# Conversion of angular momentum to linear momentum in free space

If two objects both with angular and linear velocity collide in free space, can the total linear velocity of the objects increase at the expense of a loss in angular momentum?

In other words, imagine we have two dumbbells thrown at each other. Both are moving forward and spinning.

When they collide will some of the angular momentum be converted into linear momentum or will linear and angular momentum be conserved independently?

• The conservation laws of angular and linear momentum have nothing to do with each other (the former derives from rotational, the latter from translational invariance), why would you think you can "convert" one into the other? – ACuriousMind Mar 20 '16 at 17:42
• @ACuriousMind Obviously you have never played pool. When a spinning pool ball hits a rail, it loses angular momentum and gains linear momentum. One is converted into another. The question is whether this can happen in free space, not involving a fixed object. – Ambrose Swasey Mar 20 '16 at 17:46
• That's not "converting" angular into linear momentum, the linear and angular momentum is just differently distributed among ball and rail after the collision. You just don't "see" the momentum in the rail because it is so massive compared to the ball, but total (i.e. that of all things participating in a collision) momentum and angular momentum are always conserved separately. – ACuriousMind Mar 20 '16 at 17:54
• Remember, angular momentum is not an intrinsic quantity. It just an expression of linear momentum at a distance. – ja72 Mar 20 '16 at 23:22

Conservation of linear momentum and conservation of angular momentum both follow from Noether's theorem, which shows how conservation laws arise from symmetries.

Conservation of linear momentum arises from translational invariance - that is, if you shift space in some direction, the laws of physics are the same. Conservation of angular momentum arises from rotational invariance - that is, if you rotate space in some direction, the laws of physics are the same.

Consider a particle moving in one-dimensional free space. The Lagrangian is $$L(x,\dot{x},t)=\frac{1}{2}m\dot{x}^2$$ Taking the transformation $$t'=t,\quad x'=x+\epsilon,\quad \dot{x}'=\dot{x}$$ This then gives $$\delta x=\epsilon,\quad\delta\dot{x}=0$$ Noether's theorem states that there is a conserved quantity for some generalized coordinate $q$, which is $$J=\frac{\partial L}{\partial \dot{q}}\delta q-F$$ That is, $$\frac{\mathrm{d}J}{\mathrm{d}t}=0$$ Here, it is simple to see that $q=x$, $\dot{q}=\dot{x}$. Therefore, $$J=m\dot{x}\epsilon-F$$ From here, it is easy to see that $m\dot{x}$ is a constant, that is, $p_x$ is conserved.

We can construct a similar Lagrangian for any number of bodies in any number of dimensions, and find that the total momentum of the system is conserved. Indeed, we can also do this for angular momentum. If we do so, we find that these two conservation laws are independent of one another. Any momenta is conserved, regardless of the other. Therefore, both are conserved, and neither can be "converted" to the other.

I would like to give a different argument:

Linear momentum and angular momentum have different dimensions - they are essentially different - you can't get apple from orange, nor can you add apple to orange.

Since the dimension of angular momentum is equal to dimension of linear momentum times length: $$[J]=[P]\cdot L$$

Obviously, there is no way to equal $$[J]$$ to $$[P]$$, so the answer to your question is: Nope.

The flaw in your argument is: in fact, there is no increase nor decrease of angular momentum afterward. If you sum up the changes of angular momentum of two tops after the collide, they will result in zero, as angular momentum is conserved in your thought experiment. The same for linear momentum.

I do not know mathematical proof but No, that should not happen.

Angular momentum can be a spinning body, or rotating body.

If the spinning body hits another body, the bodies may bounce from one another, but that would not change the linear momentum of the system, because, the bounce will be in opposite directions and linear momentum of system will remain unchanged.

If rotating body hits another body. Rotating means one body is orbiting another body, or both the bodies are rotating around one another. In this case, also, the angular momentum is due to the mutual attraction of bodies. And when they collide, they equally push/pull one another to merge.

The question then would be, where the kinetic energy due to angular momentum goes - That is lost in the collision, or converted into rotation of the combined body.

• What do you mean "do not have mathematical proof"? The conservation of linear and angular momentum arises from Noether's theorem because a collision is translationally and rotationally invariant. – ACuriousMind Mar 20 '16 at 18:05
• The bottom line here is that when tops collide they lose their rotational speed and increase their linear speed. – Ambrose Swasey Mar 20 '16 at 18:09
• @ACuriousMind: I meant I do not know the proof. – kpv Mar 20 '16 at 21:35
• @TylerDurden: The tops bounce in opposite directions, thus do not change linear momentum. – kpv Mar 20 '16 at 21:39

Yes, I just did a thought experiment that shows that indeed angular momentum will be converted to linear momentum.

Imagine we have two gyroscopes or tops spinning on a table. The tops each spin rapidly but move around the table slowly, i.e., they have a lot of angular momentum, but little linear momentum.

Now, imagine the two tops collide, what will happen is that both of the tops will shoot off at high speed, there angular momentum being converted to linear momentum.

It can be seen in this kenkogama video. When the tops touch they move away from each other much more rapidly than they approached.

• Incorrect I'm afraid. If the tops shoot off in opposite directions at equal speeds then the change in linear momentum has been zero. Remember that momentum is a vector so equal and opposite momenta sum to zero. – John Rennie Mar 20 '16 at 18:00
• The total linear momentum of the system can still be very small. What may happen here is that rotational kinetic energy may be transformed into translational kinetic energy, but the total angular and linear momenta will stay the same. – HDE 226868 Mar 20 '16 at 18:01
• @HDE226868 No, if two tops, moving in a straight line at 5 meters per second collide and both shoot off at 50 meters per second, then their linear momentum has increased. – Ambrose Swasey Mar 20 '16 at 18:02
• @TylerDurden: from this last comment it sounds as if you're really asking if rotational energy can be converted to translational energy, which as your example of the tops shows it can. Energy is a scalar so it is simply additive regardless of direction. It doesn't make any physical sense to take the magnitude of the momenta and add them without considering the direction. The result is not a physically meaningful quantity. – John Rennie Mar 20 '16 at 20:20
• A thought experiment never shows that something is true. The thing they do for you, assuming you ask the right questions, is illuminate the consequences of a supposition or assumption. That makes them a powerful tool, but they never substitute for data. – dmckee Mar 20 '16 at 23:23

## protected by Qmechanic♦Mar 20 '16 at 19:52

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