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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 constant added to a function changes its reference level without changing its derivative:Geometrically, it is a vertical translation of the graph. For $F(x)=x^3/3$, changing $C$ moves every plotted value by the same amount while preserving every tangent slope. This is the geometric reason that antiderivatives form a family.

When $F$ represents a physical quantity, the physical dimension of $C$ must match that of $F$. When position is obtained from velocity, $C$ is a position offset. An unknown starting position remains unknown even when the rate of change is known exactly.

When describing the family of antiderivatives on an interval, the arbitrary additive constant is the constant of integration. On a disconnected open domain, the corresponding freedom is a locally constant function, as in antiderivatives on disconnected domains, rather than necessarily one constant on the whole domain.

Angular momentum measures how a point mass's momentum passes around a chosen origin. Motion straight toward or away from that origin contributes zero; transverse momentum contributes more when it passes farther away. The picture shows the sign convention before the calculation.

Fix an origin $O$ and nonrotating axes in an inertial frame. For a nonrelativistic point mass of constant mass $m>0$, let $\mathbf{r}$ be its position from $O$ and $\mathbf{p}=m\mathbf{v}$ its mechanical momentum in that same reference frame. Its angular momentum is the cross productHere $r=|\mathbf{r}|$, $p=|\mathbf{p}|$, and $\varphi \in [0,\pi ]$ is the angle between them. The right-hand rule gives the direction: curl from $\mathbf{r}$ toward $\mathbf{p}$, and the thumb points along ${\mathbf{L}}_O$. Thus angular momentum is an axial vector, perpendicular to both position and momentum. Its International System of Units unit is $\mathrm{kg}{\mathrm{m}}^2/s$. A point mass need not follow a circle to have angular momentum.^[ref]

With $x$ rightward, $y$ upward, and $+z$ out of the page, the cross product gives $L_z=x{p}_y-y{p}_x$. TakeThen ${\mathbf{L}}_O=(0,0,6)\mathrm{kg}{\mathrm{m}}^2/s$. Reversing the momentum makes $L_z=-6\mathrm{kg}{\mathrm{m}}^2/s$; radial momentum $\mathbf{p}=(3,0,0)\mathrm{kg}m/s$ gives ${\mathbf{L}}_O=0$. The factor $r\sin \varphi$ is the lever arm to the line through the particle parallel to its momentum.

For motion in the $xy$ plane, plane polar coordinates give $\mathbf{v}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta }\hat{\mathbf{\theta }}$, with $\hat{\mathbf{r}}\times \hat{\mathbf{\theta }}=\hat{\mathbf{z}}$. ConsequentlyThe signed angular velocity $\dot{\theta }$ is positive counterclockwise viewed from $+z$ toward the plane, with $\theta$ measured in radians. Radial velocity does not enter the cross product. Under conservation of angular momentum, this relation predicts faster angular motion at smaller $r$ and constant areal velocity $\dot{A}=L_z/(2m)$ for consistently oriented swept area. That is the geometric content of Kepler's second law.

For several point masses, add their angular momenta about the same origin: ${\mathbf{L}}_O={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$. The origin matters: angular momentum under an origin shift explains why the same moving particle can have different values about two fixed points. The central axis of a momentum system and angular momentum decomposition about the centre of mass make this dependence useful.

Move the fixed origin from $O$ to $O^{\prime }=O+\mathbf{a}$, keeping the axes parallel and the inertial frame's velocity measurements unchanged. The new positions are ${\mathbf{r}}_i^{\prime }={\mathbf{r}}_i-\mathbf{a}$, so distributivity of the cross product givesHere $\mathbf{P}$ is total linear momentum. For the angular momentum example with $\mathbf{a}=(1,0,0)\mathrm{m}$, the new position is $(1,0,0)\mathrm{m}$ and $L_{O^{\prime },z}=6-3=3\mathrm{kg}{\mathrm{m}}^2/s$. If total momentum is zero, angular momentum is independent of this origin shift. A moving origin additionally contributes $-\dot{\mathbf{a}}\times \mathbf{P}$ when this expression is differentiated while retaining the original momenta; see angular momentum about the centre of mass for an important case where that term vanishes.

