Collision Response with Rotation

 Linear velociites at point p before the collision

\[ \begin{aligned} v_{ap}=v_a+(\omega_a \times r_{ap})\\ v_{bp}=v_b+(\omega_b \times r_{bp}) \end{aligned}\]

After collision:

\[ \begin{aligned} v_{ap}'=v_a'+(\omega_a' \times r_{ap})\\ v_{bp}'=v_b'+(\omega_b' \times r_{bp}) \end{aligned}\]

Recall that relative velocity when there is no rotation:

\[ \begin{aligned}  v_{ab}'=v_a-v_b \\  v_{ab}'=v_{a}'-v_{b}'  \end{aligned}\]

and with rotation:

\[ \begin{aligned}   v_{abp}= v_{ap}- v_{bp}\\  v_{abp}'=v_{ap}'- v_{bp}'     \end{aligned}\]

We can also say that:

\[ \begin{aligned}  v_{ab}'\cdot n=-e(v_{ab}\cdot n)\\ v_{ab}'\cdot t=-f(v_{ab}\cdot t)    \end{aligned}\]

In a world of no ration, impulse method describes the change in their linear velociteis by the impulse J, scaled by the inverse of their corresponding masses, m(a) and m(b). Note that below asumes J is pointing from A to B.

\[ \begin{aligned}    v_{a}'= v_{a}- \frac{J}{m_a}\\ v_{b}'= v_{b}+ \frac{J}{m_b}     \end{aligned}\]

The rotational equivalent is:

\[ \begin{aligned}    \omega_a'=\omega_a-\left( r_{ap}\times \frac{J}{I_a} \right) \\ \omega_b'=\omega_b+\left( r_{bp}\times \frac{J}{I_b} \right)      \end{aligned}\]

Impulse can be decomposed into components along the collision normal and tangent.

\[J=j_n\cdot n + j_t\cdot t\]

We can say that:

\[\omega_a' = \omega_a -\left( r_{ap}\times \frac{j_n\cdot n + j_t \cdot t}{I_a} \right)=\omega_a-\frac{j_n}{I_a}(r_{ap}\times n)-\frac{j_t}{I_a}(r_{ap}\times t)\]

Simiarly,

\[\omega_b' = \omega_b +\left( r_{bp}\times \frac{j_n\cdot n + j_t \cdot t}{I_b} \right)=\omega_b+\frac{j_n}{I_b}(r_{bp}\times n)+\frac{j_t}{I_b}(r_{bp}\times t)\]

Note that the equation has been expanded to describe the change in angular velocities, casued by the normal and tangent compoenents of the impulse.

The corresponding equations describing the linear velocity changes are:

\[\begin{aligned} v_a'=v_a-\frac{J}{m_a}=v_a-\frac{j_n}{m_a}\cdot n-\frac{j_t}{m_a}\cdot t \\ v_b'=v_b+\frac{J}{m_b}=v_b+\frac{j_n}{m_b}\cdot n+\frac{j_t}{m_b}\cdot t  \end{aligned}\]

From the above, by substitution we get:

\[\begin{aligned}v_{ap}'=\left( v_{a}-\frac{j_n}{m_a}n-\frac{j_t}{m_a}t \right)+\left( \omega_a-\frac{j_n}{I_a}(r_{ap}\times n)-\frac{j_t}{I_a}(r_{ap \times t}) \right)\times r_{ap}\\ v_{bp}'=\left( v_{b}+\frac{j_n}{m_b}n+\frac{j_t}{m_b}t \right)+\left( \omega_b+\frac{j_n}{I_b}(r_{bp}\times n)+\frac{j_t}{I_b}(r_{bp \times t}) \right)\times r_{bp}\end{aligned}    \]

Ignore the contribution from the angular velocity tangent component as immaterial.

