Selected Problems in Dynamics: Stability of a Spinning Body -and- Derivation of the Lagrange Points...
Transcript of Selected Problems in Dynamics: Stability of a Spinning Body -and- Derivation of the Lagrange Points...
Selected Problems in Dynamics:
Stability of a Spinning Body-and-
Derivation of the Lagrange PointsJohn 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:
(good for an inertial coordinate system)Problem: Inertia matrix “I” of a rotating body changes in an inertial coordinate system
Idt
dM
Moments of Torque
Inertia MatrixAngular Velocity
Euler’s Second Law (2)
In a body-fixed coordinate system:
Inertia Matrix: Describes distribution
of mass on the body
IIdt
dM body
Momentsof Torque
Inertia MatrixAngular Velocity
zzyzxz
yzyyxy
xzxyxx
III
III
III
I
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 a freely-spinning body:
0 IIdt
dbody
zz
yy
xx
I
I
I
I
00
00
00
0M
Euler’s Second Law (4)
Rigid body: “I” matrix is constant in body coordinates
Do the matrix multiplication:
0 IIdt
dbody
zzz
yyy
xxx
z
y
x
zz
yy
xx
I
I
I
I
I
I
I
00
00
00
0 I
dt
dI
Euler’s Second Law (5)
Do the cross product:
Result:
xxxyyyyx
zzzxxxxz
yyyzzzzy
zzz
yyy
xxx
z
y
x
II
II
II
I
I
I
I
0
00
00
00
xxxyyyyx
zzzxxxxz
yyyzzzzy
z
y
x
zz
yy
xx
II
II
II
dt
d
I
I
I
Euler’s Second Law (6)
As scalar equations:
0 zyyyzzx
xx IIdt
dI
0 xzzzxxy
yy IIdt
dI
0 yxxxyyz
zz IIdt
dI
0
00
00
00
xxxyyyyx
zzzxxxxz
yyyzzzzy
z
y
x
zz
yy
xx
II
II
II
dt
d
I
I
I
Small Perturbations:
Assume a primary rotation about one axis:
Remaining angular velocities are small
xx '
yy '
zz ' zyx ',','
Small Perturbations (2)
Equations:
0'''
zyyyzzx
xx IIdt
dI
0'''
xzzzxxy
yy IIdt
dI
0'''
yxxxyyz
zz IIdt
dI
Small Perturbations (3)
Linearize: Neglect products of small quantities
0'
dt
dI xxx
0''
zzzxxy
yy IIdt
dI
0''
yxxyyz
zz IIdt
dI
Small Perturbations (4)
Combine the last two equations:
z
yy
zzxxy
I
II
dt
d'
'
0''
yxxyyz
zz IIdt
dI
0''
zzzxxy
yy IIdt
dI
y
zz
yyxxz
I
II
dt
d'
'
switched
Small Perturbations (4)
Combine the last two equations:
dt
d
I
II
dt
dz
yy
zzxxy ''2
2
0''
yxxyyz
zz IIdt
dI
0''
zzzxxy
yy IIdt
dI
dt
d
I
II
dt
d y
zz
yyxxz''
2
2
Small Perturbations (4)
Combine the last two equations:
0'
' 22
2
yzzyy
yyxxzzxxy
II
IIII
dt
d
0''
yxxyyz
zz IIdt
dI
0''
zzzxxy
yy IIdt
dI
0'
' 22
2
zzzyy
yyxxzzxxz
II
IIII
dt
d
Small Perturbations (5)
Case 1: Ixx > Iyy, Izz or Ixx < Iyy, Izz:
Sinusoidal solutions for , Rotation is stable
02
zzyy
yyxxzzxx
II
IIII
y' z'
0'
' 22
2
yzzyy
yyxxzzxxy
II
IIII
dt
d
0'
' 22
2
zzzyy
yyxxzzxxz
II
IIII
dt
d
Small Perturbations (6)
