7 - 1

Non-Central Coordinate SystemIn 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

tvv tˆ

where v is the velocity magnitude and 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.

t

7 - 2

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 - 3

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 revolutions (or radians) per second, and is the axis around which the vector is rotating. The direction change is shown mathematically above.

7 - 4

ˆˆˆ

ˆˆˆ

ˆˆˆ

ez

te

ner

t

n

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:

nR

vtvnvtvtvtvva

tvv

tttttt

t

ˆˆˆˆˆˆ

ˆ2

t

n

t

n

nntttdt

d

ttnnndt

d

ˆ

ˆ

0

0ˆ

ˆ

ˆˆˆˆˆˆ

ˆˆˆˆˆˆ

7 - 5

Why would we ever want to use this coordinate system, whose origin is constantly shifting?

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…

7 - 6

Indianapolis 500 speedtrack

5/8 mile straightaway

1/8 mile

straight-

away

1/4 mile

turnGiven 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 - 7

milemile

R

mileR

159.02

1

12

ngnm

mile

milemphmphn

R

va

sms

mt ˆ4ˆ8.37

1609

1

159.0

1

1

447.0220ˆ 2

22

nR

vtva t

t ˆˆ2

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:

0tv

The acceleration of the car can be computed from:

However, assuming the car has constant velocity:

The acceleration becomes:

7 - 8

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

T

mg

nt

ˆcossinˆsincosˆ NTmgnNTtvmF t

7 - 9

nR

vtva t

t ˆˆ2

R

mvNT t

2

sincos

ˆ0ˆˆ

ˆcossinˆsincosˆ2

nR

mvtvmF

NTmgnNTtvmF

tt

t

In general, we know that the acceleration of the car is given by:

0cossin NTmg

From this, we can arrive at two relations to solve for the two unknowns, T and N

We also have the relation for force:

7 - 10

mgR

mv

mgR

mv

N

T

N

T

mgR

mv

tt

t

221

2

cossin

sincos

cossin

sincos

cossin

sincos

R

mvNT t

2

sincos

0cossin NTmg

NTmat 22

Given the 2 equations and 2 unknowns:

The solution is given as follows:

The condition for slipping becomes:

22

2

222 cossinsincos

g

R

vg

R

va tt

t

7 - 11

Case 1: Assume no angle on the bank

22

222

222

222 0cos0sin0sin0cos

gR

va

gR

vg

R

va

tt

ttt

22224 Ragv tt

Qualitatively, one can see that any acceleration (braking or accelerating) reduces the maximum speed at which you can take the turn without slipping.

Assume α = 0

7 - 12

Case 2: Assume no tangential acceleration

sincossincos

cossinsincos

22

22

ggR

v

R

v

gR

vg

R

v

tt

tt

tan1

tan

sincos

sincos

gRgRvt

1tantan1

22

2

222 cossinsincos

g

R

vg

R

va tt

t

Thus, the velocity can be expressed as:

Curiously, a banked curve can support potentially “infinite” velocity if the denominator equals zero: (i.e.)

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 - 13

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 - 14

At the Indy500, the maximum speed around a turn assuming μ=0.3 is:

mphmile

s

mv

o

o

t 1.779tan3.01

9tan3.0159.08.9

2

assuming μ=0.7

mphmile

s

mv

o

o

t 1859tan8.11

9tan8.1159.08.9

2

assuming μ=1.8

mphmile

s

mv

o

o

t 1109tan7.01

9tan7.0159.08.9

2

7 - 15

Problem 1

The driver of an automobile decreases her speed at aconstant rate from 45 to 30 mi/h over a distance of 750 ftalong a curve of 1500-ft radius. Determine the magnitudeof the total acceleration of the automobile after theautomobile has traveled 500 ft along the curve.

7 - 16

Problem 1

The driver of an automobile decreases her speed at aconstant rate from 45 to 30 mi/h over a distance of 750 ftalong a curve of 1500-ft radius. Determine the magnitudeof the total acceleration of the automobile after theautomobile has traveled 500 ft along the curve.

1. Use tangential and normal components: These componentsare used when the particle travels along a circular path. Theunit vector et is tangent to the path (and thus aligned with thevelocity) while the unit vector en is directed along the normal tothe path and always points toward its center of curvature.

