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Momentum—Downhill Dustpan

  1. Apr 11, 2017 #1
    1. The problem statement, all variables and given/known data
    A dustpan slides down a plane inclined at angle θ. Dust is uniformly dis- tributed on the plane, and the dustpan collects the dust in its path. After a long time, what is the acceleration of the dustpan? Assume there is no friction between the dustpan and plane.

    p=linear mass density of dust

    x is chosen as distance along the plane

    2. Relevant equations
    d(mv)/dt=mgsin(θ) (net force parallel to the plane)
    m(x)=px

    3. The attempt at a solution
    My solution:

    d(mv)/dt=mdv/dt+vdm/dt=mgsin(θ)

    dm/dt=pdx/dt=pv
    dv/dt=x''

    direct substitution:

    px*x''+p(x')^2=pxgsin(θ)
    xx''+(x')^2-xgsin(θ)=0

    Dimensional analysis: x(g, t, θ) must be of the form:
    x=Agt^2
    x'=2Agt
    x''=2Ag

    Substitute into DiffEq:
    2A^2g^2t^2+4A^2g^2t^2-Ag^2t^2sin(θ)=0

    cancel g^2t^2
    6A^2-Asin(θ)=0

    A=sin(θ)/6

    x''=gsin(θ)/3


    My question is, is this reasoning correct?




     
    Last edited by a moderator: Apr 11, 2017
  2. jcsd
  3. Apr 11, 2017 #2

    kuruman

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    Why must x(t) have this form?
     
  4. Apr 11, 2017 #3
    g, t, and theta are the only variables that can possibly affect position. Dimensional analysis gives me the result. (A is a dimensionless constant.)

    m=mas-2asb

    seconds must cancel: -2a+b=0, a=1 (to ensure that length dimension is 1 as it should be)

    a=1, b=2

    hence x=Agt2

    I thought that x shouldn't dimensionally depend on θ as it is dimensionless. θ will likely be part of the A (which it ended up being in my answer).
     
  5. Apr 11, 2017 #4

    kuruman

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    What about the density ρ of the sand? Should that not affect the answer? By stating that the position is x(t) = Agt2 you are asserting that the acceleration is constant. Is that obvious?
     
  6. Apr 11, 2017 #5
    Damn it. I forgot. But let me justify my work so far.

    xx''+(x')^2-xgsin(θ)=0

    This is my final differential equation, arrived at through only momentum considerations and not dimensional analysis (if my reasoning is indeed correct). No density term appears here (it cancelled out). Since any x(t) must come from this equation, it cannot contain an empirical variable that does not appear in the equation—if there was indeed a (dimension-bearing) ρ in my x(t), it would have appeared in the diffEQ. The "unknown constant" problem is solved with my treatment of the "A".

    Let me try to incorporate ρ, though. My dimensional analysis equation becomes:

    m=mas-2asbkgcm-c (the kgcm-c is mass density raised to an unknown power)

    To ensure that the final result (m) has no powers of mass (kg), we must set c to 0 as there is no other kg term. This is tantamount to saying that x is independent of ρ.
     
  7. Apr 11, 2017 #6

    kuruman

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    What happened to the mass? If you look at your starting expression,
    the second term on the left has no "m" term that cancels. Also, how do you get xx'' from mdv/dt?
     
  8. Apr 11, 2017 #7
    Starting equation which you had no problem with in this reply: mdv/dt+vdm/dt=mgsin(θ)

    Relevant equations:

    p=linear mass density along the plane (read as rho)

    v=dx/dt (definition)
    m=px (dustpan travels a distance x, picks up p kg per distance traveled)
    dm/dt=pdx/dt=pv (differentiating above expression)

    dv/dt=x''=d/dt(dx/dt)=d2x/dt2 (definition)

    mdv/dt=px*x'' (substitutions)
    vdm/dt=v*pv=pv^2 (substitute pv for d/dt)
    mgsin(THETA)=pxgsin(THETA) (substitute px for m)

    pxx''+pv^2=pxgsin(THETA)

    p cancels
    v=x' (definition)
     
  9. Apr 11, 2017 #8

    kuruman

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    OK, I would agree with this equation under the assumption that the dustpan is massless. If it had mass, say m0, then m(x) = m0 + ρx.
    BTW, you can make your equations more legible. Look at the ribbon directly below "Have something to add?" and you will find all sorts of useful stuff.

    Anyway, this diff. eq. doesn't look separable to me.
     
  10. Apr 11, 2017 #9
    I think the problem wants us to treat it as massless. I've seen other versions of the problem that treat it the same way. The diffEQ may look ugly, thats why I used dimensional analysis. Inspired by one of David Morin's solutions to a similar problem in his textbook.
     
  11. Apr 12, 2017 #10

    kuruman

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    I am not familiar with that. Are you talking about the Buckingham π theorem?
     
  12. Apr 12, 2017 #11

    haruspex

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    Yes, that works by dimensional analysis.
    If you don't trust that, you can just suppose that xt-a tends to a constant for some a. Plugging that into the DE, the powers of x that result are a and 2a-2. To make the equation work we need a=2.
    Yes.
     
  13. Apr 12, 2017 #12

    kuruman

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    I am bothered by this statement. When am I justified in just supposing that? If I had the ODE x' - bx = 0, which has an exponential solution, this assumption does not work. Or is this method of the "try it and see if it works" variety like separating variables in a PDE?
     
  14. Apr 12, 2017 #13

    haruspex

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    Yes, but in the present case the physics tells us it cannot be a higher power than 2 (or less than 0). The dust won't make it go faster.
     
  15. Apr 12, 2017 #14

    kuruman

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    Thanks.
     
  16. Apr 12, 2017 #15
    why can't it be an exponential decay then? (This is for argument's sake Im pretty sure my polynomial solution is correct)
     
  17. Apr 13, 2017 #16

    haruspex

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    I assume you mean negative exponential.
    That would imply x converges to a limit, no? Clearly that is not possible.
     
  18. Apr 13, 2017 #17
    you're right. My bad. That can't happen if the mass keeps increasing.
     
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