chapter 4. kinematics in two dimensions - gsu...

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


• 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


3

Kinematics in Two Dimensions:

Instantaneous Velocity

0t

rv

t ∆ →∆ →∆ →∆ →

∆∆∆∆====

∆∆∆∆

rrrrrrrr


4

Kinematics in Two Dimensions


5

0t

rv

t ∆ →∆ →∆ →∆ →

∆∆∆∆====

∆∆∆∆

rrrrrrrrds

vdt

====

(motion along a

line)

Kinematics in Two Dimensions:

Instantaneous Acceleration

avg

va

t

∆∆∆∆====

∆∆∆∆

rrrrrrrr

Average acceleration:

Instantaneous acceleration:


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.


7

0t

va

t ∆ →∆ →∆ →∆ →

∆∆∆∆====

∆∆∆∆

rrrrrrrr


8

A Projectile is an object that moves in two dimensions (in a

plane) under the influence of only the gravitational force.


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


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====


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====


12

Example of Projectile Motion


EXAMPLE 4.4 Don’t try this at home!

QUESTION:












Problem-Solving Strategy: Projectile Motion Problems


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,

EXAMPLE 4.8 A speeding bullet

QUESTION:






Motion in a Circle


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


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= = == = == = == = =


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====


29

Then

2

T

ππππωωωω ====

Uniform Circular Motion

Then

s vt====

tθ ωθ ωθ ωθ ω====

2

T

ππππωωωω ====

v

rωωωω ====


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 ====


31

Positive angle

Uniform Circular Motion


EXAMPLE 4.13 At the roulette wheel

QUESTIONS:








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


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


38

2a r vr

ω ωω ωω ωω ω= = == = == = == = =

EXAMPLE 4.14 The acceleration of a Ferris wheel

QUESTION:


EXAMPLE 4.14 The acceleration of a Ferris wheel


EXAMPLE 4.14 The acceleration of a Ferris

wheel


Chapter 4. Summary SlidesChapter 4. Summary Slides


Chapter 4. Summary SlidesChapter 4. Summary Slides

General Principles


General Principles


Important Concepts


Important Concepts


Applications


Applications


Applications


Applications


Chapter 4. QuestionsChapter 4. Questions


Chapter 4. QuestionsChapter 4. Questions A. slow down and curve downward.

B. slow down and curve upward.

This acceleration will cause the particle to



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.


This acceleration will cause the particle to



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?


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?


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



C. greater than 2 m from the base.

A. less than 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



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


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


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?


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 > (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.


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?


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.

The fan blade is slowing down. What are the signs of ω and α?


Α. ω is positive and α is positive.B. ωωωω is negative and αααα is positive.C. ω is positive and α is negative.D. ω is negative and α is negative.

chapter 4. kinematics in two dimensions - gsu...

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