Final velocity and total mass of train car below hopper

[oddjobmj]

Homework Statement

A cart, starting at rest, is below a hopper at t=0. The hopper drops coal into the cart at a constant rate μ. At t=0 a constant force starts pushing the cart to the right and the coal begins to fall. The force ends once the back of the cart passes the hopper.

A) Find final velocity of cart

B) Find total mass of coal delivered to the cart

Mass of empty cart = M = 3000 kg
Length of cart = L = 20 m
Mass delivery rate = μ = 507 kg/s
Force = F = 1343 N

Homework Equations

F=ma

The Attempt at a Solution

F=ma => a=[itex]\frac{F}{m}[/itex] = [itex]\frac{F}{m_0+μt}[/itex]=[itex]\frac{dv}{dt}[/itex]

Since there are no time dependent variables can I just do this?:

[itex]\int[/itex]dv=[itex]\frac{F dt}{m_0+μt}[/itex] => v=[itex]\frac{F*ln(m_0+μt)}{μ}[/itex]+C₁

When t=0, v=0 so:

0=[itex]\frac{F*ln(m_0)}{μ}[/itex]+C₁ => C₁=[itex]\frac{-F*ln(m_0)}{μ}[/itex]

v(t)=[itex]\frac{F*ln(m_0+μt)}{μ}[/itex]-[itex]\frac{F*ln(m_0)}{μ}[/itex]=[itex]\frac{dx}{dt}[/itex]

[itex]\int[/itex]dx=[itex]\int[/itex][itex]\frac{F}{μ}[/itex](ln(m₀+μt)-ln(m₀))dt =>

x(t)=[itex]\frac{F(-μt-μt*ln(m_0)-(m_0+μt)*ln(m_0+μt)+m_0*ln(m_0))}{μ^2}[/itex]+C₂

When t=0, x=0 so:

C₂=[itex]\frac{Fm_0*ln(m_0)}{μ^2}[/itex]

Plugging that into the x(t) equation, solving for t, and plugging in the knowns I get

v(t)=2.9 m/s

I guess mass delivered is just μ*t=6000 kg

Does that look right? Where did I go wrong? Thanks for your help!

 

[voko]

Assume that F = 0 and the car has some initial velocity. Will it just keep that velocity as coal is being loaded?

 

[oddjobmj]

No, it would maintain momentum but velocity would be reduced as mass is added. Great point but I'm not sure where to bring in the factor of the mass being added in that respect. Ahh, I thought I had this...

 

[voko]

Consider what happens over some short time ##dt##. The car has velocity ##v## with mass ##m##, mass ##dm##, with velocity 0, is added to the car, and all this time there is constant force ##F##, the new velocity of the car is ##v +dv##. Use the impulse/momentum equation.

 

[oddjobmj]

Oohh, right, thank you!

Well, P=mv

P=(m₀+μt)(v₀+at)

P=(m₀+μt)(at)

a=[itex]\frac{P}{t(m_0+μt)}[/itex]

Is that what you are suggesting?

then a=[itex]\frac{dv}{dt}[/itex]=[itex]\frac{P}{t(m_0+μt)}[/itex]

The integral here is a little nastier because now I do have a time dependent variable besides t to contend with. Will this work?

 

[voko]

Oohh, right, thank you!

Well, P=mv

P=(m₀+μt)(v₀+at)

No, this is not correct, because that assumes the acceleration is constant. Please read again what I wrote in #4. Forget acceleration, just write the total momentum before, total momentum after, and the impulse in between.

 

[oddjobmj]

Ah, sorry. I have very rarely work with momentum directly so I'm still acclimating myself with these relationships.

Initial momentum + Fdt=new momentum

vm+Fdt=(v+dv)(m+dm)

Closer?

 

[voko]

With variable mass systems, you typically have to deal with momentum/impulse. Your equation is correct. Simplify it now.

 

[oddjobmj]

Well, if I multiply through I get:

vm+Fdt=vm+vdm+mdv+dvdm

I can cancel the vm's and divide through by dt:

F=v dm/dt + m dv/dt + (dv/dt)(dm/dt)

I have dm/dt so I could plug that in and simplify a bit further but I'm not sure where that would get me. Also F is constant, as you pointed out, and I do know F.

F=vμt+[itex]\frac{dv}{dt}[/itex](m+μt)

 

[voko]

Well, if I multiply through I get:

vm+Fdt=vm+vdm+mdv+dvdm

I can cancel the vm's and divide through by dt:

F=v dm/dt + m dv/dt + (dv/dt)(dm/dt)

How come you have dt twice in the last term?

 

[oddjobmj]

Great question... I'm really not doing too well today, am I? Haha

F=vμt+(dv/dt)(m+μ)

 

[voko]

I do not see how that answers my question. The correct term should be ## \frac {dm dv} {dt} ##. Now observe that all the other terms are finite, while this goes to zero as ##dt \to 0##.

 

[oddjobmj]

Ah, I went with dmdv/dt and continued to simplify as suggested, sorry.

So for really small t the term is negligible? So could I do:

F=v[itex]\frac{dm}{dt}[/itex]+m[itex]\frac{dv}{dt}[/itex]

Then solve for dv and integrate to find v with respect to t?

 

[voko]

Well, you actually do not have to solve for dv just yet. Observe that the right hand side is simply the derivative of (mv).

