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.

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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).