## 2 Description of the Control Object
\begin{split}&\dot{\mathrm{\varphi}}_{1}=\frac{1}{(\mathrm{R}+ \mathrm{r})}\Big{[}\ \dot{\mathrm{x}}_{\mathrm{s}}(\cos\beta+\sin\beta)+\ \dot{\mathrm{y}}_{\mathrm{s}}(-\cos\beta+\sin\beta)-\ \dot{\mathrm{\dot{\beta}}}\big{(}\mathrm{s}_{\mathrm{x}}+\mathrm{s}_{ \mathrm{y}}\big{)}\Big{]}\\ &\dot{\mathrm{\varphi}}_{2}=\frac{1}{(\mathrm{R}+\mathrm{r})} \Big{[}\ \dot{\mathrm{x}}_{\mathrm{s}}(\cos\beta-\sin\beta)+\ \dot{\mathrm{y}}_{ \mathrm{s}}(\cos\beta+\sin\beta)+\ \dot{\mathrm{\dot{\beta}}}\big{(}\mathrm{s}_{ \mathrm{x}}+\mathrm{s}_{\mathrm{y}}\big{)}\Big{]}\\ &\dot{\mathrm{\varphi}}_{3}=\frac{1}{(\mathrm{R}+\mathrm{r})} \Big{[}\ \dot{\mathrm{x}}_{\mathrm{s}}(\cos\beta-\sin\beta)+\ \dot{\mathrm{y}}_{ \mathrm{s}}(\cos\beta+\sin\beta)-\ \dot{\mathrm{\dot{\beta}}}\big{(}\mathrm{s}_{ \mathrm{x}}+\mathrm{s}_{\mathrm{y}}\big{)}\Big{]}\\ &\dot{\mathrm{\varphi}}_{4}=\frac{1}{(\mathrm{R}+\mathrm{r})} \Big{[}\ \dot{\mathrm{x}}_{\mathrm{s}}(\cos\beta+\sin\beta)+\ \dot{\mathrm{y}}_{ \mathrm{s}}(-\cos\beta+\sin\beta)+\ \dot{\mathrm{\dot{\beta}}}\big{(}\mathrm{s}_{ \mathrm{x}}+\mathrm{s}_{\mathrm{y}}\big{)}\Big{]}\\ \end{split}  \tag{2}\]  
The system of Equation (2) calculated the angular velocity waveforms of individual wheels on the basis of the desired motion path and the velocity profile of a selected point on the robot's frame. A robot dynamics model was analyzed in [15]. Lagrange equations of the second kind were used to describe the WMR dynamics in the form  
\[\frac{\mathrm{d}}{\mathrm{dt}}\Bigg{(}\frac{\partial\mathrm{E}}{\partial \dot{\mathrm{q}}_{\mathrm{j}}}\Big{)}-\frac{\partial\mathrm{E}}{\partial \dot{\mathrm{q}}_{\mathrm{j}}}=\mathrm{Q}_{\mathrm{j}} \tag{3}\]  
where j = 1, 2,..., s; s is the number of the object's degrees of freedom, q\({}_{\mathrm{j}}\) is the value of the jth generalized coordinate, \(\ \dot{\mathrm{q}}_{\mathrm{j}}\) is the value of the jth generalized velocity, E is the kinetic energy of the solid system, and Q\({}_{\mathrm{j}}\) is the value of the jth generalized force. The solution of the inverse dynamics problem was determined using the following:  
\[\mathrm{M}\ddot{\mathrm{q}}+\mathrm{N}\big{(}\dot{\mathrm{q}}\big{)}=\tau \tag{4}\]  
where \(\tau\) is the vector of moment of force, \(\ddot{\mathrm{q}}=\ddot{\mathrm{\varphi}}\) is the vector of generalized angular accelerations, N is the vector of resistance to motion, and M is the inertia matrix. The values of matrices and vectors were introduced to Equation (4), obtaining the solution of the inverse dynamics problem for the ith mecanum wheel:  
\[\mathrm{M}\left[ \begin{array}{c}\ddot{\mathrm{q}}_{1}\\ \ddot{\mathrm{q}}_{2}\\ \ddot{\mathrm{q}}_{3}\\ \ddot{\mathrm{q}}_{4}\end{array} \right]+\left[ \begin{array}{c}\mathrm{N}_{1} \mathrm{f}_{1}\mathrm{sgn}\big{(}\dot{\mathrm{q}}_{1}\big{)}\\ \mathrm{N}_{2}\mathrm{f}_{2}\mathrm{sgn}\big{(}\dot{\mathrm{q}}_{2}\big{)}\\ \mathrm{N}_{3}\mathrm{f}_{3}\mathrm{sgn}\big{(}\dot{\mathrm{q}}_{3}\big{)}\\ \mathrm{N}_{4}\mathrm{f}_{4}\mathrm{sgn}\big{(}\dot{\mathrm{q}}_{4}\big{)}\end{array}  \right]=\left[ \begin{array}{c}\tau_{1}\\ \tau_{2}\\ \tau_{3}\\ \tau_{4}\end{array} \right] \tag{5}\]  
where \(\tau_{\mathrm{i}}\) is the moment of force acting on the ith wheel, \(\ddot{\mathrm{q}}_{\mathrm{i}}=\ddot{\mathrm{\varphi}}_{\mathrm{i}}\) is the angular acceleration of the ith mecanum wheel, \(\mathrm{N}_{\mathrm{i}}\) is the pressure force of the ith wheel, and \(\mathrm{f}_{\mathrm{i}}\) is the coefficient of rolling friction of the ith wheel. The inertia matrix M was determined using the following:  
\[\text{M}=\left[ \begin{array}{cccc}(\text{A}\,+\,\text{B}\,+\,\text{C})&(\,-\, \text{B})&\text{B}&(\text{A}\,-\,\text{B})\\ (-\,\text{B})&(\text{A}\,+\,\text{B}\,+\,\text{C})&(\,\text{A}\,-\,\text{B})& \text{B}\\ \text{B}&(\text{A}\,-\,\text{B})&(\text{A}\,+\,\text{B}\,+\,\text{C})&(\,-\, \text{B})\\ (\text{A}\,-\,\text{B})&\text{B}&(\,-\,\text{B})&(\text{A}\,+\,\text{B}\,+\, \text{C})\end{array}