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. (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 $\vec{\omega}$. 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. (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

$$ \boxed{ \vec r = \vec r' } $$

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

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:

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

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

(ii) Transformation of acceleration :-

Equation (9) describes the relation of rate of change of position vector in both reference frames S and R and this is valid for all vectors. Therefore equation (9) can also be written in terms of operator as

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

From which the operator

$$ \boxed{ \frac{d}{dt} = \frac{d'}{dt} + \vec{\omega}\times } \tag{11} $$

Therefore, the acceleration of a particle in reference frame S is

$$ \vec a = \frac{d}{dt} \left( \frac{d\vec r}{dt} \right) = \left( \frac{d'}{dt} + \vec{\omega}\times \right) \left( \frac{d'\vec r}{dt} + \vec{\omega}\times\vec r \right) $$

or

$$ \vec a = \frac{d^2\vec r}{dt^2} = \frac{d'^2\vec r}{dt^2} + 2\vec{\omega}\times \frac{d'\vec r}{dt} + \vec{\omega}\times \left( \vec{\omega}\times\vec r \right) $$

where $\vec{\omega}$ is assumed to be constant.

Therefore,

$$ \boxed{ \vec a = \vec a' + 2 \left( \vec{\omega}\times\vec v' \right) + \vec{\omega}\times \left( \vec{\omega}\times\vec r \right) } \tag{12} $$

where

$$ \vec a' = \frac{d'^2\vec r}{dt^2} = \text{Acceleration of a particle in reference frame R.} $$

$2(\vec{\omega}\times\vec v')$ = Coriolis acceleration of particle in frame R.

$\vec{\omega}\times(\vec{\omega}\times\vec r)$ = Centripetal acceleration.

From equation (12), the acceleration of a particle in reference frame R is

$$ \boxed{ \vec a' = \vec a - 2(\vec{\omega}\times\vec v') - \vec{\omega}\times (\vec{\omega}\times\vec r) } \tag{13} $$

This equation (13) describes the relation of acceleration of a particle in both reference frame S and R. Therefore it is called transformation equation for acceleration.

(iii) Transformation of force :-

Multiplying equation (13) by the mass $m$ of the particle,

$$ m\vec a' = m\vec a - 2m (\vec{\omega}\times\vec v') - m\vec{\omega}\times (\vec{\omega}\times\vec r) $$

or

$$ \boxed{ \vec F' = \vec F - 2m(\vec{\omega}\times\vec v') - m\vec{\omega}\times (\vec{\omega}\times\vec r) } \tag{14} $$

where $\vec F'$ = Force on a particle with respect to reference frame R.

$-2m(\vec{\omega}\times\vec v')$ = Coriolis force acting on a particle in frame R. This force is a fictitious force which appears to be acting on the particle moving in the reference frame R.

$-m\vec{\omega}\times(\vec{\omega}\times\vec r)$ = Centrifugal force. This force appears to be acting on a particle due to rotation of reference frame R. So it is also a fictitious force. It does not act on the particle in real sense but appears to be acting to the particle as viewed by the observer in the reference frame S.

Coriolis force :

(a) When the angle between $\vec{\omega}$ and $\vec v'$ is $\theta$, then

$$ \boxed{ \vec F_c = -2m\omega v'\sin\theta\, \hat n } $$

where $\hat n$ is a unit vector perpendicular to both $\vec{\omega}$ and $\vec v'$.

(b) When $\vec{\omega}$ and $\vec v'$ are normal to each other, then $\theta=90^\circ$ and $\sin\theta=1$.

$$ \boxed{ \vec F_c = -2m\omega v'\hat n } $$

(c) When $\vec{\omega}$ and $\vec v'$ are parallel to each other, then $\theta=0$ and $\sin\theta=0$.

$$ \boxed{ \vec F_c=0 } $$