## 3 Results
### Analytical Model Development Using Euler-Bernoulli Beam Theory  
As we outlined in the previous sections, a simple mathematical model may give some primary useful insight into the dominant system eigenmodes. The computed dynamical behaviour can be used in simulation prior to assembling the physical setup and can therefore help to redesign it. Let us suppose that we have just some principal idea about the machine construction, material, and purpose of application, and we would like to get some initial insight into its dynamical behaviour. Considering this, we include just the fundamental parts of the complex construction to the dynamical model descriptio. In addition, in the mathematical model. the arm length, the payload position, or the mass can be easily modified to visualise the impact on the dynamical behaviour of such possible changes. The whole set of models can thus be validated and investigated thoroughly. The Lagrange-Euler method [20] with the Euler-Bernoulli theory [21; 26] can be now employed. We shall start with the following formula  
\[\frac{d}{dt}\frac{\partial L}{\partial\dot{q}}-\frac{\partial L}{\partial q}= \mathcal{T}_{m}, \tag{39}\]  
where in the Lagrangian \(L=T-V\), the kinetic energy \(T\) and potential energy \(V\) are subtracted. \(\mathcal{T}_{m}\) denotes the torque vector, and \(q\) is the state co-ordinates vector. In this modelling case study, the energies of the hub and payload are taken into account as well as the energy of the arm that is divided into four elements. Therefore, the state co-ordinates vector can be established as \(q=\left[ \begin{array}{cccc}\theta&q_{1}&\ldots&q_{10}\end{array} \right]^{T}\), where \(\theta\) is the hub angle and \(q_{i}\) denotes for the \(i\)th element the displacement (if \(i\) is odd) and the angle (if \(i\) is even). Specifically, the kinetic energy of the hub is considered simply as \(T_{h}=\frac{1}{2}J_{h}\dot{\theta}^{2}\), where \(J_{h}\) is the hub inertia moment. The energy of the payload (\(T_{p}\)) and the energies of the arm elements (\(T_{i}\), \(V_{i}\)) depend on the deflection function \(u(x,t)\) outlined in (2). The payload is considered as a concentrated mass located on the last element, i.e., \(s=l\). This leads to  
\[T_{p}=\frac{1}{2}m_{z}\dot{u}^{2}(l,t)|_{j=n+1}, \tag{40}\]  
where \(m_{z}\) is the mass of the payload, \(\left.\dot{u}(l,t)\right|_{j=n+1}=\overline{N}(l,t)\vec{\mathcal{Q}}(t)=nl\dot {\theta}(t)+q_{2n+1}(t)\). Specifically, the payload mass is located at the last element, i.e., \(n=4\), and is related to the variable \(q_{9}\). The energies of the arm are considered as (12) for \(n=4\). By summing energies of all considered components, we obtain  
\[T=\sum_{i=1}^{4}T_{i}+T_{h}+T_{p},V=\sum_{i=1}^{4}V_{i}. \tag{41}\]  
Thus, we can complete the Lagrange-Euler method and derive the system of the second ordinary differential equations for the Lagrangian \(L=T-V\), where \(T\) and \(V\) are given as (41). This system can also be expressed as the mechanical equation  
\[M(q)\ddot{q}+Kq=b\tau, \tag{42}\]  
where \(M(q)\) is the mass matrix, \(K\) is the stiffness matrix, and \(\tau\in R\) is the input torque, \(b=\left[ \begin{array}{cccc}1&0&\ldots&0\end{array} \right]^{T}\). These formulas are moreover restricted by the condition on the arm attachment. When the arm is modelled with one fixed end and one free end, the variables \(q_{1},q_{2},q_{1},q_{2}\) vanish. Hence, the adjusted state co-ordinates vector becomes  
\[q=\left[ \begin{array}{cccc}\theta&q_{3}&\ldots&q_{10}\end{array} \right]^{T}. \tag{43}\]We obtain nonlinear differential equations and linearise them in the equilibrium point \(\left[ \begin{array}{cccccc}0&0&\ldots&0\end{array} \right]^{T}\). We derive the state-space model for the state vector \(z=\left[ \begin{array}{cccccc}\theta&q_{3}&\ldots&q_{10}&\dot{\theta}&q_{3}& \ldots&q_{10}\end{array} \right]^{T}\in R^{18}\) as  
\[\dot{z}=\left[ \begin{array}{cccccc}0&I\\ -M^{-1}K&0\end{array} \right]z+\left[ \begin{array}{cccccc}0\\ M^{-1}b\end{array} \right]\tau.