{
  "index": "1942-B-4",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "10. A particle moves under a central force inversely proportional to the \\( \\boldsymbol{k} \\) th power of the distance. If the particle describes a circle (the central force proceeding from a point on the circumference of the circle \\( ) \\), find \\( k \\).",
  "solution": "Solution. Choose polar coordinates with pole at the center of force and initial ray a diameter of the circular orbit of the particle. The equation of the orbit is then\n\\[\nr=A \\cos \\theta\n\\]\nwhere \\( A \\) is the diameter of the circle.\nThe equations of motion are\n\\[\n\\begin{array}{c}\n\\frac{d^{2} r}{d t^{2}}-r\\left(\\frac{d \\theta}{d t}\\right)^{2}=-\\frac{1}{m} f \\\\\nr \\frac{\\partial^{2} \\theta}{\\partial t^{2}}+2 \\frac{d r}{d t} \\frac{d \\theta}{d t}=0\n\\end{array}\n\\]\nwhere \\( m \\) is the mass of the particle and \\( f \\) is the magnitude of the central force. Since the sign is taken as negative, a positive \\( f \\) means an attractive force.\n\nAfter multiplication by \\( r \\), equation (3) can be integrated to give\n\\[\nr^{2} \\frac{d \\theta}{d t}=h\n\\]\nwhich asserts that the angular momentum of the particle is constant.\nDifferentiating (1) twice, using (4), we get\n\\[\n\\begin{aligned}\n\\frac{d r}{d t} & =-A \\sin \\theta \\frac{d \\theta}{d t}=-\\frac{A h \\sin \\theta}{r^{2}} \\\\\n\\frac{d^{2} r}{d t^{2}} & =-\\frac{A h \\cos \\theta}{r^{2}} \\cdot \\frac{d \\theta}{d t}+\\frac{2 A h \\sin \\theta}{r^{3}} \\cdot \\frac{d r}{d t} \\\\\n& =-\\frac{h^{2}}{r^{3}}-\\frac{2 A^{2} h^{2} \\sin ^{2} \\theta}{r^{5}}\n\\end{aligned}\n\\]\n\nThen substituting in (2), we obtain\n\\[\n\\begin{aligned}\n-\\frac{f}{m} & =-\\frac{h^{2}}{r^{3}}-\\frac{2 A^{2} h^{2} \\sin ^{2} \\theta}{r^{5}}-r\\left(\\frac{h}{r^{2}}\\right)^{2} \\\\\n& =-\\frac{2 h^{2}}{r^{5}}\\left(r^{2}+A^{2} \\sin ^{2} \\theta\\right)=-\\frac{2 A^{2} h^{2}}{r^{5}}\n\\end{aligned}\n\\]\n\nThus \\( f=2 m A^{2} h^{2} r^{-5} \\) and \\( k=5 \\).\nRemark. See F. R. Moulton, Introduction to Celestial Mechanics, Macmillan, 1902, for a discussion of the general inverse power law of attraction.",
  "vars": [
    "r",
    "\\\\theta",
    "t",
    "f"
  ],
  "params": [
    "A",
    "h",
    "k",
    "m"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "r": "radius",
        "\\theta": "azimuth",
        "t": "duration",
        "f": "centralforce",
        "A": "diameter",
        "h": "angmoment",
        "k": "powerindex",
        "m": "massconst"
      },
      "question": "10. A particle moves under a central force inversely proportional to the \\( \\boldsymbol{powerindex} \\) th power of the distance. If the particle describes a circle (the central force proceeding from a point on the circumference of the circle \\( \\) \\), find \\( powerindex \\).",
      "solution": "Solution. Choose polar coordinates with pole at the center of force and initial ray a diameter of the circular orbit of the particle. The equation of the orbit is then\n\\[\nradius = diameter \\cos azimuth\n\\]\nwhere \\( diameter \\) is the diameter of the circle.\nThe equations of motion are\n\\[\n\\begin{array}{c}\n\\frac{d^{2} radius}{d duration^{2}}-radius\\left(\\frac{d azimuth}{d duration}\\right)^{2}=-\\frac{1}{massconst} centralforce \\\\\nradius \\frac{\\partial^{2} azimuth}{\\partial duration^{2}}+2 \\frac{d radius}{d duration} \\frac{d azimuth}{d duration}=0\n\\end{array}\n\\]\nwhere \\( massconst \\) is the mass of the particle and \\( centralforce \\) is the magnitude of the central force. Since the sign is taken as negative, a positive \\( centralforce \\) means an attractive force.\n\nAfter multiplication by \\( radius \\), equation (3) can be integrated to give\n\\[\nradius^{2} \\frac{d azimuth}{d duration}=angmoment\n\\]\nwhich asserts that the angular momentum of the particle is constant.\nDifferentiating (1) twice, using (4), we get\n\\[\n\\begin{aligned}\n\\frac{d radius}{d duration} & =-diameter \\sin azimuth \\frac{d azimuth}{d duration}=-\\frac{diameter angmoment \\sin azimuth}{radius^{2}} \\\\\n\\frac{d^{2} radius}{d duration^{2}} & =-\\frac{diameter angmoment \\cos azimuth}{radius^{2}} \\cdot \\frac{d azimuth}{d duration}+\\frac{2 diameter angmoment \\sin azimuth}{radius^{3}} \\cdot \\frac{d radius}{d duration} \\\\\n& =-\\frac{angmoment^{2}}{radius^{3}}-\\frac{2 diameter^{2} angmoment^{2} \\sin ^{2} azimuth}{radius^{5}}\n\\end{aligned}\n\\]\n\nThen substituting in (2), we obtain\n\\[\n\\begin{aligned}\n-\\frac{centralforce}{massconst} & =-\\frac{angmoment^{2}}{radius^{3}}-\\frac{2 diameter^{2} angmoment^{2} \\sin ^{2} azimuth}{radius^{5}}-radius\\left(\\frac{angmoment}{radius^{2}}\\right)^{2} \\\\\n& =-\\frac{2 angmoment^{2}}{radius^{5}}\\left(radius^{2}+diameter^{2} \\sin ^{2} azimuth\\right)=-\\frac{2 diameter^{2} angmoment^{2}}{radius^{5}}\n\\end{aligned}\n\\]\n\nThus \\( centralforce=2 massconst diameter^{2} angmoment^{2} radius^{-5} \\) and \\( powerindex=5 \\).\nRemark. See F. R. Moulton, Introduction to Celestial Mechanics, Macmillan, 1902, for a discussion of the general inverse power law of attraction."
