Physics of gyroscope nutation

Usubamatov, R.

doi:10.1063/1.5099647

The known publications of the gyroscope property of maintaining its axis are represented by simplified mathematical models formulated on the law of kinetic energy conservation and the action of an inertial torque expressed by the change in the angular momentum. This is the reason that practice does not validate the correctness of such publications and gyroscopic effects remained for a long time as unsolved problems. The new study demonstrates that the origin of gyroscopic effects is more complex and is manifested by the simultaneous and interdependent action of the several inertial resistance and precession torques produced by the rotating mass of a spinning object. These inertial torques are represented by the action of centrifugal, Coriolis, the common inertial forces and the change in the angular momentum of the rotating mass of a gyroscope. In case of the complex gyroscope motion, the inertial torques can deactivate that represent the new property of the matter. The principles of energy conservation, the action of the system of internal and external torques on gyroscopic devices and their new mathematical models enable for the describing all gyroscopic effects. The main contribution and novelty of this paper is the mathematical modeling and describing the physics of the gyroscope nutation called galloping, which was a mostly unsolved problem in engineering.

I. INTRODUCTION

Gyroscopic effects are relayed in many engineering calculations of rotating parts and gyroscope properties enable for the functioning of the numerous gyroscopic devices in engineering. There are many publications regarding the gyroscope theory as well as many approaches and mathematical solutions that describe the gyroscope properties.^1–4 Numerous valuable manuscripts dedicated to applications of gyroscopic effects in mechanics.^5,6 However, all publications in the area of gyroscope theory describe gyroscope properties with simplifications and assumptions that cannot be confirmed by practice.^7–9 This is the reason that the gyroscope theory still attracts many researchers who continue to discover new properties among gyroscopic devices.^10–12 The nature of gyroscope effects is more complex and the known publications demonstrate the inaccuracy of gyroscope theories and unexplainable gyroscope properties.^13,14

Recent investigations in the area of the gyroscopic effects represent new mathematical models for inertial forces acting on a gyroscope.^15–18 These publications show that the external load acting on a gyroscope generates several inertial forces of the mass-elements of the spinning rotor. Inertial forces produce the resistance and precession torques that manifest the combined action of the centrifugal, Coriolis and the common inertial forces of the gyroscope mass as well as the change in the angular momentum. Inertial torques are acting simultaneously and interdependently around two axes of the gyroscope that are strictly perpendicular to each other. These inertial torques represent fundamental principles of the gyroscope theory.¹⁵ Researchers should consider them for the mathematical modeling of all gyroscopic effects.

The inertial torques express the kinetic energy of the spinning disc and always act to the opposite direction of the external or load torques applied on the gyroscope. The new approach to the gyroscope’s torques, as well as practice tests, discovered the unknown effects, which are the deactivation of the inertial forces generated by the rotor’s mass.¹⁸ This fact contradicts the principles of classical mechanics but discovers the new property of a physical matter that should be taken into account for the computing of the gyroscopic forces and motions.

Numerous gyroscopic devices manifest the gyroscopic effects of which one is the nutation processes called galloping. Vast tomes have been written on the gyroscope nutations.^19–21 However, all known explanations of the gyroscope nutation present only an intuitive approach and the relevant simplified models are not validated by practice. New mathematical models for the system of the inertial torques generated by the mass of the spinning rotor enable for describing the physics of the gyroscope nutation formulated on the law of energy conservation.

II. METHODOLOGY

The gyroscope nutation is represented in terms of machine dynamics particularly in terms of vibration analysis. The process of nutation of a gyroscope should be analytically described in details by principles of conservation of potential, kinetic energies and work of a force. The gyroscope nutation is considered as the object of the classical viscous damped vibration with the constant action of the damping forces. Nevertheless, the peculiarities of acting inertial torques demonstrate the following differences in the gyroscope nutation:

The equations of the inertial torques contain the product of ωω_x, which is constant and expresses the total (kinetic, potential, and friction losses) energy of the spinning rotor.¹⁵ The increase of the angular velocity ω of the rotor leads to the decrease the angular velocity ω_x of the gyroscope precession and the decrease of ability for the gyroscope nutation, and vice versa.
The angular velocity ω_x depends on the stand center mass M and ω. Their variations and the relationship demonstrate the amplitude of the gyroscope nutation. The big value of ω_x leads to the big amplitude of the gyroscope nutation and the small value of ω_x leads to the overdamped condition. The effect of overdamping is strong. After the action of the single input torques, the gyroscope system does not return to its original position. The gyroscope nutation does not have a critically damped condition. Its nutation with the action of the frictional forces is dynamically irreversible and presents a monotonically damping process.
The action of the gyroscopic inertial torques at the process of nutation is variable, a fact that leads to the variable process of the nutation.
The inertial torques of the spinning rotor do not possess the property of the spring’s torque and do not have the index of torsional stiffness or the spring constant that described in the theory of the vibration.

This peculiarity presents the specific properties of the action of the inertial torques and the gyroscope damped nutation is described by the terms that differ from the known classical theory for the viscously damped vibration.^19–21

The nutation process of the gyroscope has been studied on the stand of Super Precision Gyroscope, “Brightfusion LTD”, which technical data and scheme of acting forces computed.¹⁸ The mathematical model for acting forces on the gyroscope stand and motions around two axes is similar to the model for the gyroscope suspended from flexible cord.¹⁶ The difference of the equations is the action of the frictional forces on the supports and pivot. The mathematical models for the gyroscope nutation with the action of the load, inertial and frictional forces are presented by the cumbersome expressions for the damping process. The computing process of such cumbersome models is tedious and the physics of the gyroscope nutation is considered by the action of only inertial forces and the gyroscope weight. This condition and the absence of the frictional forces represents the undamped gyroscope nutation.

For the following analysis of the gyroscope nutation is necessary to express the equations of the acting torques. The mathematical model of the gyroscope nutation deals with the quantification of the amount of its potential and kinetic energies that express a function tending to the change between extremes. The acting forces and torques on the gyroscope stand are presented by the following components. The weight of the gyroscope components produces the external torque which expression is as follows:

T = (W l - s l_{s} / 2 - G a) g cos γ = M g l_{m} cos γ

(1)

where W is the gyroscope mass; s is the axle mass; G is the mass of the counter-weight; g is the gravity acceleration; l is the distance of location the center gravity of the gyroscope; l_s is the length of the axle; a is the distance of location the center gravity of the counter-weight, M = W - s - G is resulting load mass, l_m is the location of resulting load mass, γ is the angle of the gyroscope axis inclination.

