See also main wiki: 
좌표계,coordinate_system 극좌표,polar_coordinate 극좌표계,polar_coordinate_system
이하 '~계' suffix는 의도적으로 자제함.
Source:![[http]](/wiki/imgs/http.png)
kocw 운동을 기술하는 방법 (2)
좌표계,coordinate_system 극좌표,polar_coordinate 극좌표계,polar_coordinate_system이하 '~계' suffix는 의도적으로 자제함.
Source:
kocw 운동을 기술하는 방법 (2)극좌표 ¶
원통좌표의 바닥 원으로 볼 수 있음.
$\displaystyle (\rho,\phi)$
$\displaystyle \vec{r}=\rho\hat{\rho}+\phi\hat\phi$
근데 이차원에서는 뒤가 필요가 없고
$\displaystyle \vec{r}=\rho\hat{\rho}$
라고 한다?
$\displaystyle (\rho,\phi)$
$\displaystyle \vec{r}=\rho\hat{\rho}+\phi\hat\phi$
근데 이차원에서는 뒤가 필요가 없고
$\displaystyle \vec{r}=\rho\hat{\rho}$
라고 한다?
이상 좌표에서 속도 ¶
직교좌표에선
$\displaystyle \vec{r}=x\hat i+y\hat j$
$\displaystyle \vec{v}=\frac{d\vec{r}}{dt}$
극좌표에선
$\displaystyle \vec{v}=\frac{d\vec{r}}{dt}$
$\displaystyle \vec{r}=x\hat i+y\hat j$
$\displaystyle \vec{v}=\frac{d\vec{r}}{dt}$
$\displaystyle =\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}$
이게 끝이지만극좌표에선
$\displaystyle \vec{v}=\frac{d\vec{r}}{dt}$
$\displaystyle =\frac{d\rho}{dt}\hat{\rho}+\rho\frac{d\hat{\rho}}{dt}$
잠깐 이 식의 + 오른쪽을 정리한다. 2D평면에서
$\displaystyle \hat{\rho}=\hat{i}\cos\phi+\hat{j}\sin\phi$
$\displaystyle \hat{\phi}=-\hat{i}\sin\phi+\hat{j}\cos\phi$
미분하면$\displaystyle \hat{\phi}=-\hat{i}\sin\phi+\hat{j}\cos\phi$
$\displaystyle \frac{d\hat{\rho}}{dt}=-\hat{i}\sin\phi\frac{d\phi}{dt}+\hat{j}\cos\phi\frac{d\phi}{dt}$
따라서 결론은$\displaystyle =(-\hat{i}\sin\phi+\hat{j}\cos\phi)\frac{d\phi}{dt}$
여기서 놀랍게도 괄호 안이 간단히 되므로
$\displaystyle =\hat{\phi}\frac{d\phi}{dt}$
여기서 놀랍게도 괄호 안이 간단히 되므로
$\displaystyle =\hat{\phi}\frac{d\phi}{dt}$
$\displaystyle \frac{d\hat{\rho}}{dt}=\hat{\phi}\frac{d\phi}{dt}$
마찬가지로$\displaystyle \frac{d\hat{\phi}}{dt}=-\hat{\rho}\frac{d\phi}{dt}$
$\displaystyle =\frac{d\rho}{dt}\hat{\rho}+\rho\frac{d\phi}{dt}\hat{\phi}$
$\displaystyle =v_\rho\hat{\rho}+v_\phi\hat{\phi}$
따라서$\displaystyle =v_\rho\hat{\rho}+v_\phi\hat{\phi}$
$\displaystyle v_\rho=\frac{d\rho}{dt}$
$\displaystyle v_\phi=\rho\frac{d\phi}{dt}$
그리고, $\displaystyle \frac{d\phi}{dt}$ 는 $\displaystyle v_\phi=\rho\frac{d\phi}{dt}$
각속도,angular_velocity 또는 각진동수,angular_frequency이다.3D에선
$\displaystyle \vec{v}=\frac{d\rho}{dt}\hat{\rho}+\rho\frac{d\phi}{dt}\hat{\phi}+\frac{dz}{dt}\hat{k}$
$\displaystyle \vec{v}=\frac{d\rho}{dt}\hat{\rho}+\rho\frac{d\phi}{dt}\hat{\phi}+\frac{dz}{dt}\hat{k}$
가속도 ¶
위에서 유도한 다음 식을 활용.
