# Angular momentum conservation and torque between two rigid bodies under contact forces

Suppose we have two bodies with CoM $$\bf{r}_a$$ and $$\bf{r}_b$$ that interact via some force $$\bf{F}_{ab}$$ at some contact point $$\bf{r}_c$$

By conservation of linear momentum the force $$\bf{F}_{ab} + \bf{F}_{ba} = 0$$

The torque the force induces on body a is:

$$\bf{T}_a = (\bf{r}_c - \bf{r}_a) \times \bf{F}_{ba}$$

likewise:

$$\bf{T}_b = (\bf{r}_c - \bf{r}_b) \times \bf{F}_{ab}$$

Now if we demand that the net angular momentum is conserved we arrive at:

$$\bf{T}_a + \bf{T}_b = 0$$ $$(\bf{r}_c - \bf{r}_a) \times \bf{F}_{ba} + (\bf{r}_c - \bf{r}_b) \times \bf{F}_{ab} = 0$$ $$(\bf{r}_c - \bf{r}_a) \times \bf{F}_{ba} - (\bf{r}_c - \bf{r}_b) \times \bf{F}_{ba} = 0$$ $$(\bf{r}_b - \bf{r}_a) \times \bf{F}_{ba} = 0$$

Which seems absurd to me, as it is saying that the force must be along the line that joins both center of masses, otherwise the total angular momentum will change.

Most contact forces between rigid bodies are not central forces, hence they will in general have components that are orthogonal to $$\bf{r}_{ba} = \bf{r}_b - \bf{r}_a$$

Question: where is the error in my analysis and how to avoid this seemingly erroneous conclusion?

## Edit

I am trying to revise my equation by writing the momenta relative to the origin of coordinates

so the total angular momenta should be

$$\sum_{i \in a} \bf{r}_i \times \bf{p}_i + \sum_{j \in b} \bf{r}_j \times \bf{p}_j$$

if we write $$\bf{r}_i = \bf{r}_a + (\bf{r}_i - \bf{r}_a)$$ and likewise for elements of b, after a few manipulations I got

$$\bf{r}_a \times \bf{p}_a + I_a \omega_a + \bf{r}_b \times \bf{p}_b + I_b \omega_b$$

If we derive on time and use the fact the only force is $$\bf{F}_{ab} = - F_{ba}$$ this becomes

$$\bf{r}_a \times F_{ba} + \frac{ d(I_a \omega_a) }{dt} - \bf{r}_b \times F_{ba} + \frac{ d(I_b \omega_b) }{dt}$$

Hence the condition on the internal torques and the direction of the force makes a lot more sense

$$(\bf{r}_b - \bf{r}_a) \times F_{ba} = \frac{ d(I_a \omega_a) }{dt} + \frac{ d(I_b \omega_b) }{dt}$$

which means that the total internal torques will be non-zero only when the force is non-central (not parallel to the line joining both CoM)

• Angular momentum is considered about a single point (or axis in some cases). About which point are you looking at the total momentum of the system? Commented Oct 23, 2023 at 17:28
• @the torque for body A is around the CoM of body A, likewise for body B being around the CoM of body B, both torques should equate the rate of change of angular momentum of the respective body, and both changes in angular momenta should be opposite in direction and magnitude, am I missing some term? Commented Oct 23, 2023 at 17:34
• If you want to use conservation of angular momentum, you must calculate all angular momenta about the same point. By only calculating around the component's COM, you are indeed possibly missing things. A point mass never has angular momentum about itself, but it may have angular momentum of motion about some other point. Commented Oct 23, 2023 at 17:58
• but that's indeed what the is question about, why am I not allowed to compute the rate of change of angular momentum of each body relative to its center of mass here? the typical treatment in a classical mechanics textbook is doing this for a single set of particles, but here we have two rigid bodies, with their own center of mass, and since the physics is entirely symmetric is not clear why one should add references to additional arbitrary points of space except each body center-of-mass Commented Oct 23, 2023 at 18:08
• Because angular momentum is not conserved when considered about different points, any more than linear momentum is conserved when considered in different inertial frames. Commented Oct 23, 2023 at 18:25

Let's imagine this interaction, with a spring or a small explosion creating a force pair between the rigid bodies A and B. Although they are separated in the diagram, assume they were in contact immediately prior and the force pair acts from a single point on each.

The forces are equal and opposite. But the torque on B considered about B's COM is zero, while the torque on A considered about A's COM is non-zero. In your equation, $$T_a + T_b = 0$$ is false.

The force pair here causes A to rotate, but not B. But that doesn't mean that the angular momentum is not conserved. It just has to be calculated properly.

If we want to use conservation of angular momentum here, we must use a single point for the consideration of all torques/momenta.

