{
  "index": "1941-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "12. A car is being driven so that its wheels, all of radius \\( a \\) feet, have an angular velocity of \\( \\omega \\) radians per second. A particle is thrown off from the tire of one of these wheels, where it is supposed that \\( a \\omega^{2}>g \\). Neglecting the resistance of the air, show that the maximum height above the roadway which the particle can reach is\n\\[\n\\frac{\\left(a \\omega+g \\omega^{-1}\\right)^{2}}{2 g}\n\\]",
  "solution": "Solution. If a particle is thrown into motion in a gravitational field starting at height \\( h \\) and with upward component of velocity \\( v \\), it will rise to the height \\( h+\\left(v^{2} / 2 g\\right) \\). [The horizontal components of the motion do not influence the maximum height].\n\nAs long as the particle remains attached to the tire, it follows the path of a cycloid, and we may take its equations of motion as\n\\[\n\\begin{array}{l}\nx=a \\omega t-a \\sin \\omega t \\\\\ny=a(1-\\cos \\omega t)\n\\end{array}\n\\]\n\nIf the particle leaves the tire when \\( \\omega t=\\theta \\), then it starts into gravitational motion with height \\( a(1-\\cos \\theta) \\) and upward velocity component \\( y^{\\prime}= \\) \\( a \\omega \\sin \\theta \\). Hence it reaches the height\n\\[\nH=a(1-\\cos \\theta)+\\frac{a^{2} \\omega^{2}}{2 g} \\sin ^{2} \\theta\n\\]\nprovided \\( 0 \\leq \\theta \\leq \\pi \\) (to ensure that the particle starts upward).\nWe are asked to maximize \\( \\boldsymbol{H} \\) by choice of \\( \\theta \\). We set\n\\[\n\\frac{d H}{d \\theta}=a \\sin \\theta+\\frac{a^{2} \\omega^{2}}{g} \\sin \\theta \\cos \\theta=0\n\\]\nand find that the critical points are \\( 0, \\pi, \\theta_{0} \\), where \\( \\theta_{0}=\\arccos \\left(-g / a \\omega^{2}\\right) \\). (Since \\( a \\omega^{2}>g \\), there is such a \\( \\theta_{0} \\).) The corresponding values of \\( H \\) are \\( 0,2 a \\), and\n\\[\nH_{0}=a\\left(1+\\frac{g}{a \\omega^{2}}\\right)+\\frac{a^{2} \\omega^{2}}{2 g}\\left(1-\\frac{g^{2}}{a^{2} \\omega^{4}}\\right)=\\frac{1}{2 g}\\left(a \\omega+g \\omega^{-1}\\right)^{2}\n\\]\n\nSince \\( \\left(a \\omega+g \\omega^{-1}\\right)^{2}>4 a g, H_{0}>2 a \\), so the maximum value of \\( H \\) is \\( H_{0} \\). We can also find the maximum value of \\( \\boldsymbol{H} \\) by writing (1) in the form\n\\[\nH=H_{0}-\\frac{a^{2} \\omega^{2}}{2 g}\\left(\\cos \\theta+\\frac{g}{a \\omega^{2}}\\right)^{2}\n\\]\n\nRemark. If \\( a \\omega^{2} \\leq g \\), the maximum value of \\( H \\) will be \\( 2 a \\), attained for \\( \\theta=\\pi \\), which means that particles flying off never go higher than the top of the wheel.",
  "vars": [
    "x",
    "y",
    "h",
    "v",
    "t",
    "H",
    "\\\\theta",
    "\\\\theta_0"
  ],
  "params": [
    "a",
    "g",
    "\\\\omega",
    "H_0"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "horizpos",
        "y": "vertpos",
        "h": "startheight",
        "v": "upvelocity",
        "t": "timevar",
        "H": "maxheight",
        "\\theta": "angletheta",
        "\\theta_0": "anglecrit",
        "a": "wheelradius",
        "g": "gravaccel",
        "\\omega": "anglspeed",
        "H_0": "peakheight"
      },
      "question": "12. A car is being driven so that its wheels, all of radius \\( wheelradius \\) feet, have an angular velocity of \\( anglspeed \\) radians per second. A particle is thrown off from the tire of one of these wheels, where it is supposed that \\( wheelradius anglspeed^{2}>gravaccel \\). Neglecting the resistance of the air, show that the maximum height above the roadway which the particle can reach is\n\\[\n\\frac{\\left(wheelradius anglspeed+gravaccel anglspeed^{-1}\\right)^{2}}{2 gravaccel}\n\\]\n",
      "solution": "Solution. If a particle is thrown into motion in a gravitational field starting at height \\( startheight \\) and with upward component of velocity \\( upvelocity \\), it will rise to the height \\( startheight+\\left(upvelocity^{2} / 2 gravaccel\\right) \\). [The horizontal components of the motion do not influence the maximum height].\n\nAs long as the particle remains attached to the tire, it follows the path of a cycloid, and we may take its equations of motion as\n\\[\n\\begin{array}{l}\nhorizpos=wheelradius anglspeed timevar-wheelradius \\sin anglspeed timevar \\\\\nvertpos=wheelradius(1-\\cos anglspeed timevar)\n\\end{array}\n\\]\n\nIf the particle leaves the tire when \\( anglspeed timevar=angletheta \\), then it starts into gravitational motion with height \\( wheelradius(1-\\cos angletheta) \\) and upward velocity component \\( vertpos^{\\prime}= \\) \\( wheelradius anglspeed \\sin angletheta \\). Hence it reaches the height\n\\[\nmaxheight=wheelradius(1-\\cos angletheta)+\\frac{wheelradius^{2} anglspeed^{2}}{2 gravaccel} \\sin ^{2} angletheta\n\\]\nprovided \\( 0 \\leq angletheta \\leq \\pi \\) (to ensure that the particle starts upward).\nWe are asked to maximize \\( \\boldsymbol{maxheight} \\) by choice of \\( angletheta \\). We set\n\\[\n\\frac{d maxheight}{d angletheta}=wheelradius \\sin angletheta+\\frac{wheelradius^{2} anglspeed^{2}}{gravaccel} \\sin angletheta \\cos angletheta=0\n\\]\nand find that the critical points are \\( 0, \\pi, anglecrit \\), where \\( anglecrit=\\arccos \\left(-gravaccel / wheelradius anglspeed^{2}\\right) \\). (Since \\( wheelradius anglspeed^{2}>gravaccel \\), there is such a \\( anglecrit \\).) The corresponding values of \\( maxheight \\) are \\( 0,2 wheelradius \\), and\n\\[\npeakheight=wheelradius\\left(1+\\frac{gravaccel}{wheelradius anglspeed^{2}}\\right)+\\frac{wheelradius^{2} anglspeed^{2}}{2 gravaccel}\\left(1-\\frac{gravaccel^{2}}{wheelradius^{2} anglspeed^{4}}\\right)=\\frac{1}{2 gravaccel}\\left(wheelradius anglspeed+gravaccel anglspeed^{-1}\\right)^{2}\n\\]\n\nSince \\( \\left(wheelradius anglspeed+gravaccel anglspeed^{-1}\\right)^{2}>4 wheelradius gravaccel, peakheight>2 wheelradius \\), so the maximum value of \\( maxheight \\) is \\( peakheight \\). We can also find the maximum value of \\( \\boldsymbol{maxheight} \\) by writing (1) in the form\n\\[\nmaxheight=peakheight-\\frac{wheelradius^{2} anglspeed^{2}}{2 gravaccel}\\left(\\cos angletheta+\\frac{gravaccel}{wheelradius anglspeed^{2}}\\right)^{2}\n\\]\n\nRemark. If \\( wheelradius anglspeed^{2} \\leq gravaccel \\), the maximum value of \\( maxheight \\) will be \\( 2 wheelradius \\), attained for \\( angletheta=\\pi \\), which means that particles flying off never go higher than the top of the wheel.\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "meadowland",
        "y": "riverstone",
        "h": "skylighter",
        "v": "moonripple",
        "t": "windwhistle",
        "H": "cloudheight",