The nutation process considers the short time action of the single impact torque around axis ox that leads to the deactivation of all inertial torques. Such condition leads to the change in the balance of its potential and kinetic energies and the change in the values and direction of inertial torques and the angular velocities around axes. The action of the single input torque increases the angular velocity and the value of the resulting resistance inertial torques of the gyroscope. This peculiarity of the gyroscope nutation process considers the several steps of its motions. The starting condition for the short time action of the single input torque on the gyroscope is considered for its horizontal position. The action of the extra input torque leads to the increasing of ω_x of the inertial resistance torques and their kinetic energy and the gyroscope turns back. The action of inertial torques, the change in the potential and kinetic energies and the gyroscope motions are considered by consistent steps of the change in their values and directions around axes. Such analysis enables for demonstrating the changes in the nutation.

The kinetic energy action of the single input torque on the gyroscope is converted into its potential energy of the location. At the process of the gyroscope nutation, the potential energy is converted into kinetic energy stored in the rotational gyroscope that described by the moment of inertia J_x, mass M, the angular and linear velocities ω_x and v = ω_xl_m, respectively. The parameters of the gyroscope nutation are depended on the time of action of external and inertial torques and the steps of consideration. The motions up and down of the gyroscope nutation are represented by four mathematical models, which conditions described at sections below. The motion from low to the upper position is presented by two equations, which conditions defined at sections a₁ and a₂. The motion from the upper point to the down position is also presented by two equations that described at sections b₁ and b₂. Following gyroscope motions from low to upper position and back are repeated.

(a₁). Removing the short time action of the input torque which activates the action of the inertial resistance torques T_r.x and the precession torque T_p.y, the physical phenomenon is as follows. The gyroscope motion is represented by two mathematical models. The first one describes the action of surplus resistance torques T_r.x around axis ox in the clockwise direction, the action of the precession torque T_p.y (originated around axis oy but acting around axis ox), and the action of the gyroscope weight T is in the counter clockwise direction. Hence, the value of the surplus resulting resistance torques T_r.x is decreased to zero (T_r.x = 0) at some angular position of γ_e. The gyroscope passes the angle of γ_e under the action of the kinetic energy of the gyroscope center mass.
(a₂). The second equation describes the action of the resistance torque T_r.x which direction is changed in the counter clockwise and added to the action of the precession torque T_p.y. The action of the excess kinetic energy of the gyroscope center mass continues to turn up the gyroscope in the clockwise direction and to change the value of the inertial resistance torques, which act in the counter clockwise direction. The actions of the resulting mass of the gyroscope, inertial resistance torques T_r.x, the precession torque T_p.y decrease the kinetic energy of the gyroscope mass that manifest the damping of the gyroscope motion. The gyroscope stops at an upper position with the high potential energies of the gyroscope mass and the resistance inertial torque T_r.x, that act in the counter clockwise direction. The action of the precession torques T_p.y is deactivated due to stops of the gyroscope motion around axis ox.
(b₁). From the upper point of location, gyroscope turns down. At this condition, the gyroscope motion is represented by two mathematical models. The first one describes the action of the gyroscope weight, resulting in resistance torques T_r.x that acting in the counter clockwise direction, and the precession torque T_p.y that act in the clockwise direction. The action of T_p.y torques continues until the angular position γ_e where the value of the T_r.x is decreased to zero.
(b₂). The second equation describes the action of all components of the gyroscope after passing the angle γ_e. At this condition, the action of the inertial resistance torques T_r.x is changed in the clockwise direction and added to the action of the precession torques T_p.y. The action of the excess kinetic energy of the gyroscope mass continues to turn down the gyroscope in the counter clockwise direction and to change in the value of the inertial resistance torques T_r.x. The gyroscope stops at a low position with the minimal potential energy and the high value of inertial resistance torques T_r.x. The action of the precession torques T_p.y is deactivated due to stops of the gyroscope motion around axis ox.

The gyroscope starts to turn up under the action of surplus resistance inertial torques around axis ox in the clockwise direction. The following actions of external and internal torques are the same as described in sections (a₁, a₂) and (b₁, b₂). The gyroscope nutation is repeated with the continuous action of changeable inertial torques. The action of the external and internal torques on the gyroscope for the conditions described above is presented in Figure 1.

FIG. 1.

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Torques and angular velocities of the gyroscope motions around axis ox in clockwise (a) and counter clockwise (b) directions.

The gyroscope motions are presented by the corresponding equations that described in several publications.¹⁷ The equations for stages (a₁, a₂ and b₁, b₂) of the gyroscope nutation are presented below.

(a₁). The equation for the gyroscope motion from the low position to the angle where the resistance torque resets to zero (T_r.x = 0) is as:

\begin{matrix} J_{x} \frac{d ω_{x}}{d t} & = & M g l_{m} cos γ - \frac{2 π^{2}}{9} J ω ω_{x} - \frac{8}{9} J ω ω_{x} + \frac{2 π^{2}}{9} J ω ω_{y} cos γ \\ + J ω ω_{y} cos γ \end{matrix}

(2)

\begin{matrix} J_{y} \frac{d ω_{y}}{d t} & = & - \frac{2 π^{2}}{9} J ω ω_{x} cos γ - J ω ω_{x} cos γ + \frac{2 π^{2}}{9} J ω ω_{y} cos γ \\ + \frac{8}{9} J ω ω_{y} cos γ \end{matrix}

(3)

where all parameters are as specified above.

Inertial torques action around axes ox and oy express their kinetic energies that are equal and present by the following equation:

\begin{matrix} - \frac{2 π^{2}}{9} J ω ω_{x} - \frac{8}{9} J ω ω_{x} + \frac{2 π^{2}}{9} J ω ω_{y} cos γ + J ω ω_{y} cos γ \\ = - \frac{2 π^{2}}{9} J ω ω_{x} cos γ - J ω ω_{x} cos γ + \frac{2 π^{2}}{9} J ω ω_{y} cos γ + \frac{8}{9} J ω ω_{y} cos γ \end{matrix}

(4)

which simplification yields the following expression:

ω_{y} = (\frac{2 π^{2} + 8}{cos γ} - (2 π^{2} + 9)) ω_{x}

(5)

Substituting ω_y (Eq. (5) into Eq. (2) simplification and transformations¹⁷ yield the following:

J_{x} \frac{d ω_{x}}{d t} = M g l_{m} cos γ - [(2 π^{2} + 9) cos γ - \frac{8 (2 π^{2} + 8)}{9}] J ω ω_{x}

(6)

