I have two objects that collide with some velocities and mass. About each of them, I know the radius, the velocity before and after the collision, the mass, and the position. I wish to calculate their angular velocities as a result of their collision.

The collision is elastic, and friction is assumed. Everything is also two-dimensional.

The following is what I tried:

$$ \begin{align} \tau = I \alpha &= mr^2 \alpha \\ \tau = r_\perp F_\perp &= mr^2 \alpha \\ \frac{r_\perp F_\perp}{mr^2} &= \alpha \\ \int \frac{r_\perp F_\perp}{mr^2} \; \mathrm{d} t &= \Delta \omega \\ \frac{r_\perp}{mr^2} \int F_\perp \; \mathrm{d} t &= \Delta \omega \\ \frac{r_\perp}{mr^2} F_\perp \Delta t &= \Delta \omega \\ \frac{r_\perp}{mr^2} \frac{m \Delta v}{\Delta t} \Delta t &= \Delta \omega \\ \frac{r_\perp}{mr^2} m \Delta v &= \Delta \omega \\ \frac{r_\perp}{r^2} \Delta v &= \Delta \omega \end{align} $$

Apologies if my notation is (at least a little) confusing. For clarification, $r_\perp$ is the perpendicular distance from the center of mass to the component of the force that generates the torque, $F_\perp$.

First, I ask if this is anywhere near correct. The dimensions work out just fine, but $v$ is a vector and the LHS of the final result concerns me as I'm treating the angular velocity $\omega$ as a scalar, and this seems inconsistent.

Alternatively, I ask if there is an easier way to calculate the angular velocity given the parameters noted in the second sentence of this post. If not, what else would I need? Would I need to calculate $\omega$ using $L/I$ (angular momentum/moment of inertia)?

  • 1
    $\begingroup$ Are you colliding perfect balls of fixed radius r or what? Are they initally not rotating at all and have been made to spin upon their respective centre of masses due to the collision? $\endgroup$ Apr 14 at 8:39
  • $\begingroup$ @naturallyInconsistent Yes, but their radii may differ. And they may or may not be rotating to begin with. $\endgroup$ Apr 14 at 10:18
  • $\begingroup$ Ah, I see a fault in my question. No friction would not change the angular momenta if the objects happened to be circles/spheres. Say friction did exist. $\endgroup$ Apr 14 at 11:09
  • $\begingroup$ Read through this post which calculates the change in angular velocity $\Delta \omega$ in terms of an impulse $J$ and how to compute said impulse. $\endgroup$ Apr 14 at 13:15
  • $\begingroup$ It is good that you see that friction is necessary. I was phrasing my comment to get you to that conclusion. But then there is no industry standard for the instantaneous approximation of the impulse due to friction at a point. I am not sure that your problem can be analytically solved, except under assumption of this impulse. $\endgroup$ Apr 14 at 14:31

2 Answers 2


Before you try to analyze a situation you have to think of the appropriate assumptions that would allow you do idealize a situation and make it solvable, or at least describable.

When it comes to collisions there are three main assumptions that are typically used:

  1. The collision happens near instantaneously. This means we can talk about before and after a collision, but any finite force (such as gravity) acting on the bodies will have no effect on their motion during the collision. Only the collision forces matter, which are said to happen for an infinitely short time and infinite large magnitude. But the end result is a step change in velocity on each body at the moment of impact.

  2. Total momentum before and after the collision is preserved as no external forces act during the collision (see #1). This is handled with an exchange of momentum between the bodies, such that however much momentum is reduced on one body is added to the other body. The unit of currency of momentum that is exchanged is called an impulse. This impulse is a vector and acts in an equal and opposite fashion between the colliding bodies (Newton's 3rd law). Typically the direction of the impulse is known, but the magnitude isn't.

  3. The relative speed of the two bodies after the collision is a fraction of the relative speed of the two bodies before the collision. The proportionality ratio is called the "coefficient of restitution". Also, the relative velocity before acts in the opposite sense from the relative velocity before.

If finite friction is to be considered then an additional assumption is needed

  1. The total impulse has two components, one along the normal direction of the contact and one tangent to the contact due to friction. Dynamic friction between the bodies limits the magnitude of the tangent component proportionally to the normal component.

So consider an example with a general object that is rotating hits the ground, with the coefficient of friction $\mu$. In the diagram below I have separated the object from the ground for clarity but it is assumed they are in contact at this moment. The location of the contact point A from the center of mass is $\vec{r}_A$


Before the contact, the motion of the center of mass is described by the velocity vector $\vec{v}$ and rotational velocity $\vec{\omega}$. The body has mass $m$ and mass moment of inertia tensor ${\rm I}$.

