Chapter 3

Non-inertial Reference
Frames and Rotating
Coordinate Systems

3.1 Time Derivatives in a Rotating Frame

Consider a vector A of fixed length, rotating about the origin with constant
velocity ω. The rate of change of A is:
dA
=ω×A (3.1)
dt
Let us now consider two different frames: an inertial frame O and a rotating
frame O′ . Let î, ĵ, and k̂ be orthogonal unit vectors of O and let î′ , ĵ′ , and
k̂′ be orthogonal unit vectors of O′ . The unit vectors in the rotating frame
rotate rigidly with O′ such that:

dî′ dĵ′ dk̂′

= ω × î′ , = ω × ĵ′ , = ω × k̂′ . (3.2)
dt dt dt
An arbitrary vector a can be resolved into components of O or O′ :

a = ai î + aj ĵ + ak k̂ = a′i î′ + a′j ĵ′ + a′k k̂′ (3.3)

Differentiating with respect to time, we obtain:

da dai daj dak
= î + ĵ + k̂
dt dt dt dt
da′i ′ da′j ′ da′k ′
= î + ĵ + k̂ + a′i ω × î′ + a′j ω × ĵ′ + a′k ω × k̂′ (3.4)
dt dt dt
where we used Eq. 3.2 to re-write the time dependence of the basis vectors
in O′ in the last three terms.

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Let us now introduce some new notation:

Da da′ da′j ′ da′k ′

= i î′ + ĵ + k̂ (3.5)
Dt dt dt dt
We can interpret Da/Dt as the rate of change of a measured in the rotating
frame. The total rate of change corresponds to:
da Da
= + (ω × a) (3.6)
dt Dt
There is one term for the rate of change with respect to the rotating axes
and a second term arising from the change (rotation) of the axes themselves.

3.2 Newton’s Second Law and the Equation of Mo-

tion in a Rotating Frame
We are now ready to find the equation of motion of a particle when it is
measured in a frame rotating at constant angular velocity ω. Let r be the
position vector of the particle. Using Eq. 3.6 to differentiate:
dr Dr
= +ω×r (3.7)
dt Dt
Differentiating a second time:
( )
d2 r d Dr
= +ω×r
dt2 dt Dt
( )
D2 r Dr Dr
= +ω× +ω× +ω×r
Dt2 Dt Dt
D2 r Dr
= 2
+ 2ω × + ω × (ω × r) (3.8)
Dt Dt
Recalling that Newton’s second law is Ftot = md2 r/dt2 , where Ftot is the
total force acting on the particle, the equation of motion in the rotating
frame becomes:
D2 r Dr
m 2
= Ftot − 2mω × − mω × (ω × r) (3.9)
Dt Dt
The last two terms are apparent forces that arise only because we are mea-
suring positions with respect to axes that are themselves rotating (corre-
sponding to constant acceleration). The apparent forces are also known as
inertial or fictitious forces.

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 w

Equator

Figure 3.1: Coordinates for describing motion near the Earth’s surface. The
displacement of a particle x is measured from the tip of a rotating vector
R, which extends from the center of the Earth to a point on the Earth’s
surface.

3.3 Motion Near the Earth’s Surface

Let us consider the apparent forces on a particle near the Earth’s surface.
For simplicity, let us assume that the Earth is spherically symmetric so that
the weight of an object is a vector directed down towards the center of the
Earth. Let O be an inertial frame with origin at the Earth’s center and O′
be a frame rotating with the Earth at angular velocity ω, also with origin
at the Earth’s center. The total force on the particle is its weight mg plus
any other external forces F.

