r ⃗ Rotating Frame Forces

1. Rotating frame 2. Coriolis force 3. Centrifugal force

Forces

Pressure

gradient Coriolis

Rotating Frame

The earth rotates with a period about 24 hours (23h 56m; or 366 rotation per year), although we can not feel it (human love geocentric view).

Ω = 7.292 x 10^-5 rad/s 
 (directing the North star)

⃗

Ω

⃗

r

⃗

R

ϕ

Coriolis force

Newtons’ laws of motion work only a inertial frame.

However, we can still use Newtons’ law in a rotating frame in a uniform rotation, if we introduce two apparent forces:

1) Coriolis and 2) Centrifugal forces

δθ

⃗ Ω = k d θ dt

i k j

⃗ A

⃗ A′

δ A ⃗

δθ

After δt D

A ⃗

Dt =

D

A ⃗

Dt + ?

? = lim δ A ⃗

δθ

⃗ Ω = k d θ dt

i k j

⃗ A

⃗ A′

δ A ⃗

δθ

After δt D

A ⃗

Dt

=

D

A ⃗ Dt

? = lim

δt→0

δ A ⃗

δt = Ω × ⃗ A ⃗ + Ω × ⃗ A ⃗

! Ω

! r

! relationship working
 R

for any vector A

Inertial frame vs. rotation frame

Da

! A Dt = D

r

! A Dt +

! Ω ×

! A

Can be obtained through
 (1) vector calculation - Holton
 (2) intuition - Vallis


Relationship between total differentiations in


a inertial frame (DaA/Dt) and a rotating frame (DrA/Dt)

ϕ

! Ω

v ! =

! Ω × !

r

! r

! R relationship working


for any vector A

Velocity (wind)

Da

!r Dt = D

r

!r Dt +

! Ω ×!

r

Relationship between total differentiations in


a inertial frame (DaA/Dt) and a rotating frame (DrA/Dt)

Da

! A Dt = D

r

! A Dt +

! Ω × !

A

v!

a = ! vr+

! Ω ×!

r

ϕ

! Ω

v ! =

! Ω × !

r

! r

! R

Acceleration

Relationship between total differentiations in


a inertial frame (DaA/Dt) and a rotating frame (DrA/Dt)

recall

relationship working
 for any vector A

v!

a =! vr+

! Ω ×!

r Da

v!

a

Dt = D

r

v!

a

Dt +

! Ω ×!

va

Da

v!

a

Dt = Dr

v!

r+

! Ω ×!

(

)

Dt +

! Ω × !

vr+

! Ω ×!

(

)

Da

v!

a = D

r

v!

r + 2

! Ω ×!

v +

! Ω ×

! Ω ×!

(

)

Da

! A Dt = D

r

! A Dt +

! Ω ×

! A

ϕ

Acceleration

Coriolis force Da

v!

a

Dt = D

r

v!

r

Dt + 2

! Ω ×!

vr+

! Ω ×

! Ω ×!

(

)

Centrifugal force ^{(= −Ω2}

! R)

D_a! va

Dt = − 1

ρ∇p − kg *+ν∇²v

!

a

D_r! vr

Dt = −2

! Ω ×!

vr+ Ω²R

!"

− kg *−1

ρ∇p +ν∇²v

!

a

D! v Dt = −2

! Ω ×!

v− 1

ρ∇p − kg +ν∇²v

!

Momentum equation in a rotating frame 
 (with rotation rate Ω)

Handle as one term using the concept of "geoid"


(max of Ω²R is ~0.034 m/s²; 
 just slightly modify g*➜g)

viscous force is local
 cannot feel rotation va ➜ v

Equation set

Six equations with six known (u, v, w, T, p, ρ) Mathematically closed; and solvable

D! v Dt = − 1

ρ∇p − 2

! Ω ×!

v− kg +ν∇²v

!

c_p DT Dt − 1

ρ Dp

Dt = J

Dρ

Dt +ρ∇ ⋅! v= 0

p =ρRT

(momentume eq. for u, v, w)

(thermodynamic energy eq. for T)

(continuity eq. for ρ)

(equation of state for p) 3

1

1

1

6

num. 


eq.