Speed of train car when sand spilling out from it

[songoku]

Homework Statement

Please see below

Relevant Equations

Momentum

Impulse

Conservation of momentum

I have several confusion about this question:

1) Since the sand spilling out downwards, it means that there is change in momentum in vertical direction (before the sand spills and after the sand spills) but no change in momentum in horizontal direction so no net force in horizontal direction and no acceleration in horizonal direction, so the speed of the train car will be constant. Is this correct?

2) If I consider the momentum of the system (train car + sand inside train car) in horizontal direction, the momentum will decrease because the speed is constant and the mass of sand decreases. But this means that the there is change in momentum in horizontal direction so the momentum in horizontal direction is not conserved and there will be net force acting on the system so the speed can't be constant, which contradicts to what I thought in part (1)

3) I want to know how the momentum of spilling sand affects the system so I draw free body diagram for the train car. There will be 3 forces acting on it, normal force from ground acting upwards, weight of the train car and also weight of sand (both acting downwards). Initially, normal force = weight of train car + weight of sand (to prevent vertical motion).
When the sand spills out, there will be additional force due to the change in momentum of spilling sand in vertical direction, directed downwards. The weight of sand in the train car decreases so to prevent vertical motion of train car:

Normal force from ground on train car = weight of train car + new weight of sand + downwards force due to change in momentum of spilling sand

Is that correct?

And if I want to compare normal force before and after the sand spills, can I say they will be the same? Maybe because the decrease in weight of sand in the train car will be the same as the downwards force due to change in momentum (I don't really know the reason behind this, just guessing)

Thanks

 

[ergospherical]

Your first answer is correct - the train continues at constant velocity. Recall the equation of motion for variable mass systems $\mathbf{F} + \mathbf{v}_{\mathrm{rel}} \dfrac{dm}{dt} = m \dfrac{d\mathbf{v}}{dt}$. Since the sand is ejected at $\mathbf{v}_{\mathrm{rel}} = \mathbf{0}$, and frictional effects are negligible ($\implies \mathbf{F} = \mathbf{0}$), then $\mathbf{v}$ is constant.

[If you would like to derive this equation then you should consider either conserving the momentum of a system of fixed mass (e.g. train $+$ sand which is ejected in time $\delta t$), or otherwise using the Reynolds transport theorem of vector calculus.]

For the normal force $\mathbf{N}$, you just need to look at the vertical component of the variable mass equation.

 

[Ibix]

If I consider the momentum of the system (train car + sand inside train car)

Is that a closed system? If not, would you expect conservation laws to apply?

 

[Orodruin]

The way I would recommend thinking of this is through conservation of momentum in the form:

Change of momentum = external forces - momentum carried out of the system.

Since sand is flowing out of the system, momentum is being carried out at a rate of $-\dot m v$, where $v$ is the speed of the sand. If the external forces are zero, then $\dot p = \dot m v$ (note that $\dot m$ is negative).

If the speed of the dropped sand is equal to the speed of the cart, then $p = mv$ resulting in $\dot p = ma + \dot m v = \dot m v$ and therefore $a = 0$.

 

[songoku]

Your first answer is correct - the train continues at constant velocity. Recall the equation of motion for variable mass systems $\mathbf{F} + \mathbf{v}_{\mathrm{rel}} \dfrac{dm}{dt} = m \dfrac{d\mathbf{v}}{dt}$. Since the sand is ejected at $\mathbf{v}_{\mathrm{rel}} = \mathbf{0}$, and frictional effects are negligible ($\implies \mathbf{F} = \mathbf{0}$), then $\mathbf{v}$ is constant.

Is $\mathbf{v}_{\mathrm{rel}}$ = 0 because the horizontal velocity of spilling sand is the same as the velocity of the train car and the initial vertical velocity of spilling sand is zero, so the resultant of initial velocity of spilling sand will be the same as train car?

Is that a closed system? If not, would you expect conservation laws to apply?

Ah I see. The closed system should be train car + sand inside train car + spilling sand. Is this correct?

So the spilling sand will contribute to momentum of the system in horizontal direction and making the momentum in horizontal direction conserves

The way I would recommend thinking of this is through conservation of momentum in the form:

Change of momentum = external forces - momentum carried out of the system.

Since sand is flowing out of the system, momentum is being carried out at a rate of $-\dot m v$, where $v$ is the speed of the sand. If the external forces are zero, then $\dot p = \dot m v$ (note that $\dot m$ is negative).

Do I need to pay attention to the direction in which momentum is being carried out at a rate of $-\dot m v$ (I think it is downwards in this case)?

If the speed of the dropped sand is equal to the speed of the cart, then $p = mv$ resulting in $\dot p = ma + \dot m v = \dot m v$ and therefore $a = 0$.

By speed of the dropped sand, do you mean the speed in horizontal direction, or in vertical direction, or maybe something else that I miss?

$\dot p = \dot m v$ is the momentum of spilling sand in horizontal direction?