2. Determine the tangential acceleration: For a constanttangential acceleration, the acceleration can be determined from

v dv = at dxvo

v

xo

x

7 - 17

Problem 1

The driver of an automobile decreases her speed at aconstant rate from 45 to 30 mi/h over a distance of 750 ftalong a curve of 1500-ft radius. Determine the magnitudeof the total acceleration of the automobile after theautomobile has traveled 500 ft along the curve.

3. Determine the normal acceleration: For known velocity andradius of curvature, the tangential acceleration is determine by

an =

4. Determine the magnitude of the total acceleration: Themagnitude of the total acceleration is given by

a = at2 + an

2

v2

7 - 18

Problem 1 Solution

1500 ft

45 mi/h

30 mi/h

at

750 ft

Determine the tangential acceleration.

45 mi/h = 44 ft/s45 mi/h = 66 ft/s

vo

v

xo

xv dv = at dx

66

44

0

750v dv = at dx

( 442 - 662 ) = at ( 750 - 0 )21

at = - 1.613 ft/s2

7 - 19

at = -1.613 ft/s2

1500 ft

45 mi/h

v1 an

500 ft

Problem 1 Solution

The speed after the automobiletraveled 500 ft.

vo

v

xo

xv dv = at dx

66

v1

0

500v dv = -1.613 dx

21 (v1

2 - 662 ) = -1.613 ( 500 - 0 )

v1 = 52.4 ft/s

7 - 20

at = -1.613 ft/s2

1500 ft

45 mi/h

52.4 ft/s an

500 ft

Problem 1 Solution

Determine the normal acceleration.

an =v2

an = = 1.828 ft/s2( 52.4 ft/s )2

1500 ft

Determine the magnitude of the total acceleration.

a = at2 + an

2

a = ( - 1.613 ft/s2 )2 + ( 1.828 ft/s2 )2

a = 2.44 ft/s2

7 - 21

Problem 2

Knowing that the conveyor beltmoves at the constant speedv0 = 24 ft/s, determine the angle for which the sand is deposited onthe stockpile at B.

15 ftA

25 ft

v0

B

7 - 22

Problem 2

Knowing that the conveyor beltmoves at the constant speedv0 = 24 ft/s, determine the angle for which the sand is deposited onthe stockpile at B.

15 ftA

25 ft

v0

B

1. Analyzing the motion of a projectile: Consider the the verticaland the horizontal motion separately.

2. Consider the horizontal motion: When the resistance of the aircan be neglected, the horizontal component of the velocityremains constant (uniform motion). The distance, velocity, andtime are related by

x = x0 + (vx)0 t

7 - 23

Problem 2

Knowing that the conveyor beltmoves at the constant speedv0 = 24 ft/s, determine the angle for which the sand is deposited onthe stockpile at B.

15 ftA

25 ft

v0

B

3. Consider the vertical motion: When the resistance of the aircan be neglected, the vertical component of the acceleration isconstant (uniformly accelerated motion). The distance, velocity,acceleration and time are related by

y = y0 + (vy)0 t - g t212

7 - 24

Problem 2 Solution

15 ftA

25 ft

v0

B

x

y

Consider the horizontal motion.

(vx)0 = v0 cos

(vx)0 = (24 ft/s) cos

x = x0 + (vx)0 t

x = -25 ftx0 = 0

(25 ft) = 0 + (24 ft/s) cos tConsider the vertical motion

y = y0 + (vy)0 t - g t212

y = -15 ft

y0 = 0

(vy)0 = v0 sin

(vy)0 = (24 ft/s) sin

(-15 ft) = 0 + (24 ft/s) sin t - (32.2 ft/s2) t2 12

Two equations withtwo unknowns, and t.

7 - 25

15 ftA

25 ft

v0

B

x

y Problem 2 Solution

25 = 0 + 24 cos t

-15 = 0 + 24 sin t - 32.2 t2 12

Solve the equations for and t.

eliminate : sin = 1 - cos2 = 1 - ( )22524 t

-15 = 24 t 1 - ( )2 - 16.1 t2

250.21 t4 - 1059 t2 + 850 = 0

two solutions:

t = 1.048 s , = 6.09o

t = 1.729 s , = 52.9o

2524 t

= 6.09o or 52.9o