(a₂). The equation for the gyroscope motion up from the angle of the reset to zero of the resistance torque (T_r.x = 0) to the upper position is as:

\begin{matrix} J_{x} \frac{d ω_{x}}{d t} & = & M g l_{m} cos γ + \frac{2 π^{2}}{9} J ω ω_{x} + \frac{8}{9} J ω ω_{x} + \frac{2 π^{2}}{9} J ω ω_{y} cos γ \\ + J ω ω_{y} cos γ \end{matrix}

(7)

\begin{matrix} J_{y} \frac{d ω_{y}}{d t} & = & - \frac{2 π^{2}}{9} J ω ω_{x} cos γ - J ω ω_{x} cos γ + \frac{2 π^{2}}{9} J ω ω_{y} cos γ \\ + \frac{8}{9} J ω ω_{y} cos γ \end{matrix}

(8)

Following steps of the analysis is the same as presented in the section (a₁) and all comments are omitted.

\begin{matrix} \frac{2 π^{2}}{9} J ω ω_{x} + \frac{8}{9} J ω ω_{x} + \frac{2 π^{2}}{9} J ω ω_{y} cos γ + J ω ω_{y} cos γ \\ = - \frac{2 π^{2}}{9} J ω ω_{x} cos γ - J ω ω_{x} cos γ + \frac{2 π^{2}}{9} J ω ω_{y} cos γ + \frac{8}{9} J ω ω_{y} cos γ \end{matrix}

(9)

ω_{y} = - (\frac{2 π^{2} + 8}{cos γ} + 2 π^{2} + 9) ω_{x}

(10)

- J_{x} \frac{d ω_{x}}{d t} = M g l_{m} cos γ + [\frac{10 (2 π^{2} + 8)}{9} + (2 π^{2} + 9) cos γ] J ω ω_{x}

(11)

(b₁). The equation for the gyroscope motion down from the upper position to the angle of the reset to zero of the resistance torque (T_r.x = 0) is as:

\begin{matrix} J_{x} \frac{d ω_{x}}{d t} & = & M g l_{m} cos γ + \frac{2 π^{2}}{9} J ω ω_{x} + \frac{8}{9} J ω ω_{x} - \frac{2 π^{2}}{9} J ω ω_{y} cos γ \\ - J ω ω_{y} cos γ \end{matrix}

(12)

\begin{matrix} J_{y} \frac{d ω_{y}}{d t} & = & \frac{2 π^{2}}{9} J ω ω_{x} cos γ + J ω ω_{x} cos γ - \frac{2 π^{2}}{9} J ω ω_{y} cos γ \\ - \frac{8}{9} J ω ω_{y} cos γ \end{matrix}

(13)

\begin{matrix} \frac{2 π^{2}}{9} J ω ω_{x} + \frac{8}{9} J ω ω_{x} - \frac{2 π^{2}}{9} J ω ω_{y} cos γ - J ω ω_{y} cos γ = \frac{2 π^{2}}{9} J ω ω_{x} cos γ \\ + J ω ω_{x} cos γ - \frac{2 π^{2}}{9} J ω ω_{y} cos γ - \frac{8}{9} J ω ω_{y} cos γ \end{matrix}

(14)

ω_{y} = [\frac{2 π^{2} + 8}{cos γ} - (2 π^{2} + 9)] ω_{x}

(15)

J_{x} \frac{d ω_{x}}{d t} = M g l_{m} cos γ + [(2 π^{2} + 9) cos γ - \frac{8 (2 π^{2} + 8)}{9}] J ω ω_{x}

(16)

(b₂). The equation for the gyroscope motion down from the angle of the reset to zero of the resistance torque (T_r.x = 0) to the low position is as:

\begin{matrix} J_{x} \frac{d ω_{x}}{d t} & = & M g l_{m} cos γ - \frac{2 π^{2}}{9} J ω ω_{x} - \frac{8}{9} J ω ω_{x} - \frac{2 π^{2}}{9} J ω ω_{y} cos γ \\ - J ω ω_{y} cos γ \end{matrix}

(17)

\begin{matrix} J_{y} \frac{d ω_{y}}{d t} & = & \frac{2 π^{2}}{9} J ω ω_{x} cos γ + J ω ω_{x} cos γ - \frac{2 π^{2}}{9} J ω ω_{y} cos γ \\ - \frac{8}{9} J ω ω_{y} cos γ \end{matrix}

(18)

\begin{matrix} - \frac{2 π^{2}}{9} J ω ω_{x} - \frac{8}{9} J ω ω_{x} - \frac{2 π^{2}}{9} J ω ω_{y} cos γ - J ω ω_{y} cos γ \\ = \frac{2 π^{2}}{9} J ω ω_{x} cos γ + J ω ω_{x} cos γ - \frac{2 π^{2}}{9} J ω ω_{y} cos γ - \frac{8}{9} J ω ω_{y} cos γ \end{matrix}

(19)

ω_{y} = (\frac{2 π^{2} + 8}{cos γ} + 2 π^{2} + 9) ω_{x}

(20)

J_{x} \frac{d ω_{x}}{d t} = M g l_{m} cos γ - [\frac{10 (2 π^{2} + 8)}{9} + (2 π^{2} + 9) cos γ] J ω ω_{x}

(21)

where all parameters are as specified above.

The angles of the gyroscope position, the potential and kinetic energies of the gyroscope and the work of the inertial torques at some points of the nutation, are defined by the rules of the law of mechanical energy conservation. The position of the resulting gyroscope weight expresses the gravitational potential energy of a gyroscope that is a conservative force. The action of the inertial torques that depends on the angles of the gyroscope turn and the direction of action, expresses the negative or positive work. The motion of the gyroscope mass expresses the kinetic energy for the intermediate position and the potential energy for its low and upper points of location and the work of the inertial torques. The datum of energies is accepted the horizontal location of the gyroscope. The total energies of the gyroscope at any points of the nutation are equalized, and this fact is expressed by the following equation:

P_{M 1} + K_{1 . M} \pm A_{1 . T_{i}} = P_{M 2} + K_{2 . M} \pm A_{2 . T_{i}}

(22)

where P_Mi = Mgl_m sin γ_i is the potential energy of the resulting gyroscope weight M; K_i.M = (1/2)J_xω²_x + (1/2)M(ω_xl_m)² is the kinetic energy of the gyroscope turn and the linear motion of it’s the center mass of location on the radius l_m about the center of rotation. $A_{i T_{i}} = D_{i} J ω ω_{x} γ$ is the work of the resulting inertial torque on the angle γ of its turn; D_i is the factor of each resulting inertial torque;¹⁵ the sign (±) is applied for the work that coincides or opposite to the action of the gyroscope weight; other parameters are as specified above.

