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Figure 10.5: Decomposition of a vector in a rotating coordinate system.

Let us assume we are considering a vector q that is constant in the rotating coordinate

system. In a non-rotating system this vector changes with time because it co-rotates with

the rotating system. The vector q can be decomposed in a component q== along the

rotation vector and a component q? to the rotation vector. In addition, a vector b is

de ned in gure (10.5) that is perpendicular to both q? and in such a way that , q?

and b form a right handed orthogonal system.

10.4. THE CORIOLIS FORCE AND CENTRIFUGAL FORCE 113

Problem b: Show that:

q== = ^ ^ q

q? = q ^ ^ q (10.36)

;

b= ^ q

Problem c: In a xed non-rotating coordinate system, the vector q rotates, hence it

position is time dependent: q = q(t). Let us consider how the vector changes over a

time interval t. Since the component q== is at all times directed along the rotation

vector , it is constant in time. Over a time interval t the coordinate system

rotates over an angle t. Use this to show that the component of q perpendicular

to the rotation vector satis es:

q?(t + t) = cos ( t) q?(t) + sin ( t) b (10.37)

and that time evolution of q is therefore given by

q(t + t) = q(t) + (cos ( t) 1) q?(t) + sin ( t) b (10.38)

;

Problem d: The goal is to obtain the time-derivative of the vector q. This quantity can

be computed using the rule dq=dt = lim t!0 (q(t + t) q(t))= t. Use this, and

;

equation (10.38) to show that

q= b

_ (10.39)

where the dot denotes the time-derivative. Use (10.36) to show that the time deriva-

tive of the vector q is given by

q= q:

_ (10.40)

At this point the vector q can be any vector that co-rotates with the rotating coordinate

system. In this rotating coordinate system, three Cartesian basis vectors x, y and ^ can

^^ z

be used as a basis to decompose the position vector:

rrot = x^ + y^ + z^ :

xyz (10.41)

Since these basis vectors are constant in the rotating coordinate system, they satisfy

(10.40) so that:

x x

^

d^=dt=

y y

^

d^=dt= (10.42)

z ^:

z

d^=dt=

It should be noted that we have not assumed that the position vector rrot in (10.41)

rotates with the coordinate system, we only assumed that the unit vectors x, y and ^

^^ z

rotate with the coordinate system. Of course, this will leave an imprint on the velocity

and the acceleration. In general the velocity and the acceleration follow by di erentiating

(10.41) with time. If the unit vectors x, y and ^ would be xed, they would not contribute

^^ z

to the time derivative. However, the unit vectors x, y and ^ rotate with the coordinate

^^ z

system and the associated time derivative is given by (10.42).

CHAPTER 10. LINEAR ALGEBRA

114

Problem e: Di erentiate the position vector in (10.41) with respect to time and show

that the velocity vector v is given by:

v =x^ + y^ + z^ + r :

_ x _y _z (10.43)

_ x _y _z

The terms x^ + y^ + z^ is the velocity as seen in the rotating coordinate system, this

velocity is denoted by vrot . The velocity vector can therefore be written as:

v = vrot+ r: (10.44)

Problem f: Give an interpretation of the last term in this expression.

Problem g: The acceleration follows by di erentiation expression (10.43) for the velocity

once more with respect to time. Show that the acceleration is given by

a =x^ + y^ + z^+2

xyz _ x _y _z r) :

(x^ + y^ + z^) + ( (10.45)

xyz

The terms x^ + y^ + z^ in the right hand side denote the acceleration as seen in the

rotating coordinate system, this quantity will be denoted by arot . The terms x^ + y^ + z^

_ x _y _z

again denote the velocity vrot as seen in the rotating coordinate system. The left hand

side is by Newton's law equal to F=m, where F is the force acting on the particle.

Problem h: Use this to show that in the rotating coordinate system Newton's law is

given by:

marot = F;2m vrot m r) :

( (10.46)

;

vrot de-

The rotation manifests itself through two additional forces. The term ;2m

( r) describes the centrifugal force.

scribes the Coriolis force and the term ;m

Problem i: Show that the centrifugal force is perpendicular to the rotation axis and is

directed from the rotation axis to the particle.

Problem j: Air ows from high pressure areas to low pressure areas. As air ows in the

northern hemisphere from a high pressure area to a low-pressure area, is it de ected

towards the right or towards the left when seen from above?

Problem k: Compute the magnitude of the centrifugal force and the Coriolis force you

experience due to the earth's rotation when you ride your bicycle. Compare this

with the force mg you experience due to the gravitational attraction of the earth.

It su ces to compute orders of magnitude of the di erent terms. Does the Coriolis

force de ect you on the northern hemisphere to the left or to the right? Did you

ever notice a tilt while riding your bicycle due to the Coriolis force?

In meteorology and oceanography it is often convenient to describe the motion of air or

water along the earth's surface using a Cartesian coordinate system that rotates with the

e e e

earth with unit vectors pointing in the eastwards (^1 ), northwards (^2 ) and upwards (^3 ),

^ that are

see gure (10.6). The unit vectors can be related to the unit vectors ^, ' and

r^

de ned in equation (3.7) of section (3.1). Let the velocity in the eastward direction be

denoted by u, the velocity in the northward direction by v and the vertical velocity by w.

10.4. THE CORIOLIS FORCE AND CENTRIFUGAL FORCE 115

â„¦

^

e3

^

e2

Î¸

^

e1

Figure 10.6: De nition of a local Cartesian coordinate system that is aligned with the

earth's surface.

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