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					416. Rotations of a dumbbell equipped with ‘the leier constraint’
A. V. Rodnikov
Bauman Moscow State Technical University, Russia
e-mail: avrodnikov@yandex.ru
(Received: 23 September; accepted: 02 December)




Abstract. We consider a special space tethered system consisting of a dumbbell-shaped rigid body and a particle. The
particle coast along on the cable. The cable ends are placed in the dumbbell endpoints. We call such system ‘the system
with leier constraint (the Dutch term ‘leier’ means the rope with both fixed ends). We assume that the system mass center
moves along the circular orbit in the Newtonian Central Force Field. We study the dumbbell's relative motion caused by
the particle of small mass in the orbital frame of reference. We deduce a sufficient condition for librations of the dumbbell
about its stable equilibrium. We find a family of the dumbbell's asymptotic motions tending to librations about unstable
equilibrium. The surface of such asymptotic motions is an interstream separating the areas of the dumbbell's right-hand
and left-hand rotations. We deduce an equation of this surface.
Key words: space tethered system, leier constraint, asymptotic solution, circular orbit

Introduction                                                               a dumbbell, i.e. it is composed of particles with masses m1
                                                                           and m 2 connecting by weightless rod of length 2c.
    Space tethered systems are one of the most interesting
topics in dynamics. For the first time the motion of a
particle tethered to a spacecraft has been suggested in [1,
2]. Presently there are hundreds papers devoted to various
aspects of the motion of tethered satellites. In this paper we
study some generalization of the classic couple.
    We consider the system that moves in the Newtonian
Central Force Field and consists of a dumbbell-shaped
rigid body and a particle. The particle coasts along on the
cable with ends placed in the dumbbell endpoints. We call
such cable ‘a leier’. (the Dutch maritime term ‘leier’ means
the rope with both fixed ends).
    We assume the system mass center describes circular
orbit, the cable length is small in comparison with orbit
radius, the particle mass is small in comparison with the
dumbbell mass, the cable do not leave the orbit plane. We
study the dumbbell rotation caused by the small particle in
the orbital frame of reference. It is well known that the
dumbbell-shaped satellite has two types of relative
equilibria. There are the stable ‘vertical’ equilibria and
unstable ‘horizontal’ equilibria. We claim that the small
particle sufficiently influence the dumbbell relative motion                                                Fig. 1.
only if the dumbbell is initially quasi-horizontal. We prove
that if the system Jacobi’s integral less than some constant                    Without loss of generality, m2 ≥ m1 . Suppose the
then only librations about the ‘vertical’ equilibrium are                  particle m3 coast along on the cable with ends fixed to the
possible. We note that there exist a set of the dumbbell                   dumbbell endpoints (Fig.1). This cable can be called ‘a
relative motions tending to librations about the ‘horizontal’              leier’. Denote by 2a the cable length. Let C be the mass
equilibria. Factually, these asymptotic motions form the                   center for considered system and O1 be the attracting
surface being an interstream between areas of left-hand                    center. Suppose C moves along the circular orbit, i.e.
and right-hand rotations of the dumbbell. We deduce the                    O1C=r=const and the particles m1 , m 2 , m3 do not leave
equation of this interstream.
                                                                           the plane of this orbit. Moreover assume a<<r. Denote by
Designations and parameters                                                ϕ  the angle between O1C and the rod.
   Consider a mechanical system consisting of a rigid                           Evidently, the particle m3 cannot leave the ellipse with
body and a particle with mass m3 . Assume that the body is                 foci in the dumbbell endpoints. The ellipse has eccentricity

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                             VIBROMECHANIKA. JOURNAL OF VIBROENGINEERING. 2008 DECEMBER, VOLUME 10, ISSUE 4, ISSN 1392-8716
                                             416. ROTATIONS OF A DUMBBELL EQUIPPED WITH ‘THE LEIER CONSTRAINT’. A. V. RODNIKOV




