Chapter 4. Kinematics in Two Dimensions
Chapter 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
plane.
Chapter Goal: To learn to
solve problems about motion
in a plane.
Topics:
• Acceleration
• Kinematics in Two Dimensions
• Projectile Motion
• Relative Motion
• Uniform Circular Motion
• Velocity and Acceleration in Uniform
Circular Motion
• Nonuniform Circular Motion and Angular
Acceleration
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Kinematics in Two Dimensions
Average velocity:
Kinematics in Two Dimensions:
Instantaneous Velocity
r
r
∆r
vavg =
∆t
r
r
∆r
vavg =
∆t
r
r ∆r
v=
∆t
3
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∆t → 0
4
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Kinematics in Two Dimensions
Kinematics in Two Dimensions:
Instantaneous Acceleration
Average acceleration:
r
r
∆v
aavg =
∆t
Instantaneous acceleration:
ds
v=
dt
r
r ∆r
v=
∆t
r
r ∆v
a=
∆t
∆t → 0
∆t → 0
(motion along a
line)
5
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6
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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.
r
r ∆v
a=
∆t
∆t → 0
7
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8
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A Projectile is an object that moves in two dimensions (in a
A Projectile is an object that moves in two dimensions with
r r
free fall acceleration a = g
plane) under the influence of only the gravitational force.
a x = 0 - motion along x-axis – with zero acceleration –
constant velocity
v x = constant
a y = − g = −9.8
m
s2
- motion along y-axis – free fall motion
with constant acceleration
9
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10
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Example: Find the distance AB
x = v i , x t = v i t cos θ
v x = constant = v i , x = v i cos θ
a y = − g = −9.8
m
s2
v y = v i , y − gt
t2
AB
then y final = vi , y t AB − g 2
t AB =
x AB
A
B
v i2
sin 2θ
g
vi , y = vi sin θ
Point A - yi = y A = 0
Point B - y final = yB = 0
x AB =
2v i , y
g
=
v i , y t AB = g
2
t AB
2
2v i sin θ
g
then
= v i t AB cos θ =
=
2v i2
v2
sin θ cos θ = i sin 2θ
g
g
11
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12
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Example of Projectile Motion
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
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.
Or, in terms of components,
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EXAMPLE 4.8 A speeding bullet
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EXAMPLE 4.8 A speeding bullet
QUESTION:
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EXAMPLE 4.8 A speeding bullet
Motion in a Circle
26
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Kinematics in Two Dimensions: Uniform Circular Motion:
Motion with Constant Speed
Example:
What is the arc length for
The speed (magnitude of velocity) is the same
θ = 450 , 900 ,100
r = 1m
if
2π
π
rad = rad = 0.78 rad
0
360
4
π
0
0 2π
rad = rad = 1.57 rad
90 = 90
3600
2
450 = 450
Period:
T=
2π r
v
100 = 100
The position of the particle is
characterized by angle (or by arc
length)
Arc length:
Arc length:
s = rθ
s45 =
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3600
= 57.30
2π
πr
= 0.78m
4
πr
s90 =
= 1.57 m
2
where angle θ is in
RADIANS
1 rad =
2π
π
rad =
rad = 0.174 rad
3600
18
s10 =
πr
18
s = rθ
= 0.174m
27
28
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Kinematics in Two Dimensions: Uniform Circular Motion:
Motion with Constant Speed
The speed (magnitude of velocity) is the same
Uniform Circular Motion
v
r
T=
ω=
θ = ωt
2π r
v
Then
Then
ω=
x = r cos θ = r cos(ω t )
2π
T
y = r sin θ = r sin(ω t )
x 2 + y 2 = r 2 sin 2 (ω t ) + r 2 cos 2 (ω t ) = r 2
29
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30
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θ = ωt
Uniform Circular Motion
ω>0
ω<0
ω=0
Positive angle
31
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2π
T
ω=
s = vt
θ = ωt
ω=
s = vt
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v
r
EXAMPLE 4.13 At the roulette wheel
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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Uniform Circular Motion: Acceleration
r
r ∆v
a=
∆t
∆t → 0
r
r ∆r
v=
∆t
Uniform Circular Motion: Acceleration
∆t → 0
r r
∆r1 = v1∆t
r r
∆r2 = v2 ∆t
r r
r
r
r v2 − v1 ∆r2 − ∆r1
a=
=
∆t
( ∆t ) 2
ω=
centripetal
acceleration
The magnitude of acceleration:
a=
v2
= ω 2 r = vω
r
Direction of acceleration is the
same as direction of vector
→
(toward the center) CB = ∆rr2 − ∆rr1
The magnitude of acceleration:
a=
CB
∆rφ
v ∆ tω ∆ t
v2
∆r (π − 2α ) ∆rθ
=
=
=
ω
v
=
=
=
( ∆t ) 2 ( ∆ t ) 2
( ∆t )2
( ∆t ) 2
( ∆t ) 2
r 37
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EXAMPLE 4.14 The acceleration of a Ferris
wheel
38
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EXAMPLE 4.14 The acceleration of a Ferris
wheel
QUESTION:
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v
r
EXAMPLE 4.14 The acceleration of a Ferris
wheel
Chapter 4. Summary Slides
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General Principles
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General Principles
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Important Concepts
Important Concepts
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Applications
Applications
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Applications
Applications
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This acceleration will cause the particle to
Chapter 4. Questions
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A. slow down and curve downward.
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.
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This acceleration will cause the particle to
A. slow down and curve downward.
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.
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During which time interval is the particle
described by these position graphs at
rest?
A. 0–1 s
B. 1–2 s
C. 2–3 s
D. 3–4 s
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During which time interval is the particle
described by these position graphs at
rest?
A. 0–1 s
B. 1–2 s
C. 2–3 s
D. 3–4 s
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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
A. less than 2 m from the base.
B. 2 m from the base.
C. greater than 2 m from the base.
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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
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
A. less than 2 m from the base.
B. 2 m from the base.
C. greater than 2 m from the base.
A.
B.
C.
D.
E.
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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
A.
B.
C.
D.
E.
right and up, more than 100 m/s.
right and up, less than 100 m/s.
right and down, more than 100 m/s.
right and down, less than 100 m/s.
right and down, 100 m/s.
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right and up, more than 100 m/s.
right and up, less than 100 m/s.
right and down, more than 100 m/s.
right and down, less than 100 m/s.
right and down, 100 m/s.
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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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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?
Rank in order, from largest to smallest,
the centripetal accelerations (ar)a to (ar)e
of particles a to e.
A. (ar)b > (ar)e > (ar)a > (ar)d > (ar)c
B. (ar)b > (ar)e > (ar)a = (ar)c > (ar)d
C. (ar)b = (ar)e > (ar)a = (ar)c > (ar)d
D. (ar)b > (ar)a = (ar)c = (ar)e > (ar)d
E. (ar)b > (ar)a = (ar)a > (ar)e > (ar)d
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Rank in order, from largest to smallest,
the centripetal accelerations (ar)a to (ar)e
of particles a to e.
A. (ar)b > (ar)e > (ar)a > (ar)d > (ar)c
B. (ar)b > (ar)e > (ar)a = (ar)c > (ar)d
C. (ar)b = (ar)e > (ar)a = (ar)c > (ar)d
D. (ar)b > (ar)a = (ar)c = (ar)e > (ar)d
E. (ar)b > (ar)a = (ar)a > (ar)e > (ar)d
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The fan blade is slowing down. What are
the signs of ω and α?
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.
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The fan blade is slowing down. What are
the signs of ω and α?
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.
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