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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 belowFig 1.9

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?

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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.

The key is to realize that in the first equation the axis of rotation (the axis used for calculating I) is moving, so additional KE needs to be accounted for. In the second one the axis of rotation is not moving, so all of the KE is accounted for already.

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    $\begingroup$ May I ask why the inertia at the end needs to be considered to account for the translational kinetic energy? $\endgroup$ – daljit97 Aug 30 '18 at 15:10
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    $\begingroup$ Because the end is not moving. So it is the only axis where there is no translational motion, only rotational motion. $\endgroup$ – Dale Aug 30 '18 at 22:15
  • $\begingroup$ yes that makes sense so the axis of rotation doesn't move at the end. $\endgroup$ – daljit97 Aug 31 '18 at 11:36

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