I Damped oscillator with changing mass

[erobz]

I only referenced them so the OP can have them to compare with not that the problem has to be addressed exactly like these. It doesn't even look like these two do the same treatment anyway so I think there is still room for fun.

For the water bucket on a spring: I'm kind of sad they used Bernoulli's without mentioning that it is derived in an inertial frame. In the frame of reference of the cup, the pressure at the nozzle will oscillate around $\rho gz$. It will inevitably be a function of $\frac{dv}{dt}$ and $z$ itself. I would have liked to see how they neglected it professionally. I would say they hand waived it away, but actually they didn't address that at all. I'm sure it's a fine assumption for all practical purposes, but I would have liked to see it regardless for my own future reference.

 

[Rezex124]

See if these papers help;

Sand;
https://www.researchgate.net/publication/228412656_Variable_mass_oscillator/link/00b4951ad53e60ac25000000/download

Water;
http://www.scielo.org.mx/pdf/rmfe/v60n1/v60n1a3.pdf

I just had a quick glance over them and yes! They will definitely help.

Thank you very much

 

[bob012345]

I just had a quick glance over them and yes! They will definitely help.

Thank you very much

You're welcome but remember just because these papers were published does not mean they are the last word and the best way or only way to do this. Maybe you and your teacher can put together your own approach and publish it.

 

[erobz]

For the water bucket on a spring: I'm kind of sad they used Bernoulli's without mentioning that it is derived in an inertial frame. In the frame of reference of the cup, the pressure at the nozzle will oscillate around $\rho gz$. It will inevitably be a function of $\frac{dv}{dt}$ and $z$ itself. I would have liked to see how they neglected it professionally. I would say they hand waived it away, but actually they didn't address that at all. I'm sure it's a fine assumption for all practical purposes, but I would have liked to see it regardless for my own future reference.

Never mind, I see that they do mention this issue:

"At this point, we point out that the quadratic dependence of mass on time is only a motivator for the leaking oscillator problem, treated here as a purely theoretical problem. Therefore, this result should be considered within its appropriate limitations. Probably, when the bucket is moving, going up and down with the oscillations, the flow rate through the hole could be seen to change as well, deviating slightly from the results obtained here. In other words, we are ignoring the fact that the bucket, as well as the water within it, are accelerating frames. However, we can admit that the quadratic dependence of mass must work as a reasonable approximation in the case the loss of water occurs at a very low rate, and the oscillating bucket experiences smooth motions as investigated in this article"[1]

[1] http://www.scielo.org.mx/pdf/rmfe/v60n1/v60n1a3.pdf

 

[Delta2]

The way I view how LAWII applies is not that the sum of these terms is zero thus making a spurious force but that both terms are zero.

How the term $\dot m v$ can be zero (assuming that the cart has initial velocity)?. And yes we apply the law in the direction of $\vec{v}$, not in the direction of dripping where there is external force (the weight ). And no you just can't say $\dot m$ is zero because the force that causes it is perpendicular to the direction we apply the law, no sorry if the mass is varying w.r.t time then $\dot m\neq 0$ simple as that...

 

[Delta2]

I don't see why the rocket equation wouldn't apply

The rocket equation applies but this equation is derived by applying Newton's 2nd law in a system where the total mass is constant that is the rocket+the exhaust gases.

 

[bob012345]

How the term $\dot m v$ can be zero (assuming that the cart has initial velocity)?. And yes we apply the law in the direction of $\vec{v}$, not in the direction of dripping where there is external force (the weight ).

Because there is no force creating a momentum change. The second law exists and holds, but has a null effect due to the falling water. I am trying to make the point I think the $\dot{m}$ term in the equation $F= m\dot{v}+\dot{m}v=0$ has nothing to do with the $\dot{m}$ due to falling water and in this case of the cart it is null in the same way the $\dot{v}$ term is null.

 

[Delta2]

I am trying to make the point I think the m˙ term in the equation F=mv˙+m˙v=0 has nothing to do with the m˙ due to falling water

The $\dot m$ term is uniquely defined, I don't know how you say in one equation it is this, and in the other equation is that. Sorry you make no sense to me.