        "\\theta": "sunsetglow",
        "\\theta_0": "dawndrift",
        "a": "harvestmoon",
        "g": "starlattice",
        "\\omega": "silvercrest",
        "H_0": "cobaltgrove"
      },
      "question": "12. A car is being driven so that its wheels, all of radius \\( harvestmoon \\) feet, have an angular velocity of \\( silvercrest \\) radians per second. A particle is thrown off from the tire of one of these wheels, where it is supposed that \\( harvestmoon silvercrest^{2}>starlattice \\). Neglecting the resistance of the air, show that the maximum height above the roadway which the particle can reach is\n\\[\n\\frac{\\left(harvestmoon silvercrest+starlattice silvercrest^{-1}\\right)^{2}}{2 starlattice}\n\\]\n",
      "solution": "Solution. If a particle is thrown into motion in a gravitational field starting at height \\( skylighter \\) and with upward component of velocity \\( moonripple \\), it will rise to the height \\( skylighter+\\left(moonripple^{2} / 2 starlattice\\right) \\). [The horizontal components of the motion do not influence the maximum height].\n\nAs long as the particle remains attached to the tire, it follows the path of a cycloid, and we may take its equations of motion as\n\\[\n\\begin{array}{l}\nmeadowland=harvestmoon \\, silvercrest \\, windwhistle-harvestmoon \\sin silvercrest \\, windwhistle \\\\\nriverstone=harvestmoon(1-\\cos silvercrest \\, windwhistle)\n\\end{array}\n\\]\n\nIf the particle leaves the tire when \\( silvercrest \\, windwhistle=sunsetglow \\), then it starts into gravitational motion with height \\( harvestmoon(1-\\cos sunsetglow) \\) and upward velocity component \\( riverstone^{\\prime}= \\) \\( harvestmoon \\, silvercrest \\sin sunsetglow \\). Hence it reaches the height\n\\[\ncloudheight=harvestmoon(1-\\cos sunsetglow)+\\frac{harvestmoon^{2} \\, silvercrest^{2}}{2 starlattice} \\sin ^{2} sunsetglow\n\\]\nprovided \\( 0 \\leq sunsetglow \\leq \\pi \\) (to ensure that the particle starts upward).\nWe are asked to maximize \\( \\boldsymbol{cloudheight} \\) by choice of \\( sunsetglow \\). We set\n\\[\n\\frac{d \\, cloudheight}{d \\, sunsetglow}=harvestmoon \\sin sunsetglow+\\frac{harvestmoon^{2} \\, silvercrest^{2}}{starlattice} \\sin sunsetglow \\cos sunsetglow=0\n\\]\nand find that the critical points are \\( 0, \\pi, dawndrift \\), where \\( dawndrift=\\arccos \\left(-starlattice / harvestmoon \\, silvercrest^{2}\\right) \\). (Since \\( harvestmoon \\, silvercrest^{2}>starlattice \\), there is such a \\( dawndrift \\).) The corresponding values of \\( cloudheight \\) are \\( 0,2 harvestmoon \\), and\n\\[\ncobaltgrove=harvestmoon\\left(1+\\frac{starlattice}{harvestmoon \\, silvercrest^{2}}\\right)+\\frac{harvestmoon^{2} \\, silvercrest^{2}}{2 starlattice}\\left(1-\\frac{starlattice^{2}}{harvestmoon^{2} \\, silvercrest^{4}}\\right)=\\frac{1}{2 starlattice}\\left(harvestmoon \\, silvercrest+starlattice \\, silvercrest^{-1}\\right)^{2}\n\\]\n\nSince \\( \\left(harvestmoon \\, silvercrest+starlattice \\, silvercrest^{-1}\\right)^{2}>4 harvestmoon starlattice, cobaltgrove>2 harvestmoon \\), so the maximum value of \\( cloudheight \\) is \\( cobaltgrove \\). We can also find the maximum value of \\( \\boldsymbol{cloudheight} \\) by writing (1) in the form\n\\[\ncloudheight=cobaltgrove-\\frac{harvestmoon^{2} \\, silvercrest^{2}}{2 starlattice}\\left(\\cos sunsetglow+\\frac{starlattice}{harvestmoon \\, silvercrest^{2}}\\right)^{2}\n\\]\n\nRemark. If \\( harvestmoon \\, silvercrest^{2} \\leq starlattice \\), the maximum value of \\( cloudheight \\) will be \\( 2 harvestmoon \\), attained for \\( sunsetglow=\\pi \\), which means that particles flying off never go higher than the top of the wheel."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "vertical",
        "y": "horizontal",
        "h": "grounded",
        "v": "slowness",
        "t": "timeless",
        "H": "deepness",
        "\\\\theta": "straight",
        "\\\\theta_0": "straightzero",
        "a": "flatness",
        "g": "weightlessness",
        "\\\\omega": "stillness",
        "H_0": "depthstart"
      },
      "question": "12. A car is being driven so that its wheels, all of radius \\( flatness \\) feet, have an angular velocity of \\( stillness \\) radians per second. A particle is thrown off from the tire of one of these wheels, where it is supposed that \\( flatness stillness^{2}>weightlessness \\). Neglecting the resistance of the air, show that the maximum height above the roadway which the particle can reach is\n\\[\n\\frac{\\left(flatness stillness+weightlessness stillness^{-1}\\right)^{2}}{2 weightlessness}\n\\]",
      "solution": "Solution. If a particle is thrown into motion in a gravitational field starting at height \\( grounded \\) and with upward component of velocity \\( slowness \\), it will rise to the height \\( grounded+\\left(slowness^{2} / 2 weightlessness\\right) \\). [The horizontal components of the motion do not influence the maximum height].\n\nAs long as the particle remains attached to the tire, it follows the path of a cycloid, and we may take its equations of motion as\n\\[\n\\begin{array}{l}\nvertical=flatness stillness timeless-flatness \\sin stillness timeless \\\\\nhorizontal=flatness(1-\\cos stillness timeless)\n\\end{array}\n\\]\n\nIf the particle leaves the tire when \\( stillness timeless=straight \\), then it starts into gravitational motion with height \\( flatness(1-\\cos straight) \\) and upward velocity component \\( horizontal^{\\prime}= flatness stillness \\sin straight \\). Hence it reaches the height\n\\[\ndeepness=flatness(1-\\cos straight)+\\frac{flatness^{2} stillness^{2}}{2 weightlessness} \\sin ^{2} straight\n\\]\nprovided \\( 0 \\leq straight \\leq \\pi \\) (to ensure that the particle starts upward).\nWe are asked to maximize \\( \\boldsymbol{deepness} \\) by choice of \\( straight \\). We set\n\\[\n\\frac{d\\,deepness}{d\\,straight}=flatness \\sin straight+\\frac{flatness^{2} stillness^{2}}{weightlessness} \\sin straight \\cos straight=0\n\\]\nand find that the critical points are \\( 0, \\pi, straight_{0} \\), where \\( straight_{0}=\\arccos \\left(-weightlessness / flatness stillness^{2}\\right) \\). (Since \\( flatness stillness^{2}>weightlessness \\), there is such a \\( straight_{0} \\).) The corresponding values of \\( deepness \\) are \\( 0,2 flatness \\), and\n\\[\ndepthstart=flatness\\left(1+\\frac{weightlessness}{flatness stillness^{2}}\\right)+\\frac{flatness^{2} stillness^{2}}{2 weightlessness}\\left(1-\\frac{weightlessness^{2}}{flatness^{2} stillness^{4}}\\right)=\\frac{1}{2 weightlessness}\\left(flatness stillness+weightlessness stillness^{-1}\\right)^{2}\n\\]\n\nSince \\( \\left(flatness stillness+weightlessness stillness^{-1}\\right)^{2}>4 flatness weightlessness, depthstart>2 flatness \\), so the maximum value of \\( deepness \\) is \\( depthstart \\). We can also find the maximum value of \\( \\boldsymbol{deepness} \\) by writing (1) in the form\n\\[\ndeepness=depthstart-\\frac{flatness^{2} stillness^{2}}{2 weightlessness}\\left(\\cos straight+\\frac{weightlessness}{flatness stillness^{2}}\\right)^{2}\n\\]\n\nRemark. If \\( flatness stillness^{2} \\leq weightlessness \\), the maximum value of \\( deepness \\) will be \\( 2 flatness \\), attained for \\( straight=\\pi \\), which means that particles flying off never go higher than the top of the wheel."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "h": "nbmcvpra",
