About which axis we should take moment of inertia? In the formula of rotational kinetic energy $\frac{1}{2} I \omega^2$; which axis should we take the moment of inertia (about the centre of mass (COM) or about the axis of rotation)? For a pure rolling sphere, we take $I$ about the COM, which is also the axis of rotation, but for a rod hinged at one end, which $I$ should take (COM or the hinge)?
Also, in $\tau = I \alpha$ which $I$ should we take (about COM or axis about which torque is written)? 
Any help is appreciated.
 A: You should always calculate rotational kinetic energy and torque about a point about which only pure rotation takes place. Thus, for a hinged object only rotational motion exists at hinge. Hence, MOI is calculated about the hinge.
Although, in the case of torque you can choose any axis (perpendicular to the plane of rotation)only if summation of external forces equate to zero.
A: the MOI is taken about the axis of rotation. IF your body is rotating about its center of mass than that is the axis of rotation.
Example the kinetic energy of the earth if to be calculated use the axis which pass through the north and south poles of the planet
If a rod is rotating about an axis than both the rod Inertia and torque are to be calculated from the hinge
HOPE IT HELPS :)
A: Kinetic energy of any system depends upon reference frame.
In combined rotation and translation,
In lab frame,
$$K_{sys,lab}=K_{sys,cm}+K_{cm,lab}$$
$$K_{sys,lab}=\frac{1}{2}{mv_{cm}^2}+\frac{1}{2}{I_{cm}\omega^2}$$
In your scenario,it's quite clear from these equations that $I$ should be written w.r.t. $com$.
In the equation $\vec{T_{net}}=I\vec{\alpha}$, $I$ should be written w.r.t that axis about which $\vec{T_{net}}$ is calculated.
A: Let us say that here the most straightforward approach to get the job done is to compute the moment of inertia with respect to the hinge, either using Huygens-Steiner theorem:
$$ I_{AOR} = I_{COM} + md^2 $$
($m$ being the mass of the rod and $d$ the distance of the COM from the AOR) 
or computing it from scratch:
$$ I = \int_0^L \rho r^2 dr $$
$L$ being the length of the rod and $\rho$ its density.
(assuming that the rod is unidimensional)
Then you can use the $I$ that you get here in both $K={1\over 2} I \omega ^2$ and $\tau = I \alpha$, without worrying about other terms due to moving reference frames.
