Center of mass

For masses $m_i$ at positions ${\mathbf{r}}_i$, the center of mass is $\mathbf{R}=({\sum }_i{m}_i{\mathbf{r}}_i)/({\sum }_i{m}_i)$, with the analogous mass-density integral for a continuous body.

For a closed system with fixed total mass, summing the constituent Newton's second law equations makes pairwise internal forces cancel when Newton's third law applies. The remaining resultant external force is the derivative of total momentum, equal to $M{\ddot{\mathbf{r}}}_{\mathrm{CM}}$. This governs translation of the center of mass; external torque additionally governs changes of angular momentum.

The point through which the resultant gravitational force acts when that force has a single point of application. In a uniform gravitational field, it coincides with the center of mass; a nonuniform field need not give the same point.

For fixed classical particle masses, differentiating the center of mass position gives the displayed mass-weighted velocity. The total momentum is $M{\mathbf{V}}_{\mathrm{CM}}$. The relativistic zero-momentum frame instead has velocity $c^2\mathbf{P}/E$; treating rest mass as the only weight there generally fails.

For a solid cone of base radius $a$, height $h$ and density $\rho (r)={\rho }_0(r/a{)}^p$, take $p>-2$ so the axis singularity, if present, is integrable. In cylindrical coordinates, at fixed radius $r$ the vertical range is $0\le z\le h(1-r/a)$. Integrating over height first givesThe first height moment is $\pi {\rho }_0{\int }_0^a(r/a{)}^{p}r{h}^2(1-r/a{)}^{2}dr=2\pi {\rho }_0{a}^2{h}^2/[(p+2)(p+3)(p+4)]$. Dividing proves $d_p=h/(p+4)$. Thus uniform density has centroid $h/4$, while density proportional to radius has centroid $h/5$. For sums of such density components, total masses and first moments add, giving a mass-weighted centroid.

The centre of mass of an astronomical system. In an isolated two-body problem, both bodies follow orbits around their shared astronomical barycenter; the Kepler orbit of their relative displacement has a semi-major axis equal to the sum of their individual semi-major axes.

Suppose a body of mass $M_0$ has centre of mass ${\mathbf{R}}_0$, and material of mass $M_c<M_0$ with centre of mass ${\mathbf{R}}_c$ is removed. Additivity of the first mass moment gives the remaining centre of massWhen the original centre of mass is at the origin, the remaining centre of mass points opposite the removed material's first mass moment. For a constrained rigid body rotating about a fixed axis, this displaced centre of mass undergoes centripetal acceleration even if the applied axial torque is zero.

The center of mass moves as if the total external force acted on the total mass; internal forces cancel in its equation.

Center-of-mass reduction separates translational motion from internal coordinates by using the conserved or externally forced center coordinate.

In the two-body problem, these coordinates separate the common translation from the internal separation. For total mass $M$ and reduced mass $\mu$, the kinetic energy is $M{\dot{R}}^2/2+\mu {\dot{r}}^2/2$. The conjugate momenta are $P=M\dot{R}$ and $p=\mu \dot{r}$ for the ordinary Newtonian kinetic term.

For fixed particle masses and positive total mass, write ${\mathbf{x}}_i=\mathbf{R}+{\mathbf{y}}_i$ with $\mathbf{R}$ the centre of mass. The weighted relative velocities sum to zero, so expanding the kinetic energy eliminates the mixed term and yields the displayed decomposition. For a rigid body, the relative contribution is the rotational kinetic energy $I{\omega }^2/2$ about the chosen centre-of-mass axis.

For particles with masses $m_j$ and positions $r_j$, Jacobi coordinates consist of the total center of mass and successive relative coordinates ${\rho }_j=r_{j+1}-({\sum }_{l=1}^j{m}_l{r}_l)/({\sum }_{l=1}^j{m}_l)$. They separate translation from internal motion and diagonalize the kinetic energy with the corresponding reduced masses.

