PHYS 1441 – Section 501
Lecture #3
Wednesday, June 9, 2004
Dr. Jaehoon Yu
Chapter two: Motion in one dimension
•
–
•
Chapter three: Motion in two dimension
–
–
–
–
•
Free Fall
Coordinate systems
Vector and Scalar
Two dimensional equation of motion
Projectile motion
Chapter four: Newton’s Laws of Motion
Today’s homework is HW #2, due 6pm, next Wednesday, June 15!!
Wednesday, June 9, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
1
Announcements
• E-mail distribution list: 25 of you have registered
– Remember 3 extra credit points if done by midnight
tonight
– Next Wednesday is the last day of e-mail registration
– -5 extra points if you don’t register by next Wednesday
– A test message will be sent out this Friday
• Homework: You are supposed to download the
homework assignment, solve it offline and input the
answers back online.
– 44 registered
– 40 submitted
Wednesday, June 9, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
2
Kinematic Equations of Motion on a
Straight Line Under Constant Acceleration
vxf t   vxi  axt
Velocity as a function of time
1
1
Displacement as a function
xf  xi  v x t   vxf  vxi  t of velocities 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 Kinetic equations, depending on the
information given to you for specific physical problems!!
Wednesday, June 9, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
3
Falling Motion
• Falling motion is a motion under the influence of gravitational
pull (gravity) only; Which direction is a freely falling object
moving?
– A motion under constant acceleration
– All kinematic formula we learned can be used to solve for falling
motions.
• 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, June 9, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
4
Example for Using 1D Kinematic
Equations on a Falling object
Stone was thrown straight upward at t=0 with +20.0m/s
initial velocity on the roof of a 50.0m high building,
What is the acceleration in this motion?
g=-9.80m/s2
(a) Find the time the stone reaches at the maximum height.
What happens at the maximum height? The stone stops; V=0
v f vyi  ayt  20.0  9.80t  0.00m / s
Solve for t
t
20.0
 2.04s
9.80
(b) Find the maximum height.
1 2
1
yf  yi  vyit  ayt  50.0  20  2.04   (9.80)  (2.04) 2
2
2
 50.0  20.4  70.4(m)
Wednesday, June 9, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
5
Example of a Falling Object cnt’d
(c) Find the time the stone reaches back to its original height.
t  2.04  2  4.08s
(d) Find the velocity of the stone when it reaches its original height.
vyf  vyi  ayt  20.0  (9.80)  4.08  20.0(m / s)
(e) Find the velocity and position of the stone at t=5.00s.
Velocity
Position
vyf  vyi  ayt  20.0  (9.80)  5.00  29.0(m / s )
1
yf  yi  vyit  ayt 2
2
 50.0  20.0  5.00 
Wednesday, June 9, 2004
1
 ( 9.80)  (5.00) 2 27.5(m)
2
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
6
Coordinate Systems
• Makes it easy to express locations or positions
• Two commonly used systems, depending on convenience
– Cartesian (Rectangular) Coordinate System
• Coordinates are expressed in (x,y)
– Polar Coordinate System
• Coordinates are expressed in (r,q)
• Vectors become a lot easier to express and compute
+y
How are Cartesian and
Polar coordinates related?
y1
(x1,y1)=(r,q)
2
2
r
x1  r cos q r   x1  y1 
q
O (0,0)
Wednesday, June 9, 2004
x1
+x
y1  r sin q
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
tan q 
y1
x1
7
Example
Cartesian Coordinate of a point in the x-y plane are (x,y)= (-3.50,-2.50)m.
Find the polar coordinates of this point.
r
y

