Rotating frames — centrifugal force, Coriolis force: Question bank

Before you start, two words you must not confuse:

Real force — a genuine push/pull from another object (a hand, a string, gravity, the floor). Survives in every frame.
Fictitious force — a bookkeeping term we add only because our frame is spinning; it has no source object. Vanishes the moment you switch to an inertial frame.

Figure — Rotating frames — centrifugal force, Coriolis force

True or false — justify

True? A body sitting still on a spinning carousel feels centrifugal force but zero Coriolis force.

True. Coriolis is

- 2 m ω \times v_{rot}

; if you are at rest in the rotating frame then

v_{rot} = 0

so Coriolis vanishes, while centrifugal

- m ω \times (ω \times r)

depends only on position and stays.

True? In an inertial (non-spinning) frame, centrifugal and Coriolis forces both equal zero.

True. They are artefacts of analysing motion inside a rotating frame; an inertial observer needs no extra terms because

F_{real} = m a

already works there.

True? The Coriolis force can speed a particle up.

False. It is always perpendicular to

v_{rot}

, so

F_{Cor} \cdot v_{rot} = 0

and it does zero work — it only bends the path, never changes the speed.

True? The centrifugal force can do work on a moving object.

True. It points radially outward, so if the object moves radially there is a component of force along the motion, giving nonzero work — unlike Coriolis it is not forced to be perpendicular to velocity.

True? Doubling the spin rate

ω

doubles the centrifugal force.

False. Centrifugal magnitude is

m ω^{2} r_{⊥}

(with

r_{⊥}

the perpendicular distance to the axis), so it scales with

ω^{2}

— doubling

ω

makes it four times larger.

True? Doubling the spin rate

ω

(with the same

v_{rot}

) doubles the Coriolis force.

True. Coriolis magnitude is

2 mω v_{rot} sin ϕ

(with

ϕ

the angle between

ω

and

v_{rot}

), linear in

ω

, so twice the spin gives twice the force.

True? Coriolis force in the Southern Hemisphere deflects horizontally-moving air to its left.

True. There

ω

points into the ground, flipping the sign of

- 2 m ω \times v

relative to the North, so cyclones spin clockwise instead of counter-clockwise.

True? At the equator, the Coriolis deflection of horizontal motion is zero.

Mostly true — the vertical component of

ω

vanishes at the equator, so horizontal winds get no horizontal Coriolis turning there; a horizontal

ω

component still deflects vertical motion (see the dropped-ball case).

True? The factor of 2 in the Coriolis force is just a convention that could be absorbed into

ω

False. It arises physically from two distinct

ω \times v_{rot}

terms in the acceleration expansion; absorbing it would break the centrifugal term, which carries a single

ω

-squared.

Spot the error

"You feel thrown outward on a turn, so a real outward force pushes you." — where's the error?

There is no real outward force. The only real force is the door/seat pushing you inward; your body's inertia wants to continue straight, so relative to the turning car you drift outward and blame a fictitious "centrifugal" force.

"Coriolis force needs the object to be at a large radius, since it's a rotation effect." — error?

Coriolis

- 2 m ω \times v_{rot}

contains no

r

at all; it depends on velocity in the frame, not position. An object at the very centre still feels Coriolis if it is moving there.

"An object at rest in the rotating frame feels no fictitious forces, since it isn't moving." — error?

It still feels centrifugal force, which depends on position not velocity. Only the Coriolis term switches off when

v_{rot} = 0

"Since Coriolis is perpendicular to velocity, it curves the path but centrifugal (radial) never curves anything." — error?

Centrifugal does alter motion — it pushes radially outward and changes the radial part of the trajectory. "Curving without changing speed" is the special property of Coriolis, not a statement that centrifugal is inert.

F_{cf} = - m ω \times (ω \times r)

points along

r

, so it always points straight away from the origin." — error?

It points away from the rotation axis, not the origin. Only the component of

r

perpendicular to

ω

matters, giving magnitude

m ω^{2} r_{⊥}

where

r_{⊥}

is the perpendicular distance to the axis.

