Calculus of One and Several Variables

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) \equiv 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) \equiv 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) \equiv 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)\timesi(t)) + y(t)(Ω (t)\timesj (t))+ z(t)(Ω (t)\timesk (t))

and so, from the last equation for r_B^′

v (t) = vB (t)+ Ω (t)\times rB (t)+ Ω(t)\times R (t) = vB (t)+ Ω (t)\times 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) \timesv (t)+ Ω'(t) \timesr (t)+ Ω (t)\times v (t) dt2 B B

= aB(t)+ (Ω(t)\times vB (t))+ (Ω'(t)\times r (t))+ Ω (t)\times (vB (t)+ Ω (t)\times r(t))

= aB (t)+ 2(Ω (t)\times vB (t))+ (Ω'(t)\times r(t))+ Ω (t)\times (Ω (t)\times r (t))

= aB(t)+ 2(Ω (t)\times vB (t))+ (Ω'(t)\times r(t))+ Ω (t)\times (Ω (t)\times rB (t)) +Ω (t)\times (Ω (t)\times R (t)) (17.17)

' = aB (t)+ 2 (Ω (t) \timesvB (t)) +(Ω (t)\times rB(t))+ Ω(t)\times (Ω(t)\times rB(t)) + Ω(t)\times R'(t)+ Ω'(t) \timesR (t)

= aB (t) + 2(Ω(t)\times vB (t))+ (Ω'(t)\times rB (t)) d- + Ω (t)\times (Ω(t)\times rB (t))+ dt (Ω (t)\times R (t)) = aB (t) + 2(Ω(t)\times vB (t))+ (Ω'(t)\times rB (t)) + Ω (t)\times (Ω(t)\times r (t))+ R''(t) B

Therefore,

2 d-r2B(t) = aB (t)+ 2 (Ω (t)\times vB(t))+ (Ω '(t)\times rB (t))+ Ω (t)\times (Ω(t)\times 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)\times vB (t))+ (Ω'(t)\times rB(t)) +Ω (t)\times (Ω (t)\times 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.