# New Treatment of the Rotary Motion of a Rigid Body with Estimated Natural Frequency

Authors: A. I. Ismail
Journal: Advances in Astronomy
 (2020)

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## Abstract

In this paper, the problem of the motion of a rigid body about a fixed point under the action of a Newtonian force field is studied when the natural frequencyω=0.5. This case of singularity appears in the previous works and deals with different bodies which are classified according to the moments of inertia. Using the large parameter method, the periodic solutions for the equations of motion of this problem are obtained in terms of a large parameter, which will be defined later. The geometric interpretation of the considered motion will be given in terms of Euler’s angles. The numerical solutions for the system of equations of motion are obtained by one of the well-known numerical methods. The comparison between the obtained numerical solutions and analytical ones is carried out to show the errors between them and to prove the accuracy of both used techniques. In the end, we obtain the case of the regular precession type as a special case. The stability of the motion is considered by the phase diagram procedures.


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## Body

## 1. Introduction

Consider a rigid body of massM moves in an asymmetric field around a fixed point O [1]. Let us assume that the surface of its ellipsoid of inertia is optional, as well as the mass center. Let the frame OX,OY,andOZ be a fixed system in space, and the frame Ox,Oy,andOz is the main axes frame for the surface of the ellipsoid of inertia of the body which moves with the it. Initially, we consider the main axis z for the surface of the ellipsoid of inertia that makes an angle ξ0≠kπ/2;k=0,1,and2 with the fixed axis Z in space. Let the body spins with small speed angular velocity r0 about the axis z. Suppose that p,q,andr represent the components of the angular velocity vector of the body about the main axes of the ellipsoid of the inertia surface; γ,γ′,andγ″ are the directional cosines vector of the axis Z; g is the acceleration of gravity; A,B,andC are the principal moments of inertia. The point (x0,y0,z0) is the center of mass in the moving coordinate system; R¯ is the position vector of the center of attraction 01 on the fixed downward coordinate Z axis, and ρ¯ is the position vector of the element dm. Let i^¯,j^¯,k^¯, and Z¯^ be the unit vectors in the shown directions (Figure 1). Consider dF¯ is the attraction force element due to the attracting center and acted on the element dm at the point px,y,z.Figure 1
Description of motion in terms of moving and fixed frames.


## 2. Formulation of the Problem

Without a loss of generality, we choose the positive direction of both the axisz and the axis x that do not make an obtuse angle ξ0 with the direction of axis Z. Under the restriction on ξ0 and the choice of the coordinate system, we get [2](1)γ0≥0,0<γ0″<1.The differential equations of motion can be reduced to an autonomous system of two degrees of freedom and one first integral as follows [3]:(2)4p¨2+p2=4ε−2Fp2,p˙2,γ2,γ˙2,ε,γ¨2+γ2=ε−2Φp2,p˙2,γ2,γ˙2,ε,(3)γ22+γ˙22+2ε−1νp2γ2+ν2p˙2γ˙2+s21+ε−2…=γ0″−2−1,where(4)F=C1A1−1p2p˙22+x0′p˙2γ˙2−y0′a−1p2γ˙2−y0′A1−1A1+a−1γ2p˙2−z0′a−1p2−0.75νe1p2−0.25p2s11+A1b−1x0′s21+Oε−1+⋯,Φ=−1−C1A1−1p2p˙2γ˙2+x0′γ˙22−y0′γ2γ˙2−z0′b−1γ2+x0′b−1−A1−2γ2p˙22+0.75νe+e1γ2−γ2s11+1+B1p2s21+Oε−1+⋯,(5)p2=p1−ε−1e+e1γ2,γ2=γ1−ε−1νp2,q1=−A1−1p˙2+ε−1A1−1y0′a−1−e2γ˙2+⋯,r1=1+0.5ε−2s11+⋯,γ1′=γ˙2+ε−1ν2p˙2+⋯,γ1″=1+ε−1s21+ε−2s22−0.5s11+⋯,(6)p1=pcγ0″,pq,r1=rr0,γ1=γγ0″,γγ′γ″,τ=r0−1t,.≡ddτ;(7)s11=ap202−p22+bp˙202−p˙22A12−2x0′γ20−γ2+y0′γ˙20−γ˙2,s21=ap20γ20−p2γ2−bA1−1p˙20γ˙20−p˙2γ˙2,s22=aνp202−p22+eγ20−γ2+e1γ202−γ22+bA1−1−ν2p˙202−p˙22+a−1y0′γ˙20−γ˙2−e2γ˙202−γ˙22,(8)A1=C−BA,ABC,a=AC,ab,c2=MglC,ε=cγ0″r0,x0=lx0′,xyz,l2=x02+y02+z02,4A1B1=−1,eb=4x0′A1,3ν=41+B1,3e1=4z0′A1b−1−a−1,e2=e1+a−1z0′,ν2=ν−A1−1.The symbols like ABC are abbreviated equations.


