Wednesday, Feb. 13, 2008
Dr. Jae hoon 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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• 1 st term exam next Wednesday, Feb. 20
– 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
2

• 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
3

Wednesday, Feb. 13, 2008 PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
4

• 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 g
 
9.8
j r  m s 2

• a
 x
0 m s
2 and a y
 
9.8
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 ) v xf
 v x 0
– Vertical motion under constant acceleration
( g ) v
 v y 0
 a t y
 v

Dr. Jaehoon Yu
9.8
 t
5

v x
x-component y-component
 v xo
 a t x v y
 v yo
 a t y x

1
 v
2 xo
 v x
 t y

1
 v
2 yo
 v y
 t v
2 x
 v
2 xo

2 a x x v
2 y
 v
2 yo

2 a y y x
 v t xo

1
2 a t x
2
Wednesday, Feb. 13, 2008 y
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
 v t yo

1
2 a t y
2
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Show that a projectile motion is a parabola!!!
x-component v
 xi v i cos

 y-component v
 yi v i sin
 i a x
=0 a
 a i x
 a j y
  gj x f
 v xi t
 v i cos
 i t t
 v i x f cos
 i
In a projectile motion, the only acceleration is gravitational one whose direction is always toward the center of the earth (downward).
y f
 v t yi

1
2
  t
2
 v i sin
 i

1 t gt
2
2
Plug t into the above y
 f v i sin
 i

 v i x f cos
 i



1

2 g

 v i x f cos
 i


2 y f
Wednesday, Feb. 13, 2008
 x f tan
 i



2 v
2 g cos
2
 i i
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu

 x f
2
What kind of parabola is this?
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Wednesday, Feb. 13, 2008
The only acceleration in this
PHYS 1441-002, Spring 2008 motion. It is a constant!!
8

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
9

v
0

22 m s v
0 y
 
40 v
0 x v
 ox v o cos
    
17 m s v
 oy v o sin
    
14 m s
Wednesday, Feb. 13, 2008 PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
10

?
y
Wednesday, Feb. 13, 2008 a y
-9.8 m/s 2 v y
0 m/s
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu v oy
+14 m/s t
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?
y a y
-9.80 m/s 2 v y
0 v oy
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?
v
Which kinematic formula would you like to use?
2 y
 v
2 oy

2 a y y y

0
2




Solve for y
14 m s
9.8 m s

2

2 y
 v
 
10 m
2 y
2
 a y v
2 oy
Wednesday, Feb. 13, 2008 PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
Because there is no acceleration in xdirection!!
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What is the time of flight between kickoff and landing?
Wednesday, Feb. 13, 2008 PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
13

y
0 m
Wednesday, Feb. 13, 2008 a y
-9.80 m/s 2 v y
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu v oy
14 m/s ?
t
14

Now solve the kinematic equations in y direction!!
y
0 a y
-9.80 m/s 2 v y v oy
14 m/s t
?
y
 v t oy

1
2 a t y
2
Since y=0
0
 v t 1 a t oy

2 y
2 
 oy

1
2 a t y
 v
Two soultions oy

1
2 a t y t

0

Wednesday, Feb. 13, 2008
0 or
Solve for t t

 v oy
1
2 a y


2 v oy a y


9.8

2.9
s
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
15

Calculate the range R of the projectile.
x
 v t ox

1
2 a t x
2  v t
 ox
    
49 m
Wednesday, Feb. 13, 2008 PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
16

Wednesday, Feb. 13, 2008
The only acceleration in this
PHYS 1441-002, Spring 2008 motion. It is a constant!!
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• 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 v
0 y
 a t y
 v
0 sin

0
 gt
A

0
Solve for t
A t
A v
0 sin

0 g
Time to reach to the maximum height!!
Wednesday, Feb. 13, 2008 PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
18

Since no acceleration is in x direction, it still flies even if v y
=0.
R
 v t
0 x
 v
0 x
 
A

2 v
0 cos

0

 v
0 sin

0 g

 v
0
2 sin 2

0 Range
R
 g

 y
 f h
 v t
0 y

1
2
 
2  v
0 sin

0

 v
0 sin

0 g



1
2 g
 v
0 sin

0 g


2
Height y f
 h

 v
0
2 sin
2

0
2 g


Wednesday, Feb. 13, 2008 PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
19

• What are the conditions that give maximum height and range of a projectile motion?
h
 v
0
2 sin
2

0
2 g


This formula tells us that the maximum height can be achieved when  i
=90 o !!!
R
 

 v
0
2 sin 2

0 g



Wednesday, Feb. 13, 2008
This formula tells us that the maximum range can be achieved when
2  i
=90 o , i.e.,  i
=45 o !!!
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
20

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 what FORCE is?
The above statement is not entirely correct. Why?
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?
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.
F
1
F
2
Wednesday, Feb. 13, 2008
NET FORCE,
F= F
1
+F
2
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
21

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
22

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 1 st 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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