Case 2: Izz > Ixx > Iyy or Izz < Ixx < Iyy:
Hyperbolic solutions for , Rotation is unstable
02
zzyy
yyxxzzxx
II
IIII
y' z'
0'
' 22
2
zzzyy
yyxxzzxxz
II
IIII
dt
d
0'
' 22
2
yzzyy
yyxxzzxxy
II
IIII
dt
d
Numerical Solution
Ixx = 1.0Iyy = 0.5Izz = 0.7
Body Coordinates
Numerical Solution (2)
Ixx = 1.0Iyy = 0.5Izz = 0.7
Global Coordinates
Numerical Solution (3)
Ixx = 1.0Iyy = 1.2Izz = 2.0
Body Coordinates
Numerical Solution (4)
Ixx = 1.0Iyy = 1.2Izz = 2.0
Global Coordinates
Numerical Solution (5)
Ixx = 1.0Iyy = 0.8Izz = 1.2
Body Coordinates
Numerical Solution (6)
Ixx = 1.0Iyy = 0.8Izz = 1.2
Global Coordinates
Spinning Body: Summary
Rotation about axis with largest inertia:Stable
Rotation about axis with smallest inertia:Stable
Rotation about axis with in-between inertia:Unstable
Derivation of the Lagrange Points
Two-Body Problem: Lighter body orbiting around a heavier
body(Let’s stick with a circular orbit)
Actually both bodies orbit around the barycenter
x
y W
mM
D
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
W
Mx
y
m
D
Two-Body Problem (3)
Force Balance:Body “M” Body “m”
02
2 D
GMmMxM 0
22
D
GMmmxm
W
x
y
m
D
Mm xxD
xmxM(<0)
M
CentrifugalForceGravitational Force
Two-Body Problem (4)
Parameters: M – mass of large body m – mass of small body D – distance between bodies
Unknowns: xM – x-coordinate of large body xm – x-coordinate of small body W – angular velocity of coordinate system
Three equations, three unknowns
Mm xxD
02
2 D
GMmMxM
02
2 D
GMmmxm
In case anybody wants to see the math …
mmm xM
mMx
M
mxD
02
2 D
GMmMxM
02
2 D
GMmmxm
Mm xxD 0
22 D
GmxM 0
22
D
GMxm
222
Dx
GM
Dx
Gm
mM
mM x
M
x
m Mm Mxmx
DmM
Mxm
DmM
mxM
32
2
D
mMG
Dx
GM
m
Two-Body Problem (5)
Solutions:
DmM
Mxm
DmM
mxM
3
2
D
mMG
W
x
y
m
DxmxM(<0)
M
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
W
x
y
mM
(x,y)
Lagrange Points (2)
Sum of Forces: particle at (x,y) of mass “ m ”
x-direction:
y-direction:
02
23
2223
22
x
yxx
xxGm
yxx
xxGM
m
m
M
M
02
23
2223
22
y
yxx
yGm
yxx
yGM
mM
Lagrange Points (3)
Simplifications: Divide by mass “m” of particle Substitute for “W2” and divide by “G”
0
32
3222
322
D
mMx
yxx
xxm
yxx
xxM
m
m
M
M
03
23
2223
22
D
mMy
yxx
my
yxx
My
mM
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
mM
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
mM
Some Solutions: Note that “y” equation is uniquely
satisfied ify = 0 (but there may be nonzero solutions also)
03
23
2223
22
D
mMy
yxx
my
yxx
My
mM
Lagrange Points (6)
Lagrange Points (7)
Some Solutions: “x” equation when “y” is zero:
Note:
0
32
322
32
D
mMx
xx
xxm
xx
xxM
m
m
M
M
3323
2MMM xxxxxx ?