## 3. Construction of Periodic Solutions with Zeros Basic Amplitudes

In this section, we use the suggested method for constructing the aimed solutions for the autonomous system (2). Consider the condition [4](9)p20,0=p˙20,0=γ˙20,ε=0.The generating system for (2) is obtained when ε⟶∞ as follows:(10)4p¨20+p20=0,γ¨20+γ20=0.The solutions for system (10) with a period T0=4π are(11)p20=a0∗cos0.5τ,γ20=b0∗cosτ,where a0∗andb0∗ are constants.Let system (2) has periodic solutions with a period T0+α in the form [5](12)p2=a∗cosψ+∑n=1Nε−npn∗a∗,ψ+Oε−N−1,γ2=b∗cosϕ+∑n=1Nε−nγn∗a∗,ϕ+Oε−N−1.For these solutions, we let the initial conditions(13)p20,ε=a∗=a0∗+a∗ε,γ20,ε=b∗=b0∗+b∗ε,γ˙20,ε=0.Here,a∗ε,b∗ε⟶0 at ε⟶∞. Considering first integral (3) with conditions (13), we get(14)0<b0∗=1−γ0″21/2γ0″−1<∞,b∗ε=−ε−1νa0∗+a∗ε+⋯.Leta∗, ψ, and ϕ are changed with time according to(15)a˙∗=∑n=1Nε−nAn∗a∗+Oε−N−1,(16)ψ˙=0.5+∑n=1Nε−nψna∗+Oε−N−1,(17)ϕ˙=1+∑n=1Nε−nϕna∗+Oε−N−1.The following derivatives are obtained:(18)p˙2=−0.5a∗sinψ+Oε−1,γ˙2=−b∗sinϕ+Oε−1,p¨2=−0.25a∗cosψ+ε−10.25∂2p1∗∂ψ2−a∗ψ1cosψ−A1∗sinψ+ε−2A1∗∂2p1∗∂a∗∂ψ−A2∗+2A1∗ψ1sinψ+A1∗dA1∗da∗cosψ+0.25∂2p2∗∂ψ2+ψ1∂2p1∗∂ψ2−a∗ψ12+2ψ2cosψ−a∗A1∗sinψdψ1da∗+Oε−3,γ¨2=−b∗cosϕ+ε−1∂2γ1∗∂ϕ2−2b∗ϕ1cosϕ+ε−2∂2γ2∗∂ϕ2+2ϕ1∂2γ1∗∂ϕ2−b∗ϕ12+2ϕ2cosϕ+2A1∗∂2γ1∗∂a∗∂ϕ−b∗A1∗dϕ1da∗sinϕ+Oε−3.Using equations (7), (12), and (18), we get(19)s110=aa0∗2cos2ψ0−cos2ψ−0.25bA1−2a0∗2sin2ψ−2b0∗x0′cosϕ0−cosϕ+y0′sinϕ,s210=a0∗b0∗acosψ0cosϕ0−cosψcosϕ+0.5bA1−1sinψsinϕ,s220=aνa0∗2cos2ψ0−cos2ψ+eb0∗cosϕ0−cosϕ+e1b0∗2cos2ϕ0−cos2ϕ+bA1−10.25ν2a0∗2sin2ψ+a−1y0′b0∗sinϕ+e2b0∗2sin2ϕ,where ψ0 and ϕ0 are the initial values of the corresponding functions.Using (4), (12), (18), and (19), we