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Let ${\mathbf{e}}_r,{\mathbf{e}}_{\varphi },{\mathbf{e}}_z$ be the cylindrical unit vectors. The surface has an upward normal vector proportional to ${\mathbf{e}}_z-h^{\prime }(r){\mathbf{e}}_r$. If $N_z$ is the vertical component of the normal force, its radial component is $-h^{\prime }(r)N_z$. In the stated approximation, vertical acceleration and the vertical driving-force component are neglected, so vertical balance gives $N_z\simeq Mg$. The horizontal contact force is consequently $-Mg{h}^{\prime }(r){\mathbf{e}}_r$. The projected velocity and its speed areThe driving force is parallel to this velocity, with components $F\dot{r}/u$ and $Fr\dot{\varphi }/u$. Use the acceleration in polar coordinates and Newton's second law to obtainThese equations apply where $u\ne 0$; at rest the direction of an exactly velocity-aligned drive is not determined.

For a prescribed path, write $r^{\prime }=dr/d\varphi$, $r^{\prime \prime }=d^{2}r/d{\varphi }^2$. The chain rule gives $\dot{r}=r^{\prime }\dot{\varphi }$ and $\ddot{r}=r^{\prime \prime }{\dot{\varphi }}^2+r^{\prime }\ddot{\varphi }$. Subtract $r^{\prime }/r$ times the azimuthal equation from the radial equation. The driving force and $\ddot{\varphi }$ both cancel:Thus the constraint for tangentially driven motion on an axisymmetric surface isThe denominator must give a nonnegative squared speed, and the calculation uses $r>0$. At the polar origin a limiting Cartesian description is appropriate.

For the paraboloid and Archimedean spiral, putThen, for $\varphi >0$,The branch tending to $\varphi =\infty$ has positive angular velocity, henceSince $u=ϵ\dot{\varphi }\sqrt{1+{\varphi }^2}$, solve the azimuthal equation for the required force:This is positive for $\varphi >0$, consistent with a forward drive. Taking the limit gives

These are the large-time limits of the printed projected equations. Their mechanical approximation has a precise range limitation: on the exact surface, $z=h(r)$ gives $\dot{z}=(2r/\ell )\dot{r}$, which along this solution grows like $2{ϵ}^2\alpha \varphi /\ell$. At fixed positive $ϵ$, it is not small as $\varphi \to \infty$. Thus the requested limits are formal limits of that reduced model; neglect of vertical motion is not a uniform approximation to the full three-dimensional motion all the way to infinity.

For the inverse-square potential $V(r)=-Q/r$, the central force is $-V^{\prime }(r)\hat{r}=-Q\hat{r}/r^2$. The transverse component of acceleration in polar coordinates is therefore zero:Thus $\boxed{h=r^2\dot{\theta }\text{ is constant}}$; for unit mass it is the signed angular momentum perpendicular to the orbital plane. The radial component givesConsequentlyThis is the mechanical energy, since the full kinetic energy is $\frac{1}{2}({\dot{r}}^2+r^2{\dot{\theta }}^2)$ and $r^2{\dot{\theta }}^2=h^2/r^2$.

For $h\ne 0$ and $Q>0$, the effective potential goes to $+\infty$ as $r\downarrow 0$ and to zero from below as $r\to \infty$. Its derivative is $U^{\prime }(r)=(Qr-h^2)/r^3$, so it decreases up to $r_0=h^2/Q$ and then increases. Its zero and minimum areFor $Q<0$, $U=h^2/(2r^2)+|Q|/r$ is positive and strictly decreasing, with limits $+\infty$ at zero and $0^{+}$ at infinity. It has no minimum at a finite positive radius.

If $h=0$, the centrifugal term disappears. For $Q>0$ the graph is instead $-Q/r$, increasing from $-\infty$ to $0^{-}$ without a minimum; for $Q<0$ it remains positive and decreasing. The bounded-orbit calculation below uses the stipulated $h>0$, which excludes this radial collision case.

In plane polar coordinates, use ${\mathbf{e}}_r=(\cos \theta ,\sin \theta )$ and ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta )$. Their derivatives are ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$. Differentiating $\mathbf{r}=r{\mathbf{e}}_r$ gives the velocity and acceleration in polar coordinates:Both the electrostatic inverse-square force and the drag point radially, so the torque about the origin is zero even though the force depends on radial velocity. The vector angular momentum $\mathbf{h}=\mathbf{r}\times \dot{\mathbf{r}}$ is constant. For $\mathbf{h}\ne 0$, every position lies in the fixed plane perpendicular to it; if $\mathbf{h}=0$, the trajectory is radial and still lies in a plane. In the nonzero case the transverse equation is $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$, givingThe signed $h$ depends on the plane orientation. The stipulated ratio $k/|h|$ requires $h\ne 0$.

Set $u=1/r$ and use primes for differentiation with respect to $\theta$. Then $\dot{\theta }=h{u}^2$, $\dot{r}=-h{u}^{\prime }$ and $\ddot{r}=-h^2{u}^2{u}^{\prime \prime }$. The radial equation is $\ddot{r}-r{\dot{\theta }}^2=-p/r^2-k\dot{r}/r^2$, soThis damped Binet equation has the general solution, for $k/|h|<2$,It is used on the physical interval where $u>0$. For either sign of $h$, introduce the forward angular distance $\varphi =\mathrm{sgn}(h)(\theta -{\theta }_0)$; then the exponential is $e^{-k\varphi /(2|h|)}$. Thus damping acts forward in time for both orientations, not just when $h>0$.

