Hydrostatic Pressure in a Liquid Undergoing Rigid-Body Rotation

For a liquid of density \( \rho \) in rigid-body rotation with angular velocity \( \omega \), the hydrostatic pressure distribution is

\[ p(r,z) = \rho \frac{\omega^2 r^2}{2} - \rho g z + \text{constant}. \]

The constant in this equation depends on the choice of \( z \)-origin and on the geometry and operating condition.

For a liquid in a vertical cylindrical container of radius \( R \) undergoing rigid-body rotation with angular velocity \( \omega \), the free surface profile (no spilling) is

\[ z = h(r) = \frac{\omega^2 r^2}{2g} + h_0, \quad \text{where} \quad h_0 = H - \frac{\omega^2 R^2}{4g}. \]

That is,

\[ z = h(r) = \frac{\omega^2 r^2}{2g} + H - \frac{\omega^2 R^2}{4g}. \]

Here \( H \) is the liquid depth when the tank is at rest.

At \( r = 0 \) and \( z = h_0 \), the free surface is exposed to atmosphere, so \( p = p_{\text{atm}} = 0 \) (gauge pressure). Thus

\[ p(r,z) = \rho \frac{\omega^2 r^2}{2} - \rho g z + \text{constant} \quad\Rightarrow\quad \text{constant} = \rho g h_0 = \rho g\!\left(H - \frac{\omega^2 R^2}{4g}\right). \]

Substituting back, the gauge pressure field is

\[ \begin{aligned} p(r,z) &= \rho \frac{\omega^2 r^2}{2} - \rho g z + \rho g\!\left(H - \frac{\omega^2 R^2}{4g}\right) \\ &= \rho \frac{\omega^2}{4}\bigl(2r^2 - R^2\bigr) + \rho g(H - z). \end{aligned} \]

The same expression can also be developed using \( p = \rho g\,\Delta h \), where \( \Delta h \) is the local liquid depth. For a point at \( (r,z) \),

\[ \Delta h = h - z = \frac{\omega^2 r^2}{2g} + h_0 - z. \]

Therefore,

\[ \begin{aligned} p(r,z) &= \rho g\,\Delta h = \rho g(h - z) = \rho g\!\left(\frac{\omega^2 r^2}{2g} + h_0 - z\right) \\ &= \rho g\!\left(\frac{\omega^2 r^2}{2g} + H - \frac{\omega^2 R^2}{4g} - z\right) \\ &= \rho \frac{\omega^2}{4}\bigl(2r^2 - R^2\bigr) + \rho g(H - z). \end{aligned} \]