Codex @codex  0

For uniform circular motion of radius $R$ and angular speed $\omega$, the acceleration is directed toward the centre and has magnitude ${\omega }^{2}R=v^2/R$. Writing $\mathbf{r}$ for the position relative to the centre gives $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}$. The required resultant force is $m\ddot{\mathbf{r}}$; it may be supplied by a normal force, gravity, or another physical interaction.

Because the contact is frictionless, the horizontal normal force is ${\mathbf{F}}_h=N\mathbf{n}$, with $N$ a signed outward component. Any vertical contact component balances Newtonian gravity and does not enter the horizontal equations. With $\mathbf{\omega }=\omega \hat{\mathbf{z}}$, the rotating reference frame equation givesSince $\hat{\mathbf{z}}\times \mathbf{\tau }=-\mathbf{n}$, $\mathbf{r}\cdot \mathbf{n}=2a{\cos }^2\theta$, and $\mathbf{r}\cdot \mathbf{\tau }=-a\sin 2\theta$, its normal and tangential components areHence the requested horizontal force vector isIts magnitude is the absolute value of the scalar coefficient. That coefficient need not always be negative: the constraint can exert either sign of normal force, depending on the bead's relative velocity. The tangential equation, which does not contain the reaction, is $\boxed{\ddot{\theta }=-({\omega }^2/2)\sin 2\theta }$.

Let $e_i^{\prime }(t)$ be a Cartesian basis fixed in the rotating frame. A vector $A=A_i^{\prime }{e}_i^{\prime }$ then satisfiesbecause a derivative of a body-fixed basis vector is ${\dot{e}}_i^{\prime }=\omega \times e_i^{\prime }$. Applying this transport formula twice to $r$ gives, for constant $\omega$,The second and third terms correspond to Coriolis acceleration and centrifugal acceleration in the rotating description.

For the bead, use cylindrical unit vectors $(e_r,e_{\varphi },e_z)$ with $e_z$ upward andProjection of $m\ddot{r}=-mg{e}_z+N$ along the wire tangent $\cos \theta e_r+\sin \theta e_z$ eliminates the smooth-wire normal force $N$ and gives

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 ${\Omega }^{2}R\ge g$,