FF
Faculty of Engineering and Materials Science
MECHANICS II
(DYNAMICS)
Prof. Dr. E.I. Imam Morgan
Dean, Faculty of EMS
elsayed.morgan@guc.edu.eg
C7.115
(1)
► Office hours: Actually any time but Thursday
► Teaching Assistant: Engineers: ● Nihal El-Zayat(C1:212) ● Ahmed Fahmy (C1:212) )
●Abdel Rahman Hatem (C1: 212) ● Omar Medhat(C1:212) ● Veronica Ehab (C3:208)
● Mohamed Manzour (C1: 212)
► Text book:
● Beer, F.P., and Johnston, E.R.
“ VECTOR MECHANICS for ENGINEERS: Statics & Dynamics”
7th, 8th, 9th, 10th, and 12th editions.
McGraw Hill Publishing Co., New York, NY, ISBN 0-07-121828-9
●● R. C. Hibbeler
“ ENGINEERING MECHANICS – DYNAMICS “
10th, 11th ,12th, and 14th editions
Pearson, Prentice Hall, New Jersey, ISBN 0-13-122882-X
► Course Assessment:
10% ....
20% ....
30% ....
40% ....
3Assignments (no best)
3Quizzes (best two)
Mid-Term exam.
Final-Term exam.
No compensation under any conditions
► Attendance: 75% of the course must be attended
Important: The student has to attend in his/her scheduled slot (No group change)
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(2)
INTRODUCTION
►What is meant by Dynamics ?
►Basic Definitions ?
►Frame Work ?
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(3)
Dynamics
Input


F and / or M 
The System is Moving
Mechanical System
✓►One Particle
✓►Multi-Particles(fixed number)
✓ ►One Rigid Body
✓ ►Several Connected Rigid
Bodies (Machine)
System
Characteristics
● Geometric Chcs
The relation(s) between the
applied actions (forces &
moments) and the generated
motion is called:
( KINETICS )
These relations must include
the System’s Characteristics
Prof. Imam Morgan
Dean of EMS – MCTR dept.
● Inertia Chcs
center of mass (gravity)
total mass (m)
mass distribution (M.O.I)
product of inertia
Output
Motion
■ Particle:

position r

velocity v

acceleration a
■ Rigid Body :


 and  for the R.B


v A and a A for po int A
Determination of motion parameters and
how are they related as well as their
variation with time without considering
the applied forces is called:
( KINEMATICS )
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Basic Definitions
► Particle
It is a material body with infinitesimal
dimensions containing a definite quantity of
matter. When the size of a body is very small
compared with its range of motion it may be
considered as a particle (airplanes, stars and
planets have small dimensions compared with
their trajectories).
Particle P
► Rigid Body
Is supposed to represent a continuous system
of an infinite number of particles. Each
particle is considered as a point on the body.
Moreover, the body is Rigid when the
distance between any two points (particles) is
constant during its motion under the action
of any loading conditions.
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(5)
► Machine (Mechanism)
Slider-Crank Mechanism
Four Bar Mechanism
The machine (or, the mechanism) is composed of
several rigid bodies that are connected using
joints. The joints are of different types and shapes
and they are chosen according to the required
performance of the machine.
The machine is designed to transmit the motion
and/or the power.
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(6)
► Motion
The motion is defined as the change of the position
of the:
particle or rigid body
in a chosen reference space during a certain interval
of time as particle P in Fig.a. For R.B., the path of a
point A (or particle A) on the body is the locus of the
points reached by the particle during motion.
Path of P
Path of A
A
A’
B’
B
Motion of a Particle P
Prof. Imam Morgan
Dean of EMS – MCTR dept.
Motion of a Rigid Body
So, Particle can be considered as a point of
a Rigid body
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Contents
Ch. 1 …. Kinematics of Particles (Rectilinear Motion)
Ch. 2 …. Kinematics of Particles (Curvilinear Motion)
Ch. 3 …. Kinetics of Particles
Part I
● Accelerating Force
Dynamics of Particle
5 Lectures
Ch. 4 …. Kinetics of a Particle
● Impulse and Momentum
● Energy Approach
Part II
Dynamics of R.B
Prof. Imam Morgan
Dean of EMS – MCTR dept.
Ch. 5 …. Kinematics of R.B.
● Translation
● Rotation about a fixed axis ( 2D & 3D)
● General Plane Motion
● Rotation about a fixed point
(Spherical Motion)
Ch. 6 …. Simultaneous Motion (Coriolis effect)
Ch. 7 …. Kinetics of Rigid Body
● Accelerating Force
7 Lectures
Ch. 8 …. Kinetics of Rigid Body
● Energy Approach
(8)
Note: Multi-particles Systems
Out of Scope
of this course
A
►Constant No. of Particles:
A/ Fixed No.
B/ Gaining and Losing
►Variable No. of Particles:
C/ Gaining or Losing
Fixed Number of Particles
Prof. Imam Morgan
Dean of EMS – MCTR dept.
B
Gaining and Losing Particles
Fluid Stream Diverted
by Vane or Duct
Jet Engine
Prof. Imam Morgan
Dean of EMS – MCTR dept.
Fan
Helicopter
( 10 )
C
Gaining or Losing Particles
Gains mass
(Variable Mass)
Loses mass
Classical form of Newton’s law can not
be applied in these cases
Prof. Imam Morgan
Dean of EMS – MCTR dept.
( 11 )
Part I
Dynamics of a Particle
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(12)
Part I
Dynamics of Particles
Lect. (1)
Ch. 1 …. Kinematics of Particles (Rectilinear Motion)
Position
Displacement
Velocity …. Average and Instantaneous
Acceleration
Displacement & total distance travelled (path description)
Motion determination
Multi connected particles and relative motions
Ch. 2 …. Kinematics of a Particle (Curvilinear)
Ch. 3 …. Kinetics of a Particle
● Accelerating Force
Ch. 4 …. Kinetics of a Particle
● Impulse and Momentum
● Energy Approach
Prof. Imam Morgan
Head of MCTR Depart.
(13)
Chapter 1
Kinematics of Particles
I Rectilinear Motion
►Position, Velocity, and Acceleration.
►Multi Connected Particles
14
General &
Important
Notes
Kinematic
Parameters

