Transcript Monday, June 14, 2004 - UTA HEP WWW Home Page

PHYS 1441 – Section 501
Lecture #4
Monday, June 14, 2004
Dr. Jaehoon Yu
Chapter three: Motion in two dimension
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Two dimensional equation of motion
Projectile motion
Chapter four: Newton’s Laws of Motion
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Force and Mass
Newton’s Laws of Motion
Solving problems using Newton’s Laws
Friction Forces
First term exam at 6pm, next Wednesday, June 23!!
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
1
Announcements
• E-mail distribution list: 36 of you have registered
– This Wednesday is the last day of e-mail registration
– -5 extra points if you don’t register by next Wednesday, June 23
– A test message was sent out today.
• Term exam:
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–
–
–
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Date: Next Wednesday, June 26
Time: In the class, 6 – 7:50pm
Covers: Chapters 1 – 6.5 and Appendix A
Location: Classroom, SH 125
Mixture of multiple choices and numeric calculations
There will be a total of 3 exams of which two best will be chose for your grades
Missing an exam is not permitted unless approved prior to the exam by me  I
will only be in e-mail contact for this exam.
• Class Schedule:
– Dr. White will teach you on June 16 and 21
– Mr. Kaushik (Rm 133, x25700) will proctor your exam on June 23
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
2
Displacement, Velocity, and Acceleration in 2-dim
Quantity
One Dimension
Two Dimensions
Displacement
x  xf  xi
r  r f  r i
Average Velocity
xf  xi x
vx 

tf  ti t
r
r f  ri
v

t
t f  ti
r
r
r
v  lim
t 0 t
Instantaneous
Velocity
Average
Acceleration
Instantaneous
Acceleration
Monday, June 14, 2004
vx  lim
Δt 0
x
t
vxf  vxi vx a   v  v f  v i
ax 

t
t f  ti
tf  ti
t
ax  lim
Δt 0
vx
t
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
r
r
v
a  lim
t 0 t
3
2-dim Motion Under Constant Acceleration
• Position vectors in x-y plane:
• Velocity vectors in x-y plane:
Velocity vectors
in terms of
acceleration
vector
r
r
r
r i  xi i  yi j
r
r
r
vi vxi i  v yi j
r
r
r
r f  xf i  yf j
r
r
r
v f  vxf i  v yf j
v xf  vxi  axt v yf  v yi  a y t
r
r
r r r
v f   vxi  axt  i  vyi  a yt j  vi  at


• How are the position vectors written in acceleration vectors?
1
1
x f  xi  vxi t  ax t 2
y f  yi  v yi t  a y t 2
2
2
r 
r
r 
r
r
1
1
2 
2 
r f  x f i  y f j   xi  vxi t  ax t  i   yi  v yi t  a y t  j
2
2





 



r
r
r
r
r 2 ur r
1 r
1r 2
 xi i  yi j  vxi i  vyi j t  ax i  a y j t  ri  v i t  at
2
2
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
4
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, June 14, 2004
 vx 
2
  vy 
2

 40
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
2

  15  43m / s
2
5
Example for 2-D Kinematic Eq. Cnt’d
Angle of the
Velocity vector
 vy
  tan 
 vx
1

1  15 
1  3 
o

tan


21
tan




 
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, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
6
Projectile Motion
Monday, June 14, 2004
The
only acceleration in this
PHYS 1441-501, Summer 2004
Dr. Jaehoon
motion.
It isYua constant!!
7
Projectile Motion
•
A 2-dim motion of an object under the gravitational acceleration with the
assumptions
– Gravitational 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!!!
vxi  vi cos 
r
r
r
r
a  ax i  a y j   g j
1
y f  v yi t    g  t 2
2
1 2
 vi sin i t  gt
2
Monday, June 14, 2004
v yi  vi sin i
ax=0
Plug in the
t above
In a projectile motion, the only
acceleration is gravitational one whose
direction is always toward the center of
the earth (downward).
x f  vxi t  vi cos i t t 
xf
vi cos i
 xf
 1  xf
  g 
y f  vi sin  i 
 vi cos  i  2  vi cos  i

