13Jan
1 Kinematics of a Particle
Geometry of
motion
Kinematics
of a
Particle
Rectilinear
Motion
Curvilinear
Motion
Relative
Motion
1.1
Kinematics of a Particle
Rectilinear
Motion
Position
General
Definitions
Uniform
Rectilinear
Motion
Velocity
Acceleration
Str. line motion
Uniformly
Accelerated
Motion
1.6
1 KINEMATICS OF A PARTICLE
The motion of a particle along a
P
O
straight line is termed rectilinear
motion. To define the position P
x
x
of the particle on that line, we
choose a fixed origin O and a
positive direction. The distance x from O to P, with the
appropriate sign, completely defines the position of the particle
on the line and is called the position coordinate of the
particle.
Coordinate system -> define starting point
1.7
P
O
x
x
The velocity v of the particle is equal to the time derivative of
the position coordinate x,
Rate of change of position with
time
dx
v=
dt
and the acceleration a is obtained by differentiating v with
respect to t,
dv
a=
dt
or
d 2x
a= 2
dt
we can also express a as
dv
a=v
dx
1.8
P
O
-
x
+
x
dx
v=
dt
or
dv
a=
dt
d 2x
dv
a = dt 2 or a = v
dx
The velocity v and acceleration a are represented by algebraic
numbers which can be positive or negative.
A positive value for v indicates that the particle moves in the
positive direction, and a negative value that it moves in the
negative direction.
A positive value for a, however, may mean that the particle is
truly accelerated (i.e., moves faster) in the positive direction, or
that it is decelerated (i.e., moves more slowly) in the negative
direction. A negative value for a is subject to a similar
1.9
interpretation.
What is the average acceleration
of a Formula 1 race car?
1.10
What is the average acceleration
of a Formula 1 race car?
• An F1 car will do 0-100 km/h in 2.1
seconds.
• A Formula 1 car will do 4g lateral
acceleration in cornering and up to 5g
acceleration in braking
1.11
Two types of motion are frequently encountered: uniform
rectilinear motion, in which the velocity v of the particle is
constant and
x = xo + vt
and uniformly accelerated rectilinear motion, in which the
acceleration a of the particle is constant and
v = vo + at
x = x o + v ot +
1
2
at2
v2 = vo2 + 2a(x - xo )
1.13
Uniformly Accelerated Motion
- Acceleration of Free Fall
• Neglecting air resistance, free fall
motion is constantly accelerated motion
in one dimension
• Taking y drection as positive upward,
ay = -g = -9.80 m/s2
1.14
Hammer and Feather Experiment
1.15
Sometimes it is convenient to use a graphical solution for
problems involving rectilinear motion of a particle. The graphical
solution most commonly involves x - t, v - t , and a - t curves.
At any given time t,
a
v = slope of x - t curve
a = slope of v - t curve
t1
t2
t
v
t2
v2
v1
v2 - v1 =  a dt
t1
t1
t2
t
x
while over any given time interval
t1 to t2,
v2 - v1 = area under a - t curve
x2 - x1 = area under v - t curve
t2
x2
x1
x2 - x1 =  v dt
t1
t1
t2
t
1.17
20Jan
Kinematics of a Particle
Curvilinear
Motion
General
Definitions
Position
Velocity
Acceleration
Rectangular
Coordinates
Path
Coordinates
Polar
Coordinates
Projectile
Motion
Circular
Motion
Entrained &
Relative
Motion
1.18
Example – Planetary Motion
Geocentric model (Ptolemy)
vs
Heliocentric model (Copernicus)
1.19
Ptolemy
1.20
Heliocentric
Model
Copernicus
Galileo
1.21
Curvilinear Motion of a Particle
Example
Motion of a pendulum in 2D
1.22
y
v
P
r
Po
O
s
x
The curvilinear motion of a particle
involves particle motion along a
curved path. The position P of the
particle at a given time is defined by
the position vector r joining the
origin O of the coordinate system
with the point P.
The velocity v of the particle is defined by the relation
dr
v=
dt
The velocity vector is tangent to the path of the particle, and
has a magnitude v equal to the time derivative of the length s of
the arc described by the particle:
ds
v=
dt
1.25
Instantaneous
Velocity
r dr

v  lim
dt
t 0 t
1.26
y
v
dr
v=
dt
ds
v=
dt
P
r
s
Po
x
O
In general, the acceleration a
of the particle is not tangent
to the path of the particle. It
is defined by the relation
a
y
P
r
Po
O
s
x
dv
a=
dt
1.27
Instantaneous Acceleration
• The instantaneous acceleration is the
limit of the average acceleration, Δv/Δt,
as Δt approaches zero
v dv

