A very simple scheme can be formulated by assuming that the body is a planar rigid system sketched by three points A, B and; A and B, performing the two contact points of the wheels, belong to a horizontal plane and is the centre of mass of. The rigid body can lean with respect to the vertical direction and bend with respect to a fixed horizontal direction. Let O be the projection of on and take a fixed reference frame, being the upward vertical, and a body reference frame such that

is the angle, is the angle. The mutual disposal of the two frames is given by

and sketched in Figure 1.

Set: the geometrical restrictions, , lead us to consider the four lagrangian coordinates. Therefore

. The angular velocity of the rigid body is.

The Lagrangian function of the system writes

where and, and are the moments of inertia with respect to the body reference frame, which is supposed to be principal.

The only kinetic constraint we are going to consider is, whose expression in the Lagrangian coordinates is

The first kind Lagrangian equations of motion

where are the lagrangian coordinates and is the unknown multiplier, will be suitably handled if one defines the pseudovelocities (see [1] for the concepts and the method we are pursuing)

We point out the following relationships involving U, V and the real velocities:

where and

Together with the kinetic constraint, Equations (4) give the set of possible velocities in terms of:

It is known (see [1] ) that linear kinetic constraints allow to refine the equations of motion (3) in a way similar to the holonomic case: as a matter of fact, the constraints identify a subspace in the tangent space of the lagrangian coordinates, giving the virtual displacement of the system. The geometrical method consists of projecting the equations according to where is defined in (6). Joining to the kinetic constraint (2) and the definition (4) we get, dividing by suitable constants,

where

are dimensionless constants. Since the rigid body is practically plane and contained in the plane orthogonal to j, it is reasonable to assume, , so that

From here on we adopt (9).

The seven ODEs (7) contain the seven unknown quantities. With respect to the first kind system (3) they have the advantage of no exhibiting multipliers and of reducing the kinetic variables of one unity.

It is not at all worthless to explain (7) in the context of the the cardinal equations of dynamics, seeing that many models in literature (some of them are [2] -[5] ) rest on such equations: the first three equations in (7) are respectively

where L is the angular momentum and the resultant momentum of the external forces. Actually, since no friction is present, the constraint in A is along, while the constraints in B can be modelled as a force along plus a horizontal force perpendicular to (see Figure 2), acting the constraint (2):

Hence no force exists along (first equation) and (second equation), along (and not, as stated in [6] ). Finally, third equation is simply due to the fact that the only external force with a non zero momentum along is the weight force.

Notice that any of the three equations do not give rise to a conserved quantity: the only evident constant of

motion is the total energy:

Actually, even if the system is nonholonomic, the first integral can be achieved starting from

and performing the usual calculations as in the holonomic case, achieving at last.

In the same matter of integral invariants for the system, we find it not correct to claim, as in [7] , that the absence of, , from entails three constant of motion, in order that four conserved quantities (including the total energy) can be obtained: as a matter of fact, the equations of the model are not

and cyclic variable does not mean conserved quantity. Besides that, even if is written in terms of and, is not a cyclical coordinate, being implicitily in such variables. For this reason we question the validity of (Equation (14) in [7] ), which would imply an integral invariant.

With regards to the same subject, we emphasize that equation (Equation (16) in [7] ), giving rise to the conservation of, is not correct in our advice, since U does not refer to a lagrangian coordinate.

We perform now a brief analysis of (7). It is evident that the first four equations in (7) form the sub-system

The last three equations in (7) give simply, , once (12) has been solved.

Statement 1.1 System (7) admits locally one solution, for any set of data, , , , , ,.

Proof. The assigned data provide and, by means of (6). Furthermore

where we defined (see also (8) and (9))

Therefore (12) can be solved in an appropriate time interval. Finally one obtains, e from the last three equations in (7) and the rest of the given data. □

Remark 1.1 If in (7) we let, we are dealing with the simpler case of a bar on a horizontal plane with one point moving at each time along the direction of the bar: since equations reduce to

The energy conservation gives the orbits on the phase plane, namely each point of the -axis and the semi-ellipses. Moreover, , that is,.

Again for, the special case, that concerns with one typical instance in nonholonomic constraints (see for instance [1] ), cannot be recovered from the system of Remark 1.2, but it requires the definition of the pseudovelocity.

