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.

Explanation

`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)`.
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