Solving an Impossible Problem - Frictionless Train Car??

[Curl]

I think I got close to finding the answer to this one, but I'm in a lot of doubt.

The problem looks deceivingly simple, but it's impossible to solve. Any scenario I can think of seems to violate some physical law. Goes like this:

You have a train car on a frictionless track. It's filled with a fluid (say, inviscid fluid). Now there is a tiny nozzle pointing STRAIGHT DOWN (the nozzle is towards the right end of the train car). You open it and let water drip out. What happens to the water and the cart?

??

 

[rcgldr]

The trains speed is unaffected by the water flow perpendicular to the direction of the train. The water flowing downwards will results in a slight reduction of pressure at the hole, causing the train car and remaining water to be slightly lighter than if the water were not flowing.

Ignoring atmosphere effects, if the water falls onto a firctionless surface, it continues to move forwards at the same speed as the train. If the water is slowed down by the surface below, the the water slows down and the Earth speeds up.

In all cases, in the closed system of earth, train, and water, total momentum and energy are conserved.

 

[Curl]

Sorry, I forgot to state the most important fact: The nozzle is at the right end of the train car. What you said is only true if the opening is dead centered on the train car (and I didn't say that either, I guess you assumed it).

 

[berkeman]

I think I got close to finding the answer to this one, but I'm in a lot of doubt.

The problem looks deceivingly simple, but it's impossible to solve. Any scenario I can think of seems to violate some physical law. Goes like this:

You have a train car on a frictionless track. It's filled with a fluid (say, inviscid fluid). Now there is a tiny nozzle pointing STRAIGHT DOWN (the nozzle is towards the right end of the train car). You open it and let water drip out. What happens to the water and the cart?

??

Sorry, I forgot to state the most important fact: The nozzle is at the right end of the train car. What you said is only true if the opening is dead centered on the train car (and I didn't say that either, I guess you assumed it).

And you didn't think that part was important to your question? Hmm.

The COM for the system has to stay constant. The position of the nozzle will vary with time as the tanker moves to the side, as the puddle forms. Not sure how to treat a puddle on a frictionless surface...

 

[Dickfore]

if water drips vertically downwards, this does not affect the horizontal position of the cm of the train cart.

 

[berkeman]

if water drips vertically downwards, this does not affect the horizontal position of the cm of the train cart.

Even if it drips from one end of the tanker? The puddle will end up being non symmetric to the tanker's initial position...

 

[Curl]

Okay it sounds like you guys are getting started. berkeman, keep thinking, you're on the right track (no pun intended).

Basically the tank must move to the left so that it spills water in a "line", then stop. How in the world this is possible, I got no idea. One case I've considered was that the momentum of the car must be somehow related to the momentum of the fluid elements that are moving to the right (towards to nozzle to escape). When the valve is shut, they will "crash" into the right wall and cause the train to stop. This is probably BS but its the best I got.

Is there even a pressure gradient in the x-direction? Probably not since the fluid is inviscid.

 

[rcgldr]

if water drips vertically downwards, this does not affect the horizontal position of the cm of the train cart.

If the hole is off center, and assuming the puddle formed below the train moves at the same speed as the train (frictionless surface), then as mentioned before, the center of mass of train and water remains the same, with the train car moving slightly forward to compensate for the puddle forming aft of the center of the train.

 

[Dickfore]

Even if it drips from one end of the tanker? The puddle will end up being non symmetric to the tanker's initial position...

If the cart is supposed to move somehow horizontally, where would it move towards?

 

[K^2]

Woah. Deja vu.

You can't solve this problem without knowing flow rate. Knowing flow rate, it's a simple integration problem.

 

[rcgldr]

Is there even a pressure gradient in the x-direction? Probably not since the fluid is inviscid.

Yes, there's a pressure gradient in the fluid, both vertical related to gravity, and horizontal, related to the flow. Although inviscid flow is indeterminant (zero viscosity and constant density), you know that the center of mass of the train and fluid remains at constant velocity absent any external horizontal forces.

