3316 19 .3256 .9455 .3443 2.904 71 1.2392
.3491 20 .3420 .9397 .3640 2.748 70 1.2217
.3665 21 .3584 .9336 .3839 2.605 69 1.2043
.3840 22 .3746 .9272 .4040 2.475 68 1.1868
.4014 23 .3907 .9205 .4245 2.356 67 1.1694
.4189 24 .4067 .9135 .4452 2.246 66 1.1519
.4363 25 .4226 .9063 .4663 2.144 65 1.1345
.4538 26 .4384 .8988 .4877 2.050 64 1.1170
.4712 27 .4540 .8910 .5095 1.963 63 1.0996
.4887 28 .4695 .8829 .5317 1.881 62 1.0821
.5061 29 .4848 .8746 .5543 1.804 61 1.0647
.5236 30 .5000 .8660 .5774 1.732 60 1.0472
.5411 31 .5150 .8572 .6009 1.664 59 1.0297
.5585 32 .5299 .8480 .6249 1.600 58 1.0123
Cosines Sines Cotangents Tangents Degrees Radians
708 Mechanics
Radians Degrees Sines Cosines Tangents Cotangents
.5760 33 .5446 .8387 .6494 1.540 57 0.9948
.5934 34 .5592 .8290 .6745 1.483 56 0.9774
.6109 35 .5736 .8192 .7002 1.428 55 0.9599
.6283 36 .5878 .8090 .7265 1.376 54 0.9425
.6458 37 .6018 .7986 .7536 1.327 53 0.9250
.6632 38 .6157 .7880 .7813 1.280 52 0.9076
.6807 39 .6293 .7771 .8098 1.235 51 0.8901
.6981 40 .6428 .7660 .8391 1.192 50 0.8727
.7156 41 .6561 .7547 .8693 1.150 49 0.8552
.7330 42 .6691 .7431 .9004 1.111 48 0.8378
.7505 43 .6820 .7314 .9325 1.072 47 0.8203
.7679 44 .6947 .7193 .9657 1.036 46 0.8029
.7854 45 .7071 .7071 1.0000 1.000 45 0.7854
Cosines Sines Cotangents Tangents Degrees Radians
Appendices 709
APPENDIX LNOBEL PRIZE WINNERS IN PHYSICSa
1901 Wilhelm Konrad Rontgen 1845-1923 German Discovery of X-rays.
1902 Hendrik Antoon Lorentz 1853-1928 Dutch Influence of magnetism on the
Pieter Zeeman 1865-1943 Dutch phenomena of atomic radiation.
1903 Henri Becquerel 1852-1908 French Discovery of natural radioactivity
Pierre Curie 1850-1906 French and of the radioactive elements
Marie Curie 1867-1934 French radium and polonium.
1904 Baron Rayleigh 1842-1919 English Discovery of argon.
1905 Philipp Lenard 1862-1947 German Research in cathode rays.
1906 Sir Joseph John Thomson 1856-1940 English Conduction of electricity throughgases.
1907 Albert A. Michelson 1852-1931 U.S. Invention of interferometer andspectroscopic and metrologicalinvestigations.
1908 Gabriel Lippmann 1845-1921 French Photographic reproduction ofcolours.
1909 Guglielmo Marconi 1874-1937 Italian Development of wireless tele-Karl Ferdinand Braun 1850-1918 German graphy.
1910 Johannes Diderik 1837-1923 Dutch Equations of state of gases andvan der Waals fluids.
1911 Wilhelm Wien 1864-1928 German Laws of heat radiation.
1912 Nils Gustaf Dalen 1869-1937 Swedish Automatic coastal lighting.
1913 Heike Kamerlingh-Onnes 1853-1926 Dutch Properties of matter at low tem-peratures; production of liquidhelium.
1914 Max von Laue 1879-1960 German Diffraction of X-rays in crystals.
1915 Sir William Henry Bragg 1862-1942 English Study of crystal structure bySir William Lawrence 1890- English— means of X-rays.Bragg his son
1916 (No award)
1917 Charles Glover Barkla 1877-1944 English Discovery of the characteristicX-rays of elements.
1918 Max Planck 1858-1947 German Discovery of the elemental quantum.
1919 Johannes Stark 1874-1957 German Discovery of the Doppler effectin canal rays and the splittingof spectral lines in the electric field.
