PHYS 1443 – Section 004
Lecture #5
Tuesday, Sept. 9, 2014
Dr. Jaehoon Yu
•
Motion in two dimensions
•
•
•
•
•
Tuesday, Sept. 9, 2014
Coordinate system
Vector and scalars, their operations
Motion under constant acceleration
Projectile Motion
Maximum range and height
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
1
• Quiz #2
Announcements
– Beginning of the class this Thursday, Sept. 11
– Covers CH1.1 through what we learn today (CH4 – 4) plus math
refresher
– Mixture of multiple choice and free response problems
– Bring your calculator but DO NOT input formula into it!
• Your phones or portable computers are NOT allowed as a replacement!
– You can prepare a one 8.5x11.5 sheet (front and back) of
handwritten formulae and values of constants for the exam
• None of the parts of the solutions of any problems
• No derived formulae, derivations of equations or word definitions!
• No additional formulae or values of constants will be provided!
• Colloquium at 4pm tomorrow, Wednesday, in SH101, UTA
faculty expo II
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
2
Reminder: Special Project #2 for Extra Credit
• Show that the trajectory of a projectile motion
is a parabola!!
– 20 points
– Due: Thursday, Sept. 11
– You MUST show full details of your OWN
computations to obtain any credit
• Beyond what was covered in this lecture note
and in the book!
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
4
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
x1  r1cos q1 r  x12 + y12
1
1
O (0,0)
Tuesday, Sept. 9, 2014
x1
+x
y1  r1 sin q1
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
(
tan q1 =
y1
x1
æ y1 ö
q1 = tan ç ÷
è x1 ø
-1
5
)
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
=

s
(-3.50,-2.50)m
2
2
= 18.5 = 4.30(m)
x
r
(
)
(( -3.50) + ( -2.50) )
x2 + y2
q = 180 + q s
tanq s =
-2.50 5
=
-3.50 7
æ 5ö
-1
 s  tan ç ÷ =
è 7ø
\q = 180 + q s =
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
6
Vector and Scalar
Vector quantities have both magnitudes (sizes)
and directions Force, gravitational acceleration, momentum
Normally denoted in BOLD letters, F, or a letter with arrow on top
Their sizes or magnitudes are denoted with normal letters, F, or
absolute values:
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!!!
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
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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
Tuesday, Sept. 9, 2014
C
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
C: The same magnitude
but opposite direction:
C=-A:A negative vector
F: The same direction
but different magnitude
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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
Tuesday, Sept. 9, 2014
B 2A
A
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
B=2A
9
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.
Bsin60o
N
B 60o Bcos60o
r

( A + Bcosq ) + ( Bsinq )
r
20
A
2



A2  B 2 cos 2   sin 2   2 AB cos 

A2  B 2  2 AB cos

20.02  35.02  2  20.0  35.0 cos 60
 2325  48.2(km)
E
  tan
 tan
1
 tan 1
Tuesday, Sept. 9, 2014
2
B sin 60
1
A  B cos 60
Do this using
35.0 sin 60
components!!
20.0  35.0 cos 60
30.3
 38.9 to W wrt N
37.5
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
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Components and Unit Vectors
Coordinate systems are useful in expressing vectors in their components
y
Ay
(-,+)
(+,+)
A
(Ax,Ay)
Ay 

