1 6/21/04 This Week 6/21 Lecture – Chapter 3 6/22 Recitation – Spelunk Problems: 3.3, 3.29, 3.31 6/23 Lab – Kinematics in 1-D 6/24 Lecture – Chapter.

16/21/04

This Week

6/21 Lecture – Chapter 3 6/22 Recitation – Spelunk

Problems: 3.3, 3.29, 3.31 6/23 Lab – Kinematics in 1-D 6/24 Lecture – Chapter 4 6/25 Recitation – Projectile Motion

Problems: 4.7, 4.18, 4.28, 4.59

Chapter 3

Vectors

36/21/04

Vectors

In 1 dimension we could specify direction with +/-

In 2 and 3 dimensions we need more

Columbus to Harrisburg315 mi and 5o wrt East

Columbus to Nashville315 mi and 235o wrt East

Vectors have a magnitude and a direction

46/21/04

Notational Convention How do we know when we are talking

about a vector quantity?

Arrow A

Boldface A

Underline A

56/21/04

Representation of VectorsGraphical Polar Cartesian

MagnitudeDirection

x and ycomponents

Use the representation most appropriate to the problem

66/21/04

Adding Vectors (Graphically)

Vectors are added “Tail to Tip”

CBA

A

B

C

76/21/04

Trigonometry Review

Sin = opp/hyp Cos = adj/hyp Tan = opp/adj

Mnemonic: Soh Cah Toa

Note: radians = 180o

hyp

otenuse

oppositeside

adjacent side

86/21/04

Unit Vectors Carry the “direction” information

1ˆ i

z k

y j

x ˆ

i

Unit magnitude

Dimensionless

jvivv yxˆˆ

velocity vector

x component of velocity vector

unit vector in the x direction

96/21/04

Converting from Polar to Cartesian

Given:Magnitude: rDirection:

sinrrr yy

cosrrr xx

r

r

yr

xr

AAA

of Magnitude

x component of the vector

y component of the vector

106/21/04

Converting from Cartesian to Polar

Given rx, ry:

x

y

r

rtan

22yx rrr

r

r

yr

xr

Pythagorean Theorem

Geometry

Caution: Quadrant Ambiguity!

116/21/04

Example:

y

xv

vy=-3m/s

vx=-5m/s

v = (vx ,vy) = (-5,-3) m/s

Find the magnitude and direction of a vector given by:

593

5tan

8.5)3()5(

1

22

sm

sm

smv

Note that is in the -x,-y quadrant, so

= 59° +180° =239°

126/21/04

Example:

jgtiv ˆˆ5

0)5( dt

d

dt

dva xx

ggtdt

d

dt

dva yy )(

Given the following velocity, calculate the acceleration

jgjaiaa yxˆˆˆ

136/21/04

Vector Addition (Quantitative)

Given two vectors

jBiBB

jAiAA

yx

yx

ˆˆ

ˆˆ

Add the components to get the vector sum

jBAiBABA yyxxˆ)(ˆ)(

146/21/04

Vector Addition

A

B

BA

x

y

Ax Bx

By

Ay

jBAiBABA yyxxˆ)(ˆ)(

We can verify this equation graphically

156/21/04

Example:

jiB

jiA

ˆ5ˆ2

ˆ5ˆ12

mm

mm

i

ji

jiji

BAC

ˆ14

ˆ)55(ˆ)212(

ˆ5ˆ2ˆ5ˆ12

m

mmmm

mmmm

Find the vector sum of the following:

166/21/04

Scalar multiplication

Can multiply a vector by a scalar:

jcAicAAc yxˆ)(ˆ)(

jmimA

jmimA

ˆ)15(ˆ)9(3

ˆ)5(ˆ)3(

Example:

176/21/04

Properties of Vectors

cbacba

abba

bcacbac

adacadc

)(

Associative Property

Commutative Property

cba

Addition

Distributive Property

Distributive Property

186/21/04

Properties of Vectors

)CB(AC)BA(

ABBA

Commutative Property

Associative Property

196/21/04

Vector PropertiesMultiplicationby a scalar

Ac

If c is positive: Just changes the length

of the vector

If c is negative: Length of vector

changes Direction changes by

180°

206/21/04

is the angle between the vectors if you put their tails together

Dot or Scalar Product

zzyyxx BABABABABA cos

B

A

ABBA

Recall that cos() = cos(-), so

216/21/04

Dot Product: Physical Meaning Measures “how much” one vector lies along

another

)cos(cos BABABA

B

A

090

0

BA

BABA

226/21/04

Example:

Find the angle between vectors A and B:

jmimB

jmimA

ˆ)1(ˆ)3(

ˆ)2(ˆ)3(

BA

BA

BA

BA

BABA

1cos

cos

cos

First, solve for

236/21/04

Solution:

mmmB

mmmA

10)1()3(

13)2()3(

22

22

27)1)(2()3)(3( mmmmmBA

1281013

7cos

cos

21

1

mm

m

BA

BA

Plugging these into our equation for :

246/21/04

“Cross” or Vector Product

...BA

zyx

zyx

BBB

AAA

kji ˆˆˆ

det

kABBA yxyxˆ)(

jABBA zxzxˆ)( iABBA zyzy

ˆ)(

Another way to multiply vectors…

sinABBA

Magnitude only…

256/21/04

Example:

BA

01831

02312

ˆˆˆ

det

kji

k)23)31(1812(

j)0)31(012(

i)018023(

Find the cross product of A and B

kjmimB

kjmimA

ˆ0ˆ)18(ˆ)31(

ˆ0ˆ)23(ˆ)12(

km ˆ)929( 2

Only in the xy-plane

Perpendicular to the xy-plane

266/21/04

Right Hand Rule

Point your fingers in the direction of the first vector

Curl them in the direction of the second vector

Your thumb now points in the direction of the cross product

Sanity check your answers!