        "v": "rktdlsqe",
        "t": "ulwpqkzo",
        "H": "sjhdkwpq",
        "\\theta": "prqbmvns",
        "\\theta_0": "flqwpdht",
        "a": "kwmfztrn",
        "g": "msznghqp",
        "\\omega": "vnwpqzlk",
        "H_0": "gjwprkzn"
      },
      "question": "12. A car is being driven so that its wheels, all of radius \\( kwmfztrn \\) feet, have an angular velocity of \\( vnwpqzlk \\) radians per second. A particle is thrown off from the tire of one of these wheels, where it is supposed that \\( kwmfztrn vnwpqzlk^{2}>msznghqp \\). Neglecting the resistance of the air, show that the maximum height above the roadway which the particle can reach is\n\\[\n\\frac{\\left(kwmfztrn vnwpqzlk+msznghqp vnwpqzlk^{-1}\\right)^{2}}{2 msznghqp}\n\\]\n",
      "solution": "Solution. If a particle is thrown into motion in a gravitational field starting at height \\( nbmcvpra \\) and with upward component of velocity \\( rktdlsqe \\), it will rise to the height \\( nbmcvpra+\\left(rktdlsqe^{2} / 2 msznghqp\\right) \\). [The horizontal components of the motion do not influence the maximum height].\n\nAs long as the particle remains attached to the tire, it follows the path of a cycloid, and we may take its equations of motion as\n\\[\n\\begin{array}{l}\nqzxwvtnp=kwmfztrn vnwpqzlk ulwpqkzo-kwmfztrn \\sin vnwpqzlk ulwpqkzo \\\\\nhjgrksla=kwmfztrn(1-\\cos vnwpqzlk ulwpqkzo)\n\\end{array}\n\\]\n\nIf the particle leaves the tire when \\( vnwpqzlk ulwpqkzo=prqbmvns \\), then it starts into gravitational motion with height \\( kwmfztrn(1-\\cos prqbmvns) \\) and upward velocity component \\( hjgrksla^{\\prime}= \\) \\( kwmfztrn vnwpqzlk \\sin prqbmvns \\). Hence it reaches the height\n\\[\nsjhdkwpq=kwmfztrn(1-\\cos prqbmvns)+\\frac{kwmfztrn^{2} vnwpqzlk^{2}}{2 msznghqp} \\sin ^{2} prqbmvns\n\\]\nprovided \\( 0 \\leq prqbmvns \\leq \\pi \\) (to ensure that the particle starts upward).\nWe are asked to maximize \\( \\boldsymbol{sjhdkwpq} \\) by choice of \\( prqbmvns \\). We set\n\\[\n\\frac{d sjhdkwpq}{d prqbmvns}=kwmfztrn \\sin prqbmvns+\\frac{kwmfztrn^{2} vnwpqzlk^{2}}{msznghqp} \\sin prqbmvns \\cos prqbmvns=0\n\\]\nand find that the critical points are \\( 0, \\pi, flqwpdht \\), where \\( flqwpdht=\\arccos \\left(-msznghqp / kwmfztrn vnwpqzlk^{2}\\right) \\). (Since \\( kwmfztrn vnwpqzlk^{2}>msznghqp \\), there is such a \\( flqwpdht \\).) The corresponding values of \\( sjhdkwpq \\) are \\( 0,2 kwmfztrn \\), and\n\\[\ngjwprkzn=kwmfztrn\\left(1+\\frac{msznghqp}{kwmfztrn vnwpqzlk^{2}}\\right)+\\frac{kwmfztrn^{2} vnwpqzlk^{2}}{2 msznghqp}\\left(1-\\frac{msznghqp^{2}}{kwmfztrn^{2} vnwpqzlk^{4}}\\right)=\\frac{1}{2 msznghqp}\\left(kwmfztrn vnwpqzlk+msznghqp vnwpqzlk^{-1}\\right)^{2}\n\\]\n\nSince \\( \\left(kwmfztrn vnwpqzlk+msznghqp vnwpqzlk^{-1}\\right)^{2}>4 kwmfztrn msznghqp, gjwprkzn>2 kwmfztrn \\), so the maximum value of \\( sjhdkwpq \\) is \\( gjwprkzn \\). We can also find the maximum value of \\( \\boldsymbol{sjhdkwpq} \\) by writing (1) in the form\n\\[\nsjhdkwpq=gjwprkzn-\\frac{kwmfztrn^{2} vnwpqzlk^{2}}{2 msznghqp}\\left(\\cos prqbmvns+\\frac{msznghqp}{kwmfztrn vnwpqzlk^{2}}\\right)^{2}\n\\]\n\nRemark. If \\( kwmfztrn vnwpqzlk^{2} \\leq msznghqp \\), the maximum value of \\( sjhdkwpq \\) will be \\( 2 kwmfztrn \\), attained for \\( prqbmvns=\\pi \\), which means that particles flying off never go higher than the top of the wheel.\n"
    },
    "kernel_variant": {
      "question": "Consider a perfectly spherical rigid Earth of radius $R_{e}$ that rotates uniformly with angular velocity  \n\n\\[\n\\boldsymbol{\\omega}_{e}\n      =\\omega_{e}\\bigl(\\cos\\varphi\\,\\hat{\\boldsymbol{\\imath}}_{N}\n                     +\\sin\\varphi\\,\\hat{\\boldsymbol{\\imath}}_{U}\\bigr),\n\\qquad -\\dfrac{\\pi}{2}<\\varphi<\\dfrac{\\pi}{2},\n\\]\n\nwhere the right–handed orthonormal triad  \n$\\bigl(\\hat{\\boldsymbol{\\imath}}_{E},\n        \\hat{\\boldsymbol{\\imath}}_{N},\n        \\hat{\\boldsymbol{\\imath}}_{U}\\bigr)$\npoints respectively East, North and Up at the point of interest and satisfies  \n$\\hat{\\boldsymbol{\\imath}}_{E}\\times\\hat{\\boldsymbol{\\imath}}_{N}\n      =\\hat{\\boldsymbol{\\imath}}_{U}$,\n$\\hat{\\boldsymbol{\\imath}}_{N}\\times\\hat{\\boldsymbol{\\imath}}_{U}\n      =\\hat{\\boldsymbol{\\imath}}_{E}$.\n\nA high–speed train travels \\emph{due west} along the circle of latitude\n$\\varphi$ with ground speed $v>0$.\nEach wheel is an ideal thin hoop of radius $R$ that rolls without\nslipping, so its spin about the \\emph{negative} north axis is  \n\n\\[\n\\boldsymbol{\\omega}_{w}=-\\alpha\\,\\hat{\\boldsymbol{\\imath}}_{N},\n\\qquad\n\\alpha=\\dfrac{v}{R},\n\\qquad\nR\\alpha^{2}>g_{0},\n\\]\n$g_{0}$ denoting the conventional surface gravity.\n\nInside one wheel a tiny bead is soldered to a spoke at the distance\n$r=\\lambda R$ from the wheel centre ($0<\\lambda\\le1$).  \nMeasured \\emph{counter–clockwise from the downward vertical as viewed\nfrom the south}, the spoke makes the angle $0\\le\\theta\\le\\pi$.  \nExactly when the spoke reaches the value $\\theta$ the weld fails and the\nbead is released.\n\nThe subsequent motion will be analysed in the Earth–fixed (rotating)\nframe with origin at the Earth’s centre $C$.  In this frame the bead\nexperiences  \n\n$\\bullet$ the Newtonian gravity $-g_{0}\\,\\hat{\\boldsymbol{\\imath}}_{U}$;\n\n$\\bullet$ the centrifugal acceleration\n      $-\\boldsymbol{\\omega}_{e}\\times\n       (\\boldsymbol{\\omega}_{e}\\times\\mathbf r)$;\n\n$\\bullet$ the Coriolis acceleration\n      $2\\,\\boldsymbol{\\omega}_{e}\\times\\mathbf v$,\n\nwhere $\\mathbf r(t)$ and $\\mathbf v(t)$ denote position and velocity in\nthe rotating frame.  Air resistance is neglected and the wheel is\nassumed to spin so fast that the bead is ejected upwards.\n\nThroughout we keep the \\emph{two} independent dimensionless small\nparameters  \n\n\\[\n\\varepsilon=\\dfrac{\\omega_{e}}{\\alpha}\\ll1,\n\\qquad\n\\eta=\\dfrac{\\omega_{e}^{2}R_{e}}{g_{0}}\\approx3\\times10^{-3},\n\\tag{$\\star$}\n\\]\nretain every contribution of order $\\varepsilon$ or $\\eta$ and discard\nall products or squares\n$\\varepsilon^{2},\\,\\eta^{2},\\,\\varepsilon\\eta$.