Being bit nit picky here, but what is filling up the space left by the fluid as it drains from the train car? That would need to be taken into account as well.

 

[Studiot]

It's not at all clear what other forces might be involved.
For instance is the train already in motion on the track as some have assumed?
If so has is been given a push and left to roll or is something still pushing (pulling) it?
And which direction is the train traveling / pointing?

To my last question here is an associated riddle.

A train is traveling due north on a track.
One rail is worn more than the other.
Which rail and why?

 

[A_B]

A train is traveling due north on a track.
One rail is worn more than the other.
Which rail and why?

The Earth rotates towards the east, so there is a coriolis force on the train towards the east. The tracks must compensate for this by exerting a force to the west. I guess most of this counter force would come from the rail to the east, which is the one that's worn more than the other?

EDIT: This is for the northern hemisphere, it would be the other way around for the southern hemisphere

 

[maimonides]

I think I`ve seen this on http://physics.stackexchange.com/" some time ago.
My opinion: when the water leaves the tank/nozzle it also leaves the system to which conservation of momentum is applied, as far as effects to the train are of concern. It cannot interact with the car or the water in it anymore. (At least for zero viscosity). You can´t manipulate a rocket by "catching" its exhaust gases somewhere behind it and doing something to them.
So the whole thing appears to be a non-problem.

 

[Dickfore]

when the water leaves the tank/nozzle it also leaves the system to which conservation of momentum is applied, as far as effects to the train are of concern.

Essentially this^.

 

[rcgldr]

My opinion: when the water leaves the tank/nozzle it also leaves the system to which conservation of momentum is applied, as far as effects to the train are of concern. It cannot interact with the car or the water in it anymore.

If you only consider the train and the internal water, it doesn't really matter what happens after the water leaves, but it took a net internal flow in order for the water to leave the tank, and that does matter.

 

[Curl]

Train starts at rest, ZERO external horizontal forces acting on it (no wind, no friction, no birds, no rain). I don't see why anyone has problems understanding the problem statement, its an easy set up.

So most of you agree that with the nozzle on the right, the train MUST move to the left. But if the train is moving to the left, the water is being poured out from a moving nozzle, so it will have some horizontal velocity to the left. But then, after the nozzle is shut, the train must all of the sudden move to the right because if it keeps moving to the left (and the water on the ground also moving to the left) momentum is not conserved.

No scenario can make sense, they'll either violate momentum or logic or both. This question is BS, someone tell me the answer so I can sleep at night.
K^2, what is the solution?

 

[CronoSpark]

Not sure if this helps/I will play the devil's advocate here, but I'm imagining that the train would actually be moving to the right.

Given a stationary tank on frictionless rails, if you imagine the water flow from within the tank you will see that it flows to the right and down. The water molecules push against the right end of the tank, pushing the tank to the right. The water flowing through the nozzle does provide a reaction force (pushing the tank back to the left) though. The flow rate is important for the acceleration of the tank.

 

[Drakkith]

Given that the water is flowing straight down out of the nozzle and train cart, I don't see how the train can move anywhere. There shouldn't be any net force up or in any direction, or so I think.

 

[RedX]

I'm not exactly sure I understand what the problem is, so maybe I'm taking a too simplistic view of things.

I don't know anything about fluids, so just imagine a truck with a hole in the bottom of it, and a human who drops boxes out of the hole, littering the frictionless roadway until all the boxes in the truck run out.

The person starts on the left side of the truck, and pushes a box rightward towards the hole. While the person is doing this, the truck moves to the left to preserve center of mass. As soon as the person is next to the hole and holds the box directly above the hole, she stops - then the truck will stop too (friction stops her, but that means she drags the truck back rightwards). She then drops the box straight down. Since she drops it straight down the truck doesn't budge horizontally, and no one is moving. She then empties the whole truck like this. Since there was more mass on the left hand side, the end result is that the truck moves more to the left. But center of mass of the littered boxes and the truck remains the same.

