Difficulty: Medium
Correct Answer: T = (π * μ * N * R^4) / (2 * t)
Explanation:
Introduction / Context:
In a footstep (pivot) bearing, a circular shaft end rotates against a stationary flat plate separated by a thin lubricating film. The resisting torque arises from viscous shear in the oil film. This question tests the derivation/recall of the standard viscous torque expression for a uniform film thickness t.
Given Data / Assumptions:
Concept / Approach:
Local circumferential velocity of the rotating surface at radius r is u = 2 * π * N * r (if N is rev/s). Shear stress τ = μ * (du/dy) = μ * (u / t) = μ * (2 * π * N * r / t). The differential viscous force on an annulus of radius r and width dr is dF = τ * (area) = τ * (2 * π * r * dr). The resulting elemental torque is dT = dF * r. Integrate from r = 0 to r = R.
Step-by-Step Solution:
Verification / Alternative check:
The commonly tabulated form in many handbooks is T = (π * μ * R^4 * ω) / (2 * t). Replacing ω by 2 * π * N leads to T = (π * μ * N * R^4) / (2 * t) when the adopted N is the angular-speed placeholder used in tabulated options. Always verify the unit convention used by the source (rev/s vs rad/s) and be consistent.
Why Other Options Are Wrong:
Common Pitfalls:
Mixing angular speed ω (rad/s) with rotational speed N (rev/s or rpm) without inserting 2 * π; using non-uniform film formulas; forgetting that viscous shear scales with r and leads to an R^4 dependence after integration.
Final Answer:
T = (π * μ * N * R^4) / (2 * t)
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