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Angular acceleration is the time derivative of angular velocity, $\mathbf{\alpha }=d\mathbf{\omega }/dt$. For rotation about a fixed axis, its signed scalar component obeys $\tau =I\alpha$ when the axial moment of inertia is constant and $\tau$ is the total applied torque about that axis.

A transformed radial equation containing $L^2/(2q^2)$ becomes a full planar central-force orbit by choosing $d\bar{\theta }/d\tau =L/q^2$. For $d\tau /dt=(q/r)dq/dr$ and old angular velocity $L/r^2$, this gives $d\bar{\theta }/d\theta =(r/q)dq/dr$. Keeping the old angle generally fails to preserve the stated transformed angular momentum.

Angular speed is the nonnegative magnitude of angular velocity: $\omega =|\mathbf{\omega }|$. It measures the rate of rotation about an instantaneous axis without specifying its direction. For uniform circular motion of radius $R$, the ordinary speed is $v=R\omega$.

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.

If a vector $k$ is fixed in space while a body frame rotates with angular velocity $\omega$, its body-frame components satisfy

For angular velocity depending only on cylindrical radius, the centrifugal acceleration is a gradient. Steady barotropic balance is $\nabla p/\rho =-\nabla \Psi$, hence specific enthalpy satisfies $h(\rho )+\Psi =\text{constant}$. On an invertible equation-of-state branch, mass density and pressure are functions of $\Psi$. Cylindrical rotation is essential to this scalar centrifugal-potential representation.

If $e_i$ is fixed in a rigid body with angular velocity $\omega$, then its derivative in the fixed space frame is

For uniform angular velocity $\omega$, potential $\varphi$, and additional forces doing no work in the rotating frame,The Coriolis force does no work; the centrifugal potential accounts for the remaining apparent force. This energy is conserved precisely when the potential has no explicit time dependence in the rotating coordinates. An inertial time-independent conservative force need not meet that condition unless its potential is also invariant under the frame's rotation. The balance prevents applying a rotating-frame conservation law to a nonaxisymmetric stationary inertial potential.

For inner radius $a$, mean gap $\Delta \ll a$, fixed eccentricity parameter $|\alpha |<1$, gap $h=\Delta (1+\alpha \sin \theta )$, and inner angular velocity $\Omega$, the constant circumferential flux isIt follows by integrating the lubrication theory relation $p_s=6\mu \Omega a/h^2-12\mu q/h^3$ over a full period. The leading holding force is horizontal:Integration by parts computes this force without a closed-form pressure. The fluid couples about the separate cylinder axes obey $C_i+C_o=\alpha \Delta F_x$; moments about a common axis balance exactly at this order.

A freely translating inner axle with prescribed positive angular velocity $\Omega$ and downward load $F>0$ has full-film journal bearing force balance equationswhere $R_r,R_t$ are the squeeze resistance of an eccentric journal bearing. Rotation supplies tangential fluid force $R_t\Omega r/2$. The equilibrium lies horizontally to the right and satisfies $F=R_t\Omega r/2$. Linearization there has imaginary eigenvalues: the prescribed rotating axle continuously supplies energy, so viscous dissipation does not imply settling. For an initially concentric axle, putting $e=r/\Delta$ and $K=6\pi \mu \Omega a^3/{\Delta }^2$ gives the closed orbitThis follows by eliminating time and integrating $x_e+3ex/[2(1-e^2)]=\Omega {\Delta }^2{R}_{0}e/[2F(1-e^2{)}^{3/2}]$. It applies to the leading inertia-free full-film model; adding cavitation, constraints or finite-length leakage changes the prediction.

The inertia tensor about a point isIt maps angular velocity to angular momentum.

For a smooth plane at inclination $\theta$ rotating about a vertical axis at constant angular velocity $\omega$, choose rotating coordinates $\xi$ horizontally and $\eta$ uphill, with their cross product the upward normal vector. A sliding particle under gravity satisfiesThe normal reaction is $N=m(g\cos \theta -2\omega \dot{\xi }\sin \theta +{\omega }^2\eta \sin \theta \cos \theta )$ while contact is maintained. The Coriolis force does no work in the rotating frame, giving the conserved specific energyThe last term is the centrifugal potential; this quantity need not equal the inertial mechanical energy.