276/21/04

Multiplication of Unit Vectors

kj

i

ˆˆ

ˆ +

+

+

kBAjBiA yxyxˆˆˆ

kji ˆˆˆ

kij ˆˆˆ

0ˆˆ iiikj ˆˆˆ ijk ˆˆˆ

0ˆˆ jj

jik ˆˆˆ

jki ˆˆˆ

0ˆˆ kk

Can multiply by components, but it usually takes longer and is easier to make mistakes

286/21/04

Properties of the Cross Product

Cross product normal to the surface of the plane created by A and B

Antisymmetric

http://physics.syr.edu/courses/java-suite/crosspro.html

ABBA

296/21/04

Properties of Vector Products

ABBA

)()()( BcABAcBAc

)()()( CABACBA

)()()( CBCACBA

CBACBA

)()(

CBABCACBA

)()()(

Antisymmetry

Multiplication by a scalar

Distributive property

Distributive property

Triple Product

Triple Product

306/21/04

Vector Multiplication

Dot Product result is a scalar projection of one

vector onto the other

Cross Product result is a vector resultant vector is

perpendicular to both vectors

cosABBA

sinABBA

316/21/04

Where to Shoot?

a) Aim at target

b) Aim ahead of target

c) Aim behind targetvc

326/21/04

Where to Shoot?If Vc=15 m/s, Va=50 m/s,what is θ?

Vnet

vc

Va θ sinθ=Vc/Va

=15/50=0.3 θ=17.5°

vc

Va

Vnet

θ

Vnet = Vc + Va

What is Vnet?

Vnet = Va cosθ = 50 cos(17.5)° = 47.7 m/s

If the arena diameter is 60m, how long is arrow flight?

Radius is 30m, time of flight is: Ta=R/Vnet=(30 m)/(47.7 m/s)=0.63s

336/21/04

Example (Problem 3.11) A woman walks 250 m in the direction 30

east of north, then 175 m directly east. Find

a) the magnitude of her final displacement,

b) the angle of her final displacement, and

c) the distance she walks.

d) Which is greater, that distance or the magnitude of her displacement?

Chapter 4

Motion in 2-D and 3-D

356/21/04

Chapter 4: 2D and 3D Motion

Ideas from 1D kinematics can be carried over into 2D and 3D.

Our kinematic variables are now vectors.

)(),(),( tatvtr

366/21/04

Simplest Quantity: Position r(t)

Separate into components

r(t)=x(t) î + y(t) ĵ

Position of particle is specified by r(t) which is a vector depending on time

376/21/04

Displacement

In 1-D

In 2-D

jyyixxd

rrrd

ifif

if

ˆ)(ˆ)(

if xxxd

386/21/04

Velocity

Average VelocityJust like 1-D:

Instantaneous Velocity

Direction is tangent to the path

t

rvavg

dt

rd

t

rv

t

0

lim

y(t)

x(t)

396/21/04

Velocity

Two 1-D Problems

jviv

jt

rit

r

t

jrir

t

rv

yx

yx

yx

ˆˆ

ˆˆ

ˆˆ

(4m, 3m)

(5m, 2m)

ji

js

mmi

s

mmv

sm

sm ˆ)1(ˆ)1(

ˆ1

)32(ˆ1

)45(

Example: Say an object moves from point 1 to point 2 in 1 s. Find the average velocity of the object:

406/21/04

Acceleration

Average Acceleration

Instantaneous Acceleration

Just like 1-D:

t

vaavg

t

va

t

0lim

vi

vf

a

416/21/04

Treat each dimension separately

In x-direction: x, vx = dx/dt, and ax = dvx/dt

In y-direction: y, vy = dy/dt, and ay = dvy/dt

In z-direction: z, vz = dz/dt, and az = dvz/dt

Like three one-dimensional problems

426/21/04

a and v Follow From r(t)

ktzjtyitxtr ˆ)(ˆ)(ˆ)()(

kvjviv

kdt

dzj

dt

dyi

dt

dx

dt

rdv

zyxˆˆˆ

ˆˆˆ

kajaia

kdt

dvj

dt

dvi

dt

dv

dt

vda

zyx

zyx

ˆˆˆ

ˆˆˆ

v

ar

Note: r, a, and v don’t have to point in the same direction!

436/21/04

Example: (Problem 4.2)

The position vector for an electron is r = (5.0 m)i – (3.0 m)j – (2.0 m)k

a) Find the magnitude of r

b) Sketch the vector on a right handed coord system

^ ^^

446/21/04

Example: (Problem 4.9)

A particle moves so that its position (in meters) as a function of time (in seconds) is r = i + 4 t2 j + t k. Write expressions for

a) its velocity as a function of time and

b) its acceleration as a function of time.