PHYS 1441 – Section 002
Lecture #3
Wednesday, Jan. 23, 2008
Dr. Jaehoon Yu
•
•
•
•
Dimensional Analysis
Trigonometry reminder
Coordinate system, vector and scalars
One Dimensional Motion: Average Velocity;
Acceleration; Motion under constant acceleration; Free
Fall
Wednesday, Jan. 23, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
1
Dimension and Dimensional Analysis
• An extremely useful concept in solving physical problems
• Good to write physical laws in mathematical expressions
• No matter what units are used the base quantities are the
same
– Length (distance) is length whether meter or inch is used to
express the size: Usually denoted as [L]
– The same is true for Mass ([M])and Time ([T])
– One can say “Dimension of Length, Mass or Time”
– Dimensions are used as algebraic quantities: Can perform two
algebraic operations; multiplication or division
Wednesday, Jan. 23, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
2
Dimension and Dimensional Analysis
• One can use dimensions only to check the
validity of one’s expression: Dimensional
analysis
– Eg: Speed [v] = [L]/[T]=[L][T-1]
•Distance (L) traveled by a car running at the speed
V in time T
•L = V*T = [L/T]*[T]=[L]
• More general expression of dimensional analysis
is using exponents: eg. [v]=[LnTm] =[L]{T-1]
where n = 1 and m = -1
Wednesday, Jan. 23, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
3
Examples
• Show that the expression [v] = [at] is dimensionally correct
• Speed: [v] =L/T
• Acceleration: [a] =L/T2
• Thus, [at] = (L/T2)xT=LT(-2+1) =LT-1 =L/T= [v]
•Suppose the acceleration a of a circularly moving particle with
speed v and radius r is proportional to rn and vm. What are n
andm?
a
r
v
a  kr n v m
Dimensionless
constant
Length
Wednesday, Jan. 23, 2008
Speed
m
n L
1 2
LT
  L    LnmT  m
T 
m  2  m  2
n  m  n  2  1  n  1
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
2
v
a  kr 1v 2 
r
4
• Definitions of sinq, cosq and tanq
sin q 
cosq 
Length of the opposite side to q

Length of the hypotenuse of the right triangle
Length of the adjacent side to q

Length of the hypotenuse of the right triangle
ha

h
tan q 
ho
h
q
90o
ho=length of the side
opposite to the angle q
Trigonometry Reminders
ha=length of the side adjacent to the angle q
ho
Length of the opposite side to q

Length of the adjacent side to q
ha
sin q

tan q 
cos q
Wednesday, Jan. 23, 2008
ho
h
ha
h
ho

ha
Pythagorian theorem: For right triangles
h 2  ho2  ha2
h  ho2  ha2
Do Ex. 3 and 4 yourselves…
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
5
Example for estimates using trig..
Estimate the radius of the Earth using triangulation as shown in the
picture when d=4.4km and h=1.5m.
Pythagorian theorem
d=4.4km
R  h 2  d 2  R 2
R 2  2hR  h 2  d 2  R 2
R
Solving for R
d h
R
2h
2
2
4400m   1.5m 


2
2
2 1.5m
 6500km
Wednesday, Jan. 23, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
6
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, Jan. 23, 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 
7
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, Jan. 23, 2008
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
o
8
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, Jan. 23, 2008
PHYS 1441-002, Spring 2008
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, Jan. 23, 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
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, Jan. 23, 2008
B 2A
A
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
B=2A
11
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, Jan. 23, 2008
 A  B cosq 
A  B cos 60
PHYS 1441-002, Spring 2008
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
A
(-,+)
(Ax,Ay)
q
(-,-)
u
r
A 
(+,-)


Wednesday, Jan. 23, 2008
Ax
ur
A cos q
 
2
Ax 
ur
A cos q
Ay 
ur
A sin q
x
A  Ax  Ay
2
ur
A sin q
u
r 2
A
cos 2 q  sin 2 q

PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu


2
}
Components
} Magnitude
2
u
r
 A
13
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, Jan. 23, 2008
r
ur
ur
r ur
r
r
A  Ax i Ay j  A cos q i  A sin q j
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
14
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, Jan. 23, 2008
ur
D



 25  59   7.0  65(cm)
2
2
PHYS 1441-002, Spring 2008
Dr. Jaehoon Yu
2
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