1. Chapter 12 : Kinematics Of A Particle

2. Chapter Outline Introduction Rectilinear Kinematics: Continuous Motion
Rectilinear Kinematics: Erratic Motion
Curvilinear Motion: Rectangular Components
Motion of a Projectile

3. Introduction
Mechanics – the state of rest of motion of bodies subjected to the action of forces
Static – equilibrium of a body that is either at rest or moves with constant velocity
Dynamics – deals with accelerated motion of a body
1) Kinematics – treats with geometric aspects of the motion
2) Kinetics – analysis of the forces causing the motion

4. Rectilinear Kinematics: Continuous Motion
Rectilinear Kinematics – specifying at any instant, the particle’s position, velocity, and acceleration
Position
1) Single coordinate axis, s
2) Origin, O
3) Position vector r – specify the location of particle P at any instant
4) Magnitude s in metres

5. Rectilinear Kinematics: Continuous Motion
Note : - Magnitude of s (and r) = Dist from O to P
=> +ve = right of origin,
-ve = left of origin

6. Rectilinear Kinematics: Continuous Motion
Displacement
change in its position
a vector quantity
If particle moves from P to P’ :
+ve if particle’s position is right of its initial position
-ve if particle’s position is left of its initial position

7. Rectilinear Kinematics: Continuous Motion
Velocity
Average velocity,
Instantaneous velocity is defined as

8. Rectilinear Kinematics: Continuous Motion
Representing as an algebraic scalar,
Velocity is +ve = particle moving to the right
Velocity is –ve = Particle moving to the left
Magnitude of velocity is the speed (m/s)

9. Rectilinear Kinematics: Continuous Motion
Average speed is defined as total distance traveled by a particle, sT, divided by the elapsed time
The particle travels along
the path of length sT in time
=>

10. Rectilinear Kinematics: Continuous Motion
Acceleration
velocity of particle is known at points P and P’ during time interval Δt, average
acceleration is
Δv represents difference in the velocity during the time interval Δt, ie

11. Rectilinear Kinematics: Continuous Motion
Instantaneous acceleration at time t is found by taking smaller and smaller values of Δt and corresponding smaller and smaller values of Δv,

12. Rectilinear Kinematics: Continuous Motion
Velocity as a Function of Time
Integrate ac = dv/dt, assuming that initially v = v0 when t = 0.
Constant Acceleration

13. Rectilinear Kinematics: Continuous Motion
Position as a Function of Time
Integrate v = ds/dt = v0 + act, assuming that initially s = s0 when t = 0
Constant Acceleration

14. Rectilinear Kinematics: Continuous Motion
Velocity as a Function of Position
Integrate v dv = ac ds, assuming that initially v = v0 at s = s0
Constant Acceleration

15. EXAMPLE 1
The car moves in a straight line such that for a short time its velocity is defined by v = (0.9t t) m/s where t is in sec. Determine it position and acceleration when t = 3s. When t = 0, s = 0.

16. EXAMPLE 1
Solution:
Coordinate System. The position coordinate extends from the fixed origin O to the car, positive to the right.
Position. Since v = f(t), the car’s position can be determined from v = ds/dt, since this equation relates v, s and t. Noting that s = 0 when t = 0, we have

17. EXAMPLE 1
When t = 3s,
s = 10.8m

18. EXAMPLE 1
Acceleration Knowing v = f(t), the acceleration is determined from a = dv/dt, since this equation relates a, v and t.
When t = 3s,
a = 6m/s2

19. Rectilinear Kinematics: Erratic Motion
When particle’s motion is erratic, it is best described graphically using a series of curves that can be generated experimentally from computer output.
a graph can be established describing the relationship with any two of the variables, a, v, s, t
using the kinematics equations a = dv/dt, v = ds/dt, a ds = v dv

20. Rectilinear Kinematics: Erratic Motion
When particle’s motion is erratic, it is best described graphically using a series of curves that can be generated experimentally from computer output.
a graph can be established describing the relationship with any two of the variables, a, v, s, t
using the kinematics equations a = dv/dt, v = ds/dt,
a = dv dt
v = ds dt
dt = ds v
a = dv . v ds
a ds = v dv

