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?

  • $\begingroup$ Related?Relative angular velocity and acceleration $\endgroup$ – Farcher Jun 14 '19 at 5:21
  • $\begingroup$ The two particles maintain a constant distance with each other and hence act as if part of a rigid body => They share a common angular velocity. $\endgroup$ – John Alexiou Feb 14 at 20:51

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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  • $\begingroup$ That was very clear. Thank you $\endgroup$ – dhanesh vijay Jun 15 '19 at 17:20

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