1. Engineering Dynamics Module code NS 111 Engr. Dr Imran Shafi

2. DYNAMICS

3. BRANCHES OF MECHANICS Mechanics
Things that do change shape when forces are applied
Solid Mechanics
Mechanics
Things that do not change shape when forces are applied
Rigid Bodies
Deformable Bodies
Fluids
Statics
Dynamics
Stress Analysis
Compressible
Flow
Incompressible Flow
Engineering Statics

4. An Overview of Mechanics
Mechanics: the study of how bodies react to forces acting on them
Statics: the study of bodies in equilibrium
Dynamics:
1. Kinematics – related to geometric aspects of motion
2. Kinetics - related to the forces causing the motion

5. Introduction Dynamics includes:
Kinematics: study of the geometry of motion. Kinematics is used to relate displacement, velocity, acceleration, and time without reference to the cause of motion.
Kinetics: study of the relations existing between the forces acting on a body, the mass of the body, and the motion of the body. Kinetics is used to predict the motion caused by given forces or to determine the forces required to produce a given motion.
Rectilinear motion: position, velocity, and acceleration of a particle as it moves along a straight line.
Curvilinear motion: position, velocity, and acceleration of a particle as it moves along a curved line.

7. Applications
Engineering Statics
Mian Ashfaq Ali

8. KINEMATICS OF PARTICLES
APPLICATIONS
The motion of large objects, such as rockets, airplanes, or cars, can often be analyzed as if they were particles.
If we measure the altitude of this rocket as a function of time, how can we determine its velocity and acceleration?

9. APPLICATIONS (continued)
A train travels along a straight length of track.
If the train accelerates at a constant rate, how can we determine its position and velocity at some instant?

10. Kinematics of a Particle
Objectives:
◦Introduce the concepts of position, displacement, velocity, and acceleration
◦Study particle motion along a straight line and represent the motion graphically
◦Investigate particle motion along a curved path using different coordinate systems

11. KINEMATICS OF PARTICLES Kinematics of particles
Road Map
Kinematics of particles
Rectilinear motion
Curvilinear motion
Normal and tangential coordinates that move along with velocity on a track, one is normal and one is at 90 degree
x-y coord.
r- coord.
n-t coord.

12. Rectilinear Kinematics: Continuous Motion
We will begin our study of dynamics by discussing the kinematics of a particle that moves along a rectilinear or straight line
Particle: has mass, but it is of negligible size and shape
The kinematics of a particle is characterize by specifying, at any instant it’s, position, velocity and acceleration
In most problems, we are interested in bodies of finite size, rockets, projectiles or vehicles that can be taken as particle where motion is char by motion of its CG and rotation is neglected
car

13. Position is a vector quantity ?
POSITION AND DISPLACEMENT
A particle travels along a straight-line path defined by the coordinate axis s.
The position of the particle at any instant, relative to the origin, O, is defined by the position vector r, or the scalar s. Typical units for r and s are meters (m) or feet (ft).
Position is a vector quantity ?
The displacement of the particle is defined as its change in position.
Vector form:  r = r’ - r
Scalar form:  s = s’ - s
Displacement is a vector and shall be distinguished from
the total distance traveled by the particle, sT, which is a positive scalar that represents the total length of the path over which the particle travels.

14. The average velocity of a particle during a time interval t is
Velocity is a measure of the rate of change in the displacement of a particle. It is a vector quantity (it has both magnitude and direction). The magnitude of the velocity is called speed, with units of m/s or ft/s.
The average velocity of a particle during a time interval t is
vavg = r/t
If we take smaller and smaller values of ∆t, the magnitude ∆r
becomes smaller and smaller.
The instantaneous velocity is defined as
v = dr/dt
Speed is the magnitude of velocity: v = ds/dt
Average speed is the total distance traveled divided by elapsed time: (vsp)avg = sT/  t, But its average velocity is vavg = -∆s/ t

15. The instantaneous velocity is defined as
v = dr/dt
From the definition of a derivative,
e.g.,

16. The instantaneous acceleration is the time derivative of velocity.
Acceleration is the rate of change in the velocity of a particle. It is a vector quantity. Typical unit is m/s2.
The instantaneous acceleration is the time derivative of velocity.
Vector form: a = dv/dt
Scalar form: a = dv/dt = d2s/dt2
Acceleration can be positive (speed increasing) or negative (speed decreasing).
The derivative equations for velocity and acceleration can be manipulated to get: a ds = v dv
v = ds/dt a = dv/dt

17. The instantaneous acceleration is the time derivative of velocity.
Vector form: a = dv/dt
Scalar form: a = dv/dt = d2s/dt2
From the definition of a derivative,

18. SUMMARY OF KINEMATIC RELATIONS:
RECTILINEAR MOTION
• Differentiate position to get velocity and acceleration.
v = ds/dt ; a = dv/dt or a = v dv/ds
• Integrate acceleration for velocity and position.
Velocity: ò = t o v dt a dv s ds or ò = t o s dt v ds
Position:
• Note that so and vo represent the initial position and velocity of the particle at t = 0.

