# Circle-circle collision, linear and angular momenta transfer

I am trying to simulate particle collisions in two dimensions (hydrated ions). I have found the formulas I need for the transfer of linear momentum. However, I kind of suspect that angular momentum of individual particles will have to be affected if they collide off-center. This is important in my case because my circles have contents inside and it looks unnatural when they just all keep floating and bumping into one another with their contents not changing their orientation whatsoever. I asked ChatGPT about this (because my understanding of mechanics is really bad so I'm not sure which forces I'm supposed to consider and how), but considering my poor knowledge of classical mechanics, I'm not sure I'm getting the right answers from it either. Here's what I got:

The detailed formulas for simulating the collision of two balls with off-center impacts, including the resulting linear and angular motions, can be complex and depend on the specific characteristics of the collision. However, I can provide you with a simplified overview of the principles involved.

### Angular Motion:

Conservation of Angular Momentum:

• Before Collision: I11i + I22i = Total Initial Angular Momentum
• After Collision: I11f + I22f = Total Final Angular Momentum

Torque τ During Collision:

• If the collision is off-center, there will be a torque τ acting on the system during the collision.
• τ = r * F, where r is the lever arm (distance from the axis of rotation to the point of application of force) and F is the force applied during the collision.

Angular Impulse α:

• The torque τ is related to the change in angular momentum through the angular impulse.
• I * α = τ * ∆t, where I is the moment of inertia and α is the angular acceleration.
• lever arm r is the magnitude of d⊥, that is, r = |d⊥|.
• d⊥ is calculated as , where d is the vector connecting the centers of colliding circles and vrel is their relative velocities.

Are any of these formulas correct?

I am particularly concerned with the vector multiplication in the last formula that I don't quite get. Also, I am not sure if the torque will be created either way upon an off-center collision or will be a result of hydrogen-bonding between the two surfaces and will also have to be taken into account.

• what's the shape of the molecules? Do they have circular shape? If so, do circular molecules have a mechanism to transfer tangential force during a collision? Commented Jan 14 at 12:13
• @basics Yes, that's the approximation I'm going for. Commented Jan 14 at 12:29
• Elastic collision is a good model for you? Commented Jan 14 at 12:50
• @basics This is what my particle looks like. I want to assume that it's just a big circle when calculating collisions, but because it has content inside, I expect it to rotate if the collision is off-center, which, if I understand correctly, not somehting that ellastic collisions allow. Commented Jan 14 at 13:07
• Before I go into details, if your main concern is the last formula, that formula is the projection of $\vec{d}$ in the directions orthogonal to the relative velocity $\vec{v}_{rel}$, i.e. nothing more than $\vec{d}$ minus the projection along vector $\vec{v}_{rel}$. I'll add an answer below, and we'll improve it together with the things you need Commented Jan 14 at 13:08

If your concern is the last formula, let's compute the moment acting on a particle w.r.t. its center of mass $$G$$ due to a collision and thus a force $$\mathbf{F}$$ at a point $$P$$ whose relative position w.r.t. $$R$$ is $$\mathbf{r} = P - G$$. Then, we'll try to understand when the last formula could come from, and the assumptions implied in that expression.

The moment $$\mathbf{M}$$ due to $$\mathbf{F}$$ acting at $$\mathbf{r}$$ simply reads

$$$$\mathbf{M} = \mathbf{r} \times \mathbf{F} \ .$$$$

Now, as we all should know, only the components of $$\mathbf{F}$$ not aligned with $$\mathbf{r}$$ have contributions to the moment. Let's write this in math, writing $$\mathbf{F}$$ as the sum of its component aligned with $$\mathbf{r}$$ and the component orthogonal to it, using projectors

\begin{aligned} \mathbf{F} & = \mathbf{F}_r + \mathbf{F}_n = \\ & = \underbrace{\dfrac{\mathbf{r} \mathbf{r}}{r^2} \cdot \mathbf{F}}_{=\mathbf{F}_n} + \underbrace{\mathbf{F} - \dfrac{\mathbf{r} \mathbf{r}}{r^2} \cdot \mathbf{F}}_{= \mathbf{F}_n} = \\ & = \left[ \dfrac{\mathbf{r} \mathbf{r}}{r^2} \right] \cdot \mathbf{F} + \left[ \mathbb{I} - \dfrac{\mathbf{r} \mathbf{r}}{r^2} \right] \cdot \mathbf{F} \ . \end{aligned}

Now, taking the vector product to evaluate the moment, it should be clear how the first term cancels out (since it contains the vector product $$\mathbf{r} \times \mathbf{r} = \mathbf{0}$$)

\begin{aligned} \mathbf{M} & = \mathbf{r} \times \left( \left[ \dfrac{\mathbf{r} \mathbf{r}}{r^2} \right] \cdot \mathbf{F} + \left[ \mathbb{I} - \dfrac{\mathbf{r} \mathbf{r}}{r^2} \right] \cdot \mathbf{F} \right) = \\ & = \mathbf{r} \times \left[ \mathbb{I} - \dfrac{\mathbf{r} \mathbf{r}}{r^2} \right] \cdot \mathbf{F} = \\ & = \mathbf{r} \times \left( \mathbf{F} - \dfrac{\mathbf{r} \mathbf{r}}{r^2} \cdot \mathbf{F} \right) = \mathbf{r} \times \mathbf{F}_{\perp \mathbf{r}} \ . \end{aligned}

We got an expression containing the content of the round parentheses that looks like the expression provided in your question, but with different terms.

