---
title: "Torque to rotate truncated cone"
subject: "Fluid Mechanics"
chapter: 1
key: "semester_3/fluid_mechanics:truncated-cone-torque"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/truncated-cone-torque
site: BCT Engineering
---

# Torque to rotate truncated cone

Fluid Mechanics · Chapter 1: Fluid and its physical properties

- Asked in: 2076 Chaitra, 6 marks
- Video: [Mechanics of Fluids - Topic 2 - Example 6 - Viscosity - Rotating Cone in Annulus](https://www.youtube.com/watch?v=RwUEfgNo99Q) — Dato Ts Dr Mohd Zamri Yusoff
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/truncated-cone-torque
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-1

## Question

Oil of viscosity $\mu$ = 2 poise fills the small gap of thickness 0.2 mm. Determine the torque required to rotate the truncated cone at constant speed of 100 rpm. Neglect fluid stress exerted on the circular bottom. [Figure: inverted truncated cone with 60° apex angle, rotating with angular velocity $\omega$ in a conical casing; dimensions 40 mm and 60 mm marked on the height]

## Short answer

Using element at axial distance z, radius r = z tanα, shear stress τ = μ ω r / t. Integrate torque: T = (π μ ω tan³α)/(2 t cosα) (z₂⁴ − z₁⁴). With ω = 10.472 rad/s, μ = 0.2 Pa·s, α = 30°, z₁=0.04 m, z₂=0.10 m, T = 0.356 N·m.

## Answer

**Reading of the figure:** the cone has a full apex angle of $60^\circ$ (half-angle $\alpha=30^\circ$). The wetted frustum extends from 40 mm to 100 mm measured along the axis from the (imaginary) apex: 40 mm from the apex to the small end, and a frustum height of 60 mm. The gap measured normal to the surface is $t=0.2$ mm.

**Data:** $\mu=2$ poise $=0.2$ Pa·s, $N=100$ rpm, so $\omega=\dfrac{2\pi\times100}{60}=10.472$ rad/s.

```text
    \____________/   large end, z2 = 100 mm from apex
     \          /
      \        /     frustum height z2 - z1 = 60 mm
       \______/      small end, z1 = 40 mm from apex
        \    /
         \  /
          \/  apex (imaginary), 60 deg
    oil gap t = 0.2 mm between cone and casing
```

Take an element at axial distance $z$ from the apex:

- radius $r=z\tan\alpha$
- slant length $ds=\dfrac{dz}{\cos\alpha}$
- shear stress $\tau=\mu\dfrac{\omega r}{t}$

$$
dT=\tau\,(2\pi r\,ds)\,r=\frac{2\pi\mu\omega}{t}\,\frac{r^{3}}{\cos\alpha}\,dz
$$

$$
T=\frac{2\pi\mu\omega\tan^{3}\alpha}{t\cos\alpha}\int_{z_1}^{z_2}z^{3}dz=\frac{\pi\mu\omega\tan^{3}\alpha}{2\,t\cos\alpha}\left(z_2^{4}-z_1^{4}\right)
$$

Substituting ($\tan30^\circ=0.5774$, $\cos30^\circ=0.8660$, $z_1=0.04$ m, $z_2=0.10$ m):

$$
T=\frac{\pi(0.2)(10.472)(0.5774)^{3}}{2(0.0002)(0.8660)}\left(0.10^{4}-0.04^{4}\right)=0.356\ \text{N·m}
$$

**Answer: $T\approx0.356\ \text{N·m}$.** (If the figure dimensions differ, use the same formula with the correct $z_1$ and $z_2$.)

## Check yourself

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

1. For an element at radius r on the cone, what is the shear stress in the oil gap t?
   A. tau = mu omega r / t
   B. tau = mu omega / (r t)
   C. tau = mu t / (omega r)
   D. tau = mu omega r^2 / t
   Answer: A. Surface speed is omega r and the profile is linear across gap t, so tau = mu omega r/t. (Key point: Shear stress in gap)

2. For the conical surface with half-angle alpha, the area of an element is 2 pi r ds. What are r and ds in terms of axial distance z?
   A. r = z cos(alpha), ds = dz sin(alpha)
   B. r = z tan(alpha), ds = dz / cos(alpha)
   C. r = z / tan(alpha), ds = dz cos(alpha)
   D. r = z, ds = dz
   Answer: B. r = z tan(alpha) and the slant length is ds = dz/cos(alpha). (Key point: Element geometry)

3. After integrating dT from z1 to z2, how does the torque depend on z?
   A. Through (z2^2 - z1^2)
   B. Through (z2^3 - z1^3)
   C. Through (z2 - z1)
   D. Through (z2^4 - z1^4)
   Answer: D. dT is proportional to r^3 dz, i.e. z^3 dz, which integrates to (z2^4 - z1^4)/4. (Key point: Torque integral)

4. What are the correct values of mu and omega used for 2 poise and 100 rpm?
   A. mu = 2 Pa·s, omega = 100 rad/s
   B. mu = 0.2 Pa·s, omega = 10.47 rad/s
   C. mu = 0.02 Pa·s, omega = 10.47 rad/s
   D. mu = 0.2 Pa·s, omega = 100 rad/s
   Answer: B. 1 poise = 0.1 Pa·s and omega = 2 pi N/60 = 10.472 rad/s, giving T of about 0.356 N·m. (Key point: Unit conversion)

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