That's an exercise from Thorton's Classical Dynamics textbook.
Basically, as stated, a uniform rod of lenght $b$ tips over to the floor, being initially stood vertically upright on the same floor. The question is what's the angular velocity of the rod when it hits the floor.
Seemed like a pretty straightforward exercise when I looked at it. I wrote the energy conservation law for the rod:
\begin{equation}mgb=\frac{1}{2}mv_{cm}^2+I\omega^2\end{equation}
Since the center of mass won't move through this rotation, $v_{cm}=0$. The moment of inertia of the rod can be found from the definition: \begin{equation}I=\int_{0}^{b}\rho_b\ r^2dr\end{equation} Where $\rho_b$ represents a linear density; $\rho_b=\frac{m}{b}$. Follows trivially that $I=\frac{1}{3}mb^2$. Then, it's easy to see that it'd follow: \begin{equation}\omega=\sqrt{\frac{6g}{b}}\end{equation} Thing is, that does not seem to be the answer. I looked at the solutions manual, and, when writing the potential energy, he used $\frac{b}{2}$, that is, up to the center of mass. That doesn't make much sense to me. He used the moment of inertia of the body as a whole, why take the potential up to the $CM$ instead?
His development then gives:
\begin{equation}\omega=\sqrt{\frac{3g}{b}}\end{equation}
What am I missing here? Seems like a silly question, but regardless. Any help will be appreciated.