0
33
MMmm
MMmm
xxxxxxxxmMx
xxxxmDxxxxMD
Lagrange Points (8)
Oversimplification: Let “m” = 0 xM = 0
Solutions: Anywhere on the orbit of “m”
0
32
322
D
Mx
yx
Mx
0
32
322
D
My
yx
My
323
22 Dyx
0
32
3222
322
D
mMx
yxx
xxm
yxx
xxM
m
m
M
M
03
23
2223
22
D
mMy
yxx
my
yxx
My
mM
Lagrange Points (9)
Perturbation:
For the y = 0 case:
Mm 10 DxM Dxm 1
0111
113
3
DxDxDxDxx
DxDxD
DxDxD
0
33
MMmm
MMmm
xxxxxxxxmMx
xxxxmDxxxxMD
Lagrange Points (10)
For the y = 0 case (cont’d):– Lagrange Points L1,L2: near x = D:
0111
11 33
DxDxDxDxx
DxDxDDxDxD
1Dx
0111
125
255
D
DD
0
33
MMmm
MMmm
xxxxxxxxmMx
xxxxmDxxxxMD
L1, L2: y = 0, x = D(1 + )d
Collecting terms:
Multiplying things out:
01
11112
2
22
332222
3232
2221
253
57333
0111
125
255
D
DD
L1, L2 (2): y = 0, x = D(1 + )d
Balancing terms:
Linearizing:
22
332222
3232
2221
253
57333
Order of magnitude d3
Order of magnitude e
33 3
3
1
Lagrange Points (11)
Lagrange Point L3: y = 0, near x = –D
Since | e |, | d | << 1 < 2:
0112211
1122
1Dx
01211
1222
22
0111
11 33
DxDxDxDxx
DxDxDDxDxD
L3: y = 0, x = –D(1 + )d
Multiply out and linearize:
12
5
02214441
444
01211
1222
22
08844444
444
0125
Lagrange Points (12)
Summary for a Small Perturbation and y = 0:
Lagrange Points:
L1:
L2:
L3:
Mm 10 DxM Dxm 1
3
3
11 Dx
12
51Dx
3
3
11 Dx
Lagrange Points (13)
What if y is not zero? Mm 10 DxM Dxm 1
x
y
2q
1DR
Lagrange Points (14)
Polar Coordinates:
Mm 10 DxM
Dxm 1
x
y
2q
2sin,2cos DDR
0,RF
1DR
2sin1
,2cos1
D
DR
ereq
Lagrange Points (15)
Consider the forces in the R-direction:
From M:
From m:
Centrifugal force:
2
322 2sin2cos1
2cos1
DDD
DDGMFM
2
322 2sin12cos11
2cos11
DDD
DDGmFm
12DFC
R-direction Forces
Simplify: From M:
From m:
Centrifugal force:
2
3222 2sin2cos1
2cos1
D
GMFM
2
3222 2sin12cos11
2cos11
D
MGFm
1
12D
GMFC
R-direction Forces (2)
Linearize in d and e: From M:
From m:
Centrifugal force:
2cos2212cos331
2cos122
D
GM
D
GMFM
2
232 22212cos1221
2cos11
D
GM
D
MGFm
1
2D
GMFC
R-direction Forces (3)
Balance the Forces:
0122
2cos221222
D
GM
D
GM
D
GM
0122
2cos221
22
2cos2211
22
2cos23
3
2cos2122
1
Lagrange Points (16)
Consider the forces in the q-direction:
From M:
From m:
Centrifugal force is completely radial
2
322 2sin2cos1
2sin
DDD
DGMFM
2
322 2sin12cos11
2sin1
DDD
DGmFm
0CF
q-Direction Forces
Simplify:
2
322 2sin2cos1
2sin
DDD
DGMFM
2
322 2sin12cos11
2sin1
DDD
DGmFm
2
3222 2sin2cos1
2sin
D
GMFM
2
3222 2sin12cos11
2sin1
D
GMFm
q-Direction Forces (2)
Linearize in d and e:
22
2sin
2cos331
2sin
D
GM
D
GMFM
2
3222 2sin2cos1
2sin
D
GMFM
2
3222 2sin12cos11
2sin1
D
GMFm
232
232
2cos22
2sin
212cos1221
2sin1
D
GM
D
GMFm
q-Direction Forces (3)
Simplifying further:
2
2sin
D
GMFM
322
322
23
222
sin8
2sin
sin222
2sin
sincos122
2sin
D
GM
D
GM
D
GMFm
2
2sin
D
GMFM
232 2cos22
2sin
D
GMFm
q-Direction Forces (4)
Balance the two forces:
0
sin8
2sin2sin322
D
GM
D
GM
2
2sin
D
GMFM
32 sin8
2sin
D
GMFm
322 sin8
2sin2sin
D
GM
D
GM
1sin8 3
2
1sin
306
60
32
Lagrange Points (17)
Solving for our radius:
3
2cos2122
1
1DR
603
2
12
281
32
121
22
11
32cos21
22
11
DD
DR
Lagrange Points (13 redux)
What if y is not zero? Mm 10 DxM Dxm 1
x
y
602
12
281DR
Lagrange Points (18):
Summary for zero y: Between M and m: L1:
On the opposite side of m from M: L2:
On the opposite side of M from m: L3:
Mm 10 DxM Dxm 1
3
3
11 Dx
3
3
11 Dx
12
51Dx
Lagrange Points (19):
Summary for nonzero y: Ahead of m in orbit: L4:
Behind m in orbit: L5:
602
602
Mm 10 DxM Dxm 1
12
281DR
12
281DR
Summary of Lagrange Points
(courtesy of Wikipedia, “Lagrange Points”)
(What is wrongwith this picture?)
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