    },
    "descriptive_long_confusing": {
      "map": {
        "r": "sandcastle",
        "\\\\theta": "moonlight",
        "t": "pineapple",
        "f": "snowflake",
        "A": "marshmallow",
        "h": "blueberry",
        "k": "lanterns",
        "m": "waterfall"
      },
      "question": "10. A particle moves under a central force inversely proportional to the \\( \\boldsymbol{lanterns} \\) th power of the distance. If the particle describes a circle (the central force proceeding from a point on the circumference of the circle \\( ) \\), find \\( lanterns \\).",
      "solution": "Solution. Choose polar coordinates with pole at the center of force and initial ray a diameter of the circular orbit of the particle. The equation of the orbit is then\n\\[\nsandcastle=marshmallow \\cos moonlight\n\\]\nwhere \\( marshmallow \\) is the diameter of the circle.\nThe equations of motion are\n\\[\n\\begin{array}{c}\n\\frac{d^{2} sandcastle}{d pineapple^{2}}-sandcastle\\left(\\frac{d moonlight}{d pineapple}\\right)^{2}=-\\frac{1}{waterfall} snowflake \\\\\nsandcastle \\frac{\\partial^{2} moonlight}{\\partial pineapple^{2}}+2 \\frac{d sandcastle}{d pineapple} \\frac{d moonlight}{d pineapple}=0\n\\end{array}\n\\]\nwhere \\( waterfall \\) is the mass of the particle and \\( snowflake \\) is the magnitude of the central force. Since the sign is taken as negative, a positive \\( snowflake \\) means an attractive force.\n\nAfter multiplication by \\( sandcastle \\), equation (3) can be integrated to give\n\\[\nsandcastle^{2} \\frac{d moonlight}{d pineapple}=blueberry\n\\]\nwhich asserts that the angular momentum of the particle is constant.\nDifferentiating (1) twice, using (4), we get\n\\[\n\\begin{aligned}\n\\frac{d sandcastle}{d pineapple} & =-marshmallow \\sin moonlight \\frac{d moonlight}{d pineapple}=-\\frac{marshmallow blueberry \\sin moonlight}{sandcastle^{2}} \\\\\n\\frac{d^{2} sandcastle}{d pineapple^{2}} & =-\\frac{marshmallow blueberry \\cos moonlight}{sandcastle^{2}} \\cdot \\frac{d moonlight}{d pineapple}+\\frac{2 marshmallow^{2} blueberry \\sin moonlight}{sandcastle^{3}} \\cdot \\frac{d sandcastle}{d pineapple} \\\\\n& =-\\frac{blueberry^{2}}{sandcastle^{3}}-\\frac{2 marshmallow^{2} blueberry^{2} \\sin ^{2} moonlight}{sandcastle^{5}}\n\\end{aligned}\n\\]\n\nThen substituting in (2), we obtain\n\\[\n\\begin{aligned}\n-\\frac{snowflake}{waterfall} & =-\\frac{blueberry^{2}}{sandcastle^{3}}-\\frac{2 marshmallow^{2} blueberry^{2} \\sin ^{2} moonlight}{sandcastle^{5}}-sandcastle\\left(\\frac{blueberry}{sandcastle^{2}}\\right)^{2} \\\\\n& =-\\frac{2 blueberry^{2}}{sandcastle^{5}}\\left(sandcastle^{2}+marshmallow^{2} \\sin ^{2} moonlight\\right)=-\\frac{2 marshmallow^{2} blueberry^{2}}{sandcastle^{5}}\n\\end{aligned}\n\\]\n\nThus \\( snowflake=2 waterfall marshmallow^{2} blueberry^{2} sandcastle^{-5} \\) and \\( lanterns=5 \\).\nRemark. See F. R. Moulton, Introduction to Celestial Mechanics, Macmillan, 1902, for a discussion of the general inverse power law of attraction."
    },
    "descriptive_long_misleading": {
      "map": {
        "r": "angularcomp",
        "\\\\theta": "straightlen",
        "t": "spacemeasure",
        "f": "placidity",
        "A": "singularity",
        "h": "idleness",
        "k": "rootvalue",
        "m": "emptiness"
      },
      "question": "10. A particle moves under a central force inversely proportional to the \\( \\boldsymbol{rootvalue} \\) th power of the distance. If the particle describes a circle (the central force proceeding from a point on the circumference of the circle \\( ) \\), find \\( rootvalue \\).",
      "solution": "Solution. Choose polar coordinates with pole at the center of force and initial ray a diameter of the circular orbit of the particle. The equation of the orbit is then\n\\[\nangularcomp=singularity \\cos straightlen\n\\]\nwhere \\( singularity \\) is the diameter of the circle.\nThe equations of motion are\n\\[\n\\begin{array}{c}\n\\frac{d^{2} angularcomp}{d spacemeasure^{2}}-angularcomp\\left(\\frac{d straightlen}{d spacemeasure}\\right)^{2}=-\\frac{1}{emptiness} placidity \\\\\nangularcomp \\frac{\\partial^{2} straightlen}{\\partial spacemeasure^{2}}+2 \\frac{d angularcomp}{d spacemeasure} \\frac{d straightlen}{d spacemeasure}=0\n\\end{array}\n\\]\nwhere \\( emptiness \\) is the mass of the particle and \\( placidity \\) is the magnitude of the central force. Since the sign is taken as negative, a positive \\( placidity \\) means an attractive force.\n\nAfter multiplication by \\( angularcomp \\), equation (3) can be integrated to give\n\\[\nangularcomp^{2} \\frac{d straightlen}{d spacemeasure}=idleness\n\\]\nwhich asserts that the angular momentum of the particle is constant.\nDifferentiating (1) twice, using (4), we get\n\\[\n\\begin{aligned}\n\\frac{d angularcomp}{d spacemeasure} & =-singularity \\sin straightlen \\frac{d straightlen}{d spacemeasure}=-\\frac{singularity idleness \\sin straightlen}{angularcomp^{2}} \\\\\n\\frac{d^{2} angularcomp}{d spacemeasure^{2}} & =-\\frac{singularity idleness \\cos straightlen}{angularcomp^{2}} \\cdot \\frac{d straightlen}{d spacemeasure}+\\frac{2 singularity idleness \\sin straightlen}{angularcomp^{3}} \\cdot \\frac{d angularcomp}{d spacemeasure} \\\\\n& =-\\frac{idleness^{2}}{angularcomp^{3}}-\\frac{2 singularity^{2} idleness^{2} \\sin ^{2} straightlen}{angularcomp^{5}}\n\\end{aligned}\n\\]\n\nThen substituting in (2), we obtain\n\\[\n\\begin{aligned}\n-\\frac{placidity}{emptiness} & =-\\frac{idleness^{2}}{angularcomp^{3}}-\\frac{2 singularity^{2} idleness^{2} \\sin ^{2} straightlen}{angularcomp^{5}}-angularcomp\\left(\\frac{idleness}{angularcomp^{2}}\\right)^{2} \\\\\n& =-\\frac{2 idleness^{2}}{angularcomp^{5}}\\left(angularcomp^{2}+singularity^{2} \\sin ^{2} straightlen\\right)=-\\frac{2 singularity^{2} idleness^{2}}{angularcomp^{5}}\n\\end{aligned}\n\\]\n\nThus \\( placidity=2 emptiness singularity^{2} idleness^{2} angularcomp^{-5} \\) and \\( rootvalue=5 \\).\nRemark. See F. R. Moulton, Introduction to Celestial Mechanics, Macmillan, 1902, for a discussion of the general inverse power law of attraction."
