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Surface of a rotating liquid

  1. Jul 22, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-7-22_20-2-38.png

    2. Relevant equations


    3. The attempt at a solution
    If I consider an arc shaped element of the liquid, then the physical forces acting on it are 4 different tensions , normal force and gravitational force on it. Two of the tensions would be acting along tangential direction (opposite to each other) and the other two along radial direction(opposite to each other) .

    Now, in steady state, normal force and gravitational force cancels each other.
    I assume that in dynamic state, too, normal force and gravitational force cancel each other.
    Assuming tangential acceleration to be 0, the two opposite tensions along tangential direction cancel each other.

    Now, wrt. inertial frame , the two opposite radial tensions provide centripetal acceleration. But then how to decide the shape?
     
  2. jcsd
  3. Jul 22, 2017 #2
    This is likely too advanced for me, but I think it would be easier for others to help you if you told them more precisely what the question's asking for. Is it asking for a qualitative description, for example, or are you supposed to somehow define the shape quantitatively?
     
  4. Jul 22, 2017 #3
    Well, it is an example given in the book which I want to solve without looking at the solution in the book.
    So, if you feel that the information given in the question is not sufficient, I am attaching the example so that you can guide me.
     

    Attached Files:

  5. Jul 22, 2017 #4
    Since it seems to want a quantitative description, you're going to have to wait for someone smarter and better at physics than I am.
     
  6. Jul 22, 2017 #5

    haruspex

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    Do you mean surface tension? It is not related to that.
    Consider a small particle (of water) on the surface at radius x. In terms of the height function, y=y(x), what is the slope there?
    There are two forces on it, and you know the resultant acceleration.
     
  7. Jul 23, 2017 #6
    I completely disagree with this solution in your book. Certainly, the normal load at the free surface is constant, and equal to the atmospheric pressure outside the fluid (assuming negligible surface tension effects, which you are clearly expected to assume). The solution in your book is totally bogus.

    Here's how I would solve this problem. As indicated in the problem description, z is the elevation of the free surface above the lowest point of the free surface at the center of the bucket. Let h be the depth below this location. I am going to focus on a fluid Free Body situated between depth h and depth ##h+\Delta h##, between angular locations ##\theta## and ##\theta+\Delta \theta##, and radial locations r and ##r+\Delta r##. So the volume of this free body is ##(r\Delta \theta)(\Delta r)(\Delta h)##. I am going to do a radial force balance on this free body. The fluid pressure at radial location r and depth h is p(r,h). The pressure force acting on the free body at the radial location r is ##[pr\Delta \theta \Delta h]_{r}\mathbf{i}_r##, where ##\mathbf{i}_r## is the unit vector in the radial direction. The pressure force acting on the free body at radial location ##(r+\Delta r)## is ##[-pr\Delta \theta \Delta h]_{r+\Delta r}\mathbf{i}_r##. The pressure force acting on the free body at angular locations ##\theta## and ##(\theta+\Delta \theta)## are ##[p\Delta r \Delta h \mathbf{i_{\theta}}]_{\theta}## and ##[-p\Delta r \Delta h \mathbf{i_{\theta}}]_{\theta+\Delta \theta}##, respectively.

