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Homework Help: Bucket of water spinning in vertical circle

  1. Jun 12, 2016 #1
    1. The problem statement, all variables and given/known data
    The water stays in the bucket, even at the top of the circular path, as long as the speed exceeds a certain value. Explain why.

    I think i have a good answer, but not 100% sure. My answer:

    There is a centripetal force acting on the bucket and the water since they are travelling circular motion. In order for the water to remain in the bucket when the bucket is at its highest point, the reaction force of the bucket acting on the water must be greater than the weight, since the centripetal acceleration of the bucket must be greater than the acceleration due to gravity in order to prevent the water from falling out of the bucket.

    This can only be achieved above (or equal to) a certain velocity. This velocity is given by:

    v= sqrt(g x r) I obtained this formula by equating the centripetal force equation with mg.

    Is what i have said correct?

  2. jcsd
  3. Jun 12, 2016 #2


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    Overall, your explanation sounds good to me except for one statement:
    When you say "greater than the weight" I'm not sure what weight you are talking about (bucket, water, both?). But actually, the force of the bucket acting on the water does not need to be greater than the weight of any of the objects. If the speed is just at the point where the water is on the verge of losing contact with the bucket, what is the magnitude of the reaction force of the bucket on the water?
  4. Jun 12, 2016 #3
    Thanks for the reply

    Oops. I meant the reaction on the water from the bucket must be greater than the weight of the water.

    I say this because wouldnt the bucket have to be accelerating towards the centre of the rotation faster than the bucket for the water to stay in the bucket/ lose contact with the bucket?
  5. Jun 12, 2016 #4
    What would a FBD look like for the water at the top of the swing?
  6. Jun 12, 2016 #5
    Wouldnt it just be two forces, both acting towards the centre of rotation and therefore directly downwards. One being the reaction, and the other being the weight?
  7. Jun 12, 2016 #6
    Good. Now, what if the "reaction" of the bucket provided zero force ONLY at the top of the swing? Is this possible? If so, would the water still be able to go in a circular path?
  8. Jun 12, 2016 #7
    Well this would mean that the only force acting on the water is weight, so the water will no longer follow a circular path, since it is no longer being accelerated perpendicular to its motion. So it would just fall towards earth with a parabolic motion rather than circular motion?
  9. Jun 12, 2016 #8
    But at the top of the journey the acceleration and the velocity IS perpendicular. In the next instance, the situation changes. How? Is the reaction force still 0? Why did it change, if it did?
  10. Jun 12, 2016 #9
    I meant after that point, the acceleration is always directly down, because the force due to gravity acts directly down. But now i realise what you were initially asking: Reaction is zero only at the top, but is not zero again afterwards when it continues to turn.

    But in order for the reaction to be zero at the top, wouldnt the centripetal acceleration of the bucket have to be less than the acceleration due to gravity, so that the water is accelerating directly down (towards centre of rotation) faster than the bucket?

    So if we say that the reaction force just about becomes zero at the top (if centripetal acceleration was just below acceleration due to gravity) the water will attempt to travel straight again, but the bucket is still rotating, so the water will "make contact" with the water again. So the water will almost instantly go from feeling no reaction, to feeling a reaction again as the bucket rotates just past the highest point.
  11. Jun 12, 2016 #10
    I think part of your confusion is believing centripetal force is its own separate force. The centripetal force is the combination of all the forces acting on the water. This combination (or net force) is always acting towards the center of the circular path. At the top, when the velocity is at its minimum, the centripetal force is composed only of the weight of the water. In the next instance, as inertia carries the water over, the centripetal force has changed direction a little(still pointing towards the center) but is a vector composes of the weight of the water down and the reaction force of the bucket on some angle.
  12. Jun 12, 2016 #11
    I know that centripetal force isnt a real force. I understand what you say. So at either side, when the string holding the bucket is parallel to the ground mv^2/r = T (tension in string). And at all other positions mv^2/r = T+mgcos of the angle from the vertical.

    I think im finding it confusing because the bucket and water are independent objects. Im totally comfortable with working with a ball on a string, or questions in other context which involve only 1 object.
  13. Jun 12, 2016 #12
    Is this correct for the water:

    At the top, mv^2/r = mg
    All other points: mv^2/r = mgcos(a) + Rcos(a)

    Sorry if i keep getting things wrong!
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