a  lim
dt
t 0 t
1.28
Producing An Acceleration
• Various changes in a particle’s motion
may produce an acceleration
– The magnitude of the velocity vector may
change
– The direction of the velocity vector may
change
• Even if the magnitude remains constant
– Both may change simultaneously
1.29
Curvilinear Motion
Rectangular Coordinates
1.30
y
vy
P
vx
vz
j
k
r yj
xi
i
z
y
zk
x
ay
P
ax
az
j
k
z
r
i
x
Denoting by x, y, and z the rectangular
coordinates of a particle P, the
rectangular components of velocity and
acceleration of P are equal, respectively,
to the first and second derivatives with
respect to t of the corresponding
coordinates:
.
vx = x
..
ax = x
.
vy = y
..
ay = y
.
vz = z
..
az = z
The use of rectangular components is
particularly effective in the study of the
motion of projectiles.
1.31
Example of Curvilinear Motion
Projectile Motion
1.32
Projectile Motion
• An object may move in both the x and y
directions simultaneously
• The form of two-dimensional motion we
will deal with is called projectile motion
1.33
Curvilinear Motion
Path Coordinates
1.36
Curvilinear motion in Path Coordinate System
y
Given
path
+s
Path coodinate
s(t)
Origin o
P
e$t
q (t) Tangent unit vector
O
x
eˆt  cosq t iˆ  sin q t  ˆj
1.37
Velocity tangential to the path

v  seˆt  veˆt  vq = sq
Given
path
y
+s

v  seˆt
Path coodinate
s(t)
Origin o
P
e$t
q (t) Tangent unit vector
O
x
eˆt  cosq t iˆ  sin q t  ˆj
1.38
How “curved” is the path at P?
+s
s
y
P’
P
O
q
o
q  q ( s)
q'
x
1.39
How “curved” is the path at P?
Osculating circle at point P
+s
(when s0)
Center of Curvature
Radius of Curvature
q
C
r
e$n '
s P’
e$n
y
P
Let s0
O
q0
q
o
q  q ( s)
q'
x
1.40
An infinitesimal segment
of the motion path at point P
coincides with
the osculating circle
at point P
1.41
At P
+s
Circular
Segment
ds  r dq
C dq
r
$e n
y
P
ds
O
q
o
q  dq
x
1.42
Radius of curvature
+s
C
C
C
dq
w
r
O
v
ds
P
s
ds
ds / dt  v
r  Lim


q 0 q
w
dq
dq / dt
1.43
Acceleration
+s
r
y
C

at  seˆt
2

v
an  eˆn
r
P
q
o
x
1.44
y
It is sometimes convenient to resolve
C
the velocity and acceleration of a
v2
particle P into components other
an =
e
r n
than the rectangular x, y, and z
components. For a particle P moving
dv
along a path confined to a plane, we
at = dt et
attach to P the unit vectors et
P
tangent to the path and en normal to
x
O
the path and directed toward the
center of curvature of the path.
The velocity and acceleration are expressed in terms of tangential
and normal components. The velocity of the particle is
27Jan
v = vet
The acceleration is
v2
dv
a=
et +
en
r
dt
1.45
y
C
v = vet
v2
an =
e
r n
P
O
dv
at = dt et
v2
dv
a=
et +
en
r
dt
x
In these equations, v is the speed of the particle and r is the
radius of curvature of its path. The velocity vector v is directed
along the tangent to the path. The acceleration vector a
consists of a component at directed along the tangent to the
path and a component an directed toward the center of
curvature of the path,
1.46
Circular Motion
Centripetal Acceleration
• The change in the
velocity vector is
due to the change in
direction
• The vector diagram
shows Δv = vf - vi
1.47
Curvilinear Motion
Polar Coordinates
1.49
Circular Motion

r  r0eˆr
Position vector of P
y
Unit vector
êr
P
x
O
r0
constant
1.50
Unit vector describes direction
eˆr  cosq iˆ  sinq ˆj
y
q : angular coordinate
q
=
arc
êr
1
(radian)
Unit circle
O
q
1
P
x
r0
constant
360º=2
1.51
Polar coordinates:
êq
y
r0
O
P (r0 ,q )
Transverse
unit vector
q
êr
P
Radial
unit vector
x
0
eˆr  cosq iˆ  sinq ˆj

