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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.

Axis theorems. The parallel axis theorem stateswhere $I_G$ is the moment of inertia about an axis through the centre of mass, $I_O$ is about a parallel axis, $m$ is the body's total mass, and $d$ is the perpendicular separation of the axes. Expanding $\int [(x-d{)}^2+y^2]dm$ proves the result because $\int xdm=0$ at the centre of mass. The perpendicular axis theorem applies to a planar lamina in the $xy$ plane:All three axes meet at the same point, $x,y$ lie in the lamina and $z$ is normal to it; the identity follows from $x^2+y^2$ being the squared distance to the $z$ axis. It is not a theorem for an arbitrary three-dimensional body.

Cylinder integrals. Put the cylinder's midpoint at $z=0$ and its symmetry axis along $z$. Its uniform mass density is $\rho =m/(\pi r^{2}l)$. The transverse moment of inertia of a uniform solid cylinder about the $x$ axis isEquivalently each thin circular slice has diametral moment of inertia $r^{2}dm/4$, by the perpendicular axis theorem and symmetry, and shifting it by $z$ adds $z^{2}dm$. The parallel axis theorem then gives the transverse end-axis moment of inertia:The axial moment of inertia of a uniform solid cylinder is $I_{\parallel }=m{r}^2/2$. Equating it to $I_{\perp ,G}$ givesHere the source's principal axis means the cylinder's symmetry axis; its transverse centre axes are also principal axes. This equality describes a short cylinder rather than the thin-rod limit.

Hammer dynamics and forces. Treat the compact head as a point mass and neglect its intrinsic moment of inertia. Let $\theta$ be the displacement from downward vertical, and let ${\mathbf{e}}_r$ point along the rod from pivot to head, with ${\mathbf{e}}_{\theta }$ in the direction of increasing $\theta$. DefineThe two external weights $Mg$ and $mg$ act at distances $l$ and $l/2$ from the pivot. The frictionless pivot supplies a reaction $\mathbf{R}$ but no torque about its axis. The total restoring torque and rotational equation of motion areFor completeness, Newton's second law for the centre of mass gives both components of the pivot force:The negative radial component points towards the pivot. These follow from the centre of mass acceleration $-d{\dot{\theta }}^2{\mathbf{e}}_r+d\ddot{\theta }{\mathbf{e}}_{\theta }$ and the total weight $m_{T}g(\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta })$. They specify the external forces even when the oscillation is not small.

With the small-angle approximation, $\sin \theta \simeq \theta$, this is simple harmonic motion. The physical pendulum has small-amplitude oscillation periodFor release from a finite amplitude $0<{\theta }_0<\pi$, conservation of mechanical energy instead gives $\frac{1}{2}I{\dot{\theta }}^2=gA(\cos \theta -\cos {\theta }_0)$. Integrating a quarter-cycle gives the exact oscillation periodwhere $K$ is the complete elliptic integral of the first kind; $K(0)=\pi /2$ recovers $T_0$.

For $m=M/10$ and $r=l/20$, compare with the simple pendulum value $T_s=2\pi \sqrt{l/g}$:For small oscillations, the hammer's oscillation period is about $0.794\%$ shorter. The extra rod weight adds restoring torque proportionately faster than it adds pivot moment of inertia.

The free rod has the same mass $m$ and length $l$ as in part (a). There is no pivot. Put the rod's initial center of mass at $y=0$ and the impact end at $y=l/2$. At impact the combined center of mass isConservation of momentum gives the combined center of mass velocity $V_C=v_0/2$. Evaluate angular momentum about this combined center of mass; the incident ball's perpendicular lever arm is $l/4$, so its initial magnitude is $m{v}_{0}l/4$.

After the perfectly inelastic collision, the parallel axis theorem gives the combined moment of inertiaWith negligible external angular impulse, conservation of angular momentum therefore givesIts sense is that of the incident ball's angular momentum, clockwise in the source diagram. As a consistency check, the final kinetic energy isThus $m{v}_0^2/10$, or one fifth of the initial kinetic energy, is dissipated in sticking. Treating the rod's original centre as a fixed pivot would give an incorrect angular velocity.

