# Dynamics and impact in a system with unilateral constraints the relevance of dry friction

Anna Sinopoli

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Anna Sinopoli. Dynamics and impact in a system with unilateral constraints the relevance of dry friction. Meccanica, 1987, 22 (4), pp.210 - 215. 10.1007/BF01573814. hal-01403604

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# DYNAMICS AND IMPACT IN A SYSTEM WITH UNILATERAL CONSTRAINTS THE RELEVANCE OF DRY FRICTION

Anna Sinopoli

Dipartimento di Scienza e Tecnica del Restauro - Istituto Universitario di Architettura - Venezia

SOMMARIO Scopo del presente lavoro è quello di indagare sulla dinamica e sull'urto di un sistema rigido, soggetto a vincoli unilaterali. Tale problema è particolarmente interessante per la comprensione del comportamento meccanico delle strutture monumentali costituite da blocchi lapidei, in presenza di forze esterne variabili. L'analisi fornisce informazioni sulla evoluzione dinamica del sistema; tali informazioni sono indispensabili per uno studio sulla stabilità.

SUMMARY. The aim of the present work is to investigate the impact and the dynamics of a rigid system, subject to unilateral constraints. The problem is particularly interesting to understand the mechanical behaviour of stone structures subject to external forces and to give informations about the dynamical evolution; such informations are basic for a stability study.

## INTRODUCTION

In the last years particular attention has been devoted to problems with unilateral constraints.

At the same time, studies on masonry structures have also received many contributions. Experimental and numerical researches have been undertaken to understand the mechanical behaviour; moreover, analytical studies have been developed to define constitutive laws.

The problem mentioned above are strictly connected; in fact, it is well known that masonry structures are characterized by a very weak (or null) tensile strength, due to the presence and the degradation of the mortar.

In particular, stone monumental structures can be modeled as rigid-labile systems with unilateral constraints.

The possibility to define a stability degree for such systems, in the case of monodimensional structures (arches, portals, etc.) with one degree of freedom has been investigated in former works [1], [2], [3]. The most interesting result is that, due to unilateral constraints, the equilibrium configuration is stable, so that it is possible to define a stability degree. Further, the stability definition is strictly connected to the possibility of stable oscillations around the equilibrium configuration, particularly in presence of external forces; this is the case of earthquakes. Therefore, a dynamical analysis of stone structures, under the assumptions above, is mandatory. The oscillations are expected to have a peculiar character, because they are possible by means of inversion of the mechanisms. Then the problem is non linear, due to the

fact that the inversion is characterized by an impact.

The free dynamics of particular systems, both in the case of small and large amplitude oscillations has been studied [1], [2], [4], under the restrictive assumption that such systems are conservative.

The amount of energy dissipation during the impulsive phenomenon is, to the author opinion, an open problem.

The dynamical analysis of a column has been undertaken in [5], with a numerical approach, but an «a priori» assumption about the motion after the impact has been imposed.

Such assumption is the same made by Housner [6]. Its fundamental work analyses both the free and forced motion only for very slender stone structures, the behaviour of which is assumed similar to the inverted pendulum.

In this work, the dynamics and the impact of a stone element are investigated, assuming as fundamental parameter the dimensional ratio, which is allowed to vary in a very large range. As a consequence, even if the analytical expression for the solution of the free motion is the same of Housner, the difference from Housner periods increases with the dimensional ratio.

The large range of variation of the dimensional ratio permits to define the relevance of dry friction and to identify the only possible motion after the impact.

Both the dynamics and the impact are studied for two extreme cases:

a) absence of dry friction;

b) dry friction large enough to prevent sliding.

Such cases are useful to define the boundary of the behaviour for all the possible values of dry friction.

## 1. DYNAMICS WITHOUT FRICTION

Referring to Fig. 1, let us assume that:

a) the system is constituted by a rigid bloc, simply supported by a rigid ground;

b) *h*, *b* are its dimensions; *h* is the height and *b* the base length (the base are is *b* × 1);

c) (*x*, *y*) is an inertial reference system with its origin initially coincident with point *A*.

Let us consider the free dynamics of this system, under the assumption that friction is negligible. Moreover, initially, let *A* be the only point in contact with the ground. This corresponds to consider one of the two possible mechanisms of the system.

