You ask for a lot, mikado, and in not too clear a way..
Remember, that if the collision period is extremely short, "normal"-sized forces and torques contribute only negligibly to changes in linear and angular velocities (their impulses are too small).
For example, the impulse imparted by the force of gravity during the collision period can be neglected.
That is, only those forces/torques that blow up towards "infinite" magnitudes during collision need to be considered.
1. DIRECT,ECCENTRIC IMPACT
I will consider the simplest 3-D model which allows for changes in the linear and angular velocities, namely a direct, eccentric impact model (with 2 objects involved).
With a direct impact, it is meant that the impulse involved is parallell to the common vector normal at the contact point of the 2 objects.
Note that we have simplified extensively here by assuming only a single point of contact, rather than a complicated contact area!
A more complicated point-contact model is the oblique impact model, where the impulse is not assumed to be necessarily parallell to the common vector normal.
An eccentric impact means that the impulse involved is not necessarily parallell to the position vectors of the contact point, measured from the objects' respective centres of mass (C.M).
This is the feature that can generate/change angular velocities of the objects involved.
2. 1. and 2. IMPACT PERIODS:
It is advantageous to split the collision period into 2 phases/ impact periods:
1.Impact period:
This phase stretches physically up to maximal deformation of the objects
(at least on the conceptual level..
).
An exact deformation analysis is of course, extremely difficult, if not impossible.
However, by using impulse analysis, we gain a simplified, but effective model.
In particular, we make use of the highly simplifying, but patently false assumption that the two bodies remain rigid bodies throughout the collision period.
The conditions that has to be met is:
a)Change in objects' linear momenta is due to impulse (6 equations)
b)Change in objects angular momenta is due to angular impulse (6 equations)
c) The normal velocities of the 2 objects at the contact point must be equal(1 equation)
Hence, we have 13 equations in 13 unknowns:
\vec{v}_{G,1}^{(i)}, \vec{v}_{G,2}^{(i)}, \vec{\omega}_{1}^{(i)},<br />
\vec{\omega}_{2}^{(i)}, I^{(i)}
I^{(i)} is the scalar value of the impulse, since the direction has been fixed in the direct impact assumption.
The superscript (i) means that the quantities are intermediate, and not final values.
2.Impact period:
In this period, elastic deformations developed during the first impact period revert, and both objects experience a 2.impulse as a result of this.
This second impulse, \vec{I}^{(2)} is simplified to being proportional with the first, that is:
\vec{I}^{(2)}=k\vec{I}^{(i)}
k is called the restitution coefficient, and for fully elastic collisions, k=1.
This is an extremely effective simplication, since we then for the enormous system of equations to be solved in the 1.impact period only need to solve explicitly for the expression of I^{(i)}!
To see why, let's consider object 1's equations for changes in linear momentum in the two impact periods:
1.period:
I^{(i)}\vec{n}_{1}=m_{1}(\vec{v}_{G,1}^{(i)}-\vec{v}_{G,1}^{(0)})
2.period:
kI^{(i)}\vec{n}_{1}=m_{1}(\vec{v}_{G,1}^{(f)}-\vec{v}_{G,1}^{(i)})
(\vec{n}_{1} is the vector normal at the contact point)
Summing together, we have for the final velocity:
\vec{v}_{G,1}^{(f)}=\vec{v}_{G,1}^{(0)}+\frac{2I^{(i)}}{m_{1}}\vec{n}_{1}
3. SOLUTIONS:
We let the vector normal at the contact point be represented
by \vec{n}_{1}=-\vec{n}_{2}
\vec{r}_{1},\vec{r}_{2} are the respective position vectors to the objects' centres of mass.
\mathcal{I}_{G,1},\mathcal{I}_{G,2} are the inertial tensors with respect to the C.M's.
\vec{b}_{1}=\mathcal{I}_{G,1}^{-1}(\vec{r}_{1}\times\vec{n}_{1})
that is a matrix-vector product, where the matrix is the inverse of the inertia tensor.
\vec{v}_{C,1}^{(0)}=\vec{v}_{G,1}^{(0)}+\vec{\omega}_{1}^{(0)}\times\vec{r}_{1}
is the initial contact point velocity for object 1.
Hence, we have:
I^{(i)}=-\frac{\vec{v}_{C,1}^{(0)}\cdot\vec{n}_{1}+\vec{v}_{C,2}^{(0)}\cdot\vec{n}_{2}}{\frac{1}{m_{1}}+\frac{1}{m_{2}}+(\vec{b}_{1}\times\vec{r}_{1})\cdot\vec{n}_{1}+(\vec{b}_{2}\times\vec{r}_{2})\cdot\vec{n}_{2}}
hope this helps..
