Since the height of the gyroscope's barycenter does not change (and the origin of the coordinate system is located at this same point), its
gravitational potential energy is constant. Therefore its Lagrangian \mathcal{L} corresponds to its kinetic energy K only. We have \mathcal{L}=K=\frac{1}{2} \vec{\omega}^{T}I\vec\omega+\frac{1}{2} M \vec{v}_{\rm CM}^{2}, where M is the mass of the gyroscope, and \vec{v}_{\rm CM}^{2}=\Omega^2 R^2 \sin^2\delta={\rm constant} is the squared inertial speed of the origin of the coordinates of the final coordinate system (i.e. the
center of mass). This constant term does not affect the dynamics of the gyroscope and it can be neglected. On the other hand, the tensor of inertia is given by I=\begin{pmatrix} I_{1}&0&0\\ 0 & I_{2}&0\\ 0 &0 & I_{2} \end{pmatrix} and \begin{align} \vec{\omega}&=\begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\psi & \sin\psi\\ 0 & -\sin\psi & \cos\psi \end{pmatrix} \begin{pmatrix} \dot{\psi}\\ 0\\ 0 \end{pmatrix}+\begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\psi & \sin\psi\\ 0 & -\sin\psi & \cos\psi \end{pmatrix} \begin{pmatrix} \cos\alpha & \sin\alpha & 0\\ -\sin\alpha & \cos\alpha & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0\\ 0\\ \dot{\alpha} \end{pmatrix}\\ &\qquad + \begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\psi & \sin\psi\\ 0 & -\sin\psi & \cos\psi \end{pmatrix} \begin{pmatrix} \cos\alpha & \sin\alpha & 0\\ -\sin\alpha & \cos\alpha & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos\delta & 0 & -\sin\delta\\ 0 & 1 & 0\\ \sin\delta & 0 & \cos\delta \end{pmatrix} \begin{pmatrix} \cos\Phi & \sin\Phi & 0\\ -\sin\Phi & \cos\Phi & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos\Omega t & \sin\Omega t & 0\\ -\sin\Omega t & \cos\Omega t & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0\\ 0\\ \Omega \end{pmatrix}\\ &= \begin{pmatrix} \dot{\psi}\\ 0\\ 0\\ \end{pmatrix}+ \begin{pmatrix} 0\\ \dot{\alpha}\sin\psi\\ \dot{\alpha}\cos\psi \end{pmatrix}+ \begin{pmatrix} -\Omega\sin\delta\cos\alpha\\ \Omega(\sin\delta\sin\alpha\cos\psi+\cos\delta\sin\psi)\\ \Omega(-\sin\delta\sin\alpha\sin\psi+\cos\delta\cos\psi) \end{pmatrix} \end{align} Therefore we find \begin{align} \mathcal{L} &= \frac{1}{2} \left [I_{1}\omega_{1}^{2}+I_{2} \left (\omega_{2}^{2}+\omega_{3}^{2} \right ) \right ]\\ &= \frac{1}{2} I_{1} \left (\dot{\psi}-\Omega\sin\delta\cos\alpha \right )^{2} +\frac{1}{2} I_{2} \left \{ \left [\dot{\alpha}\sin\psi+\Omega(\sin\delta\sin\alpha\cos\psi+\cos\delta\sin\psi) \right ]^{2} + \left [\dot{\alpha}\cos\psi+\Omega(-\sin\delta\sin\alpha\sin\psi+\cos\delta\cos\psi) \right ]^{2} \right \} \\ &= \frac{1}{2} I_{1} \left (\dot{\psi}-\Omega\sin\delta\cos\alpha \right )^{2}+\frac{1}{2} I_{2} \left \{ \dot{\alpha}^{2} +\Omega^{2} \left (\cos^{2}\delta+\sin^{2}\alpha\sin^{2}\delta \right ) +2\dot{\alpha}\Omega\cos\delta \right \} \end{align} The Lagrangian can be rewritten as \mathcal{L}=\mathcal{L}_{1}+\frac{1}{2} I_{2}\Omega^{2}\cos^{2}\delta+\frac{d}{dt}(I_{2}\alpha\Omega\cos\delta), where \mathcal{L}_{1}=\frac{1}{2} I_{1} \left (\dot{\psi}-\Omega\sin\delta\cos\alpha \right )^{2}+\frac{1}{2} I_{2}\left (\dot{\alpha}^{2}+\Omega^{2}\sin^{2}\alpha\sin^{2}\delta \right ) is the part of the