v
Path
P

a

 
 r ... Position vector r  r t 

 
 v ...Velocity vector v  v t 
tangent to the path

 
 a ... Acceleration vector a  a t 
not tangent to the path

r
►How can we express each
observer
vector at any time ?
O
The path can be:
■ Straight line …… Rectilinear Motion
■ Plane or Space curve…. Curvilinear
Motion
Prof. Imam Morgan
Dean of EMS – MCTR dept.
►How can we derive the relation(s)
between them ?
►The physical meaning of each
vector (why ?and how ?)
(15)
Rectilinear
Motion
The aircraft shown is considered as a particle in rectilinear
motion during landing operation. Its motion relative to the
ground is characterized at any instant by its position, velocity,
and acceleration.
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(16)
1
Rectilinear Motion
t
t=0
x

v

r
Po

a
Straight line
Path
P
O
In this case,

 the Kinematic
quantities r , v ,and a are
directed along the specified
straight line.
Prof. Imam Morgan
Dean of EMS – MCTR dept.
Therefore, the magnitude and sense of
each quantity are required to be
known for complete description of
that quantity. So, scalar analysis is an
adequate approach for solution
(17)
A: Position
+ve x
►The path of the particle is defined
using a single coordinate axis x.
►The origin O is a chosen fixed
point on the path.
- ve x
►The position of the particle at any
time is given by the distance x
measured from O.
x = x(t)
For example:
x  6t 2  t 3
or, it may be given in the
form of a graph (x - t )
Prof. Imam Morgan
Dean of EMS – MCTR dept.
Note: The value of x does not
indicate the direction of
motion!!! ☺
(18)
B: Displacement
■The displacement, Δx = Δx1,2 ,
of the particle, over a time
interval Δt = Δt1,2 = t2 – t1 ,
represents the change of
position of the particle during
that interval.
■The displacement is given by:
+ve x
x1
x2
Δx1,2= x2 – x1
Δx can be: positive,
negative, or
zero.
Third
position 3
•
O
Initial
position 1
Δx1,3 – ve
Prof. Imam Morgan
Dean of EMS – MCTR dept.
Δx2,3 – ve
Second
position 2
Δx1,2 + ve
x
(19)
Example: The position of a particle is define by the relation: x  6t 2  t 3
where x in m and t in sec.
Determine Δx0,2 and Δx2,5 and Δx2,7
t=2
Δx0,2 = 16
Δx2,5 = 9
t=0
x
O
t=7
Δx2,7 = - 65
t=5
Calculate: xi (at time t = i) i = 0, 2, 5, and 7
then, xo = 0
x2 = 16 m
x5 = 25 m
x7 = – 49 m
So, Δx0,2 = x2 – xo = 16 – 0 = 16 m
Δx2,5 = x5 – x2 = 25 – 16 = 9 m
Δx2,7 = x7 – x2 = – 49 – 16 = – 65 m
Prof. Imam Morgan
Head of MCTR Depart.
(20)
C: Velocity
(average and instantaneous)
►Consider a particle which occupies
position P at time t and P’ at t +Dt ,
Average velocity 
Dx
Dt
Dx
Instantaneous velocity  lim
Dt 0 Dt
►From the definition of the derivative,
Dx dx
v  lim

