PHYS 1441 – Section 002
Lecture #6
Monday, Feb. 16, 2009
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
Today’s homework is homework #4, due 9pm, Tuesday, Feb. 25!!
Monday, Feb. 16, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
1
Announcements
• First term exam
– 1 – 2:20pm, Wednesday, Feb. 18
– Covers: CH1.1 – what we complete on Monday, Feb. 16
+ appendix A1 – A8
– Do not miss the exam
Monday, Feb. 16, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2
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
Monday, Feb. 16, 2009
yf
y0

t
yPHYS 1441-002,
y 0 Spring 2009 Dr.
Jaehoon Yu
3
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
Monday, Feb. 16, 2009
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 2009 Dr.
Jaehoon Yu
1
2
2
4
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
Monday, Feb. 16, 2009
 2
x f


PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2
What kind of parabola is this?
5
Projectile Motion
Maximum height
Monday, Feb. 16, 2009
The
only acceleration in this
PHYS 1441-002, Spring 2009 Dr.
Jaehoon
motion.
It isYu a constant!!
6
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
7
Ex. 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.
Monday, Feb. 16, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
8
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
Monday, Feb. 16, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
9
Motion in y-direction is of the interest..
y
ay
vy
voy
?
-9.8 m/s2
0 m/s
+14 m/s
Monday, Feb. 16, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
t
10
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
Monday, Feb. 16, 2009
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 2009 Dr.
Jaehoon Yu
11
Ex. The Time of Flight of a Kickoff
What is the time of flight between kickoff and landing?
Monday, Feb. 16, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
12
What is y when it reached the max range?
y
ay
0m
-9.80 m/s2
Monday, Feb. 16, 2009
vy
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
voy
t
14 m/s
?
13
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
Monday, Feb. 16, 2009
Since y=0
vy
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 2009 Dr.
Jaehoon Yu
14
Ex. 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
Monday, Feb. 16, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
15
Example for a Projectile Motion
• A stone was thrown upward from the top of a cliff at an angle of 37o
to horizontal with initial speed of 65.0m/s. If the height of the cliff is
125.0m, how long is it before the stone hits the ground?
vxi  vi cos   65.0  cos 37  51.9m / s
v yi  vi sin i  65.0  sin 37  39.1m / s
1 2
y f  125.0  v yi t  gt
2
Becomes
gt 2  78.2t  250  9.80t 2  78.2t  250  0
t
78.2 
 78.22  4  9.80  (250)
2  9.80
t  2.43s or t  10.4s
does not exist.
t Monday,
10.Sept.
4s22, 2008 Since negative timePHYS
1443-002, Fall 2008
Dr. Jaehoon Yu
16
Example cont’d
• What is the speed of the stone just before it hits the ground?
v xf  v xi  vi cos   65.0  cos 37  51.9m / s
v yf  v yi  gt  vi sin i  gt  39.1  9.80 10.4  62.8m / s
v  vxf  v yf  51.9   62.8  81.5m / s
2
2
2
2
• What are the maximum height and the maximum range of the stone?
Do these yourselves at home for fun!!!
Monday, Sept. 22, 2008
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
17
Projectile Motion
Monday, Feb. 16, 2009
The
only acceleration in this
PHYS 1441-002, Spring 2009 Dr.
Jaehoon
motion.
It isYu a constant!!
18
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
Monday, Feb. 16, 2009
v0 sin  0
tA 
g
PHYS 1441-002, Spring 2009 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


Monday, Feb. 16, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
20
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 
Monday, Feb. 16, 2009
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 2009 Dr.
Jaehoon Yu
21
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 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
Monday, Feb. 25, 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
22
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.
Monday, Feb. 16, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
23
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?
Monday, Feb. 16, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
NO!
24
Mass
Mass: A measure of the inertia of a body or quantity of matter
•
•
Independent of the object’s surroundings: The same no matter where you go.
Independent of the method of measurement: The same no matter how you
measure it.
The heavier the object, the bigger the inertia !!
It is harder to make changes of motion of a heavier object than a lighter one.
The same forces applied to two different masses result
in different acceleration depending on the mass.
m1
a2

m2
a1
Note that the mass and the weight of an object are two different quantities!!
Weight of an object is the magnitude of the gravitational force exerted on the object.
Not an inherent property of an object!!!
Weight will change if you measure on the Earth or on the moon but the mass won’t!!
kg 25
Unit of mass?
Monday, Feb. 25, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
Newton’s Second Law of Motion
The acceleration of an object is directly proportional to the
net force exerted on it and is inversely proportional to the
object’s mass.
How do weurwrite the above statement in a mathematical expression?
r
a
F
i
i
ur
r
 Fi  ma
From this
we obtain
m
Newton’s 2nd
Law of Motion
i
Since it’s a vector expression, each component must also satisfy:
F
ix
 max
i
Monday, Feb. 25, 2008
F
iy
 may
i
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
F
iz
 maz
i
26
Unit of the Force
From the vector expression in the previous page, what do
you conclude the dimension and the unit of the force are?
ur
r
 Fi  ma
i
The dimension of force is
The unit of force in SI is
[m][ a]  [ M ][ LT 2 ]
[ Force]  [m][ a]  [ M ][ LT 2 ] 
For ease of use, we define a new
derived unit called, Newton (N)
Monday, Feb. 25, 2008
m
kg
   2   kg  m / s 2
s 
1
1N 1kg  m / s  lbs
4
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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