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For the separated stationary state, restore its time factor $e^{-iEt/\hslash }$. Away from radial nodes, the quantum phase is $S=k\varphi -Et/\hslash$, up to a constant $0$ or $\pi$ where the real radial function has fixed sign. The guidance equation in plane polar coordinates therefore gives the Bohmian circulation of an angular-momentum eigenstateThe direction is $\hat{\mathbf{\varphi }}$ for $k>0$ and $-\hat{\mathbf{\varphi }}$ for $k<0$; the velocity is zero for $k=0$. Each admissible trajectory is a circle:The orbital angular momentum along the trajectory is $m{r}_0{v}_{\varphi }=\hslash k$. The origin or any zero-amplitude circle is excluded from this local formula. For the real degenerate superposition constructed above, the spatial quantum phase is constant on each nodal sector, so its Bohmian mechanics velocity is instead zero. Thus the circular motion is a conclusion about the separated angular-momentum eigenstate, not an arbitrary energy eigenstate.

The rotational invariance of a central-potential Hamiltonian allows a simultaneous eigenstate of energy and the two-dimensional orbital angular momentum $L_z=-i\hslash {\partial }_{\varphi }$. For a separated wavefunction $F(r)G(\varphi )$, the Laplacian in polar coordinates givesThe $\varphi$ term must be a constant; write $G^{\prime \prime }/G=-k^2$. The angular eigenfunctions can be chosen as $e^{ik\varphi }$. Single-valuedness under $\varphi \mapsto \varphi +2\pi$ imposes $e^{2\pi ik}=1$, hence $k\in \mathbb{Z}$. The radial equation in the separation of a two-dimensional central-potential eigenstate isFor a real central potential and the usual real self-adjoint radial boundary conditions, the radial equation admits a basis of real solutions: real and imaginary parts of a complex solution obey the same equation and boundary conditions. Choose a real normalized radial eigenfunction. Since the plane area element in plane polar coordinates is $rdrd\varphi$, the normalization isThis establishes the intended separated simultaneous eigenstate form. It is not the form of every stationary state. The radial equation depends on $k^2$, so the $k$ and $-k$ sectors have the same energy. For $k\ge 1$, their normalized superpositionis a single-valued stationary state with that energy but is not a single angular exponential. Quantum degeneracy is precisely why separation of variables selects a convenient eigenstate basis rather than all vectors in an energy eigenspace. The printed assertion needs this qualification. The printed polar-coordinate aid also labels a gradient component tuple as a divergence; the gradient used below is the two-dimensional vector $\hat{\mathbf{r}}{\partial }_r+\hat{\mathbf{\varphi }}{r}^{-1}{\partial }_{\varphi }$.

The area element in plane polar coordinates is $rdrd\theta$. Thus the axial moment of inertia isFor the removed material, the angular width is $2\pi /13$, andIts moment of inertia is therefore $I_{\mathrm{cut}}=\pi {\rho }_0{a}^5/160$, leavingThe torque equation $I\dot{\Omega }=\tau$ gives constant angular acceleration, so the time to reach the prescribed angular speed from rest is

To find the centre of mass after removing material, let ${\mathbf{e}}_1$ point along the bisector of the missing sector in the disc. The original disc has zero first mass moment about its centre. The missing sector has zero transverse first moment by reflection symmetry, while its component along ${\mathbf{e}}_1$ isbecause the radial integral is $11a^4/192$. The remaining first moment is $-Q_{\mathrm{cut}}{\mathbf{e}}_1$. Dividing by the given mass $k{\rho }_0{a}^3$ gives the body-fixed centre of massIt lies away from the missing sector. In the inertial frame, let $t_{\ast }$ be the moment the applied torque stops and let $\psi ={\psi }_{\ast }+\Omega (t-t_{\ast })$ give the orientation of that sector's bisector. Then, with $d=11a\sin (\pi /13)/(96k)$,The centre of mass has inward centripetal acceleration of magnitude ${\Omega }^{2}d$. The required real force is the constraint reaction exerted by the fixed rod and its support. Its horizontal resultant is $k{\rho }_0{a}^3\ddot{\mathbf{R}}$; a zero applied axial torque does not imply a zero resultant force.

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.

For plane polar coordinates, $x=r\cos \theta$ and $y=r\sin \theta$. The chain rule givesTheir Euclidean norms are $\Vert {\partial }_r\Vert =1$ and $\Vert {\partial }_{\theta }\Vert =r$ for $r>0$.

Motion under a central force is planar. In plane polar coordinates, the radial component of the acceleration is $\ddot{r}-r{\dot{\theta }}^2$, soThe conserved specific angular momentum is $h=r^2\dot{\theta }$. Substitution of $\dot{\theta }=h/r^2$ yieldsThe last term is the centrifugal part of the radial effective potential.