Dt 0 Dt
dt
►Instantaneous velocity may be positive or negative.
►Magnitude of velocity is referred
to as particle speed.
Note: The sign of v indicates the
direction of motion.
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(21)
x  6t 2  t 3
dx
v   12t  3t 2 → instantaneous velocity at any time
dt
Find the average velocity of the particle during the time
interval 2 < t < 7.
Dx2,7
average velocity vav 
Dt2,7
e.g.,
but,
Dx2,7   65
Dt2,7   5

vav 
 65
 13
5
m/ s
x7 = – 49
x2 = + 16
t=2
t=0
t=7
actual path
x
O
vav
Prof. Imam Morgan
Dean of EMS – MCTR dept.
Δx2,7 = - 65
(22)
D: Acceleration
(average and instantaneous)
►Consider particle with velocity v at time t
and v’ at t +Dt,
Dv
Instantaneous acceleration  a  lim
Dt 0 Dt
►From the definition of the derivative,
Dv dv d 2 x
a  lim
  2
Dt 0 Dt
dt dt
e.g. x  6t 2  t 3
v  12t  3t 2
dv
a   12  6t
dt
Prof. Imam Morgan
Dean of EMS – MCTR dept.
Note: The sign of a does not indicate
the direction of motion.
(23)
Instantaneous acceleration may be:
Positive
Negative
▪▪▪
▪
▪▪▪▪
▪▪
▪ increasing positive velocity
▪▪ decreasing negative velocity
▪▪▪ decreasing positive velocity
▪▪▪▪ increasing negative velocity.
If a = 0, then v will be kept
constant in positive or negative
direction.
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(24)
Displacement ∆x and Total distance traveled xT
D x  xT
O
t
t
xT
x
D xt , t 
D xt ,t 
D x  xT
O
xT
t
t
xT = ∆𝑥𝑡,𝑡ሗ + ∆𝑥𝑡,ሷ 𝑡ሗ
Prof. Imam Morgan
Dean of EMS – MCTR dept.
D xt , t 
t 
x
D xt ,t 
Time of zero
velocity
t
t”
●
t'
v=0
(25)
Example (1)
The position of a particle in rectilinear motion is
given by :
x = t 3 – 6t 2 ‒ 36t + 116
(m,s)
For the time interval 0 ≤ t ≤ 8 seconds, determine:
a- the displacement, and the average velocity,
b- the total distance traveled by the particle.
c- Sketch the motion.
x
O
P
x
(t )
Solution:
x  t 3  6t 2  36t  116
(m)
v  3t 2  12t  36
(m / s)
a  6t  12
(m / s 2 )
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(26)
Solution:
x  t 3  6t 2  36t  116
x
O
P
x
v  3t 2  12t  36
a  6t  12
(t )
a- The displacement, and the average velocity:
Dx0,8  x8  x0


x0  116 m and x8  44 m
Dx0,8  44  116  160
vav 
m
Dx0,8  160

 20 m / s
Dt 0,8
8
– 44
t8 = 8
●
O
116
to = 0
x
vav = – 20
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(27)
b- The total distance traveled by the particle.
x  t 3  6t 2  36t  116
v  3t 2  12t  36
a  6t  12
At first, find (if exist) the time(s) within the given
interval at which v = 0.
For v  0
3t 2  12t  36  0
 t  6  t6 and t  2 refused 
as t6 lies within the int erval :

