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Variable Mass:Sand from balloon

  1. Apr 17, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A balloon of mass M contains a bag of sand of initial mass ##m_o##. The balloon experiences a constant upthrust and is initially at rest and in equilibrium: the upthrust compensates exactly for the gravitational force. At time t=0, the sand is released at a constant rate. It is fully disposed of in time ##t_o##. Determine the velocity ##v(t)## at any time between t=0 and ##t=t_o##. Express the result in terms of $$k = \frac{m_o}{M+m_o} \frac{1}{t_o}$$

    3. The attempt at a solution

    At t=0, sand released at constant rate, ##\alpha## say. So, ##\frac{dm}{dt} = -\alpha, ##m the mass of the sand bag. Solve the above to give ##m_s(t) = m_o - \alpha t##

    The mass of the whole system (balloon + sand bag), ##M(t) = M + m_o -\alpha t##

    By NII, $$P(t+\Delta t) - P(t) = F_{ext} \Delta t \Rightarrow M(t + \Delta t)v(t + \Delta t) - M(t)v(t) = F_{ext}\Delta t$$

    Taylor expand the first two terms to give $$(M(t) + \Delta t \dot{M})(v(t) + \Delta t \dot{v}) - M(t)v(t) = [(M + m_o)g - M(t)g]\Delta t,\,\,F_{ext} = (M + m_o)g - M(t)g$$

    Collect terms ignore the (very small) ##\Delta t^2## term to get $$\dot{v} - \frac{\alpha}{M + m_o - \alpha t}v = \frac{(M+m_o)g - M(t)g}{M + m_o - \alpha t},$$ which can be solved by integrating factor.

    Does it look good so far? When I look at the solution, they end up doing just an ordinary integration.

    Many thanks.
     
    Last edited: Apr 17, 2013
  2. jcsd
  3. Apr 17, 2013 #2

    TSny

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    ##P(t+\Delta t)## should include the momentum of all of the material that is included in ##P(t)##. That would include not only the material still left in the balloon at ##t+\Delta t## but also the sand material that was released during the interval ##\Delta t##.
     
  4. Apr 17, 2013 #3

    CAF123

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    ##P(t + \Delta t)## is the momentum of the material still left in the balloon after ##\Delta t## and the momentum of the sand released in that ##\Delta t##. So I should have had something like: $$P(t + \Delta t) = M(t+\Delta t)v(t+\Delta t) - dm(v+dv),$$ dv the change in velocity as a result of disposing the sand.

    This will reduce the eqn to $$M(t)dv = (M+m_o)g - M(t)g$$ Is it better?
     
  5. Apr 17, 2013 #4

    TSny

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    Yes, that looks good (but you dropped a ##dt## somewhere).
     
  6. Apr 17, 2013 #5

    TSny

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    Maybe a minor point: For the last term ##dm(v+dv)## you could use just ##v## instead of ##v+dv## since you only need terms to first order.
     
  7. Apr 17, 2013 #6

    CAF123

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    So integrating from v(t=0) to v and from t=0 to t gives: $$\int_0^v dv = (M+m_o)g\int_0^t \frac{dt}{M+m_o - \alpha t} - gt$$

    This gives $$v(t) = \frac{(M+m_o)g}{\alpha} \ln \left(\frac{M+m_o}{M+m_o-\alpha t}\right) - gt$$

    Is there another way to express ##\alpha##?
     
  8. Apr 17, 2013 #7

    TSny

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    You should be able to express ##\alpha## in terms of ##m_0## and ##t_0##.
     
  9. Apr 17, 2013 #8

    CAF123

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    That would be ##\alpha = m_o/t_o##.
     
  10. Apr 17, 2013 #9

    TSny

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    Right.
     
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