The Study on Motion of a Rigid Body Carrying a Rotating Mass

The free motion of a rigid body carrying a rotating mass without change of the centroid (this system may be called one-rotor gyrostat) is discussed. Equations of motion are derived: first integrals as a vectorial equation which contained the right vector of an angular velocity of the given rotor with respect to the carrier body and the turn-tensor of this body; a scalar relation between rotation angle of the given rotor with respect to the carrier body and the angular velocity of the carrier body. Only two of these parameters are independent variables. To get equations and to exclude the singular points in the solutions, it is necessary to determine the turn-tensor of the carrier body in the most suitable form. To this end the representation theorem of the turn-tensor and some additional arguments are used. As a final result, we enabled to get two complicated differential equations of the first order. In particular case, the exact solution is represented. Excluding the singular points numerical solutions are determined.


Introduction
The problem about the motion of gyrostat in the elementary statement (free motion of gyrostat) and in more complex statement (motion of heavy gyrostat, motion of gyrostat in Newtonian field of gravitation) was investigated by many authors [1]- [7].But although this, it is impossible to assert that even the elementary case-free motion of gyrostat is investigated in all details.There is an analytical solution of this problem in the book [8].The similar problem for rigid body without rotors is investigated by using the turn-tensor (see, for examples, [9]).
The turn-tensor is the most suitable tool for the description of turns and rotations of rigid bodies.Therefore, the method of construction of solution of the problem essentially based on the use of the turn-tensor is stated below.The use of the modern mathematical tool allows simplifying the derivation of solution of the problem and making it more evident.The problem is reduced to a complicated system of two differential equations.It is necessary to note that, this system may have singular points with a representation of the turn-tensor.So that our main purpose is the determination of conditions for representing the turntensor in a suitable form to avoid the singular points in the solution.The exact solution is obtained only in two cases.The numerical solution is represented for some given parameters.

Statement of the Problem
Let us consider a gyrostat, which moves without effect of external moments and forces.Suppose that the given system is a carrier body with one-rotor, see Figure 1.
Specifying the following terms m − mass of the gyrostat, The relation between the tensor of inertia of carrier body in the inertial position 1  and its tensor of inertia in actual position ( )  is given by Assume that 1  is the tensor of inertia of this rotor, calculated with respect to its centre of mass.Since the rotation of rotor does not change the distribution where λ is axial moment of inertia, µ is the equatorial moment of inertia, and e is a unit vector, which determine the axis of rotation of the rotor in the initial position, E is a unit tensor.
In the actual position the tensor of inertia takes a form similar to Equation (2.1) , the respectively.The left vector of an angular velocity of the composition (1.6) can express as follows [8]: Equation of motion of the moving system: In this problem for the gyrostat Euler's second law of dynamics takes the form where 2 K is the kinetic moment of gyrostat with respect to its centre of mass.
Euler's second law of dynamics for the rotor w.r.to its centre mass the form ( ) The kinetic moment of gyrostat w.r.to its center of mass is defined by the following formula Substituting Equation (2.11) into (2.10), and using (2.8) it is easy to have the following expression where L is a constant vector, which is determine from the initial condition of the problem, and Since the moment of friction is absent, multiplying Equation (2.9) by P e ⋅ we will get Form Equations (2.7), (2.11) and (2.13) we have After substituting equations (2.14) into (2.12),we can write (2.12) as follows where ( ) The problem is reduced to the following system of equations The kinetic energy system has the following form To show the kinetic energy has a constant value: from this equation and with the help of the previous equations, it is easy to calculate Substitution (2.16) into (2.18),it's correct to write We proved that K-const., hence from Equation (2.17) the energy integral has the following form

