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The uniform rod has linear mass density $m/\ell$. Integrating squared distance from the pivot gives its moment of inertia about the axis perpendicular to the plane of motion:Measure $\theta$ from the downward vertical. The center of mass is at distance $\ell /2$ from the pivot, so gravity produces the restoring torque $-mg(\ell /2)\sin \theta$. The fixed-axis rotational equation of motion is $I_O\ddot{\theta }={\tau }_O$. Therefore the physical pendulum obeysThe pivot force has zero torque about the pivot.

The rod and disc contribute restoring torques $-mg(\ell /2)\sin \theta$ and $-Mgd\sin \theta$, respectively. Using i in the rotational equation of motion givesOnly the disc's center of mass enters its gravitational torque; its finite radius enters its moment of inertia. The formula is also regular at $M=m$, although the source happens to exclude that value.

Axis theorems. The parallel axis theorem stateswhere $I_G$ 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 $\int xdm=0$ at the centre of mass. The perpendicular axis theorem applies to a planar lamina in the $xy$ plane: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.

Cylinder integrals. Put the cylinder's midpoint at $z=0$ and its symmetry axis along $z$. Its uniform mass density is $\rho =m/(\pi r^{2}l)$. The transverse moment of inertia of a uniform solid cylinder about the $x$ axis isEquivalently each thin circular slice has diametral moment of inertia $r^{2}dm/4$, by the perpendicular axis theorem and symmetry, and shifting it by $z$ adds $z^{2}dm$. The parallel axis theorem then gives the transverse end-axis moment of inertia:The axial moment of inertia of a uniform solid cylinder is $I_{\parallel }=m{r}^2/2$. Equating it to $I_{\perp ,G}$ givesHere 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.

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 $\theta$. DefineThe 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 areFor completeness, Newton's second law for the centre of mass gives both components of the pivot force: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 $m_{T}g(\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta })$. 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 periodFor release from a finite amplitude $0<{\theta }_0<\pi$, conservation of mechanical energy instead gives $\frac{1}{2}I{\dot{\theta }}^2=gA(\cos \theta -\cos {\theta }_0)$. Integrating a quarter-cycle gives the exact oscillation periodwhere $K$ 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 $T_s=2\pi \sqrt{l/g}$: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.

The moment of inertia about a specified axis is $I=\int s^{2}dm$, where $s$ is perpendicular distance from the axis. The parallel axis theorem states $I=I_{\mathrm{cm}}+m{d}^2$ for two parallel axes separated by $d$, one passing through the centre of mass of a body of mass $m$.

It is easiest to calculate the eccentric disc by subtraction. Let $m_f$ be the mass of the unperforated disc and $m_h$ the removed mass. Uniform thickness and density give $m_h=m_f/4$ and $m_f-m_h=M$, hence $m_f=4M/3$ and $m_h=M/3$. The hole is tangent to the rim, so its centre, and therefore the shaft axis, lies $R/2$ from the original disc centre. About the shaft axis,The shaft contributes $I_{\mathrm{shaft}}=M(R/2{)}^2/2=M{R}^2/8$. ThusThe motor's torque is $P/\omega$ at nonzero angular velocity, while the resisting torque is $-\gamma \omega$. The rotational equation of motion and its equivalent energy balance areAt the steady state, input power equals dissipated power, givingFor the acceleration, use $u={\omega }^2$. Then $\dot{u}+(2\gamma /I)u=2P/I$ with $u(0)=0$, soThe approach to ${\omega }_0$ takes infinite time. This constant-power rotational acceleration idealizes the start: $P/\omega$ diverges at rest, although the energy balance has a well-defined limiting solution. A real motor has finite starting torque.