{"id":222238,"date":"2025-05-31T07:13:00","date_gmt":"2025-05-31T07:13:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222238"},"modified":"2025-05-31T07:13:02","modified_gmt":"2025-05-31T07:13:02","slug":"none-none-12-176","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/none-none-12-176\/","title":{"rendered":"None None *12-176."},"content":{"rendered":"\n

None None *12-176. The car travels around the circular track with a constant speed of 20 m\/s. Determine the car’s radial and transverse components of velocity and acceleration at the instant = \u00cf\u20ac\/4 rad. 12-177. The car travels around the circular track such that its transverse component is 0 (0.0062) rad, where t is in seconds. Determine the car’s radial and transverse components of velocity and acceleration at the instant = 4 s. O r = (400 cos 0) m<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

Let\u2019s solve Problem 12\u2013177<\/strong>:<\/p>\n\n\n\n

\n\n\n\n

Given:<\/strong><\/h3>\n\n\n\n
    \n
  • Transverse angular position:\u2003\u03b8(t) = 0.006\u202ft\u00b2 (radians)<\/li>\n\n\n\n
  • Time:\u2003t = 4 s<\/li>\n\n\n\n
  • Radius:\u2003r = 400\u202fcos(\u03b8) m<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step 1: Find \u03b8, \u03b8\u0307, and \u03b8\u0308 at t = 4 s<\/strong><\/h3>\n\n\n\n

    \u03b8 = 0.006\u202ft\u00b2 = 0.006(4\u00b2) = 0.096 rad<\/strong><\/p>\n\n\n\n

    Differentiate to get angular velocity and acceleration:<\/p>\n\n\n\n

      \n
    • \u03b8\u0307 = d\u03b8\/dt = 0.012\u202ft \u2192 \u03b8\u0307(4) = 0.012(4) = 0.048 rad\/s<\/strong><\/li>\n\n\n\n
    • \u03b8\u0308 = d\u00b2\u03b8\/dt\u00b2 = 0.012 rad\/s\u00b2 (constant)<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Step 2: Compute radius r and its derivatives at \u03b8 = 0.096<\/strong><\/h3>\n\n\n\n

      r = 400\u202fcos(\u03b8) = 400\u202fcos(0.096) \u2248 400(0.9954) \u2248 398.17 m<\/strong><\/p>\n\n\n\n

      Differentiate r with respect to time:<\/p>\n\n\n\n

        \n
      • \u1e59 = dr\/dt = dr\/d\u03b8 \u00b7 d\u03b8\/dt = d(400\u202fcos\u03b8)\/d\u03b8 \u00b7 \u03b8\u0307
        \u2003\u2003\u2003= \u2013400\u202fsin(\u03b8)\u00b7\u03b8\u0307 \u2248 \u2013400\u202fsin(0.096)(0.048) \u2248 \u20131.837 m\/s<\/li>\n\n\n\n
      • r\u0308 = d\u00b2r\/dt\u00b2
        \u2003\u2003\u2003= d\/dt(\u2013400\u202fsin(\u03b8)\u202f\u03b8\u0307)
        \u2003\u2003\u2003= \u2013400[cos(\u03b8)\u202f(\u03b8\u0307)\u00b2 + sin(\u03b8)\u202f\u03b8\u0308]
        \u2003\u2003\u2003\u2248 \u2013400[cos(0.096)(0.048)\u00b2 + sin(0.096)(0.012)]
        \u2003\u2003\u2003\u2248 \u2013400[0.9954(0.002304) + 0.0958(0.012)]
        \u2003\u2003\u2003\u2248 \u2013400[0.002294 + 0.00115] \u2248 \u2013400(0.003444) \u2248 \u20131.378 m\/s\u00b2<\/strong><\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        Step 3: Radial and Transverse Components<\/strong><\/h3>\n\n\n\n

