{
  "index": "1958-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "2. Two uniform solid spheres of equal radii are so placed that one is directly above the other. The bottom sphere is fixed, and the top sphere, initially at rest, rolls off. At what point will contact between the two spheres be \"lost\"? Assume the coefficient of friction is such that no slipping occurs.",
  "solution": "Solution. Let \\( S \\) be the rolling sphere; let \\( a \\) be its radius and \\( M \\) its mass. Recall that the moment of inertia of a uniform solid sphere is \\( I=(2 / 5) M a^{2} \\).\n\nWe assume that the motion is essentially two-dimensional. Then the state of the system is defined by the angle \\( \\theta \\) between the line of centers and the vertical as shown in the figure. We denote the time derivative of \\( \\theta \\) by \\( \\theta \\). Let \\( v \\) be the translational speed of the center of \\( S \\) and let \\( \\omega \\) be the speed of rotation of \\( S \\).\n\nAs long as contact is maintained between the two spheres, the center of \\( S \\) moves along a circle of radius \\( 2 a \\), so \\( v=2 a \\dot{\\theta} \\) and \\( \\omega=2 \\dot{\\theta} \\). The kinetic energy of \\( S \\) is given by\n\\[\n\\frac{1}{2} M v^{2}+\\frac{1}{2} I \\omega^{2}=\\frac{14}{5} M a^{2} \\dot{\\theta}^{2}\n\\]\nand the potential energy by\n\\( 2 M a g \\cos \\theta \\)\nrelative to the level of the center of the lower sphere. The total energy is constant so\n\\[\n2 M a g \\cos \\theta+\\frac{14}{5} M a^{2} \\dot{\\theta}^{2}=2 M a g\n\\]\n(The right member is the left evaluated for \\( \\theta=0 \\).) Therefore we have\n\\[\n\\frac{7}{5} a \\dot{\\theta}^{2}=g(1-\\cos \\theta)\n\\]\n\nTo keep \\( S \\) in a circular orbit of radius \\( 2 a \\), a force toward the center of magnitude \\( 2 a M \\dot{\\theta}^{2} \\) is required. This force is supplied by the component of the gravitational force along the line of centers, which is \\( M g \\cos \\theta \\). As long as this component exceeds the required centripetal force, the excess is opposed by the force of contact between the two spheres. When the component of gravity becomes inadequate to supply the necessary force, contact is lost and \\( S \\) goes into a parabolic path while rotating at constant speed. The loss of contact occurs when\n\\[\n2 a M \\dot{\\theta}^{2}=M g \\cos \\theta\n\\]\n\nCombining this with (1), we obtain\n\\[\n\\frac{7}{10} \\cos \\theta=1-\\cos \\theta\n\\]\n\nHence contact is lost when \\( \\cos \\theta=10 / 17 \\), i.e., when \\( \\theta=\\arccos (10 / 17) \\).\n\nRemarks. It is interesting to verify that contact is lost at the same point even if the two spheres are of different sizes.\n\nThe motion described is not realizabie, for equation (1) shows that the time required for the upper sphere to roll to angular position \\( \\alpha \\) would be\n\\[\n\\int_{0}^{\\alpha} \\frac{1}{\\sqrt{1-\\cos \\theta}} d \\theta\n\\]\nand this improper integral diverges since\n\\[\n\\frac{1}{\\sqrt{1-\\cos \\theta}} \\sim \\frac{\\sqrt{2}}{\\theta} \\quad \\text { as } \\theta \\rightarrow 0 .\n\\]\n\nHence it would be more accurate to phrase the question as follows: The upper sphere is displaced slightly and allowed to roll off. Find the limit of the position at which contact is lost as the initial displacement approaches zero.\n\nFor a more general treatment of this problem, see A. S. Ramsey, Dynamics, Cambridge University Press, 1929, page 210.",
  "vars": [
    "\\\\theta",
    "\\\\alpha",
    "v",
    "\\\\omega"
  ],
  "params": [
    "S",
    "a",
    "M",
    "I",
    "g"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "\\\\theta": "orbitangle",
        "\\\\alpha": "dispangle",
        "v": "centerspeed",
        "\\\\omega": "spinrate",
        "S": "rollersphere",
        "a": "sphradius",
        "M": "sphmass",
        "I": "massmoment",
        "g": "gravconst"
      },
      "question": "Two uniform solid spheres of equal radii are so placed that one is directly above the other. The bottom sphere is fixed, and the top sphere, initially at rest, rolls off. At what point will contact between the two spheres be \"lost\"? Assume the coefficient of friction is such that no slipping occurs.",