If you want to generalize this to giving the boxes a horizontal velocity when it exits the hole, that's fine. Just make the hole slightly bigger than the size of the box, and give the box a horizontal push on the way out. In this case it's possible to attain net motion of the truck when the truck is emptied, and in any direction.

Of course all the energy comes from the soda pop that she's sipping.

But what direction will the truck move? Well it depends. Place a cylinder wall around the hole. Now in order to dump the boxes through the hole, she has to lift the boxes a little higher and drop them through the cylinder. Depending on how the boxes undergo multiple bounces off the wall will determine the direction the boxes exit the truck. It could go out either way depending on how many ricochets there are. Therefore my guess for fluids is that it doesn't flow out all in one direction. It changes directions. Hence the motion is hard to predict. If the pipe is long enough (pipe corresponding to cylinder wall) my guess is that it leaks straight down, so the analogy with the boxes applies, and the cart will have moved left, but stopped once all the water had gone.

 

[Curl]

No, good try though. The "box" thing you thought about is a discreet version of this fluid problem. I already though about that, and it is much easier to solve this problem. But with fluids, since the flow is essentially continuous, the water MUST leak out off a moving train car, which screws everything up.

A better version of the "box" view is to say you have marbles sliding down an incline (fed from a hopper), hitting the wall, stopping, and falling down, one by one. What ends up happening is, the car gains equal and opposite momentum of the marble, but when the marble hits the wall, both the marble and the car stop, then marble falls. By this time, the CG of the train car has moved to the left by an equal amount as the CG of the marble, so it's all good.

But with the fluid it doesn't work like this, which is the reason I cannot figure it out. Where's K^2 with the answer?!

 

[RedX]

Well, say water drips out the hole, leaving a void of water just above the hole. This void needs to be filled up. So say water rushes in from the both the left and the right to fill this void. Then they collide, so they no longer have horizontal momentum, and just fall straight down. Say the truck is of length 4, and the hole is at 3, and the left side of the truck is at 0. Then all the water between 2 and 4 will collide at 3, and fall straight down. Then all the water between 0 and 2 will rush to the right (this actually happens simultaneously as water between 2 and 3 leaves). When all the water rushes to the right, the truck moves to the left. Then the water leaks out to the right and the truck will gain motion to the left (the wall on the left pushes the water to the right).

I know this might not be satisfactory for a variety of reasons, but real water is discrete is it not?

 

[maimonides]

When a brick moving to the right reaches the nozzle, its horizontal movement is stopped. It is deflected and goes down. This also happens simultaneously (for small enough bricks), so there is no net effect.

Fluid mechanics assumes fluids to be continous.

 

[CronoSpark]

Oh right, completely forgot about the water being pushed from the left end of the train. I guess that means that the train just doesn't move?

 

[D H]

So most of you agree that with the nozzle on the right, the train MUST move to the left.

Edit: See post #29.
[strike]Not necessarily. As specified in the original post, the fluid fills the train car. By that, I assume that you mean that the center of mass of the fluid is at the center of the train car. If this is the case the train will not move. There is no law of conservation of position in physics.[/strike]

There is however a law of conservation of momentum. Instead of having the fluid fill the train car let's have the fluid be in a tank on the left side of the car, with a pipe connecting the tank to the nozzle on the right. Start the train at rest and open the nozzle. Now the train will move to the left while the fluid is dripping from / flowing from the nozzle.

The tricky part is explaining why the car is moving to the right after the flow stops.

This is very similar to an interview problem that I throw at some candidates.

 

[rcgldr]

I thought of an alternative example, although it doesn't change the situation much.

