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A regularly parametrized curve $\gamma$ on an embedded surface $\Sigma$ is a geodesic when its covariant acceleration vanishes. Equivalently, in the ambient Euclidean space,for an affine geodesic parameter. A complete geodesic is one whose maximal affine parameter interval is all of $\mathbb{R}$.

Radiation conservation givesConsequentlyEvaluating at $t_0$, where $a=1$, shows thatPutThen $\dot{b}=2\sqrt{A+B{b}^2}$. Choosing the big bang as $t=0$ and integrating givesThereforewhere

At early times, $\sinh (\beta t)\sim \beta t$, so $a\propto t^{1/2}$, as in a radiation-dominated universe. At late times,the expected de Sitter expansion with $H\to \sqrt{\Lambda c^2/3}$.

Acceleration begins when the radiation deceleration and cosmological-constant acceleration balance:Thus $a=\alpha$, which in the exact solution means $\sinh (\beta t_{\Lambda })=1$. HenceThis is the radiation-to-cosmological-constant transition.

The Friedmann equation immediately givesUsing $P=w\rho$ in the acceleration equation and eliminating $\rho a^2$ with Friedmann givesDifferentiate $\Omega -1=k{c}^2{\mathcal{H}}^{-2}$ and substitute this equation:where ${\mathcal{H}}^2+k{c}^2=\Omega {\mathcal{H}}^2$ was used in the last step.

With $\gamma =(1-v^2/c^2{)}^{-1/2}$,Differentiation with respect to proper time givesOrthogonality $U\cdot A=0$ implies that in the instantaneous rest frame $A^{\prime \mu }=(0,a^{\prime })$. For collinear motion a Lorentz boost givesFor constant positive proper acceleration $a^{\prime }$,This hyperbola has future null asymptote $x=ct-c^2/a^{\prime }$. A right-moving signal emitted from $x_0$ at $t=0$ follows $x=ct+x_0$; if $x_0<-c^2/a^{\prime }$, it lies beyond that acceleration horizon and never intersects the worldline.

In local coordinates $u=(u^1,u^2)$ with metric $g_{ij}={\partial }_i\sigma \cdot {\partial }_j\sigma$, the energy isThe Euler–Lagrange equations are equivalentlywhere ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{k\ell }({\partial }_i{g}_{j\ell }+{\partial }_j{g}_{i\ell }-{\partial }_{\ell }{g}_{ij})$.

For the plane-section curve $\eta$, constant speed gives ${\eta }^{\prime \prime }\perp {\eta }^{\prime }$. Both ${\eta }^{\prime \prime }$ and the surface normal lie in $P$, while $P$ is spanned by ${\eta }^{\prime }$ and that normal along the intersection. Hence ${\eta }^{\prime \prime }$ is normal to the surface. Its tangential acceleration vanishes, which is the geodesic equation.

Linearizing the constraint gives $\delta \rho =-2{\rho }_0\delta a/a_0$. Since $\delta p=\delta \rho /3$, the acceleration equation gives $\delta \ddot{a}=2\Lambda \delta a$. Hence perturbations contain $e^{\sqrt{2\Lambda }t}$: the Einstein static universe is unstable.

Define $p(t)=\int x^{\prime }\rho (t,x^{\prime })d^3{x}^{\prime }$. Expanding the retarded potential for source size $d\ll r$ and speed $v\ll c$, with radiation zone $r\gg d,c/\omega$, and using charge conservation gives $B={\mu }_0\ddot{p}(t-r/c)\times \hat{x}/(4\pi rc)$. Then $S={\mu }_0|\ddot{p}\times \hat{x}{|}^2\hat{x}/(16{\pi }^{2}c{r}^2)$ and $dP/d\Omega ={\mu }_0(|\ddot{p}{|}^2-(\hat{x}\cdot \ddot{p}{)}^2)/(16{\pi }^{2}c)$. For circular motion, $P={\mu }_0{q}^2{\omega }^4{R}^2/(6\pi c)$ and $⟨dP/d\Omega ⟩={\mu }_0{q}^2{\omega }^4{R}^2(1+{\cos }^2\theta )/(32{\pi }^{2}c)$. The stated axial and equatorial polarizations follow directly. Relativistically, perpendicular acceleration gives $P={\mu }_0{q}^2{\gamma }^4{\omega }^4{R}^2/(6\pi c)$.