Let $x_G$, $y_G$ and $\theta$ be the lagrangian coordinates (respectively the two center of mass translations and the rotation angle), then the equazions of motion are:
<figure><img src="image_2.png" /><figcaption>Fig. 1</figcaption></figure>

$$
\begin{cases}
m\ddot{x}_G = 0 \\
m\ddot{y}_G = \lambda - mg \\
I_G \ddot{\theta} = -\lambda \left(\frac{b}{2} \cos \theta - \frac{h}{2} \sin \theta\right)
\end{cases}
\quad (1)
$$

where *m* is the mass of the system, *λ* is the constraint reac-
tion, *g* the gravity acceleration and *I*<sub>G</sub> the baricentric inertia
moment.

In order to solve system (1), the constraint equation
due to ground impenetrability must also be imposed; this
is a unilateral constraint:

$$
d y_A \geq 0 \tag{2}
$$

If we assume that, during the motion, point A never gets
off the ground, eq. (2) can be written as:

$$
d y_G + \left( \frac{h}{2} \sin \theta - \frac{b}{2} \cos \theta \right) d \theta = 0 \quad (3)
$$

From eqs. (2) and (3) the dynamical system becomes:

$$
\begin{cases}
\ddot{x}_G = 0 \\
\ddot{y}_G = \left(\frac{b}{2} \cos \theta - \frac{h}{2} \sin \theta\right) \ddot{\theta} - \left(\frac{b}{2} \sin \theta + \frac{h}{2} \cos \theta\right) \dot{\theta}^2 \\
\lambda = m \ddot{y}_G + mg \\
I_G \dot{\theta} + \frac{m}{4} (b^2 \cos^2 \theta + h^2 \sin^2 \theta - 2bh \sin \theta \cos \theta) \ddot{\theta} + \\
\qquad + \frac{m}{4} [bh \sin^2 \theta - bh \cos^2 \theta + (h^2 - b^2) \sin \theta \cos \theta] \dot{\theta}^2 - \\
\qquad - \frac{m}{2} g (h \sin \theta - b \cos \theta) = 0.
\end{cases}
\quad (4)
$$

It is noted that system (4) has only two degrees of freedom,
corresponding to the lagrangian coordinates x_G and θ; further,
the pertinent equations are uncoupled.

The horizontal motion of the center of mass is characteri-
zed by:

$$
\dot{x}_G(t) = \text{const.} \tag{5}
$$

Regarding to rotation, if we assume that θ is small enough,
so that sen θ ≃ θ and cos θ ≃ 1, the corresponding equation
can be written as:

$$
\begin{equation}
\begin{split}
& \left( I_G + m \frac{b^2}{4} - m \frac{bh}{2} \theta \right) \ddot{\theta} + \\
& \qquad + m \left( \frac{h^2 - b^2}{4} \theta - \frac{bh}{4} \right) \dot{\theta}^2 - mg \frac{h}{2} \theta = -mg \frac{b}{2}
\end{split}
\tag{6}
\end{equation}
$$

Let use note that in eq. (6), in spite of the small amplitude
assumption, a term proportional to ˙θ² is present. In this
respect some observations both of mathematical and physical
character must be made. We are concerned to study the small
amplitude motion around the equilibrium configuration
corresponding to θ = 0.

This configuration is stable only because of ground impe-
netrability; such constraint, introduced when the bloc
reaches the ground, cannot be taken into account by eq. (6),
but it must be imposed, by means of another relationship. In
fact, if we consider solutions with the form: θ(t) = const, eq.
(4) gives the two equilibrium configurations: θ = π arctg b/h,
which are, respectively, stable and unstable.

Regarding to motion near θ = 0, it is far from natural
equilibrium configurations of the system and in particular, it
can be considered as a large amplitude motion around the
unstable configuration. As a consequence, the angular veloci-
ty can assume finite values, so that ˙θ² cannot be neglected. In
proof of this argument, it is worth to remark that the poten-
tial energy U(θ), for θ = 0, exhibits a boundary minimum
characterized by $\frac{\partial U}{\partial \theta} \neq 0$.

Eq. (6) is a non linear equation, with variable coefficients.

