Acceleration Form

We set

$$\begin{array}{r}{x}_1=q\\ x_2=\dot{q}\end{array}$$

Giving

\begin{align} \dot{x}_{1} &= x_{2} \\ \dot{x}_{2} &= W^{-1}(x_{1})\left( \frac{\partial L}{\partial x_{1}}+Q-\frac{\partial}{\partial x_{1}}(W(x_{1})x_{2})x_{2} \right) \end{align} $$___ ## Proof Remember ![[Kinetic Energy#^f01942|^f01942]] Giving

\frac{\partial T}{\partial \dot{q}} = W(q)\dot{q}

Since we know from the [[Euler-Lagrange Equation]] ![[Euler-Lagrange Equation#^b8cacc|^b8cacc]]We get

\frac{d}{dt}(W(q)\dot{q}) = \frac{\partial L}{\partial q}+Q

$$Wethenfind$$

\frac{d}{dt}(W(q)\dot{q}) = W(q)\ddot{q} + \frac{\partial}{\partial q}\left(W(q)\dot{q}\right)\dot{q}

$$Puttingthisintotheequationofmotionweget$$

W(q)\ddot{q} + \frac{\partial}{\partial q}(W(q)\dot{q})\dot{q} - \frac{\partial L}{\partial q} = Q

Solving for $\ddot{q}$

\ddot{q} = W^{-1}(q) \left[ \frac{\partial L}{\partial q}+Q - \frac{\partial}{\partial q}(W(q)\dot{q})\dot{q} \right]

$$Ifsetset$$

\begin{align} x_{1}=q \ x_{2}=\dot{q} \end{align}

$$Wethenget$$

\begin{align} \dot{x}{1} &= x{2} \ \dot{x}{2} &= W^{-1}(x{1})\left( \frac{\partial L}{\partial x_{1}}+Q-\frac{\partial}{\partial x_{1}}(W(x_{1})x_{2})x_{2} \right) \end{align}

$$Q.E.D$$