19. at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0
Consider particle with motion given by
at t = 0, x = 0, v = 0, a = 12 m/s2
at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0
at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2
at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s2

20. CONSTANT ACCELERATION
The three kinematic equations can be integrated for the special case when acceleration is constant (a = ac) to obtain very useful equations. A common example of constant acceleration is gravity; i.e., a body freely falling toward earth. In this case, ac = g = 9.81 m/s2 downward. These equations are: t a v c o + =
yields ò dt dv 2 s
(1/2)a ds ) - (s 2a (v
ac= 3m/s2
Insert v before integration in equation 2 t a v c o + =

22. CONSTANT ACCELERATION
But what if ac= 3t2

23. Important Points

24. Engineering Statics
Mian Ashfaq Ali

25. EXAMPLE
Given: A motorcyclist travels along a straight road at a speed of 27 m/s. When the brakes are applied, the motorcycle decelerates at a rate of -6t m/s2.
Find: The distance the motorcycle travels before it stops.
Plan: Establish the positive coordinate s in the direction the motorcycle is traveling. Since the acceleration is given as a function of time, integrate it once to calculate the velocity and again to calculate the position.

26. v = ds/dt a = dv/dt a ds = v dv
EXAMPLE (continued)
Solution:
1) Integrate acceleration to determine the velocity.
a = dv / dt => dv = a dt =>
=> v – vo = -3t2 => v = -3t2 + vo ò - = t o v dt dv ) 6 (
2) We can now determine the amount of time required for the motorcycle to stop (v = 0). Use vo = 27 m/s.
0 = -3t => t = 3 s
3) Now calculate the distance traveled in 3s by integrating the velocity using so = 0:
v = ds / dt => ds = v dt =>
=> s – so = -t3 + vot
=> s – 0 = (3)3 + (27)(3) => s = 54 m ò + - = t o s dt v ds ) 3 ( 2
v = ds/dt a = dv/dt a ds = v dv
s = vt v = vo + act s = so + vot + (1/2) act v2 = (vo)2 + 2ac (s - so)

27. v = ds/dt a = dv/dt a ds = v dv
A baseball is thrown downward from a tower of height h with an initial speed vo. Determine the speed at which it hits the ground and the time of travel.
Given:
h = 50 m vo =6 m/s
Find: v, t
Solution:
30.7 m/s
2.64 s
v = ds/dt a = dv/dt a ds = v dv
Engineering Statics
Dr Riaz Mufti
s = vt v = vo + act s = so + vot + (1/2) act v2 = (vo)2 + 2ac (s - so)
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

28. v = ds/dt a = dv/dt a ds = v dv
Engineering Statics
s = vt v = vo + act s = so + vot + (1/2) act v2 = (vo)2 + 2ac (s - so)
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

29. Acceleration due to gravity is a vector quantity.
Remember…………………
Acceleration due to gravity is a vector quantity.

30. A Ball dropped from a building
The vector is oriented down.

31. A baseball tossed up in the air, halfway up the path
The vector is oriented down.

32. A baseball tossed up in the air, right before it strikes the ground
The vector is oriented down.

33. GROUP PROBLEM SOLVING
Given: Ball A is released from rest at a height of 12 m at the same time ball B is thrown upward, 1.5 m from the ground. The balls pass one another at a height of 6 m.
Find: The speed at which ball B was thrown upward.

34. sA = (sA )o + (vA)ot + (1/2)act2
GROUP PROBLEM SOLVING (continued)
Solution:
1) First consider ball A. With the origin defined at the ground, ball A is released from rest ((vA)o = 0) at a height of 12 m
((sA )o = 12 m). Calculate the time required for ball A to drop to 6 m (sA = 6 m) using a position equation.
sA = (sA )o + (vA)ot + (1/2)act2
6 m = 0m + (0)(t) + (1/2)(9.81)(t2) => t = s
2) Now consider ball B. It is throw upward from a height of 5 ft ((sB)o = 1.5 m). It must reach a height of 6 m (sB = 6 m) at the same time ball A reaches this height (t = s). Apply the position equation again to ball B using t = 1.106s.
sB = (sB)o + (vB)ot + (1/2) ac t2
6 m = (vB)o(1.106) + (1/2)(-9.81)(1.106)2
=> (vB)o = 9.49 m/s
v = ds/dt a = dv/dt a ds = v dv
s = vt v = vo + act s = so + vot + (1/2) act v2 = (vo)2 + 2ac (s - so)

