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Kane’s equations for nonholonomic systems in bond-graph-compatible velocity and momentum forms

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Abstract

The authors’ previously published results on a bond-graph-compatible and nonholonomic momentum form of Kane’s equations are extended, from scleronomic to rheonomic systems. The momentum form is found to be partially Hamiltonian, and bond-graph-compatible velocity forms are also found, of which one is partially Lagrangian. The results are derived for general particle systems and then specialized to rigid-body systems. Direct dependence of kinetic energy upon time is avoided, making the results power-conserving and suitable for use with bond graphs. A new nonholonomic IC (NIC) bond-graph element is defined, and bond graphs for the partially Hamiltonian and partially Lagrangian forms are exhibited using this element.

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Notes

  1. The IC field element implements only inertial (kinetic) energy storage, but does have a set of ports with effort, or C-port causality.

  2. Where feasible, we use terminology from [22], a revised and updated edition of the earlier exposition of Kane’s methodology [10]. In the earlier exposition the terminology used for the elements of \(\boldsymbol {\mathrm {f}}\) is “generalized speeds”. Cogent objections to the earlier terminology have been raised [18], and so we use the newer terminology “generalized velocities” from [22]. In the more general dynamics literature, “quasi-velocities” and “nonholonomic velocities” are also used as synonyms. Generalized velocities, as defined here, are to be distinguished from the coordinate derivatives \(\dot{\boldsymbol {\mathrm {q}}}\), unless \(\boldsymbol {\mathrm {Q}}\) is the identity matrix.

  3. The use of the adjectives “reduced” and “unreduced”, as describing the noun “Kane’s equations”, describes the process involved in proceeding from an expression equating generalized impressed and inertia forces (Sects. 3.2, 3.3), to several standard forms of ordinary differential equations, with associated matrix parameters (Sect. 3.4 et seq.). It is acknowledged that this usage does not appear in any previous published treatment the authors are aware of, and thus does not yet represent established terminology.

  4. \(\boldsymbol {\mathrm {D}}\) is intended to be a simplified version of \(\bar{\boldsymbol {\mathrm {D}}}\) in which (17) is satisfied while terms that always reduce to zero in both \(\bar{\boldsymbol {\mathrm {D}}}\boldsymbol {\mathrm {f}}\) and \(\bar{\boldsymbol {\mathrm {D}}}^{\mathsf{T}}\boldsymbol {\mathrm {f}}\) do not appear. In Appendix B, we derive in detail the form that \(\bar{\boldsymbol {\mathrm {D}}}\) takes for rigid bodies and show that it contains such terms.

  5. The standard 6-DOF flight equations, using body-referenced velocity components as elements of \(\boldsymbol {\mathrm {f}}\), are an example of this circumstance.

  6. The utility of rheonomic constraints, expressed here at the velocity level, is sometimes obvious, e.g., for analysis of a system driven by a shaft with imposed rotational velocity. Less obviously, a rheonomic formulation is also useful for computing joint or internal reaction forces or moments by introducing fictitious degrees of freedom to the system model and then constraining the associated generalized velocities to be zero while computing the associated generalized constraint reactions.

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Author information

Authors and Affiliations

  1. Senior Applications Engineer, Applied Dynamics International, Ann Arbor, MI, USA

    James R. Phillips

  2. Department of Orthopaedic Surgery, University of Illinois at Chicago, Chicago, IL, USA

    Farid Amirouche

  3. Vice-Chairman, Department of Orthopaedic Surgery, Northshore University HealthSystem, Skokie, IL, USA

    Farid Amirouche

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  1. James R. Phillips
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Correspondence to James R. Phillips.

