Summing linear momentum in circular rotation

[Chenkel]

TL;DR Summary

I'm wondering about this approach, and also the motivation for the definition of angular momentum.

Hello everyone!

I was wondering why can't we take a rotating body and see the linear movement that each particle moves to find the 'total linear momentum,' I imagine this quantity would be conserved, and furthermore couldn't you write the total linear momentum as a function of angular velocity? i.e, $$p = \int \dot \theta r dm = \dot \theta \int r dm$$If this is true then what is the motivation for the definition of angular momentum?

 

[Chenkel]

Just to clearify, I know what the definition of angular momentum is, but I'm wondering about the approach of looking at a body's linear momentum when rotating and setting the system to contain all elemental masses of the body, and solving for angular velocity to study the body's motion. We could apply an 'external force' instead of an external torque if we want to cause an angular acceleration. My intuition is that the angular momentum approach is better, but I'm not sure why. Something tells me that trying to analyze a rotating body in terms of linear momentum might be just plain wrong, difficult, and just incorrect, but it's hard for me to figure out why. If there's someone who can shed light on this topic it will be greatly appreciated, thank you!

 

[Filip Larsen]

The usefulness of angular momentum depends on what you model. For, say, a cloud of non-interacting particles (dust) there are no particular benefit of using angular momentum instead of simple linear momentum of the particles. But angular momentum for a collection of particles becomes really useful and practical once you start model the dynamics of rigid object where the particles maintain their relative position by exchanging internal forces, especially since the linear and rotational dynamics can become separable when using the center of mass as reference point. That does not mean you could not also model the same situation using linear momentum and forces only, just that it probably will become rather tedious to do, e.g. you have to keep track of all the internal force pairs as the object rotates.

 

[Orodruin]

couldn't you write the total linear momentum as a function of angular velocity? i.e, $$p = \int \dot \theta r dm = \dot \theta \int r dm$$

First of all, what you wrote down is not the linear momentum. Momentum is a vectorial quantity and its direction matters. Of course, nothing stops you from writing down an expression like that, but the question is whether or not it is useful or not. Both linear momentum and angular momentum are useful quantities, particularly when conserved, which is related to translational and rotational symmetry.

For, say, a cloud of non-interacting particles (dust) there are no particular benefit of using angular momentum instead of simple linear momentum of the particles.

I disagree. There is generally no ”instead of” here. Both are useful quantities in their own right. Both for the case of a cloud of particles as well as for a rigid body. Once you do things such as constrain the rigid body to rotate around a single point or axis, then the usefulness of linear momentum becomes less relevant, but that is because you then introduced an unknown force at that point and most of the analysis can be made using angular momentum only.

 

[Chenkel]

First of all, what you wrote down is not the linear momentum. Momentum is a vectorial quantity and its direction matters. Of course, nothing stops you from writing down an expression like that, but the question is whether or not it is useful or not. Both linear momentum and angular momentum are useful quantities, particularly when conserved, which is related to translational and rotational symmetry.I disagree. There is generally no ”instead of” here. Both are useful quantities in their own right. Both for the case of a cloud of particles as well as for a rigid body. Once you do things such as constrain the rigid body to rotate around a single point or axis, then the usefulness of linear momentum becomes less relevant, but that is because you then introduced an unknown force at that point and most of the analysis can be made using angular momentum only.

I see, you make a good point. My expression doesn't capture the direction for the momentum of the elemental masses.. I do wonder why angular momentum has an additional dimension of length compared to linear momentum, there are definitely some very useful things that come from the definition, such as conservation of angular momentum, torque, direction of rotation, etc, I can't think of a 'better' way to do it, but sometimes I wonder if there is a very good way to see the motivation for angular momentum's definition and it's dimensions.

 

[malawi_glenn]

I do wonder why angular momentum has an additional dimension of length

Because it is defined as (for a point particle)
angular momentum w.r.t. point A:
$\vec L_A= \vec r_A \times \vec p$ where $\vec r_A$ is the displacement vector from the point A to the particle.
The unit of L is thus unit of momentum times unit of length.

I wonder if there is a very good way to see the motivation for angular momentum's definition

Angular momentum and torque plays the rotational role as momentum and force

https://scripts.mit.edu/~srayyan/PERwiki/images/b/b3/3BallsTurnTable.png

Now when ball A, B and C hits the apparatus, there will a force of impact from each of them on the apparatus. We can determine the way the apparatus will rotate around Q, based on the angular momentum w.r.t. Q by the particles A, B and C. We can do that by using conservation of angular momentum w.r.t Q.

 

[Orodruin]

but sometimes I wonder if there is a very good way to see the motivation for angular momentum's definition and it's dimensions.

There is. It is the generator of rotations just as linear momentum generates translations. This is why they are the corresponding conserved quantities when these are symmetries.