    If we sum these forces and set them equal to the mass times acceleration of the free body, we obtain:
    $$[pr\Delta \theta \Delta h]_{r}\mathbf{i}_r-[pr\Delta \theta \Delta h]_{r+\Delta r}\mathbf{i}_r+[p\Delta r \Delta h \mathbf{i_{\theta}}]_{\theta}-[p\Delta r \Delta h \mathbf{i_{\theta}}]_{\theta+\Delta \theta}=-\rho (r\Delta \theta)(\Delta r)(\Delta h)\omega^2 r\mathbf{i}_r$$If we divide this equation by ##(\Delta h)(\Delta r)(\Delta \theta)## and take the limit as ##\Delta r## and ##\Delta \theta##, we obtain:
    $$\frac{\partial (pr)}{\partial r}\mathbf{i}_r+p\frac{\partial \mathbf{i}_{\theta}}{\partial \theta}=\rho \omega^2 r^2\mathbf{i}_r\tag{1}$$
    We know that the derivative of the unit vector in the theta direction with respect to theta is given by:
    $$\frac{\partial \mathbf{i}_{\theta}}{\partial\theta}=-\mathbf{i}_r\tag{2}$$
    If we substitute Eqn. 2 into Eqn. 1, we obtain:
    $$\frac{\partial p}{\partial r}=\rho \omega^2 r\tag{3}$$
    Integrating Eqn. 3 between r = 0 and r yields: $$p(r,h)-p(0,h)=\frac{1}{2}\rho \omega^2 r^2\tag{4}$$
    I'm going to stop here for now and give you a chance to digest what I have written.

    Chet
     
  8. Jul 23, 2017 #7
    In a private communication, Haruspex has induced me to think about the analysis in the book some more, and, based on this, I have concluded that the analysis was not bogus after all. I will, however, say that the book failed to discuss the reasons why the pressure on the sides of the free body in the figure are zero, and the reason that the net force of the surrounding fluid is normal to the free surface. The reasoning should have been that (a) the pressure on the top surface of the free body is zero, since that is in contact with the air and (b) the pressures on the sides of the free body are zero because the pressure is not varying along the free surface. So the only surrounding fluid force remaining is at the lower boundary of the free body, and the pressure must be perpendicular to this boundary (which is parallel to the free surface). So the net force of the surrounding fluid on the free body must be in the normal direction. Such a discussion would certainly have been called for (in my judgment).
     
  9. Jul 24, 2017 #8
    Considering Cylindrical coordinate system

    Let's consider a small volume element of water dV = r dr d##\theta ## dz at the position (r, ## \theta ##, z )
    Neglecting the effect of the atmospheric pressure, there are two physical forces acting on it 1) Contact forces ##\vec F_{con} ## 2) gravitational force dm ##\vec g ##.

    Now, contact force is due to the contact of this volume of water with the surrounding water particles.

    Wrt. inertial system,
    ## \vec F_{con} + m \vec g = m\{ \left ( \ddot r - r \dot {\theta}^2\right ) ~\hat r+ \left ( 2 \dot r \dot \theta +r \ddot \theta \right ) ~\hat \theta\} +\ddot z \hat z##

    How to find ## \vec F_{con} ## ?

    I can't relate the information that the bucket is rotating with constant angular velocity to this problem right now because water is non-rigid.
    How to do this?
     
  10. Jul 24, 2017 #9
    ## \vec F_{con} ## is comprised of pressure forces on your element of volume. There are hydrostatic pressure forces on the top and bottom (in the z direction) and, in the r and ##\theta## directions, there are the pressure forces that I included in post #6.

    In your equation,
    ## \vec F_{con} + m \vec g = m\{ \left ( \ddot r - r \dot {\theta}^2\right ) ~\hat r+ \left ( 2 \dot r \dot \theta +r \ddot \theta \right ) ~\hat \theta\} +\ddot z \hat z##, since r and z are not changing with time and the angular velocity is constant, the right hand side reduces to ##- mr \dot {\theta}^2\hat{r}##. The mass m is expressed in terms of the density and volume of the element, as in my post #6. The gravitational term is in the z direction, and is equal to ##-mg\hat{z}##.

    The net result of all this is that, when your force balance equation here is dotted with ##\hat{r}##, you obtain Eqn. 3 from my post #6.

    Incidentally, it doesn't matter if the water is rigid. In this problem, it is rotating as a rigid body, and, more generally, even a liquid satisfies Newton's 2nd law, at least on a differential basis.
     
  11. Jul 25, 2017 #10
    When I am told to find out the shape of the liquid, what am I supposed to calculate quantitatively?

