# Rotating and Moving Water Container

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) 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?

• @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. – Soroush khoubyarian Jul 24 '18 at 13:50