Lecture (2) - Kinematics of Particles II.pdf

Faculty of Engineering and Materials Science
MECHANICS II
(Dynamics)
4th MCTR, DPE,
MATS, and CIVIL
Lecture (2)
Prof. Dr. E. I. Imam Morgan
Faculty of Engineering and Materials
Science (EMS)
Mechatronics Department
elsayed.morgan@guc.edu.eg
C7.105
Prof. Dr. Imam Morgan
Former Dean EMS, MCTR Depart. 1

Ch. 1 …. Kinematics of Particles (Rectilinear Motion)
 Position
 Motion determination
 Multi connected particles and relative motions
Ch. 2 …. Kinematics of Particles (Curvilinear Motion)
 Position, velocity, and acceleration vectors in general forms and
their relations with respect to the path
 Applications in different frames of references (2D and 3D)
 Rectangular coordinates (2D and 3D)
 Intrinsic Coordinates
 Polar Coordinates
Ch. 3 …. Kinetics of a Particle
● Accelerating Force
Ch. 4 …. Kinetics of a Particle
● Impulse and Momentum
● Energy Approach
Part I
Dynamics of Particles
Lect. (2)

3
Chapter 2
Kinematics of Particles
Curvilinear Motion
Cartesian Coordinates
Intrinsic Coordinates
Polar Coordinates
II

Curvilinear Motion
Particle moving along a curve other than a straight line is in
curvilinear motion. Indeed, in the shown photo, each plane
performs a space curvilinear motion. We are interest to
specify, completely, this motion and to determine the
kinematic quantities that define the motion
Definition

a

r

v

P
Path
O
General
Objectives
 
 
t
r
r
vector
Position
r




 ...
 
 
path
the
to
tangent
t
v
v
vector
Velocity
v




 ...
 
 
path
the
to
angent
t
not
t
a
a
vector
on
Accelerati
a




 ...
►How can we express each
vector at any time.
►How can derive the relation(s)
between them.
►The physical meaning of each
vector (why? and how?)
Kinematic
Parameters

Curvilinear Motion
(General Discussion)
Curvilinear motion occurs when the particle moves along a curved path.
Since this path is often described in three dimensions, vector analysis
(mainly components) will be used to formulate the particle’s position,
velocity, and acceleration.
1 Position Vector, Velocity Vector, and Acceleration vector
A Position Vector
►Position vector of a particle at
time t is defined by a vector
between origin O of a fixed
reference frame and the
position occupied by particle.

 The distance between P(tj) and
P(ti) is the magnitude of the
displacement vector i.e.,
∆Ԧ
𝑟 𝑡𝑖, 𝑡𝑗
∆Ԧ
𝑟1,2 = Ԧ
𝑟2 − Ԧ
𝑟1
 The distance traveled
during the time interval (ti,tj)
is the length of the yellow
curved line shown in the
figure.
It is quite clear that the
magnitudes of the displacement
and the distance traveled over a
given time interval are different!
Displacement & Distance Traveled and Path

The average velocity: is defined by;
Ԧ
𝑣𝑎𝑣 =
∆Ԧ
𝑟
∆𝑡
The instantaneous velocity of the particle at
time t is obtained by considering
that𝑃′approaches P.
So, ∆Ԧ
𝑟 becomes shorter and shorter,
Hence, in the limit at t = 0 it tends to be in
tangent direction. (See next slide )
B Velocity Vector
displacement s
∆s
∆Ԧ
𝑟

Velocity and Path
Points P(t + ∆t/n) are
5 consecutive
Positions of a car
driving around a track.
As Δt goes to zero, the
change of position
vector becomes tangent
to the path. This implies
that …
The velocity vector is always tangent to the path
Details















dt
ds
t
s
v
dt
r
d
t
r
v
t
t
0
0
lim
lim



instantaneous velocity (vector)
instantaneous speed (scalar)
is tangent to
the path
v