"We invented fictitious forces to fix Newton's second law, so they violate it." — error?

The opposite — they restore

F = m a

inside a non-inertial frame by supplying exactly the terms the frame's own acceleration removed. Nothing is violated; we just bookkeep the frame's motion as forces.

Why questions

Why is there a centrifugal (outward) force but the real physics of circular motion needs a centripetal (inward) force?

In the inertial frame you need a real inward (centripetal) force to bend the path into a circle; in the rotating frame, that same requirement appears as an outward fictitious centrifugal force balancing the inward real force so the object seems to stand still.

Why does the Coriolis force do no work while the centrifugal force can?

Coriolis is built as

ω \times v_{rot}

, hence always

⊥ v_{rot}

, killing

F \cdot v

. Centrifugal points radially and has no such perpendicularity constraint, so it can share a direction with the velocity.

Why is the Coriolis effect negligible for a thrown baseball on Earth but decisive for hurricanes?

The acceleration

2 ω_{\oplus} v

is tiny per second (

\sim 1 0^{- 3} m/s^{2}

), so over a ball's short flight it's imperceptible; over the hours and hundreds of kilometres of a weather system it accumulates into a dominant sideways turning.

Why do we take

ω

constant when deriving the two-force result, and what changes if it isn't?

Constant

ω

(

\dot{ω} = 0

) drops a term; if the frame's spin rate changes, a third fictitious force

- m \dot{ω} \times r

(the Euler force) appears — a tangential kick you feel when a carousel speeds up or slows down.

Why does the transport theorem produce a cross product

ω \times A

rather than a plain multiple of

A

Rigid rotation only reorients a vector, sweeping its tip along a circle; the instantaneous rate of that sideways sweep is exactly

ω \times A

, which is perpendicular to

A

— a plain scalar multiple would wrongly stretch the vector along itself.

Why is the Foucault Pendulum slow plane-rotation evidence for Earth's spin rather than gravity?

Its swing plane turns because Coriolis nudges the bob sideways each swing; only a rotating frame produces that steady precession, so its measurable turning rate is a direct fingerprint of

ω_{\oplus}

Edge cases

v_{rot}

is exactly parallel to

ω

(motion straight along the spin axis), what is the Coriolis force?

Zero, because

ω \times v_{rot} = 0

when the vectors are parallel (

sin ϕ = 0

) — motion purely along the axis feels no Coriolis deflection.

If a particle sits exactly on the rotation axis (

r_{⊥} = 0

), what is the centrifugal force?

Zero, since

F_{cf}

has magnitude

m ω^{2} r_{⊥}

; on the axis there is no perpendicular distance to be flung out from.

At the North Pole, a puck slides horizontally on frictionless ice — describe its Coriolis behaviour.

There

ω

is vertical, so any horizontal velocity gets the full Coriolis deflection to the right, curving the puck into a clockwise circle (an inertial circle) as seen from the rotating Earth.

What happens to the Coriolis force if you reverse the direction of

v_{rot}

It flips sign, because

F_{Cor} = - 2 m ω \times v_{rot}

is linear in

v

— reversing your motion reverses the sideways push, which is why outgoing and incoming air deflect oppositely.

In the limit

ω \to 0

(frame stops spinning), what happens to both fictitious forces?

Both vanish — Coriolis linearly and centrifugal quadratically in

ω

— recovering the ordinary inertial-frame Newton's law

F_{real} = m a

, exactly as expected.

For a ball dropped from a tower at the equator, why does it land slightly east rather than straight down?

Its downward velocity crosses with Earth's spin so

- 2 m ω \times v

points east; the fictitious deflection nudges the fall eastward of the plumb line, a real, measured effect.

Recall One-line summary you should be able to recite

Centrifugal = position-based, outward from the axis, $\propto ω^{2} r_{⊥}$ , can do work; Coriolis = velocity-based, perpendicular to motion, $\propto ω v$ , does no work — and both vanish in an inertial frame.