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Chapter 4. Kinematics in Two DimensionsChapter 4. Kinematics in Two Dimensions
A car turning a corner, a basketball sailing toward the hoop, a planet orbiting the sun, and the diver in the photograph are examples of two-dimensional motion or, equivalently, motion in a
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
two-dimensional motion or, equivalently, motion in a plane.
Chapter Goal: To learn to
solve problems about motion
in a plane.
Topics:
• Acceleration
• Kinematics in Two Dimensions
• Projectile Motion
Chapter 4. Kinematics in Two DimensionsChapter 4. Kinematics in Two Dimensions
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
• Relative Motion
• Uniform Circular Motion
• Velocity and Acceleration in Uniform Circular Motion
• Nonuniform Circular Motion and Angular Acceleration
Kinematics in Two Dimensions
Average velocity:avg
rv
t
∆∆∆∆====
∆∆∆∆
rrrrrrrr
avg
rv
t
∆∆∆∆====
∆∆∆∆
rrrrrrrr
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3
Kinematics in Two Dimensions:
Instantaneous Velocity
0t
rv
t ∆ →∆ →∆ →∆ →
∆∆∆∆====
∆∆∆∆
rrrrrrrr
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4
Kinematics in Two Dimensions
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5
0t
rv
t ∆ →∆ →∆ →∆ →
∆∆∆∆====
∆∆∆∆
rrrrrrrrds
vdt
====
(motion along a
line)
Kinematics in Two Dimensions:
Instantaneous Acceleration
avg
va
t
∆∆∆∆====
∆∆∆∆
rrrrrrrr
Average acceleration:
Instantaneous acceleration:
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6
Instantaneous acceleration:
0t
va
t ∆ →∆ →∆ →∆ →
∆∆∆∆====
∆∆∆∆
rrrrrrrr
Kinematics in Two Dimensions: Instantaneous Acceleration
Motion along a line: acceleration results in change of speed (the magnitude
of velocity)
Motion in a plane: acceleration can change the speed (the magnitude of
velocity) and the direction of velocity.
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7
0t
va
t ∆ →∆ →∆ →∆ →
∆∆∆∆====
∆∆∆∆
rrrrrrrr
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8
A Projectile is an object that moves in two dimensions (in a
plane) under the influence of only the gravitational force.
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9
a g====r rr rr rr r
0x
a ====
9.8m
a g= − = −= − = −= − = −= − = −
- motion along x-axis – with zero acceleration –
constant velocity
constantx
v ====
- motion along y-axis – free fall motion
A Projectile is an object that moves in two dimensions with
free fall acceleration
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10
29.8
y
ma g
s= − = −= − = −= − = −= − = −
- motion along y-axis – free fall motion
with constant acceleration
29.8
y
ma g
s= − = −= − = −= − = −= − = −
,constant cos
x i x iv v v θθθθ= = == = == = == = =
Example: Find the distance AB
,cos
i x ix v t v t θθθθ= == == == =
,y i yv v gt= −= −= −= −
2
,2
AB
final i y AB
ty v t g= −= −= −= −
Point A - 0i A
y y= == == == =
Point B - 0final B
y y= == == == = then
2
,2
AB
i y AB
tv t g====
,sin
i y iv v θθθθ====
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11
2
,2 2 sini y i
AB
v vt
g g
θθθθ= == == == =
A B
then
2 2
cos
2sin cos sin 2
AB i AB
i i
x v t
v v
g g
θθθθ
θ θ θθ θ θθ θ θθ θ θ
= == == == =
= == == == =
2
sin 2i
AB
vx
gθθθθ====
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12
Example of Projectile Motion
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EXAMPLE 4.4 Don’t try this at home!
QUESTION:
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EXAMPLE 4.4 Don’t try this at home!
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EXAMPLE 4.4 Don’t try this at home!
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EXAMPLE 4.4 Don’t try this at home!
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EXAMPLE 4.4 Don’t try this at home!
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EXAMPLE 4.4 Don’t try this at home!
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Problem-Solving Strategy: Projectile Motion Problems
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Relative Motion
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Relative Motion
If we know an object’s velocity measured in one reference
frame, S, we can transform it into the velocity that
would be measured by an experimenter in a different
reference frame, S´, using the Galilean transformation of
velocity.