At the point of contact A the velocity of the object is

$$\vec{v}_A = \vec{v} + \vec{\omega} \times \vec{r} \tag{1} $$

where $\times$ is the vector cross product.

At the moment of impact, the contact normal direction is $\hat{n}$ and the contact tangent direction is $\hat{t}$. For this example, these are perpendicular and parallel to the floor.

This means that the impact speed along the contact normal is

$$v_{\rm imp} = \hat{n} \cdot \vec{v}_A \tag{2}$$

where $\cdot$ is the vector dot product.

As a result of the impact, a normal impulse $J_n$ and a tangential impulse $J_t$ develop at the point of contact and act in an equal and opposite fashion on the body and the floor. Thus the total impulse acting on the body is the combination of those two impulses.

$$ \vec{J} = J_n \hat{n} + J_t \hat{t} \tag{3}$$

We will find the impulse magnitudes later, but the important point here is that once $\vec{J}$ is known, then the response of the object to the collision is known. Each motion vector will suddenly change values according to the following changes

$$ \begin{aligned} \Delta \vec{v} &= \tfrac{1}{m} \vec{J} \\ \Delta \vec{\omega} &= {\rm I}^{-1} (\vec{r}\times \vec{J}) \end{aligned} \tag{4}$$

This means the rotational velocity after the impact would be $\vec{\omega} + \Delta \vec{\omega}$ for example.

This also means the change in velocity of the body at the point of contact A is

$$ \Delta \vec{v}_A = \Delta \vec{v} + \Delta \vec{\omega} \times \vec{r} \tag{5}$$

which is the same velocity transformation as (1), but after the collision.

Now how to calculate the impulse component magnitudes $J_n$ and $J_t$?

Use the law of contact to project the impact velocity along the contact normal

$$ \underbrace{ \hat{n} \cdot ( \vec{v}_A + \Delta \vec{v}_A)}_\text{bounce} =- \epsilon\; \underbrace{\hat{n} \cdot \vec{v}_A}_\text{impact} = \tag{6}$$

where $\epsilon$ is the coefficient of restitution.

The above can be re-arranged as $\hat{n} \cdot \Delta \vec{v}_A =-(1+\epsilon)\,v_{\rm imp}$ and use (5) to relate the change in velocity at the point of contact to the change in velocity at the center of mass. This is $\hat{n} \cdot (\Delta\vec{v}-\vec{r}\times\Delta\vec{\omega}) =-(1+\epsilon)\,v_{\rm imp}$ and now use (4) to relate to the impulse magnitudes

$$ \hat{n}\cdot\left(\tfrac{1}{m}\vec{J}-\vec{r}\times{\rm I}^{-1}\left(\vec{r}\times\vec{J}\right)\right)=-(1+\epsilon)\,v_{{\rm imp}} \tag{7a}$$

The nature of friction is such that the surfaces either stick together or slide apart. This means that along the contact tangent, the effective coefficient of restitution is zero. This is not always the case, as with super bouncy balls, which allow bouncing along the tangential direction as well as the normal direction. But for this example, I am ignoring this and formulating an expression similar to (7) but projected along the tangential direction

$$ \hat{t}\cdot\left(\tfrac{1}{m}\vec{J}-\vec{r}\times{\rm I}^{-1}\left(\vec{r}\times\vec{J}\right)\right)=-(1+\epsilon)\left(\hat{t}\cdot\vec{v}_{A}\right) \tag{7b}$$

Considering the directions $\hat{n}$ and $\hat{t}$ are unit vectors and perpendicular to each other when the impulse vector $\vec{J}$ is decomposed using (3) some simplifications occur to arrive at the following 2×2 system of equations in terms of $J_n$ and $J_t$.

$$ \begin{aligned}\left[\tfrac{1}{m}+\left(\hat{n}\times\vec{r}\right)\cdot{\rm I}^{-1}\left(\hat{n}\times\vec{r}\right)\right]J_{n}+\left[\left(\hat{n}\times\vec{r}\right)\cdot{\rm I}^{-1}\left(\hat{t}\times\vec{r}\right)\right]J_{t} & =-(1+\epsilon)\,\left(\hat{n}\cdot\vec{v}_{A}\right)\\ \left[\left(\hat{t}\times\vec{r}\right)\cdot{\rm I}^{-1}\left(\hat{n}\times\vec{r}\right)\right]J_{n}+\left[\tfrac{1}{m}+\left(\hat{t}\times\vec{r}\right)\cdot{\rm I}^{-1}\left(\hat{t}\times\vec{r}\right)\right]J_{t} & =-\left(\hat{t}\cdot\vec{v}_{A}\right) \end{aligned} \tag{8} $$

