PHYS 1443 – Section 003
Lecture #6
Monday, Sept. 13, 2004
Dr. Jaehoon Yu
1.
Motion in two dimension
•
•
Motion under constant acceleration
Projectile motion
•
•
–
2.
Heights and Horizontal Ranges
Maximum ranges and heights
Reference Frames and relative motion
Newton’s Laws of Motion
•
•
Force
Law of Inertia
Quiz Next Monday, Sept. 20!!
Monday, Sept. 13, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
1
Announcements
• Quiz #2 Next Monday, Sept. 20
– Will cover Chapters 1 – up to what we cover this Wednesday
• e-mail distribution list: 34/47 of you have subscribed so far.
– Remember -3 points extra credit if not registered by Wednesday
– A test message will be sent Thursday for verification purpose
• Remember the 1st term exam, Monday, Sept. 27, two weeks
from today
– Covers up to chapter 6.
– No make-up exams
• Miss an exam without pre-approval or a good reason: Your grade is F.
– Mixture of multiple choice and free style problems
• Homework is designed to cover the most current material
– Sometimes we don’t cover them all so you might have to go beyond
what is covered in the class.
– But this is better than covering material too old.
Monday, Sept. 13, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
2
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
i, j , k
So the vector A can
be re-written as
Monday, Sept. 13, 2004
A  Ax i  Ay j  A cos i  A sin  j
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
3
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
r  r f  r i
• Average Velocity:
r
r f  ri
v

t
t f  ti
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Monday, Sept. 13, 2004
v  lim
t 0
r
dr

t
dt
How is each of
these quantities
defined in 1-D?
v
v f  vi
a

t
t f  ti
v d v d  d r  d 2 r
a  lim


 2


t 0 t
dt dt  dt  dt
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
4
Kinematic Quantities in 1d and 2d
Quantities
1 Dimension
2 Dimension
Displacement
x  x f  xi
r  r f  r i
Average Velocity
x x f  xi
vx 

t
t f  ti
v
Inst. Velocity
x dx
v x  lim

t  0 t
dt
Average Acc.
v x v xf  v xi
ax 

t
t f  ti
r d r
v  lim

t 0 t
dt
Inst. Acc.
Monday, Sept.What
13, 2004is
a
r
r f  ri

t
t f  ti
v
v f  vi

t
t f  ti
vx dvx d 2 x
v d v d 2 r
a x  lim

 2 a  lim

 2
t 0 t
t 0 t
dt dt
dt dt
the difference
between
1D2004
and 2D quantities?
PHYS
1443-003, Fall
Dr. Jaehoon Yu
5
2-dim Motion Under Constant Acceleration
• Position vectors in x-y plane:
r
r
r
r
r
r
r i xi i  yi j
r f x f i  y f 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 acceleration vector
X-comp
v xf  vxi  axt
Y-comp
v yf  v yi  a y t
r
r
r
r r
v f   vxi  axt  i   vyi  a yt  j  vi  at
Monday, Sept. 13, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
6
2-dim Motion Under Constant Acceleration
• How are the position vectors written in acceleration
vectors?
Position vector
components
Putting them
together in a
vector form
Regrouping
above results in
Monday, Sept. 13, 2004
1 2
x f  xi  vxi t  ax t
2
r
r
r
r f  xf i  yf j 
1 2
y f yi  v yi t  a y t
2
r 
r
1
1

2 
2 
  xi  vxi t  ax t  i   yi  v yi t  a y t  j
2
2

 






r
r
r
r
r 2
1 r
 xi i  yi j  vxi i  vyi j t  ax i  a y j t
2
r
ur r
1 2
 ri  v i t  at
2
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
7
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 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 
r
r
r
r
r
Velocity vector
v(t )  vx  t  i v y  t  j   20  4.0t  i 15 j (m / s)
Compute the velocity and speed of the particle at t=5.0 s.
r
r
r
r
r
r
r
v t 5  vx ,t 5 i v y ,t 5 j   20  4.0  5.0  i 15 j  40i 15 j m / s

r
speed  v 
Monday, Sept. 13, 2004
 vx 
2
  vy 
2

 40
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
2

  15  43m / s
2
8
Example for 2-D Kinematic Eq. Cnt’d
Angle of the
Velocity vector
 vy
  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 
Monday, Sept. 13, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
9
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 ( no acceleration )
– Vertical motion under constant
acceleration ( g )
Monday, Sept. 13, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
10
Show that a projectile motion is a parabola!!!
x-component
vxi  vi cos 
v yi  vi sin  i
y-component
a  ax i  a y j   g j
ax=0
x f  vxi t  vi cos i t
t
xf
vi cos 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

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 i 
  2 g  v cos  
i 
 i
 vi cos i 

g
y f  x f tan  i  
2
2
2
v
cos
i
 i
Monday, Sept. 13, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
 2
x f


2
What kind of parabola is this?
11
Projectile Motion
Monday, Sept. 13, 2004
The
only acceleration in this
PHYS 1443-003, Fall 2004
Dr. Jaehoon
motion.
It isYua constant!!
12
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 y component, because
t 80  gt   0
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.
Monday, Sept. 13, 2004
80
 t  0 or t 
 8 sec
g
 t  8sec
x f  vxi t  20  8  160  m
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
13
Horizontal Range and Max Height
• Based on what we have learned in the previous pages, one
can analyze a projectile motion in more detail
– Maximum height an object can reach
– Maximum range
What happens at the maximum height?
At the maximum height the object’s vertical
motion stops to turn around!!
v yf  v yi  a y t
vi

Monday, Sept. 13, 2004
h
 vi sin   gt A  0
vi sin 
t A 
g
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
14
Horizontal Range and Max Height
Since no acceleration in x direction, it still flies even if vy=0.
 vi sin i 
R  vxi t  vxi  2t A   2vi cos i 

g


 vi 2 sin 2 i 


Range
R

y f  h  v yi t 
Height
g


1
 vi sin i  1  vi sin  i 
2
  g  t  vi sin i 

  g
g
g
2

 2 

 vi 2 sin 2 i 
yf  h  

2g


Monday, Sept. 13, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
15
2
Maximum Range and Height
• What are the conditions that give maximum height and
range of a projectile motion?
 vi 2 sin 2  i
h  
2g

 vi 2 sin 2 i 

R  

g


Monday, Sept. 13, 2004




This formula tells us that
the maximum height can
be achieved when i=90o!!!
This formula tells us that
the maximum range can
be achieved when
2i=90o, i.e., i=45o!!!
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
16