1443-501 Spring 2002
Lecture #3
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
Coordinate Systems
Vector Properties and Operations
2-dim Displacement, Velocity, & Acceleration
2-dim Motion Under Constant Acceleration
Projectile Motion
Coordinate Systems
• Make it easy to express locations or positions
• Two commonly used systems, depending on convenience
– Cartesian (Rectangular) Coordinate System
• Coordinates are expressed in (x,y)
– Polar Coordinate System
• Coordinates are expressed in (r,q)
• Vectors become a lot easier to express and compute
y
How are Cartesian and
y1
(x1,y1)=(r,q)
r
x  r cos q
y  r sin q
q
O (0,0)
Jan. 28, 2002
Polar coordinates related??
x1
x
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
r
x
2
1
 y1 2 
y1
tan q 
x1
2
Example 3.1
Cartesian Coordinate of a point in the xy plane are (x,y)= (-3.50,2.50)m. Find the polar coordinates of this point.
r
y
x
2
1
 y1 2  
 3.50   2.50 
2
q
qs
(-3.50,-2.50)m
Jan. 28, 2002
 18.5  4.30(m)
x
r
2
q  180  q s
 2.50 5

 3.50 7
5
qs  tan 1    35.5
7
q  180  q s  180  35.5  216
tan qs 
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
3
Vector and Scalar
Vector quantities have both magnitude (size)
and direction Force, gravitational pull, momentum
Normally denoted in BOLD letters, F, or a letter with arrow on top F
Their sizes or magnitudes are denoted with normal letters letters, F,
or absolute values: F or F
Scalar quantities have magnitude only
Can be completely specified with a value
and its unit Normally denoted in normal letters, E
Energy, heat,
mass, weight
Both have units!!!
Jan. 28, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
4
Properties of Vectors
• Two vectors are the same if their sizes and the direction
are the same, no matter where they are on a coordinate
system.
Which ones are the
same vectors?
y
D
A=B=E=D
F
A
Why aren’t the others?
B
x
E
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C
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
C: The same magnitude
but opposite direction:
C=-A:A negative vector
F: The same direction
but different magnitude
5
Vector Operations
•
Addition:
– Triangular Method: One can add vectors by connecting the head of one vector to
the tail of the other
– Parallelogram method: Connect the tails of the two vectors and extend
– Addition is commutative: Changing order of operation does not affect the results
A+B=B+A, A+B+C+D+E=E+C+A+B+D
A+B
B
A
•
A
=B
B
A+B
A+B
A
Subtraction:
– The same as adding a negative vector:A - B = A + (-B)
A
A-B
•
-B
Since subtraction is the equivalent to adding
a negative vector, subtraction is also
commutative!!!
Multiplication by a scalar is
increasing the magnitude A, B=2A
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B 2A
A
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
B=2A
6
Example 3.2
A car travels 20.0km due north followed by 35.0km in a direction 60.0o west
of north. Find the magnitude and direction of resultant displacement.
N
B 60o
r
q
 A  B cosq 2  B sin q 2
r
20
A

A2  B 2 cos 2 q  sin 2 q  2 AB cos q

A2  B 2  2 AB cos q

20.02  35.02  2  20.0  35.0 cos 60
 2325  48.2(km)
E
B sin 60
q  tan
1
 tan 1
35.0 sin 60
20.0  35.0 cos 60
30.3
 38.9 to W wrt N
37.5
 tan 1
Jan. 28, 2002


A  B cos 60
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
Find other
ways to
solve this
problem…
7
Components and Unit Vectors
• Coordinate systems are useful in expressing vectors in their components
y
Ay
(+,+)
(Ax,Ay)
A
(-,+)
(+,-)
Ay  A sin
A 
A 
q
(-,-)
}
Ax  A cos q
Ax
x

Ax
Components
 Ay
2
 A cosq    A sin q 
2
2
A
2
q
cos
2
2
q  sin 2 q   A
• Unit vectors are dimensionless vectors whose magnitude is exactly 1
• Unit vectors are usually expressed in i,j,k or i, j , k
• Vectors can be expressed using components and unit vectors
So the above vector A
can be written as
Jan. 28, 2002
A  Ax i  Ay j  A cosq i  A sin q j
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Dr. J. Yu, Lecture #3
8
Examples 3.3 & 3.4
Find the resultant vector which is the sum of A=(2.0i+2.0j) and B =(2.0i-4.0j)

 
C  A  B  2.0i  2.0 j  2.0i  4.0 j

2.0  2.0 i  2.0  4.0  j  4.0i  2.0 j m
4.0 2   2.0 2  16  4.0  20 
C 

q  tan 1
Cy
Cx
 tan 1
4.5( m )
 2.0
 27 
4.0
Find the resultant displacement of three consecutive displacements:
d1=(15i+30j +12k)cm, d2=(23i+14j -5.0k)cm, and d1=(-13i+15j)cm
D  d1  d 2  d 3

 
 
 15i  30 j  12 k  23i  14 j  5.0 k   13i  15 j

 15  23  13i  30  14  15  j  12  5.0 k
 25i  31 j  7.0 k (cm)
D 
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252
 31
2
 7.0 
2
 40(cm)
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
9
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
 r  r f  ri
• Average Velocity:
r f  ri
r
v 

t
t f  ti
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Jan. 28, 2002
v  lim
t 0
r
dr

t
dt
v f  vi
v
a 

t
t f  ti
v d v
d  d r  d 2 r
a  lim





t 0 t
dt
dt  dt  dt 2
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
10
2-dim Motion Under Constant Acceleration
• Position vectors in xy plane:
ri  xi i  yi j
rf  x f i  y f j
• Velocity vectors in xy plane:
vi  v xi i  v yi j
v f  v xf i  v yf j
v xf  v xi  a x t , v yf  v yi  a y t
v f  v xi  a x t i  v yi  a y t  j  vi  at
• How are the position vectors written in acceleration vectors?
1
1
a x t 2 , y f  yi  v yi t  a y t 2
2
2
1
1

