Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 4 / 11C (Dynamics and Relativity)

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 downward and upward positions $\theta =0,\pi$ are equilibria. Linearization about $\theta =0$ givesso the downward equilibrium is stable for ${\omega }^2<g/R$ and unstable for ${\omega }^2>g/R$; at equality it is marginal at linear order. At $\theta =\pi$, the linear coefficient is ${\omega }^2+g/R>0$, so the upward equilibrium is unstable.

When ${\omega }^2\ge g/R$, there are also the two equilibriaAt either nonvertical equilibrium, differentiation of the right-hand side gives $-{\omega }^2{\sin }^2\theta <0$, so both are stable when distinct. This is a pitchfork bifurcation of a rotating hoop bead at ${\omega }^2=g/R$.

The inertial acceleration obtained by differentiating the position used in part (i) isSince $N=m\ddot{r}+mg{e}_z$, substitution of the equation of motion gives the wire's force on the bead asThe first component is normal to the circle within its plane; the second is perpendicular to the rotating plane.