Horizontal component of the Coriolis force

[MatinSAR]

Homework Statement

Show that the horizontal component of the Coriolis force is independent of the direction of motion of the particle on Earth's surface.(The particle is moving on a horizontal plane.)

Relevant Equations

Newton's Laws in non-inertial reference frames.

The coriolis force that acts on the object moving on the Earth is: $$F_{cor}=2m(\vec v \times \vec \omega)$$$F_{cor}$ is the Coriolis force, $m$ is the mass of the object, $\vec{v}$ is the velocity of the object in the Earth frame, and $\vec{\omega}$ is the angular velocity of the Earth.

Is it true to say that the horizontal component of this force is equal to $2m \omega v \sin \theta$ where $\theta$ is the angle between $\vec \omega$ and $\vec v$?

 

[jbriggs444]

Is it true to say that the horizontal component of this force is equal to $2m \omega v \sin \theta$ where $\theta$ is the angle between $\vec \omega$ and $\vec v$?

That is the formula for the magnitude of the Coriolis force. Not for its horizontal component.

 

[MatinSAR]

That is the formula for the magnitude of the Coriolis force. Not for its horizontal component.

Thank you for pointing out my mistake. I think I need to reread this chapter before trying to solve.