This document discusses kinematics of particles using polar components. It defines the position vector of a particle as the vector from the origin to the particle's position. For curvilinear motion, the instantaneous velocity vector is defined as the limit of the displacement vector divided by the time interval as the interval approaches zero. Similarly, the instantaneous acceleration vector is defined as the limit of the change in velocity vector over the time interval. Polar components (radial and transverse) and tangential and normal components are also introduced to analyze curvilinear motion. Expressions are derived for velocity and acceleration in terms of these component directions. An example problem of a centrifuge is worked out using these concepts.

1. Engineering DYNAMICS
Lecture
9 Kinematics of Particles:
Polar Components

2. The softball and the car both undergo
curvilinear motion.
• A particle moving along a curve other than a
straight line is in curvilinear motion.
Curvilinear Motion: Position, Velocity & Acceleration

3. • The 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.
• Consider a particle which occupies position P defined by at time t
and P’ defined by at t + Dt,
r

r

Curvilinear Motion: Position, Velocity & Acceleration

4. 0
lim
t
s ds
v
t dt
D 
D
 
D
Instantaneous velocity
(vector)
Instantaneous speed
(scalar)
0
lim
t
r dr
v
t dt
D 
D
 
D
Curvilinear Motion: Position, Velocity & Acceleration

5. 0
lim
t
v dv
a
t dt
D 
D
  
D
instantaneous acceleration (vector)
• Consider velocity of a particle at time t and velocity at t + Dt,
v

v


• In general, the acceleration vector is not tangent
to the particle path and velocity vector.
Curvilinear Motion: Position, Velocity & Acceleration

6. • When 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






















• 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
Rectangular Components of Velocity & Acceleration

7. Example 2/5

8. Example 2/5

9. • Rectangular components particularly effective
when component accelerations can be integrated
independently, e.g., motion of a projectile,
0
0 





 z
a
g
y
a
x
a z
y
x 





with initial conditions,
      0
,
,
0 0
0
0
0
0
0 


 z
y
x v
v
v
z
y
x
Integrating twice yields
𝑣𝑥 = 𝑣𝑥 0 𝑣𝑦 = 𝑣𝑦 0
− 𝑔𝑡 𝑣𝑧 = 0
𝑥 = 𝑣𝑥 0𝑡 𝑦 = 𝑣𝑦 0
𝑡 −
1
2
𝑔𝑡2
𝑧 = 0
• Motion in horizontal direction is uniform.
• Motion in vertical direction is uniformly accelerated.
• Motion of projectile could be replaced by two
independent rectilinear motions.
Projectile motion

10. A projectile is fired from the edge
of a 150-m cliff with an initial
velocity of 180 m/s at an angle of
30°with the horizontal. Neglecting
air resistance, find (a) the horizontal
distance from the gun to the point
where the projectile strikes the
ground, (b) the greatest elevation
above the ground reached by the
projectile.
SOLUTION:
• Consider the vertical and horizontal motion
separately (they are independent)
• Apply equations of motion in y-direction
• Apply equations of motion in x-direction
• Determine time t for projectile to hit the
ground, use this to find the horizontal
distance
• Maximum elevation occurs when vy=0
Problem

11. SOLUTION:
Given: (v)o =180 m/s (y)o =150 m
(a)y = - 9.81 m/s2
(a)x = 0 m/s2
Vertical motion – uniformly accelerated:
Horizontal motion – uniformly accelerated:
Choose positive x to the right as shown
Problem

12. SOLUTION:
Horizontal distance
Projectile strikes the ground at:
Solving for t, we take the positive root
Maximum elevation occurs when vy=0
Substitute into equation (2) above
Substitute t into equation (4)
Maximum elevation above the ground =
Problem

13. If we have an idea of the path of a vehicle, it is often convenient to analyze
the motion using tangential and normal components (sometimes called
path coordinates).
Tangential and Normal Components

14. • The tangential direction (et) is tangent to the path of the
particle. This velocity vector of a particle is in this direction
x
y
et
en
• The normal direction (en) is perpendicular to et and points
towards the inside of the curve.
v= vt et
r= the instantaneous
radius of curvature
2
dv v
dt r
 
t n
a e e
v
 t
v e
• The acceleration can have components in both the en and et directions
Tangential and Normal Components
r
r

15. • To derive the acceleration vector in tangential
and normal components, define the motion of a
particle as shown in the figure.
• are tangential unit vectors for the
particle path at P and P’. When drawn with
respect to the same origin, and
is the angle between them.
t
t e
e



and
t
t
t e
e
e






D

D
 
 







d
e
d
e
e
e
e
e
t
n
n
n
t
t







D
D

D
D
D

D

D

D 2
2
sin
lim
lim
2
sin
2
0
0
Tangential and Normal Components
∆𝜽

16. t
e
v
v



• With the velocity vector expressed as
the particle acceleration may be written as
dt
ds
ds
d
d
e
d
v
e
dt
dv
dt
e
d
v
e
dt
dv
dt
v
d
a t
t













but
v
dt
ds
ds
d
e
d
e
d
n
t 

 
r



After substituting,
r
r
2
2
v
a
dt
dv
a
e
v
e
dt
dv
a n
t
n
t 






• The tangential component of acceleration
reflects change of speed and the normal
component reflects change of direction.
• The tangential component may be positive or
negative. Normal component always points
toward center of path curvature.
Tangential and Normal Components