      "solution": "Solution. Let \\( rollersphere \\) be the rolling sphere; let \\( sphradius \\) be its radius and \\( sphmass \\) its mass. Recall that the moment of inertia of a uniform solid sphere is \\( massmoment=(2 / 5) sphmass sphradius^{2} \\).\n\nWe assume that the motion is essentially two-dimensional. Then the state of the system is defined by the angle \\( orbitangle \\) between the line of centers and the vertical as shown in the figure. We denote the time derivative of \\( orbitangle \\) by \\( \\dot{orbitangle} \\). Let \\( centerspeed \\) be the translational speed of the center of \\( rollersphere \\) and let \\( spinrate \\) be the speed of rotation of \\( rollersphere \\).\n\nAs long as contact is maintained between the two spheres, the center of \\( rollersphere \\) moves along a circle of radius \\( 2 sphradius \\), so \\( centerspeed=2 sphradius \\dot{orbitangle} \\) and \\( spinrate=2 \\dot{orbitangle} \\). The kinetic energy of \\( rollersphere \\) is given by\n\\[\n\\frac{1}{2} sphmass centerspeed^{2}+\\frac{1}{2} massmoment spinrate^{2}=\\frac{14}{5} sphmass sphradius^{2} \\dot{orbitangle}^{2}\n\\]\nand the potential energy by\n\\( 2 sphmass sphradius gravconst \\cos orbitangle \\)\nrelative to the level of the center of the lower sphere. The total energy is constant so\n\\[\n2 sphmass sphradius gravconst \\cos orbitangle+\\frac{14}{5} sphmass sphradius^{2} \\dot{orbitangle}^{2}=2 sphmass sphradius gravconst\n\\]\n(The right member is the left evaluated for \\( orbitangle=0 \\).) Therefore we have\n\\[\n\\frac{7}{5} sphradius \\dot{orbitangle}^{2}=gravconst(1-\\cos orbitangle)\n\\tag{1}\n\\]\n\nTo keep \\( rollersphere \\) in a circular orbit of radius \\( 2 sphradius \\), a force toward the center of magnitude \\( 2 sphradius sphmass \\dot{orbitangle}^{2} \\) is required. This force is supplied by the component of the gravitational force along the line of centers, which is \\( sphmass gravconst \\cos orbitangle \\). As long as this component exceeds the required centripetal force, the excess is opposed by the force of contact between the two spheres. When the component of gravity becomes inadequate to supply the necessary force, contact is lost and \\( rollersphere \\) goes into a parabolic path while rotating at constant speed. The loss of contact occurs when\n\\[\n2 sphradius sphmass \\dot{orbitangle}^{2}=sphmass gravconst \\cos orbitangle\n\\tag{2}\n\\]\n\nCombining (1) and (2), we obtain\n\\[\n\\frac{7}{10} \\cos orbitangle=1-\\cos orbitangle\n\\]\n\nHence contact is lost when \\( \\cos orbitangle=10 / 17 \\), i.e., when \\( orbitangle=\\arccos (10 / 17) \\).\n\nRemarks. It is interesting to verify that contact is lost at the same point even if the two spheres are of different sizes.\n\nThe motion described is not realizable, for equation (1) shows that the time required for the upper sphere to roll to angular position \\( dispangle \\) would be\n\\[\n\\int_{0}^{dispangle} \\frac{1}{\\sqrt{1-\\cos orbitangle}} \\, d\\,orbitangle\n\\]\nand this improper integral diverges since\n\\[\n\\frac{1}{\\sqrt{1-\\cos orbitangle}} \\sim \\frac{\\sqrt{2}}{orbitangle} \\quad \\text { as } orbitangle \\rightarrow 0 .\n\\]\n\nHence it would be more accurate to phrase the question as follows: The upper sphere is displaced slightly and allowed to roll off. Find the limit of the position at which contact is lost as the initial displacement approaches zero.\n\nFor a more general treatment of this problem, see A. S. Ramsey, Dynamics, Cambridge University Press, 1929, page 210."
    },
    "descriptive_long_confusing": {
      "map": {
        "\\theta": "marigold",
        "\\alpha": "cupboard",
        "v": "lighthouse",
        "\\omega": "rainstorm",
        "S": "cassette",
        "a": "teaspoon",
        "M": "sandcastle",
        "I": "beanstalk",
        "g": "buttercup"
      },
      "question": "2. Two uniform solid spheres of equal radii are so placed that one is directly above the other. The bottom sphere is fixed, and the top sphere, initially at rest, rolls off. At what point will contact between the two spheres be \"lost\"? Assume the coefficient of friction is such that no slipping occurs.",
