PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
1Wednesday, June 8, 2011
PHYS 1443 – Section 001Lecture #3
Wednesday, June 8, 2011Dr. Jaehoon Yu
• One Dimensional Motion: Free Fall• Coordinate System • Vectors and Scalars and their operations• Motion in Two Dimensions: Motion under constant
acceleration; Projectile Motion; Maximum ranges and heights
• Newton’s Laws of Motion
Today’s homework is homework #2, due 10pm, Monday, June 13!!
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Announcements• Homework registration and submissions
– 27/28 registered but only 9 completed the submission– The due for homework #1, the freebee, is 10pm tonight!
• Quiz #2 coming Tuesday, June 14– Covers: CH 1.1 – what we finish on Monday, June 13
• Reading assignment– CH3.9
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Reminder: Special Problems for Extra Credit• Derive the quadratic equation for Bx2-Cx+A=0
5 points• Derive the kinematic equation
from first principles and the known kinematic equations 10 points
• You must show your work in detail to obtain full credit
• Due at the start of the class, Thursday, June 9
ifxf xxavv xxi 222
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Special Project for Extra Credit• Show that the trajectory of a projectile motion
is a parabola!!–20 points–Due: next Tuesday, June 14–You MUST show full details of your OWN
computations to obtain any credit• Much beyond what was covered in page 40 of
this lecture note!!
PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Free Fall• Free fall motion is a motion under the influence of the
gravitational pull (gravity) only; Which direction is a freely falling object moving?– A motion under constant acceleration – All kinematic formula we learned can be used to solve for falling
motions. • Gravitational acceleration is inversely proportional to the
distance between the object and the center of the earth• The magnitude of the gravitational acceleration is g=9.80m/s2
on the surface of the earth, most of the time.• The direction of gravitational acceleration is ALWAYS toward
the center of the earth, which we normally call (-y); where up and down direction are indicated as the variable “y”
• Thus the correct denotation of gravitational acceleration on the surface of the earth is g=-9.80m/s2 when +y points upward
Yes, down to the center of the earth!!
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Example for Using 1D Kinematic Equations on a Falling object (similar to Ex. 2.16)
Stone was thrown straight upward at t=0 with +20.0m/s initial velocity on the roof of a 50.0m high building,
g=-9.80m/s2
fv
fy
(a) Find the time the stone reaches at the maximum height.
What is the acceleration in this motion?
What is so special about the maximum height? V=0
t
(b) Find the maximum height.
vyi ayt 20.0 9.80t 0.00 /m s Solve for t20.09.80
2.04s
2150.0 20 2.04 ( 9.80) (2.04)2
50.0 20.4 70.4( )m yi vyit 1
2ayt2
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Example of a Falling Object cnt’d
t
vyf
yf Position
vyf Velocity
(c) Find the time the stone reaches back to its original height.
(d) Find the velocity of the stone when it reaches its original height.
(e) Find the velocity and position of the stone at t=5.00s.
2.04 2 4.08s
vyi ayt20.0 ( 9.80) 4.08 20.0( / )m s
vyi ayt20.0 ( 9.80) 5.00 29.0( / )m s
2150.0 20.0 5.00 ( 9.80) (5.00)2
yi vyit 1
2ayt2
27.5( )m
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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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, θ (r,θ)• Vectors become a lot easier to express and compute
O (0,0)
(x1,y1)r1
θ1
1x
How are Cartesian and Polar coordinates related?
1r
y1
x1
+x
+y
1y tanθ1
x1
2 y12
1y r1
r1
= (r1,θ1)
1x
θ1 tan−1 y1
x1
⎛⎝⎜
⎞⎠⎟
cosθ1
sinθ1
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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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.
y
x
(-3.50,-2.50)mr
θ
θs
r
θ 180 +θ s
tanθs
sθ
∴θ
−3.50
2
−2.50 2
18.5 4.30m
x 2 y2
−2.50−3.50
57
tan−1 5
7⎛⎝⎜
⎞⎠⎟ 35.5o
180 θs 180o 35.5o 216o
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Vector and ScalarVector quantities have both magnitudes (sizes) and directions
Scalar quantities have magnitudes onlyCan be completely specified with a value and its unit
Force, gravitational acceleration, momentumNormally denoted in BOLD letters, F, or a letter with arrow on top F
urTheir sizes or magnitudes are denoted with normal letters, F, or absolute values:
Fur
or F
Energy, heat, mass, timeNormally denoted in normal letters, E
Both have units!!!
