core33 min studyLesson 117 in Physics

Coriolis deflection depends on relative velocity

Predict Coriolis directions with a cross product, estimate short-time deflection, and distinguish coordinate curvature from an energy source.

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What you’ll understand

  • Compute Coriolis components in a clearly oriented rotating frame.
  • Explain why the Coriolis term does no instantaneous work on relative motion.
  • State the short-time approximation behind a constant-deflection estimate.

A moving object viewed from a rotating platform can curve sideways even when an inertial observer sees a simpler trajectory. The Coriolis term describes the velocity-dependent part of this coordinate effect. Its sign follows an oriented cross product, not a universal instruction that motion bends right. Define the rotation axis, the observer's axes, and the relative velocity first. These definitions make the same method work on a turntable, Earth, or another rotating reference frame.

Use an oriented basis to determine the sign

Take a horizontal platform rotating counterclockwise as seen from above, so its angular velocity points along positive z. Let x point right and y point upward on the platform diagram, making a right-handed Cartesian basis. An object moving along positive x relative to the platform has angular velocity crossed into relative velocity along positive y. The minus sign and factor two in the Coriolis expression therefore give an acceleration along negative y. Reversing either the platform rotation or the relative velocity reverses this acceleration. Reversing both preserves it.

At a given instant, the Coriolis direction is perpendicular to the relative velocity and rotation axis. A velocity parallel to the rotation axis has zero Coriolis acceleration, even though the frame rotates. The factor two arises because a moving object's displacement and the rotating basis both contribute when differentiating acceleration; it is not a convention that can be dropped. When resolving components, distinguish angular speed in radians per second from rotation frequency in cycles per second. Replacing angular speed by frequency without the factor of two pi creates a quantitative error even when the direction is correct.

aCor=−2Ω×v′,Ω=Ωz^ ⟹ (ax,ay)=(2Ωvy′,−2Ωvx′)\mathbf a_{\mathrm{Cor}}=-2\boldsymbol\Omega\times\mathbf v',\qquad \boldsymbol\Omega=\Omega\hat{\mathbf z}\ \Longrightarrow\ (a_x,a_y)=(2\Omega v_y',-2\Omega v_x')
The displayed components use right-handed x, y, z axes and relative velocity v′. Coriolis acceleration is measured in m/s² when Ω is in rad/s and velocity in m/s; it is one term in the full rotating-frame equation.↔ Scroll wide equations horizontally.

A perpendicular term changes direction without supplying power

The dot product of relative velocity with its Coriolis acceleration vanishes identically. Consequently the Coriolis term alone does no instantaneous work in the rotating-frame kinetic-energy balance. This does not imply that every object's speed is constant in a rotating frame. Gravity, contact forces, the centrifugal term, or a changing frame can alter relative kinetic energy. Energy accounting must include the full equation and any suitable effective potential. A bent path is not by itself evidence that a sideways coordinate term supplies or removes mechanical energy.

For a short interval, hold the leading relative velocity approximately constant and estimate the Coriolis contribution to transverse displacement using constant acceleration. The approximation requires the accumulated direction change to be small, typically angular speed times duration much less than one. If the launch is near the rotation axis, the centrifugal term starts small, but it develops as the object moves away. Its leading displacement correction can then occur at higher order in this short-time expansion. At longer times, integrate the full rotating equation or transform an inertial trajectory; do not extend the quadratic transverse estimate indefinitely.

v′⋅aCor=0,ΔyCor≈−Ωv0x′t2(∣Ωt∣≪1)\mathbf v'\cdot\mathbf a_{\mathrm{Cor}}=0,\qquad \Delta y_{\mathrm{Cor}}\approx-\Omega v_{0x}'t^2\quad (|\Omega t|\ll1)
The displacement estimate assumes initial motion along +x, rotation along +z, and treats the leading velocity as constant over a short interval. It is not an exact trajectory or a replacement for other forces and frame terms.↔ Scroll wide equations horizontally.

Earth's geometry adds a latitude dependence

For local horizontal motion on Earth, resolve its rotation vector into local vertical and horizontal parts. The vertical component is angular speed times the sine of latitude. Keeping only this component gives the familiar horizontal Coriolis acceleration perpendicular to horizontal velocity. It deflects toward the right of motion in the Northern Hemisphere and left in the Southern Hemisphere. Those statements require the local horizontal approximation; the full three-dimensional effect also contains vertical couplings. At the equator the simplified horizontal deflection vanishes, but the complete Coriolis vector does not vanish for every possible velocity.

