Wonders of physics

L G. AslamazovLate Professor, Moscow Technological UniversityA. A. VarlamovItalian Institute of Condensed Matter Physics

First let us consider the motion of our hero in a stationary reference

frame. Obviously the motion is circular with the linear velocity v that adds

up of the linear velocity u r of the merry-go round and his relative velocity:

v — uir + vo.

dG. G. Coriolis, (1792-1843), French civil engineer.

eSee the novel by M. Cruz-Smith for further reference.

44 The Fucault pendulum and the Baer law

The centripetal acceleration is denned by the common formula,

v v

aCT> = — = — + w 2 r + 2v0u>.

r r

According to the second Newton law the acceleration is due to the horizontal

component of the force exerted onto the man by the rotating platform, the

seat, handles etc.,

macp = Q.

Now consider the motion in the reference frame bound to the merry-goround.

Here the linear velocity is vo and the centripetal acceleration is

2

acp = ~f~- With the help of the two previous equalities we may write:

mvo ^ 2

maco = = Q — mcj r — 2mi>ow.

r

In order to apply the second Newton law in the revolving frame of reference

we must introduce the force of inertia:

F\n = – (m ui2 r + 2m v0 w) = – (Fct + Fcor),

where the minus sign indicates that it is directed away from the axis. In

the non-inertial frame the equation of motion will be:

m o^p = Q + Fi n = Q – (Fct + FCor).

It seems that the inertia force throws you of the center of the merry-goround.

However the word “seems” is not a slip. No new interactions between

the bodies appear in the rotating reference frame. The only real

forces acting onto the man are the same reactions of the seat and bars.

Their net horizontal component Q is directed towards the center. In the

stationary reference frame the force Q resulted into the centripetal acceleration

acp. In the rotating frame due to kinematical reasons the acceleration

changed to the smaller value a’cp. In order to restore the balance between

the two sides of the equation we had to introduce the force of inertia.

In our case the force Fin comes up of the two addends. The first is the

centrifugal force Fcf that increases with the frequency of rotation and with

the distance from the center. The second is the Coriolis force FQ0T named

after the person who first calculated it. This force has to be introduced

only when the body moves relative to the rotating frame. It depends not

Inertia forces in the rotating reference frame 45

on the position of the body but on it’s velocity and the angular velocity of

rotation.

If the body in the rotating frame moves not along a circle but radially,

Fig. 6.2, then, just the same, one must introduce the Coriolis force. Now

it is perpendicular to the radius unlike the previous case. One of the basic

features of the Coriolis force is that it is always perpendicular both to the

axis of rotation and to the direction of motion. It may look strange but

in the revolving frame inertia forces not only push a body away from the

center but tend to swerve it astray.

We must emphasize that the Coriolis force like all other inertia forces is

of kinematical origin and can not be related to any physical objects*. Here

is an explicit example.

Imagine a cannon set at the North pole and pointed along a meridian(the pole is chosen for simplicity). Let the target lie on the same meridian.Is it possible that the projectile hits the target? From the point of view ofexternal observer which uses the inertial frame bound to the Sun the answeris obvious: the trajectory of the projectile lies in the initial meridional planewhereas the aim revolves with the Earth. Thus the projectile will never getthe target (unless a whole number of days will elapse). But how could oneexplain the fact in the reference frame bound to the Earth? What causesthe projectile stray from the initial vertical plane? In order to restoreconsistency one has to introduce the Coriolis force that is perpendicularto the rotation axis and to the velocity of a body. This force pulls theprojectile away from the meridional plane and it misses the target.Now let us return to the precession of the oscillation plane of the Fucaultpendulum from which we have started. It comes of a quite similarreason. Suppose again that the pendulum is situated at the pole. Then fora stationary observer the oscillation plane is at rest and it is the Earth thatrotates. A denizen of the North pole will see the opposite. For him themeridional plane looks fixed whereas the oscillation plane of the pendulumperforms a full revolution every 24 hours. The only way to explain this iswith the help of the Coriolis force. Unfortunately in general the picture isnot so transparent as at the poleg.

fEven though the force of inertia is not produced by any real bodies, observers feel it asa real force, akin to gravity. Remember the centrifugal force in the turning car.gOscillation plane of a Fucault pendulum located elsewhere turns 2n sin a radians perday, where a is the latitude of the place.