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Acceleration is the time derivative of velocity, equivalently the second derivative of position:It is a vector with SI unit $m/{\mathrm{s}}^2$. A change in direction produces acceleration even if speed stays constant. For example, a car rounding a bend at steady speed needs an inward resultant force by Newton's second law. Thus acceleration does not mean only speeding up. When $v>0$, the dot product gives $dv/dt=\mathbf{v}\cdot \mathbf{a}/v$: a perpendicular acceleration changes direction without changing speed at that instant.

Take counterclockwise circular motion about the origin with radius $R=2\mathrm{m}$ and constant angular speed $\omega =0.5rad/s$. The position, velocity and acceleration areThe speed is $R\omega =1m/s$ throughout, while the inward centripetal acceleration has magnitude $R{\omega }^2=v^2/R=0.5m/{\mathrm{s}}^2$. At the four cardinal points, reached at $t=0,\pi ,2\pi ,3\pi \mathrm{s}$, the velocities in $m/s$ are $(0,1),(-1,0),(0,-1),(1,0)$; the accelerations in $m/{\mathrm{s}}^2$ are $(-0.5,0),(0,-0.5),(0.5,0),(0,0.5)$. The vectors turn together, remaining perpendicular. ^[ref]

The proof outline is just two applications of the chain rule to $\mathbf{r}(t)$; ${\sin }^2(\omega t)+{\cos }^2(\omega t)=1$ then fixes both magnitudes. The usefulness is geometric: turning requires a force even when a speed reading remains steady, and doubling speed at fixed radius quadruples the required centripetal acceleration. The same mechanism appears in a circular orbit.

Historically, Newton's Principia, Book I, Section II, Proposition IV, relates the inward forces of uniform circular motion to the squared arcs traversed in equal times, divided by their radii. In modern notation, its dependence on speed and radius is $v^2/R$; Newton presents a geometric argument rather than the coordinate derivatives above. ^[ref]

A mechanical bearing supports relative motion while transmitting constraint forces between machine components. A shaft support can permit rotation while constraining translation, with the permitted motions depending on its construction and arrangement. A ball bearing uses rolling contacts; a journal bearing supports a shaft through a sliding or fluid-film interface. The support reaction is distinct from resistance to the permitted motion: an ideal support can transmit a large transverse force with zero resisting shaft torque.^[ref]

For a simple ideal shaft supported at axial positions0 and $L$, an upward load $P$ applied at $a$ requires reactionsWith $P=200\mathrm{N}$, $L=100\mathrm{mm}$ and $a=40\mathrm{mm}$, the downward support loads are120N and80N. These reactions exist even if the shaft rotates without an applied resisting torque. In a real mechanical bearing, an illustrative frictional resisting moment $M_f=0.020\mathrm{N}\mathrm{m}$ at angular speed $100rad/s$ dissipates $M_f|\omega |=2\mathrm{W}$; force equilibrium alone does not determine that loss.

A locating bearing arrangement fixes the shaft's axial position relative to the housing, while a non-locating bearing support permits the required axial displacement, for example from thermal expansion. The manufacturer distinguishes this arrangement from adjusted and floating arrangements.^[ref] Radial support does not by itself establish which support carries axial thrust.

Selection must therefore consider permitted motion, combined loads, alignment and operating conditions. An idler support reaction determines a shaft-level load; it does not directly give each rolling contact's Hertzian contact pressure or a service-life prediction. Those require the bearing's internal geometry and a suitable load/life model.

A mechanism in which rotating weights move outward as angular speed increases, allowing their displacement to indicate or regulate a machine's speed. The steady configuration follows the balance of actual applied forces and centripetal acceleration, or equivalently of applied forces and centrifugal force in a rotating reference frame. Geometry and the restoring mechanism determine the speed-displacement relation.

