Equation of motion in a rotating frame

Suppose a rotating reference frame shares its origin with an inertial frame and rotates at constant angular velocity $\mathbf{\Omega }$. Let $\mathbf{x}$ be the relative position, with dots denoting derivatives of its rotating components. The two accelerations, expressed in the same axes, satisfyApply the derivative rule $(d\mathbf{A}/dt{)}_{\mathrm{inertial}}=(d\mathbf{A}/dt{)}_{\mathrm{rotating}}+\mathbf{\Omega }\times \mathbf{A}$ twice to the position vector to derive this relation. Newton's second law therefore gives the rotating-frame equationHere $\mathbf{F}$ contains the actual interaction forces; the extra terms are the Coriolis force and centrifugal force.^[ref]

For example, an object at rest on a rotating table still needs an inward interaction force to maintain its circular motion; in rotating coordinates it balances the outward centrifugal force. If the origin has acceleration ${\mathbf{a}}_O$, add $-m{\mathbf{a}}_O$ to the right side. If the angular velocity varies, also add the Euler force $-m\dot{\mathbf{\Omega }}\times \mathbf{x}$. These terms cannot be omitted merely because the object is instantaneously at rest in the rotating axes.

For constant density and a conservative force per mass $-\nabla \Phi$, gravity and centrifugal acceleration can be incorporated into the displayed reduced pressure. The resulting Euler equations in rotating axes are ${\mathbf{u}}_t+(\mathbf{u}\cdot \nabla )\mathbf{u}+2\mathbf{\Omega }\times \mathbf{u}=-\nabla P/\rho$. The sign of the centrifugal contribution follows from $\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{x})=-\nabla |\mathbf{\Omega }\times \mathbf{x}{|}^2/2$. Constant density is what permits combination into one pressure gradient without extra density-gradient terms.

For an isotropic harmonic oscillator with inertial frequency $2\omega$, viewed in a frame rotating at $\omega$, initial rotating coordinates $(1,0)$ and zero relative velocity produce the displayed astroid trajectory. The complex coordinate satisfies $\ddot{\zeta }+2i\omega \dot{\zeta }+3{\omega }^2\zeta =0$, with solution $\zeta =\frac{3}{4}{e}^{i\omega t}+\frac{1}{4}{e}^{-3i\omega t}$. The relative speed is $(3|\omega |/2)|\sin 2\omega t|$, attaining its first maximum at $\pi /(4|\omega |)$ for nonzero $\omega$.