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Oscillating bucket, variable mass

  1. Jun 25, 2012 #1
    1. The problem statement, all variables and given/known data
    A 2kg bucket containing 10kg of water is hanging from a vertical ideal spring of force constant 125N/m and oscillating up and down with amplitude 3cm. Suddenly the bucket springs a leak in the bottom such that water drops out of the bucket at a steady rate of 2g/s.


    2. Relevant equations
    When the bucket is half full find the period of oscillation.


    3. The attempt at a solution

    What confuses me about this problem is that book solutions manual simply find the mass of the half full bucket (7kg) and plugs to the equation T = 2pi*sqrt(m/k).

    What I initially tried to do is write the general equation of 2nd Newtons law:

    dp/dt = -kx => x''m + x'm' + kx = 0, where x' means a derivative with respect of time.

    And this is basically the equation with exact form as equation for damped oscillations. It only has replaced damping constant with m'. And it oscillates with frequency
    ω' = √[k/m-(1/2m*dm/dt)^2]

    However. the book assumes that the angular frequency is ω=√[k/m]

    The answers in both cases agree with 5 significant figures, this is because dm/dt is very small. But in the case of dm/dt is significantly big, is this is the right way to attack this kind of problem?
     
  2. jcsd
  3. Jun 25, 2012 #2

    Simon Bridge

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    Pretty much what you did was the correct approach - the book used an approximate method which, as you saw, was quite good.

    Suspect a misplaced minus sign though ... lets see:
    m(t).g - k.x(t) = v(t).dm(t)/dt + m(t).a(t): m(t) = 10 - 2t
     
  4. Jun 25, 2012 #3
    I think that the period of oscillation didn't depend on amplitude or other effects.It only depend on m and k.So that T=2π.[itex]\sqrt{\frac{m}{2k}}[/itex]
     
  5. Jun 25, 2012 #4
    Are you sure you can use the damping equation? The derivative of m(t) is a constant in this example, but m(t) is not. Isn't this required for the damping equation?
     
  6. Jun 25, 2012 #5
    If dm/dt is quite big, then the approximation used by the book will become inaccurate.

    Though the rate of mass change, from a small hole, wouldn't normally exceed lot more than that in practical situations.
     
  7. Jun 25, 2012 #6

    ehild

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    The equation is not correct. See: http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Derivation

    The momentum of the system bucket and water conserves during that very short time when the drop leaves the bucket. As the drop has the same velocity as the bucket, its falling out has no influence on the velocity of the bucket. The equation of motion is

    mx"=F(x),

    but you need to take into account that the mass is function of time.
    The external force includes both the elastic force and gravity. The change of mass will change the "equilibrium" position of the bucket, so it will move upward and oscillate with changing frequency and amplitude.
     
  8. Jun 25, 2012 #7
  9. Jun 25, 2012 #8

    ehild

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    You can do it, and you get the same equation. But in that equation, the relative velocity of the drop (with respect to to bucket) would appear. And that relative velocity is zero.

    ehild
     
  10. Jun 26, 2012 #9
    Lets say a different case... Imagine a pendulum with a hollow sphere filled with water at the end of the spring. There are two holes in a sphere such that, the water leaks out in the tangential direction of the the sphere motion.

    Will the equation of motion be the same as in the case where water cannot leak out?
     
  11. Jun 26, 2012 #10

    ehild

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    You misunderstood something. The rocket equation is

    mdV/dt+vreldm/dt=ƩFi,

    where vrel is the relative velocity of the exhaust with respect to the rocket.
    This is not the same equation as without the exhaust.

    In your problem, "leaking" means that the water drops leave the rocket with zero relative velocity. But that is not quite true, as the speed of the water pouring out from a hole depends on the height of the water level in the bucket and the phase of the oscillation. That would be a different system and a different equation, too complicated for me.

    ehild
     
  12. Jun 26, 2012 #11
    I get it now, thanks for your effort. :)
     
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