KINEMATICS OF PARTICLES
___________________________________
PROJECTILE MOTION, NORMAL
TANGENTIAL AND CYLINDRICAL
COMPONENTS
LECTURE #1
KINEMATICS OF PARTICLES
___________________________________
PROJECTILE MOTION
LECTURE #1
Objectives of Today’s Lecture
Students will be able to analyze:
 Free-flight
motion of a projectile
 Kinematic equations for projectile motion
 Particle motion along a curved path using three
coordinate systems
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Projectile Motion
• Projectile: any body that is given an initial velocity
and then follows a path determined by the effects of
gravitational acceleration and air resistance.
• Trajectory – path followed by a projectile
4
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Concept of Projectile Motion
• Projectile motion can be treated as two rectilinear
motions, one in the horizontal direction experiencing
zero acceleration and the other in the vertical direction
experiencing constant acceleration (i.e., gravity).
Consider the two balls on the left.
The red ball falls from rest, whereas the
yellow ball is given a horizontal velocity. Each
picture in this sequence is taken after the
same time interval. Notice both balls are
subjected to the same downward acceleration
since they remain at the same
elevation/height at any instant. Also, note that
the horizontal distance between successive
photos of the yellow ball is constant since the
velocity in the horizontal direction is constant.
5
Horizontal Motion is Uniform Motion
Notice that the Horizontal motion is in no way affected by the Vertical motion.
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Applications of Projectile Motion
• A kicker should know at what angle, q, and initial
velocity, vo, he must kick the ball to make a field
goal.
• For a given kick “strength”, at what angle should the
ball be kicked to get the maximum distance?
7
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Applications of Projectile Motion
• A fireman wishes to know the maximum height on
the wall he can project water from the hose. At what
angle, q, should he hold the hose?
8
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Kinematic Equations
Vertical Motion
Horizontal Motion
9
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Kinematic Equations
Horizontal Motion
Since ax = 0, the velocity in the horizontal direction remains
constant (vx = vox) and the position in the x direction can be
determined by:
x = xo + (vox)(t)
Why is ax equal to zero (assuming movement through the air)?
10
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Kinematic Equations
Vertical Motion
Since the positive y-axis is directed upward, ay = -g. Application of
the constant acceleration equations yields:
vy = voy – g(t)
y = yo + (voy)(t) – ½g(t)2
vy2 = voy2 – 2g(y – yo)
For any given problem, only two of these three equations can be used. Why?
11
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Kinematic Equations
• Horizontal and vertical components of velocity are
independent.
• Vertical velocity decreases at a constant rate due to the
influence of gravity.
12
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Mathematical Verification
Horizontal Motion
• Acceleration : ax= 0