The first step (section (a₁)) considers the equation of the equality of the gyroscope energies for the low point and angular position (γ_e) of the gyroscope that turns in the clockwise direction. Substituting Eq. (1), equations of inertial torques¹⁵ into Eq. (22) and using Eq. (6), and transformation yields the following equation:

\begin{matrix} M g l_{m} sin (- γ_{l . 1}) & = & M g l_{m} sin γ_{e} + \frac{1}{2} J_{x} ω_{x . e}^{2} + \frac{1}{2} M {(ω_{x . e} l_{m})}^{2} \\ - [(2 π^{2} + 9) cos γ_{l} - \frac{8 (2 π^{2} + 8)}{9}] J ω ω_{x . l} (γ_{e . 1} - γ_{l . 1}) \end{matrix}

(23)

where the index l and e.1 belong to the low and the angular position (for T_r.x = 0) of the gyroscope respectively, other parameters are as specified above.

The second step (section (a₂)) considers the equation of the equality of the gyroscope energies for the angular position γ_e of the reset to zero of the resistance torques (T_r.x = 0) and the upper point of location γ_u of the gyroscope (Eqs. (1) and (11)).

\begin{matrix} M g l_{m} sin γ_{u} & = & M g l_{m} sin γ_{e . 1} + \frac{1}{2} J_{x} ω_{x . e}^{2} + \frac{1}{2} M {(ω_{x . e} l_{m})}^{2} \\ - [(2 π^{2} + 9) cos γ + \frac{10 (2 π^{2} + 8)}{9}] J ω ω_{x} (γ_{u} - γ_{e . 1}) \end{matrix}

(24)

where γ_e.1 and γ_u is the angle of the gyroscope at the reset to zero of the resistance torques (T_r.x = 0) and the upper position respectively, other parameters are as specified above.

The angular velocity of the gyroscope motion down is presented by Eq. (16). The first step (section (b₁)) consider the equation of the equality of the gyroscope energies for the upper γ_u and γ_e positions of the reset to zero of the resistance torques (T_r.x = 0) that turns down the gyroscope in the counter clockwise direction (Eqs. (1) and (16)).

\begin{matrix} M g l_{m} sin γ_{u} & = & M g l_{m} sin (- γ_{e . 2}) + \frac{1}{2} J_{x} ω_{x . e}^{2} + \frac{1}{2} M {(ω_{x . e} l_{m})}^{2} \\ + [(2 π^{2} + 9) cos γ - \frac{8 (2 π^{2} + 8)}{9}] J ω ω_{x . u} (γ_{u} - γ_{e . 2}) \end{matrix}

(25)

where all parameters are as specified above.

The second step (section (b₂)) considers the equation of the equality of the gyroscope energies for the angular position (γ_e.2) of the reset to zero of the resistance torques (T_r.x = 0) and low point (γ_l.2) of location of the gyroscope that turns down in the counter clockwise direction (Eqs. (1) and (21)).

\begin{matrix} M g l_{m} sin (- γ_{l . 2}) & = & M g l_{m} sin (- γ_{e . 2}) + \frac{1}{2} J_{x} ω_{x . e}^{2} + \frac{1}{2} M {(ω_{x . e} l_{m})}^{2} \\ - [\frac{10 (2 π^{2} + 8)}{9} + (2 π^{2} + 9) cos γ_{l}] \\ \times J ω ω_{e . x . .} (γ_{e . 2} - γ_{l . 2}) \end{matrix}

(26)

where all parameters are as specified above.

The time of the gyroscope nutation is defined by the expression t = γ/ω_x. Following gyroscope nutation is repeated with the change in the numerical data of processes.

A. Case study and practical tests

The mathematical model for the nutation of the gyroscope with the counter-weight is based on the technical parameters of the test stand.¹⁸ The practical solution of the gyroscope nutation is considered for the conditions of action of the weight of the gyroscope components, single input torque, and inertial torques. The angular velocity of the spinning rotor is accepted ω = 600 rpm. The nutation process is started by the action of the single input torque which time action is t = 1.0 s and the value of the angular velocity is ω_s = 0,35 rad/s. The single input torque turns down the gyroscope on the angle: −γ_l = ω_st = 0.35 × (180/π) × 1.0 = −20.053522°, where the sign (-) means, the negative angle from the horizontal datum of the gyroscope. The geometrical parameters of the gyroscope h = 0.0569254 m, τ = 38.581266°.¹⁸ Numerical solution of the gyroscope amplitude from horizontal versus time of the motion used mathematical models for the gyroscope nutation (Eqs. (6), (11), (16), (21), (23) – (26)) and represented in Appendix, in Table I and Figure 2. In order to elucidate all particular steps of the nutation process, computations have been performed using calculus; all numerical operations have been performed using a simple calculator.

TABLE I.

Numerical solution of the gyroscope amplitude from horizontal versus time of the motion.

	First		Second		Third
Nutation	up	down	up	down	up	down
Angular location, γ°	23.81044	−23.9893	23.8453	−23.9783	23.8253	−23.9805
Time of motion, t (s)	0.570	1.417	1.423	1.424	1.421	1.423
Amplitude, γ°		47.7997		47.8236		47.8142

	First		Second		Third
Nutation	up	down	up	down	up	down
Angular location, γ°	23.81044	−23.9893	23.8453	−23.9783	23.8253	−23.9805
Time of motion, t (s)	0.570	1.417	1.423	1.424	1.421	1.423
Amplitude, γ°		47.7997		47.8236		47.8142

FIG. 2.

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The gyroscope nutation around axis ox.

Figure 2 demonstrates the gyroscope nutation process generated by the action of the surplus inertial torques of the spinning rotor. Practical observation confirms the diagram presented in Figure 2.

III. RESULTS AND DISCUSSION

The work presents mathematical models for the undamped gyroscope nutation based on the action of the gyroscope weight and the system of inertial torques generated by the rotating mass of the spinning rotor. The action of the frictional torques is not considered due to cumbersome mathematical expressions for the gyroscope motions. The model for the gyroscope nutation demonstrates the change in the potential and kinetic energies and the work of the changeable inertial torques of the spinning rotor, i.e., the physics of the nutation process. A gyroscope with a high angular velocity of the spinning rotor is represented the overdamped system that does not manifest the nutation. The relatively low angular velocity of the spinning rotor enables the manifestation of the gyroscope nutation. The mathematical model of the gyroscope nutation can be used for defining the amplitudes, frequency of the nutation, and the change in the values of resulting inertial torques. Conducted practical observations confirmed the mathematical models and represent a good example for the educational process of the physics of gyroscope nutation.

