Non-Central Coordinate System
In addition to the 1-D, 2-D, and 3-D coordinate systems
discussed in previous lectures, there is another useful coordinate
system known as normal/tangential coordinates. The main
difference in this coordinate system is that origin is no longer
fixed. Instead, this coordinate system references particle motion
by the tangential velocity along its path of motion

v  vt tˆ
where v is the velocity magnitude and tˆ is a unit vector
denoting the tangential direction of the path. As long as the
motion is primarily confined to two dimensions, this
coordinate system proves to be very useful in solving
problems.
7-1
For example, given the path trajectory s, the associated
directions of the tangential unit vector, which is parallel to the
instantaneous velocity, and normal unit vector are shown
below.
However, the normal unit vector could be drawn any number of
ways that is perpendicular to the tangential direction. Why is it
drawn this way? What significance is this normal direction, given
the radius of curvature?
7-2
Unit Vector Time Derivatives
The time derivative of a unit vector is proportional to how
quickly it changes direction. For example, suppose you have a
vector rotating about an axis with angular velocity vector

ˆ
  
where  is the frequency of rotation, denoted in
ˆ is the axis
revolutions (or radians) per second, and 
around which the vector is rotating. The direction change is
shown mathematically above.
7-3
Let’s provide a model system equivalent with cylindrical
coordinates to make the derivation easier. For example, in the
system of a particle in a stable circular orbit around a point, the
vectors can be written as follows:
rˆ  eˆn  nˆ
ˆ  eˆ  tˆ
t
ˆ
zˆ  eˆ  

v  vt tˆ
d
ˆ  nˆ  tˆ  tˆ
nˆ  nˆ  
dt
d ˆ ˆ
ˆ  tˆ  nˆ  nˆ
t  t  
dt
nˆ   0  nˆ 
    
 ˆ

ˆ
 t    0   t 
2
v
 
a  v  vt tˆ  vt tˆ  vt tˆ  vtnˆ  vt tˆ  t nˆ
R
7-4
Thus, the normal-tangential coordinate system is fundamentally
different than all the previously studied coordinate systems in that
the origin is not at a fixed location. To see why, consider the path
of a particle shown in the diagram below…
Why would we ever want to use this coordinate system, whose origin is
constantly shifting?
7-5
Indianapolis 500 speedtrack
1/8 mile
straightaway
5/8 mile straightaway
1/4 mile
turn
Given the average speed of a car for 1 lap is around 220 mph and the driver is
moving at a constant speed, what is the acceleration around the turn?
7-6
In ¼ mile, the car goes through a 90 degree turn. Thus,
the circumference of the circle is 1 mile, and its radius can
be computed from: 2R  1m ile
R
1m ile
 0.159m ile
2
The acceleration of the car can be computed from:
vt2

a  vt tˆ  nˆ
R
However, assuming the car has constant velocity:
vt  0
The acceleration becomes:

v
 0.447

ˆ
a   n   220m ph
R
 1m ph

2
t
m
s
2

1
 1m ile 


  37.8 sm2 nˆ  4 gnˆ
 0.159m ile  1609m 
7-7
With advances in automotive technology, how
fast can a car safely drive in the Indy 500?
The catch here, is how we define “safely”. According to my definition, the
word “safely” means the speed at which the car does not slip on the road.

N
ˆ

nˆ
tˆ

T

mg

ˆ
F  mvt tˆ  T cos   N sin nˆ  mg  T sin   N cos 
7-8
In general, we know that the acceleration of the car is given by:
2
t
v

a  vt tˆ  nˆ
R
We also have the relation for force:

ˆ
F  mvt tˆ  T cos   N sin  nˆ  m g  T sin    N cos 
2

m
v
ˆ
F  mvt tˆ  t nˆ  0
R
From this, we can arrive at two relations to solve for the two unknowns, T and N
m vt2
T cos   N sin   
R
mg  T sin   N cos   0
7-9
mg  T sin   N cos   0
Given the 2 equations
and 2 unknowns:
2
t
mv
T cos   N sin   
R
The solution is given as follows:
 m vt2   cos  sin    T 