    ##\vec F_{con} + m \vec g = m\{ \left ( \ddot r - r \dot {\theta}^2\right ) ~\hat r+ \left ( 2 \dot r \dot \theta +r \ddot \theta \right ) ~\hat \theta\} +\ddot z \hat z##
    Earlier I thought that r ,z and angular velocity are changing as the system is liquid. Now, I understand that in these kind of problems I have to approximate the system to a semi- rigid body,( because if I assume the water to be a rigid body, then by the definition of "rigid body " the shape of the water will remain unchanged), for which r and z are not changing with time and the angular velocity is constant.

    Now,
    ##\vec F_{con} + m g \left (-\hat z \right) = mr \dot {\theta}^2\left( -\hat r \right) ##

    For calculating ##\vec F_{con} ##,
    There will be pressure on the system due to the collision with water particles from left and right along radial direction and up and down.
    As it is said that r is constant for the system, so it is for other water particles, so no particle is moving along radial direction, no particle will impart momentum to the system along radial direction, hence there is no pressure along radial direction.

    Similar is the case with z direction.
    ???
     
  12. Jul 25, 2017 #11
    In a typical simple problem involving hydrostatics, when you apply the hydrostatic equation, the fluid is not moving in the vertical direction and yet the pressure is increasing with depth. How do you explain that?
     
  13. Jul 25, 2017 #12
    upload_2017-7-25_20-25-59.png ## \vec F_{net}= \vec F_up + \vec F_{down} - dm~g \hat y =0 ##
    ## - P_{h} A \hat y + P_{h +dh} A \hat y = \rho A dh g \hat y ##
    ## P_{h +dh} - P_{h } = \rho g dh ##
    ## dP = \rho g dh ##
    Integrating both sides,
    ## P_{h_f} -P_{h_i} = \rho g\left ( h_f - h_i \right) ##

    where h is the depth measured from the upper water surface level
    ## P_{h_f}## is the pressure at depth hf
    ##\vec F_{up} ## is the force on the volume element dV due to the upper water particles

    So, I understood that the component of ##\vec F_{con}## in ##\hat z ## cancels m##\vec g## .
    But , what creates a component of ##\vec F_{con}## in radial direction?
     
  14. Jul 25, 2017 #13
    When the fluid is rotating about the axis, from the perspective of an observer in the rotating frame of reference, the apparent centrifugal force is equivalent to gravitation in the radial direction (i.e., it appears to be a body force). So it's virtually analogous to vertical hydrostatic force, except that the gravitational acceleration radially is dependent on radial location.
     
  15. Jul 27, 2017 #14

    But, this is wrt rotating frame. I am solving it in an inertial frame.
     
  16. Jul 27, 2017 #15
    It doesn't matter whether you are solving it in an inertial frame or in a rotation frame. All that happens is the the "ma" term gets moved to the other side of the equation with a change in sign. The final answer comes out the same either way.
     
  17. Jul 27, 2017 #16
    The fluid is rotating, and the parcels of liquid have to travel in circles about the axis, so they have to be accelerated in the negative radial direction (as reckoned from an inertial frame of reference). So the wall must presses inward on the parcels further out, and the parcels further out must press inward on those closer to the axis. This is how the radial pressure gradient is established.
     
  18. Jul 27, 2017 #17

    RonL

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    I can't do math but perhaps there is a comment or link in this article that will make something clearer :smile:

    https://en.wikipedia.org/wiki/Liquid_mirror_telescope
     
  19. Jul 27, 2017 #18
    In the link ,it is said that the ##\vec F_{con} ## is perpendicular to the surface.
    ##\vec F_{con} ## cancels weight and provides centripetal acceleration. For, providing centripetal acceleration it has to have a component equal to ## mr \dot {\theta}^2## in -ve radial direction and ## -\hat r## lies in the surface. So, how can it be normal to the surface?
     
  20. Jul 27, 2017 #19

    TSny

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    ## -\hat r## is horizontal.
     
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