# Why can't we conserve angular momentum about any other point except Center of mass

A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end A of the rod with a velocity v0 in a direction perpendicular to AB. The collision in elastic. After the collision the particle comes to rest (a). Find the ratio m/M

I just want to know why the angular momentum is conserved about the center of mass. Why can't we take any other point on the rod to conserve angular momentum as no external torque is acting on the system.I am a a getting different value for angular velocity if i try to conserve angular momentum about any point except for center of mass.

Actually, You can take arbitrary point, and may calculate the angular momentum, torque, and so on. Center of mass is just one of them. Although the physical variables may be seen different with respect to your choice of point, the law of the Physics will be same, and the total angular momentum will be conserved. If something goes different, double-check your calculation.

• I took an arbitrary point and tried to conserve angular momentum. But I am not getting the same answer
– user267164
Jun 23, 2020 at 3:04
• @SpaceX May I check your solution, then? Jun 23, 2020 at 3:16
• I got the answer.I forgot to add some terms which resulted in incorrect calculations.
– user267164
Jun 23, 2020 at 3:22

I will add to K.R.Park's answer by saying that just because angular momentum is conserved about any point doesn't mean it is a trivial task so to so. We pick the center of mass of the rod because this point will not be accelerating after the collision, so we can move to an inertial frame where the center of mass of the rod is at rest. However, if we follow some other point on rod keep in mind that it is accelerating. If you want to follow some other point, you will have to include in your analysis the fact that you will be working in a non-inertial frame.

As a sort of intermediate method, you could pick a point on the rod but then stay with that location in space as the rod moves away after the collision. The calculations still aren't trivial, but at least you will be working in an inertial frame of reference.