You can work all of this out; it's quite a tractable problem. Take the configuration linked by
@vanhees71, for instance, and assume that the angle between the local vertical and the line joining the Earth's centre to the North Pole is ##\gamma## [
you may set ##\gamma = 0## if the apparatus is located at the North Pole]. Consider an Earth fixed frame ##\{ \hat{\mathbf{r}}, \hat{\mathbf{x}}, \hat{\mathbf{y}} \}## and a gyroscope fixed frame ##\{ \hat{\mathbf{r}}, \hat{\mathbf{x}}', \hat{\mathbf{y}}' \}## which are related by a rotation ##\mathsf{R}(\varphi)## around the ##\hat{\mathbf{r}}## axis$$\begin{pmatrix} \hat{\mathbf{x}}' \\ \hat{\mathbf{y}}' \end{pmatrix} = \begin{pmatrix} \cos{\varphi} & \sin{\varphi} \\ -\sin{\varphi} & \cos{\varphi} \end{pmatrix} \begin{pmatrix} \hat{\mathbf{x}} \\ \hat{\mathbf{y}} \end{pmatrix}$$Write the angular velocity at which the Earth rotates about its axis as ##\boldsymbol{\Omega} = \Omega (\cos{\gamma} \hat{\mathbf{r}} + \sin{\gamma} \hat{\mathbf{x}})##.
Let ##\psi## be the angle through which the gyroscope has rotated about its axis. Relative to an inertial (space fixed) frame, then, the angular velocity of the gyroscope takes the form \begin{align*}
\boldsymbol{\omega} &= \boldsymbol{\Omega} + (\dot{\psi} \hat{\mathbf{x}}' + \dot{\varphi} \hat{\mathbf{r}}) \\
&= (\Omega \sin{\gamma} \cos{\varphi} + \dot{\psi}) \hat{\mathbf{x}}' - \Omega \sin{\gamma} \sin{\varphi} \hat{\mathbf{y}}' + (\Omega \cos{\gamma} + \dot{\varphi}) \hat{\mathbf{r}}
\end{align*}where we have chosen ##\hat{\mathbf{x}}'## such that it lies along the axial direction of the gyroscope; the inertia tensor relative to the centre of mass of the gyroscope is hence also reduced to diagonal form ##(I_{ij}) = \mathrm{diag}(a,b,b)##. Then the angular momentum about the centre of mass relative to the inertial (space fixed) frame is nothing but ##\mathbf{L} = a (\Omega \sin{\gamma} \cos{\varphi} + \dot{\psi}) \hat{\mathbf{x}}' - b\Omega \sin{\gamma} \sin{\varphi} \hat{\mathbf{y}}' + b (\Omega \cos{\gamma} + \dot{\varphi})\hat{\mathbf{r}}##.
This should be sufficient information for you to determine the equations of motion, after writing ##\dot{\mathbf{L}} = \mathbf{G} = \alpha \hat{\mathbf{y}}'## for some ##\alpha \in \mathbb{R}##. It might also help to know that each vector in the gyroscope fixed basis satisfies the equation ##\dot{\mathbf{e}} = (\dot{\varphi} \hat{\mathbf{r}} + \boldsymbol{\Omega}) \times \mathbf{e}##.
