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If no mass is accreted, $M$ is constant and $\Sigma \sim M/R^2$. The viscous-time estimate becomes $t\propto R^{2-m+2n}$, while $J\sim M(G{M}_{\star }R{)}^{1/2}$. ThusThe central torque supplies the angular momentum that allows the fixed disk mass to spread.

At exact conjunction the planet's force is radial and has no instantaneous torque, but the approach and departure do not cancel when conjunction is displaced from apoapsis. Write the conjunction longitude as ${\lambda }_c={\varpi }_d+\pi +\delta$. At conjunction,The resonant part of the given evolution law has ${\dot{a}}_d=-8e_d{a}_{d}D\sin \varphi$. Thus a conjunction after apoapsis, $\delta >0$, gives $\sin \varphi <0$ and transfers angular momentum to the dust, increasing $a_d$. A conjunction before apoapsis removes angular momentum, while one exactly at apoapsis has zero secular transfer by symmetry.

Poynting–Robertson drag removes angular momentum, so a trapped grain requires positive resonant torque from the planet. Part (c) shows that conjunction must occur after the dust has passed apoapsis. In the planet's rotating frame, the four-lobed pattern is therefore rotated so that the relevant outer lobe lags the fixed planet in the direction of orbital motion. Equivalently,with the stable phase displaced beyond $\varphi =\pi$ rather than sitting at the torque-free value $\varphi =\pi$.

For a circular binary star, the relative orbit has speed $v=\sqrt{GM/a}$. Its orbital angular momentum is thereforeLogarithmic differentiation, using $\dot{\mu }/\mu ={\dot{M}}_1/M_1+{\dot{M}}_2/M_2-\dot{M}/M$, givesFor conservative binary mass transfer, $\dot{M}=0$, ${\dot{J}}_{\mathrm{orb}}=0$, and ${\dot{M}}_2=-{\dot{M}}_1$. HenceSince a donor has ${\dot{M}}_1<0$, the orbit widens for $M_1<M_2$ and shrinks for $M_1>M_2$.

Write the Roche lobe radius as $R_L=aF(q)$, whereThenConservative transfer has $d\ln q/d\ln M_1=1+q$, so

The donor response $R_1\propto M_1^{-n}$ has ${\zeta }_{\ast }=d\ln R_1/d\ln M_1=-n$, whereasAfter $d\ln M_1<0$, overflow decreases only if ${\zeta }_{\ast }>{\zeta }_L$. The dynamical stability of binary mass transfer criterion is thusFor $q>1$, both terms make $n_{\mathrm{crit}}<0$, which is incompatible with $n>0$. Transfer from the more massive donor is therefore dynamically unstable: mass loss shrinks its Roche lobe while the donor expands. The resulting runaway commonly produces a common envelope, followed by envelope ejection into a tighter binary or by a stellar merger.

Time and azimuthal invariance give energyand angular momentum $J=(7m/5)r^2\dot{\varphi }$. They represent mechanical energy and vertical angular momentum.

The initial transverse velocity gives $h=a(\sqrt{2k}/a)=\sqrt{2k}$. The initial mechanical energy isSince $\Phi (1/u)=-mk{u}^2$ and $m{h}^2/2=mk$, the $u^2$ terms cancel in the conserved energy equation:The initial radial velocity fixes the sign, because $\dot{r}=-h{u}^{\prime }$ gives $u^{\prime }(0)=1/a$. The continuous derivative $u^{\prime }$ cannot change sign while its square remains $1/a^2$. As $u(0)=1/a$, this givesThe future orbit is a hyperbolic spiral: it starts at $(a,0)$, turns counterclockwise and winds indefinitely towards the origin. Its radial distance decreases on every turn. The positive-$r$ continuation has $\theta >-1$, but only $\theta \ge 0$ is needed after release.

To decide whether infinitely many turns require infinite time, use the conserved angular momentum:Integrating from $0$ givesConsequentlyThe particle reaches the origin in finite time, in the sense that $r\to 0$ as $t\uparrow t_{\ast }$. Meanwhile $\theta =t/(t_{\ast }-t)\to \infty$. The central force is undefined at the origin, so these formulas describe motion up to collision. This is finite-time collision in an inverse-cube central force.

Define the classical axial spin by the axial angular velocity of a symmetric top,Since $\mathbf{e}$ is a unit vector, $\mathbf{e}\cdot \dot{\mathbf{e}}=0$. The vector triple product identity applied to $\dot{\mathbf{e}}=\mathbf{\omega }\wedge \mathbf{e}$ givesThe inertia tensor multiplies the component perpendicular to $\mathbf{e}$ by $A$ and the component along $\mathbf{e}$ by $C$. Therefore the angular momentum about $Q$ isIts time derivative isbecause $\dot{\mathbf{e}}\wedge \dot{\mathbf{e}}=0$.