q
qs
(-3.50,-2.50)m
  3.50 
2
  2.50 
2

 18.5  4.30( m)
x
r
2
2
x

y


q  180  q s
tan q s 
2.50 5

3.50 7
o
1  5 
tan
qs 
   35.5
7
q  180  q s  180o  35.5o  216o
Wednesday, June 9, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
8
Vector and Scalar
Vector quantities have both magnitude (size)
and direction Force, gravitational acceleration, momentum
Normally denoted in BOLD letters, F, or a letter with arrow on top F
Their sizes or magnitudes are denoted with normal letters, F, or
absolute values: F or F
Scalar quantities have magnitude only
Can be completely specified with a value
and its unit Normally denoted in normal letters, E
Energy, heat,
mass, weight
Both have units!!!
Wednesday, June 9, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
9
Properties of Vectors
• Two vectors are the same if their sizes and the directions
are the same, no matter where they are on a coordinate
system.
Which ones are the
same vectors?
y
D
A=B=E=D
F
A
Why aren’t the others?
B
x
E
Wednesday, June 9, 2004
C
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
C: The same magnitude
but opposite direction:
C=-A:A negative vector
F: The same direction
but different magnitude
10
Vector Operations
•
Addition:
– Triangular Method: One can add vectors by connecting the head of one vector to
the tail of the other (head-to-tail)
– Parallelogram method: Connect the tails of the two vectors and extend
– Addition is commutative: Changing order of operation does not affect the results
A+B=B+A, A+B+C+D+E=E+C+A+B+D
A+B
B
A
•
A
=
B
A+B
OR
A+B
B
A
Subtraction:
– The same as adding a negative vector:A - B = A + (-B)
A
A-B
•
-B
Since subtraction is the equivalent to adding
a negative vector, subtraction is also
commutative!!!
Multiplication by a scalar is
increasing the magnitude A, B=2A
Wednesday, June 9, 2004
B 2A
A
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
B=2A
11
Example of Vector Addition
A car travels 20.0km due north followed by 35.0km in a direction 60.0o west
of north. Find the magnitude and direction of resultant displacement.
r
Bsinq N
B 60o Bcosq
r
q1
20
A



A2  B 2 cos 2 q  sin 2 q  2 AB cosq

A2  B 2  2 AB cosq

20.02  35.02  2  20.0  35.0 cos 60
 2325  48.2(km)
E
q 1  tan 1
 tan 1
 tan 1
Wednesday, June 9, 2004
 A  B cosq 2  B sin q 2
ur
B sin 60
ur
ur
A  B cos 60
35.0 sin 60
20.0  35.0 cos 60
30.3
 38.9 to W wrt N
37.5
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
Find other
ways to
solve this
problem…
12
Components and Unit Vectors
Coordinate systems are useful in expressing vectors in their components
y
Ay
(-,+)
Ax  A cos q
(+,+)
(Ax,Ay)
A
(+,-)
Ax
x
2

}
Components
} Magnitude
ur
ur
A
cos
q

  A sin q 

A  Ax  Ay
u
r
A 
q
(-,-)
Ay  A sin q
2
2
u
r 2
A
cos 2 q  sin 2 q


u
r
 A
• Unit vectors are dimensionless vectors whose magnitudes are exactly 1
• Unit vectors are usually expressed in i, j, k or i, j , k
• Vectors can be expressed using components and unit vectors
ur
r ur
r
r
r
ur
So the above vector A
A  Ax i  Ay j  A cos q i  A sin q j
can be written as
Wednesday, June 9, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
13
2
Examples of Vector Operations
Find the resultant vector which is the sum of A=(2.0i+2.0j) and B =(2.0i-4.0j)

 
r
r
r
r
ur ur ur
C  A  B  2.0i  2.0 j  2.0i  4.0 j
r
  2.0  2.0  i
r
  2.0  4.0  j
ur
C   4.02   2.02
 16  4.0  20  4.5(m)

r
r
 4.0i 2.0 j  m 
q
2.0
tan
 tan
 27o
Cx
4.0
1
Cy
1
Find the resultant displacement of three consecutive displacements:
d1=(15i+30j +12k)cm, d2=(23i+14j -5.0k)cm, and d3=(-13i+15j)cm
ur ur ur ur
r
r
r
r
r
r
r
r
D  d 1  d 2  d 3  15i  30 j  12k  23i  14 j  5.0k  13i  15 j
r
r
r
r
r
r
 15  23  13 i   30  14  15  j  12  5.0  k  25i 59 j 7.0k (cm)

Magnitude
Wednesday, June 9, 2004
ur
D



 25  59   7.0  65(cm)
2
2
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
2
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