PHYS 1443 – Section 002
Lecture #5
Wednesday, Sept. 12, 2007
Dr. Jaehoon Yu
•
•
•
Coordinate Systems
Vectors and Scalars
Motion in Two Dimensions
– Motion under constant acceleration
– Projectile Motion
– Maximum ranges and heights
Wednesday, Sept. 12, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
1
Announcements
• E-mail distribution list: 63 of you subscribed to the
list so far
– Please subscribe to the class distribution list since this is
the primary communication tool for this class
• Two physics department seminars this week
– At 4pm today in SH101
• Dr. Sandy Dasgupta, UTA Chemistry Department chair:
Playing With Drops and Films Analytical Chemistry in Small
Places
– At 4pm Friday in SH101
• Dr. Amitava Patra: Nano Materials for Photonic Application
Wednesday, Sept. 12, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
2
Wednesday, Sept. 12, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
3
Wednesday, Sept. 12, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
4
Special Problems for Extra Credit
• Derive the quadratic equation for Bx2-Cx+A=0
 5 points
• Derive the kinematic equation v  vxi  2axx  x 
from first principles and the known kinematic
equations  10 points
• You must show your work in detail to obtain full
credit
• Due Wednesday, Sept. 19
2
xf
Wednesday, Sept. 12, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
2
f
i
5
1D Kinematic Equations of Motion on a
Straight Line Under Constant Acceleration
vxf t   vxi  axt
Velocity as a function of time
1
1
xf  xi  v x t  vxf  vxi t Displacement as a function
of velocities and time
2
2
1 2
xf  xi  vxit  axt
2
vxf  vxi  2axxf  xi 
2
2
Displacement as a function of
time, velocity, and acceleration
Velocity as a function of
Displacement and acceleration
You may use different forms of Kinematic equations, depending on
the information given to you for specific physical problems!!
Wednesday, Sept. 12, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
6
Coordinate Systems
• They make it easy and consistent to express locations or positions
• Two commonly used systems, depending on convenience, are
– Cartesian (Rectangular) Coordinate System
• Coordinates are expressed in (x,y)
– Polar Coordinate System
• Coordinates are expressed in distance from the origin ® and the angle measured
from the x-axis, q (r,q)
• Vectors become a lot easier to express and compute
+y
How are Cartesian and
Polar coordinates related?
y1
(x1,y1) =(r1,q1)
r1
x1  r1 cos q1 r   x12  y12 
1
q1
O (0,0)
Wednesday, Sept. 12, 2007
x1
+x
y1  r1 sin q1
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
y1
tan q1 
x1
 y1 

x
 1
q1  tan 1 
7
Example
Cartesian Coordinate of a point in the xy plane are (x,y)= (-3.50,-2.50)m.
Find the equivalent polar coordinates of this point.
r
y

q
qs
(-3.50,-2.50)m
2
 y2 
  3.50 
2
  2.50 
2

 18.5  4.30( m)
x
r
x
q  180  q s
tan q s 
2.50 5

3.50 7
1
5
q s  tan    35.5
7
o
o
o
q  180  q s  180  35.5  216
Wednesday, Sept. 12, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
o
8
Vector and Scalar
Vector quantities have both magnitudes (sizes)
and directions Force, gravitational acceleration, momentum
ur
Normally denoted in BOLD letters, F, or a letter with arrow on top F
Their sizes or magnitudes are denoted with normal letters, F, or
ur
absolute values: F or F
Scalar quantities have magnitudes only
Can be completely specified with a value
and its unit Normally denoted in normal letters, E
Energy, heat,
mass, time
Both have units!!!
Wednesday, Sept. 12, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
9
Properties of Vectors
• Two vectors are the same if their sizes and the directions
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
Wednesday, Sept. 12, 2007
C
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
C: The same magnitude
but opposite direction:
C=-A:A negative vector
F: The same direction
but different magnitude
10
Vector Operations
•
Addition:
– Triangular Method: One can add vectors by connecting the head of one vector to
the tail of the other (head-to-tail)
– 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
A+B
OR
A+B
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
Wednesday, Sept. 12, 2007
B 2A
A
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
B=2A
11
Example for Vector Addition
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.
r
Bsinq N
B 60o Bcosq
r
q
20
A
2

