PHYS 1443 – Section 003
Lecture #3
Wednesday, Sept. 11, 2002
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
2-D Displacement, Velocity and Speed
Projectile Motion
Uniform Circular Motion
Nonuniform Circular Motion
Relative Motion
Today’s homework is homework #4, due 1am, next Wednesday!!
Wednesday, Sept. 11, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
1
Announcements
• e-mail:24 of you have subscribed so far.
– This is the primary communication tool. So do it ASAP.
– A test message will be sent this Wednesday.
• Homework registration: 45 of you have registered (I
have 56 of you)
– Roster will be locked at the end of the day today (5:30pm),
Sept. 11
Wednesday, Sept. 11, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
2
Kinematic Equations of Motion on a
Straight Line Under Constant Acceleration
Velocity as a function of time
v t   vxi  axt
xf
1
1
xf  xi  vxt  vxf  vxi t Displacement as a function
of velocity 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. 11, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
3
Free Fall
• Free fall is a motion under the influence of gravitational pull
(gravity) only; Which direction is a freely falling object moving?
• Gravitational acceleration is inversely proportional to the
distance between the object and the center of the earth
• The gravitational acceleration is g=9.80m/s2 on the surface of
the earth, most of the time.
• The direction of gravitational acceleration is ALWAYS toward
the center of the earth, which we normally call (-y); where up
and down direction are indicated as the variable “y”
• Thus the correct denotation of gravitational acceleration on
the surface of the earth is g=-9.80m/s2
Wednesday, Sept. 11, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
4
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:
Wednesday, Sept. 11, 2002
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 2002
Dr. Jaehoon Yu
5
2-dim Motion Under Constant Acceleration
• Position vectors in x-y plane: r i
• Velocity vectors in x-y plane:
Velocity vectors
in terms of
acceleration
vector
v xf  v xi  a x t
 xi i  yi j
r f  xf i  yf j
v i  v xi i  v yi j
v f  v xf i  v yf j
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
x f  xi  v xi t  a x t 2
2
1
y f  yi  v yi t  a y t 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

 


 

r f  xi i  yi j  v xi i  v yi j t 
Wednesday, Sept. 11, 2002


1
1
a x i  a y j t 2  ri  v i t  at 2
2
2
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
6
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 yf  v yi  a y t  15m / s 
v xf  vxi  a x t  20  4.0t 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
  tan
1
vx 2  v y 2
speed  v 
  15 
1   3 


  tan 
  21
 40 
 8 

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  150(m)
2
2
Can you write down the position vector at t=5.0s?
Wednesday, Sept. 11, 2002
y f  v yi t  15  5  75 (m)


r f  x f i  y f j  150i  75 j m
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
7
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!!!
v xi  vi cos  ;
In a projectile motion, the only
acceleration is gravitational one whose
direction is always toward the center of
the earth (downward).
v yi  vi sin  i
a  ax i  a y j   g j
1
 g  t 2
2
1
 vi sin  i t  gt 2
2
y f  v yi t 
Wednesday, Sept. 11, 2002
x f  v xi t  vi cos  i t
ax=0
yf
Plug in the
t above
t
xf
vi cos i
xf
xf

 1 



 vi sin  i 

g
 v cos   2  v cos  

i 
i 
 i
 i

g
y f  x f tan  i  
2
2
 2vi cos  i
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
 2
x f


What kind of parabola is this?
8
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?
1
y f  40t   g t 2  0m
2
t 80  gt   0
Flight time is determined
by y component, because
the ball stops moving
when it is on the ground
after the flight.
80
 t  0 or t 
 8 sec
g
Distance is determined by x
component in 2-dim,
because the ball is at y=0
position when it completed
it’s flight.
Wednesday, Sept. 11, 2002
x f  v xi t
 20  8  160m 
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
9
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 What happens at the maximum height?
– Maximum range
At the maximum height the object’s
vertical motion stops to turn around!!
v yf  v yi  a y t  vi sin   gt A  0
vi
Since no
acceleration in
x, it still flies
even if vy=0

h
t A 
y f  h  v yi t 
R  v xi 2t A 
 vi sin  i 

2vi cos  i 


g


 vi 2 sin 2 i
R 

g

Wednesday, Sept. 11, 2002
vi sin 
g




1
 g t 2
2
 v sin  i
y f  vi sin  i  i
g

PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
yf
 1  vi sin  i
  g 
g
 2 
 vi 2 sin 2  i


2g





10



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


Wednesday, Sept. 11, 2002




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 2002
Dr. Jaehoon Yu
11
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   20.0  cos 30  17.3m / s
v yi  vi sin  i  20.0  sin 30  10.0m / s
y f  45.0  v yi t 
t 
20.0 
1
gt 2
2
gt 2  20.0t  90.0  9.80t 2  20.0t  90.0  0
 20 2
 4  9.80  ( 90)
2  9.80
t  2.18s or t  4.22s
•
t  4.22s
What is the speed of the stone just before it hits the ground?
vxf  vxi  vi cos  20.0  cos 30  17.3m / s
v yf  v yi  gt  vi sin  i  gt  10.0  9.80  4.22  31.4m / s
v 
v xf
2
 v yf
2
Wednesday, Sept. 11, 2002

17.32   31.4   35.9m / s
2
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
12