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Appendices

Appendix A: Proofs of results for rigid bodies

Here we provide demonstrations of the results stated above for rigid bodies, i.e., Eqs. (56)–(61). Without loss of generality, we consider only a single rigid body, with the velocity of its mass center given by \(\boldsymbol {v}_{\mathrm{c}}\) and its angular velocity given by \(\boldsymbol {\omega}\) in an inertial frame. A typical material point, with differential mass \(\mathrm{d}m\), is located at a vector radius \(\boldsymbol {\rho}\) from the mass center and has the inertial velocity

$$ \boldsymbol {v}=\boldsymbol {v}_{\mathrm{c}}+\boldsymbol {\omega}\times \boldsymbol {\rho}; $$
(64)

its partial velocity matrix is therefore

$$ \frac{\partial \boldsymbol {v}}{\partial \boldsymbol {\mathrm {f}}}= \frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}+ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\times \boldsymbol {\rho}\,. $$
(65)

The second term on the right of (64) is the derivative of the radius vector, i.e.,

$$ \dot{\boldsymbol {\rho}}=\boldsymbol {\omega}\times \boldsymbol {\rho}\,. $$
(66)

By the definition of the mass and mass center, we have the following properties:

$$ \int \mathrm{d}m=m\,, $$
(67)
$$ \int \boldsymbol {\rho}\,\mathrm{d}m=\boldsymbol {0}\,. $$
(68)

The symmetric and positive-definite rotational inertia tensor \(\boldsymbol {\mathsf {I}}\) is defined as

$$ \boldsymbol {\mathsf {I}}\equiv \int \tilde{\boldsymbol {\rho}}\cdot \tilde{\boldsymbol {\rho}}^{ \mathsf{T}}\mathrm{d}m\,, $$
(69)

where \(\tilde{\boldsymbol {\rho}}\) is the antisymmetric dual tensor corresponding to the radius vector, defined as

$$ \tilde{\boldsymbol {\rho}}\equiv \boldsymbol {\mathsf {1}}\times \boldsymbol {\rho}\,, $$
(70)

and \(\boldsymbol {\mathsf {1}}\) is the identity tensor. The inertia tensor \(\boldsymbol {\mathsf {I}}\) has the property that its components are constant when expressed with respect to a set of body-fixed basis vectors; using this property, it is easy to show that the time derivative of \(\boldsymbol {\mathsf {I}}\) is given by

$$ \frac{\mathrm{d}}{\mathrm{dt}}\boldsymbol {\mathsf {I}}=\boldsymbol {\omega}\times \boldsymbol {\mathsf {I}}- \boldsymbol {\mathsf {I}}\times \boldsymbol {\omega}\,. $$
(71)

Eq. (56)

From the definition of \(\boldsymbol {\mathrm {e}}\) in (9), using (65), we have

$$\begin{aligned} \boldsymbol {\mathrm {e}} & =\int \left ( \frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}+ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\times \boldsymbol {\rho} \right )\cdot \mathrm{d}\boldsymbol {F} \\ & =\frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}\cdot \int \mathrm{d}\boldsymbol {F}+\frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}} \cdot \int \boldsymbol {\rho}\times \mathrm{d}\boldsymbol {F} \\ & \text{$\phantom{=}$(using properties of the scalar triple product}) \\ & =\frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}\cdot \boldsymbol {F}+\frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\cdot \boldsymbol {M} \\ & \hphantom{=} (\text{by definition of}\,\boldsymbol {F}\,\text{and}\,\boldsymbol {M})\,. \end{aligned}$$

 □

Eq. (57)

From the definition of \(\boldsymbol {\mathrm {p}}\) in (20), using (64) with (65), we have

