# Conservation of linear and angular momentum

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.

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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

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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To prove the conservation of quantities, you need to be able to compute the motion of the system so that you can directly compute these quantities from the time dependent coordinates and verify that they do not change with time.

However I don't think that the motion of this system is integrable: it looks like a multiple oscillator and it is very prone to develop chaotic motion (as a double pendulum), so I'm afraid that the conservation of momenta has to be assumed.

If you were just looking for a way to write it down I suggest: $$\left\{\begin{array}{l} \frac{\rm d}{\rm dt}\Big( p_A + p_B \Big)= 0\\ \frac{\rm d}{\rm dt}\Big( L_A+L_B + L_A^0 + L_B^0 \Big)= 0 \end{array}\right.$$

which should take into account every possible motion of the system components. You can take $L_A^0$ and $L_B^0$ wrt the centre of mass or any other external fixed point.

Finally I have a comment about Fig 2. which does not represent a reasonable complete physic case. Those two misaligned forces are producing a torque out of nothing within the system! You should not be able to find such a case in nature.

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The angular momentum $L_{A/B}$ of a rigid body $A/B$ about its center of mass is

$$L_{A/B} = I_{A/B} \omega_{A/B},$$

where $I_{A/B}$ is the inertia matrix of $A/B$ about its center of mass in the world frame and $\omega_{A/B}$ is the angular velocity of $A/B$. The angular momentum $L_{A/B}^0$ of a rigid body $A/B$ about the origin of the world frame is

$$L_{A/B}^0 = L_{A/B} + x_{A/B} \times p_{A/B},$$

where $x_{A/B}$ are the coordinates of the center of mass in the world frame. Then the total angular momentum in the system with the rigid bodies $A$ and $B$ about the origin of the world frame is $L_{total}^0 = L_A^0 + L_B^0$, which is supposedly conserved. The total linear momentum is $p_{total} = p_A + p_B$.

The total linear momentum is conserved as soon as the forces $f_A$ and $f_B$ are of equal magnitude and opposite direction ($f_A = f = -f_B$):

$$\frac{\mathrm d}{\mathrm{d}t} ( p_A + p_B ) = m_A \dot{v}_A + m_B \dot{v}_B = f_A + f_B = f - f = 0,$$

where $v_{A/B}$ is the translational velocity of $A/B$ in the world frame. The derivative of the total angular momentum with respect to time is

$$\begin{split} \frac{\mathrm d}{\mathrm{d}t} ( L_A^0 + L_B^0 ) & = \frac{\mathrm d}{\mathrm{d}t} (L_A + L_B + x_A \times p_A + x_B \times p_B) = \dot{L}_A + \dot{L}_B + x_A \times \dot{p}_A + x_B \times \dot{p}_B \\ & = \tau_A + \tau_B + x_A \times f_A + x_B \times f_B = r_A \times f_A + r_B \times f_B + x_A \times f_A + x_B \times f_B \\ & = (x_A + r_A - x_B - r_B) \times f. \end{split}$$

Thus the total angular momentum is conserved if:

1. $x_A + r_A - x_B - r_B = 0$, that is the force pair acts at the same coordinates in the world frame,
2. $f = 0$, that is no force acts,
3. $(x_A + r_A - x_B - r_B)\ ||\ f$, that is the force acts along the line of connection.

Fig. 1 satisfies condition number 3 and thus conserves the total angular momentum. Fig. 2 satisfies none of the three conditions and thus does not conserve the total angular momentum.

Edit: The SO post Is angular momentum always conserved in the absence of an external torque? contains a proof for point particles, which has an analogous requirement for the conservation to hold (forces along the line of connection). The author of the proof corrected it by now.

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## protected by Qmechanic♦Mar 21 '14 at 11:27

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