dynamics - kopykitab...dynamics [containing complete unified u.g.c. syllabus for ba/b.sc. (general...

DYNAMICS[Containing complete unified U.G.C. syllabus for BA/B.Sc. (General and

Honours) Part II Mathematics]

(WITH OBJECTIVE TYPE QUESTIONS)For

* Under-graduate, Honours and Post Graduate courses of Mathematics, Physics andEngineering students of all Indian Universities/Institutions.

* Competitive examinations like P.C.S. etc.

Dr. M.D. RAISINGHANIAM.Sc., Ph.D.

Formerly Reader and Head,Department of Mathematics,

S.D. (Post-graduate) College,Muzaffarnagar, U.P.

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PREFACE TO THE FOURTH EDITIONSome important problems have been added and solved in the miscellaneous set of problem which

was introduced in last edition of this book. Some articles have been re-written to bring more clarity inthe discussion of theory. Referenced to the latest papers of various universities have been inserted atproper places.

I hope that these changes will make the book more useful.Suggestions for further improvement of the book will be highly appreciated.

AUTHOR

PREFACE TO THE FIRST EDITIONThis book on dynamics has been specially written to meet the requirements of under-graduate

and Honours students of Mathematics, Physics and Engineering, and those preparing for I.A.S., I.F.S.and other competitive examinations. This book contains complete unified U.G.C. syllabus ofmathematics recommended by the board constituted by the University Grants Commission of India.

The authors possess a very long and rich experience of teaching mathematics and have first handexperience of the problems and difficulties that students generally face.

The salient features of the book are :* The matter has been presented in a simple and lucid language, so that students themselves shall

be able to understand the solutions of the problems.* Each chapter opens with necessary definitions and complete proofs of the standard theorems

and results. These in turn are followed by solved examples which have been classified in varioustypes and methods. This classification will help the students to revise the subject matter at thetime of examination without losing any confidence.

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CONTENTSChapters Pages1. BASIC CONCEPTS OF DYNAMICS 1.1—1.13

1.1. Some basic definitions 1.11.2. Motion in a straight line with uniform acceleration 1.21.3. Vertical motion under gravity 1.21.4. Momentum of a body, force and Newton’s laws of motion 1.41.5. Reaction on the lift when a body of mass m is carried on a lift

moving with an acceleration 1.61.6. Motion of two bodies connected by a string 1.61.7. Composition and resolution of velocities 1.71.8. Relative velocity 1.7

2. WORK, POWER, ENERGY AND IMPULSE. SYSTEM OF PARTICLES AND CONSERVATION PRINCIPLES 2.1—2.16

2.1. Work. Definition 2.12.2. Measurement of work done by a constant force 2.12.3. Measurement of work done when the force is variable and

displacement takes place along its line of action 2.12.4. Measurement of work done when the force is variable and the

point of application moves along a plane curve 2.22.5. Power of an agent. Definition 2.22.6. Units of work 2.22.7. Units of power 2.32.8. Kinetic energy 2.32.9. Principle of work-energy 2.3

2.10. Energy 2.42.11. Conservative and non-conservative forces. Potential energy 2.42.12. Examples of conservative and non-conservative forces 2.42.13. The principle of conservation of energy 2.52.14. Impulse of a force 2.52.15. Impulse-momentum principle for a particle 2.52.16. Units of impulse 2.62.17. Principle of linear momentum for a system of particles 2.62.18. Principle of conservation of linear momentum 2.82.19. Principle of conservation of linear momentum for a dis-continuous motion 2.82.20. Motion of a shot and gun 2.82.21. Law of motion of mass centre 2.92.22. Principle of angular momentum for a system of particles 2.92.23. Principle of angular momentum for motion relative to mass centre 2.10

3. RECTILINEAR MOTION WITH VARIABLE ACCELERATION 3.1—3.183.1. Rectilinear Motion 3.13.2. Motion under inverse square law 3.13.3. Motion of a particle under the attraction of the earth 3.23.4. Rectilinear motion under miscellaneous laws of forces 3.7

4. SIMPLE HARMONIC MOTION (S.H.M.) 4.1—4.384.1. Simple harmonic motion 4.1


4.2. Geometrical representation of S.H.M. 4.24.3. Compounding of two simple harmonic motions of the same period

and in the same straight line 4.34.4. Some useful results 4.34.5. Elastic strings and spiral strings. Hooke’s law 4.124.6. Problems based on particle atached to a horizontal elastic sring 4.124.7. Particle suspended by an elastic string (or vertical elastic spring) 4.174.8. Forced vibrations of a horizontal oscillator 4.304.9. Damped harmonic oscillation (or resisted S.H.M.) 4.32

5. PLANE KINEMATICS 5.1—5.295.1. Cartesian rectangular components of velocity and acceledration

of a particle moving along a plane curve 5.15.2. Angular velocity and angular acceleration 5.45.3. Relation between angular velocity and linear velocity 5.45.4. Angular velocity and angular acceleration of direction of motion 5.55.5. Rate of change of unit vector in a plane 5.95.6. Radial and transverse components of velocity and acceleration 5.105.7. Tangential and normal components of velocity and acceleration 5.21

6. PROJECTILES 6.1—6.296.1. Motion of a particle in a vertical plane 6.16.2. The motion of a projectile and its trajectory 6.16.3. To find the latus rectum, vertex, focus etc. 6.26.4. To find time of flight, horizontal range, greatest height attained 6.36.5. Velocity and direction of motion of a projectile after a given time 6.36.6. Velocity at a point of the trajectory 6.46.7. Some geometrical properties of a parabola 6.46.8. To determine the possible directions in which a particle be

projected to pass through a given point 6.196.9. Range and time of flight up an inclined plane 6.20

6.10. Range and time of flight down an inclined plane 6.217. DIRECT AND OBLIQUE IMPACTS (Collision of elastic bodies) 7.1—7.18