    },
    "garbled_string": {
      "map": {
        "r": "quxmplyz",
        "\\theta": "bnavziro",
        "t": "yzpkdvos",
        "f": "latimnuw",
        "A": "molfaxen",
        "h": "qudictar",
        "k": "zarlequi",
        "m": "tebidros"
      },
      "question": "10. A particle moves under a central force inversely proportional to the \\( \\boldsymbol{zarlequi} \\) th power of the distance. If the particle describes a circle (the central force proceeding from a point on the circumference of the circle \\( ) \\), find \\( zarlequi \\).",
      "solution": "Solution. Choose polar coordinates with pole at the center of force and initial ray a diameter of the circular orbit of the particle. The equation of the orbit is then\n\\[\nquxmplyz=molfaxen \\cos bnavziro\n\\]\nwhere \\( molfaxen \\) is the diameter of the circle.\nThe equations of motion are\n\\[\n\\begin{array}{c}\n\\frac{d^{2} quxmplyz}{d yzpkdvos^{2}}-quxmplyz\\left(\\frac{d bnavziro}{d yzpkdvos}\\right)^{2}=-\\frac{1}{tebidros} latimnuw \\\\\nquxmplyz \\frac{\\partial^{2} bnavziro}{\\partial yzpkdvos^{2}}+2 \\frac{d quxmplyz}{d yzpkdvos} \\frac{d bnavziro}{d yzpkdvos}=0\n\\end{array}\n\\]\nwhere \\( tebidros \\) is the mass of the particle and \\( latimnuw \\) is the magnitude of the central force. Since the sign is taken as negative, a positive \\( latimnuw \\) means an attractive force.\n\nAfter multiplication by \\( quxmplyz \\), equation (3) can be integrated to give\n\\[\nquxmplyz^{2} \\frac{d bnavziro}{d yzpkdvos}=qudictar\n\\]\nwhich asserts that the angular momentum of the particle is constant.\nDifferentiating (1) twice, using (4), we get\n\\[\n\\begin{aligned}\n\\frac{d quxmplyz}{d yzpkdvos} & =-molfaxen \\sin bnavziro \\frac{d bnavziro}{d yzpkdvos}=-\\frac{molfaxen qudictar \\sin bnavziro}{quxmplyz^{2}} \\\\\n\\frac{d^{2} quxmplyz}{d yzpkdvos^{2}} & =-\\frac{molfaxen qudictar \\cos bnavziro}{quxmplyz^{2}} \\cdot \\frac{d bnavziro}{d yzpkdvos}+\\frac{2 molfaxen qudictar \\sin bnavziro}{quxmplyz^{3}} \\cdot \\frac{d quxmplyz}{d yzpkdvos} \\\\\n& =-\\frac{qudictar^{2}}{quxmplyz^{3}}-\\frac{2 molfaxen^{2} qudictar^{2} \\sin ^{2} bnavziro}{quxmplyz^{5}}\n\\end{aligned}\n\\]\n\nThen substituting in (2), we obtain\n\\[\n\\begin{aligned}\n-\\frac{latimnuw}{tebidros} & =-\\frac{qudictar^{2}}{quxmplyz^{3}}-\\frac{2 molfaxen^{2} qudictar^{2} \\sin ^{2} bnavziro}{quxmplyz^{5}}-quxmplyz\\left(\\frac{qudictar}{quxmplyz^{2}}\\right)^{2} \\\\\n& =-\\frac{2 qudictar^{2}}{quxmplyz^{5}}\\left(quxmplyz^{2}+molfaxen^{2} \\sin ^{2} bnavziro\\right)=-\\frac{2 molfaxen^{2} qudictar^{2}}{quxmplyz^{5}}\n\\end{aligned}\n\\]\n\nThus \\( latimnuw=2 tebidros molfaxen^{2} qudictar^{2} quxmplyz^{-5} \\) and \\( zarlequi=5 \\).\nRemark. See F. R. Moulton, Introduction to Celestial Mechanics, Macmillan, 1902, for a discussion of the general inverse power law of attraction."