$$\begin{aligned} \boldsymbol {\mathrm {p}} & =\int \left ( \frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}+ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\times \boldsymbol {\rho} \right )\cdot \left (\boldsymbol {v}_{\mathrm{c}}+\boldsymbol {\omega}\times \boldsymbol {\rho}\right )\mathrm{d}m \\ & =\int \left (\frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}+ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\cdot \tilde{\boldsymbol {\rho}}\right )\cdot \left (\boldsymbol {v}_{\mathrm{c}}+ \boldsymbol {\omega}\cdot \tilde{\boldsymbol {\rho}}\right )\mathrm{d}m \\ & \phantom{=} (\text{by definition of}\,\tilde{\boldsymbol {\rho}}) \\ & =\frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}\cdot \boldsymbol {v}_{\mathrm{c}}\int \mathrm{d}m+ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\cdot \left (\int \tilde{\boldsymbol {\rho}}\cdot \tilde{\boldsymbol {\rho}}^{\mathsf{T}}\mathrm{d}m \right )\cdot \boldsymbol {\omega} \\ & \phantom{=} ( \text{using (68), and by definition of the tensor transpose operator}) \\ & =\frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}\cdot \boldsymbol {v}_{\mathrm{c}}\,m+ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\cdot \boldsymbol {\mathsf {I}}\cdot \boldsymbol {\omega} \\ & \phantom{=} (\text{by definitions of}\,m\,\text{and}\,\boldsymbol {\mathsf {I}}) \\ & =\frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}\cdot \boldsymbol {p}+\frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\cdot \boldsymbol {h} \\ & \phantom{=} (\text{by definitions of}\,\boldsymbol {p}\,\text{and}\,\boldsymbol {h})\,. \end{aligned}$$

 □

Eq. (58)

Using (23) and (57), we have

$$\begin{aligned} \boldsymbol {\mathrm {A}} & =\frac{\partial}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\left [ \left (\frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}\right ) \cdot \boldsymbol {p}+\left (\frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}} \right )\cdot \boldsymbol {h}\right ] \\ & =\left (\frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}} \right )\cdot \frac{\partial}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\left (m \,\boldsymbol {v_{\mathrm{c}}}\right )+\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \frac{\partial}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\left (\boldsymbol {\mathsf {I}}\cdot \boldsymbol {\omega}\right ) \\ & \phantom{=} ( \text{by definition of }\,\boldsymbol {p}\,\text{and}\,\boldsymbol {h}) \\ & =m\left (\frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}} \right )\cdot \left ( \frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}} \right )+\left (\frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right ) \cdot \boldsymbol {\mathsf {I}}\cdot \left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right )\,. \end{aligned}$$

 □

Eq. (59)

We prove this by showing first that \(\dot{\boldsymbol {\mathrm {A}}}=\boldsymbol {\mathrm {D}}+\boldsymbol {\mathrm {D}}^{\mathsf{T}}\) and second that \(\bar{\boldsymbol {\mathrm {D}}}^{\mathsf{T}}\boldsymbol {\mathrm {f}}=\boldsymbol {\mathrm {D}}^{\mathsf{T}}\boldsymbol {\mathrm {f}}\) when \(\boldsymbol {\mathrm {D}}\) is given by (59). N.B. it is not claimed that \(\boldsymbol {\mathrm {D}}=\bar{\boldsymbol {\mathrm {D}}}\); what is claimed is that by satisfying these properties, \(\boldsymbol {\mathrm {D}}\) may serve in place of \(\bar{\boldsymbol {\mathrm {D}}}\) in the equations of motion. Starting first then with (58) and using the definition of \(\boldsymbol {\mathrm {C}}\) in (62), by direct differentiation we have

$$\begin{aligned} \dot{\boldsymbol {\mathrm {A}}} & =\boldsymbol {\mathrm {C}}+\boldsymbol {\mathrm {C}}^{\mathsf{T}}+\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \frac{\mathrm{d}}{\mathrm{d}t}\boldsymbol {\mathsf {I}}\cdot \left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right ) \\ & =\boldsymbol {\mathrm {C}}+\boldsymbol {\mathrm {C}}^{\mathsf{T}}+\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \left ( \boldsymbol {\omega}\times \frac{\partial \boldsymbol {h}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right )+\left ( \boldsymbol {\omega}\times \frac{\partial \boldsymbol {h}}{\partial \boldsymbol {\mathrm {f}}}\right ) \cdot \left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right ) \\ & \phantom{=} ( \text{using (71) and properties of the scalar triple product}) \\ & =\boldsymbol {\mathrm {D}}+\boldsymbol {\mathrm {D}}^{\mathsf{T}} \\ & \phantom{=} (\text{using (63)}) \\ & (\text{First part of proof.}) \end{aligned}$$

