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The moment of inertia about a specified axis measures how a mass distribution resists changes of angular velocity about that axis. For point masses $m_i$ at perpendicular distances $r_{\perp i}$, or a continuous distribution with mass density $\rho$,Its SI unit is $\mathrm{kg}{\mathrm{m}}^2$. The distance is to the entire axis, not to one point on it: moving a point mass along the axis leaves its contribution unchanged. The moment of inertia depends on both the distribution and the chosen axis; it is not determined by total mass alone. About an axis fixed in an inertial frame, a rigid body with larger moment of inertia has smaller angular acceleration under the same net axial torque, and greater rotational kinetic energy at the same angular speed.^[ref]

For a simple model, put four equal point masses of total mass $M$ on light rigid spokes at radius $a$. Their moment of inertia is $M{a}^2$. Doubling $a$ quadruples the moment of inertia without changing $M$. The illustrated placements $a=R/3$ and $a=R$ give $M{R}^2/9$ and $M{R}^2$. Actual spokes and finite-sized weights contribute their own moments of inertia; the point mass model neglects them. Repositioning weights during rotation also breaks the fixed-distribution assumption, so it requires a separate work and angular momentum calculation.

The moment of inertia of a uniform disc follows by dividing it into concentric annuli: $dm=(2M/R^2)rdr$ and $I={\int }_0^R{r}^2(2M/R^2)rdr=M{R}^2/2$. For an ideal thin ring every mass element is at $R$, so $I=M{R}^2$. A ring of finite radial thickness instead uses the moment of inertia of a uniform annulus, $I=M(R^2+b^2)/2$, with inner radius $b$. A moment of inertia is thus a compact summary of radial placement rather than a complete description of shape. For several joined pieces, their moments of inertia add only when each is calculated about the same axis.^[ref]

A physical pendulum makes the moment of inertia measurable through timing. Martín Monteiro, Cecilia Stari, Cecilia Cabeza and Arturo C. Martí used a metal bar with suspension holes $1.0\mathrm{cm}$ apart and an attached smartphone. They report bar length $1.199\mathrm{m}$, bar mass $0.2518\mathrm{kg}$ and phone mass $0.1311\mathrm{kg}$. Changing the suspension hole changes the pivot axis. Their readouts were phone gyroscope angular velocity records, analyzed by autocorrelation or a fit to a sinusoid, and an independent photogate; the logged records for the sinusoid fits lasted $10\mathrm{s}$. They observed agreement between the three timing methods and the geometry-based model, including a minimum of oscillation period as the suspension point varied. These are reported apparatus details and observations, not measurements made for this article.^[ref]

To infer a moment of inertia from such a measurement, determine the total mass $M_{\mathrm{tot}}$ including every attached part, locate the assembly's centre of mass $G$, and measure its distance $h>0$ below the horizontal pivot $O$. Record successive crossings in the same direction, or the elapsed time $t_N$ for $N$ complete oscillations, giving oscillation period $P=t_N/N$. The physical pendulum model with a fixed pivot, a uniform gravitational field, negligible friction and the small-angle approximation givesThe crossing times, masses and geometric distances are the measured inputs; $I_{O,\mathrm{inferred}}$ is an inference conditional on this model, while an integral over the assumed shape supplies an independent model prediction. The parallel axis theorem converts the inferred pivot moment of inertia to one about a parallel axis through $G$. The controls and interpretation of that comparison are described there. An axial measurement determines one scalar moment of inertia, not the complete inertia tensor needed for general rotation.

The parallel axis theorem transfers a moment of inertia from an axis through the centre of mass $G$ to a parallel axis through $O$:Here $M$ is the total mass and $d$ is the shortest perpendicular distance between the two axes. Both moments of inertia concern the same mass distribution and the same axis direction. The axis for $I_G$ must pass through the centre of mass; $I_O$ may refer to an axis outside the body. No uniform-density or planar-shape assumption is needed. Within any family of parallel axes, the centre of mass axis minimizes the moment of inertia.^[ref]

A short proof isolates why the centre of mass matters. Choose parallel axes perpendicular to an $xy$ plane, put $G$ at the origin, and place $O$ at $(d,0)$. Thenbecause $\int xdm=0$ in centre of mass coordinates. Individual mass elements can move closer to the new axis, but their weighted cross terms cancel. For arbitrary parallel axes $A$ and $B$, use $I_B-I_A=M(d_B^2-d_A^2)$, where each distance is measured from the parallel centre of mass axis; adding $M$ times the squared separation of $A$ and $B$ is generally incorrect. A displacement along the axis has $d=0$. Changing the axis direction instead requires the inertia tensor or, in its applicable planar setting, the perpendicular axis theorem.

For a uniform disc, mass $M$ and radius $R$, the central perpendicular moment of inertia is $I_G=M{R}^2/2$. Shifting that axis to the rim gives $d=R$ and $I_O=3M{R}^2/2$, three times $I_G$. In the computed $M=2\mathrm{kg}$, $R=0.30\mathrm{m}$ example, $I_G=0.090\mathrm{kg}{\mathrm{m}}^2$, $M{d}^2=0.180\mathrm{kg}{\mathrm{m}}^2$ and $I_O=0.270\mathrm{kg}{\mathrm{m}}^2$. At angular speed $10rad/s$ the rim-pivoted disc has rotational kinetic energy $13.5\mathrm{J}$. Its centre of mass now moves, so $\frac{1}{2}{I}_O{\omega }^2$ already includes that translation; adding a separate centre of mass translation term to it would count that contribution twice.

The physical pendulum measurement described under moment of inertia provides a practical check: infer $I_O=Mgh(P/2\pi {)}^2$ at several pivot positions, then subtract $M{h}^2$ to obtain $I_G$. Keep all pivot axes parallel and measure $h$ to the centre of mass of the entire moving assembly. The same assembly should yield the same $I_G$ within measurement precision. If a smartphone is attached, its mass and its own moment of inertia belong in both the geometric prediction and the centre of mass calculation. The published bar-and-phone study explicitly adds both contributions using the parallel axis theorem.^[ref]

For a careful replication, release small swings without a push, repeat timings at several small amplitudes, and check that the extrapolated oscillation period is stable as amplitude decreases. Time many complete oscillations and repeat the run; two crossings of a photogate per cycle must not be mistaken for two cycles. Inspect decay of amplitude and compare early and late timings to detect significant pivot friction or air drag. Mark the bar and phone mount to detect relative slip, tighten the attachment, and check that the pivot support does not rock or migrate between holes. Repeat with an independent photogate or video readout to check the phone's timing. These are proposed controls for the replication, not additional results reported by the authors. The ideal model excludes appreciable damping, large swings and flexing; subtraction of $M{h}^2$ also magnifies relative errors when $I_G$ is much smaller than $I_O$.

The oscillation period need not increase monotonically with pivot distance: the shift raises the moment of inertia but also increases the gravitational restoring torque. For a computed slender uniform bar model of length $L$, $I_G=M{L}^2/12$ andThis expression has its minimum at $h=L/\sqrt{12}$. With $L=1\mathrm{m}$, $h=0.50\mathrm{m}$ and stipulated $g=9.81m/{\mathrm{s}}^2$, the prediction is $P\simeq 1.638\mathrm{s}$; it is not a recorded timing or a prediction for the bar with a phone attached. The competing terms explain why a timing experiment can reveal a mass distribution rather than merely its weight.