Gyroscopic Couple

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

        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.

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 $-z$.

  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.

Data: $m = 250$ kg, $h = 0.6$ m, $r_w = 0.3$ m, $I_w = 1.0$ kg·m² (each), $I_e = 0.2$ kg·m², $G = 4$, $v = 54$ km/h $= 15.0$ m/s, $R = 50$ m.

Spin quantities

$$\omega_w = \frac{v}{r_w} = \frac{15.0}{0.3} = 50.00\ \text{rad/s}, \qquad \omega_e = 4\omega_w = 200.00\ \text{rad/s}$$ $$\sum I\omega = 2I_w\omega_w + I_e\omega_e = 2(1.0)(50.00) + 0.2(200.00) = 140.0\ \text{kg·m}^2/\text{s}$$

Precession rate about the vertical axis: $\omega_p = \dfrac{v}{R} = \dfrac{15.0}{50} = 0.30$ rad/s.

Couples (at heel angle $\theta$)

• Gyroscopic couple: $C_g = \left(\sum I\omega\right)\omega_p\cos\theta = 42.0\cos\theta$ N·m
• Centrifugal couple about the ground contact: $C_c = \dfrac{mv^2}{R}\,h\cos\theta = \dfrac{250 \times 15.0^2}{50}\times 0.6\cos\theta = 675.0\cos\theta$ N·m
• Weight couple (restoring) $mgh\sin\theta = 250 \times 9.81 \times 0.6\sin\theta = 1471.5\sin\theta$ N·m

   CG *   weight mg (down)
      |\
      | \  centrifugal force mv^2/R (outward)
   h  |  \
      |   \
  ----O----  ground contact, lean angle theta

Equilibrium

Weight couple balances the sum of the centrifugal and gyroscopic effects:

$$mgh\sin\theta = (C_c + C_g)\cos\theta$$ $$\tan\theta = \frac{C_c + C_g}{mgh} = \frac{675.0 + 42.0}{1471.5} = 0.4873$$ $$\theta = 26.0^\circ$$

Answer: The motorcycle must be heeled (leaned) by about $26.0^\circ$ towards the centre of the curve.