7.1. Direct and oblique impacts 7.17.2. Three fundamental principles of impact 7.1

I. Newton’s experimental law or Newton’s law of restitution 7.1II. Motion of two smooth bodies perpendicular to the line of impact 7.2III. Principle of conservation of momentum 7.2

7.3. Impact on a fixed plane 7.27.4. Direct impact of two spheres 7.87.5. Loss of kinetic energy by direct impact 7.97.6. Oblique impact of two spheres 7.117.7. Loss of kinetic energy by oblique impact 7.12

8. CONSTRAINED MOTION IN VERTICAL AND HORIZONTAL CIRCLES 8.1—8.418.1. Constrained motion 8.1

Part I : Constrained motion in a vertical circle 8.18.2. Motion on the outside of a smooth vertical circle 8.18.3. Motion on the inside of a smooth vertical circle 8.78.4. Motion of the bead on a smooth vertical circular wire or motion


of a particle inside a vertical circular tube 8.88.5. Motion of a particle attached to the end of a string 8.88.6. Simple pendulum 8.208.7. Formulae for gain or loss of beats (time) by a clock 8.218.8. Formulae for gain or loss of beats (time) by a clock when it is

taken to a point above or below the surface of the earth 8.22Part II : Constrained motion in a horizontal circle 8.27

8.9. Motion in a horizontal circle with uniform speed 8.278.10. Conical pendulum 8.298.11. Relative rest inside a rotating sphere 8.348.12. Motion of a bicyclist on a curved path (or nearly circular path) 8.358.13. Motion of a railway carriage (or a motor car) on a curved level track 8.368.14. Motion of a railway carriage (or a motor car) on a curved banked

up track 8.378.15. Lateral thrust of the rails on the wheels of a carriage on a banked

up track 8.37 9. CENTRAL ORBITS 9.1—9.39

9.1. Central force and central orbits 9.19.2. To prove that central orbit is always a plane curve 9.1

9.3A. Differential equation of a central orbit (polar form) 9.19.3B. Differential equation of a central orbit (pedal form) 9.2

9.4. Rate of description of the sectorial area 9.2Areal velocity 9.2

9.5. To show that v = h/p 9.29.6. Energy equation 9.39.7. Different forms of velocity at a point in a central orbit 9.39.8. Escape velocity of a particle under a central force 9.49.9. Elliptic orbit (centre of force being the focus) 9.4

9.10. Working rule to get the law of force when the central orbit is known 9.59.11. Apse, apse line, apsidal distance and apsidal angle 9.129.12. A particle moves with central acceleration µ/(distance), to find the

path and to distinguish the cases 9.139.13. Given the law of force, to find the orbit 9.149.14. Solve examples based on Art. 9.13 9.149.15. Stability of circular orbits 9.309.16. Condition for stable circular orbit 9.319.17. General case of conditions for stability of circular orbit under

central force 9.3110. THE INVERSE SQUARE LAW (PLANETARY MOTION) 10.1—10.20

10.1. Newton’s law of gravitation 10.110.2. Motion under the inverse square law 10.110.3. Kepler’s laws of planetary motion 10.210.4. Deductions from Kepler’s laws 10.210.5. Equivalence between Kepler’s laws for planetary motion and

Newton’s law of gravitation 10.210.6. Necessary modifications of Kepler’s third law 10.410.7. Some useful results of ellipse 10.410.8. Time of description of a given arc of a parabolic orbit 10.14


10.9.Time of description of a given arc of an elliptic orbit 10.1410.10.Time of description of a given arc of a hyperbolic orbit 10.1510.11.Tangential disturbing force. Disturbed elliptic motion 10.1811. CONSTRAINED MOTION IN A PLANE 11.1—11.21

11.1. Introduction 11.111.2. Motion of a heavy particle on a smooth curve in a vertical plane 11.111.3. Cycloid and related results 11.211.4. Motion on a smooth cycloid in a vertical plane 11.211.5. Cycloidal pendulum 11.311.6. Motion under constraint in various curves 11.10

12. MOTION IN A RESISTING MEDIUM. MOTION ON ROUGH CURVES MOTION WHEN THE MASS MOVING VARIES 12.1—12.37

12.1. Motion in resisting medium in a horizontal line 12.112.2. Motion in resisting medium in a vertical line 12.1

Terminal (or limiting) velocity 12.212.3. Motion in a medium whose resistance varies as the velocity 12.212.4. Motion in a medium whose resistance varies as the square of the

velocity 12.512.5. Motion of a projectile in a resisting medium in which resistance

varies as velocity 12.1212.6. Trajectory in a resisting medium whose resistance varies as

square of velocity 12.1712.7. Trajectory in a resisting medium whose resistance varies as nth

power of velocity 12.1712.8. Motion on a rough cycloid 12.1812.9. Motion of a particle on any rough plane curve 12.21

12.10. Solved examples based on motion on a smooth curve underresistance 12.25

12.11. Motion when the mass moving varies 12.2813. MOTION IN THREE DIMENSIONS 13.1—13.18

13.1. Cylindrical (polar) co-ordinates 13.113.2. To find the components of velocity and acceleration of a moving

point in terms of cylindrical co-ordinates 13.113.3. Spherical polar co-ordinates 13.213.4. To find the components of velocity and acceleration of a moving

point in terms of spherical co-ordinates 13.313.5. Method of solving problems based on motion in three dimensions

while using spherical (polar) co-ordinates 13.413.6. Solved examples based on particular case I of Art. 13.5 13.513.7. Solved examples based on particular case II of Art. 13.5 13.11

14. MOMENTS AND PRODUCTS OF INERTIA 14.1—14.4314.1. Moment of inertia (M.I.) 14.114.2. Radius of gyration 14.1

14.3A. Perpendicular axis theorem 14.114.3B. The converse of perpendicular axis theorem 14.1

