= Solution
The circular constraint is
$$
x=R\sin\theta,\qquad z=R\cos\theta.
$$
Its tangent in the $xz$-plane is $(\cos\theta,0,-\sin\theta)$, while the <normal force> is radial. Projecting the equations from part (a) onto this tangent eliminates the normal force and gives
$$
\boxed{
R\ddot\theta
=-g\sin\theta+\Omega^2R\cos\theta\sin\theta
-2\Omega\dot y\cos\theta}.
$$
Because the ramp is translation-invariant along $y$, $N_y=0$; using $\dot x=R\dot\theta\cos\theta$ gives
$$
\boxed{\ddot y=\Omega^2y+2\Omega R\dot\theta\cos\theta}.
$$
For rest in the rotating frame, $\dot\theta=\dot y=\ddot\theta=\ddot y=0$. Assuming $\Omega\ne0$, the second equation gives $y=0$, and the first gives
$$
\sin\theta(\Omega^2R\cos\theta-g)=0.
$$
On the semicircular ramp the rest points are therefore
$$
\boxed{\theta=0,\ y=0}
$$
and, when $\Omega^2R\geq g$,
$$
\boxed{\theta=\pm\arccos\frac{g}{\Omega^2R},\ y=0}.
$$
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