C Acceleration Vector
The instantaneous acceleration vector is;
Consider velocity Ԧ
𝑣 of particle at time t and
velocity 𝑣′ at t + t, then,
the average acceleration is given by:
∆ Ԧ
𝑣 =𝑣′ − Ԧ
𝑣
Ԧ
𝑎𝑎𝑣 =
∆ Ԧ
𝑣
∆𝑡
In general, acceleration vector is
directed in the concave side of the path.
dt
v
d
t
v
a
t








 0
lim
where,

On a curved path, the
acceleration has a
component toward the
concave side of the path.
Curved path, velocity
change, point accelerates.
Acceleration and Path
The acceleration vector is always in concave side of
the path (never tangent)
Details
∆ Ԧ
𝑣 =𝑣′ − Ԧ
𝑣
∆ Ԧ
𝑣

motion
of
direction
the
give
not
does
side)
concave
(in
path
the
t to
not tangen
on vector
accelerati
...
motion
of
direction
the
gives
path
the
to
tangent
vector
velocity
...
ector
position v
...
dt
v
d
a
a
dt
r
d
v
v
r









Final Conclusion
Now we are going to know
how can the vector Ԧ
𝑟 be
defined as well as the
derivative of any vector.

►In this case, the position vector of particle P is
given by its rectangular components,
k
z
j
y
i
x
r







►Velocity vector,
k
v
j
v
i
v
k
z
j
y
i
x
k
dt
dz
j
dt
dy
i
dt
dx
v
z
y
x






















The total velocity is tangent to the path.
v

1 Rectangular Components
Now, we will express each vector in terms of its components for the following
different cases;
 Rectangular Components
 Tangential and Normal components
 Radial and Transversal components

►Acceleration vector,
k
a
j
a
i
a
k
z
j
y
i
x
k
dt
z
d
j
dt
y
d
i
dt
x
d
a
z
y
x
























 2
2
2
2
2
2
a

in concave side

Example (1)
A particle is constrained to travel along the shown path.
If 𝑥 = 4𝑡4 m, where t is in seconds, determine the
magnitude and direction of the particle’s velocity and
acceleration when t = 0.5 s.
Prove that Ԧ
𝑣 is in tangent direction while Ԧ
𝑎 is in
concave side.(recommended)
Solution: Rectangular coordinates
𝑥 = 4𝑡4 𝑦2 = 16𝑡4
𝑦 = +4𝑡2
𝑣𝑥 = 16𝑡3 𝑣𝑦 = 8𝑡
𝑎𝑥 = 48𝑡2 𝑎𝑦 = 8
𝑣𝑥 = 2 𝑣𝑦 = 4
𝑎𝑥 = 12 𝑎𝑦 = 8
𝑣 = 4.47 𝑚/𝑠 𝜃𝑣 = 63.4𝑜
𝑎 = 14.42 𝑚/𝑠2 𝜃𝑎 = 33.4𝑜
𝜃𝑎
12
8
𝜃𝑣
2
4
at t = 0.5
at t = 0.5
Prove that v is in
tangent direction
Note that 𝜃𝑣 > 𝜃𝑎

Example (2)
At any instant the horizontal position of the weather
balloon is defined by x = (9t) m, where t is in second
and x in meters. 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 both
the velocity and acceleration when t = 2 s.
Solution:
Given: x = 9t and y = (x2/30)
   
  j
t
i
t
j
y
i
x
r
ˆ
30
81
ˆ
9
ˆ
ˆ
:
is
baloon
the
of
ector
position v
The
2













 
 
30
81
30
9
30
find
first
At
2
2
2
t
t
x
y
t
y
y












x (t)
y (t)
 
   
o
r
m
r
j
i
j
i
r
31
18
8
.
10
tan
99
.
20
8
.
10
18
ˆ
8
.
10
ˆ
18
ˆ
30
4
81
ˆ
2
9
1
2
2