35. v = ds/dt a = dv/dt a ds = v dv
Given: A stone is dropped from rest down a well, and in 1s another stone B is dropped from rest.
Find: Determine the Distance between the stones another second later.
Solution:
S=Si + vit + ½ ac t2
SA= ½ (9.81) (2)2
SA= m
SB = ½ (9.81) (1) 2
SB = m
∆S = = m
v = ds/dt a = dv/dt a ds = v dv
s = vt v = vo + act s = so + vot + (1/2) act v2 = (vo)2 + 2ac (s - so) 35

36. v = ds/dt a = dv/dt a ds = v dv
Given: When two cars A and B are next to one another, they are travelling in the same direction with speeds Va and Vb, respectively. If B maintains its constant speed, while A begins to decelerate at aA.
Find: Determine the distance ‘d’ between the cars at the instant once car ‘A’ stops.
Solution:
Motion of car B:
Motion of car A
SB = vBt
v = vo + ac t
SB = vBt = vB (vA/aA ) = (vA vB) / aA
0 = vA – aA t
The distance between cars A and cars B is
SBA = │SB-SA │
= │((vAvB)/ aA ) – vA2/ 2aA │
= │(2vAvB – vA2) / 2aA│
t = vA/aA
v2 = v02 + 2ac (S-S0)
0 = vA2 +2(-aA) (SA – 0)
SA =vA2 / 2aA
v = ds/dt a = dv/dt a ds = v dv
s = vt v = vo + act s = so + vot + (1/2) act v2 = (vo)2 + 2ac (s - so) 36

37. v = ds/dt a = dv/dt a ds = v dv
Given: A particle has an initial speed of 27 m/s. If it experiences a deceleration of a = (-6t)m/s2., where t is in seconds.
Find: Determine the distance traveled before it stops.
Solution:
a = -6t
dv=a dt
v = ds/dt a = dv/dt a ds = v dv
s = vt v = vo + act s = so + vot + (1/2) act v2 = (vo)2 + 2ac (s - so) 37

38. v = ds/dt a = dv/dt a ds = v dv
Given: The acceleration of a rocket travelling upward is given by
a = ( s)m/s2, where s is in meters.
Find: Determine the time needed for the rocket to reach an altitude of s=100 m. Initially v = 0 and s = 0 when t = 0.
Solution:
v = ds/dt a = dv/dt a ds = v dv 38

39. a) the velocity when t = 5 sec
Given: If the effects of atmospheric resistance are accounted for, a freely falling body has an acceleration defined by the equation a=9.81[1-v2 (10-4 )] m/s2 , where v is in m/s and the positive direction is downward. If the body is released from rest at a very high altitude,
Find: Determine
a) the velocity when t = 5 sec
b) the body’s terminal or maximum attainable velocity (as t approaches infinity) 39

40. v = ds/dt a = dv/dt a ds = v dv
Solution:
v = ds/dt a = dv/dt a ds = v dv
s = vt v = vo + act s = so + vot + (1/2) act v2 = (vo)2 + 2ac (s - so) 40

41. v = ds/dt a = dv/dt a ds = v dv
a is given.
a = const. ? No
Yes
V = Vo + at
V2 = Vo2 + 2a(S-So)
S = So + Vo + 1/2 at2
find V
and S.
is a = f(t)
a = dV/dt
V = dS/dt
is a = f(V)
‘a’ is a function of S.
dt = dV/a
find V in terms of t.
s = ∫V.dt
find S.
a.ds = V.dV
dS = V.dV/ a
V in terms of S.
find S in terms of V.
V = dS/ dt
dt = dS/ V
find S in terms of t.
KEY:
a = Acceleration
V = Velocity
S = Position
t = Time
v = ds/dt a = dv/dt a ds = v dv
e.g (a= 4)
e.g (a= 2t-1)
Integrate
e.g (a= -0.4V3)
e.g (a= 4S)

42. v = ds/dt a = dv/dt a ds = v dv
V is given.
V = Const. ? No
Yes
Is V = f (t)
a = 0
S = linear
s = vt
a = dV/dt
V = dS/dt
Integrate to find ‘S’.
Differentiate to find ‘a’.
Is V = f (S)
dS/V = dt
find S in terms of t.
V = f (a)
a.ds = V.dV
dS = V.dV/a = f (a)
KEY:
a = Acceleration
V = Velocity
S = Position
t = Time
find S in terms of ‘a’.
e.g (V= 3t2-2t)
e.g V= 5/(4+S)

