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Mechanics of Dean Drive on Frictional Ground

Christopher G. Provatidis

Department of Mechanical Engineering, National Technical University of Athens, Greece

Abstract: This paper investigates the unidirectional motion of an object forced to move on the ground due to an attached inertial drive which comprises counter-rotating eccentric masses on the horizontal plane. The study deals with the case in which the motion is generated by motors rotating at high speed, preferably constant. This inertial propulsion eventually enables the body to travel a limited maximum distance, which was found to be proportional to the square of the motor speed and inversely proportional to the coefficient of sliding (kinetic) friction. Other significant parameters such as the ratio of rotating masses over the object mass, as well as the initial position of these masses when the object is released to move and the magnitude of eccentricity, are discussed. The simulation is based on the combination of analytical formulas with the numerical solution of Single-DOF nonlinear ordinary differential equations for the unidirectional motion of the object to which the rotating masses are attached.

Keywords: inertial propulsion, Synchronization, Dean drive, friction

Received November 25, 2013; Revised December 08, 2013; Accepted December 18, 2013

Citation: Christopher G. Provatidis, Mechanics of Dean Drive on Frictional Ground. Journal of Mechanical Design and Vibration, 1 (1). 10-19. January 2014. doi:10.12691/jmdv-1-1-3

1. Introduction

Since 1959, Dean’s drive [1] has been a matter of controversy, either of rejection [2,3,4] or admiration [5]. The idea is to produce motion due to rotating lumped masses that are attached to the ends of rigid rods (eccentricities). Although the induced inertial forces are internal ones and also no mass exits the system, until the time the object is released to move, the system is open (ground forces are induced). Therefore, unidirectional motion is possible but, as we shall show, it cannot last for too long.

In more detail, using two fully synchronized counter-rotating masses, motion is conditionally performed only towards the bisector straight line between the points at which the rods are articulated. Previous studies have shown that the object is oscillating and, at the same time, moves like a projectile [6,7]. What actually happens is that, during every rotation, the initial linear momentum of the rotating masses is continuously transferred to the object and vice versa. In this way, although the object was initially at rest, the rotating masses possessed a sufficient amount of kinetic energy (and momentum), which was then transferred to the object. In other words, if the entire mechanical system was inside a closed box, it would give the impression that it is at rest, a thought that is incorrect, because the center of mass is at constant rotation.

It is worth mentioning that if there is only one rotating mass instead of two the object generally performs an oscillating oblique shot [8]. The range of the shot highly depends on the polar angle of the rods upon the release of the object, the magnitude of the angular velocity, the eccentricity and the ratio of the rotating mass over the object mass. For example, for the hypothetical conditions occurring in the old Rutherford–Bohr model of hydrogen atom, the maximum range is 72 km [8].

Although the vertical motion has been fully studied [6,7,8], the same does not hold for the horizontal one. Internet sources (even from academics) still refer to perpetual reactionless motion, and these claims inspired the author to perform this study. Within this context, this paper aims at investigating the influence of the friction, as well as other parameters such as motor speed, on the maximum distance for the object to travel. In addition, it explains the dominating mechanism that causes the progressive decrease of the mean average velocity of the object with respect to time.

2. Inertial Propulsion Mechanisms

Figure 1. Top view of the drive on the ground

In general, a “Dean drive” is a particular device (inertial drive) that consists of two exciters (rotating masses, No. 1 and No. 2), attached to an object of centroid G, as shown in Figure 1. Below we present the case according to which the rods and the masses rotate on the horizontal xy-plane on which the motion of the object (sliding) takes place as well. The two masses are fully synchronized so as to possess identical angular velocities in magnitude but opposite in direction. Due to the opposite rotation, it is evident that the motion of the object will take place entirely along the y-axis.

2.1. Equations of Motion

Referring to Figure 1, let us denote yM the position of the object (of mass M), ym the position of mass m, and (t) the time-varying polar angle between x-axis and the rotating rod. We assume that the object is at rest until the rods obtain a certain angular velocity ω. In the beginning of time (t = 0), at which the centroid of the object is at the point G₀ and the rods form angle ₀, the object is released to move on the ground as shown in Figure 1 (bottom).