 2
g

x f
y f  x f tan  i  
2
2

 2vi cos  i 
PHYS 1441-501, Summer 2004 What kind of parabola is this?
Dr. Jaehoon Yu
8



2
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, June 14, 2004
80
 t  0 or t 
 8 sec
g
 t  8sec
x f  vxi t  20  8  160  m
PHYS 1441-501, Summer 2004
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
– Maximum range
What happens at the maximum height?
At the maximum height the object’s vertical
motion stops to turn around!!
vi
Since no
acceleration in
x, it still flies
even if vy=0
v yf  v yi  a y t  vi sin   gt A  0

h
R  vxi 2t A 
 v sin i 
2vi cos i  i

g


t A 
vi sin 
g
y f  h  v yi t 
1
g t2
2
 v sin  i  1  vi sin  i 
  g 

y f  vi sin  i  i
 g  2  g 
 vi 2 sin 2 i 
PHYS 1441-501, Summer 2004
R
Monday, June 14, 2004 

g
Dr. Jaehoon Yu


 vi 2 sin 2  i 

y f  

 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


Monday, June 14, 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 1441-501, Summer 2004
Dr. Jaehoon Yu
11
Example for Projectile Motion
•
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?
o
vxi  vi cos   20.0  cos30  17.3m / s
v yi  vi sin i  20.0  sin 30o  10.0m / s
y f  45.0  v yi t 
t
20.0 
1 2
gt
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
Only
physical
answer
What is the speed of the stone just before it hits the ground?
vxf  vxi  vi cos   20.0  cos 30o  17.3m / s
v yf  v yi  gt  vi sin i  gt  10.0  9.80  4.22  31.4m / s
v  vxf 2  v yf 2  17.32   31.4 2  35.9m / s
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
12
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, June 14, 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 1441-501, Summer 2004
Dr. Jaehoon Yu
13
Force
We’ve been learning kinematics; describing motion without understanding
what the cause of the motion was. Now we are going to learn dynamics!!
FORCEs are what cause 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 in 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, June 14, 2004
NET FORCE,
F= F1+F2
When there is force, there is change of velocity.
Forces cause acceleration.
Forces are vector quantities, so vector sum of all
forces, the NET FORCE, determines the motion of
the object.
When net force on an objectis 0, it has
constant velocity and is at its equilibrium!!
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
14
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 Force?
A calibrated spring whose length changes linearly with the exerted force.
Forces are vector quantities, so addition of multiple forces must be done
following the rules of vector additions.
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
15
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 force.
Galileo’s statement on natural states of matter: Any velocity once imparted to a moving
body will be rigidly maintained as long as 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 is no
force that interferes with the motion. This tendency is called the Inertia.
A frame of reference that is moving at constant velocity is called an Inertial Frame
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
16
Mass
Mass: A measure of the inertia of a body or quantity of matter
1.
2.
Independent of the object’s surroundings: The same no matter where you go.
Independent of method of measurement: The same no matter how you
measure it.
The heavier an object gets the bigger the inertia!!
It is harder to make changes of motion of a heavier object than the lighter ones.
The same forces applied to two different masses result
in different acceleration depending on the mass.
m1
a2

m2
a1
Note that mass and weight of an object are two different quantities!!
Weight of an object is the magnitude of 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.
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
17
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 we write the above statement
in a mathematical expression?
Since it’s a vector expression, each
component should also satisfy:
F
ix
 F  ma
i
i
 max
i
F
iy
F
 may
iz
i
 maz
i
From the above vector expression, what do you conclude the dimension and
unit of force are?
2
The unit of force in SI is [ Force]  [m][a]  kg m / s
For ease of use, we define a new
derived unit called, a Newton (N)
Monday, June 14, 2004
1
1N  1kg  m / s  lbs
4
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
2
18
Example 4.2
What constant net force is required to bring a 1500kg car to rest from a speed
of 100km/h within a distance of 55m?
What do we need to know to figure out the force?
What are given? Initial speed:
Acceleration!!
vxi  100km / h  28m / s
Final speed: v xf  0m / s
Displacement: x  x f  xi  55m
This is a one dimensional motion. Which kinetic formula do we use to find acceleration?
2
vxf2  vxi2
2
2



