Rotating liquid in a vertical-axis cylinder – depth of paraboloid A cylindrical vessel of radius r is rotated steadily at angular speed ω about its vertical axis. The free surface becomes a paraboloid. The depth difference between the rim and the vertex (parabola depth) is:

Introduction:
When a liquid rotates as a forced vortex, its free surface forms a paraboloid of revolution. This question asks for the depth difference between the center and rim, which is fundamental in vortex dynamics and rotating-tank problems.

Given Data / Assumptions:

• Steady rigid-body rotation with angular speed ω.
• Cylindrical container of radius r; gravity g acts downward.
• Neglect evaporation, capillarity, and transient sloshing.

Concept / Approach:
In a forced vortex, all fluid particles have the same angular velocity ω. The free-surface profile satisfies z = (ω^2 * r^2) / (2 * g) + constant. Hence the height difference between the rim (radius r) and the center (r = 0) is Δh = ω^2 * r^2 / (2 * g).

Step-by-Step Solution:
1) Write energy balance in a rotating frame: pressure + potential + centrifugal potential is constant on the free surface.2) Along the free surface, p ≈ p_atm; thus elevation varies with r to balance centrifugal effects.3) Integrate centrifugal acceleration a_c = ω^2 * r to get head: (ω^2 * r^2) / (2 * g).4) Evaluate at r and at 0: Δh = [(ω^2 * r^2)/(2 * g)] − 0.

Verification / Alternative check:
Dimensional check: [ω^2 * r^2 / g] has units of length; the factor 1/2 arises from integrating r dr.

Why Other Options Are Wrong:

• ω^2 * r^2 / g: missing the 1/2 integration factor.
• ω^2 * r / g: incorrect dependence; should be proportional to r^2.
• r^2 / (2 * g * ω^2): inverted dependence on ω.

Common Pitfalls:
Confusing free-vortex (irrotational) with forced vortex; forgetting that the profile is parabolic (∝ r^2), not linear in r.

Final Answer:
ω^2 * r^2 / (2 * g)