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I am trying to simulate a collision between two molecules. I know the energy for every position/orientation, from which I can calculate the forces. The treatment is classical and the molecules are assumed to be rigid.

I converted this situation to the center of mass frame so I only need the relative position, and I use the mass $m_{1} \cdot m_{2} / (m_{1} + m_{2})$.

But I get stuck at the rotations. Can I use rotating particles in center of mass frame, because all I can find is examples of point particles? If I can, what do I use for the moment of inertia? If not, what's the best alternative?

EDIT: Hopefully this clarifies things: the reduced representation seems to be for point particles. I would like to know if I can use it for rotating (non-point) particles in some way. It seems to me that if I can use it at all, I would need to use a special moment of inertia for individual particles. I think that because the mass is changed, so if I use normal moment of inertia but scaled linear inertia, then rotations will be disproportionately slow or fast. Then again, maybe I can't use the reduced representation at all.

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Do you mean that you want to find a reduced representation for the collision of rotating particles, analogous to the reduced mass for two point particles? – Nanite Oct 17 '13 at 20:24
Either that or find if/how I can use non-reduced rotation in the 'normal' reduced representation – Mark Oct 18 '13 at 2:49
IF you want to do 3D rigid body collision it gets really complicated really fast. – ja72 Apr 16 '14 at 18:03

1 Answer 1

Ok it seems from your calculations you essentially chose a spot in space around which both molecules rotate--a barycenter. This means that you can use the Parallel Axis Theorem to calculate the moment of inertia of both particles as a system basically orbiting this barycenter. Which for this case is simply $I_{total}=\sum\limits_{i=1}^n m_i r_i^2$.

Note that this is not the same as the molecules rotating around their own cg, in which case $I=Cmr^2$ where $C$ is a coefficient associated with the shape of that molecule. $C$ is 1 for point masses, hoops and hollow cylinders (basically any shape where $r$ is the same for every infinitesimal mass in the object), and $<1$ for other shapes. You can derive the constant or consult a table (wikipedia List of moments of inertia)

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I want to know for the individual molecules. So do I understand correctly that I just use the same formula and mass as I would in lab frame? It seems weird because the reduced mass of the molecule is different... – Mark Oct 17 '13 at 12:34

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