mechanics of rigid bodies
DESCRIPTION
Mechanics of Rigid Bodies. 4.1. Rigid body Rigid body : a system of mass points subject to the holonomic constraints that the distances between all pairs of points remain constant throughout the motion If there are N free particles, there are 3N degrees of freedom - PowerPoint PPT PresentationTRANSCRIPT
Mechanics of Rigid Bodies
Rigid body
• Rigid body: a system of mass points subject to the holonomic constraints that the distances between all pairs of points remain constant throughout the motion
• If there are N free particles, there are 3N degrees of freedom
• For a rigid body, the number of degrees of freedom is reduced by the constraints expressed in the form:
• How many independent coordinates does a rigid body have?
4.1
ijij cr
The independent coordinates of a rigid body
• Rigid body has to be described by its orientation and location
• Position of the rigid body is determined by the position of any one point of the body, and the orientation is determined by the relative position of all other points of the body relative to that point
4.1
The independent coordinates of a rigid body
• Position of one point of the body requires the specification of 3 independent coordinates
• The position of a second point lies at a fixed distance from the first point, so it can be specified by 2 independent angular coordinates
4.1
0
The independent coordinates of a rigid body
• The position of any other third point is determined by only 1 coordinate, since its distance from the first and second points is fixed
• Thus, the total number of independent coordinates necessary do completely describe the position and orientation of a rigid body is 6
4.1
0
Orientation of a rigid body
• The position of a rigid body can be described by three independent coordinates,
• Therefore, the orientation of a rigid body can be described by the remaining three independent coordinates
• There are many ways to define the three orientation coordinates
• One common ways is via the definition of direction cosines
4.1
Direction cosines
• Direction cosines specify the orientation of one Cartesian set of axes relative to another set with common origin
• Orthogonality conditions:
4.1
333231
232221
131211
cosˆcosˆcosˆ'ˆ
cosˆcosˆcosˆ'ˆ
cosˆcosˆcosˆ'ˆ
kjik
kjij
kjii
1'ˆ'ˆ'ˆ'ˆ'ˆ'ˆˆˆˆˆˆˆ
0'ˆ'ˆ'ˆ'ˆ'ˆ'ˆˆˆˆˆˆˆ
kkjjiikkjjii
ikkjjiikkjji
Orthogonality conditions
• Performing similar operations for the remaining 4 pairs we obtain orthogonality conditions in a compact form:
4.1
'ˆ'ˆ ii)cosˆcosˆcosˆ()cosˆcosˆcosˆ( 131211131211 kjikji
'ˆ'ˆ ji
)cosˆcosˆcosˆ()cosˆcosˆcosˆ( 232221131211 kjikji
132
122
112 coscoscos 1
231322122111 coscoscoscoscoscos 0
ikl
lkli
3
1
coscos
Orthogonal transformations
• For an arbitrary vector
• We can find components in the primed set of axes as follows:
• Similarly
4.1
321ˆˆˆ GkGjGiG
3211ˆ'ˆˆ'ˆˆ'ˆ'ˆ' GkiGjiGiiGiG
3131211
2131211
1131211
ˆ)cosˆcosˆcosˆ(
ˆ)cosˆcosˆcosˆ(
ˆ)cosˆcosˆcosˆ(
Gkkji
Gjkji
Gikji
313212111 coscoscos GGG
3332321313
3232221212
coscoscos'
coscoscos'
GGGG
GGGG
Orthogonal transformations
• Therefore, orthogonal transformations are defined as:
• Orthogonal transformations can be expressed as a matrix relationship with a transformation matrix A
• With orthogonality conditions imposed on the transformation matrix A
4.2
ijijj
jiji aGaG cos ;'3
1
AGG '
ikl
lkliaa
3
1
Properties of the transformation matrix
• Introducing a matrix inverse to the transformation matrix