For three particles, choose the relative position of particles one and two, the position of their center of mass relative to particle three, and the total center of mass. The kinetic energy diagonalizes with masses

For Jacobi coordinates $\mathbf{a},\mathbf{b},\mathbf{c}$ and their corresponding reduced masses $\alpha ,\beta ,\gamma$, total angular momentum is

For three ordered angular positions $q_1,q_2,q_3$, define consecutive gapsand the collective angle $\Phi =(q_1+q_2+q_3)/3$. Then $\alpha +\beta +\gamma =2\pi$. For three equal masses on a circle of radius $r$,Thus the ignorable coordinate $\Phi$ decouples from the relative gap dynamics whenever the potential depends only on the gaps.

The reduced mass is the effective mass for the two-particle relative coordinate of two bodies with constant positive masses $m_1,m_2$:Both bodies can respond to their mutual force, so their relative acceleration is greater than if one were fixed. Consequently $0<\mu <\min (m_1,m_2)$. This explains why using either individual mass, or their sum, generally gives the wrong internal oscillation rate. The same reduced mass describes a two-body problem, a free two-mass spring, and the approximate molecular vibration of a diatomic molecule.^[ref]

For equal masses $m_1=m_2=m$, $\mu =m/2$. Two $1\mathrm{kg}$ bodies therefore have $\mu =0.5\mathrm{kg}$: in their stationary centre of mass frame a relative speed $2m/s$ means individual speeds $1m/s$ and total kinetic energy $1\mathrm{J}$. With $m_1=1\mathrm{kg}$ and $m_2=100\mathrm{kg}$, instead $\mu =100/101\mathrm{kg}\approx 0.990099\mathrm{kg}$. At the same relative speed, the individual speeds are $200/101$ and $2/101m/s$, and the total kinetic energy is $200/101\mathrm{J}$. These are calculations for stipulated masses and velocities.

For fixed $m_1$ and $q=m_2/m_1>0$, $\mu /m_1=q/(1+q)$. The heavy-partner limit $q\to \infty$ gives $\mu \to m_1$ from below; the relative error incurred by replacing $\mu$ with $m_1$ is $(m_1-\mu )/\mu =1/q$. At $q=100$ this is $1\%$. Treating the heavy body as fixed is thus a controlled approximation to internal motion, rather than an exact consequence of its being heavier.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.

Take two point masses joined by a massless spring of spring constant $k>0$ and equilibrium length $\ell >0$. Assume one-dimensional motion along the spring, no external force, no damping, and an extension $s=x_2-x_1-\ell$ satisfying Hooke's law; keep $x_2>x_1$. For a real spring this is a model within its elastic range. Its potential energy is $U=k{s}^2/2$, soHere $\omega$ is angular frequency and $\nu$ is frequency. Choose $|A|<\ell$ so the model bodies do not cross. The internal normal mode has displacements $u_1=-m_{2}s/M$ and $u_2=m_{1}s/M$ about equilibrium offsets from the centre of mass; its amplitude ratio is $|u_1|/|u_2|=m_2/m_1$. The other normal mode is collective translation with zero frequency and no spring extension. This free pair has different boundary conditions from normal modes of two equal masses between three springs.

For $k=10N/m$ and two $1\mathrm{kg}$ masses, $\omega =\sqrt{20}{\mathrm{s}}^{-1}\approx 4.47214{\mathrm{s}}^{-1}$ and $\nu \approx 0.711763\mathrm{Hz}$. Replacing the second mass by $100\mathrm{kg}$ gives $\omega =\sqrt{10.1}{\mathrm{s}}^{-1}\approx 3.17805{\mathrm{s}}^{-1}$ and $\nu \approx 0.505802\mathrm{Hz}$. Fixing the second endpoint instead would give $\sqrt{10}{\mathrm{s}}^{-1}$: the free heavy partner still moves. The free pair's angular frequency exceeds that fixed-end limit by the factor $\sqrt{1+1/q}$, about $1.004988$ at $q=100$.

