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The moment of inertia about a specified axis is $I=\int s^{2}dm$, where $s$ is perpendicular distance from the axis. The parallel axis theorem states $I=I_{\mathrm{cm}}+m{d}^2$ for two parallel axes separated by $d$, one passing through the centre of mass of a body of mass $m$.

It is easiest to calculate the eccentric disc by subtraction. Let $m_f$ be the mass of the unperforated disc and $m_h$ the removed mass. Uniform thickness and density give $m_h=m_f/4$ and $m_f-m_h=M$, hence $m_f=4M/3$ and $m_h=M/3$. The hole is tangent to the rim, so its centre, and therefore the shaft axis, lies $R/2$ from the original disc centre. About the shaft axis,The shaft contributes $I_{\mathrm{shaft}}=M(R/2{)}^2/2=M{R}^2/8$. ThusThe motor's torque is $P/\omega$ at nonzero angular velocity, while the resisting torque is $-\gamma \omega$. The rotational equation of motion and its equivalent energy balance areAt the steady state, input power equals dissipated power, givingFor the acceleration, use $u={\omega }^2$. Then $\dot{u}+(2\gamma /I)u=2P/I$ with $u(0)=0$, soThe approach to ${\omega }_0$ takes infinite time. This constant-power rotational acceleration idealizes the start: $P/\omega$ diverges at rest, although the energy balance has a well-defined limiting solution. A real motor has finite starting torque.

In a fixed point-mass Newtonian gravitational field, a test mass $m$ experiencesBecause this is a central force, its torque about the attracting centre vanishes and angular momentum is conserved. The orbit lies in a fixed plane and sweeps out equal areas in equal times. Since the force is also conservative, the specific orbital energy is constant:The last two terms form an effective potential. They show how the angular-momentum barrier competes with attraction and determines the turning radii. For nonradial motion the orbit is a conic section:Negative energy gives an ellipse, with $e<1$; zero energy gives a parabola; positive energy gives a hyperbola. The circular limit has $e=0$, $v_c=\sqrt{GM/r}$ and $\epsilon =-GM/(2r)$. For a bound ellipse of semimajor axis $a$, $\epsilon =-GM/(2a)$ and the orbital period is $T=2\pi \sqrt{a^3/(GM)}$. The local escape speed is $\sqrt{2GM/r}$. These are the point-mass, test-particle limits; extended mass distributions or noncentral perturbations change the orbit laws.

The Lorentz force determines the motion of a charged particle:At nonrelativistic speeds, $\mathbf{p}=m\mathbf{v}$. The electric field does work at rate $q\mathbf{E}\cdot \mathbf{v}$, while the magnetic part of the Lorentz force is perpendicular to velocity and does no work. A uniform electric field gives constant acceleration $q\mathbf{E}/m$; with a transverse initial velocity, the trajectory is a parabola.

In a uniform magnetic field, the parallel velocity is constant, while the perpendicular component undergoes circular motion. The angular cyclotron frequency, orbit radius and helical pitch areThe sign of $q$ determines the sense of rotation. For perpendicular uniform electric fields and magnetic fields, the guiding centre has the electric drift velocityIndeed, $\mathbf{E}+{\mathbf{v}}_D\times \mathbf{B}=0$, so the remaining motion is cyclotron rotation about that drifting centre, with possible motion along the magnetic field. The drift is independent of electric charge sign and mass. At relativistic energies, retain $\mathbf{p}=\Gamma m\mathbf{v}$ and the corresponding relativistic orbit radius rather than assuming constant nonrelativistic acceleration.

Take a representative tennis-ball mass of $m\simeq 0.06\mathrm{kg}$. The speed is nonrelativistic, so the de Broglie wavelength isThis is an estimate because the mass was not specified. The wavelength is vastly smaller than any practicable aperture or obstacle separation: a diffraction angle of order $\lambda /d$ is unresolvably small for laboratory-scale $d$. Moreover, interactions with air, thermal radiation and the apparatus destroy observable coherence between macroscopically distinct paths. Quantum mechanics still applies; the small wavelength and rapid decoherence explain why ordinary experiments reveal classical trajectories rather than a measurable interference pattern.

