## 2 The System's Dynamics Modeling  
According to [PERSON]'s simplified model of the \"rotor-torsion spring system\" [39], in the case of small deformations, a flexible joint can be regarded as a linear spring with no moment of inertia between the motor rotor and the link. The structures of rigid joints and flexible joints are shown. Figure 1a is the rigid-joint, and Figure 1b is the flexible-joint. From Figure 1a we can see that the rotor's rotation angle is equal to that of the link \(B_{i}\). Their rotation angles are all \(\theta_{i}\). However, the rotor's rotation angle is \(\theta_{i}\) and the link \(B_{i}\)'s rotation angle is \(q_{i}\) in Figure 1b. So, there is an angle error \(\sigma_{i}=\theta_{i}-q_{i}\) caused by the spring: \(k_{\theta i}\) is the spring's stiffness coefficient. The free-floating multi-flexible space manipulator system is composed of a rigid-base \(B_{0}\), n links \(B_{1}\)\(\sim\)\(B_{n}\), and n flexible-joints \(O_{1}\)\(\sim\)\(O_{n}\), as shown in Figure 2. (\(\mathbf{OXY}\)) is the inertial frame, \((O_{0}x_{0}y_{0})\) is the base frame, \((O_{i}x_{i}y_{i})\) is the coordinate frame attached to the link \(B_{i}(i=1,2,\cdots,n)\). The base's center of mass is \(O_{0}\), its position vector is \(\mathbf{r}_{0}\); \(\mathbf{r}_{i}\) is the position vector of \(B_{i}\)'s center of mass. \(C\) is the system's center of mass, its position vector is \(\mathbf{r}_{c}\). Meanwhile, \(q_{0}\) is the attitude angle of the base, which is the angle between the \(x_{0}\) axis and the X axis; \(q_{i}\) is the angle between the \(x_{i-1}\) axis and the \(x_{i}\) axis, \(\theta_{i}\) is the rotation angle of the motor rotor in the joint \(O_{i}\). Figure 1: The simple model of joint. (**a**) The rigid-joint; (**b**) The flexible-joint. The position vector \(\mathbf{r}_{i}\) is:  
\[\mathbf{r}_{i}=\mathbf{r}_{0}+\sum\limits_{j=\,0}^{i-1}l_{j}e_{j}+d_{i}e_{i} \tag{1}\]  
where, \(e_{i}(i=0,\cdots,n)\) is the unit vector for axle \(x_{i}\),\(e_{i}=\left[\sin\left(\sum\limits_{i=0}^{n}q_{i}\right)^{T}\ \ \cos\left(\sum\limits_{i=0}^{n}q_{i}\right)^{T}\right]^{T}\). \(l_{j}\) is the distance between \(O_{j}\) and \(O_{j+1}\), \(d_{i}\) is the distance between \(O_{i}\) and \(O_{ci}\). Without losing the generalization, the system satisfies the conservation of momentum and moment of momentum. The system is initially stationary, the initial momentum and momentum is zero. The system's kinetic energy \(T\) is:  
\[T=T_{\rm r}+T_{\theta} \tag{2}\]  
where, \(T_{\rm r}\) is the kinetic energy of the base and links, \(T_{\rm r}=\sum\limits_{i\,=\,0}^{n}T_{i}\), \(T_{i}=\frac{1}{2}m_{i}\tilde{r}_{i}^{2}+\frac{1}{2}l_{i}\omega_{i}^{2}\). \(T_{\theta}\) is the kinetic energy of the motor rotors, \(T_{\theta}=\sum\limits_{i\,=\,1}^{n}T_{\theta i}\), \(T_{\theta i}=\frac{1}{2}l_{\theta i}\omega_{\theta i}^{2}\). \(m_{i}\) is the mass of \(B_{i}\). \(J_{i}\) is the moment of inertia of \(B_{i}\), \(J_{\theta i}\) is the moment of inertia of the motor rotor in joint \(O_{i}\). \(\omega_{i}\) is \(B_{i}\)'s angular velocity. \(\omega_{\theta i}\) is the motor rotor's angular velocity. The space environment is microgravitational, so the gravity is ignored. The system's potential energy \(U\) is the flexible-joints' elastic potential energy:  
\[U=\frac{1}{2}\sum\limits_{i\,=\,1}^{n}k_{i}(\theta_{i}-q_{i})^{2} \tag{3}\]  
Let \(\mathbf{\theta}=\left[\theta_{1}\ \ \ \theta_{2}\ \ \cdots\ \ \theta_{n}\right]^{T}\), \(\mathbf{q}_{\theta}=\left[q_{1}\ \ \ q_{2}\ \ \cdots\ \ \ q_{n}\right]^{T}\mathbf{\mathcal{A}}=\left[q_{0}\ \ \ \mathbf{q}_{\theta}\right]^{T}\). The motor's output torque \(\tau=\left[\tau_{1}\ \ \ \tau_{2}\ \ \cdots\ \ \tau_{n}\right]^{T}\). We choose \(\mathbf{Q}=\left[\mathbf{\theta}^{T}\ \ \ \mathbf{q}^{T}\right]^{T}\) to be the system's generalized coordinate, \(F=\left[\tau^{T}\ \ \ \ 0_{1\times\left(n+1\right)}\right]^{T}\) to be the generalized force.