    },
    "kernel_variant": {
      "question": "A particle of mass $m$ and charge $q$ moves in a fixed plane $\\Pi$.  \nAt the point $O\\in\\Pi$ an attractive central force of magnitude  \n\n\\[\n\\lvert F_{c}(r)\\rvert \\;=\\;K\\,r^{-k},\n\\qquad K>0,\\;k\\in\\mathbb{R},\n\\tag{1}\n\\]\n\nacts on the particle, where $r$ denotes the distance to $O$.  \nOn $\\Pi$ there is a uniform magnetic field  \n\n\\[\n\\mathbf{B}=B\\,\\mathbf{k},\\qquad \\mathbf{k}\\ \\hbox{parallel to the positive $z$–axis},\n\\]\n\nso that only $F_{c}$ and the Lorentz force  \n\n\\[\n\\mathbf{F}_{L}=q\\,\\mathbf{v}\\times\\mathbf{B}\n\\]\n\nact on the particle.  \nThe observed trajectory is the circle  \n\n\\[\n\\Gamma:=\\bigl\\{P\\in\\Pi:\\;CP=\\ell\\bigr\\},\\qquad OC=\\ell ,\n\\tag{2}\n\\]\n\ni.e.\\ $\\Gamma$ has centre $C$ on the $x$–axis, radius $\\ell$, and its circumference passes through $O$.  \nPlane polar coordinates $(r,\\theta)$ are taken with pole $O$ and initial ray $OC$; in these  \n\n\\[\nr(\\theta)=2\\ell\\cos\\theta,\n\\qquad -\\frac{\\pi}{2}<\\theta<\\frac{\\pi}{2}.\n\\tag{3}\n\\]\n\nThroughout set  \n\n\\[\n\\mu:=\\frac{qB}{2m}.\n\\]\n\nEmpirical facts  \n(A) the geometrical areal velocity with respect to $O$, i.e.\\ $\\tfrac12 r^{2}\\dot\\theta$, is constant;  \n(B) the geometric locus of the particle is exactly $\\Gamma$.\n\nTasks  \n\n1.  Prove that (A) and (B) cannot hold simultaneously when $B\\neq0$; a perpendicular uniform magnetic field is incompatible with a constant areal velocity on the off–centred circle $\\Gamma$.\n\n2.  Put $B=0$.  Show that (1)–(3) admit a motion only when  \n\n    \\[\n        k=5,\n        \\tag{4}\n    \\]\n\n    and that the constant of areas $h:=r^{2}\\dot\\theta$ and the force\n    constant $K$ are related by  \n\n    \\[\n        K \\;=\\;2m\\,h^{2}\\,A^{2}\n          \\;=\\;8m\\,h^{2}\\,\\ell^{2},\n        \\qquad\\bigl(A=2\\ell\\bigr).\n        \\tag{5}\n    \\]\n\n3.  Let $\\varphi(t)$ be the polar angle of the particle with respect\n    to $C$.  Derive  \n\n    \\[\n        \\dot\\varphi(t)=\\omega_{0}\\sec^{2}\\theta(t),\n        \\qquad\n        \\omega_{0}:=\\frac{h}{2\\ell^{2}},\n        \\tag{6}\n    \\]\n\n    and show that $\\dot\\varphi$ attains its minimum\n    $\\omega_{\\min}=\\omega_{0}$ at $\\theta=0$\n    (point $P=C+\\ell$) and diverges as $\\theta\\to\\pm\\pi/2$.\n    Check that $\\omega_{\\min}$ coincides with the value obtained by equating the centripetal acceleration $\\ell\\omega^{2}$ to the projection of the central force on $CP$.\n\n4.  Impose at $t=0$ a small radial displacement\n    $r(0)=r_{\\Gamma}(0)+\\varepsilon$, $0<\\varepsilon\\ll\\ell$, leaving\n    $\\dot\\theta(0)$ unchanged.  Let\n    $r(t)=r_{\\Gamma}(t)+\\delta r(t)$ with $\\delta r(0)=\\varepsilon$,\n    $\\delta\\dot r(0)=0$.  \n\n    (a) Prove that, to first order in $\\delta r$,  \n\n    \\[\n       \\boxed{%\n       \\delta\\ddot r+\\Xi(t)\\,\\delta r=0},\n       \\qquad\n       \\Xi(t):=\\frac{h^{2}}{A^{4}}\\cos^{-6}\\theta(t)\n               \\Bigl(3\\cos^{2}\\theta(t)-10\\Bigr).\n       \\tag{7}\n    \\]\n\n    (b) Show that $\\Xi(t)<0$ for all $t$ and that  \n\n    \\[\n       -10\\,\\frac{h^{2}}{A^{4}}\n       \\;\\le\\;\n       \\Xi(t)\n       \\;\\le\\;\n       -7\\,\\frac{h^{2}}{A^{4}}.\n       \\tag{8}\n    \\]\n\n    (c) Conclude that the perturbation $\\delta r(t)$ grows exponentially; the orbit $\\Gamma$ is therefore \\emph{linearly unstable}.\n\n5.  Return to $B\\neq0$ with no restriction on the orbit.\n    Using the symmetric gauge\n    ${\\bf A}=\\tfrac12{\\bf B}\\times{\\bf r}$ show that the canonical\n    $z$–component of the angular momentum is  \n\n    \\[\n        L_{z}=m r^{2}\\dot\\theta+\\frac{qB}{2}\\,r^{2},\n        \\tag{9}\n    \\]\n\n    and explain why the conservation of $L_{z}$ does not imply a\n    constant geometric areal velocity when $B\\neq0$.",
      "solution": "Throughout we write  \n\n\\[\nA:=2\\ell,\n\\qquad\nr(\\theta)=A\\cos\\theta\n\\quad\\bigl(-\\tfrac{\\pi}{2}<\\theta<\\tfrac{\\pi}{2}\\bigr).\n\\tag{10}\n\\]\n\nThe polar basis vectors satisfy  \n\n\\[\n\\mathbf{e}_{r}\\times\\mathbf{k}=-\\mathbf{e}_{\\theta},\n\\qquad\n\\mathbf{e}_{\\theta}\\times\\mathbf{k}=+\\mathbf{e}_{r}.\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n1.  Equations of motion and incompatibility of $B\\neq0$  \n\nThe velocity is $\\mathbf{v}=\\dot r\\,\\mathbf{e}_{r}+r\\dot\\theta\\,\\mathbf{e}_{\\theta}$.\nHence the Lorentz acceleration is  \n\n\\[\n\\mathbf{a}_{L}=\\frac{q}{m}\\,\\mathbf{v}\\times\\mathbf{B}\n           =2\\mu\\bigl(r\\dot\\theta\\,\\mathbf{e}_{r}-\\dot r\\,\\mathbf{e}_{\\theta}\\bigr).\n\\]\n\nAdding the central acceleration\n$-(K/m)\\,r^{-k}\\mathbf{e}_{r}$ yields  \n\n\\[\n\\boxed{%\n\\begin{aligned}\n\\ddot r-r\\dot\\theta^{2}&=-\\frac{K}{m}\\,r^{-k}+2\\mu r\\dot\\theta,\\\\[1mm]\nr\\ddot\\theta+2\\dot r\\dot\\theta&=-2\\mu\\dot r.