By Newton's second law, the central force obeysDifferentiating the cross product givessince the acceleration is parallel to $\mathbf{r}$. Hence $\mathbf{n}=\mathbf{n}(0)$ is constant. Since $\mathbf{r}\cdot \mathbf{n}=0$, the entire motion lies in the fixed plane through the origin with nonzero normal vector $\mathbf{n}(0)$.

Choose polar coordinates in that plane, orienting $\theta$ so that its positive normal vector points along $\mathbf{n}(0)$. The velocity isThereforeMultiplication by $m$ identifies this with conservation of angular momentum; geometrically, the swept area per unit time is $h/2$.

Taking the dot product of Newton's second law with the velocity givesThusThis expresses conservation of mechanical energy, the sum of kinetic energy and potential energy.

Finally, for $u=1/r$ and a prime denoting a derivative with respect to $\theta$, the chain rule givesSubstitution into the conserved energy yields

In cylindrical coordinates,Differentiating the position vector $\mathbf{r}=r{\mathbf{e}}_r+z{\mathbf{e}}_z$ twice gives the velocity and accelerationThis derives acceleration in cylindrical coordinates rather than assuming it.

At a fixed time, the wire has parametrization $r\mapsto r{\mathbf{e}}_r+f(r){\mathbf{e}}_z$. Its derivative is the nonzero tangent vectorA smooth wire exerts a constraint force $\mathbf{R}$ with $\mathbf{R}\cdot \mathbf{t}=0$. Because $g$ is the gravitational acceleration, the gravitational force is $-mg{\mathbf{e}}_z$; the printed $-g{\mathbf{e}}_z$ is interpreted per unit mass. Projecting Newton's second law, $m\ddot{\mathbf{r}}=\mathbf{R}-mg{\mathbf{e}}_z$, along $\mathbf{t}$ eliminates $\mathbf{R}$:The chain rule gives $\ddot{z}=f^{\prime }(r)\ddot{r}+f^{\prime \prime }(r){\dot{r}}^2$. Rearrangement yieldsThis is the governing equation for a bead on a uniformly rotating wire. Its azimuthal acceleration contributes nothing to the projection because ${\mathbf{e}}_{\theta }\cdot \mathbf{t}=0$.

In cylindrical coordinates, $\theta =\omega t$ and $z=r^2/(2a)$. ThusThe acceleration isA tangent vector along the wire is $\mathbf{t}={\mathbf{e}}_r+(r/a){\mathbf{e}}_z$. Because the wire is smooth, its constraint force is orthogonal to $\mathbf{t}$. Taking the dot product of Newton's second law with $\mathbf{t}$ eliminates that force; gravitational weight contributes $-mgr/a$. HenceMultiplying by $a^2/m$ gives

Take the pivot as origin, with $x$ horizontal and $y$ vertically upward. For the simple pendulum, putThe velocity and acceleration areThe forces are the tension $-T{\mathbf{e}}_r$ and gravity $(0,-mg)$. Taking the dot product of Newton's second law with ${\mathbf{e}}_{\theta }$ eliminates the tension and givesNear the downward stable equilibrium $\theta =0$, the small-angle approximation $\sin \theta \simeq \theta$ gives $\ddot{\theta }=-(g/\ell )\theta$ to first order. Since $x=\ell \sin \theta \simeq \ell \theta$, its linearization isThis is simple harmonic motion with angular frequency $\sqrt{g/\ell }$; the restoring sign confirms the stable equilibrium.

In terms of the positive cylindrical radius $r=-z$, the conserved mechanical energy becomesThe effective potential $U$ tends to infinity at both ends of $(0,\infty )$ for $v>0$, and $U^{\prime \prime }(r)=3a^2{v}^2/r^4>0$. Thus its two turning points enclose the allowed radial motion. At a turning point, $\dot{r}=0$, soBesides $r=a$, the positive root of a polynomial isthe other root of a polynomial is negative. The quadratic polynomial $Q(r)=2g{r}^2-v^{2}r-a{v}^2$ is negative at zero and has just one positive root of a polynomial. Since $Q(a)=2a(ga-v^2)$, we have $r_{\ast }<a$ when $v^2<ga$, and $r_{\ast }>a$ when $v^2>ga$.

