Equilibria of a double pendulum in a circular orbit

PII: S0094-5765(99)00015-6

Acta Astronautica Vol. 44, No. 1, pp. 63±65, 1999 # 1999 Elsevier Science Ltd. All rights reserved Printed in Great Britain 0094-5765/99 $ - see front matter

ACADEMY TRANSACTIONS NOTE EQUILIBRIA OF A DOUBLE PENDULUM IN A CIRCULAR ORBIT V.A. SARYCHEV{ Universidade da Beira Interior, CovilhaÄ, Portugal (Received 11 October 1998) AbstractÐAll spatial equilibria of a double pendulum relative to orbital reference frame are determined in a circular orbit. # 1999 Elsevier Science Ltd. All rights reserved

3 3  1 l21 o20 cos2 W1 ‡ m  2 l22 o20 cos2 W2 Uˆ m 2 2 ÿ ÿm  12 l1 l2 o20 sin c1 sin W1 sin c2 sin W2

Consider the motion of three mass point system (P1 of mass m1, P of mass m3, and O2 of mass m2) connected by two rigid weightless rods P1P of length l1 and PO2 of length l2 (Fig. 1) in a circular orbit. Two rods are connected with each other by a spherical hinge at the point P. We call such system as a double pendulum.


‡ cos c1 sin W1 cos c2 sin W2 ÿ 2 cos W1 cos W2 : …2†

The orientation of axes Oizi (i = 1,2) is determined relative to the orbital reference frame OXYZ (the axis OZ directs along the radius vector connecting the Earth's mass center with the mass center O of the double pendulum; the axis OX is perpendicular to the axis OZ, lies in the orbital plane, and in the direction of the pendulum orbital motion; the axis OY is perpendicular to the orbital plane and completes the right-handed triad) by means of Euler angles ci (angle of precession) and Wi (angle of nutation). Here o0 is angular velocity of orbital motion of the double pendulum; O1 is the mass center of points m1 and m3 system;

It is easy to obtain the following expressions of kinetic energy T and force function U of a double pendulum from general relations (3.2) and (3.3) in [1]: ÿ 1 Tˆ m  1 l21 c_ 21 sin2 W1 ‡ W_ 21 ‡ o0 c_ 1 cos c1 sin 2W1 2
‡ 2o0 W_ 1 sin c1 ÿ o20 cos2 c1 sin2 W1 ÿ 1 ‡ m  2 l22 c_ 22 sin2 W2 ‡ W_ 22 ‡ o0 c_ 2 cos c2 sin 2W2 2
‡ 2o0 W_ 2 sin c2 ÿ o20 cos2 c2 sin2 W2 hÿ ‡m  12 l1 l2 c_ 1 cos c1 sin W1 ‡ W_ 1 sin c1 cos W1

m 1ˆ m  12 ˆ


ÿ ‡ o0 cos W1
c_ 2 cos c2 sin W2 ‡ W_ 2 sin c2 cos W2
ÿ
‡ o0 cos W2 ‡ c_ 1 sin c1 sin W1 ÿ W_ 1 cos c1 cos W1 ÿ
c_ 2 sin c2 sin W2 ÿ W_ 2 cos c2 cos W2 ÿ
ÿ
i ‡ sin W1 sin W2 W_ 1 ‡ o0 sin c1 W_ 2 ‡ o0 sin c2 ; …1†

m1 …m2 ‡ m3 † m2 …m1 ‡ m3 † ;m 2ˆ ; m1 ‡ m2 ‡ m3 m1 ‡ m2 ‡ m3 m1 m2 : m1 ‡ m2 ‡ m3

…3†

The dot designates the di€erentiation with respect to time t. Using expressions for T and U, it is possible to derive the equations of motion of the double pendulum for angles c1, c2, W1, W2 in Lagrange form. Equilibria of the double pendulum in the reference frame OXYZ ci0 ˆ const;

Wi0 ˆ const

…4†

are determined by equations @…T0 ‡ U† ˆ 0; @ci

{IAA Member, Section 2. On leave from the Keldysh Institute of Applied Mathematics, Russian Academy of Sciences, Moscow, Russia.

where 63

@…T0 ‡ U† ˆ 0; @Wi

…5†

64

V. A. Sarychev

Fig. 1

1  1 l21 o20 cos2 c1 sin2 W1 T0 ˆ ÿ m 2

cos W10 ˆ 0; sin c10 ˆ 0; cos W20 ˆ 0; x2 cos c20 ‡ cos c10 ˆ 0:

1  2 l22 o20 cos2 c2 sin2 W2 ÿ m 2 ÿ ‡m  12 l1 l2 o20 cos W1 cos W2

Solutions (12) exist if x2e1; the axis of symmetry O1z1 is parallel to the axis 2OY, the axis of symmetry O2z2 lies in the plane OXY.


