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On a beta plane, $f=f_0+\beta y$, the preceding vorticity equation isIn the broad ocean interior, assume a statistically steady flow, weak advection of relative vorticity, and negligible friction compared with planetary-vorticity advection. The constant-depth, constant-density assumptions of 2(a) exclude topographic and baroclinic vorticity generation contributions. More precisely, require $|{\xi }_t|,|\mathbf{u}\cdot \nabla \xi |,|r\xi |\ll |\beta v|$. Then Sverdrup balance isSmall Rossby number and slow forcing help justify these conditions in the interior; they do not justify them inside an intense western boundary current. For fluctuations, the mean eddy contribution to relative-vorticity advection must also be negligible.

The depth-integrated northward volume flux per unit zonal distance is $V=Hv$. For a basin $0<x<L$ and wind stress independent of $x$, the northward interior volume flux isA negative value denotes southward transport. The cancellation of $H$ is why this interior volume flux is insensitive to the chosen constant depth. In a closed, steady basin the net volume flux across a full latitude section is zero: a boundary return must compensate $T$. Thus the displayed formula cannot represent the total closed-basin transport including that return.

For a concrete circulation sketch, use a northern-hemisphere subtropical basin $0<y<B$ andThis represents westward trade winds at the southern edge and eastward westerlies at the northern edge. The negative wind stress curl drives southward interior flow. Define an ocean transport streamfunction by $Hu=-{\Psi }_y$, $Hv={\Psi }_x$ and impose no normal flow at the eastern wall through $\Psi (L,y)=0$. Integrating Sverdrup balance westward givesSince $V=0$ at $y=0,B$, this also gives no normal flow at the north and south walls. Interior water moves westward in the southern part, southward across the basin, and eastward in the northern part. A northward western boundary current completes the clockwise ocean gyre.

Pure Sverdrup balance cannot satisfy the western impermeability condition as well: ${\Psi }_I(0,y)=-LV(y)$ varies along that wall. Indeed imposing no normal flow on both east and west walls would require their constant streamfunction values to agree in this simply connected closed basin, whereas their difference here is the nonzero $T(y)$. The shaded return in the sketch represents a boundary region where the interior approximation breaks down. It is supplied by friction in 2(c), rather than inferred from Sverdrup balance alone. A southern-hemisphere subtropical ocean gyre has the reflected, anticlockwise circulation.

For an order-of-magnitude estimate, use ${\tau }_0=0.1\mathrm{N}{\mathrm{m}}^{-2}$, ${\rho }_0=10^3\mathrm{kg}{\mathrm{m}}^{-3}$, $\beta =2\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$, $L=5\times 10^6\mathrm{m}$ and a wind stress variation scale ${\ell }_{\tau }=10^6\mathrm{m}$. ThenHere $1\mathrm{Sv}=10^6{\mathrm{m}}^3{\mathrm{s}}^{-1}$ is a Sverdrup. The sinusoidal example has maximum $|{\tau }_{s,y}|=\pi {\tau }_0/B$; taking $B=\pi \times 10^6\mathrm{m}$ makes it agree with the chosen variation scale. Other reasonable basin widths and wind profiles change the coefficient, but retain a transport of order tens of Sverdrups.

The annual, zonal-mean absorbed sunlight falls from the tropics towards the poles much more strongly than outgoing thermal radiation. Thus Earth's top-of-atmosphere radiation budget has a tropical surplus and a polar deficit, with changes of sign around $35^{\circ }$ latitude in a basic sketch. The curves below are illustrative, rather than measured data; their area-weighted global means agree.

In a statistically stationary climate, local conservation of energy requires the surplus to be exported. If $R(\varphi )$ is the net downward radiative heat flux and $\mathcal{H}(\varphi )$ the northward total power carried across the latitude circle, thenHere $a$ is Earth's radius. The terrestrial atmosphere and ocean circulation carry this energy poleward by moving warm and cold fluid, including atmospheric transport of latent heat. Without this transport the assumed steady radiation budget could not persist: the tropics would warm and polar regions cool until the budget changed. The tropical atmospheric contribution includes the Hadley cell: rising warm air near the equator, poleward flow aloft, subtropical descent, and an equatorward lower return.