21. Rectilinear Kinematics: Erratic Motion
Given the s-t Graph, construct the v-t Graph
The s-t graph can be plotted if the position of the particle can be determined experimentally during a period of time t.
To determine the particle’s velocity as a function of time, the v-t Graph, use v = ds/dt
Velocity as any instant is determined by measuring the slope of the s-t graph

22. Rectilinear Kinematics: Erratic Motion
Slope of s-t graph = velocity

23. Rectilinear Kinematics: Erratic Motion
Given the v-t Graph, construct the a-t Graph
When the particle’s v-t graph is known, the acceleration as a function of time, the a-t graph can be determined using a = dv/dt
Acceleration as any instant is determined by measuring the slope of the v-t graph

24. Rectilinear Kinematics: Erratic Motion
Slope of v-t graph = acceleration

25. Rectilinear Kinematics: Erratic Motion
Since differentiation reduces a polynomial of degree n to that of degree n-1, then if the s-t graph is parabolic (2nd degree curve), the v-t graph will be sloping line (1st degree curve), and the a-t graph will be a constant or horizontal line (zero degree curve)

26. EXAMPLE 2
A bicycle moves along a straight road such that it position is described by the graph as shown. Construct the v-t and a-t graphs for 0 ≤ t ≤ 30s.

27. EXAMPLE 2
Solution:
v-t Graph. The v-t graph can be determined by differentiating the eqns defining the s-t graph
The results are plotted.

28. EXAMPLE 2
We obtain specify values of v by measuring the slope of the s-t graph at a given time instant.
a-t Graph. The a-t graph can be determined by differentiating the eqns defining the lines of the v-t graph.

29. EXAMPLE 2
The results are plotted.

30. Rectilinear Kinematics: Erratic Motion
Given the a-t Graph, construct the v-t Graph
When the a-t graph is known, the v-t graph may be constructed using a = dv/dt
Change in velocity
Area under a-t graph =

31. Rectilinear Kinematics: Erratic Motion
Knowing particle’s initial velocity v0, and add to this small increments of area (Δv)
Successive points v1 = v0 + Δv, for the v-t graph
Each eqn for each segment of the a-t graph may be integrated to yield eqns for corresponding segments of the v-t graph

32. Rectilinear Kinematics: Erratic Motion
Given the v-t Graph, construct the s-t Graph
When the v-t graph is known, the s-t graph may be constructed using v = ds/dt
Displacement
Area under v-t graph =

33. Rectilinear Kinematics: Erratic Motion
knowing the initial position s0, and add to this area increments Δs determined from v-t graph.
describe each of the segments of the v-t graph by a series of eqns, each of these eqns may be integrated to yield eqns that describe corresponding segments of the s-t graph

34. EXAMPLE 3
A test car starts from rest and travels along a straight track such that it accelerates at a constant rate for 10 s and then decelerates at a constant rate. Draw the v-t and s-t graphs and determine the time t’ needed to stop the car. How far has the car traveled?

35. EXAMPLE 3
Solution:
v-t Graph. The v-t graph can be determined by integrating the straight-line segments of the a-t graph. Using initial condition v = 0 when t = 0,

36. EXAMPLE 3
When t = 10s, v = 100m/s, using this as initial condition for the next time period, we have
When t = t’ we require v = 0. This yield t’ = 60 s
s-t Graph. Integrating the eqns of the v-t graph yields the corresponding eqns of the s-t graph. Using the initial conditions s = 0 when t = 0,

37. EXAMPLE 3 When t = 10s, s = 500m. Using this initial condition,
When t’ = 60s, the position is s = 3000m

38. Rectilinear Kinematics: Erratic Motion
Given the a-s Graph, construct the v-s Graph
v-s graph can be determined by using v dv = a ds, integrating this eqn between the limit v = v0 at s = s0 and v = v1 at s = s1
Area under a-s graph

39. Rectilinear Kinematics: Erratic Motion
determine the eqns which define the segments of the a-s graph
corresponding eqns defining the segments of the v-s graph can be obtained from integration, using vdv = a ds

40. Rectilinear Kinematics: Erratic Motion
Given the v-s Graph, construct the a-s Graph
v-s graph is known, the acceleration a at any position s can be determined using a ds = v dv
Acceleration = velocity times slope of v-s graph

41. Rectilinear Kinematics: Erratic Motion
At any point (s,v), the slope dv/ds of the v-s graph is measured
Since v and dv/ds are known, the value of a can be calculated

42. EXAMPLE 4
The v-s graph describing the motion of a motorcycle is shown in Fig 12-15a. Construct the a-s graph of the motion and determine the time needed for the motorcycle to reach the position s = 120 m.

