2. Rotating Frame of Reference and Coriolis Force

Let the coordinates of any point P are $(x,y,z)$ and unit vectors are $(\hat{i},\hat{j},\hat{k})$ along the axes of the stationary reference frame S. The position vector of point P is

$$ \overrightarrow{OP}=\vec r =\hat{i}x+\hat{j}y+\hat{k}z \tag{1} $$

Fig. (2.1): Rotating reference frame R

If another reference frame R which is initially at $t=0$ coincident with the frame S, is rotating with an angular velocity $\vec{\omega}$ in such a way that origins of both reference frames remain coincident. After $t$ seconds all the axes of frame R will be inclined by an angle

$$ \theta=\omega t $$

with the respective axes of frame S. In this state if the coordinates of the same point P are $(x',y',z')$ and unit vectors are $(\hat{i}',\hat{j}',\hat{k}')$ along the axes of the rotating reference frame R, then the position vector of point P is

$$ \overrightarrow{OP}=\vec r' =\hat{i}'x'+\hat{j}'y'+\hat{k}'z' \tag{2} $$

Since the directions of the axes of frame R are changing with respect to time as shown in Fig. (2.1), so the unit vectors $(\hat{i}',\hat{j}',\hat{k}')$ will be time dependent.

From equations (1) and (2), the position of particle P is given as

$$ \vec r = \hat{i}'x' +\hat{j}'y' +\hat{k}'z' \tag{3} $$

This equation is called transformation equation of position vector in rotating frame R.

(i) Transformation of velocity :-

Differentiating equation (3) with respect to $t$, we have

$$ \frac{d\vec r}{dt} = \left( \hat{i}'\frac{dx'}{dt} + \hat{j}'\frac{dy'}{dt} + \hat{k}'\frac{dz'}{dt} \right) + \left( \frac{d\hat{i}'}{dt}x' + \frac{d\hat{j}'}{dt}y' + \frac{d\hat{k}'}{dt}z' \right) \tag{4} $$

Now the velocity of a particle in reference frame S is

$$ \vec v = \frac{d\vec r}{dt} = \hat{i}\frac{dx}{dt} + \hat{j}\frac{dy}{dt} + \hat{k}\frac{dz}{dt} \tag{5} $$

Similarly the velocity of a particle in reference frame R is

$$ \vec v' = \frac{d'\vec r}{dt} = \hat{i}'\frac{dx'}{dt} + \hat{j}'\frac{dy'}{dt} + \hat{k}'\frac{dz'}{dt} \tag{6} $$

To determine the value of $\frac{d\hat{i}'}{dt}$, $\frac{d\hat{j}'}{dt}$, $\frac{d\hat{k}'}{dt}$, let us take a unit vector $\vec R$ which is moving with an angular velocity $\vec{\omega}$ just like $\hat{i}',\hat{j}',\hat{k}'$ are moving with an angular velocity $\vec{\omega}$.

Therefore,

$$ \vec V=\vec{\omega}\times\vec R $$

or

$$ \frac{d\vec R}{dt} = \vec{\omega}\times\vec R \tag{7} $$

Substituting $\hat{i}',\hat{j}',\hat{k}'$ in place of $\vec R$ in equation (7), we have

$$ \frac{d\hat{i}'}{dt} = \vec{\omega}\times\hat{i}' $$

$$ \frac{d\hat{j}'}{dt} = \vec{\omega}\times\hat{j}' $$

$$ \frac{d\hat{k}'}{dt} = \vec{\omega}\times\hat{k}' \tag{8} $$

Substituting these equations and equation (6) in equation (4),

$$ \frac{d\vec r}{dt} = \frac{d'\vec r}{dt} + \vec{\omega}\times \left( \hat{i}'x' + \hat{j}'y' + \hat{k}'z' \right) $$

But from equation (3),

$$ \hat{i}'x' + \hat{j}'y' + \hat{k}'z' = \vec r $$

Therefore,

$$ \boxed{ \frac{d\vec r}{dt} = \frac{d'\vec r}{dt} + \vec{\omega}\times\vec r } $$

Since

$$ \vec v=\frac{d\vec r}{dt} \qquad\text{and}\qquad \vec v'=\frac{d'\vec r}{dt}, $$

we obtain the transformation equation of velocity:

$$ \boxed{ \vec v = \vec v' + \vec{\omega}\times\vec r } $$