Footstep bearing with viscous lubrication If the shaft speed is doubled, how does the torque needed to overcome viscous resistance change?

Introduction:
This question examines viscous shear in a lubricated thrust (footstep) bearing. The torque required to overcome viscous resistance depends on shear stress in the oil film, which scales with relative speed and viscosity for Newtonian lubricants.

Given Data / Assumptions:

• Newtonian lubricant with constant viscosity mu.
• Uniform film thickness h under operating conditions.
• Laminar Couette-type shear between the rotating surface and stationary pad.

Concept / Approach:
Shear stress tau = mu * (du/dy). For a rotating disk over a stationary surface with film thickness h, the velocity gradient scales with omega * r / h. The elemental shear force scales with tau * area, and elemental torque scales further with radius r. Integration across the radius yields torque T proportional to mu * omega / h times a geometric factor. Thus, T ∝ omega (and ∝ speed N).

Step-by-Step Solution:

Verification / Alternative check:
Classical formulas for a flat, full-film thrust bearing give T = (pi * mu * omega * R^4) / (2 * h) for idealized geometry, explicitly linear in omega, confirming the doubling behavior.

Why Other Options Are Wrong:

• four/eight/sixteen times: imply quadratic or higher power laws; not valid for linear viscous shear.
• no change: contradicts direct proportionality to speed.

Common Pitfalls:
Confusing viscous shear scaling with turbulent skin-friction laws or bearing hydrodynamic lift scaling; this question targets the linear Newtonian-shear regime with constant film thickness.

Final Answer:
double