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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?

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Possibly related: physics.stackexchange.com/q/19724/2451 –  Qmechanic Jan 20 '12 at 19:48

1 Answer 1

up vote 2 down vote accepted

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.

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Okay. Works for me if no one's going to disagree. Would this still hold to describe the instantaneous state of the combined system if the two bodies weren't actually rigidly combined? –  JCooper Nov 23 '11 at 16:53
@JCooper: Well, a physics proposal is not correct just because nobody bothers/has time to disagree. Number 1 person that you have to convince is always yourself, and the fact that you repeatedly writes it feels wrong smells like we didn't get to the bottom of the problem. I get the impression that you are trying to describe two independently moving rigid bodies as just one rigid body, which can't be done by definition. –  Qmechanic Nov 23 '11 at 18:18
@Qmechanic I really did want to know where the velocities go if two independent bodies become a single rigid body. However, as an after-thought I wanted to know if you can talk about the instantaneous velocities of a single system composed of multiple independent rigid bodies. E.g., if a ballistic rigid body were to fracture into multiple pieces, the linear and angular momentum of the system should be conserved; so I wonder if you could add up the pieces and get a measurement of the instantaneous linear and angular velocities of the system. –  JCooper Nov 23 '11 at 18:54

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