Damped oscillator with changing mass

[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?

 

[bob012345]

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

Are the $V_j, V_b$ terms independent of the oscillation?

 

[erobz]

Are the $V_j, V_b$ terms independent of the oscillation?

They are supposed to be, but I don't know...maybe that is a poor model. They are the velocity of water in the jet, and the water in the bucket relative to the bucket.
EDIT:
The way I thought about it was what comes out the hole was a function of the pressure at the bottom of the bucket. That pressure was dependent on the acceleration of the bucket. Those velocities are indirectly dependent on the bucket acceleration, through the pressure at the bottom and Bernoulli's principle.

 

[bob012345]

They are supposed to be, but I don't know...maybe that is a poor model. They are the velocity of water in the jet, and the water in the bucket relative to the bucket.
EDIT:
The way I thought about it was what comes out the hole was a function of the pressure at the bottom of the bucket. That pressure was dependent on the acceleration of the bucket. Those velocities are indirectly dependent on the bucket acceleration, through the pressure at the bottom and Bernoulli's principle.

I still recommend solving the water flowing from the still bucket first. One thing also, water is virtually incompressible so I wonder about assuming the pressure changes with acceleration? But obviously something changes and the rate water comes out must vary.

Here is a past PF thread dealing with that question;

https://www.physicsforums.com/threads/basic-calc-problem-with-flow-out-of-a-hole-in-a-bucket.840469/

 

[erobz]

One thing also, water is virtually incompressible so I wonder about assuming the pressure changes with acceleration?

I think it does. Imagine if you were sitting at the bottom of a tank of water of depth $z$. you are going to feel pressure $P = \rho g z$. Now, put that tank on a rocket ship. You are now going to feel pressure increase proportional to the rocket acceleration.

 

[bob012345]

One thing, the water leaking from the bucket should be less than normal maybe even zero when the bucket is accelerating downward. Does you equation physically allow for that?