The gravitational torque about $Q$ is $h\mathbf{e}\wedge (-mg\mathbf{k})$, and the applied torque is $\ell \mathbf{e}\wedge (\kappa \mathbf{s})$. The support force acts at $Q$, so has zero torque about $Q$. The angular momentum equation is consequently

Write $c=\mathbf{k}\cdot \mathbf{e}$. Since $\mathbf{s}=\ell (\mathbf{k}-\mathbf{e})$, the total torque simplifies toTaking the dot product of the equation of motion with $\mathbf{e}$ gives $C\dot{n}=0$: both torque terms are perpendicular to $\mathbf{e}$, as are the other terms on the left. HenceTaking the dot product of ${\dot{\mathbf{L}}}_Q=\mathbf{\tau }$ with the fixed vector $\mathbf{k}$ givesThese are conservation of the axial spin and of the vertical component of angular momentum.

For energy, the kinetic energy and potential energy areDot the equation of motion, with $\dot{n}=0$, with $\mathbf{e}\wedge \dot{\mathbf{e}}$. The cross product identity $(\mathbf{e}\wedge \mathbf{u})\cdot (\mathbf{e}\wedge \mathbf{v})=\mathbf{u}\cdot \mathbf{v}-(\mathbf{e}\cdot \mathbf{u})(\mathbf{e}\cdot \mathbf{v})$ givesIt follows thatsoThe applied force is conservative even if $\kappa$ is negative: its potential energy is $\kappa |\mathbf{s}{|}^2/2$. This is a symmetric top with a quadratic alignment potential.

Far from the planet, the magnitude of angular momentum per unit mass is the magnitude of the cross product of position vector and velocity. The perpendicular distance from the incoming asymptote is the impact parameter $d$, soChoose $\theta =0$ at closest approach. The general solution of the differential equation in a, with $u^{\prime }(0)=0$, isThe specific orbital energy isAt infinity $\epsilon =v^2/2$, giving $e=\sqrt{1+h^2{v}^2/{\mu }^2}>1$. Thus the hyperbolic orbit hasA different zero of polar angle simply replaces $\theta$ by $\theta -{\theta }_0$.

The incoming and outgoing asymptotes have radial directions $-{\theta }_{\infty }$ and ${\theta }_{\infty }$, where $\cos {\theta }_{\infty }=-1/e$ and $\pi /2<{\theta }_{\infty }<\pi$. The incoming velocity points opposite its radial direction, so the deflection angle between incoming and outgoing velocities is $\delta =2{\theta }_{\infty }-\pi$. HencePutting $\delta =\pi /2$ gives

Use $g$ as the gravitational acceleration, so the weights are $mg$ and $\sqrt{2}mg$, consistently with the requested equations. The printed introductory phrase calling $g$ a force magnitude omits the factor $m$.

For the particle on a cone coupled to a hanging mass, write $z=-r$ and use the cylindrical coordinates unit vectors ${\mathbf{e}}_r,{\mathbf{e}}_{\theta },{\mathbf{e}}_z$. The ideal string segment from the cone particle to the hole has length $\sqrt{2}r$ and direction $(-{\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$. If $h$ is the downward length of the hanging segment and $s_0$ the fixed total length, thenThe light ideal string and smooth hole give the same tension $T$ on both sides. Applying the vertical component of Newton's second law to the hanging particle givesWrite the normal reaction on the cone particle as $N({\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$, allowing $N$ to be signed according to the side of the ideal constraining surface. The acceleration in cylindrical coordinates and Newton's second law giveThere is no azimuthal force. Multiplying the middle equation by $r/m$ proveswhere the positive angular direction is chosen along the initial velocity. This is conservation of angular momentum about the vertical axis.

Subtract the radial equation from the vertical equation to eliminate the normal reaction. Since $\ddot{r}=-\ddot{z}$, this givesSubstitute the tension and the conserved angular momentum, using $r=-z$:Multiplication by $\dot{z}$ now gives an exact time derivative:Initially $z=-a$ and $\dot{z}=0$, because the initial velocity is horizontal. ThereforeThis is also total mechanical energy divided by $m$, with an irrelevant constant removed: the two particles contribute $(1+\sqrt{2})m{\dot{z}}^2$ to the radial and vertical kinetic energy, while their combined gravitational potential energy is $-mgz$ plus a constant.

