# Linear momentum of a rotating system

So say we have a thin rod resting horizontally on a flat frictionless surface. The rod is pinned at its center, and a small mass collides perfectly inelastically with one end of the rod after moving at a speed perpendicular to the rod.

I know that angular momentum is conserved in this case and that linear momentum is not due to the force applied to the rod by the pin. My question is - how would we actually calculate the final linear momentum of the system and the force applied by the pin?

• After the collision, the system rotates about the pivot of rod. The system has some angular momentum. Which linear momentum are you referring to? – Tojrah Jul 7 '19 at 3:32
• The force applied by pin? It varies. May be you want the initial force applied by pin – Tojrah Jul 7 '19 at 3:33

how would we actually calculate the final linear momentum of the system...

The total linear momentum of a system of particles is just the sum of the momentum of each particle. This is fairly simple to do for the uniform rod and the mass stuck to the end of it.

Rod

Since the rod is uniform, each element of the rod on one side of the pin has an equal but opposite linear momentum as an element on the other side of the pin. Therefore, the total momentum of the rod is $$0$$. This makes sense, as the rod by itself would not need an external force to rotate about it's center of mass where the pin is.

Mass

So really the only net linear momentum comes from the mass that sticks to the rod. Since $$\mathbf p=m\mathbf v$$, and since $$\mathbf v=\omega R\hat\theta$$, we have the total linear momentum of the system is $$\mathbf p=m\omega R\hat\theta$$ where the direction of $$\hat\theta$$ is determined by the angular position of the mass at some point in time.

and the force applied by the pin?

Since the only external force here is exerted by the pin, we have $$\mathbf F_{\text{pin}}=\dot{\mathbf p}=m\omega R\frac{\text d\hat\theta}{\text dt}=-m\omega^2 R\hat r$$ where like before $$\hat r$$ is determined by the position of the mass at some point in time. Notice that this is just the centripetal force acting on the mass.

I leave it to you to determine the polar unit vectors as explicit functions of time.