obtain(20)F0=0.25C1A1−1a0∗3cosψsin2ψ+0.5a0∗b0∗x0′sinψsinϕ+a−1a0∗b0∗y0′cosψsinϕ+0.5A1−1A1+a−1a0∗b0∗y0′sinψcosϕ−z0′a−1a0∗cosψ−0.75νe1a0∗cosψ−0.25a0∗cosψaa0∗2cos2ψ0−cos2ψ−0.25bA1−2a0∗2sin2ψ−2b0∗x0′cosϕ0−cosϕ+y0′sinϕ+A1b−1x0′a0∗b0∗acosψ0cosϕ0−cosψcosϕ+0.5bA1−1sinψsinϕ,Φ0=0.25C1−1A1−1a0∗2b0∗sin2ψsinϕ+0.5x0′b0∗21−cos2ϕ+0.5y0′b0∗2sin2ϕ−z0′b−1b0∗cosϕ+x0′b−1−0.125A1−2a0∗2b0∗1−cos2ψcosϕ+0.75νe+0.75νe1b0∗cosϕ−aa0∗2b0∗cos2ψ0cosϕ+0.5aa0∗2b0∗1+cos2ψcosϕ+0.125bA1−2a0∗2b0∗1−cos2ψcosϕ+2x0′b0∗2cosϕ0cosϕ−x0′b0∗21+cos2ϕ+y0′b0∗2sin2ϕ+a0∗2b0∗1+B10.5bA1−1sinψsinϕ+acosψ0cosϕ0−cosψcosϕcosψ.Substituting from (12), (18), and (20) into (2) and equating coefficients of ε−1 in both sides, we get(21)∂2p1∗∂ψ2+p1∗=4a0∗ψ1cosψ+4A1∗sinψ,∂2γ1∗∂ϕ2+γ1∗=2b0∗ϕ1cosϕ,∂2p2∗∂ψ2+p2∗=4A2∗sinψ+a0∗4ψ2+0.25C1A1−1a0∗2−3.25aa0∗2−4z0′a−1−3νe1+0.125bA1−2a0∗2+2x0′b0∗cosϕ0cosψ+0.25a0∗3a−C1A1−1−0.25bA1−2cos3ψ+4aa0∗x0′A1b−1b0∗cosψ0cosϕ0+x0′a0∗b0∗1−2aA1b−1cosϕ−ψ−x0′a0∗b0∗3+2aA1b−1cosϕ+ψ+y0′a0∗b0∗2a−1−A1−1a−1sinϕ−ψ+y0′a0∗b0∗2+2a−1+A1−1a−1sinϕ+ψ,∂2γ2∗∂ϕ2+γ2∗=2ϕ2−z0′b−1+0.125A1−2a0∗2b−1+0.75νe1−aa0∗2cos2ψ0−0.5aB1a0∗2+2x0′b0∗cosϕ0b0∗cosϕ−0.5x0′b0∗2+x0′b−1+0.75νe+1+B1aa0∗2b0∗cosψ0cosϕ0cosψ−0.67x0′b0∗2cos2ϕ+1.5y0′b0∗2sin2ϕ+0.5a0∗20.25A1−21−b−aB1+A1−1b0∗b−1cos2ψ−ϕ+0.25A1−21−b−aB1−A1−1b0∗b−1cos2ψ+ϕ.Canceling singular terms from (21) as in [6], we get(22)ψ1=A1∗=ϕ1=A2∗=0,ψ2=−0.06C1A1−1a0∗2+0.81aa0∗2+z0′a−1+0.75νe1−0.02bA1−2a0∗2−0.5x0′b0∗cosϕ0,ϕ2=0.5z0′b−1−0.125A1−2a0∗2b−1−0.75νe1+aa0∗20.5B1+cos2ψ0−2x0′b0∗cosϕ0.Substituting from (22) into (15)–(17) and integrating, we obtain(23)a∗=a0∗arbitrary const.,ψ=0.5τ+0.5ε−2−0.125C1A1−1a0∗2−0.375aa0∗2+2aa0∗2+2z0′a−1+1.5νe1−0.31bA1−2a0∗2−x0′b0∗cosϕ0τ,ϕ=τ+0.5ε−2z0′b−1−0.125A1−2a0∗2b−1−0.75νe1+aa0∗21+0.5B1−2x0′b0∗τ.From the previous results, we get(24)ψ0=ψ0=0,ϕ0=ϕ0=0.From (13) and (23), we obtain a∗ from the order greater than Oε−2.The periodic solutionsp2andγ2 are obtained by substituting (22) and (23) into (21) and using (12) and (14). Finally, the periodic solutions p1,q1,r1,γ1,γ1′,andγ1″ are obtained from (5), (19), (23), and (24).


