Rotating Frame and Angular Velocity We have an equation $ \frac{dr}{dt}=\Omega \times  \bf r  \tag 1$
SPECIFICATIONS


*

*$\times$ means cross product,$\Omega$ constant angular velocity,${\bf r}$ is the postion vector of an object

*Given object has a position vector ${\bf r}$ in some non-rotating inertial reference frame

*This object is in a non-inertial reference frame which rotates with constant angular velocity  $\mbox{ $\Omega$}$ about an axis passing through the origin of the inertial frame.

*Our object appears stationary in the rotating reference frame.
In the non-rotating frame, the object's position vector ${\bf r}$ will appear to precess about the origin with angular velocity  $\mbox{$\Omega$}$


Question


*

*What will be the case when $\Omega $ is not constant? Means varying with time.Will that be the case as follows?  $ \frac{dr}{dt}=\Omega(t) \times  \bf r  \tag 2$

*In some other way imagine if I am happened to know  $ \frac{dr}{dt}$,$\bf r$ at each s  and able to find a vector $f(t)$ such that  $ \frac{dr}{dt}=f(t) \times  \bf r  \tag 3$. Then can I say r is rotating with a varying angular velocity $f(t)=\Omega(t)$ related to the non moving frame?

 A: If $\Omega$ is not constant, the equation still holds. But except in special cases, you will have to solve it by numerical integration.
If you have a certain rotation motion, and you found that $f(t)$ fits the equation $dr/dt=f(t)\times r$, it is not necessarily your angular velocity, but it will be true that $\Omega(t)=f(t)+\lambda(t) r(t)$, with $\lambda$ real. If you prescribe a rotation by giving $f(t)$, it surely defines a rotation motion, and can be recovered by numerical integration.
A: 1. What will be the case when Ω is not constant? 
If it is a function of time, the cross-product will be also a function of time, cross product is to be taken at each instant of time.
2. In some other way imagine if I happened to know dr/dt,r at each s and able to find ...
Hint: Only perpendicular vectors produce a cross product, like in:
k $ X (a i + b j + c k )  = a j - b i + 0. $
A: *

*Yes (the same expression holds).

*Yes (it is called motion synthesis).


It is worth it for you reading about differentiating vectors on rotating frames.


*

*http://envsci.rutgers.edu/~broccoli/dynamics_lectures/lect_06_dyn12_mom_eq_rot.pdf

*https://en.wikipedia.org/wiki/Rotating_reference_frame#Time_derivatives_in_the_two_frames
A: An angular velocity represented by a vector $\mathbf\Omega(t)$ consists of
an axis of rotation (parallel to $\mathbf\Omega(t)$) and a speed of rotation
(equal to the magnitude of $\mathbf\Omega(t)$) around that axis.
A single point-mass particle at coordinates $\mathbf r,$ moving at velocity 
$\frac d{dt}\mathbf r,$ could be rotating around any axis that is perpendicular
to the direction of motion and that does not pass through $\mathbf r.$
(Or any axis at all, if you allow it to have other components of motion.)
A rigid body, however, if it is rotating, has a particular axis of rotation.
There are an infinite number of vectors that solve the equation
$$\frac d{dt}\mathbf r = \mathbf f(t) \times \mathbf r,$$
but only a vector parallel to the body's axis of rotation
can be the angular velocity of that body.
A: 1
“What will be the case when Ω is not constant?“
The angular velocity vector formula for point
$$  \frac {\vec r_n \times \vec v_n} { \vert \vec r_n \vert ^2 } = \vec \omega_n \tag {a}$$
the derivative of the vector of the angular velocity of a point
$$  \frac {(\vec r_n \times \vec a_n)  \vert \vec r_n \vert ^2 - (\vec r_n \times \vec v_n)(2 r_x v_x + 2 r_y v_y  + 2 r_z v_z) } { \vert \vec r_n \vert ^4 } =  \frac {d\vec \omega_n} {dt} \tag {b} $$
Acceleration vector of a point dependent on the vector of angular velocity and angular acceleration it's derivative of your formula (1)
$$ \frac {d \vec v_{\bot}} {dt} = \vec \omega \times  \vec v  + \vec \varepsilon \times  \vec r = \vec a_{\omega \varepsilon } \tag c$$
Because there are such situations when the angular acceleration vector is zero, therefore
$$ \exists  \vec \varepsilon (0,0,0) ; [ \vec \varepsilon \times  \vec r = (0,0,0) \to \vec \omega \times  \vec v = \vec a_{\omega } ] \tag {d}$$
this is why
$$\vec \omega \times  \vec v = \vec a_{\omega }  \tag {e}$$
so the angular acceleration vector will give the following acceleration vector
$$  \vec \varepsilon \times  \vec r = \vec a_{\varepsilon \omega } - \vec a_{\omega }  = \vec a_{\varepsilon  }  \tag   {f}$$
Acceleration vector $\vec a_{\omega }$ it does not always have to correspond to reality and so for example: for orbital motion acceleration vector from the omega vector it does not have to be equal to the acceleration by gravity.


*"able to find a vector f(t)"

The acceleration vector from the angular acceleration vector is (f) times the mass
$$ \vec F_{\varepsilon} (t) = m(\vec \varepsilon (t) \times  \vec r (t))  \tag   {g}$$