Due to the fact that the counter-rotation of the two rotating masses cancels the two x-components of the inertial forces, their resultant will appear only in the y-direction and will be equal to . The external forces on the object (of mass M) consist of (i) the aforementioned d’Alembert force () plus (ii) the friction force Ff that is directed opposite to the velocity . Therefore, Newton’s second law becomes:

[picture] | (1)

Also, the relationship between ym and yM is (see Figure 1):

[picture] | (2)

Considering , by virtue of Eq (2) the second temporal derivative of ym becomes

[picture] | (3)

Substituting (3) into (1), one obtains

[picture] | (4)

Dividing both parts of (4) by (2 m + M) and introducing the parameter

[picture] | (5)

the former leads to the final equation of motion:

[picture] | (6)

2.2. Friction

2.2.1. Definitions

It is well known that the frictional forces acting between surfaces at rest with respect to each other are called forces of static friction (Fs). The maximum force of static friction (Fs, max) will be the same as the smallest force necessary to start motion. Once motion is started, the frictional forces acting between the surfaces usually decrease, so that a smaller force is necessary to maintain uniform motion. The forces acting between surfaces in relative motion are called forces of sliding (or kinetic) friction (Fk), and usually have a constant value.

In terms of the normal force N, the static friction is written as

[picture] | (7)

while the sliding friction as

[picture] | (8)

where and is the coefficient of static and sliding friction, respectively.

Usually, for a given pair of surfaces:

[picture] | (9)

Moreover, considering that the initial kinetic energy will be transformed into frictional work, for an object sliding along a straight horizontal road at speed v₀, the braking distance until it comes to a complete stop is given by

[picture] | (10)

In the case under consideration, neither the energy is conserved (the motor continuously consumes energy) nor the initial velocity is well described.

2.2.2. Friction Models

When the object moves, the induced friction equals to the sliding one (of a constant value). In contrast, when the velocity vanishes (object temporarily at rest) there are two cases depending on the instantaneous acceleration: (i) if the inertial force is smaller than the maximum static value the object does not tend to slide (therefore, the friction equals to the inertial force and the acceleration vanishes (), (ii) if the inertial force is greater than the maximum static value, the object tends to slide (therefore, the friction equals to the sliding friction and the object is accelerated (). The aforementioned model can be easily described and computer programmed using the formula:

[picture] | (11)

In Eq (11), the function is defined by

[picture] | (12)

In our case, the normal force is permanently given by

[picture] | (13)

and therefore, if the combination () never occurs, Eq (6) becomes:

[picture] | (14)

3. Constant Angular Velocity

The case of a constant angular velocity is the most straightforward one and therefore is thoroughly studied below for both cases of the frictionless and frictional grounds. Then, in either of Eq (6) and Eq (14), .

3.1. Frictionless Model

When μ = 0 (i.e. Fj = 0) the integration of Eq (6) gives the following analytical solution:

[picture] | (15)

whence the object velocity becomes (see Appendix A):

[picture] | (16)

Equation (16) dictates that the velocity of the object is a periodic function with respect to and t. When v₀ = 0 and varies, obtains values in the interval:

In the particular case that, ₀ = 0, it takes the form: , with average value .

Due to the periodicity involved in the terms and of Eq (15) and Eq (16), respectively, two important remarks have to be made:

1. The average velocity is cosine-dependent and is equal to

[picture] | (17)

2. Every full rotation (360 degrees) the object velocity becomes zero and, at the same time, the rods are oriented in the same direction . In other words, there is no phase difference between object velocity and the corresponding direction of the rods ().

3. At the end of every rotation (), the net displacement changes by the same amount, which is equal to

[picture] | (18)

According to Eq (17) and Eq (18), the maximum average velocity, as well as the maximum forward distance travelled, are achieved when , whereas the cosine term is zeroed when . It should however be clarified that, when the object is initially at rest (), Eq (15) denotes that even if the object will perform harmonic oscillations, of amplitude , about its initial position and the net displacement over time is zero.

In contrast, when , due to the fact that , the last term in Eq (15) is bounded, i.e. . Therefore, there will always exist a critical time later than which the third (linear with respect to t) term in Eq (15) will exceed . In other words, the object will never stop moving and will disappear at infinity.

A closer look on Eq (16) reveals that when the object starts from rest (), except for the particular cases or , during any rotation there are two positions at which the velocity of the object vanishes. In this paper these positions will be called “reversal points”, in the sense that at these points the object velocity changes its sign.

As an example, considering M = 5 kg, m = 1 kg, r = 0.1 m, and ω = 3000 rpm, the influence of angle is shown in Figure 2, where it is depicted that the maximum velocity is produced when (see the thickest red line). Except for the case where (magenta and cyan lines in Figure 2), under all other conditions the object will never stop travelling, provided the motor consumes energy and preserves a constant value for its speed (angular velocity, ).