 

[bob012345]

The $\dot m$ term is uniquely defined, I don't know how you say in one equation it is this, and in the other equation is that. Sorry you make no sense to me.

I realize my point is subtle. The $\dot m$ term is uniquely defined. It is defined by the existence or not of a force. As I quoted Newton above, I think he meant that a force causes a momentum change not a momentum change causes a force. I think we all can agree there is no slowing down or speeding up, no force, on the cart due to water dripping. You then say there is a paradox if we look at the cart alone and can only resolve it by considering the falling water as well so everything works out to cancel any spurious forces. That makes no sense to me. I think it is more natural to say Newton's Second Law always works in every situation and on every part of a system but understanding and applying it is sometimes tricky. Of course, I could be all wet...

 

[Delta2]

That makes no sense to me. I think it is more natural to say Newton's Second Law always works in every situation and on every part of a system but understanding and applying it is sometimes tricky

And you make no sense to me sorry, you apply the law by defining $\dot m$ as zero, when it is not zero. Sorry that's way too tricky way to apply the law...

 

[alan123hk]

In a broad sense, I agree that a system is defined by people. If we define a system that only considers the force and momentum remaining inside the cart, even though the cart is losing mass, then I wouldn't think the statement that Newton's second law does not apply to this system is false. Because the basis of Newtonian mechanics is the conservation of mass, since this system is not an isolated system, it interacts with other systems, and its mass is not conserved, we cannot apply Newton's laws to this non-isolated system.

This is certainly not a problem of Newtonian mechanics, not a problem of equation $F=m\frac{dv}{dt}+v\frac{dm}{dt}$, nor a problem of system definition, because I think one has the right to choose the parameters involved in defining the system (including non-isolated systems).

For the entire isolated system.
$$\sum F = \frac{dp_1}{dt}+\frac{dp_2}{dt}=\left(m_1\frac{dv}{dt}+v\frac{dm_1}{dt}\right) +\left(m_2\frac{dv}{dt}+v\frac{dm_2}{dt}\right)=m_0\frac{dv}{dt}+v\frac{dm_0}{dt} $$Since $~\sum F ~$ and $~\frac{dm_0}{dt}~$ are equal to zero, $\frac {dv}{dt}=0 ~~$

 

[haruspex]

Newton's Second Law always works in every situation and on every part of a system

I doubt Newton even considered systems of varying mass. We can either specify that Newton 2 only applies to constant mass systems (my preference) or specify how to adjust it to cope with them.
@Orodruin's fix (post #22) is to say that mass entering or exiting the system with its own momentum constitutes a force on the system. I have not seen that elsewhere, but maybe it is standard.

 

[erobz]

In what sense is this used? I remember my Physics professor making a point about Newtons 2nd Law actually being:

$$ \sum F = \frac{d}{dt} \left( mv \right) $$

Yet we never used it that way ( as far as I can tell, or remember )

Is it just the basis for the derivation of Impulse/Momentum?

$$ \int \sum F dt = \int d ( mv ) = m_f v_f - m_o v_o $$

 

[Orodruin]

I mean, effectively Newton's second law is just a definition of force as the rate of change of a system's momentum. By itself it does not tell you very much. Only once you start modelling the forces does it actually give you a physical model. So, we define force as the rate of change in a system's momentum. How then can a system's momentum change? We can have contact or volume forces that transfer momentum between different systems and model those - and we can have momentum leaving a system by having material flowing out of the system and taking some momentum with it. The only question is what you decide to call "an external force" on a system.

 

[bob012345]

So, we define force as the rate of change in a system's momentum.

Do we? I always thought a force has some physical basis and causes a change in momentum when it acts on some object.

 

[bob012345]

I doubt Newton even considered systems of varying mass. We can either specify that Newton 2 only applies to constant mass systems (my preference) or specify how to adjust it to cope with them.
@Orodruin's fix (post #22) is to say that mass entering or exiting the system with its own momentum constitutes a force on the system. I have not seen that elsewhere, but maybe it is standard.

