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For uniform circular motion of radius $R$ and angular speed $\omega$, the acceleration is directed toward the centre and has magnitude ${\omega }^{2}R=v^2/R$. Writing $\mathbf{r}$ for the position relative to the centre gives $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}$. The required resultant force is $m\ddot{\mathbf{r}}$; it may be supplied by a normal force, gravity, or another physical interaction.

For a weak force depending only on position, the amplitude average is a periodic total derivative and vanishes. The leading harmonic amplitude is constant, while the phase may drift. The method of multiple scales agrees with the exact conservation of energy for bounded orbits in the effective potential. A cubic position force on the right side of the oscillator equation lowers the frequency at first order when its coefficient is positive.

For any vector $\mathbf{a}$, differentiating its components and the rotating basis gives the rotating-frame derivative formulaApply it twice to the position $\mathbf{x}$, with constant angular velocity $\mathbf{\omega }$. The inertial acceleration iswhere dots denote rotating-frame derivatives. Newton's second law therefore becomesThe last terms are the Coriolis force and the force corresponding to centrifugal acceleration.

For the particle on a uniformly rotating inclined plane, choose orthonormal vectors ${\mathbf{e}}_1$ along the horizontal $\xi$ axis and ${\mathbf{e}}_2$ up the slope, and set ${\mathbf{e}}_3={\mathbf{e}}_1\times {\mathbf{e}}_2$, the upward normal vector. Then ${\mathbf{e}}_z=\sin \theta {\mathbf{e}}_2+\cos \theta {\mathbf{e}}_3$ andHere $N$ is the normal reaction. The marble is modeled as the smooth sliding particle specified by the printed equations, without an additional rolling constraint. The rotating-frame Coriolis acceleration and centrifugal acceleration are respectivelyThe normal rotating-frame acceleration is zero, so the normal reaction is not generally $mg\cos \theta$; it satisfiesProjecting the same equation onto the two tangential axes gives exactly

Multiply the equations by $\dot{\xi }$ and $\dot{\eta }$ and add. The Coriolis force terms cancel, as expected because this force does no work relative to the rotating frame. The normal reaction also does no work along the plane. The conserved quantity is rotating-frame kinetic energy plus gravitational and centrifugal potential energy:This need not be the conserved inertial energy, since the externally driven rotating plane can exchange energy with the particle.

For conservative forcing in averaged oscillator equations, $f=f(y)$, the average determining $R_T$ vanishes. If $F^{\prime }=f$, thenwhose period average is zero. Thus $R=R_0$ to this order, while a position-dependent force generally shifts the oscillation phase and frequency. The exact equation also conserves the energy $\frac{1}{2}{\dot{y}}^2+\frac{1}{2}{y}^2-\epsilon F(y)$, consistent with the absence of amplitude drift for bounded orbits.

For the cubic position force, $⟨f\cos \varphi ⟩=R^3⟨{\cos }^4\varphi ⟩=3R^3/8$. ThereforeThe leading oscillation is $R_0\cos [(1-3\epsilon R_0^2/8)t+{\theta }_0]$. For positive $\epsilon$ it is a softening cubic oscillator, obtained by weakly perturbing a harmonic oscillator. The weak-force condition is $\epsilon R_0^2\ll 1$; the constant $R$ is the leading harmonic amplitude, not a claim that the exact waveform remains sinusoidal.

Introduce the slow time $T=\epsilon t$ and treat $t,T$ independently in the method of multiple scales. WriteAt the next order,For fixed $R$ define the period average byHere $f$ inside the averages is evaluated at the leading position and velocity $R\cos \varphi ,-R\sin \varphi$. Orthogonality to both fundamental harmonics is the solvability condition in the method of multiple scales: it removes the resonant forcing that would otherwise generate a secular term. Since the squared sine and cosine averages are $1/2$, the amplitude-phase equations areEquivalently $\dot{R}=-\epsilon ⟨f\sin (t+\theta )⟩$ and $\dot{\theta }=-\epsilon ⟨f\cos (t+\theta )⟩/R$ at first order. These equations describe a bounded, weakly perturbed oscillation over $t=O({\epsilon }^{-1})$ while the amplitude remains in a range where the expansion is ordered. At zero amplitude the oscillation phase coordinate is singular; Cartesian harmonic coefficients or an equilibrium analysis should replace it.

Use an Earth-centred rotating reference frame, with Earth's constant angular velocity vector $\mathbf{\omega }$. The vectors $\mathbf{r}$, $\mathbf{v}$, and $\mathbf{a}$ are the particle's position, velocity, and acceleration measured in that rotating frame. The vector $\mathbf{g}$ is the gravitational acceleration due to the actual gravitational force, expressed in the same axes; it excludes the separate centrifugal acceleration. The remaining term $-2\mathbf{\omega }\times \mathbf{v}$ is the Coriolis acceleration. A constant rotation axis and an unaccelerated Earth-centred origin account for the absence of an Euler term or a translational-origin term.