A temporary reversal in a planet's apparent direction across the background sky, arising from the observer's changing position. For an outer planet observed from the Earth, the Earth overtakes it along its smaller, faster orbit. Different apparent reversals for two planets can produce three apparent conjunctions around one Sun-centred encounter, even when both follow prograde circular orbits.

Areal velocity measures how fast a position vector from a chosen origin sweeps signed swept area. It connects a geometric picture of motion to angular momentum: a long position vector with slow transverse motion can sweep the same area per unit time as a short one with fast transverse motion. Constancy requires a dynamical condition, given by the torque criterion for constant areal velocity.

Choose a fixed origin $O$ in an inertial frame and a fixed plane through $O$, with constant unit vector $\mathbf{n}$ normal to it. For a differentiable trajectory in that plane, let $\mathbf{r}(t)$ be position relative to $O$ and $\mathbf{v}=\dot{\mathbf{r}}$ its velocity. Positive swept area follows the right-hand rule about $\mathbf{n}$. In positively oriented Cartesian coordinates in the plane, and in plane polar coordinates with angle $\theta$ in radians,Here the dot product selects the normal component of the cross product. Motion away from or towards $O$ alone contributes no area. The signed specific angular momentum component $h_n=\mathbf{n}\cdot (\mathbf{r}\times \mathbf{v})$ obeysfor a constant mass $m$. If $\mathbf{n}$ is chosen along a nonzero conserved specific angular momentum vector, then $h_n=h=|\mathbf{r}\times \mathbf{v}|$ and the familiar positive formula is $\dot{A}=h/2$. Reversing the orientation changes the sign; the instantaneous unsigned swept-area rate is $|\dot{A}|$.

Astrometry measures the positions and motions of celestial objects. Repeated position measurements can distinguish stellar parallax, caused by the observer's motion, from proper motion of the source across the sky.

An astrometry measurement needs a date and a reference frame, as well as an angle. In an equatorial coordinate system, right ascension and declination label two viewing directions. A catalogue's reference date states when its quoted position applies; comparing entries at different dates without accounting for proper motion can make the same star appear misplaced. The observational problem is to infer a consistent history from successive incoming light directions, while accounting for the observer's changing position and measurement error.

To see the separation clearly, consider just one angular coordinate $\theta$, with $t$ the numerical elapsed time in years. Assume a simplified circular orbit of the Earth, a fixed source Euclidean distance, constant proper motion in this coordinate and no other accelerations. Choose the direction and the starting date so that the annual stellar parallax projection is the cosine termHere ${\theta }_0$ is the coordinate without the annual stellar parallax offset at the reference date; the observed $\theta (0)$ is ${\theta }_0+p$. The proper motion coefficient $\mu$ is quoted in arcseconds per year, and $p$ is the annual stellar parallax in arcseconds. This is an illustrative choice of a full-amplitude projection. General sky directions have different projected amplitudes and phases in their two coordinates; their apparent paths need not be circles. Real astrometry uses the observer's calculated position rather than imposing this single cosine. ^[ref].

Set ${\theta }_0=0$, $\mu =100\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}$ and $p=50\mathrm{mas}$, where $1\mathrm{mas}$ is one thousandth of an arcsecond. The model gives $\theta (0)=50$, $\theta (0.5)=0$ and $\theta (1)=150\mathrm{mas}$. Same-season observations one year apart differ by $100\mathrm{mas}$ because the annual stellar parallax contribution repeats. Subtracting that fitted straight trend leaves an annual signal between $-50$ and $+50\mathrm{mas}$. The existing stellar parallax distance relation then gives $20$ parsecs.

Observing only at the same annual phase would recover the proper motion trend but leave $p$ indistinguishable from a constant offset. Sampling several seasons as well as several years separates the repeating signal from the accumulating displacement. The two signatures answer different questions using the same measurements.

Astronomy studies celestial objects and phenomena, including stars, planets and galaxies. Astrometry measures their positions and motion; astrophysics uses physics to explain their properties and behaviour.

Much of astronomy begins with incoming light. An image tells us which direction the light arrived from, rather than directly providing the source's three-dimensional position. Two stars close together on the sky may be very far apart in space. The celestial sphere organizes these viewing directions; locating an object on it is a different task from measuring how far away that object is. Repeated observations turn this limitation into an opportunity: the changing direction of light contains evidence about both Euclidean distance and motion.

