Introduction
Rotating reference frames produce two main fictitious forces: centrifugal (radially outward) and Coriolis (depends on velocity in rotating frame). These explain phenomena on Earth's surface and in rotating machinery. Understanding these forces is essential for geophysics, meteorology, and navigation.
Centrifugal Force
In frame rotating with angular velocity ω, centrifugal force on mass m at distance r from axis: F_cf = mω2r radially outward. Magnitude increases with distance from rotation axis. Adds to gravity, making effective gravity weaker at equator than poles. Independent of velocity in rotating frame - acts on all objects.
Coriolis Force
F_cor = -2m(ω × v') where v' is velocity in rotating frame. Perpendicular to both rotation axis and velocity. Deflects moving objects: right in Northern Hemisphere, left in Southern Hemisphere. Zero for objects at rest in rotating frame. Magnitude: 2mωv'sinθ where θ is angle between ω and v'.
\nIntroduction
Rotating reference frames produce two main fictitious forces: centrifugal (radially outward) and Coriolis (depends on velocity in rotating frame). These explain phenomena on Earth's surface and in rotating machinery. Understanding these forces is essential for geophysics, meteorology, and navigation.
Centrifugal Force
In frame rotating with angular velocity ω, centrifugal force on mass m at distance r from axis: F_cf = mω2r radially outward. Magnitude increases with distance from rotation axis. Adds to gravity, making effective gravity weaker at equator than poles. Independent of velocity in rotating frame - acts on all objects.
Coriolis Force
F_cor = -2m(ω × v') where v' is velocity in rotating frame. Perpendicular to both rotation axis and velocity. Deflects moving objects: right in Northern Hemisphere, left in Southern Hemisphere. Zero for objects at rest in rotating frame. Magnitude: 2mωv'sinθ where θ is angle between ω and v'.
\nCombined Effects on Earth
Effective gravity: g_eff = g - ω2Rcos2λ (λ is latitude). Centrifugal effect reduces g by about 0.3% at equator (about 0.034 m/s2). Coriolis deflects winds, creates trade winds, affects ocean currents, deflects artillery shells and falling objects. Important for long-range ballistics and weather patterns.
Foucault Pendulum
Pendulum's swing plane rotates due to Coriolis effect. Rotation rate: Ω = ωsinλ (λ is latitude). At poles: rotates once per day (clockwise in North, counterclockwise in South). At equator: no rotation. Demonstrates Earth's rotation directly. Period = 24 hours/sinλ. Famous demonstration in science museums.
Applications
Weather patterns: cyclones rotate counterclockwise in Northern Hemisphere due to Coriolis. Ocean currents: gyre circulation. Long-range targeting: artillery and missile trajectories must correct for Coriolis. Flight paths: aircraft routes affected by Earth rotation. Centrifuges: artificial gravity in space stations and separation technologies.
\nSolved Example: Coriolis Deflection
A projectile is fired horizontally northward at 500 m/s from latitude 45°N. Calculate the Coriolis acceleration and deflection after traveling 100 km. Earth: ω = 7.27×10^-5 rad/s. Solution: At 45°, horizontal component of ω is ωsin(45°) = 5.14×10^-5 rad/s (vertical component is ωcos45°). For horizontal northward motion, Coriolis force is to the east (right in Northern Hemisphere). Coriolis acceleration: a_c = 2ωv sinλ = 2 × 7.27×10^-5 × 500 × sin(45°) = 2 × 7.27×10^-5 × 500 × 0.707 = 0.0514 m/s2 eastward. Time to travel 100 km: t = distance/speed = 100,000/500 = 200 s. Deflection: d = 1/2a_c t2 = 1/2 × 0.0514 × (200)2 = 0.0257 × 40000 = 1028 m ≈ 1.03 km eastward. Significant deflection for long-range artillery. Snipers at shorter ranges (1 km) have smaller deflection (about 10 cm), still measurable. Deflection is to the right in Northern Hemisphere regardless of direction (except purely vertical).
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