For a parcel moving without an axial torque, conservation of angular momentum applies in an inertial frame. The specific axial angular momentum at latitude $\varphi$ is“Initially at rest” means $u=0$ relative to the rotating surface at the equator, so $M=\Omega a^2$. Neglecting atmospheric height compared with $a$ givesWith $\Omega =7.292\times 10^{-5}{\mathrm{s}}^{-1}$ and $a=6.371\times 10^6\mathrm{m}$, the eastward velocity at $30^{\circ }\mathrm{N}$ isThis is the ideal angular-momentum-conserving upper branch of the Hadley cell. Real jet streams exchange angular momentum through eddies and friction, so this estimate is not an observational claim about the actual wind speed.

Isaac Newton made the greater contribution if the criterion is the scope and explanatory integration of mathematical natural philosophy; Galileo Galilei has the stronger claim if the criterion is the formative transformation of how terrestrial motion could be investigated. The first criterion gives a defensible overall verdict for Isaac Newton, but a serious comparison must show what each changed rather than award credit according to who worked later.

Galileo Galilei's contribution was not simply to attach numbers to familiar events. In the Third Day of Two New Sciences by Galileo Galilei, he defines uniformly accelerated motion by equal additions of speed in equal intervals of time, then derives consequences. Starting from rest, the modern notation is $v=at$; the mean speed over the interval is $at/2$, soHence distances in successive equal intervals have the ratios $1:3:5:7:\dots$. The novelty lies in making a precisely specified kind of motion the subject of mathematical demonstration, with observed falls assessed against its consequences. This is not the same as deriving the underlying physical cause of gravity.^[ref]

The reported apparatus shows how such an idealization could be investigated. A smooth bronze ball descended in a straight grooved wooden channel, raised at one end. Water discharged through a small opening into a vessel during the descent; its collected mass served as a measure of elapsed time. Repeating runs over different portions of the channel compared the predicted squared dependence of time on distance. The inclined plane slowed the event, while repetition and weighing made a temporal comparison possible without a modern stopwatch. The published report is evidence of Galileo Galilei's experimental presentation, not a surviving raw-data archive from which every reported precision can now be reconstructed.^[ref]

A necessary physical qualification is often missed. A ball in rolling without slipping is not a point particle sliding without friction. For a solid sphere on an ideal inclined plane, modern rigid body dynamics gives $a=5g\sin \alpha /7$, rather than $g\sin \alpha$. Both are constant accelerations, so both give $s\propto t^2$. The trial can therefore test the temporal form without directly establishing the free-fall coefficient. This distinction makes the experiment more intelligible: the evidence supports a specified relation under approximations, rather than every detail of a modern idealization.

The Fourth Day extends this approach to projectile motion. Combine uniform horizontal velocity $u$ with constant downward acceleration $g$, taking downward displacement positive. Then $x=ut$ and $y=g{t}^2/2$. Eliminating time givesa parabola. Actual projectiles experience air resistance, but the ideal curve organizes the effects that need separate treatment. Galileo Galilei thus gave a powerful example of how mathematical abstraction and physical observation could cooperate.^[ref]

Isaac Newton went further by connecting kinds of motion through a general account of force. Newton's second law links the change of momentum to the impressed force; for constant mass, its modern differential expression is $\mathbf{F}=m\mathbf{a}$. Newton's third law makes interaction reciprocal, enabling a consistent analysis of interacting bodies. The Philosophiae Naturalis Principia Mathematica by Isaac Newton develops consequences of central forces and then asks which laws fit astronomical phenomena. Mathematical deduction now connects measurable orbits to an underlying force law, rather than just describing each trajectory separately.^[ref]

A simple modern reconstruction shows the increase in explanatory reach. For a circular orbit of radius $r$ and orbital period $T$ about a much more massive central body, centripetal acceleration is $4{\pi }^{2}r/T^2$. Kepler's third law, $T^2\propto r^3$, therefore implies an acceleration proportional to $r^{-2}$. This circular argument is illustrative; Isaac Newton's treatment of elliptical orbits is more general. With Newton's law of universal gravitation,the two-body orbital period is tied to the semimajor axis $a$ and both masses. The same theory connects the Moon's orbital acceleration with terrestrial gravity, and supports further investigations of tides, comets and departures from a simple two-body orbit.^[ref]

This achievement depended on more than two individuals. Johannes Kepler's relations supplied crucial astronomical constraints, and both authors inherited geometry, observational records and technical practices. Nor should the Philosophiae Naturalis Principia Mathematica by Isaac Newton be described simply as the first application of symbolic calculus: its published arguments are largely geometrical. What distinguishes Isaac Newton is the demonstrative network uniting terrestrial and celestial problems, together with explicit idealizations whose failures generated further calculable questions.

