All four inertial terms checked on explicit data, and the exact return to an inertial chart

Declare a one-label model of the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N=1$, mass $m=1\mathrm{k}\mathrm{g}$, time window $I_0=\mathbb{R}$, configuration domain $Q={\mathbb{R}}^3$ and the zero force map $F:I_0\times Q\times {\mathbb{R}}^3⟶{\mathbb{R}}^3,F(t,\mathbf{x},\mathbf{v})\equiv 0$ (admissible: constant, hence of the standing regularity), in a fixed inertial frame $F$ with adapted coordinates $(t,x)$ (Inertial frames, stationary observers and the adopted Galilean transformations between frames, Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). For real parameters ${\omega }_0,\beta ,\gamma$, vectors $a_0,v_c\in {\mathbb{R}}^3$ and Moving spatial charts of an inertial frame: rotating and translating coordinates, angular velocity and inertial terms data, take $\theta$ to be a dimensionless angle in radians, with $[{\omega }_0]={\tau }^{-1}$, $[\beta ]={\tau }^{-2}$, $[\gamma ]=\mathrm{\ell }{\tau }^{-2}$, $[a_{0,j}]=\mathrm{\ell }$ and $[v_{c,j}]=\mathrm{\ell }{\tau }^{-1}$ for each spatial component $j$: $\theta (t):={\omega }_{0}t+\frac{1}{2}\beta t^2,R(t):=\left(\begin{array}{ccc}\cos \theta (t)& -\sin \theta (t)& 0\\ \sin \theta (t)& \cos \theta (t)& 0\\ 0& 0& 1\end{array}\right),c(t):=a_0+v_{c}t+\frac{1}{2}\gamma t^2{e}_1,$ consider the motion $x(t):=x_0+vt$ of the zero-force model with $x_0=(1,0,0)\mathrm{m}$, $v=(0,1,0)\mathrm{m}{\mathrm{s}}^{-1}$, its moving-chart coordinate curve $y(t)=R(t{)}^{\mathsf{T}}(x(t)-c(t))$, and the inertial-terms field $\tau (t,y,\dot{y})=-(2\omega (t)\times \dot{y}+{\omega }^{\mathrm{\prime }}(t)\times y+\omega (t)\times (\omega (t)\times y)+R(t{)}^{\mathsf{T}}{c}^{\mathrm{\prime }\mathrm{\prime }}(t))$.

Then, for the parameter values ${\omega }_0=2{\mathrm{s}}^{-1}$, $\beta =1{\mathrm{s}}^{-2}$, $\gamma =3\mathrm{m}{\mathrm{s}}^{-2}$ and $a_0=v_c=0$:

1. $R(t)\in SO(3)$ is $C^{\mathrm{\infty }}$ and $\mathrm{\Omega }=R^{\mathsf{T}}{R}^{\mathrm{\prime }}$ has the angular velocity $\omega (t)={\theta }^{\mathrm{\prime }}(t)e_3=(2+t)e_3$ and ${\omega }^{\mathrm{\prime }}(t)=e_3$, while $c^{\mathrm{\prime }\mathrm{\prime }}(t)=(3,0,0)$;
2. $y(0)=e_1$, $\dot{y}(0)=-e_2$, and the moving-chart coordinate acceleration is $\ddot{y}(0)=-3e_1-e_2$ (in units $\mathrm{m}{\mathrm{s}}^{-2}$); the inertial-chart acceleration is $\ddot{x}(t)=0$ for every $t$;
3. the four inertial terms at $t=0$ are $-2\omega \times \dot{y}=-4e_1$, $-{\omega }^{\mathrm{\prime }}\times y=-e_2$, $-\omega \times (\omega \times y)=+4e_1$ and $-R^{\mathsf{T}}{c}^{\mathrm{\prime }\mathrm{\prime }}=-3e_1$, and their sum is $\tau (0,y(0),\dot{y}(0))=-3e_1-e_2=\ddot{y}(0)$; clause 1 of The equation of motion in a moving chart: inertial terms, inertial forces, and when it restricts to an inertial chart therefore holds at $t=0$ for these data, with $m\ddot{y}=R^{\mathsf{T}}F+m\tau =m\tau$;
4. because $\omega \equiv ̸0$, clause 3 of that theorem identifies the chart as not inertial; and with the alternative parameters ${\omega }_0=\beta =\gamma =0$, $a_0=(1,0,0)$, $v_c=(0,1,0)$, the chart has constant rotation $R\equiv I_3$ and affine reference curve, hence vanishing inertial terms, and the same motion has $\ddot{y}\equiv 0$ in the chart, which is the inertial-frame description with $y(t)=x(t)-a_0-v_{c}t$.

All values are model values in the declared coordinates; the example reports no measurement and asserts nothing about any apparatus.

• Reading the terms. On this data the Coriolis term and the centrifugal term cancel ($-4e_1+4e_1=0$) and the Euler and translational terms add, so the example is a check of the individual terms rather than of any one of them. A motion with $\dot{y}$ not aligned as here separates the Coriolis contribution; the companion pendulum and bead examples do not use a moving chart at all.

• No equivalence principle. The non-inertial chart's terms are computed from its own motion data; nothing in the example compares them with a gravitational field, and the zero-force model remains force-free in every chart.