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Spinning bucket-forces

  1. Nov 25, 2008 #1
    a bucket is filled with water and tied to a rope with a length of "L=1m", the bucket is then spun in a verticle circular motion ,
    what is the minimum frequency the bucket can be spun at in order for the water to not spill?

    i realise that i need to find the centripeutal force. is it mv^2/r??

    from there i can find f=(v/L)*(1/2pi)

    the correct answer is (1/2pi)*sqrt(g/L)--- where have i gone wrong
     
  2. jcsd
  3. Nov 25, 2008 #2

    Doc Al

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    Staff: Mentor

    Yes.

    How did you get this result?
     
  4. Nov 25, 2008 #3
    using the equation for frequency in circular motion, i realise its wrong,
    do i need to compare mv^2/L to mg, saying that it will not fall out when mv^2/L=mg
    ie when v^2/L=g
    v=(sqrt(g*L)
    still not right??
     
  5. Nov 25, 2008 #4

    Doc Al

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    Staff: Mentor

    Yes, that's the minimum speed at the top to keep the water in the bucket. Use that speed to find the minimum frequency. Hint: It goes in a circle.
     
  6. Nov 26, 2008 #5
    so, using equation for frequency (f)

    [tex]\omega[/tex]=2(pi)f

    f=[tex]\frac{\omega}{2pi}[/tex]

    [tex]\omega[/tex]=[tex]\frac{v}{L}[/tex]

    f=[tex]\frac{v/L}{2pi}[/tex]


    v=[tex]\sqrt{g*L}[/tex]

    f=[tex]\frac{\sqrt{g*L}/L}{2\pi}[/tex]
     
    Last edited by a moderator: Nov 26, 2008
  7. Nov 26, 2008 #6

    Doc Al

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    Staff: Mentor

    Good. Now simplify that a bit so it looks like the given answer.
     
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