\n\nUseful abbreviation (effective gravity at latitude $\\varphi$):\n\n\\[\ng=g_{0}-\\omega_{e}^{2}R_{e}\\cos^{2}\\varphi\n  =g_{0}\\bigl(1-\\eta\\cos^{2}\\varphi\\bigr).\n\\tag{$\\dagger$}\n\\]\n\nTasks  \n\n(a)  In the local basis\n$\\bigl(\\hat{\\boldsymbol{\\imath}}_{E},\n       \\hat{\\boldsymbol{\\imath}}_{N},\n       \\hat{\\boldsymbol{\\imath}}_{U}\\bigr)$\ndetermine the bead’s initial position $\\mathbf r_{0}$ (with respect to\n$C$) and its \\emph{rotating–frame} velocity $\\mathbf v_{0}$,\nretaining all terms up to $O(\\varepsilon)$ and $O(\\eta)$ and expressing\nyour answer in $R_{e},R,\\lambda,\\theta,\\alpha,\\omega_{e}$.\n\n(b)  Let\n$L_{N}=m\\,\\hat{\\boldsymbol{\\imath}}_{N}\\!\\cdot(\\mathbf r\\times\\mathbf v)$\nbe the north component of the angular momentum in the rotating frame.\n\n(i)  Prove that  \n\n\\[\n\\dfrac{{\\mathrm d}L_{N}}{{\\mathrm d}t}\n      =m\\,g\\,r_{E}(t)\n       +O\\bigl(\\varepsilon\\,m g R\\bigr)\n       +O\\bigl(\\eta\\,m g R\\bigr),\n\\qquad\nr_{E}(t)=\\hat{\\boldsymbol{\\imath}}_{E}\\!\\cdot\\mathbf r(t),\n\\]\n\nand that over the whole flight  \n\n\\[\n\\dfrac{L_{N}(t)-L_{N}(0)}{L_{N}(0)}\n      =O\\!\\Bigl(\\dfrac{R}{R_{e}}\\Bigr)+O(\\varepsilon)+O(\\eta).\n\\]\n\n(ii)  Hence show that the trajectory departs from a single vertical\nplane by at most $O\\!\\bigl(R/R_{e}\\bigr)+O(\\varepsilon)+O(\\eta)$ and\nthat, to the same order, this plane coincides with the local meridian.\n\n(c)  Working consistently to $O(\\varepsilon)$ and $O(\\eta)$, prove that\nthe greatest height attained by the bead above the rail top is  \n\n\\[\nh(\\lambda,\\theta)=\nR-\\lambda R\\cos\\theta\n+\\dfrac{\\lambda^{2}R^{2}\\alpha^{2}\\sin^{2}\\theta}{2g}\n\\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr).\n\\]\n\n(d)  For fixed $\\lambda$ maximise $h(\\lambda,\\theta)$ with respect\nto $\\theta$ and show that  \n\n\\[\n\\boxed{\\;\n\\cos\\theta^{\\!*}\n   =-\\dfrac{g}{\\lambda R\\alpha^{2}}\n     \\Bigl[1-2\\varepsilon\\cos\\varphi+O(\\eta)\\Bigr]\\;},\n\\]\n\\[\n\\boxed{%\nh_{\\max}(\\lambda)=\nR+\\dfrac{g}{2\\alpha^{2}}\\bigl[1-2\\varepsilon\\cos\\varphi\\bigr]\n+\\dfrac{\\lambda^{2}R^{2}\\alpha^{2}}{2g}\n\\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr)}.\n\\]\n\n(e)  Finally maximise $h_{\\max}(\\lambda)$ over\n$0<\\lambda\\le1$ and prove that \\emph{every} bead satisfies  \n\n\\[\n\\boxed{%\nh\\le\n\\dfrac{\\bigl(R\\alpha+g/\\alpha\\bigr)^{2}}{2g}\n+\\varepsilon\\cos\\varphi\\,\n\\dfrac{R^{2}\\alpha^{2}-g^{2}/\\alpha^{2}}{g}\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr)},\n\\]\n\nwith equality iff $\\lambda=1$ and  \n\n\\[\n\\cos\\theta=-\\,\\dfrac{g}{R\\alpha^{2}}\n           \\Bigl[1-2\\varepsilon\\cos\\varphi+O(\\eta)\\Bigr].\n\\]\n\n(f)  Give a qualitative explanation showing why the Coriolis force\ncontributes only at $O(\\varepsilon)$ to the maximum height, whereas the\ncentrifugal term already enters at $O(\\eta)$, and discuss the relative\nmagnitude of the two effects for realistic high–speed wheels.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout the solution we strictly enforce the double–expansion rule\n$(\\star)$: every $O(\\varepsilon)$ and every $O(\\eta)$ contribution is\nretained, whereas all products\n$\\varepsilon^{2},\\ \\eta^{2},\\ \\varepsilon\\eta$ are discarded.\n\n----------------------------------------------------------------\n(a)  Initial data  \n\nPosition of the wheel centre in the rotating frame:  \n$\\mathbf r_{C}=(R_{e}+R)\\,\\hat{\\boldsymbol{\\imath}}_{U}$.\n\nBead offset (adopted $\\theta$–convention):\n\n\\[\n\\mathbf r'=\\lambda R\n\\bigl(\\sin\\theta\\,\\hat{\\boldsymbol{\\imath}}_{E}\n      -\\cos\\theta\\,\\hat{\\boldsymbol{\\imath}}_{U}\\bigr).\n\\]\n\nHence  \n\n\\[\n\\boxed{\\;\n\\mathbf r_{0}=\\mathbf r_{C}+\\mathbf r'\n=\\bigl(R_{e}+R-\\lambda R\\cos\\theta\\bigr)\\hat{\\boldsymbol{\\imath}}_{U}\n+\\lambda R\\sin\\theta\\,\\hat{\\boldsymbol{\\imath}}_{E}}\\!.\n\\tag{1}\n\\]\n\nVelocity in the rotating frame comes from three contributions.\n\n(i)  \\emph{Translation of the wheel centre}  \n(the ground–frame speed is purely westward)\n\\[\n\\mathbf v_{C}=-v\\,\\hat{\\boldsymbol{\\imath}}_{E}\n             =-\\alpha R\\,\\hat{\\boldsymbol{\\imath}}_{E}.\n\\]\n\n(ii)  \\emph{Spin of the wheel}\n\\[\n\\mathbf v_{\\text{spin}}\n   =\\boldsymbol{\\omega}_{w}\\times\\mathbf r'\n   =\\alpha\\lambda R\n      \\bigl(\\cos\\theta\\,\\hat{\\boldsymbol{\\imath}}_{E}\n           +\\sin\\theta\\,\\hat{\\boldsymbol{\\imath}}_{U}\\bigr).\n\\]\n\n(iii)  \\emph{Transformation from the inertial to the rotating frame.}  \nBecause the centre–of–mass term has already been converted in (i), only\n$\\mathbf r'$ contributes:\n\n\\[\n-\\boldsymbol{\\omega}_{e}\\times\\mathbf r'\n   =-\\omega_{e}\\lambda R\n      \\Bigl(\\cos\\varphi\\cos\\theta\\,\\hat{\\boldsymbol{\\imath}}_{E}\n           -\\sin\\varphi\\sin\\theta\\,\\hat{\\boldsymbol{\\imath}}_{N}\n           +\\cos\\varphi\\sin\\theta\\,\\hat{\\boldsymbol{\\imath}}_{U}\\Bigr).\n\\]\n\nCollecting (i)–(iii) and introducing\n$\\varepsilon=\\omega_{e}/\\alpha$ gives\n\n\\[\n\\boxed{\n\\begin{aligned}\n\\mathbf v_{0}&=\n\\alpha R\\Bigl[\n\\bigl(-1+\\lambda\\cos\\theta\n      +\\varepsilon\\lambda\\cos\\varphi\\cos\\theta\\bigr)\\,\n          \\hat{\\boldsymbol{\\imath}}_{E} \\\\\n&\\quad -\\varepsilon\\lambda\\sin\\varphi\\sin\\theta\\,\n          \\hat{\\boldsymbol{\\imath}}_{N}\n\\;+\\;\\lambda\\sin\\theta\n      \\bigl(1+\\varepsilon\\cos\\varphi\\bigr)\\,\n          \\hat{\\boldsymbol{\\imath}}_{U}\\Bigr]\n+O\\!\\bigl(\\varepsilon^{2}\\alpha R\\bigr).\n\\end{aligned}}\n\\tag{2}\n\\]\n\nNote especially the \\emph{non–vanishing} north component\n$(v_{0})_{N}=-\\varepsilon\\alpha\\lambda R\\sin\\theta\\sin\\varphi\n              =O(\\varepsilon\\alpha R)$.\n\n----------------------------------------------------------------\n(b)  Near–conservation of $L_{N}$ and quasi–planarity  \n\nDefine $L_{N}=m\\,\\hat{\\boldsymbol{\\imath}}_{N}\\!\\cdot(\\mathbf r\\times\\mathbf v)$.\nIn the rotating frame  \n\n\\[\n\\mathbf a=-g_{0}\\hat{\\boldsymbol{\\imath}}_{U}\n-\\boldsymbol{\\omega}_{e}\\times\n      (\\boldsymbol{\\omega}_{e}\\times\\mathbf r)\n+2\\,\\boldsymbol{\\omega}_{e}\\times\\mathbf v .\n\\]\n\nTherefore  \n\n\\[\n\\dfrac{{\\mathrm d}L_{N}}{{\\mathrm d}t}\n      =m\\,\\hat{\\boldsymbol{\\imath}}_{N}\\!\\cdot(\\mathbf r\\times\\mathbf a)\n      =\\tau_{g}+\\tau_{c}+\\tau_{\\text{Cor}}.\n\\tag{3}\n\\]\n\nGravitational torque:  \n\\[\n\\tau_{g}=m\\,g\\,r_{E}(t)+O(\\eta\\,m g R), \\qquad\\text{cf.\\ }(\\dagger).\n\\]\n\nCentrifugal torque:  \n$\\tau_{c}=O(\\eta\\,m g R)$.