Choose positive $x$ downhill and positive $\theta$ in the corresponding rolling direction, with $x=\theta =0$ initially. For rolling without slipping, the contact point has zero velocity, so $\dot{x}-b\dot{\theta }=0$. Integrating the rolling constraint givesLet $F$ be the uphill static friction. Along the inclined plane, Newton's second law for the centre of mass givesThe weight passes through the centre of mass, and the normal reaction has its line of action through it. Thus only static friction supplies torque about $G$. The angular velocity is along the cylinder's symmetry axis, so the inertia tensor gives angular momentum $C\dot{\theta }$ in that direction. The torque equation is thereforeEliminating static friction and inserting $C$ givesThe extra term represents the rotational inertia that must be accelerated along with the translational mass.

Write $q=|\dot{\mathbf{r}}|=\sqrt{{\dot{r}}^2+r^2{\dot{\theta }}^2}$. Differentiating the angular momentum and applying Newton's second law givesThe central force contributes no torque, because it is parallel to $\mathbf{r}$. The angular momentum direction is fixed whenever it is nonzero; the motion is consequently planar. In the plane's polar coordinates, $\mathbf{L}=mh{\mathbf{e}}_z$, soFor the mechanical energy, use the chain rule, Newton's second law, and $V^{\prime }(r)=-F_c(r)$:ThusFor a trajectory on a time interval, both $h$ and $E$ are each a constant of motion precisely when $\alpha =0$ or the velocity vanishes throughout that interval. Under the stated strictly attractive central force at a regular point $r>0$, a permanently stationary trajectory is impossible by Newton's second law. For the stated attractive motion, both are conserved if and only if $\alpha =0$. An isolated instant of zero speed only makes the derivatives vanish at that instant. A radial trajectory has $h=0$ even with quadratic drag, but its mechanical energy still decreases whenever it moves.

All forces on the first particle lie in the plane spanned by ${\mathbf{e}}_r$ and ${\mathbf{e}}_z$: this holds for gravity, tension, and the normal reaction. Hence the azimuthal component of Newton's second law vanishes:Using the product rule givesStrictly, $h$ is the axial specific angular momentum; the actual axial angular momentum is $m_{1}h$, which is conserved as well.

Differentiate the stated total energy using the chain rule and then substitute the two Newton's second law equations:The gravitational potential energy terms have canceled the power of gravity. Also $\mathbf{N}$ is proportional to ${\mathbf{e}}_r+{\mathbf{e}}_z$, whose dot product with ${\dot{\mathbf{r}}}_1$ is ${\dot{r}}_1+{\dot{z}}_1=0$. Thus the stationary smooth mechanical constraint does no work. The remaining tension powers cancel by part (b), proving $\dot{E}=0$ and conservation of total energy while the massless inextensible string remains taut and contact is maintained.

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

Using the same cylindrical-shell volume,For unit density, this is the axial angular momentum. To bound it, apply the integral Cauchy-Schwarz inequality to $f(r)=\sqrt{r}v(r)$ and $g(r)=r^{3/2}$:Multiplication by $(2\pi h{)}^2$ gives $L^2\le \pi h{a}^{4}K$. Since the chamber's volume is $V=\pi a^{2}h$, this is precisely

A normal mode has all coordinates following one fixed displacement pattern with a common time factor. Its normal frequency is that factor's oscillation frequency; normal mode coordinates are amplitudes in a basis of such patterns, making the linear equations independent. A zero-frequency mode can instead have constant velocity along a symmetry direction.

For the unit circle, each angular displacement equals its corresponding arc displacement. The kinetic energy and change in potential energy areAny common spring prestress changes only the constant $V_0$, because the sum of the three arc lengths is fixed. The Euler-Lagrange equations giveThese are the normal modes of three beads connected by springs on a circular hoop, with third mass $M=\mu m$. For $\mu >0$, their three mode vectors and squared angular frequencies areThe zero mode rotates the entire configuration without stretching a spring. The antisymmetric mode moves the equal masses oppositely while leaving the third fixed. In the symmetric mode the equal masses move together against the third, with zero total angular momentum. These vectors are orthogonal in the mass inner product with matrix $m\mathrm{diag}(1,1,\mu )$.