We are going now to investigate the stability of the system at.

For what concerns with the initial data, we can certainly assume with no loss in generality, , , hence.

Let us first check whenever is a solution of (7).

Statement 1.2 is solution of (7) if and only if U is constant and.

Proof. Set in (7), first three equations:

which entails,. We incidentally notice that if then U, V consistent with are those we discussed in Remark 1.3.

On the other hand, U constant and make us write (12), second and third equation, as

(it is physically correct to assume), which implies. □

Remark 1.2 The following statement also follows from the previous analysis: if the angle is constant (that is never changes direction), then U has to be constant and has to be zero.

Corollary 1.1 Assume,; then is solution of (7) if and only if.

Proof. If is solution, then must be zero at any time; on the other hand, the set of data, , , give univocally the solution, ,. □

Our aim points now to discuss the stability of. Thre physical problem requires (see (5)). Incidentally we notice that if is the solution of (7) starting from, then is the solution of the same equations corresponding to.

It may be helpful by the way to set in (7) in order to figure out the behaviour of the solution for short times:

that is, assuming for istance, , U and W are initially increasing, V decreasing.

Let us now show the following Proposition 1.1 The equilibrium point of (12) is unstable.

Proof. We set (12) in normal form: calling, where is computed in (13), it is easily found

so that can be calculated (see (12) for).

We now compute the Jacobian matrix of at the equilibrium: calculations lead to

where. The eigenvalues are found by solving

Since, the polynomial in brackets is such that and, so that there exists one real and positive eigenvalue and standard results in this sense (see e.g. [8] ) can be implemented. □

The linear approximation entails and

which give the equation for:

whose solution contains,. As to, the linear approximation gives

which diverges the same.

An analytical investigation can be performed directly for system (7): choosing for istance, as it is natural, , and, (16) shows that initially increases, so that P₀ enters the quarter. Furthermore, setting, one writes (7) as

with appropriate,; on the other hand, using also

which is (11) with the appropriate G, one should acquire information about the maintenance of (does not change verse), of (the transverse velocity of O is opposite to the position of with respect to the vertical direction) and of (absence of inversion with respect to). Such as analysis will be not expanded, in order not to overload the Section.

Remark 1.3 We have not imposed any constraint on the velocity of A yet: the velocity can be calculated a posteriori by means of (5) and the angle between and is such that

where.

Remark 1.4 It is sometimes assumed in literature (more or less expressly) to know U e V: in that case system (7) is obviously simpler, but such an assumption corresponds to impose the constraints, , with given U and V. Hence must appear on the right-hand side of (3), with, unknown multipliers. As a whole, we get seven equations in the seven unknown quantities, , and.

Following the approach in [6] , we add an external force in order to modify the dynamics of the system and to infer the stability of the stationary solution.

We impose a force in, where (horizontal versor perpendicular to) and in; we expect, the same for the other coefficients. Notice that a force along in would have no effects.

Computing the Lagrangian components of the vector of forces in the tangent space and taking the projection (see (6)), one can check that the term to add to the right-hand side of (7), first three equations, is

The conclusion of Statement 1.1 about existence and uniqueness is not altered, since the matrix A of (12) is still the same.

Let us investigate about the effect of stabilization by the external device in the case of the force in A only: (actually the overlap of does not change the substance). Moreover, has to vanish at the equilibrium:

It can be easily seen that the characteristic polynomial (17) changes into

where the partial derivatives are calculated in. The following Proposition sets a selection of choices for f.

Proposition 1.2 (o) If then the system is unstable.

1) for:

a) if or then the system is unstableb) if and and [resp.], then the system is unstable [resp. stable].

2) For:

c) if or, then the system is unstabled) if and, then the (real or complex) eigenvalues different from zero have negative real part.

We now consider a rigid device simulating the front wheel, adding to the body a rigid part (say the front wheel together with handlebars) hung in A and forming the angle (front steering) between the direction and a direction fixed in the body frame:

For the sake of simplicity, we may imagine as a rigid bar laying on, with no active force operating on it. We now consider the five lagrangian coordinates. The angular velocity of is hence.

