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Final velocity and total mass of train car below hopper

  1. Dec 3, 2013 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations

    F=ma

    3. 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!
     
  2. jcsd
  3. Dec 3, 2013 #2
    Assume that F = 0 and the car has some initial velocity. Will it just keep that velocity as coal is being loaded?
     
  4. Dec 3, 2013 #3
    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...
     
  5. Dec 3, 2013 #4
    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.
     
  6. Dec 4, 2013 #5
    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?
     
  7. Dec 4, 2013 #6
    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.
     
  8. Dec 4, 2013 #7
    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?
     
    Last edited: Dec 4, 2013
  9. Dec 4, 2013 #8
    With variable mass systems, you typically have to deal with momentum/impulse. Your equation is correct. Simplify it now.
     
  10. Dec 4, 2013 #9
    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)
     
  11. Dec 4, 2013 #10
    How come you have dt twice in the last term?
     
  12. Dec 4, 2013 #11
    Great question... I'm really not doing too well today, am I? Haha

    F=vμt+(dv/dt)(m+μ)
     
  13. Dec 4, 2013 #12
    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##.
     
  14. Dec 4, 2013 #13
    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?
     
  15. Dec 4, 2013 #14
    Well, you actually do not have to solve for dv just yet. Observe that the right hand side is simply the derivative of (mv).
     
  16. Dec 4, 2013 #15
    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.
     
  17. Dec 4, 2013 #16
    $$

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

    $$
     
  18. Dec 5, 2013 #17
    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
     
    Last edited: Dec 5, 2013
  19. Dec 5, 2013 #18
    Yes, that is correct.
     
  20. Dec 5, 2013 #19
    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?
     
  21. Dec 5, 2013 #20
    Why do you need acceleration if you already have velocity? Integrate velocity with respect to time, and you get displacement.
     
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