For uniform circular motion of radius $R$ and angular speed $\omega$, the acceleration is directed toward the centre and has magnitude ${\omega }^{2}R=v^2/R$. Writing $\mathbf{r}$ for the position relative to the centre gives $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}$. The required resultant force is $m\ddot{\mathbf{r}}$; it may be supplied by a normal force, gravity, or another physical interaction.

Centripetal subtraction error is the residual error left when a modeled centripetal contribution is removed from a rotating accelerometer signal. For the positive magnitude $C=r{\omega }^2$, small errors in radius and angular speed giveThe residual's sign depends on whether the correction is overestimated or underestimated. A correct model also needs synchronized speed information and the appropriate sensor-axis projection; a primary rotating-sensor experiment documents the importance of radial/tangential directions and rotating-frame effects.^[ref]

Using first-order propagation of uncertainty, the standard uncertainty satisfiesThe covariance term must be included when the estimates are correlated.^[ref] For $r=0.020\mathrm{m}$, $\omega =100rad/s$, $u_r=10\mu \mathrm{m}$ and $u_{\omega }=0.1rad/s$, the two uncorrelated contributions have magnitudes0.10 and0.40m/s², giving $u_C\approx 0.4123m/{\mathrm{s}}^2$.

A centripetal timing error can remain even with accurate individual channels. If the speed used for correction lags by $\Delta t$, then $\Delta \omega \approx -\dot{\omega }\Delta t$. At the same radius and speed, acceleration rate $\dot{\omega }=10rad/{\mathrm{s}}^2$ and lag5ms produce correction error approximately $-0.20m/{\mathrm{s}}^2$. Channel timing therefore belongs in the measurement model. These linearized formulas describe small errors; they do not replace accounting for other motion or specific force contributions.

A conical pendulum is a pendulum whose bob executes uniform horizontal circular motion beneath a fixed pivot, with its string at a constant angle $\theta$ to the downward vertical. The string sweeps a right circular conical surface; the bob follows its circular rim, without any physical cone supporting it. The ideal model has a point mass $m>0$, a taut, massless, inextensible string of length $\ell >0$, uniform gravitational acceleration of magnitude $g>0$, and negligible drag force and pivot friction. A nonzero circular motion requires $0<\theta <\pi /2$. This geometry and the corresponding forces are described in Richard Fitzpatrick, The conical pendulum, equations 262–267.

The right triangle formed by the pivot, the orbit centre, and the bob gives $r=\ell \sin \theta$ and $h=\ell \cos \theta$. Here $r$ is the orbit radius and $h$ is the vertical drop from the pivot to the bob's horizontal plane, not the bob's height above a table. Write $T$ for tension, $P$ for the oscillation period of one revolution, and $\omega =2\pi /P$ for the angular speed; then the bob's speed is $v=\omega r$. A side view compresses the horizontal circle to a line segment; the top view reveals the circle and its tangent velocity.

The two actual forces on the bob are tension toward the pivot and gravitational weight vertically downward. The vertical acceleration is zero, while the horizontal centripetal acceleration is $v^2/r={\omega }^{2}r$. Thus Newton's second law givesThe inward net force is the horizontal component of tension, rather than an extra interaction to add to these two forces. Constant speed still requires acceleration because the velocity changes direction. ^[ref]

Dividing the two component equations and using $r=\ell \sin \theta$ gives the useful collection of predictionsIf $\varphi$ is the azimuthal angle, then $\dot{\varphi }=\pm \omega$; either direction of rotation satisfies ${\dot{\varphi }}^2\cos \theta =g/\ell$. The mass cancels from the angular speed and oscillation period, but not from the tension. The motion is a relative equilibrium: in a rotating frame following the bob at angular speed $\omega$, its position is fixed. This is a concrete way to see how rotational symmetry can turn an apparently moving solution into a stationary one.

