Gyroscopic effect on an aeroplane turbine

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.