 ˆ
ˆ
ˆ
 90q) i 
eˆq eˆq cos(
q q 90j ) j
cos
 q sin
sin(
1.52
êr
Radial
unit vector
y
P
êq
Transverse
unit vector
q
r0
x
O
0
1.53
Angular
coordinate
Angular
velocity
Angular
acceleration
q
d
w  q  q
dt
2
d
  2 q  q
dt
1.54
We use q  q (t) to describe
(1) Circular motion of a particle P
(2) Rotation of a rigid disc.
1.55
Ang. acceleration

  q k̂

w  q k̂
Ang. velocity
z
Angular displacement
y
qk̂

r
o
Right hand rule
q

v
P
x
r0
1.56
Circular
motion
Angular
Coordinate
Straight-line
motion
( k̂ )
q
w  q
  q
q kˆ, w kˆ,  kˆ
Cartesian
Coordinate
( iˆ )
x
v  x
a  x
x iˆ, v iˆ, a iˆ
(All are signed magnitudes)
1.57
Circular
motion
Angular
Coordinate
Straight-line
motion
( k̂ )
q
w  q
  q
q kˆ, w kˆ,  kˆ
Cartesian
Coordinate
( iˆ )
x
v  x
a  x
x iˆ, v iˆ, a iˆ
Rotational motion about k̂ Linear motion along iˆ
1.58
Velocity of a particle in circular motion

y
  dr
v
dt

r  r0 eˆr

v
r0
O
q
êr
P
0
x
eˆr  cosq iˆ  sinq ˆj
1.59
Position vector :

r  r0 eˆr  r0 (cosq iˆ  sin q ˆj )
Velocity vector :

 dr d
deˆr
v
 r0eˆr   r0
dt dt
dt
deˆr 
 q (sinq iˆ  cosq ˆj )  w eˆq
dt
1.60
y
êq
r0
O
d
eˆr  w eˆq
êr dt
q
P
0
x
eˆr  cosq iˆ  sinq ˆj
eˆq   sinq iˆ  cosq ˆj
1.61
Velocity vector :

v  r0w eˆq  veˆq
v  wr0
êq
Signed magnitude
Direction
1.62
Use Vector Cross Product
$k
e$q
z
y
O
ˆ
k  eˆr  êq
e$r
r0
x
Right-hand rule
1.63
ˆ
w k  r0 eˆr  wr0 êq
  
w r  v
z

w  wk̂
y
O

v  wr0 eˆq

r0
r  r0eˆr
x
1.64
In
circular
motion

r is a vector
with constant magnitude
d    
r w r  v
dt
1.65
In general, for any vector with
constant magnitude in rotation
z

w

d
 
C  w C
dt
y
x

C with CONSTANT
magnitude
d 
C
dt
1.66
d   
r w r
dt

d
( ) w ( )
dt
Rate of change
due to
rotation
d

eˆr  w  êr
dt

d
eˆq  w  êq
dt
1.67
Acceleration (Tangential and Normal)
 d  d
dv
d
a  v  (veˆq )  eˆq  v (eˆq )
dt
dt
dt
dt

d
eˆq  w  êq  wêr
dt
 dv eˆ  wv (eˆ )
r
a  dt q
Tangential component + Normal component
1.68

dv
at eˆq
a  dt

Tangential component
avn (eˆr )
w
(eˆq )
dv d
 (w r )   r  at
dt dt
Normal component
(eˆr )
wv  w r  an
2
1.69

a  at eˆq 
y
an (eˆr )

at

a
êq
êr

an
0
q
P
x
1.70

ˆ
a
e
a t t
Define:

y

at

a
eˆt  eˆq
eˆn   eˆr
an (eˆn )
ên

an
q
0
êt
ên
êt
P
x
tangential unit vector
normal unit vector
1.71
 a eˆ
a t t
Tangential component
at   r
Normal component

an eˆn
(eˆt )
(eˆn )
2
v
an  vw  rw 
r
2
1.72

at  ( r )eˆt
eˆt  kˆ  eˆr

at  ( kˆ)  ( r eˆr )
 

at    r
k̂
êt
êr
z
k̂
o

at

r
y
P
x
1.73

an  (w v) eˆn
eˆn  kˆ  eˆt

an  (w kˆ)  ( v eˆt )

 
an  w  v
k̂
êt

ên
z
wk̂

v

an
  
2
 w  (w  r )   w r
o
w
P
x
1.74
    
a   r  w v



w
z
y
 

at    r
P
o
x

 
an  w  v
1.75
Physical meaning of components



a  at  an
 
 
  r  w v

Rate of change of magnitude of v

Rate of change of direction of v
1.76
Using polar coordinates rP , θ
Wire rotates about O
y
O
Pivoted at O
rP
θ
Bead slides along the rigid wire
x
It is in curvilinear motion
1.77

drP
d


( rPe$r )
vP 
dt
dt

rP  rPe$r
y

r q eˆ



ˆ
vP  rP er  rPqeˆq
q
P
rP eˆr
Physical Meaning?