14.4. Moments of inertia. Some standard cases 14.114.4A. M.I. of a rod 14.114.4B. M.I. of a rectangular lamina 14.214.4C. M.I. of a rectangular parallelopiped (or cuboid) 14.2


14.4D. M.I. of a circular wire 14.314.4E. M.I. of a circular disc 14.314.4F. M.I. of a uniform triangular lamina about one side 14.414.4G. M.I. of a spherical shell (or hollow sphere) about a diameter 14.414.4H. M.I. of a solid sphere about a diameter 14.414.4I. M.I. of an elliptic disc 14.514.4J. M.I. of an ellipsoid 14.5

14.5A. Reference table 14.614.5B. Routh’s rule 14.7

14.6. Product of inertia (P.I.) 14.714.7. Notations 14.714.8. Method of decomposition 14.714.9. Theorem of parallel axis 14.15

14.10. Moment of inertia of a body about a line 14.1814.11. Moments and products of inertia of a plane lamina about a line 14.1814.12. Some useful propositions 14.1814.13. Momental ellipsoid 14.2114.14. Momental ellipse 14.2214.15. Principal axes and principal moments of inertia 14.2514.16. Existence of principal axes 14.2514.17. General method of finding principal axes and principal moments

of inertia 14.2714.18. Equimomental (or kinetically or dynamically) equivalent bodies

or system of bodies 14.3614.19. Necessary and sufficient conditions for the two systems to be

equivalent 14.3614.20. To show that M.I. and P.I. of a triangle about any lines are the

same as the M.I. and P.I. about the same lines, of three particlesplaced at the middle points of the sides, each equal to one thirdmass of the triangle 14.36

15. D’ALEMBERT’S PRINCIPLE. THE GENERAL EQUATIONS OF MOTION 15.1—15.915.1. Introduction 15.115.2. D’Alembert’s principle 15.115.3. Working rule to solve problems by using D’Alembert’s principle 15.215.4. General equations of motion of a rigid body 15.215.5. Motion of the centre of inertia 15.215.6. Motion relative to the centre of inertia 15.315.7. Impulse of an impulsive force 15.815.8. General equations of motion under impulsive forces 15.8

16. MOTION ABOUT A FIXED AXIS 16.1—16.2516.1. Moment of the effective forces about the axis of rotation 16.116.2. Equation of motion of the body about the axis of rotation 16.116.3. Moment of momentum about the axis of rotation 16.216.4. Kinetic energy of a body rotating about a fixed axis 16.216.5. Compound pendulum 16.6

Time of a complete oscillation of a compound pendulum 16.616.6. Simple equivalent pendulum 16.7


16.7. Centre of suspension and centre of oscillation 16.716.8. To show that the centre of suspension and the centre of

oscillation of a compound pendulum are convertible 16.716.9. Minimum time of oscillation of a compound pendulum 16.8

16.10. Reaction of the axis of rotation 16.1516.11. Motion about a fixed axis. Impulsive forces 16.1916.12. Centre of percussion 16.2116.13. Centre of perecussion of a rod 16.2116.14. Working rule to find the centre of percussion of a body for a

fixed axis. General case 16.22

17. SOME ASPECTS OF RIGID BODY DYNAMICS 17.1—17.1217.1. Kinetic energy and angular momentum 17.117.2. Kinetic energy of a rigid body with a fixed point 17.217.3. Kinetic energy of a rigid body in general 17.317.4. Angular momentum of a particle and of a system of particles 17.517.5. To find an expression for angular momentum of a rigid body

turning about a fixed point and fixed axes 17.517.6. Angular momentum of a rigid body in general 17.617.7. Equations of motion of a rigid body in two dimensions 17.10

Miscellaneous problems and results on the entire book M.1–M.51Index I.1–I.3

New-Unified U.G.C. Syllabus of Dynamics for B.A./B.Sc. (General and Honours) Part II Mathematics.

Topics Chapter (s)covering the Topic

(i) Velocities and accelerations along radial and 5transverse directions, and along trangential andnormal directions

(ii) Simple harmonic motion. Elastic strings 3 and 4(iii) Motion on smooth plane curves 8 and 11(iv) Motion on rough plane curves 12(v) Motion in a resisting medium 12

(vi) Motion of particles of varying mass 12(vii) Central orbits 9

(viii) Kepler’s laws of motion 10(ix) Motion of a particle in three dimensions. 13

Acceleration in terms of different co-ordinatesystems

Reference : S.L. Loney. An elementary treatise on the dynamics of a particle and of rigid bodies,Cambridge University Press, 1956.


1Basic Concepts of Dynamics

1.1 Some basic definitions1. Dynamics. It is that branch of mechanics which deals with the motion of bodies under the

action of given forces.2. Particle. A particle is a portion of matter which is indefinitely small in size i.e., which for the

purpose of our investigation is so small that the distance between its different parts may be neglected.It may be treated as a mathematical point endowed with mass.

3. Body. A portion of matter occupying finite space, i.e., limited in every direction is called abody.

Rigid body : A rigid body is defined to be a collection of particles, the distance between any twoof which remains invariant

4. Rest and Motion. A particle is said to be at rest if it does not change its position relative toits surroundings with time. If it changes its position with time, it is said to be in motion.

5. Path. The path of a particle is the curve drawn through the successive positions of the pointduring motion. If the path is a straight line, then the motion is called rectilinear and if the path is acurve, then the motion is said to be curvilinear.

6. Position. The position of a particle at a given time is determined by the point occupied by theparticle on its path.

7. Displacement. The change of position of a particle in the given interval of time is known asits displacement in that interval. The displacement of a particle from P to Q will be denoted by thesymbol .PQ

Note that displacement is a vector quality..

8. Velocity. The rate change of displacement with respect to time is known as velocity. It is avector quantity because it has both magnitude and direction.

