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Water in a whirling tube

  1. Oct 9, 2016 #1
    1. The problem statement, all variables and given/known data
    A tube of length '##L##' is filled completely with an incompressible liquid of mass '##M##' and closed at both of the ends. The tube is then rotated in a horizontal plane about one of it's ends with a uniform angular velocity '##\omega##'. Then which of the following statements are true:
    1. The force exerted by liquid at the other end is ##\frac{1}{2}M\omega^2 L##
    2. Ratio of force at middle point and the end point of the tube will be ##4:1##
    3. The force between liquid layers linearly increases with the distance along the length of the tube.
    4. Force is constant.
    Note: The question is a multiple correct question
    2. Relevant equations
    ##\vec{F}=m\vec{a}##

    3. The attempt at a solution
    Now lets not bother with the options, what I am puzzled about is how to find the force exerted at a linear element at a distance ##x## from the rotation axis. Wouldn't the scenario be just like that of swirling massive rope. In the case of a whirling rope the tension at any point is given by ##\dfrac{M\omega^2}{2L}(L^2-x^2)##, so in this case also wouldn't it be the same. But, wait I also (just like you) doubt that I am entirely correct because in the case of the rope there was no mass or any other body which applied a force (or needed to be rotated at ##x=L##) on the differential element that we were considering at a distance ##L## from the rotational axis, that's why we could find the integral constant by putting ##F=0## as ##c=\dfrac{M\omega^2}{2L}## in the expression ##F=-\dfrac{M\omega^2x}{2}+c##. So, what point do I need to consider for finding out the integral constant in this situation or is it that the situation is not like that of the rope at all.
     
  2. jcsd
  3. Oct 9, 2016 #2

    haruspex

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    It's a bit like a rope, but reversed. With the rope, least tension is at the free end, max at the axis end. The water pressure gradient is the othe way around. It is not clear whether you can directly map the rope result across, but you can certainly use an analogous method.
    Consider an element dx distance x from the axis. What is its mass? What is its centripetal acceleration? If the pressure on it from the axis side is P(x), what is P(x+dx)?
     
  4. Oct 10, 2016 #3
    So,I searched about how the pressure works in the case of fluids and found out that on any shape of element considered in the containing vessel of the fluid it experiences fluid pressure towards it so we can safely say that same will be the case with the force it experiences, correct me if I am wrong.

    Edit:- Would it be correct to discuss the doubt I have regarding the derivation of tension in a massive whirling rope here or in some other thread, its just a very small thing.
     
  5. Oct 10, 2016 #4

    haruspex

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    I am not at all sure what you mean by that.
    Perhaps if you try to answer my question at the end of post #2 your meaning will become clear.
     
  6. Oct 10, 2016 #5
    As I have understood it, the answer to your question would be ##P(x)+dP(x)## in the direction opposite to that in which ##P(x)## acts from the side nearer to the rotational axis.
     
  7. Oct 10, 2016 #6

    haruspex

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    Yes, obviously, but connect that with the centripetal force.
     
  8. Oct 10, 2016 #7
    Okay so we get ##dF(x)=(dm)\omega^2r## (I am considering the force acting on the element instead of the pressure hence used ##F(x)## instead of ##P(x)##), so we get ##F(x)=\dfrac{M\omega^2x^2}{2L}##. Also, please read the edit that I made in post #3
     
  9. Oct 10, 2016 #8

    haruspex

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    Yes.
    I reread your OP regarding the rope, and I do not see what doubt you express for that case. Please clarify.
     
  10. Oct 10, 2016 #9
    Okay I will be posting my doubt regarding the whirling of the massive rope that I was mentioning in the post #3 edit so here goes. What bothers me about it is that why is it that the tension that tension that acts at a distance ##x=0## form the rotational axis is greatest while the one that is acting at the end is the least, i.e. 0. It doesnt quite add up for me, because the part of the rope that is attached to the thing that makes the whole of the rope rotate is, as I think of it, has that part of the rope attached (i.e the part where ##x=0##, I know I am not being very clear about it but I cant find any other words for it right now) so there is no need for any tension to act at that part.
     