      "solution": "Solution. Let \\( cassette \\) be the rolling sphere; let \\( teaspoon \\) be its radius and \\( sandcastle \\) its mass. Recall that the moment of inertia of a uniform solid sphere is \\( beanstalk=(2 / 5) sandcastle teaspoon^{2} \\).\n\nWe assume that the motion is essentially two-dimensional. Then the state of the system is defined by the angle \\( marigold \\) between the line of centers and the vertical as shown in the figure. We denote the time derivative of \\( marigold \\) by \\( \\dot{marigold} \\). Let \\( lighthouse \\) be the translational speed of the center of \\( cassette \\) and let \\( rainstorm \\) be the speed of rotation of \\( cassette \\).\n\nAs long as contact is maintained between the two spheres, the center of \\( cassette \\) moves along a circle of radius \\( 2\\,teaspoon \\), so \\( lighthouse=2\\,teaspoon \\dot{marigold} \\) and \\( rainstorm=2 \\dot{marigold} \\). The kinetic energy of \\( cassette \\) is given by\n\\[\n\\frac{1}{2} sandcastle\\, lighthouse^{2}+\\frac{1}{2} beanstalk\\, rainstorm^{2}=\\frac{14}{5} sandcastle\\, teaspoon^{2} \\dot{marigold}^{2}\n\\]\nand the potential energy by\n\\( 2\\, sandcastle\\, teaspoon\\, buttercup \\cos marigold \\)\nrelative to the level of the center of the lower sphere. The total energy is constant so\n\\[\n2\\, sandcastle\\, teaspoon\\, buttercup \\cos marigold+\\frac{14}{5} sandcastle\\, teaspoon^{2} \\dot{marigold}^{2}=2\\, sandcastle\\, teaspoon\\, buttercup\n\\]\n(The right member is the left evaluated for \\( marigold=0 \\).) Therefore we have\n\\[\n\\frac{7}{5}\\, teaspoon\\, \\dot{marigold}^{2}=buttercup\\,(1-\\cos marigold)\n\\]\n\nTo keep \\( cassette \\) in a circular orbit of radius \\( 2\\,teaspoon \\), a force toward the center of magnitude \\( 2\\,teaspoon\\, sandcastle\\, \\dot{marigold}^{2} \\) is required. This force is supplied by the component of the gravitational force along the line of centers, which is \\( sandcastle\\, buttercup \\cos marigold \\). As long as this component exceeds the required centripetal force, the excess is opposed by the force of contact between the two spheres. When the component of gravity becomes inadequate to supply the necessary force, contact is lost and \\( cassette \\) goes into a parabolic path while rotating at constant speed. The loss of contact occurs when\n\\[\n2\\,teaspoon\\, sandcastle\\, \\dot{marigold}^{2}=sandcastle\\, buttercup \\cos marigold\n\\]\n\nCombining this with (1), we obtain\n\\[\n\\frac{7}{10}\\, \\cos marigold=1-\\cos marigold\n\\]\n\nHence contact is lost when \\( \\cos marigold=10 / 17 \\), i.e., when \\( marigold=\\arccos (10 / 17) \\).\n\nRemarks. It is interesting to verify that contact is lost at the same point even if the two spheres are of different sizes.\n\nThe motion described is not realizable, for equation (1) shows that the time required for the upper sphere to roll to angular position \\( cupboard \\) would be\n\\[\n\\int_{0}^{cupboard} \\frac{1}{\\sqrt{1-\\cos marigold}} \\, d marigold\n\\]\nand this improper integral diverges since\n\\[\n\\frac{1}{\\sqrt{1-\\cos marigold}} \\sim \\frac{\\sqrt{2}}{marigold} \\quad \\text { as } marigold \\rightarrow 0 .\n\\]\n\nHence it would be more accurate to phrase the question as follows: The upper sphere is displaced slightly and allowed to roll off. Find the limit of the position at which contact is lost as the initial displacement approaches zero.\n\nFor a more general treatment of this problem, see A. S. Ramsey, Dynamics, Cambridge University Press, 1929, page 210."
    },
    "descriptive_long_misleading": {
      "map": {
        "\\theta": "straightline",
        "\\alpha": "flatness",
        "v": "stillness",
        "\\omega": "calmness",
        "S": "blockform",
        "a": "diameter",
        "M": "lightness",
        "I": "nimbleness",
        "g": "levitation"
      },
      "question": "Two uniform solid spheres of equal radii are so placed that one is directly above the other. The bottom sphere is fixed, and the top sphere, initially at rest, rolls off. At what point will contact between the two spheres be \"lost\"? Assume the coefficient of friction is such that no slipping occurs.",