Let $a$ be the center of mass acceleration downhill and $F$ the uphill static friction. Translation and rotation about the center of mass obeyThe rolling without slipping constraint is $a=r\alpha$. For the coin's moment of inertia $I=m{r}^2/2$, this gives $F=ma/2$, soThe normal force is $N=mg\cos \theta$. Static friction can enforce rolling without slipping only if $F\le \mu N$, givingThe static friction acts uphill to provide the torque; it is not automatically equal to $\mu N$, attaining that limiting value only at the threshold. For finite $\mu \ge 0$ the result lies between $0$ and $\pi /2$.

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

Substitution in the moment of inertia givesFor independent errors interpreted as standard uncertainties, logarithmic differentiation and propagation of uncertainty giveConsequentlyThis is a $0.515\%$ statistical precision requirement. If “accuracy” instead means hard bounds whose worst cases must be guaranteed, first-order absolute errors add: $0.01+0.01+4{ϵ}_R\le 0.025$, giving $\boxed{{ϵ}_R\le 0.125\%,|\delta R|\le 1.25\mathrm{mm}}$ to first order. For an exact symmetric hard bound about the nominal values, the upper extreme is slightly more restrictive: $(1.01{)}^2(1+{ϵ}_R{)}^4\le 1.025$ gives ${ϵ}_R\le 0.1199\%$, or $1.20\mathrm{mm}$. The question does not distinguish a statistical uncertainty from a worst-case tolerance, so the convention matters.

The original apparatus has two small masses at distance $L/2$ from the suspension axis. Neglecting the rod's moment of inertia, their total moment of inertia is $I=2m(L/2{)}^2=m{L}^2/2$, consistent with the supplied torsion pendulum period. Treat $m$ and $L$ as exact because no uncertainties are specified. Rearranging the period relation givesThe torsion constant is a restoring torque per unit angle. In dimensional calculations the radian is dimensionless.

Using first-order propagation of uncertainty, with the quoted period uncertainty interpreted as a standard uncertainty,Consequently

An Euler strut is a slender initially straight elastic column under axial compression. For a uniform bending stiffness $EI$ and a small transverse displacement $y(x)$, the second-order potential energy under a conservative compressive load $P$ isThe bending term is positive; shortening releases load potential. Stationarity yields $EI{y}^{(4)}+P{y}^{\prime \prime }=0$. Pinned ends impose $y(0)=y(L)=0$ and zero end bending moments, $y^{\prime \prime }(0)=y^{\prime \prime }(L)=0$. Substitution of $y=A\sin (n\pi x/L)$ givesThe straight configuration loses stability first in the longest admissible bending mode. The second moment of area $I=\int z^{2}dA$ is geometric, not a mass moment of inertia. Boundary conditions matter: a fixed-free column has effective length $2L$, while fixed-fixed ends have effective length $L/2$, changing the threshold by factors $1/4$ and $4$. Slenderness, elastic material response, alignment and the weak bending direction must be checked. Imperfections produce bending before the ideal threshold, and a short column can yield or crush first. Linear buckling theory does not predict an unlimited deflection or a complete postbuckling curve.

For a Lagrangian $\mathcal{L}=T-U$, the Euler-Lagrange equation is $d(\partial \mathcal{L}/\partial \dot{q})/dt=\partial \mathcal{L}/\partial q$ for each unconstrained generalized coordinate. For example, $\mathcal{L}=m{\dot{x}}^2/2-k{x}^2/2$ gives $m\ddot{x}+kx=0$ for a spring oscillator.

For a uniform solid sphere, slicing into disks normal to the rotation axis givesUse total dumb-bell mass $M$ and $\ell =L-r$. Since both centres lie on its spin axis, its moment of inertia about that axis is $2M{R}^2/5$. If $\theta$ measures displacement from the bottom of the track, rolling without slipping imposes spin speed $|\dot{\varphi }|=\ell |\dot{\theta }|/r$. The rolling inside a fixed circular cylinder therefore hasThe equation of motion and its small-amplitude limit areWith $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the given lengths, $\boxed{T=2\pi /\omega \simeq 80.5\mathrm{s}}$. If the centre rises by height $h$, conservation of energy at the bottom givesThe speed expression uses exact height-energy conversion within the ideal rolling constraint; only the period formula uses the small-angle approximation.