rP
O
Physical Meaning?
P
x
1.78
y

vP

ˆ
rv
e
q
P
P 'q

vrPP eˆ/ rf
Entrained
velocity

rP
O
Absolute
velocity
P
Relative
velocity
x
1.79

 
vP  rP eˆr  w  rP



vP / f + vP '
v P = SO
ABSOLUTE
velocity
Curvilinear
RELATIVE
velocity
ENTRAINED
velocity
= rectilinear + Circular
1.80
 
 
d 
rP  = rP  + w  rP 
t
dt
Absolute = Relative + Entrained

w:
Entraining angular velocity
of the moving frame f
1.81
eq
When the position of a particle moving
er
in a plane is defined by its polar
coordinates r and q, it is convenient to
r = r er
P
use radial and transverse
components directed, respectively,
along the position vector r of the
q
particle and in the direction obtained by
x
O
rotating r through 90o
counterclockwise. Unit
vectors er and eq are attached to P and are directed in the radial
and transverse directions. The velocity and acceleration of the
particle in terms of radial and transverse components is
.
.
v = rer + rqeq
..
.. . 2
..
a = (r - rq )er + (rq + 2rq)eq
1.82
eq
r = r er
er
P
.
.
v = rer + rqeq
..
.. . 2
..
a = (r - rq )er + (rq + 2rq)eq
q
O
x
In these equations the dots represent differentiation with
respect to time. The scalar components of the velocity and
acceleration in the radial and transverse directions are
therefore
.
.
vr = r
.. . 2
ar = r - rq
vq = rq
..
..
aq = rq + 2rq
It is important to note that ar is not equal to the time derivative
of vr, and that aq is not equal to the time derivative of vq.
1.83
10Feb
Kinematics of a Particle
Relative
Motion
1 dimension
2
Dimensions
Pulley and
Cable
Systems
1.84
Relative Motion
O
A
B
xB/A
xA
x
xB
When particles A and B move along the same straight line, the
relative motion of B with respect to A can be considered.
Denoting by xB/A the relative position coordinate of B with respect
to A , we have
xB = xA + xB/A
Differentiating twice with respect to t, we obtain
vB = vA + vB/A
aB = aA + aB/A
where vB/A and aB/A represent, respectively, the relative velocity
and the relative acceleration of B with respect to A.
1.85
xC
xA
xB
C
A
B
When several blocks are are connected by inextensible cords,
it is possible to write a linear relation between their position
coordinates. Similar relations can then be written between
their velocities and their accelerations and can be used to
analyze their motion.
1.86
1m/s
1.87
1.88
1m/s
F
A
xA
g
xB
B
1.89
y’
B
y
rB
rA
rB/A
x’
A
z’
z
x
For two particles A and B moving
in space, we consider the relative
motion of B with respect to A , or
more precisely, with respect to a
moving frame attached to A and
in translation with A. Denoting by
rB/A the relative position vector of
B with respect to A , we have
rB = rA + rB/A
Denoting by vB/A and aB/A , respectively, the relative velocity and
the relative acceleration of B with respect to A, we also have
vB = vA + vB/A
and
aB = aA + aB/A
1.90
Given vA0 = vB0 = 50 m/s
Both A and B in curvilinear motion
y
vA0
A
30
30
Will collision happen?
x
vB0
q =?
100 m
B
Example 1-15
1.91
What is the motion of B
relative to A ?
y
A
x’
x’ x’
x’
x’
x’
30
30
y’
y’ y’
y’
y’
y’x
100 m
B
Example 1-15
1.93
Motion path of B relative to A
should be a straight line!!!!
WHY??
x’
A
stationary
To hit A,
make sure
y’


(v B 0  v A 0 )
 
v A  v A0  gtˆj
 
vB  vB 0  gtˆj



vB / A  vB 0  v A0
= constant
B
points to A initially.
1.94



vB / A  vB 0  v A0
Must
be
along
the
line


vB 0
vB / A
150
30

v A0
vB 0 =50
1.95
Follower f Is In translation
y’
Exam 1-17
17Feb
B-x’y’
=
y
B
x’
follower f
vA/B
O
q
r
w
=
A
Point attached
translating frame
x
Body attached
translating frame
vA/B = vA/f
A relative to B-x’y’ ( f ) is rectilinear
1.97
Relative velocity
y’
vB
y
B
follower f
vA x’=
vB
vA
A
vA/B
O
qw
r
B-x’y’
vA/B/f + vB
A’
Entrained velocity
A’
x
vB = vA’
A’ on f is the entraining point for A
Entrained motion of B-x’y’ is rectilinear
1.98
vB
vA
vA =
q
vA = w r (q  90 )

vA/B
r
O
vA/B + vB
q
w
A

w
rsin
q
vA/B =
180
vB = w rcosq 90

Example 1-17
1.99