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I'm having a hard time trying to understand the conservation of angular momentum. Let's assume the following 2D experiment:

enter image description here

At the start of the experiment both rectangles have no angular momentum, but if the upper moves downwards and their corners hit, they will both start to rotate clockwise (red arrows). In my apparently flawed understanding of angular momentum this would mean that the total angular momentum has increased because two clockwise rotations would be added together and a clockwise rotation would have to be countered by a counterclockwise rotation. The math would work out if we would subtract the second rectangles angular momentum from the first's but I don't understand why and if we had more than two bodies which rotations would we have to add and which to subtract?

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Total angular momentum about a given point is conserved in the absence of external forces. When you do this, you need to add up the contributions of $\vec{r}\times\vec{p}$ for each particle, where $\vec{r}$ is the position of the centre of mass, and $\vec{p}$ is the momentum, as well as counting the angular momentum about the centre of mass for each particle. If you do this for your system, you'll see that there is some angular momentum even before the two bodies collide.

Here are the details. Label the two objects $1$ and $2$. Suppose, when they collide, that their centres of mass are at $\vec{r}_1$ and $\vec{r}_2$, relative to some fixed origin of coordinates. The point of contact is $\vec{r}_c$ and, at that point, a force acts between the two objects, changing the momentum of $1$ by $+\vec{p}_c$ and the momentum of $2$ by $-\vec{p}_c$ (Newton's third law). Then the change in total angular momentum has four parts to it:

  1. Change in angular momentum of $1$ about its COM: $(\vec{r}_c-\vec{r}_1)\times \vec{p}_c$
  2. Change in angular momentum of $2$ about its COM: $(\vec{r}_c-\vec{r}_2)\times (-\vec{p}_c)$
  3. Change in $\vec{r}\times\vec{p}$ term for $1$: $\vec{r}_1\times \vec{p}_c$
  4. Change in $\vec{r}\times\vec{p}$ term for $2$: $\vec{r}_2\times (-\vec{p}_c)$

These sum up to zero.

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  • $\begingroup$ Worth noting that while it is VERY not obvious, the angular momentum of a body is the angular momentum of the COM plus the angular momentum about the COM. That's what makes this method work. $\endgroup$ – Jahan Claes Oct 15 '18 at 22:53
  • $\begingroup$ Now it clicked for me. I thought we could just add up the rotation of all bodies around their COM and sum it up. But now I understand it much better. Thanks! That helped me a lot. $\endgroup$ – Michael Mahn Oct 16 '18 at 7:27

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