If \vec{F_{}}→F is the force generating the torque \vec{\tau_{}}→τ on the gyroscope, the precession frequency \omega_pωp is related to the angular momentum \vec{L_{}}→L as follows. The angular momentum itself is related to the moment-of-inertia II and spin frequency \omegaω.
Torque : \vec{\tau_{}} \equiv \vec{r_{}}\times\vec{F_{}}=\frac{d\vec{L_{}}}{dt}→τ≡→r×→F=d→Ldt,
Angular Momentum: \vec{L_{}}\equiv I\omega→L≡Iω
Precission Frequency: \omega_p = 1/L\frac{dL}{dt}=\tau/Lωp=1LdLdt=τL
The torque generating force is the weight ( F=MgF=Mg) of the spinning disk which acts vertically down. Since the gyroscope's axle is horizontal to the ground, \vec{r_{}}→r and \vec{F_{}}→F are perpendicular to each other.
\tau = r.F.sin90^o=r.Mg.sin90^o=Mgrτ=r.F.sin90o=r.Mg.sin90o=Mgr
Moment of inertia of a disk of mass MM and radius RR is I=\frac{MR^2}{2}I=MR22
Angular momentum L=I.\omega=1/2 MR^2\omegaL=I.ω=12MR2ω
Precessional frequency \omega_p = \tau/L = \frac{Mgr}{1/2MR^2\omega}=\frac{2gr}{\omegaR^2}ωp=τL=Mgr12MR2ω=2grωR2
Given : R=49cm=0.49mR=49cm=0.49m;\quadr=4cm=0.04m\quadg=9.8ms^{-2}
\omega=990 rev/min =33\pi rad/s,
\omega_p = \frac{2gr}{\omegaR^2}=0.03148 rad/s =0.3007 rev/min.
