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A particle of mass $m$ constrained to the cone $z=-r$ is joined through the apex by a light inextensible ideal string to a hanging particle of mass $\sqrt{2}m$. The cylindrical radius is $r>0$ and gravitational acceleration is $g$. If the total ideal string length is $s_0$, the second particle has vertical coordinate $z_2=-s_0+\sqrt{2}r$. The ideal string and normal reaction exert no azimuthal force, so conservation of angular momentum gives $r^2\dot{\theta }=h$.

The first particle has kinetic energy $m{\dot{r}}^2+m{h}^2/(2r^2)$, while the second has kinetic energy $\sqrt{2}m{\dot{r}}^2$. Their combined gravitational potential energy is $mgr$ plus a constant. Hence conservation of energy gives the displayed equation, with effective potential $U(r)=gr+h^2/(2r^2)$. For $h\ne 0$, $U$ is strictly convex and has its unique minimum at $r=(h^2/g{)}^{1/3}$. The energy level has two radial turning points unless the motion is a circular orbit at that minimum.

The model maintains a mechanical constraint on the surface. For unilateral contact with one side of the cone, a normal reaction computed with a fixed outward sign must also have the physically allowed sign. A finite length of ideal string additionally restricts $r<s_0/\sqrt{2}$ before the hanging particle reaches the hole.

Use $s$ for cylindrical radius, to distinguish it from spherical radius $r$. In cylindrical coordinates $(s,\varphi ,z)$, the azimuthal angle $\varphi$ is measured from positive $x$ in the horizontal plane:Here $0\le \varphi <2\pi$ and $z\in \mathbb{R}$.

In spherical polar coordinates $(r,\theta ,\varphi )$, take $\theta$ as the polar angle from positive $z$ and use the same azimuthal angle $\varphi$:The ranges are $r\ge 0$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. The radial direction has horizontal projection of length $r\sin \theta$, which explains the extra factor in $x,y$. The azimuth is undefined on the polar axis, and both angles are nonunique at the origin; these are coordinate degeneracies rather than geometric singularities.

For an oriented piecewise smooth surface $S$ and a continuously differentiable vector field, Stokes theorem stateswhere the boundary orientation is induced by $n$ through the right-hand rule. Choose the upward orientation of the elliptic paraboloid. Its band has cylindrical radius $1/2\le r\le 1$, and a parametrization isIts oriented surface element and scalar area element areThe requested sketch shows the open band, not a capped solid.

For the specified vector field, direct differentiation gives $\nabla \times B=(0,0,3x^2+3y^2)$. HenceTo check this by Stokes theorem, parametrize a counterclockwise circle of radius $\rho$ at height ${\rho }^2$ by $r_{\rho }(\phi )=(\rho \cos \phi ,\rho \sin \phi ,{\rho }^2)$. Along it $dz=0$, andUsing ${\int }_0^{2\pi }{\sin }^4\phi d\phi ={\int }_0^{2\pi }{\cos }^4\phi d\phi =3\pi /4$ gives $3\pi {\rho }^4/2$. The induced outer circle is counterclockwise as viewed from above, and the inner circle is clockwise. ThereforeThis verifies the direct surface integral and the Stokes flux through an annular paraboloid. Reversing the chosen surface orientation reverses the final sign.

Distinguish the cylindrical radius $R$ from the spherical radius $r=\sqrt{R^2+z^2}$. A Taylor expansion of the central gravitational potential for $|z|/R\ll 1$ givesThus the usual thin-disk expansion of the stellar gravitational potential isThe vertical acceleration is $-{\partial }_z\Phi =-{\Omega }_K^{2}z$. The first term can be omitted when considering only vertical forces at fixed $R$, but must be retained for radial dynamics.

Take $\gamma >1$ for the ideal gas. At fixed cylindrical radius, put $\mathcal{R}=k/(\mu m_p)$, where $k$ is the Boltzmann constant, $\mu$ the mean molecular weight and $m_p$ the proton mass. Also write $a=9\alpha \Omega /4$. Vertical hydrostatic equilibrium, thermal equilibrium and the ideal gas law requireConsequently, with $K=A/a$,Integrating from the midplane yieldsThe polytropic vertical structure in stellar gravity therefore has, for $|z|\le H$,whereThis is a polytropic equation of state $P\propto {\rho }^{1+1/\beta }$ with polytropic index $\beta$. For $|z|>H$, set $\rho =P=0$; temperature has no physical gas value in that vacuum. Here $H$ is the finite semi-thickness, rather than an exponential disk scale height.