IV. CONCLUSION

The gyroscope nutation was one of the most complex and intricate in terms of analytical solutions and for a long time represented the unsolvable problem. New mathematical models for the inertial torques acting on the spinning rotor finely resolved all questions of gyroscopic effects. The action of the single input torque on the gyroscope is resulting in the change in the value of the inertial torques generated by the rotating mass of the spinning rotor. The origin of inertial forces acting on the gyroscope is well known in classical mechanics but their actions have never been described analytically. The physics of the gyroscope nutation is explained by the change in potential and kinetic energies of the gyroscope and by the work of the changeable inertial torques generated by rotating mass of the spinning rotor. The presented solution can be used for describing the physics of the gyroscope nutation at the educational process.

ACKNOWLEDGMENTS

The work supported by the Kyrgyz State Technical University after I. Razzakov.

NOMENCLATURE

A _i: work of resistance torques
g: gravity acceleration
e: base of the natural logarithm
i: index for axis ox or oy
J: mass moment of inertia of a gyroscope’s rotor
J _i: mass moment of inertia of a gyroscope around axis i
K _M: kinetic energy of a gyroscope stand
l: distance between a gyroscope centre mass and a support
l _m: distance between a resulting mass of the gyroscope and a support
m: mass of a rotor’s disc
M: resulting mass of the gyroscope stand
P _M: potential energy of a resulting gyroscope weight
T: load torque
T _am.i: torque generated by a change in an angular momentum acting around axis i
T_cti, T_cr.i, T_in.i: torque generated by centrifugal, Coriolis and common inertial forces respectively, and acting around axis i
T _p.i: precession torque acting around axis i
T _r.i: resistance torque acting around axis i
t: time
W: mass of a gyroscope
γ: angle of inclination of a rotor’s axle
γ _e: angle of reset to zero of the resistance torque (T_r.x = 0)
ω: angular velocity of a rotor
ω _i: angular velocity of precession around axis i

1. Solution

The gyroscope parameters and data of the case study¹⁸ are used for several Euler’s form equations of the gyroscope undamped nutation. Solutions of gyroscope nutation are based on the action of the system of inertial torques, the gyroscope weight, and the principle of the law of energy conservation. The gyroscope nutation is defined by substituting initial gyroscope parameters¹⁸ into basic Eq. (22) that accepted for the datum.

2. First nutation

After the short time action of the single input torque, the gyroscope turns up in the clockwise direction until the upper position of the nutation. The motion of the gyroscope is considered for the four stages that presented by Eqs (6), (11), (16) and (21)) by substituting data defined above into corresponding equations:

(a₁). The first stage is the gyroscope motion up from the low position until the value of the resistance torque rests to zero (T_r.x = 0), thus according to Eq. (6) we have

\begin{matrix} 5.551045166 \times 1 0^{- 4} \frac{d ω_{x}}{d t} & = & 0.043 \times 9.81 \times 0.0039069767 cos γ \\ - [(2 π^{2} + 9) cos γ - \frac{8 (2 π^{2} + 8)}{9}] \\ \times 0.5726674 \times 1 0^{- 4} \times 600 \times (\frac{2 π}{60}) ω_{x} \end{matrix}

(A1)

Simplification and transformation of Eq. (A1) yields the following expression:

\frac{5.368063832 \times 10^{- 3}}{cos γ - 0.857959405} \frac{d ω_{x}}{d t} = \frac{0.015937525 cos γ}{cos γ - 0.857959405} - ω_{x}

(A2)

Separating variables and transformation yields the following:

\frac{d ω_{x}}{\frac{0.015937525 cos γ}{cos γ - 0.857959405} - ω_{x}} = 186.286905508 \times (cos γ - 0.857959405) d t

(A3)

The differential Eq. (A3) is presented by the integral form of defined limits and yields the following:

\begin{matrix} \int_{- 0.35}^{- ω_{x}} \frac{d ω_{x}}{\frac{0.015937525 cos γ}{cos γ - 0.857959405} - ω_{x}} & = & 186.286905508 \\ \times \int_{0}^{t} (cos γ - 0.857959405) d t \end{matrix}

(A4)

The left integral of Eq. (A4) is tabulated and represented the integral $\int \frac{d x}{a - x} = - ln |a - x| + C$ ⁠. Solving of the integral Eq. (A5) yields the following expression:

\begin{matrix} ln (\frac{0.015937525 cos γ}{cos γ - 0.857959405} - ω_{x}) |_{- 0.35}^{- ω_{x}}) \\ = - 186.286905508 (cos γ - 0.857959405) |_{0}^{t}) \end{matrix}

giving rise to the following expression:

\begin{matrix} \frac{0,015937525 cos γ}{cos γ - 0.857959405} + ω_{x} & = & [\frac{0.015938455 cos γ}{cos γ - 0.857959405} + 0.35] \\ \times e^{- 186.286905508 (cos γ - 0.857959405)} \end{matrix}

(A5)

The right side of Eq. (A5) has a small value of the high order that can be neglected. Solving Eq. (A5) yields the equation of the dependency ω_x in γ:

ω_{x} = - \frac{0.015937525 cos γ}{cos γ - 0.857959405}

(A6)

where the sign (-) means the motion in the clockwise direction.

The first derivative of Eq. (A6), transformation and the following integral equation with definite limits enables for defining the expression of ω_x

d ω_{x} = \frac{- 0.013673749 sin γ}{{(cos γ - 0.857959405)}^{2}} d γ

(A7)

\begin{matrix} \int_{- 0.35}^{- ω_{x . 1}} d ω_{x} = - 0.013673749 \int_{- 20, 053522^{○}}^{γ_{1}} \frac{sin γ}{{(cos γ - 0.857959405)}^{2}} d γ \end{matrix}

(A8)

Solving of the integral Eq. (A8) brings the following:

ω_{x} |_{0.35}^{ω_{x . e}}) = \frac{- 0.013673749}{cos γ - 0.857959405} |_{- {20.053522}^{○}}^{γ_{l}})

giving rise to the following result:

ω_{x . e} = 0.517954705 \times (\frac{cos γ - 0.884358912}{cos γ - 0.857959405})

(A9)