      "solution": "Solution. Let \\( blockform \\) be the rolling sphere; let \\( diameter \\) be its radius and \\( lightness \\) its mass. Recall that the moment of inertia of a uniform solid sphere is \\( nimbleness=(2 / 5) lightness diameter^{2} \\).\n\nWe assume that the motion is essentially two-dimensional. Then the state of the system is defined by the angle \\( straightline \\) between the line of centers and the vertical as shown in the figure. We denote the time derivative of \\( straightline \\) by \\( \\dot{straightline} \\). Let \\( stillness \\) be the translational speed of the center of \\( blockform \\) and let \\( calmness \\) be the speed of rotation of \\( blockform \\).\n\nAs long as contact is maintained between the two spheres, the center of \\( blockform \\) moves along a circle of radius \\( 2 diameter \\), so \\( stillness=2 diameter \\dot{straightline} \\) and \\( calmness=2 \\dot{straightline} \\). The kinetic energy of \\( blockform \\) is given by\n\\[\n\\frac{1}{2} lightness\\, stillness^{2}+\\frac{1}{2} nimbleness\\, calmness^{2}=\\frac{14}{5} lightness\\, diameter^{2} \\dot{straightline}^{2}\n\\]\nand the potential energy by\n\\( 2 lightness\\, diameter\\, levitation \\cos straightline \\)\nrelative to the level of the center of the lower sphere. The total energy is constant so\n\\[\n2 lightness\\, diameter\\, levitation \\cos straightline+\\frac{14}{5} lightness\\, diameter^{2} \\dot{straightline}^{2}=2 lightness\\, diameter\\, levitation\n\\]\n(The right member is the left evaluated for \\( straightline=0 \\).) Therefore we have\n\\[\n\\frac{7}{5} diameter \\dot{straightline}^{2}=levitation(1-\\cos straightline)\n\\]\n\nTo keep \\( blockform \\) in a circular orbit of radius \\( 2 diameter \\), a force toward the center of magnitude \\( 2 diameter\\, lightness \\dot{straightline}^{2} \\) is required. This force is supplied by the component of the gravitational force along the line of centers, which is \\( lightness\\, levitation \\cos straightline \\). As long as this component exceeds the required centripetal force, the excess is opposed by the force of contact between the two spheres. When the component of gravity becomes inadequate to supply the necessary force, contact is lost and \\( blockform \\) goes into a parabolic path while rotating at constant speed. The loss of contact occurs when\n\\[\n2 diameter\\, lightness \\dot{straightline}^{2}=lightness\\, levitation \\cos straightline\n\\]\n\nCombining this with (1), we obtain\n\\[\n\\frac{7}{10} \\cos straightline=1-\\cos straightline\n\\]\n\nHence contact is lost when \\( \\cos straightline=10 / 17 \\), i.e., when \\( straightline=\\arccos (10 / 17) \\).\n\nRemarks. It is interesting to verify that contact is lost at the same point even if the two spheres are of different sizes.\n\nThe motion described is not realizable, for equation (1) shows that the time required for the upper sphere to roll to angular position \\( flatness \\) would be\n\\[\n\\int_{0}^{flatness} \\frac{1}{\\sqrt{1-\\cos straightline}} d straightline\n\\]\nand this improper integral diverges since\n\\[\n\\frac{1}{\\sqrt{1-\\cos straightline}} \\sim \\frac{\\sqrt{2}}{straightline} \\quad \\text { as } straightline \\rightarrow 0 .\n\\]\n\nHence it would be more accurate to phrase the question as follows: The upper sphere is displaced slightly and allowed to roll off. Find the limit of the position at which contact is lost as the initial displacement approaches zero.\n\nFor a more general treatment of this problem, see A. S. Ramsey, Dynamics, Cambridge University Press, 1929, page 210."
    },
    "garbled_string": {
      "map": {
        "\\theta": "qzxwvtnp",
        "\\alpha": "hjgrksla",
        "v": "lkjhgfds",
        "\\omega": "poytrewq",
        "S": "mxncbvla",
        "a": "zqplkmnr",
        "M": "rskdjtug",
        "I": "vhslpqnr",
        "g": "lwtzmnrv"
      },
      "question": "2. Two uniform solid spheres of equal radii are so placed that one is directly above the other. The bottom sphere is fixed, and the top sphere, initially at rest, rolls off. At what point will contact between the two spheres be \"lost\"? Assume the coefficient of friction is such that no slipping occurs.",
      "solution": "Solution. Let \\( mxncbvla \\) be the rolling sphere; let \\( zqplkmnr \\) be its radius and \\( rskdjtug \\) its mass. Recall that the moment of inertia of a uniform solid sphere is \\( vhslpqnr=(2 / 5) rskdjtug \\, zqplkmnr^{2} \\).\n\nWe assume that the motion is essentially two-dimensional. Then the state of the system is defined by the angle \\( qzxwvtnp \\) between the line of centers and the vertical as shown in the figure. We denote the time derivative of \\( qzxwvtnp \\) by \\( \\dot{qzxwvtnp} \\). Let \\( lkjhgfds \\) be the translational speed of the center of \\( mxncbvla \\) and let \\( poytrewq \\) be the speed of rotation of \\( mxncbvla \\).