At fixed cylindrical radius, let $w=u_z$ and use the continuity equation ${\partial }_t\rho +{\partial }_z(\rho w)=0$. For the similarity solution $z=\xi \eta (t)$ and $\rho =\tilde{\rho }(\xi )/\eta$, the continuity equation reduces toHere $\eta$ is dimensionless and $\xi$ labels the initial height. Midplane symmetry sets the integration constant to zero, so the flow is a homologous vertical motion of an astrophysical disk:The hydrostatic approximation and ideal gas law becomeThe material derivatives are $D_{t}P=\tilde{P}\dot{\eta }$ and $D_t\rho =-\tilde{\rho }\dot{\eta }/{\eta }^2$, so the energy equation givesSeparation of variables requires a positive constant $K$ such thatIntegrating with $\eta (0)=1$ gives the slow cooling of an astrophysical disk similarity factorA closed set of profile equations is thereforeThese have the same polytropic vertical structure in stellar gravity as part (b). Writingwith $P_0=\mathcal{R}{\rho }_0{T}_0$, fixesReturning to physical height, the explicit profiles and evolving semi-thickness for $|z|\le H(t)$ areandOutside this moving surface $\rho =P=0$. The surface density of a disk is conserved because $\rho dz=\tilde{\rho }d\xi$. The column cools and contracts without changing its scaled profile: its midplane temperature falls as ${\eta }^2$, its mass density rises as ${\eta }^{-1}$ and its pressure falls as $\eta$.

The value $T_0$ determines $H_0$, but the mass density normalization also enters $C$. If the initial column is specifically the heated equilibrium of part (b), switching off its heating givesOtherwise $P_0$, or equivalently the conserved column mass, is additional initial data. The contraction timescale grows with $1+Ct$, so sufficiently slow initial evolution remains slow relative to the fixed vertical dynamical timescale of a disk. This is an exact solution of the reduced hydrostatic approximation, rather than the full momentum equation: its omitted vertical acceleration is $\ddot{\eta }\xi$, with $\ddot{\eta }/(\eta {\Omega }^2)=2(1+\beta )C^2/[(1+2\beta {)}^2{\Omega }^2(1+Ct{)}^2]$, which remains small and decreases if $C\ll \Omega$.

PutFor an axisymmetric vector field depending only on the cylindrical radius $R$, the solenoidal vector field condition isHence $R{B}_R$ is constant. The hypothesis $B_R\to 0$ on the axis forcesThe azimuthal and axial components of $\nabla \times \mathbf{B}=q\mathbf{B}$ are thenEliminating $B_{\varphi }=-B_z^{\prime }/q$ gives the closed ordinary differential equationOnce $B_z$ is known, $B_{\varphi }=-B_z^{\prime }/q$ and $B_R=0$ determine the full cylindrical force-free magnetic field.

In terms of the positive cylindrical radius $r=-z$, the conserved mechanical energy becomesThe effective potential $U$ tends to infinity at both ends of $(0,\infty )$ for $v>0$, and $U^{\prime \prime }(r)=3a^2{v}^2/r^4>0$. Thus its two turning points enclose the allowed radial motion. At a turning point, $\dot{r}=0$, soBesides $r=a$, the positive root of a polynomial isthe other root of a polynomial is negative. The quadratic polynomial $Q(r)=2g{r}^2-v^{2}r-a{v}^2$ is negative at zero and has just one positive root of a polynomial. Since $Q(a)=2a(ga-v^2)$, we have $r_{\ast }<a$ when $v^2<ga$, and $r_{\ast }>a$ when $v^2>ga$.

Height is $z=-r$, so its maximum occurs at the smaller positive turning point. HenceIndeed, the initial vertical acceleration isBelow the critical speed the particle first rises to $z=-r_{\ast }$; above it the particle initially descends, and its initial height is already its greatest height. At $v=\sqrt{ga}$ the turning points merge and the particle follows a horizontal circular orbit at $z=-a$. For zero initial speed the radial conservation of energy gives motion to the apex level $z=0$; if the hole excludes the apex itself, zero is the supremum of the heights before arrival at the hole.

These are the heights under the maintained mechanical constraint used in part (a). A unilateral contact with only one side of a cone would additionally require checking the sign of the normal reaction; the ideal constrained equations alone do not impose that condition.

A thin disk has vertical semi-thickness $H$ much smaller than cylindrical radius $r$. Vertical gravity is then approximately linear in height and orbital dynamics is nearly planar.

At fixed cylindrical radius $R$ with $|z|\ll R$, the central gravitational potential is harmonic in height. The vertical epicyclic frequency equals ${\Omega }_K$ in this spherical potential.