() v  v0  ac t

1 2
() x  x0  v0t  ac t
2

() v 2  v0  2ac ( s  s0 )
2
v x  (v0 ) x
x  x0  (v0 ) x t
v x  (v0 ) x
• Conclusion # 1: Horizontal velocity remains constant
• Conclusion # 2: Equal distance covered in equal time intervals
( x  x0 )
vx 
t
13
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Mathematical Verification
Vertical Motion
( ) v  v0  ac t
v y  (v0 ) y  gt
1
( ) y  y0  v0t  ac t 2
2
y  y0  (v0 ) y t 
(   ) v  v 0  2 ac ( s  s 0 )
2
2
2
2
y
y
v  v0
1 2
gt
2
 2 g ( y  y0 )
Conclusion # 1: Equal increments of speed gained in equal
increments of time
Distance increases in each time interval
14
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Projectile Motion
• Assumptions:
(1) free-fall acceleration
(2) neglect air resistance
• Choosing the y direction as positive upward:
ax = 0; ay = - g (a constant)
y
• Take x0= y0 = 0 at t = 0
• Initial velocity v0 makes an
angle q0 with the horizontal
v 0 x  v 0 cosq 0
v0
q
x
v 0 y  v 0 sin q 0
15
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Projectile Motion
Maximum Height [altitude]
At the peak of its trajectory, vy = 0.
From
v y  v0 y  gt  voy  gt  0
t1 
Time t1 to reach the peak
Substituting into:
v0 y
g
1 2
y  v0 y t  gt
2
h  ymax 
v02y
2g
16
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Projectile Angle
• The optimal angle of projection is dependent on the
goal of the activity.
• For maximal height the optimal angle is 90o.
• For maximal distance the optimal angle is 45o.
17
Projection angle = 10 degrees
10 degrees
Projection angle = 45 degrees
10 degrees
30 degrees
40 degrees
45 degrees
Projection angle = 60 degrees
10 degrees
30 degrees
40 degrees
45 degrees
60 degrees
Projection angle = 75 degrees
10 degrees
30 degrees
40 degrees
45 degrees
60 degrees
75 degrees
So angle that maximizes Range
(qoptimal) = 45 degrees
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Projectile Motion
A ball is given an initial velocity of V0 = 37 m/s at an angle of q = 53.1. Find
the position of the ball, and the magnitude and direction of its velocity, when t
= 2.00 s. Find the time when the ball reaches the highest point of its flight, and
find its height h at this point
The initial velocity of the ball has components:
v0x = v0 cos θ0 = (37.0 m/s) cos 53.1° = 22.2 m/s
v0y = v0 sin θ0 = (37.0 m/s) sin 53.1°= 29.6 m/s
a)
position
x = v0xt = (22.2 m/s)(2.00 s) = 44.4 m
y = v0yt - ½gt2
= (29.6 m/s)(2.00 s) –½ (9.80 m/s2)(2.00 s)2
= 39.6 m
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Projectile Motion
A ball is given an initial velocity of V0 = 37 m/s at an angle of q = 53.1. Find
the position of the ball, and the magnitude and direction of its velocity, when t
= 2.00 s. Find the time when the ball reaches the highest point of its flight, and
find its height h at this point
• Velocity
• vx = v0x = 22.2 m/s
• vy = v0y – gt = 29.6 m/s – (9.80 m/s2)(2.00 s) = 10.0 m/s
v  vx2  v y2 
22.2 m / s 2  (10.0 m / s) 2
 24.3 m / s
 10.0 m / s 
  arctan 0.450  24.2
q  arctan 
 22.2 m / s 
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Projectile Motion
A ball is given an initial velocity of V0 = 37 m/s at an angle of q = 53.1. Find
the position of the ball, and the magnitude and direction of its velocity, when t
= 2.00 s. Find the time when the ball reaches the highest point of its flight, and
find its height h at this point
• b) Find the time when the ball reaches the highest point of its flight, and find
its height H at this point.
1 2
v y  0  v0 y  gt1
H  v0 y t1  gt 1
2
v0 y 29.6 m / s
1
t1 

 3.02 s
2
2
2

(
29
.
6
m
/
s
)(
3
.
02
s
)

(
9
.
80
m
/
s
)(
3
.
02
s
)
g 9.80 m / s
2
 44.7 m
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Projectile Motion
A ball is given an initial velocity of V0 = 37 m/s at an angle of q = 53.1. Find
the position of the ball, and the magnitude and direction of its velocity, when t
= 2.00 s. Find the time when the ball reaches the highest point of its flight, and
find its height h at this point
c) Find the horizontal range R, (that is, the horizontal distance from the starting
point to the point at which the ball hits the ground.)
1 2
1
y  0  v0 y t 2  gt 2  t 2 (v0 y  gt 2 )
2
2
2v0 y
2(29.6 m / s )
t 2  0 and t 2 

