---
title: "Torque to overcome bearing friction"
subject: "Fluid Mechanics"
chapter: 1
key: "semester_3/fluid_mechanics:journal-bearing-torque"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/journal-bearing-torque
site: BCT Engineering
---

# Torque to overcome bearing friction

Fluid Mechanics · Chapter 1: Fluid and its physical properties

- Asked in: 2075 Asoj, 8 marks
- Video: [Problem to Determine the Power Lost in the Bearing (Sleeve)- Properties of Fluid - Fluid Mechanics 1](https://www.youtube.com/watch?v=2lhmSUnTckE) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/journal-bearing-torque
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-1

## Question

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.

## 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.

## Answer

**Data:** shaft radius $R=8$ cm $=0.08$ m, bearing radius $8.025$ cm, so the clearance $t=0.025$ cm $=0.25$ mm. $L=0.30$ m, $N=200$ rpm, $\mu=2.5$ poise $=0.25$ Pa·s.

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

Angular speed and surface velocity:

$$
\omega=\frac{2\pi\times200}{60}=20.944\ \text{rad/s},\qquad V=\omega R=1.6755\ \text{m/s}
$$

Shear stress (linear profile across the thin film):

$$
\tau=\mu\frac{V}{t}=0.25\times\frac{1.6755}{0.00025}=1675.5\ \text{N/m}^2
$$

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

$$
F=\tau A=1675.5\times0.1508=252.7\ \text{N}
$$

Torque:

$$
T=F\,R=252.7\times0.08=20.21\ \text{N·m}
$$

**Answer: $T\approx20.2\ \text{N·m}$.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Shaft radius 8 cm and bearing radius 8.025 cm. What is the oil film thickness?
   A. 0.25 mm
   B. 0.025 mm
   C. 2.5 mm
   D. 0.8 mm
   Answer: A. 8.025 - 8 = 0.025 cm = 0.25 mm. (Key point: Clearance)

2. Which formula gives the shear stress at the shaft surface?
   A. tau = mu t / (omega R)
   B. tau = mu omega R / t^2
   C. tau = mu V / t with V = omega R
   D. tau = mu V t
   Answer: C. A linear profile across the film gives tau = mu V/t with V = omega R = 1.6755 m/s. (Key point: Shear stress)

3. How is the friction torque obtained?
   A. T = tau x A x R, with A = 2 pi R L
   B. T = tau x A / R
   C. T = tau x L x R
   D. T = tau x pi R^2 x R
   Answer: A. Shear force F = tau (2 pi R L) acts at radius R, so T = F R = 20.2 N·m. (Key point: Torque from force)

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