I have seen the argument recently that Newton's Second Law in the differential form doesn't apply to systems of changing mass because of Galilean invariance. Comments?

https://physics.stackexchange.com/questions/53980/second-law-of-Newton-for-variable-mass-systems

 

[erobz]

I have seen the argument recently that Newton's Second Law in the differential form doesn't apply to systems of changing mass because of Galilean invariance. Comments?

https://physics.stackexchange.com/questions/53980/second-law-of-Newton-for-variable-mass-systems

That argument is also in the paper you linked. Section 2. Variable-mass systems and Newton’s second law

http://www.scielo.org.mx/pdf/rmfe/v60n1/v60n1a3.pdf

 

[bob012345]

That argument is also in the paper you linked. Section 2. Variable-mass systems and Newton’s second law

http://www.scielo.org.mx/pdf/rmfe/v60n1/v60n1a3.pdf

Sure but their situation is different from the cart leaking water 90 degrees to its motion.

They also make this claim

In order to correctly obtain the equation of motion, we have to apply the principle of conservation of linear momentum for the entire system, which is the basic principle behind the Newton’s second law

So, they are changing what Newton said too. Not saying it's 'wrong' just saying it's a different perspective.

 

[Orodruin]

Do we? I always thought a force has some physical basis and causes a change in momentum when it acts on some object.

You need to model a force somehow to even put it into Newton's second law. For example, what Newton's law of gravitation tells you is that the rate of momentum exchange between two bodies is given by $\pm GMm/r^2$.

In statics problems in mechanics, you are essentially modelling the forces by requiring them to be contact forces at particular points (or spread out loads) and asking what their magnitudes must be for a particular configuration to be in equilibrium.

Without a model for the forces involved, Newton's second law does not really tell you anything except that there is something called "force" that is a change in an object's momentum.

 

[erobz]

Without a model for the forces involved, Newton's second law does not really tell you anything except that there is something called "force" that is a change in an object's momentum.

"net force" correct?

 

[bob012345]

You need to model a force somehow to even put it into Newton's second law. For example, what Newton's law of gravitation tells you is that the rate of momentum exchange between two bodies is given by $\pm GMm/r^2$.

In statics problems in mechanics, you are essentially modelling the forces by requiring them to be contact forces at particular points (or spread out loads) and asking what their magnitudes must be for a particular configuration to be in equilibrium.

Without a model for the forces involved, Newton's second law does not really tell you anything except that there is something called "force" that is a change in an object's momentum.

That's what I said or meant but I wish the language used made it more clear that a physical force causes a momentum change, not the other way around.

 

[Orodruin]

That's what I said or meant but I wish the language used made it more clear that a physical force causes a momentum change, not the other way around.

A force is a momentum flow rate. This is the point.

 

[Orodruin]

"net force" correct?

Any force is a momentum change rate. What is true is that the total change rate for an object’s momentum is the sum of all individual contributions and so the net force is the total change rate of the object’s momentum.

 

[erobz]

Any force is a momentum change rate. What is true is that the total change rate for an object’s momentum is the sum of all individual contributions and so the net force is the total change rate of the object’s momentum.

You said "momentum change", not "momentum change rate" in what I quoted. Any "force" will not necessarily change the objects momentum, but a "net force" will. Not that it matters.

 

[Orodruin]

You said "momentum change", not "momentum change rate" in what I quoted.

That was just sloppiness. Force is momentum change rate, not momentum change. Momentum change is the time integral of the change rate, ie, impulse.

Momentum change is not a force.

 

[bob012345]

A force is a momentum flow rate. This is the point.

What about static forces?

 

[Orodruin]

What about static forces?

They are also flow rates of momentum. It just so happens that in a static situation the net flow of momentum into an object is zero. In the static case you have p=0 and therefore dp/dt=0. This puts constraints on the momentum flows in/out of the object - they have to add up to zero.

 

[bob012345]

They are also flow rates of momentum. It just so happens that in a static situation the net flow of momentum into an object is zero. In the static case you have p=0 and therefore dp/dt=0. This puts constraints on the momentum flows in/out of the object - they have to add up to zero.

I said to myself when I mentioned static forces that someone here would probably suggest momentum flows adding up to zero. Your answer here almost created a non-zero momentum flow between me, my chair and the floor!