Choose $z$ normal to the sheet and $x$ along the in-plane projection of the rotation axis. A uniform square has $I_x=I_y=m{a}^2/12$ and $I_z=m{a}^2/6$. Since ${\omega }_x={\omega }_z=\omega /\sqrt{2}$, the two nonzero angular-momentum components are $J_x=I_x\omega /\sqrt{2}$ and $J_z=I_z\omega /\sqrt{2}$. ThereforeAngular momentum is not parallel to the prescribed rotation axis. Multiplying the moment of inertia about that axis by $\omega$ would give the projection of angular momentum on the axis, not its magnitude.

Let $s$ denote distance from the neutral axis within a cross section. Bending to radius $R$ gives longitudinal strain $s/R$, so Hooke's law gives stress $Es/R$. Its moment isHere $E$ is Young's modulus and $I$ is the second moment of area, not a mass moment of inertia. For the stated square orientation, $I=a^4/12$.

For self-weight deflection of a square cantilever, the downward load per unit length is $q=\rho a^{2}g$. A free-end segment from $x$ to $l$ has shear-force magnitude $F(x)=q(l-x)$ and bending-moment magnitude $B(x)=q(l-x{)}^2/2$. Thus, with the given sign conventions,Using the small-slope curvature $1/R\simeq y^{\prime \prime }$ for downward deflection,The clamp fixes position and slope, and the unloaded free end has zero moment and shear:Four integrations yieldThe linear beam model requires small slopes and strains; the strong $l^4$ dependence eventually invalidates that approximation if other dimensions and material properties are fixed.

A normal mode is a motion in which all displacement coordinates oscillate at one angular frequency with fixed amplitude ratios. For small conservative oscillations, the symmetric stiffness and mass matrices give the generalized eigenvalue problem for small oscillations, $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. For positive mass and a stable equilibrium, its mass-orthogonal eigenvectors span the motion. A superposition of these modes, with phases and amplitudes fixed by initial displacements and velocities, gives the general solution; different modes need not share a phase.

Use upward end displacements. The spring compressions are $x_0-x_1$ and $x_0-x_2$, while the beam's centre rises by $q=(x_1+x_2)/2$. The potential energy isStatic force balance gives $2k{x}_0=mg$, cancelling the linear terms and leavingFor a uniform beam, its central moment of inertia in the vertical plane is $m{l}^2/12$. To first order in the tilt, $\theta =(x_2-x_1)/l$. Its translational and rotational kinetic energies areThus $\mathsf{M}=(m/6)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $\mathsf{K}=k1$. The normal modes of a rigid beam on two end springs have eigenvectorsThe first translates without tilt; the second rocks about a stationary centre. Ordinary frequencies are ${\omega }_s/(2\pi )$ and ${\omega }_a/(2\pi )$.

Assume the system had settled before the end load leaves, and put that load at end 1. Its extra static spring load is $Mg$ at that end and zero at the other. Therefore $x_1(0)=-Mg/k$, $x_2(0)=0$, and both initial velocities vanish because no impulse is imparted. Decomposing into the two normal modes givesOnly the symmetric mode contributes to the centre position:The datum $z_{\mathrm{eq}}$ is the unloaded beam's equilibrium height; no absolute height is supplied. The rocking mode is present even though it cancels out of this coordinate. These results use small tilt, $Mg/(kl)\ll 1$, ideal light vertical springs and negligible damping.

For rolling without slipping through height $h$, conservation of energy givesA uniform solid sphere has $I/(m{R}^2)=2/5$, a solid cylinder $1/2$, and a pipe $(1+r_{\mathrm{in}}^2/R^2)/2>1/2$. Thus the ball is fastest, then the solid cylinder, then the pipe. A smaller moment of inertia leaves more energy in translation. This assumes release from rest through the same height, adequate static friction and negligible rolling resistance; a thin pipe has $v^2=gh$.

