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A rod falling (rotation)

  1. Oct 14, 2016 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    Consider a rod of length ##L## that is attached to a hinge making an angle ##\theta## with the horizontal. There is a force ##\vec{F}## acting vertically on the rod that creates constant angular velocity ##-\omega##. If there is fluid under the rod with constant density, what force is the rod exerting on the hinge?

    2. Relevant equations
    Continuity. Navier-Stokes?

    3. The attempt at a solution
    So I'm not really sure where to start. I let the rod have a distance coming out of the page, call this ##b##. Then total volume of fluid in the triangular region under the rod and above the horizontal is ##V(t) = b L^2 \cos \theta \sin \theta/2 = b L^2 \sin (2 \theta) /4 \implies V'(t) = -b L^2 \omega\cos (2 \theta) /2##, which is at least negative, implying fluid is leaving the region (good).

    Now I know torque ##\vec{\tau} = \vec{r} \times \vec{F}##, so if we use polar coordinates we have ##\vec{\tau} = L \hat{r} \times F\hat{y}##. Now we know the position vector in polar coordinates is ##\vec{R} = r\cos\theta \hat{x}+r\sin\theta\hat{y} \implies \hat{r} = \cos\theta\hat{x}+\sin\theta\hat{y}##. Then ##\vec{\tau} = L \hat{r} \times F\hat{y} = L(\cos\theta\hat{x}+\sin\theta\hat{y}) \times -F\hat{y} = F\hat{y} \times L(\cos\theta\hat{x}+\sin\theta\hat{y}) = F\hat{y} \times L\cos\theta\hat{x} +F\hat{y} \times \sin\theta\hat{y} = -F L \cos\theta \hat{z}## where ##z## is out of the page. I'm not really sure how to consolidate this info. I'm thinking that the fluid would exert a sort of pressure force normal to the rod and in the positive ##\theta## direction. Then if we summed forces on the rod we would have the downward force ##-F \hat{y}##, the pressure force ##F_p \hat{\theta}## and the reacting force from the hinge to the rod ##F_h \hat{x}##. Since the hinge is not accelerating, these forces should balance, right?
     
    Last edited: Oct 14, 2016
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  3. Oct 14, 2016 #2

    haruspex

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    Yes, that all seems reasonable.
    I tried to solve it by assuming a drag force f(x)=Axn, where x is the distance from the hinge (which I take to be at the top). I got an answer, but I had to assume where the force F acts on the rod. I took that to be at the bottom.
    By the way, you titled it "rod falling", but I assume gravity is to be ignored.
     
  4. Oct 15, 2016 #3

    joshmccraney

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    Yes, I have ignored gravity, though I suppose we wouldn't necessarily have to. Would you mind posting your work, or giving me some insight into how you incorporated drag?

    Additionally, I assume to find the pressure force the fluid exerts on the rod I would have to use navier-stokes. Do you agree? I'm thinking of using Navier-Stokes in polar coordinates, although if we assume ##\theta## is small, then perhaps we could assume the horizontal motion is greater than vertical, and hence use Cartesian Navier-Stokes? What are your thoughts?

    Thanks for the reply!
     
  5. Oct 15, 2016 #4

    haruspex

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    I have no idea... I have never studied anything that advanced in fluid dynamics.
    We are not given much information - not the density, not the viscosity, not the width of the rod... - so I assumed a fairly simple treatment is expected.
    If we take the drag to be normal to the rod and proportional to some power, n, of the linear speed, we can get expressions for the net force from the drag and its net torque about the hinge. That allows us to write the usual force balance equations, except that we need to make an assumption regarding the point of application of F.
    On that basis, I got an answer in terms of F, θ and n.
     
  6. Oct 15, 2016 #5

    joshmccraney

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    Perhaps we could see what @Chestermiller thinks? This is for a high-level fluids class (well, it's from a set of notes). This makes me think we'll need the pressure exerted by the fluid, but again, I could be wrong.