The train car is "attached" near the center of a massive and huge flat plane in space such that "horizontal" motion is frictionless. The water inside the train car is being boiled (due to lack of pressure in space at the nozzle) and the steam is allowed to escape through a nozzle located at the right end of the train car, but oriented so the steam leaves the nozzle with the same "horizontal" component of speed as the train and "upwards" (away from the plane). The flat plane massive and large enough that angular momentum related effects are tiny. Eventually the train is emptied, and there's a plume of "steam" flowing away from the train. There are no external forces, so the center of mass of this system doesn't move with respect to it's original position (or velocity).

 

[Curl]

Not necessarily. As specified in the original post, the fluid fills the train car. By that, I assume that you mean that the center of mass of the fluid is at the center of the train car. If this is the case the train will not move. There is no law of conservation of position in physics.

If you can prove this you will win the Nobel Prize in Physics. But you can't:

Let's suppose you are correct, and the train doesn't move. I put a cup under the nozzle, and let some water drip until the cup fills up. Now I have fluid in the cup, and the CG of the system (the system is the train car plus the fluid that started in it) is now slightly to the right. The CG of the system moved WITH NO HORIZONTAL FORCES ON THE SYSTEM! This is not possible. We already went over this above.

 

[RedX]

If you can prove this you will win the Nobel Prize in Physics. But you can't:

Let's suppose you are correct, and the train doesn't move. I put a cup under the nozzle, and let some water drip until the cup fills up. Now I have fluid in the cup, and the CG of the system (the system is the train car plus the fluid that started in it) is now slightly to the right. The CG of the system moved WITH NO HORIZONTAL FORCES ON THE SYSTEM! This is not possible. We already went over this above.

Maybe the train starts moving to the left when the water leaks, and if this gives the outgoing water a horizontal velocity to the left, then to conserve momentum something has to be going to the right. What would be going to the right would be the rest of the water in the train as it flows from the left side towards the hole. Sooner or later all that moving water is going to crash into the right side of the edge of the hole or nozzle, thereby causing the train to move to the right and the rest of the water to fall down the hole.

I mean it could happen could it not? Whether it does I don't know, but that's good enough for me lol!

 

[D H]

Correction to post #29:

The train will move, but how much it moves and the manner of the motion depends on the scenario.

Imagine a chamber that can be isolated from the rest of the fluid in the car by one valve and whose contents can be released by another valve. I'll respectively call these the fill valve and dump valve. For a first scenario, we'll operate the system in a discrete manner. Open the fill valve, allowing the chamber to be filled. After the chamber is full that valve is closed. We wait for any transients to damp out and then open the dump valve. Once the chamber is empty we close the dump valve and repeat the process.

In this case the car will move in a jerky fashion. The car will move to the left as water flows into the chamber and will come to a stop shortly after closing the fill valve. Each release of water occurs with the train at rest, but in a slightly different location each time. If the train car is much less massive than the fluid it holds it will have moved by almost the full length car length by the time the tank is empty. Note that the center of mass of the fluid+car system does not change in this scenario.

Something very different happens if we make the flow continuous. Opening both valves simultaneously means there is a fluid flow inside the car to the left. Conservation of momentum says the car as a whole needs to be moving to the left while the fluid is flowing. However, conservation of momentum also dictates that when the flow finally stops the car will be moving to the right. Once again the center of mass of the fluid+car system does not change.

 

[maimonides]

The discussion on stackexchange is http://physics.stackexchange.com/questions/1683/mechanics-around-a-rail-tank-wagon" .
There seems to be at least one very detailed and reasonable post on it, which wasn´t there when I looked last time.

 

[D H]

The first several answers there violate conservation of momentum. Assuming a continuous (but not necessarily constant) fluid flow through a hole to the right of the center of mass, the train does move to the left initially but the final velocity of the train is necessarily to the right.

 

[Chewy0087]

The first several answers there violate conservation of momentum. Assuming a continuous (but not necessarily constant) fluid flow through a hole to the right of the center of mass, the train does move to the left initially but the final velocity of the train is necessarily to the right.