## 4. Construction of Periodic Solutions with Nonzeros Basic Amplitudes

We use the large parameter method [7] for constructing the periodic solutions with nonzeros basic amplitudes for system (2) when A<B<C or A>B>C. Consider generating system (10) has periodic solutions with a period T0=2πn as follows:(25)p20τ=Ecos0.5τ−μ,γ20τ=M3cosτ,where E=M12+M22,μ=tan−1M2/M1, and M1,M2,andM3 are constants.Let system (2) has periodic solutions with a period T0+α that reduces to generating solutions (21) when ε⟶∞, where α is a function of ε such that α∞=0. Consider the following initial conditions:(26)p20,ε=M˜1,p˙20,ε=00.5M˜2,γ20,ε=M˜3,γ˙20,ε=0.The notation∼ denotes the following substitution:(27)Mi⟶M˜i=Mi+βi,i=1,2,3,where β1,0.5β2, and β3 represent the deviations of the initial values of the required solutions from their values of the generating ones M1,M2, and M3, respectively. These deviations are functions of ε and vanish when ε⟶∞. Now, we construct the required solutions in the following forms [8]:(28)p2=E˜cosψ−μ+∑n=1Nε−npn∗E˜,ψ+Oε−N−1,γ2=M˜3cosϕ+∑n=1Nε−nγn∗E˜,ϕ+Oε−N−1,where pn∗ and γn∗ are periodic functions in ψ and ϕ, respectively. The quantity M˜3 is determined from the first integral (3). Let E˜,ψ, and ϕ are changed with time according to(29)dE˜dτ=∑n=1Nε−nEnE˜+Oε−N−1,(30)dψdτ=0.5+∑n=1Nε−nψnE˜+Oε−N−1,(31)dϕdτ=1+∑n=1Nε−nϕnE˜+Oε−N−1.Substituting initial conditions (26) into integral (3), when τ=0, we deduce that(32)0<M3=1−γ0″2γ0″<∞,β3=−ε−1νM˜1+⋯.The derivatives become(33)p˙2=dE˜dτ∂p2∂E˜+dψdτ∂p2∂ψ,γ˙2=dE˜dτ∂γ2∂E˜+dϕdτ∂γ2∂ϕ,p¨2=dE˜dτ2∂2p2∂E˜2+d2E˜dτ2∂p2∂E˜+2dE˜dτdψdτ∂2p2∂E˜∂ψ+dψdτ2∂2p2∂ψ2+d2ψdτ2∂p2∂ψ,γ¨2=dE˜dτ2∂2γ2∂E˜2+d2E˜dτ2∂γ2∂E˜+2dE˜dτdϕdτ∂2γ2∂E˜∂ϕ+dϕdτ2∂2γ2∂ϕ2+d2ϕdτ2∂γ2∂ϕ.Using equations (7), (28), and (33), we