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CrusaderSean
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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.
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.
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