Equation (23) of the equality of the gyroscope energies for the low point and the equilibrium position of the resisting torque enables for expressing the angular velocity and position of the gyroscope. Substituting the gyroscope data¹⁸ and the result of Eq. (A9) into Eq. (23) yields the following expression:

\begin{matrix} 0.001648079 sin (- 20.05352 2^{○}) & = & 0.001648079 sin γ_{e} + \frac{1}{2} \times 5.551045166 \times 1 0^{- 4} \times {[0.517954705 \times (\frac{cos γ - 0.884358912}{cos γ - 0.857959405})]}^{2} \\ + \frac{1}{2} \times 0.043 \times {[0.517954705 \times (\frac{cos γ - 0.884358912}{cos γ - 0.857959405}) \times 0.0039069767]}^{2} - [(2 π^{2} + 9) cos γ - \frac{8 (2 π^{2} + 8)}{9}] \\ \times 0.5726674 \times 1 0^{- 4} \times 600 \times (\frac{2 π}{60}) \times 0.517954705 \times (\frac{cos γ - 0.884358912}{cos γ - 0.857959405}) \times (γ + 20.05352 2^{○}) \frac{π}{180^{○}} \end{matrix}

(A10)

Simplification and transformation of Eq. (A10) brings the following transcendental equation:

\begin{matrix} - 0.604528077 & = & 1.762997863 sin γ_{e} + 0.079747175 \\ \times {(\frac{cos γ - 0.884358912}{cos γ - 0.857959405})}^{2} - (cos γ - 0.884358912) \\ \times (γ + 20.05352 2^{○}) \end{matrix}

(A11)

Numerical solution of Eq. (A11) yields the following result:

γ_{e} = 23.809 6^{○}

(A12)

Substituting Eq. (A12) into Eq. (A9) and transformation yields the following angular velocity of the gyroscope:

\begin{matrix} ω_{x . e} & = & 0.517954705 \times (\frac{cos {23.8096}^{○} - 0.884358912}{cos {23.8096}^{○} - 0.857959405}) \\ = & 0.277771610 rad / s \end{matrix}

(A13)

Defined information enables for calculating the time of the gyroscope turn up until equilibrium. The angle of the gyroscope turn in the clockwise direction is γ₁ + γ_e= |−20.053522°|+ 23.8096° = 43.863122° and the time of gyroscope motion is defined by the expression t = γ/ω_x. Substituting ω_x.e (Eq. (A13)), the value of 43.863122° into the expression t, derivation and integration giving rise to the following result:

t_{1} = \frac{{43.863122}^{○} \times (π / 180^{○})}{0.517954705 \times (\frac{cos γ - 0.884358912}{cos γ - 0.857959405})}

(A14)

d t_{1} = \frac{0.039019431 sin γ}{{(cos γ - 0.884359952)}^{2}} d γ

(A15)

\int_{0}^{t_{1}} d t_{1} = \int_{- {20.053522}^{○}}^{{23.8096}^{○}} \frac{0.039019431 sin γ}{{(cos γ - 0.884358912)}^{2}} d γ

(A16)

t_{1} |_{0}^{t_{1}}) = \frac{0.039019431}{cos γ - 0.884358912} |_{- {20.053522}^{○}}^{{23.8096}^{○}})

t_{1} = 0.568671 s .

(A17)

(a₂). The second stage is the motion of the gyroscope from the equilibrium until the upper position, (Eq. (11)):

\begin{matrix} - 5.551045166 \times 1 0^{- 4} \frac{d ω_{x}}{d t} & = & 0.043 \times 9.81 \times 0.0039069767 cos γ \\ - [\frac{10 (2 π^{2} + 8)}{9} + (2 π^{2} + 9) cos γ] \\ \times 0.5726674 \times 1 0^{- 4} \times 600 \times (\frac{2 π}{60}) ω_{x} \end{matrix}

(A18)

Following steps of solutions are similar as for the first stage presented above. For the simplification of presentation all comments are omitted:

\begin{matrix} \frac{d ω_{x}}{\frac{0.015937527 cos γ}{1.072449256 + cos γ} + ω_{x}} = - 186.286904371 (1.072449256 + cos γ) d t \end{matrix}

(A19)

\begin{matrix} \int_{- 0.277771610}^{- ω_{x}} \frac{d ω_{x}}{\frac{0.015937525 cos γ}{1.072449256 + cos γ} + ω_{x}} \\ = - 186.286904371 (1.072449256 + cos γ) \int_{0}^{t} d t \end{matrix}

(A20)

\begin{matrix} ln (\frac{0.015937525 cos γ}{1.072449256 + cos γ} + ω_{x}) |_{- 0.277771610}^{- ω_{x}}) \\ = - 186.286904371 (1.072449256 + cos γ) |_{0}^{t}) \end{matrix}

(A21)

\begin{matrix} \frac{0.015937525 cos γ}{1.072449256 + cos γ} - ω_{x} & = & (\frac{0.015937525 cos γ}{1.072449256 + cos γ} - 0.277771610) \\ \times e^{- 186.286904371 (1.072449256 + cos γ) t} \end{matrix}

(A22)

ω_{x} = \frac{0.015937525 cos γ}{1.072449256 + cos γ}

(A23)

d ω_{x} = - \frac{0.017092188 sin γ}{{(1.072449256 + cos γ)}^{2}} d γ

(A24)

ω_{x} |_{0.277771610}^{ω_{x u}}) = - \frac{0.017092188}{1.072449256 + cos γ} |_{{23.8096}^{○}}^{γ_{u}})

(A25)

ω_{x u} = 0.286372139 (\frac{1.0127640261 + cos γ}{1.072449256 + cos γ})

(A26)

Equation (24) of the equality of the gyroscope energies for the equilibrium and the upper position of the resistance torque enables for expressing the angular velocity and position of the gyroscope.

\begin{matrix} 0.001648079 sin 23.809 6^{○} + \frac{1}{2} 5.551045166 \times 1 0^{- 4} \times {(0.277771610)}^{2} + \frac{1}{2} \times 0.043 \times {(0.277771610 \times 0.0039069767)}^{2} = 0.001648079 sin γ_{u} \\ + [(2 π^{2} + 9) cos γ + \frac{10 (2 π^{2} + 8)}{9}] \times 0.5726674 \times 1 0^{- 4} \times 600 (\frac{2 π}{60}) \times 0.286372139 (\frac{1.0127640261 + cos γ}{1.072449256 + cos γ}) \frac{(γ - {23.8096}^{○}) π}{180^{○}} \end{matrix}

(A27)

\begin{matrix} 3.251982763 (0.408733697 - sin γ_{u}) & = & (1.0127640261 + cos γ) \\ \times (γ - 23.809 6^{○}) \end{matrix}

γ_{u} = 23.8104 4^{○}

(A28)

\begin{matrix} ω_{x u} = 0.286372139 (\frac{1.0127640261 + cos {23.81044}^{○}}{1.072449256 + cos {23.81044}^{○}}) = 0.277771581 \end{matrix}

(A29)

The time of the gyroscope turn from equilibrium to the upper position is determined as follows. The angle of the gyroscope turn in the clockwise direction is γ_u – γ_e= $23.8104 4^{○} - 23.809 6^{○} = 0.00084 5^{○}$ that is a small value. Then, the time of gyroscope motion (t = γ/ω_x.e) can be computed by the value of average angular velocity.

t_{2} = \frac{{0.000845}^{○} \times π / 180^{○}}{(0.277771581 + 0.277771610) / 2} = 5.309409 \times 1 0^{- 5} s .