• Let us consider a matrix element
• Orthogonality conditions
4.3
jlk
klkjaa
3
1
1AA 1ki
lliklaa
3
1
3
1
3
1k lliklkj aaa
3
1
3
1k lliklkj aaa
3
1
3
1l kklkjli aaa
3
1ljllia
3
1kkikja
jia
ija
ija
AA~1
Properties of the transformation matrix
• Calculating the determinants
• The case of a negative determinant corresponds to a complete inversion of coordinate axes and is not physical (a.k.a. improper)
4.3
1~ AA 1~ AAAA 1AA ~
AA~
AA~
AA2
A 1 1
1 A
Properties of the transformation matrix
• In a general case, there are 9 non-vanishing elements in the transformation matrix
• In a general case, there are 6 independent equations in the orthogonality conditions
• Therefore, there are 3 independent coordinates that describe the orientation of the rigid body
4.14.2
ikl
lkli
3
1
coscos1'ˆ'ˆ'ˆ'ˆ'ˆ'ˆ
0'ˆ'ˆ'ˆ'ˆ'ˆ'ˆ
kkjjii
ikkjji
333231
232221
131211
aaa
aaa
aaa
A
Example: rotation in a plane
• Let’s consider a 2D rotation of a position vector r
• The z component of the vector is not affected, therefore the transformation matrix should look like
• With the orthogonality conditions
• The total number of independent coordinates is
4 – 3 = 1
4.2
100
0
0
2221
1211
aa
aa
A
jlk
klkjaa
2
1
2,1, lj
Example: rotation in a plane
• The most natural choice for the independent coordinate would be the angle of rotation, so that
• The transformation matrix
4.2
33
212
211
'
cossin'
sincos'
xx
xxx
xxx
100
0cossin
0sincos
A
Example: rotation in a plane
• The three orthogonality conditions
• They are rewritten as
4.2
0cossinsincos
1cossin
1sincos22
22
0
1
1
22211211
22221212
21211111
aaaa
aaaa
aaaa
Example: rotation in a plane
• The 2D transformation matrix
• It describes a CCW rotation of the coordinate axes
• Alternatively, it can describe a CW rotation of the same vector in the unchanged coordinate system
4.2
100
0cossin
0sincos
A
The Euler angles
• In order to describe the motion of rigid bodies in the canonical formulation of mechanics, it is necessary to seek three independent parameters that specify the orientation of a rigid body
• The most common and useful set of such parameters are the Euler angles
• The Euler angles correspond to an orthogonal transformation via three successive rotations performed in a specific sequence
• The Euler transformation matrix is proper
4.4
Leonhard Euler(1707 – 1783)
1A
The Euler angles
• First, we rotate the system around the z’ axis
• Then we rotate the system around the x’’ axis
4.4
'
'
'
100
0cossin
0sincos
z
y
x
''' Dxx
''
''
''
cossin0
sincos0
001
z
y
x
''CxX
The Euler angles
• Finally, we rotate the system around the Z axis
• The complete transformation can be expressed as a product of the successive matrices
4.4
Z
Y
X
100
0cossin
0sincos
BXx
BXx ''BCx 'BCDx 'Ax
'Axx
The Euler angles
• The explicit form of the resultant transformation matrix A is
• The described sequence is known as the x-convention
• Overall, there are 12 different possible conventions in defining the Euler angles
4.4
coscossinsinsin
sincoscoscoscossinsincossincoscossin
sinsinsincoscossincossinsincoscoscos
BCDA
Euler theorem
• Euler theorem: the general displacement of a rigid body with one point fixed is a rotation about some axis
• If the fixed point is taken as the origin, then the displacement of the rigid body involves no translation; only the change in orientation
• If such a rotation could be found, then the axis of rotation would be unaffected by this transformation
