Question about a Rotating system with mass moving inwards

[losbellos]

TL;DR Summary

I have a question. I have a rotating tube like a line that has two end and one of them is the center of rotation (like a watch arrow just tube), and inside the tube a mass that is moving towards the center of rotation...

Dear People,
I have a question. I have a rotating tube like a line that has two end and one of them is the center of rotation (like a watch arrow just tube), and inside the tube a mass that is moving towards the center of rotation. So the masses moving along the line aka along the length of the tube. Both starts to move from the same distance from the center of rotation and has the same rotational velocity. Now I have 2 case:

1, the mass moves inwards to the center of rotation slower
2, the mass moves inwards to the center of rotation faster

Both cases the initial angular velocity is the same. Both of them reaches a specific distance which is the same in both case but of course it takes different times. So is that true that the angular velocity of the system would be the same when the mass reaches the destination which is the same for both cases. So it would not matter how fast they move inwards towards the center the final velocities would be the same at the destination? I mean this is how the kinetic energy preservation aka angular momentum conservation would work right?

Thank you!

 

[vanhees71]

I'd just derive the equations of motion for the mass inside the tube. It's a nice excercise using the Lagrange formalism!

 

[losbellos]

I'd just derive the equations of motion for the mass inside the tube. It's a nice excercise using the Lagrange formalism!

Well I am only asking it because this is how I know it. And if I would put into a solver it would be the same. So I would like to know the answer pls.

 

[Lnewqban]

...
1, the mass moves inwards to the center of rotation slower
2, the mass moves inwards to the center of rotation faster

Both cases the initial angular velocity is the same. Both of them reaches a specific distance which is the same in both case but of course it takes different times. So is that true that the angular velocity of the system would be the same when the mass reaches the destination which is the same for both cases. So it would not matter how fast they move inwards towards the center the final velocities would be the same at the destination? I mean this is how the kinetic energy preservation aka angular momentum conservation would work right?
...

I would say that the initial and final velocities would be the same.
For case 1, it would take longer to reach the final velocity than for case 2.

 

[jbriggs444]

Well I am only asking it because this is how I know it. And if I would put into a solver it would be the same. So I would like to know the answer pls.

The process is irrelevant. The fact that the final configurations are identical and the fact that angular momentum is conserved fixes the final angular velocity regardless of how the final configuration came to be achieved.

 

[losbellos]

I'd just derive the equations of motion for the mass inside the tube. It's a nice excercise using the Lagrange formalism!

The process is irrelevant. The fact that the final configurations are identical and the fact that angular momentum is conserved fixes the final angular velocity regardless of how the final configuration came to be achieved.

Obvious now. I had such a trauma this weekend some people sent me files that did something else and it was from a very famous simulation package... It doesn't conserve the energy!

 

[vanhees71]

Maybe I misunderstood the problem. So let's do the calculation. Of course it's most convenient to use polar coordinates with ##\varphi=\omega t##, with ##\omega## the angular velocity of the rotating tube. Then
$$\vec{x}=\begin{pmatrix} r \cos(\omega t) \\ r \sin(\omega t)\end{pmatrix}$$
and
$$\dot{\vec{x}}=\dot{r} \begin{pmatrix} \cos(\omega t) \\ \sin(\omega t) \end{pmatrix} + \omega r \begin{pmatrix}-\sin(\omega t) \\ cos(\omega t) \end{pmatrix}.$$
The Lagrangian thus is
$$L=\frac{m}{2} \dot{\vec{x}}^2 = \frac{m}{2} (\dot{r}^2+\omega^2 r^2).$$
The equation of motion follows from
$$p_r=\frac{\partial L}{\partial \dot{r}}=m \dot{r}, \quad \dot{p}_r=m \ddot{r} = \frac{\partial L}{\partial t} = m \omega^2 r.$$
The general solution is
$$r(t)=r_0 \cosh(\omega t) + \frac{v_0}{\omega} \sinh(\omega t).$$
Now you can plug in any initial condition you like to see, what happens.

 

[losbellos]

My solution is somehow different. My solution energy conserving and it is not outside of the generally accepted newtonien basic mechanics.

Let me share a scientific method to calculate this with time steps.
The idea here is that whatever amount of energy it has to pass it also has to loose and it is implemented with Toruqe's and precalculations.

