Linear momentum converted to angular momentum?

[Michio Cuckoo]

http://img688.imageshack.us/img688/2310/81332204.png 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?

 

[tiny-tim]

Hi Michio Cuckoo!

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

No such thing as only linear momentum.

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

 

[Michio Cuckoo]

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).

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

 

[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.

 

[Michio Cuckoo]

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.

Your explanation isn't very clear.

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

 

[Doc Al]

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

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

 

[Michio Cuckoo]

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

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.

 

[Doc Al]

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

Well, depending upon your chosen reference point, that formula shows that the angular momentum of the small ball is nonzero.

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

I suspect you are confusing total angular momentum with angular momentum about an object's center of mass.

 

[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!

 

[Doc Al]

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 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'.

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.

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.

 

[haruspex]

Your explanation isn't very clear.

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

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?

 

[jfy4]

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?

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:
<br /> L_{\text{small ball about }\mathcal{O}}+L_{\text{large ball about }\mathcal{O}}+I_{\text{large ball}}\omega=0<br />
from
<br /> 0=L_{\text{initial}}=L_{\text{final}}<br />
So that total angular momentum is conserved, but individually it need not.

 

[haruspex]

So that total angular momentum is conserved, but individually it need not.

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

 

[jfy4]

ok, cool :)

 

[Michio Cuckoo]

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:
<br /> L_{\text{small ball about }\mathcal{O}}+L_{\text{large ball about }\mathcal{O}}+I_{\text{large ball}}\omega=0<br />
from
<br /> 0=L_{\text{initial}}=L_{\text{final}}<br />
So that total angular momentum is conserved, but individually it need not.

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

 

[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.

 

[Michio Cuckoo]

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.

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

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

 

[jfy4]

Is the Earth spinning according to you?

 

[kmarinas86]

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

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.

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

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.

 

[Michio Cuckoo]

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.

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.

 

[jfy4]

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?

The question I think you are trying to ask is:

what sort of motion is always distinguishable?

or

what defines non-inertial motion?

The local gravitational field determines what motions are inertial. If you don't move along the gravitationaly defined paths then you are moving non-inertially, and it can be felt/distinguished.

But, two people in the "Graviton" (its a circus ride), certainly don't see each other spinning since they are both on the ride.

Does that help?

 

[Darwin123]

>The small ball only has linear momentum,
This is a false statement. The small ball has a nonzero
angular momentum about a axis that doesn't include
the small ball.
For example, chose an axis that passes through the center of the
large ball and points out of the page. Once the small ball is fired,
the angular momentum of the small ball around this axis is very large.
To make up for this increase in angular momentum, the large
ball has to rotate in the opposite direction.
Angular momentum varies with the axis of rotation. When one
does a problem with angular momentum, one has to choose a
convenient axis of rotation and stick with it.

 

[Michio Cuckoo]

The question I think you are trying to ask is:

or

The local gravitational field determines what motions are inertial. If you don't move along the gravitationaly defined paths then you are moving non-inertially, and it can be felt/distinguished.

But, two people in the "Graviton" (its a circus ride), certainly don't see each other spinning since they are both on the ride.

Does that help?

But I wanted you to ignore all relativistic effects.

 

[kmarinas86]

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.

Yes and yes.

 

[jfy4]

But I wanted you to ignore all relativistic effects.

I did, I stick by my answer, it's true even for Galilean relativity.

 

[kmarinas86]

[1] The local gravitational field determines what motions are inertial.

[2] If you don't move along the gravitationaly defined paths then you are moving non-inertially,

[3] and it can be felt/distinguished.

The only reason why you feel anything at all is material strain. Acceleration is only felt because application of force is uneven causing material strain, which is detected by your nervous system. If the force/mass density is evenly distributed in 3D dimensions, you don't feel a thing, regardless of how many G's you are pulling. However, to do that requires something like gravity, or alternatively, a force that is applied directly to each particle of your body. The seats people sit on don't have the ability to apply the force like that. The result is a gradient of pressure directed towards the direction of acceleration.

Therefore, non-inertial vs. inertial motion cannot be distinguished purely in terms of how something is "felt".

If in fact you are falling inertially in a non-uniform gravitational field, your nerves will still strain, because the inertial paths for each cell will be different, and they will converge and/or diverge. The tidal force is another manifestation of non-uniform gravitational fields, and it exists for any appreciable mass.

http://en.wikipedia.org/wiki/Microgravity#Free_fall

In Low Earth orbit (LEO), the force of gravity decreases upward by 0.33 μg/m. Objects which have a non-zero size will be subjected to a tidal force, or a differential pull, between the high and low ends of the object. (An extreme version of this effect is spaghettification.)