For uniform Earth density, the shell theorem says that only mass within radius $r$ contributes to the gravitational field there. Since enclosed mass is proportional to $r^3$,At depth $h$, $g_h/g_0=1-h/R$. A simple pendulum has small-amplitude period $P=2\pi \sqrt{l/g}$, so the pendulum clock below a uniform-density surface runs at the fraction $P_0/P_h=\sqrt{1-h/R}$ of its surface-calibrated rate. After a surface-clock hour, its lag isThe mine clock is slower because gravity is weaker under the stated uniform-density assumption. The expansion $\Delta t\simeq 3600h/(2R)$ gives the same result. This is a change in the mechanical pendulum period, not a relativistic clock effect; a realistic Earth density profile would require a different interior-gravity calculation.

Use distinct symbols for the dimensions of mass, length, time, temperature and amount of substance. The molar gas constant has dimensions $\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}{\mathsf{\Theta }}^{-1}{\mathsf{N}}^{-1}$, while the given $M$ is a molar mass with dimensions $\mathsf{M}{\mathsf{N}}^{-1}$. Matching dimensions in the proposed power law giveswith the amount-of-substance equation repeating $\alpha +\gamma =0$. ThereforeDimensional analysis fixes these exponents but not a dimensionless multiplier. For example, an open-tube mode number and the gas's heat-capacity ratio enter the actual resonance frequency without changing the dimensions.

The original diagram has a horizontal bottle neck, so weight does not enter the axial force balance. Gauge pressure acts on the cork's end area and supplies outward force $F_g=\pi a^2{P}_g$. The radial contact force is the contact pressure times the cylindrical contact area, $N=2\pi a(l-x)P_c$. The limiting friction is $\mu N$, so outward sliding occurs whenThe assumption of equal static and kinetic friction coefficients removes a separate force drop when slipping starts.

Initially the pressure is marginal at $x=0$, so $\pi a^2{P}_g=2\pi al\mu P_c$. During extraction the contact length decreases, while the pressure force stays constant. The net force is therefore $F(x)=\pi a^2{P}_{g}x/l$. The cork's mass is $m_c=\rho \pi a^{2}l$, giving the pressure-driven cork ejection equationFor $x(0)=0$ and $\dot{x}(0)=v_0>0$, solving the constant-coefficient equation givesThe equilibrium is unstable: as the cork moves out, friction decreases and the unbalanced force grows.

Multiplying the equation of motion by $\dot{x}$ and integrating gives ${\dot{x}}^2=v_0^2+b^2{x}^2$. The contact ends at $x=l$, soEquivalently, the net work is ${\int }_0^{l}F(x)dx=\pi a^2{P}_{g}l/2$. Exactly zero initial velocity would leave the ideal marginal configuration stationary. The exit time $b^{-1}\mathrm{arsinh}(bl/v_0)$ diverges as $v_0\to 0^{+}$, even though the limiting exit speed is finite.

The density is $500\mathrm{kg}{\mathrm{m}}^{-3}$, so $v_{\mathrm{exit}}\simeq \sqrt{400000/500}=28.3\mathrm{m}{\mathrm{s}}^{-1}$. For an ideal upward launch at this speed, conservation of mechanical energy gives height above the exitThis neglects air resistance and the small gravitational work during extraction compared with the launch energy. Actual gas pressure need not stay constant, so it is an idealized estimate.

Adding a stopper that doubles the moving mass, while preserving the same pressured area and contact friction, leaves the extraction work unchanged. The squared exit speed and the ballistic height are therefore halved:

For particles in an inertial frame, total momentum and angular momentum about a fixed origin areThe momentum derivative is the total external force when internal forces cancel pairwise. Angular momentum is conserved when the total external torque vanishes and the internal torques sum to zero, as for pairwise equal-and-opposite central forces. Equal-and-opposite internal forces alone need not cancel internal torque if their lines of action are different. An isolated system with these internal-force properties conserves both quantities; if fields carry momentum, the complete system must include them.

