Speed of horizontal rotating lamina after perpendicular impact from above A uniform square lamina of side $2l$ and mass $m$ is rotating with angular speed $\omega$ in a horizontal plane about a smooth fixed vertical axis through the centre of the lamina. A particle of mass $3m$ is held at height $l$ above the lamina. The particle is released and adheres to the lamia at a corner. Find the new angular speed of the lamina. I assumed the height $l$ makes no difference to the result since the particle's impact is perpendicular to the plane of circular motion although I cannot get the answer in the back of my book so I assume I'm missing something.

The moment of inertia of a square plate of length $2l$ and mass $m$ about an axis at its centre and perpendicular to its plane is $\frac{1}{12}m((2a)^2+(2a)^2)$. The particle has moment of inertia $3m(\sqrt2a)^2$.

Thus $$\frac23ma^2\omega = \left(\frac23ma^2 +6ma^2\right)\omega' =\frac{20}{3}ma^2\omega'$$ where $\omega'$ is the final angular velocity. Dividing both sides by $\frac{20}{3}ma^2$ gives the desired answer.