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Write the position of a mass element relative to the center of mass as $\mathbf{r}$, so its inertial-frame velocity is $\dot{\mathbf{X}}+\mathbf{\omega }\times \mathbf{r}$. The center-of-mass identity $\int \mathbf{r}dm=0$ removes all cross terms. Define the inertia tensor about the center of mass byThe total angular momentum about the origin and kinetic energy are then

Let $\rho (r)=C{r}^{-1}\sin (\pi r/R)$. Spherical symmetry makes $I=I_{0}1$. The mass normalization isso $C=M/(4R^2)$. A moment about any diameter isTherefore

Writing $x_i=(x_i,y_i,z_i)$, rotation about the $z$-axis gives particle $i$ the speed $\omega \sqrt{x_i^2+y_i^2}$. Its kinetic energy is therefore $m_i{\omega }^2(x_i^2+y_i^2)/2$, soThis is the defining mass sum for the moment of inertia about the $z$-axis.

For the uniform cuboid, the mass density is $\rho =M/(8abc)$ and

The work done along the trajectory isBy Newton's second law, $F=m\ddot{x}$, soThis is the work-energy theorem: the work equals the change in kinetic energy.

A force is conservative when its work between two points is independent of the path, equivalently when it is the negative gradient of a potential energy. Here a potential with $V(0)=0$ isConservation of energy for $x>0$ givesThus

If $v_0>v_e$, the particle escapes with asymptotic speed $\sqrt{v_0^2-v_e^2}$; the graph decreases from $v_0$ to this horizontal asymptote. At the escape velocity $v_0=v_e$, it decreases as $v_e{e}^{-x/(2\lambda )}$ and approaches zero only at infinity. If $v_0<v_e$, it reaches the turning pointwhere $v=0$, then returns on the negative branch of the same curve and oscillates symmetrically about the origin.

For oscillation put $w=1-v_0^2/v_e^2$. One quarter of the period iswhere $u=e^{-x/\lambda }$. Therefore

The kinetic energy of the two equal masses isThe extensions of the left, coupling, and right springs are respectively $x_1$, $x_2-x_1$, and $x_2$. Their potential energy is thereforeThus the mechanical Lagrangian $L=T-U$ has the requested quadratic form with the symmetric small-oscillation mass and stiffness matrices

Choose ordered particle angles $q_1,q_2,q_3$ withand put $\Phi =(q_1+q_2+q_3)/3$. Solving for the individual angles givesThe kinetic energy is thereforeThe collective angle is an ignorable coordinate and decouples from the gaps, so its term may be omitted from the relative Lagrangian. Substituting $\gamma =2\pi -\alpha -\beta$ into the potential energy gives

At step $n$, the parent and daughter rest masses areThe decay is allowed when $M\ge m$, which holds because $M=\lambda m$ with $\lambda >1$. Hence all $N$ stages occur:The initial rest mass is $NM+m_0$ and the final rest mass is $Nm+m_0$. Thusof rest mass is converted into kinetic energy, corresponding to energy $Nm(\lambda -1)c^2$.

Take the sphere to move in the positive $z$-direction and use spherical coordinates centred on it. An axisymmetric harmonic potential that decays at infinity has the dipole form $A\cos \theta /r^2$. The no-penetration condition in the laboratory frame isIt fixes $A=-U{a}^3/2$, so the potential flow around a translating sphere isThe laboratory-frame velocity components are

The fluid kinetic energy iswhere $V=4\pi a^3/3$. Thus the added mass of a sphere is $\rho V/2$.

When the sphere falls, replacing fluid of density $\rho$ by material of density ${\rho }_s$ lowers the gravitational potential energy at rateThe total kinetic energy of sphere and fluid isConservation of total energy givesFor $U\ne 0$, the acceleration of a freely falling sphere with added mass isThe same formula holds at release by continuity.

Assume an inviscid homogeneous ocean, no horizontal pressure gradient, and no wind stress. Horizontal uniformity then removes advective acceleration, and the parcel equations on a beta plane arewith $u=\dot{x}$ and $v=\dot{y}$.

