Transcript Wednesday, February 13, 2008

PHYS 1441 – Section 002
Lecture #9
Wednesday, Feb. 13, 2008
Dr. Jaehoon Yu
•
•
Motion in Two Dimensions
– Projectile Motion
– Maximum ranges and heights
Newton’s Laws of Motion
– Force
– Newton’s first law: Inertia & Mass
– Newton’s second law
Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Announcements
• 1st term exam next Wednesday, Feb. 20
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Time: 1 – 2:20pm
Place: SH103
Covers: Appendices, CH 1 – CH4.4
Style: mixture of multiple choices and essay problems
Class on Monday, Feb. 18: Jason will be here to go over
the any problems you would like to review
• Colloquium today
– UTA Physics faculty research topics
– Good opportunity to learn what UTA Physics does
– Possibly working with them if you are interested in…
Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Reminder: Special Project
• Show that a projectile motion’s trajectory is a
parabola!!
– 20 points
– Due: Wednesday, Feb. 27
– You MUST show full details of computations to
obtain any credit
Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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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 
• ax  0 m s 2 and a y  9.8m 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 ) v xf  vx 0
– Vertical motion under constant acceleration
( g ) v  v  a t  v  9.8
Wednesday, Feb. 13,
yf 2008
y0

t
Spring 2008
y PHYS 1441-002,
y0
Dr. Jaehoon Yu
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Kinematic Equations in 2-Dim
x-component
y-component
vx  vxo  axt
v y  v yo  a y t
x
1
2
 vxo  vx  t
2
y
2
xo
x  vxot  ax t
1
2
Wednesday, Feb. 13, 2008
v
yo
 vy  t
v  v  2ay y
v  v  2a x x
2
x
y
1
2
2
2
yo
y  vyot  ayt
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
1
2
2
6
Show that a projectile motion is a parabola!!!
x-component
vxi  vi cos 
v yi  vi sin i
y-component
a axi  a y j   gj
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
Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
 2
x f


2
What kind of parabola is this?
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Projectile Motion
Maximum height
Wednesday, Feb. 13, 2008
The
only acceleration in this
PHYS 1441-002, Spring 2008
Dr. Jaehoon
motion.
It isYua constant!!
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Example 6 The Height of a Kickoff
A placekicker kicks a football at and angle of 40.0 degrees and
the initial speed of the ball is 22 m/s. Ignoring air resistance,
determine the maximum height that the ball attains.
Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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First, the initial velocity components
v0  22 m s
v0 y
  40
v0 x
vox  vo cos   22m s  cos 40  17 m s
voy  vo sin    22m s  sin 40  14 m s
Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Motion in y-direction is of the interest..
y
ay
vy
voy
?
-9.8 m/s2
0 m/s
+14 m/s
Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
t
11
Now the nitty, gritty calculations…
y
ay
vy
voy
?
-9.80 m/s2
0
14 m/s
t
What happens at the maximum height?
The ball’s velocity in y-direction becomes 0!!
And the ball’s velocity in x-direction? Stays the same!! Why?
Because there is no
acceleration in xdirection!!
Which kinematic formula would you like to use?
v  v  2ay y
2
y
2
oy
y
Wednesday, Feb. 13, 2008
Solve for y
y
0  14 m s 
2  9.8 m s
2
2

v y2  voy2
2a y
 10 m
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Example 7 The Time of Flight of a Kickoff
What is the time of flight between kickoff and landing?
Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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What is y when it reached the max range?
y
ay
0m
-9.80 m/s2
Wednesday, Feb. 13, 2008
vy
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
voy
t
14 m/s
?
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Now solve the kinematic equations in y direction!!
y
ay
0
-9.80 m/s2
y  voyt  ayt
1
2
Two soultions
2
t 0
voy  12 a y t  0
Wednesday, Feb. 13, 2008
vy
Since y=0
voy
t
14 m/s
?
0  voyt  ayt  t  voy  12 a y t 
1
2
2
or
Solve
for t
voy 2voy
2 14  2.9 s