43. EXAMPLE 4
Solution:
a-s Graph Since the eqns for the segments of the v-s graph are given, a-s graph can be determined using a ds = v dv.

44. EXAMPLE 4
Time The time can be obtained using v-s graph and v = ds/dt. For the first segment of motion, s = 0 at t = 0,
At s = 60 m, t = 8.05 s

45. EXAMPLE 4
For second segment of motion,
At s = 120 m, t = 12.0 s

46. General Curvilinear Motion
Curvilinear motion occurs when the particle moves along a curved path
Position. The position of the particle, measured from a fixed point O, is designated by the position vector r = r(t).

47. General Curvilinear Motion
Displacement. Suppose during a small time interval Δt the particle moves a distance Δs along the curve to a new position P`, defined by r` = r + Δr. The displacement Δr represents the change in the particle’s position.

48. General Curvilinear Motion
Velocity. During the time Δt, the average velocity of the particle is defined as
The instantaneous velocity is determined from this equation by letting Δt 0, and consequently the direction of Δr approaches the tangent to the curve at point P. Hence,

49. General Curvilinear Motion
Direction of vins is tangent to the curve
Magnitude of vins is the speed, which may be obtained by noting the magnitude of the displacement Δr is the length of the straight line segment from P to P`.

50. General Curvilinear Motion
Acceleration If the particle has a velocity v at time t and a velocity v` = v + Δv at time t` = t + Δt. The average acceleration during the time interval Δt is

51. General Curvilinear Motion
a acts tangent to the hodograph, therefore it is not tangent to the path

52. Curvilinear Motion: Rectangular Components
Position Position vector is defined by
r = xi + yj + zk
The magnitude of r is always positive and defined as
The direction of r is specified by the components of the unit vector ur = r/r

53. Curvilinear Motion: Rectangular Components
Velocity
where
The velocity has a magnitude defined as the positive value of
and a direction that is specified by the components of the unit vector uv=v/v and is always tangent to the path.

54. Curvilinear Motion: Rectangular Components
Acceleration
where
The acceleration has a magnitude defined as the positive value of

55. Curvilinear Motion: Rectangular Components
The acceleration has a direction specified by the components of the unit vector ua = a/a.
Since a represents the time rate of change in velocity, a will not be tangent to the path.

56. EXAMPLE 5
At any instant the horizontal position of the weather balloon is defined by x = (9t) m, where t is in second. If the equation of the path is y = x2/30, determine the distance of the balloon from the station at A, the magnitude and direction of the both the velocity and acceleration when t = 2 s.

57. EXAMPLE 5
Solution:
Position When t = 2 s, x = 9(2) m = 18 m and y = (18)2/30 = 10.8 m
The straight-line distance from A to B is m
Velocity

58. EXAMPLE 5 When t = 2 s, the magnitude of velocity is
The direction is tangent to the path, where
Acceleration

59. EXAMPLE 5
The direction of a is

60. Motion of a Projectile
Free-flight motion studied in terms of rectangular components since projectile’s acceleration always act vertically
Consider projectile launched at (x0, y0)
Path defined in the x-y plane
Air resistance neglected
Only force acting on the projectile is its weight, resulting in constant downwards acceleration
ac = g = 9.81 m/s2

61. Motion of a Projectile

62. Motion of a Projectile Horizontal Motion Since ax = 0,
Horizontal component of velocity remain constant during the motion

63. Motion of a Projectile
Vertical Positive y axis is directed upward, then ay = - g

64. EXAMPLE 6
The chipping machine is designed to eject wood chips at vO = 7.5 m/s. If the tube is oriented at 30° from the horizontal, determine how high, h, the chips strike the pile if they land on the pile 6 m from the tube.

65. EXAMPLE 6
Coordinate System Three unknown h, time of flight, tOA and the vertical component of velocity (vB)y. Taking origin at O, for initial velocity of a chip,
(vA)x = (vO)x = 6.5 m/s and ay = m/s2

66. EXAMPLE 6 Horizontal Motion Vertical Motion
Relating tOA to initial and final elevation of the chips,