• Thus, any vector lying along the axis of rotation must have the same components before and after the orthogonal transformation:
4.6
RARR '
Euler theorem
• This formulation of the Euler theorem is equivalent to an eigenvalue problem
• With one of the eigenvalues
• So we have to show that the orthogonal transformation matrix has at least one eignevalue
• The secular equation of an eigenvalue problem is
• It can be rewritten for the case of
4.6
RAR 1RAR 0)( R1A
0)( R1A
1
1
0 1A 1
0 1A
Euler theorem
• Recall the orthogonality condition:
• n is the dimension of the square matrix
• For 3D case:
• It can be true only if
4.6
1AA ~
A1AAA~~~
A1A1A~~
)( A1A1A~~
)(
A1A1A~~
A1A1A~
1A
A11A~
~A11A A11A )( 1A1A
1A1A n)1(
1A1A 3)1( 1A1A
0 1A ... DEQ
Euler theorem
• For 2D case (rotation in a plane) n = 2:
• Euler theorem does not hold for all orthogonal transformation matrices in 2D: there is no vector in the plane of rotation that is left unaltered – only a point
• To find the direction of the rotation axis one has to solve the system of equations for three components of vector R:
• Removing the constraint, we obtain Chasles’ theorem: the most general displacement of a rigid body is a translation plus a rotation
4.6
0)( R1A
1A1A n)1( 1A1A Michel Chasles(1793–1880)
Infinitesimal rotations
• Let us consider orthogonal transformation matrices of the following form
• Here α is a square matrix with infinitesimal elements
• Such matrices A are called matrices of infinitesimal rotations
• Generally, two rotations do not commute
• Infinitesimal rotations do commute
4.8
RAARAA 1221
α1A
))(( 21 α1α1 2121 αα1α1α1 21 αα1
))(( 12 α1α1 1212 αα1α1α1 12 αα1
Infinitesimal rotations
• The inverse of the infinitesimal rotation:
• Proof:
• On the other hand:
• Matrices α are antisymmetric
• In 3D we can write:
• Infinitesimal change of a vector:
4.8
α1A 1
Aα1 )( ))(( α1α1 αα1αα11 1
AA~1 α1α1 ~ αα ~
0
0
0
12
13
23
dd
dd
dd
α
rrr 'd rrα1 )( αr
3
1
)(j
jiji rdr
Infinitesimal rotations
• is a differential vector, not a differential of a vector
• is normal to the rotation plane
4.8
0
0
0
12
13
23
dd
dd
dd
α
3
1
)(j
jiji rdr
12213
31132
23321
)(
)(
)(
drdrdr
drdrdr
drdrdrk
kjjijki drdr
3
1,
)(
)(drrd
)(d
)(d
dnrrd )(
dnd
)(
drdr )sin(
Infinitesimal rotations
• These matrices are called infinitesimal rotation generators
4.8
0
0
0
12
13
23
dd
dd
dd
α
d
nn
nn
nn
0
0
0
12
13
23
3
1iiind Mα
010
100
000
1M
000
001
010
;
001
000
100
32 MM
dnd
)(
3
1ikijkijji MMMMM
Example: infinitesimal Euler angles
• For infinitesimal Euler angles it can be rewritten as
4.8
coscossinsinsin
sincoscoscoscossinsincossincoscossin
sinsinsincoscossincossinsincoscoscos
A
10
1)(
01
d
ddd
dd
A α1
00
0)(
00
d
ddd
dd
α
0
0
0
12
13
23
dd
dd
dd
α
)(ˆˆ)( ddkdid
Rate of change of a vector
• Dividing by dt
4.9
AGG '
3
1
'j
jiji GaG
3
1kkijkij d
iii ddGdG ΩG'
ωGGG
dt
d
dt
d 'Ωω ddt ωGGG '
3
1
'j
ijjjiji daGdGadG
ijijijijij daa
3
1
)(j
ijjjijij daGdGda
3
1jijjjij GdG
3
1
3
1j kkijkji dGdG
3
1,kjkjijk
i
dG
dG
Example: the Coriolis effect
• Velocity vectors in the rotating and in the “stationary” systems are related as
• For the rate of change of velocity
• Rotating system: acceleration acquiresCoriolis and centrifugal components
4.10
GωGG '
Gaspard-GustaveCoriolis
(1792 - 1843)
rωvv rs
)( 2 rωωvωaa rsr
rωrr rs
srsss vωvv )()( ) (
) (rωvω
dt
rωvdr
r
r
)( 2 )( rωωvωv rrr
constω
Example: the Coriolis effect
• On the other hand
• This is the Lorentz acceleration
• What is the relationship between those two?
