{"id":233924,"date":"2025-06-13T14:52:27","date_gmt":"2025-06-13T14:52:27","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233924"},"modified":"2025-06-13T14:52:29","modified_gmt":"2025-06-13T14:52:29","slug":"your-cousin-throckmorton-skateboards-from-rest-down-a-curved-ramp-in-the-form-of-a-quarter-circle-of-radius","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/your-cousin-throckmorton-skateboards-from-rest-down-a-curved-ramp-in-the-form-of-a-quarter-circle-of-radius\/","title":{"rendered":"Your cousin Throckmorton skateboards from rest down a curved ramp in the form of a quarter circle of radius"},"content":{"rendered":"\n

Your cousin Throckmorton skateboards from rest down a curved ramp in the form of a quarter circle of radius
. Assume that the wheels of the skateboard roll without slipping and that the friction in the wheel bearings is negligible. If we treat Throcky and his skateboard as a particle of mass
, he undergoes (nonuniform) circular motion as he rides down the ramp.\\ a) When Throcky has descended half the distance to the bottom, so that the surface of the ramp is at
to the horizontal, what is his instantaneous speed?\\nb) What is the magnitude of the normal force acting on Throcky at that moment?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

We are given a problem involving circular motion along a quarter-circle ramp. Let’s solve it in two parts using principles of energy conservation<\/strong> and circular dynamics<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Given:<\/strong><\/h3>\n\n\n\n
    \n
  • Ramp is a quarter-circle (i.e., 90\u00b0 total).<\/li>\n\n\n\n
  • Radius of the circular ramp: RRR (we\u2019ll keep it general).<\/li>\n\n\n\n
  • Mass of Throcky and skateboard: mmm<\/li>\n\n\n\n
  • Throcky starts from rest.<\/li>\n\n\n\n
  • At halfway down the ramp (i.e., 45\u00b0 from the top), find:\n
      \n
    • (a) Speed<\/strong><\/li>\n\n\n\n
    • (b) Normal force<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Part (a): Speed at Halfway Down (45\u00b0)<\/strong><\/h2>\n\n\n\n

      We apply the principle of conservation of mechanical energy<\/strong>.<\/p>\n\n\n\n

        \n
      • Initial position (top of ramp):<\/strong>
        Total energy = Potential energy = Ei=mgh=mgRE_i = mgh = mgREi\u200b=mgh=mgR (since height = radius)<\/li>\n\n\n\n
      • At halfway (45\u00b0 down):<\/strong>
        Throcky has descended halfway in arc length but in height, the vertical drop is: h=R(1\u2212cos\u2061\u03b8)=R(1\u2212cos\u206145\u2218)=R(1\u221222)h = R(1 – \\cos \\theta) = R(1 – \\cos 45^\\circ) = R(1 – \\tfrac{\\sqrt{2}}{2})h=R(1\u2212cos\u03b8)=R(1\u2212cos45\u2218)=R(1\u221222\u200b\u200b) So the potential energy is now: PE=mgR(1\u221222)PE = mgR(1 – \\tfrac{\\sqrt{2}}{2})PE=mgR(1\u221222\u200b\u200b) Then by conservation of energy: Initial\u00a0Total\u00a0Energy=Kinetic\u00a0Energy+Potential\u00a0Energy\u00a0at\u00a045\u00b0\\text{Initial Total Energy} = \\text{Kinetic Energy} + \\text{Potential Energy at 45\u00b0}Initial\u00a0Total\u00a0Energy=Kinetic\u00a0Energy+Potential\u00a0Energy\u00a0at\u00a045\u00b0 mgR=12mv2+mgR(1\u221222)mgR = \\tfrac{1}{2}mv^2 + mgR(1 – \\tfrac{\\sqrt{2}}{2})mgR=21\u200bmv2+mgR(1\u221222\u200b\u200b) Cancel mmm and solve for vvv: gR=12v2+gR(1\u221222)gR = \\tfrac{1}{2}v^2 + gR(1 – \\tfrac{\\sqrt{2}}{2})gR=21\u200bv2+gR(1\u221222\u200b\u200b) 12v2=gR(22)\\tfrac{1}{2}v^2 = gR(\\tfrac{\\sqrt{2}}{2})21\u200bv2=gR(22\u200b\u200b) v=2gR(22)=gR2v = \\sqrt{2gR(\\tfrac{\\sqrt{2}}{2})} = \\sqrt{gR\\sqrt{2}}v=2gR(22\u200b\u200b)\u200b=gR2\u200b\u200b<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        Part (b): Normal Force at 45\u00b0<\/strong><\/h2>\n\n\n\n

        At 45\u00b0, Throcky is undergoing nonuniform circular motion<\/strong>. The centripetal force<\/strong> is provided by components of gravity and the normal force<\/strong>:<\/p>\n\n\n\n