e = c / a and semi-axises a and b = a 2 − c 2 . Let Oxy be
                                                                                                              1 − e 2 ( 1 − e µ cos γ )( ϕ ′ + 1 )2 +
a coordinate system with origin in the dumbbell midpoint
(see fig.1). Clearly, if x and y is the coordinates of the                                                    + 2( 1 − e 2 cos 2 γ )( ϕ ′ + 1 )γ ′ +
particle m3 the inequality
                                                                                                                    3                                 1 − e2
            x + dy − a ≤ 0 ; d = a / b
                2            2     2                             2    2
                                                                                 (1)             + 1 − e 2 γ ′2 −     ( 1 − e 2 ) sin 2γ sin 2ϕ +            ⋅
                                                                                                                    2                                   2
is valid. The motion of m3 is called the constrained one if
                                                                                                 ⋅ ( 3 cos 2ϕ cos 2γ + 1 − eµ cos γ ( 1 + 3 cos 2ϕ )) ≥ 0.
(1) is equality. In this case the coordinates of m3 can be
determine by formulae                                                                       The dumbbell rotations caused by the small particle
                x = a cos γ , y = b sin γ                       (2)
                                                                                                  Let the mass m3 be small in comparison with the
where γ is an eccentric anomaly of the mentioned ellipse.
If m3 moves inside the ellipse then we say that the motion                                  dumbbell mass, i.e. k << 1.
                                                                                                  It is well-known that there exist two types of
is the unconstrained one (or the free one).                                                 stationary motions of the dumbbell-shaped satellite. There
     Let           µ = ( m2 − m1 ) /( m2 + m1 )                and                          are ‘the vertical’ equilibria ( ϕ = 0 or ϕ = π ) and ‘the
ν = m3 /( m2 + m1 ) . Trivially, 0 < µ < 1 , 0 < e < 1 , ν > 0 ,                            horizontal’ equilibria ( ϕ = ±π / 2 ).
    It is clear that the dimensionless parameters µ, ν, e and                                     Obviously ‘the vertical’ equilibria are stable. The
the variables ϕ, γ determine the considered system                                          particle motion does not destroy these equilibria. Only
dynamics completely in the case of constrained motion.                                      some librations of the dumbbell about ‘vertical’ position
                                                                                            are possible in this case.
Lagrangian and Jacobi’s integral

    Lagrangian for relative motion of the considered
couple has a form [3,4]
                  L = L 2 + L1 + L 0             (3)
where

L2 =
     1
     2
            {
       ϕ ′ 2 + k [( 1 − 2 eµ cos γ + e 2 cos 2 γ )ϕ ′ 2 +                           }
{
+ 1 − e 2 ( 1 − 2 e µ cos γ )ϕ ′γ ′ + ( 1 − e 2 cos 2 γ )γ ′ 2                          }
           L1 = ke cos γ ( e cos γ − 2 µ )ϕ ′
                              3
                L 0 = −W = cos 2 ϕ +
                              2                                                                                            Fig. 2.
      9                3                3               
  + k  e 2 cos ϕ − e µ cos γ + e 2 cos 2γ −                                                     It can easily be checked that if the dumbbell is
      8                2                8                                                 ‘quasi-horizontal’ initially then the particle motion along
                −
                    3
                    4
                         [
                      e µ ( 1 − 1 − e 2 ) cos( 2ϕ − γ ) +
                                                                                            the leier force the upturning of the dumbbell. The further
                                                                                            motion of the dumbbell belongs to one of three types.
                [+ ( 1 + 1 − e ) cos( 2ϕ + γ )]+
                                   2
                                                                                            There are ‘the libratory motion’ about the ‘vertical’
                                                                                            equilibria, ‘the rotary motion’ about mass center, the
                +
                   3
                  16
                       [( 1 − 1 − e ) cos( 2ϕ − γ ) +
                                         2     2                                            complicated ‘tumbling motion’ consisting of libratory and
                                                                                            rotary segments.

                 [+ ( 1 + 1 − e ) cos( 2ϕ + γ )]}.
                                    2    2
                                                                                                 Let us remark that the dumbbell tends to librations
                                                                                            about its ‘horizontal’ equilibria for some singular initial
                                              ν                                             values of ( γ ,γ ′ ) .
                             k=                              .
                                  e (1 − µ 2 )
                                    2
                                                                                            A sufficient condition for the dumbbell libration
Hence we have Jacobi’s integral L2 + W = h.
      The       prime _' ’
                      ‘            denotes             the        derivative   w.r.t.             It is not hard to prove that if Jacobi’s integral constant
dimensionless time τ = G M r           1/ 2        1 / 2 −3 / 2
                                           t , where G is the                               h is smaller than h * = 3 / 8 ⋅ k ( 5 e 2 − 2 ) then only `the
gravity constant, M is the mass of the attracting center.                                   libratory motion' is possible. Consider a plot of W (Fig. 2).
                                                                                            We see a mountain country consisting of parallel ridges
The constrained motion condition                                                             ϕ = π / 2 + π k and valleys ϕ = π k , where k is integer.
                                                                                            The ridge is the sequence of ‘peaks’ γ = π k and saddle-
Note also that the constrained motion is possible only if
                                                                                            points   γ = π / 2 + π k . In the saddle-point              W = h* .
[3,4]