ωva rc
2
m
BqvaL
m
BqL
LL va
Kinetic energy of a system of particles
• Kinetic energy of a system of particles
• Introducing a center of mass:
• We can rewrite the coordinates in the center-of-mass coordinate system:
• Kinetic energy can be rewritten:
1.2
i
ii rmT 2)(2
1
ii
iii
m
rmR
Rrr ii
' Rrr ii
'
i
ii rmT 2)(2
1 i
iii RrRrm )'()'(2
1
i
ii
iii
ii RmRrmrm 22 )(2
1 )'()'(
2
1
M
rmi
ii
RMrmi
ii
Kinetic energy of a system of particles
• On the other hand
• In the center-of-mass coordinate system, the center of mass is at the origin, therefore
1.2
i
ii
iii
ii mRrmRrm 22 )(2
1 ')'(
2
1
MRrmdt
dRrm
iii
iii
22 )(2
1 ')'(
2
1
i
ii
iii
ii RmRrmrmT 22 )(2
1 )'()'(
2
1
RMrmi
ii
'' RMrm
iii
MRrmTi
ii22 )(
2
1 )'(
2
1
Kinetic energy of a system of particles
• Kinetic energy of the system of particles consists of a kinetic energy about the center of mass plus a kinetic energy obtained if all the mass were concentrated at the center of mass
• This statement can be applied to the case of a rigid body: Kinetic energy of a rigid body consists of a kinetic energy about the center of mass plus a kinetic energy obtained if all the mass were concentrated at the center of mass
• Recall Chasles’ theorem!
1.25.1
MRrmTi
ii22 )(
2
1 )'(
2
1
Kinetic energy of a system of particles
• Chasles: we can represent motion of a rigid body as a combination of a rotation and translation
• If the potential and/or the generalized external forces are known, the translational motion of center of mass can be dealt with separately, as a motion of a point object
• Let us consider the rotational part or motion
5.1
MRrmTi
ii22 )(
2
1 )'(
2
1
i
iiR rmT 2)'(2
1
Rotational kinetic energy
• Rate of change of a vector
• For a rigid body, in the rotating frame of reference, all the distances between the points of the rigid body are fixed:
• Rotational kinetic energy:
5.3
i
iiR rmT 2)'(2
1 i
iii rrm ''2
1
' )'()'( irisi rωrr
0)'( rir ')'( isi rωr
i
iii rωrωm )'()'(2
1
i
iiiR rωrωmT )'()'(2
1
i j
jijii rωrωm3
1
)'()'(2
1
i j nmnimjmn
lklikjkli rωrωm
3
1
3
1,
3
1,
''2
1
Rotational kinetic energy5.3
i j nmnimjmn
lklikjkliR rωrωmT
3
1
3
1,
3
1,
''2
1
i nmlkj
nilimkjmnjkli rrωωm3
1,,,,
''2
1 knlmnlkm
jjmnjkl
3
1
i nmlk
nilimkknlmnlkmi rrωωm3
1,,, '')(
2
1
i lkllikik
lkliki ωrrωrωm
3
1,
3
1,
22 '')'()(2
1
3
1,
2 ]'')'[(2
1
lklikikli
iilk rrrmωω
]'')'[( 2likikli
iikl rrrmI
3
1,2
1
lklklk ωIω
2
~Iωω
Inertia tensor and moment of inertia
• (3x3) matrix I is called the inertia tensor
• Inertia tensor is a symmetric matrix (only 6 independent elements):
• For a rigid body with a continuous distribution of density, the definition of the inertia tensor is as follows:
• Introducing a notation
• Scalar I is called the moment of inertia
5.3
2
~IωωRT
lkkl II
dVrrrIV
lkklkl ])[( 2
nω ω 2
~IωωRT
2
~ ωω Inn
2
2Iω
Inn~I
]'')'[( 2likikli
iikl rrrmI
Inertia tensor and moment of inertia
• On the other hand:
• Therefore
• The moment of inertia depends upon the position and direction of the axis of rotation
5.3
2
2IωTR
i
iiiR rωrωmT )'()'(2
1 i
iii rnrnmω
)'()'(2
2
i
iii rnrnmI )'()'(
Parallel axis theorem