(A30)

The total time of the gyroscope motion up is as:

t_{u} = t_{1} + t_{2} = 0.568871 + 5.309409 \times 1 0^{- 5} = 0.568 s .

(A31)

(b₁). The gyroscope turns down from the upper position to the equilibrium that leads to change in the direction of the action for the inertial torque (Eq. (16)). Substituting initial data into modified Eq. (16) yields the following equation, which solution is similar to Eq. (A1):

\begin{matrix} 5.551045166 \times 1 0^{- 4} \frac{d ω_{x}}{d t} & = & 0.043 \times 9.81 \times 0.0039069767 cos γ \\ - [\frac{8 (2 π^{2} + 8)}{9} - (2 π^{2} + 9) cos γ] \\ \times 0.5726674 \times 1 0^{- 4} \times 600 \times (\frac{2 π}{60}) ω_{x} \end{matrix}

(A32)

\frac{5.368063832 \times 10^{- 3}}{0.857959405 - cos γ} \frac{d ω_{x}}{d t} = \frac{0.015937525 cos γ}{0.857959405 - cos γ} - ω_{x}

(A33)

\begin{matrix} \frac{d ω_{x}}{\frac{0.015937525 cos γ}{0.857959405 - cos γ} - ω_{x}} = 186.286905508 (0.857959405 - cos γ) d t \end{matrix}

(A34)

\begin{matrix} \int_{0.277771581}^{ω_{x}} \frac{d ω_{x}}{\frac{0.015937525 cos γ}{0.857959405 - cos γ} - ω_{x}} \\ = \int_{0}^{t} 186.286905508 (0.857959405 - cos γ) d t \end{matrix}

(A35)

\begin{matrix} ln (\frac{0.015937525 cos γ}{0.857959405 - cos γ} - ω_{x}) |_{0.277771581}^{ω_{x}}) \\ = - 186.286905508 (0.857959405 - cos γ) |_{0}^{t}) \end{matrix}

\begin{matrix} \frac{0.015937525 cos γ}{0.857959405 - cos γ} - ω_{x} \\ = [\frac{0.401312606 (cos γ - 0.892031973)}{0.857959405 - cos γ} - 0.277771581] \\ \times e^{- 186.286905508 (cos γ - 0.857959405)} \end{matrix}

(A36)

ω_{x} = \frac{0.015937525 cos γ}{0.857959405 - cos γ}

(A37)

The first derivative of Eq. (A37), transformation and the following integral equation with definite limits enables for defining the value of ω_x

d ω_{x} = \frac{- 0.013673749 sin γ}{{(0.857959405 - cos γ)}^{2}} d γ

(A38)

\begin{matrix} \int_{0.277771581}^{ω_{x . 1}} d ω_{x} = - 0.013673749 \int_{{23.81044}^{○}}^{γ_{1}} \frac{sin γ}{{(0.857959405 - cos γ)}^{2}} d γ \end{matrix}

(A39)

ω_{x} |_{0.277771581}^{{ω_{x .}}_{e}}) = \frac{0.013673749}{0.857959405 - cos γ} |_{{23.81044}^{○}}^{γ_{1}})

ω_{x . e} = 0.517970731 (\frac{0.884358095 - cos γ}{0.857959405 - cos γ})

(A40)

Substituting of Eq. (A40) into Eq. (25) of the equality of the gyroscope energies for the upper position and the reset to zero of the resisting torque enables for expressing the angular velocity and position of the gyroscope.

\begin{matrix} 0.001648079 sin 23.8104 4^{○} & = & 0.001648079 sin (- γ_{e}) + \frac{1}{2} 5.551045166 \times 1 0^{- 4} \times {[0.517970731 (\frac{0.884358095 - cos γ}{0.857959405 - cos γ})]}^{2} \\ + \frac{1}{2} \times 0.043 \times {[0.517970731 (\frac{0.884358095 - cos γ}{0.857959405 - cos γ}) \times 0.0039069767]}^{2} + [(2 π^{2} + 9) cos (- γ_{e}) - \frac{8 (2 π^{2} + 8)}{9}] \\ \times 0.5726674 \times 1 0^{- 4} \times 600 \times (\frac{2 π}{60}) \times 0.517970731 (\frac{0.884358095 - cos γ}{0.857959405 - cos γ}) (23.8104 4^{○} + γ_{e}) \frac{π}{180} \end{matrix}

(A41)

Simplification and transformation of Eq. (A41) brings the following transcendental equation:

\begin{matrix} 0.711721383 & = & 1.762943316 sin (- γ_{e}) + 0.079749642 \\ \times {[(\frac{0.884358095 - cos γ}{0.857959405 - cos γ})]}^{2} - (0.884358095 - cos γ) \\ \times (23.8104 4^{○} + γ_{e}) \end{matrix}

(A42)

Numerical solution of Eq. (A42) yields the following result:

γ_{e} = - 23.967 3^{○}

(A43)

\begin{matrix} ω_{x . e} = 0.517970731 (\frac{0.884358095 - cos {23.9673}^{○}}{0.857959405 - cos {23.9673}^{○}}) = 0.273000658 \end{matrix}

(A44)

The angle of the gyroscope turn in counter clockwise direction is γ_u – γ_e= |−23.9673°| + 23.81044° = 47.77774° and the time of gyroscope motion is t = γ/ω_x.u.

t_{1} = \frac{{47.77774}^{○} \times (π / 180^{○})}{0.517970731 (\frac{0.884358095 - cos γ}{0.857959405 - cos γ})}

(A45)

d t_{1} = \frac{0.042499138 cos γ}{{(0.884358095 - cos γ)}^{2}} d γ

(A46)

\int_{0}^{t_{1}} d t_{1} = \int_{{28.81044}^{○}}^{- {23.9683}^{○}} \frac{0.042499138 sin γ}{{(0.884358095 - cos γ)}^{2}} d γ

(A47)

t_{1} |_{0}^{t_{1}}) = \frac{0.042499138}{0.884358095 - cos γ} |_{{23.81044}^{○}}^{- {23.9683}^{○}})

t_{1} = 1,416 s .