28
m
/
s
vxf  vxi  2ax x f  xi Acceleration ax 

 7.1m / s 2
2x f  xi 
255m 

Thus, the force needed
to stop the car is

Fx  max  1500kg   7.1m / s 2   1.1104 N



vxf2  v xi2
m v xf2  v xi2
m v xf2  v xi2
Given the force how far does
x  x f  xi 


the car move till it stops?
2a x
2max
2 Fx
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu

•Linearly proportional to
the mass of the car
•Squarely proportional
to the speed of the car
•Inversely proportional
19 brake
to the force by the
Example for Newton’s 2nd Law of Motion
Determine the magnitude and direction of acceleration of the puck whose
mass is 0.30kg and is being pulled by two forces, F1 and F2, as shown in the
picture, whose magnitudes of the forces are 8.0 N and 5.0 N, respectively.
 
F1x  F1 cos 1  8.0  cos 60o  4.0 N
Components
of F1
F1 y  F1 sin 1  8.0  sin 60o   6.9 N
F1

1
220o
F2
of F2

Fx  F1x  F2 x  4.0  4.7  8.7 N  max
Fy  F1 y  F2 y  6.9  1.7  5.2 N  ma y
F
8.7
ax  x 
 29m / s 2
m 0.3
Fy
Magnitude and
direction of
a
1  17 
o
1  y 
tan
tan



acceleration a
 
   30
29
a

x


o
F2 y  F2 sin  2  5.0  sin  20  1.7 N
Components of
total force F
Monday, June 14, 2004

o
F
cos


5
.
0

cos

20
 4.7 N
F

2
2
2
x
Components
60o

5.2
ay 

 17 m / s 2
m
0.3

Acceleration
Vector a
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
r
2
2
2
a   ax 2   a y    29  17 
 34m / s 2


r




a  ax i  ay j   29 i  17 j  m / s 2


20
Gravitational Force and Weight
Gravitational Force, Fg The attractive force exerted
on an object by the Earth
F G  ma  m g
Weight of an object with mass M is W  F G  M g  Mg
Since weight depends on the magnitude of gravitational acceleration,
g, it varies depending on geographical location.
By measuring the forces one can determine masses. This is why you
can measure mass using spring scale.
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
21
Newton’s Third Law (Law of Action and Reaction)
If two objects interact, the force, F21, exerted on object 1 by object 2
is equal in magnitude and opposite in direction to the force, F12,
exerted on object 1 by object 2.
F21
F12
F 12   F 21
The action force is equal in magnitude to the reaction force but in
opposite direction. These two forces always act on different objects.
What is the reaction force to the
force of a free fall object?
The force exerted by the ground
when it completed the motion.
Stationary objects on top of a table has a reaction force (normal force)
from table to balance the action force, the gravitational force.
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
22
Example of Newton’s 3rd Law
A large man and a small boy stand facing each other on frictionless ice. They put their
hands together and push against each other so that they move apart. a) Who moves away
with the higher speed and by how much?
F 12  F 21  F
F 12   F 21
F12
M
F21=-F12
m
b) Who moves farther while
their hands are in contact?
F 12  ma b
F 12x  mabx F
F 21  M a M
F 21x  MaMx F
F 12   F 21
F 12   F 21  F
21y
ab x 
 maby  0
 MaMy  0
F M

aMx
m m
vMxf  vMxi  aMxt  aMxt
vbxf  vbxi  abxt  abxt 
M
aMxt 
m
M
vMxf
m
 vbxf  vMxf if M  m by the ratio of the masses
Given in the same time interval, since the boy
has higher acceleration and thereby higher
speed, he moves farther than the man.
Monday, June 14, 2004
12 y
xb  vbxf t 
xb 
M
m
1
M
M
abxt 2 
vMxf t 
aMxt 2
2
m
2m
1

