# Inertial Forces and Friction in Propulsion of a Rigid Body
## 2 Problem Formulation and Mathematical Model  
The concept of a body moving due to the inertial drive is schematically illustrated in Figure 1. The mechanical system consists of a rigid main body (a box or a capsule) of mass \(M\) and two internal masses treated as material points (particles), each having mass \(m\). The box is placed on a rough horizontal plane and can move along the \(x\) axis. Its interaction with the flat ground is described with the dry friction coefficient \(\mu\). The internal masses, in turn, revolve in circles of radius \(r\) about the axes fixed to the box. Both particles move in opposite directions with a constant common angular speed \(\omega\), which is controlled by appropriate driving torques. Positions of the point masses are specified by the angle \(\alpha\). The particles in circular motion undergo the normal (radial) accelerations relative to the box, \(a_{r}=r\omega^{2}\). As a reaction, forces \(F_{1}\) and \(F_{2}\) of an inertial nature are exerted on the box. They play the crucial role in causing the robot to move. Continuously alternating directions of \(F_{1}\) and \(F_{2}\) lead to cyclic changes in the overall load on the ground (and the friction force), as well as to cyclic generation of the propelling force. The basic problem is to choose such values of the model parameters that ensure efficient and smooth motion of the whole system. Consider the circular motion of the particles. The Cartesian components of the inertial forces are given by  
\[F_{1x}=F_{2y}=mr\omega^{2}\text{sin}(\alpha),\qquad F_{1y}=F_{2x}=mr\omega^{2} \text{cos}(\alpha). \tag{1}\]  
Consequently, the total inertial force \(F\) acting on the box has the following equal components:  
\[F_{x}=F_{y}=mr\omega^{2}[\text{sin}(\alpha)+\text{cos}(\alpha)]. \tag{2}\]  
Assume that rightward motion of the mechanical system is the desired one. A positive (upward) component \(F_{y}\) reduces both the load on the ground and the friction force, which facilitates moving the robot. The opposite effect is caused by negative (downward) \(F_{y}\) that weighs down the system. When the horizontal component \(F_{x}\) is positive (rightward), it can be regarded as the propelling force that may excite forward motion of the robot, if only this force overcomes friction. Negative (leftward) \(F_{x}\), in turn, may brake the system or even move it back. Different configurations of the internal masses together with forces \(F_{1}\) and \(F_{2}\) are shown in Figure 2. The coexistence of positive \(F_{x}\) and positive \(F_{y}\) (the phase of easier forward motion), as well as negative \(F_{x}\) and negative \(F_{y}\) (the rest phase, no backward motion), is particularly advantageous for efficient motion of the robot. Let \(x(t)\) denote the displacement of the box as a function of time \(t\). The horizontal motion and the static equilibrium in the vertical direction are described by  
\[(M+2m)\ddot{x}=F_{x}-T, \tag{3a}\] \[-(M+2m)g+F_{y}+N=0, \tag{3b}\]  
where \(T\) is the friction force, \(N\) is the normal reaction of the ground, and \(g\) is the acceleration due to gravity. According to the classical [PERSON]'s law of friction,  
\[T=\mu N\text{ sgn}\big{(}\dot{x}\big{)}. \tag{4}\]  
Figure 1: The main body with two internal masses in circular motion.