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Fluid in Rigid-Body Motion

Intuition

Slam the brakes with a coffee in your cup holder: the surface tilts. Spin a bucket of water: the surface forms a bowl. In both cases the fluid moves as a rigid body — no shearing between layers — so it is still "statics", just in an accelerating frame. Pressure now has to balance gravity and supply the acceleration.

Mathematical formulation

For a fluid element \(dx\,dy\,dz\), the surface force is \(-\nabla P\, dV\) and the body force is \(-\rho g\,\hat{k}\, dV\). Newton's second law gives the governing equation:

\[\boxed{\,\nabla P + \rho g\,\hat{k} = -\rho\,\mathbf{a}\,}\]

Hydrostatics is the special case \(\mathbf{a} = 0\).

Three canonical cases

1 · Free fall (\(\mathbf{a} = -g\hat{k}\)): the equation gives \(\nabla P = 0\) — pressure is uniform. A freely falling container feels no internal pressure gradient; the bottom feels only atmospheric pressure. (Fluids in orbit behave the same way.)

2 · Uniform linear acceleration (\(\mathbf{a} = a_x\hat{i} + a_z\hat{k}\)):

\[\frac{\partial P}{\partial x} = -\rho a_x, \qquad \frac{\partial P}{\partial z} = -\rho(g + a_z)\]
\[P(x,z) = P_0 - \rho a_x x - \rho(g + a_z) z\]

Isobars (and the free surface) are inclined planes with slope

\[\frac{\Delta z_s}{\Delta x} = -\frac{a_x}{g + a_z}.\]

3 · Rotating container (angular velocity \(\omega\), cylindrical coordinates): the centripetal acceleration \(-r\omega^2\hat{r}\) gives

\[\frac{\partial P}{\partial r} = \rho r \omega^2, \qquad \frac{\partial P}{\partial z} = -\rho g\]

Setting \(dP = 0\) yields the isobar slope \(dz/dr = r\omega^2/g\), hence a paraboloid of revolution. Matching the fluid volume \(V = \pi R^2 h_0\) fixes the free surface:

\[z_s = h_0 - \frac{\omega^2}{4g}\left(R^2 - 2r^2\right)\]

Application: the liquid-mirror telescope

The Large Zenith Telescope used a slowly rotating dish of mercury: rotation makes the surface an exact paraboloid — precisely the shape a telescope mirror needs — for a tiny fraction of the cost of polished glass. The focal length is set by \(\omega\): \(f = g/(2\omega^2)\).

Common mistakes

  • Forgetting the \(g + a_z\) coupling — vertical acceleration changes the effective gravity, which also rescales buoyancy.
  • Assuming the deepest point stays put when spinning up. Fluid rises at the rim and dips at the axis by equal volume: \(z_s(0) = h_0 - \omega^2R^2/4g\).
  • Applying these results to sheared flows. Rigid-body motion means no relative motion between fluid particles — otherwise viscosity enters and you need Navier–Stokes.

Knowledge graph position

Prerequisites: Hydrostatic equilibrium. Leads to: Euler's equation.

Quiz

Q1 (computational). A tank of water accelerates horizontally at \(a_x = 4.9\ \text{m s}^{-2}\). At what angle does the free surface tilt?

Answer

\(\tan\alpha = a_x/g = 0.5 \Rightarrow \alpha \approx 26.6°\), surface tilting up toward the rear of the tank.

Q2 (conceptual). In a freely falling elevator, a cork is held at the bottom of a water bottle and released. Does it rise?

Answer

No. Free fall makes \(\nabla P = 0\) — no pressure gradient, no buoyancy. The cork just floats where it is (effective gravity is zero).

Q3 (multiple choice). Doubling \(\omega\) for a rotating bucket changes the dip of the surface at the axis by a factor of:

  • (a) 2 (b) 4 (c) \(\sqrt{2}\) (d) 8
Answer

(b). The dip is \(\omega^2 R^2/4g \propto \omega^2\).