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Properties of Vectors• Two vectors are the same if their and the
are the same, no matter where they are on a coordinate system!!
x
y
AB
E
D
C
F
Which ones are the same vectors?
A=B=E=D
Why aren’t the others?
C: The same magnitude but opposite direction: C=-A:A negative vector
F: The same direction but different magnitude
sizes directions
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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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
BA
B=A
BA+B
• Subtraction: – The same as adding a negative vector:A - B = A + (-B)
A-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
A B=2A
AB 2
A+BA+B
A-B
OR
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Example for Vector AdditionA 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.
N
E
60o
θr
r
20A
B
60cos
60sintan 1
BA
B
θ
Do this using components!!
θθθ cos2sincos 2222 ABBA
θcos222 ABBA
60cos0.350.2020.350.20 22
)(2.482325 km
60cos0.350.2060sin0.35tan 1
N W wrt to9.385.373.30tan 1
Bcos60o
Bsin60o A Bcosθ
2 Bsinθ
2
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Components and Unit VectorsCoordinate systems are useful in expressing vectors in their components
Aur
Ax2 Ay
2
(Ax,Ay)A
θ
Ay
Ax
x
y
xA
Aur
}Components
(+,+)
(-,+)
(-,-) (+,-)
yA
} Magnitude
Aur
cosθ 2
Aursinθ
2
A
ur2
cos2θ sin2θ Aur
Aur
cosθ
Aur
sinθ
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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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
ir, jr, kr
AurSo a vector A can be
expressed as yA
A
urcosθ
Aur
sinθ xA ir
ir
jr
jr
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Examples of Vector OperationsFind the resultant vector which is the sum of A=(2.0i+2.0j) and B =(2.0i-4.0j)
Cur
Find the resultant displacement of three consecutive displacements: d1=(15i+30j +12k)cm, d2=(23i+14j -5.0k)cm, and d3=(-13i+15j)cm
Dur
Cur
Dur
θ
2.0i
r 2.0 j
r 2.0i
r−4.0 j
r
2.0 2.0 ir
tan−1
Cy
Cx
tan−1 −2.0
4.0−27o
15 23−13 ir
25ir d
ur1 d
ur2 d
ur3
15i
r 30 j
r12k
r 23i
r14 j
r−5.0k
r −13i
r15j
r
25
2 59
2 7.0
2 65(cm)
4.0
2 −2.0
2
16 4.0 20 4.5m
2.0 −4.0 jr
4.0ir
−2.0 jrm
Aur Bur
30 14 15 jr
12−5.0 kr
59 jr 7.0k
rcm
Magnitude
Wednesday, June 8, 2011
Displacement, Velocity, and Acceleration in 2-dim
• Displacement: r r
• Average Velocity: v r
• Instantaneous Velocity:
v r
• Average Acceleration
a r
• Instantaneous Acceleration: a
r
How is each of these quantities defined in 1-D?
f ir rr r
rt
rf i
f i
r rt t
r r
0lim
t
rt
rdrdt
r
vt
rf i
f i
v vt t
r r
0lim
t
vt
rdvdt
rd drdt dt
r2
2
d rdt
r
PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Wednesday, June 8, 2011
initial positiono rr
final positionrr
Displacement:
2D Displacement
rr rr−r
ro
PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Wednesday, June 8, 2011
rv
Average velocity is the displacement divided by the elapsed time.
rr−r
ro
t −to
rr
t
2D Average Velocity
PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Wednesday, June 8, 2011
The instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time.
rv
limt→ 0
rr
t
PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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Wednesday, June 8, 2011
ra
rvo
rv
rv
2D Average Acceleration
rv−r
vo
t −to rv
tPHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu21
Wednesday, June 8, 2011
Kinematic Quantities in 1D and 2DQuantities 1 Dimension 2 Dimension
Displacement
Average Velocity
Inst. Velocity
Average Acc.
Inst. Acc.
x rr
vx xt
xf −xitf −ti
vr≡rr
trr
f −rri
tf −ti
vr≡lim
t→ 0
rr
tdrr
dt vx ≡lim
t→ 0
xt
dxdt
ar≡vr
tvr
f −vri
tf −tiax ≡
vxt
vxf −vxitf −ti
ar≡lim
t→ 0
vr
tdvr
dt ax ≡lim
t→ 0
vxt
dvxdt
What is the difference between 1D and 2D quantities?
x f −xi r
rf −r
ri
PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu
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