Suppose I have two rigid bodies A and B and they are connected by a spring which is attached off-center (thus possibly causing torques). Due to the spring a force $f$ acts on A and a force $-f$ acts on B (at the respective attachment points) in direction of the spring as in Fig. 1. How can I show the conservation of momentum? $\frac{\rm d}{\rm dt} p_A + p_B = 0$ (where $p_A$ and $p_B$ are the linear momenta of A and B respectively) is missing the angular part and $\frac{\rm d}{\rm dt} p_A + p_B + L_A + L_B = 0$ (where $L_A$ and $L_B$ are the angular momenta of A and B around their center of masses respectively) seems to be wrong. Is $\frac{\rm d}{\rm dt} p_A + p_B + L_A^0 + L_B^0 = 0$ (where $L_A^0$ and $L_B^0$ are the angular momenta of A and B around the origin respectively) the correct ansatz?
What if the forces are opposite but not in the direction of the spring as in Fig. 2?
Fig. 1: Opposite forces along the line between the points where the forces act.
Fig. 2: Opposite forces but not along the line between the points where the forces act.
Best Answer
The angular and linear momentum of the two masses A and B are not necessarily conserved individually; it is the momenta of the system $S_{AB}$ that is conserved. If you know the conditions of the system at any particular time $t$, draw a free body diagram and work out the momentums for the system. Knowing that these values are conserved, you can use them as conditions to help you solve for the forces on the system at any other time.