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Use base dimensions $\mathsf{M},\mathsf{L},\mathsf{T}$ for mass, length and time. The proposed dimensional analysis givessoEquating the independent dimensional powers requiresConsequentlyThe mass density drops out because there is no other mass-bearing quantity with which to combine it under the stated assumptions.

Dimensional analysis fixes the powers but leaves a dimensionless coefficient undetermined. For a linear deep-water gravity wave, the dispersion relation is ${\omega }^2=gk$, giving phase velocity $\omega /k=g/\omega$ and group velocity $d\omega /dk=g/(2\omega )$. Both obey the inferred scaling. The phrase wave speed alone does not distinguish these two velocities or fix the coefficient.

Definitions. The torque of a force $\mathbf{F}$ about an origin is $\mathbf{\tau }=\mathbf{r}\times \mathbf{F}$, where $\mathbf{r}$ runs from that origin to the force's point of application. Its component about a fixed axis changes angular momentum about that axis. The moment of inertia about that axis iswhere $s$ is perpendicular distance to the axis. It measures resistance to angular acceleration; for fixed-axis rigid rotation, ${\tau }_{\mathrm{net}}=I\alpha$. The units are $\mathrm{N}\mathrm{m}$ for torque and $\mathrm{kg}{\mathrm{m}}^2$ for moment of inertia.

Mass and axial inertia. Slice the shaft perpendicular to its common axis. A slice of radius $r(z)$ has mass $dm=\rho \pi r(z{)}^{2}dz$ and moment of inertia $dI=\frac{1}{2}r(z{)}^{2}dm$, using the moment of inertia of a uniform disc. Thus the axial moment of inertia of a solid of revolution uses the second and fourth powers of its radius:For the tapered section, take $z$ from its narrow end toward its wide end, so $r(z)=a(1+z/l)$ for $0\le z\le l$. Its contributions areThe two cylinders contribute $4\rho \pi a^{2}l$ and $\rho \pi a^{2}l$ to mass, and $8\rho \pi a^{4}l$ and $\frac{1}{2}\rho \pi a^{4}l$ to moment of inertia. Adding the three sections givesAll slice rotation axes coincide with the shaft axis, so the moments add directly.

Frictional torque. The normal load on the narrow circular end is $Mg$. With the stipulated uniform pressure, $p_N=Mg/(\pi a^2)$. A contact annulus of radius $s$ and width $ds$ has area $dA=2\pi sds$, normal force $p_{N}dA$ and tangential kinetic friction $dF=\mu p_{N}dA$. Multiplying by its lever arm $s$ and integrating gives the uniform-pressure friction torque on a circular contact:Here $G_f$ is the positive magnitude; its torque opposes the angular velocity.

Steady speed and stopping rotations. Mechanical power in rotation is $P=\tau \omega$. At steady angular velocity, the driving and frictional torques balance, so constant-power rotation against dry friction givesUsing $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $a=0.100\mathrm{m}$,Once the motor is removed, the speed-independent kinetic friction gives constant angular acceleration $-G_f/I$. Equivalently, the initial rotational kinetic energy is dissipated as frictional work. The stopping angle under constant resisting torque is thereforeThe total angular displacement corresponds toHence the shaft completesand then another $0.16668$ of a rotation. Its stopping time is $I{\omega }_{\ast }/G_f=2.05461\mathrm{s}$. Using the rounded value $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ changes the fractional rotation slightly but leaves four complete rotations.

Use a stationary wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$ solving the time-independent Schrödinger equation. For energy above the positive step potential, the positive wavenumbers areIncidence only from the left permits a right-moving incident plane wave and a left-moving reflected plane wave for $x\le 0$, but only a right-moving transmitted plane wave for $x>0$:The omitted right-incident component would describe a different scattering experiment. Wavefunction matching at a finite potential step requires continuity of both $\psi$ and $d\psi /dx$ for a finite jump and the same mass on either side:Eliminate $B=C-A$ from the second condition:The first ratio is a transmission amplitude; a transmitted probability current also depends on its different propagation velocity.

Experimental evidence. Tests of special relativity check linked predictions about light propagation, clocks and four-momentum. Several independent kinds of experiment support those predictions.

