= Solution
\b[Axis theorems.] The <parallel axis theorem> states
$$
I_O=I_G+md^2,
$$
where $I_G$ is the <moment of inertia> about an axis through the <centre of mass>, $I_O$ is about a parallel axis, $m$ is the body's total <mass>, and $d$ is the perpendicular separation of the axes. Expanding $\int[(x-d)^2+y^2]dm$ proves the result because $\int x\,dm=0$ at the <centre of mass>. The <perpendicular axis theorem> applies to a <planar lamina> in the $xy$ plane:
$$
I_z=I_x+I_y.
$$
All three axes meet at the same point, $x,y$ lie in the lamina and $z$ is normal to it; the identity follows from $x^2+y^2$ being the squared distance to the $z$ axis. It is not a theorem for an arbitrary three-dimensional body.
\b[Cylinder integrals.] Put the cylinder's midpoint at $z=0$ and its symmetry axis along $z$. Its uniform <mass density> is $\rho=m/(\pi r^2l)$. The <transverse moment of inertia of a uniform solid cylinder> about the $x$ axis is
$$
I_{\perp,G}=\rho\int_{-l/2}^{l/2}dz\int_0^r s\,ds\int_0^{2\pi}d\varphi\,(z^2+s^2\sin^2\varphi)
=\frac{ml^2}{12}+\frac{mr^2}{4}.
$$
Equivalently each thin circular slice has diametral <moment of inertia> $r^2dm/4$, by the <perpendicular axis theorem> and symmetry, and shifting it by $z$ adds $z^2dm$. The <parallel axis theorem> then gives the transverse end-axis <moment of inertia>:
$$
\boxed{I_{\perp,G}=\frac{m}{12}(l^2+3r^2),\qquad
I_{\perp,\rm end}=\frac{m}{12}(4l^2+3r^2).}
$$
The axial <moment of inertia of a uniform solid cylinder> is $I_\parallel=mr^2/2$. Equating it to $I_{\perp,G}$ gives
$$
\boxed{l^2=3r^2,\qquad l=\sqrt3\,r.}
$$
Here the source's principal axis means the cylinder's symmetry axis; its transverse centre axes are also <principal axes>. This equality describes a short cylinder rather than the thin-rod limit.
\b[Hammer dynamics and forces.] Treat the compact head as a <point mass> and neglect its intrinsic <moment of inertia>. Let $\theta$ be the displacement from downward vertical, and let $\mathbf e_r$ point along the rod from pivot to head, with $\mathbf e_\theta$ in the direction of increasing $\theta$. Define
$$
m_T=M+m,\qquad A=Ml+\frac{ml}{2},\qquad d=\frac{A}{m_T},\qquad
I=Ml^2+m\left(\frac{l^2}{3}+\frac{r^2}{4}\right).
$$
The two external weights $Mg$ and $mg$ act at distances $l$ and $l/2$ from the pivot. The frictionless pivot supplies a reaction $\mathbf R$ but no <torque> about its axis. The total restoring <torque> and <rotational equation of motion> are
$$
\tau=-gA\sin\theta,
\qquad \boxed{I\ddot\theta+gA\sin\theta=0.}
$$
For completeness, <Newton's second law> for the <centre of mass> gives both components of the pivot <force>:
$$
\boxed{R_r=-m_Td\dot\theta^2-m_Tg\cos\theta,\qquad
R_\theta=m_Td\ddot\theta+m_Tg\sin\theta.}
$$
The negative radial component points towards the pivot. These follow from the <centre of mass> <acceleration> $-d\dot\theta^2\mathbf e_r+d\ddot\theta\mathbf e_\theta$ and the total weight $m_Tg(\cos\theta\mathbf e_r-\sin\theta\mathbf e_\theta)$. They specify the external <forces> even when the <oscillation> is not small.
With the <small-angle approximation>, $\sin\theta\simeq\theta$, this is <simple harmonic motion>. The <physical pendulum> has small-amplitude <oscillation period>
$$
\boxed{T_0=2\pi\sqrt{\frac{I}{gA}}
=2\pi\sqrt{\frac{Ml^2+m(l^2/3+r^2/4)}{gl(M+m/2)}}.}
$$
For release from a finite amplitude $0<\theta_0<\pi$, <conservation of mechanical energy> instead gives $\tfrac12I\dot\theta^2=gA(\cos\theta-\cos\theta_0)$. Integrating a quarter-cycle gives the exact <oscillation period>
$$
T(\theta_0)=4\sqrt{\frac{I}{gA}}K\!\left(\sin\frac{\theta_0}{2}\right),
\qquad K(k)=\int_0^{\pi/2}\frac{d\psi}{\sqrt{1-k^2\sin^2\psi}},
$$
where $K$ is the complete <elliptic integral of the first kind>; $K(0)=\pi/2$ recovers $T_0$.
For $m=M/10$ and $r=l/20$, compare with the <simple pendulum> value $T_s=2\pi\sqrt{l/g}$:
$$
\boxed{\frac{T_0}{T_s}
=\sqrt{\frac{1+\frac1{10}(\frac13+\frac1{1600})}{1+\frac1{20}}}
=\sqrt{\frac{49603}{50400}}\simeq0.992062.}
$$
For small <oscillations>, the hammer's <oscillation period> is about $0.794\%$ shorter. The extra rod weight adds restoring <torque> proportionately faster than it adds pivot <moment of inertia>.
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