$q_r = \xi$ のとき \[ \begin{align} \frac{\partial T}{\partial \dot{\xi}} &= \frac{\partial}{\partial \dot{\xi}} \left[ \frac{1}{2}m \left\{\left( l + \xi \right)^2 \dot{\theta}^2 + \dot{\xi}^2 \right\} \right] \\ &= \frac{1}{2}m \cdot 2 \dot{\xi} \\ &=m \dot{\xi} \end{align} \] \[ \begin{align} \frac {d}{dt} \left( \frac{\partial T}{\partial \dot{\xi}} \right) &= \frac{d}{dt} (m \dot{\xi}) \\ &= m \ddot{\xi} \end{align} \] \[ \begin{align} \frac{\partial T}{\partial \xi} &= \frac{\partial}{\partial \xi} \left[ \frac{1}{2}m \left\{ \left( l + \xi \right)^2 \dot{\theta}^2 + \dot{\xi}^2 \right\} \right] \\ &= \frac{1}{2}m \cdot 2\left( l + \xi \right) \dot{\theta}^2 \\ &=m \left( l + \xi \right) \dot{\theta}^2 \end{align} \] \[ \begin{align} \frac{\partial U}{\partial \xi} &= \frac{\partial}{\partial \xi} \left\{ \frac{1}{2}k \xi^2 +mg \left( l + \xi \right) \left( 1 - \cos \theta \right) \right\} \\ &= \frac{1}{2}k \cdot 2\xi + mg \left( 1 - \cos \theta \right) \\ &= k \xi +mg \left( 1 - \cos \theta \right) \end{align} \] \[ \therefore m \ddot{\xi} - m \left( l+\xi \right) \dot{\theta}^2 + k \xi + mg \left( 1- \cos\theta \right) =0 \] $ q_r = \theta $ のとき \[ \begin{align} \frac{\partial T}{\partial \dot{\theta}} &= \frac{\partial}{\partial \dot{\theta}} \left[ \frac{1}{2}m \left\{ \left( l + \xi \right)^2 \dot{\theta}^2 + \dot{\xi}^2 \right\} \right] \\ &= \frac{1}{2}m \left( l + \xi \right)^2 \cdot 2 \dot{\theta} \\ &=m \left( l + \xi \right)^2 \dot{\theta} \end{align} \] \[ \begin{align} \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{\theta}} \right) &= \frac{d}{dt} \left\{ m \left( l + \xi \right)^2 \dot{\theta} \right\} \\ &= m \left\{ 2 \left( l + \xi \right) \dot{\xi} \dot{\theta} + \left( l + \xi \right)^2 \ddot{\theta} \right\} \end{align} \] \[ \frac{\partial T}{\partial \theta} =0 \] \[ \begin{align} \frac{\partial U}{\partial \theta} &= \frac{\partial}{\partial \theta} \left\{ \frac{1}{2} k \xi^2 + mg \left( l + \xi \right) \left( 1- \cos\theta \right) \right\} \\ &= mg \left( l + \xi \right) \sin \theta \end{align} \] \[ \therefore \left( l+\xi \right) \ddot{\theta} + 2 \dot{\xi} \dot{\theta} + g \sin \theta =0 \]