Download Selected Problems in Dynamics

Selected Problems in Dynamics:
Stability of a Spinning Body
-and-
Derivation of the Lagrange Points
John F. Fay
June 24, 2014
Outline
 Stability
of a Spinning Body
 Derivation
of the Lagrange Points
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Stability of a Spinning Body
 Euler’s
Second Law:
Moments of Torque
d
M  I 
dt
Angular Velocity
Inertia Matrix
(good for an inertial coordinate system)
Problem: Inertia matrix “I” of a rotating body
changes in an inertial coordinate system
Euler’s Second Law (2)
 In
a body-fixed coordinate system:
Moments
of Torque
d
M  I body    I 
dt
Angular Velocity
Inertia Matrix
 Inertia
Matrix:
 I xx
 Describes distribution

I   I xy
of mass on the body
 I xz

I xy
I yy
I yz
I xz 

I yz 
I zz 
Euler’s Second Law (3)
 Planes
of Symmetry:
 x-y
plane: Ixz = Iyz = 0
 x-z plane: Ixy = Iyz = 0
 y-z plane: Ixz = Ixz = 0
 For
 I xx
I   0
 0
a freely-spinning body:
M 0
d
I body    I   0
dt
0
I yy
0
0
0 
I zz 
Euler’s Second Law (4)
 Rigid
body:
 “I”
 Do
d
I body    I   0
dt
matrix is constant in body coordinates
d
I
   I   0
dt
the matrix multiplication:
 I xx
I   0
 0
0
I yy
0
0   x   I xx x 
0   y    I yy y 
I zz   z   I zz z 
Euler’s Second Law (5)
 Do
the cross product:
 x   I xx x   y I zz z   z I yy y 






  I    y    I yy y     z I xx x   x I zz z 
 z   I zz z   x I yy y   y I xx x 
 Result:
 I xx
0

 0
0
I yy
0
0   x   y I zz z   z I yy y 
d   


0   y     z I xx x   x I zz z   0
dt
I zz   z   x I yy y   y I xx x 
Euler’s Second Law (6)
 I xx
0

 0
 As
0
I yy
0
0   x   y I zz z   z I yy y 
d


0   y     z I xx x   x I zz z   0
dt
I zz   z   x I yy y   y I xx x 
scalar equations:
d x
I xx
 I zz  I yy  y z  0
dt
d y
I yy
 I xx  I zz  z x  0
dt
d z
I zz
 I yy  I xx  x y  0
dt
Small Perturbations:
 Assume
axis:
a primary rotation about one
 Remaining
angular velocities are small
x     'x
y  'y
z   ' z
 ' x ,  ' y ,  ' z  
Small Perturbations (2)
 Equations:
d ' x
I xx
 I zz  I yy  ' y  ' z  0
dt
d ' y
I yy
 I xx  I zz  ' z    ' x   0
dt
d ' z
I zz
 I yy  I xx    ' x  ' y  0
dt
Small Perturbations (3)
 Linearize:
 Neglect
products of small quantities
d ' x
I xx
0
dt
d ' y
I yy
 I xx  I zz  ' z  0
dt
d ' z
I zz
 I yy  I xx  ' y  0
dt
Small Perturbations (4)
 Combine
I yy
d ' y
the last two equations:
 I xx  I zz  ' z  0
dt
d ' z
I zz
 I yy  I xx  ' y  0
dt
d ' y

I xx  I zz 
 
 'z
dt
I yy
switched

I xx  I yy 
d ' z

'y
dt
I zz
Small Perturbations (4)
 Combine
I yy
d ' y
the last two equations:
 I xx  I zz  ' z  0
dt
d ' z
I zz
 I yy  I xx  ' y  0
dt
d 2 ' y