get(34)s110=E2acos2μ−0.5+0.25bA1−2sin2μ−0.5+0.50.25bA1−2−acosτ−2μ−2M3x0′1−cosτ+y0′sinτ,s210=M3Eacosμ+0.50.5bA1−1−acos0.5τ+μ−0.50.5bA1−1+acos1.5τ−μ,s220=E2νacos2μ−0.5−0.25bA1−1ν2sin2μ−0.5−0.5νa+0.25bA1−1ν2cosτ−2μ+0.5M32e1a+bA1−1e21−cos2τ+M3ae1−cosτ+by0′a−1A1−1sinτ.Using (4), (28), (33), and (34), we obtain(35)F0=0.25C1A1−1E3cos0.5τ−μsin20.5τ−μ+EM3sinτ0.5x0′sin0.5τ−μ+y0′a−1cos0.5τ−μ+0.5y0′A1−1A1+a−1M3Ecosτsin0.5τ−μ−Ez0′a−1+0.75νe1cos0.5τ−μ−0.25Ecos0.5τ−μE2acos2μ−0.5+0.25bA1−2sin2μ−0.5+0.50.25bA1−2−acosτ−2μ−2M3x0′1−cosτ+y0′sinτ+A1b−1x0′M3Eacosμ+0.50.5bA1−1−acos0.5τ+μ−0.50.5bA1−1+acos1.5τ−μ,Φ0=b−1x0′−0.5M32x0′+0.75νe−z0′b−1M3+0.125A1−2M3E2−0.75νe1M3+M3E2acos2μ−0.5+0.25bA1−2sin2μ−0.5−2M32x0′+0.5a1+B1M3E2cosτ−1.5M32x0′cos2τ−y0′sin2τ+1+B1M3E2acosμcos0.5τcosμ+sin0.5τsinμ+0.25E2M30.51−C1A1−1+0.25A1−21−b+a−1+B10.5A1−1+acos2μ−τ+0.25M3E2×0.5A1−1C1−1+0.25A1−21−b+a+1+B10.5bA1−1−acos2μ.Substituting from (28), (33), and (35) into initial system (2) and equating coefficients of ε−1 and ε−2 in both sides, we obtain the following:Coefficients ofε−1:(36)∂2p1∗∂ψ2+p1∗=4Eψ1cosμ−E1sinμcosψ+4Eψ1sinμ+E1cosμsinψ,∂2γ1∗∂ϕ2+γ1∗=2ϕ1M3cosϕ.We neglect the singular terms [4] to get(37)Eψ1cosμ−E1sinμ=0,Eψ1sinμ+E1cosμ=0,(38)ϕ1=0,such that determinant (37) becomes(39)Δ=Ecosμ−sinμsinμcosμ=Ecos2μ+sin2μ=E≠0.For this case, the solution of (37) becomes(40)ψ1=E1=0.The particular solutions for (36) become(41)p1∗=γ1∗=0.Coefficients ofε−2:(42)d2p2∗dτ2+0.25p2∗=E2cosμ+Eψ2sinμsin0.5τ−E2sinμ−Eψ2+0.06C1A1−1E3−z0′a−1E−0.75νe1E−0.25E3acos2μ−0.5−0.06E3bA1−2sin2μ−0.5−0.06E30.25bA1−2−a+0.5M3x0′Ecosμcos0.5τ+A1b−1x0′M3Eacosμ−0.06E3C1A1−1+0.25bA1−2−acos3μcos1.5τ+sin3μsin1.5τ+M3E0.5A1b−10.5bA1−1−ax0′cosμ+0.5a−1−0.5A1−1A1+a−1+0.5y0′sinμcos0.5τ−0.5M3EA1b−10.5bA1−1−ax0′sinμ−a−1−1+A1−1a−1+0.5y0′cosμsin0.5τ−0.5M3E0.25+0.5+A1b−1ax0′cosμ+a−1+1+A1−1a−1+0.5y0′sinμcos1.5τ−0.5M3E1.5A1b−1ax0′sinμ−a−1+1+A1−1a−1+0.5y0′cosμsin1.5τ,(43)d2γ2∗dτ2+γ2∗=x0′b−1−0.5M32+0.75νe+M32ϕ2−z0′b−1−0.125A1−2E2+0.75νe1−E2acos2μ−0.5+0.25bA1−2sin2μ−0.5+2M3x0′−0.5aE21+B1cosτ+1.5M32y0′sin2τ−x0′cos2τ+1+B1M3E2acosμcos0.5τ+sinμsin0.5τcosμ+0.125M3E21−C1A1−1+0.5A1−2−0.5bA1−2−2a−1+B1bA1−1+2acos2μcos2τ+sin2μsin2τ+0.125M3E2C1−1A1−1+0.5A1−2−0.5bA1−2−2a+1+B1bA1−1−2acos2μ.Neglecting