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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. 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.

Fig. 2: Opposite forces but not along the line between the points where the forces act.

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Since linear and angular momentum have different units the above ansatz is certainly wrong. More likely the linear and angular momenta are conserved separately: $\frac{\rm d}{\rm dt}p_A + p_B = 0$ and $\frac{\rm d}{\rm dt}L_A^0 + L_B^0 = 0$. – user1225999 Sep 10 '12 at 9:14
Conservation of linear momentum for case 1 is $\frac{\rm d}{\rm dt} p_A + p_B = f_A + f_B = f - f = 0$ and conservation of angular momentum is $\frac{\rm d}{\rm dt} L_A^0 + L_B^0 = \frac{\rm d}{\rm dt} L_A + L_B + x_A \times p_A + x_B \times p_B = \tau_A + \tau_B + x_A \times f_A + x_B \times f_B = r_A \times f - r_B \times f + x_A \times f - x_B \times f = (x_A + r_A - x_B - r_B) \times f = 0$ – user1225999 Sep 10 '12 at 10:58
However, for the second case $x_A + r_A - x_B - r_B$ is not parallel $f$ and thus it seems angular momentum is not conserved. – user1225999 Sep 10 '12 at 11:02

1 Answer

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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.

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