Stability and Stabilization of Steady Rotations of a Spherical Robot on a Vibrating Base

Abstract

This paper addresses the problem of a spherical robot having an axisymmetric pendulum drive and rolling without slipping on a vibrating plane. It is shown that this system admits partial solutions (steady rotations) for which the pendulum rotates about its vertical symmetry axis. Special attention is given to problems of stability and stabilization of these solutions. An analysis of the constraint reaction is performed, and parameter regions are identified in which a stabilization of the spherical robot is possible without it losing contact with the plane. It is shown that the partial solutions can be stabilized by varying the angular velocity of rotation of the pendulum about its symmetry axis, and that the rotation of the pendulum is a necessary condition for stabilization without the robot losing contact with the plane.

APPENDIX A. PROOF OF THE EQUIVALENCE OF TWO SYSTEMS AND DERIVATION OF THE HAMILTONIAN (2.15)

1. Due to the spherical symmetry of the shell and due to the invariance of the system under rotations about the vertical, the equations for the variables $(\boldsymbol{\omega},\boldsymbol{\gamma})$ decouple from the complete system of equations (2.1) and (2.7). Using the variables $\boldsymbol{M},\boldsymbol{\gamma}$, these equations can be represented as

$$\displaystyle\dot{\boldsymbol{M}}+\boldsymbol{\omega}\times\boldsymbol{M}=\eta(\boldsymbol{e}_{3},\boldsymbol{\gamma}\times\boldsymbol{\omega})\boldsymbol{\gamma}\times(\boldsymbol{e}_{3}\times\boldsymbol{\omega})+(\tilde{{\rm g}}-\ddot{\tilde{\xi}}(t))\boldsymbol{\gamma}\times\boldsymbol{e}_{3},$$

(A.1)

$$\displaystyle\dot{\boldsymbol{\gamma}}=\boldsymbol{\gamma}\times\boldsymbol{\omega},$$

where $\boldsymbol{\omega}={\bf D}^{-1}\boldsymbol{M}$ and the following notation has been introduced:

$$\tilde{{\rm g}}=\dfrac{{\rm g}}{\ell},\quad\tilde{\xi}(t)=\dfrac{\xi(t)}{\ell},\quad\ell=\dfrac{\hat{I}}{m\rho\tilde{I}}.$$

In this case, the dynamics of the variables $\boldsymbol{\alpha},\boldsymbol{\beta},{\bf S},{\bf r}$ is given by the quadratures (2.1), and the angular velocities of rotation of the shell $\boldsymbol{\Omega}$ on a fixed level set of the Chaplygin integral $\boldsymbol{\mathcal{K}}^{abs}$ depend on the variables $\boldsymbol{\omega},{\bf Q}$ as follows:

$$\boldsymbol{\Omega}={\bf J}^{-1}\left({\bf Q}\boldsymbol{\mathcal{K}}^{abs}+m\rho R\boldsymbol{\gamma}\times(\boldsymbol{e}_{3}\times\boldsymbol{\omega})\right).$$

As shown in [10], the reduced system of equations describing the motion of the top fastened to the shell rolling without slipping on the plane is equivalent, up to a change of parameters, to a system of equations describing the dynamics of an unbalanced dynamically asymmetric ball on an absolutely smooth plane.

It is easy to see (by direct verification of the equations of motion) that the following proposition holds.