I xx  I zz  d ' z
 
2
dt
I yy
dt

I xx  I yy  d ' y
d 2 ' z

2
dt
I zz
dt
Small Perturbations (4)
 Combine
I yy
d ' y
the last two equations:
 I xx  I zz  ' z  0
dt
d ' z
I zz
 I yy  I xx  ' y  0
dt
I xx  I zz I xx  I yy 
d 2 ' y
2

' y  0
2
dt
I yy I zz
I xx  I zz I xx  I yy 
d 2 ' z
2

 'z  0
2
dt
I yy I zz
Small Perturbations (5)
d 2 ' y
dt
2

2
I xx  I zz I xx  I yy 
I yy I zz
' y  0
I xx  I zz I xx  I yy 
d 2 ' z
2

 'z  0
2
dt
I yy I zz
 Case
1: Ixx > Iyy, Izz or Ixx < Iyy, Izz:

2
I xx  I zz I xx  I yy 
 Sinusoidal
I yy I zz
solutions for
 Rotation is stable
0
 ' y , 'z
Small Perturbations (6)
d 2 ' y
dt
2

2
I xx  I zz I xx  I yy 
I yy I zz
' y  0
I xx  I zz I xx  I yy 
d 2 ' z
2

 'z  0
2
dt
I yy I zz
 Case
2: Izz > Ixx > Iyy or Izz < Ixx < Iyy:

2
I xx  I zz I xx  I yy 
I yy I zz
0
solutions for  ' y , ' z
 Rotation is unstable
 Hyperbolic
Numerical Solution
Ixx = 1.0
Iyy = 0.5
Izz = 0.7
Body Coordinates
Numerical Solution (2)
Ixx = 1.0
Iyy = 0.5
Izz = 0.7
Global Coordinates
Numerical Solution (3)
Ixx = 1.0
Iyy = 1.2
Izz = 2.0
Body Coordinates
Numerical Solution (4)
Ixx = 1.0
Iyy = 1.2
Izz = 2.0
Global Coordinates
Numerical Solution (5)
Ixx = 1.0
Iyy = 0.8
Izz = 1.2
Body Coordinates
Numerical Solution (6)
Ixx = 1.0
Iyy = 0.8
Izz = 1.2
Global Coordinates
Spinning Body: Summary
 Rotation
about axis with largest inertia:
 Rotation
about axis with smallest inertia:
 Rotation
about axis with in-between inertia:
Stable
Stable
Unstable
Derivation of the Lagrange Points
 Two-Body
 Lighter
Problem:
body orbiting around a heavier body
(Let’s stick with a circular orbit)
 Actually
both bodies orbit around the
barycenter
y

D
M
m
x
Two-Body Problem (2)
 Rotating
reference frame:
 Origin
at barycenter of two-body system
 Angular velocity matches orbital velocity of
bodies
 Bodies are stationary in the rotating reference
frame
y

D
M
m
x
Two-Body Problem (3)
 Force
Balance:
Body “M”
Body “m”
GMm
Mx M  
0
2
D
GMm
mxm  
0
2
D
2
y
xM(<0)
2
Centrifugal
Force
Gravitational Force

xm
D
M
D  xm  xM
m
x
Two-Body Problem (4)
GMm
Mx M  
0
 Parameters:
2
D
 M – mass of large body
GMm
2
 m – mass of small body
mxm  
0
2
D
 D – distance between bodies
D  xm  xM
2
 Unknowns:
 xM
– x-coordinate of large body
 xm – x-coordinate of small body
  – angular velocity of coordinate system
 Three
equations, three unknowns
In case anybody wants to see the
GMm
math …
Mx  
0
D
2
M
2
GM
GMm
2
Gm
mx


