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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.

Dan
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  • It would help if you explained your own reasoning, and the answer you got, and the book's answer. At first glance this looks like it can be done easily with conservation of angular momentum. There are some hidden assumptions: e.g., the lamina continues to rotate about its center after the particle adheres. Also, why a "smooth" vertical axis? Just another way to say frictionless? – Michael Weiss May 07 '14 at 13:55
  • @Michael Yes smooth means frictionless. The answer in the back is $\omega/10$ and I haven't seen how to get this yet. I got $w/19$ by reasoning that the angular momentum must be equivalent before and after the impact and that since the impact is perpendicular $I_1 . \omega = I_2 . \omega_2$ where $I_1$ and $I_2$ are the moments of inertia of the structure before and after the addition of the extra weight respectively and where $w_2$ is the new angular speed. – Dan May 10 '14 at 09:54

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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.

user110503
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