 □

Then for the second part of the proof, we begin with the definition of \(\bar{\boldsymbol {\mathrm {D}}}\) in (16), finding

$$\begin{aligned} \bar{\boldsymbol {\mathrm {D}}}^{\mathsf{T}}\boldsymbol {\mathrm {f}} & =\int \frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {v}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \left ( \frac{\partial \boldsymbol {v}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right )\boldsymbol {\mathrm {f}} \,\mathrm{d}m \\ & =\int \frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {v}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \boldsymbol {v}\, \mathrm{d}m \\ & =\int \frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}+ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\times \boldsymbol {\rho} \right )\cdot \left (\boldsymbol {v}_{\mathrm{c}}+\boldsymbol {\omega}\times \boldsymbol {\rho}\right )\,\mathrm{d}m \\ & \phantom{=} (\text{using (64) and (65)}) \\ & =\frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \boldsymbol {v}_{\mathrm{c}}\int \mathrm{d}m+\int \left [ \frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \tilde{\boldsymbol {\rho}}+\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\times \left ( \boldsymbol {\omega}\cdot \tilde{\boldsymbol {\rho}}\right )\right ]\cdot \left ( \boldsymbol {\omega}\cdot \tilde{\boldsymbol {\rho}}\right )\,\mathrm{d}m \\ & \phantom{=} ( \text{using (68), (66), and (70)}) \\ & =\frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \boldsymbol {v}_{\mathrm{c}}\,m+\frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \left ( \int \tilde{\boldsymbol {\rho}}\cdot \tilde{\boldsymbol {\rho}}^{\mathsf{T}} \mathrm{d}m\right )\cdot \boldsymbol {\omega} \\ & \phantom{=} (\text{using the definition of }\,m\, \text{and properties of the cross product and tensor} \\ & \phantom{=}\text{transpose operator}) \\ & =\frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \boldsymbol {p}+\frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \boldsymbol {h} \\ & \phantom{=} (\text{using the definitions of }\,\boldsymbol {p},\,\boldsymbol {\mathsf {I}},\,\text{and}\, \boldsymbol {h})\,. \end{aligned}$$

Then using (62) and (63), we easily find that \(\boldsymbol {\mathrm {D}}^{\mathsf{T}}\boldsymbol {\mathrm {f}}\) simplifies to the same result, so that \(\bar{\boldsymbol {\mathrm {D}}}^{\mathsf{T}}\boldsymbol {\mathrm {f}}=\boldsymbol {\mathrm {D}}^{\mathsf{T}}\boldsymbol {\mathrm {f}}\), as desired (second part of proof).  □

Eq. (60)

We have already found that \(\boldsymbol {\mathrm {D}}^{\mathsf{T}}\boldsymbol {\mathrm {f}}\) simplifies to the correct result, immediately above. All that is needed is to recall that \(\boldsymbol {\mathrm {D}}^{\mathsf{T}}\boldsymbol {\mathrm {f}}=\hat{\boldsymbol {\mathrm {e}}}\) by definition from (27).  □

Eq. (61)

Let \(\boldsymbol {u}\) be an arbitrary unit vector fixed in a frame rotating with angular velocity \(\boldsymbol {\omega}\). The derivative of \(\boldsymbol {u}\), \(\dot{\boldsymbol {u}}\), is given by \(\boldsymbol {\omega}\times \boldsymbol {u}\). Using “cancellation of dots” (31) with \(\boldsymbol {u}\), we have

$$ \frac{\partial \boldsymbol {u}}{\partial q_{r}}= \frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{r}}\times \boldsymbol {u}\,. $$
(72)

