when to consider linear kinetic energy and when to use angular kinetic energy? I saw a problem about a pivoting rod online (http://dev.physicslab.org/Document.aspx?doctype=3&filename=RotaryMotion_RotationalDynamicsPivotingRods.xml), I am thinking the problem if we initially hold the rod horizontally with one end hinged on the wall without friction. When I let go, the rod rotate all the way to the vertical position (downward). If we want to consider the instant speed of the center of mass of the rod when it rotated to vertical position downward. In this case, the energy conserved since no friction in the pivot and we ignore the air resistance. So 
$$G.P.E = \frac{mv^2}{2}$$
But in that webpage, it consider the angular kinetic energy instead. But in the text, it said that I should consider both linear and angular kinetic energy at center of mass for a rotational object, $K.E. = \frac{MV_{CM}^2}{2} + \frac{I_{CM}\omega^2}{2}$. It is confusing to me because when the rod vertical, the speed is pointing to the right so the speed of the center of mass is linear. So which one is correct?
 A: In the problem given on the link, you can do either of the following two calculations:-   
1)Only angular kinetic energy considering $I$ and $\omega$ about the point, about which the rod performs pure rotational motion (i.e. only rotational and no translational), which in this case will the hinged end of the rod. Calling that point $O$:-
$$K=\frac12 I_O\omega_{O}^2=\frac12 \frac{Ml^2}{3}\omega^2=\frac{Ml^2\omega^2}{6}$$
2)Or you can calculate translational energy of centre of mass and the rotational energy about centre of mass. This method is true in all cases.i.e you can always find the total energy by summing the translational energy of the centre of mass, and the rotational energy about(considering $I$ and $\omega$ about the COM) the centre of mass. In this case ($\omega_{com}=\omega_O$, and $v_{com}=\omega l/2$):-
$$K=\frac12Mv_{com}^2+\frac12I_{com}\omega_{com}^2=\frac18 M\omega^2l^2+\frac12 \frac{Ml^2}{12}\omega^2=\frac{Ml^2\omega^2}{6}$$
Thus, any way you calculate, the net energy turns out to be the same. Here we cannot use only translational kinetic energy of com or otherwise, because in no frame of reference is the motion purely translational, but is either purely rotational (about $)$) or both rotational and translational (about COM). Therefore, if you are considering the motion about a point, which is purely rotational about that point, you need to use only rotational KE , but in the com frame, you need to use both, rotational and translational kinetic energy about and of the com respetively.
