Intuition What this page is forThe foundations gave us the two fictitious-force formulas. The visual walkthrough showed us what they look like. Now we grind through every kind of case a problem can hand you — every sign of the cross product, every quadrant of the carousel, the degenerate "standing still" case, the limiting "far from axis" case, and a real weather problem. When you finish, no exam scenario should feel new.
Before anything, let me re-anchor the two tools we keep reusing, in plain words:
Every problem in this topic is one (or a blend) of the cells below. Each worked example is tagged with the cell it covers.
Cell
Case class
Covered by
C1 Centrifugal, position-only, standard magnitude
Ex 1
C2 Coriolis, velocity radially outward (sign of z ^ × r ^
Ex 2
C3 Coriolis, velocity tangential (sign of z ^ × θ ^
Ex 3
C4 Degenerate : object at rest in frame (v rot = 0 Ex 4
C5 Both forces at once , adding as vectors, and comparing sizesEx 5
C6 Limiting behaviour : on the axis (r ⊥ → 0 Ex 6
C7 Real-world word problem : Earth, hemisphere sign, accumulation over timeEx 7
C8 Exam twist : velocity along the axis (v ∥ ω Ex 8
We fix a convention used throughout: the spin axis is z ^ ω = ω z ^ ω > 0 counter-clockwise seen from above . On the flat turntable we use polar directions r ^ θ ^ r ^
z ^ × r ^ = θ ^ , z ^ × θ ^ = − r ^ , r ^ × θ ^ = z ^ .
Look at the figure: r ^ θ ^ z ^ z ^ r ^ θ ^
Worked example Sitting on a carouselYou sit at r ⊥ = 2 m ω = 1.5 rad/s m = 60 kg
Forecast: bigger radius or faster spin → bigger fling. Guess the number before reading on.
Step 1 — pick the right tool. No velocity relative to the seat is mentioned (you sit still), so only centrifugal matters. Its magnitude is position-only:
F cf = m ω 2 r ⊥ . Why this step? Coriolis needs v rot = 0
Step 2 — plug in.
F cf = 60 × ( 1.5 ) 2 × 2 = 60 × 2.25 × 2 = 270 N . Why this step? Direct substitution; units kg ⋅ ( rad/s ) 2 ⋅ m = kg⋅m/s 2 = N
Step 3 — interpret the real force. In the inertial frame there is no outward force. The seat pushes you inward with 270 N centripetal force m ω 2 r ⊥ feels from inside.
Verify: 60 ⋅ 2.25 = 135 135 ⋅ 2 = 270 N
Worked example Throwing outwardOn the same carousel (ω = 1.5 z ^ straight outward (r ^ v = 5 m/s m = 0.2 kg
Forecast: which way does the ball seem to curve — with the spin or against it?
Step 1 — write velocity in frame directions. v rot = 5 r ^ Why this step? Coriolis cares only about the object's velocity as measured inside the rotating frame , and radial-out is exactly r ^
Step 2 — form the cross product.
F Cor = − 2 m ω × v rot = − 2 ( 0.2 ) ( 1.5 ) ( z ^ × 5 r ^ ) . z ^ × r ^ = θ ^ F Cor = − 2 ( 0.2 ) ( 1.5 ) ( 5 ) θ ^ = − 3.0 θ ^ N . Why this step? The cross product rotates the velocity direction by 90°, turning "outward" into "sideways." The minus sign then flips it to − θ ^ opposite to the spin direction .
Step 3 — picture it. From the rider's seat the ball veers backward against the spin. From outside, the ball flies straight; the rider (and turntable) rotate under it.
Verify: 2 ⋅ 0.2 ⋅ 1.5 ⋅ 5 = 3.0 N − θ ^ r ^ F Cor ⋅ v rot = 0
Worked example Moving sideways with the spinSame carousel. Now a mouse runs tangentially, in the spin direction (+ θ ^ v = 4 m/s m = 0.1 kg
Forecast: last time radial motion pushed sideways . Tangential motion should push… where?
Step 1 — velocity in frame. v rot = 4 θ ^ Why this step? Running along the spin means the velocity points along θ ^
Step 2 — cross product with the other handedness. Use z ^ × θ ^ = − r ^ F Cor = − 2 ( 0.1 ) ( 1.5 ) ( z ^ × 4 θ ^ ) = − 2 ( 0.1 ) ( 1.5 ) ( 4 ) ( − r ^ ) = + 1.2 r ^ N . Why this step? This is the other sign case in the matrix. Tangential motion produces a radial Coriolis push — here outward (+ r ^ with the spin, effectively speeding up its circling and needing more inward force than the frame provides.
Step 3 — reverse the direction, flip the sign (sanity of C3). If the mouse ran against the spin (− θ ^ − 1.2 r ^ v
Verify: 2 ⋅ 0.1 ⋅ 1.5 ⋅ 4 = 1.2 N r ^ θ ^
Worked example Standing stillA 70 kg stands still on the carousel at r ⊥ = 3 m ω = 1.5 rad/s
Forecast: does anything Coriolis happen when you don't move?
Step 1 — Coriolis. v rot = 0 F Cor = − 2 m ω × 0 = 0. Why this step? Coriolis is proportional to velocity inside the frame . Zero velocity ⇒ zero Coriolis. This is the single most common student error to avoid.
Step 2 — Centrifugal still acts.
F cf = m ω 2 r ⊥ = 70 × ( 1.5 ) 2 × 3 = 70 × 2.25 × 3 = 472.5 N , outward . Why this step? Centrifugal depends only on position , not velocity — it never switches off while you sit on the spinning platform.
