# Does the angular kinetic energy accounts for the translational kinetic energy of the centre of mass?

So I am looking at the problem 1.19 (page 33) of "Analytical Mechanics by Hand and Finch". The problem is about a stick held at a vertical angle $\theta$ on a table with its base fixed on the bottom of table as shown below

Now my professor said that the kinetic energy of the system is: $$T = \frac{1}{2}mv^2 + \frac{1}{2}Iw^2$$

where $m$ is the mass of the stick, $v$ the velocity ($\dot{\theta}$), $I$ the moment of inertia and $w$ the angular frequency. Now my confusion arises from the fact that I always though that the angular kinetic energy of a system also accounts for the translational kinetic energy of the centre of mass of the system (for example see here the answer at the following link Energy in the physical pendulum). So I thought the kinetic energy would simply be: $$T= \frac{1}{2}Iw^2$$. Is my teacher correct and if yes, why?

Actually, both formulas are correct, depending on the value used for I. If you calculate I as the moment about the center of mass then: $$T = \frac{1}{2} m v^2 + \frac{1}{2} I_{Center} \omega^2$$ If you calculate I as the moment of inertia about the end then $$T=\frac{1}{2} I_{End} \omega^2$$ Where $$I_{Center}=\frac{1}{12}mL^2$$ $$I_{End}=\frac{1}{3}mL^2$$ A little algebra should convince you that they are the same.