Ćwiczenia 30.03.09

x  =  ct cos (bt)

y  =  ct sin (bt)

x  =  ρcosφ

y  =  ρsinφ

$$\rho\ = \sqrt{x^{2} + y^{2}}\ $$

$$\text{tg}\text{\ φ} = \frac{y}{x}$$

$$\varphi = arc\ tg\frac{y}{x}$$

$$\rho\ = \sqrt{\left\lbrack \text{ct}\cos{(bt)} \right\rbrack^{2} + \left\lbrack \text{ct}\sin{(bt)} \right\rbrack^{2}\ }\text{\ \ }\int_{}^{}{\rho = ct}$$

$$\varphi = arc\ tg\ \frac{\text{ct}\sin{(bt)}}{\text{ct}\cos{(bt)}}\ $$

φ = bt, $t = \frac{\varphi}{b},\ \varphi = c\frac{\varphi}{b}$ wartość wektora…

Prędkość w układzie biegunowym

$\dot{\overrightarrow{r}}\ = \dot{\rho}\hat{\rho_{0}} + \rho\dot{\varphi}\hat{\varphi_{0}}$

$$\overrightarrow{V}\ = c\hat{\rho_{0}} + ctb\hat{\varphi_{0}}$$

$$\left| \overrightarrow{V} \right| = \sqrt{{V_{\rho}}^{2} + {V_{\varphi}}^{2}} = \sqrt{c^{2} + c^{2}t^{2}b^{2}}$$

$\overrightarrow{a}\ = \left( \ddot{\rho} - \rho\dot{\varphi^{2}}\ \right){\hat{\rho}}_{0} + \left( \rho\ddot{\varphi} + 2\dot{\rho}\dot{\varphi} \right)\hat{\varphi_{0}\ } = a_{r}\hat{\rho_{0}} + a_{\rho}\ \hat{\varphi_{0}}$

$$\overrightarrow{\left| a \right|}\ = \sqrt{{a_{r}}^{2} + {a_{\varphi}}^{2}} = \sqrt{{(0 - ctb^{2})}^{2} + {(0 + 2bc)}^{2}} = \sqrt{c^{2}b^{4}t^{2} + {4b}^{2}c^{2}}$$

a_s = …?