\n\nAs long as contact is maintained between the two spheres, the center of \\( mxncbvla \\) moves along a circle of radius \\( 2 zqplkmnr \\), so \\( lkjhgfds = 2 zqplkmnr \\dot{qzxwvtnp} \\) and \\( poytrewq = 2 \\dot{qzxwvtnp} \\). The kinetic energy of \\( mxncbvla \\) is given by\n\\[\n\\frac{1}{2} rskdjtug \\, lkjhgfds^{2}+\\frac{1}{2} vhslpqnr \\, poytrewq^{2}=\\frac{14}{5} rskdjtug \\, zqplkmnr^{2} \\dot{qzxwvtnp}^{2}\n\\]\nand the potential energy by\n\\( 2 rskdjtug \\, zqplkmnr \\, lwtzmnrv \\cos qzxwvtnp \\)\nrelative to the level of the center of the lower sphere. The total energy is constant so\n\\[\n2 rskdjtug \\, zqplkmnr \\, lwtzmnrv \\cos qzxwvtnp+\\frac{14}{5} rskdjtug \\, zqplkmnr^{2} \\dot{qzxwvtnp}^{2}=2 rskdjtug \\, zqplkmnr \\, lwtzmnrv\n\\]\n(The right member is the left evaluated for \\( qzxwvtnp = 0 \\).) Therefore we have\n\\[\n\\frac{7}{5} zqplkmnr \\, \\dot{qzxwvtnp}^{2}=lwtzmnrv\\,(1-\\cos qzxwvtnp)\n\\]\n\nTo keep \\( mxncbvla \\) in a circular orbit of radius \\( 2 zqplkmnr \\), a force toward the center of magnitude \\( 2 zqplkmnr \\, rskdjtug \\, \\dot{qzxwvtnp}^{2} \\) is required. This force is supplied by the component of the gravitational force along the line of centers, which is \\( rskdjtug \\, lwtzmnrv \\cos qzxwvtnp \\). As long as this component exceeds the required centripetal force, the excess is opposed by the force of contact between the two spheres. When the component of gravity becomes inadequate to supply the necessary force, contact is lost and \\( mxncbvla \\) goes into a parabolic path while rotating at constant speed. The loss of contact occurs when\n\\[\n2 zqplkmnr \\, rskdjtug \\, \\dot{qzxwvtnp}^{2}=rskdjtug \\, lwtzmnrv \\cos qzxwvtnp\n\\]\n\nCombining this with (1), we obtain\n\\[\n\\frac{7}{10} \\cos qzxwvtnp = 1 - \\cos qzxwvtnp\n\\]\n\nHence contact is lost when \\( \\cos qzxwvtnp = 10 / 17 \\), i.e., when \\( qzxwvtnp = \\arccos (10 / 17) \\).\n\nRemarks. It is interesting to verify that contact is lost at the same point even if the two spheres are of different sizes.\n\nThe motion described is not realizabie, for equation (1) shows that the time required for the upper sphere to roll to angular position \\( hjgrksla \\) would be\n\\[\n\\int_{0}^{hjgrksla} \\frac{1}{\\sqrt{1-\\cos qzxwvtnp}} \\, d qzxwvtnp\n\\]\nand this improper integral diverges since\n\\[\n\\frac{1}{\\sqrt{1-\\cos qzxwvtnp}} \\sim \\frac{\\sqrt{2}}{qzxwvtnp} \\quad \\text { as } qzxwvtnp \\rightarrow 0 .\n\\]\n\nHence it would be more accurate to phrase the question as follows: The upper sphere is displaced slightly and allowed to roll off. Find the limit of the position at which contact is lost as the initial displacement approaches zero.\n\nFor a more general treatment of this problem, see A. S. Ramsey, Dynamics, Cambridge University Press, 1929, page 210."
    },
    "kernel_variant": {
      "question": "A thin–walled spherical shell of mass M and radius r is gently released from rest at the north-pole of a fixed solid sphere of radius 2r that is rigidly bolted to the surface of a small, uniformly rotating asteroid.\n\n•  The asteroid spins with constant angular speed Ω about the local “vertical”, i.e. the straight line through the asteroid’s centre and the north-pole of the big sphere.  \n•  In the uniformly rotating frame that co-rotates with the asteroid the effective gravitational acceleration is a constant vector of magnitude g₀ directed toward the asteroid’s centre.  \n•  Static friction is sufficiently large that the shell rolls without slipping while contact is maintained.  \n•  The shell is released with zero initial azimuthal velocity; consequently, by assumption, the motion is confined to the initial meridian plane so that the Coriolis force does not enter.\n\nLet θ be the angle between the upward vertical and the line joining the two centres (θ = 0 at the top).  \nThe shell’s moment of inertia is I = β M r² (β = ⅔ for a thin spherical shell, but keep β symbolic).  \nIntroduce the abbreviations  \n A := 1 + β ,  L := R + r = 3r ,  F := Ω²L ⁄ g₀ .\n\n(a) Show that the shell loses contact with the large sphere when θ satisfies the quadratic  \n    (A + 1)F cos²θ + (A + 2) cos θ − [2 + (A + 1)F] = 0.  (⋆)\n\n(b) Solve (⋆) explicitly for cos θ and show that the physically admissible root is  \n\n  cos θ = { −(A + 2) + √[(A + 2)² + 4(A + 1)F(2 + (A + 1)F)] } ∕ [2(A + 1)F]. (1)\n\n(c) Obtain the first non-trivial term in the small-spin expansion (F ≪ 1):  \n    cos θ = 2/(A + 2) + A(A + 1)(A + 4)F/(A + 2)³ + 𝒪(F²).\n\n(d) Show that in the rapid-spin limit (F ≫ 1) the detachment angle obeys  \n    cos θ = 1 − A/[2(A + 1)F] + 𝒪(F⁻²).\n\n(e) Determine the shell’s angular speed ω about its own centre at the instant contact is lost, expressing the final answer solely in terms of g₀, Ω, r and β.\n\nAnswers that are fully justified, dimensionally consistent and that quote the thin-shell value β = ⅔ where appropriate will receive full credit.",