 6.04 s
2
g
9.80 m / s
R  v0 xt2  (22.2 m / s)(6.04 s) 134 m
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Projectile Motion
A ball traveling at 25 m/s drive off of the edge of a cliff 50 m
high. Where does it land?
Horizontally
25 m/s
x = x0 + (v0)x t
Initial Conditions
x = 25 *3.19 = 79.8 m
vx = 25 m/s
Vertically
vy0 = 0 m/s
v = v0-gt
a =- 9.8 m/s2
y = y0 + v0t + 1/2gt2 ….
t=0
v2 = v02 - 2g(y-y0)….
y0 = 0 m
y =- 50 m
x0 =0 m
-50 = 0+0+1/2(-9.8)t2 … t = 3.19 s
79.8 m
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Example
Given: vo and θ
Find: The equation that defines
y as a function of x.
Plan: Eliminate time from the
kinematic equations.
Solution: Using
vx = vo cos θ
We can write: x = (vo cos θ)t
or
and
t =
y = (vo sin θ)t – ½ g(t)2
vy = vo sin θ
x
vo cos θ
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Solution
Simplifying the last equation, we get:
y = (x tanq) –
(
g x2
2vo2
)
(1 + tan2q)
The above equation is called the “path equation” which describes
the path of a particle in projectile motion. The equation shows that
the path is parabolic.
ASSIGNMENT
DEADLINE

REVIEW EXAMPLES
 Example
12.11
 Example 12.12
 Example 12.13
QUESTIONS
THANK YOU
FOR YOUR
INTEREST
KINEMATICS OF PARTICLES
___________________________________
GENERAL CURVILINEAR MOTION &
NORMAL & TANGENTIAL COMPONENTS
AND CYLINDRICAL COMPONENTS
LECTURE #1
Kinematics of Particles
General Curvilinear Motion
OBJECTIVE
Students should be able to:
1. Determine the normal and tangential components of
velocity and acceleration of a particle traveling along a
curved path.
2. Determine velocity and acceleration components using
cylindrical coordinates
33
Kinematics of Particles
General Curvilinear Motion
Normal and Tangential Components I
When a particle moves along a curved path, it is sometimes convenient
to describe its motion using coordinates other than Cartesian. When
the path of motion is known, normal (n) and tangential (t) coordinates
are often used
In the n-t coordinate system, the origin is
located on the particle (the origin moves
with the particle)
The t-axis is tangent to the path (curve) at the instant considered,
positive in the direction of the particle’s motion
The n-axis is perpendicular to the t-axis with the positive direction
toward the center of curvature of the curve
34
Kinematics of Particles
General Curvilinear Motion
Normal and Tangential Components II
The positive n and t directions are
defined by the unit vectors un and ut
respectively
The center of curvature, O’, always lies
on the concave side of the curve.
The radius of curvature, r, is defined
as the perpendicular distance from the
curve to the center of curvature at that
point
The position of the particle at any instant is defined by the
distance, s, along the curve from a fixed reference point.
35
Kinematics of Particles
General Curvilinear Motion
Velocity in the n-t coordinate system
The velocity vector is always tangent
to the path of motion (t-direction)
The magnitude is determined by taking the
time derivative of the path function, s(t)
v = vut
where
v = ds/dt
Here v defines the magnitude of the velocity (speed) and
ut defines the direction of the velocity vector.
36
Kinematics of Particles
General Curvilinear Motion
Acceleration in the n-t coordinate system I
Acceleration is the time rate of change of velocity:
·
a = dv/dt = d(vu )/dt = vu
t
.
+ v ut
.
Here v represents the change in
.
the magnitude of velocity and ut
represents the rate of change in
the direction of ut.
t
After mathematical manipulation,
the acceleration vector can be
expressed as:
.
a = vut + (v2/r)un = atut + anun
37
Kinematics of Particles
General Curvilinear Motion
Acceleration in the n-t coordinate system II
There are two components to
the acceleration vector:
a = at ut + an un
The tangential component is tangent to the curve and in the direction of
increasing or decreasing velocity.
.
at = v
or
at ds = v dv
The normal or centripetal component is always directed toward the center of
curvature of the curve. an = v2/r
The magnitude of the acceleration vector is
a = [(at)2 + (an)2]0.5
38
Kinematics of Particles
General Curvilinear Motion
Special cases of motion I
There are some special cases of motion to consider
1)
The particle moves along a straight line.
r --->