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                                   VIBROMECHANIKA. JOURNAL OF VIBROENGINEERING. 2008 DECEMBER, VOLUME 10, ISSUE 4, ISSN 1392-8716
                                      416. ROTATIONS OF A DUMBBELL EQUIPPED WITH ‘THE LEIER CONSTRAINT’. A. V. RODNIKOV




Therefore if h < h* then the dumbbell ‘cannot pass through                       where σ ( τ ,γ 1 , h2 ) is T-periodic function of τ . Thus if the
the ridge’ and rotations on complete angle are impossible.                       motion is constrained then
For instance, the libratory motion is observed for any                           D = D ( γ ′′( τ , ,γ 1 , h2 ), γ ′( τ , ,γ 1 , h2 ), γ ( τ , ,γ 1 , h2 ) = D1 ( τ ) is T-
initial value of ϕ and zero initial velocities if initial value                  periodic function of τ . (Here h2 depends on ( γ 1 , γ 1 ) ).                    ′
of γ is about π / 2 . It can be shown numerically that ‘the
rotary motion’ is guaranteed only if the initial value of γ ′                    The reduced equations’ solutions
is sufficiently big.
                                                                                      Solutions of (4) can be represented in a form
The motion equations reduction for the symmetric                                 ψ ( τ ) = p( τ ) + q( τ ) , where
dumbbell
                                                                                    p( τ ) =
                                                                                           2 3
                                                                                              k
                                                                                                        (     +∞
                                                                                                  exp( τ 3 )∫τ exp( −ξ 3 )D1( ξ )dξ +                         )
      Note that ‘the tumbling motion’ is a set of right-hand
and left-hand rotations with close to 1800 angles. Factually,                   (               τ
                                                                                 + exp( −τ 3 )∫−∞ exp( ξ 3 )D1( ξ )dξ ,             )
chaotic rotations of the dumbbell are obtained.
      Consider a single rotation from this set. Let
                                                                                        q( τ ) =
                                                                                                   1
                                                                                                  2 3
                                                                                                               (
                                                                                                       C1 exp( τ 3 ) + C2 exp( −τ 3 )                     )           (7)
          ′        ′
 ( γ 1 ,γ 1 ,ϕ1 ,ϕ 1 ) be values of ( γ , γ ′ ,ϕ ,ϕ ′ ) in the beginning         From equalities
of this rotation. It is clear that ϕ ′ ≈ 0 and ϕ 1 ≈ ± π / 2 .                                                p( τ + T ) =

                                                                                                 (exp((τ + T )
(Without loss of generality it can be assumed that                                         k                                  +∞
ϕ 1 ≈ − π 2 ). It is obvious that the motion in the vicinity of                      =                                   3 )∫τ +T exp( −ξ 3 )D1( ξ )dξ +
                                                                                         2     3
                                                                                                                                                              )
`horizontal' equilibrium determine the direction of the
                                                                                                               τ +T
considered rotation. Substituting ϕ ≈ −π / 2 + kψ in the                                 + exp( −( τ + T ) 3 )∫−∞ exp( ξ 3 )D1( ξ )dξ =

                                                                                                                         (                    )
dumbbell's motion equation we obtain                                                                                 k
                         ψ ′ − 3ψ + D k = 0 ,                        (4)                                       =       exp( τ + T ) 3 ⋅
                                                                                                                   2 3
            2( 1 − e 2 cos 2 γ )γ ′ + e 2 γ ′ sin 2 γ −                                             +∞
                                                                        (5)                      ⋅ ∫τ exp( −( ζ + T ) 3 )D1( ζ + T )dζ +
                       − 3( 1 − e ) sin γ = 0
                                  2

                                                                                                                   + exp( −( τ + T ) 3 ) ⋅
where D = 1 − e γ ′′ − e 2 γ ′ sin 2γ − 3 / 2 1 − e 2 sin 2γ .
                                                                                                 (⋅ ∫                                                 )
                      2
                                                                                                    τ +T
                                                                                                    −∞        exp((ζ + T ) 3 ) D1(ζ + T )dζ =
Here we are restricted to a case of symmetric dumbbell
µ = 0 ⇔ m1 = m 2 and neglect the terms of order higher
than k .
                                                                                          =
                                                                                                 k
                                                                                               2 3
                                                                                                        (    +∞
                                                                                                   expτ 3 )∫τ exp( −ξ 3 )D1( ξ )dξ +                              )
     Note that (5) is equation of motion for the particle if
the dumbbell is fixed in the ‘horizontal’ position [3].
                                                                                                                    τ
                                                                                        + exp( −τ 3 )∫−∞ exp( ξ 3 )D1( ξ )dξ = p( τ )             )
Equality                                                                         it follows that p( τ ) is T-periodic function of τ . Constants
      ( 1 − e 2 cos 2 γ )γ ′ 2 + 3( 1 − e 2 ) cos 2 γ = h2  (6)                  C1 and C2 are defined by formulae
is the Jacobi’s integral for (5). Analyzing phase portrait of                                                   3π      
(6) we see that there exist three types of equation (5)                                    C1 = k −1 / 2  3ϕ1 +
                                                                                                                   + ϕ1  − k 1 / 2 A ,
                                                                                                                       ′
                                                                                                                         