If you really think about it, it is obvious that a body in "freefall" in a non-uniform gravitational field is made up of particles which are not in freefall.

I did, I stick by my answer, it's true even for Galilean relativity.

No. [1] and [2] are not true for Galilean relativity as an inertial path in Galilean relativity must be a straight path. [3] is wrong for both General and Galilean relativity because non-inertial acceleration does not necessarily lead to a pressure gradient, so you might not feel it, and conversely, inertial motion can coincide with it, which would cause you to feel the relative movements inside of you.

 

[jfy4]

The only reason why you feel anything at all is material strain. Acceleration is only felt because application of force is uneven causing material strain, which is detected by your nervous system. If the force/mass density is evenly distributed in 3D dimensions, you don't feel a thing, regardless of how many G's you are pulling. However, to do that requires something like gravity, or alternatively, a force that is applied directly to each particle of your body. The seats people sit on don't have the ability to apply the force like that. The result is a gradient of pressure directed towards the direction of acceleration.

Therefore, non-inertial vs. inertial motion cannot be distinguished purely in terms of how something is "felt".

You are right, but the "feeling" analogy is I think a good place to start, and I think you are being a little pedantic.

If in fact you are falling inertially in a non-uniform gravitational field, your nerves will still strain, because the inertial paths for each cell will be different, and they will converge and/or diverge. The tidal force is another manifestation of non-uniform gravitational fields, and it exists for any appreciable mass.

http://en.wikipedia.org/wiki/Microgravity#Free_fall



If you really think about it, it is obvious that a body in "freefall" in a non-uniform gravitational field is made up of particles which are not in freefall.

Absolutely, but as I said above, I think you focused too much on my use of the word "feeling"... I regret using that word now for your sake.

No. [1] and [2] are not true for Galilean relativity as an inertial path in Galilean relativity must be a straight path. [3] is wrong for both General and Galilean relativity because non-inertial acceleration does not necessarily lead to a pressure gradient, so you might not feel it, and conversely, inertial motion can coincide with it, which would cause you to feel the relative movements inside of you.

What you wrote is not in conflict with what I wrote, except for your use of the word "No" at the begining...

The reason spinning in a circle in flat space-time is non-inertial is because geodesics are stright lines... we agree... In fact the majority of your post hinged on us not considering a uniform G-field, and taking tidal effects into consdieration, whereas I was answering OPs question about local, Galilean reference frames, with uniform G-fields. you don't have to be a human to feel non-inertial motion, just be an electron, and spin in a circle. Light shows us how it feels. If you just don't like my use of the word feel, I can change it from now on.

spinning in a circle is motion that is distinguishable, you can tell you are doing it. But the frequncy of rotation isn't absolute. Just run around the outside of a merry-go-round and tell me if you ever noticed it complete even one period; then stop running.

 

[kmarinas86]

spinning in a circle is motion that is distinguishable, you can tell you are doing it. But the frequncy of rotation isn't absolute. Just run around the outside of a merry-go-round and tell me if you ever noticed it complete even one period; then stop running.

If we understand that spherically-symmetric static scalar potentials do not have a rotational component, and yet they provide a definitive force, it is clear that given a force and a distance away from the axis of revolution a certain velocity with respect to a non-rotating inertial reference frame must be attained to make sure that an object stays in a circular orbit as opposed to falling down the gradient of the potential. This is true for electrical charges, and it is true for gravitational masses. Without some kind of orbital velocity around a star, planets such as Earth and Saturn, to name a few, would simply crash into the Sun. The rate of revolution of the Earth and other planets is indeed physical and is not some arbitrary matter of one's frame of choosing.

It's also true for pizza dough:



In this case, you have vicosity instead of gravity, but the idea is mechanically analogous. Rotation is real and physical. It's not an artifact of observation.

 

[jfy4]

In this case, you have vicosity instead of gravity, but the idea is mechanically analogous. Rotation is real and physical. It's not an artifact of observation.

We agree on this, but certainly the period of rotation is not absolute, which you claimed true a few posts ago, I am primarily concerned that the OP gets this, and I hope our agreement on this will settle it.

 

[kmarinas86]

We agree on this, but certainly the period of rotation is not absolute, which you claimed true a few posts ago, I am primarily concerned that the OP gets this, and I hope our agreement on this will settle it.

If you neglect relativity, as the O.P. insists that we do, then how do you suppose that the period of rotation is not absolute? I mean, we know how long it takes for Mars to orbit the Sun. We don't say that the Martian year depends on what planet we happen to be looking from, now do we?