In order to find a solution which takes into account infini-
tesimal values of θ and finite values of ˙θ let us consider a
solution with the form:

$$
\theta(t) = \theta_{0}^{*} + \theta_{0}(t) \tag{7}
$$

that is, the angle θ is obtained as the difference between
two finite terms, where θ₀* is the unstable equilibrium con-
figuration corresponding to: θ = arctg b/h.

Substituting (7) in (6), and neglecting the first order
terms, eq. (6) becomes:

$$
\left( I_G + m \frac{b^2}{4} \right) \ddot{\theta}_0 - \frac{bh}{4} \dot{\theta}_0^2 = -mg \frac{b}{2} \quad (8)
$$

Eq. (8) can be easily transformed into a first order diffe-
rential equation, with the substitution ẛ₀ = y, so that:

$$
\theta_0(t) = - \frac{4b^2 + h^2}{3bh} \ln \cosh \left[ \frac{3b}{4b^2 + h^2} (2gh)^{\frac{1}{2}} t + C_1 \right] + C_2 \quad (9)
$$

The constants C₁ and C₂ depend on the initial conditions.

If $\dot{x}_G(0) = 0$, the motion of the center of mass is along
the vertical. At instant $t_1$, when $\theta(t_1) = 0$, the instantaneous
center of rotation is, in a reference system fixed with respect
to the bloc, coincident with the point the coordinates of
which are $(0, \frac{h}{2})$.
The motion, at the same instant, is characterized by:

$$
\left\{
\begin{aligned}
\dot{x}_G(t_1) &= 0 \\
\dot{y}_G(t_1) &= -\frac{b}{2} \dot{\theta}(t_1) \\
\dot{\theta}(t_1) &= -\left(\frac{2g}{h}\right)^{\frac{1}{2}} \tanh \left[ \frac{3b}{4b^2+h^2} (2gh)^{\frac{1}{2}} t_1 + C_1 \right].
\end{aligned}
\right.
\quad (10)
$$

For $t > t_1$, the motion is undetermined; it will depend on
the impact between the rigid bloc and the ground. This
aspect will be investigated in the following.

2. DYNAMICS WITH DRY FRICTION

Now let us assume that dry friction is present. Moreover
the amount of friction is large enough to prevent sliding of
point A in Fig. 1. Then the dynamical equations are:

$$
\left\{
\begin{aligned}
m \ddot{x}_G &= \lambda_1 \\
m \ddot{y}_G &= \lambda_2 - mg \\
I_G \ddot{\theta} &= \lambda_1 \left(\frac{b}{2} \sin \theta + \frac{h}{2} \cos \theta\right) + \lambda_2 \left(\frac{h}{2} \sin \theta - \frac{b}{2} \cos \theta\right)
\end{aligned}
\right.
\quad (11)
$$

with the constraints equations:

$$
\left\{
\begin{aligned}
dx_G + \left(\frac{b}{2} \sin \theta + \frac{h}{2} \cos \theta\right) d\theta &= 0 \\
dy_G + \left(\frac{h}{2} \sin \theta - \frac{b}{2} \cos \theta\right) d\theta &= 0
\end{aligned}
\right.
\qquad (12)
$$

where $\lambda_1$ and $\lambda_2$ represent, respectively, the horizontal and vertical constraint reactions.

The motion, in this case, is a rotations around the fixed
point A. Then, let us assume the rotation angle $\theta$, as lagran-
gian coordinate. The corresponding dynamical equation is:

$$
I_A \ddot{\theta} - \frac{mg}{2} (h \sin \theta - b \cos \theta) = 0 \quad (13)
$$

where $I_A$ is the inertia moment with respect to A.

With the assumption of small values of θ, eq. (13), on the
contrary of eq. (6), does not contain θ̇ terms, so that it can
be solved and the solution of the first order approximation
is:

$$
\theta(t) = \frac{b}{h} + C_1 \cosh(\alpha t + C_2) \quad (14)
$$

where:

$$
\alpha = \left[ \frac{3gh}{2(b^2 + h^2)} \right]^{\frac{1}{2}} .
$$

It is particularly interesting to calculate $\lambda_1/\lambda_2$, that is the
ratio between the horizontal and the vertical reaction. Such
ratio, in fact provides information about our assumption of
friction large enough to prevent sliding of point A.