Compare the relevant time and length scales before including or omitting Earth's rotation. A ball crossing a classroom in a fraction of a second usually accumulates a tiny rotation-induced deflection compared with launch uncertainty and air resistance. A large-scale flow evolving for many hours can be strongly affected. Observing a curved path alone does not isolate Coriolis acceleration: pressure gradients, contact constraints, or aerodynamic forces may dominate. A controlled platform experiment can reverse rotation while preserving launch conditions, providing a sign-reversal test that helps separate a rotation-dependent effect from a fixed laboratory bias.

Estimate a short turntable deflection

A small object starts at a turntable's axis with relative velocity 2 m s−12\,\mathrm{m\,s^{-1}} along +x. The platform rotates counterclockwise at 0.1 rad s−10.1\,\mathrm{rad\,s^{-1}}. Estimate the leading Coriolis contribution to y displacement after 0.5 s0.5\,\mathrm s, neglecting horizontal interactions.

  1. Rotation points along +z, so initial Coriolis acceleration is along −y with magnitude 2Ωv=2(0.1)(2)=0.4 m s−22\Omega v=2(0.1)(2)=0.4\,\mathrm{m\,s^{-2}}.

  2. The short-time estimate gives ΔyCor≈12(−0.4)(0.5)2=−0.05 m\Delta y_{\mathrm{Cor}}\approx\tfrac12(-0.4)(0.5)^2=-0.05\,\mathrm m. The predicted deflection is 5 cm toward −y.

  3. Check the expansion parameter: Ωt=0.05\Omega t=0.05, which is small. Centrifugal and velocity-change corrections enter the full trajectory, so the estimate is leading-order rather than an exact 5 cm prediction.

The result

The leading transverse displacement is −0.05 m. Reversing platform rotation reverses this leading sign while the outward centrifugal contribution does not reverse.

Explore the model

Interactive lab

Interactive lab · fields

Inspect the perpendicular rotation operator

Set source strength q to zero and rotation Ω to 1, then −1. Predict the arrow reversal at the same point before checking. The displayed field is Ω(−y,x), perpendicular to the position vector. If that input vector is interpreted abstractly as velocity, the same rotation operator with coefficient −2Ω gives the planar Coriolis direction. This is a vector-field analogy, not a rotating-frame trajectory solver.

Inspect the perpendicular rotation operatorThe softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Current readings: Fx at (1, 0): 0.833; Fy at (1, 0): 0.5; Field magnitude there: 0.972. All parameters are adjustable with the labeled controls below.−3.2−1.601.63.2−1.5−0.7500.751.5xyq = 1
Local field directionSource strength
Fx at (1, 0)
0.833
Fy at (1, 0)
0.5
Field magnitude there
0.972
t = 0 s

The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe planar cross product rotates a vector by a right angle; signs and factors still come from the physical equation.

Check your understanding

Which velocity gives zero Coriolis acceleration in a frame rotating about a fixed z axis?

Try it yourself

A frame rotates with Ω=0.2z^ rad s−1\boldsymbol\Omega=0.2\hat{\mathbf z}\,\mathrm{rad\,s^{-1}}. An object's instantaneous relative velocity is 3x^+4y^ m s−13\hat{\mathbf x}+4\hat{\mathbf y}\,\mathrm{m\,s^{-1}}. Find the Coriolis acceleration and verify its instantaneous power per unit mass.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the planar component formula, then take the dot product with the relative velocity.

Compare with the solution

The acceleration is (2Ωvy′,−2Ωvx′)=(1.6,−1.2) m s−2(2\Omega v_y',-2\Omega v_x')=(1.6,-1.2)\,\mathrm{m\,s^{-2}}. Its dot product with velocity is 3(1.6)+4(−1.2)=0 m2 s−33(1.6)+4(-1.2)=0\,\mathrm{m^2\,s^{-3}}. The Coriolis term therefore contributes no instantaneous kinetic-energy change, although other terms in the full equation may do so.

Take it with you

  • Coriolis acceleration follows −2Ω × v′ with explicitly oriented relative coordinates.
  • The Coriolis term is perpendicular to relative velocity and supplies no instantaneous power.
  • Latitude rules and short-time deflections are qualified approximations to the full frame transformation.

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