Let us decompose the position vector of the particle into two parts. First
we can define a vector R extending from the center of the Earth to some
point on its surface, as shown in Fig. 3.1. Then let x be a vector connecting
the tip of R to the particle. The position of the particle is then:

r=R+x (3.10)

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Note that since R is fixed in the rotating frame O′ , DR/Dt = 0 and also
D2 R/Dt2 = 0. The equation of motion (Eq. 3.9) is then:

D2 x Dx
m 2
= F + mg − 2mω × − mω × (ω × (R + x)) (3.11)
Dt Dt
To proceed we will now make an approximation that x ≪ R (note that the
radius of the Earth is around 6400km so this is a good approximation even
if the particle moves a few kilometers). With this approximation, the last
term simplifies:
ω × (ω × (R + x)) ≈ ω × (ω × R) (3.12)
We can also re-write the term involving g:
GM GM R
g=− 3 (R + x) ≈ − R3 R = −g R (3.13)
|R + x|

With these approximations, the equation of motion becomes:

D2 x Dx
m 2
= F + mg∗ − 2mω × (3.14)
Dt Dt
where we have defined the apparent gravity:
R
g∗ = −g − ω × (ω × R) (3.15)
R
The second term in the above equation, which modifies the true gravity to
give the apparent gravity, is known as the centrifugal force (an apparent
force).

We will take the latitude λ to be as shown in Fig. 3.1 (λ = 0 at the equator).

3.3.1 Apparent Gravity

The apparent gravity g∗ defines a local apparent vertical direction. If one
hangs a mass on a pendulum such that it is stationary in the rotating frame
D2 x
fixed to the Earth then DxDt = 0 and Dt2 = 0. In such case the upward
force on the pendulum bob, provided by the string, must balance exactly
the apparent gravity mg∗ . In other words, if the pendulum is stationary,
then it must hang in the direction of the apparent gravity rather than di-
rectly towards the Earth’s center.

To find the deflection angle between the apparent gravity and −R we can
use Fig. 3.2. The centrifugal term has a magnitude:

|−ω × (ω × R)| = ω 2 R cos λ (3.16)

25
 α
g

R
g * g*
R g

w 2 R cos
w x (w x R )

Figure 3.2: Determining the deflection angle between true and apparent
vertical directions at the Earth’s surface.

Applying the cosine rule to the triangle in Fig. 3.2, we have:

( )2
g ∗2 = g 2 + ω 2 R cos λ − 2gω 2 R cos2 λ (3.17)

Note that this tells us that g ∗ = g + O(ω 2 ). Applying the sine rule to the
same triangle gives:
sin α sin λ
2
= ∗ (3.18)
ω R cos λ g
where α is the deflection angle. This angle should be small such that we can
approximate sin α ≈ α. To order ω 2 , we can replace g ∗ with g, giving:

ω2R
α≈ sin λ cos λ (3.19)
g
The deflection vanishes at the equator and the poles, while it is maximum
at latitude 45◦ . The maximum deflection is given by:

ω2R 1 3.4cms−2
= = 1.73 × 10−3 rad ≈ 0.1◦ (3.20)
g 2 2g

3.3.2 Coriolis Force

The last term in Eq. 3.14 plays a role when the particle moves in the rotating
frame and is known as the Coriolis force:
Dx
−2mω × (3.21)
Dt

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Figure 3.3: Particle moving across a perfectly smooth rotating disk as seen
from an inertial frame above the disk (left), a frame rotating with the disk
(center) and a frame rotating with the disk when the disk rotates at high
angular velocity (right).

Let us stress once more that this is a fictitious force, which is only appar-
ently present when observing from a non-inertial (accelerating) frame. Note
that the Coriolis force acts at right angles to the direction of motion and is
proportional to the particle speed.

To understand the origin of the Coriolis force let us imagine a particle mov-
ing across a perfectly smooth rotating disk. Being perfectly smooth, the disk
doesn’t impart any horizontal force on the particle. To an observer watch-
ing the disk from above (an inertial reference frame) the particle moves in
a straight line across the disk, as shown by the first image in Fig. 3.3.

However, an observer moving with the disk will see the particle follow a
curve path as shown in the center image of Fig. 3.3. The observer will thus
conclude that some apparent forces act on the particle at right angles to
its velocity and this is the Coriolis force (note that the observer will also
see the apparent force mω 2 x). As the angular velocity of the disk increases
the path of the particle as seen by an observer rotating with the disk can
become rather complicated, as shown by the last image in Fig. 3.3.

