## 3 Conversion of Rotational Energy into Translational Vehicle Energy
Thus, the maximum translational velocity is as follows:  
\[v_{max}=\frac{1}{\sqrt{2}}R\omega_{0}\sqrt{\frac{M}{M^{\prime}}} \tag{13}\]  
Maximum velocity \(v_{max}\) depends on radius \(R\), angular velocity \(\omega_{0}\), and construction factor \(C=\frac{M^{\prime}}{M}\); however, \(v_{max}\) is independent of the unbalance mass \(m_{u}\). We are looking now at intermediate states of energy transformation: Energy theorem postulates that each increase in translational energy asks for an equal decrease in rotational energy, i.e.,  
\[-dE_{rot}=dE_{trans} \tag{14}\]  
\[E_{rot}=\frac{1}{4}MR^{2}\omega^{2} \tag{15}\]  
\[E_{rot}=\frac{1}{4}MR^{2}2\omega d\omega=\frac{1}{2}MR^{2}\omega d\omega \tag{16}\]  
\[E_{trans}=\frac{1}{2}M^{\prime}v^{2} \tag{17}\]  
\[\max E_{rot}=\frac{1}{2}\theta\omega_{0}^{2}=E_{0} \tag{11}\]0  
\[\max E_{rot}=\frac{1}{2}\theta\omega_{0}^{2}=E_{0} \tag{11}\]1  
Equation (19) teams our rotational magnitudes with the translational ones of interest, i.e., vehicle velocity \(v\) and its translational acceleration \(a\). The third axiom of Newton (action = reaction) tells us about acceleration \(a\) of mass: Centrifugal force \(F=m_{u}R\omega^{2}\) caused by unbalance and inertial force \(F=M^{\prime}a=M^{\prime}\frac{d^{2}x}{dt^{2}}\) acting on vehicle mass \(M^{\prime}\) must be in equilibrium, i.e.,  
\[\max E_{rot}=\frac{1}{2}\theta\omega_{0}^{2}=E_{0} \tag{11}\]2  
Thus, we obtain the acceleration \(\Big{(}a=\frac{dx^{2}}{dt^{2}}=\frac{dv}{dt}\Big{)}\) of vehicle mass \(M^{\prime}\):  
\[\max E_{rot}=\frac{1}{2}\theta\omega_{0}^{2}=E_{0} \tag{11}\]3  
From (19) follows the time-dependent change of rotation frequency:  
\[\max E_{rot}=\frac{1}{2}\theta\omega_{0}^{2}=E_{0} \tag{11}\]4  
\[\max E_{rot}=\frac{1}{2}\theta\omega_{0}^{2}=E_{0} \tag{11}\]5  
After integration, we have the following:  
\[\max E_{rot}=\frac{1}{2}\theta\omega_{0}^{2}=E_{0} \tag{11}\]6  
\[\max E_{rot}=\frac{1}{2}\theta\omega_{0}^{2}=E_{0} \tag{11}\]7  
We obtain instantaneous rotation frequency in dependence of \(x=x(t)\), which is the trajectory length depending on trust duration time \(t\):  
\[\max E_{rot}=\frac{1}{2}\theta\omega_{0}^{2}=E_{0} \tag{11}\]8  
Rotation frequency decreases exponentially with increasing \(x(t)\). This follows from the increase in translational energy and instantaneous velocity \(v\) (see Figure 4):  
\[\max E_{rot}=\frac{1}{2}\theta\omega_{0}^{2}=E_{0} \tag{11}\]9With (11) and (25), we obtain via \(E_{0}\ =\ max\ E_{rot}\ =\ \frac{1}{4}MR^{2}\omega_{0}^{2}\)  
\[\theta=\frac{1}{2}MR^{2}\]0  
\[\theta=\frac{1}{2}MR^{2}\]1  
\[\theta=\frac{1}{2}MR^{2}\]2  
\(v(x)\) is the instantaneous velocity in dependence of trajectory length \(x\). Setting \(x=0\) delivers initial velocity \(v=0\) in accordance with our takeoff scenario above. From (28), we obtain by differentiation the instantaneous translational acceleration of the rotor:  
\[\theta=\frac{1}{2}MR^{2}\]3  
Initial acceleration \(a_{0}\) follows from (29) by setting starting point condition \(x=0\):  
\[\theta=\frac{1}{2}MR^{2}\]4  
Both instantaneous acceleration \(a\) and velocity \(x\) depend on mass quotient \(\frac{m}{M}\), which presents the main steering magnitude for propulsion control. Acceleration increases proportionally with \(\frac{m_{u}}{M^{\prime}}\). Regarding the momentum and angular momentum laws during the energy conversion, the rotational momentum constantly decreases due to the reduction of the rotational frequency, while the translational momentum increases in magnitude due to the energy conversion accompanying the increase in translational velocity. However, it would be inappropriate to speak of a conversion of momentum, as these quantities cannot be reconciled due to the different physical dimensions of angular and translational momentum. There is a certain analogy to the behavior of a non-conservative system (e.g., a system with friction). Here, the law of conservation of energy applies, but not the law of conservation of momentum.