Basket centrifuge scaling: if the basket radius is halved and the rotational speed (rpm) is doubled, what happens to (i) rim linear speed and (ii) centrifugal force?

Introduction / Context:
Understanding how geometric and kinematic changes affect centrifuge performance is crucial for scale-up and troubleshooting. Two key relationships are the rim linear speed v = 2 * π * N * r and the centrifugal (centripetal) acceleration a_c = ω^2 * r, which governs the "g-force" driving separation.

Given Data / Assumptions:

• Initial radius r and rotational speed N (rpm).
• New radius r/2 and new speed 2N.
• Rigid basket; focus on kinematics and acceleration, not throughput complexities.

Concept / Approach:
Translate rpm to angular speed ω = 2πN (in rad/s). The rim linear speed is v = ω * r. The centrifugal acceleration is proportional to ω^2 * r and the corresponding "g" level determines filtration/drainage rate capabilities for a given material system.

Step-by-Step Solution:
Original v0 = 2πN * r; after change v1 = 2π*(2N)(r/2) = 2πN * r → unchanged.Original a0 ∝ (2πN)^2 * r; after change a1 ∝ (2π(2N))^2 * (r/2) = 4*(2πN)^2 * (r/2) = 2*(2πN)^2 * r → doubled.Therefore, rim speed remains the same, but centrifugal force (per unit mass) doubles.

Verification / Alternative check:
Express in dimensionless "g" terms: G = ω^2 * r / g; halving r and doubling N gives G_new = 2 * G_old.

Why Other Options Are Wrong:
Rim speed doubled/halved: incorrect; it remains unchanged.Capacity increased: not guaranteed; capacity depends on feed properties, drainage path, cake behavior, and area, not G alone.

Common Pitfalls:
Confusing linear speed with centrifugal acceleration; they scale differently when r and N change simultaneously.

Final Answer:
centrifugal force is doubled.