Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics IV / 6

All time derivatives in this part are taken relative to the rotating reference frame, so its basis is held fixed during differentiation. Put $f(\varphi )=p(\varphi )\sin \varphi$ and $q(\varphi )=p(\varphi )\cos \varphi$. The chain rule and product rule giveExplicitly,The separate Coriolis force and centrifugal force terms in the stated equation account for the motion of the basis relative to an inertial observer.

Continue with $f=p\sin \varphi$ and $q=p\cos \varphi$. The tangent vector to the hoop is ${\mathbf{r}}_{\varphi }=f^{\prime }{\mathbf{e}}_1-q^{\prime }{\mathbf{e}}_3$. Smooth contact means that the hoop's normal force has zero dot product with this tangent vector. The remaining applied force is $-mg{\mathbf{e}}_3$, whose dot product with ${\mathbf{r}}_{\varphi }$ is $mg{q}^{\prime }$.

Since $\mathbf{\omega }=\omega {\mathbf{e}}_3$, the Coriolis force is perpendicular to the hoop's plane and hence to ${\mathbf{r}}_{\varphi }$. AlsoTake the dot product of the rotating-frame Newton's second law with ${\mathbf{r}}_{\varphi }$, substitute a, and divide by $m$:DefineThen $K^{\prime }/2=f^{\prime }{f}^{\prime \prime }+q^{\prime }{q}^{\prime \prime }$ and $J^{\prime }=-{\omega }^{2}f{f}^{\prime }-g{q}^{\prime }$, soExpanding the squares makes the cancellation of the mixed terms explicit:The strict inequality follows from $p>0$. Here $J$ is the rotating-frame effective potential per unit mass. Multiplication of the equation by $\dot{\varphi }$ gives the useful first integral $\frac{1}{2}K(\varphi ){\dot{\varphi }}^2+J(\varphi )=\text{constant}$.

An equilibrium point has $\dot{\varphi }=0$ and $\boxed{J^{\prime }({\varphi }_0)=0}$. Since $K({\varphi }_0)>0$, the linearized displacement $\psi =\varphi -{\varphi }_0$ satisfiesThus $J^{\prime \prime }({\varphi }_0)>0$ gives a stable equilibrium, while $J^{\prime \prime }({\varphi }_0)<0$ gives an unstable equilibrium point. More generally, a strict local minimum of $J$ is sufficient for stability by the conserved first integral in b. When $J^{\prime \prime }({\varphi }_0)=0$, the second-derivative test alone is inconclusive, and the local shape of $J$ must be examined.

For the specified ellipse, put $D=1+e\cos \varphi >0$. Direct differentiation givesConsequently, with $\eta =g/{\omega }^2$ and $\omega \ne 0$,Two equilibria always occur at $\boxed{\varphi =0,\pi (\mathrm{mod}2\pi )}$. Any further equilibria must satisfyTo decide exactly when this is possible, view the equation as $\eta =(c+e)/(1+ec)$ with $c=\cos \varphi$. Its derivative $(1-e^2)/(1+ec{)}^2$ is positive on $[-1,1]$, and it takes endpoint values $-1$ and $1$. Since $g>0$ implies $\eta >0$, the complete list is:

These are positions of the bead at rest relative to the rotating hoop. No stability classification of the particular ellipse is needed.