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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 Mbigg[bigg(frac L2 dot thetacos theta bigg)^2 + bigg(frac L2 dot theta sin thetabigg)^2 bigg] + frac 12bigg(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: $T=dfrac 16 ML^2 dot theta ^2$