1920 Charles Edouard Guillaume 1861-1938 Swiss Discovery of the anomalies ofnickel-steel alloys.
1921 Albert Einstein 1879-1955 German Discovery of the law of the photo-elec-tric effect.
710 Mechanics
1922 Niels Bohr 1885–1963 Danish Study of structure and radiationsof atoms.
1923 Robert Andrews Millikan 1868-1953 U.S. Work on elementary electriccharge and the photoelectric effect.
1924 Manne Siegbahn 1886 Swedish Discoveries in the area of X-rayspectra.
1925 James Franck 1882–1964 German Laws governing collision between
Gustav Hertz 1887 German electron and atom.
1926 Jean Perrin 1870–1942 French Discovery of the equilibrium ofsedimentation.
1927 Arthur H. Compton 1892–1962 U.S. Discovery of the scattering ofX-rays by charged particles.
Charles T.R. Wilson 1869–1959 English Invention of the cloud chamber,a device to make visible thepaths of charged particles.
1928 Sir Owen Williams 1879–1959 English Discovery of the law known byRichardson his name (the dependency of
the emission of electrons ontemperature).
1929 Louis-Victor de Broglie 1892 French Wave nature of electrons.
1930 Sir Chandrasekhara Raman 1888 Indian Work on the scattering of lightand discovery of the effectknown by his name.
1931 (No award)
1932 Werner Heisenberg 1901 German Creation of quantum mechanics.
1933 Paul Adrien Maurice Dirac 1902 English Discovery of new fertile forms of
Erwin Schrodinger 1887–1961 Austrian the atomic theory.
1934 (No award)
1935 James Chadwick 1891 English Discovery of the neutron.
1936 Victor Hess 1883–1964 Austrian Discovery of cosmic radiation.
Carl David Anderson 1905 U.S. Discovery of the positron.
1937 Clinton Joseph Davisson 1881–1958 U.S. Discovery of diffraction of elec-
George P. Thomson 1892 English trons by crystals.
1938 Enrico Fermi 1901–1954 Italian Artificial radioactive elementsfrom neutron irradiation.
1939 E.O. Lawrence 1901–1958 U.S. Invention of the cyclotron.
1942 (No awards)
1943 Otto Stern 1888 U.S.g Work with molecular beams andmagnetic moment of proton.
1944 Isidor Isaac Rabi 1898 U.S. Nuclear magnetic resonance.
1945 Wolfgang Pauli 1900–1958 Austrian Discovery of quantum exclusionprinciple.
1946 Percy Williams Bridgman 1882–1961 U.S. High-pressure physics.
Appendices 711
1947 Sir Edward Appleton 1892 English Upper atmosphere physics anddiscovery of Appleton layer.
1948 Patrick Maynard Stuart 1897 English Discoveries in cosmicBlackett radiation and nuclear physics.
1949 Hideki Yukawa 1907 Japanese Prediction of existence of meson.
1950 Cecil Frank Powell 1903 English Photographic method of studyingnuclear processes; discoveriesabout mesons.
1951 Sir John Douglas Cockeroft 1897 English Transmutation of atomic nuclei
Ernes Thomas Sinton Walton 1903 Irish by artificially acceleratedatomic particles.
1952 Felix Bloch 1905 U.S. Measure of magnetic fields in
Edward Mills Purcell 1912 U.S. atomic nuclei.
1953 Frits Zernike 1888 Dutch Invention of phase contrastmicroscopy.
1954 Max Born 1882 Englishh Work in quantum mechanics andstatistical interpretation ofwavefunction.
Walther Bothe 1891–1957 German Analysis of cosmic radiationsusing the coincidence method.
1955 Willis E. Lamb, Jr. 1913 U.S. Fine structure of hydrogen.
Polykarp Kusch 1911 U.S. Magnetic moment of electron.
1956 John Bardeen 1908 U.S. Invention and development of
Walter H. Brattain 1902 U.S. transistor.
William B. Shockley 1910 U.S.c
1957 Chen Ning Yang 1922 Chinesed Non-conservation of parity and
Tsung Dao Lee 1926 Chinesed work in elementary particletheory.
1958 Pavel A. Cerenkov 1904 Russian Discovery and interpretation of
Ilya M. Frank 1908 Russian Cerenkov effect of radiation by
Igor Y. Tamm 1895 Russian fast charged particles in matter.
1959 Owen Chamberlain 1920 U.S. Discovery of the antiproton.
Emilio Gino Segre 1905 U.S.e
1960 Donald A. Glaser 1926 U.S. Invention of bubble chamber.
1961 Robert L. Hofstadter 1915 U.S. Electromagnetic structure of nu-cleons from high-energy elec-tron scattering.
Rudolf L. Mossbauer 1929 German Discovery of recoilless resonanceabsorption of gamma rays innuclei.
