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Find V(t) of train car with changing mass.

  1. Oct 1, 2013 #1
    1. The problem statement, all variables and given/known data
    An empty freight car of mass M starts moving from rest with a constant applied force F0. At the same time, coal runs into the car at steady rate b from a hopper above the car at rest along the track. Calculate the velocity of the car as a function of time.

    Data: M = 1900 kg; F0 = 47000 N; b = 190 kg/s; t = 6.8 s.


    2. Relevant equations
    F=m*a
    V(t)=v0+a*t

    3. The attempt at a solution
    M(t)=M+bt

    F=ma=M[itex]\frac{dv}{dt}[/itex]

    F dt=M dv

    dv=[itex]\frac{F}{M}[/itex]dt=[itex]\frac{F}{M=bt}[/itex]dt

    Integrating both sides I get:

    v=[itex]\frac{F*ln((bt/M)+1)}{b}[/itex]

    Plugging in my known values I find that v(6.8)=128.33 m/s which is incorrect. Any suggestions?

    Thank you!
     
  2. jcsd
  3. Oct 1, 2013 #2
    Newtons 2nd law says Ftot=dp/dt= dmv/dt=(dm/dt)*v+(dv/dt)m
     
  4. Oct 1, 2013 #3
    Thank you, lep11. I'm actually not sure how to solve that. I am very rusty with regard to differential equations.

    Can I multiply through by dt to get Ftotaldt=vdm+mdv ?

    Then I can integrate both sides to get Ft=2mv where m=M+bt

    Ft=2v(M+bt)

    v=[itex]\frac{Ft}{2(M+bt)}[/itex]

    I tried this and came up with 50.06 m/s which is also not correct.
     
  5. Oct 1, 2013 #4
    Hi oddjobmj...Could you please tell what is the correct answer ?
     
  6. Oct 1, 2013 #5
    I don't have the correct answer, unfortunately. It is an online based program where I have a certain number of attempts.
     
  7. Oct 1, 2013 #6

    haruspex

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    You have to be careful with that formulation. E.g. how would you represent the case where the coal is falling from a hopper which is itself moving?
    Oddjobmj, the problem with your formulation is that it treats the added coal as already having the same forward velocity as the cart. Instead, just think about the total momentum provided by the force in time t and the total momentum obtained by cart plus its load of coal at that time.
     
    Last edited: Oct 1, 2013
  8. Oct 1, 2013 #7
    oddjobmj...You should approach the variable mass system problems using momentum approach.

    P(t) = M(t)v(t)

    P(t+dt) = ( M(t)+dm )( v(t)+dv )

    dP = M(t)dv+v(t)dm

    or,dP/dt = M(t)dv/dt+v(t)dm/dt

    Now dP/dt =F , dm/dt = b and M(t)=M+bt

    F = M(t)dv/dt+bv(t)

    F-bv(t) = M(t)dv/dt

    dt/M(t) = dv/(F-bv(t))

    Integrate with proper limits and you should end up with v=Ft/(M+bt) .

    Does that mean we can apply dP/dt approach directly in variable mass problems only when the velocity of the outgoing/incoming mass(added coal) is same as that of the system(car+coal at time t) ?
     
    Last edited: Oct 1, 2013
  9. Oct 2, 2013 #8

    haruspex

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    Yes, but it's much easier than that. No calculus is needed. Just think about the total momentum that has been gained by the cart and the coal in it at time t.
    Think about what the v dm/dt term means. It says some amount of mass has been added / created at speed v. Since in reality mass does not get created out of nothing, it must have already existed with some momentum. Taking the gain in momentum to be v dm/dt assumes it had no prior momentum (within the reference frame). In the present problem that works because we are taking the added coal to be initially at rest, but it does not work in general.
     
  10. Oct 2, 2013 #9
    In theory, this approach is always valid. In practice, total momentum ## p(t) = m(t)v(t) ## only in special cases, in general ## p(t) ## is the sum (or integral) of various masses moving at various speeds. For example, if sand is ejected at relative velocity ##v_e##, then ## p(t) = \int_0^t (-m'(\tau)) (v(\tau) + v_e(\tau)) d\tau + m(t) v(t) ## and ## p'(t) = - m'(t) (v(t) + v_e(t)) + m'(t) v(t) + m(t) v'(t) = m(t) v'(t) - m'(t) v_e(t) ##.
     
  11. Oct 2, 2013 #10
    I know what you have been hinting at :) .This is an excellent way of thinking in the context of the problem.Let oddjobmj figure it out.

    Okay…So vdm/dt is the change in the momentum of the added/ejected mass ‘dm’ in inertial reference frame. For this reason ,in rocket equation we have (v-u)dm/dt instead of vdm/dt .

    Suppose the cart and hopper both would have been moving at same velocity ‘v’.In that case the term representing change in momentum of the increased mass dm, vdm/dt would have been zero.

    Is my understanding correct ?
     
  12. Oct 2, 2013 #11

    haruspex

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    Yes. In the context of the OP, but in a reference frame where the coal in the hopper has a constant nonzero velocity u and mass is transferred to the cart at rate b, you could reason that the change in the momentum for hopper+coal is u dm/dt + m du/dt = -ub. Combining that with vb+m(t)dv/dt for the cart+coal we have F = (v-u)b+m(t)dv/dt.
     
  13. Oct 3, 2013 #12
    Sorry, I was away for a few days!

    This worked, thank you! I was pretty darn close.
     
  14. Oct 4, 2013 #13

    haruspex

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    But as I mentioned, there is a much easier way. The cart and all the coal in it started at rest. If it is now mass M+bt and moving with velocity v how much momentum has it gained in total? How much momentum has the force F supplied in time t?
     
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