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A Cavendish torsion balance is a torsion balance arranged to measure gravitational torque from nearby source masses. Its suspended horizontal rod rotates about a vertical fiber. In the linear elastic regime, a twist angle $\theta$, measured from the fiber's untwisted orientation, gives a restoring torque ${\tau }_{\mathrm{fiber}}=-\kappa \theta$, where $\kappa >0$ is the torsion constant. The minus sign is essential: the fiber resists the twist. Under an approximately constant gravitational torque, the rotational equation using moment of inertia $I$ and angular acceleration $\ddot{\theta }$ isAt equilibrium, the two torques balance. The angle is a displacement from the fiber's reference orientation, not the largest excursion during an oscillation. ^[ref]

For the toy parameters in the Cavendish experiment, the predicted equilibrium twist isHere $L|{\theta }_{\mathrm{eq}}|/r\simeq 0.0107$, consistent with using a small geometric displacement in the toy model, but not an estimate of its total measurement error. Reversing the source positions ideally takes the equilibrium angle from $+{\theta }_{\mathrm{eq}}$ to $-{\theta }_{\mathrm{eq}}$: the magnitude of the change is $2|{\theta }_{\mathrm{eq}}|\simeq 0.6119^{\circ }$. A stable zero offset then cancels in the difference. A real reading may track a reflected light spot; the law of reflection doubles a mirror's rotation in the outgoing beam, so the spot's angular change must not be confused with the rod's angle. ^[ref]

How can the fiber's torsion constant be found without already knowing the gravitational force? Measure the oscillation period and calculate the moment of inertia. For the same toy rod, neglecting its mass and treating both test masses as point masses, $I=m{L}^2+m{L}^2=2m{L}^2$. With the source torque absent or treated as constant, let $\varphi =\theta -{\theta }_{\mathrm{eq}}$ be a small departure from equilibrium. Neglect damping and any change of gravitational torque with angle. The rotational equation becomes simple harmonic motion:Here $A$ is the angular amplitude, $\delta$ specifies the initial phase, and $\omega$ is the angular frequency. A softer fiber gives a larger equilibrium twist but also a longer oscillation period: greater sensitivity costs observation time. An undamped oscillation does not settle at equilibrium; damping is needed for settling. ^[ref]

The toy moment of inertia and undamped oscillation period areTiming this oscillation period would recover the assumed torsion constant. The units check explains the result: torque divided by torsion constant is an angle in radians, and $I/\kappa$ has units of time squared when radians are treated as dimensionless in SI.

Combining the equilibrium twist with the period calibration, the two-near-pair model would infer the gravitational constant asThe test mass cancels in the second expression because it enters both the gravitational torque and the toy moment of inertia. If the measured quantity is the magnitude $|\Delta \theta |$ of the change between opposite source positions, use $|{\theta }_{\mathrm{eq}}|=|\Delta \theta |/2$. Substituting the toy twist and oscillation period returns the input $G$; it is a consistency check, not independent experimental evidence.

The sensitivity of a Cavendish torsion balance also explains its practical difficulty. Air motion can exert unwanted torque; unequal temperatures can drive that motion. Cavendish discusses this effect on p. 471 and describes observing from outside a closed room to avoid disturbing the instrument. ^[ref] Mechanical vibrations can excite the same slow oscillations used to calibrate the torsion constant, while changes in the fiber or source separations alter the inferred force. Precision therefore depends on resolving the equilibrium angle, calibrating the full suspended moment of inertia and torsion constant, and modeling the actual source geometry. The two-near-pair model omits opposite-source attraction, rod contributions, finite-size effects, damping, and the variation of gravitational torque as the rod turns. ^[ref]

Three identical cylinders initially form an equilateral triangle on a horizontal floor and are released from rest. If each has moment of inertia $\beta m{R}^2$, the initial top downward acceleration is $g/(7+6\beta )$ for no-slip rough ground and smooth cylinder contacts, or $g/(7+8\beta )$ for smooth ground and no-slip rough cylinder contacts. In both cases the lower cylinders separate with outward acceleration $u=\sqrt{3}{a}_{\mathrm{top}}$. The corresponding lower-cylinder angular acceleration magnitudes are $u/R$ and $2a_{\mathrm{top}}/R$, with opposite rotation senses between the cases. Sufficient static friction is required; a finite coefficient allowing sliding changes the problem.