‡ sin c1 sin c2 sin W1 sin W2 :

cos W10 ˆ 0; x1 cos c10 ‡ cos c20 ˆ 0; cos W20 ˆ 0; sin c20 ˆ 0:

…6†

Equations (5) have the following expanded form: D1 sin c10 sin W10 ˆ 0; D2 sin c20 sin W20 ˆ 0; ÿ
D1 cos c10 cos W10 ‡ 3 sin W10 x1 cos W10 ‡ cos W20 ˆ 0; ÿ
D2 cos c20 cos W20 ‡ 3 sin W20 x2 cos W20 ‡ cos W10 ˆ 0:

sin W10 ˆ 0; x2 cos W20 ‡ cos W10 ˆ 0; cos c20 ˆ 0: …14† Solutions (14) exist if x2e1; the axis of symmetry O1z1 is parallel to the axis 2OZ, the axis of symmetry O2z2 lies in the plane OXZ. x1 cos W10 ‡ cos W20 ˆ 0; cos c10 ˆ 0; sin W20 ˆ 0:

Here D1 ˆ x1 cos c10 sin W10 ‡ cos c20 sin W20 ; D2 ˆ x2 cos c20 sin W20 ‡ cos c10 sin W10 ; m2 ‡ m3 l1
; m2 l2

x2 ˆ

m1 ‡ m3 l2
: m1 l1

…8†

Making simple but rather tiring calculations we determine all solutions of eqns (7), which correspond to equilibria of the double pendulum in a circular orbit. cos W10 ˆ 0; cos c10 ˆ 0; cos W20 ˆ 0; cos c20 ˆ 0: …9† For solutions (9) the axes of symmetry O1z1 and O2z2 coincide with the axis2OX (the tangent to the orbit).

…15† Solutions (15) exist if x1e1; the axis of symmetry O2z2 is parallel to the axis 2OZ, the axis of symmetry O1z1 lies in the plane OXZ. sin c10 ˆ 0;

…10† For solutions (10) the axes of symmetry O1z1 and O2z2 coincide with the axis 2OY (the normal to orbital plane). …11†

For solutions (11) the axes of symmetry O1z1 and O2z2 coincide with the axis 2OZ (the direction of radius vector).

sin c20 ˆ 0;

cos2 W10 ˆ P1 2Q1 ; cos2 W20 ˆ P2 3Q2 ; where P1 ˆ

…16†

ÿ 2
ÿ
8x1 ÿ 5 x1 x2 ÿ 5x21 ÿ 2 2x21 ‡ 1 p  ; Q ˆ ; 1 4x21 8x21 …x1 x2 ÿ 1†…4x1 x2 ÿ 1† …17:1†

cos W10 ˆ 0; sin c10 ˆ 0; cos W20 ˆ 0; sin c20 ˆ 0:

sin W10 ˆ 0; sin W20 ˆ 0:

…13†

Solutions (13) exist if x1e1; the axis of symmetry O2z2 is parallel to the axis 2OY, the axis of symmetry O1z1 lies in the plane OXY.

…7†

x1 ˆ

…12†

P2 ˆ

ÿ 2
ÿ
8x2 ÿ 5 x1 x2 ÿ 5x22 ÿ 2 2x22 ‡ 1  p ; Q ˆ : 2 4x22 8x22 …x1 x2 ÿ 1†…4x1 x2 ÿ 1† …17:2†

For solutions (16) the axes of symmetry O1z1 and O2z2 are in the plane OYZ. All solutions (16) exist if the following inequalities are valid: ÿ
ÿ
0 < cos2 W10 1 < 1; 0 < cos2 W20 1 < 1; …18:1†

Equilibria of a double pendulum in a circular orbit

65

Fig. 2

ÿ




0 < cos2 W10 2 < 1;

ÿ




0 < cos2 W20 2 < 1;

…x1 x2 ÿ 1†…4x1 x2 ÿ 1† > 0: Here ÿ 2
cos W10 1 ˆ P1 ‡ Q1 ;

ÿ

…18:2† …18:3†


cos2 W20 1 ˆ P2 ÿ Q2 ; …19:1†

ÿ 2
cos W10 2 ˆ P1 ÿ Q1 ;

ÿ


cos2 W20 2 ˆ P2 ‡ Q2 : …19:2†

Solutions (19.1) exist if the inequalities (18.1) and (18.3) take place; solutions (19.2) exist if the inequalities (18.2) and (18.3) take place. Consider regions determined by inequalities (18) where solutions (19) exist. The analysis of boundaries of inequalities (18) allows us to decompose the plane of parameters x1, x2 into regions (Fig. 2), where some of the solutions (19) exist. The boundaries of these regions are formed by following three hyperbolas: 4x1x2ÿx1ÿx2ÿ2 = 0, 4x1x2+x1+x2ÿ 2 = 0, 4x1x2ÿ3x1ÿ3x2+2 = 0. The symbol r(t) in Fig. 2 designates the regions, where the solutions (19.1) ((19.2)) exist. The region, where both solutions (19.1) and (19.2) exit, is designated by the symbol q. Adding in Fig. 2 the boundaries x1=1 for solutions (13) and (15) and x2=1 for solutions (12) and (14), we obtain ®nal decomposition of the

plane of parameters x1, x2 into regions where various but ®xed number of solutions exist. Thus, the result of this research is the determination of all spatial equilibria of the double pendulum in orbital reference frame and of regions of their existence. Note that a simpler problem of determination of equilibria of two arbitrary bodies connected by a cylindrical hinge in the plane of circular orbit was ®rst considered in [2,3], where solutions which are analogous to (9), (11), (14), (15) were obtained, and their stability was also investigated. The results obtained in [2,3] were considerably later repeated in [4] for a double pendulum. AcknowledgementsÐThe present work was supported by the Portugal Program PRAXIS XX1 (project 3/3.1/CTAE/ 1992/95)

REFERENCES

1. Sarychev, V. A. 1978. Problems of orientation of satellites. Itogi Nauki i Tekhniki, Ser.Space Research, Moscow. In Russian. 2. Sarychev, V.A., Cosmic Research, 1967, 5(3), 360±364. 3. Wittenburg, J., Abhandlungen der Braunschweigischen Wissenschaftlichen Gesellschaft, 1968, Band XX, 198± 278. 4. Misra, A.K., Amier, Z. and Modi, V.J., Acta Astronautica, 1988, 17(10), 1059±1068.