Lagrange vs Newton-Euler Formalisms

Section 13.8 Lagrange vs Newton-Euler Formalisms

It is of interest to relate Lagrange equations and to those obtained by applying the Fundamental Theorem of Dynamics. Recall from Section 13.4 that Lagrange equation with respect to coordinate \(q_i\) applied to rigid body \(\cB\) (of mass \(m\text{,}\) mass center \(G\)) is equivalent to the equation

\begin{equation} \{ \cD_{\cB/\cE} \} \cdot\{\cV_{\cB/\cE}^{q_i}\} = \{\cA_{\coB \to \cB}\} \cdot\{\cV_{\cB/\cE}^{q_i}\}\tag{13.8.1} \end{equation}

Hence, it is seen that Lagrange equation \(\cL^{q_i}_{\cB/\cE}\) can be obtained by “projection” (in the sense of screws) of the Fundamental Theorem of Dynamics \(\{ \cD_{\cB/\cE} \}=\{\cA_{\coB \to \cB}\}\) onto partial kinematic screw \(\{\cV_{\cB/\cE}^{q_i}\}\text{:}\)

\begin{equation} \underbrace{ \frac{\partial\bom }{\partial \dq_i} \cdot \bD_{A} + \frac{\partial\vel_{A\in \cB} }{\partial \dq_i}\cdot m \ba_G} _{\frac{d}{dt} \frac{\partial\kin}{\partial \dq_i} - \frac{\partial\kin}{\partial q_i}} = \underbrace{\frac{\partial\bom }{\partial \dq_i} \cdot \bM_{B,\coB \to \cB} + \frac{\partial\vel_{B\in \cB} }{\partial \dq_i} \cdot \bR_{\coB \to \cB}}_{\qQ_{\coB \to \cB}^{q_i}} \label{lag-to-FTD}\tag{13.8.2} \end{equation}

where \(A\) and \(B\) are two arbitrary points.

A similar equation is obtained for a system of \(p\) rigid bodies by summation of the \(p\) Lagrange equations \(\cL^{q_i}_{j/0}\text{,}\) \(j=1,...,p\text{.}\) Equation (13.8.2) demonstrates that in order to obtain equation \(\cL^{q_i}_{\cB/\cE}\) a complicated (linear) combination of equations extracted from the Fundamental Theorem of Dynamics must be devised. We see here the significant advantage offered by Lagrange equations: while an equation of motion or a first integral can readily be obtained in an automatic and systematic way with the Lagrange formalism, use of the Newton-Euler formalism requires a careful and often difficult, if not fruitless, strategy.

Example 13.8.1.

Using the system \(\Si\) of Example 13.7.6, show how the two Lagrange equations \(\cL^{\psi}_{\Si/0}\) and \(\cL^{\te}_{\Si/0}\) relate to equations extracted from the Fundamental Theorem of Dynamics.

Solution.

With the results of Example 13.7.6, Lagrange equation \(\cL^{\psi}_{\Si/0}\) is obtained as follows:

\begin{equation*} \{ \cD_{1/0} \} \cdot\{\cV_{1/0}^{\psi}\} + \{ \cD_{2/0} \} \cdot\{\cV_{2/0}^{\psi}\} = \{\cA_{\bar{1} \to 1}\} \cdot\{\cV_{1/0}^{\psi}\} + \{\cA_{\bar{2} \to 2}\} \cdot\{\cV_{2/0}^{\psi}\} \end{equation*}

with the partial kinematic screws

\begin{equation*} \{ \cV^\psi_{1/0} \} = \{ \cV_{2/0}^\psi \} = \begin{Bmatrix} \bz_0\\ \bze \end{Bmatrix}_O, \end{equation*}

we find

\begin{equation*} \bz_0 \cdot (\underbrace{\bD_{O, 1/0} + \bD_{O, 2/0}}_{\bD_{O, \Si/0}}) = \bz_0 \cdot (\underbrace{\bM_{O, \bar{1} \to 1} + \bM_{O, \bar{2}\to 2}}_{\bM_{O, \bSi \to \Si} }) \end{equation*}

Hence Lagrange equation \(\cL^{\psi}_{\Si/0}\) is equivalent to extracting from \(\{ \cD_{\Si/0} \}= \{\cA_{\bSi \to \Si}\}\) the dynamic moment equation about point \(O\) along \(\bz_0\text{.}\) Likewise, Lagrange equation \(\cL^{\te}_{\Si/0}\) is obtained as follows:

\begin{equation*} \{ \cD_{1/0} \} \cdot\{\cV_{1/0}^{\te}\} + \{ \cD_{2/0} \} \cdot\{\cV_{2/0}^{\te}\} = \{\cA_{\bar{1} \to 1}\} \cdot\{\cV_{1/0}^{\te}\} + \{\cA_{\bar{2} \to 2}\} \cdot\{\cV_{2/0}^{\te}\} \end{equation*}

with the partial kinematic screws

\begin{equation*} \{ \cV^\te_{1/0} \} = \{0\} , \qquad \{ \cV_{2/0}^\psi \} = \begin{Bmatrix} -\by_1 \\ \bze \end{Bmatrix}_O \end{equation*}

giving

\begin{equation*} \by_1 \cdot \bD_{O, 2/0} = \by_1 \cdot \bM_{O, \bar{2}\to 2} \end{equation*}

This shows that \(\cL^{\te}_{\Si/0}\) is equivalent to extracting from \(\{ \cD_{2/0} \}= \{\cA_{\bar{2} \to 2}\}\) the dynamic moment equation about point \(O\) along \(\by_1\text{.}\)

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