0
x



0
xM   2  0
m
2
D
D
D
D x x
Gm
GM
2
 

2
M
xM D
xm D 2
xm 
D
M m
m
m M
x

D
M


 mxm  MxM
M m
x M xm
2
2
m
2
m
m
M m
D  xm 
xm 
xm
M
M
GM
G M  m 
 

2
xm D
D3
2
M
Two-Body Problem (5)
 Solutions:
G M  m 
 
D3
M
xm 
D
M m
m
xM  
D
M m
2
y
xM(<0)

xm
D
M
m
x
Lagrange Points
 Points
at which net gravitational and
centrifugal forces are zero
 Gravitational
attraction to body “M”
 Gravitational attraction to body “m”
 Centrifugal force caused by rotating coordinate
system
y
(x,y)

M
m
x
Lagrange Points (2)
 Sum
of Forces: particle at (x,y) of mass “m ”
 x-direction:
 GMm x  xM 
x  x
M
2
y
2

 x  x
3
2
 y-direction:
 GMmy
x  x
M
2
y
2
Gmm x  xm 

2
y
2
Gmmy
 x  x
3
m
2
m
2
y
2

3

3
 mx 2  0
2
 my 2  0
2
Lagrange Points (3)
 Simplifications:
by mass “m” of particle
 Substitute for “2” and divide by “G”
 Divide
 M  x  xM 
x  x

2
M
y
2
 My
x  x

2
M
y
2

mx  xm 
 x  x
3
m

 x  x
3

y
2
2
2
my
2
m

2
y
2

3

3

M  m
x
0
2
D3

M  m
y
0
2
D3
Lagrange Points (4)
 Qualitative
considerations:
Far enough from the system, centrifugal force will
overwhelm everything else Net force will be away
from the origin
 Close enough to “M”, gravitational attraction to that
body will overwhelm everything else
 Close enough to “m”, gravitational attraction to that
body will overwhelm everything else

M
m
Lagrange Points (5)
 Qualitative
considerations (2)
Close to “m” but beyond it, the attraction to “m” will
balance the centrifugal force  search for a Lagrange
point there
 On the side of “M” opposite “m”, the attraction to “M”
will balance the centrifugal force  search for a
Lagrange point there
 Between “M” and “m” their gravitational attractions will
balance  search for a Lagrange point there

M
m
Lagrange Points (6)
 Some
Solutions:
 Note
that “y” equation is uniquely satisfied if
y = 0 (but there may be nonzero solutions also)
 My
x  x

2
M
y
2

 x  x
3
my
2
m

2
y
2

3

M  m
y
0
2
D3
Lagrange Points (7)
 Some
Solutions:
 “x”
equation when “y” is zero:
 M  x  xM 

mx  xm 
x  x   x  x  
Note:  x  x    x  x
2
3
2
2
M

3

M  m
x
0
D3
2
m
2
M
3
3
2
M
  x  xM 
?
 MD 3  x  xm  x  xm  mD3 x  xM  x  xM
3
 xM  m x  xm  x  xm x  xM  x  xM  0
Lagrange Points (8)
 M  x  xM 
x  x
M

M

y
 My
x  x
2
2
y
 Oversimplification:
 xM
x
 Mx
2
y
2

3
2
2
2
mx  xm 

 x  x
3
m

m

2

 x  x
3
y
my
2
2
2
y

2

2
3
2
2
x
3
M  m  0
D3

M  m
y
0
D3
Let “m” = 0
=0
M
x 3 0
D
 Solutions:
x
 Anywhere
2
y
x
2

3
2
 My
2
y
 D3
on the orbit of “m”
2

3
2
M
 y 3 0
D
Lagrange Points (9)
 MD 3  x  xm  x  xm  mD3 x  xM  x  xM
 xM  m x  xm  x  xm x  xM  x  xM  0
 Perturbation:
 For
m  M
0    1
xM  D xm  1   D
the y = 0 case:
 D 3  x  1   D  x  1   D
 D 3  x  D  x  D
 x1    x  1   D  x  1   D  x  D  x  D  0
Lagrange Points (10)
 MD 3  x  xm  x  xm  mD3 x  xM  x  xM
 xM  m x  xm  x  xm x  xM  x  xM  0
 D 3  x  1   D  x  1   D  D 3  x  D  x  D
 x1    x  1   D  x  1   D  x  D  x  D  0
 For
the y = 0 case (cont’d):
– Lagrange Points L1,L2: near x = D:
x  D1   
 D        D 1     
5
5
2
 D 1   1         1       0
5
2
L1, L2: y = 0, x = D(1 + 
 D 5        D 5 1     
2
 D 5 1   1         1       0
2
 Collecting
terms:
       1  1   1   1      
2
  1       0
2
 Multiplying
things out:
3  3 2   3  3  7  5 2   3 
       2

