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Mass flow and rolling carts (kleppner 3.11)

  1. Dec 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Material is blown into cart A from cart B at a rate b kilograms per
    second. The material leaves the chute vertically down-
    ward, so that it has the same horizontal velocity u as cart B. At the
    moment of interest, cart A has mass M and velocity v. Find dv/dt,
    the instantaneous acceleration of A.

    2. Relevant equations
    Momentum

    3. The attempt at a solution

    System studied: cart A and the chute.

    Let's put that ##M_A## is the mass of cart A (unloaded), ##m(t)## is the material's mass in cart A at time ##t##, and ## \triangle m = m(t+\triangle t) - m(t)## is a small amount of material falling from the chute into cart A in a ##\triangle t## seconds.

    At a given time ##t##, the horizontal momentum will be:
    ## P(t) = (M_A + m(t)) v(t) + \triangle m \ u(t) ##
    ## P(t+\triangle t ) = (M_A + m(t) + \triangle m) v(t+\triangle t) ##

    So that in time ##t##:
    ## \frac{dP}{dt} = (M_A + m) \frac{dv}{dt} + \frac{dm}{dt} (v - u)##

    In this system, only friction from the wheels of cart A contribute to horizontal external force :
    ##f_{ext} (t) = -\mu g (M_A +m(t)) ##

    Since at time of interest ##t_i##, we are given:
    ##M_a + m(t_i) = M##,
    ## v(t_i) = v##,
    ##u(t_i) = u##,
    ## \frac{dm}{dt} = b##,

    the acceleration should be:
    ##\frac{dv}{dt}(t_i) = \frac{1}{M_A + m(t_i)}(f_{ext}(t_i) + \frac{dm}{dt} (u - v) = -\mu g +\frac{b}{M} (u-v) ##


    Do you agree with this solution ?
     
  2. jcsd
  3. Dec 13, 2014 #2

    haruspex

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    Yes, but you did not need to separate MA from m. Since friction is not mentioned, I would have assumed it zero.
     
  4. Dec 13, 2014 #3
    Thank you haruspex!
    I did separate because it wasn't so clear at first, but now I see it would work just the same ! Thanks
     
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