Proposition 3

The reduced system of equations (A.1) $($ for the variables $\boldsymbol{M},\boldsymbol{\gamma}$ $)$ , which describes the dynamics of an axisymmetric pendulum fastened at the center of a spherical shell rolling on an absolutely rough vibrating plane, coincides, up to a change of parameters, with the reduced system describing the dynamics of a spherical top with an axisymmetric mass distribution on a smooth vibrating plane. In this case, the parameters of the top are related to the dimensionless parameters of the system (A.1) by

$$\eta=\dfrac{m_{t}a_{t}^{2}}{i_{1}^{(t)}},\quad\nu=\dfrac{i_{1}^{(t)}}{i_{3}^{(t)}},\quad\ell=\dfrac{i_{1}^{(t)}}{m_{t}a_{t}},$$

where $m_{t}$ is the mass of the top, $i_{1}^{(t)},i_{3}^{(t)}$ are its equatorial and axial moments of inertia, respectively, and $a_{t}$ is the displacement of the center of mass along the symmetry axis. The dimensionless angular momentum $\boldsymbol{M}$ (2.9) coincides, up to a constant factor, with the spherical top’s angular momentum calculated with respect to the point of contact.

This conclusion can be extended to an arbitrary mass distribution of the bodies (the top and the pendulum) for motion on a vibrating plane. In this case, the shell must be homogeneous and have a spherical tensor of inertia.

2. Next, we consider the procedure of reducing equations (A.1) to a system with one and a half degrees of freedom.

Equations (A.1) admit three integrals of motion $\mathcal{F}_{0},\mathcal{F}_{1},\mathcal{F}_{2}$ and the symmetry field (2.13). Following [19], we choose as variables of the reduced system the integrals of the symmetry field

$$k_{3}=\dfrac{1}{k}(\omega_{2}\gamma_{1}-\omega_{1}\gamma_{2})\quad\text{and}\quad\gamma_{3},$$

where

$$k=\sqrt{\dfrac{1-\gamma_{3}^{2}}{1+\eta(1-\gamma_{3}^{2})}}.$$

On the fixed level set of the integrals (2.8) and (2.10)

$$\mathcal{F}_{1}=k_{1},\quad\mathcal{F}_{2}=k_{2}$$

the equations of motion take the form

$$\begin{array}[]{c}\dot{k}_{3}=-\dfrac{k(k_{2}-k_{1}\gamma_{3})(k_{2}\gamma_{3}-k_{1})}{(1-\gamma_{3}^{2})^{2}}+k(\tilde{{\rm g}}-\ddot{\tilde{\xi}}(t)),\\ \dot{\gamma}_{3}=kk_{3}.\end{array}$$

(A.2)

To recover the dynamics of the complete system (A.1), we need to add to Eqs. (A.2) a quadrature for the angle of proper rotation

$$\dot{\varphi}=\nu k_{1}-\dfrac{\gamma_{3}(k_{2}-k_{1}\gamma_{3})}{1-\gamma_{3}^{2}}.$$

In this case, the time dependence of the angular velocities $\boldsymbol{\omega}$ and the other two components, $\gamma_{1},\gamma_{2}$, will be determined by the algebraic relations

$$\begin{array}[]{c}\omega_{1}=\dfrac{(k_{2}-k_{1}\gamma_{3})\gamma_{1}-kk_{3}\gamma_{2}}{1-\gamma_{3}^{2}},\quad\omega_{2}=\dfrac{(k_{2}-k_{1}\gamma_{3})\gamma_{2}+kk_{3}\gamma_{1}}{1-\gamma_{3}^{2}},\\ \omega_{3}=\nu k_{1},\quad\gamma_{1}=\sqrt{1-\gamma_{3}^{2}}\cos\varphi,\quad\gamma_{2}=\sqrt{1-\gamma_{3}^{2}}\sin\varphi.\end{array}$$

Thus, the reduction yields a system with one and a half degrees of freedom (A.2).

