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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
• For a rigid body, the number of degrees of freedom
is reduced by the constraints expressed in the form:
rij  cij
• How many independent coordinates does a rigid
body have?
4.1
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
0
4.1
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
0
4.1
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
iˆ'  iˆ cos 11  ˆj cos 12  kˆ cos 13
ˆj '  iˆ cos   ˆj cos   kˆ cos 
21
22
23
kˆ'  iˆ cos  31  ˆj cos  32  kˆ cos  33
• Orthogonality conditions:
iˆ  ˆj  ˆj  kˆ  kˆ  iˆ  iˆ' ˆj '  ˆj 'kˆ'  kˆ'iˆ'  0
iˆ  iˆ  ˆj  ˆj  kˆ  kˆ  iˆ'iˆ'  ˆj ' ˆj '  kˆ'kˆ'  1
4.1
Orthogonality conditions
iˆ'iˆ' 
 (iˆ cos11  ˆj cos12  kˆ cos13 )  (iˆ cos11  ˆj cos12  kˆ cos13 )
 cos 2 11  cos 2 12  cos 2 13  1
iˆ' ˆj ' 
 (iˆ cos11  ˆj cos12  kˆ cos13 )  (iˆ cos21  ˆj cos22  kˆ cos23 )
 cos 11 cos  21  cos 12 cos  22  cos 13 cos  23  0
• Performing similar operations for the remaining 4
pairs we obtain orthogonality conditions in a compact
form:
3
 cos 
l 1
li
cos  lk   ik
4.1
Orthogonal transformations

• For an arbitrary vector G  iˆG1  ˆjG2  kˆG3
• We can find components in the primed set of axes

as follows:
G '  iˆ'G  iˆ'iˆG  iˆ' ˆjG  iˆ'kˆG
1
1
2
 (iˆ cos 11  ˆj cos 12  kˆ cos 13 )  iˆG1
 (iˆ cos   ˆj cos   kˆ cos  )  ˆjG
11
12
13
2
 (iˆ cos 11  ˆj cos 12  kˆ cos 13 )  kˆG3
 cos 11G1  cos12G2  cos13G3
• Similarly
G2 '  cos  21G1  cos  22G2  cos  23G3
G3 '  cos  31G1  cos  32G2  cos  33G3
3
4.2
Orthogonal transformations
• Therefore, orthogonal transformations are defined
as:
3
Gi '   aijG j ;
aij  cos ij
j 1
• Orthogonal transformations can be expressed as a
matrix relationship with a transformation matrix A
G'  AG
• With orthogonality conditions imposed on the
transformation matrix A
3
a a
l 1
li lk
  ik
4.3
Properties of the transformation matrix
• Introducing a matrix inverse to the transformation
matrix
3
1
AA  1
a
l 1
kl
ali   ki
• Let us consider a matrix element a ij 
3
a 
k 1
kj
ki
3
3
3
  3

  3




   akj   akl ali     akj akl ali    ali   akj akl  


k 1 
 l 1
  k 1 l 1
l 1 
k 1



3
  ali jl  a ji  aij
l 1
~
1
A A
3
3
• Orthogonality conditions
a
k 1
kj
akl   jl
4.3
Properties of the transformation matrix
~
A  A 1
~
AA  AA 1
~
AA  1
• Calculating the determinants
2
~
~
AA  A A  A A  A  1  1
 A  1
• The case of a negative determinant
corresponds to a complete inversion
of coordinate axes and is not
physical (a.k.a. improper)
Properties of the transformation matrix
• In a general case, there are 9 non-vanishing
elements in the transformation matrix
 a11 a12
A  a21 a22
 a31 a32
a13 
a23 
a33 
• In a general case, there are 6 independent equations
in the orthogonality conditions
3
 cos 
l 1
li
cos  lk   ik
iˆ' ˆj '  ˆj 'kˆ'  kˆ'iˆ'  0
iˆ'iˆ'  ˆj ' ˆj '  kˆ'kˆ'  1
• Therefore, there are 3 independent coordinates that
describe the orientation of the rigid body
4.1
4.2
4.2
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
 a11 a12

