이전 글에서는 원통좌표계 및 구면좌표계에서의 벡터 연산자에 대해서 알아보았다. 그런데 이전 글에서 이미 위치벡터를 구해 놓았으므로(물론 벡터연산자를 구하기 위한 목적으로), 이제 이를 이용하여 원통좌표계 및 구면좌표계에서의 속도 및 가속도벡터에 대해 구해보도록 하자.
$$[Cylindrical\,Coordinates]\,:\,x\,=\,r\,cos\theta,\,y\,=\,r\,sin\theta,\,z\,=\,z$$
$$\hat{\mathbb{q_r}}\,=\,e_r\,=\,\frac{\frac{\partial \mathbb{r}}{\partial r}}{\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial r}\end{vmatrix}}\,=\,\frac{cos\theta\,i\,+\,sin\theta\,j}{\begin{vmatrix}cos\theta\,i\,+\,sin\theta\,j\end{vmatrix}}\,=\,cos\theta\,i\,+\,sin\theta\,j,\,h_r\,=\,\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial r}\end{vmatrix}\,=\,1$$
$$\hat{\mathbb{q_{\theta}}}\,=\,e_{\theta}\,=\,\frac{\frac{\partial \mathbb{r}}{\partial {\theta}}}{\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial {\theta}}\end{vmatrix}}\,=\,\frac{-r\,sin\theta\,i\,+\,r\,cos\theta\,j}{\begin{vmatrix}-r\,sin\theta\,i\,+\,r\,cos\theta\,j\end{vmatrix}}\,=\,-sin\theta\,i\,+\,cos\theta\,j,\,h_{\theta}\,=\,\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial \theta}\end{vmatrix}\,=\,r$$
$$\hat{\mathbb{q_z}}\,=\,e_z\,=\,\frac{\frac{\partial \mathbb{r}}{\partial z}}{\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial z}\end{vmatrix}}\,=\,\frac{k}{\begin{vmatrix}k\end{vmatrix}}\,=\,k,\,h_z\,=\,\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial z}\end{vmatrix}\,=\,1$$
$$\mathbb{r}\,=\,x\,i\,+\,y\,j\,+\,z\,k\,=\,rcos\theta\,i\,+rsin\theta\,j\,+\,z\,k\,=\,r\,e_r\,+\,z\,e_z$$
$$[Cylindrical\,Coordinates]\,\begin{cases} \mathbb{r}\,= & re_r\,+\,ze_z \\ \mathbb{v}\,= & \frac{d\mathbb{r}}{dt}\,=\,\frac{d}{d\,t}\,
(re_r\,+\,ze_z) \\ & =\,\frac{d\,r}{d\,t}e_r\,+\,r\,\frac{d\,e_r}{d\,t}\,+\frac{d\,z}{d\,t}\,e_z \\ & =\,\dot{r}\,e_r\,+\,r\,\dot{\theta}\,e_{\theta}\,+\dot{z}\,e_z \\ \mathbb{a}\,= & \frac{d\,\mathbb{v}}{d\,t} \\ & =\,\frac{d}{d\,t}\,[\dot{r}\,e_r\,+\,r\,\dot{\theta}\,e_{\theta}]\,+\,\frac{d}{d\,z}[\dot{z}\,e_z] \\ & =\,\ddot{r}\,e_r\,+\,\dot{r}\,\frac{d\,e_r}{d\,t}\,+\,\dot{r}\,\dot{\theta}\,e_{\theta}\,+\,r\,\ddot{\theta}\,e_{\theta}\,+r\,\dot{\theta}\,\cdot\,\dot{\theta}\,(-e_r)\,+\,\frac{d}{d\,t}\,(\frac{d\,z}{d\,t})e_z \\ & =\,[\ddot{r}\,-\,r({\dot{\theta}}^2)]e_r\,+\,[\dot{r}\,\ddot{\theta}\,+\,2\,\dot{r}\,\dot{\theta}]e_{\theta}\,+\,\ddot{z}\,e_z \end{cases} $$
$$[Spherical\,Coordinates]\,:\,x\,=\,\rho\,cos\theta\,sin\,\phi,\,y\,=\,\rho\,sin\theta\,sin\phi,\,z\,=\,\rho\,cos\phi$$
$$\hat{\mathbb{q_{\rho}}}\,=\,e_{\rho}\,=\,\frac{\frac{\partial \mathbb{r}}{\partial \rho}}{\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial \rho}\end{vmatrix}}\,=\,\frac{cos\theta\,sin\,\phi\,i\,+\,sin\theta\,sin\phi\,j\,+\,cos\phi\,k}{\begin{vmatrix}cos\theta\,sin\,\phi\,i\,+\,sin\theta\,\,sin\phi\,j\,+\,cos\phi\,k\end{vmatrix}}\,=\,cos\theta\,sin\,\phi\,i\,+\,sin\theta\,\,sin\phi\,j\,+\,cos\phi\,k,\,h_{\rho}\,=\,\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial r}\end{vmatrix}\,=\,1$$
