Introduction

1. k1 Introduction Mechanics is that branch of science which deals with the state of rest or motion of bodies under the action of forces. The subject of Mechanics is logically divided into two parts.Statics which concerns the equilibrium of bodies under the action of forces,and Dynamics concerns the motion of Bodies. Dynamics is again subdivided into a. Kinematics b. Kinetics

2. k2 Study of Kinematics concerns with the motion of a body, without referring to the forces causing the motion of that body. Study of Kinetics concerns with the motion of the body considering the forces causing the motion. Terms and definitions space: is the geometric region occupied by bodies whose positions are described by linear and angular measurements relative to a coordinate system.

3. k3 Time is the measure of succession of events and is basic quantity in dynamics. Mass is a measure of the inertia of a body, which is its resistance to a change of velocity Particle. A body of negligible dimensions is called a particle. In the mathematical sense a particle is a body whose dimensions approach to zero so that it may be analyzed as point mass. Rigid body. A body is considered rigid when the relative movements between its parts are negligible for the purpose at hand.

4. k4 Newton’s Laws of Motion First Law. A particle remains at rest or continues to move in a straight line with a uniform velocity if there is no unbalanced force acting on it. Second Law. The acceleration of a particle is proportional to the resultant force acting on it and is in the direction of this force. Third Law. The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear.

5. k5  Newton’s second law forms the basis for most of the analysis in dynamics. As applied to a particle of mass m, it may be stated as F=ma Where F is the resultant force acting on the particle and a is the resulting acceleration. This equation is a vector equation. Rectilinear Motion: It is the motion of a particle along a straight line.

6. Rectilinearmotion with uniform acceleration: k6 O X M N ∆s S t ∆t   A particle moves along a straight line OX as shown in the figure. The moving particle is in position M at any time ‘t’ and it covers a distance ‘s’ from ‘O’. The particle moves to N, through a distance ∆s in a small interval of time ∆t. The velocity v at the instant when the particle is at certain point M, at time ‘t’ is the rate of change of displacement ∆s as the increment of time ∆t approaches to zero as limit is known INSTANTANEOUS VELOCITY and is given by   Lt v = ∆t  0 ∆s/∆t = dS/dt

7. k7 Average Velocity : It is the uniform velocity with which the particle may be considered to be moving in order to cover the total distance s in a total time t. (i) When the particle moves with uniform velocity   s = u.t (ii) When particle moves with variable velocity

8. k8 (iii) When particle moves with initial velocity u and constant acceleration a, its velocity changes to v, then vav = (u+v)/2 s = (u+v)/2 x t s = distance covered in time ‘t’ (iv) If the distances moved by the particle from start are s1 in t1, s2 in t2, then average velocity may also be found by vavg = (s2 - s1) / (t2-t1)

9. k9 Acceleration : If the velocity of a particle is v at M and it changes by ∆v, in a small interval of time ∆t then the acceleration of moving particle, a at the instant at which particle is at M i. e. the instantaneous acceleration is given by A particle may move in a straight line with constant acceleration or with variable acceleration.

10. k10 Uniform Acceleration : If the velocity of a body changes by equal amounts in equal intervals of time, the body is said to move with uniform acceleration. Variable Acceleration: If the velocity of a body changes by unequal amounts in equal intervals of time, the body is said to move with variable acceleration. Note: When the velocity is increasing the acceleration is reckoned as positive, when decreasing as negative (retardation or deceleration) .

11. k11 Displacement – Time Variations: Displacement Time Fig. 1 In fig (1)The graph is parallel to the time axis indicating that the displacement is not changing with time. The slope of the graph is zero. The body has no velocity and is at rest.

12. k 12 ∆x Displacement ∆t Time Fig.2 In fig (2)The displacement increases linearly with time. The displacement increases by equal amount in equal intervals of time. The slope of the graph is constant.