FEA calculate the tube torque requirement for reference acceleration. ra
element's mass center
arml = distance x,y
TQtube = TQtube +ra/(r/arml) * arml * element's mass ;

Same for the mass inside at every time step and reference the ra to the mass center of the mass. This means the mass elements away from center will have ra/(mc/element distance) multiplied otherwise divided values.

for inwards motion
TQX
(vt+( TQX/TQtube *ra*ts))/(vm-( TQX/TQmass *ra*ts)) == r/rn(mass center's distance from origo)
Now solve for TQX and you will use it
vt = (vt+( TQX/TQtube *ra*ts))
vm = (vm-( TQX/TQtube *ra*ts))

The resulting velocities always be in tune with the system so the energy too and this tell this too. This also solves that the velocity of the mass inwards direction doesnt matter only the location...

This produces very accurate and fast result with time steps.

 

[jbriggs444]

My solution energy conserving

Energy is not conserved in this scenario. We have some unspecified agency pumping in the energy that draws the two masses closer together in the tube.

 

[losbellos]

Validate your solver it is good!

 

[vanhees71]

My solution is somehow different. My solution energy conserving and it is not outside of the generally accepted newtonien basic mechanics.

Let me share a scientific method to calculate this with time steps.
The idea here is that whatever amount of energy it has to pass it also has to loose and it is implemented with Toruqe's and precalculations.

FEA calculate the tube torque requirement for reference acceleration. ra
element's mass center
arml = distance x,y
TQtube = TQtube +ra/(r/arml) * arml * element's mass ;

Same for the mass inside at every time step and reference the ra to the mass center of the mass. This means the mass elements away from center will have ra/(mc/element distance) multiplied otherwise divided values.

for inwards motion
TQX
(vt+( TQX/TQtube *ra*ts))/(vm-( TQX/TQmass *ra*ts)) == r/rn(mass center's distance from origo)
Now solve for TQX and you will use it
vt = (vt+( TQX/TQtube *ra*ts))
vm = (vm-( TQX/TQtube *ra*ts))

The resulting velocities always be in tune with the system so the energy too and this tell this too. This also solves that the velocity of the mass inwards direction doesnt matter only the location...

This produces very accurate and fast result with time steps.

What I posted above IS the accepted Newtonian dynamics. Since the Lagrangian is not explicitly time dependent, "energy" is conserved. The Hamiltonian is
$$H=p_r \dot{r}-L=\frac{m}{2} \dot{r}^2 -\frac{m \omega^2}{2} r^2=\text{const}.$$

 

[losbellos]

Energy is not conserved in this scenario. We have some unspecified agency pumping in the energy that draws the two masses closer together in the tube.

Thats a wrong understanding of the energy preservation! you are talking about the full value but we are talking about the mass and the tube circumferential velocities and energies.... The two must be the same plus your energy input is perpendicular so if you add that vector there it would not change the value's length to the radius-perpendicular aka velocity vector...

 

[weirdoguy]

kinetic energy preservation aka angular momentum conservation

Not "aka" because these are two different things.

 

[losbellos]

Sorry for my language, so the energy preservation and the conservation of angular momentum is the same in the sense of preserving the systems total energy or angular momentum. Because both mean velocity and masses. But its ok, better to be strict.

 

[jbriggs444]

Sorry for my language, so the energy preservation and the conservation of angular momentum is the same in the sense of preserving the systems total energy or angular momentum. Because both mean velocity and masses. But its ok, better to be strict.

Conservation of angular momentum does not entail conservation of energy. Consider the case where, as here, the moment of inertia of the system can change due to an internal reconfiguration.

The classic example is a spinning skater who pulls in her arms and increases her rotation rate as a result. It takes effort to pull in the arms. That effort manifests as increased mechanical energy.

 

[losbellos]

Conservation of angular momentum does not entail conservation of energy. Consider the case where, as here, the moment of inertia of the system can change due to an internal reconfiguration.

The classic example is a spinning skater who pulls in her arms and increases her rotation rate as a result. It takes effort to pull in the arms. That effort manifests as increased mechanical energy.

why do you have to include that into this while I even stated do not! Please read it through!

 

[jbriggs444]

why do you have to include that into this while I even stated do not! Please read it through!

Because your claim that mechanical energy is conserved in this scenario is incorrect.

 

[vanhees71]

There's energy conservation in this scenario. See #11.

 

[jbriggs444]

There's energy conservation in this scenario. See #11

Apparently I had trouble with the problem description. So we have a tube anchored to a pivot at one end and a mass projected up the tube rather than a mass drawn up the tube.

 

[vanhees71]

Ok, then I don't understand the problem right either. Of course, in the setup I consider, i.e., a rotating tube with a frictionless moving point mass inside, you have only a centrifugal force outwards, and that's what the Lagrange formalism applied above indeed gives.