9. Speed. The mangnitude of velocity is called speed.10. Average speed. Average speed of a particle in a given interval of time is given by the ratio of

the total distance travelled by it to the total time taken.Thus, average speed = (Total distance)/(Total time)11. Uniform velocity. A particle is said to move with uniform velocity if it moves in a constant

direction and covers equal distances in equal intervals of time, however small these intervals may be.12. Acceleration and retardation. The rate of change of velocity with respect to time is called

acceleration. It is a vector quantity as it has both magnitude and direction. If a particle moves suchthat its velocity decreases, then the acceleration is negative and is called retardation. Clearly, thevelocity of a particle is uniform when its acceleration is zero.

13. Uniform acceleration. When velocity of a particle moving in a straight line changes byequal amounts in equal intervals of time, however small, the particle is said to move with uniformacceleration.

14. Expression for velocity at a point. Let a particle move along the line OX, where O is origin.Let P be the position of the particle at any time t such that OP = s. Then,

v = velocity at time t = ds/dtThe velocity at any time t is positive or negative according as

the particle is moving towards right (i.e., away from origin O) ortowards left (i.e., towards the origin O).

Expressions for acceleration at a point. Let a particle move along the line OX, where O is origin.1.1


1.2 Dynamics

Let P be the position of the particle at any time t and let v be its velocity at P. Then, if f is theacceleration of the particle at any time t at P, then

f =2

2, and d d d sf fdt ds dtv v

v

Instantaneous rest. A particle is said to be at instantaneous rest at any time if and only if v = 0but f 0.1.2 Motion in a straight line with constant acceleration

A particle moves along a straight line from a fixed point in it with initial velocity u and constantacceleration f, in its direction of motion. Let s be the distance of the particle from the origin at timet and v be the velocity attained at that time. Then

(i) v = u + f t (ii) s = ut + (1/2) f t2 (iii) v2 = u2 + 2fs.[Note : Replace f be f if f is retardation in any problem].

(iv) Distance travelled in the nth second = u + (1/2) (2n 1) f.(v) The average velocity is the mean of the initial and the final velocities and is equal to the

velocity at the middle of the interval i.e.,average velocity = (u + v)/2 = u + f . (t/2)

ILLUSTRATIVE SOLVED EXAMPLESEx. 1. If a body starting from rest, moving with uniform acceleration describes 1000 cm in 10

seconds, then the acceleration with which it moves will be(a) 20 cm/sec2, (b) 25 cm/sec2, (c) 30 cm/sec2, (d) 35 cm/sec2. [I.A.S. (Prel) 1993]Sol. Ans. (a) s = ut + (1/2)f t2 1000 = 0 + (1/2) f 102 f = 20Ex. 2. A train starts from a station with constant acceleration for 2 minutes and attains a

constant speed. It then runs for 11 minutes at this speed and retards uniformly during the next 3minutes and stops at the next station which is 9 km off. The maximum speed (in km/hour) attainedby the train is (a) 30 (b) 35 (c) 40 (d) 45. [I.A.S. (Prel.) 1995]

Sol. Ans. (c). Let the particle move from A to B withuniform acceleration f1 for 2 minutes. Then, using v = u +ft and s = ut + (1/2) ft2, we get V = f1 2 ands1 = 0 + (1/2)f1 22 so that s1 = V. Motion from B to C iswith uniform max. velocity V, so s2 = 11V. Next, for motion from C to D with uniform retardation f2 for3 minutes, using v = u ft and s = ut (1/2)f t2, we get 0 = V 3f2 and s3 = 3V (1/2) f2 32 so that s3= (3/2) V. Now, 9 = AD = s1 + s2 + s3 = V + 11V + (3/2)V = (27/2)V so that

V = (2/3) km/minute = (2/3) 60 km/hour = 40 km/hour.Ex. 3. If a particle starting with a velocity u and subject to a uniform acceleration ‘a’ travels for

‘n’ seconds, then the distance travelled by it in nth second is(a) u + (a/2) (2n – 1), (b) u + (a/2)(2 n), (c) u + (n/2) (2a – 1), (d) u + (n/2) (2a + 1).

[I.A.S. (Prel.) 98]Sol. Ans. (a). Refer result (iv) of Art 1.2. Here f = a.Ex. 4. A bus starts from rest with an acceleration of 1 m/s2. A man, who is 48 m behind the bus,

starts running towards it with a uniform velocity of 10 m/s. He will be able to catch the bus in(a) 6 s (b) 7 s (c) 8 s (d) 9s. [I.A.S. (Prel.) 2002]Sol. Ans. (c). Let M be the position of the man when the bus starts

from B. Let the man be able to catch the bus at C after t sec. Then, forman, MC = 10t and for bus, BC = 0 + (1/2) 1 t2. Form figure MC = MB+ BC so that 10t = 48 + t2/2 or t2 – 20t + 96 = 0 or t = 8 or 12. Requiredtime is t = 8 when for the first time man catches bus at C. [Note that t = 12 corresponds to time whenman and bus will be together at a later stage of further motion].1.3 Vertical motion under gravity. When a body is let fall in vacuum towards the earth, it will movevertically downwards with an acceleration g (known as acceleration due to gravity). Its value in F.P.S.

/ 2dx dt at b


Basic Concepts of Dynamics 1.3

system is 32 ft/sec2, in C.G.S. system is 981 cm/sec2 and in M.K.S. system is 9.8 m/sec2.Downward motion. If a particle is projected vertically downwards from a point A with initial

velocity u and h be the distance of the particle from A at time t and v be the velocity attained at thattime, then

(i) v = u + gt (ii) h = ut + (1/2) gt2 (iii) v2 = u2 + 2gh(iv) Distance travelled in the nth sec = u + (g/2) (2n 1).In particular, if the body starts from rest or is simply let fall or dropped so that u = 0, then the

above formulas become (i) v = gt, (ii) h = (1/2) gt2, (iii) v2 = 2gh or v = (2gh)1/2, (iv) distance travelledin the nth sec. = (g/2) (2n 1).