(A48)

(b₂). The next step is to define the angle of the gyroscope location at the lower point and the time of the turn down. Substituting initial data into Eq. (21) and transformation yields the following equation:

\begin{matrix} 5.551045166 \times 1 0^{- 4} \frac{d ω_{x}}{d t} & = & 0.043 \times 9.81 \times 0.03674418 cos γ \\ - [\frac{10 (2 π^{2} + 8)}{9} + (2 π^{2} + 9) cos γ] \\ \times 0.5726674 \times 1 0^{- 4} \times 600 \times (\frac{2 π}{60}) ω_{x} \end{matrix}

(A49)

The solution of Eq. (A49) is similar to Eq. (A19)

\begin{matrix} \frac{d ω_{x}}{\frac{0.015937525 cos γ}{1.072449256 + cos γ} - ω_{x}} = 186.286904371 (1.072449256 + cos γ) d t \end{matrix}

(A50)

\begin{matrix} \int_{0.272969538}^{ω_{x}} \frac{d ω_{x}}{\frac{0.015937525 cos γ}{1.072449256 + cos γ} - ω_{x}} \\ = 186.286904371 (1.072449256 + cos γ) \int_{0}^{t} d t \end{matrix}

(A51)

\begin{matrix} ln (\frac{0.015937525 cos γ}{1.072449256 + cos γ} - ω_{x}) |_{0.273000658}^{ω_{x}}) \\ = - 186.286904371 (1.072449256 + cos γ) |_{0}^{t}) \end{matrix}

(A52)

\begin{matrix} \frac{0.015937525 cos γ}{1.072449256 + cos γ} - ω_{x} & = & (\frac{0.015937525 cos γ}{1.072449256 + cos γ} - 0.273000658) \\ \times e^{- 186.286904371 (1.072449256 + cos γ) t} \end{matrix}

(A53)

ω_{x} = \frac{0.015937525 cos γ}{1.072449256 + cos γ}

(A54)

d ω_{x} = - \frac{0.017092188 sin γ}{{(1.072449256 + cos γ)}^{2}} d γ

(A55)

ω_{x} |_{0.273000658}^{ω_{x u}}) = - \frac{0.017092188}{1.072449256 + cos γ} |_{- {23.9673}^{○}}^{γ_{u}})

ω_{x u} = 0.281606014 (\frac{1.011753863 + cos γ}{1.072449256 + cos γ})

(A56)

d ω_{x} = - \frac{- 0.017092184 sin γ}{{(1.072449256 + cos γ)}^{2}} d γ

(A57)

\begin{matrix} \int_{0,273000658}^{ω_{x . l}} d ω_{x} = - 0.017092184 \int_{- {23.9673}^{0}}^{γ_{l}} \frac{\sin γ}{{(1.072449256 + \cos γ)}^{2}} d γ \end{matrix}

(A58)

ω_{x} |_{0,273000658}^{ω_{x . l}}) = - \frac{0,017092184}{1,072449256 + \cos γ} |_{- 23, 9673^{0}}^{γ_{l}})

ω_{x} = 0.281606012 (\frac{1.011753489 + \cos γ}{1.072449256 + \cos γ})

(A59)

Equation (A59) of the equality of the gyroscope energies for the equilibrium and the low position of the resisting torque enabled for expressing the angular velocity and position of the gyroscope.

\begin{matrix} 0.001648079 sin (- γ_{l}) = 0.001648079 sin (- 23.967 3^{○}) + \frac{1}{2} 5.551045166 \times 1 0^{- 4} \times {[0.281606012 (\frac{1.011753489 + cos γ}{1.072449256 + cos γ})]}^{2} \\ + \frac{1}{2} \times 0.043 \times {[0.281606012 (\frac{1.011753489 + cos γ}{1.072449256 + cos γ}) \times 0.0039069767]}^{2} - [\frac{10 (2 π^{2} + 8)}{9} + (2 π^{2} + 9) cos γ] \\ \times 0.5726674 \times 1 0^{- 4} \times 600 \times (\frac{2 π}{60}) \times 0.281606012 (\frac{1.011753489 + cos γ}{1.072449256 + cos γ}) [γ_{l} - (- 23.967 3^{○}] \frac{π}{180} \end{matrix}

(A60)

\begin{matrix} 3.242661731 sin (- γ_{l}) & = & - 1.317218469 + 0.043818482 \\ \times {(\frac{1.011753489 + cos γ}{1.072449256 + cos γ})}^{2} \\ - (1.011753489 + cos γ) [γ_{l} - (- 23.967 3^{○}] \end{matrix}

(A61)

γ_{l} = - 23.989 3^{○}

(A62)

\begin{matrix} ω_{x} & = & 0.281606012 (\frac{1.011753489 + cos {23.9893}^{○}}{1.072449256 + cos {23.9893}^{○}}) \\ = & 0.273000031 rad / s \end{matrix}

(A63)

The time of the gyroscope turn from equilibrium to the upper position is determined as follows. The angle of the gyroscope turn in the clockwise direction is γ_u – γ_e= −23.9893° − (−23.9673°) = 0.0220° and the time of gyroscope motion is t = γ/ω_x.u.

t_{2} = \frac{{0.0220}^{○} \times π / 180^{○}}{(0.273000031 + 0.273000658) / 2} = 0.001406 s .

(A64)

The total time of the gyroscope motion up is as:

t_{u} = t_{1} + t_{2} = 1,416 + 0,001 = 1.417 s .

(A65)

The first nutation of the gyroscope is terminated with the surplus kinetic energy of the inertial torques. The value of the angular velocity of the gyroscope is less than the angular velocities of the initial condition of gyroscope’s nutation. It means the gyroscope will continue the nutation. The computing parameters of the following nutation demonstrate the same results as presented above by Eqs. (A1) – (A65).

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2019

Author(s)

Physics of gyroscope nutation

I. INTRODUCTION

II. METHODOLOGY

A. Case study and practical tests

III. RESULTS AND DISCUSSION

IV. CONCLUSION

ACKNOWLEDGMENTS

NOMENCLATURE

1. Solution

2. First nutation

REFERENCES

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Physics of gyroscope nutation

I. INTRODUCTION

II. METHODOLOGY

A. Case study and practical tests

III. RESULTS AND DISCUSSION

IV. CONCLUSION

ACKNOWLEDGMENTS

NOMENCLATURE

1. Solution

2. First nutation

REFERENCES

Citing articles via

Submit your article

Sign up for alerts

Related Content

Resources

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