Differentiating this expression again with respect to \(q_{s}\), we find

$$ \frac{\partial ^{2}\boldsymbol {u}}{\partial q_{s}\partial q_{r}}= \frac{\partial ^{2}\boldsymbol {\omega}}{\partial q_{s}\partial \dot{q}_{r}} \times \boldsymbol {u}+\frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{r}} \times \frac{\partial \boldsymbol {u}}{\partial q_{s}}\,. $$

Noting that the left side of the equation is symmetric in the indices \(s\), \(r\), we may equate the right sides of the equation with reversed indices, i.e.,

$$ \frac{\partial ^{2}\boldsymbol {\omega}}{\partial q_{s}\partial \dot{q}_{r}} \times \boldsymbol {u}+\frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{r}} \times \frac{\partial \boldsymbol {u}}{\partial q_{s}}= \frac{\partial ^{2}\boldsymbol {\omega}}{\partial q_{r}\partial \dot{q}_{s}} \times \boldsymbol {u}+\frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{s}} \times \frac{\partial \boldsymbol {u}}{\partial q_{r}}\,, $$

from which we find

$$ \frac{\partial ^{2}\boldsymbol {\omega}}{\partial q_{s}\partial \dot{q}_{r}} \times \boldsymbol {u}= \frac{\partial ^{2}\boldsymbol {\omega}}{\partial q_{r}\partial \dot{q}_{s}} \times \boldsymbol {u}+\frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{s}} \times \frac{\partial \boldsymbol {u}}{\partial q_{r}}- \frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{r}}\times \frac{\partial \boldsymbol {u}}{\partial q_{s}}\,. $$

Using (72) again in the right side, this becomes

$$ \frac{\partial ^{2}\boldsymbol {\omega}}{\partial q_{s}\partial \dot{q}_{r}} \times \boldsymbol {u}= \frac{\partial ^{2}\boldsymbol {\omega}}{\partial q_{r}\partial \dot{q}_{s}} \times \boldsymbol {u}+\frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{s}} \times \left (\frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{r}} \times \boldsymbol {u}\right )- \frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{r}}\times \left ( \frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{s}}\times \boldsymbol {u} \right )\,. $$

Using the vector identity \(\boldsymbol {A}\times (\boldsymbol {B}\times C)-\boldsymbol {B}\times (\boldsymbol {A}\times \boldsymbol {C})=( \boldsymbol {A}\times \boldsymbol {B})\times \boldsymbol {C}\) on the right side, this becomes

$$ \frac{\partial ^{2}\boldsymbol {\omega}}{\partial q_{s}\partial \dot{q}_{r}} \times \boldsymbol {u}= \frac{\partial ^{2}\boldsymbol {\omega}}{\partial q_{r}\partial \dot{q}_{s}} \times \boldsymbol {u}+\left ( \frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{s}}\times \frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{r}}\right )\times \boldsymbol {u}\,. $$

Since this is true for an arbitrary unit vector \(\boldsymbol {u}\), we therefore must have

$$ \frac{\partial ^{2}\boldsymbol {\omega}}{\partial q_{s}\partial \dot{q}_{r}}= \frac{\partial ^{2}\boldsymbol {\omega}}{\partial q_{r}\partial \dot{q}_{s}}+ \left (\frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{s}}\times \frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{r}}\right )\,. $$

Now contracting both sides of the equation with respect to \(\dot{q}_{s}\), we find

$$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{r}}\right )= \frac{\partial \boldsymbol {\omega}}{\partial q_{r}}+\left (\boldsymbol {\omega} \times \frac{\partial \boldsymbol {\omega}}{\partial \dot{q}_{r}}\right )\,. \end{aligned}$$

 □

Appendix B: Derivation of \(\bar{\boldsymbol {\mathrm {D}}}\) for rigid bodies

Here we present a derivation of the exact result for \(\bar{\boldsymbol {\mathrm {D}}}\) in the case of rigid bodies, showing in detail how it differs from \(\boldsymbol {\mathrm {D}}\).

