17.6 Coriolis Force and Centripetal Force
Let p
be a point which has constant components relative to the moving coordinate
system described above
. For example, it could be a single point on the
rotating earth. Letting
i ∗ , j ∗ , k ∗ be a typical rectangular coordinate system fixed
in space and let
R be the position vector of
p from the origin fixed in
space. In the case of the earth, think of the origin as the center of the earth.
Thus the components of
R with respect to the moving coordinate system are
constants. Let
r B be the position vector from this point
p to some other
point.
rB (t) ≡ x(t) i(t)+ y(t)j(t)+ z (t)k(t)
The acceleration perceived by an observer moving with the moving coordinate system
would then be
′′ ′′ ′′ ′′
rB (t) ≡ aB (t) = x (t)i(t)+ y (t)j(t) +z (t)k (t)
and the perceived velocity would be r B ′
≡ v B .
vB (t) ≡ x′(t)i(t)+ y′(t)j(t)+ z′(t)k(t)
Let r
≡ R +
r B . Then, since
R has constant components relative to the
moving coordinate system,
v (t) = R ′(t)+ r′ (t), r′(t) = v (t)+ x (t)i′(t)+ y (t)j′(t)+ z(t)k′(t)
B B B
= vB (t)+ x(t)(Ω(t)×i(t)) + y(t)(Ω (t)×j (t))+ z(t)(Ω (t)×k (t))
and so, from the last equation for r B ′
,
v (t) = vB (t)+ Ω (t)× rB (t)+ Ω(t)× R (t)
= vB (t)+ Ω (t)× r(t)
Now take a further derivative to find the total acceleration. Using what was just shown,
it equals
a(t) = R ′′(t) + d2rB(t) = a (t)+ Ω (t) ×v (t)+ Ω′(t) ×r (t)+ Ω (t)× v (t)
dt2 B B
= aB(t)+ (Ω(t)× vB (t))+ (Ω′(t)× r (t))+ Ω (t)× (vB (t)+ Ω (t)× r(t))
= aB (t)+ 2(Ω (t)× vB (t))+ (Ω′(t)× r(t))+ Ω (t)× (Ω (t)× r (t))
= aB(t)+ 2(Ω (t)× vB (t))+ (Ω′(t)× r(t))+ Ω (t)× (Ω (t)× rB (t))
+Ω (t)× (Ω (t)× R (t)) (17.17)
′
= aB (t)+ 2 (Ω (t) ×vB (t)) +(Ω (t)× rB(t))+ Ω(t)× (Ω(t)× rB(t))
+ Ω(t)× R′(t)+ Ω′(t) ×R (t)
= aB (t) + 2(Ω(t)× vB (t))+ (Ω′(t)× rB (t))
d-
+ Ω (t)× (Ω(t)× rB (t))+ dt (Ω (t)× R (t))
= aB (t) + 2(Ω(t)× vB (t))+ (Ω′(t)× rB (t))
+ Ω (t)× (Ω(t)× r (t))+ R′′(t)
B
Therefore,
2
d-r2B(t) = aB (t)+ 2 (Ω (t)× vB(t))+ (Ω ′(t)× rB (t))+ Ω (t)× (Ω(t)× rB (t))
dt
where recall that a B
is the perceived acceleration relative to the moving coordinate
system. Solving for this yields
2
d-rB-(t)− (2(Ω (t)× vB (t))+ (Ω′(t)× rB(t)) +Ω (t)× (Ω (t)× rB (t))) = aB (t)
dt2
The part of the acceleration on the left depending on the relative velocity is called the
Coriolis acceleration. The rest of it is sometimes called centrifugal acceleration. It is felt
by the observer by regarding the moving coordinates as fixed. On the earth, this force is
small enough to be neglected. However, when v B is large, one can get a significant
contribution from the Coriolis force.