TITLE: Relative angular velocity
QUESTION [0 upvotes]: Two bodies are in circular motion with A at one diameter and having velocity tangentially downwards and B at other end of diameter tangentially upwards. Find angular velocity of particle A wrt B if angular velocity of particle A wrt centre of circle is $\omega.$
I tried to do vector subtraction of velocity of A and Velocity of B with angle between them as $\pi$ and got the answer to be $2V.$ Then I tried to apply $V=R\omega$ and got the relative angular velocity as $2\omega.$ But the book shows $\omega$ as answer. Where have I done wrong and how to do it?

REPLY [2 votes]: If you choose a frame of reference where B is steady, then in that
frame A's velocity is $2V$, as you said. Distance of A from B is $2R$
and angular velocity is $(2V)/(2R) = V/R = \omega$.
Actually this is a theorem in kinematics of rigid bodies: if you
choose several reference frames differing one from another only for a
translational motion velocities do change, but angular velocity
always stays the same.
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