(-,-)
Tuesday, Sept. 9, 2014
(+,-)
Ax 
Ax
x
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
A = Ax2 + Ay2
}
Components
} Magnitude
11
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
So a vector A can be
expressed as
Tuesday, Sept. 9, 2014
Ax Ay
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
12
Examples of Vector Operations
Find the resultant vector which is the sum of A=(2.0i+2.0j) and B =(2.0i-4.0j)
( 4.0) + ( -2.0)
2
2
= 16 + 4.0 = 20 = 4.5(m)
q=
tan
-1
Cy
Cx
=
Find the resultant displacement of three consecutive displacements:
d1=(15i+30j +12k)cm, d2=(23i+14j -5.0k)cm, and d3=(-13i+15j)cm
Magnitude
Tuesday, Sept. 9, 2014
( 25) + (59) + (7.0)
2
2
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
2
= 65(cm)
13
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
• Average Velocity:
How is each of
these quantities
defined in 1-D?
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
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2D Displacement
Displacement:
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
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2D Average Velocity
Average velocity is the
displacement divided by
the elapsed time.
t - to
Tuesday, Sept. 9, 2014
=
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
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The instantaneous velocity indicates how fast the car
moves and the direction of motion at each instant of time.
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
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2D Average Acceleration
t - to
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
=
18
Kinematic Quantities in 1D and 2D
Quantities
1 Dimension
Displacement
Dx = x f - xi
Average Velocity
Inst. Velocity
Average Acc.
Inst. Acc.
2 Dimension
Dx x f - xi
vx =
=
Dt
t f - ti
Dx dx
vx º lim
=
Dt ®0 Dt
dt
Dvx vxf - vxi
ax º
=
Dt
t f - ti
ax º lim
Dt®0
Dvx
Dt
=
dvx
dt
What is the difference
between 1D and 2D quantities?
PHYS 1443-004, Fall 2014
Tuesday, Sept. 9, 2014
Dr. Jaehoon Yu
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A Motion in 2 Dimension
This is a motion that could be viewed as two motions
combined into one. (superposition…)
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
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Motion in horizontal direction (x)
(v
)
vx = vxo + axt
x=
v = v + 2ax x
x = vxot + ax t
2
x
2
xo
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
1
2
+
v
t
xo
x
1
2
2
21
Motion in vertical direction (y)
v y = v yo + a y t
y=
1
2
(v
)
+
v
t
yo
y
v = v + 2a y y
2
y
2
yo
y = v yot + a y t
1
2
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
2
22
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.
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
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Kinematic Equations in 2-Dim
x-component
y-component
vx = vxo + axt
v y = v yo + a y t
x=
1
2
(v
)
+
v
t
xo
x
2
y
2
xo
Dx = vxot + ax t
1
2
Tuesday, Sept. 9, 2014
(v
)
+
v
t
yo
y
v = v + 2a y y
v = v + 2ax x
2
x
y=
1
2
2
2
yo
Dy = v yot + ayt
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
1
2
24
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.
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
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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 in time,
remember that the final velocity of one time segment is
the initial velocity for the next.
6. Keep in mind that there may be two possible answers to
a kinematics problem.
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
26
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
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
27
First, the motion in x-direciton…
x
ax
vx
vox
t
?
+24.0 m/s2
?
+22 m/s
7.0 s
Dx = vox t + ax t
(
1
2
)(
2
) ( 24m s )(7.0 s)
= 22m s 7.0 s +
2
1
2
vx = vox + ax t
(
) (
= 22m s + 24m s
Tuesday, Sept. 9, 2014
2
2
= +740 m
)(7.0 s) = +190m s
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
28
Now, the motion in y-direction…
y
ay
vy
voy
t
?
+12.0 m/s2
?
+14 m/s
7.0 s
Dy = voy t + a y t
(
)(
1
2
2
) (
= 14m s 7.0 s + 12m s
1
2
v y  voy + a y t
(
) (
= 14m s + 12m s
Tuesday, Sept. 9, 2014
2
2
)(7.0 s)
2
= +390 m
)(7.0 s) = +98m s
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
29
The final velocity…
v
v y = 98m s
q
vx = 190 m s
(
) + ( 98m s)
(98 190) = 27
v  190 m s
q = tan
-1
2
= 210 m s
A vector can be fully described when
the magnitude and the direction are
given. Any other way to describe it?
Yes, you are right! Using
components and unit vectors!!
Tuesday, Sept. 9, 2014
2
m s
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
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If we visualize the motion…
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
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2-dim Motion Under Constant Acceleration
• Position vectors in x-y plane:
• Velocity vectors in x-y plane:
Velocity vectors in terms of the acceleration vector
X-comp
v xf  vxi  axt
Tuesday, Sept. 9, 2014
Y-comp
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
v yf  v yi  a y t
32
2-dim Motion Under Constant Acceleration
• How are the 2D position vectors written in
acceleration vectors?
Position vector
components
1 2
x f  xi  vxi t  ax t
2
1 2
y f  yi  v yi t  a y t
2
Putting them
together in a
vector form
Regrouping
the above
Tuesday, Sept. 9, 2014
r
 ri
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
2D problems can be
interpreted as two 1D
problems in x and y
33
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 the 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 
Velocity vector
Compute the velocity and the speed of the particle at t=5.0 s.

Tuesday, Sept. 9, 2014
 40
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
2
  15  43m / s
2
34
Example for 2-D Kinematic Eq. Cnt’d
Angle of the
Velocity vector
 vy
  tan 
 vx
1

1  15 
1  3 
o

tan

tan


21



 
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?
Tuesday, Sept. 9, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
35