On breadth, quantitative unification and the inference from motion to force, Isaac Newton's contribution was greater. Galileo Galilei's work supplied an exemplary way to mathematize terrestrial motion and test its idealized relations. Giving the former the overall verdict should therefore acknowledge the latter's distinct foundational achievement, not portray the Philosophiae Naturalis Principia Mathematica by Isaac Newton as a creation without predecessors.

Write dimensional variables as their scales times dimensionless variables, with positive scales $S_0,Q_0,m_0,N_0,t_0$. For compactness defineA consistent explicit choice isThese identities give $t_0{m}_0/({\rho }_i{S}_0)=1$, $K{t}_0{N}_0^3=1$, $f{Q}_0^2/(\sin \alpha S_0^{8/3})=1$ and ${\rho }_{w}g{t}_0{Q}_0/(S_L{N}_0)=1$. Substitution therefore leaves the unit coefficients in all four balances, and the sole supply parameter is $\mu =M_L{S}_L/Q_0$. Expanding its scales yieldsAfter dropping scale hats, eliminate the instantaneous quantities to obtain the useful two-dimensional systemThese scales assume a positive slope, positive friction coefficient and nonzero lake area; they nondimensionalize exactly the hydraulic law printed in the source.

To obtain a power-law river kinematic wave, let $P_b$ be the wetted perimeter and $I$ the small bed slope. A uniform-flow force balance per unit river length is $\rho gAI={\tau }_b{P}_b$. With a Chézy friction law ${\tau }_b=\rho f{u}^2$, it gives $u=(gI/f{)}^{1/2}(A/P_b{)}^{1/2}$. Thus the hydraulic radius determines the power of area.

For a channel family with $P_b=P_0{A}^q$, where $q<1$, we have $u=C{A}^{(1-q)/2}$, with $C=(gI/(f{P}_0){)}^{1/2}$. Set $m=(1-q)/2>0$ and $c=(m+1)C$. ThenFor example a broad fixed-width rectangular channel has approximately constant wetted perimeter, giving $m=1/2$; a geometrically similar channel family has $P_b\propto A^{1/2}$, giving $m=1/4$. The exponent is therefore a geometry-dependent consequence of the friction balance, not a universal arbitrary value fixed by Chezy's law alone.

Let $B$ be the constant canal width, large compared with water depth. Then $A\simeq Bh$ and $l\simeq B$, so friction is $f{u}^2/h$. Choose a reference uniform water depth $H$, normal-flow speed $U$ satisfying $f{U}^2/H=gS$, and scalesThe continuity equation becomes ${\hat{h}}_{\hat{t}}+(\hat{h}\hat{u}{)}_{\hat{s}}=0$. Dividing momentum by $gS$ and dropping hats givesThe Froude number compares velocity to long gravity-wave speed. At the reference state, $F<1$ is subcritical flow, $F=1$ critical flow, and $F>1$ supercritical flow. In dimensional variables the wave characteristics travel at $u\pm \sqrt{gh}$: one can propagate upstream in subcritical flow, while both move downstream in supercritical flow. Away from the reference state the local Froude number is $F|u|/\sqrt{h}$.

These are the one-dimensional shallow water equations with bed slope and friction. The coordinate $s$ measures distance downstream, $t$ is time, $A(s,t)$ is the wetted cross-sectional area, $u(s,t)$ is section-averaged downstream velocity, and $h(s,t)$ is water water depth above the bed. Thus $Au$ is water volume flux and the first equation expresses mass conservation.

In the momentum equation, $u_t+u{u}_s$ is the material acceleration, $g$ is gravitational acceleration, and $S$ is the positive downward bed slope, so $gS$ drives the flow. The wetted perimeter is $l$ and $f$ is a dimensionless friction coefficient in the stated convention. The frictional deceleration is $fl{u}^2/A$ for downstream flow, while $-g{h}_s$ is the hydrostatic pressure contribution from surface-height variation relative to the sloping bed. The formula $u^2$ assumes the chosen positive-flow direction; a bidirectional friction model uses $u|u|$.

