Question

The rotor of an aeroplane engine has a mass moment of inertia 1.0 kg m\(^2\). The engine rotates at a speed of 500 RPM in the clockwise direction if viewed from the front of the aeroplane. If the aeroplane while flying at 1200 km/hr turns with a radius of 2 km at same elevation, then the magnitude of the gyroscopic moment exerted by the rotor on the aeroplane structure in N m is

• A. 8.73
• B. 17.46
• C. 4.37
• D. 26.19

Hint:

Find the spin angular velocity and the turn rate (precession) separately, then multiply with the inertia.

Answer 1

The Correct Option is A

A spinning rotor forced to change the direction of its spin axis pushes a reaction couple onto its supports, the gyroscopic couple, equal to the spin angular momentum times the rate at which the spin axis turns.

1. Spin angular velocity: $\omega_s = 2\pi N/60 = 2\pi(500)/60 = 52.36$ rad/s, from the given engine speed of 500 RPM.
2. Precession angular velocity: the aeroplane turning at radius $R = 2000$ m with speed $V = 1200/3.6 = 333.33$ m/s sweeps its heading at $\omega_p = V/R = 333.33/2000 = 0.1667$ rad/s.
3. Gyroscopic couple: $C = I\omega_s\omega_p = 1.0 \times 52.36 \times 0.1667 = 8.73$ N m, the moment the rotor pushes onto the airframe as it is forced to precess.

None of the other three values match this product, so 8.73 N m, option A, is correct.