=>
an =
v2/r
=0
=>
.
a = at = v
The tangential component represents the time rate of change in the
magnitude of the velocity.
39
Kinematics of Particles
General Curvilinear Motion
Special cases of motion II
There are some special cases of motion to consider
2)
The particle moves along a curve at constant speed.
.
at = v = 0
=>
a = an = v2/r
The normal component represents the time rate of change in the
direction of the velocity.
40
Kinematics of Particles
General Curvilinear Motion
Special cases of motion III
There are some special cases of motion to consider
3) The tangential component of acceleration is constant, at = (at)c.
In this case,
s = so + vot + (1/2)(at)ct2
v = vo + (at)ct
v2 = (vo)2 + 2(at)c(s – so)
As before, so and vo are the initial position and velocity of the
particle at t = 0
41
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Normal and Tangential Components
• To investigate particle motion along a curved path
“Curvilinear Motion” using three coordinate systems
– Rectangular Components
• Position vector
• Velocity
• Acceleration
r=xi+yj+zk
v = vx i + vy j + vz k
a = ax i + ay j +az k
(tangent to path)
(tangent to hodograph)
– Normal and Tangential Components
– Polar & Cylindrical Components
42
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Normal and Tangential Components
• If the path is known i.e.
– Circular track with given radius
– Given function
y
1 2
x
60
y  15 ln(
x
)
80
• Method of choice is
normal and tangential components
43
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Planer Motion
• From the given geometry and/or given
function
• More emphasis on radius of curvature
velocity and acceleration
• At any instant the origin is located at the particle
itself.
• The t-axis is tangent to the curve at P and is +ve in
the direction of increasing s.
• The normal axis is perpendicular to t-axis and
directed toward the center of curvature O’.
• un is the unit vector in the normal direction while ut is
a unit vector in the tangential direction.
44
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Radius of Curvature
• For the Circular motion :
(r) = radius of the circle
• For y = f(x):

1  (dy / dx) 
r
2 3/ 2
d 2 y / dx 2
45
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Velocity
• The particle velocity is always tangent to the path.
• Magnitude of velocity is the time derivative of path
function s = s(t)
ds

 s
dt
v   ut
– From constant tangential acceleration
  o  (at ) c t
– From time function of tangential acceleration
dv
at 
dt

 dv   a dt
t
– From acceleration as function of distance
v dv  at ds 
 vdv   f (s)ds
46
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Acceleration
Acceleration is time derivative of velocity
v   ut
a  v   u t   u t

u t  u n
r
2
a   u t  u n
r
2
an 
r
at  
a  at u t  an u n
a  at  an
2
2
47
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Types of Acceleration
3 types of acceleration:
– linear
– radial (centripetal)
– angular
• Linear acceleration: is a
change in speed without
change in direction (increase in
thrust in straight-and-level
flight)
• Radial (or centripetal)
acceleration : when there is a
change in direction (turn, dive)
• Angular acceleration: when
body speed and direction are
changed (tight spin)
48
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Special Case
1- Straight line motion
r 
an  0
a  at  
2- Constant speed curve
motion (centripetal
acceleration)
at    0
2
a  an 
r
49
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Centripetal Acceleration
• Recall that acceleration is defined as a change
in velocity with respect to time.
• Since velocity is a vector quantity, a change in
the velocity’s direction , even though the
speed is constant, represents an acceleration.
• This type of acceleration is known as
Centripetal acceleration
2
ac 
r
at    0
50
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Sample Problem
• A truck is traveling a long a circular path having a radius of 50 m at a
speed of 4 m/s. For a short distance from s = 0, its speed is increased
by   (0.05s) m / .s 2Where s is in meters. Determine its speed and the
magnitude of its acceleration when it moved s = 10 m.
v
10
 vdv   0.05s ds
4
at  0.05 s  0.05(10)  0.5 m / s 2
0
1 2  0.05 2 
v  
s 
2 4
2
0
(4.58) 2
an 

