Transcript Wednesday, February 6, 2008

PHYS 1441 – Section 002
Lecture #7
Wednesday, Feb. 6, 2008
Dr. Jaehoon Yu
•
Motion in Two Dimension
– Motion under constant acceleration
– Vector recap
– Projectile Motion
– Maximum ranges and heights
Wednesday, Feb. 6, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
1
Announcements
• 1st term exam on Wednesday, Feb. 20
– Time: 1 – 2:20pm
– Place: SH103
– Covers: Appendices, CH 1 – what we learn till next
Wednesday, Feb. 13
– Class on Monday, Feb. 18: Jason will be here to go over
the any problems you would like to review
• Colloquium
– Today at 4pm in SH101, following the refreshment at
3:30pm in SH108
– Please be sure to sign in the sign-in sheet
Wednesday, Feb. 6, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
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Wednesday, Feb. 6, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
3
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.
6, 2008is
What
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 between
1441-002, Spring
the difference
1D2008
and 2D quantities?
Dr. Jaehoon Yu
r r
r f  ri
4
Kinematic Equations
v  vo  at
x
1
2
 vo  v  t
v  v  2ax
2
2
o
x  vot  at
1
2
Wednesday, Feb. 6, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
2
5
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
6
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
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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
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8
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
9
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
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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. 6, 2008
x1
+x
y1  r1 sin q1
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
y1
tan q1 
x1
 y1 

x
 1
q1  tan 1 
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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
2.50 5

3.50 7
tan q s 
1
5
q s  tan    35.5
7
o
o
o
q  180  q s  180  35.5  216
Wednesday, Feb. 6, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
o
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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
Wednesday, Feb. 6, 2008
C
PHYS 1441-002, Spring 2008
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
Wednesday, Feb. 6, 2008
B 2A
A
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
B=2A
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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
Bsin60oN
B 60o Bcos60o
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. 6, 2008
 A  B cosq 
A  B cos 60
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
Find other
ways to
solve this
problem…
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