 






r
18 m
10.8 m
θr

► Velocity:
   
 
j
v
i
v
j
t
i
j
y
i
x
v
y
x
ˆ
ˆ
ˆ
30
162
ˆ
9
ˆ
ˆ
:
is
baloon
the
of
vector
velocity
The











 


o
x
y
v
y
x
v
v
s
m
v
s
m
v
s
m
v
t
2
.
50
tan
/
058
.
14
8
.
10
9
/
8
.
10
/
9
:
.
sec
2
at
1
2
2


















► Acceleration:
   
j
a
i
a
j
j
y
i
x
a
y
x
ˆ
ˆ
ˆ
30
162
0
ˆ
ˆ
:
is
baloon
the
of
on vector
accelerati
The











 




o
x
y
a
y
x
a
a
s
m
a
s
m
a
a
t
90
tan
/
4
.
5
4
.
5
0
/
4
.
5
0
:
.
sec
2
at
1
2
2

















Prove that v is in
tangent direction
Find components of
a in direction and
perpendicular to v.
   j
t
i
t
r ˆ
30
81
ˆ
9 2




Projectile
Path
Vertical
Plane
Reference
Line
P
Firing (initial)
Velocity
Reference Plane
Projectile
Motion
►An important application
of Cartesian coordinates is
that for study of motion
of a particle that moves in
constant acceleration
field: (Projectile Motion)
►Basic Assumptions:
● g = constant (near
surface of the earth)
● Neglect air resistance
● The projectile is
considered as a
particle.
Important application
g = constant

vo
vx= ?
vy= ?
y(t)=?
x(t)=?
α
Basic Relations
Given: vo ≡ initial velocity
α ≡ inclination of vo
w.r.to horizontal
Required:
Position ……… x (t) , y (t)
Velocity ………vx (t) , vy (t)
Equation of the path y = y (x)
 
   
2
1
0
t
cos
v
x
cos
v
v
x
o
o
x
..




  
   
4
2
1
3
2
gt
t
sin
v
y
gt
sin
v
v
g
y
o
o
y
..








Eliminating t from (2)
and (4) we get the path
x
y
O
 
 
5
cos
2
tan 2
2
2
x
v
g
x
y
o 
 

Parabolic
Path

Important Note
►The equations (1) through (5) can be used to solve any projectile’s
motion if the following conditions are satisfied:
● The origin O is taken at the firing point.
● x-axis is horizontal (in direction of firing).
● y-axis is vertically upward.
x
y
α = 0
x
y
α = -15o
4 km
1 km x
y
x
y

For the shown projectile, find in terms of vo and α the following:
a- The time of flight from A to C.
b- The range R.
c- The maximum height Hmax
C
B
O ≡ A
Hmax
R
x
y
vo
α
Example (3)

a- The time of flight from A to C.
this time is denoted as T = tA→C .
 
 
g
sin
v
T
gT
T
sin
v
T
t
and
y
substitute
and
;
equation
use
y
:
C
at
o
o


2
2
1
0
0
4
0
2






b- The range R.
you can use equation (2) by
substituting x = R and t = T.
 
 
 
g
sin
v
R
g
sin
v
cos
v
T
cos
v
R
o
o
o
o




2
2
2



C
B
A
vo
α
R
Hmax
y
x
 
   
2
cos
1
cos
t
v
x
v
v
o
o
x




 
   
4
2
1
sin
3
sin
2
gt
t
v
y
gt
v
v
o
o
y






 
 
5
cos
2
tan 2
2
2
x
v
g
x
y
o 
 


vo
α = 45o
Rmax
The maximum rang
Rmax is obtained when
α = 45o in condition
that vo is kept constant
g
v
R o
max
2

 
g
v
R o 
2
sin
2


c- The maximum height Hmax .
at B we have the condition that vy = 0
So, use equation (3) to get the time of flight from A to B.
 
g
v
H
g
v
g
g
v
v
H
in
substitute
g
v
t
t
g
v
o
o
o
o
o
B
A
B
A
o
2
sin
sin
2
1
sin
sin
;
4
sin
sin
0
2
2
max
2
max



























