= Solution
For a thin <gyrocompass> flywheel with rotational <symmetry>, the <perpendicular axis theorem> gives $I_1=I_x+I_y=2I_2$. This does not require the mass to be concentrated at the rim. Convert the specified spin to <angular velocity> and use one <sidereal day> for <Earth's rotation>:
$$
\omega=5000\frac{2\pi}{60}=523.6\ \mathrm{rad\,s^{-1}},\qquad
\Omega_E=\frac{2\pi}{86164\ \mathrm s}=7.2921\times10^{-5}\ \mathrm{rad\,s^{-1}}.
$$
At the given latitude,
$$
\omega_{\rm shm}=\sqrt{2(523.6)(7.2921\times10^{-5})\cos52.2^\circ},\qquad
\boxed{\omega_{\rm shm}\simeq0.216\ \mathrm{rad\,s^{-1}}.}
$$
The associated <oscillation period> is $T=2\pi/\omega_{\rm shm}\simeq29.0\ \mathrm s$. This uses the stated neglect of the support assembly's contribution to the <moment of inertia>; a substantial additional vertical <moment of inertia> would lengthen the <oscillation period>.
Back to article page
