Stability analysis of an acted asymmetric rigid body by a gyrostatic moment and a constant body-fixed torque

Abstract

This work focuses on the dynamical rotatory motion of an asymmetric rigid body (RB) subjected to a constant body-fixed torque and a vector of a gyrostatic moment (GM). The motion is considered in the absence of the first two components of the GM. The novelty of the current work is the conversion of the stability analysis of this body from a two dimensional phase plane to three ones phase space. Specifically, two scenarios of stationary torques are considered, the first one is directed on the minor or major axis, while the latter is presumed to act on the middle axis. Therefore, three dimensional phase space routes are generated. In both scenarios some novel analytical and simulation outcomes are provided regarding equilibrium manifolds, periodic solutions or non-periodic ones, separator surfaces, as well as the extreme periodic solutions. The significance of this work is due to its great applications, especially those that use the gyro’s theory.

Introduction

In the last few decades, many researchers have focused their work on studying the RB’s motion when the applied external torques have various types.^1–10 The main factor of this progress is due to its broad range of uses in diverse fields, including satellites, spacecraft, and the like. An investigation of an asymmetrical satellite’s motion under the action of pressure forces and resistive moments is presented in Ref. 2 and Ref. 3 when the motion is regarded to be with the satellite’s center of mass. After averaging the controlling system, the authors examined the results according to the Euler-Poisson motion. More contributions to this problem are discussed in Ref. 4 when the satellite is subjected to both drag and gravitational force. The authors considered that the drag torque is linearly related to the satellite’s angular velocity. The issue of estimating the solution of the average time of a slightly perturbed dynamical system that is a stationary domain of phase variables is examined in. Ref. 5.

The averaging approach ¹¹ is utilized in Refs. 6–8 to obtain a corresponding averaging system of the controlling one for the rotatory motion of a symmetric rigid body (RB). The motion is acted upon by a Newtonian field of force (NFF), a gyrostatic moment (GM), and a perturbing one. In addition to these forces and moments, the action of an electromagnetic field is considered in Ref. 7 and. Ref. 8. The achieved averaging system is solved according to several applications, in which the gained outcomes have been graphed to show the significant effect of various values of the used parameters on the body’s motion. The gained results generalized the works in Ref. 9 and Ref. 10 when the gravitational field’s influence and perturbing torques are taken into consideration.

A general rotatory motion of a charged RB has been discussed in Ref. 12 under the action of the GM when the body initially has high speed around one of its principal axes. The method of Krylov–Bogoliubov–Mitropolski and its adjustment have been used to obtain the solutions of the governing system of motion. This motion was studied in Ref. 13 to obtain the periodic solutions according to the conditions of Bobylev–Steklov when the body is acted upon by the GM and NFF. The stability of the obtained results has been checked in light of the phase portraits of these results. The influence of other moments like perturbing and restoring, besides the aforementioned moment and force, has been investigated in. Ref. 14 Some of the applications have been looked over and analyzed.

The homotopy method (HM), which is employed in Ref. 15 and Ref. 16 to treat with the vibrating movements of dynamical systems, is regarded as a significant perturbation method that may be utilized to the solutions of numerous differential equations. ¹⁷ These solutions are contrasted with the numerical ones to show the precise accuracy of this method. These solutions have been compared with the numerical ones to show the precise accuracy of this method. The motion of a connected pendulum with a controlled wheel was studied in Ref. 18 while in Ref. 19 the authors examined the solution and stability of a planar rotating pendulum model, while the linearized stability of a hybrid oscillator of Rayleigh-Van der Pol-Duffing was achieved near the equilibrium points in. Ref. 20. The modified HPM is used in Ref. 21 to obtain the solution of string axial vibrations to overcome the shortcoming of the HPM, in which the stability criterion is carried out and the frequency equation is determined. Li-He’s method is used in Ref. 22 to gain the formulation of frequency amplitude for a microelectromechanical system.

The rotatory motion of a RB for the displacement case of the body’s center of mass was examined in Ref. 23 when the body’s motion is considered to be acted by a gravitational field only. The method of small parameter ¹¹ was used to gain approximate solutions. The extension of this work may be found in Ref. 24 and Ref. 25 when the body suffers from the full existence of GM and the presence of NFF besides the two components of GM, respectively. Moreover, the approximate solutions of an irrational frequency case for the RB motion are acquired in Ref. 26. The numerical outcomes are gained using Runge–Kutta algorithms of fourth order, and they are compared with approximate analytical ones to show the agreement between them. The stability of the acquired data is explored using several phase plane plots, which demonstrate the motion’s lack of chaos. The angles of Euler have been used to assess the geometric description of motion to pinpoint the body’s orientation at every given moment.

Exact solutions and perturbed ones for a large range of special instances have been produced through several scientific works. In Ref. 27, the authors proposed a perturbation approach for an asymmetric RB in arbitrary torqued motion. An Encke-type perturbation approach was used to formulate the problem with regard to the analytic approximate solution in the absence of torque. Computer processing durations for these and two additional motion prediction algorithms were compared in. Ref. 28. In Ref. 29, the authors provided a precise solution for a dual-spin spacecraft to move without experiencing torque. With a constant spin rate and traversing torques, a closed-form solution to the linearized equations of rotation was investigated in Ref. 30. In Refs. 31–34, the authors established the approximations for the motion of a near-symmetric RB exposed to body-fixed torques (BFT) that are both stationary and varying with time around the three main axes. Euler’s equations have been decoupled to obtain the solution of the RB motion through several decoupling processes in Ref. 35. This body is subject to a GM in addition to another external time-dependent harmonic moment about the body’s main axes, in which a few applications are presented theoretically.

This paper’s major outcomes can be interpreted in the manner shown below. A collection of one dimension (D) separated curves were displayed on a 2D phase plane in Ref. 1 to study the stability of an asymmetric RB subjected to a constant body-fixed torque along one of its main axes. The phase plane’s coordinates are somewhat physically illogical, and the analysis is only performed for a single set of a time’s initial conditions. The main innovation of this study is the conversion of the stability analysis from a 2D phase plane to a 3D phase space, which makes the current work novel. The dimensionless angular velocity of the coordinate system of the body is represented by the 3D phase space coordinates. The generated coordinate system is the same as the one that was applied to the stability analysis of a spinning body without torque. Consequently, it lends itself more easily to practical interpretation. The separators are graphed as 2D surfaces in the 3D phase space, allowing for the capture of all potential initial conditions with just a single graph. The conversion of this body’s stability analysis will be from a 2D to a 3D phase space.

