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Homework Help: Vertical Circular Motion Question - Why does the water stay in the bucket?

  1. May 18, 2006 #1
    Hello all,

    This is (I think) one of the more popular introductory physics questions. Someone ties a bucket to a rope, puts some water in it and swings it around in a vertical circle with relatively constant speed. When the bucket is directly above the person's head, the water stays in the bucket. I, however, can't seem to figure out why this happens.

    I've drawn the Free Body Diagram of the water a number of times, and this is what I get:
    F(g) = mg [down]
    F(n) [down]
    Thus the centripital force is down, as it should be, into the centre of the circle.

    I also drew the FBD of the bucket, here are the forces I have:
    F(g)= mg[down]
    T = [down]
    F(c) is therefore also down.

    I realize that there is a centripital velocity to the bucket, which is perpendicular to the circle. If I square this velocity, devide it by the radius (r) and the mass (m), I should get a centripital force F=ma, a=v^2/r, F=m(v^2/r).

    From what I understand, this vector is paralel to the velocity (vc). However, that would also make it perpendicular to the gravitational and tension forces, meaning that it does not affect them. So, in the x-hat direction, I have this (centrifugational?) force, and in the y-hat I have the Tension and the gravitional force. However, I still cannot figure out why the water does not fall out. The forces should be orthogonal and thus there is no upwards force or any component of a force in the positivie y-hat direction to keep it up...

    If someone could provide me with an explanation, I would greatly appretiate it,


    EDIT: Appologies if the problem is somewhat muddled, I've become rather confused trying to solve it :D. I think that the description is clear enough, though...
  2. jcsd
  3. May 18, 2006 #2

    Doc Al

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    This is good. Now you just need to understand what normal force means. For the water to stay in the bucket it must be pressed against the bucket. What does that tell you about the normal force?

    OK. Call the velocity a tangential velocity, not a centripetal velocity. (The word "centripetal" means "towards the center".)

    The key is the normal force. (And once you see that connection you'll be able to calculate the minumum speed at the top of the motion that will keep the water in the bucket.) Realize that the normal force is not fixed: it depends on the speed.
  4. May 18, 2006 #3


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    Firstly, I really must point one thing out because it is something that really bugs me: There is no such thing as the centrifugal force, it arises if one views circular motion from a rotating i.e. non-inertial reference frame.

    At the top of circle you basically have two forces acting. As you corrctly said: Gravity (downwards) and the normal force (downwards), which is equal to the tension in the string. No other forces are acting. Therefore together, these two forces must equal the centripetal force for the water to reamain in the bucket;

    [tex]F_{g} + F_{n} = F_{c}[/tex]

    Think of it like this, these forces are used to provide the centripetal force, any additional force would used to 'pull' the water out of the bucket. Now, using newton's second law; [itex]F = ma[/itex] and the fact that centripetal acceleration is given by;

    [tex]a = \frac{v^2}{r}[/tex]

    We can say that;

    [tex]F = \frac{mv^2}{r}[/tex]

    Therefore, using the origonal equation we can say that;

    [tex]mg + T = \frac{mv^2}{r}[/tex]

    Does that make sense?


    Edit: Sorry Doc, I was typing for a while.
  5. May 18, 2006 #4
    I think I understand it now...

    For the bucket, F(c)=F(g)+T, which, as Hootenanny said, can be written as F(g)+T=m*v^2/r. At the minimum speed, however, the rope will be "loose", meaning that there is no tension. Thus, T=0 and mg=m*v^2/r, g=v^2/r

    For the water inside the bucket, F(n)+F(g)=F(c)=m*v^2/r. Just like the bucket, at minimum speed, the water does not push against the bucket (thus there is no F(n) back on the water). Therefore, mg=M*v^2/r. This makes sense because the speed (v) of the bucket has to be equal to the speed of anything inside of it (the water).

    If the above is correct, then the speed is what keeps the water pressed to the bucket. Since, if there is no speed, there normal force equals mg, as it did when the bucket was at the bottom, before the acceleration was applied. If the bucket is at the top and v=0, then it also makes sense that the Fn=mg, since the bucket will fall with acceleration of g, thus pushing the water with a force of mg.

    Is this correct?
  6. May 18, 2006 #5


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    Sounds about right :smile: I'm impressed, it took me a while to get my head around circular motion (lots of questions about eskimo's sliding down the roof of their igloos!).

  7. May 18, 2006 #6

    Doc Al

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    Just a slight correction. For the water to be pressed into the bucket, the bucket must exert a force on the water. That force is the normal force. To find the minimum speed, solve for the speed where the normal force just becomes zero. (Any less, and water will start spilling.) So, at the top of the motion, when the water just barely stays in the bucket the normal force is zero (not mg!).

    If the speed were zero at the top, there would be no circular motion at all, and since the only force on the water would be gravity (certainly no normal force!), you'd have F = mg, making the acceleration = g (as you would expect for a falling body). Since both bucket and water fall with acceleration g (free fall), the normal force of bucket on water is zero, not mg.
  8. Aug 31, 2008 #7
    A bucket of water is swung in a vertical circle of radius r in such a way that the bucket is upside down when it is at the top of the circle. What is the minimum speed that the bucket may have at this point if the water is to remain in it?
  9. Aug 31, 2008 #8

    Doc Al

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    The solution to the problem is discussed in this thread. Show what you've done so far. Where are you stuck?
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