The observed partial-drag coefficient agrees with this first-order prediction rather than simple Galilean addition of $u$. It tests a different consequence of the Lorentz transformation from a particle-lifetime measurement.^[ref]

Atmospheric muons and the two frame descriptions. Atmospheric cosmic rays create unstable particles whose decay chains produce muons. A stopped-muon experiment measures the time between a stopping signal and the subsequent decay electron, obtaining the proper mean lifetime ${\tau }_0\simeq 2.2\mu \mathrm{s}$. The associated light-travel distance is only $c{\tau }_0\simeq 660\mathrm{m}$, yet substantial numbers of energetic atmospheric muons reach detectors after travelling several kilometres.

For an ideal selected cohort produced height $h$ above a detector and moving downward at constant speed $v$, the flight time in the terrestrial frame is $h/v$. Time dilation predicts a laboratory mean lifetime $\gamma {\tau }_0$, hence survival fractionWithout time dilation, the predicted exponent would be $-h/(v{\tau }_0)$ and the surviving population much smaller. Altitude comparisons constitute a cosmic-ray muon time-dilation experiment: select comparable momenta or energies and account for detector acceptance, atmospheric energy loss and backgrounds before interpreting the count ratio.

For illustration, not as a claimed measurement, take $h=10\mathrm{km}$ and $v=0.998c$. Then $\gamma =15.82$, the laboratory flight time is $33.4\mu \mathrm{s}$, and the predicted surviving fraction is about $0.38$. Omitting time dilation would give only about $2.5\times 10^{-7}$ on the same idealized constant-speed model.

In the muon rest frame, its decay still follows $e^{-\Delta \tau /{\tau }_0}$. The atmosphere and ground move upward. Their separation, measured simultaneously in that frame, is shortened by length contraction:The same survival fraction follows. The terrestrial description uses a longer-lived moving particle; the muon-frame description uses a shorter atmospheric flight path. They describe the same production and detection events. This consistency illustrates both time dilation and length contraction, while the survival measurement is not two independent experimental tests merely because it can be described in two frames.

Energy-momentum invariant. For a particle with rest mass $m$,Since $E=m{c}^2+K$, substitution and cancellation of $m^2{c}^4$ giveThis relativistic energy-momentum relation reduces to $K\simeq p^2/(2m)$ when $K\ll m{c}^2$.

Laboratory beam energies. If both final muons are at rest in the laboratory, their total momentum is zero and their total energy is $2m_{\mu }{c}^2$. Four-momentum conservation therefore requires equal and opposite incoming momenta. Because the electron and Positron have equal rest masses, they then have equal total energies:Subtract each particle's own rest energy:Using the usual handbook rest energies gives approximately $105.1\mathrm{MeV}$ for each beam. These are the ideal kinematic threshold values. Just above threshold, the pair has small, equal and opposite momenta in this laboratory frame.

Positron momentum. Apply the relativistic energy-momentum relation to $E_{+}=m_{\mu }{c}^2$ and use its stipulated negative-$x$ direction:The electron has the opposite momentum, $p_{e,x}=+p_0$ with $p_0=c\sqrt{m_{\mu }^2-m_e^2}$.

Boost to the electron rest frame. The electron's velocity is $v=p_{e,x}{c}^2/E_{-}=p_0/m_{\mu }>0$. Thus $S^{\prime }$ moves in the positive-$x$ direction, and its Lorentz factor satisfiesThe positron has the same laboratory total energy as the electron, so $v{E}_{+}/c^2=v{m}_{\mu }=p_0$. The supplied Lorentz transformation then givesAs a sign check, the same boost gives $p_{e,x}^{\prime }=\gamma (p_0-v{E}_{-}/c^2)=0$, as required for the electron rest frame.

The original diagram shows two equal balls, each of mass $m$, attached to the fixed upper pivot and light sliding collar by four equal rods. Each ball describes a horizontal circle of radius $r=L\sin \theta$. The pivot-to-collar distance is $2L\cos \theta$, so the upward collar displacement compresses the spring byThus Hooke's law gives a downward spring force $F_s=k\delta$ on the collar. Let $T_u,T_l$ be the tensions in the upper and lower rods attached to one ball. Vertical force balance on the light collar and then on either ball givesThe radial force on one ball supplies its centripetal acceleration:For $\theta \ne 0$, adding the vertical-balance results and eliminating the rod tensions givesFor the given centrifugal governor, $\cos 30^{\circ }=\sqrt{3}/2$ andHenceThe factor of two in the collar balance matters: both lower rods support the same light collar. The spring force is not an extra downward force of this full magnitude on each ball; its effect on each ball is transmitted through its own lower rod. The formula is the steady spring-loaded centrifugal governor balance.