Upward motion. If a particle is projected vertically upwards from O with velocity u and h bethe distance of the particle from O at time t and v be the velocity attained at that time, then

(i) v = u gt (ii) h = ut (1/2) gt2 (iii) v2 = u2 2gh(iv) Greatest height attained during upward motion = u2/2g(v) Time to reach the greatest height = u/g

(vi) Time of flight (i.e., the total time taken by the particle to reach the greatest height and thento return to the starting point again) = 2u/g.

(vii) Time of ascent = time of descent.ILLUSTRATIVE SOLVED EXAMPLES

Ex. 1. If a particle moves along a straight line according to the law 2 2 2s at bt c where sis the displacement at any instant, then how does the acceleration vary with displacement ?

(a) as 1/s (b) as 1/s2 (c) as 1/s3 (d) as 1/s4 [I.A.S. (Prel.) 2005]Sol. Ans. (c). We have 2 1/ 2( 2 )s at bt c ... (1)From (1) , v = ds/dt = (at + b) / 2 1/ 2( 2 )at bt c ... (2)

acceleration = f = 2 1 / 2 2 2 1/ 2

2( 2 ) ( ) ( 2 )

2a at bt c at b at bt cd

dt at bt c

v

2 2 2

2 3/ 2 3( 2 ) ( )

,( 2 )

a at bt c at b ac bat bt c s

using (1)

Ex. 2. Two particles of m1 and m2 gms are projected vertically upwards such that the velocity ofprojection of m1 in double that of m2. If the maximum height to which m1 and m2 rise be h1 and h2respectively, then (a) h1 = 2h2 (b) 2h1 = h2 (c) h1 = 4h2 (d) 4h1 = h2. [I.A.S. (Prel.) 1996]

Sol. Ans. (c). Let velocity of m2 be u. Then by problem the velocity of m1 is 2u. Now, byproblem h1 = (2u)2/2g and h2 = u2/2g so that h1 = 4 (u2/2g) = 4h2.

Ex. 3. A particle is thrown vertically upwards with a velocity 30 m/sec. Find its greatest height.[Pune 2010]

Sol. Greatest height 2 30 30 45.91m

2 2 9.8u

g

Ex. 4. If 2x at bt c , find velocity when t = 2 [Agra 2009]

Sol. By definition, velocity = / 2dx dt at b Hence, when t = 2, the required velocity = 4a + bEx. 5. The velocity of a jet projected vertically upwards to reach a height of 4 km is

(g = 9.8 m/sec2) (a) 28.0 km/s (b) 2.8 km/s (c) 0.28 km/s (d) 0.028 km/s.[I.A.S. (Prel.) 2001]

Sol. Ans. (c). Max. height H = u2/2g u2 = 2gH = 2 9.8 4 1000 so that u = 280 m/sec= (280/1000) km/sec. = 0.28 km/sec.

Ex. 6. An object was thrown vertically downwards with the initial velocity v0. If during the fifthsecond of its fall, it travels (3/2) times the distance it had travelled during the third second then thevalue of v0 is given by (a) 2g m/sec (b) (3g/2) m/sec (c) g m/sec (d) g/2 m/sec.

[I.A.S. Prel. 2002]

/ 2dx dt at b


1.4 Dynamics

Sol. Ans. (b). Since distance travelled in nth sec = u + (g/2) (2n – 1), here v0 + (g/2) (2 5 1) =(3/2) [v0 + (g/2) (2 3 1)] so v0 = (3/2) g.

Ex. 7. A stone falling vertically from rest travels half of its total path in the last second of its fall.The total time of its fall is (a) less than 2 seconds (b) equal to 2 seconds (c) greater than 2 secondsbut less than 3 seconds (d) greater than 3 seconds. [I.A.S. Prel. 2003]

Sol. Ans. (c). Let total time of fall be n seconds then by problem, (g/2) (2n 1) = (1/2) (1/2)gn2 so that n2 4n + 2 = 0 giving n = 1 2 and n = 1 2. But (1 2) gives –ve value of time.So reject this value. Then the required time = 1 2 which is > 2 seconds but < 3 seconds.1.4 Momentum of a body, force and Newton’s laws of motion.

Momentum of a body. If m be the mass of the body moving with velocity v, then the momentumof body = mv. It is a vector quantity.

Force. Force is an action which changes or tends to change the state of rest or of uniform motionof a body in a straight line.

Impressed forces. The external forces acting upon a body are known as the impressed forces.Internal forces. The mutual forces of action and reaction are known as internal forces.Newtons laws of motion. They are stated as follows :First law. Every body (or particle) continues in its state of rest or of uniform motion in a

straight line unless it is compelled by impressed forces to change that state.Second law. The rate of change of momentum is directly proportional to the impressed force

and takes place in the direction of the force.Third law. To every action, there is an equal and opposite reaction.Same useful results and definitions(i) Inertia and law of inertia. A body has no tendency of itself to change its state of rest or

motion if it is kept free from the action of external forces. The inability of a body to change itsposition by itself is known as inertia and hence first law is also known as the law of inertia.

(ii) Equation of motion or fundamental equation of dynamics is given by P = mf, when P isthe impressed force acting on a body (or particle) of mass m and f is its acceleration. Formula P = mfis remembered as follows :

Force causing motion in absolute units = mass accelerationIn vector notation P = mf is written as P = mf.(iii) Deduction of first law from second law. [Pune 2010]From Newton’s law II, d(mv)/dt = kP, ...(1)

where m is the mass of the particle, v is the velocity, P is the external force and k is a constant. If thereis no force acting on the body, then P = 0 and so (1) gives

d(mv)/dt = 0 so that mv = c or v = c/m = u, a constant. ...(2)where c is constant of integration. Now (2) shows that the velocity of the particle is constant andhence the particle continues to move in a straight line with constant speed. Again, if u = o, then (2)shows that v = o i.e., the particle continues to be at rest. This is nothing but a statement of the first law.The first law of motion can thus be deduced from the second law.