A uniform disk of radius $R$ and area mass density $\rho$ has a circular hole of radius $r_o$, whose centre H is displaced by $r_o$ from the original centre C. Set $D=\pi \rho R^2$, $h=\pi \rho r_o^2$ and $W=D-h$. The parallel axis theorem gives $I_H=D{R}^2/2+D{r}_o^2-h{r}_o^2/2$. Removing the material shifts the remaining centre of mass G away from H by $CG=h{r}_o/W$, so $HG=D{r}_o/W$ and $I_G=I_H-D^2{r}_o^2/W$. Thus the rotational component of free rigid-body kinetic energy is smaller than fixed-axis kinetic energy about H at the same angular speed, although total rolling kinetic energy also includes translation.

Two equal point masses can move around the same fixed circle of radius $a$ in diametrically opposite positions. Their stationary centre of mass is the circle's center. A separation $2a$ gives gravitational force $G{m}^2/(4a^2)$ on each mass, and Newton's second law equates it to $ma{\omega }^2$. The common angular speed is therefore fixed by the displayed relation. Opposite tangential velocities produce the same sense of angular rotation.

A gear is a toothed mechanical element that transmits motion and torque through contact with another gear or a rack and pinion. The teeth maintain a prescribed relation between the motions: unlike a smooth friction drive, the ideal gear ratio is set by geometry and tooth counts. The gear pitch circle supplies an imaginary rolling geometry for describing that relation; it is not the circle through the tooth tips.^[ref]

For two external gears on fixed parallel axes, let $N_1,N_2$ be their tooth counts and let signed angular velocities ${\omega }_1,{\omega }_2$ use the same orientation. Each tooth on one gear must pass a tooth space on the other, so counting tooth passages givesThus a 20-tooth input driving a 40-tooth output makes the output turn half as fast in the opposite direction. An input of $1200$ revolutions per minute gives an output of $600$ revolutions per minute in magnitude. These signs describe an external pair; a rack and pinion instead converts rotation into translation.

The useful trade is between angular speed and torque. In an ideal steady transmission, conservation of energy equates input and output power, $|{\tau }_1{\omega }_1|=|{\tau }_2{\omega }_2|$. The same 20-tooth/40-tooth pair therefore turns an input torque of $2\mathrm{N}\mathrm{m}$ into an output torque of $4\mathrm{N}\mathrm{m}$ while halving angular speed. This does not create energy: more torque accompanies less angular displacement per input revolution. Real gears dissipate energy through friction, and backlash permits a small lost motion on reversal; tooth strength and bearings, rather than the gear ratio alone, determine allowable load.

Gear base pitch is the spacing of successive involute tooth profiles referred to the gear base circle. For base diameter $d_b$, tooth count $N$, gear module $m$ and reference gear pressure angle $\alpha$,It also gives the spacing between successive corresponding tooth contacts measured along the gear line of action. The gear circular pitch uses the reference circle instead, giving $p=\pi m$ and $p_b=p\cos \alpha$.^[ref]

For $m=2\mathrm{mm}$ and $\alpha =20{}^{\circ }$, $p=6.2832\mathrm{mm}$ but $p_b\approx 5.9043\mathrm{mm}$. A second gear with the same module but $25{}^{\circ }$ reference angle has $p_b\approx 5.6945\mathrm{mm}$. Matching reference tooth spacing alone therefore does not make these the same specified mating geometry.

Changing the mounting separation of a compatible involute pair does not change its cut profiles, base circles or gear base pitch. In operating-circle notation, $p_w=2\pi r_w/N$ changes with mounted radius, whilestays fixed. This links the mounted operating pressure angle to an invariant tooth-profile spacing.

For a tooth rotating at angular speed $\omega$, base-circle tangential speed is $r_b|\omega |$. Dividing by gear base pitch gives the gear mesh frequency,A 30-tooth gear at $600$ revolutions per minute gives $f_m=300\mathrm{Hz}$. Finally, the gear contact ratio divides usable gear path of contact length by this base pitch, so profile spacing and available contact length jointly determine overlap between successive tooth engagements.