Figure 2. Displacement along y-axis of an object sliding on frictionless horizontal ground for several characteristic initial polar angles (₀=-180, -90,0 and 90 deg) when it is released (ω = 3000 rpm)

It is worth mentioning that since for it is for every polar angle, the object velocity never becomes negative, and it becomes zero at the ends of every rotation only for . Equivalently, the object is always moving forward, as shown in Figure 3 for the first two rotations.

Figure 3. Displacement and velocity of an object sliding on frictionless horizontal ground for the first two rotations (ω = 3000 rpm, period T = 4 s, ₀ = 0)

On the contrary, when the initial angle is different than zero (e.g. ), the velocity given by Eq (16) changes its sign, being negative in () and positive in (), while it vanishes at . An example for is shown in Figure 4 (top), where one can notice that at the end of the period the displacement decreases and the velocity becomes slightly negative. In a similar way, supposing that , the displacement will decrease in the beginning of the period where the velocity will become slightly negative (Figure 4, bottom).

Figure 4. Displacement and velocity of an object sliding on frictionless horizontal ground for the first rotation (ω = 3000 rpm, period T = 0.02 s, TOP: ₀ = π/ 6, BOTTOM: ₀ = -π/ 6)

3.2. Frictional Model

When the coefficient of friction is not equal to zero, it is not further possible to identify a unique expression for the analytical solution of Eq (6) or Eq (14). It is however possible to split the time in successive intervals in which the direction of the object velocity does not change its sign. Therefore it becomes necessary to determine the aforementioned “reversal points” at which vanishes (it passes from positive to negative values and vice versa). Between two successive reversal points, the first being at the time instance , we can still derive a unique analytical expression, different only in terms of the sign before as follows:

3.2.1. Forward Sliding

[picture] | (19)

whence the object velocity becomes:

[picture] | (20)

3.2.2. Backward Sliding

[picture] | (21)

whence the object velocity becomes:

[picture] | (22)

It is noted that , as the latter corresponds to the current reversal point, which is calculated using a standard Newton-Raphson algorithm by setting (forward motion, Eq (20)) or (backward motion, Eq (22)). Obviously, in Eqs(19)-(22) the quantities are variable, and take their current value accordingly.

Due to the complexity involved in the aforementioned analytical procedure, concerning the proper choice of initial guess values necessary for the Newton-Raphson algorithm, particularly when the number of rotations is high, we prefer to solve Eq (6) in a purely numerical way. For example, we can use standard Runge-Kutta algorithms of variable time step, such as the efficiently implemented function “ode45” in MATLAB. Having said this, it should be noted that particular care must be paid on selecting a properly small tolerance to ensure convergence.

3.3. Critical Angular Velocity

For an arbitrary angle , the condition for ensuring sliding (subscript “sl”) becomes , whence the critical minimum value is

[picture] | (23)

and by virtue of Eq (5) obtains its normalized form,

[picture] | (24)

It is evident that for every there is a particular critical speed for ensuring start of sliding. Therefore, the critical speed is a function of every angle , and has a minimum value for , whence the minimum critical speed is given by

[picture] | (25)

For the particular example (M = 5 kg, m = 1 kg, r = 0.1 m), for several coefficients of friction the critical sliding speed is given in Table 1.

Table 1. Influence of the coefficient of friction to the initiation of sliding

Table 2. Influence of the coefficient of friction to the distance travelled

3.4. Displacement Decay

3.4.1. The Role of Varying Angle at Reversal Points

Based on Eq (24), the initial angle to start sliding is given by

[picture] | (26)

Considering again the same speed ω = 3000 rpm, the transient phenomenon of the decay is shown in Figure 5 for several coefficients of friction.

Figure 5. Displacement decay of an object sliding on frictional horizontal ground for a large number of rotations (ω = 3000 rpm, period T = 0.02 s, ₀ = 0, M = 5 kg, m = 1 kg, r = 0.1 m)

In a first attempt to correlate the above-calculated maximum distance travelled with the distance obtained using the formula provided by Eq (10), applied to the center of mass, the comparison is shown in Table 2, where it is noted that the simulated value is about three times bigger than what would be expected using Eq (10).

The above difference is explained as follows.

