Solution Example I

We first consider the vector transformation from the coordinate system K to the co-rotational system K'. The pointer vector ${\bf r}$ is represented by the linear combination of the unit vector ${\bf e_i}$($i=x,y,z$) in K:

$${\bf r} = \sum\limits_i r_i {\bf e_i}.$$ (1)

Equally one can write

$${\bf r} = \sum\limits_i r_i^{'} {\bf e_i}^{'},$$ (2)

in the co-rational system K'. The time differential of the unit vectors in K' is given by

$${d{\bf e_i}^{'}\over dt}= {\bf\omega}\times {\bf e_i}^{'},$$ (3)

where $\omega$ is the angler velocity vector. When the differential operator $d/dt$ is applied in equation (2), the relation (3) above gives

$${d{\bf r}\over dt} = \sum\limits_i {dr_i^{'}\over dt}{\bf e_i}^{'} + {\bf\omega}\times \sum\limits_i r_i^{'}{\bf e_i}^{'}.$$ (4)

Here the first term in the right hand side above is the velocity seen by the observer in the co-rational system K'. Consequently the equation (4) can read

$${\bf v} = {\bf v}^{'} + {\bf\omega}\times {\bf r},$$ (5)

where ${\bf v}$ is the velocity vector.

The acceleration ${\bf a}$ in K is then obtained by the time differential of the equation (4) as

$${d^2{\bf r}\over dt^2} = \sum\limits_i {d\over dt}\left({dr_i^{'}\over dt}{\bf e_i}^{'}\ri... ... + {d\over dt}\left({\bf\omega}\times \sum\limits_i r_i^{'}{\bf e_i}^{'}\right)$$

$$= \sum\limits_i {d^2r_i^{'}\over dt^2}{\bf e_i}^{'}+ {\bf\omega}\ti... ...er dt}{\bf e_i}^{'} +\dot{{\bf\omega}}\times \sum\limits_i r_i^{'}{\bf e_i}^{'}$$

$$+ {\bf\omega}\times\sum\limits_i {dr_i^{'}\over dt}{\bf e_i}^{'} ... ...f\omega}\times \left({\bf\omega}\times\sum\limits_i r_i^{'}{\bf e_i}^{'}\right)$$

$$= \sum\limits_i {d^2r_i^{'}\over dt^2}{\bf e_i}^{'}+ 2{\bf\omega}\t... ...\omega}\times \left({\bf\omega}\times\sum\limits_i r_i^{'}{\bf e_i}^{'}\right).$$ (6)

Here we used the relation (3). The first term in the right hand side in the equation above represents the accelerator for the observer in K'. We, therefore, finally get

$${\bf a}={\bf a}^{'}+2\omega\times {\bf v}^{'}+\dot{{\bf\omeg... ... r} +{\bf\omega}\times \left({\bf\omega}\times {\bf r}\right).$$ (7)

The equation described above and the Newtonian equation of the motion

$${\bf F}= m{\bf a}$$ (8)

then lead to the following final equation of the motion for the observer in k';

$$m{\bf a}^{'}= {\bf F}-2m\left(\omega\times {\bf v}^{'}\right... ...t) -m{\bf\omega}\times \left({\bf\omega}\times {\bf r}\right).$$ (9)

The terms in the right hand side imply the external force, the coliori Force, the apparent force due to the angular (de-)acceleration, and the centrifugal force, respectively.