Question

The mass of the turbine rotor of a ship is 20 tonnes and has a radius of gyration of 0.60 m. Its speed is 2000 r.p.m. The ship pitches 6° above and 6° below the horizontal position. A complete oscillation takes 30 seconds and the motion is simple harmonic. Determine Maximum gyroscopic couple.

(a) 11.185 kN-m

(b) 22.185 kN-m

(c) 33.185 kN-m

(d) 44.185 kN-m

I had been asked this question during an internship interview.

This is a very interesting question from Effect of Gyroscopic Couple on an Aeroplane in division Gyroscopic Couple and Precessional Motion of Machine Dynamics

Answer 1

Correct option is (c) 33.185 kN-m

To elaborate: Given : m = 20 t = 20 000 kg ; k = 0.6 m ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s; φ = 6° = 6 × π/180 = 0.105 rad ; tp = 30 s

We know that mass moment of inertia of the rotor,

I = m.k^2 = 20 000 (0.6)^2 = 7200 kg-m^2

and angular velocity of the simple harmonic motion,

ω1 = 2π / tp = 2π/30 = 0.21 rad/s

Maximum angular velocity of precession,

ωPmax = φ.ω1 = 0.105 × 0.21 = 0.022 rad/s

We know that maximum gyroscopic couple,

Cmax = I.ω.ωPmax = 7200 × 209.5 × 0.022 = 33 185 N-m

                = 33.185 kN-m