Pendulum with a uniform rod and an endpoint mass ($T=2\pi \sqrt{l(m+M_r/3)/[g(m+M_r/2)]}$)

A uniform rigid rod of length $l$ and mass $M_r$, pivoted at one end with a point mass $m$ at the other, has pivot moment of inertia $I=l^2(m+M_r/3)$. Its gravitational potential energy is $-gl(m+M_r/2)\cos \theta$. The planar Euler-Lagrange equation is $I\ddot{\theta }+gl(m+M_r/2)\sin \theta =0$, giving the displayed small-amplitude oscillation period. The limit $M_r=0$ gives a simple pendulum, whereas $m=0$ gives the uniform-rod physical pendulum. Thinness does not specify a rod's mass, so both contributions must be stated or explicitly neglected.