In this study, the next contributions are organized: First, scaling of the equations of motion (EOM) to remove their dependency on the inertia characteristics for the scenario of a variation of time for the torque fixed body; Second, the EOM’s dependence on both the inertia qualities and the moment’s magnitude should be eliminated in the situation of constant body-fixed torque; Third, the dimensionless system’s equilibrium manifolds are determined; Fourth, development of the characteristic equation, the linearized EOM, and the stability characteristics of the dimensionless system; Fifth, for the scenario of a constant moment on the minor/major axis, evaluation of the motion integrals and the two separators surfaces in a 3D phase space; and Sixth, assessment of the motion integrals and the two common separators surfaces in a 3D phase space for the scenario of a stationary torque on the median axis.

Problem formulation

This section aims to provide further information regarding the description of the problem under consideration. Therefore, we consider the rotatory motion of a RB around a fixed point $O$ , which can be considered the origin of two systems. One of them $XYZ$ , is fixed in space, while the other $xyz$ , is fixed in the body and moving with it, see Figure 1. Some forces and moments are considered to be influencing the body which are represented by the gyrostatic moment vector $\underset{–}{\ell }\equiv \left({\ell }_1,{\ell }_2,{\ell }_3\right)$ ; ${\ell }_j\left(j=\mathrm{1,2,3}\right)$ are its projections on the gyrostat’s principal axes of inertia, and other additional constant torques $M_j$ , in which they are guided along the same axes of the gyrostat. Let $\left(I_1,I_2,I_3\right);I_1>\hspace{0.33em}{I}_2>I_3$ and $\left(p,q,r\right)$ be the principal moments of inertia and the angular velocities, respectively, which are directed about the body’s inertia principal axes. According to the basic equation of the body’s angular momentum, we can write the EOM when ${\ell }_1={\ell }_2=0$ and ${\ell }_3\ne 0$ , as follows ¹

$I_1\dot{p}+\left(I_3-I_2\right)qr+{\ell }_{3}q=M_1,$

$I_2\dot{q}+\left(I_1-I_3\right)rp-{\ell }_{3}p=M_2,$

$I_3\dot{r}+\left(I_2-I_1\right)pq=M_3.$

(1)

Figure 1.

The dynamical motion.

Let us consider the following parameters

$k_{\nu }=\frac{\left(I_{\zeta }-I_{\xi }\right)}{I_{\nu };}\hspace{1em}\left[\nu ,\zeta ,\xi \left(=\mathrm{1,2,3}\right),\nu \ne \zeta ,\nu \ne \xi ,\zeta <\xi \right],k=k_1{k}_2{k}_3,\tau =\sqrt{k}t,$

$u_j=\frac{M_j}{I_j},\hspace{1em}{x}_j=\frac{{\omega }_j}{\sqrt{k_j}},\hspace{1em}{\mu }_j=\frac{u_j}{\sqrt{k{k}_j}};\hspace{1em}\left(j=1,2,3\right),\hspace{1em}{\omega }_j=\left(p,q,r\right).$

(2)

Equation (1) can be rewritten without reference to the rotating body’s inertia characteristics as follows

$\dot{p}-k_{1}qr+\frac{{\ell }_{3}q}{I_1}=u_1,$

$\dot{q}-k_{2}rp-\frac{{\ell }_{3}p}{I_2}=u_2,$

$\dot{r}-k_{3}pq=u_3.$

(3)

Further, removing the dependency of the previous equations (3) on the magnitude of the projections $M_j$ of the torque vectors is achievable in the case of constant BFT.

Considering

$u=\frac{\sqrt{k_2{k}_3{u}_1^2+k_3{k}_1{u}_2^2+k_1{k}_2{u}_3^2}}{k},$

(4)

and redefining $\tau ,x_j,$ and ${\mu }_j$ as follows

$\tau =\sqrt{ku}t,\hspace{1em}{x}_j=\frac{{\omega }_j}{\sqrt{u{k}_j}},\hspace{1em}{\mu }_j=\frac{u_j}{u\sqrt{k{k}_j}};\hspace{1em}\left(j=\mathrm{1,2,3}\right).$

(5)

Substituting (5) into the systems of equation (1) yields

$\frac{d{x}_1}{d\tau }-x_2{x}_3+a{x}_2={\mu }_1,$

$\frac{d{x}_2}{d\tau }+x_3{x}_1-b{x}_1={\mu }_2,$

$\frac{d{x}_3}{d\tau }-x_1{x}_2={\mu }_3,$

(6)

where $\tau ,x_j,$ and ${\mu }_j$ are, respectively, the dimensionless quantities of time, angular velocity, and constant torque vectors, also $a={\ell }_3/I_1{k}_1\sqrt{k_3}$ and $b={\ell }_3/I_2{k}_2\sqrt{k_3}$ .

Making use of (4) and (5), we get

${\mu }_1^2+{\mu }_2^2+{\mu }_3^2=1$

(7)

Using equation (5), the previous equation can be easily verified. As cleared in equation (5), the quantities $\tau ,x_j,$ and ${\mu }_j$ can be used in the rest of this work.

The equations of motion (EOM) at steady state become

$\left(-X_2\left(X_3-a\right),X_1\left(X_3-b\right),-X_1{X}_2\right)=\left({\mu }_1,{\mu }_2,{\mu }_3\right)$

(8)

where $X_j\left(j=\mathrm{1,2,3}\right)$ are the points of equilibrium. Equation (8) produces ${\mu }_1{\mu }_2{\mu }_3=X_1^2{X}_2^2\left(X_3-a\right)\left(X_3-b\right)$ . This means that the points of equilibrium exist if and only if ${\mu }_1{\mu }_2{\mu }_3\ge 0$ . There are eight equilibrium locations specified by a dimensionless torque vector, as follows.