…… (6)
The equations for T, R, and
Hmax are valid only for this
Figure.
Important
 
   
2
cos
1
cos
t
v
x
v
v
o
o
x




 
   
4
2
1
sin
3
sin
2
gt
t
v
y
gt
v
v
o
o
y






 
 
5
cos
2
tan 2
2
2
x
v
g
x
y
o 
 

C
B
A
vo
α
R
Hmax
y
x

C
B
O ≡ A
x
y
y
E
D
vE
vD
Important Note
►The projectile will have the same velocity, as it passes through the same
height y: vD= vE . Moreover, the velocity at y can be obtained from:
y
g
v
v o 2
2
2

 Proof is highly
recommended
vo
α

A projectile is fired from the edge of a 150 m cliff with an initial velocity
vo = 180 m/s as shown.
a) Determine the horizontal distance (R) from A to C.
b) Calculate the time of flight from A to C.
c) Find the maximum height (hmax) above the lower horizontal plane.
R
hmax
C
B
A
R
hmax
C
B
A
y
x
vo= 180 m/s
α = 30o
Example (4)

a) Determine the horizontal distance (R)
from A to C.
 
   
 
 
m
.
R
refused
m
.
R
,
or
m
.
R
R
.
R
.
R
cos
.
tan
R
y
and
R
x
:
C
at
where
,
path
the
of
equation
Use
7
3097
8
239
7
3097
0
150
577
0
10
019
2
30
180
2
81
9
30
150
150
5
2
4
2
2
2
















◄
b) Calculate the time of flight from A to C.
 
   
 
sec
.
T
refused
sec
.
T
,
or
sec
.
T
.
T
.
T
T
.
T
sin
y
substitute
equation
In
91
19
54
1
91
19
0
6
30
37
18
81
9
2
1
30
180
150
150
4
2
2













◄
R
hmax
C
B
A
y
x
vo= 180 m/s
α = 30o
Eq. (2) can be used
with x = 3097.7 m
 
   
2
cos
1
cos
t
v
x
v
v
o
o
x




 
   
4
2
1
sin
3
sin
2
gt
t
v
y
gt
v
v
o
o
y






 
 
5
cos
2
tan 2
2
2
x
v
g
x
y
o 
 


c) Find the maximum height (hmax) above the lower horizontal plane.
   
 
m
h
.
sin
h
:
before
obtained
is
axis
x
above
H
height
The
v
:
B
at
max
max
max
y
563
413
150
81
9
2
30
180
150
0
2
2









◄
R
hmax
C
B
A
y
x
vo= 180 m/s
α = 30o

Example (5)
A projectile is fired with an initial
velocity of 800 m/s at a target B
located 2000 m above the gun A at
horizontal distance of 12000 m.
Neglecting air resistance, determine
the value of the firing angle α.
Solution:
We can use directly equation of
the path:
2
2
2
cos
2
tan x
v
g
x
y
o 
 

Where, at B:
x = 12000 m, y = 2000 m
800 m/s
m
m
800 m/s
m
m
y
x
O

 
 2
2
2
12000
cos
800
2
2
.
32
tan
12000
2000

 

Knowing that:



2
2
2
tan
1
sec
cos
1



then, 2000 = 12000 tan α – 3622.5 [ 1 + tan2α ]
3622.5 tan2α – 12000 tan α + 5622.5 = 0
tan α1 = 0.565 or, tan α2 = 2.748
α1 = 29.5o or, α2 = 70o
So, the target will be hit if either of these two firing
angles is used (two possible solutions)
Recommended:
Calculate the time of flight
from A to B in each angle.