The tunnel is taken thin enough that removal of its material does not significantly disturb spherical symmetry. For $a\le r\le 2a$, the enclosed mass isBy the spherical shell theorem, the inward gravitational acceleration has magnitudeThe released gravitational potential energy is the positive work done by gravity while the mass falls from radius $2a$ to $a$:By conservation of mechanical energy,Inside the cavity, the spherical shell theorem gives zero gravitational field, so the velocity stays constant. The original figure places B and C at opposite ends of a cavity diameter, a distance $2a$ apart. ThereforeThe quantity $G\rho$ has units ${\mathrm{s}}^{-2}$, so this is a time. There are no numerical values for $a$ or $\rho$ in the paper. The transit time is independent of $a$ and the falling mass $m$. The cavity's gravitational potential is constant, not necessarily zero; only its gradient vanishes. The calculation is a hollow-sphere gravitational transit.

Translation and rotation. For particles constituting a rigid body, the total linear momentum isThus the total external force governs the motion of the centre of mass. About that centre of mass, the angular momentum and applied torque arewhere primes denote relative positions and velocities. The derivative of the vector is taken relative to nonrotating axes. The inertia tensor relates angular momentum to angular velocity, ${\mathbf{L}}_{\mathrm{CM}}=\mathsf{I}\mathbf{\omega }$. For rotation about a fixed unit direction $\mathbf{n}$, the moment of inertia is $I_{\mathbf{n}}=\int r_{\perp }^{2}dm=\mathbf{n}\cdot \mathsf{I}\mathbf{n}$; it measures the resistance to changing that rotation and gives rotational kinetic energy $I_{\mathbf{n}}{\omega }^2/2$. The simpler vector relation $\mathbf{L}=I\mathbf{\omega }$ applies when the rotation axis is a principal axis, not for every orientation of an arbitrary rigid body. These translational and rotational balances must be applied together; a nonzero torque does not replace the net-force equation.

Gyroscope precession. A rapidly spinning gyroscope has a large angular momentum approximately along its spin axis. The torque component parallel to this vector changes its magnitude, while the perpendicular component turns its direction. If the vector undergoes precession with angular velocity $\mathbf{\Omega }$, then over $dt$ its change is the infinitesimal rotation $d\mathbf{L}=(\mathbf{\Omega }dt)\times \mathbf{L}$. HenceThis describes direction-changing precession at constant $|\mathbf{L}|$; any spin-up contribution must be added separately. For perpendicular torque and angular momentum, the precessional component perpendicular to $\mathbf{L}$ is ${\mathbf{\Omega }}_{\perp }=\mathbf{L}\times \mathbf{G}/L^2$. The familiar tendency of a supported gyroscope to turn sideways is this vector balance, not the absence of gravitational force.

Wheel-lift threshold without the flywheel. Let $h=0.50\mathrm{m}$ be the centre of mass height, $a=1.0\mathrm{m}$ half the wheel separation, and $R_b=10\mathrm{m}$ the bend radius. The required inward force is $F=M{v}^2/R_b$, provided by static friction. Let $N_o,N_i$ be the summed outer and inner wheel normal forces. Vertical balance gives $N_o+N_i=Mg$. Moment balance about the centre of mass givesThe inner wheels lift when $N_i=0$. Thus the vehicle rollover threshold isThe stated mass cancels. This assumes the rigid geometry and sufficient static friction specified in the paper; the threshold is set by loss of contact, not by sliding.

Engine flywheel. The moment of inertia of a uniform disc givesIts spin angular speed, spin angular momentum, and turn angular speed at $40\mathrm{km}{\mathrm{h}}^{-1}$ areThe transverse flywheel axis is perpendicular to the vertical turn axis, soTo specify its direction, use right-handed axes $+x$ forward, $+y$ to the car's left, and $+z$ upward, and consider a left turn. Then $\mathbf{\Omega }=+\Omega \hat{\mathbf{z}}$. Choose the flywheel angular momentum to the car's right, ${\mathbf{L}}_f=-L_f\hat{\mathbf{y}}$. The car must applyto obtain the required precession. This gyroscopic roll torque in a turning vehicle opposes the outward roll tendency in a left turn. By the right-hand rule, the flywheel must turn counterclockwise viewed from the car's right side toward its centre, opposite to the rotation of wheels rolling forward. Reversing flywheel spin would aggravate the outward roll. Reversing the turn reverses both roll and gyroscopic torques, so the same flywheel spin remains helpful.