    Regarding density, we can assume the fluid has constant density ##\rho##, and viscous/surface tension is negligible. No viscosity implies zero drag I think (though perhaps the fluid drag is not the drag you're referring to?). I suppose we can assume the rod has no mass (though it would be cool, once solving this, to see how the mass would impact the solution, along with say, friction at the hinge.)

    Either way, let's assume I am able to find the pressure force of the fluid. Then would we simply have ##F_h = F_p \sin \theta## and then ##F_p \cos \theta = F##?
     
  7. Oct 15, 2016 #6

    haruspex

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    I meant it in the general sense of whatever the fluid does to impede the rod.
     
  8. Oct 15, 2016 #7

    joshmccraney

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    Ohhhhh, so you're saying whether it's pressure or something else like drag? Also, I edited post #5. I have a final question there I was hoping you could address?
     
  9. Oct 15, 2016 #8

    haruspex

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    No. You do not know the direction of the hinge reaction. You will also need to consider torque.
     
  10. Oct 15, 2016 #9

    joshmccraney

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    Ok, that's actually something I was wondering. So would we apply conservation of angular momentum, which for this case, is sum of torque's is zero since the rod is moving at constant angular velocity?
     
  11. Oct 15, 2016 #10

    haruspex

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    Yes.
     
  12. Oct 15, 2016 #11

    joshmccraney

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    Ok cool. So we would have the torque I explained in my initial post, which is the torque resulting from the downward force. Then we would also have the pressure torque. Now if I can come up with an expression for pressure as a function of position, how would we go about this torque since it's acting on all of the rod? Would it by an integral of some sorts, say ##\iint_S p \hat{n}\, dS## where ##\hat{n}## is unit-normal to the plane (so ##\hat{\theta}##)?
     
  13. Oct 15, 2016 #12

    haruspex

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    Yes, but I took it as a single integral over the length of the rod. Since the force is normal to the rod, this is quite easy.
     
  14. Oct 15, 2016 #13

    joshmccraney

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    Right, if we had pressure ##p## then $$\vec{F_p} = \int_0^b \int_0^L p \hat{\theta} \,dl \, dw = b \int_0^L p \hat{\theta} \, dl $$ right? This gives us a force, but torque is ##\vec{r} \times \vec{F}##. What would the ##\vec{r}## be; not ##L## right, since ##\vec{F_p}## is distributed over all of ##L##, right?
     
  15. Oct 15, 2016 #14

    haruspex

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    Just put a factor l in the integrand.
     
  16. Oct 15, 2016 #15
    I really don't understand the geometry. Can you please draw a diagram.
     
  17. Oct 15, 2016 #16

    joshmccraney

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    Oooooh ok, so you're saying sum all the torques from pressure, like $$\tau_p = b\int_0^L l \hat{r} \times p \hat{\theta} \, dl = b\int_0^L l p \hat{z} \, dl$$ where ##\tau_p## is the torque due to the pressure.

    And Chet, I've attached the picture for your consideration.
     

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  18. Oct 15, 2016 #17

    haruspex

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    yes. And I assumed p=Aln, for some A, n.
     
  19. Oct 15, 2016 #18

    joshmccraney

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    Gotcha, this makes a lot of sense! So if gravity were to be present would we do a similar analysis as we did on pressure (only ##\hat{r}## would be crossed with ##-\hat{y}## (I am calling ##y## the vertical direction so ##z## is out of the page)? At any rate, if we know the pressure then the torque balance would give us ##\tau_p + \tau_f+\tau_r = 0## where we have the pressure torque, the force torque, and the reaction torque respectively. Then we could solve for the reaction torque ##\tau_r##, which would be in the ##z## direction. But how would we get a force from this?

    Hey Chet, any idea how to compute the pressure?
     
  20. Oct 15, 2016 #19

    haruspex

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    It's a hinge.
     
  21. Oct 15, 2016 #20

    joshmccraney

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    Sorry, hinge torque then? But do you agree with the gravity assessment?
     
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