---
title: "Gyroscopic effect on an aeroplane turbine"
subject: "Theory of Mechanism and Machine II"
chapter: 3
key: "semester_7/theory_of_mechanism_and_machine_ii:aeroplane-turbine-gyroscopic-couple"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/aeroplane-turbine-gyroscopic-couple
site: BCT Engineering
---

# Gyroscopic effect on an aeroplane turbine

Theory of Mechanism and Machine II · Chapter 3: Gyroscopic Couple

- Asked in: year not given, 6 marks
- Video: [DETERMINING THE EFFECT OF GYROSCOPIC COUPLE](https://www.youtube.com/watch?v=NhRQVGN8iqc) — Bhanu Tej J v
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/aeroplane-turbine-gyroscopic-couple
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-3

## Question

The rotor of the turbine of an aeroplane has a mass of 500 kg and a radius of gyration of 0.3 m. It rotates at 3000 rpm clockwise when viewed from the rear of the aeroplane. The aeroplane flies at 360 km/h and turns to the right in a curve of 200 m radius. Find the gyroscopic couple on the aeroplane and state its effect on the aeroplane.

## Answer

**Data:** $m = 500$ kg, $k = 0.3$ m, $N = 3000$ rpm, $v = 360$ km/h, $R = 200$ m.

#### Quantities
$$
I = mk^2 = 500 \times 0.3^2 = 45.0\ \text{kg·m}^2
$$

$$
\omega = \frac{2\pi \times 3000}{60} = 314.16\ \text{rad/s}
$$

$$
v = \frac{360}{3.6} = 100\ \text{m/s}, \qquad \omega_p = \frac{v}{R} = \frac{100}{200} = 0.50\ \text{rad/s}
$$

#### Gyroscopic couple
$$
C = I\omega\omega_p = 45.0 \times 314.16 \times 0.50 = 7069\ \text{N·m}
$$

#### Direction of the reaction
Take x forward, y to the left, z upward. A rotor turning clockwise as seen from the rear has its spin vector along $+x$ (forward). A right turn is a clockwise yaw seen from above, so the precession vector is along $-z$.

```text
  spin H : forward  (+x)
  precession : down (-z)  [right turn]
  active couple on rotor : axis -y (to right)
  reactive couple on aircraft : axis +y (to left)
```
The reactive couple on the aeroplane has its axis pointing to the left, which turns the nose downward (pitching the nose down and the tail up).

**Answer:** Gyroscopic couple $= 7069$ N·m ($\approx 7.07$ kN·m). The reactive couple tends to **dip the nose** of the aeroplane (raise the tail) in the right turn.

## Check yourself

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

1. How is the angular velocity of precession found for the turning aeroplane?
   A. w_p = v / R
   B. w_p = v R
   C. w_p = v^2 / R
   D. w_p = R / v
   Answer: A. w_p = 100/200 = 0.5 rad/s with v = 360 km/h = 100 m/s. (Key point: Precession speed)

2. The rotor's moment of inertia is:
   A. m k = 150 kg.m
   B. m k^2 = 45 kg.m^2
   C. 1/2 m k^2 = 22.5 kg.m^2
   D. m k^2 / 2 pi
   Answer: B. I = m k^2 = 500 x 0.3^2 = 45 kg.m^2. (Key point: Moment of inertia)

3. The gyroscopic couple is closest to:
   A. 3535 N.m
   B. 7069 N.m
   C. 14138 N.m
   D. 707 N.m
   Answer: B. C = I w w_p = 45 x 314.16 x 0.5 = 7069 N.m. (Key point: Magnitude of couple)

4. For a clockwise rotor viewed from the rear and a right turn, the reactive couple tends to:
   A. Raise the nose
   B. Dip the nose and raise the tail
   C. Roll the plane to the left
   D. Yaw further to the right
   Answer: B. The reactive couple axis points to the left, which pitches the nose down. (Key point: Effect on aeroplane)

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