 

r f   xi  v xi t  a x t 2 i   yi  v yi t  a y t 2  j
2
2

 

1
 ri  vt  at 2
2
x f  xi  v xi t 
Jan. 28, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
11
Example 4.1
A particle starts at origin when t=0 with an initial velocity v=(20i-15j)m/s.
The particle moves in the xy plane with ax=4.0m/s2. Determine the
components of velocity vector at any time, t.
v xf  v xi  a x t  20  4.0t m / s 
v yf  v yi  a y t  15m / s 
v (t )  20  4.0t i  15 j m / s
Compute the velocity and speed of the particle at t=5.0 s.




v  20  4.0  5.0i 15 j m / s  40i 15 j m / s
  15 
1   3 

  tan 
  21
 40 
 8 
q  tan 1 
speed  v 

vx 2  v y 2
402   152
 43m / s
Determine the x and y components of the particle at t=5.0 s.
x f  v xi t 
1
1
a x t 2  20  5   4  52  150m, y f  v yi t  15  5  75m
2
2


rf  x f i  y f j  150i  75 j m
Jan. 28, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
12
Projectile Motion
•
A 2-dim motion of an object under the gravitational acceleration with the
assumptions
– Free fall acceleration, -g, is constant over the range of the motion
– Air resistance and other effects are negligible
•
A motion under constant acceleration!!!!  Superposition of two motions
– Horizontal motion with constant velocity and
– Vertical motion under constant acceleration
Show that a projectile motion is a parabola!!!
a  ax i  a y j   g j
v xi  vi cos q , v yi  vi sin q i
x f  v xi t  vi cos q i t
1
 g t 2
2
1
 vi sin q i t  gt 2
2
y f  v yi t 
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t
xf
vi cos q i
xf
xf

 1 
  g 
y f  vi sin q i 
v
cos
q
2  vi cos q i
i 
 i

 2
g
x f
 x f tan q i  
2
2

 2vi cos q i 
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
13



2
Example 4.2
A ball is thrown with an initial velocity v=(20i+40j)m/s. Estimate the time of
flight and the distance the ball is from the original position when landed.
Which component determines the flight time and the distance?
Flight time is determined
by y component, because
the ball stops moving
when it is on the ground
after the flight.
Distance is determined by x
component in 2-dim,
because the ball is at y=0
position when it completed
it’s flight.
Jan. 28, 2002
1
y f  40t   g t 2  0m
2
t 80  gt   0
80
 t  0 or t 
 8 sec
g
x f  v xi t
 20  8  160m 
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
14
Horizontal Range and Max Height
• Based on what we have learned previously, one can
analyze a projectile motion in more detail
– Maximum height an object can reach
– Maximum range
At the maximum height the object’s
vertical motion stops to turn around!!
vi
Since no
acceleration
in x, it still
flies at vy=0
Jan. 28, 2002
v yf  v yi  a y t  vi sin q  gt A  0
q
h
t A 
vi sin q
g
y f  h  v yi t 
R  v xi 2t A 
 vi sin q i 

2vi cos q i 


g


 vi 2 sin 2q i 




g


1
 g t 2
2
 vi sin q i 
1  vi sin q i 


 vi sin q i 

g




g
2
g




2
 vi sin 2 q i 




2
g


1443-501 Spring 2002
Dr. J. Yu, Lecture #3
15
2
Maximum Range and Height
• What are the conditions that give maximum height and
range in a projectile motion?
 vi 2 sin 2 q i
h  
2g

 vi 2 sin 2q i 

R  

g


Jan. 28, 2002




This formula tells us
that the maximum
hieght can be achieved
when qi=90o!!!
This formula tells us that
the maximum hieght can
be achieved when
2qi=90o, i.e., qi=45o!!!
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
16
Example 4.5
•
A stone was thrown upward from the top of a building at an angle of 30o to
horizontal with initial speed of 20.0m/s. If the height of the building is 45.0m,
how long is it before the stone hits the ground?
v xi  vi cos q  20.0  cos 30   17.3m / s
v yi  vi sin q i  20.0  sin 30   10.0m / s
1
gt 2
2
 20.0t  90.0  9.80t 2  20.0t  90.0  0
y f  45.0  v yi t 
gt 2
 20 2
 4  9.80  ( 90)
2  9.80
t  2.18 s or t  4.22 s
 t  4.22 s
t 
•
20.0 
What is the speed of the stone just before it hits the ground?
v xf  v xi  vi cos q  20.0  cos 30  17.3m / s
v yf  v yi  gt  vi sin q i  gt  10.0  9.80  4.22  31.4m / s
v 
v xf
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2
 v yf
2

17.32   31.4 
2
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
 35.9m / s
17
Uniform Circular Motion
• A motion with a constant speed on a circular path.
– The velocity of the object changes, because the direction
changes
– Therefore, there is an acceleration
r
q
r1
r2 q r
vi
vj
vj
v
vi
Angle
is q
The acceleration pulls the object inward: Centripetal Acceleration
Average
Acceleration
Instantaneous
Acceleration
a
vf  vi
v

t f  ti
t
q 
v
v

r
v r
, v  v
, a
r
r
t r
r v
v
v2
ar  lim a  lim
 v 
t  0
t  0 t r
r
r
r
Is this correct in
dimension?
What story is this expression telling you?
Jan. 28, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #3
18