A  a21 a22
 0
0
3
• With the orthogonality conditions
a
k 1
0

0
1
kj
akl   jl
j , l  1,2
• The total number of independent
coordinates is
4–3=1
4.2
Example: rotation in a plane
• The most natural choice for the independent
coordinate would be the angle of rotation, so that
x1 '  x1 cos   x2 sin 
x2 '   x1 sin   x2 cos 
x3 '  x3
• The transformation matrix
 cos 
A   sin 
 0
sin 
cos 
0
0
0
1
4.2
Example: rotation in a plane
• The three orthogonality conditions
a11a11  a21a21  1
a12a12  a22a22  1
a11a12  a21a22  0
• They are rewritten as
cos 2   sin 2   1
sin 2   cos 2   1
cos  sin   sin  cos   0
4.2
Example: rotation in a plane
 cos 
• The 2D transformation matrix
A   sin 
• It describes a CCW rotation of
 0
the coordinate axes
• Alternatively, it can describe
a CW rotation of the same
vector in the unchanged
coordinate system
sin 
cos 
0
0
0
1
4.4
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
A 1
Leonhard Euler
(1707 – 1783)
4.4
The Euler angles
• First, we rotate the system around the z’ axis
 cos  sin  0  x'

 
x' '  Dx'   sin  cos  0  y '
 0
0
1  z ' 
• Then we rotate the system around the x’’ axis
0
1
X  Cx' ' 0 cos 

0  sin 
0   x' '



sin    y' '
cos    z ' ' 
4.4
The Euler angles
• Finally, we rotate the system around the Z axis
 cos

x  BX   sin 
 0
sin 
cos
0
0  X 



0  Y 
1  Z 
• The complete transformation can be expressed as a
product of the successive matrices
x  BX  BCx' '  BCDx'  Ax '
x  Ax '
4.4
The Euler angles
• The explicit form of the resultant transformation
matrix A is
A  BCD 
 cos cos   cos  sin  sin 
  sin  cos   cos  sin  cos

sin  sin 
cos sin   cos  cos  sin 
 sin  sin   cos  cos  cos
 sin  cos 
sin  sin  
cos sin  
cos  
• The described sequence is known as the xconvention
• Overall, there are 12 different possible conventions
in defining the Euler angles
4.6
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:
R '  AR  R
4.6
Euler theorem
AR  R
( A  1)R  0
AR  1R
• This formulation of the Euler theorem is equivalent
to an eigenvalue problem
( A  1)R  0
• With one of the eigenvalues
 1
• So we have to show that the orthogonal
transformation matrix has at least one eignevalue
• The secular equation of an eigenvalue problem is
A  1  0
• It can be rewritten for the case of
A 1  0
 1
 1
4.6
Euler theorem
~
• Recall the orthogonality condition: A  1 AA  1
~
~
~ ~
~
~
~
( A  1) A  1  A
AA  A  1  A
( A  1) A  1  A
~
~
~
~
A 1 A  1 A
A 1 A  1 A
A 1  1 A
~
A 1  1 A
A 1  1 A
A  1   (A  1)
A  1  (1) A  1
n
• n is the dimension of the square matrix
• For 3D case:
A  1  (1)3 A  1
• It can be true only if A 1  0
A 1   A 1
Q.E.D.
4.6
Euler theorem
• For 2D case (rotation in a plane) n = 2:
A  1  (1) A  1
n
A 1  A 1
Michel Chasles
(1793–1880)
• 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:
( A  1)R  0
• Removing the constraint, we obtain Chasles’
theorem: the most general displacement of a rigid
body is a translation plus a rotation
4.8
Infinitesimal rotations
• Let us consider orthogonal transformation matrices
of the following form
A  1 α
• Here α is a square matrix with infinitesimal elements
• Such matrices A are called matrices of infinitesimal
rotations
• Generally, two rotations do not commute
A1A2R  A2 A1R
• Infinitesimal rotations do commute
(1  α1 )(1  α 2 )  1  α11  1α 2  α1α 2 1  α1  α 2
(1  α 2 )(1  α1 )  1  α 2 1  1α1  α 2α1  1  α 2  α1
4.8
Infinitesimal rotations
• The inverse of the infinitesimal rotation: A 1  1  α
• Proof:(1  α ) A  (1  α )(1  α )  1  α1  1α  αα  1
~
• On the other hand: A 1  A
~ α
~  α
1α  1 α
• Matrices α are antisymmetric
• In 3D we can write:
 0
α   d3
 d 2
d3
0
 d1
 d 2 
d1 
0 
• Infinitesimal change of a vector:
dr  r'r  (1  α )r  r  αr
3
(dr )i    ij rj
j 1
4.8
Infinitesimal rotations
3
(dr )i    ij rj
j 1
(dr )1  r2 d3  r3d 2
(dr ) 2  r3d1  r1d3
d3  d 2 
 0


