homework and exercises - Rotational velocity of tethered shape after falling

My solution to the above question involves equating the potential energy to the to the kinetic energy at the point at which the wire tightens as: $$ \frac{1}{2}mv^2 = mgh $$ However, I am having trouble finding the initial rotational velocity of the object, I initially thought that it was, if l is length: $$ \omega = \frac{2}{l} v $$ However, seeing a this is a full (past) exam question, I think the solution if not as straightforward. My second thought was that perhaps the radius of rotation is the distance from the connection point to the centroid of the shape and v the velocity component perpendicular to this.
But I am not sure that all of the velocity after falling is converted into rotational velocity at that instant, or if the point of connection accelerates to the right at the point as the string becomes taut.