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Coriolis and Centrifugal Forces

We are now going to analyze a very simplephysical situation from a less simple point ofview.

Lets say we have a body which moves atconstant velocity with respect to an inertial

frame. No net force acts on it. Now suppose welook at this body from a frame of reference whichis accelerating. In general, we no longer observe

the body to be moving with constant velocity. Itappears to accelerate, and from our point of viewit looks as though there were a force acting on it.

Such a force is called an effective or ficti-tious force. The acceleration due to such a forceis caused solely by the motion of the observer.

A common example of this effect is when youjump up and down, or spin around on your fancydesk chair. Everything appears to bounce up and

down or go around and around, as the case maybe.

In this section, we will be concerned with theparticular case in which we are observing theuniformly-moving body from a frame of refer-ence which is rotating at a constant rate. Our

first example will involve a kid on a merry-go-round.

1. A kid on a merry-go-round

A kid is sitting on the edge of a merry-go-round,and he drops a ball. We will be interested in

what happens in the horizontal plane - that is,when the whole thing is observed from above.Never mind the vertical motion of the ball.

If we are attached to the earth, viewing the wholething from above, the ball just goes sailing off ina straight line as shown in this top view:

kid and ballstart here

kid endshere

ball endshere

Here is a movie illustrating the balls path.

The question is, what path does the ball follow,according to the kid? Heres the answer:

http://../MOVIES/FRGROUND.MOVhttp://../MOVIES/FRGROUND.MOVhttp://../MOVIES/FRGROUND.MOV

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kid stayshere

If you dont believe it, work it out! Its easiest tosee if you realize that the kid has to keep turninghis head, in order to see the ball. Heres a movieshowing the motion seen by the kid.

According to the kid, there appears to be a forcewith one component pointing outwards andanother directed around. How large are thesecomponents?

The calculation of the effective force seen by thekid is rather involved, and we wont attempt to

reproduce it here. However, the result is remark-ably simple, especially when written in vectornotation. There are only two terms:

FPeff

= mP H P HrP 2mP HvP

centrifugal coriolis,

where vP is the velocity as seen in the rotating

frame (that is, as seen by the kid). The angularvelocity vector of the rotating frame is P . In our

case, it points vertically out of the merry-go-round:

P =z .

Lets work out the centrifugal force. In planepolar coordinates, the body is located at

rP = r r .

The various vectors are shown in the next figure:

P

=

z

r

Using the relations

zHr = and zH = r ,

we find

FPcentrifugal

= m2r r.

Hence, the centrifugal force is directed outwards

from the center of rotation, and is exactlyopposite to the centripetal force discussed earli-er.

http://../MOVIES/FRMTTBLE.MOVhttp://rotation.pdf/http://rotation.pdf/http://rotation.pdf/http://rotation.pdf/http://rotation.pdf/http://../MOVIES/FRMTTBLE.MOV

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What about the coriolis force? If the velocity asseen in the rotating frame is

vP = r vr

+ v

,

then the coriolis force is

FPcor

= 2m r v vr .

Notice that this has two parts, one in the radialdirection and the other in the angular direction.

In our present example, at the instant the ball isreleased both components of the velocity arezero. The only apparent force is the centrifugalforce, and this causes the radial component of the

velocity to become nonzero. Therefore, the kidsees the ball come off the merry-go-round at

right angles as seen by the kid. Then, since the

radial component of the velocity is positive, the

coriolis force points in the direction and thekid starts to see the ball go around.

Fcoriolis

P

Fcentrifugal

P

Fcentrifugal

P

Here is another movie of the motion observed bythe kid, except this time the coriolis and centrifu-gal forces are shown.

Lets make an interesting and useful modification

to the above situation. Suppose the kid throwsthe ball towards the center of the merry-go-round. What path will the kid observe the ball tofollow?

If the kid defines straight ahead to meandirectly towards the center of the merry-go-

round, then it turns out that he will see the balldeflected to the right. Here is a movie showing

what the kid sees, and another showing the viewfrom an observer fixed to the earth. The latterview makes it clear that the deflection observedby the kid is entirely due to his rotational motion.

http://../MOVIES/CORIOLIS.MOVhttp://../MOVIES/CORIOLIS.MOVhttp://../MOVIES/C2BODY.MOVhttp://../MOVIES/C2BODY.MOVhttp://../MOVIES/C2FIXED.MOVhttp://../MOVIES/C2FIXED.MOVhttp://../MOVIES/C2FIXED.MOVhttp://../MOVIES/C2BODY.MOVhttp://../MOVIES/CORIOLIS.MOV

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2. Motion relative to the earth

We are now going to talk about the fictitiousforces due to the rotation of the earth. This iswhat we were referring to in an earlier section,when we hinted that the surface of the earth isnot really an inertial frame. You will see that theideas of this section are just souped-up versionsof the kid on the merry-go-round. All you haveto do is look down from above the north pole.

Lets concentrate on the coriolis force. Suppose abody is located at (,) in spherical coordinates,and that its velocity as seen on the earth is given

by

vP = r vr

+ v

+ v

.

The earths angular velocity vector is

P =z .

We have to work out the cross product

FPcoriolis

= 2mP HvP ,

so we will need to know the cross products of thevarious unit vectors shown in the followingfigure:

P

r

Using the relations

zHr = sin , zH = cos

and

zH = r sin cos ,

it is easy to show that the coriolis force is

FPcor

= 2m:;