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The Centrifugal and Coriolis Forces
The “proof” given in class of the general expression (in two dimensions) is my own, and
diﬀerent from what is in the textbook, or in other books. Here it is:
First, look at motion in a straight line as measured in polar coordinates. Suppose that
the velocity of a point mass
m
is
±v
=
v
ˆ
i
as shown in the diagram, superposed on the drawing
of a
stationary
disk. Imagine that the motion is along a straight line drawn on the disk
parallel to the
x
axis, not through the centre.
r
w
¡
¡
@
@
@
@I
φ
⇒
v
¡
¡
¡±
@
@
@I
ˆ
φ
ˆ
r
Then, the velocity of the mass has components
v
r
=
v
cos
φ
and
v
φ
=

v
sin
φ
so that
±v
=
v
cos
φ
ˆ
r

v
sin
φ
ˆ
φ
The angle
φ
is written instead of the usual
θ
, because we will need
θ
for when the disk is
rotating. The unit vectors ˆ
r
and
ˆ
φ
are as indicated.
Notice that the acceleration corresponding to this
±v
is zero, because
±a
=

v
sin
φ
˙
φ
ˆ
r
+
v
cos
φ
˙
φ
ˆ
φ

v
cos
φ
˙
φ
ˆ
φ
+
v
sin
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 Fall '07
 Harris
 Force

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