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  -  . 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  . The similar problem for rigid body without rotors is investigated by using the turn-tensor (see, for examples,  ). 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 turn- tensor 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.
2. 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
mass of the gyrostat,
mass of the carrier body,
mass of the rotor,
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 ,
velocity of the point ,
velocity of the point
velocity of the point .
The relation between the tensor of inertia of carrier body in the inertial position and its tensor of inertia in actual position is given by
Assume that 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
Figure 1. One rotor gyrostat.
of masses in the change the distribution of masses in the gyrostat, can represent as followed
where is axial moment of inertia, is the equatorial moment of inertia, and is a unit vector, which determine the axis of rotation of the rotor in the initial position, is a unit tensor.
In the actual position the tensor of inertia takes a form similar to Equation (2.1)
where is the tensor of the rotor. The full turn of rotor around its symmetrical axis in initial position and the turn-tensor . It is easy to prove that the turn-tensor can be represented in the following form of composition of turns around and around orthogonal to
where is defined as a tensor of inclination of axis . Hence the full turn of rotor may be taken the following form
From the Equations (2.4) and (2.5) we get
where is the angle of turn of rotor with respect to carrier body. Let the angular velocity and correspond to turn tensor , the respectively. The left vector of an angular velocity of the composition (1.6) can express as follows  :
Equation of motion of the moving system:
In this problem for the gyrostat Euler’s second law of dynamics takes the form
where 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
where the kinetic moment of gyrostat w.r.to is its centre of mass; is a moment, which acts on the rotor form the sides of the carrier body.
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 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 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
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
3. 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 may be introduced by
Substituting Equations (2.16), (3.1) into equation (2.19), we have
Using Zhilin’s theorem  to represent the turn-tensor . The theorem of representation of the turn-tensor can be formulated: Let there be given two arbitrary unit vectors and . Any turn-tensor can be represented in the form composition of turns around vectors #Math_75# ,#Math_76# and
where are called angles of precession, nutation, and own rotation, respectively.
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 and . Moreover, the unsuccessful choice of vectors and may lead to appearance of singular points in the solution. For choosing and we will use the fact: gives the motion of the kinetic 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
It’s obvious, that is a good choice, then
Since is a symmetrical tensor, it may be represented as
where , , are unit vectors, which depend on this tensor. Then we can write
The vectors and can be represented as
The choice of the vector is not unknown. This choice may be , , or other vector. The solution does not contain singular points, if following inequality is valid.
i.e. the angle belongs to , where the angle is the angle between and .
4. 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 . This hodograph is a circle, as shown in Figure 2.
Since vector is perpendicular to the plane of hodograph, then it is suitable to represent the turn tensor of the carrier body in the form:
The corresponding angular velocity of :
The right angular velocity is given by:
Figure 2. Hodograph of .
Substituting (4.3), (4.4) into equation (2.16) to get:
Rewriting Equation (4.5) as:
From the vectorial Equation (4.7) it follows that:
Equations (4.2), (4.3), (4.4), (4.9) and from which given in (2.16), give the exact solution of the problem.
Case (2): when
Equation (3.2) may be represented in the following form:
Inserting (3.1) into (4.10), we get:
The intersection of plane (4.12) with the unit Sphere gives a hodograph of the vector as shown in Figure 3.
Figure 3. A horizontal hodograph of .
Substituting (4.14) into (4.13) to get:
5. System of Differential Equations Describing the Rotation
Putting in Equation (3.3) and by using , we can drive
Substituting Equations (2.16), (3.1) into (5.1), we get
By using Equations (3.2), and (5.2). It is can be written
From the Equation (3.5), we have
Then we can get
Multiplying Equation (5.2) by scalar product, and using Equations (5.3), and (5.4) we get the system of first order non linear differential equation
Making use Equations (5.3), (5.4) and substituting them into Equation (5.2) after multiplying it by scalar product. We can get
6. 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 give a circle:
The intersection of surface (3.2) with is given by the equation of ellipse:
The distance between the origin and a point on the ellipse (6.3) may given by
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 and . Solving the system:
Four points are verifying the solution of the previous system (6.5). At these points may be evaluated. The smallest real value of them is and the largest real value is . If or valid, then the condition:
will be satisfied. It is easy to see that the conditions
(a) (b) (c) (d)
Figure 4. (a) Intersection between the surface (6.3) and sphere (6.2); (b) Intersection of (6.3), (6.2) with ; (c) Oscillating nature of Θ; (d) Monotonically variation of Φ.
(a) (b) (c) (d)
Figure 5. (a) Intersection between the surface (6.3) and sphere (6.2); (b) Intersection of (6.3), (6.2) with ; (c) Oscillating nature of Θ; (d) Oscillating nature of Φ.
7. 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,
The angle precession is varying monotonically but the variation of angle of nutation has oscillating nature. The angle of own rotation is varying monotonically if the unit vector will be inside the hodograph of the vector , but if it will be outside this hodograph, then the variation of angle has an oscillating nature. The turn-tensor may be represented by another vector except when the condition or is invalid. Particularly, the exact solution is found when and when . Excluding the singular points under the conditions or the numerical solution is represented graphically for some given parameters.