α   d3
0
d1 
 d 2  d1
0 
3
(dr )i    ijk rj dk
(dr )3  r1d 2  r2 d1
j , k 1
  
dr  r  (d)

• (d) is a differential vector, not a
differential of a vector
• From the diagram: dr  (r sin  )d

  
dr  (r  n )d


nd  (d)
• (d) is normal to the rotation plane
4.8
Infinitesimal rotations
 d 2   0
 n


0
d1   3
 d1
0   n2
0 0 0
3


α  d  ni M i
M1  0 0 1
i 1
0  1 0
 0

α   d3
 d 2
d3
n3
0
 n1
 n2 
n1  d
0 


(d)  nd
0 0  1
 0 1 0
M 2  0 0 0 ; M 3   1 0 0
1 0 0 
 0 0 0
• These matrices are called infinitesimal
3
rotation generators
M i M j  M j M i    ijk M k
i 1
4.8
Example: infinitesimal Euler angles
A
 cos cos   cos  sin  sin 
  sin  cos   cos  sin  cos

sin  sin 
cos sin   cos  cos  sin 
 sin  sin   cos  cos  cos
 sin  cos 
sin  sin  
cos sin  
cos  
• For infinitesimal Euler angles it can be rewritten as
1
d  d

A   (d  d )
1

0
 d
0
d   1 α
1 
0
d  d


α   (d  d )
0

0
 d
0

d 
0 
 0
α   d3
 d 2

d3
0
 d1
 d 2 
d1 
0 
(d)  iˆd  kˆ(d  d )
4.9
Rate of change of a vector
G'  AG
3
Gi '   aijG j
j 1
aij   ij  daij   ij   ji
dGi '   aij dG j  G j daij    ( ij  daij )dG j  G j daij 
3
3
j 1
j 1
3


   ij dG j  G j ij   dGi    G j   ijk d k   dGi 
j 1 
k 1

j 1
3
   ijkG j d k
3
j ,k 1
dGi '  dGi  G  dΩi
 ij    ijk d k
3
3
k 1
• Dividing by dt
dG ' d G

 G ω
dt
dt
ωdt  dΩ
 ' G
  G ω
G
4.10
Example: the Coriolis effect
 G
 ' ω  G
G

ω  const
• Velocity vectors in the rotating and in the
“stationary” systems are related as
   
rs  rr  ω  r
   
vs  vr  ω  r
• For the rate of change of velocity
 


   d (vr  ω  r ) 
   
(vs ) s  (vs )r  ω  vs  
  ω  (vr  ω  r )
dt

r

    
 (vr ) r  2ω  vr  ω  (ω  r )
 
    
ar  as  2ω  vr  ω  (ω  r )
• Rotating system: acceleration acquires
Coriolis and centrifugal components
Gaspard-Gustave
Coriolis
(1792 - 1843)
Example: the Coriolis effect
 

ac  vr  2ω
• On the other hand


 qB
aL  v 
m
• This is the Lorentz acceleration
  
aL  v  L


qB
L 
m
• What is the relationship between those two?
1.2
Kinetic energy of a system of particles
1
2