From eq. (11), (12) and (13), it can be found, at the zero

order approximation, that:

$$
\begin{align}
\lambda_1 &= mg \left( \frac{3}{4} \frac{bh}{h^2 + b^2} - \frac{b \dot{\theta}^2}{2g} \right) \notag \\
\lambda_2 &= mg \left( \frac{b^2 + 4h^2}{4(h^2 + b^2)} - \frac{h \dot{\theta}^2}{2g} \right). \tag{15}
\end{align}
$$

If $f_a$ represents the limit value of the dry friction coefficient, during the motion, point A does not slide only if:

$$
\frac{\lambda_1(t)}{\lambda_2(t)} \le f_a. \quad (16)
$$

Neglecting first order terms, eq. (16) corresponds to:

$$
(b - h f_a) \dot{\theta}^2(t) \ge \frac{g}{2(h^2 + b^2)} [3bh - f_a(b^2 + 4h^2)] \quad (17)
$$

Inequality (17) has been investigated by means of para-
metric analysis on $\frac{b}{h}$ and $f_a$. The results are reported in fig. 2,
where the lines $f_a = 0,75$ and $f_a = \frac{b}{h}$, useful for algebraic
considerations, have been drawn. It can be said that, in region
for which $\frac{b}{h} \ge f_a$ and $f_a > 0,75$, enequality (17) is verified
for any initial conditions and at any instant.

On the contrary, for $\frac{b}{h} \ge f_a$ and $f_A \le 0,75$, (17) is satisfied
only for particular values of the ratio $\frac{b}{h}$.

In tab. 1 such values are reported in correspondence to
different values of f<sub>a</sub>.

<table><caption>Table 1.</caption>
  <thead>
    <tr>
      <th>f<sub>a</sub></th>
      <th>range of values for <sup>b</sup>&frasl;<sub>h</sub></th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>0.1</td>
      <td><sup>b</sup>&frasl;<sub>h</sub> &le; 0.13 and <sup>b</sup>&frasl;<sub>h</sub> &ge; 29.9</td>
    </tr>
    <tr>
      <td>0.2</td>
      <td><sup>b</sup>&frasl;<sub>h</sub> &le; 9.27 and <sup>b</sup>&frasl;<sub>h</sub> &ge; 14.7</td>
    </tr>
    <tr>
      <td>0.3</td>
      <td><sup>b</sup>&frasl;<sub>h</sub> &le; 0.42 and <sup>h</sup>&frasl;<sub>b</sub> &ge; 9.58</td>
    </tr>
    <tr>
      <td>0.4</td>
      <td><sup>b</sup>&frasl;<sub>h</sub> &le; 0.57 and <sup>h</sup>&frasl;<sub>b</sub> &ge; 6.92</td>
    </tr>
    <tr>
      <td>0.5</td>
      <td><sup>b</sup>&frasl;<sub>h</sub> &le; 0.76 and <sup>h</sup>&frasl;<sub>b</sub> &ge; 5.23</td>
    </tr>
    <tr>
      <td>0.6</td>
      <td><sup>b</sup>&frasl;<sub>h</sub> &le; 1.00 and <sup>h</sup>&frasl;<sub>b</sub> &ge; 4</td>
    </tr>
    <tr>
      <td>0.65</td>
      <td><sup>b</sup>&frasl;<sub>h</sub> &le; 1.16 and <sup>h</sup>&frasl;<sub>b</sub> &ge; 3.46</td>
    </tr>
    <tr>
      <td>0.70</td>
      <td><sup>b</sup>&frasl;<sub>h</sub> &le; 1.37 and <sup>h</sup>&frasl;<sub>b</sub> &ge; 2.91</td>
    </tr>
    <tr>
      <td>0.75</td>
      <td><sup>b</sup>&frasl;<sub>h</sub> = 2</td>
    </tr>
  </tbody>
</table>

For $f_a < 0.75$ and $\frac{b}{h} > f_a$.

In order to verify inequality (17) in the region where
$\frac{b}{h} < f_a$, initial conditions of the motion must be specified.
It we consider ones corresponding to $\theta(0) = \theta_0$ and $\dot{\theta}(0) = 0$,
no contraints must be imposed to verify (17).
The region where the friction is not sufficient to prevent sliding has been dashed in Fig. 2.

