How is momentum conserved in this collision with loosely-connected spheres?

Question

There are three identical spheres. B and C are loosely connected with a light inextensible string, and A approaches B directly at some initial velocity $v$. There is no angular momentum.

A and B have a perfectly elastic collision and A comes to rest, B moving away with velocity $v$.

Some time later, the string becomes taut, and spheres B and C experience a force directed towards each other, roughly perpendicularly to $v$.

This produces a torque on the combined B+C system, and these spheres will enter a sort of mutual orbit with an overall motion away from A.

Why does this not violate conservation of angular momentum?

Answer 1

There is no angular momentum.

Nothing is rotating, but that does not mean there is no angular momentum. For angular momentum, you need to pick a point (or in some cases, an axis) about which to calculate. The angular momentum of a particle about that point depends on the distance that the line of motion has from the point.

http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html#amp

Why does a particle moving with a constant linear velocity that's is offset to the center have a constant angular momentum?

You could pick a point where the angular momentum of the initial setup is zero. For instance, we could use the starting position of $B$. As $A$ is moving directly to that point, its angular momentum is zero.

But after the collision, you will find that during the rotation, at all times the angular momentum of $B$ and $C$ sum to zero about that point.

You are probably noticing that the $B-C$ system is rotating clockwise, and that's going to contribute a positive angular momentum. But the system as a whole is moving upward as well. That motion too may contribute to the angular momentum and needs to be accounted for.

Answer 2

It does not violate angular momentum because sphere A has angular momentum about the origin that equals

$$ \boldsymbol{L}_A = \boldsymbol{r}_A \times \boldsymbol{p}_A $$

where $\boldsymbol{p}_A = m_A \boldsymbol{v}_A$ is (translational) momentum of A and $\boldsymbol{v}_A$ is the velocity vector of A.

After the first collision, if sphere A is motionless, the entirety of this angular momentum is transferred into sphere B

$$ \boldsymbol{L}_A = \boldsymbol{L}_B = \boldsymbol{r}_B \times \boldsymbol{p}_B $$

And later, then two spheres B and C start to interact, via an exchange of impulses along the string, and because these impulses are offset from the coordinate system, they also exchange some of their angular momentum such that

$$ \boldsymbol{L}_B^{\text{before}} = \boldsymbol{L}_B^{\text{after}} + \boldsymbol{L}_C^{\text{after}} $$

where again

$$\begin{aligned} \boldsymbol{L}_B^{\text{after}} = \boldsymbol{r}_B \times \boldsymbol{p}_B^{\text{after}} \\ \boldsymbol{L}_C^{\text{after}} = \boldsymbol{r}_C \times \boldsymbol{p}_C^{\text{after}} \\ \end{aligned}$$

you can confirm this since the momentum of sphere C is purely due to the impulse from the string

$$\begin{aligned} \boldsymbol{p}_B^{\text{after}} = \boldsymbol{p}_B^{\text{before}} - \boldsymbol{J}_{\text{string}} \\ \boldsymbol{p}_C^{\text{after}} = + \boldsymbol{J}_{\text{string}} \\ \end{aligned} $$

If you do an accounting of momentum you get

$$\begin{array}{c|lll|c} \text{Momentum} & \text{Sphere A} & \text{Sphere B} & \text{Sphere C} & \text{Total Momentum}\\ \hline \text{Stage 1} & \boldsymbol{p}_{A}=m_{A}\boldsymbol{v}_{A} & \boldsymbol{p}_{B}=0 & \boldsymbol{p}_{C}=0 & m_{A}\boldsymbol{v}_{A}\\ \text{Stage 2} & \boldsymbol{p}_{A}=m_{A}\boldsymbol{v}_{A}-\boldsymbol{J}_{AB} & \boldsymbol{p}_{B}=\boldsymbol{J}_{AB} & \boldsymbol{p}_{C}=0 & m_{A}\boldsymbol{v}_{A}\\ \text{Stage 3} & \boldsymbol{p}_{A}=m_{A}\boldsymbol{v}_{A}-\boldsymbol{J}_{AB} & \boldsymbol{p}_{B}=\boldsymbol{J}_{AB}-\boldsymbol{J}_{BC} & \boldsymbol{p}_{C}=\boldsymbol{J}_{BC} & m_{A}\boldsymbol{v}_{A} \end{array}$$

Each $\boldsymbol{J}_{AB}$ and $\boldsymbol{J}_{BC}$ represents in impulse vector describing the impact, and the tightening of the string.

And more importantly an accounting of angular momentum

$$\begin{array}{c|lll|c} \text{Angular Mom.} & \text{Sphere A} & \text{Sphere B} & \text{Sphere C} & \text{Total Ang. Mom.}\\ \hline \text{Stage 1} & \boldsymbol{L}_{A}=\boldsymbol{r}_{A}\times\boldsymbol{p}_{A} & \boldsymbol{L}_{B}=0 & \boldsymbol{L}_{C}=0 & \boldsymbol{r}_{A}\times\boldsymbol{p}_{A}\\ \text{Stage 2} & \boldsymbol{L}_{A}=\boldsymbol{r}_{A}\times\left(\boldsymbol{p}_{A}-\boldsymbol{J}_{AB}\right) & \boldsymbol{L}_{B}=\boldsymbol{r}_{B}\times\boldsymbol{J}_{AB} & \boldsymbol{L}_{C}=0 & \boldsymbol{r}_{A}\times\boldsymbol{p}_{A}\\ \text{Stage 3} & \boldsymbol{L}_{A}=\boldsymbol{r}_{A}\times\left(\boldsymbol{p}_{A}-\boldsymbol{J}_{AB}\right) & \boldsymbol{L}_{B}=\boldsymbol{r}_{B}\times\left(\boldsymbol{J}_{AB}-\boldsymbol{J}_{BC}\right) & \boldsymbol{L}_{C}=\boldsymbol{r}_{C}\times\boldsymbol{J}_{BC} & \boldsymbol{r}_{A}\times\boldsymbol{p}_{A} \end{array}$$

Note the condition that $\left(\boldsymbol{r}_{A}-\boldsymbol{r}_{B}\right)\times\boldsymbol{J}_{AB}=0$ which requires the impulse $\boldsymbol{J}_{AB}$ to be inline with the line connecting the centers of sphere A and B. Similarly, $\left(\boldsymbol{r}_{B}-\boldsymbol{r}_{C}\right)\times\boldsymbol{J}_{BC}=0$ for the second impulse $\boldsymbol{J}_{BC}$ due to the string.

Answer 3

Why does this not violate conservation of angular momentum?

It doesn't. The angular momentum of the center of mass (COM) of the three sphere system depends on the location of the reference point (or axis) about which it is calculated, which is arbitrary. Since the three sphere system is isolated, that angular momentum must be conserved, regardless of the chosen reference point.

Hope this helps.