\n\\end{aligned}}\n\\tag{11}\n\\]\n\nFrom the $\\theta$–equation  \n\n\\[\n\\frac{{\\rm d}}{{\\rm d}t}\\bigl(r^{2}\\dot\\theta\\bigr)\n      =-2\\mu r\\dot r\n      =-\\mu\\frac{{\\rm d}}{{\\rm d}t}(r^{2}),\n\\]\n\nwhence  \n\n\\[\n\\boxed{r^{2}\\dot\\theta+\\mu r^{2}=h=\\hbox{const}},\n\\quad\n\\dot\\theta=\\frac{h}{r^{2}}-\\mu.\n\\tag{12}\n\\]\n\nThe areal velocity with respect to $O$ is  \n\n\\[\n\\frac12 r^{2}\\dot\\theta=\\frac{h}{2}-\\frac{\\mu}{2}\\,r^{2}.\n\\tag{13}\n\\]\n\nBecause $r(\\theta)$ varies along $\\Gamma$, constancy of\n$\\tfrac12 r^{2}\\dot\\theta$ forces $\\mu=0$, i.e.\\ $B=0$.  \nTasks (A) and (B) cannot both hold when $B\\neq0$, completing Task 1.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n2.  Determination of $k$ and of the relation between $h$ and $K$  \n\nSet $B=\\mu=0$.  Equation (11) reduces to  \n\n\\[\n\\ddot r-r\\dot\\theta^{2}=-\\frac{K}{m}\\,r^{-k}.\n\\tag{14}\n\\]\n\nSince $h=r^{2}\\dot\\theta$ is constant,\n\n\\[\n\\dot\\theta=\\frac{h}{r^{2}},\n\\qquad\n\\dot r=\\frac{{\\rm d}r}{{\\rm d}\\theta}\\dot\\theta\n      =-\\frac{h\\sin\\theta}{A\\cos^{2}\\theta},\n\\tag{15}\n\\]\n\nand\n\n\\[\n\\ddot r=\\frac{{\\rm d}}{{\\rm d}\\theta}\\bigl(\\dot r\\bigr)\\dot\\theta\n       =-\\frac{h^{2}}{A^{3}}\n         \\Bigl(2\\cos^{-5}\\theta-\\cos^{-3}\\theta\\Bigr).\n\\tag{16}\n\\]\n\nMoreover  \n\n\\[\nr\\dot\\theta^{2}=A\\cos\\theta\\frac{h^{2}}{A^{4}\\cos^{4}\\theta}\n               =\\frac{h^{2}}{A^{3}}\\cos^{-3}\\theta.\n\\tag{17}\n\\]\n\nSubstituting (16)–(17) and $r=A\\cos\\theta$ into (14) gives  \n\n\\[\n-\\frac{2h^{2}}{A^{3}}\\cos^{-5}\\theta\n     =-\\frac{K}{m}\\,A^{-k}\\cos^{-k}\\theta,\n\\]\n\nwhich holds for all $\\theta$ only if  \n\n\\[\nk=5,\n\\qquad\nK=2m\\,h^{2}\\,A^{2}\n   =8m\\,h^{2}\\,\\ell^{2}.\n\\]\n\nThis establishes (4)–(5) and completes Task 2.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n3.  Angular velocity with respect to $C$  \n\nCartesian coordinates give  \n\n\\[\nx=r\\cos\\theta=A\\cos^{2}\\theta,\n\\qquad\ny=r\\sin\\theta=A\\cos\\theta\\sin\\theta.\n\\]\n\nHence  \n\n\\[\n\\boldsymbol{\\rho}:=\\overrightarrow{CP}\n=(x-\\ell,\\;y)\n=(\\ell\\cos2\\theta,\\;\\ell\\sin2\\theta),\n\\quad\\lvert\\boldsymbol{\\rho}\\rvert=\\ell,\n\\]\n\nso that $\\varphi=2\\theta$.  Therefore  \n\n\\[\n\\dot\\varphi=2\\dot\\theta\n           =\\frac{2h}{r^{2}}\n           =\\frac{2h}{A^{2}\\cos^{2}\\theta}\n           =\\omega_{0}\\sec^{2}\\theta,\n\\qquad\n\\omega_{0}=\\frac{h}{2\\ell^{2}}.\n\\]\n\nThe minimum $\\dot\\varphi_{\\min}=\\omega_{0}$ is reached at $\\theta=0$,\nwhile $\\dot\\varphi\\to\\infty$ when $\\theta\\to\\pm\\pi/2$.  \nAt $\\theta=0$: $r=A$, $\\ell=A/2$, $v=h/A$ and the radial force gives  \n\n\\[\n\\ell\\omega^{2}\n=\\frac{K}{m}\\,r^{-5}\n=\\frac{2h^{2}A^{2}}{A^{5}}\n=\\frac{h^{2}}{4\\ell^{3}},\n\\]\nso $\\omega=h/(2\\ell^{2})=\\omega_{0}$, confirming the result.  \nTask 3 is finished.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n4.  Linear radial (in)stability of $\\Gamma$  \n\nWrite $r(t)=r_{\\Gamma}(t)+\\delta r(t)$ with\n$\\lvert\\delta r\\rvert\\ll\\ell$, $\\delta r(0)=\\varepsilon$,\n$\\delta\\dot r(0)=0$.  \nBecause $k=5$ the exact radial equation is  \n\n\\[\n\\ddot r-h^{2}r^{-3}=-\\frac{K}{m}\\,r^{-5},\n\\qquad\nK=2m h^{2}A^{2}.\n\\tag{18}\n\\]\n\nExpanding (18) to first order in $\\delta r$ gives the variational\nequation\n\n\\[\n\\delta\\ddot r\n+\\Bigl(3h^{2}r_{\\Gamma}^{-4}-5K\\,m^{-1}r_{\\Gamma}^{-6}\\Bigr)\\delta r\n=0.\n\\tag{19}\n\\]\n\nWith $r_{\\Gamma}=A\\cos\\theta(t)$ and (5) this becomes the Hill–type\nequation\n\n\\[\n\\boxed{%\n\\delta\\ddot r+\\Xi(t)\\,\\delta r=0},\n\\qquad\n\\Xi(t)=\\frac{h^{2}}{A^{4}}\\cos^{-6}\\theta(t)\n       \\bigl(3\\cos^{2}\\theta(t)-10\\bigr),\n\\tag{20}\n\\]\n\nestablishing (7).  \n\nBecause $0<\\cos^{2}\\theta(t)\\le1$ we have  \n\n\\[\n-10\\;\\le\\;3\\cos^{2}\\theta(t)-10\\;\\le\\;-7,\n\\]\n\nso  \n\n\\[\n-10\\,\\frac{h^{2}}{A^{4}}\n\\;\\le\\;\n\\Xi(t)\n\\;\\le\\;\n-7\\,\\frac{h^{2}}{A^{4}}\n<0,\n\\tag{21}\n\\]\n\nwhich is precisely (8).\n\nInequality (21) implies  \n$\\Xi(t)\\le-\\lambda^{2}$ with\n$\\lambda:=\\sqrt{7}\\,h/A^{2}>0$.  \nSturm comparison (or a standard Grönwall argument) now shows that any\nnon–trivial solution satisfies  \n\n\\[\n\\lvert\\delta r(t)\\rvert\\;\\ge\\;\n\\lvert\\varepsilon\\rvert\\cosh\\bigl(\\lambda t\\bigr),\n\\]\n\nso $\\lvert\\delta r(t)\\rvert$ grows at least exponentially.  The orbit\n$\\Gamma$ is therefore linearly \\emph{unstable}.  \nThis completes Task 4.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n5.  Canonical angular momentum for $B\\neq0$  \n\nWith the symmetric gauge\n${\\bf A}=\\tfrac12{\\bf B}\\times{\\bf r}$ the Lagrangian is  \n\n\\[\nL=\\tfrac12 m\\bigl(\\dot r^{2}+r^{2}\\dot\\theta^{2}\\bigr)\n  +\\frac{qB}{2}\\,r^{2}\\dot\\theta\n  -\\frac{K}{4}\\,r^{-4}.\n\\]\n\nBecause $\\theta$ is cyclic,  \n\n\\[\n\\boxed{%\nL_{z}=m r^{2}\\dot\\theta+\\frac{qB}{2}\\,r^{2}}\\;=\\text{const}.\n\\tag{22}\n\\]\n\nThe first summand is the mechanical angular momentum, the second the\nfield contribution.  Since the second term is proportional to $r^{2}(t)$, the\nconstancy of $L_{z}$ does \\emph{not} imply that\n$r^{2}\\dot\\theta$ (and hence the areal velocity) is constant whenever\n$B\\neq0$.  \nTask 5 (and the whole problem) is finished. \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.399493",
