Four persons are placed at the vertices of ABCD square frame of side a mounted on xy-plane having its center as the origin. The square frame is revolving with angular velocity $\vec \omega = \omega \hat k$. With what velocity vector should A throw a light ball with respect to himself when the square frame has all its sides parallel to x and y axis with point A lying in the second quadrant, so that the ball just falls downward as seen from the ground as if it were free falling?
Solution
The square frame is rotating anti-clockwise. The velocity of person A w.r.t. ground = $\frac {a\sqrt 2}{2}\omega (-cos 45^\circ \hat i - sin 45^\circ \hat j)$
= $-\frac {a\omega}{2} (\hat i + \hat j)$
The person should throw the ball at velocity = $+\frac {a\omega}{2} (\hat i + \hat j)$
This way, the initial velocity of ball would be 0 w.r.t. the ground and it will just fall downward like a free fall.
Note that whether ABCD is labelled clockwise or anti-clockwise on square, it does not matter.