A flywheel weighs \(\frac{{981}}{\pi }\) kgf and has a radius of gyration of 100 cm. It is given a spin of 100 r.p.m. about its horizontal axis. The whole assembly is rotating about a vertical axis at 6 rad/s. The gyroscopic couple experienced will be

Concept:

Gyroscopic Couple = I × ω × ω_p

Where I is the moment of inertia of the flywheel, ωp is the angular velocity of spin and ω is the rotational speed of the flywheel

Calculation:

Given, W = \(\frac{{981}}{\pi }\) kgf ⇒ m =  \(\frac{{981}}{\pi }\)kg, radius of gyration k = 100 cm = 1m, ω = 6rad/s, N = 100 r.p.m,

\(I = m{k^2} = \frac{{981}}{{\pi }} × {\left( 1 \right)^2}\) kg-m²

\(I = \frac{{981}}{\pi }\;kg-\;{m^2}\)

\(ω_p = \frac{{2\pi N}}{{60}} = \frac{{2\pi × 100}}{{60}} = \frac{{20\pi }}{6}\;rad/sec\)

\(T = Iω \;{ω _p} = \frac{{981}}{\pi } × \frac{{20\pi }}{6} × 6= 19620 \ kg \ m^2/s^2 = 19620\;Nm\)

1 kgf = 9.81 N

⇒ T = 2000 kgf-m