So, I was going to ask you to find at least one academic or published source for that concept of static momentum flows but I found some myself. Both papers are from the same source and I found no others so while this is an approach that has been published, I doubt if it is standard fare in engineering statics courses.

http://www.physikdidaktik.uni-karlsruhe.de/download/statics_momentum_currents.pdf

https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.572.3896&rep=rep1&type=pdf

Essentially, if I understand it, the concept tries to do away with forces altogether and replace them with an electric current analog.

 

[Orodruin]

Essentially, if I understand it, the concept tries to do away with forces altogether and replace them with an electric current analog.

There is no need to replace anything. It is just a question of writing down the continuity equation for momentum. This is commonly done in, for example, fluid dynamics in the Euler equations.

Edit: Or more generally, the Cauchy momentum equations. These describe non-relativistic momentum transport in any continuum.

So, I was going to ask you to find at least one academic or published source for that concept of static momentum flows but I found some myself.

It is a simple matter of looking at the continuity equation for momentum in any static situation. This is not in any way controversial or strange but certainly not the way that introductory mechanics courses would look at things simply because you don’t need to. The momentum current is the stress tensor and the momentum density is zero. With momentum being conserved, it is inevitable that you obtain a corresponding continuity equation on the form change = amount in - amount out + amount sourced.

Another way of seeing this is that even in a static situation any force acting on an object will give an impulse over time. This is a transfer of momentum by definition. That the object does not start moving is because the opposite impulse was transferred to the object somewhere else (or equivalently, by the third law, the impulse was passed on to another object).

What you do need to do in an introductory class is to model your forces or Newton’s second law will not help you.

 

[erobz]

Interesting stuff. A bit over my head. Is there a "mechanics of momentum flow" where the "fluid type" model is utilized regularly?

Never mind. I see your edit. Definitely over my head!

 

[bob012345]

There is no need to replace anything.

Great. So, getting back to the problem at hand, first, how would you attack the dripping cart problem? There seems to be a consensus among others here that Newton's Second Law simply does not apply to the cart and remaining water it carries because the mass is changing.

 

[Orodruin]

Great. So, getting back to the problem at hand, first, how would you attack the dripping cart problem? There seems to be a consensus among others here that Newton's Second Law simply does not apply to the cart and remaining water it carries because the mass is changing.

That’s not really true. Newton’s second law does apply. You just have to apply it correctly and include all momentum flows - including those resulting from momentum being transported out of the system by the mass lost. This is also included in the Cauchy momentum equations in the continuity form by the term $\rho \vec u \otimes \vec u$ in the momentum current.

I have already described in this thread how that would apply to a mass loss with a general ejected velocity (which is significantly simpler than solving the Cauchy momentum equations but still requires somehow modelling the ejected velocity).

 

[pnachtwey]

You need to solve two differential equations simultaneously. One models the water flowing from the cup and the other models the position of the cup. Use Runge-Kutta to solve numerically.

 

[bob012345]

That’s not really true. Newton’s second law does apply. You just have to apply it correctly and include all momentum flows - including those resulting from momentum being transported out of the system by the mass lost. This is also included in the Cauchy momentum equations in the continuity form by the term $\rho \vec u \otimes \vec u$ in the momentum current.

I have already described in this thread how that would apply to a mass loss with a general ejected velocity (which is significantly simpler than solving the Cauchy momentum equations but still requires somehow modelling the ejected velocity).

Ok, thanks. I thought it did. But don't we agree the cart will not speed up or slow down at all as the water falls out? It just gets lighter as the water leaks but maintains its velocity. We model the water leakage rate independent of the cart dynamics.

This is - as many times - only a problem if one misinterprets that equation. With $F$ being the rate of momentum change in the system, the case of just dripping does not satisfy $F=0$ because momentum is leaking out of the system with the mass at a rate of $-\dot m u$, where $u$ is the velocity of the leaking mass. For $u=v$ we obtain $\dot v=0$.

For $u=v+u_0$, we do end up with an acceleration of $-[d\ln (m/m_0)/dt] u_0$.

In this post #22, you appear to be saying the loss of momentum as water drips will cause an acceleration of the cart?