• Kinetic energy of a system of particles T  2  mi (ri )
i


mi ri
• Introducing a center of mass:   mi ri


R i
 i

M
 mi ri  MR
 mi
i
i
• We can rewrite the coordinates in the center-ofmass coordinate system:
  
ri  ri ' R
  
ri  ri ' R
• Kinetic energy can be rewritten:
   
1
2 1
T   mi (ri )   mi (ri ' R)  (ri ' R)
2 i
2 i
  1
1
2
  mi (ri ' )   mi (ri 'R)   mi ( R ) 2
2 i
2 i
i
1.2
Kinetic energy of a system of particles
  1
1
2
2


T   mi (ri ' )   mi (ri 'R)   mi ( R)
2 i
2 i
i

 1  2
1
2
  mi (ri ' )  R   mi ri '  ( R)  mi
2 i
2
i
i
 d
 1  2
1
2
  mi (ri ' )  R   mi ri '  ( R) M
2 i
dt i
2
• On the other hand


 mi ri  MR
i


 mi ri '  MR'
i
• In the center-of-mass coordinate
system, the center of mass is at the
origin, therefore
1
1  2
2
T   mi (ri ' )  ( R) M
2 i
2
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
1  2
2
T   mi (ri ' )  ( R) M
2 i
2
1.2
5.1
5.1
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
1
TR   mi (ri ' ) 2
2 i
1
1  2
2
T   mi (ri ' )  ( R) M
2 i
2
5.3
Rotational kinetic energy
 
 
 
1
1
1
2
TR   mi (ri ' )   mi ri 'ri '   mi (ω  ri ' )  (ω  ri ' )
2 i
2 i
2 i


 
• Rate of change of a vector (r ' )  (r ' )  ω  r '
i
s
i
r
i
• For a rigid body, in the rotating frame of reference,
all the distances between the points of the rigid body

 
are fixed: 
(ri ' ) r  0
(ri ' ) s  ω  ri '
• Rotational kinetic energy:
3
 
 
1
 
 
1
TR   mi (ω  ri ' )  (ω  ri ' )   mi  (ω  ri ' ) j  (ω  ri ' ) j
2 i
2 i
j 1
3 
 3
 3

1
  mi      jkl ωk ri 'l    jmnωm ri 'n  


2 i
j 1   k ,l 1
m
,
n

1


5.3
Rotational kinetic energy


    jkl jmn 
1
TR   mi      jkl ωk ri 'l    jmnωm ri 'n  
j 1


2 i
j 1   k ,l 1
 m,n 1
    km l n   lm kn
3
1
   mi jkl jmnωk ωm ri 'l ri 'n
2 i j ,k ,l ,m,n1
3
1
   mi ( km l n   lm kn )ωk ωm ri 'l ri 'n
2 i k ,l ,m,n1
3
 3

1
2
2
  mi   (ωk ) (ri 'l )   ωk ri 'k ri 'l ωl 
2 i
k ,l 1
 k ,l 1

3
~ Iω
1 3
1
ω
2
  ωk ωl  mi [( ri ' )  kl  ri 'k ri 'l ]   ωk I kl ωl 
2 k ,l 1
2 k ,l 1
2
i
3
3
3
I kl   mi [( ri ' )  kl  ri 'k ri 'l ]
2
i
3
5.3
Inertia tensor and moment of inertia
~ Iω
ω
TR 
2
I kl   mi [( ri ' ) 2  kl  ri 'k ri 'l ]
i
• (3x3) matrix I is called the inertia tensor
• Inertia tensor is a symmetric matrix (only 6
independent elements):
I kl  I lk
• For a rigid body with a continuous distribution of
density, the definition of the inertia tensor is as
follows:
I  [( r ) 2   r r ]dV
kl

V
• Introducing a notation
kl
k l
2
~
~
Iω
ωIω ωn In ω

TR 

ω  ωn
2
2
2
~In
I

n
• Scalar I is called the moment of inertia
5.3
Inertia tensor and moment of inertia
Iω 2
TR 
2
• On the other hand:
ω2
 
