## 3 Results and Discussion
Using these initial conditions, we find:  
\[C_{1}=\frac{\lambda_{2}\cdot(T+Q)-S\cdot\omega}{\lambda_{1}-\lambda_{2}},\]  
and  
\[C_{2}=\frac{-\lambda_{1}\cdot(T+Q)+S\cdot\omega}{\lambda_{1}-\lambda_{2}}. \tag{42}\]  
So, with the general solution of Equation (30), which satisfies the specified initial conditions, will have the form:  
\[ \begin{array}{l}x=\frac{\lambda_{2}\cdot(T+Q)-S\cdot\omega}{\lambda_{1}- \lambda_{2}}\cdot e^{\lambda_{1}t}-\frac{\lambda_{1}\cdot(T+Q)-S\cdot\omega}{ \lambda_{1}-\lambda_{2}}\cdot e^{\lambda_{2}t}+\\ +S\cdot\sin\omega t+T\cdot\cos\omega t+Q.\end{array}  \tag{43}\]  
Expression (43) is the law of relative motion of the haulm particle \(M\) along the thrower blade at an arbitrary moment of time \(t\). \[ \begin{array}{l}\dot{x}=\frac{\lambda_{1}\cdot\lambda_{2}\cdot(T+Q)-S\cdot \lambda_{1}\omega}{\lambda_{1}-\lambda_{2}}\cdot e^{\lambda_{1}t}-\frac{ \lambda_{2}\cdot\lambda_{1}\cdot(T+Q)-S\cdot\lambda_{2}\cdot\omega}{\lambda_ {1}-\lambda_{2}}\cdot e^{\lambda_{2}t}+\\ +S\cdot\omega\cdot\cos\omega t-T\cdot\omega\cdot\sin\omega t.\end{array}  \tag{44}\]  
For further simulation of the process of movement of the haulm particle \(M\) along the cylindrical and rectilinear parts of the casing of the loading mechanism, after it leaves the thrower blade, it is necessary to calculate the relative speed of the particle leaving the thrower blade. If the length of the blade is given, namely, \(AB=l\), then from dependence (43) it is possible to calculate the time \(t_{1}\) for the leaves of the haulm particle to leave the blade at \(x=l\). Such a calculation can be done by means of a PC. By substituting the obtained value of time \(t_{1}\) into Expression (44), one can find the relative value \(\dot{x}_{1}=\dot{x}(t_{1})\) of the haulm particle \(M\) leaving the end of the thrower blade. Let us consider an important partial case when the blades of the thrower are arranged radially. In this case \(\psi_{1}=0\), \(\psi=0\) and \(\psi_{0}=\pi\cdot 2^{-1}\). From Expressions (29), we obtain:  
\[K=g;\ L=-f\cdot g;\ C=\cos\gamma+f\cdot\sin\gamma;\ D=f;\ r_{0}=0 \tag{45}\]The differential Equation (30) for the movement of a haulm particle along the blade in this case is greatly simplified and has the following form:  
\[ \begin{array}{l}\ddot{x}+\left(2f\cdot\omega+\frac{k}{m}\right)\cdot\dot{x}- \omega^{2}\cdot x=g\cdot\cos\omega t-\\ -g\cdot f\cdot\sin\omega t+\frac{k}{m}\cdot V_{n}\cdot(\cos\gamma+f\cdot\sin \gamma).\end{array}  \tag{46}\]  
The general solution of Equation (46), which satisfies the above initial conditions, can be obtained from the general solution (43); however, the coefficients \(S\), \(T\) and \(Q\), taking into account (45), will already be determined from the following expressions, obtained on the basis of (38)-(40):  
\[Q=-\frac{(\cos\gamma+f\cdot\sin\gamma)\cdot k\cdot V_{n}}{m\cdot\omega^{2}} \tag{47}\]  
\[S=\frac{4g\cdot f\cdot\omega+\frac{k\cdot g}{m}}{4\omega^{3}+\omega\cdot\left( 2f\cdot\omega+\frac{k}{m}\right)^{2}} \tag{48}\]  
\[T=\frac{2g\cdot f^{2}\cdot\omega+\frac{k\cdot f\cdot g}{m}-2g\cdot\omega}{4 \omega^{3}+\omega\cdot\left(2f\cdot\omega+\frac{k}{m}\right)^{2}} \tag{49}\]  
So the general solution of the differential Equation (46), when substituting Expressions (47)-(49) into (43), will be written in the following form:  
\[