For the size estimates, Earth's angular speed isWith drop height $H=20\mathrm{m}$, unperturbed free fall gives the fall timeand final speed $gT=20{\mathrm{m}\mathrm{s}}^{-1}$. Thus the gravitational term has size $10{\mathrm{m}\mathrm{s}}^{-2}$, the Coriolis acceleration is at most about $2\omega gT\approx 2.8\times 10^{-3}{\mathrm{m}\mathrm{s}}^{-2}$, and the centrifugal acceleration is at most ${\omega }^2{R}_0\approx 2.9\times 10^{-2}{\mathrm{m}\mathrm{s}}^{-2}$. These are characteristic sizes $10^1$, $10^{-3}$, and $10^{-2}$ in acceleration units; geometric factors can reduce them, and the Coriolis term is zero at release.

For a suitable approximation to the rotational deflection of a falling body, take $\mathbf{g}$ constant over the short fall, start at rest relative to Earth, and use the zeroth-order velocity ${\mathbf{v}}^{(0)}(t)=\mathbf{g}t$ inside the small Coriolis term. In the centrifugal term replace $\mathbf{r}$ by ${\mathbf{R}}_0$, because $H/R_0\ll 1$. Integrating the resulting acceleration twice, with initial position ${\mathbf{r}}_{\mathrm{i}}$, givesSince ${\mathbf{r}}_{\mathrm{i}}+\mathbf{g}{T}^2/2={\mathbf{R}}_0$, evaluating at the unperturbed fall time yields the printed first-iterate estimate

The approximation keeps the leading Coriolis contribution and the leading centrifugal contribution enhanced by Earth's large radius. Feeding the Coriolis velocity correction back into the equation produces terms of size ${\omega }^{2}g{T}^4$, smaller than the retained centrifugal displacement by a factor of order $H/R_0$. Replacing $\mathbf{r}$ by ${\mathbf{R}}_0$ in that term has the same relative error. Here $\omega T\approx 1.4\times 10^{-4}$ and $H/R_0\approx 3.3\times 10^{-6}$, so both approximations are well controlled. Strictly, the displayed vector is the position at time $T$: its small upward component is removed by the slightly longer actual fall time. That arrival-time correction changes the retained horizontal displacements only at higher order. Thus the horizontal components of the printed estimate give the landing point on the locally flat surface; its vertical component should not be interpreted as a landing above the ground.

Let ${\hat{\mathbf{e}}}_E,{\hat{\mathbf{e}}}_N,{\hat{\mathbf{e}}}_U$ point east, north, and vertically upwards. They form a right-handed basis, with ${\hat{\mathbf{e}}}_N\times {\hat{\mathbf{e}}}_U={\hat{\mathbf{e}}}_E$. At latitude $\lambda$, in the spherical-Earth approximation,The required cross products areAt $\lambda =45^{\circ }$, the northerly and easterly displacements are thereforeThe negative northerly component means a southward displacement. With the supplied values, these are approximately $2.9\mathrm{cm}$ south and $1.3\mathrm{mm}$ east. The eastward effect comes from Coriolis acceleration, while the leading southward effect here comes from centrifugal acceleration.

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.

Assume the particle masses $m_i$ are fixed and their total mass $M={\sum }_i{m}_i$ is positive. The centre of mass and the relative position vectors areTheir weighted sum is zero, so differentiation gives ${\sum }_i{m}_i{\dot{\mathbf{y}}}_i=0$. Expanding the kinetic energy with ${\dot{\mathbf{x}}}_i=\dot{\mathbf{R}}+{\dot{\mathbf{y}}}_i$ makes the mixed term vanish:This kinetic-energy decomposition about the centre of mass separates translation from internal motion without assuming any particular force law.

For the rigid body, its relative velocity is ${\dot{\mathbf{y}}}_i=\omega \mathbf{n}\times {\mathbf{y}}_i$. The cross product identity for a unit vector gives $|\mathbf{n}\times {\mathbf{y}}_i{|}^2=|{\mathbf{y}}_i{|}^2-(\mathbf{n}\cdot {\mathbf{y}}_i{)}^2$, the squared perpendicular distance to the rotation axis. Hence the moment of inertia of the particle is $I_i=m_i[|{\mathbf{y}}_i{|}^2-(\mathbf{n}\cdot {\mathbf{y}}_i{)}^2]$, and summing the rotational kinetic energy yields

The Lorentz force and Newton's second law giveWriting $\Omega =qB/m$, the component equations areThe initial velocity therefore givesand integration with the initial position givesThis is helical motion in a uniform magnetic field: a helix of radius $v/\Omega$ around the line $x=1+v/\Omega$, $y=0$, with its axis parallel to the magnetic field.