Write $e=E^{\prime }(1)>0$, $q_1=q^{\prime }(1)>0$ and introduce perturbations $(H,U,C,S)e^{\lambda t+ikx}$ to water depth, velocity, concentration and bed height. Take the leading quasi-steady approximation $ϵ=0$ for the fast water and suspension response, retaining the small friction parameter $\delta$. For $k\ne 0$, the linearized equations areHere surface height is $H+S$, and linearization of $1-u^2/h$ gives $H-2U=-3U$. Eliminating $H,C,U$ gives the dispersion relationIn particular, putting $B=F^2-1\ne 0$ and expanding at large $|k|$,Thus the temporal growth rate tends to a finite constant, while the phase velocity isFor supercritical flow $F>1$, the waves propagate upstream. They grow if $\beta e(F^2-1)>3\delta q_1$; at leading order in small $\delta$, this holds for fixed positive $\beta ,e$ away from critical flow. They are antidunes, upstream-migrating bedforms coupled to the free surface. Indeed $H=-U$ and $S\simeq -(F^2-1)U$, so surface height $H+S\simeq -F^{2}U$ is in phase with the bed perturbation in this regime.

For clarity, neglecting $\delta$ completely gives the exact leading-order expressionsThese also identify the missing qualification in the source: the unstable upstream branch requires $F>1$. For $F<1$ this leading model gives decay and downstream propagation. The near-critical case $F^2-1\simeq 0$ needs a different balance; it is not covered by the expansion. Small nonzero $ϵ$ perturbs the slow bedform branch and adds fast water modes, so the stated result concerns the leading slow morphodynamic branch. This is the high-wavenumber antidune dispersion calculation.

For a smooth surface the friction force vanishes. The ${\mathbf{e}}_{\theta }$ component of Newton's second law then gives $r^2\dot{\theta }=h$, constant. Along the constraint $z=H(r)$, the chain rule givesThe acceleration in cylindrical coordinates isTake the dot product of Newton's second law with the tangent vector ${\mathbf{e}}_r+H^{\prime }{\mathbf{e}}_z$. The normal force disappears, while the gravity term is $-mg{H}^{\prime }$. Dividing by $m$ givesUsing $\dot{\theta }=h/r^2$ yields

The free-body diagram shows a meridional section of the right circular cone. On the first particle, gravity acts vertically downwards, tension points along the straight line to the apex, and the normal reaction is perpendicular to the surface. The smooth mechanical constraint supplies no tangential friction. The second particle has only its downward gravitational weight and the upward tension.

Adding the forces on each particle and applying Newton's second law givesAt $z_1=-r_1$, a normal vector is ${\mathbf{e}}_r+{\mathbf{e}}_z$. Its sign depends on the supporting face of the cone; either sign has zero dot product with every permitted velocity. The diagram chooses the reaction towards the upper side of the surface.

Apply the general linear momentum equation $d\mathbf{p}/dt=\sum \mathbf{F}$ to the whole rigid body. Its center of mass has acceleration $\ddot{x}{\mathbf{e}}_1$, and each rail contributes its own static friction and normal force. HencesoHere $N$ is the normal force at each rail, not the combined support.

Take torques about the center of mass, with positive angular velocity along ${\mathbf{e}}_3$. The thread leaves the lower side shown in the source diagram: its contact position vector relative to the center isThus $\mathbf{q}\times \mathbf{T}=aT{\mathbf{e}}_3$. Each contact point has relative position vector $-R{\mathbf{e}}_2$, so each friction torque is $-RF{\mathbf{e}}_3$. The gravitational weight acts through the center of mass and the normal forces have zero torque about it. The angular momentum equation becomesThe condition of rolling without slipping makes the contact point instantaneously stationary:Integrating and using $x(0)=\varphi (0)=0$ gives $x=-R\varphi$. Thus rightward translation corresponds to clockwise rotation. Eliminating $F$ from the two dynamical equations with $\ddot{x}=-R\ddot{\varphi }$ yieldsand therefore

Put $D=m{R}^2+{\mathcal{I}}_{33}=m(R^2+a^2/2)$. The angular acceleration is constant. The initial rest conditions and $x(0)=\varphi (0)=0$ giveFor the upward rightward pull in the diagram, $0\le \alpha \le \pi /2$, the sign of $x(t)$ is the sign of $R\cos \alpha -a$. Consequently it rolls towards the horizontal pull when $0\le \alpha <\arccos (a/R)$. At $\alpha =\arccos (a/R)$ it remains at rest, with balanced forces and torques. For $\arccos (a/R)<\alpha \le \pi /2$, it rolls to the left. In particular a horizontal pull, $\alpha =0$, makes it roll to the right, since $R>a$. If other pulling angles are allowed, the general condition for motion in the same horizontal direction as the pull is $\cos \alpha (R\cos \alpha -a)>0$.