(iv) The physical independence of forces. Suppose a force P1 acting on a particle of mass m,produces an acceleration f1 in it. Similarly if a force P2 acting on the same particle, produces anacceleration f2 in it. Then,

mf1 = P1 and mf2 = P2On adding these, m (f1 + f2) = P1 + P2,

showing that a force P1 + P2 acting on the particle produces an accelerationf1 + f2. Hence we see that the additional force P2 produces an additional acceleration f2, as it would doif acting alone on the particle. This result is known as the principle of physical independence offorces.



(v) Unit of forcePoundal. It is the absolute unit of force in the F.P.S. (Foot-Poundal-Second) system. Thus, a

poundal is the force which produces an acceleration of 1 ft/sec2 in a mass of one pound.Dyne. It is the absolute unit of force in the C.G.S. (Centimeter-Gramme-Second) system. Thus, a

dyne is the force which produces an acceleration 1 cm/sec2 in a mass of one gramme.Newton. It is the absolute units of force in the M.K.S. (Meter-Kilogram-Second) system. Thus,

a Newton is the force which produces an acceleration of 1 m/sec2 in a mass of one kilogram. ANewton in briefly written as N.

Note 1. One Newton = 105 dynesNote 2. The above units of force are called absolute units because their values are the same

everywhere and do not depends on the g, the acceleration due to gravity.Gravitational units of force. Units of force which depend on the value of g are called gravitational

units of force. In this system in F.P.S., it is called a pound weight or lb. wt, in C.G.S., it is called gramweight or gm. wt and in M.K.S., it is called kilogram weight or kg. wt.

Connection between absolute and gravitational units of force1 lb. wt = g poundals = 32 poundals

1 gm. wt = g dynes = 981 dynes1 kg. wt = g Newtons = 9.8 Newtons = 9.8 N.

Note. While using the formula P = mf, P is always measured in absolute units, i.e., in poundalsor dynes or Newtons.

ILLUSTRATIVE SOLVED EXAMPLESEx. 1. Which one of the following statements is correct ? (a) Newton’s three laws of motion are

independent (b) first law of motion is contained in the second law as a special case (c) second law ofmotion can be deduced from the third law (d) third law of motion can be deduced from the secondlaw. [I.A.S. (Prel.) 1997]

Sol. Ans. (b) Refer result (iii) of Art 1.4.Ex. 2. Newton’s second law of motion is given by (a) velocity = mass/force (b) acceleration =

moving mass/force, (c) acceleration = moving force/mass (d) None of the above. [I.A.S. (Prel.) 1997]Sol. Ans. (c) Since P = mf so f = P/m.Ex. 3. Newton’s third law says (a) for every action there is an unequal and opposite reaction (b)

to every action there is an equal and opposite reaction (c) to every action there is an equal and samereaction (d) None of the above. [I.A.S. (Prel.) 1998]

Sol. Ans. (b). Refer Art 1.4, third law.Ex. 4. If a body of mass M kg at rest is acted upon by a constant force of W kg.wt, then in T

seconds it moves through a distance of (a) gTW/2M meters (b) gTW2/2M meters (c) g2TW/2M meters(d) gT2W/2M metres. [I.A.S. (Prel.) 1996]

Sol. Ans. (d). From P = mf, Wg = Mf so that f = Wg/M. Using s = ut + (1/2) ft2 we get, s = 0 + (1/2) (Wg/M)T2 = gT2W/2M.

Ex. 5. If a body of (1/2) kg is subjected to a force (i + 2j + k) N and if the body is initially at rest,then its velocity after 4 second will be (a) 2 6 m/sec (b) 4 6 m/sec (c) 8 6 m/sec(d) 16 6 m/sec. [I.A.S. (Prel.) 1998]

Sol. Ans. (c). Magnitude of force = | i + 2j + k | = (l2 + 22 + l2)1/2 = 6 Newton. Now P = mf 6 = (1/2) f f = 2 6 m/sec2. Then by,, v = u + ft, v = 0 (2 6) 4 8 6 m/sec.

Ex. 6. A body of mass 50 gm is acted upon by a constant force F = 100 dynes. The time requiredto move the body through a distance 25 cm from rest is (a) 10 s, (b) 5.5 s, (c) 5 sec, (d) 3.5 sec.

[I.A.S. (Prel.) 2001]Sol. Ans. (c). P = mf 100 = 50 f f = 2. Then

s = ut + (1/2) ft2 25 = 0 + (1/2) 2 t2 t = 5 sec.


1.6 Dynamics

Ex. 7. The magnitude of a force which is acting on a body of mass 1 kg for 5 sec. to producevelocity of 1 m/sec in it, is (a) 1000 dynes, (b) 2000 dynes, (c) 10,000 dynes, (d) 20,000 dynes.

[I.A.S. (Prel.) 1999]Sol. Ans. (d). v = u + ft 100 = f 5 f = 20. P = mf = 1000 20 = 20,000 dynes.

1.5 Reaction on the lift when a body of mass m is carried on a lift moving with an acceleration f

Fig. (i) Fig. (ii)

Case I Lift moving up Case II. Lift moving downR = reaction = m (g + f) R = reaction = m (g f)

Note 1. The reaction does not depend upon the velocity but it depends on the acceleration.When f = 0, R = mg i.e., when the lift is at rest or moves with constant velocity, the reaction is equalto the weight of the body.

