# Energy in the physical pendulum

I am considered the problem of the physical pendulum, in particular the derivation of the energy of the physical pendulum. I am slightly confused by one point.

Looking at the derivation given in this link,

it seems to be that they only considered the gravitational potential energy and the rotational kinetic energy. However the bod is not rotating about its centre of mass, therefore should we not also account for the translational kinatic energy of the centre of mass?

If we did this, then the final expression for energy would have an additional factor of $$\frac{1}{2}Mv^2=\frac{1}{2}Ml^2\omega^2=\frac{1}{2}I\omega^2$$; so then the final expression for energy would be

$$mg\frac{d}{2}(1-cos\theta) +I\omega^2$$? Where am i going wrong here?

• The extra translational KE of the CM is already accounted for in the rotational KE. This extra energy is exactly the reason why the parallel axis theorem has a $+md^2$ term. Commented Feb 25, 2017 at 19:37

The extra translational KE of the CM is already accounted for in the rotational KE. This extra energy is exactly the reason why the parallel axis theorem has a term $+md^2$.
So $I_s=I_{cm}+m {l_{cm}}^2$ (see text after eqn 24.2.6) then the total rotational KE is $\frac12I_s\omega^2$.