 
1
TR   mi (ω  ri ' )  (ω  ri ' ) 
2
2 i
 
 
 mi (n  ri ' )  (n  ri ' )
i
• Therefore
 
 
I   mi (n  ri ' )  (n  ri ' )
i
• The moment of inertia depends upon the position
and direction of the axis of rotation
5.3
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:

 
  
   
I 0   mi (n  ri )  (n  ri )   mi (n  (ri ' R))  (n  (ri ' R))
i

i


  2
 
 
ri  ri ' R
  mi (n  ri ' )  2 mi (n  ri ' )  (n  R)
i
  2
  mi (n  R)
i
i
 I CM


 
  2
 2(n   mi ri ')  (n  R)  (n  R) M
i
5.3
Parallel axis theorem
I 0  I CM
  2
 M (n  R)
• 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
5.1
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 R  R  R
2
1





 
• Then ( R2 ) s  ( R1 ) s  ( R ) s  ( R1 ) s  ( R ) r  ω1  R



• Similarly ( R1 ) s  ( R2 ) s  ( R ) s


 
 ( R2 ) s  ( R) r  ω2  R


 
( R2 ) s  ( R1 ) s  ω1  R

 
(ω1  ω2 )  R  0


 
( R1 ) s  ( R2 ) s  ω2  R
5.1
Parallel axis theorem

 
(ω1  ω2 )  R  0
• 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:


ω1  ω2
• The angular velocity vector
is the same for all coordinate
systems fixed in the body
5.3
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
I kl   [( r ) 2  kl  rk rl ]dV
V
a a a
a a a
0 0 0
0 0 0
I11      [( r ) 2  r1r1 ]dr1dr2 dr3      [( r2 ) 2  (r3 ) 2 ]dr1dr2 dr3
5
2
2

a
2
Ma
 I 22  I 33
 a   [( r2 ) 2  (r3 ) 2 ]dr2 dr3 

3
3
0 0
a a
5.3
Example: inertia tensor of a
homogeneous cube
I kl   [( r ) 2  kl  rk rl ]dV
V
a a
a a a
a 5
Ma 2
I12      [r1r2 ]dr1dr2 dr3   a   [r1r2 ]dr1dr2  

4
4
0 0 0
0 0
I12  I 21  I13  I 31  I 23  I 32
 2
 3
 1
I  Ma 2 
 4
 1
 4
1

4
2
3
1

4
1
 
4
1
 
4
2 
3 
5.1
Angular momentum of a rigid body
• Angular momentum of a system of particles is:

 
L   mi (ri  ri )
i


 
• Rate of change of a vector (ri ) s  (ri ) r  ω  ri
• For a rigid body, in the rotating frame of reference,
all the distances between the points of the rigid body

 

are fixed:
 (r )  ω  r
(ri ) r  0
i s
i

  
• Angular momentum of rigid body: L   mi (ri  (ω  ri ))
i
 3
 3

L j   mi    jkl rik    lmnωm rin  
 k ,l 1

i
m
,
n

1


3

    jkl lmnrik rinωm mi
i
k ,l , m, n 1
5.1
Angular momentum of a rigid body
Lj  
i

i
3
 (
k , m, n 1
3

k ,l , m , n 1
 r r ωm mi
jkl lmn ik in
   jn km )rik rinωm mi
jm kn
3
3
  ωk  mi [( ri )  jk  rij rik ]   I jk ωk
2
k 1
k 1
i
L  Iω
• Rotational kinetic energy:
~
~
~
Lω
ωIω
ωL

TR 

2
2
2
5.5
5.6
Free rigid body
• For a free rigid body, the Lagrangian is:
3
1
1
1
1
2
2
2



ω
I
ω

(
R
)
M

L  T   mi (ri ' )  ( R) M

k kl l
2 k ,l 1
2
2 i
2
• Recall ωi dt  dΩi
1 3 
 T
• Then L    k I kl 
l
CM
2 k ,l 1
• We separate the Lagrangian into two independent
parts and consider the rotational part separately
• Then, the equations of motion for rotation
d  LR  LR