  11. Oct 10, 2016 #10

    haruspex

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    If there were no tension there, what would keep the rope attached to the axis?
     
  12. Oct 10, 2016 #11
    Hmm...your reply made me think in a different direction then what I was initially thinking, so what I thought of was that the tension is needed at that point to rotate the center of mass with an angular speed ##\omega## and as the center of mass is situated at ##x=L/2##, so the tension needed there is ##F(x=0)=\dfrac{M\omega^2L}{2}##. Now, the same thing can also be considered for the whirling tube filled with water but I don't see any force acting at ##x=0##, so what keeps the water together (I wanted to point out the analogy of "connected" as that in the rope with that of the filled tube).
     
  13. Oct 10, 2016 #12

    haruspex

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    It's needed to restrain the whole rope. As it happens that the centripetal acceleration is proportional to the radius, yes, it is equivalent to the whole mass being at L/2, but you should not just assume that.

    The water is under compression, not tension, with the maximum compression at the distal point.
    You get the same with a simple gravitational model. A rope suspended from the ceiling has maximum tension at the top, none at the bottom, while a standing pole has max compression at the base, none at the top. The difference in the rotational model is that the 'gravitational field' is non-uniform.
     
  14. Oct 10, 2016 #13
    Okay after a lot of thought I got the point that you were seeking my attention to, please correct me if I still didn't get your point. So what you were trying to tell me is that that the tension is max at x=0 because it needs to keep the rope from being detached from the rotational axis.

    But the force that we need to keep the rope from detaching from the rotational axis can also be calculated by assuming that whole of mass is present at center of mass, then why is my assumption wrong.

    I cant relate the gravitational model to the rotating tube filled fully model and how the water is compressed.
     
  15. Oct 10, 2016 #14
    @decentfellow Have you drawn a free body diagram for the parcel of fluid in the tube between locations x and x + dx, or, do you feel that you have advanced to the point where you no longer need to use free body diagrams? If you have drawn a free body diagram, what are the forces in the x direction acting on the parcel. What is its mass, and what is its acceleration?

    Chet
     
  16. Oct 10, 2016 #15
    I see...so essentially the centrifugal force was creating the pressure and hence the pressure was maximum at the maximum distance i.e. ##x=L##. And now I got the analogy too what @haruspex was trying to tell me was that just like a standing pole has maximum compression due to the weight above it. Am I correct? And sorry for not uploading the free body diagram of the differential element I was still int the process of writing the code for it in Texmaker, but as I have just started it takes me some time to write it. I dont mean to sound rude but do you want me to still upload the free body diagram.
     
  17. Oct 10, 2016 #16
    No. please just show us your force balance equation for the parcel of fluid between x and x + dx .
     
  18. Oct 10, 2016 #17
  19. Oct 10, 2016 #18
    @Chestermiller Oh I am apologise I saw your post after uploading the pic. From the FBD we can see that
    $$dF=(dm)\omega^2x=(\dfrac{M}{L}\omega^2x)dx \implies F=\dfrac{M}{2L}\omega^2x^2+c$$

    Now after such a long thread I have totally understood what I have to do here, which is to consider the force acting at ##x=0## is ##F=0##.

    So we get ##c=0##, hence ##F=\dfrac{M\omega^2x^2}{2L}##
     
  20. Oct 10, 2016 #19
    Excellent. Now, the only other thing to add is P(x) = F(x)/A.
     
  21. Oct 10, 2016 #20
    No need the question asked for the force only and it has not mentioned the cross-sectional area. And my doubt was only regarding why do have ##F=0## at ##x=0##. I cannot thank you enough for clearing my doubt with fast responses to my queries, @haruspex also.
     
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