Chapter 3: Motion in a Plane

• Vector Addition

• Velocity

• Acceleration

• Projectile motion

• Relative Velocity

• CQ: 1, 2.

• P: 3, 5, 7, 13, 21, 31, 39, 49, 51.

Two Dimensional Vectors

• Displacement, velocity, and acceleration each have (x, y) components

• Two methods used:

• geometrical (graphical) method

• algebraic (analytical) method

• /

2

Graphical, Tail-to-Head

3

Addition Example

• Giam (11)

Order Independent (Commutative)

5

Subtraction, tail-to-tail

6

Subtraction Example

• Giam (19)

Algebraic Component Addition• trigonometry & geometry

• “R” denotes “resultant” sum

• Rx = sum of x-parts of each vector

• Ry = sum of y-parts of each vector

8

Vector Components

9

Examples

• Magnitude || (g4-5) Notation, Example

• Component Example Animated

• Phet Vectors

10

Trigonometry

h

osin

h

acos

a

otan

o

a

h

11

Using your Calculator: Degrees and Radians

5.030sin

866.060sin

866.030cos

5.060cos

Check this to verify your calculator is working with degrees

12

Example:

o

a

h

Given:

= 10°, h = 3

Find o and a.

310sin

o

52.0

10sin3

o

o

310cos

a

95.2

10cos3

a

a

13

Inverse Trig

• Determine angle from length ratios.

• Ex. o/h = 0.5:

• Ex. o/a = 1.0:

14

30)5.0(sin 1

45)0.1(tan 1

Pythagorean Theorem

o

a

h

Example: Given,

o = 2 and a = 3

Find h

222 hao

605.313

13

94

32

2

2

222

h

h

h

h

15

Azimuth: Angle measured counter-clockwise from +x direction.

Examples:

East 0°, North 90°, West 180°, South 270°.

Northeast = NE = 45°

16

Check your understanding:

Note: All angles measured from east.17

What are the Azimuth angles?

A:

B:

C:

180°

60°

110°

70°

Components: Given A = 2.0m @ 25°, its x, y components are:

mAAx 81.125cos0.2cos

mAAy 85.025sin0.2sin

Check using Pythagorean Theorem:

0.29996.185.081.1 2222 yx AAA

18

Vector Addition by Components:

BAy

BAx

BAR

BAR

sinsin

coscos

22yx RRR

),.(180tan 1 IIIIIquadsR

R

x

y

19

R = (10cm, 0°) + (10cm, 45°):

45sin100sin10

45cos100cos10

yyy

xxx

BAR

BAR

07.707.70

07.1707.710

y

x

R

R

cmRRR yx 5.1807.707.17 2222 20

Example Vector Addition

),.(180tan 1 IIIIIquadsR

R

x

y

5.220

07.17

07.7tan 1

(cont) Magnitude, Angle:

21

axisx above 5.22at 5.18 cmR

General Properties of Vectors

• size and direction define a vector

• location independent• change size and/or direction when multiplied by

a constant• Vector multiplied by a negative number changes

to a direction opposite of its original direction.

• written: Bold or Arrow

22

these vectors are all the same

23

Multiplication by Constants

A

-A0.5A

-1.2A

24

Projectile Motion• time = 0: e.g. baseball leaves fingertips

• time = t: e.g. baseball hits glove

• Horizontal acceleration = 0

• Vertical acceleration = -9.8m/s/s

• Horizontal Displacement (Range) = x

• Vertical Displacement = y

• Vo = launch speed

• o = launch angle

Range vs. Angle

26

Example 1: 6m/s at 30

vo = 6.00m/s o = 30°

xo = 0, yo = 1.6m; x = R, y = 0

smvv ooox /20.530cos00.6cos

smvv oooy /00.330sin00.6sin

27

Example 1 (cont.)

2

221

221

9.436.1

)8.9(30sin66.1

tt

tt

tatvy yoy

06.139.4 2 tt

Step 1

28

Quadratic Equation

02 cbxaxa

acbbx

2

42

06.139.4 2 tt

6.1

3

9.4

c

b

a

29

a

acbbx

2

42

Example 1 (cont.)