\n\nCoriolis torque:\n$\\tau_{\\text{Cor}}\n       =O(\\varepsilon\\,m g R)$\nbecause $|\\mathbf v|=O(\\alpha R)$.\n\nCollecting:\n\n\\[\n\\boxed{\\dfrac{{\\mathrm d}L_{N}}{{\\mathrm d}t}\n      =m\\,g\\,r_{E}(t)\n       +O\\!\\bigl(\\varepsilon\\,m g R\\bigr)\n       +O\\!\\bigl(\\eta\\,m g R\\bigr)}.\n\\]\n\nFlight time $T$ is of order  \n$T\\sim2(v_{0})_{U}/g=O(\\alpha\\lambda R/g)$.  \nHence\n\n\\[\nL_{N}(t)-L_{N}(0)\n      =O\\!\\bigl((\\varepsilon+\\eta)\\,m g R T\\bigr)\n      =O\\!\\bigl((\\varepsilon+\\eta)\\,m\\alpha R^{2}\\bigr).\n\\]\n\nFrom (1)–(2)\n\n\\[\nL_{N}(0)=m\\bigl(r_{E}v_{U}-r_{U}v_{E}\\bigr)_{t=0}\n        =m\\alpha R R_{e}\n          \\Bigl[1-\\lambda\\cos\\theta+O(\\varepsilon)\\Bigr],\n\\]\nso that\n\n\\[\n\\boxed{\\;\n\\frac{L_{N}(t)-L_{N}(0)}{L_{N}(0)}\n          =O\\!\\Bigl(\\dfrac{R}{R_{e}}\\Bigr)+O(\\varepsilon)+O(\\eta)}.\n\\]\n\nSince $(v_{0})_{N}=O(\\varepsilon)$, the initial departure\nfrom the meridian plane is already $O(\\varepsilon)$ and, by the above\nbound, does not grow beyond\n$O(R/R_{e})+O(\\varepsilon)+O(\\eta)$ during the flight, proving (ii).\n\n----------------------------------------------------------------\n(c)  Maximum height for fixed $(\\lambda,\\theta)$  \n\nThe Coriolis force does no work.  \nThe centrifugal force is conservative, its potential being\n$\\tfrac12\\omega_{e}^{2}r_{\\perp}^{2}$, so the vertical dynamics is the\nusual one but with the \\emph{effective} gravity $g$ of $(\\dagger)$.\nHence mechanical energy in the vertical direction is conserved to\n$O(\\varepsilon)$ and $O(\\eta)$.\n\nInitial height above the rail:\n$z_{0}=R-\\lambda R\\cos\\theta$.\n\nUpward velocity component from (2):\n\n\\[\n(v_{0})_{U}=\\alpha\\lambda R\\sin\\theta\n            \\bigl(1+\\varepsilon\\cos\\varphi\\bigr).\n\\]\n\nTherefore  \n\n\\[\n\\dfrac{(v_{0})_{U}^{2}}{2g}\n        =\\dfrac{\\lambda^{2}R^{2}\\alpha^{2}\\sin^{2}\\theta}{2g}\n          \\bigl[1+2\\varepsilon\\cos\\varphi\\bigr].\n\\]\n\nAdding the initial height gives\n\n\\[\n\\boxed{%\nh(\\lambda,\\theta)=\nR-\\lambda R\\cos\\theta\n+\\dfrac{\\lambda^{2}R^{2}\\alpha^{2}\\sin^{2}\\theta}{2g}\n\\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr)}.\n\\tag{4}\n\\]\n\n----------------------------------------------------------------\n(d)  Maximisation with respect to $\\theta$  \n\nDifferentiate (4) and set to zero:\n\n\\[\n0=\\lambda R\\sin\\theta\n   \\Bigl[1+\\dfrac{\\lambda R\\alpha^{2}}{g}\n         \\bigl(1+2\\varepsilon\\cos\\varphi\\bigr)\\cos\\theta\\Bigr].\n\\]\n\nThe non–trivial root is  \n\n\\[\n\\boxed{\\;\n\\cos\\theta^{\\!*}\n   =-\\dfrac{g}{\\lambda R\\alpha^{2}}\n     \\Bigl[1-2\\varepsilon\\cos\\varphi+O(\\eta)\\Bigr]\\;}.\n\\tag{5}\n\\]\n\nInsert (5) into (4).  A straightforward algebraic expansion yields  \n\n\\[\n\\boxed{%\nh_{\\max}(\\lambda)=\nR+\\dfrac{g}{2\\alpha^{2}}\\bigl[1-2\\varepsilon\\cos\\varphi\\bigr]\n+\\dfrac{\\lambda^{2}R^{2}\\alpha^{2}}{2g}\n\\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr)}.\n\\tag{6}\n\\]\n\n----------------------------------------------------------------\n(e)  Maximisation with respect to $\\lambda$  \n\nFrom (6)\n\n\\[\n\\dfrac{{\\mathrm d}h_{\\max}}{{\\mathrm d}\\lambda}\n   =\\dfrac{\\lambda R^{2}\\alpha^{2}}{g}\n     \\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]>0,\n     \\qquad(0<\\lambda\\le1),\n\\]\nso $h_{\\max}$ is strictly increasing and attains its absolute maximum at\n$\\lambda=1$.  Substituting $\\lambda=1$ into (6) gives  \n\n\\[\nh_{\\text{top}}\n    =R+\\dfrac{g}{2\\alpha^{2}}\\bigl[1-2\\varepsilon\\cos\\varphi\\bigr]\n     +\\dfrac{R^{2}\\alpha^{2}}{2g}\n      \\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]\n     =\\dfrac{(R\\alpha+g/\\alpha)^{2}}{2g}\n      +\\varepsilon\\cos\\varphi\\,\n       \\dfrac{R^{2}\\alpha^{2}-g^{2}/\\alpha^{2}}{g}.\n\\tag{7}\n\\]\n\nBecause\n$h_{\\max}(\\lambda)\\le h_{\\text{top}}$ for every\n$0<\\lambda\\le1$, we obtain the bound stated in part (e):\n\n\\[\n\\boxed{%\nh\\le\n\\dfrac{\\bigl(R\\alpha+g/\\alpha\\bigr)^{2}}{2g}\n+\\varepsilon\\cos\\varphi\\,\n\\dfrac{R^{2}\\alpha^{2}-g^{2}/\\alpha^{2}}{g}\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr)}.\n\\]\n\nEquality holds iff $\\lambda=1$ and $\\theta=\\theta^{\\!*}$, i.e.\\\n\\[\n\\cos\\theta=-\\,\\dfrac{g}{R\\alpha^{2}}\n           \\Bigl[1-2\\varepsilon\\cos\\varphi+O(\\eta)\\Bigr].\n\\]\n\n----------------------------------------------------------------\n(f)  Discussion of rotational corrections  \n\nThe Coriolis force is always orthogonal to the instantaneous velocity\nand therefore does \\emph{no} work.  Its influence on the maximum height\ncan arise only indirectly through tiny plane–change effects and is\nconsequently suppressed to $O(\\varepsilon)$.  \n\nThe centrifugal force, on the other hand, derives from the potential\n$\\tfrac12\\omega_{e}^{2}r_{\\perp}^{2}$ and its vertical component\nconstitutes a genuine reduction of the effective gravity by a fraction\n$\\eta\\cos^{2}\\varphi$.  Since $\\eta\\approx3\\times10^{-3}$ whereas\n$\\varepsilon=\\omega_{e}/\\alpha\\lesssim10^{-6}$ for realistic\nhigh–speed wheels ($\\alpha\\gtrsim10^{2}\\,\\mathrm{s}^{-1}$),\ncentrifugal corrections exceed Coriolis corrections by three orders of\nmagnitude and thus provide the first measurable rotational effect on the\nmaximum height.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.388866",
        "was_fixed": false,
        "difficulty_analysis": "1.  Three–dimensional setting.  While the core idea (a bead detaching from a rolling wheel) is retained, the motion now occurs on a rotating spherical Earth, demanding full vector treatment instead of plane kinematics.\n2.  Additional forces.  Besides gravity the solver must handle centrifugal and Coriolis terms and decide which ones influence the height to a given order.\n3.  Small–parameter perturbation.  A systematic expansion in ε = ωₑ/α is required; this calls for perturbative reasoning rather than purely algebraic manipulation.\n4.  Conservation laws in non-inertial frames.  The candidate must know which inertial forces do or do not perform work and exploit that to keep the calculation manageable.\n5.  Two layers of optimisation.  First θ must be chosen to maximise the altitude for each λ, then λ itself must be optimised.  Because λ enters both algebraically and through the root of a transcendental equation, careful order‐keeping in ε is essential.\n6.  Geometric insight.  Showing that the path lies in a tilted plane (part (b)) demands recognition of an almost-conserved component of angular momentum, a concept absent from the original problem.\n\nAll these elements go well beyond the planar cycloid of the original exercise, making the enhanced variant significantly harder and richer in mathematical structure."