θ
α
ሷ
𝑥 = 0 ሷ
𝑦 = −g
𝑣𝑥 = 𝑣𝑜𝑐𝑜𝑠𝛽 𝑣𝑦 = 𝑣𝑜𝑠𝑖𝑛𝛽 − 𝑔𝑡
𝑥 = (𝑣𝑜𝑐𝑜𝑠𝛽)𝑡 𝑦 = (𝑣𝑜𝑠𝑖𝑛𝛽)𝑡 −
1
2
𝑔𝑡2
ሷ
𝑥 = −𝑔𝑠𝑖𝑛𝜃 ሷ
𝑦 = −𝑔 𝑐𝑜𝑠𝜃
𝑣𝑥 = 𝑣𝑜𝑐𝑜𝑠 ∝ − 𝑔𝑠𝑖𝑛𝜃 𝑡 𝑣𝑦 = 𝑣𝑜𝑠𝑖𝑛 ∝ − 𝑔 𝑐𝑜𝑠𝜃 𝑡
𝑥 = (𝑣𝑜𝑐𝑜𝑠 ∝)𝑡 −
1
2
𝑔𝑠𝑖𝑛𝜃 𝑡2
𝑦 = (𝑣𝑜𝑠𝑖𝑛 ∝)𝑡 −
1
2
(𝑔𝑐𝑜𝑠𝜃)𝑡2
Horizontal – Vertical Frame x-y
Along and normal to the incline Frame x-y
x
y
x
y
𝛽 = 𝛼 + 𝜃
Important
Different Frames
Best used
to find hmax

This type of coordinates is used
whenever the path of the particle
is known (well defined).
2 Tangential and Normal Components
(Intrinsic or,
Natural Coordinates)
or
Sharp turn has a
large normal
acceleration, often
equal to several g.

Cars traveling along a
clover-leaf interchange
experience an acceleration
due to a change in velocity
as well as due to a change
in direction of the velocity.
If the car’s speed is
increasing at a known rate as
it travels along a curve, how
can we determine the
magnitude and direction of
its total acceleration?
►
►
Why would you care about the total acceleration of
the car? …. The answer will be clarified in Kinetics

s = s(t)
Position
The position of the particle on the shown curved plane path is
defined by the curvilinear distance s covered by the particle
during time t.
The reference point 𝑂′
is considered as a point on the path.
Plane
Path
Ƹ
𝑒𝑛 Ƹ
𝑒𝑡
𝑠(𝑡)
ሖ
𝑂
P
 Two perpendicular unit
vectors; Ƹ
𝑒𝑡 and Ƹ
𝑒𝑛 are
attached to the particle.
 Ƹ
𝑒𝑡 is directed along the
tangent to the path and
pointing in the direction
of motion, while Ƹ
𝑒𝑛 is
always pointing toward
the concave side of the
path.

Velocity
Plane
Path
Ƹ
𝑒𝑛 Ƹ
𝑒𝑡
𝑠(𝑡)
ሖ
𝑂
P
Ԧ
𝑣
The speed of the particle is defined before
as the rate of change of the covered
distance.
Therefore the speed is given by:
𝑣 =
𝑑𝑠
𝑑𝑡
= ሶ
𝑠
It is known that the velocity
vector is always directed
tangent to the path,
Therefore,
Ԧ
𝑣 = 𝑣 Ƹ
𝑒𝑡

  n
t e
v
e
s
dt
v
d
a ˆ
ˆ
2
















tangential
acceleration
( at )
normal
acceleration
( an )

2
v
a
s
v
dt
dv
a
n
t



 


Tangent to the path.
with or against v.
Always towards center
of curvature.
concav
in
always
.......
a
a
a n
t
tot
2
2

 In concave side
Acceleration
Plane
Path
Ƹ
𝑒𝑛
Ƹ
𝑒𝑡
𝑠(𝑡)
ሖ
𝑂
P
Ԧ
𝑎𝑛
Ԧ
𝑎𝑡
Ԧ
𝑎
C
ρ
C .... Center of curvature
ρ .... Radius of curvature (PC)

an1
ρ1
ρ2
an2
ρ = ∞
v1
v2
Each point on the path has its own
radius of curvature depending on the
shape of the path. In any position;
Physical meaning
of at and an
Acceleration
vectors from
A to B.
𝑎𝑛 =
𝑣2
𝜌