  B sin q 
2


A2  B 2 cos 2 q  sin 2 q  2 AB cosq

A2  B 2  2 AB cosq

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
Wednesday, Sept. 12, 2007
 A  B cosq 
A  B cos 60
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
Find other
ways to
solve this
problem…
12
Components and Unit Vectors
Coordinate systems are useful in expressing vectors in their components
y
(+,+)
Ay
A
(-,+)
(Ax,Ay)
q
(-,-)
u
r
A 

(+,-)

Ax
ur
A cos q
 
2
Ax 
ur
A cos q
Ay 
ur
A sin q
x
2
ur
A sin q
u
r 2
A
cos 2 q  sin 2 q
Wednesday, Sept. 12, 2007
A  Ax  Ay

PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu


2
}
Components
} Magnitude
2
u
r
 A
13
Unit Vectors
• Unit vectors are the ones that tells us the
directions of the components
• Dimensionless
• Magnitudes are exactly 1
• Unit vectors are usually expressed in i, j, k or
r r r
i, j, k
So the vector A can
be re-written as
Wednesday, Sept. 12, 2007
r
ur
ur
r ur
r
r
A  Ax i Ay j  A cos q i  A sin q j
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
14
Examples of Vector Operations
Find the resultant vector which is the sum of A=(2.0i+2.0j) and B =(2.0i-4.0j)

 
r
r
r
r
ur ur ur
C  A  B  2.0i  2.0 j  2.0i  4.0 j
r
  2.0  2.0  i
ur
C 
 4.0   2.0
2
r
  2.0  4.0  j

r
r
 4.0i 2.0 j  m 
2
 16  4.0  20  4.5(m)
q
2.0
tan
 tan
 27o
Cx
4.0
1
Cy
1
Find the resultant displacement of three consecutive displacements:
d1=(15i+30j +12k)cm, d2=(23i+14j -5.0k)cm, and d3=(-13i+15j)cm
ur ur ur ur
r
r
r
r
r
r
r
r
D  d 1  d 2  d 3  15i  30 j  12k  23i  14 j  5.0k  13i  15 j
r
r
r
r
r
r
 15  23  13 i   30  14  15  j  12  5.0  k  25i 59 j 7.0k (cm)

Magnitude
Wednesday, Sept. 12, 2007
ur
D



 25  59   7.0  65(cm)
2
2
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
2
15
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
• Average Velocity:
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Wednesday, Sept. 12, 2007
r r
r
r  r f  r i
r
r
r
r
r f  ri
r

v
t f  ti
t
r
r
r
r
dr
v  lim

t  0 t
dt
r
r
r
r
v f  vi
v

a
t f  ti
t
How is each of
these quantities
defined in 1-D?
r
r
r
r
2
r
v dv d  d r   d r
a  lim


2


dt
dt
t 0 t
dt
dt


PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
16
Kinematic Quantities in 1D and 2D
Quantities
1 Dimension
2 Dimension
r
r 
x  x f  xi
Displacement
Average Velocity
vx 
r
r
r
r r r f  r i
v

t
t f  ti
x f  xi
x

t
t f  ti
Inst. Velocity
x dx
v x  lim

t  0 t
dt
Average Acc.
v x v xf  v xi
ax 

t
t f  ti
Inst. Acc.
What
is the
Wednesday, Sept.
12, 2007
r r
r f  ri
r
r
r
r d r
v  lim

t 0 t
dt
r
r
r
r
v v f  vi
a

t
t f  ti
r
r
r
2
vx dvx d x r
v dv d r
a x  lim

 2 a  lim

 2
t 0 t
t 0 t
dt dt
dt dt
2
difference
between
1D2007
and 2D quantities?
PHYS
1443-002, Fall
Dr. Jaehoon Yu
17