# About what point is angular momentum conserved?

When we have a collision of two rigid bodies: about what point is angular momentum conserved? I had a specific example with a point mass colliding with the end of a rod and noticed that only if I calculated the conservation about the total centre of mass did I get the correct answer for the angular velocity after collision. However, I do not understand the physics behind why angular momentum is only conserved about certain points and would be grateful for an explanation. Also, if we have a collision when both spin about the own axis and rotation about another point is involved: what applies then?

I had a specific example with a point mass colliding with the end of a rod and noticed that only if I calculated the conservation about the total centre of mass did I get the correct answer for the angular velocity after collision.

Assuming that there are no external forces/torques acting then angular momentum is conserved about any point.

Here is an analysis of your example where the incoming mass sticks to a rod.

If there had been a hinge involved this would mean that there was an external force/torque applied on the rod and incoming mass system.
In such a case making the point about which to find the angular momentum the hinge would mean that angular momentum would be conserved because the force exerted by the hinge would exert no torque about the hinge.

• Thank you! This made me realise that my mistake was not to include a separate orbital and spin component! So if I get it right: If there is a point on which no external force has a torque, then the sum of all the orbital and spin components of the angular velocity is conserved? – Jhonny Apr 18 '17 at 10:32

You can do the calculation about any axis, but the axis under consideration needs to not be accelerating if it is not at the COM, unless you want to complicate your calculation.

Was the other point you considered at a particular $x,y$ distance from the COM of the collision in some frame, or was it at a certain point on the rod, which probably was accelerating at different times? Showing the calculations and why you thought they didn't work could be helpful.

• Thank you, this really made some things clear! Yes, I used a point on the rod and forgot to include the orbital angular velocity. Out of curiosity: How comes the point we calculate about cannot be accelerating for our equations to hold? – Jhonny Apr 18 '17 at 10:33
• Because in the accelerating reference frame, fictitious acceleration forces appear that act through the COM. You probably did not include them. – BowlOfRed Apr 18 '17 at 17:56

Let's consider two different stationary coordinate systems O and O', with $\vec{O'O} = \vec{r_0}$ and write equations for angular momentum for the system of N particles in both of them (with $[a, b]$ standing for cross-product) $$\vec{M} = \sum_{i = 1}^{i = N} m_i[\vec{r_i}, \vec{v_i}]$$ $$\vec{M} = \sum_{i = 1}^{i = N} m_i[\vec{r_i'}, \vec{v_i'}] = \sum_{i = 1}^{i = N} m_i[\vec{r_i+r_0}, \vec{v_i}] = \sum_{i = 1}^{i = N} m_i[\vec{r_i}, \vec{v_i}] + \sum_{i = 1}^{i = N} m_i[\vec{r_0}, \vec{v_i}]$$ Now recall that $\sum_{i = 1}^{i = N} m_i\vec{v_i} = \vec{P_{0}}$ is the momentum of whole system. So you get that $\vec{M'} = \vec{M} + [\vec{r_0}, \vec{P_0}]$. Now you can see that under assumption of $\vec{P_0}$ being constant and $\vec{M}$ being conserved, $\vec{M'}$ is conserved as well.

Note, that you should take into the account all parts of the system to get the correct answer.