      }
    },
    "original_kernel_variant": {
      "question": "Consider a perfectly spherical rigid Earth of radius $R_{e}$ that\nrotates uniformly with angular velocity  \n\n\\[\n\\boldsymbol{\\omega}_{e}\n      =\\omega_{e}\\bigl(\\cos\\varphi\\,\\hat{\\boldsymbol{\\imath}}_{N}\n                     +\\sin\\varphi\\,\\hat{\\boldsymbol{\\imath}}_{U}\\bigr),\n\\qquad -\\dfrac{\\pi}{2}<\\varphi<\\dfrac{\\pi}{2},\n\\]\n\nwhere the right–handed orthonormal triad\n$\\bigl(\\hat{\\boldsymbol{\\imath}}_{E},\n        \\hat{\\boldsymbol{\\imath}}_{N},\n        \\hat{\\boldsymbol{\\imath}}_{U}\\bigr)$\npoints respectively East, North and Up at the point of interest and\nsatisfies\n$\\hat{\\boldsymbol{\\imath}}_{E}\\times\\hat{\\boldsymbol{\\imath}}_{N}\n      =\\hat{\\boldsymbol{\\imath}}_{U}$,\n$\\hat{\\boldsymbol{\\imath}}_{N}\\times\\hat{\\boldsymbol{\\imath}}_{U}\n      =\\hat{\\boldsymbol{\\imath}}_{E}$.\n\nA high–speed train travels {\\em due west} along the circle of latitude\n$\\varphi$ with ground speed $v>0$.\nEach wheel is an ideal thin hoop of radius $R$ that rolls without\nslipping, so its spin about the {\\em negative} north axis is  \n\n\\[\n\\boldsymbol{\\omega}_{w}=-\\alpha\\,\\hat{\\boldsymbol{\\imath}}_{N},\n\\qquad\n\\alpha=\\dfrac{v}{R},\n\\qquad\nR\\alpha^{2}>g_{0},\n\\]\n$g_{0}$ denoting the conventional surface gravity.\n\nInside one wheel a tiny bead is soldered to a spoke at the distance\n$r=\\lambda R$ from the wheel centre ($0<\\lambda\\le1$).  \nMeasured {\\em counter–clockwise from the downward vertical as viewed\nfrom the south}, the spoke makes the angle $0\\le\\theta\\le\\pi$.  \nExactly when the spoke reaches the value $\\theta$ the weld fails and the\nbead is released.\n\nThe subsequent motion will be analysed in the Earth–fixed (rotating)\nframe with origin at the Earth’s centre $C$.  In this frame the bead\nexperiences\n\n$\\bullet$ the Newtonian gravity $-g_{0}\\,\\hat{\\boldsymbol{\\imath}}_{U}$;\n\n$\\bullet$ the centrifugal acceleration\n      $-\\boldsymbol{\\omega}_{e}\\times\n       (\\boldsymbol{\\omega}_{e}\\times\\mathbf r)$;\n\n$\\bullet$ the Coriolis acceleration\n      $2\\,\\boldsymbol{\\omega}_{e}\\times\\mathbf v$,\n\nwhere $\\mathbf r(t)$ and $\\mathbf v(t)$ denote position and velocity in\nthe rotating frame.  Air resistance is neglected and the wheel is\nassumed to spin so fast that the bead is ejected upwards.\n\nThroughout we keep the {\\em two} independent dimensionless small\nparameters  \n\n\\[\n\\varepsilon=\\dfrac{\\omega_{e}}{\\alpha}\\ll1,\n\\qquad\n\\eta=\\dfrac{\\omega_{e}^{2}R_{e}}{g_{0}}\\approx3\\times10^{-3},\n\\tag{$\\star$}\n\\]\nretain every contribution of order $\\varepsilon$ or $\\eta$ and discard\nall products or squares\n$\\varepsilon^{2},\\,\\eta^{2},\\,\\varepsilon\\eta$.\n\nUseful abbreviation (effective gravity at latitude $\\varphi$):\n\n\\[\ng=g_{0}-\\omega_{e}^{2}R_{e}\\cos^{2}\\varphi\n  =g_{0}\\bigl(1-\\eta\\cos^{2}\\varphi\\bigr).\n\\tag{$\\dagger$}\n\\]\n\nTasks  \n\n(a)  In the local basis\n$\\bigl(\\hat{\\boldsymbol{\\imath}}_{E},\n       \\hat{\\boldsymbol{\\imath}}_{N},\n       \\hat{\\boldsymbol{\\imath}}_{U}\\bigr)$\ndetermine the bead’s initial position $\\mathbf r_{0}$ (with respect to\n$C$) and its {\\em rotating–frame} velocity $\\mathbf v_{0}$,\nretaining all terms up to $O(\\varepsilon)$ and $O(\\eta)$ and expressing\nyour answer in $R_{e},R,\\lambda,\\theta,\\alpha,\\omega_{e}$.\n\n(b)  Let\n$L_{N}=m\\,\\hat{\\boldsymbol{\\imath}}_{N}\\!\\cdot(\\mathbf r\\times\\mathbf v)$\nbe the north component of the angular momentum in the rotating frame.\n\n(i)  Prove that  \n\n\\[\n\\dfrac{{\\mathrm d}L_{N}}{{\\mathrm d}t}\n      =m\\,g\\,r_{E}(t)+O\\bigl(\\varepsilon\\,m g R\\bigr),\n\\qquad\nr_{E}(t)=\\hat{\\boldsymbol{\\imath}}_{E}\\!\\cdot\\mathbf r(t),\n\\]\n\nand that over the whole flight  \n\n\\[\n\\dfrac{L_{N}(t)-L_{N}(0)}{L_{N}(0)}\n      =O\\!\\Bigl(\\dfrac{R}{R_{e}}\\Bigr)+O(\\varepsilon).\n\\]\n\n(ii)  Hence show that the trajectory departs from a single vertical\nplane by at most $O\\!\\bigl(R/R_{e}\\bigr)+O(\\varepsilon)$ and that, to the\nsame order, this plane coincides with the local meridian.\n\n(c)  Working consistently to $O(\\varepsilon)$ and $O(\\eta)$, prove that\nthe greatest height attained by the bead above the rail top is  \n\n\\[\nh(\\lambda,\\theta)=\nR-\\lambda R\\cos\\theta\n+\\dfrac{\\lambda^{2}R^{2}\\alpha^{2}\\sin^{2}\\theta}{2g}\n\\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr).\n\\]\n\n(d)  For fixed $\\lambda$ maximise $h(\\lambda,\\theta)$ with respect\nto $\\theta$ and show that  \n\n\\[\n\\boxed{\\;\n\\cos\\theta^{\\!*}\n   =-\\dfrac{g}{\\lambda R\\alpha^{2}}\n     \\Bigl[1-2\\varepsilon\\cos\\varphi+O(\\eta)\\Bigr]\\;},\n\\]\n\\[\n\\boxed{%\nh_{\\max}(\\lambda)=\nR+\\dfrac{g}{2\\alpha^{2}}\\bigl[1-2\\varepsilon\\cos\\varphi\\bigr]\n+\\dfrac{\\lambda^{2}R^{2}\\alpha^{2}}{2g}\n\\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr)}.\n\\]\n\n(e)  Finally maximise $h_{\\max}(\\lambda)$ over\n$0<\\lambda\\le1$ and prove that {\\em every} bead satisfies  \n\n\\[\n\\boxed{%\nh\\le\n\\dfrac{\\bigl(R\\alpha+g/\\alpha\\bigr)^{2}}{2g}\n+\\varepsilon\\cos\\varphi\\,\n\\dfrac{R^{2}\\alpha^{2}-g^{2}/\\alpha^{2}}{2g}\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr)},\n\\]\n\nwith equality iff $\\lambda=1$ and  \n\n\\[\n\\cos\\theta=-\\,\\dfrac{g}{R\\alpha^{2}}\n           \\Bigl[1-2\\varepsilon\\cos\\varphi+O(\\eta)\\Bigr].