The particularly satisfying combination is stellar parallax with proper motion. The first supplies a geometric Euclidean distance, while the second measures how rapidly the viewing direction changes after accounting for the observer's annual motion. Together they give a transverse speed, connecting tiny angles on an image to kilometres travelled each second. This bridge from a flat sky map to moving stars is one reason astrometry matters to the study of the Milky Way.

Consider a stipulated ideal cart modeled as a point mass of mass $2\mathrm{kg}$ on a horizontal track in a laboratory inertial frame. Its initial position is $x(0)=0$ and its initial velocity is $v_x(0)=1.5m/s$ to the right. Independent devices exert horizontal forces $+6\mathrm{N}$ and $-6\mathrm{N}$ on this same cart. Its upward support force balances its downward gravitational weight; no other forces contribute. Rotation and deformation are excluded from this model.After $4\mathrm{s}$, the cart is at $6\mathrm{m}$, still moving right at $1.5m/s$. Starting the same model at rest instead would leave it at rest. Force balance determines the zero acceleration, while the initial velocity determines whether the cart moves.

For the plotted model, both horizontal devices are switched off simultaneously at $t=2\mathrm{s}$, with no impulse. The support and gravitational weight remain balanced. The individual horizontal forces change, but their sum stays zero on both sides of the switch, so the velocity stays continuous and unchanged. Switching off just one device would be a different model with a nonzero net external force.

The Born-Oppenheimer approximation separates electronic and nuclear motion. At leading order, fixed nuclear positions and electric charges specify an electronic potential energy curve, on which the atomic nuclei move with their own masses.

A real centered Gaussian wave packet with initial position variance $1/(2\alpha )$ and momentum variance ${\hslash }^2\alpha /2$ evolves under a quantum harmonic oscillator byThe covariance matrix obeys $\Sigma (t)=M\Sigma (0)M^{\mathsf{T}}$ and $\det \Sigma ={\hslash }^2/4$. Thus the Robertson-Schrödinger uncertainty relation stays saturated, although the position-momentum variance product can exceed ${\hslash }^2/4$ because of nonzero quantum covariance. The widths are stationary only for $\alpha =m\omega /\hslash$; otherwise they breathe with frequency $2\omega$.

Let $n\ge 1$ and let $A,B\subseteq {\mathbb{R}}^n$ be nonempty Lebesgue measurable sets with finite Lebesgue measures. Assume that $A+B$ is also a Lebesgue measurable set; its measure may be infinite. ThenHere ${\lambda }_n$ denotes $n$-dimensional Lebesgue measure in Euclidean space. Nonempty compact sets and nonempty open sets of finite measure satisfy the measurability assumptions. Measurability of arbitrary summands alone does not guarantee measurability of their Minkowski sum. The theorem does not require convexity. See R. J. Gardner, The Brunn-Minkowski Inequality (2002), Theorem 4.1 and the following remarks.

The $n$th root converts a volume into a quantity that scales like a length: uniform dilation gives ${\lambda }_n(tA{)}^{1/n}=t{\lambda }_n(A{)}^{1/n}$ for $t>0$. In the Euclidean plane the theorem compares square roots of areas, not the areas themselves. For the two rectangles above it reads $3\ge \sqrt{2}+\sqrt{2}=2\sqrt{2}$. Equivalently their sum has area $9\ge 8$. For the two disks, it reads $3\sqrt{\pi }=\sqrt{\pi }+2\sqrt{\pi }$, an equality. These are exact geometric examples, not measured data.

Applying the theorem to the scaled sets gives, whenever that sum is a Lebesgue measurable set,For convex bodies, this means the root-volume of the interpolated shape lies above the straight line between the input root-volumes. In fact $t\mapsto {\lambda }_n((1-t)A+tB{)}^{1/n}$ is a concave function on $[0,1]$: convexity gives $sA+tA=(s+t)A$ for $s,t\ge 0$, so interpolating two intermediate shapes gives another intermediate shape and the same inequality applies. The beauty is that no particular boundary shape appears in the bound: possible pointwise displacements force a minimum amount of space.

For convex bodies $A,B$, equality holds exactly when they are homothetic convex bodies, meaning $B=v+sA$ for some vector $v$ and $s>0$. This equality qualification is part of the Brunn-Minkowski theorem for convex bodies. ^[ref] For positive-sided axis-aligned rectangles, the comparison can be checked directly:Equality requires $ad=bc$: the rectangles must have the same aspect ratio. Our $2\times 1$ and $1\times 2$ rectangles do not. All positive-radius disks are homothetic convex bodies, explaining their equality.

