Rotating Liquid – Pressure Rise at Drum Rim For a drum of radius r completely filled with a liquid of density rho and rotating at angular speed omega (rad/s), the increase in pressure between the center and the outer edge equals Δp = (1/2) * rho * omega^2 * r^2.

Introduction:
Liquids in rigid-body rotation develop a radial pressure gradient that increases outward. Knowing the exact pressure rise is vital for centrifuges, rotating tanks, and turbomachinery cavities.

Given Data / Assumptions:

• Rigid-body rotation at constant omega.
• Incompressible Newtonian liquid; no free surface curvature effects on pressure difference along a radius.
• Neglect gravity along the radial balance.

Concept / Approach:

Radial equilibrium in the rotating frame gives dp/dr = rho * omega^2 * r. Integrating from r = 0 to r = R yields Δp = (1/2) * rho * omega^2 * R^2. This is independent of axial position for a completely filled drum (no free surface pressure variation along a radius).

Step-by-Step Solution:

Verification / Alternative check:

Experiments with manometer taps at different radii confirm quadratic pressure variation with radius in rigid-body rotation.

Why Other Options Are Wrong:

rho * omega^2 * r^2 and 2 * rho * omega^2 * r^2 overestimate the integral. Terms linear in omega or missing squared dependence are dimensionally inconsistent for pressure.

Common Pitfalls:

Mixing this result with the free-surface parabolic shape relation z = (omega^2 * r^2) / (2g); both share the 1/2 factor but represent different phenomena.

Final Answer:

(1/2) * rho * omega^2 * r^2