time scale
t0 = 0
xT  Dx0 ,6  Dx6 ,8
 x6  x0  x8  x6
↑
x0  116
t6 = 6
↑
∆t0,8
t8 = 8
↑
x6  100
x8  44

xT   100  116   44   100 
 272
m
–100
●
t6 = 6
v=0
Prof. Imam Morgan
Dean of EMS – MCTR dept.
t8 = 8
– 44
●
O
116
to = 0
(28)
x  t 3  6t 2  36t  116
c- Sketch the motion.
v  3t 2  12t  36
a  6t  12
x6 = – 100
x0 = + 116
t6 = 6
●
v6 = 0
a6 = +ve
t0 = 0
O
t8 = 8
v8 = –ve
x
v0 = –ve
x8 = – 44
a0 = –ve
∆x0,8 = – 160
xT  Dx0,6  Dx6,8
 x6  x0  x8  x6

xT   100  116   44   100
 272
Prof. Imam Morgan
Dean of EMS – MCTR dept.
x0  116
x6  100
x8  44
m
(29)
2
Determination of Rectilinear Motion
Two Types of Problems:
Given
x = x (t)
?
?
v = v (t)
?
?
a = a (t)
Given
Direct
differentiation
Direct
Integration
(I. C.)
Typically, conditions of motion are
specified by the type of acceleration
experienced by the particle.
a = a (t)
a = a (x)
a = a (v)
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(30)
►Acceleration given as a function of time, a = a(t):
a  at 
dv
 at 
dt
v
t
v0
0
t
 dv   at  dt
v  vt 
dx
 vt 
dt
x
t
x0
0
vt   v0   at  dt
0
t
 dx   vt  dt
xt   x0   vt  dt
0
x  xt 
►►Acceleration given as a function of position, a = a(x):
a  a x 
dv
a
or
dt
dv
a  v  a x 
dx
v  v x 
dx
 v x 
dt
Prof. Imam Morgan
Dean of EMS – MCTR dept.
v
x
vo
xo
 vdv   a x dx
x
dx t
 v x    dt
xo
0
x  xt 
(31)
►►►Acceleration given as a function of velocity, a = a(v):
a  av 
v t 
dv
 av 
dt

v0
dv t
  dt
av  0
v  vt 
dx
 vt 
dt
x
t
xo
o
 dx   vt dt
x  xt 
or :
a  av 
v
v dv x
 av    dx
vo
xo
dv
v  av 
dx
v  v x 
dx
 v x 
dt
Prof. Imam Morgan
Dean of EMS – MCTR dept.
x
dx t
 v x    dt
xo
0
x  x(t )
(32)
vo
Example (2)
Brake mechanism used to reduce gun
recoil consists of piston attached to barrel
moving in fixed cylinder filled with oil.
As barrel recoils with initial velocity vo ,
piston moves and oil is forced through
orifices in piston, causing piston and
barrel to decelerate at rate proportional to
its velocity.
Determine v (t), x (t), and v (x).
x
a  kv
SOLUTION:
■ Integrate a = dv/dt = -kv to find v(t).
vt 
t
dv
 v  k  dt
v0
0
dv
a
 kv
dt
ln
vt 
 kt
v0
vt   v0 e kt
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(33)
■■ Integrate v (t) = [dx/dt ] to find x (t):
vt  
xt 
dx
 v0 e kt
dt
t
t
 1

xt   v0  e kt 
 k
0
kt
 dx  v0  e dt
0
0

v
xt   0 1  e kt
k

■■■ Integrate a = v [dv/dx ]= - kv to find v (x).
dv
a  v  kv
dx
dv  k dx
v
x
v0
0
 dv  k  dx
v  v0  kx
v  v0  kx
Prof. Imam Morgan
Dean of EMS – MCTR dept.
(34)
Or, one can use an alternative solution:
with
and
then

v0
xt  
1  e kt
k

vt 
vt   v0e kt or e kt 
v0
v  vt  

xt   0 1 
k 
v0 
v  v0  kx
Note: Uniformly accelerated Rectilinear Motion
v  vo  at
v 2  vo2  2ax
1 2
x  vot  at
2
Prof. Imam Morgan
Dean of EMS – MCTR dept.
a = const.
t=0
vo
t
v
x
(35)
End of
Lecture
Thank you: Imam Morgan
Prof. Imam Morgan
Dean of EMS - MCTR Depart.
(36)