<figure><img src="image_3.png" /><figcaption>Fig. 2.</figcaption></figure>

### 3. THE IMPACT PROBLEM

The problem of the impact, which occurs when the rotation angle is zero, is difficult to solve. In fact, using only the fundamental equations of the impulsive phenomenon, the motion after the impact is undetermined, due to the unilaterality of the constraints and to the finite extension of the surface involved.

Moreover, the experimental law of Newton cannot be used, because the impact is not central and direct.

An alternative approach to solve the problem is the one which refers to the symbolic equation of impulsive motion and uses some principle of minimum. The symbolic inequality of impulsive motion for discrete systems with unilateral constraints is:

$$ \sum_{i=1}^{n} (I_i - m_i \Delta v_i) \times \delta P_i \le 0 \quad (18) $$

where $I_i$ is the external impulse applied to the i-th mass point, $\Delta v_i$ is the corresponding velocity variation and $\delta P_i$ the infinitesimal displacement congruent with the constraints.

For our system, the symbolic inequality, when the bloc has reached the ground, imposes a restriction on the velocities distribution.

But, in the same time, it allows an infinite number of motions after the impact, congruent with the unilateral constraint.

The determination of the unique natural motion is possible by using a varied form of the Gauss principle of least performance, furnished by the constraint reactions.

This varied form due to Robin for systems with bilateral constraints, states that the unknown motion after an impact is the one which minimizes the $G$ function between any motion congruent with the constraint system.

In the case of null external impulse such $G$ function is the kinetic energy due to the sudden variations of the velocities.

The validity of the Robin principle for systems with unilateral constraints has been verified by Mayer [9], so that it is used to study the impact in both the cases with and without dry friction.

#### 3.1. The impact without friction

Let us consider again the system of Fig. 1. At the instant $t_1$, when $\theta(t_1) = 0$, the absolute instantaneous center of rotation is coincident with point $H$ (Fig. 3).

Then, the motion before the impact is characterized by:

$$ \begin{cases} \dot{x}_G(t_1) = \text{const.} \\ \dot{y}_G(t_1) = \frac{b}{2} \dot{\theta}^{-} \\ \dot{\theta}^{-}(t_1) < 0. \end{cases} \quad (19) $$

The unilateral constraint introduced by the impact is:

$$ \delta y_p \ge 0 \quad (20) $$

where $P$ is any point of the bloc.

Eq. (20) can be written as:

$$ \delta y_G + (x_P - x_G) \delta \theta \ge 0 \quad 0 \le x_P \le b \quad (21) $$

that is:

$$ \delta y_G \ge (x_G - x_P) \delta \theta \quad 0 \le x_P \le b \quad (22) $$

Eq. (22), as function of the velocities implies that the permitted motion are only those for which:

$$ \dot{y}_G^+ \ge (x_G - x_P) \dot{\theta}^+ \quad 0 \le x_P \le b \quad (23) $$

A consequence of inequality (23) is:

$$ \dot{y}_G^+ \ge \frac{b}{2} |\dot{\theta}^+| \quad (24) $$

because $x_P$ can assume any value between zero and $b$. In virtue of the Robin principle, the natural motion is the one which minimizes the kinetic energy, $G$, due to the sudden

<figure><img src="image_4.png" /><figcaption>Fig. 3.</figcaption></figure>
velocities variations. In our case the expression of G is:

$$
G = \frac{1}{2} m [(\dot{x}_G^+ - \dot{x}_G^-)^2 + (\dot{y}_G^+ - \dot{y}_G^-)^2] + \frac{1}{2} I_G (\dot{\theta}^+ - \dot{\theta}^-)^2 \quad (25)
$$

of which we must find the minimum value according to (24).

Regarding to eq. (25) we can note that:

a) $\dot{x}_G$ is an independent parameter of the motion;

b) from (24), the minimum value for G is the one cor-
responding to $\dot{y}_G^+ = \frac{b}{2} |\dot{\theta}^+|$, because $\dot{\theta}^- < 0$ so that $-\dot{y}_G^- > 0$
and $\dot{y}_G^+ \ge 0$.