2
2 2
3
3
 3  5  2     

  1  2  2   2  2   2


L1, L2 (2): y = 0, x = D(1 + 
 Balancing
terms:
Order of magnitude 3
3  3 2   3  3  7  5 2   3 
       2

2
2 2
3
3
 3  5  2     

  1  2  2   2  2   2


Order of magnitude 
 Linearizing:
3  
3
1
  
3
3
Lagrange Points (11)
 D 3  x  1   D  x  1   D  D 3  x  D  x  D
 x1    x  1   D  x  1   D x  D  x  D  0
 Lagrange
Point L3: y = 0, near x = –D
x  D1   
2      2       1     1    
 1   1   2      2     1     1    
 Since
| |, | | << 1 < 2:
2     2   1     2
2
2
 1   1   2      1       0
0
L3: y = 0, x = –D(1 + 
2        1     
2
2
 1   1   2      1       0
2
 Multiply
out and linearize:
2
4  4  4   
 1     4  4  4 1  2  2   0
4  4  4   
 4  4  4  4  4  8  8   0
5  12  0
5
 
12
Lagrange Points (12)
 Summary
for a Small Perturbation and
m  M 0    1
xM  D xm  1   D
y = 0:
 Lagrange
 L1:
Points:


1
x  D1  3  
3 

 L2:
 L3:
5 

x   D 1   
 12 


1
x  D1  3  
3 

Lagrange Points (13)
 What
if y is not zero? m  M 0    1
xM  D xm  1   D
R  D1   
y
2q
x
Lagrange Points (14)
 Polar
m  M
0    1
xM  D
xm  1   D
R  D1   
Coordinates:
eq
R  D cos2q , D sin2q 
y
er
FR ,0
  R  D1    cos2q ,



 D1   sin2q 

2q
x
Lagrange Points (15)
 Consider
 From
FM 
 From
Fm 
the forces in the R-direction:
M:
 GMm D1     D cos2q 
D1     D cos2q   D sin2q  
2
2
3
2
m:
Gmm  D1     D1    cos2q 
 D1     D1   cos2q   D1   sin2q  
2
 Centrifugal
force:
2
3
2
FC  m 2 D1   
R-direction Forces
 Simplify:
 From
M:
FM 
 From
Fm 


 GMm 1     cos2q 
D 1     cos2q    sin 2q 
2
2
2
2

2
m:
 GMm 1    1    cos2q 
D 1    1    cos2q   1   sin 2q 
2

3
 Centrifugal
2
force:
2
3
GM 1   m
1   
FC 
2
D
R-direction Forces (2)
 Linearize
in  and :
 From
M:
 From
m:
 GMm 1     cos2q   GMm
1  2  2 cos2q 
FM 

2
2
D 1  3  3 cos2q 
D
Fm 
 GMm 1    1    cos2q 
D 1  2  21      cos2q   1  2  2
3
2
 Centrifugal
force:
 GMm


2
2 2D
GMm
1     
FC 
2
D
R-direction Forces (3)
 Balance
the Forces:
GMm
GMm
GMm
1  2  2 cos2q  

