PHYS 1441 – Section 002
Lecture #6
Monday, Sept. 27, 2010
Dr. Jaehoon Yu
•
•
•
•
•
Components of the 2D Vector
Understanding the 2 Dimensional Motion
2D Kinematic Equations of Motion
Projectile Motion
Maximum Range and Height
Today’s homework is homework #4, due 10pm, Tuesday, Oc. 5!!
Monday, Sept. 27, 2010
PHYS 1441-002, Fall 2010
Dr. Jaehoon Yu
1
2D 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,  (r )
• Vectors become a lot easier to express and compute
+y
How are Cartesian and
Polar coordinates related?
y1
(x1,y1) =(r1 1)
r1
r  x 12  y 12
x  r cos 
1
1
O (0,0)
Monday, Sept. 20, 2010
x1
+x
1
1
y1  r1 sin 1
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
1

tan 1 
y1
x1
 y1 
1  tan  
 x1 
1
3

Components and Unit Vectors
Coordinate systems are useful in expressing vectors in their components
y
Ay
(+,+)
A
(-,+)
(Ax,Ay)

(-,-)
ur
A 

Monday, Sept. 27, 2010
(+,-)

ur
A cos
x
Ax
Ax 
ur
A cos 
Ay 
ur
A sin 
ur
A  Ax2  Ay2
 
2

ur
A sin 
ur 2
A cos2   sin 2 


PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
}
Components
} Magnitude

2
ur
 A
4
Unit Vectors
• Unit vectors are the ones that tells us the
directions of the components
• Dimensionless
• Magnitudes these vectors are exactly 1
• Unit vectors are usually expressed in i, j, k or
r r r
i, j, k
So a vector A can be
expressed as
Monday, Sept. 27, 2010
ur
ur
ur
r
r
r
r
A  Ax i Ay j  A cos i  A sin  j
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
5
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
r
r
r
 2.0  2.0 i  2.0  4.0 j  4.0i 2.0 j m
ur
C 

 
4.0  2.0
2


2
 16  4.0  20  4.5(m)

tan
1
Cy
 tan 1
Cx
2.0
 27o
4.0
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  d1  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
Monday, Sept. 27, 2010


 
ur
D

 


25  59  7.0  65(cm)
2
2
2
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
6
2D Displacement
r
ro  initial position
r
r  final position
Displacement: rr  rr  rr
o
Monday, Sept. 27, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
7
2D Average Velocity
Average velocity is the
displacement divided by
the elapsed time.
r r
r
r r  ro r
v

t  to
t
Monday, Sept. 27, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
8
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
Monday, Sept. 27, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
9
2D Average Acceleration
r
v
r
v
r
vo
Monday, Sept. 27, 2010
r r
r
r v  vo
v
a

t  to
t
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
10
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
• Average Velocity:
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Monday, Sept. 27, 2010
r r
r
r  r f  r i
r
r
r
rf
r

v
tf
t
r
r
r
lim
v  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, Fall 2010 Dr. Jaehoon
Yu
11
Kinematic Quantities in 1D and 2D
Quantities
Displacement
Average Velocity
Inst. Velocity
Average Acc.
Inst. Acc.
Monday, Sept. 27,
2010 is
What
1 Dimension
x  x f  xi
x x f  xi
vx 

t
t f  ti
x
vx  lim
t 0 t
vxf  vxi
vx
ax 

t
t f  ti
vx
ax  lim
t0 t
2 Dimension
r
r
r
f  ri
r
r
r
r
r 
r
r
r f  ri
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,
Fall 20101D
Dr. Jaehoon
the difference
between
and 2D quantities?
Yu
12
A Motion in 2 Dimension
This is a motion that could be viewed as two motions
combined into one. (superposition…)
Monday, Sept. 27, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
13
Motion in horizontal direction (x)
v

vx  vxo  ax t
x
v  v  2ax x
x  vxot  ax t
2
x
2
xo
Monday, Sept. 27, 2010
1
2
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
xo
 vx t
1
2
2
14
Motion in vertical direction (y)
v y  v yo  a y t
y
1
2
v
yo

 vy t
v  v  2a y y
2
y
2
yo
y  v yot  a y t
1
2
Monday, Sept. 27, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
2
15
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.
Monday, Sept. 27, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
16
Kinematic Equations in 2-Dim
x-component
y-component
vx  vxo  ax t
v y  v yo  a y t
x
1
2
v


v
t
xo
x
2
y
2
xo
x  vxot  axt
1
2
Monday, Sept. 27, 2010
v


v
t
yo
y
v  v  2a y y
v  v  2ax x
2
x
y
1
2
2
2
yo
y  v yot  a y t
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
1
2
17
2
Ex. A Moving Spacecraft
In the x direction, the spacecraft in zero-gravity zone 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, Sept. 27, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
18
How do we solve this problem?
1. Visualize the problem  Draw a picture!
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, Sept. 27, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
19
Ex. continued
In the x direction, the spacecraft in a zero gravity zone 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, Sept. 27, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
20
First, the motion in x-direciton…
x
ax
vx
vox
t
?
+24.0 m/s2
?
+22 m/s
7.0 s
x  vox t  ax t

1
2
2

 24 m s 7.0 s  740 m
 22 m s 7.0 s 
2
1
2
v x  vox  ax t


 22m s  24m s
Monday, Sept. 27, 2010
2
2
7.0 s 190m s
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
21
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
 12 m s 7.0 s  390 m
 14 m s 7.0 s 
2
1
2
v y  voy  a y t


 14m s  12m s
Monday, Sept. 27, 2010
2
2
7.0 s 98m s
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
22
The final velocity…
v
v y  98m s

vx  190 m s

  98m s
98 190 27
v  190 m s
  tan
1
Yes, you are right! Using
components and unit vectors!!
Monday, Sept. 27, 2010
2
2
o
 210 m s
A vector can be fully described when
the magnitude and the direction are
given. Any other way to describe it?


r
r
r
r
r
v  vx i  v y j  190 i  98 j m s
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
23
If we visualize the motion…
Monday, Sept. 27, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
24
What is a 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 s2
• ax  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  v
x0
xf
– Vertical motion under
constant
acceleration ( g )
Monday, Sept. 27, 2010


v yf  v y0  PHYS
a y t 1441-002,
 v y0 Fall
 2010
9.8
t
Dr. Jaehoon
25