Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV / 6

In cylindrical coordinates, $\theta =\omega t$ and $z=r^2/(2a)$. ThusThe acceleration isA tangent vector along the wire is $\mathbf{t}={\mathbf{e}}_r+(r/a){\mathbf{e}}_z$. Because the wire is smooth, its constraint force is orthogonal to $\mathbf{t}$. Taking the dot product of Newton's second law with $\mathbf{t}$ eliminates that force; gravitational weight contributes $-mgr/a$. HenceMultiplying by $a^2/m$ gives

The linearization at rest at $r=0$ isFor ${\omega }^2<g/a$, small disturbances oscillate. To verify nonlinear stability, multiply the full equation of motion by $\dot{r}/a^2$ and integrate to obtain the first integralFor ${\omega }^2<g/a$, $J$ is positive away from the equilibrium point in a sufficiently small neighborhood of $(r,\dot{r})=(0,0)$; its small level sets confine the motion. ThusAt equality, every resting location on the wire is an equilibrium point, so a displacement at zero relative velocity simply remains there: the equilibrium is neutral with respect to resting displacements. It is unstable with respect to general perturbations in phase space, however. An arbitrarily small outward initial velocity $v_0>0$ gives $\dot{r}=v_0/\sqrt{1+r^2/a^2}>0$ from the first integral, eventually taking the bead outside any prescribed small neighborhood. Thus strict phase space stability holds only for ${\omega }^2<g/a$.

Put $\lambda ={\omega }^2-g/a>0$. The zero value of the first integral in b givesAt the endpoint $r=a$, this gives ${\dot{r}}^2=\lambda a^2/2$. The absolute speed includes radial, azimuthal and vertical components. Since $\dot{z}=\dot{r}$ there,Hence the limiting exit speed isThere is a small initial condition qualification in the wording. The exact initial conditions $r=\dot{r}=0$ give the stationary solution forever, by Picard-Lindelöf theorem. The displayed zero-first integral moving branch instead approaches that unstable equilibrium as $t\to -\infty$.

For a precise small disturbance, release the bead with zero relative velocity at $r=ϵ>0$. Its first integral is $J=-\lambda {ϵ}^2/2$, givingThe bead initially accelerates outward and reaches the end. Letting $ϵ\to 0^{+}$ recovers exactly the stated exit speed. The conserved $J$ is a Jacobi integral associated with the imposed rotation; ordinary mechanical energy need not be conserved because the apparatus driving the wire can do work.