13. k13 ∆x2 Displacement ∆t2 ∆x1 ∆t1 Time Fig. 3 In fig(3) The displacement is not changing by equal amounts in equal intervals of time. The slope of the graph is different at different times.The velocity of the body is changing with time. The motion of the body is accelerated.

14. k14 Velocity – Time Variations: ∆v Constant velocity ∆t velocity O Time Fig. (a) In fig. (a): The velocity of the body increases linearly with time. The slope of the graph is constant i.e. the velocity changes by equal amounts in equal intervals of time. (acceleration of the body is constant and at t = 0, the velocity is finite. Thus the body moving with a finite initial velocity, and has constant acceleration).

15. k15 Velocity – Time Variations: Uniform acceleration velocity ∆v ∆t O Time Fig. (b) In fig. (b): The body has a finite initial velocity. As time passes, the velocity decreases linearly with time until its final velocity becomes zero i.e it comes to rest. Thus the body at a constant deceleration , since the slope of the graph is negative.

16. k16 Variable acceleration ∆v2 velocity ∆t2 ∆v1 ∆t1 O Time Fig. (c) In fig. (c): The velocity –Time graph is a curve. The slope is, therefore, different at different times. In other words, the velocity is not changing at constant rate. The body does not have a uniform acceleration since acceleration is changing with time.

17. k17 Equations of Motion Under Uniform Acceleration  1. Equation of motion (Relation between v,u, a & t)  If we assume a body starts with an initial velocity u and uniform acceleration a. After time t, it attains a velocity v. Therefore the change in velocity in t seconds is (v – u) Change in velocity / sec. = v – u / t = a v = u + at -----(1)

18. k18 2. Equation of Motion: (Relation between s, u, a and t)  Let a body moving with an initial uniform velocity u be accelerated with a uniform acceleration afor time t. If v is the final velocity, the distance s which the body travels in time t is determined as follows. Now since acceleration is uniform it is obvious that the average velocity = (u + v) /2  Distance traveled = vav x t = (u + v)/2 x t = (u + (u + at))/2 x t (Substituted from 1) s = ut + ½ at2-----(2)

19. k19 3. Equation of motion: (Relation between u, v, a and s) s = average velocity x time = (u + v)/2 x t = (u + v)/2 x (v - u)/a for t = (v – u)/a therefore s = (v2 - u2)/2a v2= u2 +2as-----------(3)

20. k20 Motion under Gravity It has been observed that bodies falling to the earth (through distances which are small as compared to the radius of the earth) experience entirely unrestricted increase in their velocity by 9.81 m/s for every second during their fall. This acceleration is called the acceleration due to gravity and as conventionally denoted by g. For downward motion: a = +g v = u + gt h = ut + ½ gt2 v2 = u2 + 2gh For upward motion: a = – g v = u – gt h = ut – ½ gt2 v2 = u2 – 2gh h = Distance moved in vertical direction

21. Comparison between equations of motion under uniform acceleration and variable acceleration: k21 Equations  Acceleration 1 2 3 Uniform v = u + at S = ut + ½ at2 v2 = u2 + 2as t t s Variable v = u+∫ a(t)dt S = ut + 1/2∫ a (t2)dt v2 – u2 = 2 ∫ a (s)ds 0 0 0 Graphical Representation: v = ds/dt s-t curve S ds dt t2 t t1 The slope ds/dt at any point gives the velocity vat that point.

22. k22 v dv v – t curve dt v t t1 dt t2 The slope dv/dt at any point gives the acceleration a at that point. The shaded area under v – t curve shown above gives the incremental displacement ds during the small interval of time dt.

23. k23 dv=adt a t2 t1 t dt The shaded area under a – t curve shown above gives the incremental velocity dv during the small interval of time dt.

24. PROJECTILES k24 INTRODUCTION Assumptions: • Mass of the projectile is not considered. • Air resistance is neglected. • The trajectory of the particle is in the vertical plane.