$\left(X_1,X_2,\left(X_3-a\right)\left(X_3-b\right)\right)=\left(\pm {\sqrt{{\mu }_2{\mu }_3/\mu }}_1,\pm \sqrt{{\mu }_3{\mu }_1/{\mu }_2},\pm \sqrt{{\mu }_1{\mu }_2/{\mu }_3}\right).$

(9)

Specifying the terms of the angular velocity perturbation as follows

$\mathrm{\Delta }{x}_j=x_j-X_j;\hspace{1em}\left(j=\mathrm{1,2,3}\right).$

(10)

Then, maintaining ${\mu }_j$ constants, we can get the motion’s linearized equations as follows

$\frac{d}{d\tau }\left[\begin{array}{c}\mathrm{\Delta }{x}_1\\ \mathrm{\Delta }{x}_2\\ \mathrm{\Delta }{x}_3\end{array}\right]=\left[\begin{array}{ccc}0& \left(X_3-a\right)& X_2\\ -\left(X_3-b\right)& 0& -X_1\\ X_2& X_1& 0\end{array}\right]\left[\begin{array}{c}\mathrm{\Delta }{x}_1\\ \mathrm{\Delta }{x}_2\\ \mathrm{\Delta }{x}_3\end{array}\right].$

(11)

The resulting characteristic equation is derived as follows

${\mathit{\lambda }}^{\mathbf{3}}+\left[{\mathit{X}}_{\mathbf{1}}^{\mathbf{2}}-{\mathit{X}}_{\mathbf{2}}^{\mathbf{2}}+\left({\mathit{X}}_{\mathbf{3}}-\mathit{a}\right)\left({\mathit{X}}_{\mathbf{3}}-\mathit{b}\right)\right]\mathit{\lambda }+{\mathit{X}}_{\mathbf{1}}{\mathit{X}}_{\mathbf{2}}\left(\mathbf{2}{\mathit{X}}_{\mathbf{3}}-\mathit{a}-\mathit{b}\right)=\mathbf{0}$

(12)

Table 1 contains a comprehensive list of all feasible combinations of angular velocities, constant torque, and resulting requirements for the presence and stability of the characteristic equation’s roots (12).

Table 1.

Linear stability of various equilibrium points.

Case	Equilibrium	Characteristic formula	Stability	Torque	Notes
1	$\left(\mathrm{0,0,0}\right)$	$s\left(s^2+ab\right)=0$	Unstable	$\left(\mathrm{0,0,0}\right)$	None
2	$\left(X_{1,},\mathrm{0,0}\right)$	$s\left(s^2+X_1^2+ab=0\right)$	Stable	$\left(\mathrm{0,0,0}\right)$	None
3	$\left(0,X_2,0\right)$	$s\left(s^2-X_2^2+ab\right)=0$	Unstable	$\left(\mathrm{0,0,0}\right)$	None
4	$\left(\mathrm{0,0},X_3\right)$	$\begin{array}{l}s[s^2+\left(X_3-a\right)\\ \times \left(X_3-b\right)]=0\end{array}$	Stable	$\left(\mathrm{0,0,0}\right)$	None
5	$\left(0,X_2,X_3\right)$	$\begin{array}{l}s[s^2+\left(X_3-a\right)\\ \times \left(X_3-b\right)-X_2^2]=0\end{array}$	Stable for $\left|X_2\right|<\left|X_3\right|$	$\left(-{\mu }_1,\mathrm{0,0}\right)$	$\begin{array}{l}-{\mu }_1=X_2\left(X_3-a\right),\\ \Vert {\mu }_1\Vert =1\end{array}$
6	$\left(X_1,0,X_3\right)$	$\begin{array}{l}s[s^2+X_1^2+\left(X_3-a\right)\\ \times \left(X_3-b\right)]=0\end{array}$	Stable	$\left(0,{\mu }_2,0\right)$	$\begin{array}{l}{\mu }_2=X_1\left(X_3-b\right),\\ \Vert {\mu }_2\Vert =1\end{array}$
7	$\left(X_1,X_2,0\right)$	$\begin{array}{l}s\left(s^2+X_1^2-X_2^2+ab\right)\\ -\left(a+b\right)X_1{X}_2=0\end{array}$	Stable for $\left|X_1\right|>\left|X_2\right|$	$\left(\mathrm{0,0},-{\mu }_3\right)$	$\begin{array}{l}-{\mu }_3=X_1{X}_2,\\ \Vert {\mu }_3\Vert =1\end{array}$
8	$\left(X_1,X_2,X_3\right)$	Equation (12)	Unstable	$\left({\mu }_1,{\mu }_2,{\mu }_3\right)$	$\begin{array}{l}{\mu }_1{\mu }_2{\mu }_3>0,\\ \Vert \mu \Vert =1\end{array}$

Steady torque on the minor/major axis

Major axis

Along the major axis $\left({\mu }_1,{\mu }_2,{\mu }_3\right)=\left(\mathrm{1,0,0}\right),$ equations (6) for a positive constant torque easily become

$\frac{d{x}_1}{d\tau }-x_2{x}_3+a{x}_2=1,$

$\frac{d{x}_2}{d\tau }+x_3{x}_1-b{x}_1=0,$

$\frac{d{x}_3}{d\tau }-x_1{x}_2=0.$

(13)

It is clear that the previous equations (13) constitute with a hyperbola $-X_1\left(X_3-a\right)=1$ . Moreover, they are stable and unstable at $\left|X_3-a\right|>1$ and $\left|X_2\right|>1$ , respectively. Inserting a new variable ${\phi }_1$ according to ¹

${\phi }_1\left(\tau \right)=\underset{0}{\overset{\tau }{{\displaystyle \int }}}{x}_1\left(\zeta \right)d\zeta ,$

(14)

or $\frac{d{\phi }_1}{d\tau }=x_1\left(\tau \right)$ where ${\phi }_1\left(0\right)=0$ . Making use of (14) into (13) to get

$\frac{d^2{\phi }_1}{d{\tau }^2}-x_2\left(x_3-a\right)=1,$

$\frac{d{x}_2}{d{\phi }_1}+\left(x_3-b\right)=0,$

$\frac{d{x}_3}{d{\phi }_1}-x_2=0.$

(15)

The last two equations of system (15) have the following solutions

$x_2=A\mathit{cos}\left({\phi }_1+\eta \right),x_3=A\mathit{sin}\left({\phi }_1+\eta \right)$

(16)

Here, $A=\sqrt{x_2^2\left(0\right)+x_3^2\left(0\right)}$ and $\eta ={\mathit{tan}}^{-1}\left[x_3\left(0\right)/x_2\left(0\right)\right]$ are known, respectively, by the amplitude and the phase angle of the solutions (16), in which $\left|\eta \right|\le \pi$ and we have

$x_2^2+x_3^2=A^2$

(17)

Based on the ${\phi }_1$ range, a full or a part of the circle is formed on the plane $x_2{x}_3$ by the tip’s projection for the dimensionless angular velocity vector $\left(x_1,x_2,x_3\right)$ .

