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Coriolis & Centrifugal Forces Reference cheat sheet - grade 11-12

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Physics Grade 11-12

Coriolis & Centrifugal Forces Reference Cheat Sheet

A printable reference covering rotating-frame forces, Coriolis acceleration, centrifugal acceleration, Earth deflection rules, and applications for grades 11-12.

TopicsPhysicsForcesAccelerationRotationVectorsNewton's Laws
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This cheat sheet covers the apparent forces that appear when motion is described from a rotating frame, such as Earth or a spinning platform. Students need it because Coriolis and centrifugal effects are common in weather, ocean currents, satellites, and rotating machines. The reference helps connect vector formulas to the physical directions of the forces.

It also separates real inertial forces from forces that appear because the observer is rotating.

The core formulas use angular velocity Ω⃗\vec{\Omega}, position r⃗\vec{r}, and velocity relative to the rotating frame v⃗rot\vec{v}_{\text{rot}}. Coriolis acceleration is a⃗cor=−2Ω⃗×v⃗rot\vec{a}_{\text{cor}} = -2\vec{\Omega} \times \vec{v}_{\text{rot}}, and centrifugal acceleration is a⃗cent=−Ω⃗×(Ω⃗×r⃗)\vec{a}_{\text{cent}} = -\vec{\Omega} \times (\vec{\Omega} \times \vec{r}). On Earth, Coriolis effects depend on latitude ϕ\phi through f=2Ωsin⁡ϕf = 2\Omega \sin\phi.

In the Northern Hemisphere moving objects deflect to the right, while in the Southern Hemisphere they deflect to the left.

Key Facts

  • In a rotating frame, Newton's second law is written as ∑F⃗real+F⃗fict=ma⃗rot\sum \vec{F}_{\text{real}} + \vec{F}_{\text{fict}} = m\vec{a}_{\text{rot}}.
  • The Coriolis acceleration is a⃗cor=−2Ω⃗×v⃗rot\vec{a}_{\text{cor}} = -2\vec{\Omega} \times \vec{v}_{\text{rot}}, so it acts perpendicular to both the rotation axis and the object's rotating-frame velocity.
  • The Coriolis force is F⃗cor=ma⃗cor=−2mΩ⃗×v⃗rot\vec{F}_{\text{cor}} = m\vec{a}_{\text{cor}} = -2m\vec{\Omega} \times \vec{v}_{\text{rot}}.
  • The centrifugal acceleration is a⃗cent=−Ω⃗×(Ω⃗×r⃗)\vec{a}_{\text{cent}} = -\vec{\Omega} \times (\vec{\Omega} \times \vec{r}), and for distance r⊥r_{\perp} from the rotation axis its magnitude is acent=Ω2r⊥a_{\text{cent}} = \Omega^2 r_{\perp}.
  • The centrifugal force is F⃗cent=mΩ2r⊥ r^⊥\vec{F}_{\text{cent}} = m\Omega^2 r_{\perp}\,\hat{r}_{\perp}, directed outward from the rotation axis in the rotating frame.
  • Earth's Coriolis parameter is f=2Ωsin⁡ϕf = 2\Omega \sin\phi, where Ω≈7.29×10−5 rad/s\Omega \approx 7.29 \times 10^{-5}\ \text{rad/s} and ϕ\phi is latitude.
  • For horizontal motion on Earth, the Coriolis acceleration magnitude is acor=fv=2Ωvsin⁡ϕa_{\text{cor}} = fv = 2\Omega v\sin\phi.
  • Coriolis acceleration changes direction of motion but does no work when it remains perpendicular to velocity, since P=F⃗⋅v⃗=0P = \vec{F}\cdot\vec{v} = 0.

Vocabulary

Rotating frame
A reference frame that spins with angular velocity Ω⃗\vec{\Omega} relative to an inertial frame.
Coriolis force
An apparent force F⃗cor=−2mΩ⃗×v⃗rot\vec{F}_{\text{cor}} = -2m\vec{\Omega} \times \vec{v}_{\text{rot}} that acts on moving objects observed in a rotating frame.
Centrifugal force
An apparent outward force in a rotating frame with magnitude Fcent=mΩ2r⊥F_{\text{cent}} = m\Omega^2 r_{\perp}.
Angular velocity
A vector Ω⃗\vec{\Omega} that gives the rotation rate and points along the rotation axis by the right-hand rule.
Coriolis parameter
The latitude-dependent value f=2Ωsin⁡ϕf = 2\Omega\sin\phi used to describe horizontal Coriolis effects on Earth.
Inertial frame
A non-accelerating reference frame in which Newton's laws apply without adding fictitious forces.

Common Mistakes to Avoid

  • Treating Coriolis and centrifugal forces as real interaction forces is wrong because they arise from using a rotating reference frame, not from a physical contact or field.
  • Using acor=2Ωva_{\text{cor}} = 2\Omega v everywhere on Earth is wrong for horizontal motion because the correct magnitude is acor=2Ωvsin⁡ϕa_{\text{cor}} = 2\Omega v\sin\phi.
  • Reversing the deflection direction is wrong because horizontal motion deflects right in the Northern Hemisphere and left in the Southern Hemisphere.
  • Pointing centrifugal force away from the center of Earth is often wrong because it points outward from the rotation axis, not necessarily directly away from Earth's center.
  • Applying Coriolis force to an object at rest in the rotating frame is wrong because F⃗cor\vec{F}_{\text{cor}} depends on v⃗rot\vec{v}_{\text{rot}} and becomes 0⃗\vec{0} when v⃗rot=0⃗\vec{v}_{\text{rot}} = \vec{0}.

Practice Questions

  1. 1 A plane flies due north at v=250 m/sv = 250\ \text{m/s} at latitude ϕ=45∘\phi = 45^{\circ}. Using Ω=7.29×10−5 rad/s\Omega = 7.29 \times 10^{-5}\ \text{rad/s}, find the horizontal Coriolis acceleration magnitude acor=2Ωvsin⁡ϕa_{\text{cor}} = 2\Omega v\sin\phi.
  2. 2 A rider is r⊥=4.0 mr_{\perp} = 4.0\ \text{m} from the axis of a rotating platform with angular speed Ω=2.5 rad/s\Omega = 2.5\ \text{rad/s}. Find the centrifugal acceleration magnitude acent=Ω2r⊥a_{\text{cent}} = \Omega^2 r_{\perp}.
  3. 3 At latitude ϕ=30∘\phi = 30^{\circ}, calculate the Coriolis parameter f=2Ωsin⁡ϕf = 2\Omega\sin\phi using Ω=7.29×10−5 rad/s\Omega = 7.29 \times 10^{-5}\ \text{rad/s}.
  4. 4 Explain why a long-range projectile has a noticeable Coriolis deflection but a short classroom projectile usually does not.