极坐标系下的速度与加速度(含结论...

物理
极坐标系下的速度与加速度(含结论与推导过程)

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坍缩态的薇汾 更新于2026-7-18 08:51:02

${ (平面)极坐标系}$

(别误会,空间里的叫柱坐标系和球坐标系)

$\textbf{点坐标:}$ $P(r,\theta)$ $\quad $r$: 极径,$ $\theta: 极角$

基矢:

径向基矢: $\hat{r}: \text{沿位矢方向} \\$

角向基矢 $\hat{\theta}: \text{垂直 } \hat{r} \text{ 沿 } \theta \text{ 增大方向}$

$\textbf{注意:} $ $\hat{r}, \hat{\theta}$ 会随时间改变

$\textbf{运动方程:} $

$$\vec{r} = \vec{r}(t) \Rightarrow \begin{cases} r = r(t) \\ \theta = \theta(t) \end{cases}$$

$\textbf{位矢:}$ $\vec{r} = r\hat{r}$

$\textbf{速度:}$ $\vec{v} = v_r\hat{r} + v_\theta\hat{\theta}$

$$v_r = \dfrac{dr}{dt} = \dot{r} $$

 $$v_\theta = r\dfrac{d\theta}{dt} = r\dot{\theta}$$

$\textbf{加速度:}$ $\vec{a} = a_r\hat{r} + a_\theta\hat{\theta}$

$$a_r = \dfrac{d^2r}{dt^2} - r\left(\dfrac{d\theta}{dt}\right)^2 =\ddot{r} - r\dot{\theta}^2 $$

$$a_\theta = 2\dfrac{dr}{dt}\cdot\dfrac{d\theta}{dt} + r\dfrac{d^2\theta}{dt^2} = 2\dot{r}\dot{\theta} + r\ddot{\theta}$$


$*{推导:}$

$\textbf{① 速度推导:}$

$$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(r\hat{r}) = \frac{dr}{dt}\hat{r} + r\frac{d\hat{r}}{dt}$$ 

$d\hat{r}$: $\begin{cases} \text{大小:} d\theta \\ \text{方向:沿 } \hat{\theta} \text{ 方向} \end{cases}$

$$☆ \Rightarrow d\hat{r} = d\theta\hat{\theta} \longrightarrow \text{\textbf{基矢变化规律}}$$ 

$$\vec{v} = \underbrace{\frac{dr}{dt}\hat{r}}_{\text{径向长度变化}} + \underbrace{r\frac{d\theta}{dt}\hat{\theta}}_{\text{角度变化}}$$ 

$\textbf{② 加速度推导:}$

$$(uvw)' = (uv)'w + uvw' = u'vw + uv'w + uvw'$$ 

$$\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}\left(\frac{dr}{dt}\hat{r} + r\frac{d\theta}{dt}\hat{\theta}\right)$$ 

$$= \frac{d^2r}{dt^2}\hat{r} + \frac{dr}{dt}\frac{d\hat{r}}{dt} + \frac{dr}{dt}\frac{d\theta}{dt}\hat{\theta} + r\frac{d^2\theta}{dt^2}\hat{\theta} + r\frac{d\theta}{dt}\frac{d\hat{\theta}}{dt}$$ 

由 $d\hat{r} = d\theta\hat{\theta}$,可得 $d\hat{\theta} = -d\theta\hat{r}$

$$\Rightarrow \vec{a} = \frac{d^2r}{dt^2}\hat{r} + 2\frac{dr}{dt}\frac{d\theta}{dt}\hat{\theta} + r\frac{d^2\theta}{dt^2}\hat{\theta} - r\left(\frac{d\theta}{dt}\right)^2\hat{r}$$ 

$$= \left[\frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2\right]\hat{r} + \left[2\frac{dr}{dt}\cdot\frac{d\theta}{dt} + r\frac{d^2\theta}{dt^2}\right]\hat{\theta}$$ 

$$= \underbrace{(\ddot{r} - r\dot{\theta}^2)}_{\begin{subarray}{c} \uparrow \\ \text{长度不均匀变化} \end{subarray}}\hat{r} + \underbrace{\underbrace{(2\dot{r}\dot{\theta}}_{\begin{subarray}{c} \uparrow \\ \text{径向角度同时变化} \\ \text{科里奥利} \end{subarray}} + \underbrace{r\ddot{\theta})}_{\begin{subarray}{c} \rightarrow \\ \text{角度不均匀变化} \end{subarray}}}_{\text{向心/角度变化}}\hat{\theta}$$ 

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