For a straight rocket, let its instantaneous remaining mass be $M$ and let $dM<0$ be ejected during a velocity increase $dv$. The exhaust's inertial velocity is $v-v_e$ to first order. Conservation of momentum givesIntegrating from initial mass $m+m_f$ to dry mass $m$ yields the rocket equationIt is the rocket-plus-exhaust system that is isolated; the rocket alone is a variable-mass open system.

For the firework, the original diagram places one tangentially firing point rocket at the end of each radial rod. The parallel axis theorem changes each rod's central inertia to $m{a}^2/12+m(a/2{)}^2=m{a}^2/3$. Adding all rods and fueled rockets givesWhen fuel is exhausted, $I_f=4m{a}^2$.

Let $M_f$ now denote the total remaining fuel in all three rockets, so $I=4m{a}^2+a^2{M}_f$. Fuel at the rim co-rotates before ejection. Its backward tangential exhaust velocity is $a\omega -v_e$, and its angular momentum per unit mass about the axle is $a(a\omega -v_e)$. For an infinitesimal ejection,Since $dI=a^{2}d{M}_f$, the co-rotation terms cancel and leave $Id\omega =-a{v}_{e}d{M}_f$. The tangential-exhaust rotor with a fixed dry inertia therefore obeysNeglecting the exhaust's initial tangential motion would give an incorrect angular-momentum balance. The rod mass remains in the denominator throughout the burn.

After the burn, the opposing constant axle torque $G>0$ acts on the fixed inertia $I_f$. Its work over the remaining turn equals the initial rotational kinetic energy:

Expand the cloud's potential energy about a stable equilibrium: $U(z)=U(0)+U^{\prime }(0)z+U^{\prime \prime }(0)z^2/2+\cdots$. Equilibrium gives $U^{\prime }(0)=0$, stability gives $U^{\prime \prime }(0)=k>0$, and reflection symmetry also removes odd terms. For small displacements, $U-U(0)\simeq k{z}^2/2$ and $F=-dU/dz\simeq -kz$.

Including the electric and damping forces, Newton's second law is $m\ddot{z}=qE(t)-b\dot{z}-kz$. Dividing by $m$ gives the Lorentz oscillator modelThe specified convention uses $2\gamma$ as the damping coefficient per unit mass.

For the steady response, represent the field by $E_0{e}^{i\omega t}$ and the displacement by $\tilde{z}{e}^{i\omega t}$, taking real parts for physical quantities. Substitution givesDefine the nonnegative amplitude $z_0=|\tilde{z}|$ and $\delta =atan2(2\gamma \omega ,{\omega }_0^2-{\omega }^2)$. ThenFor positive $q$, $\varphi =-\delta$. For electrons $q<0$, the phase of a negatively charged driven oscillator is $\varphi =\pi -\delta$: the low-frequency displacement is opposite to the field, rather than in phase with it. A signed amplitude could absorb this extra $\pi$, but the phase convention must then be stated. The induced dipole $p=qz$ has phase $-\delta$ for either charge sign. Transient free oscillations have been omitted because the question asks for steady state.

For the electron cloud, $q/m=-e/m_e$ andThe visible range corresponds to $\omega =2\pi c/\lambda$ between $2.35\times 10^{15}$ and $4.71\times 10^{15}\mathrm{rad}{\mathrm{s}}^{-1}$, below this resonance. The resonant wavelength is about $180\mathrm{nm}$, in the ultraviolet.

The acceleration amplitude is ${\omega }^2{z}_0$. At fixed incident field amplitude, the scattered intensity is therefore proportional toWell below resonance and outside strong damping corrections, $z_0$ is nearly frequency-independent and this becomes the Rayleigh scattering wavelength dependence $I_{\mathrm{sc}}\propto {\omega }^4\propto {\lambda }^{-4}$. The shortest visible wavelengths are consequently scattered much more strongly into directions away from the direct solar beam than the longest ones, producing a blue sky. The simple fourth-power approximation gives a factor 16 between 400 and 800 nm at equal incident intensity. Retaining the undamped denominator with the given resonance raises that factor to about 22.7, since the upper end is not extremely far below resonance. The exact ratio needs the unspecified damping and incident solar spectrum; the stronger scattering of the blue end is the robust conclusion here.