1962 Len D. Landau 1908 Russian Theory of condensed matter;phenomena of superfluidity andsuperconductivity.
1963 Eugene B. Wigner 1902 U.S.f Contributions to theoreticalatomic and nuclear physics.
712 Mechanics
Maria Goeppert-Mayer 1906 U.S.g Shell model theory and magic
J.H.D. Jensen 1907 German numbers for the atomic nucleus.
1964 C.H. Townes 1915 U.S. Invention of the maser and theory
Nikolai Basov 1922 Russian of coherent atomic radiation.
Aleksandr Prokhorov 1916 Russian
1965 Richard Feynman 1918 U.S. Development of quantum electro-
Julian Schwinger 1918 U.S. dynamics.
Shin-Ichiro Tomonaga 1906 JapaneseaSee Nobel : The Man and His Prizes, by Schuck et al., Elsevier, N.Y.bBorn in Germany; naturalized British citizen.cBorn in England; naturalized U.S. citizen.dBoth have permanent U.S. resident status.eBorn in Italy; naturalized U.S. citizen.fBorn in Hungary; naturalized U.S. citizen.gBorn in Germany; naturalized U.S. citizen.
Appendices 713
APPENDIX MTHE GAUSSIAN SYSTEM OF UNITS
Much of the literature of physics is written, and continues to be written, in the Gaussian systemof units. In electromagnetism many equations have slightly different forms depending on whether itis intended, as in this book, that mks variables be used or that Gaussian variables be used.Equations in this book can be cast in Gaussian form by replacing the symbols listed below under“rationalized mks” by those listed under “Gaussian.” For example, Eq. 37–26,
B = µ0 (H + M)
becomes Bc
=4
42π
πc
cc
+
. H M
or B = H + 4πM
in Gaussian form. Symbols used in this book that are not listed below remain unchanged. Thequantity c is the speed of light.
Quantity Rationalized mks Gaussian
Permittivity constant ∈0 1/4πPermeability constant µ0 4π/c2
Electric displacement D D/4πMagnetic induction B B/c
Magnetic flux ΦB ΦB/c
Magnetic field strength H cH/4πMagnetization M cM
Magnetic dipole moment µ cµ
In addition to casting the equations in the proper form it is of course necessary to use aconsistent set of units in those equations. Below we list some equivalent quantities in mks andGaussian units. This table can be used to transform units from one system to the other.
Quantity Symbol Mks system Gaussian system
Length l 1 meter 102 cm
Mass m 1 kg 103 gm
Time t 1 sec 1 sec
Force F 1 newton 105 dynes
Work or Energy W, E 1 joule 107 ergs
Power P 1 watt 107 ergs/sec
Charge q 1 coulomb 3 × 109 statcoul
Current i 1 ampere 3 × 109 statamp
Electric field strength E 1 volt/meter13
10 4× − statvolt / cm
Electric potential V 1 volt1
300statvolt
Electric polarization P 1 coul/meter2 3 × 105 statcoul/cm2
714 Mechanics
Electric displacement D 1 coul/meter2 12π × 105 statvolt/cm
Resistance R 1 ohm19
× 10 sec cm11 1− −
Capacitance C 1 farad 9 × 1011 cm
Magnetic flux Φ B 1 weber 108 maxwells
Magnetic induction B 1 tesla ≡ 1 weber/ 104 gauss
meter2
Magnetic field strength H 1 amp-turn/meter 4π × 10– 4 oersted
Magnetization M 1 weber/meter2 1/4π × 10– 3 gauss
Inductance I 1 henry19
10 11× −
All factors of 3 in the above table, apart from exponents, should be replaced by (2.997925 ± 0.000003)for accurate work; this arises from the numerical value of the speed of light. For example the mksunit of capacitance (= 1 farad) is actually 8.98758 × 1011 cm rather than 9 (= 32) × 1011 cm as listedabove. This example also shows that not only units but also the dimensions of physical quantitiesmay differ between the two systems. In the mks system (see Appendix F) the dimensions of capacitanceare M– 1L– 2T2Q2; in the Gaussian system they are simply L, the Gaussian standard unit of capacitancebeing 1 cm.
The student should consult Classical Electromagnetism, p. 611, by J.D. Jackson (John Wiley and Sons,1962) for a fuller treatment of units and dimensions.