$$\hat{\mathbb{q_{\theta}}}\,=\,e_{\theta}\,=\,\frac{\frac{\partial \mathbb{r}}{\partial {\theta}}}{\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial {\theta}}\end{vmatrix}}\,=\,\frac{-\rho\,sin\theta\,sin\phi\,i\,+\,\rho\,cos\theta\,sin\phi\,j}{\begin{vmatrix}-\rho\,sin\theta\,sin\phi\,i\,+\,\rho\,cos\theta\,sin\phi\,j\end{vmatrix}}\,=-\,sin\theta\,i\,+\,cos\theta\,j,\,h_{\theta}\,=\,\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial \theta}\end{vmatrix}\,=\,\rho\,sin\phi$$
$$\hat{\mathbb{q_{\phi}}}\,=\,e_{\phi}\,=\,\frac{\frac{\partial \mathbb{r}}{\partial \phi}}{\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial \phi}\end{vmatrix}}\,=\,\frac{\rho\,cos\theta\,cos\,\phi\,i\,+\,\rho\,sin\,\theta\,cos\phi\,j\,-\,\rho\,sin\phi\,k}{\begin{vmatrix}\rho\,cos\theta\,cos\,\phi\,i\,+\rho\,sin\,\theta\,cos\phi\,j\,-\rho\,sin\phi\,k\end{vmatrix}}\,=\,cos\theta\,cos\,\phi\,i\,+\,sin\,\theta\,cos\phi\,j\,-\,sin\phi\,k,\,h_z\,=\,\begin{vmatrix}\frac{\partial \mathbb{r}}{\partial z}\end{vmatrix}\,=\,\rho$$
$$[Spherical\,Coordinates]\,\begin{cases} \mathbb{r}\,= & \rho\,e_{\rho} \\ \mathbb{v}\,= & \frac{d\,\rho}{d\,t}\,e_{\rho}\,+\,\rho\,\frac{d\,e_{\rho}}{d\,t} \\ & =\,\dot{\rho}\,e_{\rho}\,+\,\rho(\frac{\partial e_{\theta}}{\partial \theta}\,\frac{d\,\theta}{d\,t}\,+\,\frac{\partial e_{\phi}}{\partial \phi}\,\frac{d\,\phi}{d\,t}) \\ & =\,\dot{\rho}\,e_{\rho}\,+\,\rho\,\dot{\theta}\,sin\,\phi\,e_{\theta}\,+\,\rho\,\dot{\phi}\,e_{\phi} \\ \mathbb{a}\,= & \frac{d\,\mathbb{v}}{d\,t} \\ & \ddot{\rho}\,e_{\rho}\,+\,\dot{\rho}(\dot{\theta}\,sin\,\phi\,e_{\theta}\,+\,\dot{\phi}\,e_{\phi})\,+\dot{\rho}\,\dot{\theta}\,sin\,\phi\,e_{\theta}\,+\,\rho\,\ddot{\theta}\,sin\,\phi\,+e_{\theta}\,+\,\rho\,\dot{\theta}\,cos\,{\phi}\,e_{\theta}\,-\,\rho\,\dot{\theta}\,sin\,\phi\,(sin\,\phi\,e_{\rho}\,+\,cos\,\phi\,e_{\phi})\,\dot{\theta}\,+\dot{\rho}\,\dot{\phi}\,e_{\phi}\,+\rho\,\ddot{\phi}\,e_{\phi} \\ & \,+\rho(\ddot{\phi}\,e_{\phi}\,+(-\dot{\phi}\,e_{\rho}\,+\,\dot{\theta}\,cos\,\phi\,e_{\theta})) \\ & =\,[\ddot{\rho}\,-\,\dot{\rho}\,{\dot{\phi}}^2\,-\,\rho\,\dot{\theta}\,sin^2\,\phi]e_{\rho}\,+\,[2\,\dot{\rho}\,\dot{\theta}\,sin\,\theta\,+\,2\,\rho\,\dot{\phi}\,\dot{\theta}\,cos\,\phi\,+\,\rho\,{\dot{\theta}}^2\,sin\,\phi]e_{\theta}\,+\,[\rho\,\ddot{\phi}\,+\,2\,\dot{\rho}\,\dot{\phi}\,-\,\rho\,{\dot{\theta}}^2\,sin\,\phi\,cos\,\phi]\,e_{\phi} \end{cases} $$
계산은 매우 길었다. 허나 가장 중요한 대원칙 둘만 제대로 상기하면 긴 계산도 깔끔하게 틀리지 않고 한 번에 유도해낼 수 있다. 바로 전미분 공식과 카테시안 기저벡터는 상수벡터라는 사실. 이렇게 구해놓은 계산식은 동역학을 공부하는 기계공학 전공의 공돌이들이 줄기차게 써먹을 식이다(...)