PHYS 1441 – Section 002
Lecture #5
Wednesday, Feb. 11, 2009
Dr. Jaehoon Yu
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•
•
•
•
Wednesday, Feb. 11,
2009
Coordinate systems
Vectors and Vector Operations
Motion in Two Dimensions
Motion under constant acceleration
Projectile Motion
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
1
Announcements
• E-mail distribution list: 58 of you subscribed to the list
– Please be sure to subscribe to the list as soon as
possible.
• First term exam
– 1 – 2:20pm, Wednesday, Feb. 18
– Covers: CH1.1 – what we complete on Monday, Feb. 16
+ appendix A1 – A8
• Physics Department colloquium scheduled at 4pm
today in SH101
– There is a double extra credit for colloquium today
Wednesday, Feb. 11,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2
Wednesday, Feb. 11,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
3
Reminder: Special Problems for Extra Credit
• Derive the quadratic equation for yx2-zx+v=0
 5 points
• Derive the kinematic equation v  v0  2a  x  x0 
from first principles and the known kinematic
equations  10 points
• You must show your work in detail to obtain the
full credit
• Due next Monday, Feb. 16
2
Wednesday, Feb. 11,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2
4
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, Feb. 11,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
5
Coordinate Systems
• They make it easy and consistent to express locations or positions
• Two commonly used systems, depending on convenience, are
– Cartesian (Rectangular) Coordinate System
• Coordinates are expressed in (x,y)
– Polar Coordinate System
• Coordinates are expressed in distance from the origin ® and the angle measured
from the x-axis, q (r,q)
• Vectors become a lot easier to express and compute
+y
How are Cartesian and
Polar coordinates related?
y1
(x1,y1) =(r1,q1)
r1
x1  r1 cos q1 r   x12  y12 
1
q1
O (0,0)
Wednesday, Feb. 11,
2009
x1
+x
y1  r1 sin q1
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
y1
tan q1 
x1
 y1 

x
 1
q1  tan 1 
6
Example
Cartesian Coordinate of a point in the xy plane are (x,y)= (-3.50,-2.50)m.
Find the equivalent polar coordinates of this point.
r
y

q
qs
(-3.50,-2.50)m
2
 y2 
  3.50 
2
  2.50 
2

 18.5  4.30( m)
x
r
x
q  180  q s
tan q s 
2.50 5

3.50 7
1
5
q s  tan    35.5
7
o
o
o
q  180  q s  180  35.5  216
Wednesday, Feb. 11,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
o
7
Vector and Scalar
Vector quantities have both magnitudes (sizes)
and directions Force, gravitational acceleration, momentum
ur
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
ur
absolute values: F or F
Scalar quantities have magnitudes only
Can be completely specified with a value
and its unit Normally denoted in normal letters, E
Energy, heat,
mass, time
Both have units!!!
Wednesday, Feb. 11,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
8
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, Feb. 11,
2009
C
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
C: The same magnitude
but opposite direction:
C=-A:A negative vector
F: The same direction
but different magnitude
9
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, Feb. 11,
B 2A
2009
A
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
B=2A
10
Example for 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
q
20
A
2

  B sin q 
2


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
B sin 60
q  tan
1
 tan 1
35.0 sin 60
20.0  35.0 cos 60
30.3
 38.9 to W wrt N
37.5
 tan 1
Wednesday, Feb. 11,
2009
 A  B cosq 
A  B cos 60
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
Find other
ways to
solve this
problem…
11
Components and Unit Vectors
Coordinate systems are useful in expressing vectors in their components
y
(+,+)
Ay
A
(-,+)
(Ax,Ay)
q
(-,-)
u
r
A 

Wednesday, Feb. 11,
2009
(+,-)

Ax
ur
A cos q
 
2
Ax 
ur
A cos q
Ay 
ur
A sin q
x
2
ur
A sin q
u
r 2
A
cos 2 q  sin 2 q

A  Ax  Ay

PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu

2
}
Components
} Magnitude
2
u
r
 A
12
Unit Vectors
• Unit vectors are the ones that tells us the
directions of the components
• Dimensionless
• Magnitudes are exactly 1
• Unit vectors are usually expressed in i, j, k or
r r r
i, j, k
So the vector A can
be re-written as
Wednesday, Feb. 11, 2009
r
ur
ur
r ur
r
r
A  Ax i Ay j  A cos q i  A sin q j
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
13
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
ur
C 
 4.0   2.0
2
r
  2.0  4.0  j

r
r
 4.0i 2.0 j  m 
2
 16  4.0  20  4.5(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, Feb. 11,
2009
ur
D



