Chapter 4

Motion in 2-D and 3-D

26/24/04

Chapter 4: Review

Our kinematic variables are now vectors:

)(),(),( tatvtr

Similarly to 1-D, we have

dt

rdv

2

2

dt

rd

dt

vda

Treat each dimension separately Like multiple 1-D problems

36/24/04

Review

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

46/24/04

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

^ ^^

56/24/04

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.

66/24/04

Projectile Motion

Free fall with horizontal motion

Conventional coordinate system:x horizontal y vertical (take "up“ as positive direction)z is not relevant (motion is 2-D)

ay = -g ax = 0

Gravity is the source of acceleration: a = -g ĵ

76/24/04

Projectile Motion (Continued)

In both directions the acceleration is constant

ax = 0 ay = -g

vx = v0x constant

x = x0 + v0xt

vy = v0y - gt

y = y0 + v0yt - ½gt2

86/24/04

Projectile Motion (Continued)

x and y motions can be treated independently

Connected by time:

x = x0 + v0xt

y = y0 + v0yt - ½gt2

Since y(t) is a parabola andx is linear in time:y(x) is a parabola too

96/24/04

Motion in x and y direction Independent

x = x0 + v0x t

ax = 0

vx(t) = v0x

y = y0 + v0y t – ½ gt2

ay = -g

vy(t) = v0y – gt

vxvx=0

106/24/04

Example: Kicking a Field Goal

40 yds

=35o

v0=21 m/s10 ft

1)Does the ball make it over the goal post? (assuming it starts from ground level)

2)How long does it take?

116/24/04

Solution:

36.6 m

=35o

v0=21 m/s 3.0 m

1) Convert to standard units

2) Get x-y components

v0x = 21 m/s cos35° = 17.2 m/s

v0y = 21 m/s sin35° = 12.0 m/s

3) Tabulate known and unknown quantities

Quantity Initial Final

time 0 s ???

x 0 m 36.6m

y 0 m ???

vx 17.2 m/s ???

vy 12.0 m/s ???

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Solution:4) Write down expressions

x(t) = x0 + v0xt

y(t) = y0 + v0yt – ½gt2

Note:

x0 = y0 = 0

The kick is good!

5) Solve for missing quantities

Quantity Initial Final

time 0 s ???

X 0 m 36.6m

y 0 m ???

vx 17.2 m/s ???

vy 12.0 m/s ???s

m

v

txt

sm

x

12.22.17

6.36)(

0

m

ssms

gttvty

sm

y

42.3

)12.2)(8.9()12.2)(12(

)(22

21

221

0

136/24/04

Basic Equations of Projectile Motion

Horizontal - constant velocity (neglect air resistance)

tvxtx

vtv

x

xx

00

0

)(

)(

Vertical - constant acceleration g downward

221

00

0

)(

)(

gttvyty

gtvtv

y

yy

146/24/04

Breaking Velocity into Components

Take components of v0

jvivv yxˆˆ

000

v0

v(t) can also be written in components

000

000

sin

cos

vv

vv

y

x

θ0

V0x

V0y

156/24/04

Writing y as a Function of x

y(x) is a parabola!

xv

txt

0

)(

2

021

00)(

xxy v

xg

v

xvxy

221

0)( gttvty y 200

2

0cos2

tan)(

v

gxxxy

tvxtx x00)(

For x0 =0:

Since y0 =0, we have:

Substituting in t = x/v0x

200

2

00

00

)cos(2cos

sin)(

v

gx

v

vxxy

166/24/04

Range Equation

200

0

200

2

0

cos2tan

cos2tan0

v

gxx

v

gxx

200

2

0cos2

tan)(

v

gxxxy

To find the range, we find where y = 0

We have the trivial solution x=0

Or this term equals 0

176/24/04

Range Equation (Continued)

g

vx

g

v

g

vx

v

gx

v

gx

)2sin(

cossin2cos2

cos

sin

cos

sintan

cos2

0cos2

tan

20

0020

200

0

0

0

002

00

200

0

To obtain a non-trivial solution, we solve:

186/24/04

Range Equation (Continued)

g

vR 0

20 2sin

Caution: Only valid when ‘launch’ and ‘landing’ are on the same level!

For given v0,

maximum range is at 45° angle.

0

50

100

0 30 60 90

Theta

Ran

ge

(m)

196/24/04

Evel Knievel: Famous Jumps

Fountains of Caesar’s Palace,Las Vegas (151 feet) in coma for thirty days afterwards

Jumping cars 13 in 1970; 19 in 1971 Injured jumping 13 Pepsi trucks Jumped 50 Cars (stacked) in 1973

at LA coliseum (35,000 spectators) Retired after accident while jumping a

tank of live sharks in 1976. Listed in Who’s Who and Guinness

Book of World Records for having broken 35 bones.

206/24/04

Caesar’s Palace Jump

151 feet

46 m0

Choose top of start ramp as the origin.x axis is horizontal, y axis is vertical

x

y

y = 0

216/24/04

So how fast was Evel going?

Assume he took Physics 211, and used a 45° ramp:

g

v

g

v

g

vR

20

200

20 90sin2sin

mph4.47

sm2.21

sm8.9m46 20

gRv

What direction is he going?

226/24/04

Another Problem

Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 15o above the horizontal. Ball 2 has twice the initial speed of ball 1. The first ball travels a distance D1 before being caught. How far does ball 2 travel before it is caught?

a) D2=2D1 b) D2=4D1 c) D2=8D1

236/24/04

Demo: Shoot Kenny

Magnet releases when ball is fired

Where do we aim?

