= Solution
For the general <Lagrangian> just obtained, the <Euler-Lagrange equation> gives
$$
I\ddot\theta+A\sin\theta=0.
$$
Using the <small-angle approximation> $\sin\theta\simeq\theta$, this is <simple harmonic motion> with $\omega_0^2=A/I$. Thus
$$
\boxed{T=2\pi\sqrt{\frac{l(m+M_r/3)}{g(m+M_r/2)}}\quad\longrightarrow\quad
T=2\pi\sqrt{\frac lg}\ \text{for a negligible-mass rod}.}
$$
The endpoint <mass> cancels in the light-rod limit. A concrete comparison shows why the finite rod cannot silently be included: the same specified endpoint $m$ with $M_r=m$ gives $T=2\pi\sqrt{8l/(9g)}$, distinct from the negligible-rod result. This is a <physical pendulum> when the rod contributes, and a <simple pendulum> in the light-rod limit. Both expressions require small oscillation amplitude; the full sine equation remains nonlinear.
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