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Or, in terms of components,
EXAMPLE 4.8 A speeding bullet
QUESTION:
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EXAMPLE 4.8 A speeding bullet
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EXAMPLE 4.8 A speeding bullet
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Motion in a Circle
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26
Kinematics in Two Dimensions: Uniform Circular Motion:
Motion with Constant Speed
The speed (magnitude of velocity) is the same
Period:2 r
Tv
ππππ====
The position of the particle is
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27
Arc length:
where angle is in
RADIANS
s rθθθθ====
θθθθ
The position of the particle is
characterized by angle (or by arc
length)
00360
1 57.32
radππππ
= == == == =
Example:
What is the arc length for if
0 0
0
245 45 0.78
360 4rad rad rad
π ππ ππ ππ π= = == = == = == = =
0 0 045 ,90 ,10θθθθ ==== 1r m====
0 0
0
290 90 1.57
360 2rad rad rad
π ππ ππ ππ π= = == = == = == = =
0 0
0
210 10 0.174
360 18rad rad rad
π ππ ππ ππ π= = == = == = == = =
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28
Arc length: s rθθθθ====
0360 18
450.78
4
rs m
ππππ= == == == =
901.57
2
rs m
ππππ= == == == =
100.174
18
rs m
ππππ= == == == =
Kinematics in Two Dimensions: Uniform Circular Motion:
Motion with Constant Speed
The speed (magnitude of velocity) is the same
Then
s vt====
tθ ωθ ωθ ωθ ω====v
rωωωω ====
2 rT
v
ππππ====
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29
Then
2
T
ππππωωωω ====
Uniform Circular Motion
Then
s vt====
tθ ωθ ωθ ωθ ω====
2
T
ππππωωωω ====
v
rωωωω ====
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30
Then
cos cos( )x r r tθ ωθ ωθ ωθ ω= == == == =
sin sin( )y r r tθ ωθ ωθ ωθ ω= == == == =
2 2 2 2 2 2 2sin ( ) cos ( )x y r t r t rω ωω ωω ωω ω+ = + =+ = + =+ = + =+ = + =
tθ ωθ ωθ ωθ ω====
0ωωωω >>>>0ωωωω <<<<
0ωωωω ====
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31
Positive angle
Uniform Circular Motion
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EXAMPLE 4.13 At the roulette wheel
QUESTIONS:
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EXAMPLE 4.13 At the roulette wheel
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EXAMPLE 4.13 At the roulette wheel
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EXAMPLE 4.13 At the roulette wheel
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0t
va
t ∆ →∆ →∆ →∆ →
∆∆∆∆====
∆∆∆∆
rrrrrrrr
Uniform Circular Motion: Acceleration
0t
rv
t ∆ →∆ →∆ →∆ →
∆∆∆∆====
∆∆∆∆
rrrrrrrr
1 1r v t∆ = ∆∆ = ∆∆ = ∆∆ = ∆r rr rr rr r
2 2r v t∆ = ∆∆ = ∆∆ = ∆∆ = ∆r rr rr rr r
2 1 2 1v v r r
a− ∆ − ∆− ∆ − ∆− ∆ − ∆− ∆ − ∆
= == == == =
r r r rr r r rr r r rr r r rrrrr
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37
2 1 2 1
2( )
v v r ra
t t
− ∆ − ∆− ∆ − ∆− ∆ − ∆− ∆ − ∆= == == == =
∆ ∆∆ ∆∆ ∆∆ ∆
rrrr
Direction of acceleration is the
same as direction of vector
(toward the center)2 1
CB r r→→→→
= ∆ − ∆= ∆ − ∆= ∆ − ∆= ∆ − ∆r rr rr rr r
The magnitude of acceleration:2
2 2 2 2 2
( 2 )
( ) ( ) ( ) ( ) ( )
CB r r r v t t va v
t t t t t r
φ π α θ ωφ π α θ ωφ π α θ ωφ π α θ ωωωωω
∆ ∆ − ∆ ∆ ∆∆ ∆ − ∆ ∆ ∆∆ ∆ − ∆ ∆ ∆∆ ∆ − ∆ ∆ ∆= = = = = = == = = = = = == = = = = = == = = = = = =
∆ ∆ ∆ ∆ ∆∆ ∆ ∆ ∆ ∆∆ ∆ ∆ ∆ ∆∆ ∆ ∆ ∆ ∆
Uniform Circular Motion: Acceleration
The magnitude of acceleration:
22v
a r vω ωω ωω ωω ω= = == = == = == = =
v
rωωωω ====
centripetal
acceleration
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38
2a r vr
ω ωω ωω ωω ω= = == = == = == = =
EXAMPLE 4.14 The acceleration of a Ferris wheel
QUESTION:
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EXAMPLE 4.14 The acceleration of a Ferris wheel
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EXAMPLE 4.14 The acceleration of a Ferris
wheel
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Chapter 4. Summary SlidesChapter 4. Summary Slides
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Chapter 4. Summary SlidesChapter 4. Summary Slides
General Principles
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General Principles
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Important Concepts
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Important Concepts
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Applications
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Applications
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Applications
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Applications
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Chapter 4. QuestionsChapter 4. Questions
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Chapter 4. QuestionsChapter 4. Questions A. slow down and curve downward.
