= Solution
{c}
For a fixed attachment ratio $\delta>0$, making the disc much heavier than the rod gives
$$
\omega^2\longrightarrow\frac g\ell\frac{\delta}{\delta^2+\alpha^2/2}
=\frac{gd}{d^2+a^2/2}.
$$
If the disc is also small relative to the attachment distance, $\alpha/\delta=a/d\to0$, its <moment of inertia> about its own center is negligible compared with $Md^2$. Thus
$$
\boxed{\omega^2\longrightarrow\frac gd,\qquad T\longrightarrow2\pi\sqrt{\frac dg}.}
$$
This is a <simple pendulum> with a <point mass> at the disc's center and effective length $d$; it is length $\ell$ only if $d=\ell$. Even before the <small-angle approximation>, its equation tends to $\ddot\theta=-(g/d)\sin\theta$.
If $\delta$ is allowed to tend to zero together with the other parameters, the bare conditions $\alpha\ll1$ and $\mu\gg1$ do not alone justify that reduction. The needed dominance conditions are $\mu\delta\gg1$, $\mu\delta^2\gg1$, and $\alpha/\delta\ll1$, so that the rod contributes negligible gravitational <torque> and <moment of inertia>, and the disc behaves as a <point mass>. They follow automatically in the combined limit with fixed $\delta>0$.
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