Note 2. In case II, if f = g, then R = 0, i.e., reaction vanishes and the contact becomes loose.Again, in case II, if f > g, R is negative. In this case the body is left behind and moves down freelyunder gravity.

Note 3. The pressure on the body is called the apparent weight of the body. Clearly, in case I,the apparent weight > actual weight whereas in case II, the apparent weight < actual weight.

ILLUSTRATIVE SOLVED EXAMPLESEx. 1. A person weighing 80 kg is standing on a lift. The lift moves upwards with a uniform

acceleration 4.9 m/s2. The apparent weight (in kg) of the person is (a) 160 (b) 120 (c) 80 (d) 40.[I.A.S. (Prel.) 1994]

Sol. Ans. (b). R = m (g + f) = 80 (4.9 + 9.8) = 120 g = 120 kg-wt.Ex. 2. If a body of mass m kg is carried by a lift moving with upward acceleration f, then the

pressure on the plane of the lift is(a) mf mg (b) mg mf (c) mg + mf (d) (mg) (mf). [I.A.S. (Prel.) 1999]Sol. Ans. (c). Refer case I of Art 1.5. Pressure = Reaction.Ex. 3. A person having a mass of 98 kg is descending in a lift with an acceleration of 2 m/s2. The

thrust of his feet on the lift while descending, is nearly equal to (take g = 9.8 m/s2) (a) 76.4 106

dynes, (b) 96.0 106 dynes, (c) 115.6 166 dynes (d) 196 106 dynes. [I.A.S. (Prel.) 2001]Sol. Ans. (a). By case II of Art. 1.5, Thrust = Reaction = m (g f) = 98 (9.8 2) = 98 7.8 N = 98

7.8 105 dynes = 76.4 106 dynes (nearly).1.6 Motion of two bodies connected by a string

Case I. Two masses m1 and m2 (m1 > m2) are suspended by a light inextensibleand flexible string over a smooth, fixed, small, light pulley (or a peg). Then,

(i) Acceleration = f = (m1 m2)g/(m1 + m2)(ii) Tension in string = T = 2m1m2g/(m1 + m2)

(iii) Pressure on pulley = 2T = 4m1m2g/(m1 + m2)Case II. Two masses m1, m2 are connected by an inelastic string; m2 is

placed on a horizontal table and the string passes over a light smooth pulleyat the edge of the table and m1 is hanging freely. Then,

(i) Acceleration = f = m1g/(m1 + m2)(ii) Tension in string = T = m1m2g/(m1 + m2)

(iii) Pressure on pulley = 2T = m1m2g 2 /(m1 + m2)



Example. Two masses of 5 kg and 9 kg are fastened to ends of a cord passing over a frictionlesspulley. The acceleration of resulting motion is

(a) 2.8 m/sec2 (b) 2.5 m/sec2 (c) 5.6 m/sec2 (d) 5.0 m/sec2. [I.A.S. (Prel.) 1993]Sol. Ans. (a). From case I of Art 1.6, f = [(9 5) 9.8]/(9 + 5) = 2.8 m/sec2.

1.7 Composition and resolution of velocitiesResultant velocity and components of velocity. When a particle possesses a number of velocities,

then the single velocity which is equivalent to the given velocities is called their resultant. The givenvelocities are known as the components of their resultant.

Resolved parts of a velocity. If a velocity be decomposed into two components at right angles toeach other, the components are known as the resolved parts of the velocity.

Parallelogram law of velocities. If a moving point possessessimultaneously two velocities, which are represented in magnitudeand direction by the sides of a parallelogram, drawn from a point,then the resultant velocity is represented in magnitude and directionby the diagonal of the parallelogram passing through that point.

Let the two velocities u and v acting at an angle be representedin magnitude, direction and sense by the two side OA and OB of the parallelogram OACB, then theirresultant velocity V is represented by the diagonal OC. Let OC make an angle with OA. Then wehave

V = (u2 + v2 + 2uv cos )1/2, tan = (u sin )/(u + v cos ) ...(1)Some particular cases of composition of velocities.Case I. When u and v are perpendicular to each other, then = 90°(1) V = (u2 + v2)1/2 and tan = u/v.Case II. When u and v are in the same direction, then = 0.(1) V = u + v and = 0.Case III. When u and v are in the opposite directions, then = .If u > v, (1) V = u v and V has the direction of u.If v > u, (1) V = v u and V has the direction v.Case IV. If v = u. Then (1) V = 2u cos (/2) and = /2. Thus, the direction of resultant

velocity bisects the angle between the two component velocities.Components u and v of a velocity V in the directions making angles and

with it on either side.u = (V sin )/sin ( + ), v = (V sin )/sin ( + ).

Resolved parts u and v of a velocity V in two directions at right angles to eachother, when u makes an angle with V.

u = V cos and v = V sin Note. The sum of the resolved parts of a number of velocities in a given direction

is equal to the resolved part of their resultant in the same direction.1.8 Relative velocity. Relative velocity of B with respect to a moving point A is obtained as theresultant of the absolute (actual) velocity of B and the reversed absolute (actual) velocity of A.

In vector notation, if u and v are the absolute velocity vectors of two points A and B respectively,then the relative velocity of B relative to A, is v u.

Ex. 1. The velocity of a boat relative to water is 2i + 2j and that of water relative to earth isi – 2j, where i, j represent velocities of one km an hour due to east and north respectively. Thevelocity of boat relative to earth is (a) 3 km an hour due east, (b) 2 21/2 + 51/2 km an hour at 45°north of east, (c) (17)1/2 km an hour at tan1 4 north of east, (d) 171/2 km an hour at tan1 4 duenorth.

