Trommel critical speed (theoretical): express the critical rotational speed N (rps) in terms of g and r for a cylindrical trommel.

Difficulty: Medium

Correct Answer: N = (1 / (2π)) * (g / r) ^ 0.5

Explanation:


Introduction / Context:
“Critical speed” for a rotating drum (trommel, mill) is the angular speed at which particles just cling to the wall and begin to centrifuge rather than tumble. This theoretical speed provides an upper bound for practical operation in screening or grinding equipment.



Given Data / Assumptions:

  • g = 9.81 m/s^2 (acceleration due to gravity).
  • r = radius of the drum (metres) measured to the particle locus.
  • Neglect lift and frictional effects; consider a single particle at the wall.



Concept / Approach:
At critical speed, centripetal acceleration equals gravitational acceleration at the top of the rotation: ω^2 r = g. Thus ω = (g/r)^0.5 (radians per second). Convert to revolutions per second using N = ω / (2π).



Step-by-Step Solution:
Set centripetal = gravitational: ω^2 * r = g.Solve for ω: ω = (g / r) ^ 0.5.Convert to rps: N = ω / (2π) = (1 / (2π)) * (g / r) ^ 0.5.



Verification / Alternative check:
Engineering handbooks also give the empirical rpm form N_rpm ≈ 42.3 / √D_m (for mills), which follows from the same derivation after unit conversions and diameter D = 2r substitution.



Why Other Options Are Wrong:
Dividing by 2r (option b) or changing the π factor (options c and d) stems from algebra/unit mistakes and does not match the physics balance.



Common Pitfalls:
Mixing diameter and radius, or rpm vs. rps; forgetting the 2π factor when converting angular velocity to revolutions per second.



Final Answer:
N = (1 / (2π)) * (g / r) ^ 0.5

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