 

[malawi_glenn]

Translational invariance -> linear momentum conservation
Rotational invariance -> angular momentum conservation
by Noethers theorem

https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/07:_Symmetries_Invariance_and_the_Hamiltonian/7.04:_Rotational_invariance_and_conservation_of_angular_momentum

 

[jbriggs444]

I imagine this quantity would be conserved,

So we have this skater on the ice. She pulls her arms in and speeds up. We already know that angular momentum is conserved. That is, we know that $\sum m \omega r^2$ is conserved.

You speculate that $\sum m \omega r$ is also conserved. But the two conservation laws lead to two different predictions.

Let us idealize the skater. 100% of her mass is in her arms. They are extended to a radius of one meter. She pulls them into a 1/2 meter radius. Our prediction is that her rotation rate increases by a factor of four. Your prediction is that her rotation rate increases by a factor of two.

We cannot both be right.

 

[malawi_glenn]

Let us idealize the skater. 100% of her mass is in her arms.

That is not an ideal body of a skater

We cannot both be right.

Same order of magnitude though

 

[Chenkel]

So we have this skater on the ice. She pulls her arms in and speeds up. We already know that angular momentum is conserved. That is, we know that $\sum m \omega r^2$ is conserved.

You speculate that $\sum m \omega r$ is also conserved. But the two conservation laws lead to two different predictions.

Let us idealize the skater. 100% of her mass is in her arms. They are extended to a radius of one meter. She pulls them into a 1/2 meter radius. Our prediction is that her rotation rate increases by a factor of four. Your prediction is that her rotation rate increases by a factor of two.

We cannot both be right.

Hmm, I tried thinking about this and I wonder why the rotation rate is different, obviously for some reason my equation can't predict the angular velocity...

 

[jbriggs444]

Hmm, I tried thinking about this and I wonder why the rotation rate is different, obviously for some reason my equation can't predict the angular velocity...

Yes indeed. There is a reason for that.

Your quantity is not a conserved quantity.

 

[pbuk]

Same order of magnitude though

I know that was meant to be a joke, but I think it could be confusing: to be clear, it is definitely NOT the same order of magnitude: one is linear (proportional to $r$) and the other quadratic ($r^2$).

 

[Orodruin]

it is definitely NOT the same order of magnitude: one is linear (proportional to r) and the other quadratic (r2).

Yes it is. Two and four are the same order of magnitude and that is what was being discussed.

 

[Chenkel]

Yes indeed. There is a reason for that.

Your quantity is not a conserved quantity.

That's hard for me to wrap my brain around, if you consider all the point masses of a body, and the magnitude of each mass's linear momentum about the center of mass, $r \dot \theta dm$, as the object rotates, in the absence of external force, shouldn't the magnitude of total linear momentum be conserved?

 

[Chenkel]

That's hard for me to wrap my brain around, if you consider all the point masses of a body, and the magnitude of each mass's linear momentum about the center of mass, $r \dot \theta dm$, as the object rotates, in the absence of external force, shouldn't the magnitude of total linear momentum be conserved?

If we take each elemental mass's momentum vector about the cm, and sum it, we will get the momentum vector for the system, but I think one problem with my summation of $r \dot \theta dm$ is that I'm taking the magnitude of all these little momentum vectors, and saying the sum total is equal to the magnitude of the momentum vector for the entire system, which isn't true.

 

[Chenkel]

If we take each elemental mass's momentum vector about the cm, and sum it, we will get the momentum vector for the system, but I think one problem with my summation of $r \dot \theta dm$ is that I'm taking the magnitude of all these little momentum vectors, and saying the sum total is equal to the magnitude of the momentum vector for the entire system, which isn't true.

But if we take magnitude of each little momentum vector about cm and sum it I have a hard time imagining that quantity wouldn't be conserved across time, but it might not be conserved as we change the geometry of the object, still, finding out if it's conserved across geometric changes is harder for me to figure out

 

[Orodruin]

But if we take magnitude of each little momentum vector about cm and sum it I have a hard time imagining that quantity wouldn't be conserved across time, but it might not be conserved as we change the geometry of the object, still, finding out if it's conserved across geometric changes is harder for me to figure out

If you sum all the contributions to linear momentum then you get total linear momentum. This may or may not be conserved for an object depending on the external force on the object.

 

[jbriggs444]

But if we take magnitude of each little momentum vector about cm and sum it I have a hard time imagining that quantity wouldn't be conserved across time, but it might not be conserved as we change the geometry of the object, still, finding out if it's conserved across geometric changes is harder for me to figure out

That is why you should consider the skater. Her geometry changes under purely internal forces.

 

[Chenkel]

If you sum all the contributions to linear momentum then you get total linear momentum. This may or may not be conserved for an object depending on the external force on the object.