17. A motorist is traveling on a curved
section of highway of radius 2500 ft
at the speed of 60 mi/h. The motorist
suddenly applies the brakes, causing
the automobile to slow down at a
constant rate. Knowing that after 8 s
the speed has been reduced to 45
mi/h, determine the acceleration of
the automobile immediately after the
brakes have been applied.
SOLUTION:
• Define your coordinate system
• Calculate the tangential velocity and
tangential acceleration
• Determine overall acceleration magnitude
after the brakes have been applied
• Calculate the normal acceleration
Problem

18. SOLUTION: • Define your coordinate system
et
en
• Determine velocity and acceleration in
the tangential direction
• The deceleration constant, therefore
• Immediately after the brakes are applied,
the speed is still 88 ft/s
2 2 2 2
2.75 3.10
n t
a a a
   
Problem

19. In 2001, a race scheduled at the Texas Motor Speedway was
cancelled because the normal accelerations were too high and
caused some drivers to experience excessive g-loads (similar to
fighter pilots) and possibly pass out. What are some things that
could be done to solve this problem?
Some possibilities:
Reduce the allowed speed
Increase the turn radius
(difficult and costly)
Have the racers wear g-suits
Practical Problem

20. SOLUTION:
• Define your coordinate system
• Calculate the tangential velocity and
tangential acceleration
• Determine overall acceleration
magnitude
• Calculate the normal acceleration
The tangential acceleration of the
centrifuge cab is given by
where t is in seconds and at is in
m/s2. If the centrifuge starts from
fest, determine the total acceleration
magnitude of the cab after 10
seconds.
2
0.5 (m/s )
t
a t

Problem

21. In the side view, the tangential
direction points into the “page”
Define your coordinate system
et
en
en
Top View
Determine the tangential velocity
0.5
t
a t

2 2
0
0
0.5 0.25 0.25
t t
t
v t dt t t
  

 
2
0.25 10 25 m/s
t
v  
Determine the normal acceleration
 
2 2
2
25
78.125 m/s
8
t
n
v
a
r
  
Determine the total acceleration magnitude
𝑎𝑚𝑎𝑔 = 𝑎𝑛
2
+ 𝑎𝑡
2
= 78.1252 + 52 2
78.285 m/s
mag
a 
𝑎𝑡 = 0.5𝑡 = 0.5 10 = 5 𝑚/𝑠
Problem

22. a) The accelerations would remain the same
b) The an would increase and the at would decrease
c) The an and at would both increase
d) The an would decrease and the at would increase
Notice that the normal
acceleration is much higher than
the tangential acceleration.
What would happen if, for a
given tangential velocity and
acceleration, the arm radius was
doubled?
Critical thinking!

23. By knowing the distance to the aircraft and the
angle of the radar, air traffic controllers can
track aircraft.
Fire truck ladders can rotate as well as extend;
the motion of the end of the ladder can be
analyzed using radial and transverse
components.
Polar Components (Radial and Transverse)

24. • The position of a particle P is
expressed as a distance r from the
origin O to P – this defines the
radial direction er. The transverse
direction e is perpendicular to er
r
v r e r e

 
• The particle velocity vector is
• The particle acceleration vector is
   
2
2
r
a r r e r r e
  
   
r
e
r
r



Polar Components (Radial and Transverse)

25. • We can derive the velocity and acceleration
relationships by recognizing that the unit
vectors change direction.
r
r e
d
e
d
e
d
e
d 










dt
d
e
dt
d
d
e
d
dt
e
d r
r 

 





dt
d
e
dt
d
d
e
d
dt
e
d
r




 





r
e
r
r



Polar Components (Radial and Transverse)

26. • We can derive the velocity and acceleration
relationships by recognizing that the unit vectors
change direction.
r
r e
d
e
d
e
d
e
d 










dt
d
e
dt
d
d
e
d
dt
e
d r
r 

 





dt
d
e
dt
d
d
e
d
dt
e
d
r




 





𝑣 =
𝑑
𝑑𝑡
𝑟𝑒𝑟 =
𝑑𝑟
𝑑𝑡
𝑒𝑟 + 𝑟
𝑑𝑒𝑟
𝑑𝑡
=
𝑑𝑟
𝑑𝑡
𝑒𝑟 + 𝑟
𝑑𝜃
𝑑𝑡
𝑒𝜃
𝑣 = 𝑟𝑒𝑟 + 𝑟𝜃𝑒𝜃
• The particle velocity vector is
• Similarly, the particle acceleration vector is
    











e
r
r
e
r
r
dt
e
d
dt
d
r
e
dt
d
r
e
dt
d
dt
dr
dt
e
d
dt
dr
e
dt
r
d
e
dt
d
r
e
dt
dr
dt
d
a
r
r
r
r

















2
2
2
2
2
2

















r
e
r
r



Polar Components (Radial and Transverse)

27. Concept Quiz
If you are travelling in a perfect
circle, what is always true about
radial/transverse coordinates and
normal/tangential coordinates?
a) The er direction is identical to the en direction.
b) The e direction is perpendicular to the en direction.
c) The e direction is parallel to the er direction.