We could have compute the vector product switching the terms $$\mathbf{r}$$ and $$\mathbf{F}$$, remembering the minus sign

$$$$\mathbf{M} = \mathbf{r} \times \mathbf{F} = - \mathbf{F} \times \mathbf{r} = - \mathbf{F} \cdot \left( \mathbf{r} - \dfrac{\mathbf{F} \mathbf{F}}{F^2} \cdot \mathbf{r} \right) = - \mathbf{F} \times \mathbf{r}_{\perp \mathbf{F}} \ .$$$$ Now, the vector $$\mathbf{r} = \frac{\mathbf{d}}{2}$$ for spherical molecules (assuming the collision occurs when two spheres come into contact), and thus the component of the lever arm orthogonal to the force reads

\begin{aligned} \mathbf{r}_{\perp \mathbf{F}} & = \left( \mathbf{r} - \dfrac{\mathbf{F} \mathbf{F}}{F^2} \cdot \mathbf{r} \right) = \\ & = \left( \dfrac{\mathbf{d}}{2} - \dfrac{\mathbf{F} \mathbf{F}}{F^2} \cdot \dfrac{\mathbf{d}}{2} \right) \ . \end{aligned}

This last expression really looks like the one provided in your question, except for the arm lever to be $$\mathbf{r} = \frac{\mathbf{d}}{2}$$ and not $$\mathbf{d}$$. Now, if the force is proportional to (aligned with) the relative velocity (you need to check this assumption, if it's ok for your model; if you have an expression for $$\mathbf{F}$$, I'd directly use that expression in $$\mathbf{M} = \mathbf{r} \times \mathbf{F} = \frac{\mathbf{d}}{2} \times \mathbf{F}$$), $$\mathbf{F} = k \mathbf{v}_{rel}$$, you get

$$$$\mathbf{r}_{\perp \mathbf{F}} = \left( \dfrac{\mathbf{d}}{2} - \dfrac{\mathbf{v}_{rel} \mathbf{v}_{rel}}{v_{rel}^2} \cdot \dfrac{\mathbf{d}}{2} \right)$$$$

Conclusions. To cut a long story short, if you have the expression of $$\mathbf{F}$$ I'd implement the expression $$$$\mathbf{M} = \dfrac{\mathbf{d}}{2} \times \mathbf{F} \ .$$$$

• Thanks for such a detailed answer! Don't have much backround in mechanics, besides, high school, but if I understood you correctly: 1) $M = \frac{d}{2}\times F$ (I'm going with $F = m \cdot \vec{a}$) The vector product here is en.wikipedia.org/wiki/Cross_product? 2) $\Delta \theta = \frac{M}{I}$ where I is $I = mr^2$ As for the linear momentum, I can just use the same formulas as for elastic collisions (although I would technically be breaking the law of conservation of energy?) Commented Jan 14 at 14:07
• Try to have a look at TenMinutePhysics on youtube. There may be some useful tutorial for what you're doing, even with some math summary. He's a very nice guy from Nvidia, providing implementations in Java of some stuff about simulations Commented Jan 14 at 14:11
• Few more comments: 1. try to compute a good approximation of the inertia of the molecules (anyway, this is just a number so you can set a value in first stages of implementation and then evaluate it better); 2. try to use a physical model of collisions in terms of energy: at least, the numerical scheme should not add energy to the system, since such a simulation would be physically inconsistent and it's likely to explode in few collisions. Commented Jan 14 at 14:14

Changing orientation is not really the right way to think about it.

Suppose you are driving a truck with a tire on the back down a road. You hit a bump and the tire falls off and bounces. Earth is a really big circle, so this is a collision like the one you want.

The tire is moving forward. There is a lot of friction, so the road slows the bottom of the tire. The tire begins to spin. The maximum spin would be if the road slows the bottom of the tire to a stop. (Think of a jet landing on a runway.)

How much spin depends on how much friction. Suppose you were driving on an icy road. The road would not slow the bottom of the tire and the tire would not begin to spin.

So you need to decide how much friction to put into your simulation.

• After having given it a lot of though, I think I phrased the question poorly, which shows my poor understanding of classical mechanics. So I think that the most realistic way to do this is: 1) have to particles collide and linear momenta be redistributed in an elastic fashion. 2) the surfaces of the particles are somewhat adhesive, so instead of separating they stay together 3) since the two parts of the new objects have different velocities, it introduces rotation. 4) once the difference in velocities of the two objects is greater than the adhesion force, they separate. Commented Jan 14 at 17:02
• But this approach raises other questions, such as: 1) what if the two objects never separate? I guess it's just the question of calibrating the velocities to adhesion force ratio. 2) What if the two objects, while they are still sticked together, are hit with a third object? Whatever the answers are, I'm afraid I have no idea where to begin since molecular simulations is not my area of expertise Commented Jan 14 at 17:05