25. k25 A Projectile is a particle moving in space under the action of gravity. The velocity with which the particle is projected into space has horizontal and vertical components. The combined effect of both components is to move the particle along a parabolic path. The parabolic path traced by the projectile is known as Trajectory of the Projectile. The horizontal component remains constant ( as air resistance is ignored) while the vertical component of motion is always subjected to acceleration due to gravity.

26. k26 DEFINITIONS: • Velocity of Projection is the velocity with which a body is projected into space. • Angleof projection is the angle which the initial velocity vector makes with the horizontal, or the angle at which a projectile is projected with respect to horizontal. • Range is the distance along the reference plane between the point of projection and the point at which it strikes the plane. • Time of Flightisthe total time during which the particle remains in motion. • Maximum Height is the maximum vertical distance covered by the projectile from the point of projection.

27. MOTION OF A PROJECTILE k27 y u trajectory P(x,y) Q  h x O M R Consider a particle thrown upwards from a point O, with an initial velocity u, at an angle  with the horizontal as shown in the figure above. After attaining maximum height h, it descends and finally hits the reference plane.

28. Equation of the Trajectory k28 From the figure above x = ux . t = u (Cos ) t t = x / (u Cos ) --------------(1) y = uy. t – ½ gt2 =u (Sin  )t – ½ gt2------------------(2) Sub (1) in (2) we get y = x tan  - gx2/ (2u2 Cos2 ) This is an equation for a parabola. Hence the path of the projectile is a parabola.

29. k29 The horizontal distance covered by the projectile is known as Range of Projectile denoted byR. Resolving the initial velocity u into horizontal and vertical components ux = u Cos  uy = uSin  (constant) Time of Flight: We know that vy = uy + at At Q vy = 0 0 = uy - gt tm = uy / g = u Sin  / g  Time of flight T = 2tm = 2 (u Sin  /g) Where tm is the time taken in seconds to reach maximum height.

30. k30 Maximum height attained (h): vy2 - uy2 =- 2gh (upward motion) vy = 0 h = uy2/2g ; h = u2(Sin2) / 2g Range: R= ux x T (time of flight) = 2u2(SinCos) / g R = u2Sin2/g (Sin2 = 2SinCos)

31. k31 • From the above equation it is clear that range will be maximum if Sin2 = 1 • i.e. 2 = 90 or  = 45 Rmax = u2/g • Projectile will cover a maximum range when it is directed at an angle of 45°. • Two angles of Projections for a given range: • We know that • Range R =u2Sin2/g • Sin = Sin ( - ) • u2Sin ( - 2)/g = u2Sin (2 1)/g (say) • where 21= ( - 2)

32. k32 Thus for same range R = u2Sin2/g = u2Sin ( - 2)/g = u2Sin21/g Which shows that the horizontal range remains the same when  is replaced by 1. Or 1=  - 2  =(  / 2 )-  2

33. k33 Projection on an Inclined Plane u Q B  P R R(Sin) P1   O R(cos) R= Range along incline; α=angle of projection; β =angle of inclined plane Sx=u x+ ½ axt2

34. k34 Projection on an Inclined Plane R(cosβ)=u(cosα)(t)+(0) ; ax =0 t= Rcosβ / ucosα…..(1) Sy=uy + ½ ayt2 R sinβ=u sinα (t)- ½ gt2 Substituting eqn.(1) and simplifying we get The Range along the inclined plane R={2u2cos2 α [sin(α- β)] } / gcos2β ..(2)

35. k35 Projection on an Inclined Plane: To get maximum range on the incline, Differentiating R w.r.t α and equating it to zero we get α= π/4 + β/4 Substituting this value of α in eqn.(2) We get maximum Range Rmax= u2/g(1+sin β) To find the time of flight: using the relation v=u + at

36. Projection on an Inclined Plane: k36 0=u x sin(α- β)-gx(cos β) x t t= {usin(α- β)}/ g(cos β) a=(g cos β) is the acceleration due to gravity along inclined plane t=time taken by the projectile to reach Q where QP is perpendicular distance to the incline plane Time of Flight T= 2xt T={2 u sin(α- β)} / g(cos β)