This circle’s radius $A$ (or part of it) is the motion’s first integral of a rotating RB, where it depends on the starting value of the second and third projections of the dimensionless angular velocity. The substitution of the solutions (16) into the first equation of the system (15) produces

$\frac{d^2{\phi }_1}{d{\tau }^2}-A\hspace{0.17em}\mathit{cos}\left({\phi }_1+\eta \right)\left[A\mathit{sin}\left({\phi }_1+\eta \right)-a\right]=1.$

(18)

Let

$\varphi =2\left({\phi }_1+\eta \right).$

Therefore, we may rewrite equation (18) as follows

$\frac{d^2\varphi }{d{\tau }^2}-A^2\mathit{sin}\varphi +2Aa\mathit{cos}\left(\frac{\varphi }{2}\right)=2.$

(19)

At the angle $\varphi *$ of equilibrium, we get $\mathit{sin}\varphi *=-2/A^2$ . There is undoubtedly no point of equilibrium for $A<\sqrt{2}$ . Consider a new variable $x=\frac{d\varphi }{d\tau }$ to analyze the system’s stability characteristics for the scenario $A\ge \sqrt{2}$ . Therefore, equation (19) should be rewritten in the $\left(\varphi ,x\right)$ phase plane as follows

$\frac{d\varphi }{d\tau }=x.$

(20)

Consequently, equation (19) has the form

$\frac{dx}{d\tau }=A^2\mathit{sin}\varphi -2Aa\mathit{cos}\left(\frac{\varphi }{2}\right)+2.$

(21)

Dividing (21) by (20) to get

$\frac{dx}{d\varphi }=\frac{A^2\mathit{sin}\varphi -2Aa\mathit{cos}\left(\frac{\varphi }{2}\right)+2}{x}.$

(22)

Observing that for a specific trajectory, the first integral of the motion $A$ does not change. Separation of the variables and the integration of (21) yields

$\left(\frac{x^2}{2}\right)+A^2\mathit{cos}\varphi +4Aa\mathit{sin}\left(\frac{\varphi }{2}\right)-2\varphi =E.$

(23)

It is noted that the point $\left({\varphi }^{*},x^{*}\right)=\left[{\mathit{sin}}^{-1}\left(\frac{-2}{A^2}\right),0\right]$ in the phase plane $\left(\varphi ,x\right)$ , as seen in Figure 2, determines the first equilibrium point which is located on the left of the origin, in which $\left|{\mathit{sin}}^{-1}\psi \right|\le \frac{\pi }{2};\forall \psi$ . The center points are situated at $\pi -{\varphi }^{*}\pm 2n\pi ;\left(n=\mathrm{1,2},...,\infty \right)$ , whereas the saddle points are situated at ${\varphi }^{*}\pm 2n\pi$ . A saddle point ${\varphi }_s$ that is related to the energy constant is represented as follows

$E_s=A^2\mathit{cos}{\varphi }_s+4Aa\mathit{sin}\left(\frac{{\varphi }_s}{2}\right)-2{\varphi }_s.$

(24)

Additionally, the separator equation through ${\varphi }_s$ has the form

$E_s=\left(\frac{x^2}{2}\right)+A^2\mathit{cos}{\varphi }_s+4Aa\mathit{sin}\left(\frac{{\varphi }_s}{2}\right)-2{\varphi }_s.$

(25)

Figure 2.

Depicts finite and infinite tracks for a constant major torque axis at distinct values of $\left[x_1\left(0\right),x_2\left(0\right),x_3\left(0\right)\right]$ .

Making use of (24) and (25), to obtain

$x=\pm \sqrt{2}A\sqrt{\left(\mathit{cos}{\varphi }_s-\mathit{cos}\varphi \right)+2a\left[\frac{\mathit{sin}\left(\frac{{\varphi }_s}{2}\right)-\mathit{sin}\left(\frac{\varphi }{2}\right)}{A}\right]+\left({\varphi }_s-\varphi \right)\mathit{sin}{\varphi }_s}.$

(26)

Remembering that $\varphi \left(0\right)=2\eta$ where $\left|\eta \right|\le \pi$ , the only relevant two separators for stability analysis are the left one (which passes via the left saddle point at ${\varphi }_{S_L}={\varphi }^{*}$ and wraps the left center ${\varphi }_{C_L}=-{\varphi }^{*}-\pi$ ), and the right one (that is passes via the right saddle point at ${\varphi }_{S_R}={\varphi }^{*}+2\pi$ and wraps the right center at ${\varphi }_{C_R}=-{\varphi }^{*}+\pi$ ). The right and left separators (RLS) for $A=\sqrt{9.25}$ , are graphed in Figure 3, which are ${\varphi }_{V_R}$ and ${\varphi }_{V_L}$ which, respectively, correspond to the vertical crossing of the RLS.

Figure 3.

Describes RLS with finite and infinite tracks for a constant major torque axis at distinct values of $\left[x_1\left(0\right),x_2\left(0\right),x_3\left(0\right)\right]$ .

Considering $x=2x_1,\mathit{cos}\varphi =\frac{1-{\mathit{tan}}^2\left(\frac{\varphi }{2}\right)}{1+{\mathit{tan}}^2\left(\frac{\varphi }{2}\right)}=\frac{x_2^2-x_3^2}{A^2}$ and $\varphi =2{\mathit{tan}}^{-1}\frac{x_3}{x_2}$ in the second integral equation (23) to yield

$2x_1^2+x_2^2-x_3^2+4a{x}_3-4{\mathit{tan}}^{-1}\left(\frac{x_3}{x_2}\right)=E,$

(27)

which depicts an energy constant trajectory in 3D. Taking into account that $\mathit{cos}{\varphi }_s=\sqrt{\frac{\left(A^4-4\right)}{A^2}}$ , then from equations (25) and (27) one obtains

$2x_1^2+x_2^2-x_3^2+4{ax}_3-4{\mathit{tan}}^{-1}\left(\frac{x_3}{x_2}\right)-\sqrt{{\left(x_2^2+x_3^2\right)}^2-4}-2{\mathit{Sin}}^{-1}\frac{2}{\left(x_2^2+x_3^2\right)}\pm \frac{4a}{\sqrt{2}}\sqrt{x_2^2+x_3^2-\sqrt{{\left(x_2^2+x_3^2\right)}^2-4}}=0,$

(28)

which is an illustration of the left separator (LS) in 3D. By adjusting the right side of equation (28) from $0$ to $-4\pi$ , the appropriate right separator (RS) can be produced.