For comparison, the outward inertial roll moment at the stated speed is $Mh{v}^2/R_b=6173\mathrm{N}\mathrm{m}$, and the limiting gravitational restoring moment is $Mga=9810\mathrm{N}\mathrm{m}$. The gyroscopic reaction is about $30\%$ of the former and $19\%$ of the latter, so it is appreciable for this large flywheel. If this ideal model is extended to include the favorable gyroscopic reaction in the wheel-lift condition, it gives $Mh{v}^2/R_b-L_{f}v/R_b=Mga$, or about $56.8\mathrm{km}{\mathrm{h}}^{-1}$ rather than $50.4\mathrm{km}{\mathrm{h}}^{-1}$. This quantifies a useful but finite increase in the rollover threshold.

Choose three mutually perpendicular axes through the same point, and label coordinates $x,y,z$. The moments of inertia of any rigid body areSubtracting givesFor a thin planar lamina lying in $z=0$, the right-hand side vanishes, proving the perpendicular axis theorem:The two in-plane axes and the perpendicular axis must intersect at one point; this need not be the centre of mass. For an extended three-dimensional mass distribution, the equality generally fails. The general moment-of-inertia triangle inequalities arewith equality in the first precisely when all mass lies in the plane perpendicular to its $z$ axis, and cyclically for the others. None of these identities requires principal axes.

Take $+x$ along the upper ship's initial velocity, and take $+y$ from the lower ship to the upper ship at closest approach. The upper and lower ships then have initial velocities $v_1\hat{\mathbf{x}}$ and $-v_2\hat{\mathbf{x}}$. Their equal masses put each at distance $l/2$ from the centre of mass, soBy momentum conservation, the centre of mass keeps moving with constant velocityIn that frame the ships have opposite velocities of magnitude $u=(v_1+v_2)/2$. Their initial angular momentum about the centre of mass is nonzero even though their trajectories were straight:The tether exerts central internal forces, so conservation of angular momentum givesEach ship follows uniform circular motion of radius $l/2$ around the translating centre of mass, clockwise viewed from $+z$. More explicitly, put the initial centre at the origin and defineThe kinetic-energy decomposition about the centre of mass verifies conservation of mechanical energy directly:At closest approach the relative radial velocity is already zero, so connecting the tether requires no dissipative radial impulse. Subsequently its tension is perpendicular to each ship's velocity relative to the centre of mass and does no net work on the pair at fixed $l$.

The mass and surface area of the spherical droplet are $m=(4\pi /3)\rho r^3$ and $A=4\pi r^2$. Constant surface mass accretion rate $\beta$ meansTherefore $\dot{r}=\beta /\rho =\mu$, and the initial radius givesHere $[\beta ]=\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$, $[\rho ]=\mathrm{kg}{\mathrm{m}}^{-3}$, and $[\mu ]=\mathrm{m}{\mathrm{s}}^{-1}$. This surface-fed growth of an accreting sphere law assumes constant droplet mass density.

Choose downward velocity $v$ positive, and assume the collected material is initially at rest in the cloud's inertial frame. This assumption is required for the equation in the question. During $dt$, momentum conservation for the droplet plus the material it collects givesup to second-order increments. Thus the variable-mass system satisfiesSince $m\propto r^3$, $\dot{m}/m=3\dot{r}/r=3\mu /(r_0+\mu t)$, givingThe notation $dv/dt$ is appropriate because $v$ here depends only on time. The extra term accounts for the momentum needed to accelerate newly collected mass; using $m\dot{v}=mg$ would omit it. For incoming material with downward velocity $u$, the more general equation is $m\dot{v}=mg+(u-v)\dot{m}$.

Introduce the dimensionless quantities $\tau =\mu t/r_0$ and $w=\mu v/(g{r}_0)$. ThenThe Taylor series at $t=0$ givesThus the initial slope is $g$, as for free fall. At late times,The oblique asymptote is $v=gt/4+g{r}_0/(4\mu )$, approached from below. Direct differentiation givesThe speed increases monotonically with decreasing slope, from $g$ towards $g/4$. It does not reach a terminal velocity in this drag-free model. Without mass accumulation, and still without drag, a drop released from rest has $v=gt$: its acceleration remains $g$, whereas the growing drop continually shares its momentum with newly acquired material.

