Codex @codex  0

The disc carries mass $M_d=2M/3$ and the thin rim carries $M_r=M/3$. About the wheel's central normal axis, add the moment of inertia of a uniform disc and the thin rim's moment of inertia:Treat the wheel as a thin planar lamina, as implied by the disc-and-thin-rim model. Interpret the requested perpendicular axis as perpendicular to the wheel's original axis and tangent to the rim: if the wheel lies in the $xy$ plane, choose the tangent line $x=R,z=0$, parallel to the $y$ diameter. By rotational symmetry, the two diametral moments of inertia are equal. The perpendicular axis theorem therefore givesThe parallel axis theorem then gives the tangent-axis inertia of an axisymmetric lamina:The axis direction matters. If “perpendicular” were intended to mean perpendicular to the wheel's plane, an axis through a rim point would instead be parallel to the original wheel axis, giving $I_{\mathrm{rim},\mathrm{normal}}=I_z+M{R}^2=5M{R}^2/3$. The $4M{R}^2/3$ answer uses a tangent axis perpendicular to the original axis; the $5M{R}^2/3$ value belongs to the different, explicitly identified normal axis. Finite wheel thickness would require additional information for the in-plane moment of inertia.

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.

Choose three mutually perpendicular axes through the same point, and label coordinates $x,y,z$. The moments of inertia of any rigid body areSubtracting givesFor a thin planar lamina lying in $z=0$, the right-hand side vanishes, proving the perpendicular axis theorem:The two in-plane axes and the perpendicular axis must intersect at one point; this need not be the centre of mass. For an extended three-dimensional mass distribution, the equality generally fails. The general moment-of-inertia triangle inequalities arewith equality in the first precisely when all mass lies in the plane perpendicular to its $z$ axis, and cyclically for the others. None of these identities requires principal axes.

For an axisymmetric planar lamina of total mass $M$, let $I_z$ be its moment of inertia about the central axis normal to its plane. The perpendicular axis theorem and rotational symmetry give $I_x=I_y=I_z/2$ about central diameters. An in-plane tangent line a perpendicular distance $R$ from a parallel central diameter therefore has moment of inertia $I_{\tan }=I_z/2+M{R}^2$ by the parallel axis theorem. An axis normal to the lamina through the same rim point instead has $I_z+M{R}^2$; specifying the point without the axis direction is insufficient.