We have:
- Position as $\vec r$
- Velocity as $\vec v=(\vec r)'$
- Acceleration as $\vec a=(\vec v)'$
- Angular velocity as $\vec \omega = \frac{\vec r × \vec v}{r^2}$
- Angular acceleration as $\vec \alpha = (\vec \omega)'$
Cross multiplying $\vec r$ and then differentiating with respect to time on both sides in (3), we get,
$\vec v - (\vec v · \hat r)\hat r =\vec \omega × \vec r$
$ \Rightarrow \vec v= (\vec v· \hat r)\hat r + \vec \omega × \vec r $
$\Rightarrow \vec a = ((\vec v·\hat r)\hat r)' + \vec \omega × \vec v + \vec \alpha × \vec r$
Now, in rotational frame of reference, $\vec \omega × \vec v$ corresponds to centrifugal force, $\vec \alpha × \vec r $ corresponds to Euler force. So, $((\vec v · \hat r)\hat r)'$ should correspond to Coriolis force and we should get $((\vec v · \hat r)\hat r)'$ equal to $ 2 \vec \omega × (\vec v · \hat r)\hat r$. Now, expanding first term we get,
$((\vec v · \hat r)\hat r)'$
$=(\frac{(\vec v · \vec r)\vec r}{r^2})'$
$=\frac{((\vec v · \vec r)\vec r)' r^2 - 2(\vec v · \vec r)(\vec v · \vec r) \vec r}{r^4}$
$=\frac{((\vec v · \vec r)\vec v + (\vec v ·\vec r)' \vec r ) }{r^2} - \frac{2(\vec v · \vec r)^2 \vec r}{r^4}$
$=\frac{((\vec v · \vec r)\vec v + (\vec a ·\vec r)\vec r + v^2 \vec r) }{r^2} - \frac{2(\vec v · \vec r)^2 \vec r}{r^4}$
And, expanding second term we get,
$2\vec \omega × (\vec v · \hat r)\hat r$
$=\frac{2(\vec r × \vec v ) × (\vec v · \vec r )\vec r}{r^2 × r^2}$
$=\frac {2(\vec v · \vec r)((\vec r × \vec v ) × \vec r)}{r^4}$
$=\frac {-2(\vec v · \vec r)((\vec r × (\vec r × \vec v)}{r^4}$
$=\frac {-2(\vec v · \vec r)((\vec r· \vec v)\vec r - r^{2} \vec v)}{r^4}$
$= \frac {2(\vec v · \vec r)\vec v}{r^2} - \frac{2(\vec v· \vec r)^2 \vec r}{r^4}$
Clearly, we are not getting them same. So, is the assumption that $((\vec v · \hat r)\hat r)'$ corresponds to Coriolis force incorrect? If not then how can we show that both the terms are equal?