The Lagrangian function of the whole system is, where is the same as (1) and

where is a new lagrangian coordinate and and are respectively the mass of and the central inertial momentum of with respect to the direction. The equation of motion with respect to is simply

Equation (24) gives

where is the angle between and.

Remark 2.1 If no further constraints are enjoined, the system is unstable the same: actually, changements in (7) are not essential: still keeping, is again an equilibrium point for the system, where

and (13), (17) are replaced respectively by

which has in the same way one real positive solution.

We now add the kinetic constraint of no skidding of at: which gives

A possible way to face the problem is to neglect the mass of the anterior part, so that the Lagrangian function is the same as (1). However, a complication is, in our point of view, the role of, which does not appear in, but only in the constraint (25).

This is a nontrivial point for the theory-building of the correct equations of motion: the way we are going to follow is not to neglect the mass and consider as the Lagrangian function. Even more, if we think of the problem as a “bicycle'” model, the front mass is not at all unsignificant for the overall frame.

We now consider the set of Lagrangian coordinates: system (3) of the first kind Lagrangian equations is now replaced by the seven equations

where, , , , , and are the seven unknown quantities. Concening with the initial conditions for (26) we can choose, with no loss in generality:

Let us change for the sake of convenience into the variable

so that. The velocities are not independent, because of (2), (25): if on the one hand and are arbitrary, on the other hand once has been fixed the initial velocities

are imposed. In order to reduce (26) and eliminate the multipliers, we define, similarly to (4), the pseudovelocities

Joining (29) with (2) and (25), the lagrangian velocities are written in terms of the parameters:

where and, we recall,.

The equations of motion replacing (7) are now, together with (2), (25) and (29): straightforward computations lead to

where

and, as before, , , , ,.

Even in this case the equations of motion can be led back to the cardinal equations: indeed, calling the rigid part containing A, B and (30), the second cardinal equation of the whole system using B for calculating the momenta and projecting along writes

where is the momentum of the external forces of the whole system. Since the constraints are smooth, the force in realizing the kinetic constraint (25) can be modelled as

so that (see also (10)). On the other hand, the first cardinal equation for the whole system along is Carring out all the computations, here omitted, one gets exactly (33). The second equation in (30) is again ascribable to the momentum balance of the system along, similarly to what discussed in Remark 1.1:

Finally, the fourth equation in (30), namely (24), is simply the second cardinal equation written only for and with respect to the point, where all the momenta of the external forces vanish.

As we already remarked in Section 1, the overview of the system in the frame of the cardinal equations does not determine any conserved quantity: the only evident one is the energy conservation

System (30), (31) consists of eight ODEs for the eight variables, , , , , , ,. The five equations (30) form a sub-system for together with the initial conditions

while the constant value Y is deduced from (24), (27) and (28)):

Once (30) has been solved, (31) and (28) allow to solve, ,.

As in the case of Section 1, we incidentally remark that, if and Y as in (35) is solution of (30), then, is solution of the same system, as we expect.

Statement 2.1 For any set of data (34), (35) system (30) admits one solution.

Proof. System (30) can be concisely written as

with and

the normal form being with

and

It is evident that the unique solution of (30) corresponding to the initial data, , , is and W, identically zero. We notice that no other solutions such that the plane is vertical are possible:

Statement 2.2 If, then, ,. Conversely, if, and, , , then.

Proof. Set in (30) and call:

The first two equations give; on the other hand, eliminating from second and third equations yields, hence and. But is consistent with the second equation only if and the third equation leads to.

Conversely, if, , are replaced in (30), one gets that, together with the initial conditions, gives. □

Still concerning with the initial data assignment, we notice that if (which is a reasonable condition for) we get from (35): we wonder whether solutions such that (meaning and constant) are possible. From (30), fourth equation, one gets,. If we clearly have, and for. If on the other hand, by replacing and in the first two equations of (30) we achieve the first integrals

so that only, can be a solution. Substituting in the first integrals we see that, , that is the stationary solution.

We are going to investigate the stability of the stationary solution.

Proposition 2.1 The equilibrium point, of (36) is unstable.

Proof. The Jacobian matrix at the equilibrium is

whose eigenvalues are, ,. The linearized system

gives

with, ,. □

We remark that, as bigger is as longer is the time when the planar body falls to the ground.