Can I neglect rotation (angular inertia) in inelastic collisions, in the overdamped approximation?

Question

In an overdamped system, "inertia", is neglected. See e.g., When is "Inertia" Negligible?

Does this mean that in an completely inelastic collision, for example when two disks collide in two dimensions and stick to each-other, I can also neglect the rotation of the joint body due to angular momentum?

Answer 1

Not at all. In an overdamped situation, it is assumed velocities change slowly and thus accelerations are almost negligible. But in plastic collisions velocities do change quite abruptly regardless if they are elastic or plastic.

There is just no equivalency here between the two situations. In the equations of motion, you can ignore the mass moment of inertia of a body only if it has fixed (zero or constant) rotational velocity. In the context of collisions, this means the change in angular velocity must be negligible also.

Specifically, in the equation below describing the change in rotational velocity due to an impact with impulse $\vec{J} $, when would it be ok to ignore ${\rm I}$?

$$ \Delta \vec{\omega} = {\rm I}^{-1} ( \vec{r} \times \vec{J} ) $$

You can't assume ${\rm I} \approx 0$ (point particle) because the resulting rotation will be infinite. Maybe the assumption ${\rm I} \rightarrow \infty$ (infinite size body) can be made if the geometry of the problem permits it (bullet hitting the earth for example).

The fact that the collision is inelastic, plastic or elastic really only affects the magnitude of impulse, ranging from one value for inelastic collisions to twice that value for elastic collisions.

If you are suggesting that angular momentum plays little part in calculating the impulse and only the mass is important, this is in generally incorrect and only under specific circumstances might apply. For example when two particles collide, or when the contact direction goes through or near the center of mass.

Consider two bodies, each with mass $m_1$ and $m_2$, and mass moment of inertia $I_1$ and $I_2$ that are in contact. The perpendicular distance between the contact point and each center of mass is $c_1$ and $c_2$ respectively, then this is how you resolve the contact (simplified 2D version).

$$ \begin{array}{r|l} \text{impact speed} & v_{\rm imp} = v_1 - v_2 \\ \text{reduced mass} & m^\star = \frac{1}{\tfrac{1}{m_1} + \tfrac{1}{m_2} + \tfrac{c_1^2}{I_1} + \tfrac{c_2^2}{I_2} } \\ \hline \text{impulse} & J = (1+\epsilon) \,m^\star \, v_{\rm imp} \\ \hline \text{step in motions} & \Delta v_1 = -\tfrac{1}{m_1} J \\ & \Delta v_2 = +\tfrac{1}{m_2} J \\ & \Delta \omega_1 = -{\rm I}_1^{-1} c_1 J \\ & \Delta \omega_2 = +{\rm I}_2^{-1} c_2 J \end{array} $$

where $\epsilon$ is the coefficient of restitution. You are proposing that when $\epsilon = 0$ then the moment of inertia can be ignored. This is not supported by the math since there is nowhere above where $\epsilon$ is a coefficient of ${\rm I}$ or ${\rm I}^{-1}$ which would cancel out for inelastic collisions.

For the impulse $J$ calculation you see this $\frac{1}{\frac{1}{m}+\frac{c^2}{I}}$ type of function and you can argue that when $\frac{1}{m} \gg \frac{c^2}{I}$ then the inertial terms can be ignored. But this is a function of the problem geometry and not if the system is elastic or inelastic.

_{See related post on the response to impulse on free floating rod.}