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Torque to overcome bearing friction

Short answer

Using a linear velocity profile, shear stress τ = μ (ωR)/t = 1675.5 Pa acts on area 2πRL = 0.1508 m², giving force 252.7 N. Torque = F R = 20.2 N·m.

A stationary bearing of length 30 cm and internal radius 8.025 cm has been used to provide lateral stability to a 8 cm radius shaft rotating at a constant speed of 200 rpm. The space between the shaft and bearing is filled with a lubricant having viscosity 2.5 poise. Find the torque required to overcome the friction in bearing. Take the velocity profile as linear.

Data: shaft radius R=8R=8 cm =0.08=0.08 m, bearing radius 8.0258.025 cm, so the clearance t=0.025t=0.025 cm =0.25=0.25 mm. L=0.30L=0.30 m, N=200N=200 rpm, μ=2.5\mu=2.5 poise =0.25=0.25 Pa·s.

   bearing  |<-t->| shaft
   (fixed)  |     | R = 8 cm, 200 rpm

Angular speed and surface velocity:

ω=2π×20060=20.944 rad/s,V=ωR=1.6755 m/s\omega=\frac{2\pi\times200}{60}=20.944\ \text{rad/s},\qquad V=\omega R=1.6755\ \text{m/s}

Shear stress (linear profile across the thin film):

τ=μVt=0.25×1.67550.00025=1675.5 N/m2\tau=\mu\frac{V}{t}=0.25\times\frac{1.6755}{0.00025}=1675.5\ \text{N/m}^2

Shear force over the shaft surface A=2πRL=2π(0.08)(0.30)=0.1508 m2A=2\pi RL=2\pi(0.08)(0.30)=0.1508\ \text{m}^2:

F=τA=1675.5×0.1508=252.7 NF=\tau A=1675.5\times0.1508=252.7\ \text{N}

Torque:

T=F R=252.7×0.08=20.21 N⋅mT=F\,R=252.7\times0.08=20.21\ \text{N·m}

Answer: T≈20.2 N⋅mT\approx20.2\ \text{N·m}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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