The 3D surfaces of the RLS in the $\left(x_1,x_2,x_3\right)$ phase space are depicted in Figure 4. The inspection of the present figure shows that the $\left(x_1,x_2,x_3\right)$ space is divided by two surfaces into three infinite regions: two of them are stable, and they are separated by the unstable one. If the motion has initial circumstances that are contained within one of these zones, it will remain within them. A spin-up maneuver will arise from any motion besides its initial conditions in the unstable domain, but every motion that has initial conditions in one of the stable domains will result in a closed and constrained trajectory that is contained in that region and rotates around its corresponding center. For instance, in the right stable zone, a trajectory will revolve around a center at $\left(0,-\sqrt{A^2-\sqrt{A^4-4}/2},\sqrt{A^2+\sqrt{A^4-4}/2}\right)$ .

Figure 4.

Shows the surfaces of RLS for a constant major torque axis.

For initial circumstances inside the unstable zone, the tip’s projection of the dimensionless vector $\left(x_1,x_2,x_3\right)$ of angular velocity on the $\left(x_2,x_3\right)$ plane represents an entire circle, and for starting circumstances inside either of the stable zones, it represents a portion of a circle. This circle’s radius A can be as large as possible, and the motion it produces will depend on whether it’s stable or unstable. The segment of the plane $\left(x_2,x_3\right)$ contained in the unstable zone will be crossed by any route with initial conditions $\left[x_1\left(0\right)<0,x_2\left(0\right),x_3\left(0\right)\right]$ in the lower unstable region. In other words, the gap that separated between the surfaces of two separators, located at a distance of $A={\left[{{\displaystyle \sum }}_{j=2}^3\hspace{0.17em}{x}_j^2\left(0\right)\right]}^{1/2}$ from the origin of the coordinate system $\left(x_1,x_2,x_3\right)$ . The distance between the surfaces of two separators reaches $\pm \infty$ along the $x_2$ -axis because it is assumed to be arbitrarily large. Furthermore, it is noted that every equilibrium point located within the stable domain, which has been demonstrated to be Lyapunov stable for the linearized system, is also Lyapunov stable for the nonlinear system.

A contour map of the upper portion $\left(x_1>0,x_2,x_3\right)$ of the RLS surfaces is shown in Figure 5. It is observed that the cylindrical coordinates are presented, in which $A={\left[{{\displaystyle \sum }}_{j=2}^3\hspace{0.17em}{x}_j^2\left(0\right)\right]}^{1/2}$ and $\eta ={\mathit{tan}}^{-1}\frac{x_3\left(0\right)}{x_2\left(0\right)}$ , inside the circle of radius $A=\sqrt{2}$ there are no points of equilibrium, the system’s points of equilibrium are located on the two branches of the hyperbola $x_2{x}_3=-1$ , and the line $x_2+x_3=0,$ divided the hyperbola branches into Lyapunov stable and unstable domains.

Figure 5.

Represents the RLS surfaces contour plots of $x_1$ at a stationary torque of the major axis.

It is possible to determine that the intersection of the two separators surfaces and the plane $\left(x_2,x_3\right)$ represents the solution that can be obtained from (26) to the transcendental equations given by

$\left(\mathit{cos}{\varphi }_s-\mathit{cos}\varphi \right)+2a\left[\frac{\mathit{sin}\left(\frac{{\varphi }_s}{2}\right)-\mathit{sin}\left(\frac{\varphi }{2}\right)}{A}\right]+\left({\varphi }_s-\varphi \right)\mathit{sin}{\varphi }_s=0.$

(29)

It must be mentioned that ${\varphi }_s=\varphi$ is considered a solution to (29), which produces

$\mathit{sin}{\varphi }_s=\mathit{sin}\varphi =\frac{2\mathit{tan}\frac{\varphi }{2}}{1+{\mathit{tan}}^2\frac{\varphi }{2}}=\frac{2x_2{x}_3}{{\displaystyle \sum }_{j=2}^3{x}_j^2}.$

(30)

Since $\mathit{sin}{\varphi }_s=-2/{{\displaystyle \sum }}_{j=2}^3\hspace{0.17em}{x}_j^2$ , then $x_2{x}_3=-1$ must be chosen to satisfy the previous equation (30). Here, the unstable region of the equilibrium points is located on the hyperbola $x_2{x}_3=-1;\left(\left|x_2\right|>1\right)$ i.e., at the point where the $x_2$ plane and the RLS surfaces meet. Equation (29) can be solved numerically to obtain the intersection of the two separators surfaces and the plane $\left(x_2,x_3\right)$ at $\left|x_3\right|>1$ .

Remember that each separator phase plane map, including Figure 3, was created for a specific value of $A$ , while Figures 4 and 5 show every conceivable value for $0\le A\le \infty$ . A circle $A=\sqrt{9.25}$ will meet the saddle, center, and vertical crossing (SCVC) curves which are shown in Figure 5 at the SCVC points connected to the LRS of Figure 3.

At $A=\sqrt{2}$ one can find that the SCVC that are related to the left dividing surface all breakdown into a single point at $\left(x_1,x_2,x_3\right)=\left(\mathrm{0,1},-1\right)$ , while the SCVC that is related to the right dividing surface all breakdown into $\left(x_1,x_2,x_3\right)=\left(0,-\mathrm{1,1}\right)$ . Consequently, it is possible to create the saddle, center, and crossing curves shown in Figure 5 by beginning at $A=\sqrt{2}$ in Figure 3 and the polar coordinates $\left(A,\eta =\varphi /2\right)$ that the saddle, center, and crossing points fall inside as $A$ is raised. An initial motion that is out from the separator surface will be contained by that surface and will asymptotically converge to the point of equilibrium at the intersection of $x_2^2+x_3^2=A^2$ with the unstable branch of the hyperbola $x_2{x}_3=-1$ , which is confined within that surface. For the situation of stationary main axis torque, for instance, the motion initiated from the right separators will converge to $\left(x_1,x_2,x_3\right)=\left(0,-\sqrt{\frac{A^2+\sqrt{A^4-4}}{2}},\sqrt{\frac{A^2-\sqrt{A^4-4}}{2}}\right)$ .

Analytical formulas for the extrema of the different components of the dimensionless angular speed along a periodic solution are provided in Table 2.

Table 2.

Represents the extreme values of $x_j\hspace{0.17em}\left(j=\mathrm{1,2,3}\right)$ at $\mu =\left(\mathrm{1,0,0}\right)$ as Figure 5.

