Rotating Frame and Angular Velocity

Question

We have an equation $ \frac{dr}{dt}=\Omega \times \bf r \tag 1$

SPECIFICATIONS

1. $\times$ means cross product,$\Omega$ constant angular velocity,${\bf r}$ is the postion vector of an object
2. Given object has a position vector ${\bf r}$ in some non-rotating inertial reference frame
3. 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.
4. 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

1. 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$
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?

Answer 1

1. Yes (the same expression holds).
2. 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

Answer 2

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.

Answer 3

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. $

Answer 4

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.

Answer 5

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.

2. "able to find a vector f(t)"

The acceleration vector from the angular acceleration vector is (f) times the mass

$$ \vec F_{\varepsilon} (t) = m(\vec \varepsilon (t) \times \vec r (t)) \tag {g}$$