Understanding Angular Velocity: Rotations and Accelerated Frames

[CrusaderSean]

i'm doing accelerated frames and rotations in class right now. I'm not sure if i understand angular velocity correctly so i hope someone can correct me.

Lets say there is an inertial frame and rotating frame.
r' = distance from origin of inertial to rotating frame
P = point in rotating frame
x = P measured from rotating frame
r = P measured from fixed frame

to describe velocity of P from inertia frame, it's
$$\frac{dr}{dt} = \frac{dr'}{dt} + \omega \times x$$

the way my textbook defined is $$\omega = \frac{d \phi}{dt} \hat{n}$$ , it points in normal direction of rotation axis. this normal direction is measured from inertial frame right? if you define the following as operator and apply it to omega:
$$\frac{d}{dt} = \frac{d}{dt} + \omega \times$$
$$\frac{d \omega}{dt} = \frac{d \omega'}{dt}$$
omega is observed from inertial while omega' from rotating frame. does this mean angular acceleration has same magnitude and direction in both frames, but angular velocity does not necessarily have to be the same?... kind of an odd question i guess, but i can't see how omega would have the same direction as if you measure it from different frames.

my textbook only emphasized omega is uniform for rotating (rigid) body because it does not depend on where the origin is in the rotating frame.

 

[marlon]

You have one frame that is 'fixed' and one frame that rotates along with the object (this does NOT necessarily have to be the case ofcourse). So when looking in the fixed frame, the object rotates but when looking in the rotating frame, the object does not move. The omega expresses the rotation of the object but the clue is that the rotating frame has the exact same angular velocity omega, because it rotates along with the object. This way of working is especially usefull when calculating the moments of inertia for rigid rotators (ie not pointlike objects but solid objects like an apple). Why ? Well, because when you are calculating this tensor in the rotating frame (the object does not move here) this tensor will reduce to a diagonal matrix.

marlon

 

[marlon]

Besides, there is only one omega which is the instantaneous rotation vector of the rotating frame wtr to the 'fixed' inertial frame. This omega really connects the base-vectors of these two frames.

marlon

 

[CrusaderSean]

Besides, there is only one omega which is the instantaneous rotation vector of the rotating frame wtr to the 'fixed' inertial frame. This omega really connects the base-vectors of these two frames.

marlon

i see. that makes more sense then. thanks for the clarification.