Zadanie 2

Z wykorzystaniem metody Hamiltona

dane są warunki:

$m_{2} = {\frac{1}{2}m}_{1}$ $m_{3} = {\frac{1}{3}m}_{1}$

zapisanie poszczególnych energii:

$$E = \frac{1}{2}I_{1}{\dot{\varphi}}_{1}^{2} + \frac{1}{2}m_{2}{\dot{x}}_{2}^{2} + \frac{1}{2}m_{3}{\dot{x}}_{3}^{2}$$

U = m₃gx₃ + m₂gx₂sinα

szukamy warunków geometrycznych:

x₂ = φ₁r₁ = x₃

$$I_{1} = \frac{1}{2}m_{1}r_{1}^{2}$$

upraszczamy równania energii:

$$E = \frac{1}{4}m_{1}r_{1}^{2}{\dot{\varphi}}_{1}^{2} + \frac{1}{4}m_{1}{\dot{\varphi}}_{1}^{2}r_{1}^{2} + \frac{1}{6}m_{1}{\dot{\varphi}}_{1}^{2}r_{1}^{2} = \frac{2}{3}m_{1}r_{1}^{2}{\dot{\varphi}}_{1}^{2}$$

$$U = {\frac{1}{3}m}_{1}g\varphi_{1}r_{1} + {\frac{1}{2}m}_{1}g\varphi_{1}r_{1}\sin\alpha = m_{1}gr_{1}\left( \frac{1}{3} + \frac{1}{2}\sin\alpha \right)\varphi_{1}$$

następnie obliczamy:

$$p_{\varphi_{1}} = \frac{\text{dE}}{d{\dot{\varphi}}_{1}} = \frac{4}{3}m_{1}r_{1}^{2}{\dot{\varphi}}_{1}$$

$${\dot{\varphi}}_{1} = \frac{3p_{\varphi_{1}}}{4m_{1}r_{1}^{2}}$$

$$\tilde{E} = \frac{2}{3}m_{1}r_{1}^{2}\left( \frac{3p_{\varphi_{1}}}{4m_{1}r_{1}^{2}} \right)^{2}$$

$$\tilde{E} = \frac{18}{48}\frac{p_{\varphi_{1}}^{2}}{m_{1}r_{1}^{2}}$$

równanie Hamiltona:

$$H = \tilde{E} + U$$

$$H = \frac{18}{48}\frac{p_{\varphi_{1}}^{2}}{m_{1}r_{1}^{2}} + m_{1}gr_{1}\left( \frac{1}{3} + \frac{1}{2}\sin\alpha \right)\varphi_{1}$$

$$\left\{ \begin{matrix} \frac{dp_{\varphi_{1}}}{\text{dt}} = - \frac{\text{dH}}{d\varphi_{1}} + Q_{i} = - m_{3}g + m_{2}g\sin\alpha = - {\frac{1}{3}m}_{1}g + {\frac{1}{2}m}_{1}g\sin\alpha \\ \frac{dp_{\varphi_{1}}}{\text{dt}} = \frac{4}{3}m_{1}r_{1}^{2}{\ddot{\varphi}}_{1} \\ \end{matrix} \right.\ $$

$$- {\frac{1}{3}m}_{1}g + {\frac{1}{2}m}_{1}g\sin\alpha = \frac{4}{3}m_{1}r_{1}^{2}{\ddot{\varphi}}_{1}$$

$$\frac{4}{3}m_{1}r_{1}^{2}{\ddot{\varphi}}_{1} + {\frac{1}{3}m}_{1}g - {\frac{1}{2}m}_{1}g\sin\alpha = 0$$

przyjmujemy dane w celu wykreślenia wykresów:

m₁ = 0.1kg r₁ = 0.5m α = 15