                                                                                                                2       
solutions. Solutions of the first type correspond to
librations about ± π / 2 . They are periodic functions of                                                            3π      
                                                                                                C2 = k −1 / 2  3ϕ1 +
                                                                                                                        − ϕ1  − k 1 / 2 B ,
                                                                                                                            ′
                                                                                                                              
τ with period                                                                                                         2
                                                                                                                             
                                           dγ
                               T =∫                 .                            where
                                      γ ′( γ , h2 )                                       +∞                                    ξ
                                                                                    A = ∫0 e −ξ         3
                                                                                                            D1( ξ )dξ , B = ∫− ∞ eξ     3
                                                                                                                                            D1( ξ )dξ .
Solutions of the second type correspond to the asymptotic
motions tending to γ = 0 or γ = π . It can easily be
checked that in this case the cable weakens. Let us remark                       A surface of asymptotic motions as an interstream
that such effect is also observed for the motions in some
                                                                                     Clearly, if C1>0 then the dumbbell will turn
vicinity of the separatrix. Solutions of the third type
                                                                                 counterclockwise and if C1<0 then the dumbbell will turn
correspond to rotations about the dumbbell. In this case
                                                                                 clockwise. Certainly, this criterion is valid only for the
derivative of γ w.r.t. τ is the periodic function with period
                                                                                 constrained motion.
                                    π
                                           dγ                                        If C1=0 then the dumbbell remain in the vicinity of
                               T =∫                 .
                                    0γ  ′( γ , h2 )                              horizontal equilibrium, i.e. we have the dumbbell
Moreover, solutions of (5) can be represented in a form                          asymptotic motion tending to librations about ϕ = −π / 2
                                     π                                           (or ϕ = π / 2 ). Clearly, this asymptotic motion is unstable.
                              ′
γ = γ ( τ , γ 1 , h2 ( γ 1 ,γ 1 )) = τ + σ ( τ ,γ 1 , h2 )
                                     T                                           Thus the equation


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                               VIBROMECHANIKA. JOURNAL OF VIBROENGINEERING. 2008 DECEMBER, VOLUME 10, ISSUE 4, ISSN 1392-8716
                                416. ROTATIONS OF A DUMBBELL EQUIPPED WITH ‘THE LEIER CONSTRAINT’. A. V. RODNIKOV




                   3π           +∞
                                                                              The similar interstream for ϕ 1 = − 90 ° 3′ ; ϕ1 = 0 is
          3ϕ1 +       +ϕ ′1= k ∫0 e −ξ     3
                                               D1( ξ )dξ      (8)          depicted in Figure 4. Here also e=1/2.
                   2
define a surface of asymptotic motions in the four-                           In figures 2 and 3 the shadowed area corresponds to the
                                                                           motion with the weakened cable.
                               ′       ′
dimensional space of ( γ 1 ,γ 1 ,ϕ1 ,ϕ 1 ) . In other words, (8)
is the equation of an original interstream dividing the                    On integral A computation
space of initial values into the areas of rotations clockwise
and rotations counterclockwise. Note also that if C1= C2=0                     Finally note that the infinite integral A is reduced up to
then we have the dumbbell periodic motion about                            definite. It follows from equalities
                                                                                                 +∞
horizontal equilibrium.                                                                     A = ∫0 exp( −ξ 3 )D1( ξ )dξ =
                                                                                                  ∞ ( n +1 )T
                                                                                            =∑          ∫ exp( −ξ 3 )D1( ξ )dξ =
                                                                                                n =0   nT
                                                                                            ∞ T
                                                                                      = ∑ ∫ exp( −( ζ + nT ) 3 )D1( ζ + nT )dξ =
                                                                                        n =0 0
                                                                                       T                                       ∞
                                                                                     = ∫ exp( −ζ 3 )D1( ζ )dζ ∑ exp( − nT 3 ) =
                                                                                       0                                      n =0
                                                                                                                      T
                                                                                                   1
                                                                                       =                    ∫ exp( −τ 3 )D1( τ )dτ
                                                                                            1 − exp( −T 3 ) 0