Item	Analytic statement	Qualitative value
General	$\begin{array}{l}A=\sqrt{x_2^2\left(0\right)+x_3^2\left(0\right)},\\ 2x_1^2\left(0\right)+x_2^2\left(0\right)-x_3^2\left(0\right)+4a{x}_3\left(0\right)-4{\tan }^{-1}\left[x_3\left(0\right)/x_2\left(0\right)\right]=E\end{array}$	$\begin{array}{l}x\left(0\right)=\left(\mathrm{1,0.5,3}\right),A=5\\ x\left(0\right)=\left(\mathrm{1,0.5,3}\right),x_3\left(0\right).\end{array}$
$x_{1_{\max }},x_{1_{\min }}$	$\begin{array}{l}{x}_1=\pm \sqrt{\frac{E+4\pi -4a{x}_3+\sqrt{A^2-4}}{2}-2{\tan }^{-1}\left(\frac{A^2+\sqrt{A^2-4}}{2}\right)},\\ x_2=-\sqrt{\frac{A^2-\sqrt{A^4-4}}{2}},x_3=\sqrt{\frac{A^2+\sqrt{A^4-4}}{2}}\end{array}$	$x=\left(\mathrm{1.04,0},-0.58\right)$
$x_{2_{\max }},x_{3{\text{local}}_{\min }}$	$\begin{array}{l}{x}_1=0,\\ 2x_2^2-4{\tan }^{-1}\left(\frac{\sqrt{A^2-x_2^2}}{x_2}\right)=E+4n\pi +A^2\end{array}$	$x=\left(\mathrm{0,0.97,2.88}\right),n=0$
$x_{2_{\min }},x_{3_{\min }}$	$x_3=\sqrt{A^2-x_2^2},n=0,\pm 1$	$x=\left(0,-\mathrm{1.65,2.44}\right),n=1$
$x_{3_{\max }}$	$x_1=\pm \sqrt{\frac{E+A^2+4aA+4\pi }{2}},x_2=0,x_3=A$	$x=\left(\pm \mathrm{1.26,0,3.44}\right)$

Minor axis

Due to the specific form of equations (6), the solution for the scenario in which there is an operating stationary torque along the minor axis may not be derived by swapping the indexes 3 and 1 in the previous derivation.

A stationary torque on the middle axis

For the scenario of a stationary positive torque that is directed on the middle axis, one can write $\left({\mu }_1,{\mu }_2,{\mu }_3\right)=\left(\mathrm{0,1,0}\right)$ . Therefore, equations (6) become

$\frac{d{x}_1}{d\tau }-x_2{x}_3+a{x}_2=0,$

$\frac{d{x}_2}{d\tau }+x_3{x}_1-b{x}_1=1,$

$\frac{d{x}_3}{d\tau }-x_1{x}_2\hspace{1em}=0.$

(31)

Here, the system’s equilibrium points generate the following hyperbola, which corresponds to the case 6 in Table 1.

$X_1(X_3-b)=1.$

(32)

As a result, Lyapunov stability exists at the points of equilibrium for the linearized system that is associated with the previous nonlinear system (31). Inserting an additional parameter ${\varphi }_2$ , as follows ¹

${\varphi }_2\left(\tau \right)=\underset{0}{\overset{\tau }{{\displaystyle \int }}}{x}_2\left(\xi \right)d\xi .$

(33)

According to this definition, one can write equations (31) as follows

$\frac{d{x}_1}{d{\varphi }_2}-x_3+a=0,$

$\frac{d^2{\varphi }_2}{d{\tau }^2}+x_3{x}_1-b{x}_1=1,$

$\frac{d{x}_3}{d{\varphi }_2}-x_1=0.$

(34)

The first and third equations in (34) have the following solutions

$x_1=x_1\left(0\right)\mathit{cos}h{\varphi }_2+x_3\left(0\right)\mathit{sin}h{\varphi }_2,$

$x_3=a+x_1\left(0\right)\mathit{sin}h{\varphi }_2+x_3\left(0\right)\mathit{cos}h{\varphi }_2.$

(35)

These solutions satisfy the following form

$x_1^2-x_3^2=x_1^2\left(0\right)-x_3^2\left(0\right)-a\left\{a+2\left[x_1\left(0\right)\mathit{sin}h{\varphi }_2+x_3\left(0\right)\mathit{cos}h{\varphi }_2\right]\right\}.$

(36)

The substitution of the solutions (35) into the second equation in (31) yields

$\frac{d^2{\varphi }_2}{d{\tau }^2}=1-\left(A\mathit{sin}h2{\varphi }_2+B\mathit{cos}h2{\varphi }_2\right)-\left(a-b\right)\left[x_1\left(0\right)\mathit{cos}h{\varphi }_2+x_3\left(0\right)\mathit{sin}h{\varphi }_2\right];$

$A=\frac{x_1^2\left(0\right)+x_3^2\left(0\right)}{2},B=x_1\left(0\right)x_3\left(0\right).$

(37)

Paths of $\left|x_1\left(0\right)\right|\ne \left|x_3\left(0\right)\right|$

In the present scenario, $A>B$ , and then equation (37) can be amended to produce

$\frac{d^2{\varphi }_2}{d{\tau }^2}=1-\sqrt{A^2-B^2}\mathit{sin}h\left(2{\varphi }_2+\eta \right)-\left(a-b\right)\sqrt{2C}\mathit{sin}h\left({\varphi }_2+\eta \right),$

(38)

where

$C=\frac{\left|x_1^2\left(0\right)-x_3^2\left(0\right)\right|}{2},\eta ={\mathit{tan}h}^{-1}\left(\frac{B}{A}\right).$

Let us consider the following notations

$\mathrm{\Theta }=2{\varphi }_2+\eta ,A^2=B^2+C^2,$

(39)

To rewrite equation (38) as follows

$\frac{d^2\mathrm{\Theta }}{d{\tau }^2}=2\left[1-C\mathit{sin}h\mathrm{\Theta }-\left(a-b\right)\sqrt{2C}\mathit{sin}h\left(\frac{\mathrm{\Theta }+\eta }{2}\right)\right].$

(40)

The point equilibrium for the defined conservative system by equation (40) has the form

${\mathrm{\Theta }}^{*}={\mathit{sin}h}^{-1}{\left[C+\left(a-b\right)\sqrt{2C}\right]}^{-1},$

(41)

where $C$ is the motion’s first integral.