 

[oddjobmj]

I'm not sure how to use that, actually. The only thing I can think of is to re-write it:

F=[itex]\frac{dmv}{dt}[/itex]

Fdt=dmv

Not sure how to deal with the dmv now.

 

[voko]

$$

F = \frac { d(mv) } {dt}
\to Fdt = d(mv)
\to \int\limits_0^tFdt = \int\limits_0^{(mv)} d(mv)
\to \ ?

$$

 

[oddjobmj]

Right, setting that up is very familiar. I've simply forgotten how to deal with d(two values). I may be over-thinking it but is this correct?:

Ft=mv

 

[voko]

Yes, that is correct.

 

[oddjobmj]

Great, thank you!

That means:

v=[itex]\frac{Ft}{m}[/itex]

Plugging in the initial conditions doesn't get me anywhere so perhaps I can figure the acceleration:

dv/dt=a

a=[itex]\frac{Ftdt}{m}[/itex] but m depends on t:

a=[itex]\frac{Ftdt}{m_0+tμ}[/itex]

a=-[itex]\frac{Ftμ}{(m_0+tμ)^2}[/itex]+[itex]\frac{F}{m_0+tμ}[/itex]+C

When t=0, a=0 so C=[itex]\frac{-F}{m}[/itex] so:

a=-[itex]\frac{Ftμ}{(m_0+tμ)^2}[/itex]+[itex]\frac{F}{m_0+tμ}[/itex]-[itex]\frac{F}{m}[/itex]

Am I on the right track?

 

[voko]

Why do you need acceleration if you already have velocity? Integrate velocity with respect to time, and you get displacement.

 

[oddjobmj]

That would be what I would do if I were to have properly thought through what I was about to do... :

v=[itex]\frac{dx}{dt}[/itex]=[itex]\frac{Ft}{m_0+μt}[/itex]

[itex]\int[/itex]dx=[itex]\int[/itex][itex]\frac{Ftdt}{m_0+μt}[/itex]

x=[itex]\frac{F(μt-m_0*ln(m_0+μt))}{μ^2}[/itex]+C

C=[itex]\frac{Fm*ln(m)}{μ^2}[/itex]

x=[itex]\frac{F(μt-m_0*ln(m_0+μt))}{μ^2}[/itex]+[itex]\frac{Fm*ln(m)}{μ^2}[/itex]

I'm not sure of a way to simplify this without solving numerically by plugging in my known values.

t≈15 seconds

Plugging t into our equation for v:

v≈ 1.9 seconds

The result seems consistent with the original interpretation. It would be a little slower when considering the effect of mass with 0 velocity being added.

 

[voko]

Because the dependence of x on t is a little nasty, you could instead try to find the dependence of v on x. You used the momentum/impulse approach to get the time dependency. You could use the work/energy approach to find v(x), then, because v(t) is relatively simple, solve v(t) = v(x) for t.

 

[oddjobmj]

Would you agree that the solution I worked through is at least reasonably accurate? If I don't end up finishing the work/energy solution in time I just want to make sure I understand the final stretch of the momentum/impulse relationship. With that said:

Work Total=Energy Initial - Energy Final

For a small change in x:

Fdx=.5(m+dm)(v+dv)²-.5mv²

Hmm, this is also pretty nasty. Any suggestions?

 

[voko]

You do not have to do the work/energy part. I am not sure whether 15 seconds is correct, but, in principle, you can solve the equation relating x to t numerically, that is a valid approach.

 

[oddjobmj]

Well, I am here to learn. I would love to continue; I much prefer general results to numerical results but beggars can't be choosers. On the other hand you have helped me quite a bit already. I greatly appreciate that, thank you!

If you did choose to proceed in helping me work through the work/energy relation I'm sure we could draw in what we have already covered.

 

[voko]

Initially I thought that the entire work of force would be equal to the kinetic energy of the car with the coal, but that is wrong because the collision of coal with the car is not elastic. Ignoring the vertical motion of coal, what we have here is basically a collision of a piece of coal ##m## at zero velocity with the car ##M## at some velocity ##u##. After the collision, the coal must have velocity ##v##, and the car velocity ##V##, so ##Mu = mv + MV##. But in this case the coal stays in the car, so ##v = V = {Mu \over (m + M)}##. The total energy of the system before the collision was ## Mu^2 \over 2 ##, after the collision it is ## {(m + M)v^2 \over 2} = {M \over (m + M)} {Mu^2 \over 2} < {Mu^2 \over 2} ##, thus energy is lost as coal is loaded, so the work/energy theorem is not applicable here.

To compensate for my misleading you, I have checked your numbers and found that the time and velocity that you found are correct. Well done!

 

[haruspex]

Although work is not conserved, you can still use the a = vdv/dx = vv' trick to eliminate time.
##(M+\mu t)v = Ft##
##t = \frac{Mv}{F-μv}##
Differentiating:
##(F-μv)^2 = (F-μv)M\dot v + Mvμ\dot v = FM\dot v = FMvv'##
##x = \int\frac{FMv }{(F-\mu v)^2}dv = \frac{Fm}{μ^2}\left(y-ln(1+y)\right)##, where y = μv/(F-μv).
Still need to use approximations to solve it, of course.

 

[oddjobmj]

Thank you both for your explanations. That definitely makes sense, Voko. You didn't have to check those but I do appreciate it! That is an interesting trick, haruspex, thank you!