 

[Orodruin]

You need to solve two differential equations simultaneously. One models the water flowing from the cup and the other models the position of the cup. Use Runge-Kutta to solve numerically.

More specifically, two coupled non-linear differential equations in the typical case.

 

[Orodruin]

But don't we agree the cart will not speed up or slow down at all as the water falls out? It just gets lighter as the water leaks but maintains its velocity. We model the water leakage rate independent of the cart dynamics.

That depends on how the water leaves the cart. If it leaves the cart with the same velocity as the cart, then Newton’s second law would tell you that $ma + \dot m v = \dot m v$, ie, $a=0$. If it leaves at any other velocity it is that velocity that enters the RHS giving rise to $ma = \dot m(u-v)$.

In this post #22, you appear to be saying the loss of momentum as water drips will cause an acceleration of the cart?

Which it will in the general case. The expression quoted is just the Tsiolkovsky rocket equation based on ejecting mass at a constant relative speed.

 

[erobz]

More specifically, two coupled non-linear differential equations in the typical case.

Just to be clear, you are referring to the "leaking bucket oscillator" problem in the OP

 

[bob012345]

That depends on how the water leaves the cart. If it leaves the cart with the same velocity as the cart, then Newton’s second law would tell you that $ma + \dot m v = \dot m v$, ie, $a=0$. If it leaves at any other velocity it is that velocity that enters the RHS giving rise to $ma = \dot m(u-v)$.

I see. I was assuming it leaves with the same velocity as a limiting case of slowly dripping.

Which it will in the general case. The expression quoted is just the Tsiolkovsky rocket equation based on ejecting mass at a constant relative speed.

Yes, the component of the velocity of the water leaving along the carts motion would act as a rocket.

 

[vanhees71]

Don't you still have to choose a direction? In the case of the cart dripping water from the bottom, the total momentum ( cart + drop) in the direction of motion remains constant. That is to say the mass flowrate in the direction of motion is zero. Sure, the cart is losing mass, but its not going to change velocity.

For the draining cup the mass flowrate in the direction of motion is non-zero, and there will be an impulse in the direction of motion from the fluid jet.

And something seems to be mis applied in taking

$$ \sum F = \frac{dM}{dt} v + M \frac{dv}{dt} $$

The Rocket Equation is derived from impulse in a stationary frame of reference $O$ :

$$ \sum F = \frac{dM}{dt} v_{e/R} + M \frac{dv}{dt} $$

The $v$ in the first would be with respect to $O$: i.e. $v = v_{e/O}$

If in doubt go back to first principles. To derive equations of motion in mechanics it's safe to say that the fundamentals are the conservation laws based on the symmetries of the Galilei (or Poincare for special-relativistic mechanics) group. The equations of motion follow from the conservation of momentum (i.e., symmetry of the physical laws under spatial translations) for a closed system.

For the rocket equation, see, e.g.,

https://www.planetary.org/articles/20170428-the-rocket-equation-part-1

In the case of the cart the relative-velocity component of the dripping water in the direction of the cart's motion is 0 and thus the velocity of the cart doesn't change, as discussed already above.

 

[erobz]

I would like to actually try to set up the equation for the leaking bucket on a spring, neglecting the fluid jet thrust and drag. I'm not fully understanding the result that was linked and was hoping someone would guide me through it.

Firstly, where do we take the coordinate from $x$ from? Does it matter? I have it shown from the ceiling the top of the cup.

The equilibrium position $x_o$ is a function of time:

Let

$M(t) = m_w (t) + m_b$

$$ x_o (t) = \frac{ \left( m_b+m_w(t) \right) g}{k} =\frac{ M(t) g}{k} $$

Where:

$m_b$ is the mass of the bucket
$k$ is the spring constant
$m_w$ is the mass of the water in the bucket at time $t$

Ignoring the thrust from the jet:

Is (1) the proper differential equation?

$$ M(t) \frac{d^2}{dt^2} \left( x- x_o(t) \right) = M(t) g - k \left( x- x_o(t) \right) \tag{1} $$

 

[Delta2]

Is (1) the proper differential equation?

In my opinion it is proper only if we consider that the water exits the bucket with negligible relative velocity, i.e just dripping. Because then (1) will be what we get from the rocket equation.