Then, eq. (25) can be written as:

$$
G = \frac{1}{2} m \left[ (\dot{x}_G^+ - \dot{x}_G^-)^2 + \left( |\dot{\theta}^+| - |\dot{\theta}^-| \right)^2 \frac{b^2}{4} \right] + \\
\qquad + \frac{1}{2} I_G (\dot{\theta}^+ - \dot{\theta}^-)^2. \tag{26}
$$

The minimum of G can be found requiring that:

$$
\left\{
\begin{array}{l}
\displaystyle \frac{\partial G}{\partial \dot{x}_G^+} = 0 \\[1em]
\displaystyle \frac{\partial G}{\partial \dot{\theta}^+} = 0.
\end{array}
\right.
\qquad (27)
$$

The first of eqs. (27) corresponds to:

$$
\dot{x}_G^+ = \dot{x}_G^- \tag{28}
$$

so that the horizontal motions before and after the impact
are the same. On the contrary, the presence of an absolute
value for ˙θ⁺, in the second of eqs. (27), requires that two
cases must be considered, respectively, for ˙θ⁺ < 0 and
˙θ⁺ ≥ 0.

The consequent conditions are:

$$
\left( I_G + m \frac{b^2}{4} \right) \dot{\theta}^+ = \left( I_G - m \frac{b^2}{4} \right) \dot{\theta}^- \quad \text{for } \dot{\theta}^+ < 0 \qquad (29)
$$

and

$$
\dot{\theta}^{+} = \dot{\theta}^{-} \qquad \text{for } \dot{\theta}^{+} \ge 0 \qquad (30)
$$

Eq. (29) implies that:

$$
\dot{\theta}^{+} = \frac{h^2 - 2b^2}{4b^2 + h^2} \dot{\theta}^{-} \qquad (31)
$$

where for $I_G$ has been assumed the expression:
$I_G = \frac{1}{12} m (b^2 + h^2)$.

Because both $\dot{\theta}^+$ and $\dot{\theta}^-$ are negative, eq. (31) is valid only
if $\frac{h}{b} > \sqrt{2}$. In this case, the motion after the impact is a
rototranslation charaterized by an angular rate reduced
in value but with the same sign.

Regarding to eq. (30), we can observe that is implies
an absurdity; in fact, if $\dot{\theta}^+ \ge 0$, it is impossible that
$\dot{\theta}^+ = \dot{\theta}^- < 0$. Nevertheless, some considerations can be made.
The G function does not have a minimum characterized
by $\frac{dG}{d\dot{\theta}^+} = 0$, but because it is an increasing function of
$\dot{\theta}^+$ (for $\dot{\theta}^+ \ge 0$) its minimum is a boundary minimum cor-

responding to $\dot{\theta}^+ = 0$.

Then, for $\frac{h}{b} \le \sqrt{2}$, the motion after the impact can be
a pure translation; its occurence depends on the value of
$\dot{x}_G^+ = \dot{x}_G^-$. In fact, if $\dot{x}_G^- = \dot{x}_G(0) = 0$, after the impact the
system is at rest.

It is interesting to note that, if $\dot{x}_G(0) = 0$, for any value of $\frac{h}{b}$, the kinetic energy depends only on the angular rate $\dot{\theta}$, so that the amount of dissipated energy can be evaluated. With reference to the ratio $e$ between the values of kinetic energy after and before the impact, it can be found that:

$$
e = \frac{\dot{\theta}^{+2}}{\dot{\theta}^{-2}} = \left[ \frac{h^2 - 2b^2}{4b^2 + h^2} \right]^2 \quad \text{for} \quad \frac{h}{b} > \sqrt{2}
$$

$$
e = 0 \qquad \text{for } \frac{h}{b} \le \sqrt{2}
$$

The diagram of ratio *e* is reported in Fig. 4 as function
of *h/b*.

3.2. The impact in presence of dry friction

Let us consider the same system in presence of dry friction; let the value of $\frac{b}{h}$ and $f_a$ be the ones (Fig. 2) which permit to point A to act as an hinge.