To study further the Coriolis force, we will adopt the set of axes on the
Earth’s surface illustrated in Fig. 3.4. The apparent vertical axis ẑ is an-
tiparallel to the apparent gravity, that is, ẑ ∥ −g∗ . We will chose the x̂ axis
as pointing due East and the third unit vector ŷ = ẑ × x̂. In this coordinate

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 w

z
y
x

Equator

Figure 3.4: Coordinate system on the Earth’s surface

system, the equations of motion (Eq. 3.14) can be written in components:

( )
D2 x Dz Dy
m 2 = Fx − 2mω cos λ − sin λ (3.22)
Dt Dt Dt
D2 y Dx
m 2 = Fy − 2mω sin λ (3.23)
Dt Dt
D2 z Dx
m 2 = Fz − mg ∗ + 2mω cos λ (3.24)
Dt Dt

3.3.3 Effects of the Coriolis Force on Free-Fall

In free-fall a particle experiences only gravity, such that the equation of
motion can be written:
D2 x Dx
= g∗ − 2ω × (3.25)
Dt2 Dt
Let us consider only the first order corrections to the motion caused by the
Coriolis force, that is, let us work in this section to O(ω) and approximate

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g∗ by g. While we could work in terms of components, as in Eqs. 3.22-3.24,
it is in this case easier to proceed with the equation of motion in vector form
(as written in Eq. 3.25.

Integrating Eq. 3.25 with respect to time we obtain:

Dx
= v + gt − 2ω × (x − a) (3.26)
Dt
where the constants v and a represent the initial conditions of the problem,
namely the velocity v when the particle is at the point a at t = 0. Since
we are working only to first order in ω, we can substitute the zeroth order
solution, x = a + vt + gt2 /2 into the cross product term, giving:
( )
Dx 1 2
= v + gt − 2ω × vt + gt (3.27)
Dt 2
Integrating again with respect to time, we obtain:
( )
1 2 1 3
x = a + vt + gt − ω × vt + gt2
(3.28)
2 3
This is the general solution for the effect of the Coriolis force on a particle in
free-fall (up to first order in ω). Let us now consider some specific examples.

Dropping a particle from a tower. Consider a particle dropped from

rest from a tower of height h. The initial conditions are:
   
0 0
v=  
0 , a=  0  (3.29)
0 h

Using ω × g = −ωg cos λx̂, Eq. 3.28 becomes:

       
x 0 0 1
1 1
 y  =  0  − gt2  0  + ωgt3 cos λ  0  (3.30)
2 3
z h 1 0
√
The particle hits the ground when z = 0 at t = 2h/g. At this time we can
see that the particle’s displacement in the Eastward direction is:
( )1/2
1 8h3
x(z = 0) = ω cos λ (3.31)
3 g
Two views of this are shown in figure 3.5. On the left is the view from a
frame rotating with the Earth. The particle is observed to land a little to
the East of the tower due to the Coriolis force. On the right is the view from
an inertial frame, in which the Earth is observed to rotate. The particle is

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Figure 3.5: Two views of a particle dropped from the top of a tall tower
fixed to the rotating Earth. Left: rotating frame fixed to the Earth. Right:
As seen from an inertial frame in which the Earth spins on its axis.

seen to be projected from the top of the tower. The angular momentum
of the particle around the center of the Earth must however be conserved,
meaning that the angular velocity of the particle must increase as it falls.
Therefore, the particle gets slightly ahead of the tower as it falls.

