# Coriolis Force in Physics

A bird of mass 3 kg is flying at 20 m/s in latitude 45 degree, heading west. Find the horizontal and vertical components of the Coriolis force acting on it.

The figure is below. The angular rotation speed of earth at 45 degree  latitude is

$Omega =omega_0*begin{pmatrix} 0\ cos(45)\ sin(45) end{pmatrix}$   where $ω_0=2π/(24*3600)=7.27*10^{-5} rad/sec$
$ω_0$ is the rotational speed of the Earth at the equator.

The speed of the bird heading west has the components:

$v=begin{pmatrix} -20\ 0\ 0 end{pmatrix}$

The Coriolis acceleration is by definition:

$a_C=-2(Omega times v)=-2omega_0*begin{pmatrix} i &j &k \ 0 &cos(45) &sin(45) \ -20 &0 &0 end{pmatrix}=2omega_0begin{pmatrix} 0\ 20sin(45)\ -20cos(45)end{pmatrix}$

Therefore $a_C=begin{pmatrix} 0\ 2.056*10^{-3}\ -2.056*10^{-3}end{pmatrix} (m/s^2)$

The Coriolis force is thus

$F_C=m*a_C=begin{pmatrix} 0\ 6.169*10^{-3}\ -6.169*10^{-3}end{pmatrix} (N)$

The horizontal component of the Coriolis force is $F_{C y}=6.169*10^{-3} N$ and the vertical component of the Coriolis force is $F_{C z}=-6.169*10^{-3} N$  (it is downwards).