A gear ratio states the relation between the rotational motions of connected gears. Because conventions differ, specify the input, output and direction. Here the positive reduction ratio isso $i>1$ means that the output has lower angular speed. For a single external pair on fixed axes, tooth-passage counting givesThe minus sign assumes the same positive rotation convention on both parallel axes. KHK uses the driven-to-driver tooth-count ratio for its speed-reduction examples.^[ref]

Ratios multiply through successive stages of a gear train. Consider a 12-tooth input driving a 36-tooth gear, with a 15-tooth gear rigidly attached to that intermediate rotating body and driving a 45-tooth output. Each stage reduces angular speed by three, so the combined reduction ratio is nine. An input of $1800$ revolutions per minute yields an output of $200$ revolutions per minute. Two external meshes reverse direction twice, so input and output rotate in the same direction.

An intermediate gear used only to pass rotation onwards gives a different result. If a 12-tooth input drives a 36-tooth intermediate gear that itself drives a 48-tooth output, the ratio is $(36/12)(48/36)=4$: the intermediate tooth count cancels. The previous ninefold reduction required two distinct tooth counts rigidly sharing the intermediate rotation.

For steady ideal transmission, conservation of energy gives $|{\tau }_{\mathrm{out}}|=i|{\tau }_{\mathrm{in}}|$. If gear efficiency is $\eta =P_{\mathrm{out}}/P_{\mathrm{in}}$, the corresponding relation is $|{\tau }_{\mathrm{out}}|=\eta i|{\tau }_{\mathrm{in}}|$. With $i=9$, $\eta =0.90$ and input torque $2\mathrm{N}\mathrm{m}$, the output torque is $16.2\mathrm{N}\mathrm{m}$. This power balance assumes steady operation: accelerating rotating components can temporarily store kinetic energy. A measured gear ratio establishes kinematics, while torque capacity and losses require additional mechanical information.

Gear tangential force is the pitch-circle-equivalent force that transmits torque in a gear mesh. At operating pitch radius $r_w$,For an external spur gear pair without tooth friction, it is the component of tooth-normal force along the common tangent to the operating pitch circles: $F_t=F_n\cos {\alpha }_w$. Together with the gear separating force, this resolves the mesh load in fixed shaft-loading directions. It is not a claim that the moving contact point always has that same local radial/tangential frame.^[ref]

The relation follows geometrically from the gear line of action. Its distance from a gear center is the gear base circle radius $r_b$, so the moment of a normal force is $|\tau |=F_n{r}_b$. Since $r_b=r_w\cos {\alpha }_w$, substitution yields $|\tau |=F_t{r}_w$.

For an original example, a 20-tooth input with operating radius $20\mathrm{mm}$ transmits $4\mathrm{N}\mathrm{m}$. Its gear tangential force is $200\mathrm{N}$. Driving a 60-tooth output at operating radius $60\mathrm{mm}$ ideally gives $12\mathrm{N}\mathrm{m}$: the pitch-circle-equivalent tangential load is the same, while its moment arm is three times larger. At input angular speed $50rad/s$, pitch speed is $1m/s$ and transmitted power is $F_{t}v=200\mathrm{W}$, agreeing with $\tau \omega$.

The calculation assumes steady ideal force transmission. Gear efficiency affects the input/output power balance, while inertia can contribute additional torque during acceleration. A gear tangential force is also not the entire contact load: at ${\alpha }_w=20{}^{\circ }$, $F_t=200\mathrm{N}$ corresponds to normal load about $212.84\mathrm{N}$ and separating load about $72.79\mathrm{N}$.

For a conducting disc of radius $a$ rotating in a uniform axial magnetic field $B$, the shaft-to-rim motional electromotive force is ${\int }_0^a\omega rBdr=\omega B{a}^2/2$. A current $I$ between shaft and rim gives the opposing torque $IB{a}^2/2$, making electromagnetic conversion power $\mathcal{E}I$ equal to mechanical torque times angular speed.