But I'm summing all magnitudes of linear momentum, and that loses the information for direction. I understand the way of getting total linear momentum, you take the vector sum of all momentum vectors. I'm wondering if the sum of all $r \dot \theta dm$ is conserved.

 

[jbriggs444]

That's hard for me to wrap my brain around, if you consider all the point masses of a body, and the magnitude of each mass's linear momentum about the center of mass, $r \dot \theta dm$, as the object rotates, in the absence of external force, shouldn't the magnitude of total linear momentum be conserved?

If you are talking about $\sum m |\vec{r}| \cdot |\vec{p}|$ for a rigid object in stable rotation (like a phonograph disc on a turn table) then yes. That figure will be conserved.

A phonograph disk that shrinks (upon cooling, perhaps) or which sheds a small fragment will fail to conserve that quantity, however.

If you are contemplating a non-rigid object, the notion of $\dot \theta$ may even cease to be meaningful.

 

[Chenkel]

That is why you should consider the skater. Her geometry changes under purely internal forces.

So if you take two arms where half the mass is distributed in each hand, each radius of 1 meter from the body then you would get for $q = \int r \dot \theta dm$ the value $q = \dot \theta$, if the arms come in by 50% you get $q = \frac 1 2 \dot \theta$ this quantity is conserved if $\dot \theta$ doubles for when she brings in her hands, but it does not, I understand that experimentation shoes that the quantity $\int {r^2} \dot \theta dm$ is conserved. I.e when the quantity $\int r^2 dm$ doubles the angular velocity halves, is that because bringing the arms inward requires no torque? I'm just trying to see what the theoretical deduction is for angular momentum and how that deduction matches reality.

 

[Orodruin]

I'm wondering if the sum of all rθ˙dm is conserved.

Then no. Physics simply does not work like that.

 

[Orodruin]

I'm just trying to see what the theoretical deduction is for angular momentum and how that deduction matches reality.

Consider a set of masses $m_i$. Each mass has position $\vec x_i$ and momentum $\vec p_i = m_i \dot{\vec x}_i$. Total angular momentum is given by
$$
\vec L = \sum_{i} \vec x_i \times \vec p_i.
$$
Taking the time derivative:
$$
\dot{\vec L} = \sum_i [\dot{\vec x}_i \times \vec p_i + \vec x_i \times \dot{\vec p}_i].
$$
The first term in the sum is zero because velocity and momentum for a given mass are parallel, meaning the cross product evaluates to zero. Furthermore, $\dot{\vec p}_i = \vec F_i$, the force acting on mass $i$. The second term in the sum is therefore the torque $\vec\tau_i$ on mass $i$ and consequently
$$
\dot{\vec L} = \sum_i \vec\tau_i \equiv \vec\tau,
$$
the total torque. If the total torque is zero, it then follows that $\vec L$ is conserved.

 

[Chenkel]

Consider a set of masses $m_i$. Each mass has position $\vec x_i$ and momentum $\vec p_i = m_i \dot{\vec x}_i$. Total angular momentum is given by
$$
\vec L = \sum_{i} \vec x_i \times \vec p_i.
$$
Taking the time derivative:
$$
\dot{\vec L} = \sum_i [\dot{\vec x}_i \times \vec p_i + \vec x_i \times \dot{\vec p}_i].
$$
The first term in the sum is zero because velocity and momentum for a given mass are parallel, meaning the cross product evaluates to zero. Furthermore, $\dot{\vec p}_i = \vec F_i$, the force acting on mass $i$. The second term in the sum is therefore the torque $\vec\tau_i$ on mass $i$ and consequently
$$
\dot{\vec L} = \sum_i \vec\tau_i \equiv \vec\tau,
$$
the total torque. If the total torque is zero, it then follows that $\vec L$ is conserved.

Thank you, that looks clear to me, so the theory is that when the ice skater brings their arms in the momentum quantity will be conserved (because there is no external torque) which means the angular velocity will increase as the moment of inertia decreases. But I think it should be noted that momentum will not be conserved during the time when the ice skater first goes into the spin (because the the ice skater needs to apply a torque to do this.)

 

[jbriggs444]

Thank you, that looks clear to me, so the theory is that when the ice skater brings their arms in the momentum quantity will be conserved (because there is no external torque) which means the angular velocity will increase because the moment of inertia stay the same.

No.

The "momentum quantity" that you have invented will not be conserved.
Angular momentum will be conserved because there is no relevant external torque.
Angular velocity will increase because the moment of inertia has decreased.

 

[Chenkel]

No.

The "momentum quantity" that you have invented will not be conserved.
Angular momentum will be conserved because there is no relevant external torque.
Angular velocity will increase because the moment of inertia has decreased.

Thank you for the correction, I modified my post to keep those things in mind.