        "was_fixed": false,
        "difficulty_analysis": "•  Additional interacting forces:  the presence of the Lorentz force makes the\n   angular-momentum integral non-standard and forces the use of canonical\n   (rather than merely kinematic) conservation laws.\n\n•  Higher technical level:  determining k now requires handling three different\n   θ-dependent powers and proving their compatibility; a single cancellation\n   does not work, so a systematic power comparison is unavoidable.\n\n•  Deeper theory:  Step 4 uses linearisation of a non-autonomous Hill equation\n   and the effective potential method to discuss stability—concepts that do not\n   appear in the original exercise.\n\n•  Multi-part structure:  the candidate must (i) establish the canonical integral,\n   (ii) build and manipulate non-trivial θ-dependent expressions, (iii) perform\n   a stability analysis, and (iv) interpret electromagnetically the conserved\n   quantity—all of which go far beyond the purely geometric trigonometric\n   manipulations of the original problem.\n\n•  Independence checks:  the solution has to recover the classical k = 5 result\n   when the magnetic field is switched off, and to link all pieces consistently\n   through equation (17).  Any shortcut or pattern-matching attempt fails\n   because each step hinges on cancelling *three* θ-dependent powers instead\n   of one.\n\nHence the enhanced kernel variant is significantly more complex, demands a\nbroader mathematical toolkit (differential equations, canonical mechanics,\nlinear stability theory and electromagnetic dynamics) and requires more\nsubstantial explicit computation than both the original problem and the\nsimple kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "A particle of mass $m$ and charge $q$ moves in a fixed plane $\\Pi$.  \nAt the point $O\\in\\Pi$ an attractive central force of magnitude  \n\n\\[\n\\lvert F_{c}(r)\\rvert \\;=\\;K\\,r^{-k},\n\\qquad K>0,\\;k\\in\\mathbb{R},\n\\tag{1}\n\\]\n\nacts on the particle, where $r$ denotes the distance to $O$.  \nOn $\\Pi$ there is a uniform magnetic field  \n\n\\[\n\\mathbf{B}=B\\,\\mathbf{k},\\qquad \\mathbf{k}\\ \\hbox{parallel to the positive $z$–axis},\n\\]\n\nso that only $F_{c}$ and the Lorentz force  \n\n\\[\n\\mathbf{F}_{L}=q\\,\\mathbf{v}\\times\\mathbf{B}\n\\]\n\nact on the particle.  \nThe observed trajectory is the circle  \n\n\\[\n\\Gamma:=\\bigl\\{P\\in\\Pi:\\;CP=\\ell\\bigr\\},\\qquad OC=\\ell ,\n\\tag{2}\n\\]\n\ni.e.\\ $\\Gamma$ has centre $C$ on the $x$–axis, radius $\\ell$, and its circumference passes through $O$.  \nPlane polar coordinates $(r,\\theta)$ are taken with pole $O$ and initial ray $OC$; in these  \n\n\\[\nr(\\theta)=2\\ell\\cos\\theta,\n\\qquad -\\frac{\\pi}{2}<\\theta<\\frac{\\pi}{2}.\n\\tag{3}\n\\]\n\nThroughout set  \n\n\\[\n\\mu:=\\frac{qB}{2m}.\n\\]\n\nEmpirical facts  \n(A) the geometrical areal velocity with respect to $O$, i.e.\\ $\\tfrac12 r^{2}\\dot\\theta$, is constant;  \n(B) the geometric locus of the particle is exactly $\\Gamma$.\n\nTasks  \n\n1.  Prove that (A) and (B) cannot hold simultaneously when $B\\neq0$; a perpendicular uniform magnetic field is incompatible with a constant areal velocity on the off–centred circle $\\Gamma$.\n\n2.  Put $B=0$.  Show that (1)–(3) admit a motion only when  \n\n    \\[\n        k=5,\n        \\tag{4}\n    \\]\n\n    and that the constant of areas $h:=r^{2}\\dot\\theta$ and the force\n    constant $K$ are related by  \n\n    \\[\n        K \\;=\\;2m\\,h^{2}\\,A^{2}\n          \\;=\\;8m\\,h^{2}\\,\\ell^{2},\n        \\qquad\\bigl(A=2\\ell\\bigr).\n        \\tag{5}\n    \\]\n\n3.  Let $\\varphi(t)$ be the polar angle of the particle with respect\n    to $C$.  Derive  \n\n    \\[\n        \\dot\\varphi(t)=\\omega_{0}\\sec^{2}\\theta(t),\n        \\qquad\n        \\omega_{0}:=\\frac{h}{2\\ell^{2}},\n        \\tag{6}\n    \\]\n\n    and show that $\\dot\\varphi$ attains its minimum\n    $\\omega_{\\min}=\\omega_{0}$ at $\\theta=0$\n    (point $P=C+\\ell$) and diverges as $\\theta\\to\\pm\\pi/2$.