Rotating Liquid in a Cylindrical Vessel — Radial Pressure Variation When a cylindrical vessel containing liquid is rotated steadily about its vertical axis, how does the static pressure vary with radial distance r from the axis?

Introduction:
Rigid-body rotation of a liquid about a vertical axis (no relative shear in the rotating frame) generates a parabolic free surface and a radial pressure gradient. This principle is important for centrifuges, rotating tanks, and free-vortex approximations.

Given Data / Assumptions:

• Steady rotation at angular speed omega about a vertical axis.
• No radial/axial flow; liquid rotates as a solid body.
• Gravity acts downward with acceleration g.

Concept / Approach:
Radial equilibrium gives dp/dr = rho * omega^2 * r. Integrating from the axis (r = 0) yields p(r) - p(0) = (1/2) * rho * omega^2 * r^2. Hence pressure increases with r^2. The free surface forms a paraboloid: z = (omega^2 * r^2) / (2g) + constant.

Step-by-Step Solution:
Write force balance in radial direction: dp/dr = rho * omega^2 * r.Integrate: p(r) = p(0) + (rho * omega^2 / 2) * r^2.Conclude: pressure increases quadratically with radial distance.

Verification / Alternative check:
Observation of the free surface curvature in a rotating beaker matches the quadratic profile predicted by the pressure distribution.

Why Other Options Are Wrong:
Linear increase contradicts dp/dr proportional to r; decrease is physically impossible for outward centrifugal effect; “varies as square” is ambiguous about direction—pressure specifically increases with r^2.

Common Pitfalls:
Confusing free-vortex (v ∝ 1/r) with forced-vortex (rigid-body) behavior; neglecting hydrostatic vertical variations when comparing points at different elevations.

Final Answer:
increases as the square of the radial distance