Bi-Rotor with Payload

For generalized coordinates $q={\left[\begin{array}{cccc}y& z& \theta & \varphi \end{array}\right]}^{\top }$ the system’s kinetic and potential energy are $$\begin{array}{rl}T& =\frac{1}{2}\left(m_Q({\dot{y}}^2+{\dot{z}}^2)+I_Q{\dot{\theta }}^2+m_P({\dot{y}}_P^2+{\dot{z}}_P^2)\right)\\ V& =g{m}_{Q}z+g{m}_P(z-l\cos (\varphi ))\end{array}$$ where $${\dot{y}}_P=\dot{y}+l\dot{\varphi }\cos (\varphi )+\dot{l}\sin (\varphi ),{\dot{z}}_P=\dot{z}+l\dot{\varphi }\sin (\varphi )-\dot{l}\cos (\varphi ).$$ The two forces acting on the system and their action arms are $$F_{1,2}=u_{1,2}\left[\begin{array}{c}-\sin (\theta )\\ \cos (\theta )\end{array}\right],r_{1,2}=\left[\begin{array}{c}y\mp a\cos (\theta )\\ z\mp a\sin (\theta )\end{array}\right].$$ We will also consider disturbances acting on the system $$D_{1,2}=d{C}_{1,2}\left[\begin{array}{c}-1\\ 0\end{array}\right],s_1=\left[\begin{array}{c}y\\ z\end{array}\right],s_2=\left[\begin{array}{c}y+l\sin (\varphi )\\ z-l\cos (\varphi )\end{array}\right].$$ The individual terms of the manipulator equations are then $$\begin{array}{rl}M& =\left[\begin{array}{cccc}{m}_Q+m_P& 0& 0& l{m}_P\cos (\varphi )\\ 0& m_Q+m_P& 0& l{m}_P\sin (\varphi )\\ 0& 0& I_Q& 0\\ l{m}_P\cos (\varphi )& l{m}_P\sin (\varphi )& 0& l^2{m}_P\end{array}\right]\\ c& =\left[\begin{array}{c}-l{m}_P\sin (\varphi ){\dot{\varphi }}^2+2\dot{l}{m}_P\cos (\varphi )\dot{\varphi }\\ l{m}_P\cos (\varphi ){\dot{\varphi }}^2+2\dot{l}{m}_P\sin (\varphi )\dot{\varphi }\\ 0\\ 2m_{P}l\dot{l}\dot{\varphi }\end{array}\right]\\ {\tau }_p& =\left[\begin{array}{c}0\\ -g(m_P+m_Q)\\ 0\\ -gl{m}_P\sin (\varphi )\end{array}\right]\\ B& =\left[\begin{array}{cc}-\sin (\theta )& -\sin (\theta )\\ \cos (\theta )& \cos (\theta )\\ -a& a\\ 0& 0\end{array}\right]\\ E& =\left[\begin{array}{c}-C_1-C_2\\ 0\\ -C_{2}l\cos (\theta )\\ 0\end{array}\right]\end{array}$$

The presence of time-variant parameters $p$ results in an additional term in $$c(q,\dot{q};p,\dot{p})=\frac{{\partial }^{2}T}{\partial \dot{q}\partial q}\dot{q}+\frac{{\partial }^{2}T}{\partial \dot{q}\partial p}\dot{p}-\frac{\partial T}{\partial q}.$$