t
9.8
1
a
y
ay
2
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Ex. 8 The Range of a Kickoff
Calculate the range R of the projectile.
2
1
v
t

a
t
x  ox 2 x  vox t  17 m s  2.9 s   49 m
Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Projectile Motion
Wednesday, Feb. 13, 2008
The
only acceleration in this
PHYS 1441-002, Spring 2008
Dr. Jaehoon
motion.
It isYua constant!!
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Horizontal Range and Max Height
• Based on what we have learned in the previous lecture, 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  v0 y  a y t  v0 sin 0  gt A  0
Solve for tA
Wednesday, Feb. 13, 2008
v0 sin  0
tA 
g
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
Time to reach to the
maximum height!!
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Horizontal Range and Max Height
Since no acceleration is in x direction, it still flies even if vy=0.
 v0 sin  0 
R  v0 xt v0 x 2t A   2v0 cos 0 

g


 v0 2 sin 20 
Range
R

y f  h  v0 y t 
Height
g


1
 v0 sin 0
2
  g  t  v0 sin 0
g
2

 1

 2
 v0 sin  0 
g

g


 v0 2 sin 2 0 
yf  h  

2g


Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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2
Maximum Range and Height
• What are the conditions that give maximum height and
range of a projectile motion?
 v0 sin 0 
h

2g


2
2
 v02 sin 20 
R

 g 
Wednesday, Feb. 13, 2008
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 1441-002, Spring 2008
Dr. Jaehoon Yu
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Force
We’ve been learning kinematics; describing motion without understanding
what the cause of the motion is. Now we are going to learn dynamics!!
FORCE is what causes an object to move.
Can someone tell me
The above statement is not entirely correct. Why?
what FORCE is?
Because when an object is moving with a constant velocity
no force is exerted on the object!!!
FORCEs are what cause any changes to the velocity of an object!!
What does this statement mean?
What happens if there are several
forces being exerted on an object?
F1
F2
Wednesday, Feb. 13, 2008
NET FORCE,
F= F1+F2
When there is force, there is change of velocity!!
What does force cause? It causes an acceleration.!!
Forces are vector quantities, so vector sum of all
forces, the NET FORCE, determines the direction of
the acceleration of the object.
When the net force on an object is 0, it has
constant velocity and is at its equilibrium!!
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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More Force
There are various classes of forces
Contact Forces: Forces exerted by physical contact of objects
Examples of Contact Forces: Baseball hit by a bat, Car collisions
Field Forces: Forces exerted without physical contact of objects
Examples of Field Forces: Gravitational Force, Electro-magnetic force
What are possible ways to measure strength of the force?
A calibrated spring whose length changes linearly with the force exerted .
Forces are vector quantities, so the addition of multiple forces must be
done following the rules of vector additions.
Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Newton’s First Law and Inertial Frames
Aristotle (384-322BC): A natural state of a body is rest. Thus force is required to move an
object. To move faster, ones needs larger forces.
Galileo’s statement on natural states of matter: Any velocity once imparted to a moving
body will be rigidly maintained as long as the external causes of retardation are removed!!
Galileo’s statement is formulated by Newton into the 1st law of motion (Law of
Inertia): In the absence of external forces, an object at rest remains at rest and
an object in motion continues in motion with a constant velocity.
What does this statement tell us?
•
•
•
When no force is exerted on an object, the acceleration of the object is 0.
Any isolated object, the object that do not interact with its surroundings, is
either at rest or moving at a constant velocity.
Objects would like to keep its current state of motion, as long as there are no
forces that interfere with the motion. This tendency is called the Inertia.
A frame of reference that is moving at a constant velocity is called the Inertial Frame
Is a frame of reference with an acceleration an Inertial Frame?
Wednesday, Feb. 13, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
NO!
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