Sorry I don't understand where this equation comes from: $\dot p = ma + \dot m v$.
Another thing I don't understand is: I thought $-\dot m v$ is the rate of change of momentum in vertical direction but $a$ is the acceleration in horizontal direction so why can we equate quantities in different direction?

Thanks

 

[kuruman]

The way I would recommend thinking of this is through conservation of momentum in the form:

Change of momentum = external forces - momentum carried out of the system.

Since sand is flowing out of the system, momentum is being carried out at a rate of $-\dot m v$, where $v$ is the speed of the sand. If the external forces are zero, then $\dot p = \dot m v$ (note that $\dot m$ is negative).

If the speed of the dropped sand is equal to the speed of the cart, then $p = mv$ resulting in $\dot p = ma + \dot m v = \dot m v$ and therefore $a = 0$.

I am not sure I understand this explanation.

If the system is just the train car, the horizontal forces exerted by the sand and the ground are zero, so its horizontal velocity does not change as OP pointed out in #1.

If the system is train car + sand (all the sand before any of it starts leaking), then there is no momentum carried out of the system because the sand is moving horizontally with velocity v until it hits the ground. If instead of falling on the ground the sand fell through a gap between the tracks and kept going forever, the horizontal momentum of the train + sand system would be conserved.

 

[Ibix]

Ah I see. The closed system should be train car + sand inside train car + spilling sand. Is this correct?

Yes. So if the car had absurdly large wheels so that the first sand didn't hit the floor until the last sand leaked out of the car then (neglecting air resistance) the falling column of sand would be always underneath the hole. Momentum is then conserved.

Alternatively you can give the Earth a very large but finite mass and include it, and the friction forces that stop the sand, in your calculations.

 

[Orodruin]

I am not sure I understand this explanation.

If the system is just the train car, the horizontal forces exerted by the sand and the ground are zero, so its horizontal velocity does not change as OP pointed out in #1.

If the system is train car + sand (all the sand before any of it starts leaking), then there is no momentum carried out of the system because the sand is moving horizontally with velocity v until it hits the ground. If instead of falling on the ground the sand fell through a gap between the tracks and kept going forever, the horizontal momentum of the train + sand system would be conserved.

No. The system is car + sand in the car. Momentum is flowing out of this system by the system boundary being redrawn by sand carrying momentum with it exiting the system.

 

[Orodruin]

Do I need to pay attention to the direction in which momentum is being carried out at a rate of −m˙v (I think it is downwards in this case)?

It is not downward. Only the horizontal components matter for the speed of the cart. The rate is $-\dot m v$ because $\dot m$ (the mass change of the cart+sand) is negative and the negative of this quantity is the mass carried out of the cart+sand system.

By speed of the dropped sand, do you mean the speed in horizontal direction, or in vertical direction, or maybe something else that I miss?

Horizontal direction.

p˙=m˙v is the momentum of spilling sand in horizontal direction?

No. $-\dot m v$ is the momentum of the spilling sand in the horizontal direction since $-\dot m$ is the mass rate of sand falling out of the system.

Sorry I don't understand where this equation comes from: p˙=ma+m˙v.

This comes from just differentiating $p = mv$ with respect to time.

Another thing I don't understand is: I thought −m˙v is the rate of change of momentum in vertical direction but a is the acceleration in horizontal direction so why can we equate quantities in different direction?

Again. Both are in the horizontal direction.

 

[kuruman]

No. The system is car + sand in the car. Momentum is flowing out of this system by the system boundary being redrawn by sand carrying momentum with it exiting the system.

I think I see what you mean. You choose to define a system with variable mass. OK, say I look at the system at some time t₀ after an amount of sand has fallen out and then again at t = t₀+dt. In time dt, mass dm which is part of the system has fallen out of the car and is moving with horizontal speed v. Are you saying that this dm is no longer part of the system because it fell out of a hole and the boundary of the system has been "redrawn" without it? Why is that necessary?

Variable mass systems are fraught with pitfalls. If one must include the sand as a system component, I think it is simpler to include all the sand. Then in the horizontal direction, there is friction force $f=-\dot m_{\text{sand}}v$ acting on the component of the system that hits the ground. One could write in the horizontal direction $P_{\text{sys}} =m_{\text{car}}v + m_{\text{sand}}v$, in which case $$\dot P_{\text{sys}}=F_{\text{net}}=0+(-\dot m_{\text{sand}}v).$$The time derivative of the car's momentum is zero because there is no unbalanced force acting on the "car" component of the system.

The bottom line is that there is no horizontal momentum exchange between car and sand, only between sand and ground. Therefore defining just the car as the system is sufficient to answer the original question.

 

[Orodruin]

Why is that necessary?

It is not necessary. It is a way of looking at things that can sometimes help different individuals.

In time dt, mass dm which is part of the system has fallen out of the car and is moving with horizontal speed v.

Yes, but it is better to define it as $-dm$ to avoid sign issues when integrating the system mass. Since the system loses mass, $-dm$ is positive.

