What is the formula of the force in an partially inelastic collision

Question

How can i find the force at a specific moment in a partially inelastic collision as a function of the distance between both bodies centers (d), the some of their radius (r), the coefficient of restitution (e) and their relative velocities (v)?

Obs: I want to be able to calculate the force during the collision at every moment given information about both bodies only in the very previous moment, not necessarly the inicial moment. I would like to know the energy loss in heat too.

Answer 1

Simple linear spring damper model

1. Problem Definition

   Two point bodies (no rotation considered) in contact with mass $m_1$ and $m_2$ respectively. The motion is in one dimensional (the x-axis for example) and they come in contact at exactly $t=0$.

   Each body center has location $x_1$ and $x_2$ and problem is parametrized using the combined center of mass location $x_c$ and overlap $e$

   $$x_c = \frac{m_1 x_1 + m_2 x_2}{m_1+m_2} \\ e = (x_1+r_1) - (x_2-r_2)$$

   _{Here $r_1$ and $r_2$ are the radii of the objects.}

2. Contact Force Model

   The contact is a simple spring/damper system where the force $F$ depends on the overlap $e$ and the overlap speed $\dot{e}$

$$ \begin{cases} F = k\, e + d\, \dot{e} & e\ge 0 \\ F =0 & e<0 \end{cases} $$

3. Body Equations of Motion

   During the contact the contact force is applied at equal and opposite measure on the two bodies

   $$ \begin{aligned} m_1 \ddot{x}_1 &= -F = -k\,e - d\,\dot{e} \\ m_2 \ddot{x}_2 & = F = k\,e + d\,\dot{e}\end{aligned} $$

   The accelerations of the two bodies are expressed in terms of the center of mass acceleration $\ddot{x}_c$ and the overlap acceleration $\ddot{e}$.

   $$ \begin{aligned} \ddot{x}_1 & = \ddot{x}_c + \frac{m_2}{m_1+m_2} \ddot{e} \\ \ddot{x}_2 & = \ddot{x}_c - \frac{m_1}{m_1+m_2} \ddot{e} \end{aligned}$$

   Use the accelerations in the equations of motion and solve for $\ddot{x}_c$ and $\ddot{e}$

   $$ \begin{aligned} \ddot{x}_c & = 0 \\ \ddot{e} & = - \frac{m_1+m_2}{m_1 m_2} (k\,e + d\,\dot{e}) \end{aligned} $$

4. Overlap Equations of Motion

   So besides the obvious that the center of mass velocity remains constant $\dot{x}_c = \mbox{(const)}$ the overlap obeys the 1D damped vibration equation

   $$ \mu\, \ddot{e} = -k\, e - d\,\dot{e} $$

   Where $\mu = \frac{m_1 m_2}{m_1+m_2}$ is the reduced mass of the system.

5. Damped Harmonic Solution

   We need to specify the initial conditions at $t=0$ as $e=0$ and $\dot{e}=v$, where $v = \dot{x}_1 - \dot{x}_2$ is the relative speed at impact. The general solution of the equations of motion is

   $$ e(t) = \frac{v}{\omega} \exp(-\beta t) \sin(\omega t) $$

   Using the substitution $k = \mu \omega_n^2$ and $d = 2 \zeta \mu \omega_n$ the above solves the equations of motion when $\omega = \omega_n \sqrt{1-\zeta^2}$ and $\beta = \zeta \omega_n$

   $$ e(t) = \frac{v}{\omega_n} \left( \frac{ \exp(-\zeta \omega_n t) \sin (\omega_n t \sqrt{1-\zeta^2})}{\sqrt{1-\zeta^2}} \right) $$

   The individual object positions are thus

   $$\begin{aligned} x_1 (t) & = \dot{x}_c\,t + \frac{m_2}{m_1+m_2} (e(t)-r_1-r_2) \\ x_2 (t) & = \dot{x}_c\,t - \frac{m_1}{m_1+m_2} (e(t)-r_1-r_2) \\ \end{aligned} $$

6. Parametrized Solution

   $$\begin{cases} v & = \dot{x}_1 - \dot{x}_2 & & \mbox{relative speed at }t =0 \\ \mu & = \frac{m_1 m_2}{m_1+m_2} & & \mbox{reduced mass} \\ \omega_n & = \sqrt{ \frac{k}{\mu} } & & \mbox{natural frequency} \\ \zeta & = \frac{d}{2 \mu \omega_n} & & \mbox{damping ratio} \\ \omega & = \omega_n \sqrt{1-\zeta^2} & & \mbox{damped frequency} \\ \varphi &= \frac{t}{\omega_n} & & \mbox{phase angle} \end{cases} $$

   $$\boxed{ \begin{aligned} e & = \frac{v}{\omega} \exp(-\zeta \varphi) \sin\left( \frac{\omega}{\omega_n} \varphi \right) \\ \dot{e} & = v\, \exp(-\zeta \varphi) \left( \cos\left( \frac{\omega}{\omega_n} \varphi \right) - \frac{\zeta \omega_n}{\omega} \, \sin\left( \frac{\omega}{\omega_n} \varphi \right) \right) \end{aligned} } $$

7. Force vs. Time

   Using the solution from above the force is evaluated from $F = k e + d \dot{e} = (\mu \omega_n^2) e + (2 \zeta \mu \omega_n) \dot{e}$ as

   $$ F = \mu v \exp(-\zeta \varphi) \left( 2 \zeta \omega_n \cos \left( \frac{\omega}{\omega_n} \varphi\right) + \frac{\omega_n^2 (1-2\zeta^2)}{\omega} \sin \left( \frac{\omega}{\omega_n} \varphi\right) \right) $$

   Note that at critical damping with $\zeta = 1$ the overlap and force equations are $$e = \frac{v \varphi \, {\rm e}^{-\varphi} }{\omega_n}$$ $$F = \mu v \omega_n (2-\varphi) \, {\rm e}^{-\varphi}$$

8. Energy Loss due to Damping

   Since the center of mass moves with constant velocity the change in kinetic energy is purely because of losses during the impact.

   $$ \Delta K = \frac{1}{2} \mu v^2 - \frac{1}{2} \mu \dot{e}^2 $$

   The end of the contact is when $e=0$, or specifically that the phase angle $\varphi = \frac{\pi}{\sqrt{1-\zeta^2}}$

   $$\Delta K = \frac{1}{2} \mu v^2 \left(1 - \exp\left( - \frac{2 \pi \zeta}{\sqrt{1-\zeta^2}} \right) \right) $$

   So now if you want to target specific energy loss ratio $\eta = \frac{\Delta K}{K}$ you need to use damping ratio

$$ \zeta = - \frac{ \ln(1-\eta) }{\sqrt{ 4\pi^2 + \ln(1-\eta)^2}} $$

8. Coefficient of Restitution

   Look at the velocity at the end of the contact when $\varphi = \frac{\pi}{\sqrt{1-\zeta^2}}$ as a fraction of the initial velocity $v$ defines the coefficient of restitution $\epsilon$

   $$ \epsilon = -\frac{\dot{e}}{v} = \exp \left( \frac{\pi \zeta}{\sqrt{1-\zeta^2}} \right) $$

   Or in reverse, if one wants to find the damping needed for a specific COR

   $$ \zeta = - \frac{\ln(\epsilon)}{\sqrt{\pi^2 + \ln(\epsilon)^2}} $$