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 area element in plane polar coordinates is $rdrd\theta$. Thus the axial moment of inertia isFor the removed material, the angular width is $2\pi /13$, andIts moment of inertia is therefore $I_{\mathrm{cut}}=\pi {\rho }_0{a}^5/160$, leavingThe torque equation $I\dot{\Omega }=\tau$ gives constant angular acceleration, so the time to reach the prescribed angular speed from rest is

To find the centre of mass after removing material, let ${\mathbf{e}}_1$ point along the bisector of the missing sector in the disc. The original disc has zero first mass moment about its centre. The missing sector has zero transverse first moment by reflection symmetry, while its component along ${\mathbf{e}}_1$ isbecause the radial integral is $11a^4/192$. The remaining first moment is $-Q_{\mathrm{cut}}{\mathbf{e}}_1$. Dividing by the given mass $k{\rho }_0{a}^3$ gives the body-fixed centre of massIt lies away from the missing sector. In the inertial frame, let $t_{\ast }$ be the moment the applied torque stops and let $\psi ={\psi }_{\ast }+\Omega (t-t_{\ast })$ give the orientation of that sector's bisector. Then, with $d=11a\sin (\pi /13)/(96k)$,The centre of mass has inward centripetal acceleration of magnitude ${\Omega }^{2}d$. The required real force is the constraint reaction exerted by the fixed rod and its support. Its horizontal resultant is $k{\rho }_0{a}^3\ddot{\mathbf{R}}$; a zero applied axial torque does not imply a zero resultant force.

Put $D=m{R}^2+{\mathcal{I}}_{33}=m(R^2+a^2/2)$. The angular acceleration is constant. The initial rest conditions and $x(0)=\varphi (0)=0$ giveFor the upward rightward pull in the diagram, $0\le \alpha \le \pi /2$, the sign of $x(t)$ is the sign of $R\cos \alpha -a$. Consequently it rolls towards the horizontal pull when $0\le \alpha <\arccos (a/R)$. At $\alpha =\arccos (a/R)$ it remains at rest, with balanced forces and torques. For $\arccos (a/R)<\alpha \le \pi /2$, it rolls to the left. In particular a horizontal pull, $\alpha =0$, makes it roll to the right, since $R>a$. If other pulling angles are allowed, the general condition for motion in the same horizontal direction as the pull is $\cos \alpha (R\cos \alpha -a)>0$.

The required static friction at each rail follows from the translational equation:For $0\le \alpha \le \pi /2$ it is positive, agreeing with the leftward force convention. The assumed rolling without slipping is physically possible only if the rails can supply this force while maintaining contact. With coefficient of static friction ${\mu }_s$, the necessary contact inequalities areFor the usual pulling range, these becomeIf either fails, the proposed motion requires lift-off or more static friction than the rails provide. Equality $T\sin \alpha =mg$ is also incompatible here: it gives $N=0$ but a nonzero required friction. The question supplies no coefficient of static friction, so it determines the required friction exactly but gives no numerical slip threshold. If $F$ is regarded as a prescribed force rather than an available contact response, it must equal $F_{\mathrm{req}}$ for the rolling constraint to hold.

Assume the cylinders are released from rest. In each case, “there is friction” is interpreted as sufficient static friction for rolling without slipping at the rough contacts. Without a coefficient or a no-slip assumption, sliding cases would not have a unique specified acceleration. On release the two lower cylinders separate; their mutual contact must not be retained as a rigid constraint.

Let the top cylinder's initial downward acceleration be $w$, and each lower cylinder's outward horizontal acceleration be $u$. The top cylinder initially has no horizontal acceleration. Differentiating the fixed upper-to-lower centre separation twice at the initial zero velocity givesBy reflection symmetry the top cylinder has zero angular acceleration, while the lower cylinders rotate in opposite senses.

