ランダウ・リフシッツ力学 §9 問題 1

ランダウ・リフシッツの力学(増補第3版) §9 問題 1 のメモです。 \begin{align} M_x & = m(y \dot{z} – z \dot{y}) = m (r \sin \varphi \dot{z} – z(\dot{r} \sin \varphi + r \dot{\varphi}\cos \varphi ) ) \\ & = m (r \sin \varphi \dot{z} – z(\dot{r} \sin \varphi + r \dot{\varphi}\cos \varphi ) ) \\ & = m (r\dot{z} – z\dot{r} )\sin \varphi – m r z\dot{\varphi}\cos \varphi \\ M_y & = m(z \dot{x} – x \dot{z}) = m (z \dot{r} \cos \varphi – zr \dot{\varphi} \sin \varphi – r\dot{z} \cos \varphi ) \\ & = m (z \dot{r} – r\dot{z} ) \cos \varphi – mrz \dot{\varphi} \sin \varphi \\ M_z & = m(x \dot{y} – y \dot{x}) = m r\cos\varphi ( \dot{r} \sin \varphi + r \dot{\varphi} \cos \varphi ) – r \sin \varphi (\dot{r} \cos \varphi – r \dot{\varphi} \sin \varphi ) \\ & = m r^2\varphi \\ M^2 & = m^2 \sin^2\varphi (r\dot{z} – z\dot{r} )^2 + m^2r^2z^2 \dot{\varphi}^2 \cos^2\varphi – 2m^2 sin\phi \cos\phi (r\dot{z} – z\dot{r} ) r z \dot{\varphi} \\ & + m^2 \cos^2\varphi ( z\dot{r} -r\dot{z} )^2 + m^2r^2z^2 \dot{\varphi}^2 \sin^2\varphi – 2m^2 sin\phi \cos\phi ( z\dot{r} – r\dot{z} ) r z \dot{\varphi} \\ & + m^2 r^4 \dot{\varphi}^2 \\ & = m^2 (r\dot{z} – z\dot{r} )^2 + m^2r^2z^2 \dot{\varphi}^2 + m^2 r^4 \dot{\varphi}^2 \\ & = m^2 (r\dot{z} – z\dot{r} )^2 + m^2r^2 \dot{\varphi}^2(z^2 + r^2) \\ \end{align}