That's fascinating... any hints as to why? I'm trying to think of the reason...

 

[Studiot]

I am concerned about the idea that a train can propel itself along a frictionless track, starting from rest.

These are the stated conditions.

Since the train starts from rest it posesses zero inital momentum.

 

[D H]

That's fascinating... any hints as to why? I'm trying to think of the reason...

First off, it should be obvious that if at the end everything is moving to the left we have indeed violated conservation of momentum. At the end, if the bulk of the exhausted fluid is moving to the left, the car can only be moving to the right. So how to get there?The discussion so far has focused on the train car and its contents. What about the liquid that has left the train car? It helps to make the train car + liquid a horizontally-isolated system. One way to do this is to have the liquid leaving the car fall through the rails onto a frictionless surface situated well beneath the rails. With this construction, the train car + liquid system does not interact horizontally with the environment. Moreover, because the surface onto which the liquid falls is well beneath the rails, the train car does not interact with the liquid once it leaves the car.

The momentum of the fluid that has left the car between time t=0 when the flow first starts to time t and the momentum of the car plus its fluid contents are given by

$$\aligned \vec p_{\mbox{ex}} &= \int_0^t \dot m(\tau)\vec v(\tau)\, d\tau \\[4pt] \vec p_{\mbox{car}} &= (M+m_f(t))\vec v(t) + \vec p_{\mbox{flow}} \endaligned$$

where

• $\vec p_{\mbox{ex}}$ is the momentum of the fluid that has left the car from the persective of an inertial observer,
• $\vec p_{\mbox{car}}$ is the momentum of the car plus its fluid contents from the persective of this inertial observer,
• $\vec p_{\mbox{flow}}$ is the horizontal momentum of fluid flow inside the car from the persective of an observer fixed with respect to the car,
• $\dot m(t)$ is the (positive) rate at which mass flowing from the car at time t,
• $\vec v(t)$ is the velocity of the car at time t,
• $M$ is the (empty) mass of the car, and
• $m_f(t)$ is the mass of the fluid contents of the car.

In steady-state flow conditions, the horizontal momentum flow of the fluid inside the car is given by

$$\vec p_{\mbox{flow}} = \dot m(t) \vec o$$

where $\vec o$ is the horizontal component of the vector from the center of the tank to the outlet.

To conserve mass and to conserve horizontal momentum, we must have

$$\aligned dm_f(t)/dt &= -\dot m \\[4pt] \vec p_{\mbox{car}} &= -\vec p_{\mbox{ex}} \endaligned$$

Let's start simple: constant flow rate, no transients, and we'll close the outlet valve before all the fluid has left the tank. In this case, while the fluid is flowing from the car, the car will have a constant velocity given by

$$\vec v = -\frac{\dot m}{M+m_0}\vec o$$

where $m_0$ is the initial mass of the fluid content of the car. By ignoring transients we have given the car an impulsive change in momentum of

$$\delta p_0 = -\dot m \vec o$$

at time t=0. How does this arise? It arises from the transients that we have so conveniently ignored. It takes a bit of time for the fluid flow to build up inside the car. There is a transfer of momentum from the car itself to the fluid within the car.

When we close the value (remember that there is still fluid in the car) we will get a reversal of this momentum transfer. The initial and final changes in momentum of the car+contents subsystem have the same magnitudes but opposite directions. However, the car+contents subsystem has lost mass in the interim. The final change in velocity is greater than the initial change in velocity: The vehicle ends up moving to the right.

Now let's make the vehicle drain under its own pressure, and keep draining until it is completely empty. We'll get a similar startup transient as above. Now the momentum flow of the liquid within the car is no longer constant. It is instead decreasing in magnitude as fluid drains from the car. This means momentum is being transferred from the fluid to the car, and this transfer is toward the right. This will make the car slow down, come to a momentary stop, and then start moving to the right.

 

[Chewy0087]

That is a really great problem, and answer, thank you very much sir!