\n    Check that $\\omega_{\\min}$ coincides with the value obtained by equating the centripetal acceleration $\\ell\\omega^{2}$ to the projection of the central force on $CP$.\n\n4.  Impose at $t=0$ a small radial displacement\n    $r(0)=r_{\\Gamma}(0)+\\varepsilon$, $0<\\varepsilon\\ll\\ell$, leaving\n    $\\dot\\theta(0)$ unchanged.  Let\n    $r(t)=r_{\\Gamma}(t)+\\delta r(t)$ with $\\delta r(0)=\\varepsilon$,\n    $\\delta\\dot r(0)=0$.  \n\n    (a) Prove that, to first order in $\\delta r$,  \n\n    \\[\n       \\boxed{%\n       \\delta\\ddot r+\\Xi(t)\\,\\delta r=0},\n       \\qquad\n       \\Xi(t):=\\frac{h^{2}}{A^{4}}\\cos^{-6}\\theta(t)\n               \\Bigl(3\\cos^{2}\\theta(t)-10\\Bigr).\n       \\tag{7}\n    \\]\n\n    (b) Show that $\\Xi(t)<0$ for all $t$ and that  \n\n    \\[\n       -10\\,\\frac{h^{2}}{A^{4}}\n       \\;\\le\\;\n       \\Xi(t)\n       \\;\\le\\;\n       -7\\,\\frac{h^{2}}{A^{4}}.\n       \\tag{8}\n    \\]\n\n    (c) Conclude that the perturbation $\\delta r(t)$ grows exponentially; the orbit $\\Gamma$ is therefore \\emph{linearly unstable}.\n\n5.  Return to $B\\neq0$ with no restriction on the orbit.\n    Using the symmetric gauge\n    ${\\bf A}=\\tfrac12{\\bf B}\\times{\\bf r}$ show that the canonical\n    $z$–component of the angular momentum is  \n\n    \\[\n        L_{z}=m r^{2}\\dot\\theta+\\frac{qB}{2}\\,r^{2},\n        \\tag{9}\n    \\]\n\n    and explain why the conservation of $L_{z}$ does not imply a\n    constant geometric areal velocity when $B\\neq0$.",
      "solution": "Throughout we write  \n\n\\[\nA:=2\\ell,\n\\qquad\nr(\\theta)=A\\cos\\theta\n\\quad\\bigl(-\\tfrac{\\pi}{2}<\\theta<\\tfrac{\\pi}{2}\\bigr).\n\\tag{10}\n\\]\n\nThe polar basis vectors satisfy  \n\n\\[\n\\mathbf{e}_{r}\\times\\mathbf{k}=-\\mathbf{e}_{\\theta},\n\\qquad\n\\mathbf{e}_{\\theta}\\times\\mathbf{k}=+\\mathbf{e}_{r}.\n\\]\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n1.  Equations of motion and incompatibility of $B\\neq0$  \n\nThe velocity is $\\mathbf{v}=\\dot r\\,\\mathbf{e}_{r}+r\\dot\\theta\\,\\mathbf{e}_{\\theta}$.\nHence the Lorentz acceleration is  \n\n\\[\n\\mathbf{a}_{L}=\\frac{q}{m}\\,\\mathbf{v}\\times\\mathbf{B}\n           =2\\mu\\bigl(r\\dot\\theta\\,\\mathbf{e}_{r}-\\dot r\\,\\mathbf{e}_{\\theta}\\bigr).\n\\]\n\nAdding the central acceleration\n$-(K/m)\\,r^{-k}\\mathbf{e}_{r}$ yields  \n\n\\[\n\\boxed{%\n\\begin{aligned}\n\\ddot r-r\\dot\\theta^{2}&=-\\frac{K}{m}\\,r^{-k}+2\\mu r\\dot\\theta,\\\\[1mm]\nr\\ddot\\theta+2\\dot r\\dot\\theta&=-2\\mu\\dot r.\n\\end{aligned}}\n\\tag{11}\n\\]\n\nFrom the $\\theta$–equation  \n\n\\[\n\\frac{{\\rm d}}{{\\rm d}t}\\bigl(r^{2}\\dot\\theta\\bigr)\n      =-2\\mu r\\dot r\n      =-\\mu\\frac{{\\rm d}}{{\\rm d}t}(r^{2}),\n\\]\n\nwhence  \n\n\\[\n\\boxed{r^{2}\\dot\\theta+\\mu r^{2}=h=\\hbox{const}},\n\\quad\n\\dot\\theta=\\frac{h}{r^{2}}-\\mu.\n\\tag{12}\n\\]\n\nThe areal velocity with respect to $O$ is  \n\n\\[\n\\frac12 r^{2}\\dot\\theta=\\frac{h}{2}-\\frac{\\mu}{2}\\,r^{2}.\n\\tag{13}\n\\]\n\nBecause $r(\\theta)$ varies along $\\Gamma$, constancy of\n$\\tfrac12 r^{2}\\dot\\theta$ forces $\\mu=0$, i.e.\\ $B=0$.  \nTasks (A) and (B) cannot both hold when $B\\neq0$, completing Task 1.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n2.  Determination of $k$ and of the relation between $h$ and $K$  \n\nSet $B=\\mu=0$.  Equation (11) reduces to  \n\n\\[\n\\ddot r-r\\dot\\theta^{2}=-\\frac{K}{m}\\,r^{-k}.\n\\tag{14}\n\\]\n\nSince $h=r^{2}\\dot\\theta$ is constant,\n\n\\[\n\\dot\\theta=\\frac{h}{r^{2}},\n\\qquad\n\\dot r=\\frac{{\\rm d}r}{{\\rm d}\\theta}\\dot\\theta\n      =-\\frac{h\\sin\\theta}{A\\cos^{2}\\theta},\n\\tag{15}\n\\]\n\nand\n\n\\[\n\\ddot r=\\frac{{\\rm d}}{{\\rm d}\\theta}\\bigl(\\dot r\\bigr)\\dot\\theta\n       =-\\frac{h^{2}}{A^{3}}\n         \\Bigl(2\\cos^{-5}\\theta-\\cos^{-3}\\theta\\Bigr).\n\\tag{16}\n\\]\n\nMoreover  \n\n\\[\nr\\dot\\theta^{2}=A\\cos\\theta\\frac{h^{2}}{A^{4}\\cos^{4}\\theta}\n               =\\frac{h^{2}}{A^{3}}\\cos^{-3}\\theta.\n\\tag{17}\n\\]\n\nSubstituting (16)–(17) and $r=A\\cos\\theta$ into (14) gives  \n\n\\[\n-\\frac{2h^{2}}{A^{3}}\\cos^{-5}\\theta\n     =-\\frac{K}{m}\\,A^{-k}\\cos^{-k}\\theta,\n\\]\n\nwhich holds for all $\\theta$ only if  \n\n\\[\nk=5,\n\\qquad\nK=2m\\,h^{2}\\,A^{2}\n   =8m\\,h^{2}\\,\\ell^{2}.\n\\]\n\nThis establishes (4)–(5) and completes Task 2.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n3.  Angular velocity with respect to $C$  \n\nCartesian coordinates give  \n\n\\[\nx=r\\cos\\theta=A\\cos^{2}\\theta,\n\\qquad\ny=r\\sin\\theta=A\\cos\\theta\\sin\\theta.\n\\]\n\nHence  \n\n\\[\n\\boldsymbol{\\rho}:=\\overrightarrow{CP}\n=(x-\\ell,\\;y)\n=(\\ell\\cos2\\theta,\\;\\ell\\sin2\\theta),\n\\quad\\lvert\\boldsymbol{\\rho}\\rvert=\\ell,\n\\]\n\nso that $\\varphi=2\\theta$.  