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Coriolis and Centrifugal Forces
We are now going to analyze a very simple
physical situation from a less simple point of 1. A kid on a merry-go-round
view.
Let’s say we have a body which moves at A kid is sitting on the edge of a merry-go-round,
constant velocity with respect to an inertial and he drops a ball. We will be interested in
frame. No net force acts on it. Now suppose we what happens in the horizontal plane - that is,
look at this body fr

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Coriolis and Centrifugal Forces
Weare now going to analyze a very simplephysical situation from a less simple point of view. Let’s say we have a body which moves atconstant velocity with respect to an inertialframe. No net force acts on it. Now suppose welook at this body from a frame of reference whichis
accelerating
. In general, we no longer observethe 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 you jump up and down, or spin around on your fancydesk chair. Everything appears to bounce up anddown 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. Ourfirst 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 inwhat 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 endshereHere is amovie illustrating the ball’s path.
The question is, what path does the ball follow,according to the kid? Here’s the answer:
kid stayshereIf you don’t believe it, work it out! It’s easiest tosee if you realize that the kid has to keep turninghis head, in order to see the ball. Here’s amovieshowing 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 won’t attempt toreproduce it here. However, the result is remark-ably simple, especially when written in vectornotation. There are only two terms:
F
P
eff
= −
m
ω
P H
á
ω
P H
r
P
é
−
2
m
ω
P H
v
P
centrifugal coriolis,where
v
P
is the velocity as seen in the rotatingframe (that is, as seen by the kid). The angularvelocity vector of the rotating frame is
ω
P
. In ourcase, it points vertically out of the merry-go-round:
ω
P
=
z
ˆ
ω
.Let’s work out the centrifugal force. In planepolar coordinates, the body is located at
r
P
=
r
ˆ
r
.The various vectors are shown in the next figure:
11111111110000000000
ω
P
=
z
ˆ
ω
θ
ˆ
r
ˆ
Using the relations
z
ˆ
H
r
ˆ
= θ
ˆ and
z
ˆ
H
θ
ˆ
= −
r
ˆ
, we find
F
P
centrifugal
=
m
ω
2
r r
ˆ .Hence, the centrifugal force is directed
outwards
from the center of rotation, and is exactlyopposite to the centripetal force discussedearli-er.
What about the coriolis force? If the velocity asseen in the rotating frame is
v
P
=
r
ˆ
v
r
+ θ
ˆ
v
θ
,then the coriolis force is
F
P
cor
=
2
m
ω
ä ã
r
ˆ
v
θ
− θ
ˆ
v
r
ë í
.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 thevelocity to become nonzero. Therefore, the kidsees the ball come off the merry-go-round
at right angles
as seen by the kid. Then, since theradial component of the velocity is positive, thecoriolis force points in the
− θ
ˆ
direction and thekid starts to see the ball go around.
111111111111111111111111000000000000000000000000
F
coriolis
P
F
centrifugal
P
F
centrifugal
P
Here is anothermovie of the motion observed by
the kid, except this time the coriolis and centrifu-gal forces are shown.Let’s make an interesting and useful modificationto 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 mean“directly towards the center of the merry-go-round”, then it turns out that he will see the ball
deflected to the right
. Here is amovie showing
what the kid sees, andanother showing the view
from an observer fixed to the earth. The latterview makes it clear that the deflection observedby the kid is
entirely due to his rotational motion
.
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 anearlier 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.Let’s concentrate on the coriolis force. Suppose abody is located at (
θ
,
φ
) inspherical coordinates,and that its velocity
as seen on the earth
is givenby
v
P
=
r
ˆ
v
r
+ θ
ˆ
v
θ
+ φ
ˆ
v
φ
.The earth’s angular velocity vector is
ω
P
=
z
ˆ
ω
.We have to work out the cross product
F
P
coriolis
= −
2
m
ω
P H
v
P
,so we will need to know the cross products of thevarious unit vectors shown in the followingfigure:
ω
P
θ φ φ
ˆ
θ
ˆ
r
ˆ
Using the relations
z
ˆ
H
r
ˆ
= φ
ˆ sin
θ
,
z
ˆ
H
θ
ˆ
= φ
ˆ cos
θ
and
z
ˆ
H
φ
ˆ
= −
r
ˆ sin
θ − θ
ˆ cos
θ
, it is easy to show that the coriolis force is
F
P
cor
=
2
m
ω
: ; <
r
ˆ
v
φ
sin
θ + θ
ˆ
v
φ
cos
θ − φ
ˆ
ä ã
v
r
sin
θ +
v
θ
cos
θ
ë í B C D
.
The easiest way to make sense of this is to drawdiagrams for some special cases:

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