Since $-\infty <\mathit{sinh}x<\infty$ , there is only one associate point of equilibrium for each set of the starting conditions $\left[x_1\left(0\right),x_3\left(0\right)\right]$ . Therefore, each set of these conditions has a specific point of equilibrium, which is defined by equation (41). Additionally, if our trajectory is started at the system’s point of equilibrium, we get ${\mathrm{\Theta }}^{*}={\varphi }_2\left(\tau \right)={\varphi }_2\left(0\right)=\eta$ , which produces

${\left[C+\left(a-b\right)\sqrt{2C}\right]}^{-1}=\mathit{sin}h\eta =\frac{\mathit{tan}h\eta }{\sqrt{1-{\mathit{tan}h}^2\eta }}=B{\left[C+\left(a-b\right)\sqrt{2C}\right]}^{-1},$

(42)

which demonstrates

$B=X_1\left(X_3-b\right)=1,$

(43)

which is the same as in equation (32). Based on the right side of equation (40), which represents a conservative field of force, the system has a constant energy that can be determined. For this purpose, equation (40) can be rewritten as follows

$\frac{d\mathrm{\Theta }}{d\tau }=y.$

(44)

Making use of equations (40) and (44) to obtain

$\frac{dy}{d\tau }=2\left[1-C\hspace{0.17em}\mathit{sin}h\hspace{0.17em}\mathrm{\Theta }-\left(a-b\right)\sqrt{2C}\mathit{sin}h\left(\frac{\mathrm{\Theta }+\eta }{2}\right)\right]$

(45)

Based on the above two equations, one can get

$\frac{dy}{d\mathrm{\Theta }}=\frac{2}{y}\left[1-C\hspace{0.17em}\mathit{sin}h\mathrm{\Theta }-\left(a-b\right)\sqrt{2C}\mathit{sin}h\left(\frac{\mathrm{\Theta }+\eta }{2}\right)\right].$

(46)

Separating the variables in (46) and integrating the result to yield

$\frac{y^2}{2}-2\left[\mathrm{\Theta }-C\mathit{cos}h\mathrm{\Theta }+2\left(a-b\right)\sqrt{2C}\mathit{cos}h\left(\frac{\mathrm{\Theta }+\eta }{2}\right)\right]=E_1,$

(47)

where $E_2$ represents the energy constant (not its actual energy) in its dimensionless form and expresses the motion’s second integral of the system. On the other side, the system’s dimensionless potential energy (not its actual potential energy) is expressed as

$V=-2\left[\mathrm{\Theta }-C\mathit{cos}h\mathrm{\Theta }+2\left(a-b\right)\sqrt{2C}\mathit{cos}h\left(\frac{\mathrm{\Theta }+\eta }{2}\right)\right].$

(48)

Therefore, one can directly obtain

$\frac{d^{2}V}{d{\mathrm{\Theta }}^2}=2C\mathit{cos}h\mathrm{\Theta }-\left(a-b\right)\sqrt{2C}\mathit{cos}h\left(\frac{\mathrm{\Theta }+\eta }{2}\right)>0$

(49)

According to the right-hand side of equation (49), the constraints on the principal values of inertia, and the expressions of $a$ and $b$ , one can write $\frac{d^{2}V}{d{\theta }^2}>0$ .

It is clear that the point of equilibrium ${\mathrm{\Theta }}^{*}$ is stable. Accordingly, each beginning condition for $\left|x_1\left(0\right)\right|\ne \left|x_3\left(0\right)\right|$ generates a closed periodic trajectory with an oval shape centered on a particular Lyapunov stable point of equilibrium. In light of the fact that any point on a path can also be thought of as a set of beginning circumstances for this path, the relations $y=2x_2$ allow us to formulate (47) as follows

$2x_2^2+\left(x_1^2+x_3^2\right)\left[1-2\sqrt{\frac{2}{C}}\left(a-b\right)\right]-2{\mathit{tan}h}^{-1}\left(\frac{2x_1{x}_3}{x_1^2+x_3^2}\right)=E_2.$

(50)

This equation describes a closed path with the constant $E_2$ of energy in 3D, in which it is obtained for the case of stationary torque along the middle axis besides the beginning circumstances that meet with $\left|x_1\left(0\right)\right|\ne \left|x_3\left(0\right)\right|$ . It is obvious that any motion started on this constrained path will stay contained within it.

Stable and unstable separator

Stable separator

The scenario $x_1\left(0\right)=x_3\left(0\right)$ leads to a pure spin up around the middle axis, as observed from (31). At $x_1\left(0\right)=x_3\left(0\right)\ne 0$ , equations (35) lead to

$x_1=x_1\left(0\right)e^{{\varphi }_2},$

$x_3=a+x_1\left(0\right)e^{{\varphi }_2}.$

(51)

Then $x_1=x_3-a$ and consequently $C=0$ , where $C$ is the motion’s first integral. Since $A=B$ in this situation, equation (37) produces

$\frac{d^2{\varphi }_2}{d{\tau }^2}=1-A{e}^{2{\varphi }_2}-\left(a-b\right)x_1\left(0\right)e^{{\varphi }_2}.$

(52)

At equilibrium, there is

$x_1^{*}=x_1\left(0\right)e^{{\varphi }_2^{*}}=x_1\left(0\right)\Rightarrow {\varphi }_2^{*}=0,$

$x_3^{*}=a+x_3\left(0\right)e^{{\varphi }_2^{*}}=a+x_3\left(0\right)\Rightarrow {\varphi }_2^{*}=0.$

(53)

Combining this equation with equation (37) produces

$A=B=x_1\left(0\right)x_3\left(0\right)=X_1{X}_3=1.$

(54)

According to an analysis of equation (51) any motion initiated from the surface $x_1\left(0\right)=x_3\left(0\right)$ won’t alter the sign; depending upon its original circumstances, it will keep in either the upper or lower section of this area during the entire time $\tau$ .

$\frac{d{\varphi }_2}{d\tau }=x_2,$

(55)

$\frac{d{x}_2}{d\tau }=1-x_1^2\left(0\right)e^{2{\varphi }_2}-\left(a-b\right)x_1\left(0\right)e^{{\varphi }_2}.$

(56)

Using equations (55) and (56) to obtain

$\frac{d{x}_2}{d{\varphi }_2}=\frac{1-x_1^2\left(0\right)e^{2{\varphi }_2}-\left(a-b\right)x_1\left(0\right)e^{{\varphi }_2}}{x_2}.$

(57)

As a result, separating the variables and integrating the result to obtain

$\frac{x_1^2}{2}+\frac{x_2^2}{2}-\mathit{ln}{x}_1+\left(a-b\right)x_1=E,$

(58)

where the motion’s second integral in a dimensionless form is denoted by the symbol $E$ .

