 Solution

The circular constraint isIts 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 givesBecause the ramp is translation-invariant along $y$, $N_y=0$; using $\dot{x}=R\dot{\theta }\cos \theta$ gives

For rest in the rotating frame, $\dot{\theta }=\dot{y}=\ddot{\theta }=\ddot{y}=0$. Assuming $\Omega \ne 0$, the second equation gives $y=0$, and the first givesOn the semicircular ramp the rest points are thereforeand, when ${\Omega }^{2}R\ge g$,