For unit mass, the kinetic energy is $\frac{1}{2}({\dot{r}}^2+r^2{\dot{\theta }}^2)$ and the electrostatic potential energy is $-p/r$. ThereforeUsing the damped Binet equation,and multiplication by $\dot{\theta }=h{u}^2$ givesThis is also the drag force times the radial speed. Energy is nonincreasing in time, strictly decreasing while radial motion occurs; a circular trajectory already has constant energy. The sign of $dE/d\theta$ alone would be misleading for negative $h$.

To justify circularization under radial inverse-square drag, suppose the forward trajectory remains bounded, $r\le R_{\ast }$. Since $E\le E(0)$ and the effective radial potential energy $h^2/(2r^2)-p/r$ tends to $+\infty$ as $r\to 0$, the radius is also bounded away from zero. Velocity then stays bounded, so the solution continues for all forward time without collision. Moreover $\dot{\varphi }=|h|/r^2\ge |h|/R_{\ast }^2$, so $\varphi \to \infty$. The decaying general solution gives $u\to p/h^2$ and $u^{\prime }\to 0$. ConsequentlyThe tangential motion remains, so the limiting orbit is circular. The convergence is asymptotic; boundedness is needed because a physical solution could otherwise reach $u=0$ and escape rather than complete indefinitely many turns.

The example uses $p=h=1$, $k=0.6$ and $u(\theta )=1+0.35e^{-0.3\theta }\cos (\sqrt{0.91}\theta )$, which stays positive. Radial oscillations decay while angular momentum stays fixed.

A central force has no component in the angular direction. The angular component of the acceleration in polar coordinates therefore gives $2\dot{r}\dot{\theta }+r\ddot{\theta }=0$, orFor unit mass, $h=r^2\dot{\theta }$ is the signed angular momentum perpendicular to the orbital plane, so its magnitude $|h|$ is conserved. Equivalently, the torque $\mathbf{r}\times \mathbf{F}$ vanishes, which also fixes that plane.

The given potential energy is $k/r$, whose radial force is $k/r^2$: positive $k$ is repulsive and negative $k$ attractive. The kinetic energy in polar coordinates is $({\dot{r}}^2+r^2{\dot{\theta }}^2)/2$, so conservation of energy givesFor $h\ne 0$ and $k>0$, the effective potential is positive, strictly decreasing from $+\infty$ to $0$, and has no stationary point. For $k=-\kappa <0$, it still tends to $+\infty$ at zero, crosses zero at $r=h^2/(2\kappa )$, and has its unique minimum atIt then approaches zero from below. The minimum corresponds to a stable circular orbit. If $h=0$, the centrifugal barrier disappears; the attractive effective potential is simply $-\kappa /r$ and has no minimum.

The perpendicular unit vectors represented by $e^{i\theta }$ and $i{e}^{i\theta }$ point radially outwards and transversely in the increasing-angle direction. Reading off their real coefficients in part (b) gives the signed componentsThese are velocity in polar coordinates and acceleration in polar coordinates. In particular $r\dot{\theta }$ is a transverse speed component; $\dot{\theta }$ alone is an angular velocity. The moving basis, rather than any extra force assumption, accounts for the additional acceleration terms. At the origin the polar directions themselves are undefined, although the Cartesian velocity and acceleration can remain well defined.

Let $x$ be signed displacement along the rails, measured from the centre. The diagram in the original paper shows a guide extending across the centre, so the signed coordinate can pass through zero. Hooke's law gives the radial force $-kx$. In the rotating reference frame, the radial centrifugal force is $+m{\Omega }^{2}x$. The Coriolis force is perpendicular to the rails and is balanced by their reaction; it has no radial component. Therefore Newton's second law givesEquivalently, the radial component of acceleration in polar coordinates in the inertial frame is $\ddot{x}-{\Omega }^{2}x$. The radial spring oscillator on a rotating guide has simple harmonic motion only when $k/m-{\Omega }^2>0$, withThe angular frequency is in $\mathrm{rad}{\mathrm{s}}^{-1}$ and the frequency in $\mathrm{Hz}$. The critical angular speed isStrictly this is an upper bound, not an attained maximum for oscillations: at equality $\ddot{x}=0$, so a displaced object released from rest stays at its displaced position relative to the rails; above it, the central equilibrium is unstable.

A path along a circle can have variable areal velocity. As an illustrative controlled trajectory, keep a point mass on a circle of radius $R$ with angleIts acceleration in polar coordinates and the required net force arewhere ${\mathbf{e}}_r$ and ${\mathbf{e}}_{\theta }$ are the radial and transverse unit vectors. A circular guide can supply the radial force, while a tangential drive supplies $mR\alpha$. HenceFor $R=m={\omega }_0=\alpha =1$ in illustrative units, the three equal time intervals $[0,0.5]$, $[0.5,1]$ and $[1,1.5]$ sweep areas $0.3125$, $0.4375$ and $0.5625$. Their difference comes from the transverse force's positive torque, even though the radius and path shape stay fixed. The area under each areal velocity graph segment equals the corresponding swept area.

For a differentiable path in plane polar coordinates with $r>0$, the moving unit vectors are $\hat{\mathbf{r}}=(\cos \theta ,\sin \theta )$ and $\hat{\mathbf{\theta }}=(-\sin \theta ,\cos \theta )$. Differentiating the position $\mathbf{r}=r\hat{\mathbf{r}}$ and using $\dot{\hat{\mathbf{r}}}=\dot{\theta }\hat{\mathbf{\theta }}$ proves the displayed velocity. Its signed radial and transverse vector components are $\dot{r}$ and $r\dot{\theta }$. A second differentiation gives the acceleration in polar coordinates. The complex representatives of the two moving unit vectors are $e^{i\theta }$ and $i{e}^{i\theta }$.