At instant $t_1$, the motion of the system, before the impact,
is:

$$
\left\{
\begin{array}{l}
\dot{x}_G^- = -\frac{h}{2} \dot{\theta}^- > 0 \\[1em]
\dot{y}_G^- = \frac{b}{2} \dot{\theta}^- < 0 \\[1em]
\dot{\theta}^- < 0.
\end{array}
\right.
\qquad (32)
$$

Now it is espected that eq. (20) is satisfied as an equality
for a particular point P*, belonging to the base AB, for
which:

$$
\delta x_{P*} = 0 \qquad \dot{\theta}^{+} \neq 0 \tag{33}
$$

<figure><img src="image_5.png" /><figcaption>Fig. 4.</figcaption></figure>
Eq. (33), for any point P, is equivalent to:

$$ \dot{x}_G^+ = (y_{P^+} - y_G) \dot{\theta}^+ = -\frac{h}{2} \dot{\theta}^+ \quad (34) $$

Then, from (23) and (24), the Robin function, for $\dot{\theta}^+ \neq 0$, becomes:

$$ G = \frac{1}{2} m \left[ \frac{h^2}{4} (-\dot{\theta}^+ + \dot{\theta}^-)^2 + \frac{b^2}{4} (|\dot{\theta}^+| - |\dot{\theta}^-|^2) + \frac{1}{2} I_G (\dot{\theta}^+ - \dot{\theta}^-)^2 \right] \quad (35) $$

where, even in this case, we have assumed that: $\dot{y}_G^+ = \frac{b}{2} |\dot{\theta}^+|$. The minimum value of G is the one corresponding to:

$$ \dot{\theta}^+ = \frac{2h^2 - b^2}{2(h^2 + b^2)} \dot{\theta}^- \quad \text{for: } \dot{\theta}^+ < 0 \quad (36) $$

and

$$ \dot{\theta}^+ = \dot{\theta}^- < 0 \quad \text{for: } \theta^+ > 0. \quad (37) $$

Regarding eq. (36), if $\frac{h}{b} > \frac{\sqrt{2}}{2}$, the motion is a rotation around point B, with reduced angular velocity. In this case it can be easily shown that the dry friction is sufficient to prevent sliding of point B, also after the impact.

On the contrary, for $\dot{\theta}^+ > 0$, eq. (37) implies as in the previous section that the function cannot have a minimum with $\frac{dG}{d\dot{\theta}} = 0$.

Let us consider again eq. (35); for $\dot{\theta}^+ > 0$, G is an increasing function and it approaches its minimum value when $\dot{\theta}^+ \to 0^+$.

But, for $\dot{\theta}^+ = 0$, the formulation of G becomes:

$$ G = \frac{1}{2} m \left[ (\dot{x}_G^+ - \dot{x}_G^-)^2 + \frac{b^2}{4} (|\dot{\theta}^+| - |\dot{\theta}^-|^2) + \frac{1}{2} I_G (\dot{\theta}^+ - \dot{\theta}^-)^2 \right] \quad (38) $$

In this case, in virtue of the same arguments used above, in order to find the minimum it must be imposed that:

$$ \begin{cases} \dot{x}_G^+ = \dot{x}_G^- \\ \dot{y}_G^+ = 0 \\ \dot{\theta}^+ = 0. \end{cases} $$

Then, for $\frac{h}{b} \le \frac{\sqrt{2}}{2}$ the motion after the impact is a translation.

Even in this case we can evaluate the amount of dissipated energy. For $\frac{h}{b} > \frac{\sqrt{2}}{2}$ the expression of e is:

$$ e = \left(\frac{\dot{\theta}^+}{\dot{\theta}^-}\right)^2 = \left[\frac{2h^2 - b^2}{2(h^2 + b^2)}\right]^2 \quad (39) $$

On the contrary, for $\frac{h}{b} \le \frac{\sqrt{2}}{2}$, it becomes:

$$ e = \frac{m \dot{x}_G^+}{I_A \dot{\theta}^-} = \frac{3}{4} \left[ \frac{h^2}{h^2 + b^2} \right]^2 \quad (40) $$

The diagram of this coefficient as function of $\frac{h}{b}$ can be seen in Fig. 4.

Surprisingly enough, it is worthwile to remark that, with the same initial conditions, the impact dissipates more energy in the case without friction than in the case with friction.

## REFERENCES

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