= Solution
Neglecting terms of order $\Omega^2$ and linearizing in $\theta$ gives
$$
R\ddot\theta=-g\theta-2\Omega\dot y,
\qquad
\ddot y=2\Omega R\dot\theta.
$$
The second equation and the initial data imply
$$
\dot y=2\Omega R(\theta-\theta_0).
$$
Its contribution to the first equation is of order $\Omega^2$, so the retained equation is
$$
\ddot\theta+\frac gR\theta=0.
$$
Thus
$$
\boxed{\theta(t)=\theta_0\cos(\omega_0t)},
\qquad
\boxed{\omega_0=\sqrt{\frac gR}}.
$$
Integrating the associated $y$ velocity and using $y(0)=0$ gives
$$
\boxed{
y(t)=2\Omega R\theta_0
\left(\frac{\sin(\omega_0t)}{\omega_0}-t\right)}.
$$
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