singular terms from (42) and (43) yields [4](44)E2=0.125Esin2μ0.25C1A1−1E2−4z0′a−1−3νe1−E2acos2μ−0.5−0.25E2bA1−2sin2μ−0.5−0.25E20.25bA1−2−a+2M3x0′,ψ2=0.25cos2μ−0.25C1A1−1E2+4z0′a−1+3νe1+E2acos2μ−0.5+0.25bE2A1−2sin2μ−0.5+0.25E20.25bA1−2−a−2M3x0′,ϕ2=0.5z0′b−1+0.125A1−2E2−0.75νe1+E2acos2μ−0.5+0.25bA1−2sin2μ−0.5−2M3x0′+0.5a1+B1E2.Substituting from (38), (40), and (44) into (29) and (30) and integrating, we get(45)2E˜=2E−ε−2Esin2μ−0.25C1A1−1E2+4z0′a−1+3νe1+E2acos2μ−0.5+0.25bE2A1−2sin2μ−0.5+0.25E20.25bA1−2−a−2M3x0′τ+⋯,2ψ=τ+0.5ε−2−0.25C1A1−1E2+4z0′a−1+3νe1+E2acos2μ−0.5+0.25b·E2A1−2sin2μ−0.5+0.25E20.25bA1−2−a−2M3x0′cos2μτ+⋯,ϕ=τ+0.25ε−22z0′b−1+0.25A1−2E2−1.5νe1+E22acos2μ−0.5+0.5bA1−2sin2μ−0.5−4M3x0′+a1+B1E2τ+⋯.Substituting (44) into (42) and (43) and solving the resulted equations, we get p2∗andγ2∗. The periodic solutions p2andγ2 are constructed using (28), (32), (41), and (45). Using (5) and (34), we get the first terms of the required solutions as follows:(46)p1=M1cos0.5τ+M2sin0.5τ−ε−1x0′bB1−e1M3cosτ+⋯,q1=0.5A1−1M1sin0.5τ−M2cos0.5τ+ε−1y0′aA1+e2A1−1M3sinτ+⋯,r1=1+0.25ε−22aM12−E2+0.5bA1−2M22−0.5E2+0.25bA1−2−aM12−M22cosτ+2M1M2sinτ−4M3x0′1−cosτ+y0′sinτ+⋯,γ1=M3cosτ+ε−1ν−M1cosτ+M1cos0.5τ+M2sin0.5τ+⋯,γ1′=−M3sinτ+ε−1νM1sinτ+0.5ν2−M1sin0.5τ+M2cos0.5τ+⋯,γ1″=1+ε−1M3Eacosμ+0.5bωA1−1−acos0.5τ−μ−0.25bA1−1+2acos1.5τ−μ+ε−2M31−a−1x0′+0.5M32z0′a−ba+b−1+M31−b−1y0′sinτ−M31−a−1x0′cosτ−0.5M32z0′a−bcos2τa+b−1+E2νacos2μ−0.5−0.25bA1−1ν2sin2μ−0.5−0.1254νa+bA1−1ν2cos20.5τ−μ−0.5E2acos2μ−0.5+0.25bA1−2sin2μ−0.5+0.125bA1−2−4cos20.5τ−μ+⋯.The correction of the period is(47)αε=ε−2πn2M3x0′−2z0′−0.125A1−2E2−E2acos2μ−0.5+0.25bA1−2sin2μ−0.5−0.5aE21+B1+⋯.


