5Rotating frames

IA Dynamics and Relativity

5.3 The Coriolis force

The Coriolis force is a more subtle force. Writing

v

=

dr

dt

S

0

, we can write the

force as

F = −2mω ×v.

Note that this has the same form as the Lorentz force caused by a magnetic

field, and is velocity-dependent. However, unlike the effects of a magnetic field,

particles do not go around in circles in a rotating frame, since we also have the

centrifugal force in play.

Since this force is always perpendicular to the velocity, it does no work.

Consider motion parallel to the Earth’s surface. We only care about the

effect of the Coriolis force on the horizontal trajectory, and ignore the vertical

component that is tiny compared to gravity.

So we only take the vertical component of

ω

,

ω sin λ

ˆ

z

. The horizontal velocity

v = v

x

ˆ

x + v

y

ˆ

y generates a horizontal Coriolis force:

−2mω sin λ

ˆ

z × v = −2mω sin λ(v

y

ˆ

x − v

x

ˆ

y).

In the Northern hemisphere (0

< λ < π/

2), this causes a deflection towards

the right. In the Southern Hemisphere, the deflection is to the left. The effect

vanishes at the equator.

Note that only the horizontal effect of horizontal motion vanishes at the

equator. The vertical effects or those caused by vertical motion still exist.

Example.

Suppose a ball is dropped from a tower of height

h

at the equator.

Where does it land?

In the rotating frame,

¨

r = g − 2ω ×

˙

r − ω × (ω × r).

We work to first order in ω. Then

¨

r = g − 2ω ×

˙

r + O(ω

2

).

Integrate wrt t to obtain

˙

r = gt − 2ω × (r − r

0

) + O(ω

2

),

where

r

0

is the initial position. We substitute into the original equation to obtain

¨

r = g − 2ω × gt + O(ω

2

).

(where some new

ω

2

terms are thrown into

O

(

ω

2

)). We integrate twice to obtain

r = r

0

+

1

2

gt

2

−

1

3

ω × gt

3

+ O(ω

2

).

In components, we have g = (0, 0, −g), ω = (0, ω, 0) and r

0

= (0, 0, R + h). So

r =

1

3

ωgt

3

, 0, R + h −

1

2

gt

2

+ O(ω

2

).

So the particle hits the ground at

t

=

p

2h/g

, and its eastward displacement is

1

3

wg

2h

g

3/2

.

This can be understood in terms of angular momentum conservation in the

non-rotating frame. At the beginning, the particle has the same angular velocity

with the Earth. As it falls towards the Earth, to maintain the same angular

momentum, the angular velocity has to increase to compensate for the decreased

radius. So it spins faster than the Earth and drifts towards the East, relative to

the Earth.

Example.

Consider a pendulum that is free to swing in any plane, e.g. a weight

on a string. At the North pole, it will swing in a plane that is fixed in an inertial

frame, while the Earth rotates beneath it. From the perspective of the rotating

frame, the plane of the pendulum rotates backwards. This can be explained as a

result of the Coriolis force.

In general, at latitude λ, the plane rotates rightwards with period

1 day

sin λ

.