Transformation of the Energy Integral
In general the turn-tensor can be expressed through three parameters.The energy integral gives a relation, which superposed on these parameters.Therefore, only two of them are independent variables, and the free rotations of the body are two-parameter movements.Thus it's necessary to find the general from of a two-parameter turn-tensor conserving the energy.The unit vector m may be introduced by Substituting Equations (2.16), (3.1) into equation (2.19), we have Using Zhilin's theorem [9] to represent the turn-tensor P .The theorem of representation of the turn-tensor can be formulated: Let there be given two arbitrary unit vectors û and v .Any turn-tensor can be represented in the form composition of turns around vectors û , v and ω ( ) The success of solution depends on the appropriate of representation of the turn-tensor i.e. the equation of solution will be in complicated from with the unsuccessful choice of vectors û and v .Moreover, the unsuccessful choice of vectors û and v may lead to appearance of singular points in the solution.
For choosing û and v we will use the fact: T P gives the motion of the ki- netic moment with respect to the carrier body (see Formula (3.1)) and energy integral (3.2), which determine the path of the motion of vector of the kinetic moment with respect to the carrier body.Substituting (3.3) into (3.1),we get Since *  is a symmetrical tensor, it may be represented as The vectors m and e can be represented as The choice of the vector v is not unknown.This choice may be 1 d , 2 d , 3 d or other vector.The solution does not contain singular points, if following inequality is valid.
i.e. the angle ( ) , where the angle α is the angle be- tween v and L .

The Exact Solution in Particular Cases
Case (1): when In this case Equation (3.2) is reduced to: ( ) The intersection of the plane Equation (4.1) with the unit sphere ( ) gives a hodograph of vector m .This hodograph is a circle, as shown in Figure 2.
Since vector e is perpendicular to the plane of hodograph, then it is suitable to represent the turn tensor of the carrier body P in the form: The corresponding angular velocity of P : ( ) The right angular velocity is given by: Rewriting Equation (4.5) as: From the vectorial Equation (4.7) it follows that: Then: where 12) The intersection of plane (4.12) with the unit Sphere ( ) gives a ho- dograph of the vector m as shown in Figure 3. Since: ( ) ( ) where ( ) Substituting (4.14) into (4.13) to get:

System of Differential Equations Describing the Rotation
3) and by using Substituting Equations (2.16), (3.1) into (5.1),we get From the Equation (3.5), we have Then we can get

Numerical Solution
The system of Equations (5.5), (5.6) may be solved numerically under the necessary condition Considering this condition to avoid the singular points during calculations.The common points of sphere (3.1) with the horizontal plane 0 m z = give a circle: The intersection of surface (3.2) with 0 m z = is given by the equation of el- lipse: ( ) The distance between the origin and a point on the ellipse (6.3) may given by , r x y x y = + (6.4) To get the minimum and maximum distance between the origin and a point on the ellipse (6.3) the following steps may be done.Suppose them are min ( ) Four points are verifying the solution of the previous system (6.5).At these points ( )

Discussion and Conclusion
To understand and predict the behaviour of gyrostat it is necessary to know the suitable form of the turn-tensor of the carrier body, ( ˆˆ, ; if 1, 1 or 1, 1 The angle precession ψ is varying monotonically but the variation of angle of nutation θ has oscillating nature.The angle of own rotation ϕ is varying monotonically if the unit vector i d will be inside the hodograph of the vector m , but if it will be outside this hodograph, then the variation of angle ϕ has an oscillating nature.The turn-tensor ( ) P t may be represented by another vector except ( ) when the condition ( ) invalid.Particularly, the exact solution is found when ( )

1 m 2 m 1 c 2 c, 1 τ − vector 1 cc , 2 τ − vector 2 cc , 1 cv − velocity of the point 1 c , 2 cv
− mass of the carrier body, − mass of the rotor, c − centre of mass of gyrostat, − centre of mass of the carrier body, − centre of mass of rotor, which rotates with respect to carrier body around axis e − velocity of the point 2 c c v − velocity of the point c .
of gyrostat w.r.to is its centre of mass; ext M * is a moment, which acts on the rotor form the sides of the carrier body.

3 d
are unit vectors, which depend on this tensor.Then we can write

Figure 3 .
Figure 3.A horizontal hodograph of m .
By using Equations (3.2), and (5.2).It is can be written
is represented graphically for some given parameters.