Free-surface shape in a rotating cylinder: how does the rise of liquid at the wall compare to the depression at the rotation axis?

Introduction / Context:When a cylindrical vessel of liquid rotates steadily about its vertical axis, the free surface forms a paraboloid of revolution. Engineers use this fact to predict head distribution in rotating equipment and mixing tanks.

Given Data / Assumptions:

• Rigid-body rotation (no sloshing), steady angular speed.
• Incompressible liquid, negligible evaporation and capillarity.
• Container is completely filled to leave a free surface exposed to atmosphere.

Concept / Approach:

The free surface adjusts until pressure is hydrostatic in a rotating frame: p/ρ + g z − (ω^2 r^2)/2 = constant. This gives a parabolic surface z = (ω^2 r^2)/(2g) + constant. Conservation of volume requires that the volume rising near the wall equals the volume depressed near the centreline, so the rise at the wall equals the depression at the axis measured from the original horizontal level.

Step-by-Step Solution:

Verification / Alternative check:

Integrating z over the radius yields zero net change in volume relative to the original plane surface, confirming symmetry of rise and depression magnitudes.

Why Other Options Are Wrong:

(a) and (b) violate volume conservation in steady rotation; (d) viscosity affects transient approach, not the final static shape; (e) the relationship is deterministic, not random.

Common Pitfalls:

Confusing transient waves with steady rigid-body rotation; forgetting the reference is the original mean level.

Final Answer:

equal to the depression at the axis