ENT252 - DYNAMICS
CHAPTER 12
INTRODUCTION TO DYNAMICS
Chapter Objectives
To introduce the concepts of position, displacement, velocity and acceleration
To study particle motion along a straight line and represent this motion graphically
To investigate particle motion along a curved path using different coordinate
systems
To present an analysis of dependent motion of two particles
To examine the principles of relative motion of two particles using translating axes
12.1 Introductions

Mechanics
–
branch of the physical science that is concerned with the state of
rest or motion of bodies subjected to the action of the forces

Mechanics of rigid body - divided into statics and dynamics

Statics - concerned with the equilibrium of the body that is either at the rest or
moves with constant velocity

Dynamics - concerned with the accelerated motion of a body. Presented in 2 parts:
a) Kinematics – geometric aspect of motion
b) Kinetics – analysis of the force causing the motion
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12.2 Rectilinear Kinematics: Continuous Motion
Rectilinear Kinematics – at any given instant, the particles position, velocity and
acceleration.
Position – the straight line path of a particle. From the origin (o), position vector r specify
the location of the particle (p).
r
P
s
O
s
Position
Convenient (r) represent by (s)
Displacement – the change in its position
Eg : If the particle moves from P to P’, the displacement is Δr = r’- r
Δs = s’ – s
r’
r
r
P
O
P’
s
s
s
s’
Displacement
 Δs is positive – particles final position is to the right of its initial position, ie :
s’>s.
 Displacement of a particle – vector quantity
 Distance traveled is a positive vector.
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{Velocity}
If the particle moves through a displacement Δr, from P-P¹ during the time
interval Δt, the average velocity
Vavg = Δr
Δt
,
V = dr : instantaneous velocity
dt
V as an algebraic scalar, V = ds
dt
v
P’
P
O
s
s
 Δt or dt always positive:
1. particle moving the right, velocity is positive
2. particle moving to the left velocity is negative.
 The magnitude of the velocity is known as the speed.
( units : m/s )
vavg = st
Δt
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{ Acceleration }
Provided the velocity of the particle is known at two point P² P¹, the average
acceleration
aavg =
v
t
a
P
P’
O
s
v’
v
 v – difference in the velocity during the time interval v = v ¹- v
Acceleration: a =
a=
a
dv
dt

 acceleration
d 2s
dt 2

deceleration
a
 deceleration – when the particle is slowing down
- speed decreasing
- v  v 1  v is negative
 acceleration is zero – when velocity is constant.
- v  v  v  o
( unit = m/s 2 )
 a ds = v dv
a=
dv
dt
& v=
ds
dt
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 constant acceleration – each of three kinematics equations a c =
dv
dt
a = ac
ds
, a c ds = vdv
dt
maybe integrated to obtained formula
that related a,v,s,t.
v=
 The three formulas of constant acceleration :
1) Velocity as a function of time
v
t
vo
o
 dv   a
c
dt
v = vo + act
2) Position as a function of time
+

s
o
t
ds   (vc + a c t) dt
v  ds dt  vo  ac t
o
s = so + vo t + 1 ac t2
2
3) Velocity as a function of position

v.dv = a c .ds
o
vo
s
vdv   a c ds
so
= v 2 = v 02 + 2a c ( s-s o )
This formula only useful when the acceleration is constant and when t = o ,
s = so , v = vo
e.g.– a body fall freely toward the earth.
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See Example:
-
12.1
-
12.2
-
12.3
-
12.4
-
12.5
Exercise : 12.1
-
12.2
-
12.3
-
12.4
-
12.5
0194579207
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12.3 Rectilinear Kinematics: Erratic Motion
When particles motion during a time is erratic, may best be describes graphically using a
series of curves.
Using the kinematics equations:
a
dv
dt
v
ds
dt
a.ds  v.dv
a) Given s-t graph, construct the v-t equations
s
By experimentally, if the position can
be determined during the time of
period, graph s-t can be plotted.
By v = ds dt , the graph v-t can be
plotted.
( “ slope of s-t graph = velocity” ).
v = ds dt ,
t
b)
v  t graph  a  t graph
a  dv
dt
(“slope of v  t graph = acceleration”)
See example 12.6, page 19
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a  dv
dt
c) a  t graph , v  t graph
using a  ds dt ,
 v   adt
( change in velocity = area
under a  t )
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graph. )
ENT252 - DYNAMICS
d) v  t graph  s  t graph
v  ds dt
  s   vdt
(displacement = area under v  t
graph)
See example 12.7, page 21
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e) a  s graph  v  s graph.
a.ds  v.dv
between the limits v  vo at s  so
s
v  v1 at s  s1
1
 1 ( v12  v02 ) =  a.ds
2
so
= area under a  s graph
f) v  s graph  a  s graph
a.ds  v.dv
a  v(
dv
)
ds
acceleration = velocity times slope of v  s graph.
See Example 12.8.
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Exercise:
-
12.42
12.43
12.44
12.45
12.46
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12.4 General Curvilinear Motion
- curvilinear motion occurs when the particle moves along a curved path.
Path
- position- considered a particle located
at point p on a space curve defined by
the path function s.
P
r
s
O


s
position vector r = r ( t )
magnitude and direction change as the
particle moves along the curve.
Position
P’
- displacement- during small line  t, the
particle moves a distance  s along the
curves.
s
r
P
r’
r’ = r +  r
r
O