Projectile Motion. A projectile is fired North with speed v and elevation

angle π/4. If we take the origin as the point from which the projectile is
fired from, then the initial conditions are:
   
0 0
v   
v= √ 1 , a= 0  (3.32)
2 1 0
Let us first evaluate the cross product of ω with v:
ωv
ω × v = √ (cos λ − sin λ) x̂ (3.33)
2
Substituting into Eq. 3.28:
       
x 0 0 2 1
 y = √ vt   1 2   ωvt
1 − gt 0 − √ (cos λ − sin λ) 0 
2 2 2
z 1 1 0
 
1
1
+ ωgt cos λ 0 
3  (3.34)
3
0
√
From the z component of the result, √ z = vt/ 2 − gt2 /2, we deduce that the
projectile hits the ground at tf = 2v/g. The value of x at impact is then
given by: √
2ωv 3
xf = (3 sin λ − cos λ) (3.35)
3g 2

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Note that the latitude λ has a strong effect on whether the deflection will
be positive (to the East) or negative (to the West). If 3 sin λ > cos λ then
the deflection will be to the East. This occurs at λ > tan−1 (1/3) = 18.4◦ .

In some cases the Eastward deflection may even change sign during the
trajectory, since there are both terms quadratic and cubic in time in the
x component of Eq. 3.34. These terms have opposite signs if λ < 45◦ .
Figure 3.6 shows the typical path of a projectile at latitude λ = 20◦ .

400

100
y HmL
z HmL
200
50

0
-20 0
0
x HmmL

Figure 3.6: Deflection of a projectile viewed from a non-inertial frame ro-

tating with the Earth. The projectile is fired at elevation angle π/4 with
speed 75ms−1 at latitude λ = 20◦ . Note that the x-axis scale is in mm.

3.3.4 The Foucault Pendulum

If we were to set up a pendulum at the North pole and start it swinging in
a plane (as viewed from an inertial frame), then according to an observer
rotating with the Earth, the plane of oscillation would rotate backwards at
angular velocity −ω.

We can expect a modified effect at lower latitudes. To describe it let us again

make use of the coordinates that we defined in Fig. 3.4, where ẑ = −g∗ /g ∗ .
Let us work to first order in ω, such that g ∗ ≈ g. Let us consider a pen-
dulum of length l, which is free to swing in any direction. We will neglect
air resistance, assumming that the pendulum is long and heavy, and able to

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swing for hours.

Taking the origin as the base of the pendulum swing, the equation of motion
of the system is given by Eqs. 3.22-3.24, where F is the force of the rope
carrying the pendulum bob. Making an approximation of small oscillations,
we can ignore terms in z compared to those in x and y. Then, Fx ≈ −mgx/l
and Fy ≈ −mgy/l. Equations 3.22 and 3.23 become:

D2 x Dy
2
= −ω02 x + 2ω sin λ (3.36)
Dt Dt
D2 y Dx
= −ω02 y − 2ω sin λ (3.37)
Dt2 Dt
where we defined ω02 = g/l (ω0 is the natural frequency of the pendulum).

To solve this system of coupled differential equations, let us define the com-
plex coordinate α = x + iy. This allows us to re-write the differential
equations for x and y as a single equation in α:

D2 α Dα
+ 2iω sin λ + ω02 α = 0 (3.38)
Dt2 Dt
Let us consider a trial solution of the form α = AeiΩt . Substituting into the
differential equation, we find that we have a solution if:
√
Ω = −ω sin λ ± ω02 + ω 2 sin2 λ (3.39)
≈ −ω sin λ ± ω0 (3.40)

where we made use of the fact that the oscillation frequency of the pendulum
is much larger than the rotation frequency of the Earth, that is, ω0 ≫ ω.
The general solution can be written as:
( )
α = Aeiω0 t + Be−iω0 t e−iω sin(λ)t (3.41)

Considering the initial condition x = x0 , y = 0, this can be simplified:

α = x0 cos(ω0 t)e−iω sin(λ)t (3.42)

The cos(ω0 t) term describes the usual swing of the pendulum with frequency
ω0 . The e−iω sin(λ)t term describes the rotation of the plane of oscillation
with angular velocity ω sin λ. The rotation of the pendulum swing was first
reported by Jean Foucault in Paris in 1851.

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