The angular velocity of a particle moving in a circle realative to the center of the circle is equal to omega. Find the angular velocity of the particle relative to a point on the circular path.

The angular velocity of a particle moving in a circle realative to the

1.	The angular velocity of a particle moving in a circle realative to the center of the circle is equal to omega. Find the angular velocity of the particle relative to a point on the circular path.
Answer» Solution :`omega_(B A) = (v_(B A)_\|_)/(A B)` where `v_(B A)_\|_ = v COS THETA "and" AB = AC cos theta` because `ABC` is a right - angled triangle. Then, `omega_(B A) = (v)/(A C) = (v)/(2 R)` Substituting `(v)/( R) omega_(B O)`, we have `omega_(B A) = (1)/(2) omega_(B O) (-(1)/(2) omega))` Alternative procedure : `phi = 2 theta` Then, `(d phi)/(d t) (- omega_(B O) = 2(d theta)/(d t)` where `(d theta)/( dt) = omega_(B A)` This gives `omega_(B O) = 2 omega_(B A)`. .