The model describes frequency-dependent scattering, not the detailed perceived color after solar-spectrum weighting and human visual response. It also explains why treating the electrons as free at visible frequencies would miss the bound-charge fourth-power limit.

At the first two specified times, the displacement profiles areThe second curve is the first shifted right by $\pi /(2k)=\lambda /4$. More generally a point of fixed wave phase satisfies $\omega t-kx=\text{constant}$, so it moves along $x=(\omega /k)t+\text{constant}$. Hence the wave travels in the positive $x$ direction withThis phase velocity is distinct from the local transverse velocity $\partial y/\partial t$ of a string element.

For the force derivation take a short segment between $x$ and $x+\Delta x$. Let the tangent angles to the positive $x$ axis be ${\theta }_L$ and ${\theta }_R$. The left tension pulls along the negative tangent and the right tension along the positive tangent. With small slopes and approximately constant tension magnitude $T$, the horizontal components cancel to leading order andThe segment mass is $\rho \Delta x$, where $\rho$ is mass per unit length, not volume density. Newton's second law therefore givesThe approximation neglects bending stiffness, damping and changes in tension caused by finite-amplitude stretching.

Converting the supplied density gives $\rho =0.020\mathrm{kg}{\mathrm{m}}^{-1}$, soFor the pulse, define $g(x)=A{e}^{-a{x}^2/2}$ with $a>0$. The given initial velocity is $-c{g}^{\prime }(x)$, exactly the velocity associated with a right-moving displacement $g(x-ct)$. ThereforeIt moves right without changing shape in this nondispersive model. With the same initial displacement but zero initial transverse velocity, D'Alembert's formula instead gives the Gaussian pulse splitting from zero initial velocityThe two half-amplitude pulses travel in opposite directions. Their initial velocities cancel, while their initial displacements add to the prescribed Gaussian. The two initial-value problems have different kinetic energies; splitting into half-height pulses does not contradict energy conservation. These forms assume a string long enough that reflections from its ends have not returned.

A charge $q$ moving nonrelativistically in prescribed fields obeys the Lorentz force lawThe magnetic part is perpendicular to velocity, so $d(m{v}^2/2)/dt=q\mathbf{E}\cdot \mathbf{v}$: the electric field can change kinetic energy, while the magnetic field alone changes direction without doing work.

In a uniform electric field with no magnetic field, acceleration is constant: $\mathbf{v}={\mathbf{v}}_0+(q\mathbf{E}/m)t$ and $\mathbf{r}={\mathbf{r}}_0+{\mathbf{v}}_{0}t+q\mathbf{E}{t}^2/(2m)$. The trajectory is a parabola when there is initial motion perpendicular to the field, and a line when motion is purely along it. Reversing the charge reverses the electric acceleration.

In a uniform magnetic field with no electric field, resolve velocity into parallel and perpendicular components. The parallel component is constant; the perpendicular component rotates at the cyclotron frequencyThe resulting path is a helix, or a circle if $v_{\parallel }=0$, with pitch $2\pi v_{\parallel }/{\omega }_c$. Charge sign determines its sense of rotation. These relations underlie magnetic bending and mass/charge analysis. At relativistic energies, use momentum $\mathbf{p}=\gamma m\mathbf{v}$; at fixed energy the cyclotron frequency becomes $|q|B/(\gamma m)$.

For uniform crossed fields $\mathbf{E}\perp \mathbf{B}$, the velocity ${\mathbf{v}}_D=\mathbf{E}\times \mathbf{B}/B^2$ satisfies $\mathbf{E}+{\mathbf{v}}_D\times \mathbf{B}=0$. The orbit is a gyromotion about a guiding center drifting withindependent of mass and charge sign. A velocity selector transmits particles with the appropriate direction and speed $E/B$ without deflection. If the electric field has a component along the magnetic field, that component accelerates the parallel motion instead. Nonuniform fields can add other drifts; the simple helix and uniform crossed-field expressions require the stated uniform-field assumptions.