Since $\dot{u}=f(y)\dot{y}$, integration from the stated initial data givesThis is conservation of the parcel's zonal canonical momentum: its zonal speed records the meridionally accumulated Coriolis acceleration. Multiplying the two momentum equations by $u$ and $v$ and adding givesThus kinetic energy and speed are constant:Eliminating $\dot{x}$ gives the required ordinary differential equation

For $f_0,\beta >0$, the equator is at $y_e=-f_0/\beta$. ThereThe parcel must encounter a meridional turning point before reaching the equator if it is to remain strictly in the Northern Hemisphere. The energy relation therefore requiresor $\beta V/f_0^2<1/2$. Equality is the limiting trajectory that reaches the equator with zero meridional speed.

The kinetic energy and squared angular-momentum magnitude areDifferentiate and substitute Euler's equations. In $\dot{T}$, the coefficient of ${\omega }_1{\omega }_2{\omega }_3$ isThe analogous weighted sum in $d{L}^2/dt$ also cancels, so both quantities are conserved.

The one-particle partition function iswhere $\lambda =(2\pi {\hslash }^2\beta /m{)}^{1/2}$. ThusCanonical differentiation now givesandThereforeThe factor $5/2$ consists of $3/2$ from kinetic energy and $1$ from gravitational potential energy.

$L^2=J^2-(q\beta {)}^2=J^2{\sin }^2\alpha$. Splitting kinetic energy into radial and angular parts therefore gives $E=\frac{1}{2}m{\dot{r}}^2+J^2{\sin }^2\alpha /(2m{r}^2)+V(r)$, proving the stated effective potential.

With the proton fixed, $[-{\hslash }^2{\nabla }^2/(2m_{\mu })-e^2/(4\pi {\epsilon }_{0}r)]\psi =E\psi$: kinetic energy plus Coulomb attraction, assuming nonrelativistic point particles and neglecting spin, radiation, and proton motion (or replacing $m_{\mu }$ by the reduced mass). Substitution of the $2s$ form gives $E=-{\hslash }^2{\alpha }^2/(8m_{\mu })$ and $\alpha =m_{\mu }{e}^2/(4\pi {\epsilon }_0{\hslash }^2)$. Direct differentiation verifies it. The curve starts positive, crosses at $r=2/\alpha$, then approaches zero from below; ordinary hydrogen has the same shape but the smaller electron mass gives a larger radial scale.

The Gaussian transform isThereforeA sufficient weak-scattering regime is kinetic energy much larger than $|V_0|$ (more intrinsically $m|V_0|r_0^2/{\hslash }^2\ll 1$).

The radial effective potential is $V_{\mathrm{eff}}=mgr\cot \alpha +5J^2/(14m{r}^2)$. Initially $r=8a$, $J=(7m/5)(8a)v_0$. Comparing the initial energy with $V_{\mathrm{eff}}(2a)$ giveswhich is exactly $v_0^2\ge (4/7)ga\cot \alpha$. Larger speed raises the centrifugal barrier at the neck faster than it raises the initial kinetic energy.

With $x=\ell \sin \theta$ and $y=-a{t}^2/2-\ell \cos \theta$, discard a total derivative from the kinetic energy to obtainThus $\ddot{\theta }+(g-a)\sin \theta /\ell =0$. If $a=g$, then $\ddot{\theta }=0$: the pendulum is weightless in the lift and rotates uniformly (or remains fixed).

If transverse kinetic energy is dissipated at the downstream axis, the remaining specific orbital energy is $v^2/2-GM/b$. Binding requires $b<2GM/v^2$, corresponding to incident impact parameter $d<2GM/v^2$. Integrating the captured mass flux gives the Bondi--Hoyle--Lyttleton accretion rate.

When two massive particles coalesce without additional outgoing radiation, four-momentum conservation makes the final four-momentum their sum. For collinear incident velocities, its rest mass is the positive square root of the displayed invariant, and its final velocity equals the center-of-momentum frame velocity $c^2{p}_{\mathrm{total}}/E_{\mathrm{total}}$. Relative kinetic energy can become internal rest energy, so the final rest mass is generally larger than the sum of the incident rest masses.

This tensor records the ordered kinetic energy of the streaming velocity $u_i=⟨v_i⟩$. It excludes random motion; its vanishing is compatible with a nonzero stellar pressure-energy tensor.

This tensor integrates the local pressure tensor, where ${\sigma }_{ij}=⟨(v_i-u_i)(v_j-u_j)⟩$ is the velocity covariance. Half its trace is the random kinetic energy. It enters the tensor virial theorem with coefficient one.