For the particular case and mentioned before, for frictionless ground the velocity of the object will become zero in the beginning and at the end of every full rotation ( etc). In other words, whenever the velocity of the object vanishes the rods will always be in the horizontal position.

In contrast, when there are frictional ground forces, by virtue of Eq (20) and Eq (22) it is evident that the velocity of the object vanishes when the condition is satisfied. Obviously this is a repeated pattern for every full rotation of the rods.

In more detail, by sweeping the time values from t = 0 to infinity, and referring to an arbitrary reversal point, it should become clear that:

1). A short time after the previous equation is satisfied, the object goes forward and therefore the friction is negative:

[picture] | (27)

2). Also, a short time before the previous equation is satisfied, the object goes backward and therefore the friction is positive:

[picture] | (28)

For a high coefficient such as μ = 1, the variation of the angle (at which the object velocity vanishes) is shown in Figure 6, where one can notice that its upper limit is 90 degrees (). The latter gives the value , for which both Eq (19) and Eq (21) depict an oscillating motion of amplitude 2λ about a fixed point. In one sense, angle could be considered as a “phase difference” between the counter-rotating exciters. Similar behavior appears also for all other cases of the coefficient μ considered in this paper.

Figure 6. Displacement decay of an object sliding on frictional horizontal ground (coefficient of friction μ = 1) for a large number of rotations (ω = 3000 rpm, period T = 0.02 s, ₀ = 0, M = 5 kg, m = 1 kg, r = 0.1 m)

3.4.2. Further Insight in the Nonlinear Phenomenon

Although the above numerical results explain the reasons for which the Dean drive moves and then stops moving, a rigorous explanation is not yet clear. Without loss of generality, let us assume that ₀ = 0 and μ = 1. In this case the corresponding results are shown in Figure 7 where one can notice that although the object continuously performs a forward net movement, its velocity gradually decreases and at the same time the initial angle (phase difference) increases.

Figure 7. Displacement and velocity decay of an object sliding on frictional horizontal ground (coefficient of friction μ = 1) for the first twenty rotations (ω = 3000 rpm, period T = 0.02 s, ₀ = 0, M = 5 kg, m = 1 kg, r = 0.1 m)

Furthermore, in contrast to the frictionless motion in conjunction with ₀ = 0, where only one intersection per rotation appears in the beginning and at the end of every period (cf. Figure 3), Figure 7 depicts that, for the same condition, during every rotation the time axis () intersects the velocity graph at two points.

A deeper explanation, based again on the case μ = 1 and time period T = 0.02 s, is as follows (see Figure 8a).

•  The first reversal point B (after the initial point A) is calculated at , where the corresponding angle is , that is degrees (< 360), equivalently -12 degrees before the end of the first rotation (below the horizontal level). By virtue of Eq. (20), the aforementioned point B lies at the intersection of the graph of the function with the straight line , both of them defined in the interval . For the object moves forward (plus sign in function f₂) but its velocity progressively decreases until it vanishes at .

•  The second reversal point C appears at , where the corresponding angle is , that is degrees, equivalently only 11.2 degrees after the end of the first rotation (beyond the horizontal level). In more details, after an almost infinitesimal amount of time after the reversal point B (t > 0.0193s) the object moves backward (minus sign in function f₂) and the next reversal point C will be produced as the new intersection of the function with the straight line , both of them defined in the interval .

•  The third reversal point D (not shown in Figure 8) appears at , where the corresponding angle is , that is degrees, equivalently -16.4 degrees before the end of the second rotation (beyond the horizontal level). This point is produced by similar functions f₁ and f₂ as those used for the first reversal point, and so on.

A representative sketch for the above procedure is shown in Figure 8b, where all the aforementioned functions and have been represented by two unique “bell” (or cosine)-shaped curves reduced in the normalized interval of the first 540 degrees. Concerning the first rotation shown in Figure 8a, we start from point A (axis origin) and then follow the straight line AB of positive slope equal to . In sequence we follow the straight line BC of negative slope and this completes the first rotation. Point C at the end of the first rotation is identical with point A' where evidently segment CA' is parallel to the horizontal axis. Therefore, the second rotation starts from the known point A', then we continue with points B' and C' (not shown), and so on, until the entire bell-shaped regions of Figure 8a are fully filled (maximum distance travelled). Indicatively, the case of 100 rotations is shown in Figure 8b.

Figure 8. Filling of the reference bell-shaped regions for (a) the first and (b) the first hundred rotations (μ = 1, ω = 3000 rpm, period T = 0.02 s, ₀ = 0, M = 5 kg, m = 1 kg, r = 0.1 m)

Based on the above nonlinear analytical procedure, a series expansion solution is given in Appendix B.