R

  sin   cos   N 
 mg 
 T   cos 
 N    sin  
  
2
2




m
v
m
v
t
t
sin  
cos   sin  







R
R
cos   m g   sin   cos    m g 




1
The condition for slipping becomes:
ma t2  T 2  N
2
2
2




v
v
2
2
t
t
at   cos   g sin      sin    g cos 
R

R

2
7 - 10
Case 1: Assume no angle on the bank
Assume α = 0
2
v


2 v
a   cos0   g sin 0     sin 0   g cos0 
R

R

2
t
2
t
v
a  
R
2
t
2
t
2
t
2

   2 g 2


2

v   g a R
4
t
2
2
2
t
2
Qualitatively, one can see that any acceleration (braking or
accelerating) reduces the maximum speed at which you can
take the turn without slipping.
7 - 11
Case 2: Assume no tangential acceleration
2
v


2 v



a   cos   g sin      sin    g cos 
R

R

2
t
2
t
2
t
2
 vt2

 vt2

 cos   g sin      sin    g cos 
R

R

vt2
vt2
cos    sin    g cos   g sin  
R
R
Thus, the velocity can be expressed as:
 cos   sin 
  tan 
vt  gR
 gR
cos    sin 
1   tan 
Curiously, a banked curve can support potentially “infinite” velocity if the
denominator equals zero: (i.e.)
1   tan   tan    1
Given typical values for the friction coefficient, this would require that the
angle of the bank exceeds 45 degrees, which would be too scary to drive on!
 
 
7 - 12
With advances in automotive technology, how
fast can a car safely drive in the Indy 500?

So, we have arrived at 2 qualitative conclusions… The first is that when taking
a steep curve, you should maintain constant velocity to prevent slipping. The
second is that the maximum speed you can take the curve depends on the
friction coefficient and the bank angle. If you were on the NASCAR team, you
would be well-advised to know these parameters in designing the car. In the
INDY 500, the bank angle around the turns is 9o 12’. Assuming a friction
coefficient of 0.3, what is the maximum speed you can take around the turn?
7 - 13
At the Indy500, the maximum speed
around a turn assuming μ=0.3 is:
 