• Ok I edited the question by reworking the derivation including the orbital momenta, the final condition makes more sense to me (net internal torques will be non-zero if and only if pairwise force is not central), does that make sense to you? Commented Oct 24, 2023 at 14:17

Your mistake is that you are not using a common point to sum $$\bf{T}_a$$ and $${\bf T}_b$$. You are using the CoM of a for the first and the CoM of b for the second.

In the same way you cannot add torques specified at different locations, you cannot add the moment of momentum vectors when specified (summed) at different points.

To do this correctly you can sum up the momentum torques about the contact point to get $$\bf{T}_{ab} = \bf{T}_{ab} = 0$$

Or more commonly we sum up angular momentum about the origin and to designate impulse as $${\bf J}_c$$. Then changes to momentum $${\bf p}$$ and angular momentum $${\bf h}$$ are summarized below:

\begin{aligned}\Delta{\bf p}_{a} & =-{\bf J}_{c} & \Delta{\bf h}_{a} & =-{\bf r}_{c}\times{\bf J}_{c}\\ \Delta{\bf p}_{b} & =+{\bf J}_{c} & \Delta{\bf h}_{b} & =+{\bf r}_{c}\times{\bf J}_{c}\\ \sum\Delta{\bf p} & =0 & \sum\Delta{\bf h} & =0 \end{aligned}

Now momentum is defined in terms of motion, as well as change in momentum in terms of change in motion.

\begin{aligned}{\bf p}_{a} & =m_{a}{\bf v}_{a} & {\bf h}_{a} & ={\bf I}_{a}{\bf \omega}_{a}+{\bf r}_{a}\times{\bf p}_{a}\\ {\bf p}_{b} & =m_{b}{\bf v}_{b} & {\bf h}_{b} & ={\bf I}_{b}{\bf \omega}_{b}+{\bf r}_{b}\times{\bf p}_{b}\\ \Delta{\bf p}_{a} & =m_{a}\Delta{\bf v}_{a}=-{\bf J}_{c} & \Delta{\bf h}_{a} & ={\bf I}_{a}\Delta{\bf \omega}_{a}+{\bf r}_{a}\times\Delta{\bf p}_{a}=-{\bf r}_{c}\times{\bf J}_{c}\\ \Delta{\bf p}_{b} & =m_{b}\Delta{\bf v}_{b}=+{\bf J}_{c} & \Delta{\bf h}_{b} & ={\bf I}_{b}\Delta{\bf \omega}_{b}+{\bf r}_{b}\times\Delta{\bf p}_{b}=+{\bf r}_{c}\times{\bf J}_{c}\\ \Delta{\bf v}_{a} & =-\tfrac{1}{m_{a}}{\bf J}_{c} & \Delta{\bf \omega}_{a} & =-{\bf I}_{a}^{-1}\left({\bf r}_{c}-{\bf r}_{a}\right)\times{\bf J}_{c}\\ \Delta{\bf v}_{b} & =+\tfrac{1}{m_{b}}{\bf J}_{c} & \Delta{\bf \omega}_{b} & =+{\bf I}_{b}^{-1}\left({\bf r}_{c}-{\bf r}_{b}\right)\times{\bf J}_{c} \end{aligned}

See this answer for more details on how to handle contacts with rotations.

I want to add a bit of insight here. Angular momentum is a quantity that encodes where the axis of momentum is in space. In the case of a contact, this axis goes through the contact point and is along the contact normal.

The conservation of angular momentum is a direct result of Newton's 3rd law since the axis by which the contact impulse is applied on the two bodies is the same, and only the magnitude is of equal and opposite sense.

Given an impulse $${\bf J}_c$$ and the corresponding angular momentum $${\bf h}_c$$ summed up about the origin then I can tell you where the contact axis is

• Direction $${\bf n} = \frac{{\bf J}_c}{\| {\bf J}_c \|}$$
• Point on Line $${\bf r}_c = \frac{ {\bf J}_c \times {\bf h}_c }{ \| {\bf J}_c \|^2}$$

As you can see, the only place angular momentum enters the geometric interpretation is where the impulse acts through.

You can derive the above with the following steps

\require{cancel} \begin{aligned}{\bf h}_{c} & ={\bf r}_{c}\times{\bf J}_{c}\\ {\bf J}_{c}\times{\bf h}_{c} & ={\bf J}_{c}\times\left({\bf r}_{c}\times{\bf J}_{c}\right)\\ & ={\bf r}_{c}\left({\bf J}_{c}\cdot{\bf J}_{c}\right)-{\bf J}_{c}\left(\cancel{{\bf r}_{c}\cdot{\bf J}_{c}}\right)\\ {\bf r}_{c} & =\frac{\left({\bf J}_{c}\times{\bf h}_{c}\right)}{\left({\bf J}_{c}\cdot{\bf J}_{c}\right)} \end{aligned}