3.4.3. Curve Fitting

The numerical results shown in Figure 5 depict an exponential decay. Unfortunately the term appearing in Eq (14) does not allow for a simple analytical solution of closed form. Therefore, in order to better understand the results, we resort to a simplified model using an “equivalent” Newtonian damping; in that case an analytical solution is available (Appendix C).

Extensive numerical experimentation has shown that a curve of the form , can successfully approximate the numerical solution shown in Figure 5, when the damping factor, c, is proportional to and the final (maximum) distance travelled, a, is proportional to .

Comparing the above experimental findings with the analytical solution given by Eq (C-4) in Appendix C, we can see that starting from the above finding , then the denominator becomes proportional to ω:

where the term has been ignored, as being very small with respect to . Therefore, the expression (C-4) becomes:

[picture] | (29)

In Eq (29), variables c₁ and c₂ are correcting factors to be determined. Trial-and-error calculations have concluded the following formula:

[picture] | (30)

The excellent quality of Eq (30) in representing the numerical solution according to Eq (30) for the first 2000 rotations is shown in Figure 9.

Figure 9. Curve fitting (red line, according to Eq (30)) for the numerically calculated position of the object (blue line) versus time

4. Energy Consumption

Due to the symmetric arrangement of the two rotating masses so as, with respect to the inertial reference frame (fixed to the ground), both velocities have the same measure given by , the kinetic energy of the entire mechanism will be:

[picture] | (31)

During the sliding of the mass M over the horizontal ground, there are two sources of external action: the first (Wf) concerns the sliding friction, while the other (Wm) concerns the mechanical work given by the electric motor to the system; actually it is the electric energy multiplied by the efficiency.

Let us now take as reference at the beginning of time, , where . At this time instance, the initial kinetic energy will be:

[picture] | (32)

Moreover, the elementary work done by friction is , which by virtue of Eq (8) and Eq (13) becomes .

After integration with respect to yM, we obtain:

[picture] | (33)

Obviously, the integral in Eq (33) is equal to the total traveled distance (not the net one: ).

Since the principle of energy conservation demands that the total external work equals to change in kinetic energy, it can be written:

[picture] | (34)

Substituting Eq (31), Eq (32) and Eq (33) into Eq (34), the latter is solved in terms of to determine the work done by the motor.

Figure 10. Total kinetic energy (Ekin), and cumulative works done by the friction (Wf) and the motor (Wmotor), for μ = 1

Figure 11. Cumulative total kinetic energy (Ekin), and works done by the friction (Wf) and the motor (Wmotor), for μ = 1 (first five rotations)

Based on the above procedure, the variation of the kinetic energy and the cumulative work done by the motor is shown in Figure 10, when the coefficient of friction is μ = 1. It can be observed that while the kinetic energy progressively decreases (always oscillating within the blue zone of Figure 10), the motor continuously consumes energy thus producing mechanical work (green zone of Figure 10), while the friction continuously consumes energy (smooth red line of Figure 10). The maximum traveled distance that corresponds to Figure 10 is about 12 m and it was achieved in 400 oscillations. Finally, the same results for the first five periods are shown in Figure 11 (period T = 0.02 s).

5. Discussion

Although the numerical estimation of the reversal points is possible through an iterative Newton-Raphson algorithm, when the number of rotations is very large their relative position progressively changes. Therefore, if the guess values are not properly chosen, some reversal points may be lost.

Due to the above fact, we preferred to trust the time integration of the ODE. Nevertheless, for a large number of rotations the application of numerical time-integration methods is again problematic. In more details, in their default mode, three different algorithms (ode45, ode23s, ode15s) lead to quite different solutions after a certain large number of rotations. In contrast, when reducing the tolerance (e.g. to 10⁻⁸ or, in the limit, to 10⁻¹⁴) all of them converge to very similar solutions. A similar behavior was previously noticed for the integration of Euler and Lagrange equations for the motion of the spinning top [9].

The most important finding of this work is perhaps the fact that, the phase difference between the velocity of the object and the polar angle of the rods continuously slips from ₀ to . Since the performance in distance traveled is maximized when , below we discuss this case only. In more details, the initial velocity of the object at t = 0 is zero, and at the same time rod’s position (initial angle ₀) is also zero. When, after an almost full rotation, the velocity of the object becomes again equal to zero, however the new polar angle is not equal to ₀ (unless the ground is frictionless). Therefore, after a large number of rotations, the polar angle of the rods tends to ; in the latter case, the object merely oscillates about the same (final) position.

