珠子在转动的杆上 Euler-Lagrange equation

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这份力学讲义 A bead slides on a wire rotating at constant angular speed ω in a horizontal plane ▶ Polar coordinates $\underline{\mathbf{v}}=\dot{r} \underline{\hat{\mathbf{r}}}+r \dot{\theta} \underline{\hat{\theta}}$ ▶ $L=T-U$ with $U=0$ ▶ $L=\frac{1}{2} m \dot{r}^{2}+\frac{1}{2} m r^{2} \omega^{2}$ ▶ Single variable $q_{k} \rightarrow r$ ▶ E-L $\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{r}}\right)=\frac{\partial L}{\partial \dot{r}}$ $\frac{\partial L}{\partial \dot{r}}=m \dot{r} \rightarrow \frac{d}{d t}\left(\frac{\partial L}{\partial \dot{r}}\right)=m \ddot{r}$ $\frac{\partial L}{\partial r}=m r \omega^{2}$ ▶ E-L $\rightarrow m \ddot{r}-m r \omega^{2}=0$ Central force $F_{c e n t r a l}=m \omega^{2} r$ ▶ $r=A e^{\omega t}+B e^{-\omega t}$ What happens if the angular speed is now a free coordinate? ▶ $L=\frac{1}{2} m \dot{r}^{2}+\frac{1}{2} m r^{2} \dot{\theta}^{2}$ ▶ Two variables $q_{k} \rightarrow r, \theta$ ▶ $r$ variable: as before $\rightarrow m \ddot{r}-m r \dot{\theta}^{2}=0$ ▶ $\theta$ variable: $\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{\theta}}\right)=\frac{\partial L}{\partial \theta}$ $\frac{\partial L}{\partial \dot{\theta}}=m r^{2} \dot{\theta}$ $\frac{\partial L}{\partial \theta}=0$ E-L $: m r^{2} \ddot{\theta}=\frac{d}{d t}\left(m r^{2} \dot{\theta}\right)=0$ $\rightarrow$ Conservation of angular momentum