 

[erobz]

In my opinion it is proper only if we consider that the water exits the bucket with negligible relative velocity, i.e just dripping. Because then (1) will be what we get from the rocket equation.

Agreed, but I just wanted to make sure I set the simplest version up correctly, before trying to add the complexity of the fluid jet which will have implications in the thrust force, and the rate of change of mass. Also, if we fully consider it, I think we have to account for how the acceleration of the oscillator impacts the jet and also then the mass flow rate out of the bucket. I think that is where the "coupled" differential equations come in that @Orodruin mentioned.

 

[Delta2]

BTW with what software did you make the figure of post #90, looks pretty decent!

 

[erobz]

BTW with what software did you make the figure of post #90, looks pretty decent!

PowerPoint. Its gotten a little more diagram friendly over the years with the addition of some "object snapping"

 

[pnachtwey]

OK, you have the acceleration of the bucket. You still need another differential equation for how the water flows out the orifice on the bottom. The water will flow out faster when the bucket is accelerating upwards and flow out slower when the bucket is decelerating. You still need two differential equations. One for the position of the bucket, which you have and another for the change in mass of the bucket do to water flowing out

 

[erobz]

OK, you have the acceleration of the bucket. You still need another differential equation for how the water flows out the orifice on the bottom. The water will flow out faster when the bucket is accelerating upwards and flow out slower when the bucket is decelerating. You still need two differential equations. One for the position of the bucket, which you have and another for the change in mass of the bucket do to water flowing out

I'm planning on working my way up to that. I had to mow the grass first!

I'm taking it you think the effect of the acceleration of the oscillator on the mass flow rate through the hole is larger than the effect of the thrust from the fluid jet?

 

[bob012345]

You could try letting the mass oscillate side to side yet dripping from the bottom first as an easier case.

 

[erobz]

Focusing in on the draining water. $z$ is measured down from the free surface ( assumed to be at $P_{atm}$ ). Neglecting viscous losses, apply Bernoulli's across the hole:

$$ \frac{P}{\rho g} + \frac{V_b^2}{2g} + z_b = \frac{P_{atm}}{\rho g} + \frac{V_j^2}{2g} + z_h $$

$z_b \approx z_h$

$A_j V_j = A_b V_b \implies V_b = \frac{A_j}{A_b} V_j = \tau V_j$

We are left with:

$$ V_j = \sqrt{ \frac{2 P}{ \rho \left( 1 - \tau^2 \right) }} $$

Furthermore, the rate of change in oscillator mass $M$ is the rate of change of the liquid mass

$$ \frac{dM}{dt} = - \rho A_j \sqrt{ \frac{2 P}{ \rho \left( 1 - \tau^2 \right) }} $$

In the accelerated frame of reference of the bucket I get the following result for the pressure at the base of the bucket:

$$ P = - \rho \left( \frac{d^2x}{dt^2} - g\right) z $$

And using the total mass of the system $M$ to eliminate $z$ we have that:

$$ z = \frac{M - m_b}{ \rho A_b} $$

So the final result for the mass flow rate is given by:

$$ \frac{dM(t)}{dt} = - A_j \sqrt{ \frac{ - 2 \rho \left( \frac{d^2x}{dt^2} - g\right) }{ \left( 1 - \tau^2 \right) } \frac{M(t) - m_b}{ A_b} } \tag{2} $$

$$ M(t) \frac{d^2}{dt^2} \left( x- x_o(t) \right) = M(t) g - k \left( x- x_o(t) \right) \tag{1} $$

If I have it correct, (1) and (2) are a nasty pair of coupled non-linear second order ODE's?

 

[bob012345]

It's not clear what all of your notation is. Could you please describe what's what? Thanks.

Also, you might solve for the dripping water first with a fixed bucket to see the form of that.

 

[erobz]

It's not clear what all of your notation is. Could you please describe what's what? Thanks.

Where do you encounter the first issue that needs clarification? Between the two posts ( and accompanying diagrams ) I honestly thought I covered everything.

Also, you might solve for the dripping water first with a fixed bucket to see the form of that.

What problem do you want me to solve? A stationary bucket dripping water?