 25  59   7.0  65(cm)
2
2
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2
14
2 Dimensional Kinematic Quantities
r
ro  initial position
r
r  final position
r r r
r  r  ro  displacement
Wednesday, Feb. 11,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
15
2D Average Velocity
Average velocity is the
displacement divided by
the elapsed time.
r r
r
r r  ro r
v  t t 
t
o
Wednesday, Feb. 11,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
16
The instantaneous velocity indicates how fast the car
moves and the direction of motion at each instant of time.
r
r
r
v  lim
t 0 t
Wednesday, Feb. 11,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
17
2D Instantaneous Velocity
r
r
r
v  lim
t 0 t
Wednesday, Feb. 11,
2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
18
2D Average Acceleration
r
v
r
v
r
vo
Wednesday, Feb. 11,
2009
r r
r
v
r v  vo

a
t  to
t
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
19
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
• Average Velocity:
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Wednesday, Feb. 11,
2009
r r
r
r  r f  r i
r
r
r
rf
r

v
tf
t
r
r
r
v  lim
t  0 t
r
r
r
vf
v

a
tf
t
r
 ri
 ti
How is each of
these quantities
defined in 1-D?
r
 vi
 ti
r
r
v
a  lim
t  0 t
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
20
Kinematic Quantities in 1D and 2D
Quantities
Displacement
Average Velocity
1 Dimension
x  x f  xi
vx 
x f  xi
x

t
t f  ti
Inst. Velocity
x
vx  lim
t  0  t
Average Acc.
v x v xf  v xi
ax 

t
t f  ti
Inst. Acc.
Wednesday, Feb.
11, 2009
What
is
v x
ax  lim
t  0  t
2 Dimension
r
r 
r
r
r
r r r f  r i
v

t
t f  ti
r
r
r
v  lim
t 0 t
r
r
r
r
v v f  vi
a

t
t f  ti
r
r
v
a  lim
t 0 t
PHYS 1441-002,
Spring
2009
Dr. 2D quantities?
the difference
between
1D
and
Jaehoon Yu
r r
r f  ri
21
A Motion in 2 Dimension
This is a motion that could be viewed as two motions
combined into one.
Wednesday, Feb. 6, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
22
Motion in horizontal direction (x)
 vxo  vx  t
vx  vxo  axt
x
v  v  2a x x
x  vxot  ax t
2
x
2
xo
Wednesday, Feb. 6, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
1
2
1
2
2
23
Motion in vertical direction (y)
v y  v yo  a y t
y
1
2
v
yo
 vy  t
v  v  2ay y
2
y
2
yo
y  vyot  ayt
1
2
Wednesday, Feb. 6, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
2
24
A Motion in 2 Dimension
Imagine you add the two 1 dimensional motions on the left.
It would make up a one 2 dimensional motion on the right.
Wednesday, Feb. 6, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
25
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. 6, 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
26
Ex. A Moving Spacecraft
In the x direction, the spacecraft has an initial velocity component of +22
m/s and an acceleration of +24 m/s2. In the y direction, the analogous
quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x and vx,
(b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s.
Monday, Feb. 11, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
27
How do we solve this problem?
1. Visualize the problem  Make a drawing.
2. Decide which directions are to be called positive (+) and
negative (-).
3. Write down the values that are given for any of the five
kinematic variables associated with each direction.
4. Verify that the information contains values for at least
three of the kinematic variables. Do this for x and y
separately. Select the appropriate equation.
5. When the motion is divided into segments, remember
that the final velocity of one segment is the initial velocity
for the next.
6. Keep in mind that there may be two possible answers to
a kinematics problem.
Monday, Feb. 11, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
28
Ex. continued
In the x direction, the spacecraft has an initial velocity component of +22
m/s and an acceleration of +24 m/s2. In the y direction, the analogous
quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x and vx,
(b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s.
x
ax
vx
vox
t
?
+24.0 m/s2
?
+22.0 m/s
7.0 s
y
ay
vy
voy
t
?
+12.0 m/s2
?
+14.0 m/s
7.0 s
Monday, Feb. 11, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
29
First, the motion in x-direciton…
x
ax
vx
vox
t
?
+24.0 m/s2
?
+22 m/s
7.0 s
2
1
v
t

a
t
x  ox 2 x
  22m s  7.0 s  
1
2
 24m s  7.0 s 
2
vx  vox  axt
  22 m s    24 m s
Monday, Feb. 11, 2008
2
2
 740 m
  7.0 s   190 m s
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
30
Now, the motion in y-direction…
y
ay
vy
voy
t
?
+12.0 m/s2
?
+14 m/s
7.0 s
y  voy t  a y t
1
2
2
 14m s  7.0 s  
1
2
12m s  7.0 s 
2
v y  voy  a y t
 14 m s   12 m s
Monday, Feb. 11, 2008
2
2
 390 m
  7.0 s   98 m s
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
31
The final velocity…
v
v y  98 m s
q
vx  190 m s
190 m s    98 m s   210 m s
A vector can be fully described when
1
q  tan  98 190   27 the magnitude and the direction are
v
2
2
given. Any other way to describe it?
Yes, you are right! Using
components and unit vectors!!
Monday, Feb. 11, 2008


r
r
r
r
r
v  vx i  v y j  190i  98 j m s
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
32
If we visualize the motion…
Monday, Feb. 11, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
33