Question:
A rigid, uniform rod of length $L$ and mass $M$ is pivoted to the origin $O$ at one end, and is then left to swing freely in a vertical plane.
If the angle the rod makes with the vertical is $\theta$, show that the kinetic energy of the rod is $$T = \frac 16 ML^2{\dot \theta} ^2$$
Attempt:
In terms of $\theta$, the position of the center of mass is
$$(x,y) = \bigg(\frac L2 \sin \theta, -\frac L2 \cos \theta \bigg)$$
In a previous part of the question, I have shown that the moment of inertia of the rod about one of its endpoints is $$I = \frac 13 ML^2$$
Thus, the total kinetic energy of the rod should be
\begin{align} T & = \frac 12 M(\dot x^2 + \dot y^2) + \frac 12 I \omega ^2 \\ & = \frac 12 M\bigg[\bigg(\frac L2 \dot \theta\cos \theta \bigg)^2 + \bigg(\frac L2 \dot \theta \sin \theta\bigg)^2 \bigg] + \frac 12\bigg(\frac 13 ML^2 \bigg) \dot \theta ^2 \\ & = \frac 18 ML^2 \dot \theta ^2 + \frac 16 ML^2 \dot \theta ^2 \\ & \neq \frac 16 ML^2 \dot \theta ^2 \end{align}
Is this perhaps not the right way to do this question?
If you are including the motion of the CM is because you are considering the motion of each part of the rod as composition of two motions: the motion of each part around the CM and the motion of the CM with respect to our frame (in wich $O$ is at rest). In this case, the moment of inertia has to be around the CM: $I = \dfrac 1{12} ML^2$ leading to
$$T=\frac 18 ML^2 \dot \theta ^2 + \frac 1{24} ML^2 \dot \theta ^2=\frac 16 ML^2 \dot \theta ^2$$
We have two contributions: kinetic energy of rotation around the CM and kinetic energy of translation of the CM.
If we choose to consider the parts of the rod as simply moving around $O$ we have to use the moment of inertia around $ O$ because there are no composition of motions for any part and there is only energy of rotation: $T=\dfrac 16 ML^2 \dot \theta ^2$