 

dt  
i
i 
d 3


  I ik  k   0
dt  k 1

d 3

  I ikk   0
dt  k 1

Free rigid body
d 3

  I ikk   0
dt  k 1

dLi
0
dt
5.5
5.6

dL
0
dt
• 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:


 dL   dL 
 
      L  0
 dt   dt 
 s  r
3
I
k 1
ik
 k 
d 3
 3
  I ikk     iklk Ll  0
dt  k 1
 k ,l 1
3

k ,l , m 1
I kl  0
ikl ml
5.4
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:  I1 0 0 
I   0
 0
I2
0
0 
I 3 
• To diagonalize the inertia tensor, we have to find the
solutions of a secular equation
I11  I
I12
I13
I 21
I 22  I
I 23
I 31
I 32
I 33  I
0
5.4
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
3
look like
I i i 

j , k 1
 j k I k  0
ijk
• After diagonalization of the inertia tensor, the
rotational kinetic energy a rigid body looks like
1 3
2
TR   I i ωi
2 i 1
5.4
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
Lk  I k ωk kn
• In this case, the rotational kinetic energy has only
one term
2
3
I n ωn
1
2
TR    in I i ωi 
2 i 1
2
5.6
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):

  1iˆ1   2iˆ2   3iˆ3
• Then, the equations of motion I i i 
I11  23 ( I 3  I 2 )  0
I 2 2  13 ( I1  I 3 )  0
I 3 3  12 ( I 2  I1 )  0
3

j , k 1
 j k I k  0
ijk
I11   2 3 ( I 3  I 2 )  0
I 22  1 3 ( I1  I 3 )  0
I 33  1 2 ( I 2  I1 )  0
5.6
Stability of a free rotational motion
1  0
I11   2 3 ( I 3  I 2 )  0
I 22  1 3 ( I1  I 3 )  0
I1  I 3
2  1 3
0
I2
I 33  1 2 ( I 2  I1 )  0
I 2  I1
3  1 2
0
I3
1  const
2(3)  2(3)
1
2
I 2 I3
( I1  I 2 )( I1  I 3 )  0
2(3)  2(3) K  0
K
12
I 2 I3
( I1  I 2 )( I1  I 3 )
5.6
Stability of a free rotational motion
2(3)  2(3) K  0
K
12
I 2 I3
( I1  I 2 )( I1  I 3 )
• The behavior of solutions of this equation depends
on the relative values of the principal moments of
2
inertia


   0
I1  I 2 ; I1  I 3
I1  I 2 ; I1  I 3
2(3)
K 0
2(3)
K  2
 2(3)  A2(3) cos( t   2(3) )
• Always stable
I 3  I1  I 2
I 2  I1  I 3
2


 2 ( 3)    2 ( 3)  0
K 0
• Exponentially unstable
K   2
 2(3)  A2(3)et
 2(3)  A2(3)et
Classification of tops
• Depending on the relative values of the principle
values of inertia, rigid body can be classified as
follows:
• Asymmetrical top: I1  I 2  I 3
• Symmetrical top: I1  I 2  I 3
• Spherical top: I1  I 2  I 3
• Rotator: I1  I 2  0; I 3  0
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:
 2
 3
 1
I  Ma 2 
 4
 1
 4
1

4
2
3
1

4
1
 
4
1
 
4
2 
3 
• The secular equation:
2Ma 2
I
3
Ma 2

4
Ma 2

4
Ma 2

4
2
2Ma
I
3
Ma 2

4
Ma 2

4
Ma 2

0
4
2
2Ma
I
3
2
2

11Ma
  2Ma
 M 2 a 4 Ma 2  2Ma 2


 I  
 I  

 I   0

8
4  3
 12
  3


2
Example: principal axes of a uniform
cube
2
11Ma 2
  2Ma 2
 M 2 a 4 Ma 2  2Ma 2


 I  
 I  

 I   0

8
4  3
 12
  3


2
2
2
11
Ma
Ma
11Ma
I2 
; I3 
I1 
12
6
12
• To find the directions of the principal axes we have
to find the directions for the eigenvectors
Ma 2
• Let us consider I 3 
6
Iω3  I 3 1ω3
13 


