ランダウ・リフシッツ力学 §5 問題 1

質点\(m_1\)については

\begin{align} T_1 = & \cfrac{1}{2} m_1 l_1^2 \dot{\varphi}_1^2 \\ U_1 = & – m_1 g l_1 \cos \varphi_1 . \end{align}

一方、質点\(m_2\)については

\begin{align} x_2 = & l_1 \sin \varphi_1 + l_2 \sin \varphi_2 \\ y_2 = & l_1 \cos \varphi_1 + l_2 \cos \varphi_2 .\\ \end{align}

時間の微分は

\begin{align} \dot{x}_2 = & l_1 \dot{\varphi}_1 \cos \varphi_1 + l_2 \dot{\varphi}_2 \cos \varphi_2 \\ \dot{y}_2 = & – l_1 \dot{\varphi}_1 \sin \varphi_1 – l_2 \dot{\varphi}_2 \sin \varphi_2 \\ \end{align}

となります。これを代入します:

\begin{align} T_2 = & \cfrac{m_2}{2} ( \dot{x}_2^2 + \dot{y}_2^2) \\ = & \cfrac{m_2}{2} ( (l_1 \dot{\varphi}_1 \cos \varphi_1 + l_2 \dot{\varphi}_2 \cos \varphi_2)^2 + (l_1 \dot{\varphi}_1 \sin \varphi_1 + l_2 \dot{\varphi}_2 \sin \varphi_2)^2) \\ = & \cfrac{m_2}{2} ( l_1^2 \dot{\varphi}_1^2 \cos^2 \varphi_1 + l_2^2 \dot{\varphi}_2^2 \cos^2 \varphi_2 + 2 l_1 l_2 \dot{\varphi}_1 \dot{\varphi}_2 \cos \varphi_1 \cos \varphi_2) \\ & + \cfrac{m_2}{2} ( l_1^2 \dot{\varphi}_1^2 \sin^2 \varphi_1 + l_2^2 \dot{\varphi}_2^2 \sin^2 \varphi_2 + 2 l_1 l_2 \dot{\varphi}_1 \dot{\varphi}_2 \sin \varphi_1 \sin \varphi_2 ) \\ = & \cfrac{m_2}{2} ( l_1^2 \dot{\varphi}_1^2 + l_2^2 \dot{\varphi}_2^2 + 2 l_1 l_2 \dot{\varphi}_1 \dot{\varphi}_2 \cos (\varphi_1 – \varphi_2)) \\ \end{align} \begin{align} U_2 & = – m_2 (l_1 \cos \varphi_1 + l_2 \cos \varphi_2) g \end{align}

最終的に

\begin{align} L = & T_1 – U_1 + T_2 – U_2 \\ = & \cfrac{1}{2} m_1 l_1^2 \dot{\varphi}_1^2 + \cfrac{m_2}{2} ( l_1^2 \dot{\varphi}_1^2 + l_2^2 \dot{\varphi}_2^2 + 2 l_1 l_2 \dot{\varphi}_1 \dot{\varphi}_2 \cos (\varphi_1 – \varphi_2)) + m_1 g l_1 \cos \varphi_1 + m_2 (l_1 \cos \varphi_1 + l_2 \cos \varphi_2) g \\ = & \cfrac{1}{2} (m_1 + m_2) l_1^2 \dot{\varphi}_1^2 + \cfrac{m_2}{2} l_2^2 \dot{\varphi}_2^2 + m_2 l_1 l_2 \dot{\varphi}_1 \dot{\varphi}_2 \cos (\varphi_1 – \varphi_2) + (m_1 + m_2 )g l_1 \cos \varphi_1 + m_2 g l_2 \cos \varphi_2 . \end{align}