An exact planar example makes the off-diagonal entries tangible. Join four equal point masses $m$ with negligible-mass rigid connectors at $a(2,1,0)$, $a(-2,-1,0)$, $a(1,-1,0)$ and $a(-1,1,0)$, where $a>0$ is a length. Their centre of mass is $O=(0,0,0)$. Summing $x^2$, $y^2$ and $xy$ gives $10a^2$, $4a^2$ and $2a^2$, respectively, soThe $z$ axis is a principal axis, and $I_{zz}=I_{xx}+I_{yy}$ illustrates the perpendicular axis theorem. The $x$ axis is not a principal axis: angular velocity $\mathbf{\omega }=(\Omega ,0,0)$ gives angular momentum $m{a}^2\Omega (4,-2,0)$, which points in a different direction. A bearing can enforce fixed-axis rotation despite this difference, supplying the necessary transverse torque.

For an in-plane unit vector $\mathbf{n}=(\cos \theta ,\sin \theta ,0)$, the planar model hasIts extrema are the principal moments of inertia $(7-\sqrt{13})m{a}^2$ and $(7+\sqrt{13})m{a}^2$; the remaining principal moment of inertia is $14m{a}^2$. These values follow from the two-by-two characteristic polynomial $(4-\lambda )(10-\lambda )-4$, so they can be checked without trusting a numerical eigenvalue routine. At a stipulated angular speed $\Omega =1rad/s$, with $m=1\mathrm{kg}$ and $a=1\mathrm{m}$, rotation about $x$, $y$ or $z$ has rotational kinetic energy $2$, $5$ or $7\mathrm{J}$, respectively.

A truly three-dimensional example adds two more equal point masses at $a(0,0,1)$ and $a(0,0,-1)$. Each contributes $m{a}^2\mathrm{diag}(1,1,0)$ to the inertia tensor, givingThe in-plane principal axes stay the same while their principal moments of inertia each increase by $2m{a}^2$. The $z$ moment of inertia stays unchanged because the added point masses lie on that axis. This comparison changes the total mass from $4m$ to $6m$; it holds the original four point masses fixed rather than redistributing a fixed total mass.

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.

Choose an inertia unit $I_{\ast }$ and an angular speed unit ${\Omega }_{\ast }>0$, so physical principal moments of inertia are $I_{\ast }(1,2,3)$, physical angular velocity is ${\Omega }_{\ast }\mathbf{\omega }$, and normalized time is $t={\Omega }_{\ast }{t}_{\mathrm{physical}}$. The Euler equations for a torque-free rigid body becomeThis is a physically admissible ideal planar mass distribution: $I_3=I_1+I_2$ is the perpendicular axis theorem. A book of finite thickness need not have these exact ratios.

For an original comparison, start with angular velocity $(1,\epsilon ,0)$, $(\epsilon ,1,0)$ or $(\epsilon ,0,1)$, respectively. With $\epsilon =0.01$, the transverse Euclidean norm stays small about axes one and three, but becomes order one about axis two. For the intermediate-axis start, its linearization givesThe hyperbolic functions contain a growing exponential function with rate ${\lambda }_2=1/\sqrt{3}$. Reducing $\epsilon$ tenfold delays reaching the same small transverse amplitude by approximately $\sqrt{3}\log 10\simeq 3.99$ normalized time units, once the growing term dominates. That delays the instability. Once the transverse angular velocity is comparable to the main component, the linearization ceases to describe the motion.

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.

Choose the cylinder centre's initial position as the coordinate origin, with downhill motion to the right and $0\le \alpha <\pi /2$. For rolling without slipping, the clockwise angular speed is $\omega =v/r$. A point initially vertically above the axis therefore hasA different origin adds constant coordinates without changing the following velocity test. Setting $\theta =vt/r$ gives $\dot{x}/v=\cos \alpha +(a/r)\cos \theta$ and $\dot{y}/v=-\sin \alpha -(a/r)\sin \theta$.