• For a constrained rigid body, the rotation may occur not around the center of mass, but around some other point 0, fixed at a given moment of time
• Then, the moment of inertia about the axis of rotation is:
5.3
Rrr ii
'
i
iii rnrnmI )()(0
i
iii RrnRrnm ))'(())'((
ii
iii
iii
Rnm
Rnrnmrnm
2
2
)(
)()'(2)'(
CMI
MRnRnrmni
ii2)()()'(2
Parallel axis theorem
• Parallel axis theorem: the moment of inertia about a given axis is equal to the moment of inertia about a parallel axis through the center of mass plus the moment of inertia of the body, as if concentrated at the center of mass, with respect to the original axis
5.3
20 )( RnMII CM
Parallel axis theorem
• Does the change of axes affect the ω vector?
• Let us consider two systems of coordinates defined with respect to two different points of the rigid body: x’1y’1z’1 and x’2y’2z’2
• Then
• Similarly
5.1
RRR
12
sss RRR )( )()( 12 RωRR rs
11 )( )(
sss RRR )( )()( 21
RωRR rs
22 )( )(
RωRR ss
112 )()(
RωRR ss
221 )()(
0)( 21 Rωω
Parallel axis theorem
• Any difference in ω vectors at two arbitrary points must be parallel to the line joining two points
• It is not possible for all the points of the rigid body
• Then, the only possible case:
• The angular velocity vector is the same for all coordinate systems fixed in the body
5.1
0)( 21 Rωω
21 ωω
Example: inertia tensor of a homogeneous cube
• Let us consider a homogeneous cube of mass M and side a
• Let us choose the origin at one of cube’s corners
• Then
5.3
dVrrrIV
lkklkl ])[( 2
a a a
drdrdrrrrI0 0
321
0
112
11 ])[( a a a
drdrdrrr0 0
321
0
23
22 ])()[(
a a
drdrrra0
32
0
23
22 ])()[(
3
2 5a
3
2 2Ma 3322 II
Example: inertia tensor of a homogeneous cube
5.3
dVrrrIV
lkklkl ])[( 2
a a a
drdrdrrrI0 0
321
0
2112 ][
3
2
4
1
4
14
1
3
2
4
14
1
4
1
3
2
2MaI
a a
drdrrra0
21
0
21 ][4
5a
4
2Ma
322331132112 IIIIII
Angular momentum of a rigid body
• Angular momentum of a system of particles is:
• Rate of change of a vector
• For a rigid body, in the rotating frame of reference, all the distances between the points of the rigid body are fixed:
• Angular momentum of rigid body:
5.1
i
iii rrmL )(
irisi rωrr )()(
0)( rir
isi rωr )(
i
iii rωrmL ))((
i lk nminmlmnikjklij rωrmL
3
1,
3
1,
i nmlk
iminiklmnjkl mωrr3
1,,,
Angular momentum of a rigid body
• Rotational kinetic energy:
5.1
i nmlk
iminiklmnjklj mωrrL3
1,,,
i nmk
iminikkmjnknjm mωrr3
1,,
)(
3
1
2 ])[(k
ikijjkii
ik rrrmω
3
1kkjkωI
IωL
2
~Lω
2
~ωL
2
~IωωRT
Free rigid body
• For a free rigid body, the Lagrangian is:
• Recall
• Then
• We separate the Lagrangian into two independent parts and consider the rotational part separately
• Then, the equations of motion for rotation
5.55.6
MRrmTLi
ii22 )(
2
1 )'(
2
1 MRωIωlk
lklk2
3
1,
)(2
1
2
1
ii ddtω
CMlk
lklk TIL
3
1,2
1
i
R
i
R LL
dt
d
0
3
1
kkikI
dt
d 03
1
k
kikωIdt
d
Free rigid body
• Angular momentum of a free rigid body is constant
• In the system of coordinates fixed with the rotating rigid body, the tensor of inertia is a constant – it is often convenient to rewrite the equations of motion in the rotating frame of reference:
5.55.6
03
1,
3
1
llk
kiklk
kik LωωIdt
d
0dt