Remember !!!
From Mathematics: If the path of the particle is defined by
the function y = y(x), then, the radius of curvature ρ at certain
point A(x,y) is given by:
Path:
y = y(x)
A
y
x
ρ
C
ρ …. Radius of curvature
C …. Center of curvature
𝜌 =
1 + ൗ
𝑑𝑦
𝑑𝑥
2
3
2
ൗ
𝑑2𝑦
𝑑𝑥2

B
t
A
t = 0
vA
vB
at = const.
s
s
a
v
v
t
a
t
v
s
t
a
v
v
t
A
B
t
A
t
A
B
2
2
1
2
2
2






Negative sign for
deceleration
Motion with constant tangential acceleration
Special Case
Path

A motorist is traveling on curved section of
highway at 108 km/h. The motorist applies
brakes causing a constant deceleration rate.
Knowing that after 8 s the speed has been
reduced to 36 km/h, determine the acceleration
of the automobile immediately after the brakes
are applied.
vA=108 km/h
750 m
an=1.2 m/s2
at=2.5 m/s2
 
 
   
o
A
n
t
t
t
A
B
.
.
.
tan
s
/
m
.
.
.
a
s
/
m
.
v
a
:
A
at
,
Then
s
/
m
.
a
a
t
a
v
v
:
on
decelerati
t
tan
cons
For
6
25
5
2
2
1
78
2
2
1
5
2
2
1
750
30
5
2
8
30
10
1
2
2
2
2
2
2
2






















s
m
h
km
v
s
m
h
km
v
B
A
/
10
/
36
/
30
/
108




Example (1)

Think over !!!!
vA
vB
θ
C
D
B
A
ρ
𝑎𝑡 = −𝑘𝑡
Given:
vA = 108 km/h
at = − kt where k is known constant
ρ = 200 m
Find:
Part I: for k = 6;
a)Is it possible to reach B?
b)Determine the location of the
car when it comes to
instantaneous stop.
c)Find the Cartesian components
of acceleration at that location.
Part II :
Determine the maximum value of k
just to reach point B.

Recommended
If the motion starts from
A, determine the position
(location) of the car at this
instant.
Example (2)
A racer car C travels around the horizontal
circular track that has a radius of 300 m as
shown in figure. If the car increases its speed at
a constant rate of 1.5 m/s2, starting from rest,
determine the time needed for it to reach an
acceleration of 2 m/s2. What is its speed at this
instant?
𝑣 = 𝑣𝑜 + 𝑎𝑡 𝑡 ∴ 𝑣 = 1.5𝑡
𝑎𝑛 =
𝑣2
𝜌
=
1.5𝑡 2
300
= 0.0075𝑡2 m/s2 and, 𝑎𝑡 = 1.5 m/s2
𝑎 = 𝑎𝑡
2
+ 𝑎𝑛
2
2 = 1.5 2 + 0.0075𝑡2 2
Solving for t: t = 13.28 s
The speed at this instant is given by:
𝑣 = 1.5 13.28 = 19.9 𝑚/𝑠
m
A
●

Example (3)
When the skier reaches point A along the
parabolic path in the figure show, he has a
speed of 6 m/s which is increasing at 2 m/s2.
Determine the direction of his velocity and
the direction and magnitude of his
acceleration at this instant.
Solution:
► By definition, the velocity is directed
tangent to the path.
  axis
x
with
angle
an
makes
v
A
at
dx
dy
m
x
at
x
dx
dy
So
x
y
Since
o
45
1
tan
,
1
/
10
10
1
/
,
20
1
1
2










Therefore, vA = 6 m/s
45o
vA

Acceleration
● The tangential acceleration is given: at = 2 m/s2
● To find the normal acceleration (an) , the radius of
curvature at A must be determined:
 
