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