The required static friction at each rail follows from the translational equation:For $0\le \alpha \le \pi /2$ it is positive, agreeing with the leftward force convention. The assumed rolling without slipping is physically possible only if the rails can supply this force while maintaining contact. With coefficient of static friction ${\mu }_s$, the necessary contact inequalities areFor the usual pulling range, these becomeIf either fails, the proposed motion requires lift-off or more static friction than the rails provide. Equality $T\sin \alpha =mg$ is also incompatible here: it gives $N=0$ but a nonzero required friction. The question supplies no coefficient of static friction, so it determines the required friction exactly but gives no numerical slip threshold. If $F$ is regarded as a prescribed force rather than an available contact response, it must equal $F_{\mathrm{req}}$ for the rolling constraint to hold.

The relevant claim concerns a machine that repeats its complete operating cycle while delivering a continuing net output of work without a corresponding supply of energy. A machine cannot supply net work in a cycle without taking in energy or depleting an energy store. This follows from conservation of energy, an experimentally supported physical law rather than a theorem of pure logic.

Let $E$ include the machine's internal energy, kinetic energy and potential energy. With $Q$ the net heat received and $W$ the net work delivered, the first law of thermodynamics givesFor a complete cycle all stores return to their initial values, so $\Delta E=0$ and $W=Q$. If there is neither heat input nor another energy input, $W=0$. A descending mass, unwinding spring, consumed fuel or discharged battery supplies energy, but it cannot be omitted from the accounting while calling the result perpetual motion.

Changing an overbalanced wheel's geometry does not evade the argument. A conservative force gives zero net work around a closed configuration cycle, while friction dissipates usable mechanical energy. Resetting a mechanism or returning a raised weight costs work that must be included.

Indefinitely continuing motion without useful output is a different possibility: an ideal frictionless oscillator or freely moving body need not stop. Connecting a load removes energy and changes that situation. A cyclic machine supplied with heat from only one reservoir and converting all that heat into work is excluded instead by the Kelvin-Planck statement of the Second law of thermodynamics. The first-law and second-law objections address different proposed energy balances.

friction force microscopy is suitable. Contact scanning twists the AFM cantilever; the lateral photodetector difference measures friction and can distinguish the supported graphene and glass even if their heights are close. Acquire both trace and retrace and use half their lateral-signal difference to suppress slope-induced cross-talk. Keep the normal load low enough to avoid dragging or damaging the flake, control humidity and contamination, and calibrate torsional sensitivity if a quantitative coefficient of friction is required. An electrical connection is unnecessary.

Three suitable approaches give complementary contrast. tapping-mode atomic force microscopy supplies surface topography and an oscillation-phase image. It reduces lateral drag and is often effective for soft block copolymer domains. Its phase contrast depends on elastic response, adhesion, dissipation and operating setpoint together: a bright phase domain is not automatically the stiffer polymer. Check contrast as amplitude and setpoint change and use a known reference.

quantitative nanomechanical mapping fits the repulsive portions of many AFM force curves and separately maps effective elastic modulus, adhesion and deformation. It directly addresses the stiffness difference and limits maximum force. Its quantitative result depends on tip radius, force calibration, contact model, loading rate and film thickness; domains narrower than the contact footprint are mixed and the substrate can dominate a thin film. Use small indentation, substrate controls and a suitable viscoelastic interpretation.

friction force microscopy can reveal domains with different frictional and adhesive response, using torsional deflection during contact scans. It is simple and sensitive to surface chemistry, but continuous shear may deform or displace soft domains, and slope/topography can contaminate the lateral signal. Acquire trace–retrace pairs at several low normal loads and check reproducibility. Use a mechanical-modulus map to test the stiffness assignment; phase or friction contrast alone does not uniquely identify stiffness or chemical composition.

friction force microscopy records cantilever torsion caused by lateral tip–sample force. Choose the detector sign so that friction on a flat region contributes positively during a left-to-right trace. On a common laboratory-position axis, write the two signals schematically asHere $B$ is an offset and $G$ represents a slope-related lateral-force or detector cross-talk contribution. A reversed electronic convention reverses all plotted signs without changing the physical conclusions.