dLi 0dt
Ld
0
Lω
dt
Ld
dt
Ld
rs
03
1,,
3
1
lkmlmlk
iklk
kik ωωIωI
03
1
k
kikωIdt
d
Principal axes of inertia
• Inertia tensor is a symmetric matrix
• In a general case, such matrices can be diagonalized – we are looking for a system of coordinates fixed to a rigid body, in which the inertia tensor has a form:
• To diagonalize the inertia tensor, we have to find the solutions of a secular equation
5.4
3
2
1
00
00
00
I
I
I
I
0
333231
232221
131211
IIII
IIII
IIII
Principal axes of inertia
• Coordinate axes, in which the inertia tensor is diagonal, are called the principal axes of a rigid body; the eigenvalues of the secular equations are the components of the principal moment of inertia
• After diagonalization of the inertia tensor, the equations of motion for rotation of a free rigid body look like
• After diagonalization of the inertia tensor, the rotational kinetic energy a rigid body looks like
5.4
03
1,
kkjkj
ijkii IωωωI
3
1
2
2
1
iiiR ωIT
Principal axes of inertia
• To find the directions of the principal axes we have to find the directions for the eigenvectors ω
• When the rotation occurs around one of the principal axes In, there is only one non-zero component ωn
• In this case, the angular momentum has only one component
• In this case, the rotational kinetic energy has only one term
5.4
knkkk ωIL
22
1 23
1
2 nn
iiiinR
ωIωIT
Stability of a free rotational motion
• Let us choose the body axes along the principal axes of a free rotating rigid body
• Let us assume that the rotation axis is slightly off the direction of one of the principal axes (α - small parameter):
• Then, the equations of motion
5.6
332211ˆˆˆ iiiωω
03
1,
kkjkj
ijkii IωωωI
0)(
0)(
0)(
122133
313122
233211
IIωωωI
IIωωωI
IIωωωI
0)(
0)(
0)(
122133
313122
233211
IIωI
IIωI
IIωI
Stability of a free rotational motion5.6
0)(
0)(
0)(
122133
313122
233211
IIωI
IIωI
IIωI
0
0
0
3
12213
2
31312
1
I
IIω
I
IIω
ω
0))(( 312132
21
)3(2)3(2
1
IIIIII
ω
constω
0)3(2)3(2 K ))(( 312132
21 IIIIII
ωK
Stability of a free rotational motion
• The behavior of solutions of this equation depends on the relative values of the principal moments of inertia
• Always stable
• Exponentially unstable
5.6
0)3(2)3(2 K ))(( 312132
21 IIIIII
ωK
0K 2K3121 ; IIII
3121 ; IIII
0)3(22
)3(2
)cos( )3(2)3(2)3(2 tA
0K 2K213 III
312 III
0)3(22
)3(2 teA )3(2)3(2
teA )3(2)3(2
Classification of tops
• Depending on the relative values of the principle values of inertia, rigid body can be classified as follows:
• Asymmetrical top:
• Symmetrical top:
• Spherical top:
• Rotator:
321 III
321 III
321 III
0;0 321 III
Example: principal axes of a uniform cube
• Previously, we have found the inertia tensor for a uniform cube with the origin at one of the corners, and the coordinate axes along the edges:
• The secular equation:
3
2
4
1
4
14
1
3
2
4
14
1
4
1
3
2
2MaI
03
2
483
2
12
11 2242222
I
MaMaaMI
MaI
Ma
0
3
2
44
43
2
4
443
2
222
222
222
IMaMaMa
MaI
MaMa
MaMaI
Ma
Example: principal axes of a uniform cube
• To find the directions of the principal axes we have to find the directions for the eigenvectors
• Let us consider
03
2
483