For strictly upward motion, $\dot{x}=0$ and $\dot{y}>0$. The first condition fixes $\cos \theta =-(r/a)\cos \alpha$; choose the negative sine branch to maximize $\dot{y}$. Positivity then requiresAt $a=r$ the relevant point is momentarily stationary, not moving upward. Thus a material point inside or on an ordinary cylinder cannot move exactly vertically upward in this motion. The mathematical extension $a>r$ describes the looping trochoid of a point beyond the rolling radius, illustrated in the figure.

Write $k=GM$ and $r=|\mathbf{r}|$. The gravitational central force has zero torque, so the specific angular momentumis constant. For a nonradial orbit $h=|\mathbf{h}|>0$, motion lies in the plane normal to $\mathbf{h}$, and plane polar coordinates give $r^2\dot{\theta }=h$. With $u=1/r$ and primes denoting $\theta$ derivatives,The radial equation $\ddot{r}-r{\dot{\theta }}^2=-k/r^2$ is therefore the Binet equationIts general solution, after choosing the polar origin of angle, isThe origin is a focus of this conic section. To identify which conic section occurs, use the conserved specific energyA bound, nonradial orbit has $\epsilon <0$, hence $0\le e<1$ and is an ellipse. For example, putting $x=r\cos \theta$, $y=r\sin \theta$ in $r+ex=p$ givesan explicit ellipse equation. This is the intended low, bound Kepler orbit. The inverse-square equation alone does not imply an ellipse: an initially tangential speed larger than $\sqrt{2GM/r}$ gives $\epsilon >0$, $e>1$ and a hyperbolic Kepler orbit. Zero angular momentum instead gives radial motion. Those qualifications are necessary for the elliptic conclusion.

In a circular orbit, centripetal acceleration balances the gravitational acceleration:Here $\Omega$ is the positive angular speed; either sense of circulation has this period.

With the stated drag force, use the full angular momentum $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$. Gravitational torque is still zero, while the drag torque isTherefore $\boxed{\mathbf{L}(t)=\mathbf{L}(0)e^{-At/m}}$ for constant $A>0$. Its direction remains fixed, and its magnitude has exponential decay with time constant $m/A$; the orbit remains planar although its shape is no longer a fixed ellipse.

A field in a circulation plane has components crossing circular streamlines. Magnetic flux freezing carries those lines with fluid elements at different angular speeds: the lines become spirals, their azimuthal components are stretched, and neighboring turns acquire increasingly small separations. This is a nonaxisymmetric winding and phase-mixing problem.

For illustration, in the ideal limit take the initial horizontal field along $x$ and fix a horizontal plane. With $\chi =\varphi -\omega (s,z)t$, the induction equations giveThe first term describes advection around a circle; the term proportional to $t{\partial }_s\omega$ is the stretching by differential rotation. Spatial variation of $\omega$ also produces phase gradients proportional to $t\nabla \omega$. A rigidly rotating region merely turns the horizontal field, whereas a shearing region progressively winds it up.

Even small magnetic diffusivity becomes important once these gradients are large. Locally a phase mode of azimuthal order $m$ has damping rate of order $\eta m^2|\nabla \omega {|}^2{t}^2$, leading to attenuation of order $\exp [-\eta m^2|\nabla \omega {|}^2{t}^3/3]$. The associated magnetic phase mixing under differential rotation time isFor a horizontal uniform seed, $m=1$; if $|\nabla \omega |\sim \Omega /\ell$, this is ${\Omega }^{-1}{R}_m^{1/3}$, much shorter than the ordinary diffusion time at large $R_m$. Resistive smoothing cancels the closely spaced oppositely directed field and, in persistent circulation regions with an imposed surrounding field, drives flux expulsion from the interiors toward their boundaries. The detailed final distribution depends on the geometry and boundary forcing. Winding alone does not supply a self-sustaining poloidal–toroidal regeneration loop, and magnetic feedback can again limit the kinematic process.