PHYS 1443 – Section 002
Lecture #5
Wednesday, September 10, 2008
Dr. Mark Sosebee
•
•
•
•
Free Fall
Coordinate System
Vectors and Scalars
Motion in Two Dimensions
• Motion under constant acceleration
• Projectile Motion
• Maximum ranges and heights
Announcements
• The first term exam is to be next Wednesday, Sept. 17
– Will cover CH1 – what we finish today (likely to be CH3) +
Appendices A and B
– Mixture of multiple choices and essay problems
– In class, from 1 – 2:20pm, SH103
– Jason will conduct a review in the class Monday, Sept. 15
• There will be a department colloquium this afternoon
at 4pm in SH101
– Extra credit
– Be sure to sign in
– Refreshment at 3:30pm in SH108, the Physics Lounge!
Wednday, Sept. 10, 2008
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
2
Wednday, Sept. 10, 2008
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
3
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)
Wednday, Sept. 10, 2008
x1
+x
y1  r1 sin q1
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
y1
tan q1 
x1
 y1 

x
 1
q1  tan 1 
4
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.
y

q
qs
(-3.50,-2.50)m
2
 y2 
  3.50 
2
  2.50 
2

 18.5  4.30( m)
x
r
x
r
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
Wednday, Sept. 10, 2008
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
o
5
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!!!
Wednday, Sept. 10, 2008
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
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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
Wednday, Sept. 10, 2008
C
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
C: The same magnitude
but opposite direction:
C=-A:A negative vector
F: The same direction
but different magnitude
7
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
Wednday, Sept. 10, 2008
B 2A
A
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
B=2A
8
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
Wednday, Sept. 10, 2008
 A  B cosq 
A  B cos 60
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
Find other
ways to
solve this
problem…
9
Components and Unit Vectors
Coordinate systems are useful in expressing vectors in their components
y
(+,+)
Ay
A
(-,+)
(Ax,Ay)
q
(-,-)
u
r
A 
(+,-)


Wednday, Sept. 10, 2008
ur
A cos q
Ax
 
2
Ax 
ur
A cos q
Ay 
ur
A sin q
x
A  Ax  Ay
2
ur
A sin q
u
r 2
A
cos 2 q  sin 2 q

PHYS1443-002-Fall 2008
Dr. Jaehoon Yu


2
}
Components
} Magnitude
2
u
r
 A
10
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
Wednday, Sept. 10, 2008
r
ur
ur
r ur
r
r
A  Ax i Ay j  A cos q i  A sin q j
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
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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
Wednday, Sept. 10, 2008
ur
D



 25  59   7.0  65(cm)
2
2
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
2
12
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
• Average Velocity:
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Wednday, Sept. 10, 2008
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


PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
13
Kinematic Quantities in 1D and 2D
Quantities
Displacement
Average Velocity
1 Dimension
2 Dimension
r
r 
x  x f  xi
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
Wednday, Sept.
10, 2008
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
the difference
between 1D2008
and 2D quantities?
PHYS1443-002-Fall
Dr. Jaehoon Yu
14
2-dim Motion Under Constant Acceleration
• Position vectors in x-y plane:
r
r
r
rf  x f i  y f j
r
r
r
ri  xi i  yi j
• Velocity vectors in x-y plane:
r
r
r
vi  vxi i  v yi j
r
r
r
v f  vxf i  v yf j
Velocity vectors in terms of the acceleration vector
X-comp
v xf  vxi  axt
Y-comp

v yf  v yi  a y t
 

r
r
r
r
r
r
r
v f   vxi  axt  i   v yi  a yt  j  vxi i  v yi j  ax i  a y j t 
r r
 vi  at
Wednday, Sept. 10, 2008
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
15
2-dim Motion Under Constant Acceleration
• How are the 2D position vectors written in
acceleration vectors?
Position vector
components
Putting them
together in a
vector form
Regrouping
the above
Wednday, Sept. 10, 2008
1 2
x f  xi  vxi t  ax t
2
r
r
r
rf  x f i  y f j 
1 2
y f  yi  v yi t  a y t
2
r 
r
1
1

2
2
  xi  vxit  axt  i   yi  v yit  a yt  j
2
2







r 2
r
r
r
r
1 r
 xi i  yi j  vxi i  v yi j t   ax i  a y j  t
2
r r
1r 2
 ri vit  at
2D problems can be
2
interpreted as two 1D
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
problems in x and y
16
Example for 2-D Kinematic Equations
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 the velocity vector at any time t.
vxf  vxiaxt  20 4.0t  m / s  v yf  v yi  a y t  15 0t  15  m / s 
Velocity vector
r
r
r
r
r
v  t   vx  t  i v y  t  j   20  4.0t  i 15 j ( m / s )
Compute the velocity and the speed of the particle at t=5.0 s.
r
r
r
r
r
r
r
vt 5  vx,t 5i v y ,t 5 j   20  4.0  5.0  i 15 j  40i  15 j m / s

r
speed  v 
Wednday, Sept. 10, 2008
 vx 
2
  vy 
2

 40
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
2

  15  43m / s
2
17
Example for 2-D Kinematic Eq. Cnt’d
Angle of the
Velocity vector
 vy
q  tan 
 vx
1

1  15 
1  3 
o

tan

tan


21

 


40
 8 



Determine the x and y components of the particle at t=5.0 s.
1 2
1
x f  vxi t  ax t  20  5   4  52  150( m)
2
2
y f  v yi t  15 5  75 (m)
Can you write down the position vector at t=5.0s?
r
r
r
r
r
r f  x f i  y f j  150i 75 j  m 
Wednday, Sept. 10, 2008
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
18
Projectile Motion
• A 2-dim motion of an object under
the gravitational acceleration with
the following assumptions
– Free fall acceleration, g, is constant
over the range
of the motion
r
r
• g  9.8 j  m s 2 
– Air resistance and other effects are
negligible
• A motion under constant
acceleration!!!!  Superposition
of two motions
– Horizontal motion with constant
velocity ( no acceleration )
– Vertical motion under constant
acceleration ( g )
Wednday, Sept. 10, 2008
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
19
Show that a projectile motion is a parabola!!!
x-component
vxi  vi cos q
v yi  vi sin qi
y-component
a axi  a y j   gj
ax=0
x f  vxi t  vi cos qi t
t
xf
vi cosq i
In a projectile motion,
the only acceleration is
gravitational one whose
direction is always
toward the center of the
earth (downward).
1 2
1
2

v
sin
q
t

gt
y f  v yi t    g  t
i
i
2
2
Plug t into
the above
xf

xf

 1 
y f  vi sin qi 
  2 g  v cos q 
i 
 i
 vi cos qi 

g
y f  x f tan q i  
2
2
2
v
cos
qi
 i
Wednday, Sept. 10, 2008
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
 2
x f


2
What kind of parabola is this?
20
Projectile Motion
Wednday, Sept. 10, 2008
ThePHYS1443-002-Fall
only acceleration
in this
2008
Dr. Jaehoon
motion.
It isYua constant!!
21
Example for Projectile Motion
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?
1
y f  40t    g  t 2  0m
2
Flight time is determined
by the y component,
t 80  gt   0
because the ball stops
moving when it is on the
ground after the flight.
So the possible solutions are…
Distance is determined by the x
component in 2-dim, because
the ball is at y=0 position when it
completed it’s flight.
Wednday, Sept. 10, 2008
80
 t  0 or t 
 8 sec
g
Why isn’t 0
 t  8sec
the solution?
x f  vxi t  20  8  160  m
PHYS1443-002-Fall 2008
Dr. Jaehoon Yu
22