2
12
11 2242222
I
MaMaaMI
MaI
Ma
12
11 2
1
MaI
6;
12
11 2
3
2
2
MaI
MaI
6
2
3
MaI
33
23
13
3
ω
ω
ω
ω333 1ωIω I
Example: principal axes of a uniform cube
13
2
33
2
23
2
13
2
6443
2ω
Maω
Maω
Maω
Ma
23
2
33
2
23
2
13
2
643
2
4ω
Maω
Maω
Maω
Ma
33
2
33
2
23
2
13
2
63
2
44ω
Maω
Maω
Maω
Ma
12
33
23
33
13 ω
ω
ω
ω
12
33
23
33
13 ω
ω
ω
ω
233
23
33
13 ω
ω
ω
ω
2313 ωω
3313 ωω 332313 ωωω
Free symmetrical top
• For a free symmetrical top:
5.6
03
1,
kkjkj
ijkii IωωωI 0)(
0)(
0)(
122133
313122
233211
IIωωωI
IIωωωI
IIωωωI
321 III
0
0)(
0)(
33
313121
133211
ωI
IIωωωI
IIωωωI
constω
I
IIωωω
I
IIωωω
3
1
31312
1
31321
)(
)(
2
1
31311
)(
I
IIωωω 2
1ω tAω
tAω
sin
cos
2
1
Motion of non-free rigid bodies
• How to tackle rigid bodies that move in the presence of a potential or in an open system with generalized forces (torques)?
• Many Lagrangian problems of such types allow separation of the Lagrangians into two independent parts: the center-of-mass and the rotational
• For the non-Lagrangian (open) systems, we modify the equations of motion via introduction of generalized forces (torques) N:
5.5
ikkjkj
ijkii NIωωωI
3
1,
Heavy symmetrical top with one point fixed
• For this problem, it is convenient to use the Euler angles as a set of independent variables
• Let us express the components of ω as functions of the Euler angles
• The general infinitesimal rotation associated with ω can be considered as consisting of three successive infinitesimal rotations with angular velocities
5.74.9
ωωω ;;
ωωωω
Heavy symmetrical top with one point fixed
4.9
0
0
A
z
y
x
'''' AxBCDxBCxBXx
0
0
coscossinsinsin
sincoscoscoscossinsincossincoscossin
sinsinsincoscossincossinsincoscoscos
A
cos
sincos
sinsin
z
y
x
Heavy symmetrical top with one point fixed
4.9
0
0
B
z
y
x
'''' AxBCDxBCxBXx
0
0
0
sin
cos
z
y
x
100
0cossin
0sincos
B
Heavy symmetrical top with one point fixed
4.9
'''' AxBCDxBCxBXx
0
0
z
y
x
0
0
0
sin
cos
z
y
x
cos
sincos
sinsin
z
y
x
Heavy symmetrical top with one point fixed
4.9
z
y
x
0
0
0
sin
cos
z
y
x
cos
sincos
sinsin
z
y
x
cos
sinsincos
cossinsin
Heavy symmetrical top with one point fixed
• The Lagrangian:
• Using the Euler angles
5.7
VTL
RotationnTranslatio TTT 22
)( 233
22
211 II
dVgrV dVrg
RMrmi
ii
MRg
cosgRMV
21 II
Heavy symmetrical top with one point fixed
5.7
cosgRMV
cos
sinsincos
cossinsin
2
)cos( 23
I
2
)cossinsin()sinsincos( 22
1
I
2
)(
2
22
211
233
IIT
2
sin
2
)cos( 222
1
2
3
II
Heavy symmetrical top with one point fixed
• The Lagrangian is cyclic in two coordinates
• Thus, we have two conserved generalized momenta
5.7
cos
2
sin
2
)cos( 222
1
2
3 gRMIIL
0;0
LL
aIconstIL
p
bIconstIIL
p
13
12
12
3
)cos(
)sin()coscos(
Heavy symmetrical top with one point fixed
• The Lagrangian does not contain time explicitly
• Thus, the total energy of the system is conserved
• To solve the problem completely, we need three additional quadratures
• We will look for them, using the conserved quantities
5.7
cos
2
sin
2
)cos( 222
1
2
3 gRMIIL
0t
L
constgRMIIE
cos
2
sin
2
)cos( 222
1
2
3
Heavy symmetrical top with one point fixed
• One variable only: we can find all the quadratures!