 0.42 m / s 2
r
50
1 2
0.05
v 8 
(10) 2  0
2
2
a  (0.42) 2  (0.5) 2  0.653 m / s 2
v
10
v2
v  4.58 m / s
51
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Components
• Particle motion along a curved path “Curvilinear Motion” using
three coordinate systems
– Rectangular Components
• Position vector r = x i + y j + z k
• Velocity
v = vx i + vy j + vz k
• Acceleration a = ax i + ay j +az k
(tangent to path)
(tangent to hodograph)
– Normal and Tangential Components
• Position (particle itself)
• Velocity
v = u ut
2
a   u t  u n
• Acceleration
r
1  (dy / dx) 
r
2 3/ 2
d 2 y / dx 2
(tangent to path)
(normal & tangent)
– Polar & Cylindrical Components
52
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Curvilinear Motion : Cylindrical Components
• Observed and/or guided from origin or from the
center
• Cylindrical component
r , q , and z
• Polar component “plane motion”
r and q
53
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Application
• Circular motion but observed and/or controlled from
the center
54
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Polar Coordinates
•
•
•
•
•
•
Radial coordinate r
Transverse coordinate
q and r are perpendicular
Theta q in radians
1 rad = 180o/p
Direction ur and uq
55
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Position, Velocity and Acceleration
• Position vector
• r = r ur
• Instantaneous velocity
= time derivative of r
r  r ur
v  r  r u r  r u r
u r  q u q
v  r u r  r q u q
v   r u r  q u q
Where
 r  r and q  r q
56
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Position, Velocity and Acceleration
• Magnitude of velocity
  (r) 2  (rq) 2
• Angular velocity q
• Tangent to the path
• Angle = q  d
1 q
d  tan ( )
r
57
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Position, Velocity and Acceleration
• Instantaneous acceleration = time derivative of v
v  r u r  rq u q
a  v  r u r  r u r  rq u q  rq u q  rq u q
u r  q u q
u q  q u r
a  (r  rq 2 )u r  (r q  2 r q)u q
ar  r  rq 2
aq  r q  2 r q
a  ar u r  aq u q
58
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Position, Velocity and Acceleration
•
Angular acceleration
q  d 2q / dt 2  d / dt (dq / dt )
•
Magnitude
a  (r  rq 2 ) 2  (rq  2rq) 2
•
•
Direction “Not tangent”
Angle q  f
aq
f  tan ( )
ar
1
59
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Cylindrical Coordinates
• For spiral motion
cylindrical coordinates
is used r, q, and z.
• Position
rp  r u r  z u z
• Velocity
v  r u r  rq u q  z u z
• Acceleration
a  (r  rq 2 ) u r  (rq  2rq) u q  z u z
60
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Time Derivatives
r, r, q, and q
• If r = r(t) and q  qt)
r  4t 2
q  (8t 3  6)
r  8t
q  24 t 2
r  8
q  48 t
• If r = f(q use chain rule
r  5q
2
r  10 q q
r  10 [ (q) q  q (q) ]
r  10 q 2  10 q q
61
Kinematics of Particles
Projectile Motion, Normal Tangential and Cylindrical Components
Three-Dimensional Motion
• For• spatial
For spatial
motion
motion
required
required
threethree
dimension.
dimension.
• Binomial
• Binomial
axis baxis
which
b which
is perpendicular
is perpendicular
to ut to ut
and uand
un is used
n is used
• ub=• uut bx=uunt x un
What we have learned today?
 Analyzed the free-flight motion of a projectile
 Studied the kinematic equations for projectile motion
 Particle motion along a curved path with respect to three
coordinate systems:
 Rectangular Components
 Normal and Tangential Components
 Polar and Cylindrical Components
Next Lecture

Kinetics of particles


Force
Acceleration
ASSIGNMENT
DEADLINE
NEXT LECTURE
STUDY ALL THE EXAMPLES till
12.20
QUESTIONS
THANK YOU
FOR YOUR
INTEREST