 Solution

For a fixed attachment ratio $\delta >0$, making the disc much heavier than the rod givesIf the disc is also small relative to the attachment distance, $\alpha /\delta =a/d\to 0$, its moment of inertia about its own center is negligible compared with $M{d}^2$. ThusThis 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 $\alpha \ll 1$ and $\mu \gg 1$ do not alone justify that reduction. The needed dominance conditions are $\mu \delta \gg 1$, $\mu {\delta }^2\gg 1$, and $\alpha /\delta \ll 1$, 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 $\delta >0$.