Zadanie 1

Z wykorzystaniem metody Hamiltona

dane są warunki:

I₂ = 2I₁ $I_{3} = \frac{3I_{1}}{2}$ $r_{2} = \frac{3r_{1}}{2}$

zapisujemy energie:

$$E_{k} = \frac{I_{1}{\dot{\varphi}}_{1}^{2}}{2} + \frac{I_{3}{\dot{\varphi}}_{3}^{2}}{2} + \frac{I_{2}{\dot{\varphi}}_{2}^{2}}{2} + \frac{m_{4}{\dot{x}}_{4}^{2}}{2}$$

U = m₄gx₄

następnie szukamy warunków geometrycznych, które nie są znane:

$$\varphi_{3} = \frac{2\varphi_{1}r_{1}}{l}$$

$$\varphi_{2} = \frac{\varphi_{1}r_{1}}{r_{2}} = \frac{2\varphi_{1}r_{1}}{3r_{1}} = \frac{2\varphi_{1}}{3}$$

x₄ = φ₂r₂ = φ₁r₁

sprowadzamy równania energii do najprostszej postaci

$$E_{k} = \frac{I_{1}{\dot{\varphi}}_{1}^{2} + \frac{3I_{1}}{2}\left( \frac{2{\dot{\varphi}}_{1}r_{1}}{l} \right)^{2} + 2I_{1}\left( \frac{2{\dot{\varphi}}_{1}}{3} \right)^{2} + m_{4}\left( {\dot{\varphi}}_{1}r_{1} \right)^{2}}{2} = = \frac{I_{1}{\dot{\varphi}}_{1}^{2} + 6I_{1}\left( \frac{{\dot{\varphi}}_{1}r_{1}}{l} \right)^{2} + \frac{8I_{1}{\dot{\varphi}}_{1}^{2}}{3} + m_{4}\left( {\dot{\varphi}}_{1}r_{1} \right)^{2}}{2}$$

U = m₄gx₄

następnie obliczamy:

$$p_{\varphi_{1}} = \frac{\text{dE}}{d{\dot{\varphi}}_{1}} = I_{1}{\dot{\varphi}}_{1} + 6I_{1}{\dot{\varphi}}_{1}\left( \frac{r_{1}}{l} \right)^{2} + \frac{8I_{1}{\dot{\varphi}}_{1}}{3} + m_{4}{{\dot{\varphi}}_{1}\left( r_{1} \right)}^{2}$$

$${\dot{\varphi}}_{1} = \frac{p_{\varphi_{1}}}{I_{1} + 6I_{1}\left( \frac{r_{1}}{l} \right)^{2} + \frac{8I_{1}}{3} + m_{4}\left( r_{1} \right)^{2}}$$

$$\tilde{E} = \frac{I_{1} + 6I_{1}\left( \frac{r_{1}}{l} \right)^{2} + \frac{8I_{1}}{3} + m_{4}\left( r_{1} \right)^{2}}{2} \bullet \left( \frac{p_{\varphi_{1}}}{I_{1} + 6I_{1}\left( \frac{r_{1}}{l} \right)^{2} + \frac{8I_{1}}{3} + m_{4}\left( r_{1} \right)^{2}} \right)^{2} = = \frac{\frac{1}{2}p_{\varphi_{1}}^{2}}{I_{1} + 6I_{1}\left( \frac{r_{1}}{l} \right)^{2} + \frac{8I_{1}}{3} + m_{4}\left( r_{1} \right)^{2}}$$

równanie Hamiltona:

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

$$H = \frac{\frac{1}{2}p_{\varphi_{1}}^{2}}{I_{1} + 6I_{1}\left( \frac{r_{1}}{l} \right)^{2} + \frac{8I_{1}}{3} + m_{4}\left( r_{1} \right)^{2}} + m_{4}gx_{4}$$

$$\frac{dp_{\varphi_{1}}}{\text{dt}} = - \frac{\text{dH}}{d\varphi_{1}} + M = - m_{4}g + Mr_{1}$$

$$\frac{dp_{\varphi_{1}}}{\text{dt}} = I_{1}{\ddot{\varphi}}_{1} + 6I_{1}{\ddot{\varphi}}_{1}\left( \frac{r_{1}}{l} \right)^{2} + \frac{8I_{1}{\ddot{\varphi}}_{1}}{3} + m_{4}{{\ddot{\varphi}}_{1}\left( r_{1} \right)}^{2}$$

$$I_{1}{\ddot{\varphi}}_{1} + 6I_{1}{\ddot{\varphi}}_{1}\left( \frac{r_{1}}{l} \right)^{2} + \frac{8I_{1}{\ddot{\varphi}}_{1}}{3} + m_{4}{{\ddot{\varphi}}_{1}\left( r_{1} \right)}^{2} + m_{4}g = Mr_{1}$$

przyjmujemy dane w celu wykreślenia wykresów:

I₁ = 0.1kgm² r₁ = 0.5m l = 1m m₄ = 0.4kg M = 25sin(t) Nm