宗旨:由$a_{n} $ $= $ $\frac{v^{2} }{\rho } $ $\Rightarrow $ $\rho $ $= $ $\frac{v^{2} }{a_{n} } $

1.直角坐标系中

已知: $\vec{r} = \vec{r} (t)$ ， 即$x= x(t)$ , $y= y(t)$

 $\vec{a}_x + \vec{a}_y = \vec{a} = \vec{a}_\tau + \vec{a}_n$

$a_n = a_x \sin\theta - a_y \cos\theta = \frac{a_x v_y - a_y v_x}{v} = \frac{|\vec{a} \times \vec{v}|}{v}$

提示:

$$\cos\theta = \frac{v_x}{v}, \quad \sin\theta = \frac{v_y}{v}$$ 

 $$\rho = \frac{v^2}{a_n} = \frac{v^3}{|a_x v_y - a_y v_x|} = \frac{(\dot{x}^2 + \dot{y}^2)^{\frac{3}{2}}}{|\dot{x}\ddot{y} - \ddot{y}\dot{x}|}$$ 

其中：

 $$v_x = \dot{x},\quad v_y = \dot{y},\quad a_x = \ddot{x},\quad a_y = \ddot{y},\quad v = \sqrt{v_x^2 + v_y^2} = \sqrt{\dot{x}^2 + \dot{y}^2}$$ 

令 $v_x = 1$，则 $\dot{x} = 1$，$\ddot{x} = 0$

 $$\dot{y} = \frac{dy}{dt} = \frac{dy}{dx},\quad \ddot{y} = \frac{d\dot{y}}{dt} = \frac{dy'}{dx} = y'' = \frac{d^2y}{dx^2}$$ 

 $$\Rightarrow \rho = \frac{(1+y'^2)^{\frac{3}{2}}}{|y''|}$$ 

2.极坐标系中

已知 $\vec{r} = \vec{r}(t)$，即 $r = r(t)$，$\theta = \theta(t)$

 $$a_n = a_\theta \cos\alpha - a_r \sin\alpha = \frac{a_\theta v_r - a_r v_\theta}{v}$$ 

 $$\cos\alpha = \frac{v_r}{v}, \quad \sin\alpha = \frac{v_\theta}{v}$$ 

 $$\rho = \frac{v^2}{a_n} = \frac{v^3}{|a_\theta v_r - a_r v_\theta|} = \frac{(\dot{r}^2 + r^2\dot{\theta}^2)^{\frac{3}{2}}}{\left|(2\dot{r}\dot{\theta} + r\ddot{\theta})\dot{r} - (\ddot{r} - r\dot{\theta}^2)r\dot{\theta}\right|}$$ 

极坐标系中的速度与加速度分量：

$v_r = \dot{r} \\$

$v_\theta = r\dot{\theta}$

$a_r = \ddot{r} - r\dot{\theta}^2 \\$

$a_\theta = 2\dot{r}\dot{\theta} + r\ddot{\theta}$

若已知：$r = r(\theta)$，令 $\theta = t$，则 $\dot{r} = r'(\theta) = r'$，$\ddot{r} = r''$

$$\rho = \frac{(r^2 + r'^2)^{\frac{3}{2}}}{|2r'^2 - rr'' + r^2|}$$ 

(随后只要代公式就好)

例：对 $r = r_0 e^{k\theta}$

$$r' = r_0 k e^{k\theta}, \quad r'' = r_0 k^2 e^{k\theta}$$ 

$$\rho = \frac{(r_0^2 e^{2k\theta} + r_0^2 k^2 e^{2k\theta})^{\frac{3}{2}}}{\left|2r_0^2 k^2 e^{2k\theta} - r_0^2 k^2 e^{2k\theta} + r_0^2 e^{2k\theta}\right|} = \frac{(1+k^2)^{\frac{3}{2}} r_0^3 e^{3k\theta}}{(1+k^2) r_0^2 e^{2k\theta}} = \sqrt{1+k^2}\, r$$