Angular-linear momentum conservation question

1. Sep 16, 2009

GluteusMedius

Consider a frictionless plane. A particle of known mass and velocity (say a lump of clay) strikes a uniform rod (for simplicity let the rod be stationary in the lab frame) and sticks to it somewhere other than at the rod's center of mass. I wish to describe the linear and angular velocity of the particle-rod system after the collision.

In the absence of an external force, the center of mass of the particle-rod system continues with the same velocity despite the collision.

I presume then that the particle-rod system will begin to rotate about its center of mass with angular momentum equal to r x p, with p being the initial momentum of the particle and r the moment arm from the cm to point of contact.

I'm almost certain this can't be right since angular momentum of the system about its cm ought to be conserved also--and in this case that initial angular momentum is zero. Further, it would seem that I've made the kinetic energy of the post collision system dependent on the point the particle strikes the rod.

I think I'm confused with something very fundamental here. Any help would be greatly appreciated.

Last edited: Sep 17, 2009
2. Sep 17, 2009

tiny-tim

Welcome to PF!

Hi GluteusMedius! Welcome to PF!
No, the initial angular momentum of the particle-rod system is not zero, because the particle has angular momentum about the (combined) centre of mass.

3. Sep 17, 2009

GluteusMedius

Thanks for the response.

I can see the initial angular momentum about the combined cm now but it bothers me that the angular momentum about the combined cm is still dependent on the particle's actual trajectory and not just on its momentum vector while the particle-rod system's linear momentum is only dependent on it. In other words, I think it's a bit contradictory to say that simply aiming the particle to collide with the rod cm on a trajectory normal to it makes the angular momentum about the combined cm go away.

Or more likely I'm not determining the angular momentum about the combined cm correctly?

4. Sep 17, 2009

tiny-tim

What do you mean by "a trajectory normal to it"?

The line of the velocity must go through the centre of mass if there's to be zero angular momentum.

If it misses the (combined) centre of mass by a perpendicular (ie shortest) distance d, then the magnitude of the angular momentum is the ordinary momentum times d.

5. Sep 17, 2009

GluteusMedius

I think things just came into focus for me. It had bothered me that the kinetic energy after the collision was dependent on the point of contact between the rod and particle but I realized that since the collision was inelastic to begin with, this isn't a problem. Thanks for the response; I was missing something fundamental after all.