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?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ 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 '20 at 20:51
Add a comment
|
$\begingroup$
$\endgroup$
1
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.
