[Physics] How to combine two rigid bodies into one

angular momentummomentumrigid-body-dynamics

With respect to some fixed frame of reference, given the inertial tensors, positions, orientations, and angular and linear velocities of two rigid bodies, how do you combine them to make a single rigid body?

Positions (center of mass in global frame): $x_1$, $x_2$
Orientations (rotation from canonical orientation in global frame): $R_1$, $R_2$
Inertial tensors (in body's frame) $I_1$, $I_2$
Total masses (scalar): $m_1$, $m_2$
Linear velocities (global frame): $v_1$, $v_2$
Angular velocities around center of mass (global frame): $\omega_1$, $\omega_2$

The new position and mass are easy, of course:

$m_f = m_1 + m_2$
$x_f = \frac{x_1m_1 + x_2m_2}{m_f}$

The canonical orientation for the combined body isn't really defined; so we can just make it the identity matrix:

$R_f = \left| \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\0 &0&1\end{array}\right|$

I can then combine the inertial tensors into the new local frame of reference:

$t_1 = x_f – x_1$ (translation of $I_1$)
$J_1 = \left| \begin{array}{ccc}
-(t_{1y}^2+t_{1z}^2)&t_{1x}t_{1y} &t_{1x}t_{1z} \\
t_{1x}t_{1y} &-(t_{1x}^2+t_{1z}^2)&t_{1y}t_{1z} \\
t_{1x}t_{1z} &t_{1y}t_{1z} &-(t_{1x}^2+t_{1y}^2)
\end{array} \right|$ (unscaled change)
And likewise for $J_2$

$I_f = (R_1I_1R_1^\intercal+ m_1J_1)+ (R_2I_2R_2^\intercal+m_2J_2)$

I think it's mostly right up to there. How do I find $\omega_f$ and $v_f$ so that all of the energy is accounted for?

Attempting to answer my own question:

Can I treat the two bodies as point masses and then combine their velocities according to conservation of momentum? It feels wrong.

$v_f = \frac{m_1v_1 + m_2v_2}{m_f}$

It feels even more wrong to find the joint angular momentum:

$L_f = I_1\omega_1+t_1\!\!\times\!\!(m_1v_1)+I_2\omega_2+t_2\!\!\times\!\!(m_2v_2) = I_f\omega_f$
$\omega_f = I_f^{-1}L_f$

Is that it?

Best Answer

I don't see anything wrong with what you've done.

Conservation of momentum, and conservation of total angular momentum will hold exactly (as you assumed). But what you're doing is an "inelastic collision" so energy is not typically conserved. So the "unaccounted for" energy appears as heat.

Related Question