Then perform the dot product on both sides by n.

\[\begin{aligned} v_{ap}'\cdot n=\left[ \left( v_{a}-\frac{j_n}{m_a}n-\frac{j_t}{m_a}t \right)+\left( \omega_a-\frac{j_n}{I_a}(r_{ap}\times n) \right)\times r_{ap}\ \right]\cdot n     \\ v_{bp}'\cdot n= \left[\left( v_{b}+\frac{j_n}{m_b}n+\frac{j_t}{m_b}t \right)+\left( \omega_b+\frac{j_n}{I_b}(r_{bp}\times n) \right)\times r_{bp}   \right]\cdot n      \end{aligned}\]

Which can be expanded to:

\[\begin{aligned} v_{ap}'\cdot n=(v_a\cdot n)-\frac{j_n}{m_a}+(\omega_a\times r_{ap})\cdot n-\frac{j_n}{I_a} \left[ \left( r_{ap}\times n\right) \times r_{ap} \right]\cdot n \\ v_{bp}'\cdot n=(v_b\cdot n)+\frac{j_n}{m_b}+(\omega_b\times r_{bp})\cdot n+\frac{j_n}{I_b} \left[ \left( r_{bp}\times n\right) \times r_{bp} \right]\cdot n   \end{aligned}\]

Let's look at the last part of this expression, and to make the analysis simpler, call \(D= r_{ap}\times n\). Utilizing the scalar triple product identity \((D\times E)\cdot F = D\cdot (E\times F)\), we can say that:

\[\left[ (r_{ap}\times n)\times r_{ap} \right]\cdot n =(D \times r_{ap})\cdot n=D\cdot(r_{ap}\times n)\]

which means that;

\[\left[ (r_{ap}\times n)\times r_{ap} \right]\cdot n =(r_{ap}\times n)\cdot (r_{ap}\times n)=\left\| r_{ap}\times n\right\|^2\]

so,

\[\begin{equation} \begin{split} v_{ap}'\cdot n=(v_a\cdot n)-\frac{j_n}{m_a}+(\omega_a\times r_{ap})\cdot n-\frac{j_n}{I_a}\left\| r_{ap}\times n \right\|^2\\ =(v_a+\omega_a\times r_{ap})\cdot n-\frac{j_n}{m_a}-\frac{j_n}{I_a}\left\| r_{ap}\times n \right\|^2 \end{split} \end{equation}\]

but recall that:

\[v_{ap}=v_a+(\omega_a\times r_{ap})\]

so

\[v_{ap}'\cdot n=v_{ap}\cdot n -\frac{j_n}{m_a}-\frac{j_n}{I_a}\left\| r_{ap}\times n \right\|^2\]

similarly,

\[v_{bp}'\cdot n=v_{bp}\cdot n +\frac{j_n}{m_b}+\frac{j_n}{I_b}\left\| r_{bp}\times n \right\|^2\]

Subtracting the above two equations:

\[(v_{ap}'-v_{bp}')\cdot n=(v_{ap}-v_{bp})\cdot n-j_n\left( \frac{1}{m_a}+\frac{1}{m_b} \right)-\frac{\left\| r_{ap}\times n \right\|^2}{I_a}+\frac{\left\| r_{bp}\times n \right\|^2}{I_b}\]

Since \(v_{abp}'=-e(v_{abp}\cdot n)\):

\[-e(v_{abp}\cdot n)=v_{abp}\cdot n-j_n\left[\frac{1}{m_a}+\frac{1}{m_b}+\frac{\left\| r_{ap}\times n \right\|^2}{I_a}+\frac{\left\| r_{bp}\times n \right\|^2}{I_b}  \right]\]

Collecting terms we get:

\[j_n=\frac{(1+e)(v_{abp}\cdot n)}{\left( \frac{1}{m_a}+ \frac{1}{m_b}+\frac{\left\| r_{ap}\times n \right\|^2}{I_a}  ++\frac{\left\| r_{bp}\times n \right\|^2}{I_b}  \right)}\]