Considering first a single rigid body, by definition \(\bar{\boldsymbol {\mathrm {D}}}\) is given by Eq. (16), where \(\partial \boldsymbol {v}/\partial \boldsymbol {\mathrm {f}}\) is given by Eq. (65). The derivative of \(\partial \boldsymbol {v}/\partial \boldsymbol {\mathrm {f}}\) is therefore

$$ \frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {v}}{\partial \boldsymbol {\mathrm {f}}}\right )= \frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}\right )+ \frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\times \boldsymbol {\rho}+\left (\frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}} \right )\times \left (\boldsymbol {\omega}\times \boldsymbol {\rho}\right )\,. $$
(73)

Substituting this into (16), we find

$$ \bar{\boldsymbol {\mathrm {D}}}=\int \mathrm{d}m\,\left [ \frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}}+ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\times \boldsymbol {\rho} \right ]\cdot \left [\frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {v}_{\mathrm{c}}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}} \right )+\frac{\mathrm{d}}{\mathrm{d}t}\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right ) \times \boldsymbol {\rho}+\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right ) \times \left (\boldsymbol {\omega}\times \boldsymbol {\rho}\right )\right ]; $$

then using the defining properties of the center of gravity and inertia tensor, this reduces to

D ¯ = m ( ∂ v c ∂ f ) ⋅ d d t ( ∂ v c ∂ f T ) + ( ∂ ω ∂ f ) ⋅ I ⋅ d d t ( ∂ ω ∂ f T ) + ∫ d m [ ∂ ω ∂ f × ρ ] ⋅ [ ( ∂ ω ∂ f T ) × ( ω × ρ ) ] ,

or, referring to the definition of \(\boldsymbol {\mathrm {C}}\) in (62),

$$ \bar{\boldsymbol {\mathrm {D}}}=\boldsymbol {\mathrm {C}}^{\mathsf{T}}+\int \mathrm{d}m\,\left [ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\times \boldsymbol {\rho} \right ]\cdot \left [\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right ) \times \left (\boldsymbol {\omega}\times \boldsymbol {\rho}\right )\right ]\,. $$
(74)

Expanding the vector triple product in this result, we have

$$ \bar{\boldsymbol {\mathrm {D}}}=\boldsymbol {\mathrm {C}}^{\mathsf{T}}+\int \mathrm{d}m\,\left [ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\times \boldsymbol {\rho} \right ]\cdot \left [\boldsymbol {\omega}\left (\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right ) \cdot \boldsymbol {\rho}\right )-\boldsymbol {\rho}\left (\boldsymbol {\omega}\cdot \left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right ) \right )\right ]\,, $$

which immediately reduces to

$$ \bar{\boldsymbol {\mathrm {D}}}=\boldsymbol {\mathrm {C}}^{\mathsf{T}}+\int \mathrm{d}m\,\left [ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\times \boldsymbol {\rho} \right ]\cdot \left [\boldsymbol {\omega}\left (\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right ) \cdot \boldsymbol {\rho}\right )\right ]\,. $$
(75)

Rearranging the scalar triple product, we have

$$ \bar{\boldsymbol {\mathrm {D}}}=\boldsymbol {\mathrm {C}}^{\mathsf{T}}-\int \mathrm{d}m\,\left [ \boldsymbol {\omega}\times \boldsymbol {\rho}\right ]\cdot \left [ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\left (\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right ) \cdot \boldsymbol {\rho}\right )\right ]\,, $$

or, using the tensor dyadic \(\boldsymbol {\rho} \boldsymbol {\rho}\),

$$ \bar{\boldsymbol {\mathrm {D}}}=\boldsymbol {\mathrm {C}}^{\mathsf{T}}-\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \left [ \boldsymbol {\omega}\times \int \mathrm{d}m\boldsymbol {\rho} \boldsymbol {\rho}\right ]\, \cdot \left [ \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right ]. $$
(76)