B. slow down and curve upward.
This acceleration will cause the particle to
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B. slow down and curve upward.
C. speed up and curve downward.
D. speed up and curve upward.
E. move to the right and down.
A. slow down and curve downward.
B. slow down and curve upward.
This acceleration will cause the particle to
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B. slow down and curve upward.
C. speed up and curve downward.
D. speed up and curve upward.
E. move to the right and down.
A. 0–1 s
B. 1–2 s
During which time interval is the particle described by these position graphs at rest?
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C. 2–3 s
D. 3–4 s
A. 0–1 s
B. 1–2 s
During which time interval is the particle described by these position graphs at rest?
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C. 2–3 s
D. 3–4 s
A. less than 2 m from the base.
B. 2 m from the base.
A 50 g ball rolls off a table and lands 2 m from the base of the table. A 100 g ball rolls off the same table with the same speed. It lands at a distance
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B. 2 m from the base.
C. greater than 2 m from the base.
A. less than 2 m from the base.
B. 2 m from the base.
A 50 g ball rolls off a table and lands 2 m from the base of the table. A 100 g ball rolls off the same table with the same speed. It lands at a distance
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B. 2 m from the base.
C. greater than 2 m from the base.
A. right and up, more than 100 m/s.
B. right and up, less than 100 m/s.
A plane traveling horizontally to the right at 100 m/s flies past a helicopter that is going straight up at 20 m/s. From the helicopter’s perspective, the plane’s direction and speed are
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C. right and down, more than 100 m/s.
D. right and down, less than 100 m/s.
E. right and down, 100 m/s.
A. right and up, more than 100 m/s.
B. right and up, less than 100 m/s.
A plane traveling horizontally to the right at 100 m/s flies past a helicopter that is going straight up at 20 m/s. From the helicopter’s perspective, the plane’s direction and speed are
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C. right and down, more than 100 m/s.
D. right and down, less than 100 m/s.
E. right and down, 100 m/s.
A particle moves cw around a circle at constant speed for 2.0 s. It then reverses direction and moves ccw at half the original speed until it has traveled through the same angle. Which is the particle’s angle-versus-time graph?
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A particle moves cw around a circle at constant speed for 2.0 s. It then reverses direction and moves ccw at half the original speed until it has traveled through the same angle. Which is the particle’s angle-versus-time graph?
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Rank in order, from largest to smallest, the centripetal accelerations (ar)a to (ar)e
of particles a to e.
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A. (ar)b > (a
r)e > (a
r)a > (a
r)d > (a
r)c
B. (ar)b > (a
r)e > (a
r)a = (a
r)c > (a
r)d
C. (ar)b = (a
r)e > (a
r)a = (a
r)c > (a
r)d
D. (ar)b > (a
r)a = (a
r)c = (a
r)e > (a
r)d
E. (ar)b > (a
r)a = (a
r)a > (a
r)e > (a
r)d
Rank in order, from largest to smallest, the centripetal accelerations (ar)a to (ar)e
of particles a to e.
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A. (ar)b > (a
r)e > (a
r)a > (a
r)d > (a
r)c
B. (ar)b > (ar)e > (ar)a = (ar)c > (ar)d
C. (ar)b = (a
r)e > (a
r)a = (a
r)c > (a
r)d
D. (ar)b > (a
r)a = (a
r)c = (a
r)e > (a
r)d
E. (ar)b > (a
r)a = (a
r)a > (a
r)e > (a
r)d
The fan blade is slowing down. What are the signs of ω and α?
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Α. ω is positive and α is positive.B. ω is negative and α is positive.C. ω is positive and α is negative.D. ω is negative and α is negative.