## 5. Geometric Interpretation of Motion

In this section, we describe the body motion using Euler’s anglesξ,ζ,andη which come from the obtained solutions (Figure 2). Replacing the time t by t+t0 where t0 is an arbitrary interval, the periodic solutions remain periodic since the initial system is autonomous [9]. For this case, we obtain from (32),(48)η0=0.5π+r0−1t0+⋯,(49)ξ0=tan−1M3,where η0areξ0 are arbitrary initial angles.Figure 2
The rotational planes in terms of Euler’s angles.Making use of (46) and (49) when τ=r0−1t, we find Euler’s angles as follows:(50)ξ=ξ0−ε−1Eξ1t+t0−ξ1t0−ε−2ξ2t+t0−ξ2t0+⋯,ζ=ζ0+0.5MgℓC−1r0cos2ξ0Q10t+0.5ε−1secξ0ζ1t+t0−ζ1t0+0.5ε−2cosξ0ζ2t+t0−ζ2t0+⋯,η=η0+r0−1−0.5MgℓC−1r0cos3ξ0h10t−0.5ε−1cotξ0η1t+t0−η1t0−0.5ε−2cos2ξ0η2t+t0−η2t0+⋯,where(51)ξ1t=0.50.5bA1−1−acost2r0+μ−0.50.5bA1−1+acos3t2r0−μ,ξ2t=y0′a−1A1−1sintr0+b−1B1−1x0′costr0−0.5tanξ0z0′a−ba+b−1cos2tr0−0.5E2cotξ0aν−0.5+0.25bA1−1ν2+0.5A1−1costr0−2μ,ζ1t=η1t=0.671+0.5A1−1M1sin3t2r0−M2cos3t2r0+2−A1−1M2cost2r0+M1sint2r0,ζ2t=Q11+Q13+Q16sintr0−Q11′+Q13′−Q16′costr0+0.5Q12sin2tr0−Q12′cos2tr0+2Q14sint2r0+Q14′cost2r0+0.67Q15sin3t2r0−Q15′cos3t2r0,η2t=h11sintr0−h11′costr0+0.5h12sin2tr0−h12′cos2tr0+h13sintr0−h13′costr0+2h14sint2r0+h14′cost2r0+0.67h15sin3t2r0−h15′cos3t2r0+h16sintr0+h16′costr0+0.34h17sin3tr0−h17′cos3tr0.


