Is the angular acceleration of a rigid body independent of the reference point? I'm solving this particular question (refer to image) in which a rod is released from equilibrium after one of the two springs to which it is initially connected is disconnected. In order to calculate the angular acceleration, I have used the torque and moment of inertia about (a) the centre of mass and (b) the pivot point (the second spring). Technically, shouldn't both turn out to be the same angular acceleration? Why am I getting different values?
Can someone please explain why it is incorrect to calculate torque and moment of inertia to find angular acceleration about any arbitrary axis? Why does it have to be the centre of mass?
(Additionally, I would also like to know which answer is correct)
A proof of the logic would be much appreciated.
https://i.stack.imgur.com/HVynj.jpg
 A: No, angular acceleration, angular momentum, torque, etc. are not independent of the reference point. For example, torque is defined as $\boldsymbol\tau=\mathbf r\times\mathbf F$, where the vector $\mathbf r$ is the vector pointing from the reference point to the point of application of the force; angular momentum of a point particle is $\mathbf L=\mathbf r\times\mathbf p$, where the vector $\mathbf r$ is the vector pointing from the reference point to the location of the particle$^*$. If you change the reference point you change $\mathbf r$, and in general you change these angular quantities.

Can someone please explain why it is incorrect to calculate torque and moment of inertia to find angular acceleration about any arbitrary axis? Why does it have to be the centre of mass?

It is not incorrect to choose some other reference point that is not at the center of mass. Often calculations are much easier when the center of mass is used, but this is not true always. You are allowed to choose any reference point you want. But once you do, you need to make sure you stay consistent with this point when doing calculations involving multiple quantities that rely on that reference point. For example, the "rotational analog" of Newton's second law gives us $\boldsymbol\tau=\text d\mathbf L/\text d t$; the reference point for $\boldsymbol\tau$ needs to be the same reference point for $\mathbf L$.

$^*$The angular momentum of a rigid body can be built up from this definition.