Use $N$ turns of mean radius $a\simeq 1.05\mathrm{mm}$, including the wire radius, on a light nonmagnetic rod. With wire radius $r_w$, length $l$ and density $\rho$, neglecting helical pitch givesThe last expression is the transverse central moment of inertia of the uniformly distributed turns. The magnetic torque is $-m_B{B}_h\sin \theta$, so the small-angle magnetic dipole oscillation obeys $J\ddot{\theta }+m_B{B}_h\theta =0$. HenceThe number of turns cancels. Taking the turn radius as $1.00\mathrm{mm}$ instead gives $4.04\mathrm{s}$, consistent at the accuracy of the thin-winding model. The rod's inertia, damping and lead-wire torque are neglected.

Use positive normal stress for tension. On a small rectangular material element, ${\tau }_i$ is the normal force per unit area on the face perpendicular to axis $i$, and $e_i=\Delta {\ell }_i/{\ell }_i$ is the corresponding normal strain. Young's modulus $E$ is the ratio of axial normal stress to axial strain in uniaxial stress; Poisson's ratio $\sigma$ is minus the transverse-to-axial strain ratio in the same test. Thus isotropic linear elasticity gives $E{e}_1={\tau }_1-\sigma {\tau }_2-\sigma {\tau }_3$ and cyclic permutations. The stress symbols here follow the question; $\sigma$ denotes Poisson's ratio, not a stress component.

For $e_2=e_3=0$, the two transverse compliance equations imply ${\tau }_2={\tau }_3=\sigma {\tau }_1/(1-\sigma )$. Substitution into the first givesThis uniaxial strain modulus is greater than the uniaxial-stress Young's modulus for the positive Poisson's ratio of interest, because the lateral constraints prevent free Poisson contraction.

For a small twist $\varphi$, the bonded faces' relative tangential displacement at radius $r$ is $r\varphi$. The layer's shear strain is $r\varphi /h$, so the shear stress is $Gr\varphi /h$. Integrating the torque givesThe minus sign denotes a restoring torque; the question quotes its magnitude. This uses the disk's polar second moment of area.

For tilt about H, choose $y$ in the plate perpendicular to H. The vertical displacement of the bonded top face is $y\theta$ to first order, giving normal strain $y\theta /h$. The thin layer is constrained laterally by bonding, so its leading normal traction is $My\theta /h$, not $Ey\theta /h$. Integrating its moment about H givessince the diametric second moment of area is $\pi a^4/4$. This is the specified thin-layer approximation to the torsion and tilt stiffness of a bonded elastic disk. Relaxation near the free cylindrical edge is neglected; for a nearly incompressible layer it can limit quantitative accuracy unless the thinness is sufficient.

The thin square plate's moment of inertia about V is $I_V=2m{b}^2/3$. Its two perpendicular in-plane central moments are equal and sum to $I_V$, so the diagonal H has $I_H=m{b}^2/3$. Dividing the two angular stiffnesses by the appropriate moments of inertia givesThe geometry factors cancel in their ratio. Using the supplied relation between shear modulus and Young's modulus,The result lies just below the incompressible isotropic limit $1/2$, as expected for a rubber with a large compression stiffness compared with its shear modulus. It is an inference within the bonded-layer model, rather than an independently measured material constant.

Faraday's law for the moving loop is $\mathcal{E}=-d{\Phi }_B/dt$, equivalently $\mathcal{E}=\oint (\mathbf{E}+\mathbf{v}\times \mathbf{B})\cdot d\mathbf{\ell }$ for its moving circuit. Lenz's law fixes its orientation: the induced current opposes the change in linked magnetic flux. The minus sign represents this opposition, not opposition to the existing field in every phase of motion.

Unroll the small loop's motion locally into translation to the right at speed $v\simeq b\omega$, where $b$ is the radius marked in the source figure. Take the imposed field into the page. On entry the linked into-page flux increases, so the current is counterclockwise. The leading vertical side carries current upward and feels a leftward magnetic force $iBa$. The in-field parts of the horizontal sides feel equal opposite vertical forces; the trailing side is outside the field. On exit the current is clockwise. The trailing vertical side is now the one in the field and again carries upward current, producing leftward drag; the horizontal forces reverse and still cancel. When the entire loop is within the uniform region, its flux is constant, so $i=0$ and every side force is zero under the stated negligible-self-inductance approximation.