43. e.g S= (t3-3t2+2) S is given. Is S= f (t) ? Yes No Is S= f (V)
V = dS/dt
a = dV/dt
Differentiate to get ‘V’ and ‘a’.
dS/V = dt
Find V in terms of t.
S = f (a)
a.dS = V.dV
Find V in terms of a.
KEY:
a = Acceleration
V = Velocity
S = Position
t = Time
e.g S= (t3-3t2+2)

44. Engineering Dynamics Problem Sheet 1 Chapter 12
Problem with Odd no 1,3,…25
Due date (on exam day)
Engineering Dynamics
Problem Sheet 1
Chapter 12
Problems 12-1 to 12-26
Engineering Statics
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

45. APPLICATION
In many experiments, a velocity versus position (v-s) profile is obtained.
If we have a v-s graph for the rocket car, can we determine its acceleration at position s = 300 meters ?
How?

46. Piston kinematics
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

47. GRAPHING
Graphing provides a good way to handle complex motions that would be difficult to describe with formulas. Graphs also provide a visual description of motion and reinforce the calculus concepts of differentiation and integration as used in dynamics.
The approach builds on the facts that slope and differentiation are linked and that integration can be thought of as finding the area under a curve.

48. S-T GRAPH
Plots of position vs. time can be used to find velocity vs. time curves. Finding the slope of the line tangent to the motion curve at any point is the velocity at that point (or v = ds/dt).
Therefore, the v-t graph can be constructed by finding the slope at various points along the s-t graph.

49. V-T GRAPH
Plots of velocity vs. time can be used to find acceleration vs. time curves. Finding the slope of the line tangent to the velocity curve at any point is the acceleration at that point (or a = dv/dt).
Therefore, the a-t graph can be constructed by finding the slope at various points along the v-t graph.
Also, the distance moved (displacement) of the particle is the area under the v-t graph during time t.
v dt= ds

50. A-T GRAPH
Given the a-t curve, the change in velocity (v) during a time period is the area under the a-t curve.
So we can construct a v-t graph from an a-t graph if we know the initial velocity of the particle.
v = vo + act

51. A-S GRAPH
A more complex case is presented by the a-s graph. The area under the acceleration versus position curve represents the change in velocity
(recall ò a ds = ò v dv ).
a-s graph
½ (v1² – vo²) = = area under the ò s2 s1
a ds
This equation can be solved for v1, allowing you to solve for the velocity at a point. By doing this repeatedly, you can create a plot of velocity versus distance.

52. V-S GRAPH
Another complex case is presented by the v-s graph. By reading the velocity v at a point on the curve and multiplying it by the slope of the curve (dv/ds) at this same point, we can obtain the acceleration at that point.
a = v (dv/ds)
Thus, we can obtain a plot of a vs. s from the v-s curve.

53. 40 EXAMPLE Given: v-t graph for a train moving between two stations
Find: a-t graph and s-t graph over this time interval 40
Think about your plan of attack for the problem!

54. EXAMPLE (continued)
Solution: For the first 30 seconds the slope is constant and is equal to:
a0-30 = dv/dt = 40/30 = 4/3 m/s2
Similarly, a30-90 = and a = -4/3 m/s2 4 -4 3
a(m/s2)
t(s)

55. EXAMPLE (continued)
The area under the v-t graph represents displacement.
Ds0-30 = ½ (40)(30) = 600 m
Ds30-90 = (60)(40) = 2400 m
Ds = ½ (40)(30) = 600 m
600
3000
3600 30 90
120
t(s)
s(m)

56. A car starting from rest moves along a straight track with an acceleration as shown. Determine the time t for the car to reach a speed of 50m/s and construct the v-t graph that describes the motion until the time t.
Given: v = 50 m/s, a1 = 8 m/s2 , t1 = 10 s
Solution:
For 0 < t < 10s.
At t= 10s.
At t > 10s.

57. The s-t graph for a train has been experimentally determined
The s-t graph for a train has been experimentally determined. From the data, construct the v-t and a-t graph for the motion; 0< t < 40s. 0< t < 30s, the curve is s= (0.4t2)m, and then it becomes straight for t > 30s.
600
360
Solution:
0< t < 30s 30 40
30< t < 40

58. v-t graph
600
360 v 30 40 t
a-t graph a t

59. The a-s graph for a train travelling along a straight track is given for the first 400m of its motion. Plot the v-s graph. v=0 s=0.
Solution: 2
0< s < 200
200
400
At s=200, v= 20 m/s
200 < s < 400; a=2
At s = 400m.

60. 2
200
400
v-s graph v s

61. Engineering Dynamics Module code ME 123 Lecture 07 Mian Ashfaq Ali