5.7
aII 13 )cos(
bIII 12
12
3 )sin()coscos(
cos313 IaII
bIIIaII 12
12
312
3 )sin()coscos()cos(
ba 2sincos
2sin
cosab )(1 f
cos)(13
1 faI
I )(2 f
cos
2
sin)(
2
))(cos)(( 2221
1
221
3 gRMf
Iff
IE
Heavy symmetrical top with one point fixed
• We have an equivalent 1D problem with an effective potential!
5.7
cos
2
sin)(
2
))(cos)(( 2221
1
221
3 gRMf
Iff
IE
cos
2
sin)(
2
))(cos)((
2
2211
2213
21 RMg
fIffIIE
cos2
sin)(
2
))(cos)(()(
2211
2213 RMg
fIffIVeff
aII 13 )cos(
cossin2
)cos(
2
)(2
21
3
21 gRM
abI
I
aI
cos
sin2
)cos()('
2
21 gRM
abIVeff
Heavy symmetrical top with one point fixed
• In the most general case, the integration involves elliptic functions
• Effective potential is a function with a minimum: motion in θ is bound by two values
5.7
)('2
'2
1 effV
IE
1
2 )](''[2
I
VE
dt
d eff
)](''[/2)(
1
effVEI
dt
Heavy symmetrical top with one point fixed
• When θ is at its minimum, we have a precession
• Otherwise, the top is bobbing: nutation
• The shape of the nutation trajectory depends on the behavior of the time derivative of φ
5.7
2sin
cosab
Charged rigid body in an electromagnetic field
• Let us consider the following vector potential (C – constant vector)
• How is magnetic field related to vector C?
5.9
iiiii
ziyixii Arqrqrrrm
L )()(2
)( 2,
2,
2,
AB
3
1,
)(kj
kjijkii rCrCA rCA
nn AB )(
m
i
imnmi r
A
3
1,
3
1,,, mkji m
kjijknmi r
rC
3
1,,
)(mkj m
kjmjnkmknj r
rC
Charged rigid body in an electromagnetic field
• Constant magnetic field
5.9
3
1,,
)(mkj m
kjmjnkmknjn r
rCB
3
1
3
1 m m
nm
m m
mn r
rC
r
rC
3
1
3m
mnmn CC nC2 nB
2
BC
2
rBA
i
iiiii
ziyixii rBrqrq
rrrmL
2)(
2
)( 2,
2,
2,
Charged rigid body in an electromagnetic field
• Let us assume a uniform charge/mass ratio
• Recall rotational kinetic energy
5.9
i
iiiii
ziyixii rBqrrq
rrrmL
2)(
2
)( 2,
2,
2,
)()( acbcba
i
iii
rBqr
2
iiii
i
i mrrm
Bq)(
2
i
iii mrrm
Bq)(
2
L
m
Bq
2
2
~LωRT
2
~)/( LBmq
Radius of gyration
• FYI: radius of gyration is
5.4
M
IR 0
20MRI
2
2IωTR 2
)( 20ωRM