Magnet

246/24/04

Shoot Kenny

If there is no gravity,shoot right at him!

Kenny

xB(t)=v0x t

yB(t)=v0y t

xK(t)=x0K

yK(t)=y0K

To hit Kenny,

Ball

x=0y=0

y0K

θ

x0K

K

K

x

y

0

0tan

256/24/04

Shoot Kenny

x=0y=0

yK

θ = ?

xK

With gravity?

221

0

0

)(

)(

gttvty

tvtx

yB

xB

221

0

0

)(

)(

gtyty

xtx

KK

KK

Ball

Kenny

sin

cos

00

00

vv

vv

y

x

Breaking v0 into components:

v0

v0x

v0y

266/24/04

Shoot KennyAt what time does xB=xK?

cos

cos

0

0

00

00

v

xt

xtv

xtv

K

K

Kx

At what time does yB=yK?

221

02

21

0 gtygttv

yy

Ky

KB

sin0

0

0

0

00

v

y

v

yt

ytv

K

y

K

Ky

To hit Kenny, these times must be equal:

K

K

KK

x

y

v

y

v

x

0

0

0

0

0

0

tancos

sin

sincos

Same result as before!

276/24/04

Shoot Kenny

Why? The ball and Kenny fall with exactly the same acceleration

221

0)( gttvty yB 221

0)( gtyty KK

286/24/04

Example:

v0 A ball rolls off a table of height h. The ball has horizontal velocity v0 when it leaves the table.

How far away does it strike the ground?

How long does it take to reach the ground?

h

296/24/04

Example: Gerbil on the Fly

What is his maximum height?

How far is the total distance (horizontal range)?

What is his velocity on impact?

A daredevil gerbil, shot from ground level, is observed to have velocity:when he is 9.1 meters above a flat field.

smjiv )ˆ1.6ˆ6.7(

9.1 m v

306/24/04

Gerbil: Maximum HeightSet x = 0 and y = 0 where he starts from the groundSet t = 0 when y = 9.1 m. Then:

At maximum height:

20

20 )(2 yyy vvyya

g

vy

a

vyy y

y

y

22

20

0

20

0

vy = 0 m/say = -g

msm

smmy 11

)/8.9(2

)/1.6(1.9

2

2

y0 = 9.1 mv0y = 6.1 m/s

316/24/04

Gerbil: Horizontal RangeTotal time in air = twice the time to fall 11 m

On landing: x = v0xtair = (7.6)(3) = 23 meters

At maximum height,

11 m

xg

yt

gttvyy y

max

221

00

2

The time in the air will then be:

326/24/04

Gerbil: Impact Velocity

vx = v0x = 7.6 m/s (does not change)

vy = 0 at the top vimpac

t

Impact velocity is:

Initial velocity was:

gtvv yy 21

0

smjivimpact /)ˆ7.14ˆ6.7(

smjivinitial /)ˆ7.14ˆ6.7(

sm

ssmgtv airy

/7.14

)3)(/8.9( 221

21

336/24/04

Uniform Circular Motion

Period (the time to make one revolution):

a

v

R R

va

2

Velocity has magnitude and direction, so changing direction also causes an acceleration.

Consider the special case of constant speed around a circular path. The magnitude of the “centripetal” acceleration is then,

pointing towards the center of the circle of radius R

v

RT

2

346/24/04

Example:

A runner takes 12 seconds round a 180º circular curve at one end of a race track. The distance covered on the curve is 100 meters.

What is her centripetal acceleration?

356/24/04

SolutionHer speed will be given by:

sms

m

t

dv /33.8

12

100

The circumference of half a circle is:

mmd

R

Rd

8.31100

)2(21

R=?

v = ?

Finally we can calculate the centripetal acceleration:

222

/2.28.31

)/33.8(sm

m

sm

R

va

366/24/04

Relative Velocity

vobject/observer = vobject/in-frame + vref. frame

Example: A kid running upthe down escalator

vescalator

vboy

Observed velocity depends on velocity of observer!

If your reference frame (coordinate system) is moving relative to a stationary observer with velocity = vref. frame and an object in your reference frame has vobject/in-frame, then according to observer, the object has velocity

376/24/04

Example: Toy in Car

vtoy/in-car

vcar

vcar

vtoy/in-car

vtoy/observer

You are standing on the side of the road, and a man drives by with a young passenger:

From what you observe,she has quite an arm!

386/24/04

Passing a signpost

x is position in road frame

x’ is position in car frame

x’=x-ut

u

396/24/04

Relative Velocity: Rowing a Boat

Michael can row a boat at vrow= 3 m/s, and he wants to go straight across a river which flows with vriver = 2 m/s. At what angle should he row?

vboat = vrow + vriver

vrivervrow

vboat

y

x

Must have vboat only in y-direction to go straight across

406/24/04

Rowing a Boat (Continued)

Note: If vriver > vrow then he cannot go straight across

vboat = vrow + vriver

vrow cos = vriver

He needs vrow,x = - vriver,x

48/3

/2coscos 11

sm

sm

v

v

row

river

vriver

vboat

vrow

416/24/04

Example:There is a moving sidewalk in the airport. John does not use it, and takes 150 seconds to walk the length of the terminal. Paul stands on the moving sidewalk, and takes 70 seconds. Ringo walks on the sidewalk at the same pace that John walked beside it.

How long will it take Ringo to walk the length of the terminal?

a) 150 s – 70 s b) 150 s + 70 s c) Something else