6
$\begingroup$

The coefficient of restitution characterizes a collision in one dimension by relating the initial and final speeds of the particles involved,

$$C_R = -\frac{v_{2f} - v_{1f}}{v_{2i} - v_{1i}}$$

In a 2D collision, the velocity can be split into components parallel and perpendicular to the "plane of collision" (the plane tangent to the two objects' surfaces at their contact point), and the equation above applies to the perpendicular components of the velocities only. One could write

$$\begin{pmatrix}v_{2f\shortparallel} - v_{1f\shortparallel} \\ v_{2f\perp} - v_{1f\perp}\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & -C_R\end{pmatrix}\begin{pmatrix}v_{2i\shortparallel} - v_{1i\shortparallel} \\ v_{2i\perp} - v_{1i\perp}\end{pmatrix}$$

My question is, is it useful (i.e. does it produce a more accurate description of realistic 2D collisions) to generalize that matrix? Perhaps by allowing the top left element to be unequal to 1, or allowing nonzero off-diagonal elements? Or is there some nonlinear relation that works better?

As usual, if you know of any relevant published research, references would be much appreciated.

$\endgroup$

2 Answers 2

4
$\begingroup$

To be simple, consider an object hitting a heavy plane. If there is coefficient of friction $C_F$ (and restitution $C_R$) between the bodies, parallel velocity is modified as well: $\Delta p_{\parallel} = F_{\parallel} \Delta t = -C_F F_{\perp} \Delta t = -C_F \Delta p_{\perp} = -C_F(1+C_R) p_{\perp}$

Thus obtaining $$ \left[ \begin{array}{c} v_{\parallel}^{out} \\ v_{\perp}^{out} \end{array} \right] = \underbrace{ \left[ \begin{array}{cc} 1 & - C_F(1+C_R) \\ 0 & - C_R \end{array} \right]}_{A} \left[ \begin{array}{c} v_{\parallel}^{in} \\ v_{\perp}^{in} \end{array} \right]. $$

Note that it works as long as $v_{\parallel}^{in}\geq C_F(1+C_R) v_{\perp}^{in}$. If there were a general (working for all input velocities) matrix $A$, it should allow dissipation, but not creation, of energy i.e. $||A||_2\leq 1$ (the condition is not fulfilled in the above example). I means that either the matrix need to have other entries as well or that the problem intrinsically needs restriction of the initial conditions.

However, bear in mind coefficient of restitution is only a effective parameter (as it has been already mentioned by jalexiou).

Anyway, it may be a nice experimental problem for an undergraduate student to find the all coefficients (and check for which conditions they work). I am curious of the results.

$\endgroup$
3
  • 1
    $\begingroup$ Thanks, I figured it might be something like that. It would indeed be great to have some experimental data about how well this works, although I guess that would be a rather tedious experiment so I'm not too optimistic about finding someone to do it ;-) $\endgroup$
    – David Z
    Nov 21, 2010 at 7:35
  • $\begingroup$ The required $\frac{v_{\|}^{in}}{v_{\perp}^{in}} \geq C_F (1 + C_R)$ is a fairly large region. For example, if $C_R=1$ and $C_F = \frac{1}{2}$, then we must have $v_{\|}^{in} \geq v_{\perp}^{in}$. We also run into trouble for direct collisions (where $v_\|^{in}=0$). Of course, this doesn't really pose any problems in practice. We just set $v_\|^{out}=0$ when the inequality is not satisfied. Interestingly, we can make $A$ diagonal with elements $[C_F,-C_R]$ if we set the tangential friction to be the dry-viscous hybrid $F_\| = \frac{v_{\|}^{in} (C_F-1) }{v_{\perp}^{in} (1 + C_R)} F_\perp$. $\endgroup$ Jul 29, 2017 at 6:21
  • $\begingroup$ Also, it is not too hard to generalize to rotating bodies. The result is $v_\|^{out} = v_\|^{in} - v_\perp^{in} (1 + C_R) \frac{C_F(M + \Omega_\|) + \Omega}{M + \Omega_\perp + C_F \Omega}$ where $M = \frac{1}{m_1} + \frac{1}{m_2}$ is the the inverse reduced mass, and the angular terms $\Omega_{\,\bullet} = \frac{r_{1\bullet}^2}{I_1} + \frac{r_{2\bullet}^2}{I_2}$ and $\Omega = \frac{r_{1\perp}r_{1\|}}{I_1} + \frac{r_{2\perp}r_{2\|}}{I_2}$ are inversely proportional to the moments of inertia $I$ (a scalar in 2D). Note the equivalence in the irrotational limit $I \rightarrow \infty$. $\endgroup$ Jul 29, 2017 at 6:22
1
$\begingroup$

What happens during a true collision is that the contact forces are split into normal and tangential components. The normal forces arise from the elastic compression and expansion of the material near the surface, and the tangential from friction, stiction or viscocity in the slip direction. Those are two different completely effects and trying to group them together into one modeling construct will result in loss of detail. It is an gross approximation to just say there is % loss of energy in the normal direction due to hysteresis losses on the contact deflection, that lumping that with an artificial % loss of energy in the tangential direction due to friction is going to result in completely unrealistic results in general.

I recommend spliting the problem up in normal and tangential components and handling them each with their own modeling techniques.

Good luck.

$\endgroup$
1
  • $\begingroup$ Yes, of course I know it's much more complicated than that in reality ;-) I'm just wondering if there has been any work done on a slightly less crude approximation than the CoR. e.g. if you know about any of those modeling techniques that could be used to analyze the normal and/or tangential components of the collision, I might find that information helpful. $\endgroup$
    – David Z
    Nov 20, 2010 at 5:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.