Take rightward translation as positive and counterclockwise angular acceleration $\alpha$ as positive. Let $f$ denote signed ground friction, positive to the right. The source's lower-tangent winding makes the tension exert counterclockwise torque $Tr$. For a horizontal pull, translation and rotation obeyThe final relation is the rolling without slipping constraint. Eliminating $f$ and $\alpha$ givesThus the requested magnitude of static friction isIt acts leftward. Released from rest, the spool accelerates rightward, toward the pulling hand, and rotates clockwise. There is no contradiction between leftward friction and rightward motion: static friction prevents slipping of the contact point, and the horizontal tension is larger than $|f|$ because $r<R$.

Let $t=0$ be the start of deposition and $I_0=M{R}^2/2$ be the original moment of inertia of a uniform disc. Assume that all deposited sand sticks and remains at radius $r$, with negligible thickness and no losses. The sand's incoming angular momentum about the vertical axis is zero. Neither the bearing nor gravity supplies an axial torque, so conservation of angular momentum givesConsequently the angular velocity and angular acceleration areThe rotational kinetic energy isThus, for $\alpha r^2>0$,This is mass-loaded rotation: the sticking process dissipates mechanical energy even though angular momentum is conserved. Indeed $\dot{K}=-\alpha r^2{\omega }^2/2$. If $r=0$ or $\alpha =0$, no additional axial moment of inertia is deposited and the nonzero initial rotational kinetic energy never halves.

Definitions. The torque of a force $\mathbf{F}$ about an origin is $\mathbf{\tau }=\mathbf{r}\times \mathbf{F}$, where $\mathbf{r}$ runs from that origin to the force's point of application. Its component about a fixed axis changes angular momentum about that axis. The moment of inertia about that axis iswhere $s$ is perpendicular distance to the axis. It measures resistance to angular acceleration; for fixed-axis rigid rotation, ${\tau }_{\mathrm{net}}=I\alpha$. The units are $\mathrm{N}\mathrm{m}$ for torque and $\mathrm{kg}{\mathrm{m}}^2$ for moment of inertia.

Mass and axial inertia. Slice the shaft perpendicular to its common axis. A slice of radius $r(z)$ has mass $dm=\rho \pi r(z{)}^{2}dz$ and moment of inertia $dI=\frac{1}{2}r(z{)}^{2}dm$, using the moment of inertia of a uniform disc. Thus the axial moment of inertia of a solid of revolution uses the second and fourth powers of its radius:For the tapered section, take $z$ from its narrow end toward its wide end, so $r(z)=a(1+z/l)$ for $0\le z\le l$. Its contributions areThe two cylinders contribute $4\rho \pi a^{2}l$ and $\rho \pi a^{2}l$ to mass, and $8\rho \pi a^{4}l$ and $\frac{1}{2}\rho \pi a^{4}l$ to moment of inertia. Adding the three sections givesAll slice rotation axes coincide with the shaft axis, so the moments add directly.

Frictional torque. The normal load on the narrow circular end is $Mg$. With the stipulated uniform pressure, $p_N=Mg/(\pi a^2)$. A contact annulus of radius $s$ and width $ds$ has area $dA=2\pi sds$, normal force $p_{N}dA$ and tangential kinetic friction $dF=\mu p_{N}dA$. Multiplying by its lever arm $s$ and integrating gives the uniform-pressure friction torque on a circular contact:Here $G_f$ is the positive magnitude; its torque opposes the angular velocity.

Steady speed and stopping rotations. Mechanical power in rotation is $P=\tau \omega$. At steady angular velocity, the driving and frictional torques balance, so constant-power rotation against dry friction givesUsing $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $a=0.100\mathrm{m}$,Once the motor is removed, the speed-independent kinetic friction gives constant angular acceleration $-G_f/I$. Equivalently, the initial rotational kinetic energy is dissipated as frictional work. The stopping angle under constant resisting torque is thereforeThe total angular displacement corresponds toHence the shaft completesand then another $0.16668$ of a rotation. Its stopping time is $I{\omega }_{\ast }/G_f=2.05461\mathrm{s}$. Using the rounded value $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ changes the fractional rotation slightly but leaves four complete rotations.

