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The Coriolis acceleration is the Coriolis force divided by the particle's mass:It is therefore independent of mass for a specified relative velocity and angular velocity. This is one contribution to the acceleration measured in a rotating reference frame. For a fixed common origin with an inertial frame and constant $\mathbf{\Omega }$, the full equation of motion in a rotating frame isAll vectors here are expressed in the rotating axes; ${\mathbf{F}}_{\mathrm{real}}$ is the sum of actual forces. An accelerating origin adds a translational apparent force, and time-dependent angular velocity also adds the Euler acceleration.

The factor two follows directly from the rotating-frame derivative formula. First ${\mathbf{v}}_{\mathrm{lab}}={\dot{\mathbf{r}}}_{\mathrm{rot}}+\mathbf{\Omega }\times \mathbf{r}$. Differentiate again in the inertial frame: differentiating ${\dot{\mathbf{r}}}_{\mathrm{rot}}$ contributes one $\mathbf{\Omega }\times {\dot{\mathbf{r}}}_{\mathrm{rot}}$, and differentiating $\mathbf{\Omega }\times \mathbf{r}$ contributes another. Move both terms to the rotating side of Newton's second law to obtain the minus sign and the factor two. The centrifugal acceleration comes from the remaining double cross product.

A mass $m$ in a rotating reference frame with varying angular velocity experiences this apparent force. Dividing it by $m$ gives the Euler acceleration.

Use $D=(d/dt{)}_S$, $D^{\prime }=(d/dt{)}_{S^{\prime }}$, and ${\mathbf{v}}^{\prime }=D^{\prime }\mathbf{r}$. By the rotating-frame derivative formula, $D\mathbf{r}={\mathbf{v}}^{\prime }+\mathbf{\omega }\times \mathbf{r}$. Apply the product rule for the cross product and the same rotating-frame derivative formula again:We have $D{\mathbf{v}}^{\prime }=D^{\prime }{\mathbf{v}}^{\prime }+\mathbf{\omega }\times {\mathbf{v}}^{\prime }$ and $D\mathbf{\omega }=D^{\prime }\mathbf{\omega }$, since $\mathbf{\omega }\times \mathbf{\omega }=0$. ThereforeRearranging this identity into an equation of motion in a rotating frame produces the apparent Coriolis acceleration, Euler acceleration and centrifugal acceleration, with signs opposite to their corresponding terms on the right of this inertial acceleration identity. No translational term is needed because the two origins coincide.

Let $\Omega$ be the Earth's angular velocity magnitude. Use the local east, north and upward unit vectors ${\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. At latitude $\lambda$, the Earth's angular velocity and the train's ground-relative velocity areThus the east component of the Coriolis force isFor a north–south track, the train has no eastward acceleration relative to Earth. Its path curvature lies in the local north–vertical plane. The centrifugal acceleration and Newtonian gravity also lie in that plane, and the Earth's constant angular velocity gives no Euler acceleration. Hence the sideways force from the track must cancel the eastward Coriolis force:Its magnitude is $2m\Omega V\sin \lambda$, and its direction is west in the Northern hemisphere. The eastward tendency is the apparent Coriolis force; the requested track force has the opposite direction.

The rotating-frame derivative formula is $(d\mathbf{A}/dt{)}_S=(d\mathbf{A}/dt{)}_{S^{\prime }}+\mathbf{\omega }\times \mathbf{A}$. Applying it twice to the position vector, with constant angular velocity, givesSolving for ${\mathbf{a}}_{S^{\prime }}$ identifies the displayed terms: its left side is the acceleration relative to the rotating reference frame, the first right-hand term is the inertial acceleration, $-2\mathbf{\omega }\times {\mathbf{v}}_{S^{\prime }}$ is the Coriolis acceleration, and $-\mathbf{\omega }\times (\mathbf{\omega }\times \mathbf{r})$ is the outward centrifugal acceleration.

For a particle of mass $m$, replace ${\mathbf{a}}_S$ by the actual force divided by $m$ using Newton's second law. The remaining terms then appear as apparent forces in $S^{\prime }$. There is no Euler acceleration because $\mathbf{\omega }$ is constant and no translational acceleration term because the common origin is fixed in the inertial frame.

If $\mathbf{r}$ is measured in the rotating frame, its inertial velocity is $\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. The masses producing the gravitational field are stationary in this frame, so their potential energy is a time-independent function $V(\mathbf{r})$. Thus kinetic minus potential energy gives the stated LagrangianPut $\mathbf{v}=\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. ThenSubstitution in the Euler-Lagrange equation yieldsThe extra terms are the Coriolis acceleration, Euler acceleration, and centrifugal acceleration of a rotating reference frame.

The canonical momentum and Hamiltonian areAccordingly, Hamilton's equations are

If a frame rotates with angular velocity $\mathbf{\Omega }$ relative to an inertial frame, the derivatives of any vector $\mathbf{a}(t)$ satisfyApplying this formula twice to a position gives the Coriolis acceleration, Euler acceleration, and centrifugal acceleration terms.

The Technische Universität Darmstadt demonstration uses a steel ball launched outward from the centre of a rotating disc. A fixed lecture-room camera records a straight path, while a camera carried above the disc records a curved path. The instructions specify launching toward the fixed camera so that the laboratory motion is clearly visible. The rotating iPod transmits its view wirelessly through Apple TV; two vertical support rods of length $1\mathrm{m}$ are specified for the slow-motion arrangement. The paired video views are the readout: one physical motion is represented in two reference frames.^[ref]

A useful preparation control appears in the Universität Freiburg instructions: with the disc stationary, check that the ball travels radially toward its edge. A camera is mounted on supports attached to the disc; turn the disc by hand as the ball leaves its centre and observe the camera view.^[ref]

The inference is a dependence of the observed trajectory on the rotating axes; a curved camera trace alone does not isolate the Coriolis force. To compare with the constant-angular velocity free particle in rotating coordinates, measure the disc angular velocity, calibrate lengths in the camera image, and compare launches at rest and with opposite rotation directions. These are checks of the ideal model, not reported measurements from the manuals. A rolling ball can exchange forces with the surface through friction, and hand rotation need not have constant angular velocity; an Euler acceleration then enters too. The ideal free-ball drawing deliberately omits those effects. Neither manual supplies a numerical uncertainty or a precision measurement of the Coriolis acceleration.