Writing $\mathbf{\theta }=Q_0{v}_0+Q_A{v}_A+Q_S{v}_S$ gives independent equations ${\ddot{Q}}_0=0$, ${\ddot{Q}}_A+{\omega }_A^2{Q}_A=0$, ${\ddot{Q}}_S+{\omega }_S^2{Q}_S=0$. In particular $Q_0=A+Bt$, retaining possible slow common rotation. For $\mu =1$, the two positive frequencies coincide at $\sqrt{3k/m}$; their two-dimensional space is all patterns with ${\theta }_1+{\theta }_2+{\theta }_3=0$.

Label the equal masses $1,2$ and the mass $2m$ by $3$, in cyclic order. Let their equilibrium angles be $0,2\pi /3,4\pi /3$, with angular displacements $q_1,q_2,q_3$. Because the hoop has unit radius, its three arc lengths areThe springs themselves are constrained to the hoop, so these are arc lengths. If their common natural length is ${\ell }_0$, Hooke's law gives $V=\frac{k}{2}{\sum }_i({\ell }_i-{\ell }_0{)}^2$. Since the arc lengths have fixed sum $2\pi$, all terms linear in the $q_i$ cancel. Dropping the constant,Thus the unspecified common natural length does not affect the motion. The mass and restoring matrices in the generalized eigenvalue problem areExchange of the two equal masses separates symmetric and antisymmetric normal modes. Direct multiplication gives the complete setThe first is rigid rotation of all three beads: every arc length stays fixed, giving a zero mode. In the second, the equal masses move oppositely while the heavy bead stays fixed; the net force on that bead cancels. In the third, the equal masses move together and the heavy bead moves oppositely with equal angular amplitude. Both nonzero normal modes have zero total angular momentum. These are the normal modes of three beads connected by springs on a circular hoop with $M=2m$.

Use the mode shapes from part (b), and put $\omega =\sqrt{2k/m}$. The initial velocity decomposes asThere is no antisymmetric component because the initial state treats the equal masses identically. Initial displacements are zero, so the zero mode contributes a linear drift and the nonzero normal mode contributes a sine:In terms of actual angular positions,The conserved angular momentum is initially $2m\Omega$ and the total moment of inertia is $4m$, so the common angular velocity must be $\Omega /2$, agreeing with the drift. The relative displacements remain of order $\Omega /\omega$, even when the common angle becomes large. Hence small $\Omega /\omega$ preserves the cyclic ordering and small spring extensions at arbitrarily late times. In the ideal arc-spring model the displayed energies are exactly quadratic in the relative angles, so this motion remains exact while that ordering persists.

In a uniform gravitational field, the centre of gravity is the point $G$ through which the resultant gravitational weight acts. It coincides with the center of mass:Indeed the sum of the elemental gravitational torques about any origin isReplacing the distributed gravitational forces by the resultant at $G$ therefore preserves both total force and torque. A strongly nonuniform field need not permit the same identification.

For equilibrium, all external forces and torques must balance. A freely suspended rigid body supported at a pivot requires $G$ on the vertical through that pivot. With $G$ below the pivot, a small angular displacement raises its gravitational potential energy and gives a restoring torque; with $G$ above it, the equilibrium is unstable. A body supported on a horizontal base can remain at rest under its weight only if the vertical projection of $G$ lies within the region that the support forces can sustain. At an edge the overturning torque becomes decisive.

For motion, the center-of-mass equation isThe centre moves as a particle carrying the total mass under the resultant external force, while rotation about it is governed by the external torque and angular momentum. This separates translation from rotation; it does not determine a rigid body's full orientation from its centre's trajectory alone. In uniform gravity and negligible air resistance, a thrown rotating body has a parabolic centre-of-mass path even while its parts follow more complicated paths.

Rolling graphene into a single-walled carbon nanotube quantizes the circumferential wave vector, leaving a set of one-dimensional longitudinal subbands. At each subband edge the group velocity tends to zero, producing a Van Hove singularity in the density of states. For a simple gapped Dirac subband, the contribution scales as $|E|/\sqrt{E^2-{\Delta }_j^2}$ outside its edge. At low temperature the valence states below the Fermi level are occupied and the conduction states above it are empty. A semiconducting tube has a gap; an ideal metallic tube has gapless channels as well as higher-subband singularities.

For light polarized along the tube, the strongest allowed interband transitions connect corresponding subbands while approximately conserving axial wave vector and circumferential angular momentum. The large joint density of states makes distinct absorption resonances, conventionally labelled $E_{11},E_{22},\dots$. Strong electron–hole interactions produce bound excitons, so the observed optical resonances differ from the noninteracting band separations. Defects, doping, bundling and dielectric screening change peak energies and oscillator strengths. The original schematic below depicts the subband-edge structure, not measured spectra.