Height is $z=-r$, so its maximum occurs at the smaller positive turning point. HenceIndeed, the initial vertical acceleration isBelow the critical speed the particle first rises to $z=-r_{\ast }$; above it the particle initially descends, and its initial height is already its greatest height. At $v=\sqrt{ga}$ the turning points merge and the particle follows a horizontal circular orbit at $z=-a$. For zero initial speed the radial conservation of energy gives motion to the apex level $z=0$; if the hole excludes the apex itself, zero is the supremum of the heights before arrival at the hole.

These are the heights under the maintained mechanical constraint used in part (a). A unilateral contact with only one side of a cone would additionally require checking the sign of the normal reaction; the ideal constrained equations alone do not impose that condition.

Let the uniform string have constant tension $\tau >0$ and linear mass density $\mu >0$. Assume small transverse motion with $|u_x|\ll 1$, negligible longitudinal motion, and no external transverse force. If the string makes angle $\beta$ with the horizontal, then $\tan \beta =u_x$, and the small-angle approximation gives $\sin \beta \simeq u_x$ and $ds\simeq dx$ to first order.

The net upward force on the segment from $x$ to $x+\Delta x$ is thereforeBy Newton's second law, its mass $\mu \Delta x$ times its transverse acceleration equals this force, to the order retained in the small displacement model. Divide by $\Delta x$ and let $\Delta x\to 0$ to obtainThus the wave equation on a string iswhere $c$ is the wave speed. The horizontal force balance makes the horizontal component of tension constant; to first order it equals $\tau$.

For polar coordinates, the rotating unit vectors satisfy ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$. Applying the product rule twice to $\mathbf{r}=r{\mathbf{e}}_r$ gives the velocity and accelerationNewton's second law for a central force now givesMultiplying the azimuthal force equation by $r/m$ yieldsThis is conservation of specific angular momentum. Here $F(r)$ is the signed radial component of the force; a negative value points inward.

When $h\ne 0$, $\theta$ is a valid local parameter along the trajectory. Put $u=1/r$ and let primes denote derivatives with respect to $\theta$. The chain rule yieldsSubstituting into the radial Newton's second law equation and dividing by $-m{h}^2{u}^2$ proves the Binet equation:

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

Take $a>0$ and the intended horizontal launch around a parallel of the surface of revolution: $\dot{r}(0)=\dot{z}(0)=0$, $a\dot{\theta }(0)=v$. Then $h=av$ and $E=v^2/2+gf(a)$. Since $\dot{z}=f^{\prime }(r)\dot{r}$, the conserved energy becomesHere $U$ is the radial effective potential on a surface of revolution. Rearrangement givesFor a nonconstant bounded radial motion with $v>0$, there must be another finite radial bound $b\ne a$ satisfying $D(b)=0$, equivalentlyThe interval between the radial bounds must be dynamically accessible: $D\ge 0$ inside, and a simple endpoint must have a forbidden region $D<0$ beyond it. Equivalently the motion must lie in a bounded part of $\{r>0:U(r)\le E\}$ separated from escape by a potential barrier. When $f$ stays finite near zero and $v>0$, the centrifugal barrier already excludes $r=0$; an outer barrier is the additional requirement. An arbitrary second algebraic root on a disconnected allowed branch would not establish boundedness.

The initial radial acceleration, obtained by projecting Newton's second law along ${\mathbf{e}}_r+f^{\prime }(a){\mathbf{e}}_z$, isThus $v^2/g>a{f}^{\prime }(a)$ starts the particle outward, and $v^2/g<a{f}^{\prime }(a)$ starts it inward. Equality gives the exact circular orbit $r=a$ and needs no second distinct radial bound. A double root at a different radius can instead be approached asymptotically: this separatrix is bounded, although the radial motion is not periodic. If $v=0$, the centrifugal barrier disappears and the particle may pass through the axis, still with bounded $r$.

The source leaves a direction ambiguity at points where $f^{\prime }(a)=0$: every horizontal tangent direction is then possible, so “projected horizontally” alone does not imply $h=av$. The displayed inequality uses the azimuthal launch specified above; at $f^{\prime }(a)\ne 0$ that direction is forced by horizontality and tangency.