the displacement  r represent the
change in the particle’s position.
s
 r = r’ - r
Displacement
v
P
- velocity – during the time t , the average
velocity.
r
V avg =
O

s
r
, V = dr
dt
t
dr will be tangent to the curve at p, the
direction of V is also tangent to the
curve.
Velocity
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The magnitude of v, called ‘speed’.
V=
V
avg 
ds
.
dt
,
v
t
where  v = v 1 - v
Instant a new acceleration, t  0
a
dv
dt
a
d 2r
dt 2
 Velocity vector is always directed
tangent to the path.
 a tangent to the hodograph, not tangent
to the path of motion.
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12.5 Curvilinear Motion: Rectangular Components
Displacement
r = x i + y j + zk
magnitude of r always positive
r=
(x 2 + y 2 + z 2 )
unit vector ur = (1/r)r
Velocity
v = vx i  v y j  v2 k
v=
v =
dr
=
dt
d
( xi ) + d
( y j ) + d ( zk )
dt
dt
dt
dr
= vx i + vy j + vz k,
dt
.
where :
vx = x
.
vy = y
.
vz = z
The velocity has a magnitude defined as the positive value of
v=
vx2 +vy
2
+ vz2
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Acceleration :
a=
dv
= ax i + ay j + az k
dt
..
.
where :
a x = vx = x
.
..
a y = vy = y
.
..
a z = vz = z
The acceleration has a magnitude defined by the positive value of
a=
ax
2
+ay
2
+ az
2
See Example 12.9 and 12.10.
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12.6 Motion of a Projectile
The free-flight motion of a projectile – studied in terms of its rectangular
components. The projectiles acceleration always act in the vertical direction.
Projectile launched at point ( x o , y o ) , initial velocity is V o , having two components
( V o )x and ( V o )y . The projectile has a constant downward acceleration,
a c = g = 9.81 m s 2 .
Horizontal motion : Since a x = 0 ;

v = vo + ac t ;
x = xo+ vo t + 1
vx = ( vo ) x
2
at 2 ;
v 2 = vo 2 + 2ac ( s-so ) ;
x = xo + ( vo ) x t
vx = ( vo ) x
First and last equation indicated that the horizontal component of velocity always remains
constant during the motion.
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Vertical motion : Since ay = -g.
+ ↑ v = vo + ac t ;
y = yo + vo t + 1
2
vy = ( vo )y – gt.
ac t 2 ;
v 2 = vo + 2 ac ( y- y 2 )
;
y = yo + (vo) yt - 1 g
2
vy 2 = (vo ) 2 -2g (y)
Only two of the above three equations are independent of one another.
 Problems involving the motion of projectile can have at most three unknowns
since only three independent equations can be written.
- one equations in the horizontal direction.
- two equations in the vertical direction.
Once vx and vy are obtained, the resultant velocity v which is always tangent to the
path.
See Example:
- 12.11
- 12.12
- 12.13
Exercise:
- 12.71
- 12.72
- 12.73
- 12.74
- 12.75
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12.7 Curvilinear motion: Normal and tangential components
When the path along which a particle is moving is known, it is convenient to
describe the motion using n and t components (normal and tangent) to the path, and at the
instant considered here their origin located at the particle.
Planer motion : ( at instant considered )
o’ - center of curvature.
s - radius of curvature.
 t–axis - tangent to the curve at P.
 n-axis - perpendicular to the t- axis,
directed from P towards the center of
curvature.
 Positive direction , will be designated
by the unit vector, u n ( normal ) and u t
( tangent ).
Velocity :Since the particle moving , s is a function of time. The particle velocity v has a
direction that is always tangent to the path, and the magnitude that is determined by
taking the time derivative of the path function s = s(t) .
v = ds

dt
v = vu t
.
where v = s
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Acceleration :The acceleration of the particle is the time rate of the change of the velocity.
.
.
.
a = v = vut+v ut
by formulation ,
.
v
s
u t = ou n= u n = u n
s
s
.
.
substitute to the above equation
a = at u t + an un
.
where a t = v
or
a t ds = v.dv
and
v2
an =
s
magnitude of acceleration is the
positive value of a =
at  a n
2
2
Two special cases of motion :
1) The particle moves along a straight line , s   .
.
 an  o , a  at = v
The tangential components of acceleration represents the time rate of change in the
magnitude of the velocity.
2) The particle moves along a curve with a constant speed then
.
at  v  o, a  an  v 2
s.
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12.8 Relative – motion analysis of two particles using Translating Axes
Position :
The axes of this frame are only persuitted to
translate relative to the fixed frame . The
relative position of “B with respect to A”
is designate by a relative position vector
rB A .
rB = rA + rB
Velocity :
A
An equation that related the velocities of the particle can be determined by
taking the time derivative.
vB = vA + vB A
where
vB =
vA =
vB A=
drB
dt
vB/A
vB
drA
dt
drB
vA
A
dt
v B and v A - refer to absolute velocities
- observed from the fixed frame.
vB A
- relative velocity
- observed from the translating frame.
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Acceleration :
aB/A
aB  a A  aB A
aA
a B A is the acceleration of B as seen by the observer located at
aB
A and translating with
the x’,y’,z’ reference frame.
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