The time-independent Schrödinger equation is the energy eigenvalue equation $\hat{H}\psi =E\psi$ for a particle of mass $m$. The Hamiltonian operator is the sum of the kinetic operator $-({\hslash }^2/2m)d^2/d{x}^2$ and the prescribed potential energy $V(x)$. Here $\hslash$ is the reduced Planck constant. A corresponding stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. The Born rule identifies $|\Psi {|}^2=|\psi {|}^2$ as probability density for a state with wavefunction normalization $\int |\Psi {|}^{2}dx=1$; integrating it over an interval gives a position probability. A plane wave scattering state is not a normalizable wavefunction on the whole line, but its probability currents give the reflection and transmission fractions of an incident wave packet.

There is a notation discrepancy in the sources: the original PDF prints an unbarred $h^2$ in the kinetic term, whereas the local TeX uses ${\hslash }^2$. The standard physical equation uses the reduced Planck constant $\hslash =h/(2\pi )$, where $h$ is the Planck constant; that convention is used below. Reading the printed $h$ literally as the Planck constant would scale all lengths in this model by $2\pi$ relative to the standard answers, but would leave the amplitude ratios unchanged.

For $E>V_0$, both sides allow oscillatory wavefunctions. A classical particle always crosses the potential step and slows because its kinetic energy falls from $E$ to $E-V_0$. In quantum mechanics there is generally both reflection and transmission even though the particle is above the step. Define the positive wavenumbersWith no wave incident from the right, writeA finite potential step with constant mass makes both the wavefunction and its first derivative continuous. The two boundary conditions at $x=0$ areSolving gives the reflected and transmitted wavefunctionsFor the specified electron energies, $q/k=\sqrt{10/50}=1/\sqrt{5}$, so the relative probability amplitudes areThe transmitted probability amplitude exceeding one is compatible with conservation of quantum probability, since the outgoing speed is smaller. In fact the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ givesThese are the quantum reflection probability and transmission probability, which must be distinguished from the requested relative amplitudes.

The string's no-slip condition gives $v=R\omega$. Conservation of mechanical energy from release to the stated height drop yieldsSolving for the spool's moment of inertia,The axle does no work because its point of application is fixed, and the ideal string and axle introduce no losses. The spool mass $M$ is not needed to infer $I$ from this energy measurement.

At closest approach the velocity is tangential. The conserved specific orbital energy and specific angular momentum areThe negative energy gives an elliptic orbit. Its semi-major axis isSince $r_p+r_a=2a$,This treats the star as fixed, appropriate to $m\ll M$. Equivalently, equate the energy at the two turning points using $v_a=\ell /r_a$. The planet mass cancels; a circular-orbit relation would be inappropriate here.

The inclined parcel has tension $T$ up the slope, weight $mg$ vertically downward, normal reaction $N=mg\cos \theta$ and friction $\mu mg\cos \theta$ down the slope. The hanging mass has weight downward and tension upward. The ideal string makes the two acceleration magnitudes equal.

The velocity slope is $a=2.4/1.5=1.60\mathrm{m}{\mathrm{s}}^{-2}$, up the track for the parcel and downward for the hanging mass. The free-body diagrams show gravity itself as one vector; its slope-parallel and normal components are used in the equations rather than counted as additional forces.

Let $m_p=0.250\mathrm{kg}$ and let $m_w$ be the hanging mass. Taking up-slope and downward respectively as positive,Eliminate the common tension to obtainHere $\theta =10^{\circ }$, $\mu =0.1$ and $a=1.60\mathrm{m}{\mathrm{s}}^{-2}$. The pulley changes the direction of the string force; in the ideal model it does not change its magnitude.

The required constant deceleration follows from $v^2=u^2+2as$:Both gravity and sliding friction oppose ascent, so $g(\sin \theta +0.1\cos \theta )=5.76$. WriteThe physical incline solution is $\theta =30.0^{\circ }$ more precisely $30.0389^{\circ }$. The combined mass cancels from the equation. The top of the track is consequently $H=2\sin \theta =1.00117\mathrm{m}$ above its base.

Let $x=0.400\mathrm{m}$ be the compression and $m_t=0.550\mathrm{kg}$ the combined mass. Starting from rest at the top, it falls through $H+x$ before maximum compression. HenceThe gravitational drop during compression must be included. Equating $kx$ to the weight would instead describe a static equilibrium, not the turning point after a fall.

The central rod occupies the holes in the discs, so the disc masses must be annular rather than counted as solid discs overlapping the rod. Define $C=L_1{\rho }_1=2L_2{\rho }_2$. ThenThe axial moment of inertia isThus $I=\frac{1}{2}M{r}_2^2$, exactly the axial inertia of a uniform disc of outer radius $r_2$ and mass $M$. The cancellation uses the given density–length relation.