      "solution": "Geometric data R = 2r ⇒ L = R + r = 3r.  \nKeep A = 1 + β throughout.\n\n─────────────────\n1. Kinematics and rolling constraint  \nThe only generalised coordinate is θ(t).  \nSpeed of the shell’s centre: v = L θ̇.  \nNo-slip condition on the outer surface: v = r ω ⇒ ω = (L/r) θ̇. (2)\n\n─────────────────\n2. Kinetic energy  \nT = ½ M v² + ½ I ω²  \n  = ½ M L² θ̇² + ½ β M r²(L²/r²) θ̇²  \n  = ½ M L²(1 + β) θ̇² = ½ M L²A θ̇². (3)\n\n─────────────────\n3. Potential energy in the co-rotating frame  \n\n(i) Gravitational (zeroed at θ = 0): U_g = −M g₀ L(1 − cos θ).  \n\n(ii) Centrifugal (zeroed at θ = 0): distance from spin axis ρ = L sin θ  \n  U_c = −½ M Ω² ρ² = −½ M Ω² L² sin²θ.\n\nTotal potential U(θ) = −M g₀ L(1 − cos θ) − ½ M Ω² L² sin²θ. (4)\n\n─────────────────\n4. Energy integral  \nReleased from rest ⇒ total mechanical energy initially zero, hence\n\n ½ M L²A θ̇² = M g₀ L(1 − cos θ) + ½ M Ω² L² sin²θ. (5)\n\nDivide by ½ M and rearrange:\n\n v² = L² θ̇² = (L/A)[2g₀(1 − cos θ) + Ω²L sin²θ]. (6)\n\n─────────────────\n5. Radial force balance  \nTake the outward normal to the big sphere as positive.  \n\nForces along the line of centres  \n• Normal reaction N (outward)  \n• Centrifugal    MΩ² L sin²θ (outward)  \n• Weight component −M g₀ cos θ (inward)\n\nThe centre of mass follows a circle of radius L, so the inward normal acceleration is v²/L. Newton’s second law along the radial line is  \n\n N + MΩ² L sin²θ − M g₀ cos θ = −M v²/L. (7)\n\nAt detachment N = 0, giving  \n\n v² = L[g₀ cos θ − Ω² L sin²θ]. (8)\n\n─────────────────\n6. Detachment condition – part (a)  \nSubstitute (6) into (8), set L = 3r and insert F = Ω²L/g₀, sin²θ = 1 − cos²θ.  \nAfter straightforward algebra one obtains\n\n (A + 1)F cos²θ + (A + 2) cos θ − [2 + (A + 1)F] = 0, which is (⋆). ✓\n\n─────────────────\n7. Closed-form root – part (b)  \nLet x := cos θ; (⋆) is quadratic in x with positive discriminant for every F ≥ 0.  \nThe physically admissible solution (0 < x ≤ 1) is  \n\n x = [−(A + 2) + √{(A + 2)² + 4(A + 1)F(2 + (A + 1)F)}] ⁄ [2(A + 1)F],  \n\nexactly equation (1). ✓\n\n─────────────────\n8. Small-spin expansion – part (c)  \nPut x = x₀ + x₁F + …, insert into (⋆):\n\n (A + 2)x₀ − 2 = 0 ⇒ x₀ = 2/(A + 2). (9)\n\nCollect the 𝒪(F) terms:\n\n (A + 1)x₀² + (A + 2)x₁ − (A + 1) = 0  \n ⇒ x₁ = [(A + 1)(1 − x₀²)]/(A + 2)  \n    = (A + 1)A(A + 4)/(A + 2)³. (10)\n\nHence  \n\n cos θ = 2/(A + 2) + A(A + 1)(A + 4)F/(A + 2)³ + 𝒪(F²). ✓\n\nRemark.  x₁ is strictly positive for every β ≥ 0 (including the point-mass limit β → 0, where x₁ = 10/27).  Rotation of the asteroid therefore always delays detachment.\n\n─────────────────\n9. Rapid-spin asymptotics – part (d)  \nWrite x = 1 − ε with ε ≪ 1 for F ≫ 1.  Insert in (⋆) and retain only the leading orders:\n\n (A + 1)F(1 − 2ε) + (A + 2)(1 − ε) − [2 + (A + 1)F] = 0  \n ⇒ −2(A + 1)F ε − (A + 2) ε + A = 0.\n\nThe term proportional to F dominates, so\n\n ε = A/[2(A + 1)F] [1 + 𝒪(F⁻¹)],  \n\nwhence\n\n cos θ = 1 − A/[2(A + 1)F] + 𝒪(F⁻²). ✓\n\nDetachment occurs ever closer to the pole as the spin increases, with an algebraic approach ε ∝ F⁻¹.\n\n─────────────────\n10. Angular speed about the shell’s centre – part (e)  \nBecause v = r ω, combine (8) with L = 3r:\n\n ω² = v²/r² = [3r g₀ (cos θ − F(1 − cos²θ))]/r²  \n    = (3g₀/r)[cos θ − F(1 − cos²θ)]. (11)\n\nEquation (11), together with cos θ from (1) (or with either asymptotic form from parts (c) or (d) when appropriate), gives ω solely in terms of g₀, Ω, r and β, as required.\n\nConsistency check (Ω → 0, i.e. F = 0).  Then cos θ → 2/(A + 2) and\n\n ω₀² = (3g₀/r) · [2/(A + 2)] = 6g₀/[r(A + 2)]. (12)\n\nFor a thin spherical shell (β = ⅔ ⇒ A = 5/3) this yields\n\n ω₀ = √(18 g₀ / 11 r),  \n\nin full agreement with classical textbooks.  All formulae are dimensionally consistent (ω in s⁻¹).\n\n─────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.502087",
        "was_fixed": false,