One could write in the horizontal direction Psys=mcarv+msandv, in which case

Just to note: Here you have split your system in two implicitly by splitting the momentum.

 

[kuruman]

Just to note: Here you have split your system in two implicitly by splitting the momentum.

Sure. The object of interest in this question is only the cart. One can define only it as "the system" or one can add anything to it that does not exchange horizontal momentum with it to some approximation, e.g. the sand, the Earth, the surrounding air, the Moon, etc. By the same argument, one can split out these addenda, without changing the answer to this question, until only the cart is left. This situation is no different from the one with two skaters holding hands and moving as one until they decide to open their hands and let go of each other. Skater A keeps going, while skater B brakes and comes to rest. If we are interested in the final speed of skater A, why consider the variable-mass, two-skater system?

I don't see a disagreement between us other than the optimum explanation to a third party. My personal preference is the one with the single-component system because it is simpler.

 

[willem2]

The speed of the rail car actually changes a little. if the railcar starts with 0 speed, then it will have moved

$$\frac {L \ln { \frac {m_{full}}{m_{empty}} }} {2}$$ when all the sand is out. It will end up with the same velocity at the end as at the start.​

​

where L is the length of the rail car.​

​

The sand is moving within the car to get to the exit. This can only happen if there is a force on it. This force can only come from the car. the last of the moving sand will give its momentum back to the car, so it will end up with the same momentum it started with.​

With a very light railcar the displacement can be arbitrarily high, as will the maximum speed.​

The speed of the car will be proportional to L and to the mass loss rate divided by the total mass. You only need to integrate this from the full mass to the empty mass, to get this result.​

 

[Halc]

The speed of the rail car actually changes a little. if the railcar starts with 0 speed, then it will have moved

$$\frac {L \ln { \frac {m_{full}}{m_{empty}} }} {2}$$ when all the sand is out. It will end up with the same velocity at the end as at the start.​

​

where L is the length of the rail car.​

​

The sand is moving within the car to get to the exit. This can only happen if there is a force on it. This force can only come from the car. the last of the moving sand will give its momentum back to the car, so it will end up with the same momentum it started with.​

With a very light railcar the displacement can be arbitrarily high, as will the maximum speed.​

The speed of the car will be proportional to L and to the mass loss rate divided by the total mass. You only need to integrate this from the full mass to the empty mass, to get this result.​

The formula assumes the drain hole is at the rear, as drawn. If it's in the middle, the internal forces are balanced and no such motion results. It needs to be generalized for the position of the hole along (L).

Now consider the system in the frame of the car. It is stationary until the sand starts draining. Since the car is gaining speed, so will the sand being dropped. Momentum to the left is leaving the system, leaving the system with a net momentum to the right. So once it empties, the car will have the remaining net momentum to the right and it will abruptly turn around and return to (and pass) its origin.

 

[Orodruin]

Now consider the system in the frame of the car. It is stationary until the sand starts draining. Since the car is gaining speed, so will the sand being dropped. Momentum to the left is leaving the system, leaving the system with a net momentum to the right. So once it empties, the car will have the remaining net momentum to the right and it will abruptly turn around and return to (and pass) its origin.

One cannot go into this detail of analysis without actually making a lot of extra assumptions about how the sand flows, which are clearly not intended to be part of the given problem. The sand gains speed, yes, but it also needs a speed in the other direction relative to the car in order to reach the hole at the rear.

 

[Leo Liu]

Homework Statement:: Please see below
Relevant Equations:: Momentum

Impulse

Conservation of momentum

2) If I consider the momentum of the system (train car + sand inside train car) in horizontal direction, the momentum will decrease because the speed is constant and the mass of sand decreases. But this means that the there is change in momentum in horizontal direction so the momentum in horizontal direction is not conserved and there will be net force acting on the system so the speed can't be constant, which contradicts to what I thought in part (1)

You forgot to take the initial horizontal velocity of the sand into account. If air resistance is negligible, then the momentum of the sand+train cart system is conserved before the sand hits the ground, and thus gets slowed down by kinetic friction.

Homework Statement:: Please see below
Relevant Equations:: Momentum

Impulse

Conservation of momentum

Normal force from ground on train car = weight of train car + new weight of sand + downwards force due to change in momentum of spilling sand

I don't think the sand ejected has any initial velocity in the vertical direction. It falls from the train car because the Earth exerts gravitational force on it. So the sand is interacting with the earth, not with the train.

 

[jbriggs444]

the sand hits the ground, and thus gets slowed down by kinetic friction.

Which has no effect on the rail car. So it makes no difference to the solution of the problem at hand. We should ignore the interaction between sand and ground entirely.

We can pay attention to the momentum flow as sand leaves the system of interest, carrying momentum with it. But we can ignore what happens to the transferred momentum thereafter.

 

[songoku]

Thank you very much for all the help and explanation ergospherical, Ibix, Orodruin, kuruman, willem2, Halc, Leo Liu, jbriggs444