# Relative angular velocity

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?

• – Farcher Jun 14 at 5:21

## 1 Answer

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.

• That was very clear. Thank you – dhanesh vijay Jun 15 at 17:20