Kinetic energy of the compound pendulum

Question

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 $$T = \frac 16 ML^2{\dot \theta} ^2$$

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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?

Answer 1

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: $T=\dfrac 16 ML^2 \dot \theta ^2$