        "difficulty_analysis": "1.  Extra dynamical ingredient: the large sphere now spins, so analysis must be carried out in a non-inertial frame, introducing centrifugal and Coriolis terms and an effective potential.  \n2.  Generalised inertia: the parameter β forces the solver to treat an arbitrary moment of inertia instead of substituting a memorised value.  \n3.  Mixed forces: gravitational, inertial (centrifugal), normal and frictional forces interact; the student must project centrifugal acceleration onto the oblique radius to the contact point, a step absent from the classical problem.  \n4.  Coupled constraints: combining the non-holonomic rolling constraint with energy conservation and radial force balance leads to a quadratic whose coefficients contain both β and the dimensionless spin parameter F = Ω²L/g₀.  Manipulating this algebra without losing terms is appreciably more involved than the single linear relation of the original problem.  \n5.  Sanity check requirement: the solver must show that the rotational limit Ω → 0 reproduces the classical answer, providing an essential self-consistency test.  \n\nThese additional layers (rotating frame mechanics, variable moment of inertia, oblique projections and algebraic complexity) make the enhanced kernel variant substantially more demanding than both the textbook original and the earlier “asteroid” kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "A thin–walled spherical shell of mass M and radius r is gently released from rest at the north-pole of a fixed solid sphere of radius 2r that is rigidly bolted to the surface of a small, uniformly rotating asteroid.\n\n•  The asteroid spins with constant angular speed Ω about the local “vertical”, i.e. the straight line through the asteroid’s centre and the north-pole of the big sphere.  \n•  In the uniformly rotating frame that co-rotates with the asteroid the effective gravitational acceleration is a constant vector of magnitude g₀ directed toward the asteroid’s centre.  \n•  Static friction is sufficiently large that the shell rolls without slipping while contact is maintained.  \n•  The shell is released with zero initial azimuthal velocity; consequently, by assumption, the motion is confined to the initial meridian plane so that the Coriolis force does not enter.\n\nLet θ be the angle between the upward vertical and the line joining the two centres (θ = 0 at the top).  \nThe shell’s moment of inertia is I = β M r² (β = ⅔ for a thin spherical shell, but keep β symbolic).  \nIntroduce the abbreviations  \n A := 1 + β ,  L := R + r = 3r ,  F := Ω²L ⁄ g₀ .\n\n(a) Show that the shell loses contact with the large sphere when θ satisfies the quadratic  \n    (A + 1)F cos²θ + (A + 2) cos θ − [2 + (A + 1)F] = 0.  (⋆)\n\n(b) Solve (⋆) explicitly for cos θ and show that the physically admissible root is  \n\n  cos θ = { −(A + 2) + √[(A + 2)² + 4(A + 1)F(2 + (A + 1)F)] } ∕ [2(A + 1)F]. (1)\n\n(c) Obtain the first non-trivial term in the small-spin expansion (F ≪ 1):  \n    cos θ = 2/(A + 2) + A(A + 1)(A + 4)F/(A + 2)³ + 𝒪(F²).\n\n(d) Show that in the rapid-spin limit (F ≫ 1) the detachment angle obeys  \n    cos θ = 1 − A/[2(A + 1)F] + 𝒪(F⁻²).\n\n(e) Determine the shell’s angular speed ω about its own centre at the instant contact is lost, expressing the final answer solely in terms of g₀, Ω, r and β.\n\nAnswers that are fully justified, dimensionally consistent and that quote the thin-shell value β = ⅔ where appropriate will receive full credit.",
      "solution": "Geometric data R = 2r ⇒ L = R + r = 3r.  \nKeep A = 1 + β throughout.\n\n─────────────────\n1. Kinematics and rolling constraint  \nThe only generalised coordinate is θ(t).  \nSpeed of the shell’s centre: v = L θ̇.  \nNo-slip condition on the outer surface: v = r ω ⇒ ω = (L/r) θ̇. (2)\n\n─────────────────\n2. Kinetic energy  \nT = ½ M v² + ½ I ω²  \n  = ½ M L² θ̇² + ½ β M r²(L²/r²) θ̇²  \n  = ½ M L²(1 + β) θ̇² = ½ M L²A θ̇². (3)\n\n─────────────────\n3. Potential energy in the co-rotating frame  \n\n(i) Gravitational (zeroed at θ = 0): U_g = −M g₀ L(1 − cos θ).  \n\n(ii) Centrifugal (zeroed at θ = 0): distance from spin axis ρ = L sin θ  \n  U_c = −½ M Ω² ρ² = −½ M Ω² L² sin²θ.\n\nTotal potential U(θ) = −M g₀ L(1 − cos θ) − ½ M Ω² L² sin²θ. (4)\n\n─────────────────\n4. Energy integral  \nReleased from rest ⇒ total mechanical energy initially zero, hence\n\n ½ M L²A θ̇² = M g₀ L(1 − cos θ) + ½ M Ω² L² sin²θ. (5)\n\nDivide by ½ M and rearrange:\n\n v² = L² θ̇² = (L/A)[2g₀(1 − cos θ) + Ω²L sin²θ]. (6)\n\n─────────────────\n5. Radial force balance  \nTake the outward normal to the big sphere as positive.  \n\nForces along the line of centres  \n• Normal reaction N (outward)  \n• Centrifugal    MΩ² L sin²θ (outward)  \n• Weight component −M g₀ cos θ (inward)\n\nThe centre of mass follows a circle of radius L, so the inward normal acceleration is v²/L. Newton’s second law along the radial line is  \n\n N + MΩ² L sin²θ − M g₀ cos θ = −M v²/L. (7)\n\nAt detachment N = 0, giving  \n\n v² = L[g₀ cos θ − Ω² L sin²θ]. (8)\n\n─────────────────\n6. Detachment condition – part (a)  \nSubstitute (6) into (8), set L = 3r and insert F = Ω²L/g₀, sin²θ = 1 − cos²θ.  \nAfter straightforward algebra one obtains\n\n (A + 1)F cos²θ + (A + 2) cos θ − [2 + (A + 1)F] = 0, which is (⋆). ✓\n\n─────────────────\n7. Closed-form root – part (b)  \nLet x := cos θ; (⋆) is quadratic in x with positive discriminant for every F ≥ 0.  \nThe physically admissible solution (0 < x ≤ 1) is  \n\n x = [−(A + 2) + √{(A + 2)² + 4(A + 1)F(2 + (A + 1)F)}] ⁄ [2(A + 1)F],  \n\nexactly equation (1). ✓\n\n─────────────────\n8. Small-spin expansion – part (c)  \nPut x = x₀ + x₁F + …, insert into (⋆):\n\n (A + 2)x₀ − 2 = 0 ⇒ x₀ = 2/(A + 2). (9)\n\nCollect the 𝒪(F) terms:\n\n (A + 1)x₀² + (A + 2)x₁ − (A + 1) = 0  \n ⇒ x₁ = [(A + 1)(1 − x₀²)]/(A + 2)  \n    = (A + 1)A(A + 4)/(A + 2)³. (10)\n\nHence  \n\n cos θ = 2/(A + 2) + A(A + 1)(A + 4)F/(A + 2)³ + 𝒪(F²). ✓\n\nRemark.  x₁ is strictly positive for every β ≥ 0 (including the point-mass limit β → 0, where x₁ = 10/27).  Rotation of the asteroid therefore always delays detachment.\n\n─────────────────\n9. Rapid-spin asymptotics – part (d)  \nWrite x = 1 − ε with ε ≪ 1 for F ≫ 1.  Insert in (⋆) and retain only the leading orders:\n\n (A + 1)F(1 − 2ε) + (A + 2)(1 − ε) − [2 + (A + 1)F] = 0  \n ⇒ −2(A + 1)F ε − (A + 2) ε + A = 0.\n\nThe term proportional to F dominates, so\n\n ε = A/[2(A + 1)F] [1 + 𝒪(F⁻¹)],  \n\nwhence\n\n cos θ = 1 − A/[2(A + 1)F] + 𝒪(F⁻²). ✓\n\nDetachment occurs ever closer to the pole as the spin increases, with an algebraic approach ε ∝ F⁻¹.\n\n─────────────────\n10. Angular speed about the shell’s centre – part (e)  \nBecause v = r ω, combine (8) with L = 3r:\n\n ω² = v²/r² = [3r g₀ (cos θ − F(1 − cos²θ))]/r²  \n    = (3g₀/r)[cos θ − F(1 − cos²θ)]. (11)\n\nEquation (11), together with cos θ from (1) (or with either asymptotic form from parts (c) or (d) when appropriate), gives ω solely in terms of g₀, Ω, r and β, as required.\n\nConsistency check (Ω → 0, i.e. F = 0).  Then cos θ → 2/(A + 2) and\n\n ω₀² = (3g₀/r) · [2/(A + 2)] = 6g₀/[r(A + 2)]. (12)\n\nFor a thin spherical shell (β = ⅔ ⇒ A = 5/3) this yields\n\n ω₀ = √(18 g₀ / 11 r),  \n\nin full agreement with classical textbooks.  All formulae are dimensionally consistent (ω in s⁻¹).\n\n─────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.420386",
        "was_fixed": false,
        "difficulty_analysis": "1.  Extra dynamical ingredient: the large sphere now spins, so analysis must be carried out in a non-inertial frame, introducing centrifugal and Coriolis terms and an effective potential.  \n2.  Generalised inertia: the parameter β forces the solver to treat an arbitrary moment of inertia instead of substituting a memorised value.  \n3.  Mixed forces: gravitational, inertial (centrifugal), normal and frictional forces interact; the student must project centrifugal acceleration onto the oblique radius to the contact point, a step absent from the classical problem.  \n4.  Coupled constraints: combining the non-holonomic rolling constraint with energy conservation and radial force balance leads to a quadratic whose coefficients contain both β and the dimensionless spin parameter F = Ω²L/g₀.  Manipulating this algebra without losing terms is appreciably more involved than the single linear relation of the original problem.  \n5.  Sanity check requirement: the solver must show that the rotational limit Ω → 0 reproduces the classical answer, providing an essential self-consistency test.  \n\nThese additional layers (rotating frame mechanics, variable moment of inertia, oblique projections and algebraic complexity) make the enhanced kernel variant substantially more demanding than both the textbook original and the earlier “asteroid” kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}