An aircraft cruising at 360 kmph takes a right turn on an arc of 100 m radius. The turbines and propellers have a total mass of 500 kg with radius of gyration of 25 cm. The engine rotates at 2000 r.p.m. The magnitude of the gyroscopic couple generated is

Concept:

Gyroscopic Couple \( = I × \omega × {\omega _p}\)

Where, I is moment of inertia of the ship, ωp is the angular velocity of precision and ω is the rotational speed of the ship

Calculation:

Given V = 360 kmph = 100 m/s, m = 500 kg, k = 25 cm = 0.25 m, N = 2000 r.p.m.   

∴ moment of inertia I = mk² = 500 × 0.25² = 31.25 kg-m2

and \(\omega = \frac{{2\pi N}}{{60\;}} = \frac{{2 \times \pi \times 2000}}{{60}} = 209.43\;rad/s\)

\({\omega _p} = \frac{v}{r} = \frac{100}{{100}} = 1\; rad/s\)

Gyroscopic Couple = 31.25 × 209.43 × 1 = 6544.98 Nm = 6.55 kN m