Do not apply that equality criterion without its hypotheses. If $A=\{a\}$, its Lebesgue measure is zero and $A+B$ is a translation of $B$, so equality holds even for nonconvex $B$. Also, changing a summand on a null set can change the measure of its Minkowski sum: in one dimension, $A=[0,1]\cup \{3\}$ and $B=[0,1]$ both have measure $1$, but $A+B=[0,2]\cup [3,4]$ has measure $3$. Removing the single extra point restores equality with sum measure $2$. Thus merely being equal almost everywhere to homothetic convex bodies is not a sufficient equality test for arbitrary sets.

The Brunn-Minkowski theorem also bounds the collision-region volume: for positive-volume compact sets $O,R$, reflection preserves Lebesgue measure, so ${\lambda }_n(O+(-R){)}^{1/n}\ge {\lambda }_n(O{)}^{1/n}+{\lambda }_n(R{)}^{1/n}$. In the toy collision example this gives area at least $(2+\sqrt{2}{)}^2=6+4\sqrt{2}\approx 11.657$, compared with the exact area $12$. A lower bound alone does not identify safe reference positions; the actual Minkowski sum supplies those.

Calculus connects local change with accumulated change. A derivative asks how fast a function changes near one input; an integral adds its contributions across an interval. The fundamental theorem of calculus explains why these operations fit together: accumulating a continuous rate produces a quantity whose derivative is that rate.

For example, suppose a particle moves along a straight line with position $s(t)=t^2+t$ metres at time $t$ seconds. Its velocity is $s^{\prime }(t)=2t+1$ metres per second. Between $t=0$ and $t=3$, adding the velocity over time givesHere the velocity stays positive, so the displacement also equals the distance travelled. If the particle reverses direction, the integral of its signed velocity still gives displacement, while the integral of its speed gives total distance.

This is useful because local rules are often easier to specify than whole trajectories. A differential equation can prescribe a particle's velocity or a population's growth rate; integration then reconstructs its evolution, subject to an initial condition. In the other direction, a derivative extracts local information from a known function, and optimization uses that information to locate or approach promising inputs. The same two operations therefore connect measurement, prediction and design.

The visual connection is developed in derivative, integral and fundamental theorem of calculus. The definitions depend on limits, which turn progressively finer finite measurements into exact mathematical statements. A graph with a corner need not have a derivative there, and the conditions permitting integration or interchange of limits matter; calculus is more than a collection of symbolic rules.

Historically, Isaac Newton described changing quantities and their rates in the language of fluents and fluxions; his manuscript History of the Method of Fluxions records that approach.^[ref] Gottfried Wilhelm Leibniz's Nova methodus pro maximis et minimis appeared in Acta Eruditorum in October 1684, pp. 467–473; the original article is preserved in the HKUST Library collection.^[ref] The modern limit definitions used here should not be read back into those seventeenth-century formulations.

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.

The charge-to-mass ratio is a body's electric charge $q$ divided by its mass $m$. Its SI unit is the coulomb per kilogram, $\mathrm{C}\mathrm{k}{\mathrm{g}}^{-1}$. In a prescribed electric field and magnetic field, nonrelativistic acceleration depends on $q/m$ through the Lorentz force. Two bodies with the same initial position, velocity, and charge-to-mass ratio therefore have the same trajectory in this ideal model. For an electron, write $q=-e$ with $e>0$ the elementary charge: $q/m=-e/m$, whereas the positive number $e/m$ is the magnitude of the charge-to-mass ratio.