  2 1       0
2
2
D
D
2 2D
 1  2  2 cos2q  

2 2
 1       0
1      1  2  2 cos2q  
3    2 cos2q  

2 2

2 2
 1

 
 1  2 cos2q 
2 2
3
Lagrange Points (16)
 Consider
 From
FM 
 From
Fm 
the forces in the q-direction:
M:
GMmD sin 2q 
D1     D cos2q   D sin2q  
2
m:
2
3
2
 GmmD1   sin 2q 
 D1     D1   cos2q   D1   sin2q  
2
 Centrifugal
force is completely radial
2
3
2
FC  0
q-Direction Forces
FM 
Fm 
GMmD sin 2q 
D1     D cos2q   D sin2q  
2
 D1     D1   cos2q   D1   sin2q  
2
2
2
 GmmD1   sin 2q 
2
 Simplify:
FM 
D
2
GMm sin 2q 
2
1      cos2q    sin2q  
2
3
2
 GMm 1   sin 2q 
Fm 
D
2
3
1     1   cos2q   1   sin2q  
2
2
3
2
2
3
q-Direction Forces (2)GMm sin2q 
FM 
1      cos2q    sin2q  
2
1     1   cos2q   1   sin2q  
2
D
Fm 
D
 Linearize
2
2
2
2
 GMm 1   sin 2q 
2
2
in  and :
GMm sin 2q 
GMm sin 2q 
FM  2


2
D 1  3  3 cos2q 
D
Fm 

3
 GMm 1   sin 2q 
D 1  2  21      cos2q   1  2  2
 GMm sin 2q 
3
2
D 2 2  2 cos2q  2
3

3
q-Direction Forces (3)
Fm 
 Simplifying
further:
GMm sin 2q 
FM 

2
D
 GMm sin 2q 
D 2 2  2 cos2q  2
3
GMm sin 2q 
FM 

2
D
 GMm sin 2q 
Fm 

3
D 2 2 2 1  cos 2 q   sin 2 q   2
 GMm sin 2q 
 GMm sin 2q 



3
2
3
2
2
8 D sin q 
D 2 2 2 sin q  2



q-Direction Forces (4)
 Balance
the two forces:
GMm sin 2q 
FM 

2
D
 GMm sin 2q 
Fm 

2
3
8 D sin q 
GMm sin 2q 
 GMm sin 2q 

 0
2
2
3
D
8D sin q 
GMm sin 2q 
GMm sin 2q 


2
2
3
D
8D sin q 
8 sin 3 q   1
1
sin q   
2

q    30
6

2q    60
3
Lagrange Points (17)
 Solving
for our radius:
R  D1   
 1

 
 1  2 cos2q 
2 2
3

2q    60
3
  1
 
R  D1  
 1  2 cos2q  
 3
 2 2
 8 2 
  1
1 
 D1  
 1  2     D1 
 
2 3
12
 2 2


Lagrange Points (13 redux)
 What
if y is not zero? m  M 0    1
xM  D xm  1   D
 8 2 
R  D1 
 
12


y
2q  60
x
Lagrange Points (18):
m  M 0    1
xM  D xm  1   D
 Summary
for zero y:
 Between
 On
M and m: L1:


1
x  D1  3  
3 

the opposite side of m from M: L2:


1
x  D1  3  
3 

 On
the opposite side of M from m: L3:
5 

x   D 1   
 12 
Lagrange Points (19):
m  M 0    1
xM  D xm  1   D
 Summary
 Ahead
for nonzero y:
of m in orbit: L4:
2q  60
 Behind

m in orbit: L5:
2q  60

 8 2 
R  D1 
 
12


 8 2 
R  D1 
 
12


Summary of Lagrange Points
(What is wrong
with this picture?)
(courtesy of Wikipedia, “Lagrange Points”)
Potential Energy Contours
Stability of Lagrange Points
 Requires
saving higher-order terms in
linearized equations
 Results:
– L1, L2, L3 unstable
 Body
perturbed from the point will move away
from the point
– L4, L5 neutrally stable
 Body
perturbed from the point will drift
around the point
Conclusions
 Stability
of Spinning Bodies
 Derivation
of Lagrange Points