At the same controlled normal contact force $N$, the flat-region contribution is approximately $F_f=\mu N$. The lower perovskite structure plateaux give a larger trace–retrace separation than the upper inorganic surface, because their coefficient of kinetic friction is larger. Genuine friction reverses sign with scan direction, so a larger friction magnitude means a higher positive trace and a more negative retrace in this convention.

Sloped edges produce additional peaks or shoulders. The normal reaction to a slope has a lateral component, and the finite tip flank can contact the rim, so edge contrast is not a direct map of the local friction coefficient. With the two scans registered and load control maintained,The half-difference therefore emphasizes friction while the half-sum diagnoses common geometrical contributions. Real contact mechanics and imperfect registration can prevent exact cancellation. The bottom panel of the figure shows these two kinds of contrast separately.

For the rolling disk with a rim point mass, impose rolling without slipping and take the clockwise angular displacement $\theta$ from the configuration with the extra mass at the bottom. The disk centre has $x_C=R\theta$ and $y_C=R$. The extra mass hasThe disk's translational and rotational kinetic energies sum to $3M{R}^2{\dot{\theta }}^2/2$. Including the rim mass givesThe exact Euler-Lagrange equation isTo first order, $3M{R}^2\ddot{\theta }+MgR\theta =0$. ThereforeAt the bottom the extra mass is instantaneously at the contact point and has no first-order velocity. The increased effective inertia comes from the moving disk. Static friction enforces the rolling constraint without doing work at the stationary contact.

Rough ground, smooth cylinder contacts. Let $N$ be either upper-to-lower normal force, and $f_g$ the magnitude of the inward ground friction force on each lower cylinder. Pure rolling on the ground requires $|\alpha |=u/R$. The torque equation about a lower cylinder's centre isThe outward translation equation is $mu=N/2-f_g$, giving $N=2m(1+\beta )u$. The top cylinder's vertical equation is $mw=mg-\sqrt{3}N$. Using $u=\sqrt{3}w$ yieldsFor uniform solid cylinders, $\beta =1/2$ and $a_{\mathrm{top},\mathrm{down}}=g/10$. The left lower cylinder rotates counterclockwise while rolling left; the right rotates clockwise while rolling right.

The assumed no-slip ground contact is possible if its coefficient of static friction satisfies ${\mu }_g\ge \beta \sqrt{3}/(10+9\beta )$ at release. This follows from $f_g=\beta m\sqrt{3}w$ and ground normal force $N_g=mg+\sqrt{3}N/2=mw(10+9\beta )$.

Use spherical coordinates about the fixed pivot, with fixed rod length $l$, polar angle $\theta$ measured from the downward vertical, and azimuth $\varphi$. The bob's position iswhen the Cartesian $z$ axis points upward. Differentiation gives kinetic energy $T=m{l}^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)/2$. Taking zero gravitational potential energy at pivot height gives $V=-mgl\cos \theta$. The spherical pendulum's Lagrangian is thereforeThe rod is light, rigid and of fixed length; pivot friction and air drag are neglected. A rigid rod, unlike a string, may sustain either tensile or compressive constraint force.

Let $\theta$ be the displacement from the upward vertical and $v$ the speed of the ball's centre of mass. Its centre follows a circle of radius $R+r$. The rolling without slipping constraint is $v=r\omega$, with $\omega$ the ball's spin angular velocity. The static contact friction does no work on the ball because the instantaneous contact point is stationary. Thus conservation of energy givesProject Newton's second law toward the hemisphere's centre. The inward component of gravity is $Mg\cos \theta$ and the normal force $N$ points outward, soAt contact loss $N=0$. Eliminating $v$ gives the ideal sphere rolling off a hemisphere resultThe radius cancels, so replacing $R+r$ by $R$ under the stated small-ball approximation leaves the same answer.

Finite-friction qualification. This is an ideal rolling-constraint result. It does not describe uninterrupted pure rolling to detachment with any finite coefficient of static friction. The tangential force and torque equations giveso $f=(2/7)Mg\sin \theta$ acts uphill, whereasConsequently the required ratio isFor a concrete example, a coefficient of static friction of one allows rolling only up to about $47.54^{\circ }$, where $17\cos \theta -10=2\sin \theta$, before sliding begins. A subsequent physical contact-loss angle needs a sliding-friction model. The boxed $53.97^{\circ }$ answers the source's expressly imposed no-slip idealization.