 
m
0.3  tan 9 o

vt   9.8 2 0.159 mile 
 77.1mph
o
s 
1  0.3 tan 9

assuming μ=0.7
 
 
m
0.7  tan 9 o

vt   9.8 2 0.159 mile 
 110 mph
o
s 
1  0.7 tan 9

assuming μ=1.8
 
 
o
m
1
.
8

tan
9


vt   9.8 2 0.159 mile 
 185 mph
o
s 
1  1.8 tan 9

7 - 14
Problem 1
The driver of an automobile decreases her speed at a
constant rate from 45 to 30 mi/h over a distance of 750 ft
along a curve of 1500-ft radius. Determine the magnitude
of the total acceleration of the automobile after the
automobile has traveled 500 ft along the curve.
7 - 15
Problem 1
The driver of an automobile decreases her speed at a
constant rate from 45 to 30 mi/h over a distance of 750 ft
along a curve of 1500-ft radius. Determine the magnitude
of the total acceleration of the automobile after the
automobile has traveled 500 ft along the curve.
1. Use tangential and normal components: These components
are used when the particle travels along a circular path. The
unit vector et is tangent to the path (and thus aligned with the
velocity) while the unit vector en is directed along the normal to
the path and always points toward its center of curvature.
2. Determine the tangential acceleration: For a constant
tangential acceleration, the acceleration can be determined from
v
x
v dv = at dx
vo
xo
7 - 16
Problem 1
The driver of an automobile decreases her speed at a
constant rate from 45 to 30 mi/h over a distance of 750 ft
along a curve of 1500-ft radius. Determine the magnitude
of the total acceleration of the automobile after the
automobile has traveled 500 ft along the curve.
3. Determine the normal acceleration: For known velocity and
radius of curvature, the tangential acceleration is determine by
an =
v2
r
4. Determine the magnitude of the total acceleration: The
magnitude of the total acceleration is given by
a=
at2 + an2
7 - 17
Problem 1 Solution
45 mi/h
45 mi/h = 44 ft/s
45 mi/h = 66 ft/s
at
Determine the tangential acceleration.
30 mi/h
v
x
v dv = at dx
vo
1500 ft
44
xo
750
v dv = at dx
66
0
1 ( 442 - 662 ) = a ( 750 - 0 )
t
2
at = - 1.613 ft/s2
7 - 18
Problem 1 Solution
45 mi/h
at = -1.613 ft/s2
v1
The speed after the automobile
traveled 500 ft.
an
v
x
v dv = at dx
vo
1500 ft
v1
xo
500
v dv = -1.613 dx
66
0
1 (v 2 - 662 ) = -1.613 ( 500 - 0 )
2 1
v1 = 52.4 ft/s
7 - 19
45 mi/h
at = -1.613 ft/s2
52.4 ft/s
Problem 1 Solution
Determine the normal acceleration.
an
1500 ft
v2
an = r
( 52.4 ft/s )2
an =
= 1.828 ft/s2
1500 ft
Determine the magnitude of the total acceleration.
a=
at2 + an2
a=
( - 1.613 ft/s2 )2 + ( 1.828 ft/s2 )2
a = 2.44 ft/s2
7 - 20
Problem 2
v0

A
15 ft
B
Knowing that the conveyor belt
moves at the constant speed
v0 = 24 ft/s, determine the angle 
for which the sand is deposited on
the stockpile at B.
25 ft
7 - 21
Problem 2
v0

A
15 ft
B
25 ft
Knowing that the conveyor belt
moves at the constant speed
v0 = 24 ft/s, determine the angle 
for which the sand is deposited on
the stockpile at B.
1. Analyzing the motion of a projectile: Consider the the vertical
and the horizontal motion separately.
2. Consider the horizontal motion: When the resistance of the air
can be neglected, the horizontal component of the velocity
remains constant (uniform motion). The distance, velocity, and
time are related by
x = x0 + (vx)0 t
7 - 22
Problem 2
v0

A
15 ft
B
25 ft
Knowing that the conveyor belt
moves at the constant speed
v0 = 24 ft/s, determine the angle 
for which the sand is deposited on
the stockpile at B.
3. Consider the vertical motion: When the resistance of the air
can be neglected, the vertical component of the acceleration is
constant (uniformly accelerated motion). The distance, velocity,
acceleration and time are related by
y = y0 + (vy)0 t - 1 g t2
2
7 - 23
Problem 2 Solution
y
v0
Consider the horizontal motion.

A
x
x = x0 + (vx)0 t
15 ft
B
x = -25 ft
x0 = 0
(vx)0 = v0 cos 
25 ft
Consider the vertical motion
y = y0 + (vy)0 t - 1 g t2
2
(vx)0 = (24 ft/s) cos 
(25 ft) = 0 + (24 ft/s) cos  t
y = -15 ft
(vy)0 = v0 sin 
y0 = 0
(vy)0 = (24 ft/s) sin 
Two equations with
two unknowns,  and t.
(-15 ft) = 0 + (24 ft/s) sin  t - 1 (32.2 ft/s2) t2
2
7 - 24
Problem 2 Solution
y
25 = 0 + 24 cos  t
v0

A
x
-15 = 0 + 24 sin  t -
15 ft
B
1
32.2 t2
2
Solve the equations for  and t.
25 ft
eliminate :
sin =
1 - cos2 =
25
1 - ( 24
t
)2
25 2
-15 = 24 t
1 - ( 24 t ) - 16.1 t2
250.21 t4 - 1059 t2 + 850 = 0
two solutions:
t = 1.048 s ,
 = 6.09o
t = 1.729 s ,
 = 52.9o
 = 6.09o or 52.9o
7 - 25