Lagrangian responsible for the dynamics of the system. Then, since \partial \mathcal{L}_1/\partial\psi = 0, we find L_{x}\equiv\frac{\partial \mathcal{L}_1}{\partial\dot{\psi}}=I_1 \left (\dot{\psi}-\Omega\sin\delta\cos\alpha \right )=\mathrm{constant}. Since the angular momentum \vec L of the gyrocompass is given by \vec L=I\vec\omega, we see that the constant L_x is the component of the angular momentum about the axis of symmetry. Furthermore, we find the equation of motion for the variable \alpha as \frac{d}{dt} \left(\frac{\partial \mathcal{L}_{1}}{\partial\dot{\alpha}}\right)=\frac{\partial \mathcal{L}_{1}}{\partial\alpha}, or \begin{align} I_{2}\ddot{\alpha} &=I_{1}\Omega \left (\dot{\psi}-\Omega\sin\delta\cos\alpha \right )\sin\delta\sin\alpha+\frac{1}{2} I_{2} \Omega^{2}\sin^{2}\delta\sin2\alpha\\ &=L_{x}\Omega\sin\delta\sin\alpha+\frac{1}{2} I_{2} \Omega^{2}\sin^{2}\delta\sin2\alpha \end{align}
Particular case: the poles At the poles we find \sin\delta=0, and the equations of motion become \begin{align} L_{x} &=I_{1}\dot{\psi}=\mathrm{constant}\\ I_{2}\ddot{\alpha}&=0 \end{align} This simple solution implies that the gyroscope is uniformly rotating with constant
angular velocity in both the vertical and symmetrical axis.
The general and physically relevant case Let us suppose now that \sin\delta\neq0 and that \alpha\approx0, that is the axis of the gyroscope is approximately along the north-south line, and let us find the parameter space (if it exists) for which the system admits stable small oscillations about this same line. If this situation occurs, the gyroscope will always be approximately aligned along the north-south line, giving direction. In this case we find \begin{align} L_{x}&\approx I_{1} \left (\dot{\psi}-\Omega\sin\delta \right )\\ I_{2}\ddot{\alpha}&\approx \left (L_{x}\Omega\sin\delta+I_{2} \Omega^{2}\sin^{2}\delta \right) \alpha \end{align} Consider the case that L_{x} and, further, we allow for fast gyro-rotations, that is \left |\dot{\psi} \right |\gg\Omega. Therefore, for fast spinning rotations, L_x implies \dot\psi In this case, the equations of motion further simplify to \begin{align} L_{x} &\approx -I_{1} \left |\dot{\psi} \right | \approx \mathrm{constant}\\ I_{2}\ddot{\alpha} &\approx -I_{1} \left |\dot{\psi} \right |\Omega \sin\delta\alpha \end{align} Therefore we find small oscillations about the north-south line, as \alpha\approx A\sin(\tilde\omega t+B), where the angular velocity of this harmonic motion of the axis of symmetry of the gyrocompass about the north-south line is given by \tilde\omega=\sqrt{\frac{I_{1}\sin\delta}{I_{2}}}\sqrt{\left |\dot{\psi} \right |\Omega}, which corresponds to a period for the oscillations given by T=\frac{2\pi}{\sqrt{\left |\dot{\psi} \right |\Omega}}\sqrt{\frac{I_{2}}{I_{1}\sin\delta}}. Therefore \tilde\omega is proportional to the geometric mean of the Earth and spinning angular velocities. In order to have small oscillations we have required \dot{\psi}, so that the North is located along the right-hand-rule direction of the spinning axis, that is along the negative direction of the X_7-axis, the axis of symmetry. As a side result, on measuring T (and knowing \dot{\psi}), one can deduce the local co-latitude \delta. == See also ==