---
title: "Derivation of gyroscopic couple"
subject: "Theory of Mechanism and Machine II"
chapter: 3
key: "semester_7/theory_of_mechanism_and_machine_ii:gyroscopic-couple-derivation"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/gyroscopic-couple-derivation
site: BCT Engineering
---

# Derivation of gyroscopic couple

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

- Asked in: year not given, 6 marks
- Video: [Gyroscopic Couple - Derivation - Precessional Motion](https://www.youtube.com/watch?v=9hXKHiQjfKs) — Dr. Ravi's Academy
- 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/gyroscopic-couple-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-3

## Question

Define precessional angular motion. Derive an expression for the gyroscopic couple acting on a rotating disc whose axis of spin is made to precess at a uniform rate, and explain how the direction of the reactive couple is found.

## Short answer

Precession is rotation of the spin axis about a perpendicular axis. Gyroscopic couple C = I ω ω_p, obtained from rate of change of angular momentum. The direction of the reactive couple on the bearings is opposite to the active couple, found by applying the right-hand rule to the spin vector's rotation.

## Answer

**Precession** is the angular motion of the axis of a spinning body about another axis (the precession axis) perpendicular to the spin axis. The spin axis keeps turning in a plane at right angles to the precession axis.

#### Derivation
```text
        spin axis OX (at t)
   O ------------> X
    \  d(theta)
     \----> X' (at t+dt)
 precession axis: vertical through O
```
Let a disc of moment of inertia $I$ spin about OX with angular velocity $\omega$. The angular momentum vector is $\vec H = I\omega$ along OX. In a small time $\delta t$ the axis turns through a small angle $\delta\theta$ about the perpendicular axis, so the vector moves to OX'.

Change in angular momentum (vector change, perpendicular to OX):

$$
\delta H = I\omega\,\delta\theta
$$

Rate of change of angular momentum:

$$
\frac{dH}{dt} = I\omega\,\frac{d\theta}{dt} = I\omega\,\omega_p
$$

where $\omega_p = \dfrac{d\theta}{dt}$ is the angular velocity of precession. By Newton's law, this rate of change of angular momentum requires an applied couple:

$$
C = I\omega\omega_p
$$

If the spin and precession axes are not at $90^\circ$ but at angle $\alpha$, then $C = I\omega\omega_p\sin\alpha$.

#### Direction of the reactive couple
1. Draw the spin vector $\vec\omega$ using the right-hand screw rule.
2. Turn this vector about the precession axis in the sense of the precession. The vector change $\delta\vec H$ points in the direction of the **active couple** that must be applied to the spinning disc (by the bearings) to make it precess.
3. The disc exerts an equal and opposite **reactive gyroscopic couple** on the supporting body (bearings, aeroplane, ship, vehicle). Its vector is opposite to $\delta\vec H$.

Short rule: the spin vector tends to turn towards the active-couple vector; the supporting body feels the reverse.

## Check yourself

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

1. Precession is:
   A. Rotation of a body about its own spin axis
   B. Angular motion of the spin axis about another axis perpendicular to it
   C. Linear motion of the centre of the disc
   D. Oscillation of a body about equilibrium
   Answer: B. The spin axis keeps turning in a plane at right angles to the precession axis. (Key point: Precession)

2. When the spin axis turns through a small angle d(theta), the change in angular momentum is:
   A. I w along the spin axis
   B. I w d(theta), perpendicular to the spin axis
   C. I d(theta) only
   D. I w / d(theta)
   Answer: B. The vector H = I w keeps its magnitude but turns, so dH = I w d(theta) perpendicular to OX. (Key point: Change of angular momentum)

3. The applied couple, by Newton's law, is:
   A. I w w_p
   B. I w / w_p
   C. I w_p^2
   D. I w^2 w_p
   Answer: A. C = dH/dt = I w (d(theta)/dt) = I w w_p when axes are at 90 degrees. (Key point: Gyroscopic couple)

4. If spin and precession axes are at an angle alpha, the couple becomes:
   A. I w w_p cos(alpha)
   B. I w w_p sin(alpha)
   C. I w w_p tan(alpha)
   D. unchanged
   Answer: B. Only the component of the spin vector perpendicular to the precession axis matters, so C = I w w_p sin(alpha). (Key point: Non-perpendicular axes)

5. The reactive gyroscopic couple on the supporting body is:
   A. Along the spin vector
   B. Same as the active couple
   C. Opposite to the active couple (opposite to dH)
   D. Zero in steady precession
   Answer: C. The disc exerts an equal and opposite couple on the bearings and body compared with the active couple that causes the change dH. (Key point: Direction of reactive couple)

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