On the equations of motion for a heavy body with a fixed point

Kharlamov, P. V.

doi:10.1016/0021-8928(63)90188-6

Cited by 6 publications

(6 citation statements)

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“…Kharlamov succeeded in reducing the problem of a gyrostat in the uniform field of gravity to a system of two differential equations of the first order [18]. His method depends on the elimination of the Poisson variables and cannot be generalized neither to more general potentials nor to cases when gyroscopic forces are present.…”

Section: Reduced Equation In Rigid Body Dynamicsmentioning

confidence: 97%

New solutions of classical problems in rigid body dynamics

Yehia

Saleh

Megahid

2015

Mechanics Research Communications

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Section: Reduced Equation In Rigid Body Dynamicsmentioning

confidence: 97%

New solutions of classical problems in rigid body dynamics

Yehia

Saleh

Megahid

2015

Mechanics Research Communications

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“…h is bounded since any E h is a compact set. To find the actual boundaries of the curves segments included in h we need to investigate the intersections 1 ∩ 2 corresponding to the real solutions of (1), (2). The equations of transversal intersections considered on either surface are of degree 5 in h and 12 in s. Even though they can be written down and numerically solved, there is no criterion for the corresponding points to belong to the image of J h .…”

Section: Bifurcation Surfacesmentioning

confidence: 99%

“…In the work [1] the complete Liouville integrability of such system (1) has been proved. Despite this fact, for the case λ = 0 the only explicit integrations or qualitative investigations of (1) known up-to-date [2][3][4] deal with the axially symmetric force (β ≡ 0). We suppose that α × β = 0, r 1 × r 2 = 0.…”

Section: Introductionmentioning

confidence: 99%

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Periodic motions of the Kowalevski gyrostat in two constant fields

Kharlamov¹

2008

J. Phys. A: Math. Theor.

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The case of the gyrostat motion found by A G Reyman and M A Semenov-Tian-Shansky is known as the Liouville integrable Hamiltonian system with three degrees of freedom without symmetry groups. We find the set of points at which the integral map has rank 1. This set consists of special periodic motions generating the singular points of bifurcation diagrams on iso-energetic surfaces. For such motions, all phase variables are expressed as algebraic functions of one auxiliary variable satisfying the differential equation integrable in elliptic functions of time. It is shown that the corresponding points in three-dimensional space of the integral constants belong to the intersection of two sheets of the discriminant surface of the Lax curve.

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“…En el capítulo dos se hace un estudio breve del planteamiento del problema del cuerpo rígido con un punto fijo en un campo de gravedad constante y se estudian los casos integrables conocidos [5,17,18] .…”

Section: Capítulo 1 Introducciónunclassified

Caos en la dinámica del cuerpo rígido

Ramírez

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Capítulo 2. Breve estudio de la cinemática y dinámica del cuerpo rígido La matriz inversa de la matriz de rotación es su matriz traspuesta. Estas transformaciones lineales que mantienen invariante la norma se llaman transformaciones ortogonales. Las rotaciones son trasformaciones lineales ortogonales que además dejan invariante la propiedad de la base de vectores. Un conjunto ordenado de vectores ortonormales b 1 , b 2 y b 3 , se dice que forman una base cuando el determinante de la matriz que tienen por columnas primera, segunda y tercera, respectivamente, a los vectores b 1 , b 2 y b 3 es igual a 1: |(b 1 b 2 b 3)| = 1, (2.7) donde las barras verticales en laúltima ecuación significan que hay que realizar la operación de tomar el determinante. Para el caso cuando este toma el valor de-1, la base se llamara izquierda. El producto escalar es invariante ante una rotación. Por esta razón una base ortonormal se transformará mediante alguna rotación en otra base ortonormal. Veremos a continuación la consecuencia de que una base derecha se rote en una base derecha. Primeramente consideramos el determinante de la matriz de los tres vectores rotados. |(Rb 1 Rb 2 Rb 3)| = 1, (2.8) el cual tiene el valor uno debido a que R es una rotación. De estaúltima ecuación se puede sacar la matriz de rotación como factor común de la matriz de vectores (b 1 b 2 b 3): (Rb 1 Rb 2 Rb 3) = R(b 1 b 2 b 3) (2.9) ahora usamos la propiedad de que el determinante de un producto matricial es igual al producto de los determinantes de las matrices factores |(Rb 1 Rb 2 Rb 3)| = |R(b 1 b 2 b 3)| = |R||(b 1 b 2 b 3)| = 1.

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On the equations of motion for a heavy body with a fixed point

Cited by 6 publications

References 1 publication

New solutions of classical problems in rigid body dynamics

New solutions of classical problems in rigid body dynamics

Periodic motions of the Kowalevski gyrostat in two constant fields

Caos en la dinámica del cuerpo rígido

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