6.1

3

9.4

c

b

a

)9.4(2

)6.1)(9.4(4)3(3 2 t

)9.4(2

353.63t

954.0

342.0

t

t

End of Step 1 30

Example 1 (cont.)

tvtatvx oxxox 221Step 2

(ax = 0)

mtvx oo 96.4)954.0(30cos6cos

“Range” = 4.96m

End of Example31

Relative Motion

• Examples: • people-mover at airport• airplane flying in wind• passing velocity (difference in velocities)• notation used:

velocity “BA” = velocity of B – velocity of A

32

Summary

• Vector Components & Addition using trig

• Graphical Vector Addition & Azimuths

• Projectile Motion• Relative Motion

33

R = (2.0m, 25°) + (3.0m, 50°):

50sin0.325sin0.2

50cos0.325cos0.2

yyy

xxx

BAR

BAR

14.330.284.0

74.393.181.1

y

x

R

R

mRRR yx 88.414.374.3 2222

34

mRRR yx 88.414.374.3 2222

),.(180tan 1 IIIIIquadsR

R

x

y

0.400

74.3

14.3tan 1

(cont) Magnitude, Angle:

35

PM Example 2:

vo = 6.00m/s o = 0°

xo = 0, yo = 1.6m; x = R, y = 0

smvv ooox /00.60cos00.6cos

smvv oooy /00sin00.6sin

36

PM Example 2 (cont.)

2

221

221

9.406.1

)8.9(0sin66.1

t

tt

tatvy yoy

571.09.4

6.1

6.19.4 2

t

t

Step 1

37

PM Example 2 (cont.)

tvtatvx oxxox 221Step 2

(ax = 0)

mtvx oo 43.3)571.0(0cos6cos

“Range” = 3.43m

End of Step 238

PM Example 2: Speed at Impact

st 571.0

tavv xoxx tavv yoyy

smtvx /6)0(6 sm

vy

/59.5

571.0)8.9()0(

smvvv yx /20.8)59.5()6( 2222

39

v1

0

1

2

3

4

5

6

0 2 4 6 8 10 12 14

x(m)

y(m)

1. v1 and v2 are located on trajectory.

1v

2v

va

40

Q1. Given 1v2v

v

locate these on the trajectory and form v.

0

1

2

3

4

5

6

0 2 4 6 8 10 12 14

x(m)

y(m)

1v

2v 41

Kinematic Equations in Two Dimensions

tavv xoxx tvvx xox )(2

1 2

21 tatvx xox

xavv xoxx 222

tavv yoyy

tvvy yoy )(21

221 tatvy yoy

yavv yoyy 222

* many books assume that xo and yo are both zero.42

Velocity in Two Dimensions

• vavg // r

• instantaneous “v” is limit of “vavg” as t 0

t

rvavg

43

Acceleration in Two Dimensions

t

vaavg

• aavg // v

• instantaneous “a” is limit of “aavg” as t 0

44

Conventions

• ro = “initial” position at t = 0

• r = “final” position at time t.

45

Displacement in Two Dimensions

ro

r

r

orrr

rrr o

46

Acceleration ~ v change

• 1 dim. example: car starting, stopping

47

Acceleration, Dv, in Two Dimensions

48

Ex. Vector Addition

• Add A = 3@60degrees azimuth, plus B = 3@300degrees azimuth.

• Find length of A+B, and its azimuth. Sketch the situation.

Ex.2:

• 10cm@10degrees + 10cm@30degrees

• Length and azimuth?

Calculate F3

• F1 = 8@60, F2 = 5.5@-45• F1 + F2 + F3 = 0• F3 = -(F1 + F2)• Rx = 8cos60+5.5cos(-45)=7.89• Ry = 8sin60+5.5sin(-45)=3.04• R = 8.46• Angle = tan-1(3.04/7.89) = 21 deg above +x axis• Answer book wants is 180 off this angle!

Addition by Parts (Components)

52