The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration of 0.45 m and a speed of 3000 rpm clockwise when looking from stern. If the ship is steering to the left on a curve of 100 m radius at a speed of 36 km/h, then the gyroscopic couple is

Concept:

Gyroscopic couple for any rotor system is given by 

C_g = Iωω_p 

Where, I = Mass moment of inertia in kg-m², ω = Rotational speed of rotor in rad/s

ω_p = Angular velocity of the precession of the axis of spin or the speed of rotation of the axis of spin about the axis of precession.

Calculation:

Given:

mass of rotor (m) = 3500 kg, the radius of gyration (k) = 0.45 m, Speed of rotor (N) = 3000 rpm

radius of curve (R) = 100 m, Velocity of ship = 36 km/hr.

We know that the mass moment of inertia of the disc, about an axis through its center of gravity and perpendicular to the plane of the disc,

I = mk² 

(where, k = Radius of gyration in m)

I = 3500 × 0.45² = 708.75 kg-m² 

Now we know that the speed of the rotor can be written as,

\(\omega =\frac{2\pi N}{60}\;\) 

\(\omega =\frac{2\pi × 3000}{60}\;\)

ω = 314.16 rad/s

Now precessional speed is given by,

\(\omega_p =\frac{V}{R}\;\)

\(\omega_p =\frac{36× 10^3}{3600× 100}\;\)    

ω_p = 0.1 rad/s

Now can easily calculate gyroscopic couple by,

Cg = Iωωp 

C_g = 708.75 × 314.16 × 0.1

C_g = 22266.04 \(\frac{kg-m^2}{s^2}\;or\; Nm\)

C_g = 22.26 kN-m