        Velocity Components:<\/strong><\/p>\n\n\n\n

          \n
        • Radial velocity:\u2003v_r = \u1e59 = \u20131.837 m\/s<\/li>\n\n\n\n
        • Transverse velocity:\u2003v_\u03b8 = r\u202f\u03b8\u0307 = 398.17 \u00d7 0.048 \u2248 19.93 m\/s<\/strong><\/li>\n<\/ul>\n\n\n\n

          Acceleration Components:<\/strong><\/p>\n\n\n\n

            \n
          • Radial acceleration:\u2003a_r = r\u0308 \u2013 r(\u03b8\u0307)\u00b2
            \u2003= \u20131.378 \u2013 398.17(0.048)\u00b2 \u2248 \u20131.378 \u2013 0.916 \u2248 \u20132.294 m\/s\u00b2<\/strong><\/li>\n\n\n\n
          • Transverse acceleration:\u2003a_\u03b8 = r\u202f\u03b8\u0308 + 2\u202f\u1e59\u202f\u03b8\u0307
            \u2003= 398.17(0.012) + 2(\u20131.837)(0.048) \u2248 4.778 \u2013 0.176 \u2248 4.60 m\/s\u00b2<\/strong><\/li>\n<\/ul>\n\n\n\n
            \n\n\n\n

            Final Answers (at t = 4 s):<\/strong><\/h3>\n\n\n\n
              \n
            • Radial velocity:\u2003\u20131.84 m\/s<\/strong><\/li>\n\n\n\n
            • Transverse velocity:\u200319.93 m\/s<\/strong><\/li>\n\n\n\n
            • Radial acceleration:\u2003\u20132.29 m\/s\u00b2<\/strong><\/li>\n\n\n\n
            • Transverse acceleration:\u20034.60 m\/s\u00b2<\/strong><\/li>\n<\/ul>\n\n\n\n
              \n\n\n\n

              Explanation <\/strong><\/h3>\n\n\n\n

              This problem involves analyzing a car’s motion along a non-uniform circular path where both the radius and angular motion change with time. The car\u2019s position is defined by \u03b8(t) = 0.006t\u00b2 and r = 400cos(\u03b8). We are asked to find the radial and transverse components of velocity and acceleration at t = 4 s.<\/p>\n\n\n\n

              To begin, we calculate the angular position \u03b8 and its first and second derivatives, \u03b8\u0307 and \u03b8\u0308. These represent the angular velocity and angular acceleration. At t = 4 s, we find \u03b8 = 0.096 rad, \u03b8\u0307 = 0.048 rad\/s, and \u03b8\u0308 = 0.012 rad\/s\u00b2.<\/p>\n\n\n\n

              The radius r at this \u03b8 is computed using the given function, resulting in r \u2248 398.17 m. Since the radius is a function of \u03b8, which itself depends on time, we apply the chain rule to find \u1e59 and r\u0308. These values reflect how the radial distance changes over time, essential for determining radial components of motion.<\/p>\n\n\n\n

              Velocity in polar coordinates splits into two parts: radial (\u1e59) and transverse (r\u202f\u03b8\u0307). Similarly, acceleration includes a radial component (r\u0308 \u2013 r\u202f\u03b8\u0307\u00b2) and a transverse component (r\u202f\u03b8\u0308 + 2\u202f\u1e59\u202f\u03b8\u0307). Substituting the calculated values yields the final answers.<\/p>\n\n\n\n

              The car\u2019s transverse velocity is large (~20 m\/s), while its radial velocity is negative, indicating it\u2019s moving slightly inward at that instant. The radial acceleration is also negative, combining centripetal and radial effects, and the transverse acceleration is positive, meaning the car is speeding up angularly. These detailed components help describe the motion in curvilinear coordinates effectively.<\/p>\n\n\n\n

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

              None None *12-176. The car travels around the circular track with a constant speed of 20 m\/s. Determine the car’s radial and transverse components of velocity and acceleration at the instant = \u00cf\u20ac\/4 rad. 12-177. The car travels around the circular track such that its transverse component is 0 (0.0062) rad, where t is in […]<\/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-222238","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222238","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=222238"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222238\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222238"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222238"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222238"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}