3. Reduction to Hamiltonian form.

We transform from the variables $k_{3},\gamma_{3}$ to the nutation angle $\vartheta$ and its derivative $\dot{\vartheta}$ by making the change of variables

$$\gamma_{3}=\cos\vartheta,\quad k_{3}=-\sqrt{1+\eta\sin^{2}\vartheta}\dot{\vartheta}.$$

Now Eqs. (A.2) can be written in the Lagrangian form

$$\begin{array}[]{c}\left(\dfrac{\partial\mathcal{L}}{\partial\dot{\vartheta}}\right)^{\!\!\boldsymbol{\cdot}}-\dfrac{\partial\mathcal{L}}{\partial\vartheta}=0,\\ \mathcal{L}=\dfrac{1}{2}(1+\eta\sin^{2}\vartheta)\dot{\vartheta}^{2}-\dfrac{1}{2}\dfrac{(k_{2}-k_{1}\cos\vartheta)^{2}}{\sin^{2}\vartheta}-\tilde{{\rm g}}\cos{\vartheta}+\sin\vartheta\dot{\vartheta}\dot{\tilde{\xi}}(t).\end{array}$$

Making a standard Legendre transformation, we can represent the resulting equations in Hamiltonian form with a canonical Poisson bracket and the Hamiltonian

$$\mathcal{H}=\dfrac{1}{2}\dfrac{\Big{(}p_{\vartheta}-\dot{\tilde{\xi}}(t)\sin\vartheta\Big{)}^{2}}{1+\eta\sin^{2}\vartheta}+\dfrac{1}{2}\dfrac{(k_{2}-k_{1}\cos\vartheta)^{2}}{\sin^{2}\vartheta}+\tilde{{\rm g}}\cos\vartheta.$$

(A.3)

Here, the momentum $p_{\vartheta}$ is related to the derivative of the nutation angle by

$$p_{\vartheta}=(1+\eta\sin^{2}\vartheta)\dot{\vartheta}+\dot{\tilde{\xi}}(t)\sin\vartheta.$$

Remark 6

The Hamiltonian (A.3) differs from the total energy (2.12) by a constant value depending on the values of the first integrals, and by the explicit function of time

$$\mathcal{H}=\left(\mathcal{E}-\dfrac{1}{2\tilde{I}}(\boldsymbol{\mathcal{K}}^{abs})^{2}-\dfrac{1}{2}{{\mathcal{K}}_{\gamma}^{2}}\left(\dfrac{1}{I}-\dfrac{1}{\tilde{I}}\right)-\dfrac{i_{3}\nu k_{1}^{2}}{2}-m\rho(R+\xi(t))-\dfrac{1}{2}m\left(\dot{\xi}(t)\right)^{2}\right)\dfrac{\tilde{I}}{\hat{I}}.$$

We note that the momentum $p_{\vartheta}$ has been calculated in the fixed coordinate system. However, it is sometimes more convenient to use a (relative) momentum calculated in a coordinate system attached to an oscillating plane. The transformation from $p_{\vartheta}$ to the new momentum is given by the following relation:

$$\tilde{p}=p_{\vartheta}-\dot{\tilde{\xi}}(t)\sin\vartheta.$$

In the new variables the Hamiltonian of the system takes the form

$$\mathcal{H}=\dfrac{\tilde{p}^{2}}{2(1+\eta\sin^{2}\vartheta)}+\dfrac{(k_{2}-k_{1}\cos\vartheta)^{2}}{2\sin^{2}\vartheta}+(\tilde{{\rm g}}+\ddot{\tilde{\xi}}(t))\cos\vartheta.$$

(A.4)

We now introduce a dimensionless time variable $\tau$ by making the following transformations:

$$t=\dfrac{1}{\Omega}\tau,\quad k_{1}=\Omega\kappa_{1},\quad k_{2}=\Omega\kappa_{2},\quad\tilde{p}=\Omega p,\quad\tilde{{\rm g}}=\Omega^{2}\gamma,$$

where $\gamma$ is the dimensionless free-fall acceleration. After these transformations and substitution of the law of oscillations of the plane (2.14), the Hamiltonian (A.4) takes the form (2.15).