A position-dependent force is a conservative force when its work depends only on endpoints. Equivalently, its work around any closed allowed path vanishes, and one can define a potential energy $U$ satisfyingThe potential is defined up to an additive constant. A vanishing curl, $\nabla \times \mathbf{F}=0$, is a local test; it guarantees a global single-valued potential on a simply connected region with suitable regularity. Holes or excluded singularities require checking the global path condition.

The work-energy theorem gives $dK/dt=\mathbf{F}\cdot \mathbf{v}$. For a static potential this implies $d(K+U)/dt=0$. If $U$ depends explicitly on time, then instead $d(K+U)/dt=\partial U/\partial t$ under that force alone. Friction and other nonconservative forces add their own power and can convert mechanical energy into heat.

Examples include the gravitational potential energy $-GMm/r$ outside a spherical mass and the electrostatic energy $U=qV$, where $\mathbf{E}=-\nabla V$. For a point source charge $Q$, $U=qQ/(4\pi {ϵ}_{0}r)$, giving attraction or repulsion through the sign of $qQ$. A spring has $U=k{x}^2/2$ and force $-kx$. An equilibrium has $\nabla U=0$; a strict local minimum is stable against small displacements when the quadratic curvature is positive. Expanding about that minimum explains the harmonic restoring force used in many small-oscillation models.

Magnetic force on a moving point charge does no work, but it is velocity-dependent and is not generally a position-only force $-\nabla U$. Likewise a time-varying magnetic flux induces a circulating electric field with $\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$, so that electric field cannot globally be represented solely by an electrostatic scalar potential. These distinctions prevent treating every field force as conservative simply because an energy description exists for the complete system.

Use dimensions $\mathsf{M},\mathsf{L},\mathsf{T}$ for mass, length and time. The reduced Planck constant, gravitational constant and speed of light have dimensionsFor $l_p={\hslash }^{\alpha }{G}^{\beta }{c}^{\gamma }$ to have the dimensions of a length, dimensional analysis requiresThe first equation gives $\beta =\alpha$; the third gives $\gamma =-3\alpha$. Substitution in the second gives $2\alpha =1$. HenceWith $\hslash =1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$, $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$, the Planck length isThe proposed monomial fixes the dimensionless prefactor to one; dimensional analysis on its own cannot select an arbitrary dimensionless multiplier.

Powered vertical flight. Take upward velocity $v$ as positive, and let $v_e>0$ be the downward exhaust speed relative to the rocket. During $dt$ the rocket's mass changes by $dm<0$, and exhaust mass $-dm$ leaves with Earth-frame velocity $v-v_e$. Compare the momentum of the rocket plus this newly expelled exhaust before and after the interval:Discard the second-order product $dmdv$. The result is $mdv+v_{e}dm=-mgdt$, orThe positive rocket thrust is $-v_e\dot{m}$. Integrating with $v(0)=0$ and $m(0)=m_0$ gives the rocket equation with gravitational loss:Using $d(mv)/dt$ for the rocket alone without the exhaust momentum flux would give the wrong equation for this open system.

Orbital speed under the stated gravity assumption. A circular orbit of radius $r_o=R_E+h_o=7.90\times 10^6\mathrm{m}$ requires centripetal acceleration $v_o^2/r_o$. If the magnitude of gravitational acceleration is the stipulated constant $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ even at the orbit,This treats gravity as centrally directed for the orbital calculation, while retaining its constant magnitude. The subsequent flat-surface approximation treats its direction as locally vertical.

Burnout height. During the unpowered projectile motion, horizontal velocity is constant. At the top of the trajectory the vertical component vanishes, so the horizontal component must equal the orbital speed $v_o$. At burnout, therefore,These follow either by integrating $\ddot{y}=-g$ or by the vertical work-energy theorem. Substituting the preceding constant-$g$ orbital result givesThe vertical component is initially positive, as required to reach the orbit after burnout.

Alternative inverse-square interpretation. A real Earth's exterior Newtonian gravitational field is not constant over a $1500\mathrm{km}$ altitude change. If $9.8\mathrm{m}{\mathrm{s}}^{-2}$ instead means the surface value $g_0$, then Newton's law of universal gravitation gives $GM=g_0{R}_E^2$ andUsing this orbital speed but the paper's constant-$g$ flat projectile motion approximation for the coast would giveThis alternative changes the global constant-gravity assumption; it is not the same model as the $183\mathrm{km}$ result. An exact curved Kepler orbit coast would also abandon the prescribed flat trajectory. The printed assumptions support the constant-magnitude answers above, while the inverse-square values explain the different answer obtained from the usual physical Earth-orbit formula.