A chirped Gaussian wave packet adds a quadratic quantum phase to the zero-mean Gaussian wave packet already evaluated above. Here $\sigma >0$ and $\kappa$ is a real dimensionless parameter. Its probability density function $|{\psi }_{\sigma ,\kappa }(x){|}^2$ is unchanged for every $\kappa$, so $⟨x⟩=0$ and $(\Delta x{)}^2={\sigma }^2$. Nevertheless, applying the momentum operator givesThe sign of the quantum covariance follows directly, without assigning a classical trajectory to the wavefunction:Here $r$ is a normalized symmetrized quantum covariance, analogous to a correlation coefficient. Positive $\kappa$ gives positive $C_{xp}$, and reversing $\kappa$ reverses $C_{xp}$ while leaving both variances unchanged. In the negative-exponent Fourier transform convention used above, the momentum density isConsequently the quantum uncertainty product and the covariance matrix determinant tell different stories:Every member of this pure state family saturates the Robertson-Schrödinger uncertainty relation. Only $\kappa =0$ saturates the Heisenberg uncertainty relation for this fixed pair of position and momentum axes. Thus being a Gaussian wave packet alone does not imply a minimum $\Delta x\Delta p$.

For a concrete comparison, choose a reference length $L$, put $X=x/L$, $P=pL/\hslash$, and set $\sigma =L$. The choices $\kappa =0,+2,-2$ all have $(\Delta X{)}^2=1$, but give $(\Delta P{)}^2=1/4,5/4,5/4$ and $C_{XP}=0,+1,-1$, respectively. The two nonzero cases have $r=\pm 2/\sqrt{5}$ and product $\sqrt{5}/2$; all three covariance matrices have determinant $1/4$.

The phase space curves in the figure satisfy $z^T{\Sigma }^{-1}z=1$ in the dimensionless coordinates, where $z=(X,P{)}^T$. They display second moments, not observed simultaneous position and momentum pairs. Their common area is $\pi \sqrt{\det \Sigma }=\pi /2$, although their coordinate projections differ.

This distinction is useful for predicting focusing. Under free particle evolution with mass $m$, $x(t)=x(0)+tp(0)/m$, soA negative chirp initially narrows the position distribution; positive chirp initially broadens it. For $\kappa <0$, the minimum occurs at $t_{\ast }=-2m{\sigma }^2\kappa /[\hslash (1+{\kappa }^2)]>0$ and has variance ${\sigma }^2/(1+{\kappa }^2)$. At that time the quantum covariance vanishes, and the unchanged momentum spread gives $\Delta x(t_{\ast })\Delta p=\hslash /2$. The extra initial product was a usable position-momentum correlation, not extra measurement backaction.

Classical mechanics describes the motion of bodies by assigning positions and velocities and specifying how their interactions change those quantities. A point mass is the simplest model: its size and internal structure are ignored, leaving a mass and a moving position. A rigid body retains orientation and rotational kinetic energy. Choosing a model means deciding which features affect the question being asked; a sliding block may be treated as a point mass, whereas predicting its rolling without slipping requires its moment of inertia.

Three complementary questions organize the subject. Newton's second law asks how the net force determines the instantaneous acceleration. The work-energy theorem asks how that force, acting through a displacement, changes kinetic energy. Conservation of mechanical energy asks which combinations of position and velocity are accessible when the interactions admit a time-independent potential energy and no other force supplies net power. These descriptions are compatible, but answer different questions: an energy calculation may find a maximum speed without first solving for position at every time.

The harmonic oscillator below makes this unity visible. A restoring force produces a repeating position trace; the corresponding phase plane has a closed ellipse; kinetic energy and potential energy exchange while their sum stays fixed. Lagrangian mechanics and Hamiltonian mechanics provide further ways to organize the same dynamics, especially when coordinates or constraints make individual forces awkward to track.

The scope of classical mechanics also matters. Newtonian mechanics approximates motion at speeds small compared with the speed of light; special relativity changes that description when this approximation fails. Quantum mechanics is needed for phenomena for which a classical assignment of simultaneous position and momentum is inadequate. These limitations do not prevent classical mechanics from being a useful predictive model within its regime. ^[ref]

If the relative position is $\mathbf{d}(t)={\mathbf{d}}_0+t\mathbf{v}$ with $\mathbf{v}\ne 0$, completing the square givesThis proves the minimum distance over all real times. If time is restricted to an interval, project $t_{\ast }$ onto that interval; if $\mathbf{v}=0$, the distance is constant. Both particles are evaluated at the same time, unlike the two independent parameters in the distance between skew lines.

A line defined by an approximately coincident detector pair in annihilation-photon tomography. An event constrains a possible annihilation location to that line within instrumental and physical uncertainties. Time-of-flight data can add a position constraint along it; attenuation, scattering and random coincidences change the forward model.

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.