[I.A.S. (Prel.) 2003]


1.8 Dynamics

Sol. Ans. (a). Let u, v and w be absolute velocity of boat, water and earth respectively. Then,given that

u – v = 2i + 2j and v – w = i – 2jAdding these, (u – v) + (v – w) = 2i + 2j + i – 2j Velocity of boat relative to earth = u – w = 3i.Ex. 2. To a man walking at 4 km per hour along a road due west the wind appears to blow

from the south while to a cyclist travelling in the same direction at 8 km per hour it appears tocome from the south-west. What is the true direction and velocity of the wind? [I.F.S. 1998]

Sol. Let v be the actual velocity of wind and let it blow in a direction making an angle southof west.

N

O E

S

W 4

v

Fig. (i)

N

OE

S

W8

v

Fig. (ii)

45°

135° u

A

Situation I. v is the resultant of the velocity (= 4 km/hr) of the person along OW and theapparent velocity of wind along OS (see figure (i)). Resolving velocities along OW, we have

cos 4 v ... (1)Situation II. In this case v is the resultant of the velocity (= 8 km/hr) of the person along OW

and apparent velocity (u, say) of wind along OA making 45° with OE (see figure (ii)). Resolvingvelocities along OW and ON, we have

cos 8 cos135 8 / 2u u v and sin sin135 / 2u u vIn order to eliminate u, adding the above equations, we get

cos sin 8 v v or sin 4, v using (1) ... (2)

Squaring and adding (1) and (2), we get 2 2 16 v 4 2v km/hrDividing (2) by (1), we get tan = 1 = 45°

Hence wind blows with velocity 4 2 km/hr from north-west.MISCELLANEOUS PROBLEMS

Ex. 1. The distance travelled by a particle, falling freely from rest in first, second and thirdseconds are in the ratio(a) 1 : 2 : 3 (b) 1 : 3 : 5 (c) 9 : 4 : 1 (d) 1 : 4 : 9. [M.S. Univ. T.N. 2007]

Sol. Ans (b). We know that, distance travelled in the nth sec = u + (g/2) × (2n –1)Here u = 0 so the ratio of distances travelled in 1st, 2nd and 3rd seconds

21 11 : (2 2 1)2 2

g g : 1 (2 3 1) 1: 3 : 52

g

Ex.2. Two balls are projected simultaneously with the same velocity from the top of a tower,one vertically upwards and the other vertically downwards. If they reach the ground in time t1and t2, then the height of the tower is (1/2) × g t1 t2 (Garhwal 2010)

Sol. Let u be the common velocity of projection for both the balls and h the height of thetower. Then for the first ball, and the second ball, we have

21 1(1/ 2)h ut gt ... (1)



and 22 2(1/ 2)h ut gt ... (2)

Multiplying (1) by t2 and (2) by t1 and then adding, we geth(t1 + t2) = (1/2) × g t1 t2 (t1 + t2) or h = (1/2) × g t1 t2

Ex. 3. Two motor cars are moving along two roads perpendicular to each other, towardspoint of intersection of two roads. Their velocities at a particular time are v1 and v2 respectivelywhile their distances from the crossing are s1 and s2 respectively. If the accelerations of the twocars be f1 and f2 respectively, then they shall avoid collision, if

(a) 21 2 2 1 2 1 1 2 2 1 1 2( ) 2( ) ( )s f s f f f s s v v v v (b) 2

1 2 2 1 1 2 2 1 2 1 1 2( ) 2( )( )s s f s f s f f v v v v

(c) 2 1 1 2 1 2 2 1 1 2 2 1( ) 2( ) ( )f f s s f s f s v v v v (d) 21 1 2 2 1 2 2 1 2 1 1 2( ) ( ) ( ) 0s s f s f s f f v v v v

[I.A.S. 2004]Sol. Ans. (a) If possible, let the two cars collide after time t at the point of intersection. Then,

using s = ut + (1/2) × ft2, for both the cars we have2

1 1 1(1/ 2)s t f t v or 21 1 12 2 0f t t s v ... (1)

and 22 2 (1/ 2)s t f t v or 2

2 2 22 2 0f t t s v ... (2)Solving (1) and (2) for t2 and t by cross-multiplication, we get

2

1 2 2 1 1 2 2 1 1 2 2 1

14( ) 2( ) 2( )

t ts s s f s f f f

v v v v

2 2 21 2 2 1 1 2 2 1 2 1 1 2 1 2 2 1( ) / 2( ) ( ) /( )t s s f f s f s f f f v v v v v v

21 2 2 1 2 1 1 2 2 1 1 2( ) ( ) ( )s f s f f f s s v v v v ... (3)

Hence to avoid collision, the condition is given by2

1 2 2 1 2 1 1 2 2 1 1 2( ) ( ) ( )s f s f f f s s v v v v

Ex. 4. If time t be regarded as a function of velocity v, prove that the rate of decrease of

acceleration is given by 3 2 2( / ),f d t d v f being the acceleration.

[Kanpur 2006, 07; Rohilkhand 2001]Sol. Let t = f (v), where f is an arbitrary function ... (1)Differentiating (1) w.r.t. ‘t’, we get 1 ( ) ( / )d dt v v

Acceleration = / 1/ ( )f d dt v v ... (2)

Differentiating (2) w.r.t. ‘t’, we get 21 ( )

{ ( )}df ddt dt

vv

v

or 32

( ) ( ),{ ( )}

df dv fdt dt

vv

v using (2) ... (3)

Now, (2) / ( )dt dv v 2 2/ ( )d t dv v ... (4) From (3) and (4), df/dt = – f 3 × (d2t/dv2),

which is negative. Hence the rate of decrease of acceleration is f 3 × (d2t/dv2).

Ex. 5. A point moves with uniform acceleation. If 1 2 3, ,v v v be average velocities in threesuccessive intervals of time t1, t2, t3, prove that (v1 – v2)/(v2 – v3) = (t1 + t2)/(t2 + t3).

[Pune 2010; Rohilkhand 2003]


Dynamics

Publisher : SChand Publishing ISBN : 9788121926492 Author : M D Raisinghania

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