Now it is well known, and could be derived from Eq. (69), that the inertia tensor can be expressed as

I=∫dm ( 1 | ρ | 2 − ρ ρ ) ,

so we have

−∫dmρρ=I−1∫dm | ρ | 2 ,

or

−∫dmρρ=I−m ρ 2 ‾ 1,
(77)

where \(\overline{\rho ^{2}}\) is the mass-weighted mean-square radius of the body. Substituting (77) into (76), we find

D ¯ = C T + ( ∂ ω ∂ f ) ⋅ [ ω × ( I − m ρ 2 ‾ 1 ) ] ⋅ [ ∂ ω ∂ f T ] ,

or

$$ \bar{\boldsymbol {\mathrm {D}}}=\boldsymbol {\mathrm {C}}^{\mathsf{T}}+\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \left ( \boldsymbol {\omega}\times \frac{\partial \boldsymbol {h}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right )-\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot m \overline{\rho ^{2}}\left (\boldsymbol {\omega}\times \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right )\,. $$
(78)

Rearranging the order of the final scalar triple product, we finally have

$$ \bar{\boldsymbol {\mathrm {D}}}=\boldsymbol {\mathrm {C}}^{\mathsf{T}}+\left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\right )\cdot \left ( \boldsymbol {\omega}\times \frac{\partial \boldsymbol {h}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right )+m \overline{\rho ^{2}}\boldsymbol {\omega}\cdot \left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\times \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right )\,. $$
(79)

This can also be written as

$$ \bar{\boldsymbol {\mathrm {D}}}=\boldsymbol {\mathrm {D}}+m\overline{\rho ^{2}}\boldsymbol {\omega}\cdot \left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}}\times \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right )\,. $$
(80)

The final term in (79) and (80) is an obviously asymmetric matrix. Thus we have

$$\begin{aligned} \bar{\boldsymbol {\mathrm {D}}}+\bar{\boldsymbol {\mathrm {D}}}^{\mathsf{T}} = & \boldsymbol {\mathrm {C}}+\boldsymbol {\mathrm {C}}^{ \mathsf{T}}+\left (\frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}} \right )\cdot \left (\boldsymbol {\omega}\times \frac{\partial \boldsymbol {h}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right )+\left ( \boldsymbol {\omega}\times \frac{\partial \boldsymbol {h}}{\partial \boldsymbol {\mathrm {f}}}\right ) \cdot \left ( \frac{\partial \boldsymbol {\omega}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}}\right ) \\ = & \dot{\boldsymbol {\mathrm {A}}}\,, \end{aligned}$$

as expected.

Generalizing Eq. (80) to multiple rigid bodies subscripted by \(n\) and using the summation convention, we have

$$ \bar{\boldsymbol {\mathrm {D}}}=\boldsymbol {\mathrm {D}}+m_{n}\overline{\rho _{n}^{2}}\boldsymbol {\omega}_{n} \cdot \left (\frac{\partial \boldsymbol {\omega}_{n}}{\partial \boldsymbol {\mathrm {f}}} \times \frac{\partial \boldsymbol {\omega}_{n}}{\partial \boldsymbol {\mathrm {f}}^{\mathsf{T}}} \right )\,. $$
(81)

Now we see that the final term in the equation always reduces to zero when the equation is postmultiplied by \(\boldsymbol {\mathrm {f}}\) or premultiplied by \(\boldsymbol {\mathrm {f}}^{\mathsf{T}}\).

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Phillips, J.R., Amirouche, F. Kane’s equations for nonholonomic systems in bond-graph-compatible velocity and momentum forms. Multibody Syst Dyn 59, 45–68 (2023). https://doi.org/10.1007/s11044-023-09919-3

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