If it so happens that $\vec{r}$ and $\hat{n}$ are parallel, then the above simplifies to

$$\begin{aligned}\left[\tfrac{1}{m}\right]J_{n} & =-(1+\epsilon)\,\left(\hat{n}\cdot\vec{v}_{A}\right)\\ \left[\tfrac{1}{m}+\left(\hat{t}\times\vec{r}\right)\cdot{\rm I}^{-1}\left(\hat{t}\times\vec{r}\right)\right]J_{t} & =-\left(\hat{t}\cdot\vec{v}_{A}\right) \end{aligned}$$

which is directly solvable.

Solve the system to get $J_n$ and $J_t$ assuming the contact sticks. But if finite friction is available then $| J_t | \leq \mu | J_n |$ which means $J_t$ needs to be capped at this value

$$ J_t = {\rm sign}(J_t) \; {\rm arg min}( | J_t|, \; \mu |J_n| ) \tag{9}$$

where ${\rm sign}(x)$ returns -1 or +1 depending if the argument $x$ is negative or positive, and ${\rm argmin}(x,y)$ return the minimum of $x$ or $y$.

Once the tangential impulse is capped (if needed) then use (3) to find the change in motion

$$ \begin{aligned}\Delta\vec{v} & =\tfrac{1}{m}\left(J_{n}\hat{n}+J_{t}\hat{t}\right)\\ \Delta\vec{\omega} & ={\rm I}^{-1}\left(\vec{r}\times\left(J_{n}\hat{n}+J_{t}\hat{t}\right)\right) \end{aligned} \tag{10}$$

Note that the above analysis is entirely impulse-based, and forces or time do not play a role here. But if you had an estimate for the time slice $\Delta t$ in which the collision occurs, then you can do a very crude approximation of the forces involved.

Assuming the contact force is almost proportional to the deformation, then the shape of the force is going to be that of half a cosine with time

$$ F = F_{\rm max} \cos \left( \frac{ \pi t}{\Delta t} \right) $$

The above function provides a positive force from the time between $-\Delta t/2 $ to $\Delta t$.

The impulse along the contact normal can be equated to the area under this curve

$$ J_n = \int_{-\Delta t/2}^{\Delta t/2} F_{\rm max} \cos \left( \frac{ \pi t}{\Delta t} \right) \,{\rm d}t = F_{\rm max} \frac{2 \Delta t}{\pi} $$

In reverse, the estimated peak force is then

$$ F_{\rm max} = \frac{\pi J_n}{2 \Delta t} \tag{11}$$

Appendix I

For the 2D case where $\vec{r}=(r_x,\,r_y)$, $\hat{n} = (n_x,\,n_y)$ and $\hat{t} = (-n_y,\,n_n)$ the 2×2 system in (8) is states in matrix form as

$$\small \begin{vmatrix} \tfrac{1}{m} + \tfrac{(n_x r_y - n_y r_x)^2}{I} & \tfrac{ (n_x r_x + n_y r_y) ( n_y r_x - n_x r_y)}{I} \\ \tfrac{ (n_x r_x + n_y r_y) ( n_y r_x - n_x r_y)}{I} & \tfrac{1}{m} + \tfrac{(n_x r_x + n_y r_y)^2}{I} \end{vmatrix} \begin{vmatrix} J_n \\ J_t \end{vmatrix} = - \begin{vmatrix} (1+\epsilon)( n_y (v_y + \omega r_x)-n_x ( -v_x + \omega r_y)\\ n_x ( v_y + \omega r_x) + n_y (-v_x + \omega r_y) \end{vmatrix} $$

where $\vec{v} = (v_x,\,v_y)$


enter image description here you have this situation

ball 1 start at position $~\mathbf R_1~$ with the velocity $~\mathbf u_1~$ , ball 2 start at position $~\mathbf R_2~$ with the velocity $~\mathbf u_2~$.

what happens after the balls collide ?

at the collision point $~c~$ we put a local coordinate system $~\mathbf t~,\mathbf n~$ where $~\mathbf n~$ is the normal vector (connection line between center of masses ) and $~\mathbf t~$ is perpendicular the normal $~\mathbf n~$

The angular velocities $~(\omega_i)~$ written in the local system

$$I_{\rm CM}\,\dot \omega_1=F_\mu\,r\\ I_{\rm CM}\,\dot \omega_2=-F_\mu\,r$$

where $~F_\mu~$ is the friction force, this means that if it is "no friction" the balls will not rotate after the collision


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.