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I'm working on a 2D physics engine somulator and I want it to be very accurate on a Physics point of view. Currently I'm making some researchs about how rigid body collisions may be calculated, and I've found this page, that presents the following equation:

enter image description here

Where $w$ is the angular velocity, $j$ is the impulse during the collision, $r$ is the radius and $\hat{n}$ is (I'm not sure but I am guessing) a normalized vector.

My question is about what exacly is $I^{-1}$ .It says it's an inertia tensor, but for what I know it is used for 3D objects. Since my engine is 2D, may I simply use this formula:

enter image description here

Which is presented here as the moment of inertia of a rectangle. If yes, should I still use the (-1) power?

Btw, I would also accept a link to some pdf explaining how collisions change angular and linear momentum in 2D... this wikipedia page was the best I could find.

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  • $\begingroup$ I know it has been a while since this was asked, but according to Wikipedia( en.wikipedia.org/wiki/Moment_of_inertia ): "For bodies constrained to rotate in a plane, only their moment of inertia about an axis perpendicular to the plane, a scalar value, matters." in other words; if constrained to a 2D XY plane then the Inertia tensor 3x3 matrix can simply be replaced by the inertia about the axis of rotation. $\endgroup$ – Nikolaj Jun 11 '18 at 9:58
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Yes the I must have the ^-1 exponent, otherwise the unit would not end up in $s^-1$ (the unit for angular velocity). $\hat n$ is the unit vector in the direction of exit after collision.

Moment of inertia of a 2D or 3D object is the same as long as they have the same cross section from the perspective of the dimension you want to ignore (for example thickness of the cuboid). E.g, moment of inertia of a rectangle rotating about it's center in its own plane has the same formula you listed. Note this is true only because the cross sections are the same in this particular axis of rotation--and of course the mass is assumed the same.

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