ω3  23 
33 
Example: principal axes of a uniform
cube
2Ma 2
Ma 2
Ma 2
Ma 2
13 
23 
33 
13
3
4
4
6
213
33
23

1
33
Ma 2
2Ma 2
Ma 2
Ma 2

13 
23 
33 
23
4
3
4
6
13 223


1
33 33
Ma 2
Ma 2
2Ma 2
Ma 2

13 
23 
33 
33
4
4
3
6
13  23

2
33 33
13  23
13  33
13  23  33
5.6
Free symmetrical top
I i i 
3

j , k 1
 j k I k  0
ijk
I11  23 ( I 3  I 2 )  0
I 2 2  13 ( I1  I 3 )  0
I 3 3  12 ( I 2  I1 )  0
• For a free symmetrical top:
I1  I 2  I 3
1  2
3 ( I 1  I 3 )
 2  1
I1
I11  23 ( I 3  I1 )  0
I1 2  13 ( I1  I 3 )  0
I 3 3  0
3 ( I 1  I 3 )
3  const
I1
2
 3 ( I 1  I 3 ) 
  1 2
1  1 
I1


1  A cos t
2  A sin t
5.5
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:
I i i 
3

j , k 1
 jk I k  Ni
ijk
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




      
  ;   ;   
5.7
4.9
4.9
Heavy symmetrical top with one point
fixed
x  BX  BCx' '  BCDx'  Ax'
x 
0
0 
 
0
0 


A
 y
 
 

z 


 
 
 
   sin  sin 
x
y   cos sin 
z   cos 
 cos cos   cos  sin  sin 
A   sin  cos   cos  sin  cos

sin  sin 
cos sin   cos  cos  sin 
 sin  sin   cos  cos  cos
 sin  cos 
sin  sin  
cos sin  
cos  
4.9
Heavy symmetrical top with one point
fixed
x  BX  BCx' '  BCDx'  Ax'
x 
 
 
 
 
 


B
0
0
y




 
z 
0
0
 
 
 
   cos
x
y   sin 
 cos
B   sin 
 0
sin 
cos
0
0
0
1
z  0
4.9
Heavy symmetrical top with one point
fixed
x  BX  BCx' '  BCDx'  Ax'
0
0
 
 
 x  0
 y  0
x   cos
 z  
z  0
y   sin 
x   sin  sin 
y   cos sin 
z   cos 
4.9
Heavy symmetrical top with one point
fixed




      
 sin  sin    cos 
 


   cos sin    sin  
 cos  





 x  0
 y  0
x   cos
 z  
z  0
y   sin 
x   sin  sin 
y   cos sin 
z   cos 
5.7
Heavy symmetrical top with one point

fixed

 m r  MR
i i
L  T V
2
2
2
T  TTranslation  TRotation  I1 (1  2 )  I 33
2
2
 
 
 
V    r  gdV   g   r dV   g  RM
• The Lagrangian:
• Using the Euler angles
V  gRM cos 
I1  I 2
i
5.7
Heavy symmetrical top with one point
 sin  sin    cos 


fixed
 

   cos sin    sin  


 cos  
2 

2
2
2
I 33
I1 (1  2 ) I 3 ( cos   )

T

2
2
2
( cos sin    sin  ) 2  ( sin  sin    cos ) 2
 I1
2
( cos   ) 2
2 sin 2    2
 I3
 I1
2
2
V  gRM cos 
5.7
Heavy symmetrical top with one point
fixed
( cos   ) 2
2 sin 2    2
L  I3
 I1
 gRM cos 
2
2
• The Lagrangian is cyclic in two coordinates
L
L
 0;
0