Verify: 70 ⋅ 2.25 = 157.5 157.5 ⋅ 3 = 472.5 N = 0 = 0
Worked example Which one dominates?A 0.5 kg radially outward at v = 6 m/s r ⊥ = 1 m ω = 2 rad/s
Forecast: at 1 2
Step 1 — centrifugal (radial, outward, + r ^
F cf = m ω 2 r ⊥ = 0.5 × ( 2 ) 2 × 1 = 2.0 N ( + r ^ ) . Why this step? Position-based, always outward along r ^
Step 2 — Coriolis (radial-out velocity → tangential force). With v rot = 6 r ^ z ^ × r ^ = θ ^ F Cor = − 2 ( 0.5 ) ( 2 ) ( 6 ) θ ^ = − 12 θ ^ N . Why this step? Same C2 machinery as Ex 2; here it's a much bigger number because ω v
Step 3 — add as perpendicular vectors. + 2.0 r ^ − 12 θ ^ ∣ F total ∣ = 2. 0 2 + 1 2 2 = 4 + 144 = 148 ≈ 12.17 N . Why this step? r ^ ⊥ θ ^
Verify: 2 ⋅ 0.5 ⋅ 2 ⋅ 6 = 12 4 + 144 = 148 ≈ 12.166
Worked example The edges of the formulaA bug crawls radially outward at v = 1 m/s across a turntable with ω = 3 rad/s m = 0.01 kg on the axis r ⊥ = 0 r ⊥ = 10 m
Forecast: which force disappears at the centre, and which one is the same at both places?
Step 1 — centrifugal at r ⊥ = 0
F cf = m ω 2 ⋅ 0 = 0. Why this step? Centrifugal is ∝ r ⊥ no outward direction, so it vanishes. This is the "on the axis" limiting cell.
Step 2 — centrifugal at r ⊥ = 10 m
F cf = 0.01 × 9 × 10 = 0.9 N . Why this step? Centrifugal grows linearly with distance — far out it can be large even for a small mass.
Step 3 — Coriolis at both places.
F Cor = 2 mω v = 2 ( 0.01 ) ( 3 ) ( 1 ) = 0.06 N ( same at r ⊥ = 0 and 10 m ) . Why this step? Coriolis has no r ⊥ — it depends on velocity, not position. So it is unchanged as the bug moves outward. That contrast (centrifugal grows, Coriolis constant) is the whole point of this cell.
Verify: 0.01 ⋅ 9 ⋅ 10 = 0.9 N 2 ⋅ 0.01 ⋅ 3 ⋅ 1 = 0.06 N = 0
Worked example Why Northern-Hemisphere storms spin counter-clockwiseAir flows toward a low-pressure centre at v = 10 m/s ω ⊕ = 7.3 × 1 0 − 5 rad/s local vertical component of ω ω up out of the ground in the Northern Hemisphere. Find the Coriolis acceleration and explain the sense of rotation. (See Coriolis Effect in Weather .)
Forecast: does the inflowing air bend to its left or right?
Step 1 — Coriolis acceleration magnitude.
a Cor = 2 ω ⊕ v = 2 ( 7.3 × 1 0 − 5 ) ( 10 ) = 1.46 × 1 0 − 3 m/s 2 . Why this step? We want the per-mass effect (acceleration), so divide the force formula by m a Cor = 2 ω v
Step 2 — direction. With ω − 2 ω × v right . Right-deflected inflow curls the incoming streams into a counter-clockwise swirl around the centre.
Why this step? Right-hand rule on z ^ × v
Step 3 — why "tiny" still wins. Over a distance L = 100 km = 1 0 5 m t = L / v = 1 0 5 /10 = 1 0 4 s Δ v = a Cor t = 1.46 × 1 0 − 3 × 1 0 4 = 14.6 m/s , Why this step? Coriolis is weak locally but accumulates over large distances and long times — that is the 80/20 idea behind Coriolis Effect in Weather and the Foucault Pendulum .
Verify: 2 ⋅ 7.3 × 1 0 − 5 ⋅ 10 = 1.46 × 1 0 − 3 1.46 × 1 0 − 3 ⋅ 1 0 4 = 14.6 m/s
Worked example The trap questionA rocket moves straight up along the rotation axis , v rot = v z ^ v = 200 m/s ω = 0.8 z ^ m = 50 kg r ⊥ = 0
Forecast: it's moving fast — surely Coriolis is huge?
Step 1 — cross product of parallel vectors.
F Cor = − 2 m ω × v rot = − 2 ( 50 ) ( 0.8 ) ( z ^ × 200 z ^ ) . z ^ × z ^ = 0 F Cor = 0. Why this step? The cross product magnitude carries sin ( angle ) parallel to ω = 0 sin 0 = 0 no Coriolis , no matter how fast. Only the component of v perpendicular to the axis produces Coriolis.
Step 2 — centrifugal here. On the axis r ⊥ = 0 F cf = m ω 2 r ⊥ = 0 Why this step? Both fictitious forces happen to vanish in this configuration — the "gotcha" is that a big velocity number tempts you into a big answer.
Verify: z ^ × z ^ = 0 ⇒ F Cor = 0 r ⊥ = 0 ⇒ F cf = 0
Recall Quick self-test across the matrixCoriolis is zero when the object is at rest in the frame ::: True (Ex 4) — v rot = 0 ω sin 0 = 0 r ⊥ = 0 r ⊥ v