APPENDIX B. DERIVATION OF AN EXPRESSION FOR THE COEFFICIENT $$c$$

To reduce the Hamiltonian (3.12) to the form (3.13), we define the generating function, which depends on the old coordinates $q$, new momenta $p_{1}$ and time $\tau$, as

$$S(q,p_{1},\tau)=qp_{1}+a_{1}(\tau)q^{4}+a_{2}(\tau)q^{3}p_{1}+a_{3}(\tau)q^{2}p_{1}^{2}+a_{4}(\tau)qp_{1}^{3}+a_{5}(\tau)p_{1}^{4},$$

where $a_{i}(\tau)$ ($i=1\ldots 5$) are unknown periodic functions of time. Then the change of variables has the form

$$\displaystyle q=q_{1}-a_{2}(\tau)q_{1}^{3}-2a_{3}(\tau)q_{1}^{2}p_{1}-3a_{4}(\tau)q_{1}p_{1}^{2}-4a_{5}(\tau)p_{1}^{3},$$

$$\displaystyle p=p_{1}+4a_{1}(\tau)q_{1}^{3}+3a_{2}(\tau)q_{1}^{2}p_{1}+2a_{3}(\tau)q_{1}p_{1}^{2}+a_{4}(\tau)p_{1}^{3},$$

where we have discarded terms higher than degree 4 in $q_{1}$, $p_{1}$. Expressing the Hamiltonian (3.12) in terms of the new variables and reducing it to the form (3.13), we obtain a system of five differential equations for the coefficients $a_{i}(\tau)$. To find an explicit form of the change of variables, it is necessary to integrate all five equations. However, only three of these five equations contain the coefficient $c$. Therefore, to search for the coefficient $c$, it suffices to consider a system of only three equations:

$$\displaystyle a^{\prime}_{1}(\tau)-\lambda a_{2}(\tau)-\dfrac{\eta}{2}n_{11}^{2}n_{21}^{2}-\dfrac{B}{24}n_{11}^{4}=\dfrac{1}{4}c,$$

$$\displaystyle a^{\prime}_{3}(\tau)3\lambda(a_{2}(\tau)-a_{4}(\tau))-\dfrac{\eta}{2}(n_{11}^{2}n_{22}^{2}+4n_{11}n_{12}n_{21}n_{22}+n_{12}^{2}n_{21}^{2})-\dfrac{B}{4}n_{11}^{2}n_{12}^{2}=\dfrac{1}{2}c,$$

$$\displaystyle a^{\prime}_{5}(\tau)+\lambda a_{4}(\tau)-\dfrac{\eta}{2}n_{12}^{2}n_{22}^{2}-\dfrac{B}{24}n_{12}^{4}=\dfrac{1}{4}c,$$

where $B=\kappa-\dfrac{3}{4}\zeta+\delta\cos\tau$. Multiplying these equations, respectively, by 24, 8 and 24 and adding up, we obtain one equation:

$$\displaystyle 8(3a^{\prime}_{1}(\tau)+a^{\prime}_{3}(\tau)+3a^{\prime}_{5}(\tau))-4\eta\left(3(n_{11}n_{21}+n_{12}n_{22})^{2}+(n_{11}n_{22}-n_{12}n_{21})^{2}\right)-B\left(n_{11}^{2}+n_{12}^{2}\right)^{2}=16c.$$

Let us integrate this equation over $\tau$, noting that the coefficients $a_{i}(\tau)$ are periodic functions. Then the integral (over the period) of the first term containing the derivatives $a^{\prime}_{i}(\tau)$ is zero. Expressing the coefficient $c$, we obtain

$$c=-\dfrac{1}{32\pi}\int\limits_{0}^{2\pi}B\left(n_{11}^{2}+n_{12}^{2}\right)^{2}+4\eta\left(3(n_{11}n_{21}+n_{12}n_{22})^{2}+(n_{11}n_{22}-n_{12}n_{21})^{2}\right)d\tau.$$

Substituting into this equation the expressions for $n_{11},n_{12},n_{21},n_{22}$, written in terms of $\mu_{1},\mu_{2},\nu_{1},\nu_{2}$, we obtain (3.14).