 
2
2
2
2
3
2
2
3
2
2
2
/
273
.
1
28
.
28
6
)
10
(
28
.
28
10
1
10
1
1
10
1
/
1
10
1
/
s
m
v
a
x
at
m
x
dx
dy
dx
y
d
Since
n 

























  o
s
m
a
5
.
57
273
.
1
2
tan
/
37
.
2
273
.
1
2
1
2
2
2







Therefore, aA = 2.37 m/s2 12.5o
aA
Determine Ƹ
𝑒𝑡 and Ƹ
𝑒𝑛 at A in
terms of Ƹ
𝑖 and Ƹ
𝑗 and hence
find the Cartesian
components of Ԧ
𝑣 and Ԧ
𝑎.

n
t
n
t
t
e
d
e
d
j
i
e
but
j
i
d
e
d
j
i
e
ˆ
ˆ
ˆ
cos
ˆ
sin
ˆ
ˆ
cos
ˆ
sin
ˆ
ˆ
sin
ˆ
cos
ˆ


















i
ˆ
θ
θ
t
e

n
e

y
x
ĵ
θ
►Velocity vector of particle is tangent to path
of particle. In general, acceleration vector
is not. Aim: is to express the acceleration
vector in terms of tangential and normal
components.
Therefore; differentiating of a rotating unit vector Ƹ
𝑒𝑡 with respect to the angle of
rotation θ leads to another unit vector Ƹ
𝑒𝑛 that is perpendicular to Ƹ
𝑒𝑡 and in sense of
increasing ofθ.
Differentiation
of rotating unit vector
𝑑 Ƹ
𝑒𝑛
𝑑𝜃
= − Ƹ
𝑒𝑡
and,

After substituting,
 Tangential component of acceleration
reflects change of speed and normal
component reflects change of direction of
the speed.
 Tangential component may be positive or
negative. Normal component always points
toward center of path curvature.
 With the velocity expressed as vector Ԧ
𝑣 = 𝑣 Ƹ
𝑒𝑡
the particle acceleration can be written as:
Ԧ
𝑎 =
𝑑
𝑑𝑡
𝑣 Ƹ
𝑒𝑡 =
𝑑𝑣
𝑑𝑡
Ƹ
𝑒𝑡 + 𝑣
𝑑 Ƹ
𝑒𝑡
𝑑𝑡
=
𝑑𝑣
𝑑𝑡
Ƹ
𝑒𝑡 + 𝑣
𝑑 Ƹ
𝑒𝑡
𝑑𝜃
𝑑𝜃
𝑑𝑠
𝑑𝑠
𝑑𝑡
𝑑 Ƹ
𝑒𝑡
𝑑𝜃
= Ƹ
𝑒𝑛 , 𝑑𝑠 = 𝜌𝑑𝜃 and 𝑣 =
𝑑𝑠
𝑑𝑡
Ԧ
𝑎 =
𝑑𝑣
𝑑𝑡
Ƹ
𝑒𝑡 +
𝑣2
𝜌
Ƹ
𝑒𝑛
𝑎𝑛
𝑎𝑡



2
2
v
a
dt
dv
a
e
v
e
dt
dv
a n
t
n
t 






■ Relations for tangential and normal acceleration also
apply for particle moving along space curve.
■ Plane containing tangential and normal unit vectors is
called the osculating plane.
n
t
b e
e
e





■ Normal to the osculating
plane is found from
binormal
e
normal
principal
e
b
n




■ Acceleration has no component along binormal.
If the particle moves along a space curve, then at
a given instant the t axis is uniquely specified;
however, an infinite number of straight lines can
be constructed normal to the tangent axis.
t
n
the positive n axis directed toward the path’s center of
curvature. This axis is referred to as the principle
normal to the curve. With the n and t axes so defined,
Motion in Space (3D)

End of
Lecture
Thank you: Imam Morgan

Lecture (2) - Kinematics of Particles II.pdf

More Related Content

Similar to Lecture (2) - Kinematics of Particles II.pdf

Recently uploaded

Lecture (2) - Kinematics of Particles II.pdf