Directions. Choose $+x$ along the horizontal spindle from the pivot to the disk, and $+z$ vertically upward. With lever arm $\mathbf{r}=l\hat{\mathbf{x}}$ and weight ${\mathbf{F}}_g=-mg\hat{\mathbf{z}}$, the torque about the pivot isClockwise precession viewed from above has $\mathbf{\Omega }=-\Omega \hat{\mathbf{z}}$. The relation $\mathbf{\tau }=\mathbf{\Omega }\times \mathbf{L}$ therefore requires the spin angular momentum to point along $-x$, from the disk toward the pivot. The disk spins clockwise when viewed from its outer face looking along the spindle toward the pivot. Looking from the pivot reverses that apparent rotation. The right-hand rule fixes these signs unambiguously.

The drawing shows an instantaneous top view; the disk's angular momentum and the horizontal spindle turn clockwise together. The pivot force has zero torque about the pivot.

Rate of precession. For steady horizontal motion the spin contribution has magnitude $L=I\omega$ and turns through $d\varphi =\Omega dt$ in time $dt$. Its change is perpendicular to itself and has magnitude $dL=Ld\varphi$. HenceThis is the usual slow steady gyroscope precession relation. Any constant vertical angular momentum associated with steady precession has zero time derivative; the horizontal spin contribution supplies the changing component here. Transient nutation is excluded by the prescribed horizontal steady motion.

Clock design. Convert both rotation rates to angular frequencies:For the moment of inertia of a uniform disc, $I=m{R}^2/2$, so the mass cancels:

Spin friction. At fixed geometry, reducing $\omega$ increases $\Omega \propto 1/\omega$. The precession period, $T_p=2\pi /\Omega$, therefore decreases. This trend holds while slow steady precession remains a suitable description; it cannot be extrapolated to a stopped disk.

Counterweight. To oppose the disk's gravitational torque, place the counterweight on the other side of the pivot, at $-l_c\hat{\mathbf{x}}$. The net torque isA horizontal axle can remain fixed when this vanishes, givingFor the fixed spin direction already found, counterbalanced gyroscope precession has signed rateMoving the counterweight outward from the pivot first slows the clockwise motion, eliminates its driving torque at balance, and then reverses the steady motion to counterclockwise. A counterweight on the disk's side would add to the gravitational torque. These statements neglect axial spin moment of inertia of the counterweight and assume the stated horizontal motion, without a separately imposed precession at balance.

Momentum identity. From the Lorentz factor, $v^2/c^2=1-{\gamma }^{-2}$. The relativistic three-momentum therefore satisfiesTogether with relativistic energy $E=\gamma m{c}^2$, this gives the energy–momentum relation $E^2-p^2{c}^2=m^2{c}^4$.

Parent in the laboratory. The rest energy is $m_0{c}^2=1\mathrm{GeV}$, while the total relativistic energy is $3\mathrm{GeV}$. ConsequentlyIts momentum magnitude is $p_{0}c=\sqrt{3^2-1^2}\mathrm{GeV}=\sqrt{8}\mathrm{GeV}$.

Daughters in the zero-momentum frame. The centre-of-momentum frame is initially the parent's rest frame. Conservation of momentum gives ${\mathbf{p}}_1^{\ast }=-{\mathbf{p}}_2^{\ast }$. Equal daughter rest masses then imply equal relativistic energies. Conservation of relativistic energy givesTheir speeds are thereforeThe two velocities are opposite, and $p_X^{\ast }c=\sqrt{(1/2{)}^2-(1/3{)}^2}\mathrm{GeV}=\sqrt{5}\mathrm{GeV}/6$. Their total kinetic energy is $m_0{c}^2-2m_X{c}^2=m_0{c}^2/3$.