If, however, we wish to force the object to repeat a similar travel the only thing we need to do is to ask for external intervention to stop the object and reposition the rods of the propulsion system at their initial value, ₀, when the object’s instantaneous velocity is zero. Caution should be paid for the object to be fully released when the polar angle is again equal to ₀. Smaller values of initial angle will cause decrease of the range according to Figure 2.

6. Conclusion

The findings of this research suggest that, an inertial drive, which consists of two counter-rotating exciters attached to an object, is capable of providing uninterrupted net forward motion on a frictionless ground. The latter is due to the initial linear momentum and does not depend on whether the motor continues to absorb energy or is switched off. This endless motion strongly depends on the cosine of the angle formed by the rods at the time the object is released to move. In contrast, frictional ground reduces the maximum distance travelled the latter being inversely proportional to the coefficient of friction. Due to the oscillation produced by a motor operating at a constant high speed and due to the alternation in the direction of the frictional forces, the distance travelled is a few times higher than the monotonic forward sliding determined for a point-like mass, for the same initial velocity of the center of mass.

References

APPENDIX A

Integration of Equations of Motion

In the absence of friction forces and irrespectively to the variation law of the angle (t) in time, Eq (6) can be written as:

[picture] | (A-1)

or, equivalently

[picture] | (A-2)

Once integrating (A-2) we obtain

[picture] | (A-3)

where

[picture] | (A-4)

It is worth mentioning that, setting and , (A-4) coincides with Eq (16).

Moreover, once more integrating (A-3), we obtain:

[picture] | (A-5)

It is again worth mentioning that, setting and , (A-5) coincides with Eq (15).

APPENDIX B

Analytical Solution in Series Expansion

We refer to Figure 8, which clearly shows that each rotation approximately corresponds to a couple of forward and backward motions. Let n, n + 1 and n + 2 represent the ascending numbers of the three reversal points in a certain rotation.

The first two rotations consist of the reversal points t₀, t₁, t₂, t₃, t₄, that correspond to polar angles ₀, ₁, ₂, ₃, ₄, velocities v₀, v₁, v₂, v₃, v₄ (equal to zero) and position coordinates y₀, y₁, y₂, y₃, y₄.

The vanishing object velocities are given by:

[picture] | (B-1)

The corresponding displacements are given by:

[picture] | (B-2)

Adding the first two and then the first three equations involved in (B-1) by parts, and eliminating all those terms twice appearing in both left and right hand sides, one obtains:

[picture] | (B-3)

and

[picture] | (B-4)

Repeating similar additions for the first two displacements of (B-2), where v₁ has been substituted by the first of (B-1), after the reduction of similar terms, the displacement at the end of the first rotation is given by:

[picture] | (B-5)

In (B-5) we recognize that the first term in brackets equals the displacement on frictionless ground, while the subtrahend term denotes the influence of the oscillating frictional force.

Repeating similar additions for the first four displacements of (B-2), and progressively eliminating v₃ and then v₂ by (B-3), the displacement at the end of the second rotation is given by:

[picture] | (B-6)

In (B-6) we recognize again the frictional displacement and the influence of the oscillating friction in the first and the rest brackets, respectively. The signs alternate with a strictly defined order.

Finally, it is worth mentioning that in the case of an object forward sliding (only in the positive y-direction), all signs in (B-6) have to be taken positive. Then, writing , by virtue of the well-known identity concerning the square on the sum of four terms appearing on the right term, the finally obtained friction dependent term in (B-6) will be equal to . Therefore, in case of a forward motion the subtrahend term is much higher (in absolute value) than what the corresponding in oscillation is (alternating plus and minus). It is evident that the latter is in full accordance to the findings in Table 2.

APPENDIX C

Newtonian Damping

In order to consider the law describing the maximum distance travelled, we temporarily consider the simpler case at which the sliding friction is not a constant but proportional to the velocity of the body (). Then Eq (6) becomes:

[picture] | (C-1)

The analytical solution of (C-1) is:

[picture] | (C-2)

where:

[picture] | (C-3)

Neglecting the last terms of (C-2), we obtain:

[picture] | (C-4)

The function appearing in (C-4) is an average smooth line about which the accurate solution slightly oscillates.

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