46.1: Introduction to the Coriolis Force

Imagine you're on a rotating merry-go-round, and you throw a ball to another person who's on the opposite side of the merry-go-round. If you aim directly at the other person, you'll miss them - the ball will travel in a straight line relative to the ground, but the merry-go-round will have rotated during the time the ball is in the air. Relative to the merry-go-round, the ball will appear to move along a curved path. You can attribute this curvature to a "fictitious force" called the Coriolis force. The Coriolis force is not a real force-it's just an artifact of viewing the ball's motion in a rotating reference frame. The ball really moves in a straight line relative to the ground.

So in the rotating reference frame of the merry-go-round, you'll see the ball move in a curved path, which can't happen unless there is a "force" present. We can compute the magnitude of this Coriolis force by considering the following situation. Suppose you're at the center of the merry-go-round, and throw a ball outward with velocity $v$ while the merry-go-round is rotating with an angular velocity $\Omega$. After a time $t$, the ball will have moved a radial distance $r=v t$. At time $t$, a point on the merry-go-round a distance $r$ from the center will have moved through an arc length

$$s =r \theta$$
$$=r(\Omega t)$$
$$=(v t) \Omega t$$
$$=\Omega v t^{2} .$$

But under a constant acceleration $a_{c}$, we know

$$s=\frac{1}{2} a_{c} t^{2} .$$

Comparing Eq. $\PageIndex{4}$ with Eq. $\PageIndex{5}$, we deduce that the Coriolis acceleration $a_{c}$ is given by

$$a_{c}=2 \Omega v \text {. }$$

More generally, in terms of vectors, the Coriolis acceleration vector $\mathbf{a}_{c}$ is given by

$$\mathbf{a}_{c}=-2(\boldsymbol{\Omega} \times \mathbf{v})$$

From Newton's second law, the corresponding Coriolis force $\mathbf{F}_{c}$ on a body of mass $m$ is then

$$\mathbf{F}_{c}=-2 m(\boldsymbol{\Omega} \times \mathbf{v})$$