 M
v
t

aMx t 2  
xM
 Mxf
2

 m
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
23
Some Basic Information
When Newton’s laws are applied, external forces are only of interest!!
Why?
Because, as described in Newton’s first law, an object will keep its
current motion unless non-zero net external force is applied.
Normal Force, n:
Reaction force that reacts to gravitational
force due to the surface structure of an object.
Its direction is perpendicular to the surface.
Tension, T:
The reactionary force by a stringy object
against an external force exerted on it.
Free-body diagram
Monday, June 14, 2004
A graphical tool which is a diagram of external
forces on an object and is extremely useful analyzing
forces and motion!! Drawn only on an object.
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
24
Free Body Diagrams and Solving Problems
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Free-body diagram: A diagram of vector forces acting on an object
A great tool to solve a problem using forces or using dynamics
Select a point on an object in the problem
Identify all the forces acting only on the selected object
Define a reference frame with positive and negative axes specified
Draw arrows to represent the force vectors on the selected point
Write down net force vector equation
Write down the forces in components to solve the problems
No matter which one we choose to draw the diagram on, the results should be the same,
as long as they are from the same motion
FN
M
FN
FG  M g
FT
Gravitational force
Me
m
A force supporting the object exerted by the floor
The force pulling the elevator (Tension)
What about the box in the elevator?
Monday, June 14, 2004
F GB  mg
FG  M g
Which one would you like to select to draw FBD?
What do you think are the forces acting on this elevator?
Gravitational force
FN
FG  M g
Which one would you like to select to draw FBD?
What do you think are the forces acting on this object?
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
Gravitational
force
Normal
force
FT
FG  M g
FN
25
F BG  m g
Applications of Newton’s Laws
Suppose you are pulling a box on frictionless ice, using a rope.
M
What are the forces being
exerted on the box?
T
Gravitational force: Fg
n= -Fg
Free-body
diagram
n= -Fg
Normal force: n
T
Fg=Mg
T
Fg=Mg
Monday, June 14, 2004
Tension force: T
Total force:
F=Fg+n+T=T
If T is a constant
force, ax, is constant
 Fx  T  Ma x
F
y
ax 
  Fg  n  Ma y  0
T
M
ay  0
v xf  vxi  axt  v xi  
T 
t
M 
1 T  2
v
t

x

x


t
x  f i xi
2M 
PHYS 1441-501, What
Summer
2004 to the motion in y-direction?
happened
Dr. Jaehoon Yu
26
Example w/o Friction
A crate of mass M is placed on a frictionless inclined plane of angle .
a) Determine the acceleration of the crate after it is released.
y
n
x

F  Fg  n  ma
n
Fx  Ma x  Fgx  Mg sin 
Free-body
Diagram

Fg
y
Supposed the crate was released at the
top of the incline, and the length of the
incline is d. How long does it take for the
crate to reach the bottom and what is its
speed at the bottom?
ax  g sin 
x
F= -Mg F  Ma  n  F  n  mg cos   0
y
y
gy
1 2 1
v
t

a x t  g sin  t 2  t 
d  ix
2
2
v xf  vix  axt g sin 
2d
g sin 
2d
 2dg sin 
g sin 
 vxf  2dg sin 
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
27
Forces of Friction
Resistive force exerted on a moving object due to viscosity or other types
frictional property of the medium in or surface on which the object moves.
These forces are either proportional to velocity or normal force
Force of static friction, fs: The resistive force exerted on the object until
just before the beginning of its movement
Empirical
Formula
f s  s n
What does this
formula tell you?
Frictional force increases till it
reaches to the limit!!
Beyond the limit, there is no more static frictional force but kinetic
frictional force takes it over.
Force of kinetic friction, fk
fk  k n
Monday, June 14, 2004
The resistive force exerted on the object
during its movement
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
28
Example w/ Friction
Suppose a block is placed on a rough surface inclined relative to the horizontal. The
inclination angle is increased till the block starts to move. Show that by measuring
this critical angle, c, one can determine coefficient of static friction, s.
y
n
n
fs=kn
x
F= -Mg


Fg
Free-body
Diagram
Net force
F  Ma  Fg n f s
x comp.
Fx  Fgx  f s  Mg sin   f s  0
f s   s n  Mg sin  c
y comp.
Fy  Ma y  n  Fgy  n  Mg cosc  0
n Fgy  Mg cos c
Mg sin  c
Mg sin  c
 tan  c
s 

Mg cos  c
n
Monday, June 14, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
29