A liquid rotates as a rigid body with constant angular velocity about a vertical axis. How does pressure vary with radial distance from the axis?

Introduction / Context:
In forced vortex motion (rigid-body rotation), every fluid particle rotates with the same angular velocity. The free surface becomes parabolic and pressure increases radially outward. Recognizing the correct pressure–radius relation is essential for problems involving rotating tanks and centrifuges.

Given Data / Assumptions:

• Angular velocity = ω (constant).
• Axis of rotation is vertical; gravity acts downward.
• Negligible viscous shear in the bulk (idealized rigid-body rotation).

Concept / Approach:
Radial equilibrium requires the pressure gradient to balance centripetal acceleration: (1/ρ) * dp/dr = ω^2 * r. Integrating from the axis (r = 0) gives p(r) = p(0) + 0.5 * ρ * ω^2 * r^2, showing a quadratic dependence on r.

Step-by-Step Solution:
Set radial equilibrium: dp/dr = ρ * ω^2 * r.Integrate: p(r) − p(0) = ∫0^r ρ ω^2 r dr = 0.5 ρ ω^2 r^2.Therefore, pressure increases proportional to r^2.

Verification / Alternative check:
The parabolic free surface z = (ω^2 r^2)/(2g) has the same r^2 dependence, consistent with the pressure field in the rotating liquid.

Why Other Options Are Wrong:
Linear increase: would correspond to a constant radial body force, not centripetal acceleration proportional to r.Inverse altitude statement is unrelated to radial pressure variation.Decrease with r^2 contradicts equilibrium.

Common Pitfalls:

• Mixing up free-vortex (irrotational) and forced-vortex (rigid-body) behaviours.
• Forgetting that p varies both with r and z; here the question isolates radial dependence.

Final Answer:
Increases as the square of the radial distance