Similar stability analysis to that accomplished for $x_1\left(0\right)\ne x_3\left(0\right)$ , indicates that every set of initial requirements that fulfils $x_1=x_3>0$ leads to an oval-shaped closed periodic solution that is constrained to the plane $x_1=x_3>0$ and envelops the point of equilibrium $\left(\mathrm{1,0,1}\right)$ located in such a plane. In contrast, every set of initial requirements satisfying $x_1=x_3<0$ leads to a closed periodic solution with an oval shape that is contained within $x_1=x_3<0$ and surrounds the equilibrium point $\left(-\mathrm{1,0},-1\right)$ within. For the scenario of constant middle axis torque with $x_1=x_3>0$ , analytical formulas for the extreme of the various dimensionless projections of the angular velocity are provided in Table 3.

Table 3.

Represents the extreme value of $x_j\hspace{0.17em}\left(j=\mathrm{1,2,3}\right)$ at $\mu =\left(\mathrm{0,1,0}\right)$ and $x_1\left(0\right)=x_3\left(0\right)\ne 0$ .

Item	Analytic expression	Numerical value
General	$x_1=x_3,E=\frac{x_1^2\left(0\right)+x_2^2\left(0\right)-\ln x_1^2\left(0\right)+\left(a-b\right)x_1}{2}$	$\begin{array}{l}x\left(0\right)=\left(\mathrm{2,1,2}\right),\\ E=1.8069\end{array}$
$x_{1_{\max }},x_{1_{\min }}$	$\frac{x_1^2}{2}-\ln x_1+\left(a-b\right)x_1=E,x_1=x_3,x_2=0$	$\begin{array}{l}x\left(0\right)=\left(\mathrm{0.17,0,0.17}\right),\\ A=\left(2.30,0,2.30\right)\end{array}$
$x_{2_{\max }},x_{3{\text{local}}_{\min }}$	$x_1=1,x_2=\pm \sqrt{2\left(E+a+b\right)-1},x_3=1$	$\begin{array}{l}x=\left(\mathrm{1,0},\pm 1.62,1\right),\\ A=\left(2.30,0,2.30\right)\end{array}$

Unstable separator

At $x_1\left(0\right)=-x_3\left(0\right)$ , equations (35) give

$x_1=x_1\left(0\right)e^{-{\varphi }_2},$

$x_3=a-x_1\left(0\right)e^{-{\varphi }_2}.$

(59)

Hence $x_3=a-x_1$ . Since $A=-B$ in this scenario, equation (37) produces

$\frac{d^2{\varphi }_2}{d{\tau }^2}=1-A{e}^{-2{\varphi }_2}-\left(a-b\right)x_1\left(0\right)e^{-{\varphi }_2}.$

(60)

It is obvious that there are no equilibrium points in this situation. A closer look at equation (60) demonstrates that

$\underset{\tau \to \infty }{\mathit{lim}}\frac{d^2{\varphi }_2}{d{\tau }^2}=1.$

(61)

The substitution of equation (61) into equation (59) yields

$\underset{\tau \to \infty }{\mathit{lim}}{x}_1=\underset{\tau \to \infty }{\mathit{lim}}{x}_3=0.$

(62)

In the case where $x_1\left(0\right)=-x_3\left(0\right)$ , a steady torque on the middle axis produces a motion that quickly converges to a pure spin-up maneuver with a steady angular acceleration.

Torque around major/minor via middle axes

The significant changes in sign between the equations (15) for minor/major axis torques and those for the middle axis equations (34) are due to the features of the paths related to major/minor axis torqueing via the middle axis.

Since their related formulations of equations (15) and (34), respectively, have different signs, major or minor axis torqueing and middle axis torques have different path properties. The distinct linear homogeneous part of the equations of motion is caused by these sign changes, and this leads to the harmonic solution of the equations (16) for the major or minor axis torqueing in comparison to the hyperbolic solution of the equations (35) for the middle axis torqueing. The circle’s radius (or part of a circle) on which the $\left(x_2,x_3\right)$ projection of the harmonic solution is located corresponds to the initial integral of motion in the minor or major axis torqueing issue, while the initial integral of motion in the intermediate axis torqueing issue is the $\left(x_1,x_3\right)$ projection of the hyperbolic solution, which is on a rectangular hyperbola with those axes as its traverse (and conjugate) axes. The motion’s second integral is formulated through the substitution of the harmonic and hyperbolic solutions into the related (nonlinear, nonhomogeneous) differential equations (15) and (34), which results in differing stability qualities for the two situations.

As was previously mentioned, the left and right separators shown in Figures 4 and 5 are the result of the minor or major axis torqueing. The two curved separator surfaces, which do not intersect, divide the $\left(x_1,x_2,x_3\right)$ state space into three infinite areas, each of which consists of two stable areas split by an unstable area. In the scenario of the middle axis torqueing, the separator surfaces can be made up of two perpendicular planes that cross on the $x_2$ axis and divide the $\left(x_1,x_2,x_3\right)$ space into four different (and infinite) areas. Each of these locations produces a motion that has a steady, localized periodic path. The only instability in this scenario comes on paths starting from the unstable separator $x_1=-x_3$ and quickly convergent to a spin-up maneuver on the middle axis.

Conclusions

Under the impact of a GM vector and a steady body-fixed torque, the problem of the rotatory dynamical motion of a generic RB has been examined. The current work’s main contribution is the transition of the stability analysis of this body from a 2D phase plane to a 3D phase space. For the situation of a time-varying body stationary torque, Euler’s rotational EOM were modified to remove their dependency on inertia constraints. These equations have been rewritten in dimensionless forms for the situation of stationary body-fixed torque to get rid of their dependency on both the inertia parameters and the torque’s magnitude. The fundamental aspects like separators, equilibrium, and the integrals of motion that related to the system’s paths were converted from their 2D into 3D phase space. To demonstrate these aspects and learn more about how they interact in this fundamental yet extremely nonlinear situation, the dimensionless equations and the brand-new 3D phase space visualizations were used. Additionally, the extremes of the dimensionless periodic solution are also thoroughly analyzed.