# Rotating circular pan of liquid

1. Sep 10, 2009

1. The problem statement, all variables and given/known data
A circular pan of liquid with density $\rho$ is centered on a horizontal turntable rotating with angular speed $\omega$. Atmospheric pressure is $P_{a}$. Find expressions for (a) the pressure at the bottom of the pan and (b) the height of the liquid surface as functions of the distance $r$ from the axis, given that the height at the center is $h_{0}$.

2. Relevant equations
In a flow tube
$$P+\frac{1}{2}\rho v^2 + \rho g h = C$$
where C is a constant

3. The attempt at a solution
I boiled up a few equations using the above equation
$E_{d}$ is the energy density
at the center and at height $h_0$:
$$E_{d} = P_{a} + \rho g h_{0}$$

at the center and at the bottom:
$$E_{d} = P_{b}(0)=P_{a}+\rho g h_{0}$$
where $P_{b}(r)$ is the pressure at the bottom measured r from the center

r from the center and at the surface of the liquid (height h)
$$E_{d}=P_{a} +\frac{1}{2}\rho r^2 \omega ^2 + \rho g h$$

r from the center and at the bottom
$$P_{b}(r) + \frac{1}{2}\rho r^2 \omega ^2 = P_{b}(0)$$

Mixing all these together out pops
$$P_{b}(r) = P_{a} + \rho g h_{0}-\frac{1}{2}\rho r^2 \omega ^2$$
and
$$h=h_{0} - \frac{1}{2g}r^2 \omega ^2$$
but the answer has a plus sign where I have my negative signs. Pretty sure I am wrong but would have to fudge the equations to make it work.

I made a couple of assumptions:
For a given vertical segment of water, it is all travelling at the same angular speed $\omega$ and the atmopsheric pressure at the different liquid heights at the surface is the same.