A uniform rod of length L and mass m is pivoted about a frictionless pin through one end such that the rod rotates due to gravity from rest in a horizontal position through a vertical position. Let $\theta$ be the angle of the rod from the horizontal such that $\theta=0$ corresponds to the rod being horizontal and $\theta=\pi/2$ to the rod being vertical. If I let the axis of rotation be the pivot point, then straightforward calculations of conservation of energy reveal that at the vertical position the speed of the free end of the rod is $v=\sqrt{3gL}$. However, as a challenge I sought to reproduce this result using the center of mass of the rod as the axis of rotation, where now the pin (still at one end) would do work as the torque is nonzero about the center of mass. First, my interpretation of conversation of energy for a generalized axis of rotation:  
$$\frac{1}{2}mv^2_{AOR} + \frac{1}{2}I\omega^2=mgh + w_{AOR}$$
where $v_{AOR}$ is the velocity of the axis of rotation, $\omega$ is the angular velocity of the rod *emphasized text*about the axis of rotation, h is the change in height of the center of mass, and w is the work done to the axis of rotation by the pin. Note that if the axis of rotation is taken to be the pivot point, then $v_{AOR}$=0, $I=\frac{1}{3}mL^2$, $\omega =\frac{v}{L}$ and $w_{AOR}=0$, and solving for v yields $v=\sqrt{3gL}$

Now for the axis of rotation as the center of mass:

I reasoned $v_{AOR}$ ($v_{cm}$) to be v/2, as both the free end and the center of mass have the same angular velocities, yet the axis of rotation is half the distance from the pivot point as the free end. 

The moment of inertia is $\frac{1}{12}mL^2$

I believe $\omega = \frac{v}{L/2}$. Initial intuition is that the free end has velocity v by definition (at least in the inertial reference frame), and the distance between the axis of rotation and the free end is L/2 so by $v=r\omega$ (and of course the vectors being orthogonal) $\omega = \frac{v}{L/2}$. I got the same result when parametrizing both the free end and the axis of rotation as $\overrightarrow{r_2}=\left\langle  Lcos(\omega t), -Lsin(\omega t) \right\rangle$ and $\overrightarrow{r_1}=\left\langle  L/2cos(\omega t), -L/2sin(\omega t) \right\rangle$, taking the difference of these two as  $\overrightarrow{r}$, taking the derivative of $\overrightarrow{r}$ and then finding the magnitude of that quantity. 

For $w_{AOR}$ I got $\frac{1}{8}mgL$. For brevity I will just post summary steps, but feel free to ask for more work. I found the centripetal acceleration of the center of mass as a function of $\theta$ to be $\frac{3}{2}gsin(\theta)$ and the tangential acceleration to be $\frac{3}{4}gcos(\theta)$. Converting to Cartesian coordinates
$$\overrightarrow{a} = \frac{3}{2}gsin(\theta)\left\langle  -cos(\theta), sin(\theta), 0 \right\rangle +  \frac{3}{4}gcos(\theta)\left\langle  -sin(\theta), -cos(\theta), 0 \right\rangle$$ Then, by Newton's second law $$\left\langle 0,-mg,0 \right\rangle + \overrightarrow{F} = m\overrightarrow{a}$$ where $\overrightarrow{F}$ is the force exerted by the pin. After solving the equation above for the force, I find the torque produced by this force on the axis of rotation by taking the cross product between the moment arm and this pin force: $$\left\langle -\frac{L}{2}cos(\theta), \frac{L}{2}sin(\theta), 0 \right\rangle \times \left\langle \frac{-9mgsin(\theta)cos(\theta)}{4}, \frac{mg(9sin^2(\theta)+1)}{4}, 0 \right\rangle$$$$=\left\langle 0,0, \frac{-mgLcos(\theta)}{8}  \right\rangle$$ Taking the integral from $\theta =0$ to $\theta = \pi/2$ of this torque $d\overrightarrow{\theta}$ yields $w_= \frac{1}{8}mgL$

However, when putting this all together: 
$$\frac{1}{2}m(v/2)^2 + \frac{1}{2}(\frac{1}{12}mL^2)(\frac{v}{L/2})^2 = \frac{1}{2}mgL + \frac{1}{8}mgL$$ and solving for the velocity of the free end at the vertical position yields $v=\sqrt{\frac{15}{7}gl}$, not the correct answer of $v=\sqrt{3gL}$. Please help me address where I went wrong