When to consider rotating body as point mass?

I’m not asking you to solve this question, but here it is anyways for clarification purposes:

A rod of mass $m$ and length $2R$ can rotate about an axis passing through $O$ in vertical plane. A disc of mass $m$ and radius $R/2$ is hinged to the other end $P$ of the rod and can freely rotate about $P$. When disc is at lowest point both rod and disc has angular velocity $omega$. If rod rotates by maximum angle $theta = 60°$ with downward vertical, then find $omega$ in terms of $R$ and $g$ (all hinges are smooth).

(Answer: $omega = sqrt{frac{9g}{16R}}$ )

Now, what I concluded is that the disc spin isn’t going to complicate the problem as no force is opposing it’s rotation and it’s going to be rotating with same angular velocity throughout.

So, my flow of solution was to calculate torque about upper most $O$ point at an arbitrary angle $theta$ for the entire system (for the entire system because that way I don’t need to deal with hinge forces on either bodies), find angular velocity as a function of $g$ and $R$.

BUT, for calculating torque, I need their moment of inertia about $O$. And that’s where the problem lies.

If I calculate $Io$ as

$Io = I(rod,,about,,O) + [ I(disc,,about,,its,,COM) + m(2R)^2]$, where $m$ is the mass of disc and $2R$ is its distance from $O$, I don’t get the answer.

IF, however, I ignore the I(disc about its COM) and treat disc as a point mass at $P$, the solution works out. Why is that the case? Is it just a coincidence or can we actually do this? And why can we do this?

My solution is posted below.

And I’ve verified the answer via solving it by energy conservation, so the answer is correct.

Sorry for the fact that in one step, I’ve randomly used $L$ in place of $R$ and also forgot a minus sign. Thankfully the mistake doesn’t propagate forward in the solution.