I. Work and Energy

Operate on Law of Motion

i p i p i pF v m A v

i i pi pd vm v

dt

12i

i p i p i pdF v m v vdt

kinetic energy T

ii p d TF v T

dt

Integrate

2 2

1 1

t t

i p

t t

d TF v dt dt

dt

2 1W T T True for ANY system

D1

Integrals of the Motion

Another form of Law of

Motion yields one scalar

differential equation

Work

Concept takes on additional significance in a special subset of problems, i.e.,

conservative systems If system is conservative, then each force is derivable from a

potential function U

2 2

1 1

t r

i p i

t r

W F v dt F dr

1 1 2 2 3 3 1 1 2 2 3 3 i F e F e F e dr e dr e dr eF dr

1 1 2 2 3 3F dr F dr F dr

If force is derivable form a potential function U, then

1 2 31 2 3

U U UF F F

r r r

So, 1 2 31 2 3

i U U UF dr dr dr dr dUr r r

2 2

1 1

r U

i

r U

W F dr dU

2 1W U U True for any conservative system

So, if any system is conservative

2 1 2 1W T T U U

C2 D2

assume working frame is inertial

for this discussion, e.g., e

Define V (potential energy) = - U

2 1 2 1 2 1W T T U U V V

2 2 1 1T V T V E Total Mechanical Energy

T V E Principle of Conservation of

Mechanical Energy

Examples of Conservative Forces

uniform gravity F m g

V m g h

inverse-square gravity 2

G M mF

r

G M mV

r

linear spring F k x 2

2

k xV

D3 D3

II. Impulse and Momentum

A. Linear Impulse and Linear Momentum

2 2

1 1

t t

i p

iit t

m A dtF dt

linear impulse

2

1

t i i p

i

t

d vm dt

dt

2

1

ti

i p

it

m d v

2

1

2 1( ) ( )

t

i p i p

it

m v t m v tF dt

Let i pp m v

2

1

2 1

t

it

p pF dt

Note: If no forces act in a given direction, linear momentum in that

direction is conserved (constant)

potentially 3 constants

D4

principle of linear impulse and

linear momentum

II. Impulse and Momentum

B. Angular Impulse and Linear Momentum

i pr F r m A

i i OO d HM

dt form previously introduced

2 2

1 1

t t i i O

O

iit t

d HdtM dt

dt

2

1

2 1( ) ( )

t

i O i OO

it

H t H tM dt

angular impulse

Note: If angular impulse is zero in a given direction, angular

momentum in that direction is conserved (constant)

potentially 3 constants

D5

principle of angular impulse

and angular momentum

Example: Earth-Orbiting Spacecraft

Consider Motion of spacecraft (satellite) moving in orbit about the Earth

Now - to set up problem, derive EOM, and solve? must add to diagram and choose variables of interest

but which ones? how many?

For most general motion, might set up as 3D and define spherical

coordinates

But, motion might actually be 2D in which case can use radial/transverse

coordinates for 2D (cylindrical with z = 0)

How to make a justifiable choice?

D6

Assume that Earth is a particle and generates an inverse square gravitational

force

2

sg

Gm mF

r

distance between masses

Initially consider the possibility of constants or integrals of the motion

==> may offer/suggest insight

Requires FBD same whether problem 2D or 3D

1. Energy (list ALL forces)

only gravity

2

sg r

Gm mF u

r

if conservative, should be able to define appropriate potential

function U

sGm m

Ur

2

sg r r

Gm mUF u u

r r

U exists total energy conserved

2. Linear Momentum

Forces in FBD zero in any or all inertial directions? NO

Linear Momentum NOT conserved

D6

Define

:e inertial :u moving with s

such that rur

2

sg r

Gm mF u

r

D7

3. Angular Momentum

Moments (resulting from FBD forces) zero in any or all inertial

directions?

0M r F

0e ed H

Mdt

constant eH tells us that the motion is 2D why?

Model problem as 2D

D8

eH is constant in magnitude and direction

Define

:e inertial :u moving with s such that

rur

e u zu

z ru u u

Kinematics

srr ru

e srv ru r u

2 2e s rA u ur r r r

e sF m A

2 2 : sr sGm m

u m r rr

: 02su m r r

2nd-order, coupled, nonlinear differential equations

The complete solution ( ), ( )r t t is not trivial to produce But we can still get a lot of information without the complete solution

Recall known constants of motion ,iH E

Can write expressions for each; if they are true constants / integrals of

the motion, then EOMs should appear when we differentiate expressions

for constants

D9 D9

:HC Angular Momentum

e s e ss r s rH r m v ru m ru r u

2 e

s zH m r u

2 02e e

s z

d Hm urr r

dt

02e e

s z

d Hm r ur r

dt

zero (EOM)

:EC Energy

E T V where 2 e s zH m r u 1

2e s e s

sT m v v

22 2 22

sm G m mr r

r

2 2 2 ssGm mdE

m rr r r r rdt r

sub: 2r r

2 22

2s sdE Gm

m r m rr r rdt r

22

0sdE Gm

m r r rdt r

EOM

D10

V = - U

Complete solution to 2nd-order, coupled, nonlinear DE

not available can we get some state information from constant? yes, specific states at given instants

Given: @ t1, satellite at 400 km altitude

velocity directed as shown above with magnitude 5km/se sv

Find: velocity vector @ t2 when altitude is 500 km

Define scalar quantities at times t1, t2

t1 t2

1 1( )s rr t r u

2 2 ( )s

rr t r u

1 1 1 1 ( )e s rv t r u r u

2 2 2 2

( )e s rv t r u r u

D11

means that there are two places along

path with altitude 500 km

:HC Total Angular Momentum Conserved (magnitude and direction)

2 e

s zH m r u

2 constante sm rH

2 constante

s

Hh r

m

2 2

1 1 2 2h r r 2 2 4.27 km/sr

:EC Total Energy Conserved (scalar)

2 2 2 constant2

s sm Gm mE r rr

s

E=

m constantss 2 2 2

Gm mm= =r +r

2 rE

1 1 1 2 2 21 2

1 12 2

2 2 2 2 2 2Gm Gm= =r +r r +r r r

E

Note: 53.986 10 Gm km

3/s

2

2 2.25 r km/s

2 ( ) 2.25 4.27

e s

rv t u u km/s

D12