37. k37 Projection on an Inclined Plane: Maximum Height attained h, (PQ) using the relation v2-u2=2as 0-u2 sin2(α- β)=-2 gcos β x h h={ u2 sin2(α- β)} / 2gcos β Vertical height P1Q=h/ cos β ={u2 sin2(α- β)} / 2gcos2β

38. k38 Problem: 1 Linear motion The distance between two stations is 3km. A Locomotive starting from one station gives the train an acceleration of 1/3 m/s2 until the speed reaches 72kmph.Then the speed is maintained until the brakes are applied giving a retardation of 1m/ s2 and the train is brought to rest at the next station. Find the time taken to perform the journey and distances covered during acceleration,constant speed and retardation

39. k39 solution: a. Using V-T curve v D B S2 S1 S3 t1 t t2 t3 a1=1/3m/s2; v=72kmph=20m/s Therefore t1= v/a1 =60s (v/t=a) Similarly t3=20s

40. solution: k40 S1=1/2 x v x t1 = 600m s3=1/2 x v x t3 = 200m S2= 3000-200-600=2200m t2= 2200/20=110s

41. t1 =60 s k41 Alternate method A B C D s3 ,t3 s1 ,t1 s2 ,t2 s1+s2 +s3=3000m From A to B, vB=20=0+1/3t1 (v = u+at)

42. k42 From C to D, 0 = 20- 1x t3 t3 = 20s s3 = 20 x 20-0.5x 1 x 202 s3 = 300m Therefore s2 =3000-600-200 s2= 2200m; t2= 2200/20 = 110s

43. k43 Problem 2 A burgler’s car had a start with an acceleration of 2m/s2. A police vigilance party came after 5 seconds and continued to chase the burgler’s car with a uniform velocity of 20m/s. Find the time taken ,in which the police will overtake the car. Solution: Let the police party overtake the burgler’s car in t seconds, after the instant of reaching the spot. Distance travelled by the burglar’s car in t1 seconds is s1

44. k44 Solution: Bugler’s car u=0, a = 2m/ s2 Time, t1 = (5 + t) secs. s = ut + 0.5at2 s1= 0 +0.5 x 2 x (5+t) 2 = (5+t) 2 Distance travelled by the police party, s2= v x t ( police moves with uniform velocity ) a=2m/s2 police v=20m/s

45. k45 At the time of overtaking, the distances covered should be equal s1= s2 (5+t)2 = 20t Solving the above quadratic equation we get t=5s

46. k46 Problem 3 Two trains A and B leave the same station on parallel lines.A starts with uniform acceleration of 0.15m/s2 and attains a speed of 24kmph when the steam is reduced to keep speed constant. B leaves 40 seconds later with uniform acceleration of 0.30m/s2 to attain a maximum speed of 48km/hr. When will B overtake A ?

47. k47 Motion of train A: Uniform Acceleration, a1= 0.15 m/s2 Initial Velocity , u1= 0 Final Velocity , v1 = 24kmph = 20/3 m/s Let t1 be the time taken to attain this velocity(in seconds)

48. k48 Using the relation: v = u + at 20/3 = 0 + 0.15 x t1 Therefore, t1 = 20/3 x 0.15 = 44.4 s Also, distance travelled during this interval,

49. k49 s1= u1t1 + ½ a1t12 = 0 + ½ x 0.15 x 44.42 = 148 m. Motion of Train B: u2 = 0 a2 = 0.3 m/s2 v2 = 48 kmph= 40/3 m/s

50. k50 Let t2 be the time taken to travel this distance,say s2, Using the relation: v = u + at 40/3 = 0 + 0.3 x t2 Therefore, t2 = 40/3x0.3 = 44.4 s. And s2= u2t2 + ½ a2t22 =0+ ½ x0.3x(44.4)2=296m