NoWorry said:
Firsly, if there has been no rotation in the non-reference frame the equation would have been like this, I call it as an accelerated reference frame (remember no rotating),
m\boldsymbol{\ddot{r}}=\boldsymbol{F}-m\boldsymbol{\ddot{R}}_{0}
which is very easy to interprete this equation. This is the equation for the 2. law in the non-inertial frame reference. The first term to right is the total force in the inertial reference frame while the last term, namely
-m\boldsymbol{\ddot{R}}_{0}
is what makes the non-inertial reference frame to accelate with a translational acceleration \boldsymbol{\ddot{R}}_{0} in one direction straightly. If it transforms to the inertial reference frame, there would be no fictitious force this time.
Secondly, the rotation reference frame without the acceleration would have been like this (i)
m\boldsymbol{\ddot{r}}=\boldsymbol{F}+ \underbrace{2m\left (\boldsymbol{\omega} \times \boldsymbol{\dot{r}} \right )}_\text{Coriolis Force}+\underbrace{m\left [\boldsymbol{\omega} \times \left (\boldsymbol{\omega} \times \boldsymbol{r} \right ) \right ]}_\text{Centrifugal Force}
(This equation is according to the Classical Mechanics by John R. Taylor p. 343)
Then IF this rotation reference frame is about to accelerate, there should have been like this (ii)
m\boldsymbol{\ddot{r}}=\boldsymbol{F}-m\boldsymbol{\ddot{R}}_{0} - \underbrace{m\left (\boldsymbol{\dot{\omega}} \times \boldsymbol{r} \right ) }_\text{“Whatever” Force} - \underbrace{2m\left (\boldsymbol{\omega} \times \boldsymbol{\dot{r}} \right )}_\text{Coriolis Force} -\underbrace{m\left [\boldsymbol{\omega} \times \left (\boldsymbol{\omega} \times \boldsymbol{r} \right ) \right ]}_\text{Centrifugal Force}
(This equation is according to the Classical Mechanics by R. Douglas Gregory p. 477).
Now we can see that the third term is added "mysteriously". I don't even understand why there are positive terms in (i) while there are negative terms in (ii). That’s why I wanted to know how to interprete this (third) or other terms more specific without doubting.
I appreciate your answers though.
The following might help:
In general, a vector \boldsymbol{A} can be written as a linear combination of basis vectors. Let \boldsymbol{e'_x}, \boldsymbol{e'_y}, \boldsymbol{e'_z} be the basis vectors of a rotating coordinate system. Then we can write:
\boldsymbol{A} = A^x \boldsymbol{e'_x} + A^y \boldsymbol{e'_y} + A^z \boldsymbol{e'_z}
Now, if we take the time derivative of \boldsymbol{A}, there will be two sources of time-dependency: (1) The components A^x, A^y, and A^z may change with time, and (2) The basis vectors \boldsymbol{e'_x}, \boldsymbol{e'_y} and \boldsymbol{e'_z} can change with time. So let the operator \dfrac{D}{Dt} represent the complete time derivative, taking into account both effects, and let \dfrac{d}{dt} be the operator that only takes into account the changes in the components. Then we can write:
\dfrac{D}{Dt}\boldsymbol{A} = \dfrac{d}{dt} \boldsymbol{A} + A^x \dfrac{D}{Dt}\boldsymbol{e'_x} + A^y \dfrac{D}{Dt}\boldsymbol{e'_y} + A^z \dfrac{D}{Dt}\boldsymbol{e'_z}
In the special case of a rotating coordinate system, we can express the time dependence of the basis vectors as follows:
\dfrac{D}{Dt}\boldsymbol{e'_i} = \boldsymbol{\omega} \times e'_i
where i is either x, y, or z.
For this special case, we can summarize the total time derivative this way:
\dfrac{D}{Dt}\boldsymbol{A} = \dfrac{d}{dt} \boldsymbol{A} + \boldsymbol{\omega} \times \boldsymbol{A}
Let's let \dot{\boldsymbol{A}} = \dfrac{d}{dt} \boldsymbol{A}. Then we can write:
\dfrac{D}{Dt}\boldsymbol{A} = \dot{\boldsymbol{A}} + \boldsymbol{\omega} \times \boldsymbol{A}
Now, let's specialize to the case of the position vector:
\dfrac{D}{Dt}\boldsymbol{r} = \dot{\boldsymbol{r}} + \boldsymbol{\omega} \times \boldsymbol{r}
Taking another derivative gives:
\dfrac{D^2}{Dt^2}\boldsymbol{r} = \dfrac{D}{Dt}\dot{\boldsymbol{r}} + (\dfrac{D}{Dt} \boldsymbol{\omega}) \times \boldsymbol{r} + \boldsymbol{\omega} \times \dfrac{D}{Dt} \boldsymbol{r}
= \ddot{\boldsymbol{r}} + \boldsymbol{\omega} \times \dot{\boldsymbol{r}} + \dot{\boldsymbol{\omega}} \times \boldsymbol{r} + (\boldsymbol{\omega} \times \boldsymbol{\omega}) \times \boldsymbol{r} + \boldsymbol{\omega} \times \dot{ \boldsymbol{r}} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})
This simplifies to:
\dfrac{D^2}{Dt^2}\boldsymbol{r} = \ddot{\boldsymbol{r}} + 2 \boldsymbol{\omega} \times \dot{\boldsymbol{r}} + \dot{\boldsymbol{\omega}} \times \boldsymbol{r} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})
Now, here's where the negative signs come in: Solve the above for \ddot{\boldsymbol{r}} to get:
This simplifies to:
\ddot{\boldsymbol{r}} = \dfrac{D^2}{Dt^2}\boldsymbol{r} - 2 \boldsymbol{\omega} \times \dot{\boldsymbol{r}} - \dot{\boldsymbol{\omega}} \times \boldsymbol{r} - \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})
Now, multiply through by the mass m to get
m \ddot{\boldsymbol{r}} = m\dfrac{D^2}{Dt^2}\boldsymbol{r} - 2m \boldsymbol{\omega} \times \dot{\boldsymbol{r}} - m\dot{\boldsymbol{\omega}} \times \boldsymbol{r} - m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})
The last move is to use Newton's law to rewrite:
m\dfrac{D^2}{Dt^2}\boldsymbol{r} = \boldsymbol{F}
So in terms of the force \boldsymbol{F} we have
m \ddot{\boldsymbol{r}} = \boldsymbol{F} - 2m \boldsymbol{\omega} \times \dot{\boldsymbol{r}} - m\dot{\boldsymbol{\omega}} \times \boldsymbol{r} - m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})
At this point, people sometimes define the "inertial force" \boldsymbol{F}_{inertial} to be the extra terms
\boldsymbol{F}_{inertial} = - 2m \boldsymbol{\omega} \times \dot{\boldsymbol{r}} - m\dot{\boldsymbol{\omega}} \times \boldsymbol{r} - m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r})
So the equations of motion look like
m \ddot{\boldsymbol{r}} = \boldsymbol{F} + \boldsymbol{F}_{inertial}
