# Linear momentum converted to angular momentum?

1. May 27, 2012

### Michio Cuckoo

http://img688.imageshack.us/img688/2310/81332204.png [Broken]

Originally the big ball is at rest and not moving at all. It has a cannon attached to it.

The cannon fires a smaller ball.

The small ball only has linear momentum, BUT the big ball now rotates and has both linear and angular momentum.

Does this mean that linear momentum can be converted to angular momentum? Where did the angular momentum come from?

Last edited by a moderator: May 6, 2017
2. May 27, 2012

### tiny-tim

Hi Michio Cuckoo!
No such thing as only linear momentum.

Everything (that moves) has angular momentum (except about an axis that its velocity passes through).

3. May 27, 2012

### Michio Cuckoo

So the small ball that was ejected also has angular momentum?

4. May 27, 2012

### haruspex

Angular momentum is measured about a reference point. If you take moments about the cannon the small ball has no moment, but then neither does the cannonball.
If you take moments about the centre of the cannonball then the small ball has a moment.

5. May 27, 2012

### Michio Cuckoo

There are only 3 objects, the small ball, big ball, and cannon, which is attached to the big ball

6. May 27, 2012

### Staff: Mentor

Sure. $\vec{L} = \vec{r}\times\vec{p}$

7. May 27, 2012

### Michio Cuckoo

Thats the formula, but how does it show that the small ball has angular momentum?

It seems as though the small ball only has linear momentum.

8. May 27, 2012

### Staff: Mentor

Well, depending upon your chosen reference point, that formula shows that the angular momentum of the small ball is nonzero.
I suspect you are confusing total angular momentum with angular momentum about an object's center of mass.

9. May 27, 2012

### jfy4

Please correct me if I'm wrong, but I think the OP is talking about the the fact that the big ball is spinning, not just translating about some reference point $\mathcal{O}$. OP is worried that there isn't angular momentum conservation because only one of the objects is spinning.

I think it's ok, conservation of angular momentum applies in the absence of external torques. In this case though, I think there was an external torque from the firing, in which case we shouldn't expect it to be conserved. Like if I roll a disc down an inclined plane, it certainly isn't spinning to begin with, but at the end, it has $I\omega$. It's because there are external torques.

That said, I've been wrong before!

10. May 27, 2012

### Staff: Mentor

I think you are correct, which is why I mentioned the OP's confusing total angular momentum with angular momentum about an object's center of mass (i.e., spinning). The key point is that it is total angular momentum that is conserved, not 'spinning'.
It's the total angular momentum of the entire system--big ball + little ball--that is conserved, not the angular momentum of either one taken separately. Taken as a single system, the cannon firing is an internal force.

11. May 27, 2012

### haruspex

Pick a reference point - let's take the centre of the cannonball.
After firing, the cannonball is rotating so has a moment about its centre.
What moment does the small ball have about our reference point?
Is it equal and opposite to the moment of the cannonball about the same point?

12. May 27, 2012

### jfy4

I think though that for this scenario, that is not the case. This is what me and Doc are saying. That it actually looks like:
$$L_{\text{small ball about }\mathcal{O}}+L_{\text{large ball about }\mathcal{O}}+I_{\text{large ball}}\omega=0$$
from
$$0=L_{\text{initial}}=L_{\text{final}}$$
So that total angular momentum is conserved, but individually it need not.

13. May 27, 2012

### haruspex

We're saying the same thing. I'm only looking at the total angular momentum.

14. May 27, 2012

### jfy4

ok, cool :)

15. May 30, 2012

### Michio Cuckoo

I don't exactly understand this equation though. What is position \mathcal{O}}+I

16. May 30, 2012

### jfy4

As you noticed, there are two things going on here, the first thing (the thing you noticed), is that one of the balls is spinning, and the other isn't. The other thing (the thing this thread is talking about), is that the angular momentum about a fixed origin changes depending on the choice of origin, but the total physics doesn't change, i.e. total angular momentum is conserved.

You are free to draw a coordinate system anywhere on the picture and place the origin $\mathcal{O}$ wherever you would like. Then you preform $\vec{L}=\vec{r}\times\vec{p}$ to both objects to get their "orbital" angular momentum. Now you can change where you put $\mathcal{O}$ and do the calculation over again.

But! those numbers, the $\vec{L}$s, won't do it for you, you will notice they don't cancel. That is because besides the "orbital angular momentum (the one as a result of choosing $\mathcal{O}$'s location), the big ball has "spin" angular momentum. This needs to get added into your equation for total angular momentum. $L_{initial}=L_{final}$ for the whole system.

Once you add those numbers up, you will find angular momentum is conserved.

Hope this helps.

17. Jun 4, 2012

### Michio Cuckoo

Ignoring all quantum and relativistic effects, is "spin" angular momentum an absolute quantity?

That is, its value is the same for all observers.

18. Jun 4, 2012

### jfy4

Is the earth spinning according to you?

19. Jun 5, 2012

### kmarinas86

If you consider that planets orbit around the sun and that obey Newton's laws to some approximation and that they must move at certain speeds with respect to the sun to have they orbits they do, then yes, spin angular momentum is an absolute quantity.

If the observer rotates, does that make any contribution to the angular momentum of the Andromeda galaxy? There is no such thing as an angular momentum "for an observer" that has any physical consequence.

20. Jun 5, 2012

### Michio Cuckoo

Maybe an observer who is present but does not interact with anything else in the universe.

What I'm trying to say is, the centripetal acceleration experienced by an object is absolute, isn't it? Which means its frequency of rotation is also absolute?

Ignoring relativistic and quantum effects of course.