For a loop crossing a magnetic field boundary, overlap changes at rate $av$, henceFor approximately constant speed this gives two negative rectangular force pulses of duration $a/v$, separated by a short zero-force interval while the loop is fully inside. Their time integral isThe quoted impulse is its magnitude; this impulse of magnetic braking opposes the motion. In fact its independence of speed remains exact within this locally translational sharp-boundary model even if speed changes: each crossing contributes $-(B^2{a}^2/R)\int vdt=-(B^2{a}^2/R)\int dx$ over distance $a$.

Write the system's moment of inertia as $\mathcal{I}$. Since $d{K}_{\mathrm{rot}}/dt=\tau \omega =F_{x}b\omega =F_{x}v$, the local energy balance during overlap isThus rotational energy lost is exactly resistive heat in this model. At nearly constant $v$, each pass givesFor a finite speed reduction the exact relation is the integral of $i(t{)}^{2}R$, rather than initial speed times impulse; the equality of work and heating still holds. Fringing fields, loop curvature and finite self-inductance smooth the ideal pulses. A practical user control is to vary the electromagnet current, hence $B$: braking scales as $B^2$.

Take $\zeta$ as signed distance from the neutral axis within the beam cross-section. Bending into a large radius $R$ changes the length of a fibre proportionally to $\zeta /R$, so linear elasticity gives $\sigma =E\zeta /R$, up to the curvature sign convention. The moment magnitude isHere $I$ is the second moment of area about the neutral axis, with units of length to the fourth power, not the mass moment of inertia. A homogeneous cross-section has its neutral axis through the centroid under pure bending.

For the straight pin-ended beam, the compressive dead load produces bending moment $-Fy$ at transverse offset $y$. In the small-slope approximation, curvature is $y^{\prime \prime }$ with the corresponding sign convention, givingThe stated endpoints fix positions rather than slopes: this interpretation is consistent with the illustrated end forces and the later sinusoidal unstressed shape. Truly clamped endpoints would have a different threshold. A nonzero solution $y=A_n\sin (n\pi x/L)$ requires $F_n=n^2{\pi }^{2}EI/L^2$. Thus the first Euler load is $F_c={\pi }^{2}EI/L^2$. Below it the perfect straight state is stable, at it the first mode is marginal in linear theory, and above it the straight state is unstable. The linear equation selects thresholds but cannot set the amplitude or describe a finite postbuckled equilibrium.

For the initially curved compressed beam, elastic stress responds to the change from its stress-free curvature. With $y=y_0+\eta$, the moment is $EI(y^{\prime \prime }-y_0^{\prime \prime })=EI{\eta }^{\prime \prime }$, so force balance givesPut $q=\pi /L$ and seek the branch continuously connected to the unloaded shape. Since $y_0=Y\sin qx$, substitute $\eta =C\sin qx$ to obtain $-EI{q}^{2}C=-F(Y+C)$, orAway from eigenvalues, the homogeneous solution $A\sin (\sqrt{F/(EI)}x)+B\cos (\sqrt{F/(EI)}x)$ vanishes under the two end conditions, so this particular solution is the unique linear equilibrium. For $0\le F<F_c$ and $Y>0$, the maximum additional displacement is ${\eta }_{\max }=YF/(F_c-F)$ and maximum total displacement is $y_{\max }=Y/(1-F/F_c)$, both at midspan.

This sinusoidal imperfection amplification of a column produces deflection at every positive load and strong amplification near $F_c$. The apparent divergence signals loss of the small-slope model before infinite displacement; nonlinear geometry is needed there. The formal continuation above the first threshold is not a stable physical loading branch. Both total and additional displacement are distinguished in the graph.

Separate a ball's orbital motion from its spin. For a rolling ball between concentric races, the ball centre lies $2a$ from the axle axis. Let its tangential centre velocity be $v$ in the direction of axle motion and let its signed spin angular velocity be $\Omega$, positive in the same sense as $\omega$. Rolling without slipping at the moving inner contact and stationary outer contact requiresHenceThe spin is opposite to the axle's rotation, while the centre orbits in the same sense. The total angular momentum about the axle is the sum of its orbital angular momentum and its angular momentum about the centre of mass:For a uniform solid spherical ball, $I_{\mathrm{cm}}=2m{a}^2/5$, soThe solid-ball assumption supplies the usual moment of inertia; the preceding expression in terms of $I_{\mathrm{cm}}$ also states the answer for a different internal mass distribution.