$\displaystyle a_{\rho}=\frac{d^2\rho}{dt^2}-\rho\left(\frac{d\phi}{dt}\right)^2$
$\displaystyle a_\phi=2\frac{d\rho}{dt}\frac{d\phi}{dt}+\rho\frac{d^2\phi}{dt^2}$
$\displaystyle a_z=\frac{d^2z}{dt^2}$
$\displaystyle \frac{d\hat{\rho}}{dt}=\hat{\phi}\frac{d\phi}{dt}$
$\displaystyle \frac{d\hat{\phi}}{dt}=-\hat{\rho}\frac{d\phi}{dt}$
$\displaystyle \vec{a}=\frac{d^2\rho}{dt^2}\hat{\rho}+\frac{d\rho}{dt}\frac{d\hat{\rho}}{dt}+\frac{d\rho}{dt}\frac{d\phi}{dt}\hat{\phi}+\rho\frac{d^2\phi}{dt^2}\hat{\phi}+\rho\frac{d\phi}{dt}\frac{d\hat{\phi}}{dt}$$\displaystyle \frac{d\hat{\phi}}{dt}=-\hat{\rho}\frac{d\phi}{dt}$
$\displaystyle =\frac{d^2\rho}{dt^2}\hat{\rho}+\rho\frac{d\phi}{dt}\frac{d\hat{\phi}}{dt}+\frac{d\rho}{dt}\frac{d\hat{\rho}}{dt}+\frac{d\rho}{dt}\frac{d\phi}{dt}\hat{\phi}+\rho\frac{d^2\phi}{dt^2}\hat{\phi}$
$\displaystyle =\left[\frac{d^2\rho}{dt^2}-\rho\left(\frac{d\phi}{dt}\right)^2\right]\hat{\rho}+\left[2\frac{d\rho}{dt}\frac{d\phi}{dt}+\rho\frac{d^2\phi}{dt^2}\right]\hat{\phi}$
그래서$\displaystyle =\left[\frac{d^2\rho}{dt^2}-\rho\left(\frac{d\phi}{dt}\right)^2\right]\hat{\rho}+\left[2\frac{d\rho}{dt}\frac{d\phi}{dt}+\rho\frac{d^2\phi}{dt^2}\right]\hat{\phi}$
$\displaystyle a_{\rho}=\frac{d^2\rho}{dt^2}-\rho\left(\frac{d\phi}{dt}\right)^2$
$\displaystyle a_\phi=2\frac{d\rho}{dt}\frac{d\phi}{dt}+\rho\frac{d^2\phi}{dt^2}$
$\displaystyle a_z=\frac{d^2z}{dt^2}$
원통좌표 (=3D 극좌표? 극좌표에 z만 더한 것?) ¶
$\displaystyle (\rho,\phi,z)$
$\displaystyle \vec{r}=\rho\hat{\rho}+z\hat{k}$
$\displaystyle \vec{r}=r\hat{r}$
r방향의 단위벡터는
$\displaystyle \hat{r}=\hat{\rho}\sin\theta+\hat{k}\cos\theta$
$\displaystyle \hat{\theta}=\hat{\rho}\cos\theta-\hat{k}\sin\theta$
$\displaystyle \hat{\phi}=-\hat{i}\sin\phi+\hat{j}\cos\phi$
$\displaystyle \hat{r}=\hat{\rho}\sin\theta+\hat{k}\cos\theta$
$\displaystyle =(\hat{i}\sin\theta\cos\phi+\hat{j}\sin\theta\sin\phi+\hat{k}\cos\theta)$
θ방향의 단위벡터는$\displaystyle \hat{\theta}=\hat{\rho}\cos\theta-\hat{k}\sin\theta$
$\displaystyle =\hat{i}\cos\theta\cos\phi+\hat{j}\cos\theta\sin\phi-\hat{k}\sin\theta$
φ방향의 단위벡터는$\displaystyle \hat{\phi}=-\hat{i}\sin\phi+\hat{j}\cos\phi$
