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.
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Arho * omega^2 * r^2
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B2 * rho * omega^2 * r^2
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C(1/2) * rho * omega^2 * r^2
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Drho * omega * r
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E(1/2) * rho * omega * r^2
Answer
Correct Answer: (1/2) * rho * omega^2 * r^2
Explanation
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:
1) Write equilibrium: dp/dr = rho * omega^2 * r.2) Integrate: p(R) − p(0) = rho * omega^2 * ∫ r dr = (1/2) * rho * omega^2 * R^2.3) Therefore the pressure rise to the rim is (1/2) * rho * omega^2 * r^2.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