## 6. The Numerical Solutions

In this section, we assume numerical values data for the parameters of a rigid body, and we achieve a computer program to solve the quasilinear system using the fourth order Runge–Kutta method [7]. We make another program to represent the analytical solutions numerically in a period t between 0 and 300 (Table 1). We use the initial values from Table 1 for obtaining the numerical solutions represented in Table 2. The comparison between the obtained numerical solutions and analytical ones is presented to know the difference between them. The numerical and analytical solutions are in good agreement with others which proves the accuracy of used methods and obtained results.Table 1
The analytical solutionsp2, γ2, and their derivatives.
tp2γ2p˙2γ˙201.511.066020.81649668.60977E − 05102.0184438.2791880.6022027−7.342405202.3610991.3222970.3354627−10.98657302.498126−6.3005180.03950729−9.096887402.417591−10.74971−0.259889−2.625185502.126507−9.784271−0.53665145.168785601.650225−3.890493−0.766676710.35926701.0302243.962964−0.929931810.33185800.32050049.82036−1.0121995.1003390−0.417135310.73134−1.006313−2.700216100−1.1184436.237032−0.9127875−9.140733110−1.722344−1.398863−0.7397668−10.97719120−2.176246−8.330236−0.5023198−7.284575130−2.440619−11.06585−0.22112530.07719428140−2.492437−8.2278050.079326747.400064150−2.327188−1.2455980.372870510.9956160−1.9592636.3639040.63394099.052767170−1.42070610.767870.8398012.550172180−0.75841899.7481080.9725226−5.236892190−0.03008123.8182821.020547−10.386132000.7008771−4.0347860.9796914−10.303932101.370795−9.8555320.8535145−5.0317012201.921331−10.712080.65300442.7749972302.304537−6.1729760.39562439.1839752402.487041.4754970.103789310.96712502.4529468.380901−0.19708567.2261892602.20522411.06502−0.4807954−0.15448212701.7654498.175891−0.7226327−7.4573642801.1719211.168729−0.9015357−11.004082900.4763283−6.427075−1.001924−9.008176300−0.2607465−10.78558−1.015054−2.474989Table 2
The numerical solutionsp2, γ2, and their derivatives.
tp2γ2p˙2γ˙201.511.066020.81649668.60977E − 05102.0184418.2796260.6022035−7.341455202.3610961.3242630.3354649−10.9858302.498125−6.2974890.03951085−9.098105402.417592−10.74771−0.2598841−2.629244502.126513−9.78573−0.53664585.163152601.650237−3.896276−0.766671210.3552701.0302423.954753−0.929927510.33263800.32052419.813907−1.0121975.10715590−0.417108110.73098−1.006314−2.68967100−1.1184146.244486−0.9127917−9.131646110−1.722318−1.38609−0.739775−10.97505120−2.176226−8.318236−0.5023316−7.291935130−2.440607−11.06137−0.22114030.06275124140−2.492436−8.2344570.079309947.385382150−2.3272−1.2613250.37285310.9887160−1.9592896.3465040.63392449.058519170−1.42074410.758120.83978742.567106180−0.75846759.7522950.9725135−5.216549190−0.030136613.8358331.020544−10.372962000.7008187−4.0117650.9796954−10.305832101.37074−9.838780.8535257−5.0492772201.921283−10.711190.65302282.7494762302.304503−6.1900030.39564849.1632642402.4870251.447830.103817210.962562502.4529528.356145−0.19705547.241632602.20525211.05635−0.4807664−0.12553942701.7654998.189053−0.7226074−7.4290352801.1719881.198392−0.9015169−10.991242900.4764102−6.395274−1.001913−9.018668300−0.2606581−10.76818−1.015054−2.504935


## 7. Conclusion

The solutions (46) and the correction of the period (47) are obtained using the large parameter method, which had never been used for solving this kind of problem in the presence of the new assumptions for motion (the weak oscillations of the body about the minor or the major axis of the ellipsoid of inertia instead of the strong oscillations in the previous works). The advantage of this method is that the energy motion of the body is assumed to be sufficiently small instead of sufficiently large with other techniques [10–12]. Also, the obtained solutions treat a singular situation for the natural frequency which was excluded from previous works [13, 14].Equations (50) and (51) describe the rotation of the body at any time and show that this motion depends on four arbitrary constants ξ0,ζ0,η0,andr0, such that r0 is sufficiently small. The obtained solutions give special cases of motions when M1=M2=0 and when M1=0,M2≠0, or M2=0,M1≠0. Also, the obtained solutions give many gyroscopic motions, which depend on the values of the moments of inertia and the initial position of the body center of gravity. In the end, we obtain the case of regular precession [10] as a special case.The analytical solutions (46) are represented indefinite intervals of time through computer programs (Table 1). The numerical solutions are obtained using the fourth order Runge–Kutta method in terms of another program (Table 2). Tables 1 and 2 give in detail the obtained results of both the analytical solutions and numerical ones. These results show that the analytical solutions are in full agreement with the numerical ones which proves the accuracy of the considered techniques and results. This case of study is considered as a general case of such ones studied in [5]. The stability phase diagrams of the solutions p2 and γ2 are given (Figures 3 and 4). From these diagrams, we note that the stability for both the analytical and the numerical solutions in full agreement. This gives the validity of the obtained solutions and the considered procedures. The considered procedures and results are very useful for the general reader’s concern with the new applications dealing with the use of functionally graded materials in such structures based on the recent works [15].Figure 3
The stability of the analytical and numerical solutionsp˙2andp2.Figure 4
The stability of the analytical and numerical solutionsγ˙2andγ2.



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