How Does a Leaking Tank-Wagon Move and What Is Its Final Velocity?

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The discussion centers on the dynamics of a rail tank wagon leaking water through a nozzle, questioning how the wagon moves and its final velocity. Participants debate the effects of water ejection on the wagon's motion, noting that if the water jet is directed vertically down, the wagon's movement may remain unaffected due to the conservation of momentum. However, the complexity arises from the water's motion inside the tank and the potential recoil effect as water is expelled. The final velocity of the wagon is suggested to depend on the flow rate and the geometry of the tank, leading to the conclusion that a general solution may be elusive without specific parameters. Ultimately, the problem illustrates the challenges of applying theoretical physics to real-world scenarios.
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This problem sounds like a homework problem but its not and it is really hard.

Consider a rail tank wagon filled with liquid, say water.

Suppose that at some moment t=0 there opens a nozzle at left side of the tank at the bottom. The water jet from the nozzle is directed vertikally down.

Question:

How will the wagon's movement be while the water is running out?
What is the final velocity of the rail tank wagon after emptying?Simplifications and assumptions:

Rail tracks lie horizontally, there is no rolling(air) friction, the speed of the water jet from the nozzle is subject to the Torricelli's law, alternately one can assume constant rate of r litre/sec, the horizontal cross-section of the tank is a constant, the water surface inside the tank remains horizontal.Data given:

M(mass of the wagon without water)
m(initial mass of the water)
S(horizontal cross-section of the tank)
S>>s(cross sectional area of the nozzle)
ρ(density of the water)
l(horizontal distance from the nozzle to the centre of the mass of the wagon with water)
g(gravitational acceleration)One thing is obvious: If l=0 then the wagon will not move at all.

For an inconclusive discussion see here:
http://physics.stackexchange.com/questions/1683/mechanics-around-a-rail-tank-wagon
 
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If the jet is directed vertically down, motion of the tank is unaffected.
 
With no offense intended, how can we possibly believe that this is not a homework question when you specify that there is no friction or air resistance? That does not apply to reality.
 
K^2 said:
If the jet is directed vertically down, motion of the tank is unaffected.

Yes, but there will be motion of the water insie the tank..
 
Danger said:
With no offense intended, how can we possibly believe that this is not a homework question when you specify that there is no friction or air resistance? That does not apply to reality.

What do you mean, is it inconceivable to have a tank in a vacuum? that must blow your mind..
Is there air everywhere in the universe, or is "reality" somewhere else ?

Also, if its homework you should be able to solve it, but you cant..
 
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kuktryne said:
Yes, but there will be motion of the water insie the tank..
Good point. That's a little more interesting, then.

The CM of the water-tank system will still remain in place, since there are no external forces. Water can be assumed to be deposited at location of the nozzle.

There has to be another variable, then. Center of mass of water in the tank. For simplification, I would assume it does not shift horizontally as function of water level.

So let me call l distance from nozzle to CoM of tank, and d the distance from CoM of water to CoM of tank.

Suppose CoM of tank is located at x initially.

Equation for center of mass of water + tank.

\frac{M x + (x+d)m}{M+m} = \frac{M(x+dx) + (x+d+dx)(m-dm) + (x+l+dx)dm}{M+m}

dx dm terms go to zero, of course, giving us a differential equation.

0 = M dx + m dx - (x+d)dm + (x+l)dm

(d-l)dm = (M+m)dx

\int_m^0 \frac{d-l}{M-m}dm = x_f-x_i

Variable shift.
\int_{m+M}^M \frac{l-d}{m} dm = x_f-x_i

x_f - x_i = (l-d)ln\left(\frac{M}{M+m}\right)

How's that?

@Danger, I don't think anyone would ask such a problem for homework. It's too abstract for that.
 
kuktryne said:
What do you mean, is it inconceivable to have a tank in a vacuum? that must blow your mind..
Is there air everywhere in the universe, or is "reality" somewhere else ?

Also, if its homework you should be able to solve it, but you cant..

Is your tank in a zero-gravity environment within a vacuum? Think carefully before answering, because neither zero-G nor a perfect vacuum are physically possible within the known universe.
 
Danger, that's not how problems are solved. Saying there is always friction is not constructive. There are always factors preventing you from solving a problem exactly. That doesn't mean you can't get a good estimate. Neglecting friction is a very common estimate that usually gives good results. When you have to get better results, you build a better model that to some degree models friction.
 
K^2 said:
Danger, that's not how problems are solved. Saying there is always friction is not constructive. There are always factors preventing you from solving a problem exactly. That doesn't mean you can't get a good estimate. Neglecting friction is a very common estimate that usually gives good results. When you have to get better results, you build a better model that to some degree models friction.

Quite true, but I have never in my history on PF read a question that suggested negating real-life factors such as gravity, friction, air resistance, etc. that was not pulled from a textbook. I might very well be wrong in this case, but I hope that you can understand from where my suspicion arises.
 
  • #10
This really doesn't look like a textbook problem. It's not useful as a textbook problem. It could be a problem from some competition, but not from a textbook.
 
  • #11
K^2 said:
This really doesn't look like a textbook problem. It's not useful as a textbook problem. It could be a problem from some competition, but not from a textbook.

Okay... I must admit that I haven't seen a textbook in almost 40 years. I was just going by past experience of having people try to have their homework done for them despite the PF Guidelines by pretending that it wasn't homework.
If Kuktryne is legitimate, I express my apologies to him/her.
 
  • #12
The wagon does nothing in this process and therefore would not move...

Water flows due to the gravitational force. The only way to lift the car is to have the water pushing against the inner surface at the top of the tank which is not true I assume in this case.
 
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  • #14
Their solution is not accounting for acceleration of the wagon causing change in forces. Mine does, because I go straight for momentum conservation.

dx dm terms go to zero because they are infinitely smaller than terms containing only dx or only dm. That's basic calculus.
 
  • #15
K^2 said:
Good point. That's a little more interesting, then.

The CM of the water-tank system will still remain in place, since there are no external forces. Water can be assumed to be deposited at location of the nozzle.

This doesn't take into account the momentum of the water that is emitted after the tank starts to move.
 
  • #16
willem2 said:
This doesn't take into account the momentum of the water that is emitted after the tank starts to move.
Damn, you are right. So then the whole thing may depend on rate of flow and there is going to be a recoil. The final state will be the tanker rolling in the opposite direction, and if this final state depends on flow rate, which will depend on water level and tank geometry, then there is no general solution to this problem.

I shall think on it some more.

Edit: Actually, final momentum shouldn't depend on flow rate. So while exact position might be unknown, final velocity should be available. I'll get back to you on that.
 
  • #17
This problem too hard, this forum is mostly for highscool-homework I think
 
  • #18
I think the wagon will move towards left because water will hit the wall at the left side where nozzle is present. So, the wagon will accelerate in left direction. But the wall of tank will give opposite reaction to water due to which a water current in right direction will be setup and the wall on the right side will experience a force. So the wagon will slow down in the left direction and water current will undergo damped oscillations and so will the acceleration of wagon affected.
 
  • #19
It will initially move in the direction opposite to the location of the nozzle, but it will then reverse direction. The fact that it reverses direction is the real kicker of this problem and what makes it so difficult.

The solution is a differential equation that has flow rate as one of parameters. If flow rate depends on motion of the tank, in general, this problem cannot be solved. If we estimate flow rate as constant, a model solution can be found, but it's not going to be a particularly interesting one.
 
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