                                                                               Further, using dτ = γ ′( τ , ,γ 1 , h2 )dγ we can change the
                                                                           variable in the last integral. For instance, consider the area
                                                                           of the particle ‘positive rotations’. In this area
                                                                           h2 > 3( 1 − e 2 ) and γ ′ > 0 . Here using (6) we get

                            Fig. 3.                                                                      γ ′ = r( h2 ,γ ) ,                                   (9)
                                                                           where
                                                                                                                h2 − 3( 1 − e 2 )
                                                                                                    r( h2 ,γ ) =                     ,
                                                                                                                1 − e 2 cos 2 γ
                                                                                      h2 = ( 1 − e 2 cos 2 γ 1 )γ 1 2 + 3( 1 − e 2 ) cos 2 γ 1 .
                                                                                                                  ′

                                                                           From (9) it follows that

                                                                                                                  γ          dγ
                                                                                                            T = ∫γ
                                                                                                                   1      r ( h 2 ,γ )
                                                                           and
                                                                                                                  π          dγ
                                                                                                            T = ∫0                     .
                                                                                                                          r ( h2 , γ )
                                                                           Hence
                                                                                                                          1
                            Fig. 4.                                                                          A=                   .
                                                                                                                  1 − exp( −T 3 )
Examples of interstreams
                                                                                        π +γ1               γ
                                                                                                                   dξ        D ( h2 ,γ )
                                                                                     ⋅ ∫γ          exp  − 3 ∫                           dγ ,
    The right side of (8) depends only on γ 1 ,γ 1 and left
                                                    ′                                       1                              
                                                                                                            γ 1 r ( h2 ,γ )  r ( h2 ,γ )
                                                                                                       
                               ′
side depends only on ϕ1 ,ϕ1 . Therefore the interstreams
can be depicted in the plane ( γ 1 ,γ 1 ) for fixed values of
                                       ′                                   where
ϕ1 ,ϕ1′ .
    In particular, if the dumbbell is precisely horizontal                                             D 2 ( h2 ,γ ) = − e 2 sin 2γ ⋅
( ϕ1 = − π / 2 ; ϕ1 = 0 at the beginning of considered
                    ′
rotation then (8) is reduced up to the equality A=0. The                         ⋅
                                                                                                   (
                                                                                   1 − e 2 3 sin 2 γ + r ( h ,γ )
                                                                                                             2
                                                                                                                   +
                                                                                                                                      )                 
                                                                                                                                           r ( h2 , γ )  .
corresponding interstream is depicted in Fig. 3 for e=1/2.                               2( 1 − e cos γ )
                                                                                                   2   2                                                
                                                                                                                                                       
In this figure the areas of right-hand and left-hand rotations
are marked by the circular arrows.

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                           VIBROMECHANIKA. JOURNAL OF VIBROENGINEERING. 2008 DECEMBER, VOLUME 10, ISSUE 4, ISSN 1392-8716
                               416. ROTATIONS OF A DUMBBELL EQUIPPED WITH ‘THE LEIER CONSTRAINT’. A. V. RODNIKOV




Conclusions                                                               References

    In this paper the space tethered system consisting of the
dumbbell-shaped rigid body and the particle of small mass                 [1] Beletsky, V. V. and Novikova, E. T. On the Relative Motion
is considered. The particle moves along the cable with                        of Two Tethered Bodies on an Orbit. //Cosmic Research, v.7,
ends fixed in the body. The dumbbell rotations caused by                      No 3, pp.377-384. (1969).
the particle are studied. The sufficient condition of the                 [2] Beletsky, V. V. On the Relative Motion of Two Tethered
                                                                              Bodies on an Orbit II. //Cosmic Reseach ,v.7,No 6. (1969).
dumbbell librations about its stable equilibrium is                       [3] Rodnikov, A. V. Equilibrium Positions of a Weight on a
obtained. The family of asymptotic motions tending to                         Cable Fixed to a Dumbbell-Shaped Space Station moving
librations about unstable equilibria is found. This family                    along a circular geocentric orbit.// Cosmic Reseach,v. 44,No
forms the interstream separating the area of the dumbbell                     1, pp. 58-68. (2006).
rotations clockwise from the area of rotations                            [4] Rodnikov A. V. The algorithms for capture of the space
counterclockwise. The equation of the interstream is                          garbage using ‘leier constraint’.: Regular and Chaotic
deduced.                                                                      Dynamics, v.11, 4, pp. 483-489. (2006).

Acknowledgment

The author thanks V.V.Beletsky, Yu.F.Golubev,
I.I.Kosenko and V.V.Sazonov for useful discussions.




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