이상을 시간에 대해 미분하면
$\displaystyle \frac{d\hat{r}}{dt}=\hat{i}\cos\theta\cos\phi\frac{d\theta}{dt}-\hat{i}\sin\theta\sin\phi\frac{d\phi}{dt}+\hat{j}\cos\theta\sin\phi\frac{d\theta}{dt}+\hat{j}\sin\theta\cos\phi\frac{d\phi}{dt}-\hat{k}\sin\theta\frac{d\theta}{dt}$
$\displaystyle \frac{d\hat{\phi}}{dt}=-\hat{\rho}\frac{d\phi}{dt}$
$\displaystyle =\hat{\theta}\frac{d\theta}{dt}+\hat{\phi}\sin\theta\frac{d\phi}{dt}$
$\displaystyle \frac{d\hat{\theta}}{dt}=-\hat{r}\frac{d\theta}{dt}+\hat{\phi}\cos\theta\frac{d\phi}{dt}$$\displaystyle \frac{d\hat{\phi}}{dt}=-\hat{\rho}\frac{d\phi}{dt}$
$\displaystyle \vec{r}=r\hat{r}$
$\displaystyle \vec{v}=\frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{dt}$
$\displaystyle v_\theta=r\frac{d\theta}{dt}$
$\displaystyle v_\phi=r\sin\theta\frac{d\phi}{dt}$
$\displaystyle \vec{v}=\frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{dt}$
$\displaystyle =\frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}+r\sin\theta\frac{d\phi}{dt}\hat{\phi}$
$\displaystyle v_r=\frac{dr}{dt}$$\displaystyle v_\theta=r\frac{d\theta}{dt}$
$\displaystyle v_\phi=r\sin\theta\frac{d\phi}{dt}$
// 운동을 기술하는 방법 (3) 42분
아무튼 결론은
아무튼 결론은
$\displaystyle \vec{r}=\rho\hat\rho$
$\displaystyle \vec{v}=\frac{d\rho}{dt}\hat{\rho}+\rho\frac{d\phi}{dt}\hat\phi$
$\displaystyle \vec{a}=\left[\frac{d^2\rho}{dt^2}-\rho\left(\frac{d\phi}{dt}\right)^2\right]\hat{\rho}+\left[2\frac{d\rho}{dt}\frac{d\phi}{dt}+\rho\frac{d^2\phi}{dt^2}\right]\hat{\phi}$
인데 이것을 dot notation으로 나타내면$\displaystyle \vec{v}=\frac{d\rho}{dt}\hat{\rho}+\rho\frac{d\phi}{dt}\hat\phi$
$\displaystyle \vec{a}=\left[\frac{d^2\rho}{dt^2}-\rho\left(\frac{d\phi}{dt}\right)^2\right]\hat{\rho}+\left[2\frac{d\rho}{dt}\frac{d\phi}{dt}+\rho\frac{d^2\phi}{dt^2}\right]\hat{\phi}$
$\displaystyle \vec{v}=\dot\rho\hat\rho+\rho\dot\phi\hat\phi$
$\displaystyle \vec{a}=(\ddot\rho-\rho\dot\phi^2)\hat\rho+(2\dot\rho\dot\phi+\rho\ddot\phi)\hat\phi$
$\displaystyle \vec{a}=(\ddot\rho-\rho\dot\phi^2)\hat\rho+(2\dot\rho\dot\phi+\rho\ddot\phi)\hat\phi$
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