A bucket of water is tied to a rope and swung in a vertical circle. The distance from the bucket centre to the axis of rotation is $2.08m$. Calculate the angular velocity (in $rad s^{-1}$) of the swing, to prevent the water from falling out of the bucket. [emphasis mine]

My answer $\sqrt{5g/l} = 4.856$ refers to the minimum initial angular velocity at the base of the circle (such that the water will remain in the bucket even when it reaches the top of the circle), while the marking scheme of the test which this problem came from has $\sqrt{g/l} = 2.172$ instead, referring to the angular velocity at the top of the circle.

I was marked wrong for this problem, and my lecturer tells me that I have over-complicated the problem. (The disagreement is basically over the interpretation of the problem, specifically its last sentence.) So learned ones, I seek your input: was I right to supply the velocity at the start of the motion, or was I over-complicating the problem?

  • $\begingroup$ My initial reading of the problem lead me to presume that the average angular velocity would be of interest - because it is the only metric that can be measured easily. Given the wording any (or preferably all) angular velocity metric would be valid IMO if specified clearly in the answer, although the point at the top would be the most appropriate in light of the scope of the problem. $\endgroup$ Apr 30 '12 at 16:17
  • $\begingroup$ Ryan, are you sure your quantity for the angular velocity at the base of the circle is correct? $\endgroup$ Apr 30 '12 at 16:34
  • $\begingroup$ @AlanSE Oh. When I asked the question, I had thought that everyone here would naturally agree with my interpretation. It's interesting to see that so far, I'm the only person to see it that way. LOL At least we can agree that the problem should have been better-phrased. Thanks for your input, Alan. $\endgroup$
    – ryang
    Apr 30 '12 at 16:35
  • $\begingroup$ I am positively certain. The physics/maths of this problem isn't challenging. But interpreting which angular speed exactly that it requires is. $\endgroup$
    – ryang
    Apr 30 '12 at 16:36
  • $\begingroup$ Ok, never mind. I got the $5$ now. That should be right. $\endgroup$ Apr 30 '12 at 16:43

I probably would have calculated the angular velocity at the top of the circle. But the question is clearly ambiguous, and I don't think it was unreasonable of you to take "the angular velocity of ... the swing" to mean the initial velocity of the swing, starting from the bottom. The question should have said "calculate the angular velocity at the top of the swing", in order to make it clear what's being asked for. Most likely the person setting the question didn't actually consider that the angular velocity would be changing throughout the bucket's motion. I've never marked a test, but if you started by calculating the angular velocity at the top of the swing and wrote down $\sqrt{g/l}$ then I probably would have given you the marks for that (since it's the correct answer according to the answer scheme) and just ignored the extra part.

  • $\begingroup$ Yes, unfortuantely, it's a computer-marked quiz, so it only recognises a set value of the final answer. My lecturer was unconvinced when I explained to him that the question was ambiguous. He knew that my working included all that was required to answer the quiz (incl. the required final value), but he would not give me even partial credit. Thanks for your input-- and I will stop pursuing this now, since no one here has said that they would actually have supplied the answer that I did. I'm genuinely surprised though that I seem to be the only one naturally reading the problem this way.:) $\endgroup$
    – ryang
    Apr 30 '12 at 17:00
  • $\begingroup$ So perhaps my lecturer is right and that I am over-complicating the problem. :) $\endgroup$
    – ryang
    Apr 30 '12 at 17:07
  • 1
    $\begingroup$ I don't think you are - the reason I think I'd have answered with the angular velocity at the top is that I don't think I would have thought about the fact that the angular velocity has to vary throughout the motion. If I was your lecturer I wouldn't want to mark you down for noticing something that others might have missed. $\endgroup$
    – N. Virgo
    Apr 30 '12 at 17:16

At the base of the circle, the water stays in the bucket without any extra work. The reason is that it is attracted to the Earth by its gravity. Try to get some water into a mug and you may experimentally observe that it will stay there if the mug is oriented in the usual "breakfast mode". (If it is oriented in the opposite way, the water can't stay in the swinging bucket whatever its acceleration is.)

The relevant point where it's hard to keep the water in the bucket is clearly the top of the circle. At that point, the water is accelerated towards the Earth i.e. out of the bucket by the acceleration $g$. One needs to compensate it by the centrifugal acceleration $l\omega^2$. So the minimal value of $\omega$ that keeps the water in the bucket is $$ l\omega^2 = g,\qquad \omega=\sqrt{\frac gl} \approx 2.17\,\,{\rm rad/s}$$ As you can see, I would also give you 0 points because you seem to be confused about the direction of the gravitational force or the direction of the centrifugal force which is a pretty serious bug. The only conceivable thing you could get a fraction of the credit for is the fact that your answer was dimensionally correct. Aside from that, I can't imagine a justification of a nonzero credit for your answer that couldn't be used to reward any wrong answer. This was really an easy question and your doing complicated things that lead you to the square roots of five indicates that you don't understand these simple things.

BTW what users may post on Stack Exchange are "questions", i.e. sentences terminated with a question mark whose purpose is to learn the right answer out of a larger of set of a priori conceivable candidate answers. The purpose of this server is not to "verify" things, especially not someone's predetermined wrong propositions. Your expectation that we would verify that your lecturer was wrong wasn't rationally justified, I think. Even statistically, I am confident that in most of such situations, the students are wrong and the lecturers are right. At any rate, your example belongs to this category.

  • $\begingroup$ Please hold your horses; I am not confused about the problem. My question is essentially: when the quiz says to "Calculate the angular velocity of the swing" does it mean the minimum initial angular velocity required at the base of the circle (such that the water remains in the bucket at the top) OR do you think it is asking for the minimum angular velocity at the top of the circle. Your $\omega$ clearly refers to the speed at the TOP of the circle, but is this really what the quiz is asking for? $\endgroup$
    – ryang
    Apr 30 '12 at 15:29
  • $\begingroup$ I don't claim that you are confused about the formulation. My claim is that you are confused about the right solution i.e. about elementary physics itself. The question is totally unambiguous. It asks you about the angular velocity that prevents the water from falling out. The first step in the solution is to realize that the water may only fall out at the top of the circle. I am confident that most kids in the kindergarten would solve this first step correctly but you have not so you deserve 0 points. The problem doesn't have to give you a hint that the top is the issue; it can be derived. $\endgroup$ Apr 30 '12 at 15:33
  • $\begingroup$ Realising that the tension of the string at the top of the circle has to be positive was my FIRST step!! It was using this speed at the top of the circle that I managed to find the speed at the base of the circle. Would you please actually read my question properly? I.e. I completely KNOW how to get the 2.17 answer because I had to get it the same way you did. But I feel that this is a partial answer because that is just the angular speed at the TOP of the circle. Isn't the quiz asking for the angular speed at the base of the circle? (P.S. There are two ways to get this final answer... $\endgroup$
    – ryang
    Apr 30 '12 at 15:36
  • $\begingroup$ ....either using the Law of Conservation of Mech Energy, or by integration (more tedious). Anyway I'm giving too much detail. $\endgroup$
    – ryang
    Apr 30 '12 at 15:39
  • 2
    $\begingroup$ Presumably you're thinking that "the angular velocity of the swing" is the angular velocity you need at the start of the swing, i.e. at the bottom, to make sure the bucket goes all the way round without any spillage. I can see the scope for confusion, but I would have calculated the angular velocity at the top not the bottom. $\endgroup$ Apr 30 '12 at 15:41

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