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I'm trying to solve the following problem, but I'm getting a different answer than the one in the book. I can't understand what I'm doing wrong. (The question is from Hibbeler's fluid dynamics)

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So since the fluid is rotating, there should be a centripetal force that makes the liquid accelerate towards the centre of rotation. This force is provided by the differences in the radial pressure. I consider an infinitesimal ring at the top of the container. The ring has an infinitely small height $dh$ and thickness $dr$. So:

$\dfrac{\partial p}{\partial r} \times dr \times 2\pi r \times dr \times dh = $Radial force exerted by pressure

This force should be equal to: $dm \times \omega^2 \times r$

$\therefore (2\pi r \times dr \times dh \times \rho) \times \omega^2 r = \dfrac{\partial p}{\partial r} \times dr \times 2\pi r \times dr \times dh$

$\dfrac{\partial p}{\partial r} = \rho \omega^2 r$

$\therefore p = \dfrac{\rho \omega^2 r^2}{2} + C$

This $C$ should be equal to $0$, since there is a whole at the centre of the container at the top, meaning that the gauge pressure is $0$ when $r=0$. So the pressure at the top of the container as a function of $r$ will be:

$p(r) = \dfrac{\rho \omega^2 r^2}{2}$

Since the maximum pressure is at the edges of the bottom of the container, we have to first calculate the pressure at the edges of the top, then use the $\rho gh$ to yield the maximum pressure. The pressure at the top edge of the container is:

$p(1) = \dfrac{1.94 \times 100 \times 1^2}{2} = 97_{lb/ft^2}$

Since the container is moving upwards at an acceleration of $6_{ft/s^2}$ , it's as if instead of $g$ being equal to $32.2_{ft/s^2}$, it is equal to $38.2_{ft/s^2}$. So:

$p_{max} = 97 + 1.94 \times 38.2 \times 3 = 319.324_{lb/ft^2} = 2.22_{psi}$

But the answer to this question is $3.52_{psi}$. Where did I make a mistake?

Thanks for your help.

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  • $\begingroup$ when there is a hole in the container the atmospheric pressure may be taken into consideration.pl. see a similar problem in the following reference- by.genie.uottawa.ca/~mcg3335/AdditonalNotes/… $\endgroup$ – drvrm Jul 24 '18 at 10:01
  • $\begingroup$ @drvrm Here I'm just using the gauge pressure, and since the whole is at the centre, the pressure at the centre is $0_{psi}$. But still I don't see how my solution is wrong. $\endgroup$ – Soroush khoubyarian Jul 24 '18 at 13:50

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