\n\\]\n\n(f)  Give a qualitative explanation showing why the Coriolis force\ncontributes only at $O(\\varepsilon)$ to the maximum height, whereas the\ncentrifugal term already enters at $O(\\eta)$, and discuss the relative\nmagnitude of the two effects for realistic high–speed wheels.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we enforce the double–expansion rule $(\\star)$, i.e.\\\nall $O(\\varepsilon)$ and all $O(\\eta)$ terms are retained whereas\n$\\varepsilon^{2},\\,\\eta^{2}$ and $\\varepsilon\\eta$ are discarded.\n\n----------------------------------------------------------------\n(a)  Initial data  \n\nPosition of the wheel centre in the rotating frame:  \n$\\mathbf r_{C}=(R_{e}+R)\\,\\hat{\\boldsymbol{\\imath}}_{U}.$\n\nBead offset (adopted $\\theta$–convention):\n\n\\[\n\\mathbf r'=\\lambda R\n\\bigl(\\sin\\theta\\,\\hat{\\boldsymbol{\\imath}}_{E}\n      -\\cos\\theta\\,\\hat{\\boldsymbol{\\imath}}_{U}\\bigr).\n\\]\n\nHence  \n\n\\[\n\\boxed{\\;\n\\mathbf r_{0}=\\mathbf r_{C}+\\mathbf r'\n=\\bigl(R_{e}+R-\\lambda R\\cos\\theta\\bigr)\\hat{\\boldsymbol{\\imath}}_{U}\n+\\lambda R\\sin\\theta\\,\\hat{\\boldsymbol{\\imath}}_{E}}\\!.\n\\tag{1}\n\\]\n\nVelocity in the rotating frame.  \nThree contributions are added:\n\n(i)  {\\em Translation of the wheel centre}  \n(the ground–frame speed is purely westward)\n\\[\n\\mathbf v_{C}=-v\\,\\hat{\\boldsymbol{\\imath}}_{E}\n             =-\\alpha R\\,\\hat{\\boldsymbol{\\imath}}_{E}.\n\\]\n\n(ii)  {\\em Spin of the wheel}\n\\[\n\\mathbf v_{\\text{spin}}\n   =\\boldsymbol{\\omega}_{w}\\times\\mathbf r'\n   =\\alpha\\lambda R\n      \\bigl(\\cos\\theta\\,\\hat{\\boldsymbol{\\imath}}_{E}\n           +\\sin\\theta\\,\\hat{\\boldsymbol{\\imath}}_{U}\\bigr).\n\\]\n\n(iii)  {\\em Transformation from the inertial to the rotating frame.}  \nBecause the centre–of–mass term has already been converted in (i), only\nthe offset $\\mathbf r'$ contributes:\n\n\\[\n-\\boldsymbol{\\omega}_{e}\\times\\mathbf r'\n   =-\\omega_{e}\\lambda R\n      \\Bigl(\\cos\\varphi\\cos\\theta\\,\\hat{\\boldsymbol{\\imath}}_{E}\n           -\\sin\\varphi\\sin\\theta\\,\\hat{\\boldsymbol{\\imath}}_{N}\n           +\\cos\\varphi\\sin\\theta\\,\\hat{\\boldsymbol{\\imath}}_{U}\\Bigr).\n\\]\n\nCollecting (i)–(iii) and introducing\n$\\varepsilon=\\omega_{e}/\\alpha$ gives\n\n\\[\n\\boxed{\n\\begin{aligned}\n\\mathbf v_{0}&=\n\\alpha R\\Bigl[\n\\bigl(-1+\\lambda\\cos\\theta\n      +\\varepsilon\\lambda\\cos\\varphi\\cos\\theta\\bigr)\\,\n          \\hat{\\boldsymbol{\\imath}}_{E} \\\\\n&\\quad -\\varepsilon\\lambda\\sin\\varphi\\sin\\theta\\,\n          \\hat{\\boldsymbol{\\imath}}_{N}\n\\;+\\;\\lambda\\sin\\theta\n      \\bigl(1+\\varepsilon\\cos\\varphi\\bigr)\\,\n          \\hat{\\boldsymbol{\\imath}}_{U}\\Bigr]\n+O\\!\\bigl(\\varepsilon^{2}\\alpha R\\bigr).\n\\end{aligned}}\n\\tag{2}\n\\]\n\nNote especially the {\\em non–vanishing} north component\n$(v_{0})_{N}=-\\varepsilon\\alpha\\lambda R\\sin\\theta\\sin\\varphi=O(\\varepsilon\\alpha R)$.\n\n----------------------------------------------------------------\n(b)  Near–conservation of $L_{N}$ and quasi–planarity  \n\nDefine $L_{N}=m\\,\\hat{\\boldsymbol{\\imath}}_{N}\\!\\cdot(\\mathbf r\\times\\mathbf v)$.\nIn the rotating frame  \n\n\\[\n\\mathbf a=-g_{0}\\hat{\\boldsymbol{\\imath}}_{U}\n-\\boldsymbol{\\omega}_{e}\\times\n      (\\boldsymbol{\\omega}_{e}\\times\\mathbf r)\n+2\\,\\boldsymbol{\\omega}_{e}\\times\\mathbf v .\n\\]\n\nTherefore  \n\n\\[\n\\dfrac{{\\mathrm d}L_{N}}{{\\mathrm d}t}\n      =m\\,\\hat{\\boldsymbol{\\imath}}_{N}\\!\\cdot(\\mathbf r\\times\\mathbf a)\n      =\\tau_{g}+\\tau_{c}+\\tau_{\\text{Cor}}.\n\\tag{3}\n\\]\n\nGravitational torque:  \n$\\tau_{g}=m\\,g\\,r_{E}(t)+O(\\eta\\,m g R)$, cf.\\ $(\\dagger)$.\n\nCentrifugal torque:  \n$\\tau_{c}=O(\\eta\\,m g R)$.\n\nCoriolis torque:\n$\\tau_{\\text{Cor}}=\nm\\,\\mathbf r\\times(2\\boldsymbol{\\omega}_{e}\\times\\mathbf v)\n       =O(\\varepsilon\\,m g R)$\nbecause $|\\mathbf v|=O(\\alpha R)$.\n\nThus  \n\n\\[\n\\dfrac{{\\mathrm d}L_{N}}{{\\mathrm d}t}\n      =m\\,g\\,r_{E}(t)+O\\!\\bigl(\\varepsilon\\,m g R\\bigr),\n\\]\nestablishing (i).  Over the characteristic flight time\n$T=2(v_{0})_{U}/g=O(\\alpha\\lambda R/g)$ the accumulated change obeys\n\n\\[\nL_{N}(t)-L_{N}(0)\n          =O\\!\\bigl(mgRT\\bigr)\n          =O\\!\\bigl(m\\alpha R^{2}\\bigr).\n\\]\n\nFrom (1)–(2)\n\n\\[\nL_{N}(0)=m\\bigl(r_{E}v_{U}-r_{U}v_{E}\\bigr)_{t=0}\n        =m\\alpha R R_{e}\n          \\Bigl[1-\\lambda\\cos\\theta+O(\\varepsilon)\\Bigr],\n\\]\nso that\n\n\\[\n\\frac{L_{N}(t)-L_{N}(0)}{L_{N}(0)}\n          =O\\!\\Bigl(\\dfrac{R}{R_{e}}\\Bigr)+O(\\varepsilon),\n\\]\nas claimed.  Since $(v_{0})_{N}=O(\\varepsilon)$, the initial departure\nfrom the meridian plane is already $O(\\varepsilon)$ and does not grow\nbeyond that order during the flight, proving (ii).\n\n----------------------------------------------------------------\n(c)  Maximum height for fixed $(\\lambda,\\theta)$  \n\nThe Coriolis force does no work.  \nThe centrifugal force is conservative, its potential being\n$\\tfrac12\\omega_{e}^{2}r_{\\perp}^{2}$, so the vertical dynamics is the\nusual one but with the {\\em effective} gravity $g$ of $(\\dagger)$.