Elasticity describes recoverable deformation. For small displacements $\mathbf{u}(\mathbf{x})$, the infinitesimal strain tensor is the symmetric part of the displacement gradient,Its diagonal entries describe extension, while its off-diagonal entries describe shear. The engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$; the antisymmetric part of the displacement gradient is a small rigid rotation, not elastic strain.

The stress tensor specifies the force per unit area on each oriented plane: $\mathbf{t}(\hat{\mathbf{n}})=\mathbf{\sigma }\hat{\mathbf{n}}$. Positive normal stress is tensile in the convention used here; tangential traction gives shear stress. In the absence of internal couples, angular-momentum balance makes the stress tensor symmetric. Force balance gives $\nabla \cdot \mathbf{\sigma }+\mathbf{f}=0$ in static equilibrium, where $\mathbf{f}$ is body force per unit volume.

Linear elasticity relates stress and strain. A homogeneous isotropic solid needs two independent elastic constants:Here $G$ is the shear modulus and $\lambda$ is the other Lamé parameter. For uniaxial loading, Young's modulus is $E={\sigma }_{xx}/{ϵ}_{xx}$, with lateral contraction described by Poisson's ratio $\nu =-{ϵ}_{yy}/{ϵ}_{xx}$. The bulk modulus is $K=-p/(\Delta V/V)$ for a small hydrostatic compression. The relations between these constants areStable isotropic elastic strain energy density requires $G>0$ and $K>0$, implying $-1<\nu <1/2$. The incompressible limit is approached as $\nu \to 1/2$. Anisotropic crystals generally need more constants; the isotropic relations should not be imposed on arbitrary directions in such a crystal.

The elastic energy density is $w=\mathbf{\sigma }:\mathbf{ϵ}/2$. In the isotropic linear model this becomes $\lambda (\mathrm{tr}\mathbf{ϵ}{)}^2/2+G\mathbf{ϵ}:\mathbf{ϵ}$. The quadratic form explains why reversible loading stores energy and unloading can return it, unlike plastic deformation or a dissipative hysteresis cycle.

For a slender beam described by the Euler-Bernoulli beam equation, plane sections approximately remain plane. A fibre a distance $y$ from the neutral axis has axial strain $-y\kappa$, where $\kappa$ is curvature, and stress $-Ey\kappa$. Integrating its moment about the neutral axis gives, with consistent signs,$I_A$ is a second moment of area, with units of length to the fourth power; it is not a mass moment of inertia. For a rectangular section of width $b$ and thickness $h$, $I_A=b{h}^3/12$, explaining the strong dependence of bending stiffness on thickness. For small transverse deflection $w(x)$, curvature is approximately $w^{\prime \prime }$, and the Euler-Bernoulli beam equation is $E{I}_A{w}^{\prime \prime \prime \prime }=q(x)$ with a suitable load-sign convention. A cantilever of length $\ell$ under end load $F$ has tip deflection $F{\ell }^3/(3E{I}_A)$ and largest bending stress of magnitude $F\ell c/I_A$, where $c$ is the furthest distance from the neutral axis.

In torsion of a solid circular shaft, a cross-section rotated by angle $\theta (x)$ has shear strain $r{\theta }^{\prime }(x)$. ThereforeHere $J$ is the polar second moment of area. A uniform shaft of length $\ell$ under constant torque twists by $\mathcal{T}\ell /(GJ)$, and its largest shear stress is $\mathcal{T}a/J$. Noncircular sections can warp, so replacing their torsional stiffness by $G$ times their polar area moment is not generally valid. Both bending and torsion formulas presume small, elastic deformations; exceeding the elastic limit can produce permanent deformation or fracture.

For the uniform solid cylinder, the axial moment of inertia isThe rotational kinetic energy at $\Omega =300\mathrm{rad}{\mathrm{s}}^{-1}$ isThis includes the shaft's bulk steady rotation; the small twist does not appreciably alter its mass moment at this order.