• Thus, we have two conserved generalized momenta
L
2
2



p    I 3 ( cos    cos  )  I1 ( sin  )  const  I1b

L
p 
 I 3 ( cos    )  const  I1a

5.7
Heavy symmetrical top with one point
fixed
( cos   ) 2
2 sin 2    2
L  I3
 I1
 gRM cos 
2
2
L
• The Lagrangian does not contain time explicitly
0
t
• Thus, the total energy of the system is conserved
( cos   ) 2
2 sin 2    2
E  I3
 I1
 gRM cos   const
2
2
• To solve the problem completely, we need three
additional quadratures
• We will look for them, using the conserved
quantities
5.7
Heavy symmetrical top with one point
fixed
I 3 ( cos   )  I1a
I 3  I1a  I 3 cos 
I 3 ( cos 2   cos  )  I1 ( sin 2  )  I1b
I 3 ( cos 2  )  ( I1a cos   I 3 cos 2  )  I1 ( sin 2  )  I1b
a cos   sin 2   b
I1
  a  f1 ( ) cos   f 2 ( )
I3
b  a cos 


sin 2 
 f1 ( )
( f1 ( ) cos   f 2 ( )) 2
f1 ( ) 2 sin 2    2
E  I3
 I1
 gRM cos 
2
2
• One variable only: we can find all the quadratures!
5.7
Heavy symmetrical top with one point
fixed
( f1 ( ) cos   f 2 ( )) 2
f1 ( ) 2 sin 2    2
E  I3
 I1
 gRM cos 
2
2
I1 2 I 3 ( f1 ( ) cos   f 2 ( )) 2 I1 f1 ( ) 2 sin 2 
E


 RMg cos 
2
2
2
• We have an equivalent 1D problem with an effective
potential!
I 3 ( f1 ( ) cos   f 2 ( )) 2 I1 f1 ( ) 2 sin 2 
Veff ( ) 

 RMg cos 
2
2
( I1a) 2 I1 (b  a cos  ) 2


 gRM cos 
2I3
2
I1 (b  a cos  ) 2
Veff ' ( ) 
 gRM cos  I ( cos   )  I a
3
1
2
5.7
Heavy symmetrical top with one point
fixed
I1 2
E' 
 Veff ' ( )
2
d
t ( )  
2 / I1[ E 'Veff ' ( )]
2[ E 'Veff ' ( )]
 d 
  
I1
 dt 
2
• 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
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
b  a cos 


sin 2 
5.9
Charged rigid body in an
electromagnetic field
 mi (ri , x 2  ri , y 2  ri , z 2 )
 


L  
 qi (ri )  qi (ri  A) 


2
i 

• Let us consider the following vector potential (C –
3
constant
 
  vector)

A  Cr
Ai  (C  r )i 

j , k 1
ijk
C j rk
• How is magnetic field related to vector C?


B   A

Bn  (  A)n 
rk
   nmi ijkC j

rm
i , j , k , m 1
3
Ai
 nmi

rm
m ,i 1
3
rk
( nj mk   nk mj )C j

rm
j , k , m 1
3
5.9
Charged rigid body in an
electromagnetic field
3
rk
Bn   ( nj mk   nk mj )C j
rm
j , k , m 1
3
rm 3
rn
 3Cn   Cm mn  2Cn  Bn
  Cn
  Cm
rm m1
rm
m 1
m 1

 
 B
 Br
C
A
2
2
3
• Constant magnetic field
 mi (ri , x  ri , y
L  

2
i 
2
2
 
 ri , z )
  B  ri  


 qi (ri )  qi  ri 

2


2
5.9
Charged rigid body in an
electromagnetic field
 
2
2
2
 mi (ri , x  ri , y  ri , z )



q
B

r
i 
L  
 qi (ri )  ri  i

2
2 
i 

 
 qi B  ri
qi B  
i ri  2  i 2m  (ri  ri mi )
i
• Let us assume a uniform charge/mass ratio


~
 
qB  (q / m)B
qB
L
L 

  (ri  ri mi ) 
2m
2m i
2
• Recall rotational kinetic energy
~L
ω
TR 
2
  
  
a  (b  c )  b  (c  a )
5.4
Radius of gyration
• FYI: radius of gyration is
I
R0 
M
I  MR0
2
Iω 2 M ( R0ω) 2
TR 

2
2
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