General solution of a ball elastically colliding with a spinning rod

Question

I am working on finding the general solution to a disc colliding with a thin spinning rod in two dimensions floating in free space. The collision is perfectly elastic. The width of the rod is negligible.

Shown below is the situation and known variables. Omega is the angular velocity of the rod spinning about its center. Theta is the orientation of the rod with respect to horizontal. l is the length of the rod and d is the diameter of the disc. r is the coordinate for the center of mass of each object.

There are 5 variables that change after the instantaneous collision: Vx1, Vy1, Vx2, Vy2, and omega, so there must be 5 equations to completely determine the final state. I can only think of 4: 1) conservation of energy, 2) conservation of linear momentum x and 3) y, and 4) conservation of angular momentum. I think the 5th equation has to do with the orientation of the ball and rod like how far the ball strikes from the rod's center and what the rod's current angle is, but I cannot think of it. What is the 5th equation to make it generally solvable? Am I missing something somewhere else?

Context: I am working on a computer simulation and need to solve the equation generally to work with any conditions. I am fully aware that the equations will be disgusting.

Answer 1

here are the equations :

$$m_1\left(v_{1xf}-v_{1xi}\right)=p\,\sin(\alpha)$$ $$m_1\left(v_{1yf}-v_{1yi}\right)=-p\,\cos(\alpha)$$ $$m_2\left(v_{2xf}-v_{2xi}\right)=-p\,\sin(\alpha)$$ $$m_2\left(v_{2yf}-v_{2yi}\right)=p\,\cos(\alpha)$$ $$ I\,\left(\omega_f-\omega_i\right)=-p\,r $$ The velocity towards the normal vector

$$\sin(\alpha)\left(v_{2xf}-v_{1xf}\right)+\omega_f\,r-\cos(\alpha)\left(v_{2yf}-v_{1yf}\right)+\\ \sin(\alpha)\left(v_{2xi}-v_{1xi}\right)+\omega_i\,r-\cos(\alpha)\left(v_{2yi}-v_{1yi}\right) =0$$

6 equations for 6 unknowns

$~v_{1xf}~,v_{1yf} ~,v_{2xf}~,v_{2yf} ~,\omega_f~,p$

• $v~$ velocity
• $\omega~$ angular velocity
• $p~$ linear momentum
• $ I=\frac 12 m_2\,l^2~$ moment of intertia
• index f final
• index i initial

Edit

you start with the equation of motion at the collision time

$$ m\,\frac{dv_x}{dt}=f_c\,\sin(\alpha)\\ m\,\frac{dv_y}{dt}=-f_c\,\cos(\alpha)$$

where $f_c ~$ is the constraint force.

multiply with dt and integrating you obtain $$ m\,(v_{xf}-v_{xi})=\underbrace{\int f_c\,dt}_{=p}\,\sin(\alpha)\\ m\,(v_{yf}-v_{yi})=-\int f_c\,dt\,\cos(\alpha)$$

analog writing the equation of motion for the rotation .

energy conservation

$$\frac 12 m_1\, \vec v_{i1}\cdot \vec v_{i1} + \frac 12 m_2\, \vec v_{i2}\cdot \vec v_{i2}+ \frac 12 I\,\omega_i^2 =\frac 12 m_1\,\vec v_{f1}\cdot \vec v_{f1}+ \frac 12 m_2\,\vec v_{f2}\cdot \vec v_{f2}+ \frac 12 I\,\omega_f^2$$

linear momentum conservation

$$m_1\left(v_{1xf}+v_{1yf}\right)+ m_2\left(v_{2xf}+v_{2yf}\right)= m_1\left(v_{1xi}+v_{1yi}\right)+ m_2\left(v_{2xi}+v_{2yi}\right)$$

Answer 2

In a complex collision, you frequently need to know something about the conditions after the collision. That reduces the number of unknowns, and appears to the case in this situation. In this case, you might make the (questionable) assumption that the angle of reflection for the disk is the same as the angle of incidence. But then again, if a disk undergoes a collision at an angle, it is likely to come away with a spin (and that would change the angle and introduce another unknown).