Near a stable equilibrium bond length $r_e$, a Taylor expansion of a diatomic molecule's potential energy gives $U(r)\simeq U(r_e)+k(r-r_e{)}^2/2$, with $k=U^{\prime \prime }(r_e)>0$. Treat the internal stretch as a harmonic oscillator of reduced mass, initially omitting molecular rotation. If replacing an isotope leaves this local potential energy curvature unchanged, thenHere $c$ is the speed of light and ${\tilde{\nu }}_{\mathrm{harm}}$ a spectroscopic wavenumber; use consistent length units for $c$ and ${\tilde{\nu }}_{\mathrm{harm}}$. The quantum harmonic oscillator gives adjacent energy eigenvalues separated by $h\nu$, with $h$ the Planck constant. Consequently a vibrational photon absorption can probe the same frequency scale, subject to electric-dipole selection rules.

For an explicitly approximate isotope example, assign the atomic masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, ${}^{35}\mathrm{Cl}$, and ${}^{37}\mathrm{Cl}$ the mass-number values $1u,2u,35u,37u$, respectively, with $u$ the unified atomic mass unit. These are stipulated approximations to individual atomic masses, not measured masses of molecules or natural-abundance averages. They giveThe symbol $\mathrm{D}$ means deuterium, ${}^2\mathrm{H}$. Doubling the light partner's approximate atomic mass therefore produces a large predicted shift, whereas changing the already heavy chlorine partner by two mass units produces a much smaller one. A quantitative comparison with infrared spectroscopy requires actual isotope-specific atomic masses, fitted band parameters, and uncertainties, rather than identifying any arbitrary peak with ${\tilde{\nu }}_{\mathrm{harm}}$.

An actual infrared spectroscopy route is to compare gaseous hydrogen chloride containing ordinary hydrogen with its deuterium counterpart. A documented teaching experiment uses a stopcock-equipped gas cell with potassium bromide windows and a Nicolet Avatar instrument for Fourier transform spectroscopy. A nitrogen-purged flask and drying tube supply the mixture; the sample also contains sulfur dioxide. An empty sample-chamber background precedes the filled-cell scan. The absorbance trace and peak-picked spectroscopic wavenumbers are the readout. The analysis assigns and indexes ${}^1{\mathrm{H}}^{35}\mathrm{Cl}$ and ${}^2{\mathrm{H}}^{35}\mathrm{Cl}$ transitions separately and propagates transition uncertainties. The instructions distinguish resolvable chlorine isotope peaks from adequate precision for the intended parameter analysis. ^[ref]

A separate published teaching dataset reports a Nicolet Nexus 670 instrument, a $10\mathrm{cm}$ gas cell at $20\mathrm{Torr}$, and chlorine isotope peaks resolved to baseline. These apparatus details describe that dataset, not the preceding experiment. ^[ref]

The inference connects several models. Fit correctly assigned transitions to separate molecular rotation from molecular vibration, then compare the inferred vibrational scales using isotope scaling of diatomic vibration. Agreement supports the approximate separation between a nearly unchanged chemical bond stiffness and a changed nuclear mass. It does not make a real molecule an exact classical spring: anharmonicity shifts the fundamental transition away from the local harmonic oscillator scale, and coupling to molecular rotation shifts individual lines.

For an independent test, calculate the reduced masses from the specified isotopes and compare the inferred values $k=\mu (2\pi c{\tilde{\nu }}_{\mathrm{harm}}{)}^2$. Agreement within uncertainty tests the common-curvature assumption; choosing separate $k$ values just to fit each frequency would not test it. The documented background addresses instrument and ambient contributions, but does not by itself subtract window absorption, identify contaminating species, or calibrate the spectroscopic wavenumber scale. Those require separate controls. Finite spectral resolution, blended peaks, incorrect isotope assignments, and departure from the harmonic approximation near a potential minimum limit the comparison. No numerical measured line positions are assumed here.