9.5 Gyroscopic Couple & Gyroscopic Effects

Angular Momentum and Precession

A disc of mass moment of inertia $I$ spinning at angular velocity $\omega$ about its own axis has angular momentum

$$H = I\omega$$

directed along the spin axis, with its sense given by the right-hand rule: curl the fingers in the direction of spin and the extended thumb points along $H$.

Now suppose the spin axis is itself made to rotate, in a plane containing the axis, at angular velocity $\omega_p$. This second rotation is called precession. The magnitude of $H$ is unchanged, but its direction changes, so angular momentum changes and a couple must be acting.

For a small precession through angle $\delta\theta$, the change in the angular momentum vector has magnitude $H\,\delta\theta$, directed perpendicular to the original $H$. Dividing by $\delta t$:

$$C = \frac{dH}{dt} = H\,\omega_p = I\,\omega\,\omega_p$$

This is the gyroscopic couple.

$$\boxed{C = I\,\omega\,\omega_{p}}$$

If the spin and precession axes are inclined at angle $\theta$ rather than perpendicular, the couple becomes $C = I\omega\omega_p\cos\theta$.

Active Versus Reactive Couple

This distinction produces more errors than the formula itself.

• The active gyroscopic couple is the couple that must be applied to the rotor to force it to precess.
• The reactive gyroscopic couple is the equal and opposite couple that the rotor exerts on its bearings and frame.

Engineering questions almost always ask about the reactive couple, because that is what loads the structure. When a question asks for the effect on the ship, the vehicle or the bearings, use the reactive couple.

Direction rule

A reliable procedure:

1. Draw the spin vector $\vec{\omega}$ along the axis using the right-hand rule.
2. Draw the precession vector $\vec{\omega}_p$ likewise.
3. The active couple vector is $\vec{C} = \vec{\omega}_p \times \vec{H}$, that is, rotate the angular momentum vector towards the direction it is being turned.
4. The reactive couple on the frame is opposite to this.

A quicker statement of the same rule: the axis of spin tries to align itself with the axis of precession, turning in the same sense as the precession.

Application 1: Ships

For a ship, three motions are named:

With a propeller shaft running fore and aft, the spin axis is longitudinal.

• Steering produces precession about the vertical axis, so the gyroscopic couple acts about the transverse axis, raising or lowering the bow.
• Pitching produces precession about the transverse axis, so the couple acts about the vertical axis, tending to swing the ship off course.
• Rolling produces precession about the same axis as the spin. Since the spin and precession axes coincide, no gyroscopic couple arises during rolling. This last point is the single most examined item in the topic.

For pitching modelled as simple harmonic motion of amplitude $\phi$ and angular frequency $\omega_1 = 2\pi/t_p$, the maximum angular velocity of precession is

$$\omega_{p,\max} = \phi\,\omega_1$$

and the maximum gyroscopic couple follows as $I\omega\phi\omega_1$.

Application 2: Aircraft

For a propeller-driven aeroplane taking a turn, the same analysis applies with the spin axis along the fuselage. The effect depends on both the direction of propeller rotation and the direction of turn:

The symmetry is worth noting rather than memorising four rows: reversing either the spin direction or the turn direction reverses the effect, and reversing both restores it.

Application 3: Four-Wheeled Vehicles

When a vehicle rounds a curve of radius $R$ at speed $v$, the wheels precess about the vertical axis at

$$\omega_p = \frac{v}{R}$$

and each wheel spins at $\omega = v/r$, where $r$ is the wheel radius. The total gyroscopic couple from four wheels, plus the engine rotating parts, is

$$C = \left(4I_w + G\,I_e\right)\frac{v^2}{R\,r}$$

where $I_w$ is the moment of inertia of each wheel, $I_e$ that of the engine rotating parts, and $G$ the gear ratio between engine and wheels. The engine term carries a positive sign when the engine rotates in the same sense as the wheels and a negative sign when it rotates oppositely.

Load transfer

The reactive gyroscopic couple tends to lift the inner wheels and press down the outer wheels — the same sense as the load transfer caused by centrifugal force. The change in reaction at each wheel, for a track width $b$, is

$$\Delta P_{\text{gyro}} = \frac{C}{2b}$$

The centrifugal contribution, for a centre of gravity at height $h$, is

$$\Delta P_{\text{cent}} = \frac{M v^2 h}{2 R b}$$

The vehicle overturns when the total transfer equals the static reaction, that is when the inner wheel reaction reaches zero:

$$\frac{Mg}{4} = \Delta P_{\text{gyro}} + \Delta P_{\text{cent}}$$

This condition sets the limiting speed on a curve. For heavy earth moving machinery with a high centre of gravity, the centrifugal term dominates, which is precisely why haul-road curve radii and speed limits are specified so conservatively in opencast mines.

Application 4: Two-Wheelers

A motorcycle taking a turn must lean inward at angle $\theta$ to the vertical. Both centrifugal and gyroscopic effects act to overturn it outward, and the leaning moment of the weight opposes them. Balancing:

$$\tan\theta \approx \frac{v^2}{gR}\left[1 + \frac{(2I_w + GI_e)}{M r R}\cdot\frac{R}{h}\right]$$

In most problems the simpler form $\tan\theta = v^2/(gR)$ is used first, with the gyroscopic term added as a correction. The essential physical point is that gyroscopic action adds to the required lean rather than reducing it.

Worked Example

A disc of mass moment of inertia 5 kg m squared spins at 300 rad/s. Its axis is made to precess at 2 rad/s in a perpendicular plane. Find the gyroscopic couple.

$$C = I\omega\omega_p = 5 \times 300 \times 2 = 3000 \text{ N m}$$

The reactive couple of 3000 N m acts on the bearings in the opposite sense — a substantial load from a modest precession rate, which is why the effect cannot be neglected in the design of turbine and propeller mountings.