Answer to Question #119980 in Classical Mechanics for Dheeraj

Question

The magnitude of Coriolis accelaration for a particle of mass m moving with a velocity 100m/s horizontally along eath's surface at the north pole is about?

Answer 1

We have given mass of the particle is $m$ ,velocity $v$ with respect to earth's frame (Non-inertial frame) is $v=100m/s$ along horizontal direction at north pole.

Let, $\Omega$ is the angular velocity of earth about axis passing through center and north pole and direction is along north pole,thus $v\bot\Omega$ .

Now, we know that,

Coriolis force is given by

$$\overrightarrow{F}_{coriolis}=-2m( \overrightarrow\Omega\times \overrightarrow v)\\ \implies|\overrightarrow{F}_{coriolis}|=2m|( \overrightarrow\Omega\times \overrightarrow v)|\\ \implies a_{coriolis}=2|( \overrightarrow\Omega\times \overrightarrow v)|$$

In this case.

$$a_{coriolis}=2|( \overrightarrow\Omega\times \overrightarrow v)|=2|\overrightarrow\Omega|\cdot| \overrightarrow v|\\ =200\Omega\\ =200\times7.2921159 × 10^{-5}=14.584\times 10^{-3}m/s^2$$