          \n
        • Centripetal force required: Fc=mv2RF_c = \\frac{mv^2}{R}Fc\u200b=Rmv2\u200b<\/li>\n\n\n\n
        • The radial component of gravity: mgcos\u2061\u03b8=mgcos\u206145\u2218=mg\u22c522mg\\cos \\theta = mg\\cos 45^\\circ = mg \\cdot \\tfrac{\\sqrt{2}}{2}mgcos\u03b8=mgcos45\u2218=mg\u22c522\u200b\u200b<\/li>\n<\/ul>\n\n\n\n

          Apply Newton\u2019s second law in radial direction: N+mgcos\u2061\u03b8=mv2RN + mg\\cos \\theta = \\frac{mv^2}{R}N+mgcos\u03b8=Rmv2\u200b<\/p>\n\n\n\n

          Substitute v2=gR2v^2 = gR\\sqrt{2}v2=gR2\u200b and cos\u206145\u2218=22\\cos 45^\\circ = \\tfrac{\\sqrt{2}}{2}cos45\u2218=22\u200b\u200b: N+mg\u22c522=m(gR2)R=mg2N + mg \\cdot \\tfrac{\\sqrt{2}}{2} = \\frac{m(gR\\sqrt{2})}{R} = mg\\sqrt{2}N+mg\u22c522\u200b\u200b=Rm(gR2\u200b)\u200b=mg2\u200b<\/p>\n\n\n\n

          Solve for NNN: N=mg2\u2212mg\u22c522=mg(2\u221222)=mg\u22c522N = mg\\sqrt{2} – mg \\cdot \\tfrac{\\sqrt{2}}{2} = mg\\left(\\sqrt{2} – \\tfrac{\\sqrt{2}}{2}\\right) = mg \\cdot \\tfrac{\\sqrt{2}}{2}N=mg2\u200b\u2212mg\u22c522\u200b\u200b=mg(2\u200b\u221222\u200b\u200b)=mg\u22c522\u200b\u200b<\/p>\n\n\n\n

          \n\n\n\n

          Final Answers:<\/strong><\/h3>\n\n\n\n

          (a)<\/strong> Speed:\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003\u2003 v=gR2v = \\sqrt{gR\\sqrt{2}}v=gR2\u200b\u200b<\/p>\n\n\n\n

          (b)<\/strong> Normal Force Magnitude: N=mg\u22c522N = mg \\cdot \\tfrac{\\sqrt{2}}{2}N=mg\u22c522\u200b\u200b<\/p>\n\n\n\n

          \n\n\n\n

          Explanation (300 words)<\/strong><\/h3>\n\n\n\n

          This physics problem examines the motion of a skateboarder descending a quarter-circle ramp, a common real-world example of nonuniform circular motion. The problem is solved using energy conservation and Newton\u2019s laws. Initially, the skateboarder is at rest at the top of a circular arc, with all energy in gravitational potential form. As the skateboarder descends, potential energy is converted into kinetic energy.<\/p>\n\n\n\n

          At halfway down the ramp\u2014corresponding to a 45\u00b0 angle from the horizontal\u2014the skateboarder\u2019s vertical height has decreased by R(1\u2212cos\u206145\u2218)R(1 – \\cos 45^\\circ)R(1\u2212cos45\u2218). Using the principle of energy conservation, the kinetic energy gained is equal to the potential energy lost. We derive his speed at this point using the equation v=gR2v = \\sqrt{gR\\sqrt{2}}v=gR2\u200b\u200b, reflecting the amount of energy converted into motion.<\/p>\n\n\n\n

          Next, we analyze the normal force<\/strong>, which is the support force from the ramp surface acting perpendicular to the curve. Because the skateboarder is moving along a circular path, there must be a centripetal force<\/strong> directed toward the center of the circle. This force is partially provided by the radial component of the skateboarder\u2019s weight and partially by the ramp\u2019s normal force. By applying Newton\u2019s second law in the radial direction, we determine the normal force at 45\u00b0 to be N=mg\u22c522N = mg\\cdot \\frac{\\sqrt{2}}{2}N=mg\u22c522\u200b\u200b.<\/p>\n\n\n\n

          This setup assumes the skateboard rolls without slipping and neglects friction in the wheels, allowing for a clean conservation of mechanical energy. The approach demonstrates how classical mechanics can be used to understand motion along curved surfaces<\/p>\n\n\n\n

          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          Your cousin Throckmorton skateboards from rest down a curved ramp in the form of a quarter circle of radius. Assume that the wheels of the skateboard roll without slipping and that the friction in the wheel bearings is negligible. If we treat Throcky and his skateboard as a particle of mass, he undergoes (nonuniform) circular […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-233924","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233924","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=233924"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233924\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233924"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233924"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233924"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}