Invariant of the complete system. Before the decay, in the centre-of-momentum frame,After the relativistic two-body decay,In the laboratory, four-momentum conservation fixes the summed daughter energy to $3\mathrm{GeV}$ and their summed momentum to the parent's ${\mathbf{p}}_0$. Thus the same calculation isThe invariant is formed after summing the daughter four-momenta; adding their separate squared rest energies would omit the relative-motion contribution to the invariant mass of the pair.

Axial moment of inertia. In the original diagram $R_1$ is the outer radius and $R_2$ the inner radius. For uniform mass density $\rho$, a coaxial annulus of thickness $dr$ has $dm=2\pi \rho Lrdr$. Therefore the axial moment of inertia of a hollow cylinder isSince $M=\pi \rho L(R_1^2-R_2^2)$, factorizing the fourth-power difference givesFor a solid cylinder $R_2\to 0$, giving $I=M{R}_1^2/2$. For a thin cylindrical shell with fixed mass and $R_1,R_2\to R$, $I=M{R}^2$. The latter limit agrees with every mass element being at radius $R$.

Undamped rolling oscillations. Let $x$ be axle displacement from the spring equilibrium. Rolling without slipping imposes $|{\omega }_{\mathrm{roll}}|=|\dot{x}|/R$. The sum of translational and rotational kinetic energy isThe potential energy is $U=k{x}^2/2$. Differentiating $T+U$ in the absence of damping, or applying the force and torque equations below, givesFor the thin shell $I=m{R}^2$, the effective inertia is $2m$, and the angular frequency of the simple harmonic motion isStatic contact friction supplies the required torque but does no work when the contact point is instantaneously stationary.

Damped force balance. Take rightward force as positive and counterclockwise rotation as positive. Write $f$ for the signed static friction at the bottom contact. ThenEliminate $f=-I\ddot{x}/R^2$ and use $I=m{R}^2$:Here $b$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, so $B$ has units ${\mathrm{s}}^{-1}$ and $C$ units ${\mathrm{s}}^{-2}$. The stated drag force is interpreted as acting through the axle, with no additional viscous torque. An oil bath could also exert such a torque physically, but its magnitude is not specified by this force-only model.

Light damping. Substitution of $z=A{e}^{-pt}$ into the differential equation gives $p^2-Bp+C=0$. In the underdamped regime $B^2<4C$,Taking real linear combinations gives the general damped harmonic oscillator displacementThe constants are fixed by the initial displacement and velocity. For example, release from rest at $x_0$ gives $A_1=x_0$ and $A_2=b{x}_0/(4m{\omega }_d)$. Light damping means $b^2\ll 8mk$, in which case ${\omega }_d\simeq {\omega }_0$ and the envelope decays slowly compared with an oscillation.

Equation and interpretation. The supplied equation is the time-independent Schrödinger equation for a nonrelativistic particle with mass $m$ in a prescribed potential energy $V(x)$. Equivalently,It is an eigenvalue problem for the Hamiltonian operator. A stationary energy state has full wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. By the Born rule, $|\psi (x){|}^{2}dx$ is the probability of finding the particle in a small interval $dx$, with wavefunction normalization ${\int }_{-\infty }^{\infty }|\psi {|}^{2}dx=1$.

Infinite barriers. Infinite exterior potential energy excludes any finite-energy probability outside the interval. The infinite square well consequently has $\psi =0$ outside and the hard-wall boundary conditionsThe interior wavefunction must meet the zero exterior amplitude continuously. Its derivative need not be continuous across an infinite wall; imposing both zero value and zero derivative would eliminate every nonzero interior solution.

Ground state. Inside the well $V=0$, so ${\psi }^{\prime \prime }+k^2\psi =0$, with $k=\sqrt{2mE}/\hslash$. Its general solution is $A\sin kx+B\cos kx$. The left wall gives $B=0$, and the right wall gives $ka=n\pi$ for $n=1,2,\dots$. Thus the discrete energy eigenvalues are $E_n=n^2{\pi }^2{\hslash }^2/(2m{a}^2)$. The smallest, $n=1$, is the ground state. Sinceits normalized wavefunction, choosing an irrelevant overall phase to make it positive inside, is