A rotary encoder converts shaft rotation into a readout of angle or changes in angle. The PS-2120A uses interruptions of a light beam to encode shaft motion; its product specification gives 4000 divisions per revolution. That count is a resolution specification, not a guarantee of the precision of an inferred angular acceleration or moment of inertia. ^[ref]

PASCO's PASPORT Rotary Motion Sensor Experiment Guide, Experiment 1, specifies a PS-2120A sensor, a rod carrying two equal-radius weights, a three-step drive wheel, a thread and hanging mass, a clamp-on guide wheel, a support, a balance and calipers. The thread leaves the drive wheel tangentially. The sensor records angular velocity against time; a linear regression over the accelerating interval supplies angular acceleration $\alpha$. The guide measures the drive radius with calipers and repeats the run with the weights removed to subtract the apparatus's own moment of inertia. Its tables are blank: it prescribes a measurement and does not report an observed numerical agreement. ^[ref]

With hanging mass $m_h$ including its hanger, effective drive radius $b$, no thread slip and negligible guide-wheel inertia, Newton's second law gives thread tension $F=m_h(g-b\alpha )$. The axial torque is $bF$. Under negligible friction, the inference is thereforeRepeat the inference for the unloaded assembly and subtract: $I_{\mathrm{weights}}=I_{\mathrm{loaded}}-I_{\mathrm{empty}}$. Omitting the $m_h{b}^2$ term incorrectly treats thread tension as the hanging weight. Comparing with $(M_1+M_2)R^2$ tests the point mass approximation; finite-sized weights also have their own moments of inertia about their centres of mass. ^[ref]

Useful proposed checks go beyond that ideal inference. Verify the rotary encoder's angle scale by turning the shaft through a marked full revolution; check the recorded time against an independent clock and the balance against a known mass. Measure $b$ at the thread's running surface, including thread thickness; avoid overlapping wraps that change $b$. Release without a push, inspect the angular velocity record for a straight interval, and stop before the hanger lands or the thread runs out. Repeat with several hanging masses, weight radii and rotation directions. A consistent inferred moment of inertia across drive torques supports the model; curvature, direction dependence or drift indicates a limitation rather than a new principal moment of inertia.

For approximately constant resisting torque ${\tau }_f\ge 0$ in one rotation direction, an original calibration model isFit the left-hand quantity against $\alpha$ across several drives: the slope estimates the axial moment of inertia and the intercept estimates ${\tau }_f$. This calibration is conditional on friction remaining constant over the fitted angular speeds; a freely slowing run can expose speed dependence. Sensor differentiation noise, guide-wheel moment of inertia, thread slip, flexing connectors and uncertain weight radii limit the inference. Removing weights can also change bearing loads, so empty-apparatus subtraction does not automatically cancel friction. These are proposed controls and a model extension, not results claimed by PASCO.

A fixed-axis readout measures ${\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$, not the full inertia tensor. In a proposed generalization, repeat about $x$, $y$, $z$ and the three diagonal directions $({\mathbf{e}}_x+{\mathbf{e}}_y)/\sqrt{2}$, $({\mathbf{e}}_x+{\mathbf{e}}_z)/\sqrt{2}$, $({\mathbf{e}}_y+{\mathbf{e}}_z)/\sqrt{2}$, all through the same $O$. For example,Six independent scalar readings then determine the six entries of a symmetric matrix. Each mounting needs its own apparatus correction, and different reference points require the tensor parallel-axis theorem before the readings can be combined.

A rigid body's axial moment of inertia $I$ and a constant resisting torque ${\tau }_f>0$ give constant angular acceleration $-{\tau }_f/I$. Hence $t_{\mathrm{stop}}=I{\omega }_0/{\tau }_f$ and the angular displacement is the displayed expression. Equivalently, the lost rotational kinetic energy $I{\omega }_0^2/2$ equals the resisting work ${\tau }_f\Delta \varphi$. The total number of revolutions is $\Delta \varphi /(2\pi )$; the number of whole revolutions completed is its floor function.