Use the dimensionless variablesThe stated $R>a/2$ gives $0<A<1$. Since $f(r)=4Rx(1-x)$, the inequality from b factorizes asThe two additional possible turning points areThey are real exactly when $\beta +A\le 1$. This exposes the intended critical valuebut the ordering of the initial radius $A$ relative to these roots must also be checked.

First suppose $0<A<1/3$. Then $A<(1-A)/2$. If $\beta <1-A$, the allowed branch starting at $A$ is trapped either between $x_{-}$ and $A$ or between $A$ and $x_{-}$, according to its initial direction. At the coincident-root value ${\beta }^2=4A(1-2A)$ it is exactly a circular orbit. For $\beta =0$, the inner radial bound is the axis rather than a nonzero turning point.

At $\beta =1-A$, the additional roots coalesce at $x_{\ast }=(1-A)/2>A$. The particle starts outward and approaches $x_{\ast }$, orwithout reaching it in finite time. Indeed $D$ is proportional to $(x-A)(x-x_{\ast }{)}^2$ and the radial energy equation gives $dt/dx$ proportional to $1/|x-x_{\ast }|$ near $x_{\ast }$. Its integral diverges. This is a bounded separatrix approaching an unstable circular orbit. For $\beta >1-A$, there is no outer root and the particle escapes. Thus for $0<A<1/3$, boundedness holds exactly when $\beta +A\le 1$, with the equality case included. At $A=1/3$, the same inequality characterizes boundedness, but equality is the exact circular orbit at the starting radius.

The source's weaker assumption $A<1$ is insufficient to make this single threshold a general criterion. For $1/3<A<1/2$, the initial radius lies beyond the coalescing-root position. The transition is instead the unstable circular orbit atBelow this speed the initial radial acceleration is inward and the particle remains in the inner allowed interval. At equality it stays exactly circular; above it the initial acceleration is outward and the accessible branch extends to infinity, even if $\beta +A<1$. For $A>1/2$, the surface slopes downward outward and every such launch escapes. At $A=1/2$, only $\beta =0$ remains at rest on the summit circle.

Consequently the complete classification for the intended azimuthal launch, including zero launch speed and exact equilibria, isFor a concrete counterexample to the unqualified proposed criterion, take $A=3/4$ and $\beta =1/10$. Then $\beta +A=17/20<1$, but $f^{\prime }(a)<0$ and the outward motion has no accessible outer barrier. This satisfies the printed radius assumption and escapes. A sufficient additional radius restriction for the intended threshold is $a\le 2R/\sqrt{3}$, with the equality and circular orbit qualifications above.

Use the ideal transverse-motion model of the question: constant tension, small slopes, no bending stiffness, and no transverse body-force term. On a short segment away from $x=0$, the vertical components of the endpoint tensions are, to first order in $y_x$, $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$, since $\rho$ is linear mass density. Newton's second law givesDividing by $\Delta x$ and taking its limit proves the wave equation on a string on each side of the mass:The string on the right pulls the point mass with vertical component $T{y}_x(0^{+},t)$, while the string on the left contributes $-T{y}_x(0^{-},t)$. HenceThe fixed supports and attachment impose the boundary conditionsTogether with the mass equation, these are the endpoint and matching conditions. The slope can jump at $x=0$: the jump supplies the mass's vertical acceleration. Requiring the slope to be continuous would incorrectly remove its dynamics.

Rewrite the complex position as $z(t)=2+3i/(t-i)$. Differentiating gives the complex velocity and accelerationFor real $t$, $|t-i|=\sqrt{t^2+1}$. Their requested magnitudes are thereforeThere is no pole for real time. The magnitude is the Euclidean length of the two Cartesian components, not the modulus of either component separately.

The perpendicular unit vectors represented by $e^{i\theta }$ and $i{e}^{i\theta }$ point radially outwards and transversely in the increasing-angle direction. Reading off their real coefficients in part (b) gives the signed componentsThese are velocity in polar coordinates and acceleration in polar coordinates. In particular $r\dot{\theta }$ is a transverse speed component; $\dot{\theta }$ alone is an angular velocity. The moving basis, rather than any extra force assumption, accounts for the additional acceleration terms. At the origin the polar directions themselves are undefined, although the Cartesian velocity and acceleration can remain well defined.