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For a smooth horizontal circular hoop of radius $a$, rotated at constant angular velocity $\omega$ about a vertical axis through a rim point, measure $\theta$ at that pivot from the diameter toward the bead. In rotating Cartesian coordinate system coordinates, $\mathbf{r}=a(1+\cos 2\theta ,\sin 2\theta ,0)$. The tangential component of the centrifugal acceleration gives the displayed equation; the Coriolis acceleration is normal to the hoop. The farthest point $\theta =0$ is stable for $\omega \ne 0$, with small oscillation angular frequency $|\omega |$. The pivot itself is unstable. This geometry differs from a vertical rotating hoop and has no tangential gravitational term.

For constant density ${\rho }_0$ and body force ${\Omega }^2(3x-2qz){\mathbf{e}}_x$, dotting the incompressible flow momentum equation with ${\rho }_0\mathbf{u}$ gives the kinetic energy flux $(K+P)\mathbf{u}$ and source ${\rho }_0{\Omega }^2(3x-2qz)u_x$, where $K={\rho }_0|\mathbf{u}{|}^2/2$. Coriolis acceleration does no work. Absorbing ${\Phi }_t=-3{\Omega }^2{x}^2/2$ gives the displayed balance. For nonzero $q$, the remaining force has curl $-2{\Omega }^{2}q{\mathbf{e}}_y$, so it cannot be incorporated into a scalar mechanical potential. The model can exchange energy with its maintained vertical-shear background; at $q=0$ ordinary Jacobi energy in a shearing sheet conservation is recovered under zero boundary flux.

For ideal magnetohydrodynamics at constant mass density $\rho$, a shearing sheet with velocity $-q\Omega x{\mathbf{e}}_y+{\mathbf{v}}_h(z,t)$ and magnetic field ${\mathbf{B}}_h(z,t)+B_z{\mathbf{e}}_z$ obeysHorizontal invariance and vanishing vertical velocity eliminate nonlinear advection. These equations retain Coriolis acceleration, orbital shear and magnetic tension. The vertical pressure adjusts to balance gravity and magnetic pressure.

For constant isothermal sound speed, time-independent shearing-sheet tidal potential ${\Phi }_t$, and inviscid flow, define $E=\Sigma u^2/2+c_s^2\Sigma \ln (\Sigma /{\Sigma }_{\mathrm{ref}})+\Sigma {\Phi }_t$. The barotropic energy density and kinetic energy balances combine to give ${\partial }_{t}E+\nabla \cdot [(E+P)\mathbf{u}]=0$. Coriolis acceleration does no work. Integral conservation requires vanishing net boundary energy flux.

The time-independent particle Lagrangian in a shearing sheet has conserved rotating-frame energyThe velocity-linear Coriolis acceleration term cancels from this expression. Up to the reference-orbit constant, it is the second-order expansion of inertial specific orbital energy minus ${\Omega }_0$ times inertial specific angular momentum. Its negative radial tidal term allows inelastic collisions to lower the total energy while increasing the radial extent of a ring.

The material-boundary and vorticity conditions give $\alpha =S/[r(r-1)]$ and $\beta =Sr/(r-1)$. A pressureless particle subject to linear drag, Coriolis acceleration and tidal gravity has the drag-polynomial stability for all positive stopping rates with $K=2\Omega (2\Omega -S)$, $L=\Omega (\alpha +\beta -S)$ and $P=\alpha \beta$. In Keplerian shear, $K/{\Omega }^2=1$, $L/{\Omega }^2=3(r+1)/[2r(r-1)]$ and $P/{\Omega }^2=9/[4(r-1{)}^2]$. The exact criterion is $r\ge 3$, including the boundary $K=L>P$ at $r=3$. Attraction is local to the centre under the prescribed interior velocity field; a trajectory leaving the patch cannot be followed using that field alone.

Expand a unit-mass particle's Lagrangian about an axisymmetric circular orbit of radius $r_0$ and frequency ${\Omega }_0$, using $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Circular-orbit balance removes the linear radial term; total-time-derivative invariance of a Lagrangian removes $r_0{\Omega }_0\dot{y}$. To second order,Here $q$ is the orbital shear parameter and ${\Omega }_z$ is the vertical epicyclic frequency. The Euler-Lagrange equations retain the local Coriolis acceleration and tidal gravity.

A pressureless test particle feels tidal gravity, Coriolis acceleration and linear drag toward the prescribed gas velocity. Its planar equations areThere is no gas-pressure force on this test particle. For $(x,y)=(X,Y)e^{\lambda t}$, the coefficient matrix isNontrivial solutions require a zero determinant. Expansion givesEach root has a nonzero modal vector in the matrix kernel; real motions are obtained from conjugate pairs. Repeated roots may also supply polynomially weighted exponential solutions. Negative real parts make the centre linearly asymptotically attracting for trajectories governed by this interior linear flow.