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Ball rolling down an incline gaining mass

  1. May 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A snowball, initially of mass m, slides down a slope inclined at an angle φ with respect to the horizontal. As it moves, the mass of additional snow Δm = αx that it accumulates is proportional to the distance travelled x. Write the differential equation
    of translational motion for the snowball, ignoring rotation and friction

    2. Relevant equations

    Conservation of momentum [itex]mv_before =mv_after [\itex]

    3. The attempt at a solution

    ##dp= (m+dm)(v+dv)-mv##
    ##dp= mdv +vdm ##
    [itex]\frac{dp}{dt}= m\frac{dv}{dt}+v\frac{dm}{dt} [/itex]
    [itex]mg sin(φ) = m\frac{dv}{dt}+vαx [/itex]

    But this doesn't seem right to me

    Many thanks :)
     
  2. jcsd
  3. May 18, 2016 #2

    BvU

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    What happended to ##dm\over dt## ? Ah, sorry ##v\alpha x## is that term ?
    So what is it that makes you feel uneasy ?
    What makes me feel uneasy is ##dm = \alpha \, x## but effectively you replaced it with ##\alpha\, dx##.
    Or did you ? because I don't see ##dm\over dt##, only ##dm##
     
  4. May 18, 2016 #3
    Ah yes, sorry I forgot to update this thread

    So dm= ##\alpha\, x##
    so [itex] \frac {dm}{dt}=\alpha\ \frac {dx}{dt} [/itex]

    and the rest follows from there. I think I have done it now :) Thank you though
     
  5. May 18, 2016 #4

    haruspex

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    Just a word of caution using Fdt=dp=vdm+mdv. In the sense in which that equation follows directly from dp=d(mv), it treats mass as though it can appear out of nowhere, which is not physically possible. The equation does work in practice, but only where the mass being added arrives with no initial velocity of its own. Similarly, if losing mass, it only works if the shed mass retains no velocity. E.g. for water leaking from a moving cart, vdm is negative, but no force is required to prevent it getting faster.
     
  6. May 18, 2016 #5
    Is that not the case here? I'm assuming the snow is initially at rest on the plane? Had it not been, I would have to take this into account.

    Thank you for your input :)
     
  7. May 18, 2016 #6

    Ray Vickson

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    As the ball slides down the ramp, the snow it picks up goes from an initial velocity of 0 to the velocity of the ball.

    Variable-mass dynamics can be quite tricky, and until fairly recently, several of the published accounts have been found to be erroneous in some respects. See, eg.,
    http://adsabs.harvard.edu/full/1992CeMDA..53..227P
    for a post-1990 account. In particular, blind use of dp/dt = m dv/dt + v dm/dt = F can lead to serious errors.
     
    Last edited: May 18, 2016
  8. May 18, 2016 #7

    haruspex

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    Yes, it works in this case, but as Ray confirms it is not really simply a result of applying the product rule of calculus to p=mv.
     
  9. May 20, 2016 #8
    I didn't did I? I thought I derived it from the conservation of momentum? I may be wrong :)
     
  10. May 21, 2016 #9

    haruspex

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    You wrote, and used, dp=vdm+mdv. Consider a cart containing water rolling smoothly on a horizontal track. Water leaks out at some steady rate ρ. dm=-ρdt, but no horizontal forces act on the cart, so you might write dp=mdv-vρdt=0, and conclude that the cart is accelerating.
    The problem lies in the definition of m. If you define it as the mass of the cart and contents at time t, that is not a closed system. The cart does not merely become less massive; that mass goes somewhere and may carry momentum away with it.
    Similarly in your snowball problem, if you define m as the mass of the snowball at time t then the accumulating mass has to come in from somewhere, and potentially arrives with its own initial momentum. In this case you get away with it because the additional snow had been at rest.
     
  11. May 21, 2016 #10
    Sorry to be a pain, but I only got that expression from considering dp=(m+dm)(v+dv)-mv-dm(u), (i.e. the total change in momentum) where u is the initial speed of 'dm'. But since u=0 and dmdv is almost zero this simply reduced to the usual formula of dp=vdm+mdv . I'm struggling to see how this is different from what you are saying?
     
  12. May 22, 2016 #11

    haruspex

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    But that is not what you originally posted. You just wrote dp=(m+dm)(v+dv)-mv. That is a correct equation if p represents just the momentum of the snowball. But you did not define p. Is it intended as just that, or is it the momentum of the system? Since m is just the mass of the snowball, the obvious guess is that p is intended as the momentum of just that. If you are then going to write dp =ΣF.dt then it had better be the total system that F acts on in time dt.
    To appreciate the subtlety, consider the reference frame of a skier going down the mountain at constant speed w. This changes v to v' =v-w everywhere, giving you dp=mdv+v'dm, yet the force is the same. With your updated formula, the -dm(u) term is +dm(w)=dm(v-v'), giving the right dp for the system.
     
  13. May 22, 2016 #12
    I thought it was since the first line of my working was : dp= (m+dm)(v+dv)-mv, i.e the change in momentum and the u=0 was implied, (and yes, p=momentum of the system, not the snowball- I should have explicitly said this, sorry). Maybe I should have made this more clear. But its an interesting point, and also highlights the importance of clarifying notation. Thanks, I appreciate the help :)
     
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