Therefore  \n\n\\[\n\\dot\\varphi=2\\dot\\theta\n           =\\frac{2h}{r^{2}}\n           =\\frac{2h}{A^{2}\\cos^{2}\\theta}\n           =\\omega_{0}\\sec^{2}\\theta,\n\\qquad\n\\omega_{0}=\\frac{h}{2\\ell^{2}}.\n\\]\n\nThe minimum $\\dot\\varphi_{\\min}=\\omega_{0}$ is reached at $\\theta=0$,\nwhile $\\dot\\varphi\\to\\infty$ when $\\theta\\to\\pm\\pi/2$.  \nAt $\\theta=0$: $r=A$, $\\ell=A/2$, $v=h/A$ and the radial force gives  \n\n\\[\n\\ell\\omega^{2}\n=\\frac{K}{m}\\,r^{-5}\n=\\frac{2h^{2}A^{2}}{A^{5}}\n=\\frac{h^{2}}{4\\ell^{3}},\n\\]\nso $\\omega=h/(2\\ell^{2})=\\omega_{0}$, confirming the result.  \nTask 3 is finished.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n4.  Linear radial (in)stability of $\\Gamma$  \n\nWrite $r(t)=r_{\\Gamma}(t)+\\delta r(t)$ with\n$\\lvert\\delta r\\rvert\\ll\\ell$, $\\delta r(0)=\\varepsilon$,\n$\\delta\\dot r(0)=0$.  \nBecause $k=5$ the exact radial equation is  \n\n\\[\n\\ddot r-h^{2}r^{-3}=-\\frac{K}{m}\\,r^{-5},\n\\qquad\nK=2m h^{2}A^{2}.\n\\tag{18}\n\\]\n\nExpanding (18) to first order in $\\delta r$ gives the variational\nequation\n\n\\[\n\\delta\\ddot r\n+\\Bigl(3h^{2}r_{\\Gamma}^{-4}-5K\\,m^{-1}r_{\\Gamma}^{-6}\\Bigr)\\delta r\n=0.\n\\tag{19}\n\\]\n\nWith $r_{\\Gamma}=A\\cos\\theta(t)$ and (5) this becomes the Hill–type\nequation\n\n\\[\n\\boxed{%\n\\delta\\ddot r+\\Xi(t)\\,\\delta r=0},\n\\qquad\n\\Xi(t)=\\frac{h^{2}}{A^{4}}\\cos^{-6}\\theta(t)\n       \\bigl(3\\cos^{2}\\theta(t)-10\\bigr),\n\\tag{20}\n\\]\n\nestablishing (7).  \n\nBecause $0<\\cos^{2}\\theta(t)\\le1$ we have  \n\n\\[\n-10\\;\\le\\;3\\cos^{2}\\theta(t)-10\\;\\le\\;-7,\n\\]\n\nso  \n\n\\[\n-10\\,\\frac{h^{2}}{A^{4}}\n\\;\\le\\;\n\\Xi(t)\n\\;\\le\\;\n-7\\,\\frac{h^{2}}{A^{4}}\n<0,\n\\tag{21}\n\\]\n\nwhich is precisely (8).\n\nInequality (21) implies  \n$\\Xi(t)\\le-\\lambda^{2}$ with\n$\\lambda:=\\sqrt{7}\\,h/A^{2}>0$.  \nSturm comparison (or a standard Grönwall argument) now shows that any\nnon–trivial solution satisfies  \n\n\\[\n\\lvert\\delta r(t)\\rvert\\;\\ge\\;\n\\lvert\\varepsilon\\rvert\\cosh\\bigl(\\lambda t\\bigr),\n\\]\n\nso $\\lvert\\delta r(t)\\rvert$ grows at least exponentially.  The orbit\n$\\Gamma$ is therefore linearly \\emph{unstable}.  \nThis completes Task 4.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \n5.  Canonical angular momentum for $B\\neq0$  \n\nWith the symmetric gauge\n${\\bf A}=\\tfrac12{\\bf B}\\times{\\bf r}$ the Lagrangian is  \n\n\\[\nL=\\tfrac12 m\\bigl(\\dot r^{2}+r^{2}\\dot\\theta^{2}\\bigr)\n  +\\frac{qB}{2}\\,r^{2}\\dot\\theta\n  -\\frac{K}{4}\\,r^{-4}.\n\\]\n\nBecause $\\theta$ is cyclic,  \n\n\\[\n\\boxed{%\nL_{z}=m r^{2}\\dot\\theta+\\frac{qB}{2}\\,r^{2}}\\;=\\text{const}.\n\\tag{22}\n\\]\n\nThe first summand is the mechanical angular momentum, the second the\nfield contribution.  Since the second term is proportional to $r^{2}(t)$, the\nconstancy of $L_{z}$ does \\emph{not} imply that\n$r^{2}\\dot\\theta$ (and hence the areal velocity) is constant whenever\n$B\\neq0$.  \nTask 5 (and the whole problem) is finished. \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.342023",
        "was_fixed": false,
        "difficulty_analysis": "•  Additional interacting forces:  the presence of the Lorentz force makes the\n   angular-momentum integral non-standard and forces the use of canonical\n   (rather than merely kinematic) conservation laws.\n\n•  Higher technical level:  determining k now requires handling three different\n   θ-dependent powers and proving their compatibility; a single cancellation\n   does not work, so a systematic power comparison is unavoidable.\n\n•  Deeper theory:  Step 4 uses linearisation of a non-autonomous Hill equation\n   and the effective potential method to discuss stability—concepts that do not\n   appear in the original exercise.\n\n•  Multi-part structure:  the candidate must (i) establish the canonical integral,\n   (ii) build and manipulate non-trivial θ-dependent expressions, (iii) perform\n   a stability analysis, and (iv) interpret electromagnetically the conserved\n   quantity—all of which go far beyond the purely geometric trigonometric\n   manipulations of the original problem.\n\n•  Independence checks:  the solution has to recover the classical k = 5 result\n   when the magnetic field is switched off, and to link all pieces consistently\n   through equation (17).  Any shortcut or pattern-matching attempt fails\n   because each step hinges on cancelling *three* θ-dependent powers instead\n   of one.\n\nHence the enhanced kernel variant is significantly more complex, demands a\nbroader mathematical toolkit (differential equations, canonical mechanics,\nlinear stability theory and electromagnetic dynamics) and requires more\nsubstantial explicit computation than both the original problem and the\nsimple kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}