28. • When particle position is given in cylindrical
coordinates, it is convenient to express the
velocity and acceleration vectors using the unit
vectors .
and
,
, k
e
eR




• Position vector,
𝑟 = 𝑅𝑒𝑅 + 𝑧𝑘
• Velocity vector,
k
z
e
R
e
R
dt
r
d
v R











 

• Acceleration vector,
    k
z
e
R
R
e
R
R
dt
v
d
a R



















 


 2
2
Polar Components (Radial and Transverse)
𝑟 = 𝑅𝑐𝑜𝑠𝜃𝑖 + 𝑅𝑠𝑖𝑛𝜃𝑗 + 𝑧𝑘
Rectangular Coordinates
Polar Coordinates

29. Rotation of the arm about O is defined
by  = 0.15t2 where  is in radians and t
in seconds. Collar B slides along the
arm such that r = 0.9 - 0.12t2 where r is
in meters.
After the arm has rotated through 30o,
determine (a) the total velocity of the
collar, (b) the total acceleration of the
collar.
SOLUTION:
• Evaluate time t for  = 30o.
• Evaluate radial and angular positions,
and first and second derivatives at
time t.
• Calculate velocity and acceleration in
cylindrical coordinates.
Problem

30. SOLUTION:
• Evaluate time t for  = 30o.
s
869
.
1
rad
524
.
0
30
0.15 2





t
t

• Evaluate radial and angular positions, and first
and second derivatives at time t.
2
2
s
m
24
.
0
s
m
449
.
0
24
.
0
m
481
.
0
12
.
0
9
.
0









r
t
r
t
r



2
2
s
rad
30
.
0
s
rad
561
.
0
30
.
0
rad
524
.
0
15
.
0










 t
t
Problem

31. • Calculate velocity and acceleration.
  
r
r
r
v
v
v
v
v
r
v
s
r
v





1
2
2
tan
s
m
270
.
0
s
rad
561
.
0
m
481
.
0
m
449
.
0














 0
.
31
s
m
524
.
0 
v
  
     
r
r
r
a
a
a
a
a
r
r
a
r
r
a







1
2
2
2
2
2
2
2
2
tan
s
m
359
.
0
s
rad
561
.
0
s
m
449
.
0
2
s
rad
3
.
0
m
481
.
0
2
s
m
391
.
0
s
rad
561
.
0
m
481
.
0
s
m
240
.
0



























 6
.
42
s
m
531
.
0 
a
Problem
𝑣 = 𝑟𝑒𝑟 + 𝑟𝜃𝑒𝜃

32. SOLUTION:
• Define your coordinate system
• Calculate the angular velocity after
three revolutions
• Determine overall acceleration
magnitude
• Calculate the radial and transverse
accelerations
The angular acceleration of the
centrifuge arm varies according to
where  is measured in radians. If the
centrifuge starts from rest, determine the
acceleration magnitude after the gondola
has travelled two full rotations.
2
0.05 (rad/s )
 

Problem

33. In the side view, the transverse
direction points into the “page”
Define your coordinate system
e
er
er
Top View
Determine the angular velocity
Evaluate the integral
2
0.05 (rad/s )
 

𝜃𝑑𝜃 = 𝜃𝑑𝜃
Acceleration is a function
of position, so use:
2(2 )
2 2
0 0
0.05
2 2


 

(2)(2 )
0 0
0.05 d d
 
   

 
 
2
2
0.05 2(2 )
 

Problem
𝑎𝑑𝑠 = 𝑣𝑑𝑣
Similar to the expression

34. er
Determine the angular velocity
Determine the angular acceleration
2.8099 rad/s
 
 
2
2
0.05 2(2 )
 

2
0.05 = 0.05(2)(2 ) 0.6283 rad/s
  
 
Find the radial and transverse accelerations
   
   
2
2
2
2
0 (8)(2.8099) (8)(0.6283) 0
63.166 5.0265 (m/s )
r
r
r
a r r e r r e
e e
e e



  
   
   
  
 
2
2 2 2
( 63.166) + 5.0265
mag r
a a a
    2
63.365 m/s
mag
a 
Magnitude:
Problem

35. You could now have additional acceleration terms. This might
give you more control over how quickly the acceleration of the
gondola changes (this is known as the G-onset rate).
What would happen if you
designed the centrifuge so
that the arm could extend
from 6 to 10 meters?
r
   
2
2
r
a r r e r r e
  
   
Problem