Four uniform rods, each of mass $m$ and length $2l$, are joined rigidly together to form a square frame $ABCD$ of side $2l$. The frame is placed with all four sides at rest on a smooth horizontal table. An inextensible string has one end attached at the corner of $A$. A particle of mass $4m$ is tied to the other end of the string. The particle, initially at $A$, is projected with speed $u$ in the direction $DA$. Given that the speed of the particle immediately after the string becomes taught is $V$, show that the initial angular speed of the square frame about an axis through it's centre of gravity perpendicular to the plane of the frame is $\omega$ where $\omega$ is $\frac{2V - u}{l}$.
I tried to solve this by equating angular moments. I took the angular moments of the body before and after the exchange of momentum as angular moments about the centre of the square (i.e. multiplying the linear moments by $l$). I calculated the moment of inertia of the square frame as $\frac{16}{3}ml^2$. By equating I got, $4mlu + \frac{16}{3}ml^2\omega = 4mlV + \frac{16}{3}ml^2\omega^\prime$ where $\omega^\prime$ is the final angular speed. I solved by substituting $l\omega^\prime=V$. This did not give me $\omega = \frac{2V - u}{l}$ so clearly my thinking is wrong somewhere. Can anybody get it?