\nHence mechanical energy in the vertical direction is conserved to\n$O(\\varepsilon)$ and $O(\\eta)$.\n\nInitial height above the rail:\n$z_{0}=R-\\lambda R\\cos\\theta$.\n\nUpward velocity component from (2):\n\n\\[\n(v_{0})_{U}=\\alpha\\lambda R\\sin\\theta\n            \\bigl(1+\\varepsilon\\cos\\varphi\\bigr).\n\\]\n\nTherefore  \n\n\\[\n\\dfrac{(v_{0})_{U}^{2}}{2g}\n        =\\dfrac{\\lambda^{2}R^{2}\\alpha^{2}\\sin^{2}\\theta}{2g}\n          \\bigl[1+2\\varepsilon\\cos\\varphi\\bigr].\n\\]\n\nAdding the initial height gives\n\n\\[\n\\boxed{%\nh(\\lambda,\\theta)=\nR-\\lambda R\\cos\\theta\n+\\dfrac{\\lambda^{2}R^{2}\\alpha^{2}\\sin^{2}\\theta}{2g}\n\\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr)}.\n\\tag{4}\n\\]\n\n----------------------------------------------------------------\n(d)  Maximisation with respect to $\\theta$  \n\nDifferentiate (4) and set to zero:\n\n\\[\n0=\\lambda R\\sin\\theta\n   \\Bigl[1+\\dfrac{\\lambda R\\alpha^{2}}{g}\n         \\bigl(1+2\\varepsilon\\cos\\varphi\\bigr)\\cos\\theta\\Bigr].\n\\]\n\nThe non–trivial root is  \n\n\\[\n\\boxed{\\;\n\\cos\\theta^{\\!*}\n   =-\\dfrac{g}{\\lambda R\\alpha^{2}}\n     \\Bigl[1-2\\varepsilon\\cos\\varphi+O(\\eta)\\Bigr]\\;}.\n\\tag{5}\n\\]\n\nInsert (5) into (4).  A straightforward algebraic expansion yields  \n\n\\[\n\\boxed{%\nh_{\\max}(\\lambda)=\nR+\\dfrac{g}{2\\alpha^{2}}\\bigl[1-2\\varepsilon\\cos\\varphi\\bigr]\n+\\dfrac{\\lambda^{2}R^{2}\\alpha^{2}}{2g}\n\\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr)}.\n\\tag{6}\n\\]\n\n----------------------------------------------------------------\n(e)  Maximisation with respect to $\\lambda$  \n\nFrom (6)\n\n\\[\n\\dfrac{{\\mathrm d}h_{\\max}}{{\\mathrm d}\\lambda}\n   =\\dfrac{\\lambda R^{2}\\alpha^{2}}{g}\n     \\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]>0\n     \\quad(0<\\lambda\\le1),\n\\]\nso $h_{\\max}$ is strictly increasing and attains its absolute maximum at\n$\\lambda=1$.  Substituting $\\lambda=1$ into (6) gives  \n\n\\[\nh_{\\text{top}}\n    =R+\\dfrac{g}{2\\alpha^{2}}\\bigl[1-2\\varepsilon\\cos\\varphi\\bigr]\n     +\\dfrac{R^{2}\\alpha^{2}}{2g}\n      \\bigl[1+2\\varepsilon\\cos\\varphi\\bigr]\n     =\\dfrac{(R\\alpha+g/\\alpha)^{2}}{2g}\n      +\\varepsilon\\cos\\varphi\\,\n       \\dfrac{R^{2}\\alpha^{2}-g^{2}/\\alpha^{2}}{2g}.\n\\]\n\nBecause\n$h_{\\max}(\\lambda)\\le h_{\\text{top}}$ for every\n$0<\\lambda\\le1$, we obtain\n\n\\[\n\\boxed{%\nh\\le\n\\dfrac{\\bigl(R\\alpha+g/\\alpha\\bigr)^{2}}{2g}\n+\\varepsilon\\cos\\varphi\\,\n\\dfrac{R^{2}\\alpha^{2}-g^{2}/\\alpha^{2}}{2g}\n+O\\!\\bigl(\\varepsilon^{2},\\eta^{2}\\bigr)}.\n\\]\n\nEquality holds iff $\\lambda=1$ and $\\theta=\\theta^{\\!*}$, i.e.\\\n\\[\n\\cos\\theta=-\\,\\dfrac{g}{R\\alpha^{2}}\n           \\Bigl[1-2\\varepsilon\\cos\\varphi+O(\\eta)\\Bigr].\n\\]\n\n----------------------------------------------------------------\n(f)  Discussion of rotational corrections  \n\nThe Coriolis force is always orthogonal to the instantaneous velocity\nand therefore does {\\em no} work.  Its influence on the maximum height\ncan arise only indirectly through tiny plane–change effects and is\nconsequently suppressed to $O(\\varepsilon)$.  \nThe centrifugal force, on the other hand, derives from the potential\n$\\tfrac12\\omega_{e}^{2}r_{\\perp}^{2}$ and its vertical component\nconstitutes a genuine reduction of the effective gravity by a fraction\n$\\eta\\cos^{2}\\varphi$.  Since $\\eta\\approx3\\times10^{-3}$ whereas\n$\\varepsilon=\\omega_{e}/\\alpha\\lesssim10^{-6}$ for realistic\nhigh–speed wheels ($\\alpha\\gtrsim10^{2}\\text{ s}^{-1}$),\ncentrifugal corrections exceed Coriolis corrections by three orders of\nmagnitude and thus provide the first measurable rotational effect on the\nmaximum height.\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.334989",
        "was_fixed": false,
        "difficulty_analysis": "1.  Three–dimensional setting.  While the core idea (a bead detaching from a rolling wheel) is retained, the motion now occurs on a rotating spherical Earth, demanding full vector treatment instead of plane kinematics.\n2.  Additional forces.  Besides gravity the solver must handle centrifugal and Coriolis terms and decide which ones influence the height to a given order.\n3.  Small–parameter perturbation.  A systematic expansion in ε = ωₑ/α is required; this calls for perturbative reasoning rather than purely algebraic manipulation.\n4.  Conservation laws in non-inertial frames.  The candidate must know which inertial forces do or do not perform work and exploit that to keep the calculation manageable.\n5.  Two layers of optimisation.  First θ must be chosen to maximise the altitude for each λ, then λ itself must be optimised.  Because λ enters both algebraically and through the root of a transcendental equation, careful order‐keeping in ε is essential.\n6.  Geometric insight.  Showing that the path lies in a tilted plane (part (b)) demands recognition of an almost-conserved component of angular momentum, a concept absent from the original problem.\n\nAll these elements go well beyond the planar cycloid of the original exercise, making the enhanced variant significantly harder and richer in mathematical structure."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}