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Leaking bucket

  1. Oct 2, 2006 #1


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    Lets say that there's a bucket swinging on a string (the top of the bucket is connected to the string) and there's a small hole in the bottom of the bucket, will the frequency get bigger or smaller? i think that it will get smaller because as the bucket leaks the water gets closer to the bottom of the bucket so the center of the the swinging mass gets further away from the spinning axis - in effect it's as if the string got longer. am i correct?
    -this isn't a homework problem.
  2. jcsd
  3. Oct 2, 2006 #2


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    Sounds good to me.
  4. Oct 2, 2006 #3
    Be careful, you cannot forget about conservation of angular momentum.
    The bucket will swing with an angular momentum of:

    L = I*w = m*r^2*w

    where r is the distance to CM
    As your mass decreases it is true that r will increase.
    However, keep in mind that this is counterbalanced to some extent by the fact that m decreases.
    Whether or not w increases or decreases depends on which factor (r^2 or m) is changing faster.
    A tall skinny bucket would probably result in a smaller frequency w but a wide fat bucket would have the opposite effect.
  5. Oct 2, 2006 #4
    Are you completely certain about that? :wink:
  6. Oct 2, 2006 #5
    I think I see your point cesium:
    As water is flung out of the bucket kinetic energy goes with it.
    Consequently you experience a loss of angular momentum equal to:
    [itex]\frac{dL}{dt}=\rho(t)\ r^2\omega[/itex]
    where rho is your water mass flow rate out of the bucket.
    Hence Angular momentum is not conserved.
    All the same I think it still may be possible to to have your frequency increase if the speed and quantity of water doesn't leave too fast. However, I cannot intuitively visualize an answer and I am still digging through setting up an equation to check.
    Last edited: Oct 2, 2006
  7. Oct 2, 2006 #6
    Not visualising a pendulum?
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